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ENGINEERIN
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1882
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THEORETICAL MECHANICS.
MECHANICS OF ENGINEERING.
THEORETICAL MECHANICS,
WITH AN
INTRODUCTION TO THE CALCULUS.
DESIGNED AS A TEXT-BOOK
FOR TECHNICAL SCHOOLS AND COLLEGES, AND FOR THE USE
OF ENGINEERS, ARCHITECTS, ETC.
BY
JULIUS WEISBACH, PH.D.,
OBERBERGRATH AND PROFESSOR AT THE ROYAL MINING ACADEMY AT FREI-
AERG; MEMBER OF THE IMPERIAL ACADEMY OF SCIENCES AT ST.
PETERSBURG, ETC.
Translated from the Fourth Augmented and Improved German Edition by
ECKLEY B. COXE, A.M.,
MINING ENGINEER.
SIXTH AMERICAN EDITION.
ILLUSTRATED WITH 902 WOOD-CUTS IN THE TEXT.
NEW YORK :
D. VAN NOSTRAND, PUBLISHER,
23 MURRAY AND 27 WARREN STREET.
1882.
¡
Entered, according to Act of Congress, in the year 1875, by
D. VAN NOSTRAND,
In the office of the Librarian of Congress, at Washington, D. C.
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PREFACE TO THE FIRST EDITION.
IT
T is not without apprehension that I give to the public my
elementary treatise upon the Mechanics of Engineering and
of the Construction of Machines. Although I can say to myself
that, in preparing this manual, I have gone to work with all pos-
sible care and attention, yet I fear that I have not been able to
satisfy the wishes of every one. The ideas, wishes and require-
ments of the public are so various, that it is not possible to do
So. Some may find the treatment of a particular subject toɔ
detailed, others perhaps too short; some will desire a more
scientific discussion of certain subjects, while others would prefer
one more popular. Many years of study, much experience in
teaching and very varied observations and experiments have led
me to adopt, as most suitable to the object in view, the method,
according to which this work has been arranged. My principal
effort has been to obtain the greatest simplicity in enunciation
and demonstration, and to treat all the important laws, in their
practical applications, without the aid of the higher mathematics.
If we consider how many subjects a technical man must master in
order to accomplish any thing very important in his profession,
we must make it our business as teachers and authors for techni-
cal men to facilitate the thorough study of science by simplicity
of diction, by removing whatever may be unnecessary, and by em-
ploying the best known and most practicable methods. For this
reason I have entirely avoided the use of the Calculus in this
work. Although at the present time the opportunities for ac-
quiring a knowledge of it are no longer rare, yet it is an unde-
niable fact that, unless we are constantly making use of it, we
soon lose that facility of calculation, which is indispensable; for
this reason so many able engineers can no longer employ the Cal-
vi
PREFACE TO THE FIRST EDITION.
As I do not
As I do not agree with
enunciate without proof
culus which they learned in their youth.
those authors, who in popular treatises
the more difficult laws, I have preferred to deduce or demon-
strate them in an elementary, although sometimes in a somewhat
roundabout, manner.
Formulas without proof will therefore seldom be found in this
work. We will assume that the reader has a general knowledge
of certain principles of natural philosophy and a thorough knowl-
edge of the elements of pure mathematics. My attention has
been especially directed to preserving the proper mean between
generalization and specialization. Although I appreciate the ad-
vantages of generalization, yet it is my opinion that in this work,
as in all elementary treatises, too much generalization is to be
avoided. The simple is oftener met with in practice than the
complex. It is also undeniable that in considering the general
case we often fail to attain a more profound knowledge of the
special one, and that it is often easier to deduce the complex from
the simple than the simple from the complex. The reader must
not expect to find in this work a treatise upon the construction
of machines, but only an introduction to or preparation for it.
Mechanics should bear the same relation to the construction of
machines that Descriptive Geometry does to Mechanical Drawing.
When the pupil has acquired sufficient knowledge of Mechanics
and of Descriptive Geometry, it appears better to combine the
course of Construction of Machines with that. of Mechanical
Drawing.
It may be doubted whether it was advisable to divide my sub-
ject into two parts, theoretical and applied. If we remember that
this work is intended to give instruction upon all the mechanical
relations of the construction and of the theory of machines, the
advantage, or rather, the necessity, of such a division becomes
evident. In order to judge of a structure or of a machine, we
must have a knowledge of mechanical principles of a very varied
character, E.G., those of friction, strength, inertia, impact, efflux,
&c.; the material for the mechanical study of a structure or of a
machine must, therefore, be gathered from almost all the divis-
ions of mechanics. Now, since it is better to study all the me-
chanical principles of a machine at once than to collect them from
all the different parts of mechanics, the advantage of such a di-
vision is apparent.
Having practical application always in view, I have endeav-
ored, in preparing my work, to illustrate the principles laid down
PREFACE TO THE FIRST EDITION.
vii
in it by examples taken from every-day life. I am justified in
asserting that this work contrasts favorably with any other of the
same character in the number of appropriate examples, which are
solved in it. I also hope that the great number of carefully-pre-
pared figures will contribute to the object in view. My thanks
are due to the publishers for having given the book in all respects
the best appearance. Particular care has been taken to have the
calculations correct; generally every example has been calculated
three times, and not by the same person. It is, therefore, im-
probable that any gross errors will be found in them. In the ex-
amples, as in the formulas, I have employed the Prussian weights
and measures, as they are probably familiar to the majority of my
readers. The printing (in this case so difficult) is open to little
complaint. The mistakes in copying, or of impression, which
have been observed, are noted at the end of the book.
I do not think that many additions to this list need be made.
An attentive examination of the illustrations will show that they
have been prepared with care. The larger illustrations, particu-
larly those representing bodies in three dimensions, are drawn
according to the method of Axonometric Projection, first treated by
me (see Polytechn. Mittheilungen Band I. Tubingen, 1845).
This method of drawing possesses all the advantages of Isometric
Projection, while in addition the pictures, which it furnishes, are
not only more beautiful in themselves, but more easily awaken in
us distinct conceptions of the objects represented. The drawings
in this work are made in such a way that the dimensions of the
width or depth appear but one-half as large as those of the height
and length of the same size. I cannot omit thanking Mr. Ernest
Röting, student at the academy in Freiberg, whose revision has
essentially contributed to the accuracy of the work
It is necessary to inform the reader that he will find much
new matter, which is peculiar to the author. Without stopping to
mention many small articles, which occur in almost every chapter,
I would call attention to the following comprehensive discussions:
A general and easy determination of the centre of gravity of plane
surfaces and of polyhedra, limited by plane surfaces, will be found
in paragraphs 107, 112, and 113; an approximate formula for the
catenary in paragraph 148; additional remarks upon the friction
of axles in paragraphs 167, 168, 169, 172, and 173. Important
additions to the theory of impact have been made, particularly
in paragraphs 277 and 278; for heretofore the impact of imper-
fectly elastic bodies has been too little considered, and that of a
viii
PREFACE TO THE FIRST EDITION.
perfectly elastic with an imperfectly elastic body has not been
treated at all. Very important additions, and in some cases en-
tirely new laws, will be found in the chapter upon hydraulics, a
subject to which I have for a number of years devoted special
study. The laws of incomplete contraction, first observed by the
author, will be found for the first tinie in a manual of mechanics.
The author has also incorporated in it the principal results, so
important in practice, of his experiments upon the efflux of water
through oblique short pipes, elbows, curved and long pipes, etc.,
although the third number of his "Untersuchungen im Gebiete
der Mechanik und Hydraulik" has not yet appeared. The chap-
ter upon running water, upon hydrometry and upon the impact.
of water contains some original matter. The theories of the re-
action of water discharging from a vessel and of the impact of
water, which are treated according to the principle of mechanical
effect, are original.
I cannot, however, conceal from the reader that, since the vol-
ume has been finished, I have wished that some few subjects had
been treated differently; but I must add that as yet I have ob-
served no great imperfections. If at times the reader should
miss something, he is referred to the second volume, which will
supply both the accidental and the intentional omissions, as has
been noted in many places in the first volume.
The printing of the second volume will now go on without in-
terruption, so that we may expect the complete work to be in the
hands of the reader before the end of the year. The pocket-book,
the "Ingenieur," cited in the Mechanics, which contains a collec-
tion of formulas, rules and tables of arithmetic, geometry and
mechanics, will soon appear.
It will be a source of great pleasure and satisfaction to me, if
I have accomplished the purposes for which this work has been
undertaken, namely, to give to the practical man a useful coun-
sellor in questions of application, to the teacher of practical
mechanics a serviceable text-book for instruction, and to the stu-
dent of engineering a welcome aid in the study of mechanics.
JULIUS WEISBACH.
FREIBERG, March 19th, 1846.
PREFACE TO THE SECOND EDITION.
THE
E present (second) edition of the Mechanics of Engineering
and of the Construction of Machines has undergone no es-
sential alterations either in method or arrangement. The inter-
nal construction of the work has been changed in many places,
and its size has been considerably increased. The author has
also endeavored, as much as possible, to correct the errors and
omissions of the first edition. The great increase in size is
mainly due to three additions. The first consists of a condensed
Introduction to the Calculus, which has been made as popular as
possible, and has been prefixed to the main work. The object of
introducing it was to avoid too complicated and too artificial de-
monstrations by means of the lower mathematics, and also to
render the reader more independent in his study of mechanics,
and to place him upon a higher stand-point in this important
branch of science. By making use of the principles explained in
the Introduction, it was possible to discuss many subjects of great
practical importance, which previously we could not treat at all,
or, at least, only imperfectly with the aid of elementary algebra
and geometry. In order to avoid interruptions to those who
have not made themselves familiar with the Elements of the Cal-
culus, prefixed to the work, all the paragraphs, in which it is ap-
plied, are designated by a parenthesis ().
The second addition consists of a new chapter on Hydrostatics,
in which the molecular action of water is treated. Since a knowl-
edge of the molecular forces (capillarity) is of importance in ex-
periments and observations in hydraulics and pneumatics, the
author has thought it advisable to treat the fundamental princi-
ples of these forces in a separate chapter. Finally, a chapter has
been added to the work in the form of an appendix, which treats
X
PREFACE TO THE SECOND EDITION.
of oscillation and wave motion. The author found himself com-
pelled to do this in consequence of the importance to the engineer
of a more accurate knowledge of the theory of oscillation. The
great influence of vibration upon the working and durability of
machines is a subject to which too much attention cannot be
given. It is also to observations of oscillations that we owe the
latest determination of the modulus of elasticity, which is of such
importance in practice. I have mentioned in the Appendix the
magnetic force, principally because it is of great use to the engi-
neer in determining directions in mines, where the access to day-
light is not easy. The theory of water-waves, which closes the
volume, is a part of hydraulics; its presence in this work requires,
therefore, no explanation. Unfortunately, it is far from complete.
The changes in the other parts of the work are the following:
the chapter upon elasticity and strength has been much extended
and altered, the subject of hydraulics has been treated more at
length, and some modifications in it have been made, in conse-
quence of the continued experiments of the author.
I trust that the present edition will be received with the same
favor as the last, by which the author was encouraged to continue
his preparation of the work.
FREIBERG, May 15th, 1850.
JULIUS WEISBACH.
PREFACE TO THE THIRD EDITION.
THE third edition of the first volume of my Mechanics of En-
gineering and of the Construction of Machines, which I now
give to the public, has, compared with its predecessors, not only
been improved, but also augmented and completed. The changes
are due principally to the advance of science, and in some cases
to the results of more recent investigations. When not withheld
by some good reason, I have endeavored, so far as possible, to
satisfy the wishes which have been communicated to me from
different quarters in regard to the work. From the extraordi-
nary favor, with which it has been received both in and out of
Germany, on this as well as on the other side of the Atlantic, I
flatter myself that it has suited both in method and size the
greater portion of the public for whom it was intended, and my
efforts in preparing the new edition have been naturally directed to
removing any errors or omissions, that have been observed, and
to incorporating in it the latest experiments, treated in the same
manner and as concisely as possible. I am sorry to be obliged to
remark that the work has been subjected to unjust criticism.
Thus, E.G., Professor Wiebe, of Berlin, in a remark upon pages
245 and 246 of his work upon “die Lehre von der Befestigung
der Machinentheile," (Berlin, 1854), states that I have given
coefficients of torsion for square shafts in my Mechanics (first
edition), as well as in the "Ingenieur," 16 times greater than
those given by Morin. The Professor has here committed an
oversight; for in my formulas, as is expressly stated in both
works, the fourth power of the half length of the side occurs,
while the formulas of Morin and Wiebe, as well as those of my
second edition, contain the fourth power of the whole length of
the side of the cross-section. Now since 2' is equal to 16, the
xii
PREFACE TO THE THIRD EDITION.
error observed by Professor Wiebe proceeds from a mistake on
his part.
I shall make no reply to the partial criticism contained in
Grunert's Archiv der Mathematik, as I do not wish to enter upon
a useless controversy here. Besides, Professor Grunert has
already printed in his Archiv enough nonsense about Physics
and Practical Mechanics (as I can easily prove) to demonstrate
his unfitness for criticising works on those subjects.
It would have been easier for me to have given my book a
more scientific form; but it would then have met with less favor,
as it is intended for practical men.
From another stand-point also the book can easily and with
equal injustice be found fault with. Any one, who has had some
practical experience, will have observed how little theory is made
use of, and how often it is put in the back-ground and looked
upon with disfavor by practical men. The fault of this is no
doubt due in great measure to that method of instruction, which
condemns the study of science for the sake of its applications.
This edition, which has been augmented principally by the
revision of the theory of elasticity and strength, and by the in-
troduction of the latest hydraulic experiments, excels its prede-
cessors not only in substance, but also in appearance, all the
illustrations being new. The printing of the second volume will
continue uninterruptedly.
FREIBERG, July, 1856.
JULIUS WEISBACH.
PREFACE TO THE FOURTH EDITION.
THE
HE fourth edition of my Mechanics of Engineering and of
the Construction of Machines has undergone no change either
in method or arrangement. As three large editions have been ex-
hausted in a comparatively short time, as two have been published
in the English language, one in England and one in North Amer-
ica, and as the work has been translated into Swedish, Polish, and
Russian, I may well hope that this manual has met the wishes and
needs of that great practical public for whom it is intended. I
have, therefore, in preparing this edition, endeavored simply to
remove any errors or omissions, which may have been observed,
and to introduce the results of the latest practically important
experiments, together with the newest developments of theory.
Thus, E.G., in the chapter upon friction I have included the results
of the latest experiments by Bochet, and the section upon elasti-
city and strength has been rewritten in accordance with the
present stand-point of science, in doing which I have made use of
the recent works of Lamé, Rankine, Bresse, etc. The section
upon hydraulics has been augmented, improved and completed.
The later researches of the author are here discussed. I will men-
tion more particularly the experiments upon the efflux of water
under great and very great pressures, as well as upon the heights
of jets, those upon the efflux of air, and the comparative experi-
ments upon the impact of streams of air and water. The chapter
upon the efflux of air has been entirely rewritten, as the author is
of the opinion that the ordinary formulas for the efflux of air
under high pressures do not represent the law of efflux. The
formulas obtained are very simple, since, without materially affect-
ing its accuracy, I have substituted in the well-known formula
for heat
xiv
PREFACE TO THE FOURTH EDITION.
1 + δ τι
Τι
1 + 6т
(3)
0,42
0,50 instead of the exponent 0,42, by which I obtain
δ τι
1 + 8 T₁ = √¹ (see § 461).
1 + δτ
Y₁
1
γ
The practical value of a formula does not depend upon its cor-
rectness even at extreme limits, but rather upon the fact that,
within given limits, it furnishes values which agree sufficiently
well with the results of experiment.
Several new paragraphs, in which Phoronomics and Aerosta-
tics are treated with the aid of the Calculus, have been added. In
hydraulics the pressure of water flowing through pipes, on account
of its practical importance, has been treated separately in two new
paragraphs (§ 439 and § 440). In the chapter upon the force and
resistance of water I have treated the theory of the simple reaction
wheel, as well as its application as an instrument for proving the
theory of the impact and resistance of water. The more recent
gas and water meters are also discussed, since these instruments
are set in motion by the reaction of the issuing fluid, the intensity
of which can easily be determined by the foregoing theory.
Finally, the Appendix has been slightly augmented by the in-
troduction of the report of the recent experiments of Geh. Ober-
baurath Hagen upon waves of water.
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In answer to the criticism, which has been made in some
quarters, that a more scientific treatment of the subject, based
upon the Calculus, would have been more in accordance with the
object of the book, I would state that my book is intended for
the use of practical men, who often do not possess either the
requisite knowledge of the Calculus or sufficient facility in the use
of it. Having labored during upwards of thirty years as instructor
in a technical institution, during which time I have been engaged
in practical works of various kinds and have made many journeys
for the purpose of technical studies, I can confidently give an
opinion upon this subject.
As I consider my reputation as an author of much more
importance than any mere pecuniary advantage, it is always a
pleasure to me to find my "Mechanics" made use of in works of
a similar character; but when writers avail themselves of it with-
out the slightest acknowledgment, I can only appeal to the judg-
ment of the public.
FREIBERG, May, 1863.
JULIUS WEISBACH.
TRANSLATOR'S PREFACE.
THE favor, with which both the English and American editions
of the Mechanics of Engineering and of the Construction of Ma-
chines were received, would sufficiently justify the appearance of a
new one, even if the original work had undergone no change. But
as the first two volumes of the last (fourth) German edition contain
more than twice as much matter as those of the first, and since a
third volume of about fifteen hundred pages has been added, the
translator feels not only that the work may be considered a new
one, but also that, in offering it to the public, he is supplying a
real want. The text of this edition has been, to a great extent,
rewritten and rearranged, and the translation is entirely original.
Weisbach's Mechanics is now so well known, wherever that sci-
ence is taught, that any eulogy on our part would be superfluous.
A large number of typographical errors, observed in the German
edition, have been corrected with the approbation of the author,
who has also communicated to the translator some slight modifica
tions in the text. The work of translation was begun with the
author's approval, while the translator was a student of the Mining
Academy at Freiberg, but the work was delayed by his professional
engagements. He hopes that it will now appear without interrup-
tion.
At the suggestion of the author, an Appendix has been added
containing an account of the articles upon the subjects treated in
this volume, which have been published by him since the appear-
ance of the last German edition.
1
Ivi
TRANSLATOR'S PREFACE.
All the tables, formulas, examples, etc., in which the Prussian
weights and measures occur, have been transformed so as to be ap-
plicable to the English system. Where the metrical system was
employed in the original work, it has been retained in the transla-
tion, as the meter is now much used both in England and America.
The "Ingenieur," which is so often quoted in this work, has,
unfortunately, not been translated into English, but all the refer-
ences to it have been preserved, as the work is a valuable one, even
to those who have little or no knowledge of German, and perhaps
an English edition of it may be published.
A list of errors and omissions observed in this volume will be
given in the succeeding one, and the translator will be glad to be
informed of any typographical errors.
He would call attention to the illustrations, which are printed
from electrotype copies of the wood-cuts prepared for the German
edition, and his thanks are due to the publisher and stereotypers
for the excellent appearance of the work.
ECKLEY B. COXE.
CONTENTS.
ARTICLE
INTRODUCTION TO THE CALCULUS.
1-4 Functions. Laws of Nature...
5- 6 Differential. Position of tangent..
7— 8 Rules for differentiating.
9-10 The function y = x²
·
11-12 Straight line, ellipse, hyperbola....
13-14 Course of curves, maximum and minimum.
15 McLaurin's series, binomial series.
· •
16-18 Integral, Integral Calculus.
19-23 Exponential and logarithmic functions.
24-27 Trigonometrical and circular functions.
28 Integration by parts...
29-31 Quadrature of curves.
32 Rectification of curves.
•
33-34 Normal and radius of curvature.
35 Function y =
0
0
36 Method of the least squares.
37 Method of interpolation.
·
PAGB
33
38
40
44
49
53
57
60
63
70
78
93
95
89 & & & ZI
76
85
87
SECTION I.
PHORONOMICS, OR THE PURELY MATHEMATICAL THEORY OF
MOTION.
CHAPTER I.
SIMPLE MOTION.
§ 1 Rest and motion..
2-3 Kinds of motion.
4-6 Uniform motion.
•
7-9 Uniformly variable motion..
10-13 Uniformly accelerated motion.
14 Uniformly retarded motion.
105
105
106
107
109
113
•
xviii
TABLE OF CONTENTS.
§ 15-18 The free fall and vertical ascension of bodies..
19 Variable motion in general. . . .
20 Differential and integral formulas of phoronomics.
21 Mean velocity..
22-26 Graphical representation of the formulas of motion.
CHAPTER II.
COMPOUND MOTION.
27-29 Composition of motions....
30 Parallelogram of motions.
31-33 Parallelogram of velocities.
34 Composition and decomposition of velocities.
35 Composition of accelerations.
36 Composition of velocities and accelerations.
37-38 Parabolic motion...
39 Motion of projectiles.
40 Jets of water
•
•
41-43 Curvilinear motion in general.
44 Application of the Calculus.
45-46 Relative motion..
•
PAGE
113
•
117
·
119
121
•
122
.. 126
. 127
128
•
131
•
•
132
132
134
136
138
141
145
149
SECTION II.
MECHANICS, OR THE PHYSICAL SCIENCE OF MOTION IN
GENERAL.
CHAPTER I.
FUNDAMENTAL PRINCIPLES AND LAWS OF MECHANICS.
47 Mechanics, phoronomics, cinematics.
48 Force, gravity
49 Equilibrium, statics, dynamics...
50 Classification of the forces, motive forces, resistances, etc.
51-52 Pressure, traction, equality of forces.
53 Matter, material bodies.
54 Unit of weight, gram, pound.
55 Inertia..
56 Measure of forces
57-59 Mass, heaviness
60-61 Specific gravity, table of specific gravities.
62 State of aggregation..
63 Classification of the forces..
•
64 Forces, how determined..
65 Action and reaction..
66 Division of mechanics..
154
154
155
•
155
156
156
157
157
158
158
161
162
163
163
164
164
TABLE OF CONTENTS.
xix
CHAPTER II.
MECHANICS OF A MATERIAL POINT.
§ 67 Material point……….
68-69 Simple constant force..
70-73 Mechanical effect or work done by a force.
74-75 Principle of the via viva..
76 Composition of forces..
77 Parallelogram of forces.
78 Decomposition of forces.
79-80 Forces in a plane.
81 Forces in space.
•
82-83 Principle of virtual velocities.
84 Transmission of mechanical effect
85 Work done in curvilinear motion.
SECTION III.
STATICS OF RIGID BODIES.
CHAPTER I.
GENERAL PRINCIPLES OF THE STATICS OF RIGID BODIES.
PAGE
165
166
168
171
174
177
•
179
180
182
185
187
189
86-87 Transference of the point of application.
88-89 Statical moment....
90-91 Composition of forces in the same plane,
92 Parallel forces.
93-95 Couples.....
96 Centre of parallel forces.
97 Forces in space..
98-102 Principle of virtual velocities....
CHAPTER II.
THE THEORY OF THE CENTRE OF GRAVITY.
103-104 Centre of gravity, line of gravity, plane of gravity…..
105-106 Determination of the centre of gravity.
107-108 Centre of gravity of lines.
· •
109-114 Centre of gravity of plane figures..
•
115 Determination of the centre of gravity by the Calculus.
116 The centre of gravity of curved surfaces.
117-133 Centre of gravity of bodies....
124 Applications of Simpson's rule.
192
193
•
195
199
200
205
207
209
213
214
216
218
226
•
227
228
237
125 Determination of the centre of gravity of solids of rotation, etc.. 239
123-128 Properties of Guldinus..
•
341
XX
TABLE OF CONTENTS.
CHAPTER III.
EQUILIBRIUM OF BODIES RIGIDLY FASTENED AND SUPPORTED.
§ 129 Method of fastening.
130 Equilibrium of supported bodies.
131 Stability of a suspended body.
132-133 Pressure upon the points of support of a body..
•
134 Equilibrium of forces around an axis
135-137 Lever, mathematical and material.
138-139 Pressure of bodies upon one another.
140-141 Stability..
142-143 Formulas for stability.
144 Dynamical stability.
•
•
•
145 Work done in moving a heavy body.
146 Stability of a body upon an inclined plane.
147 Theory of the inclined plane..
PAGE
247
248
249
•
250
•
254
255
261
263
266
269
•
271
•
•
272
•
274
•
275
277
148 Application of the principle of virtual velocities.
149 Theory of the wedge....
CHAPTER IV.
•
EQUILIBRIUM IN FUNICULAR MACHINES.
150 Funicular machines, funicular polygon.
151-153 Fixed and movable knots.
154-156 Equilibrium of a funicular polygon.
157 The parabola as catenary .
158-160 The catenary.
• ·
161-162 Equation of the catenary.
163-164 The pulley, fixed and movable.
•
165-166 Wheel and axle, equilibrium of the same.
CHAPTER V.
280
281
286
291
293
299
303
305
THE RESISTANCE OF FRICTION AND THE RIGIDITY OF CORDAGE.
167-168 Friction..
169 Kinds of friction, sliding and rolling.
309
310
170 Laws of friction.
171 Coefficient of friction...
172 Angle of friction and cone of friction.
173 Experiments on friction.
174 Friction tables...
311
312
314
315
·
318
175 Latest experiments on friction..
320
•
176-177 Inclined plane, friction upon an inclined plane...
323
178 The theory of the equilibrium of bodies with reference to the
friction...
328
179-180 Wedge, friction on the wedge.
$29
181-185 Coefficients of friction of axles, friction of axles..
233
TABLE OF CONTENTS.
XXI
PAGE
§ 186 Poncelet's theorem...
341
187 Lever, axial friction of the lever.
343
188 Friction of a pivot.....
345
189 Friction on conical pivots.
347
190 Anti-friction pivots....
349
191 Friction on points and knife-edges.
352
192 Rolling friction...
853
193-194 Friction of cords and chains.
356
361
363
195 Rigidity of chains.
196-200 Rigidity of cordage.
SECTION IV.
THE APPLICATION OF STATICS TO THE ELASTICITY AND
STRENGTH OF BODIES.
CHAPTER I.
ELASTICITY AND STRENGTH OF EXTENSION, COMPRESSION
AND SHEARING.
201 Elasticity of rigid bodies..
371
202 Elasticity and strength...
372
203 Extension and compression.
374
204 Modulus of elasticity....
376
205 Modulus of proof strength, modulus of ultimate strength.
379
206 Modulus of resilience and fragility..
382
207 Extension of a body by its own weight..
381
208 Bodies of uniform strength....
387
209 Experiments upon extension and compression.
391
•
213 Strength of shearing.
•
210 Experiments upon extension....
211 Elasticity and strength of iron and wood.
212 Numbers determined by experiment..
CHAPTER II.
ELASTICITY AND STRENGTH OF FLEXURE OR BENDING.
393
397
401
406
214 Flexure of a rigid body
215 Moment of flexure (W)
216-217 Elastic curve.....
218 More general equation of the elastic curve.
219-222 Flexure produced by two parallel forces..
223 A uniformly loaded girder...
224-225 Reduction of the moment of flexure.
226 Moment of flexure of a strip....
227 Moment of flexure of a parallelopipedical girder.
228 Hollow, double-webbed or tubular girders...
409
412
•
414
419
422
430
432
435
436
•
437
xxii
TABLE OF CONTENTS.
229 Triangular girders..
230 Polygonal girders.
•
231 Cylindrical or elliptical girders.
• •
232 Application of the calculus to the determination of W.
233-231 Beams with curvilinear cross-sections.
235 Strength of flexure....
236 Formulas for the strength of bodies...
237 Difference in the moduli of proof strength.
238 Difference in the moduli of ultimate strength
239 Experiments upon flexure and rupture.
240 Moduli of proof and ultimate strength.
241 Relative deflection.
242 Moments of proof load.
243 Cross-section of wooden girders..
244 Hollow and webbed girders.
215 Eccentric loads....
246-248 Girders supported in different ways.
249-250 Girders not uniformly loaded.
251-252 Cross-section of rupture..
253-254 Bodies of uniform strength..
•
255 Flexure of bodies of uniform strength.
256 Deflection of metal springs...
•
·
CHAPTER III.
PAGE
439
441
443
445
447
450
453
457
460
463
466
469
472
474
477
480
484
491
494
498
504
506
THE ACTION OF THE SHEARING ELASTICITY IN THE BENDING
AND TWISTING OF BODIES.
257 The shearing force parallel to the neutral axis..
258 The shearing force in the plane of the cross-section.
259 Maximum and minimum strain....
510
513
•
•
. 515
260 Influence of the strength of shearing upon the proof load of a
girder...
... 519
elastic curve.
261 Influence of the elasticity of shearing upon the form of the
262 Elasticity of torsion.
263 Moment of torsion or twisting moment.
522
523
524
•
264 Resistance to rupture by torsion..
528
CHAPTER IV.
ON THE PROOF STRENGTH OF LONG COLUMNS, OR THE RESIST-
ANCE TO CRUSHING BY BENDING OR BREAKING ACROSS.
265–266 Flexure and proof load of long pillars...
267 Bodies of uniform resistance to breaking across.
•
268 Hodgkinson experiments.
269 More simple determination of the proof load.
532
539
542
544
TABLE OF CONTENTS.
xxiii
CHAPTER V.
COMBINED ELASTICITY AND STRENGTH.
§ 270 Combined elasticity and strength.
271 Eccentric pull and thrust.
272-273 Oblique pull or thrust.
•
•
274-275 Flexure of girders subjected to a tensile force.
276 Torsion combined with a tensile or compressive force..
277 Flexure and torsion combined..
278 Bending forces in different planes..
•
SECTION V.
DYNAMICS OF RIGID BODIES.
CHAPTER I.
THE THEORY OF THE MOMENT OF INERTIA.
PACE
547
551
553
559
563
567
570
279 Kinds of motion.
280 Rectilinear motion.
281 Motion of rotation
573
574
·
575
282 Moment of inertia
•
283 Reduction of the mass.
284 Reduction of the moments of inertia.
576
578
580
·
285 Radius of gyration..
286 Moment of inertia of a rod.
288 Prism and Cylinder....
289 Cone and Pyramid..
•
290 Sphere....
291 Cylinder and Cone.
293 Parabola and Ellipse.
•
581
582
287 Rectangle and Parallelopipedon (moments of inertia of)..
583
585
587
588
539
•
292 Segments...
294 Solids and surfaces of revolution.
590
592
593
295-296 Accelerated rotation of a wheel and axle..
595
297 Atwood's machine......
599
298-299 Accelerated motion of a system of pulleys or tackle..
300 Rolling motion of a body on an inclined plane....
601
605
CHAPTER II.
THE CENTRIFUGAL FORCE OF RIGID BODIES.
301 Normal force.....
302 Centripetal and centrifugal forces...
606
·
608
xxiv
TABLE OF CONTENTS.
§ 303-304 Mechanical effect of the centrifugal force...
305-308 Centrifugal force of masses of finite dimensions.
309-311 Free axes, principal axes....
312 Action upon the axis of rotation.
313 Centre of percussion..
CHAPTER III.
PAGE
610
•
614
624
629
634
OF THE ACTION OF GRAVITY UPON BODIES DESCRIBING
PRESCRIBED PATHS.
314-318 Sliding upon an inclined plane....
319 Rolling motion upon an inclined plane.
320 Circular pendulum…….
321-323 Simple pendulum.
324 Cycloid..
•
• •
325-326 Cycloidal pendulum.
327 Compound pendulum.
328 Kater's pendulum..
329 Rocking pendulum.
CHAPTER IV.
THE THEORY OF IMPACT.
330-331 Impact in general...
332 Central impact.
333 Elastic impact..
334 Particular cases of impact.
335 Loss of energy by impact.
336 Hardness of a body....
337 Elastic-inelastic impact.
338 Imperfectly elastic impact.
•
•
639
646
648
649
•
•
655
656
•
661
664
665
. 667
669
671
672
674
676
678
680
· 339-340 Oblique impact.....
682
341 Friction of impact, friction during impact.
685
342 Impact of revolving bodies.
343 Impact of oscillating bodies.
344 Ballistic pendulum..
345 Eccentric impact.
•
346 Application of the force of impact.
347 Pile driving..
348 Absolute strength of impact.
349 Relative strength of impact..
350 Strength of torsion in impact.
688
690
693
695
696
•
•
698
702
•
705
707
TABLE OF CONTENTS.
XXV
SECTION VI.
STATICS OF FLUIDS
CHAPTER I.
OF THE EQUILIBRIUM AND PRESSURE OF WATER IN VESSELS.
§ 351 Fluids ....
352 Principle of equal pressure.
•
353 Pressure in the water.
354 Surface of water.
•
355 Pressure upon the bottom.
356 Lateral pressure.
•
357-359 Centre of pressure.
•
•
360 Pressure in a given direction.
361 Pressure upon curved surfaces.
•
362 Horizontal and vertical pressure in water.
363 Thickness of pipes and boilers...
CHAPTER II.
EQUILIBRIUM OF WATER WITH OTHER BODIES.
364-366 Buoyant effort or upward pressure..
367-368 Depth of floatation...
371 Inclined floating...
369-370 Stability of a floating body.
•
373 Hydrometers, Areometers.
372 Specific gravity..
374 Equilibrium of liquids of different densities.
CHAPTER III.
OF THE MOLECULAR ACTION OF WATER.
375 Molecular forces
376 Adhesion plates.
377 Adhesion to the sides of a vessel.
378-379 Tension of the surface of the water.
380 Curve of the surface of water..
381 Parallel plates...
382-383 Capillary tubes.
•
CHAPTER IV.
OF THE EQUILIBRIUM AND PRESSURE OF THE AIR.
384 Tension of gases..
385 Pressure of the atmosphere.
PAGE
712
718
715
718
721
724
725
731
734
736
738
742
746
•
750
754
756
758
761
762
762
763
765
767
770
773
776
777
xxvi
TABLE OF CONTENTS.
§ 386 Manometer.
•
387 Mariotte's law.
£88 Work done by compressed air....
389 Pressure in the different layers of air. Barometric measure-
ments of heights...
390 Stereometer and volumeter
•
391 Air pump.
· •
392 Gay Lussac's law...
393 Heaviness of the air.
394 Air manometer
•
3.5 Buoyant effort of the air.
•
SECTION VII.
DYNAMICS OF FLUIDS.
CHAPTER I.
THE GENERAL THEORY OF THE EFFLUX OF WATER FROM
VESSELS.
PAGE
778
780
783
787
788
790
793
795
796
797
396 Efflux. Discharge.
•
397 Velocity of efflux..
398 Velocities of influx and efflux.
•
399 Velocities of efflux, pressure and heaviness
400 Hydraulic head....
401 Efflux through rectangular lateral orifices..
402 Triangular and trapezoidal lateral orifices.
403 Circular orifices...
404 Efflux from a vessel in motion.
CHAPTER II.
800
801
803
804
. 808
•
. 810
. 813
815
817
OF THE CONTRACTION OF THE VEIN OR JET OF WATER, WHEN
ISSUING FROM AN ORIFICE IN A THIN PLATE.
405 Coefficient of velocity.
406 Coefficient of contraction.
•
• •
• •
407 Contracted vein of water.
408 Coefficient of efflux..
409 Experiments upon efflux.
•
410 Rectangular lateral orifices, Efflux through them.
411 Overfalls......
412 Maximum and minimum contraction..
413 Scale of contraction....
414 Partial or incomplete contraction.
415 Imperfect contraction..
416-417 Efflux of moving water
418-419 Lesbros' experiments..
•
820
821
823
824
825
828
•
833
834
836
837
810
842
846
TABLE OF CONTENTS.
xxvii
CHAPTER III.
OF THE FLOW OF WATER THROUGH PIPES.
§ 420 Short tubes..
421 Short cylindrical tubes.
422 Coefficient of resistance.
•
423 Inclined short tubes or ajutages..
424 Imperfect contraction...
·
425-426 Conical short tubes or ajutages.
427-429 Resistance of the friction of water.
430 Motion of water in long pipes
431 Motion of water in conical pipes..
432 Conduit pipes.
433 Jets of water.
434 Height of jets of water.
435 Piezometer..
•
PAGE
852
853
855
857
$58
831
863
•
869
872
874
876
€ 878
. 881
CHAPTER IV.
RESISTANCE TO THE MOTION OF WATER WHEN THE CONDUIT
IS SUDDENLY ENLARGED OR CONTRACTED.
436 Sudden enlargement..
437 Contraction
438 Influence of imperfect contraction..
•
439 Relations of pressure in cylindrical pipes.
440 Relations of pressure in conical pipes.
441 Elbows, resistance of.
442 Bends
443-444 Valve gates, cocks, valves.
445 Valves....
446 Compound vessels...
CHAPTER V.
883
$85
887
. 888
•
891
894
896
900
904
907
OF THE EFFLUX OF WATER UNDER VARIABLE PRESSURE.
447 Prismatic vessels.
448-449 Communicating vessels.
• •
450 Notch in the side.....
910
911
. 914
. 916
•
919
921
922
924
926
•
451 Wedge-shaped and pyramidal vessels
452 Spherical and obelisk-shaped vessels.
453 Irregularly shaped vessels....
454 Simultaneous influx and efflux.
455 Locks and sluices...
·
456 Apparatus for hydraulic experiments..
xxviii
TABLE OF CONTENTS.
CHAPTER VI.
OF THE EFFLUX OF AIR AND OTHER FLUIDS FROM VESSELS
AND PIPES.
PAGE
§ 457 Efflux of mercury and oil.
980
458 Velocity of efflux of air.
952
459 Discharge..
933
460 Efflux according to Mariotte's law.
934
· ·
461 Work done by the heat.
926
:
462 Efflux of air, when the cooling is taken into consideration...
939
463 Efflux of moving air....
464 465 Coefficients of efflux of air.
466 Coefficients of friction of air.
467 Motion of air in long pipes.
•
468 Efflux when the pressure diminishes.
CHAPTER VII.
OF THE MOTION OF WATER IN CANALS AND RIVERS,
941
944
949
950
952
469 Running water..
470 Different velocities in a cross-section.
471 Mean velocity of running water..
472-474 Most advantageous profile..
475 Uniform motion of water.
476 Coefficients of friction.
•
477-478 Variable motion of water
479 Floods and freshets...
955
956
957
959
965
966
969
973
CHAPTER VIII.
HYDROMETRY, OR THE THEORY OF MEASURING WATER.
481-483 Regulators of efflux...
480 Gauging, or the measurement of water in vessels..
484 Prony's method..
·
485 Water inch.....
976
977
982
983
486 Methods of causing a constant efflux.
985
487 Hydrometric goblet.
986
•
488 Floating bodies..
989
489 Determination of the velocity and of the cross-section.
490-491 Woltman's mill or tachometer...
990
992
998
492 Pitot's tube...
493 Hydrometrical pendulum.
999
1001
494 Rheometer
CHAPTER IX.
OF THE IMPULSE AND RESISTANCE OF FLUIDS.
495-496 Reaction of water.
497 Impulse and resistance of water.
•
1002
1006
TABLE OF CONTENTS.
xxix
§ 498-500 Impact of an isolated stream
501 Impact of a bounded stream.
502 Oblique impact.
503 Impact of water in water.
•
504-505 Experiments with reaction wheels.
506 Water-meters.
507-508 Gas-meters
• •
509 Action of unlimited fluids.
510 Theory of impact and resistance..
511 Impulse and resistance against surfaces.
512 Impulse and resistance against bodies.
513 Motion in resisting media.
•
514 Projectiles...
PAGE
1006
1011
11012
. 1014
€ 1015
1020
1023
1029
1030
1031
1033
1035
1038
APPENDIX.
THE THEORY OF OSCILLATION.
1- 2 Theory of Oscillation.
3-4 Longitudinal vibrations
5 Transverse vibrations
•
6 Vibrations due to torsion.
7 Density of the earth...
8- 9 Magnetism.
•
1042
1045
1048
1050
1051
1053
10 Oscillations of a magnetic needle.
1055
•
11-13 Law of magnetic attraction...
1056
13 Determination of the magnetism of the earth.
1059
14-15 Wave motion..
1061
16 Velocity of propagation of waves.
1064
體
​17 Period of a vibration ...
1067
18 Determination of the modulus of elasticity.
1069
19 Transverse vibrations of a string
1070
20-21 Transverse vibrations of a rod.
22 Resistances to vibration..
23 Oscillation of water
❤
24 Elliptical oscillations.
25-28 Water waves.
Translator's Appendix.
Index..
1072
1077
1079
1081
1084
1092
1105
THEORETICAL MECHANICS.
INTRODUCTION
TO
THE CALCULUS.
ø
ART. 1. The dependence of a quantity y upon another quan-
tity is expressed by a mathematical formula: E.G., y = 3x², or
y = a x",
a x", etc. We write y = f (x) or z = (y) etc., and we call y a
function of x, and z a function of y. The symbols ƒ and o, etc., in-
dicate in general that y is dependent upon 2, or z upon y, but leave
the dependence of these quantities upon one another entirely un-
determined, and do not give the algebraical operation by which y
can be deduced from x, or z from y. A function y = f(x) is an
indeterminate equation; it gives an unlimited number of values for x
and y, which correspond to it. If one of them (r) is given, the other
(y) is determined by the function, and if one of them is changed, the
other also undergoes a change. Therefore the indeterminate quan-
tities x and y are called VARIABLES, or variable quantities; and the
quantities which are given, or are to be regarded as given, and in-
dicate the operation by which y is to be deduced from a, are called
CONSTANTS, or constant quantities. That one of the variables
which can be chosen at pleasure is called the independent variable,
and the other, which is determined by means of a given operation
from the first, is called the dependent variable. In y=a", a and
m are constants, a is the independent and y the dependent va-
riable.
The dependence of z upon two other quantities, x and y, is ex-
34
[ART. 2.
INTRODUCTION TO THE CALCULUS.
pressed by the equation z=ƒ (x, y). In this case z is at the same
time a function of x and y, and we have here two independent
variables.
ART. 2. Every dependence of a quantity y upon another quan-
tity x, expressed by a function or formula y = ƒ (x) can be repre-
sented by means of a curve, A P Q, Fig. 1 and Fig. 2.

FIG. 1.
P
R
FIG. 2.

P
R
T
X
A
X
A
M N
T
M N
The different values of the independent variable x answer to the
abscissas A M, A N, etc., and the different values of the dependent
variable to the ordinates M P, N Q, etc., of the curve.
The co-or-
dinates (abscissas and ordinates) represent then the two variables
of the function.
The graphic representation of a function, or the referring of the
same to a curve, presents several advantages. It furnishes us in
the first place with a general view of the connexion between the
two variable quantities; secondly, it replaces a table or summary of
every two values of the function belonging together; and thirdly, it
affords us a knowledge of the different properties and relations of the
function. If with the radius CA CB=r we describe a circle
ADB (Fig. 3), corresponding to the function y 2 r x-x²
where x and y indicate the cc-ordinates A M,
MP, this curve affords us not only a general
view of the different values that the function
can assume, but also makes us acquainted
with other peculiarities of this function, for
the properties of the circle have also their
meaning in the function. We know, E.G.,
without farther research, that y becomes equal
to zero, not only when x = 0 but also when
P
FIG. 3.
D

A
M
C
B
x= 2 r, and that y is a maximum and =r when x = r.
ART. 3.]
35
INTRODUCTION TO THE CALCULUS.
ART. 3. The Laws of Nature can generally be expressed by
functions between two or more quantities, and are therefore in
most cases capable of a graphic representation.
FIG. 4.
3
v
(1) When a body falls freely in vacuo, we have for the ve-
locity y, which corresponds to the height of fall x, y = 2gx,
but this formula corresponds to the equation y = Vpx of the para-
bola, when the parameter (p) of the latter
is made equal to the double acceleration
(2 g) of gravity. We can therefore repre-
2 sent graphically the laws of the free fall
of a body by the parabola A P Q (Fig. 4),
whose parameter p = 2 g.
The abscissas
AM, AN, of this curve are the space
traversed by the body in its fall, and the

2
P
1 M 4
N
9
ordinates MP, and NQ, the corresponding velocities.
(2) If a is a certain volume of air under the pressure of one
atmosphere, we have according to Marriotte's Law, the volume of
the same mass of air under a pressure of x atmospheres, y
a
and we have, for x=1, y = a; for x = 2, y for x = 4, y
2'
مح
a
4'
α
for x=10, y=
10; for x=100, y=
a
100'
for x=∞,y=0.
We see in this manner that the volume becomes smaller as the ten-
sion becomes greater, and that if the law of Marriotte were correct
for all tensions an infinitely great tension would correspond to an
infinitely small volume.
Further, for x = 1, we have y= 2a; for x=1, we have y=4a;
y=10a; “x=0,
y = ∞ a ;
X = 109
CC
so that the smaller the tension, the greater the volume becomes;
and if the tension is infinitely small the volume is infinitely
great.
The curve which corresponds to this law is drawn in Fig. 5.
A M, A N, are the tensions or abscissas x, MP, NQ, the corre-
sponding volumes or ordinates y. We see that this curve ap-
proaches gradually the axes A X and A Y without ever reaching
them.
(3) The dependence of the expansive force of saturated steam
36
[ART. 3.
INTRODUCTION TO TIIE CALCULUS.
upon its temperature x can be expressed, at least within certain
limits, by the formula
and by experiment we
and m-6. If we put
y = (a + x)
(out)
Y
have within certain limits a = 75, b = 175,
4
2
FIG. 5.
P
1
1
3
Y
- (75 +α)
Y
FIG. 6.

15

A
S
ՄՐ
P
1
1 M
10
2 N 3
-X
4
-75
A
M 100
N
200
for x =
100°, y
x =
50°, y
Y
and assume the formula to be correct without limit, we obtain
1,000 atmosphere,
0,133
175.
125
C
175
75
x =
O", y =
0,006
175
66
X
-75°, Y
(15)*
0,000
CC
X =
X =
120°, y = (195) = 1,914
G
150°, y = (225) = 4,517
9
x = 200°, y = (275) °
66
=
15,058
CC
PQ, Fig. 6, presents to the eye the corresponding curve. It
75 from the origin of co-ordinates
passes at a distance A 0 =
ART. 4.]
37
INTRODUCTION TO THE CALCULUS.
A through the axis of abscissas and at a distance AS 0,000
cuts the axis of ordinates; an abscissa A M< 100 corresponds to
an ordinate M P < 1, and an abscissa A N > 100 belongs to an
ordinate NQ> 1; and we can also see that not only y augments
as x increases to infinity, but also that the curve becomes steeper
and steeper as x becomes greater.
ART. 4. A function z = f (xy) with two independent varia-
bles can be represented by means of a curved surface B C' D, Fig.
7, in which the independent variables x and y are given by the
abscissas AM and A N on the axes AX and Y, and the de-
pendent variable z by the ordinate OP of a point P in the surface
ABC. If for a definite value of x we give different values to y, the
values of z deduced furnish us with the ordinates of the points of a
curve EPF parallel to the co-ordinate plane YZ; if on the contra-
ry for a given value of y we take different values of x, we determine
the ordinates z of the points of a curve GP H parallel to the co-or-
dinate plane X Z. We can consequently consider the whole curved
surface B CD as the union of a series of curves parallel to the co-or-
a (1+dy)
dinate planes. The law of Marriotte and Gay-Lussac z = X
by means of which we can calculate the volume z of a mass of air
from the pressure and the temperature y, is graphically repre-
sented by the curved surface CKP H, Fig. 8. A Mis the pres-
H
Z
FIG. 7.
A
B
M
X
Ο
T
E
N
Y
FIG. 8.
Z


H
B
K
M
sure x, A N or MO the temperature y, and O P the correspond-
ing volume z: the co-ordinates of the curve P G H give the vol-
umes for a temperature A N=y, and those of the right line KP
the volumes for the same pressure A M = x.
38
[ART. 5.
INTRODUCTION TO THE CALCULUS.
ART. 5. When we increase the independent variable of a func-
tion or the abscissa A M= x (Fig. 9 and Fig. 10) of the correspond-
ing curve an infinitely small quantity M N, which we will in future
designate by d x, the corresponding dependent variable or ordinate
MP = y becomes NQy', being increased by an infinitely small
quantity RQNQ- MP, to be designated by dy. Both these
increments dx and dy of x and y are called the Differentials of the
Variables or Co-ordinates x and y, and our principal problem now
"is to determine for the functions that most commonly occur the
differentials, or rather the ratio of the differentials of the varia-
bles x and y belonging together. If in the function y = f (x),
where x represents the abscissa A M, and y the ordinate MP, we
substitute, instead of x, x + dx = AM + MN AN, we obtain,
instead of y, y + dy = MP + RQNQ; therefore
y+dy = f (x +dx),
≈
and subtracting the first value of y from it, the differential of the
variable y remains, i. e.
dy
FIG. 9.
dƒ (x) = ƒ (x + d x) − ƒ (x)
FIG. 10.


P
R
P
R
T
A
M N
X A
T
M N
Y
This is the general rule for the determination of the differential of a
function, which when applied to different functions furnishes sev-
eral rules more or less general: E.G., if y
x², we have
d y = (x + dx)' — x²
(x + d x)²
x² + 2 x d x + d x²
d
Y
= 2 x d x + d x²
(2 x + d x) d x ;
and more simply since dx, being infinitely small compared to 2 x,
disappears, or since 2 x is not sensibly changed by the addition
of d x, and the latter can therefore be disregarded,
d y = d (x)² = 2 x d x.
ART. 6.]
39
INTRODUCTION TO THE CALCULUS.
N
D
A
FIG. 11.
PQ
The formula y=x corresponds to the contents of a square,
ABCD, Fig. 11, whose side is A BAD = x,
and we see from the figure that, by the addition to
the side of BM=D N = d x, the square is in-
creased by two rectangles B O and D P = 2 x d x,
and by a square (d x²), so that by an infinitely
small increase dx of x the square y x² is in-
creased by the differential quantity 2 x d x.

C
B M
ART. 6. The right line, TP Q, Fig. 9 and
Fig. 10, passing through two points P and Q of
the curve, which are at an infinitely small distance from each other,
is called the Tangent to this curve, and determines the direction
of the curve between these two points. The direction of the tan-
gent is given by the angle P T Ma at which the axes of abscis-
sas AX is cut by the line. When the curve is concave,`as A P Q,
Fig. 9, the tangent lies beyond the curve and the axis of abscissas;
but when it is convex, as A P Q, Fig. 10, the line lies between the
curve and the axis of abscissas.
In the infinitely small right-angled triangle PQR (Fig. 9 and
Fig. 10), with the base PR dx, and the altitude R Q = dy, the
angle QPR is equal to the tangential angle PTM a, and we
have
whence
tang. QPR =
Q R
PR'
dy
tang. a =
d x
therefore the ratio or quotient of the two differentials dy and dr
gives the trigonometrical tangent of the tangential angle; E.G., for
the parabola whose equation is y²=px we have, putting y²=px=z,
dz = (y+dy)² — y² = y² + 2y dy + dy — y² = 2y dy + dy',
or as dy vanishes before 2 y dy, or what is the same thing, dy
before 2 y,
and also
dz = 2ydy,
d z = p(x + dx) −— px,
therefore 2 y dy = pdx, whence for the tangential angle of the
parabola we have
dy
p
y'
Y
tang. a =
d x
2y
QxY
2 x
40
[ART. 7.
INTRODUCTION TO THE CALCULUS.
The definite portion P T of the tangent between the point of
tangency P and the point T where it cuts the axes of abscissas
FIG. 12.
FIG. 13.


Ꭲ
A
P
R
P
R
M N
-X A-
T
M N
X
is generally called the Tangent, and the projection T M of the
same upon the axes of abscissas the Sub-tangent; hence we have,
PM cot. P TM
subtang.
d x
= y cot. a =
У
d y
2x
Y
2 x.
E.G., for the parabola, subiang.=y
The subtangent is therefore equal to the double abscissa, and
from it the position of the tangent for any point P of the para-
bola is easily found.
For the curved surface B C D, Fig. 7, the angles of inclina-
tion a and ẞ of the tangents P T and PU at a point P are
determined by the formulas:
d z
tang. a =
tang. B=
d x
d z
dy
The plane P T U passing through P T and PU is the tan-
gent-plane of the curved surface.
ART. 7. For a function y = a + mf (x) we have
i. c.
I.)
•
mf
dy= [a+mf (x + dx)] - [a + mƒ (x)];
=α
a
a + mf (x + dx) — mf (x
= m[f(x + dx) — ƒ (x)];
d [a + mf (x)] = mdf (x),
E.G., d (5 + 3x²) = 3 [(x + dx)' x'] = 3. 2x dx = 6 x dx.
In like manner:
—
d (4 − 1 x³), = − 1 d (x)³ = − ¦ [(x + d x)³ — x³]
2
= √(x² + 3x dx + 3x dx² + d x² - x³)
= 1.3x² d x = x² dx.
ART. 7.]
41
INTRODUCTION TO THE CALCULUS.
Hence we can establish the following important rule: The con-
stant member (a, 5) of a function disappears by differentiation, and
the constant factors remain unchanged.
The correctness of this rule can be graphically represented.
For the curve A P Q, Fig. 14, whose co-ordinates in one case are
FIG. 14.
FIG. 15.

P
R
M
A
N
ΑΞ
Μι N₁
Q1

S
P
R1
R
N
M
·
A M = x and MP = y = f (x), and in the other A, M₁ = x and
M₁ P = a + y = a + f(x), we have P Rdx and R Q = d y =
df (x) and also = d (a + y) = d [a + f (x)]; and for the curves
A P, Q, and. A P Q, Fig. 15, whose corresponding ordinates MP,
and MP as well as N Q, and N Q have a certain relation to one
another, the relation between the differentials R, Q₁ = N Q₁—
MP, and RQNQ-MP is the same; for if we put MP₁ = m.
MP and N Q₁ = m. NQ, it follows that R, Q₁ = NQ,
(NQ - MP) = m. Q R.
i.c.
If y
1
d [m f (x)] = mdf (x).
Q1
1
MP₁ = m.
= u + v, or the sum of two variables u and v, we have
dy= u + du + v + d v ~ (u +), i. e., according to Art. 5.
II.)
d (uv) = du + dv, and in like manner,
df
d [ƒ (x) + ¢ (x)] = dƒ (x) + dọ (x).
The differential of the sum of several functions is then equal
to the sum of the differentials of the separate function; E.G.
d (2x + 3x² - 1x) = 2 dx+6xdx-xdx=(2+6x-3x²) d x.
The correctness of this formula can also be made evident by the
consideration of the curve A PQ, Fig. 15. If M P = f (x) and
PP₁ = (x) we have
MP₁ = y=f(x) + (x) and
¢
dy= R₁ Q₁ = R, S + SQ₁ = R Q + SQ₁ = df (x) + d p (x);
42
[ART. 8.
INTRODUCTION TO THE CALCULUS.
1
for P₁ S can be drawn parallel to P Q, and therefore we can put
R, SRQ and QS P P₁.
= 1.
ART. 8. If y = u v or the product of two variables, E.G. the
contents of the rectangle ABCD, Fig. 16, with the variable sides
AB: = u and B C v, we have
dy = (u + du) (v +
ud v + v du +
A
FIG. 16.
C
P
B M
d v) u v = u v + u dv+vdu + du dv—uv,
du dv=ud v + (v + d v) d u.
But in v + d v, dv is infinitely small com-
pared to v, and we can put
v + d v = v, and (v + d v) du = vdu,
and also
u d v + (v + d v) duud v + vdu,

so that
III.) .
it follows therefore that
d (uv) = ud v + vdu,
d[f(x). (x)] = f (x) d p (x) + (x) df (x).
The differential of the product of two variables is then equal to
the sum of the products of each variable by the differential of the
other.
=
When the sides of the rectangle ABCD are increased by
B M = du and D 0 = d v its contents y = A B × A D — u v is aug-
mented by the rectangles CO udv and CM v du and CP
= du dv, the latter, being infinitely small, compared with the oth-
ers, disappears; the differential of this surface is only equal to the
sum udv+vdu of the contents of the two rectangles CO and C M.
In conformity with this rule we have for y = x (3 x² + 1):
dy=xd (3x² + 1) + (3x² + 1) dx = 3 x d (x²) + (3x² + 1) dx
3 x.2 x d x + 3 x² d x + d x (9 x² + 1) ḍ x.
Further, if w be a third variable factor, we have
d (u vw) = ud (vw) + v w du,
or since d (vw) = vd w + wd v,
d (u v w) ⇒u v d w+u w d v + v w du, and in like manner
d (u v w z) = u v w dz+u v z dw+u w zd v + v w z du;
if w=v=w=z, it follows that d (u')=4 u³ d u, and in general
IV.). d (x")=m x”—¹ dx, if m is a positive integer, E.G.
d (x)=7x° dx, d3x-6x dx.
4
If y = x¯”, m being again a positive integer, we have also
xm
yx
1 and d (y x™)
=
yd (x)+x" dy = 0, and therefore
dy
y d (x²)
=
20
= 0, i. e.
x˜” m x¹-¹ dx
-1
1
M X
dx,
X"
ART. 8.]
43
INTRODUCTION TO THE CALCULUS.
or, if we put
m = n,
d (x") = n x-1dx.
The Rule IV. applies also to powers, whose exponents are neg
ative whole numbers, as E.G.,
3 d x
d (x³)=-3x dx = -
and
¹
2
-3
d (3 x² + 1)−² = − 2 (3 x² + 1) −³ d (3 x²) =
m m
If in y = x
n
• N
1 2 x d x
(3x+1)*
is a fraction whose denominator n and whose
numerator m are integers, we have also y"=" and d (y")=d (x"), I.E.,
1
n y¹→¹ dy = mxm-¹ dx, therefore
M X
dy
ጎ
m-1 d x
yr-1
1
m xπ-¹ d x M
M
x¯¯`dx.
m
n
n
xm- R
If we put
M
N
p, it follows that
-1
dy= d (x²) = p x²-¹ dx, which agrees with Rule IV., which
can now be considered as general.
Also d (ur) p up-du, when u denotes any function de-
pendent upon x.
Hence we have, E.G., d ( √ x³)=d (x²)=3x dx=3 √x dx,
d √2rx-x=d√ u=d (u) = } u
du
= 1 d (2 r x − x²)
2 rdx-2xd x
(r-x) dx
ບ
2 N u
√2rx-x²
И
we put u =
In order to find the differential of a quotient y =
vy, whence du = vdy + yd v, and
=
d u
du-y d v
V
dy
V.) a (1) =
v du-udv
v2
According to this Rule, E.G.,
บ
И
V
d v
I.E.,
X 1 (x + 2) d (x² - 1) — (x²-1) d (x + 2)
d
x+2
(x + 2)²
(x² + 4x + 1 ) d x
dx.
_ (x + 2). 2 x d x − (x² - 1). d x
(x + 2)²
(x + 2)²
44
[ART. 9.
INTRODUCTION TO THE CALCULUS.
We have also:
d
2'
a ( C ) = — a d¹³ ; E. c, d ( )
4
4 d (x²)
8 d x
E.G.,
X³
1
2
2
ααν
v²
important in the
When we give the
1
±1
ART. 9. The function y = 2" is the most
whole analysis, for we meet it in all researches.
exponent n all possible values, positive and negative, whole and
fractional, etc., it furnishes the different kinds of curves, which are
represented in Fig. 17. A is here the point of origin of the co-ordi-
nates, XY the axis of abscissas, and that of the ordinates.
If on both sides of the co-ordinate axes at the distances x =
and y = ± 1 from the point A we draw the parallels X₁ X1, X2 X2,
Y₁ Y₁, F₂ Y₂ to the axes, and join the points P₁, P2, P3, and P₁,
where they cut each other, by means of the diagonals Z Z, Z, Z₁, we
obtain a diagram which contains all the curves, given by the equa-
tion y=x". For every point on the axis of abscissas XX we have
Y 0, and for every point on the axis of ordinates II, x =
and for the points in the axes X, X and Y, X, y=1, and
for the points in the axes F, F, and F, Y,
If in the equation y = ∞" we put x = 1, we obtain for all
possible values of n, y = 1, and for certain values of n, also
y= 1; consequently all the curves belonging to the equa-
tion y pass through the point P₁, whose co-ordinates are
A M1 and A N = 1. If we take n 1 we have y = and we
obtain the right line Z A Z, which is equally inclined to the two
axes XX and Y Y, and which rises on one side of 1 at an angle
1
2
±1.
0 ;
of 45°), and on the other side dips at the same angle. On the
contrary, for y=x we obtain the right line Z, 4 Z, which dips
on one side of A at an angle of 45°, and rises on the other side at
the same angle.
x>
If, however, n > 1, Y x" becomes smaller for x < 1, and for
> 1 greater, than 2, and when n < 1, y=2" is greater for x < 1
and smaller for ≈ 1 than x. The first case (n > 1) corresponds
to convex curves, which run in the beginning under, and from P,
over the right line (Z A Z), and the second case (n < 1) to concave
curves, where the reverse takes place.
When, in the first case, we take n smaller and smaller until at
last it disappears, or becomes equal to zero, the ordinates approach
ART. 9.]
45
INTRODUCTION TO THE CALCULUS.
1
the constant value y = 2º = 1 and the corresponding curve ap-
proaches more and more to the broken line A N P₁ X₁; if, on the
contrary, in the second case, n becomes greater and greater, the
values of the ordinates approach the limit У = x² = x² = ∞,
X
∞, and
FIG. 17.

1
IV
2
Y₂ -2
Y
1
N
X₁
P
1
1
2 Y 3 2
لذات
Ꮓ
T
04/09 110-150
1
X
X
M
IM
_2
_X2
P
<-18
X 9
1
18
[[]
-2
3 Y₂
_Y
Y
clai
Z1
II
those of the abscissas, on the contrary, approach the value x=y°=1,
and the corresponding curve approximates more and more to the
broken line A M P₁ Y₁.
If we take n=
<= ∞,
1
1
x²
1, whence y = x¹= for x = 0, we have y
∞, and for x = ∞, Y
discussed in Art. 3, and drawn in Fig. 5 (1 Pī); it approaches
on one side the axes of ordinates, and on the other the axes of ab-
scissas without ever reaching them.
O and we obtain curve, which has been
If the exponent (-n) of the function y = x
11
1
222 |
is a proper
46
[ART. 9.
INTRODUCTION TO THE CALCULUS.
A
1
x²
x
fraction, for ≈ < 1, we have y < and on the contrary for ≈ >1,
x
:
1
y >-, and if this exponent is greater than unity, we have on the con-
x
trary for < 1, y
> and for x > 1, y < 1. The curve corre-
1
x²
X
sponding to y = x¯”, according as n is greater or smaller than unity,
runs in the beginning below or above, and from P, above or below,
the curve y = x
While those curves, which correspond
1
X
to the positive values of n, are placed in the beginning below,
and from P, on above, the right line X, X, the curves of the nega-
1
tive exponents (-n) run first above, and from P, on below, X₁₁.
For the former curves we have, for y 0, x = 0, and for x =
Y
=
co,
0,
0, y =∞, and for x =
more and more from the co-or-
∞, and for the latter, for x
y 0. While the former diverge
dinate axes I and II, the farther we follow them from the
origin A₁, the latter approach more and more on one side the axis
IX, and on the other axis IF, without ever reaching them.
If in y
X
3
1
1
The last system of curves approach nearer and nearer the
broken line FN P, X, or the broken line Y, P, M X as the expo-
nent approaches nearer and nearer the limit n = 0 or n = ∞.
x±m,m (1,3,5,
m is an entire uneven number (1, 3, 5, 7 . . .), y
and a have the same sign. Positive values of ≈ correspond to positive
values of y, and negative values of x to negative values of y. If on
the contrary m is an entire even number (2, 4, 6, etc.), y becomes
positive for all values of x, positive or negative. Therefore the
curves in the first case, as E.G., (3 P₁ A P; 3) or (1 P, 1, 1 P, 1),
run on one side of the axis of ordinates above, and on the other side
below, the axis of abscissas X ; on the contrary the curves in
the second case, as E.G., (2 P₁ A P₁ 2) or (2 P, 2, 2 P, 2), are placed
above the axis of abscissas only, and are contained in the first and
fourth quadrants; the former corresponds for m = ±∞ to the limit-
ing lines Y, M A M₁ Y, and X M Y₁, X M₁ F, the latter on the
contrary to the limiting lines Y, MA M, Y, and Y MY
XM Y
1
If we have y
x have the same
positive value of
X
1
1
4
1
n being an entire uneven number, y and
signs, and if n is an entire even number, every
gives two equal values for y, one of which
ART. 10.]
47
INTRODUCTION TO THE CALCULUS.
3
1
3
is positive and the other negative, and on the contrary for every
negative value of x, y is imaginary or impossible. The curves,
as E.G. ( P, A P ), which correspond to the first case, are found
only.in the first and third quadrants, and the curves of the second
case, as E.G. ( P₁ A P. 1), only in the first and second quad-
rants: the former become form the limiting lines X, N
A N₁ X, and X₁ N Y, X, N, Y, and the latter the limiting lines
X₁ NA N₁ X and X, N Y, X½ N₁ Ÿ.
1
1
Since y = x
I
n
2
2
involves x = y", it follows, that the latter sys-
}
tem of curves | y = x
24
:)
differs from the former (y =
x) in its
position only, and that by causing them to revolve, the curves of
one system may be made to coincide with those of the other.
Since y = x
m
1
(x4)* = (2x)=
1
(x")" we can always give from what
has gone before the general course of a curve. E.G., the curve
for
У
x3 = (x} ) 2
3
(√)
has, for both positive and negative values of x, positive ordinates;
on the contrary, the curve for
3
y = x² = (x!)³ = (√ √ x)°
W
has, for positive values of ≈ only, real ordinates, and they are equal
in magnitude, but with opposite signs. Further, for the curve
5
=
Y = x3
3
(Vx)
y and ≈ have the same sign, since neither the fifth root nor the
cube causes a change of sign.
Finally, the curves, which correspond to the equation y
Th
M
-x", differ from those of the equation y=x" only by their reversed
position in regard to the axis of abscissas II, and they form the
symmetrical halves of a complete curve.
ART. 10. From the important formula d (x") = n x”-' d x we
obtain the formula for the tangential angle of the corresponding
curves represented in Fig. 18. It is
tang. a =
dy
d x
= n x²,
48
[ART. 10.
INTRODUCTION TO THE CALCULUS.
i
and therefore we have the subtangent of these curves
d x
xn
X
= y
Y d y
n x²-
n
Hence, for the so-called parabola of Neil, the equation of which
X
is
a y²
=
x²³ or y =
a,
we have
tang. a =
1 d(x)
1
Na d x
Na
α
mint
32
and the subtangent
Farther, for the curve already discussed y =
x.
يم
α
a²
18
a² x-¹,
tang. a = a²
d (x−¹)
d x
a²
x²
(金​)
818
X
and the subtangent
X.
(See Fig. 5.)
-1
IV
2
Y₂ -2
_Z₁
233
FIG. 18.
1
2
-
1
IN
Xi
P₁
-2
X
M
_1
P
_X
-X2
N
1100
2
_Z
III
3 Y₂ -1 -Y 1
2
со
IM
3

2
Z
1
alma H
ని/లు.
X,
112
-X
2
-X
2
Z₁
12
II
ART. 11.]
49
INTRODUCTION TO THE CALCULUS.
Consequently, we have for x = 0, tang. a =
∞ and a =
90°,
1 and a 135°
for x = a, tang. a =
and for x = ∞, tang. a = 0 and a =
0°, etc.
ART. 11. When a right line A 0, Fig. 19, cuts the axis of ab-
scissas at an angle 0 A X = a, and is at a distance C K = n from
the origin of co-ordinates C, the equation between the co-ordinates
NP x and C N = M P
y of a point in the same is
y cos. a x sin. a = n, since n = MR M L, M R = y cos. a
and M L = x sin. a.
C M
Պ
For x =
0, y becomes C B = b =
therefore we have n=
>
cos. a
b cos. α,
and
y cos. a
x sin. a = b. cos. a or
y = b b + x tang. a.
Generally the lines CA and C B, which measure the distances
FIG. 19.
Y
P
from the points where the line cuts
the co-ordinate axes CX and CY
to the origin of co-ordinates, are
called the parameters of the line,
and are designated by the letters a
and b. According to the figure
a, therefore

X
N
R
B
Y
K
CA =
b
L
1
C B
Ъ
A la
tang. a =
X
M
CA
α
the straight line becomes
b
Y
=
b
a
and consequently the equation of
x, or + = 1. (See Ingenieur, page 164.)
X y
a b
When a curve approaches more and more a line, which is sit-
uated at a finite distance from the origin of co-ordinates, without
ever attaining it, the line is called the ASYMPTOTE OF THE
CURVE.
The asymptote can be considered as the tangent to a point
of the curve situated at an infinite distance. Its angle of inclina-
tion to the axis of abscissas can be determined by
d y
tang, a =
d x²
and its distance n from the origin of co-ordinates by the equation
y cos. a x sin. a = (y — x tang. a) cos. a
N
༩/
x tang. a
√1 + (tang. a)²
=(y⋅
d y
X
:
1 +
d x
(dy)
50
[ART. 12.
INTRODUCTION TO THE CALCULUS.
as well as by the formula n=(y cotg. a—x) sin. a =
=(3
d x
d y
X
):
'd x
1 +
dy
(d)
when we substitute x and y = ∞ in them.
y cotg. a
VI+(cotg. a)
X
In order that a tangent to a point infinitely distant can be an
asymptote, it is necessary, that for x or y = ∞, y — x tang. a or
Y cos, a ≈ shall not become infinitely great.
—
For a curve whose equation is y = x¯
1
= X" =
Xm
tang. a =
M
xn+1
and y
x tang. a = xπ +
m
хт
m + 1
хт
and also y cotg. a — X =
X
772
х
X =
(m + 1)
therefore
>
M
=
0, tang. a = 0, y
0,
1) for x = ∞, y
x, tang. a = 0 and n
and 2) for y = ∞, x = 0, tang. a = ∞, y cotg. x = 0 and n = 0.
The axis of abscissas
S
corresponds to the conditions tang. a
∞ and n = 0, the axis of ordinates Y Y to the conditions
tang. a = 0 and n = 0; therefore these axes are the asymptotes
of the curve, corresponding to the equation y=x". (Compare
the curves 1 P, 1, 2 P, 2, and P, in Fig. 18, page 48.)
ན
1
1
ART. 12. The equation of an ellipse A D A₁ D₁, Fig. 20, can
be deduced from the equation
FIG. 20.
Y
B
DI
P
D
B.
L
KM A T
X
x² + y² 1
a²
a
of the circle A B A, B₁, whose ra-
dius is CA – C B = C P
and whose co-ordinates are CM
=x and M P = y₁, when we
consider, that the ordinate MQ
=y of the ellipse is to the ordi-
nate M P = y₁ of the circle, as the
lesser semi-axis C D b of the el-
lipse is to the greater semi-axis,
which is equal to the radius of the
circle C B = a. We have then

a²
1
y
b
a
Y₁
α
a²
whence y₁ =
b
y and x² + y²
a³, I.E.
b2
ART. 12.]
51
INTRODUCTION TO THE CALCULUS.
x²
y
+
a²
1, the equation of the ellipse.
If we substitute in this equation for + b², — b², we obtain the
equation
x³
y²
1,
a²
b2
which is that of the hyperbola formed by the two branches PA Q
and P₁ A, Q1, Fig. 21.
When in the formula
b
У
√ x² - α²
a
deduced from the latter equation we take x infinitely great, a' dis-
appears before x², and we have
b
У
a
b x
a
√ x² = ± = ±x tang. a,
the equation of two right lines CU and CV passing through the
origin of co-ordinates C.
Since the ordinates
b
b
b
2
±
α
X
V x² and
Vx²
a²
a
a
FIG. 21.
U

-V
Y
Ni
P
B
B
iP
K
E
E
K
M
M.
-X
-X-
D₁
D
-Ÿ
tend to become equal as x becomes greater, it follows that the right
lines CU and CV are the asymptotes of the Hyperbola.
If we take C Aa, the perpendicular A B = + b and
AD =
b, we can determine the two asymptotes; for the tan-
a, formed by the asymptotes with the axis of
gent of the angle
abscissas, is
tang. A CB
=
A B
CA'
b
I.E. tang, a =
and
a
in like manner
A D
b
tang. A C D =
CA'
I.E. tang. (— a)
α
If we take the asymptotes U U and VV as axes of co-ordi-
52
[ART. 12.
INTRODUCTION TO THE CALCULUS.
nates, and put the abscissa or co-ordinate C N in the direction of
the one axis = u, and the ordinate or co-ordinate NP in the di-
rection of the other v, we have, since the direction of u varies
from the axis of abscissas by the angle a, and that of v by the
angle a
CM = x = C N cos. a +
MP = y = C N sin. a
NP cos. a
(u + v) cos. a, and
NP sin. a
(u — v) sin. a.
If we designate the hypothenuse C B =
we have
and consequently
e
√ a² + b² by e,
α
b
cos.α =
and sin. a =
е
cos. a
sin. a
1
and
a
b
>
е
xx
y²
(u³ + 2 u v + v²)
( u²
· 2 u v + v²)
cos.² a
sin.' a
a²
b²
a²
b2
u² + 2 u v + v²
ге
Q u v + v²
4 u v
= 1.
e²²
e²
e²
From the latter we obtain what is known as the equation of the
hyperbola referred to its asymptotes
U v =
€³
4
e²
or v=
4 u
According to this it is easy to draw the hyperbola between the
two given asymptotes.
The co-ordinates of the vertex A are CEE A
e
and
FIG. 22,
U

Y
Ni
P
B₁
K
E1
E
K
M
A
M
-X
-X-
-Y
the co-ordinates for the point K are C B = e and B K =
ther, for the abscissas 2 e, 3 e, 4 e, etc., the ordinates are
С
118 etc.
4'
; fur-
4
e
e
, 4
ART. 13.]
53
INTRODUCTION TO THE CALCULUS.
dy
i x
ART. 13. If in the ratio of the differentials or in the for-
mula for the tangent tang. a of the tangential angle, we substitute
successively the different values of x, we obtain all the different po-
sitions of the tangent to the corresponding curve. If we take ≈ =: 0,
we obtain the tangent of the tangential angle at the origin of co-
ordinates, and if on the contrary we take x = co, we have the same
for a point infinitely distant. The most important points are those
where the tangent to the curve runs parallel to one or other of the
co-ordinate axes, because here one or other of the co-ordinates ≈ and
y have their greatest or smallest value, or, as we say, is a maximum
or minimum. When the curve is parallel to the axis of abscissas we
have a = 0, and tang. a = 0; when parallel to the axis of ordinates
a = 90°, or tang. a = ∞, whence we deduce the following Rule:
To find the values of the abscissa or independent variable x,
which correspond to the maximum or mini-
mum value of the ordinate or dependent va-
riable y, we must put the ratio of the differ-
and resolve the result-
FIG. 23.
P
entials
d Y
d x
0, or
ing equation in regard to x; E.G., for the
equation y 6 x
x²+x, which corre-
sponds to the curve A P Q R in Fig. 23.
-X
A
M
N
d y
d x
=
3 (1-x) (2
x) (2 − x);
consequently, in placing
d
y
d x
69x+3x² = 3 (2-3x + x^) =
= 0, we have
1
X
x = 0 and 2
x = 0,
I.E. x 1 and x 2.
Substituting these values in the formula
Y 6 x § x² + x³,
we have the maximum value of y, M P = 6 —
the minimum value, N Q = 12
18+ 8 = 2.
3 + 1
5, and
Farther, for the curve KO P Q R, Fig. 24, whose equation is
1)², we have
Y = x +
2/
d
d y
d x
૨૭
=
tang. a = 1 + 2/3 (x
-
1)=1+
3 V
X
1
૭
which becomes
0, for
1, I.E. for A M = x = 1
3 V x − 1
(3)'= 19 = 0.7037, and on the contrary ∞, for 4 N = x = 1.
7
!

54
[ART. 14.
INTRODUCTION TO THE CALCULUS.
The first case corresponds to the maximum value,
3
MP = Ym = 1 − ()' + (3)² = 24
-
읽
​1.148,
and the last to the minimum value, N Qy, = 1.
FIG. 24.
R
Y
Max.
Р
Min.
K
X
- 2 _1
A MN
+2
n
We have also A 0=y
=1 for x- 0, and y =0
for the abscissa A K = x,
corresponding to the cubic
equation x + x² - 2 x + 1,
2³
whose value is x = - 2.148.
ART. 14. Since in the
equation of a curve which
starts from the origin of
co-ordinates A, and rises
above the axis of abscissas,
y increases with x, dy is
always positive, and since

when the curve on the contrary descends towards that axis, y de-
creases when x increases, dy becomes negative. Finally at the
point where the curve runs parallel to the co-ordinate axis A X,
dy becomes equal to zero, and the differentials of the ordinates,
corresponding to the equal differentials dx = M N N O P S
=
QT of the abscissas, are
S Q = P S tang. Q P S, I.E., d Yi = d
TR= QT tang. R Q T, I.E., d y
dx tang. a₁,
d x tang. a,, etc.
The tangential angles a1, a2, etc., also increase for a convex
curve A P R, Fig. 25, and decrease for a concave curve A P R,
FIG. 25.
R
P
S
FIG 26.
K
T
P
S
T
FIG. 27.



R
T
P
S
X
X
A
M NO
X
A
两个
​O
N
A
M
NO
Fig. 26; consequently in the first case
d (tang. a)
=
d
X
() is positive,
and in the second d (tang. a)
= d
(12) is negative, and for the
points of inflexion Q, Fig. 27, I.E. for the places where the con-
ART. 14.]
55
INTRODUCTION TO THE CALCULUS.
dy'
(2/2)
= 0.
vexity changes into concavity, or where the contrary takes place,
we have SQ TR, and therefore d (tang. a) = d
Hence we have the following Rule:
If the differential of the tangential angle is positive, the curve is
convex, if it is negative, the curve is concave, and if it is equal to
zero we have a point of inflexion of the curve to deal with. From
the foregoing we can easily make the following deductions:
The place, where the curve runs parallel with the axis of abscis-
sas and for which tang. a = 0, corresponds either to a minimum or
to a maximum, or to a point of inflexion of the curve, according as
the curve is convex, concave, or neither, I.E., as d (tang. a) is pos-
itive, negative, or equal to zero. On the contrary, the point, where
the curve runs parallel with the axis of ordinates and for which we
have tang. a∞, corresponds to a minimum, or maximum, or to a
point of inflexion of the curve, according as the latter is concave,
convex, or in part concave, or in part convex: I.E., as d (tang. a)
is negative or positive on each side of this point, or has a different
sign on different sides of it.
=
A portion of a curve with a point of inflexion of the first kind
is shown in Fig. 28, and a curve with one of the second kind in
Fig. 29. We perceive that the corresponding ordinate NQ is nei-
ther a maximum nor a minimum, for in this case both of the
neighboring ordinates MP and OR are larger or smaller than
N Q. In Geometry, Physics, Mechanics, etc., the determination
FIG. 28.
Y
R
P Q
FIG. 29
FIG. 30.
R
A
P
X
A
M NO
A M NO
of the maximum and minimum, or the so-called eminent, values of a
function, is often of the greatest importance. Since in the course.
of this work various determinations of such values of functions will
be met with, we will here treat only the following geometrical
problem.
To determine the dimensions of a circular cylinder 4 N, Fig.
30, which for a given contents V has the smallest surface, let us
:


56
[ART. 14.
INTRODUCTION TO THE CALCULUS.
designate the diameter of the base of the cylinder by x and the
height of the same by y; here we have
V
π
√ =
x² y
4
and the surface or the area of the two bases plus that of the
curved portion
2 π x²
4
+ пху,
but from the first equation we have
4 V
пу
ог π x y = 4 √ x−¹
x²
substituting this value of π x y, we obtain
O
π x²
2
+ 4 √x²,
and since we can treat O and x as the co-ordinates of a curve, we have
tang. a =
d O
d x
π X 4 V x ².
Putting this quotient equal to zero, we obtain the equation of con-
dition
π X =
4 V
ха
oг π x³ = 4 V.
Resolving the equation in reference to x, we have
3
4 V
x
and
>
π
3
y =
4 V
π x²
64 V³
T
π-3
π²
16 V¹
3
4 V
= X.
π
8 V
Since d (tang. a) = (· π + ³T)
da is positive, the value found
furnishes the required minimum. We can employ the same
method when we wish to determine the dimensions of a cylindri-
cal vessel which for a given contents will need the smallest amount
of material. They are already determined directly when the vessel
besides its circular bottom is to have a circular cover, but when
the latter is not needed we have
0:
0 =
Пх
2
x =
π X²
+4 Va, consequently
4
4 V
,
V
2
π
whence it follows that
and y =
3
173 π
3/ V
V
1 x.
3
π
π
19-
ART. 15.]
57
INTRODUCTION TO THE CALCULUS.
While in the first case we must make the height equal to the
width of the cylinder, in the second we must make it but one-half
the width of the latter.
ART. 15. By successive differentiations of a function y = f(x),
we obtain a whole series of new functions of the independent va-
riable x, which are
dy
d f (x)
f(x) =
d x
d x
d f₁ (x)
ƒ₂ (x) =
d x
E.G., for
f2
fa (x) = d f₂ (x), etc..
y = f(x) = x, we have
d x
10
§
f(x) = x³, f: (x) = ¹º xs, ƒ, (x) = — 19 x, etc.
For a function which is developed according to a series of the
ascending powers of a
y = f (x)
=
1
A。 + Á₁ x ÷ Á½ x² + ¸ ó + Á₁ x² + etc., we have
f(x) = A₁ + 2 A₂ x + 3 A3 x² + 4 A¸ x³ + etc.
f(x)= 2 A +2.3 4x + 3.4 A, x² + etc.
?
B
4
f(x) = 2.3 4 + 2.3.4 44 x + etc.
Substituting in these series x = 0 we obtain a series of expres-
sions suitable for the determination of the constants A, A1, A2…… viz.
ƒ (0) A。, fi (0) = 1 A₁, f: (0) = 2 A, fs (0) 2.3. A3,
etc., whence we deduce these co-efficients themselves.
=
=
0
A。ƒ (0), A, = f (0), A, = ƒ½ (0), A₂ =
½
3
=
1
2 . 3 fs (0),
1
A₁
2.3.4.f (0) etc.
Thus we can develop a function into the following series, known
as McLaurin's.
ƒ (x) = ƒ (0) + ƒ₁ (0)
fi
X
·
+ f (0)
1
x²
1.2
+ fs (0) ·
x³
1.2.3
+ ƒ₁ (0) •
+
1.2.3.4
For the binomial function y = ƒ (x)
· ƒ (x) = (1 + x)" we have
f(x) = n(1 + x)"¹, ƒ₂ (x) = n (n − 1) (1 + x)n−²
₤3 (x) = n (n - 1) (n-2) (1 + x), etc.
When we put x = 0, we obtain
ƒ (0) = 1,f,(0) = n, f₂ (0) = n (n − 1)
fs (0) = n (n − 1) (n 2), etc., whence the binomial series.
58
[ART. 15.
INTRODUCTION TO THE CALCULUS.
N
I.) (1 + x)" = 1 +
x +
1
n (n−1)
1.2
x² +
n (n - 1) (n-2)
1.2.3
x³ + etc.
We have also
(1 − x)” — 1
-
as well as
N
1.
n (n
x +
1)
x²
1
1.2
n (n − 1) (n − 2)
1.2.3
x³ + etc.,
n
(1 + x)−” = 1
n
x +
n (n + 1)
1.2
x²
n (n + 1) (n + 2)
x³ + ...
1.2.3
1
Farther, putting 1 + x = (1
(1 − z)−¹
(1 + x)” − (1 − z)—” −1+nz+
we have z =
1 — z
X
1+x
and
n (n + 1)
z² +
1.2
n(n + 1)(n+2)
1.2.3
23... I.E.
n
X
II.) (1 + x)" = 1 +
n (n + 1)
X
1
1 + x
1.2
1 + x
X
+
+ ..
1.2.3
1 + x
n (n + 1) (n + 2)
The series I. is finite for entire positive values of n, and the
series II. for entire negative values of the same.
E.G., (1 + x)³ 1 + 5 x + 10 x² + 10 x³ + 5 x¹ + x³, and
(1 + x)−³ — 1 — 5
Since a + x = a
ت)
8!+
1 + x
+10
a)
+
(1 + 2),
it
an
(a + x) = a² (1 +
+
n
10 (1
х
+ X
х
Ang
х
10 (
1 +
C
XC
5 (₁ = ) - ( - ).
(1
+
1 +
follows also that
2) = œ [1 +1()
" a"
x
n (n - 1) (2) '+...] L.B.
1.2
a
I.E.
a
n
n (n
∙1)
III.) (a + x)" = a” +
x)”
an-1 x +
1
1.2
n (n
1) (n
2)
+
1.2.3
an
1
a²-³ x³ +...
1
ART. 15.]
59
INTRODUCTION TO THE CALCULUS.
E.G.,
=
3
√ 1009² =
(1000+9)= 100 (1+ 0,009)}
100 (1 + 3.0,009 +
(
= 100 (1 + 0,006
We have also
2
1)
3
(0,009)² + ...
2
)
0,000009) 100,5991.
n (n
1)
(x + 1)" = x² + n x²-1 +
1.2
and approximately for very great values of x,
(x + 1)” = x² + N x²-¹.
x²-² + ...etc.
From this it follows that
(x + 1)"
Xn
xr-1 =
further
and finally
(x − 1)^-¹
(x-2)"-1
n
Xr
(x
A g
(x-3)n-1
=
(x
n
1)"
―
N
1)"
(x
2)"
(x — 2)" — (x − 3)"
27 - 1"
1- =
;
ጎ
22
adding the two members of these equations together, we have
1
x²-1 + (x − 1)”—¹ + (x − 2)”−¹ + (x − 3)¹¹ +
(x + 1)" - 1"
ጎ
+ 1
or, putting n 1 m, and writing the series in the reversed
order, we have
1″ + 2″ + 3m + ... + (x
1m 2m
(x + 1)m+1 - 1
1) " + 2 =
m + 1
Now since x is very great, or properly infinitely great, we can
put (x + 1)+1 2m+¹, and we then obtain the sum of the powers
of the natural series of numbers.
IV.)
1m + 2m + 3” +
+ 20m
m + 1'
E.G.,
V1 +
1º + √ ½³ + √ 3ª +
√ 42
3
4³ + . . . + 1000 approximately
1000
01/09
✓ 1000°
1000° 60000.+
10/09
534
60
[ART. 16.
INTRODUCTION TO THE CALCULUS.
ART. 16. The ordinate O Py, Fig. 31, corresponding to
the abscissa A O = x, can be considered as composed of an infinite
number of unequal elements dy, as
which cor-
B
G
FIG. 31.
D
TH
E
K
H
A
FL M M N
1
P
of
F B, G C, HD, KE... .,
respond to the equal differentials dx =
A F, = F L L M = M N
the abscissa. If therefore d y = ¢ (x).
dx were given, we could determine y
by summing all the values of d y, which
we obtain, by substituting successively
in o (x) d x for x, d x, 2 d x, 3 d x ...
O to n d x = x. This summing is indi-
cated by the so-called sign of integra-
tion, which is placed before the general expression of the differ-
ential to be summed. Thus we write, instead of
y = [p (dx) + p (2 d x) + 4 (3 d x) + . . . + ¢ (x)] d x,
y = So (x) d x.
In this case we call y the integral of p (x) d x, and ø (x) d x the
differential of y. Sometimes we can obtain the integral / (x) d x,
by really summing up the series ø (d x), 4 (2 d x), 4 (3 d x), etc.;
but it is always simpler in the determination of an integral to em-
ploy one of the Rules of what is known as the Integral Calculus,
which will be the next subject treated.
If is the number of differentials d x of x, we have x = n d x

X
or d x = and we can put
12
N
/ ¢ (x) d x =
Φ
a x = [ † ( * ) + •
= +
(² ²)
2 x
+Ф
22
(3,₂ a)
+...+ $
N
(9)]%2
X
n°
Y =
Thus for the differential d y = a x d x, we have
•
Sax d x = a dx (dx + 2 d x + 3 d x + • + n d x)
(1 + 2 + 3 + + n) a d x²,
or since according to Art. 15, IV., for n = ∞ we have the sum of
the natural series of numbers
x²
1 + 2 + 3 + 4 + 5 ... + N. —
n² and d x²
2
N
x²
y = faxdx = { n²
a
1/2 α x².
12
22
In a similar way we find, if x =
elements d x,
n
dx
d x or if x is composed of n
„²x² dx =
+(nda)'] dæ
y=S ¢(x) dx= ƒx² dx = [ (dx)² + (2 dx)² + (3 dx)² + ….. +(ndx)
α
= (1 + 2² + 3² + .... + n²) d x²
α
a
x
ART. 17.]
61
INTRODUCTION TO THE CALCULUS.
But from § 15, IV., for n = ∞, we have
1 + 2² + 3² + .... + n²
по
3
3'
whence it follows that
Sx d
x² d x
a
n³ d x³
3
( n d x)³
X³
α
За
3 a
ART. 17. From the formula d (a + m ƒ (x)) = m d f (x), we
obtain by inversion
Smd f (x) = a + m f
(x) = a + mf d f (x), or putting
df (x) =
S ¢ (x) d x,
(x). d x
I.)
S m p (x) d x = a + m
and hence it follows that the constant factor m remains, in the In-
tegration as in the Differentiation, unchanged, and that a constant
member such as a can not be determined by mere integration;
the integration furnishes only an indefinite integral.
In order to find the constant member, a pair of corresponding
values of x and y = p(x) d x must be known. If for x = c, y=k,
and we have found y S ¢ (x) d x = a + f (x) then we must
also have k = a + ƒ (c), and by subtraction we obtain y
f - k
ƒ (x) − ƒ (c); therefore in this case we have
y = £ $ (x) d x = k + f (x) − ƒ (c) = ƒ (x) + k − ƒ (c),
· S
and the constant factor a k — ƒ (c).
When, E.G., we know that the indefinite integral y = x d x =
gives, for a 1, y
x²
x =
2
3
14
تن
==
3 we have the necessary constant a =
, and therefore the integral
y = S x d x = a +
x²
2
5 + x2
2
Even the determination of the constant leaves the integral still
indefinite, for we can assume any value for the independent varia-
ble x; but if we wish to have the definite value k, of the integral
corresponding to the definite value c, of x, we must substitute this
value in the integral which we have found, or, h₁=k+f (c₁)—f (c).
5 + x²
2
E.G., Y = S x d x =
gives, for a 5, y = 15.
x =
=
Generally the value of x for which y becomes 0 is known;
in this case we have k = 0, and the indefinite integral of the form
62
[ART. 18.
INTRODUCTION TO THE CALCULUS.
f
√ 4 x (d x) = ƒ (x) leads to the definite one k₁ = ƒ (c₁) — ƒ (c),
which can also be found by substituting in the expression ƒ (x)
of the indefinite integral the two given limits, c, and c, of x, and
by subtracting the values found from one another. In order to
1
indicate this we write instead of ƒ ☀ (x) d x, ƒˆ✨ (x) d x,
1
if, E.G., √ 4 d x = 22, *' p (x) d x =
/
S'o
By the inversion of the differential formula
2
C₁ c²
2
d [f(x) + (x)] = d f (x) + d (x) we obtain the integral
$
formula [d f (x) + d p (x)] = ƒ (x) + (x), or putting
/
df (x) = (x) d x and do (x) = x (x) d x,
4
II.)
4
S 4
L [4 (x) d x + x (x) d x] = £ ¥ (x) d x + S x (x) d x.
Therefore the integral of the sum of several differentials is equal
to the sum of the integrals of each of the differentials.
E.G. (35 x) d x = 3 d x + 5 x d x = 3 x + § x².
S
S S
ART. IS. The most important differential formula, IV., Art. 8,
d (x') = . n x²-1dx, gives by inversion an integral formula which
is equally important.
-1
It is / n x¹ d x = x², or n S x”-¹ d x = X", whence
Xh
ƒ x²-¹ d x =
;
N
"2
=
substituting 1 m, and n = m + 1, we obtain the following
important integral:
Sx d x =
Xm+1
m + 1'
which is employed at least as often as all the other formulas
together.
The form of this integral shows that it corresponds to the sys-
tem of curves treated in Art. 9 and represented in Fig. 17.
From it we have, E.G.,
/ 5 x³ d x = 5 x d x = § x*;
/
3
V x¹ d x = S x³ d x = x² = 3
3
37
Vx;
(4-6x+5x¹) d x = f4dx-6x dx + 5 x d x
= 4 dx 6 / x d x + 5 x dx=4x-223+25; farther,
/ − ƒ
S
d u
putting 3 x 2 = u,
3 d x = du, or d x =
3
>
we have
ART. 19.]
63
INTRODUCTION TO THE CALCULUS.
I'v
2.dx=
d u
3
V x
/ √3 x = 2. d x = fu² = "" = } √ @ = ; √ (3 z − 2) ' ;
and finally, substituting 2 x²
138
1 = u and 4 x d x = du or
d u
x d x =
we have
,
4
S
5 x d x
3
可
​ل
2 x²
1
5
5
uૐ
= ƒ³ du = √ u d u = 4 2 ²
3
4 V u
2
Su³du
15
8
3
³√ u² =
15 3
3/
(2 x² — 1)².
By the substitution of the limits the indefinite integral can be
changed into a definite one.
E.G.
=
5 x³ d x = § (2' 1') 5. (161) 183.
525
1
v d x
So z N x
→6
2
—
= √ √4 = 1
[ "$ √ 3 x − 2 . d x = {} ( √ 16³ -- √T³) = § (64 − 1) = 14.
If E.G. (4 6 x² + 5 x¹) d x = 7, for x = 0 we would have,
–
in general,
S (4-6 x² + 5 x') d x = 7 + 4 x − 2 x³ + 25.
-
ART. 19. The so-called exponential function y a, which
consists of a power with a variable exponent, can be developed as
follows into a series by means of McLaurin's Theorem, and its dif-
ferential can then be found.
Putting a² = A。 + A₁ x + А₂ x² + A3 x³ +
1
we have, for
we have
x = 0, a² a" =
་་་
1, whence A。
1;
½ ó +
From a = 1 + A₁ x + ½ x² +
1
a¹² = A¸
1 + A₁ d x + A‚ d x²
= artis
d (a³)
= a² (A₁ d x + A₂ d x²
= aª (Â₁ + A, d x +
2
+ A 3 d x² + .... and also
a² = a* (ad≈
1)
+ A3 d x³ + . . . .)
•
.) d x = A₁ a² d x.
Hence, by successive differentiation of the series, we have
a+
f(x) = α = 1 + A, x + A₂ x² + A3 x² + ...
£i (x) = d(aº)
d x
= А A₁ α* =
1
A₁ + 2 A½ x + 3 A3 x² +...
64
[ART. 19
INTRODUCTION TO THE CALCULUS.
fr (x)
₤3 (x)
d (A₁ a*)
d x
d (A,' a*)
2
A‚² a² = 2 A₂ + 2 . 3 . A¸¤ + ..
3
=
Putting x
=
1
2
d x
0, it follows that
A₁ = A, 2 A, A,, 2.3. A₂ = A‚ª
=
1
A‚³ a² = 2.3 . A½ +
1
whence A2
1
1.2
2
A₁², A₁ =
3
1
1.2.3
1
3
19
A₁³, A4 =
A₁', &c.
1.2.3.4
and the exponential series takes the form
I.
a² = 1 + 4,₁
X
1
2
x²
1.2
+ A₁³
3
X3
1.2.3
+ A₁₁
X¹
1.2.3.4
+
1
The constant coefficient A, is of course a definite function of
the constant base, as the latter is a function of the former. If one
of the two numbers be given, the other is then determined. The
most simple, or the so-called natural series of powers, whose base
(a) will be designated hereafter by e, is obtained by putting A, = 1.
Then we have,
II.) e² = 1 + +
x²
X³
X¹
+
1.2 1.2.3 1.2.3.4
+
+
X
1
1
and if we put x 1 we obtain the base of the natural series of
powers,
e¹ − e − 1 + 1 + ¦¿ + d + a + ...
= =
1
24
= 2,7182828.
1
la, which is the Nape-
If we put e = a”, or a
a", or a = cm
em, we have
m
rian or hyperbolic Logarithm of a, and
1
1
III.) a* =
(em)
3
еть
1 +
1
()
1
+
1.2
(黒​)
81
+
1
1.2.3
3 m
+..
Since this series corresponds in its form to that of I, we have
1
also A₁
and.
M
IV.) d (a) = A, a dx =
V.) d (e) e' d x.
E.G.
a² d x
la. a* dx, as well as
M
d (e³*+1) == e³* + d (3 x + 1) = 3 e³+1 d x.
ART. 20.]
INTRODUCTION TO THE CALCULUS.
65
If we put y
The number m
z
a* = e™ we have, on the contrary,
X
X = log, y and = 1 y.
m
log y = m l y, and, on the contrary,
1
1 y, or log, y = loga Y.
M
is called the modulus of the system correspond-
ing to the base a. By means of it we can transform the Naperian
logarithm into any artificial one, or one of the latter into the
former. For Brigg's system of Logarithms the base is a =
1
m
whence = 7 10 = 2,30258, and, on the contrary, m ==
0,43429.
We have also log y = 0,43429 1 y, and
l y = 2,30258 log y.
(See Ingenieur, page 81, etc.)
1
7 10
10,
ART. 20. The course of the curves which correspond to the
exponential functions ye, and y = 10', is represented by Fig.
32. For x = 0, we have in both cases y eº = aº
a" = 1. Hence
both curves OQ S and O Q, S, pass through the same point (0)
of the axis of ordinates A For x = 1 we have,
Y.
1
2,718, and
y = e²
Y
10* =
10,
x = 2 gives
y = e²
e² = 2,718² = 7,389, and
y = 10² = 10º = 100, &c.
Both curves rise on the positive side of the axis of abscissas very
steeply, particularly the latter.
For x= - ex
1 we have e² = e−¹ =
e-
1
2,718 ...
0,368 .., and
10* = 10-¹
0,1;
farther, for x =
2, we have
1
ex
2,7182
0,135,
and
10 = 10-2 = 0,01;
for both equations give
x=
∞
5
66
[ART. 21.
INTRODUCTION TO THE CALCULUS.
1
1
0,
е
α
FIG. 32.
12
10
7,4
2,7
Y
S1
Q1
Ꭱ
R
S
The two curves approach
nearer and nearer this axis of
abscissas on the negative side
of the axis of abscissas, the
last more quickly than the first,
but they never really meet this
axis.
Since we deduce from the

equation
y = e², x = ly
and also from
y = a*, x =
loga Y
the abscissas of these curves
furnish a scale for the Nape-
rian and common logarithms;
for the abscissas are the loga-
rithms of the ordinates.
E.G. we have,
A M = 1 M P
=
log, M P₁, etc.
From the differential for-
mula IV of the last article the
tangential angle of the expo-
nential curve is determined by
the simple formula,
d y
tang. a =
d x
a² d x
m d x
-X
X
a²
2 -1 TAM 1
2
Y
= y
1 a.
m
M
Consequently for the curve O P Q S₁, Fig. 32, the subtan-
gent
=Y
y cotg. a = m, that is, is constant; and for the curve
O P Q S it is always = 1, E.G., for the point Q, 4 1, 1 for the
point R, 12 = 1, etc.
ART. 21. If x = a', we have also
=
•
dx = d (a")
a dy
m
and by inversion,
d y
m d x
αι
m d x
X
ART. 22.]
67
INTRODUCTION TO THE CALCULUS.
But y = log. x, that is, to the logarithm of the variable power
x with the constant base a; therefore we have the following differ-
ential formula for the logarithmic functions,
y =
log. x and y = 1x:
m d x
1 dx
I.) d (log, x)
X
l a X
d x
II.) d ( x)
X
If a is the tangential angle of the curve corresponding to the equa
M
log, x, we have tang. a = and the subtangent
tion y =
ху
X
= y
cotg. a = or proportional to the area x y of the rectangle con-
M
,
structed with the sides x and y.
By means of the differential formulas I. and II. we obtain
2
d V x
d (x)
x d x
d x
1) d (1x)
12
or also
=
√x
Xi
x}
2 x
dx
x
d (1 l x) = ¦ d (l x) = ! ·
2) d¹² + * = d [1 (2 + x) − 1x³]
dl
x²
[l
= dl (2 + x) – dl (x")
d x
2 + x
d x
2
X
(4 + x) d x
x (2 + x) *
d
3) a (10² = 1)
— d [? (e² − 1)]
− d [¹ (e² + 1)]
+
d (e)
=
ex - 1
d (e²)
e² + 1
c² d x
e* — 1
e² d x
e² + 1
2 e² d x
1
ART. 22. If we reverse the differential formulas of the fore-
going article, we obtain the following important integral formulas.
a² d x
From d (a)
it follows that a d
d x
= α², I.E.,
M
M
I.)
a²
S a* d x = m a² = a² : la, and therefore
II.)
S
ƒ e² d x = e².
Farther, from d (log, x)=
m d x
it follows that
X
Smd 2
x
= log. x, I.R
68
[ART. 23.
INTRODUCTION TO THE CALCULUS.
III.)
fdx
1
m
d x
log。 x = 1x, which is also given by the for-
mula d ( x) =
X
By their aid we can easily calculate the following examples:
Se³z¹ d x = { c³¹ d (5 x − 1) = } e¹¹.
e5x-1
√
3 d x
7x+2
+
pd (7 x + 2)
7 x + 2
= 3 1 (7 x + 2).
S ( ± 1 ) d x = √(x + 1 + = 21)dx
d (x
X
= Sx dx + S d x + 2
Sa
1)
X · 1
a
x.
x²
2
+ x + 21 (x − 1).
xm+1
leaves
xm d x =
ART. 23. The first integral formula Sa
the last integral undetermined; for putting m =
Sdx = Sx¹ax=
that fa
if we put x = 1 + u, and d x = d u, we have
d x
х
d
d u
1 + и
S
=
d u
m + 1
1, it follows
u + u² — u³ + U"
- u³ + u• — …….. ) du; and therefore
хо
+ a constant = ∞ + constant, but
0
-
(1 − u + u²
= √(1
S' (1 − u + u²
u³ + u¹ — ...
.) du
S u³ du +
sa x = √ √ + u
X
1
S du
Sudu + Su² d u
3
U² W
= И
+
too
2
3
4
u²
we can therefore also put 7 (1+u) = u
+
2
3
И
+ . . ., or
4
IV.) 1x = (x − 1)
(x − 1)²
2
(x
1)³
(x
1)*
+
+ .
3
4
With the aid of this series we can calculate the logarithm of all
numbers which differ very little from 1; but if we require the
logarithm of large numbers we must adopt the following method.
Taking u negative in the foregoing formula, we have
262
28
7 (1 − u)
И
2
3
U:
4
and subtracting one series from the other, we have
ART. 23.]
69
INTRODUCTION TO THE CALCULUS.
7 (1 + u) — 7 (1 − u) = 2 (u +
+
1 (1 ± 2) = 2 (u
-
1 +
3
3
=
+
X 1
-
U³
5
из
+...)
+ +
3 5
+ ...) or putting
we have
1 + u
x', or u =
1.
и
x + 1
V.)
1 x =
= 2 [
x — 1
X
X
1\5
+
+ 1/1/13
+
X +1
x + 1
x +
This formula is to be employed for the determination of the
logarithm of such numbers as differ sensibly from 1, since
X
1
x + 1
is always less than 1.
We have also 7 (x + y) − 1 x = 1
=
Y
X
Y
+
3
(1)
etc.
+
y + 3 ( 2
x
Y
+
y) + +
2
[2x² + y
2 [ 2 x
[2
.. Y
VI) 1 (x + y) = 2x +
Y
+
(+2) = 1 (1+2)
)+...] whence
(
Y
2 x +
Y
5
+ 1 ( 2 2 + y) + ...])
x
This formula is used to calculate from one logarithm, that
of a somewhat greater number
E.G.,
-
12 = 2 [ 2 + 1 + 1 - ( + 1
) + ...]
2 1/
= 2 (1 + 1·27 +213 + ...)
5
11
0,33333
0,01234
2
— 2 . 0,34656 = 0,69312,
0,00082
0,00007
more exactly = 0,69314718.
=
Hence 1872 3122,0794415, and according to the last
formula, 7 10 = 7 (8 + 2)
=
= 18 + 2 [16
2
[ 16 2 + 2 + 3 ( 16 24+
+ 2
= 2,0794415 +0,2231436 = 2,302585.
70
* [ART. 24.
INTRODUCTION TO THE CALCULUS.
1
1 S
We can also put 12 = 11 + 2 [ 2 + 1 + 1 (2 + 1) + ....)]
?
2
(3 +
1
•
1
+ 3/3 + 3/1/10 +
33
36
farther, 7 5 = 7 (4 + 1) = 2 1 2 + 2 ( § +
we can put 7 10 = 72 + 75.
(Compare Art. 19.)
..
) = 0,693147;
1
+
93
...),
), and finally
ART. 24. The trigonometrical and circular functions, whose
differentials will now be determined, are of practical importance.
The function of the sine,y sin. x, gives for x = 0, y = 0;
π
for x=
4
3,1416
4
=
0,7854..., y = √1 = 0,7071,
π
Y =
66
x=
2
CC
X
3 π, y =
1, for x = π, Y = 0 ;
· 1, for a = 2, y = 0, etc.
Taking a as the abscissa A 0, and y as the corresponding ordi-
nate O P, we obtain the serpentine curve (A PBT C2 π), Fig. 33,
which continues to infinity on both sides of A.
FIG. 33.
Y K
E
M
G

+1
Q
IB
LA
1
П
T
IC
-Y
F
L
H
N
2 π
1
ART. 24.]
71
INTRODUCTION TO THE CALCULUS.
The function of the cosine, y = cos. x, gives, for x = 0, y = 1;
π
π
Y - 1; for
for x= 1,y= √1; for x= , y = 0; for x = π, y
4
ん
​x=3π, y = 0; for 2, y = 1, etc.; it corresponds to exactly
X
x =
the same serpentine line (+
זה
3 п
1 P D + 1) as the function of
2 2
the sine, but it is always a distance=1,5708 behind or in
front of the curve of the sine.
The curves, corresponding to the function of the tangent or co-
tangent,y=tang. a and y = cotang. a, are, however, of an entirely
different form.
x,
If we substitute in y = tang. x, x
tang. x, x = 0, π, π, we obtain y = 0,
1, ∞, and therefore a curve (A QE) which approaches more and
more, without ever attaining it, a line parallel to the axis of ordi-
nates A Y, and cutting the axis of abscissas A X at a distance
Π
2
T
2
2. π, § π,
2*
from the origin of the co-ordinates. Now if we put x =
we obtain y ∞, 0, +∞, and therefore a curve (Fπ G), which
continually approaches the parallel lines, passing through (7)
and
(3), and for which these parallel lines are asymptotes. (See
Art. 11.)
If we increase ≈ still more, the same values of y are repeated,
and therefore the function y tang. x corresponds to a series of
curves which are separated from each other in the direction of the
axis of abscissas by a distance = 3,1416. On the contrary, the
0,
function y = cot. x gives for x =
π π
4 2
, y = ∞, 1, 0, — ∞, and
therefore corresponds to a curve
tangential curve only by its position; it is also easy to perceive
that an infinite number of branches of the curve, as, E.G. (M
(A QTL) which differs from the
correspond to this function.
3 п
(µ³™ N)
While the curve of the Sine and Cosine forms a continuous,
unbroken whole, the curve of the Tangent as well as that of the
Cotangent is formed of separate branches; for the ordinates for
certain values of a change from positive to negative infinity, in
consequence of which the curve naturally loses its continuity.
72
[ART. 25.
INTRODUCTION TO THE CALCULUS.
ART. 25. The differentials of the trigonometrical lines or
functions are given by the consideration of Fig. 34, in which
CACPCQ 1, arc A P= x, P Q = d x,
=
P M = sin. x, C M = cos. x, A S = tang. x;
O Q = N Q MP = sın. (x + d x) sin. x = d sin. x,
OP= (CN-CM) cos. (x+dx) + cos. x =— d cos. x, and
STATA S = tang. (x + d x) - tang. x = d tang. x.
−
Since the elementary arc P Q is perpendicular to the radius
CP, and since the angle P C A between the two lines CP and
C' A is equal to the angle P Q O between the two perpendicular to
them, P Q and O Q, the triangles CP M and Q P O are similar,
and we have
O Q
P Q
C M
C' p'
I.E.
d sin. x
d x
COS. X
whence
,
1
I.)
d (sin. x)
= cos. x. d x, and in like manner,
OP
P Q
PM
CP'
I.E.
d cos. x
d x
sin. x
1
whence
"
II.) d (cos. x) =
FIG. 34.
R
T
S
C
A
N M
sin. x d x.
We see from this, that the influence of
errors in the arc or angle upon the sine
increases as cos. x becomes greater, or as
the arc or angle becomes smaller, while on
the contrary their influence upon the co-
sine increases as sine x becomes greater,
that is, the more the arc approaches to

π
2'
and that finally the differential of the co-
sine has the opposite sign from that of the
arc, for we know that an increase of x
causes a decrease of cos. x, and a decrease
of x an increase of cos. x.
Letting fall a perpendicular S R upon
CT we form a triangle SRT which is
similar to the triangle C P M, since the angle RTS is equal to
CQ N or C P M, and we have
ST
I.E.
SR
C M'
CP d tang. x
S R
1
but we have also
cos. X
SR
P Q
I.E. SR
CS
C P'
CS.dx
1
and
ART. 26.]
73
INTRODUCTION TO THE CALCULUS.
1
d x
CS secant. x =
whence S R
and
cos. x'
COS. X
III.) d (tang. x) =
d x
(cos. x)²°
π
If instead of x we substitute
x, and instead of d r,
2
d("
(5 — ~)
I
dx, we obtain
d x
d tang. ( − x)
IV.) d (cotang. x)
*
d x
(sin. x)²
[cas. ( - )]
I.E.,
By inversion this formula gives for the differential of the arc
d sin. x
d x =
COS. X
d sin. x
d cos. x
sin. x
=
(cos. x) d tang. x
(sin. x) d cotang. x, or
d tang. x
as well as
d cotang. x
dx =
√1 — (sin. x)”
1 + (tang. x)*'
d cos. x
d x =
√1 - (cos. x)
1 + (cotang. x)²°
If we designate sin. x by y, and x by sin.-' y, we have
V.)
d sin.¹ y =
dy
√ 1 — y²
and in the same manner we find
d y
VI.)
d cos.¹ y
√1 — y²
d y
VII.)
d tang.¹ y =
1 + y²
d y
1
+ y²°
VIII.) d cotang.¹ y =—
ART. 26. By inversion the latter differential formulæ give
S cos. x d x = sin. x,
f sin. x d x =
I.)
II.)
cos. x',
III.)
S
d x
cos." x
= tang. x,
IV.)
S
d x
sin.2 x
cotang. x.
* Sin.-¹y, tang.-¹ y, etc., designate the arc whose sine is y, whose tangent
is y, etc.-TR.
74
[ART. 26.
INTRODUCTION TO THE CALCULUS.
V.)
S
d x
-
sin.-¹ x =
1
COS.
x, and
1/1
X"
VI.)
S
d x
√1 + x²
tang.¹ x =
cotang.-¹ x.
From the above, since we have d (l sin. x):
cotg. x. d x, we can easily deduce
=
d sin. x
sin. x
cos. x. d x
sin. x
VII.)
S cotg. x d x = 1 sin. x, and also
VIII.) S tang. x d x =
l cos. x; further
d (l tang. x)
=
d tang, x
tang. x
d x
cos. x² tang. x
d x
d (2x)
sin. x cos. X
sin. 2 x
d x
whence
d (1 tang. ½ x)
and
sin. x'
I
d x
3C
IX.)
S
sin. x
d x
cos. x
= 1 tang⋅ 2
T
=2 trang. ( tế = cot (3)
+
X.)
1
a
Now pytting
1
x²
+
1 + x 1-
—
b
X
4
a (1 − x) + b (1 + x)
(1 + x) (1-x)
0, or x =
- X = = 0,
we have 1 - — a (1 − x) + b (1 + x), and taking 1 + x = = a
– 1, we obtain 1 = a(1 + 1) whence a = , and putting 1
or x = 1, we obtain 1 = 2 b or b =
π
- 1
1
2/
, whence
1.
1- x²
12
++;
+
1 + x 1.
; and finally
X
S=
d x
S
d x
1-x²
= √ √
d x
d x
1 + x
ι
+
/1
1
XL) S²² = 1 / (1 + 2)
XI.)
1.
x
2
x²
x
padding
XII) // d² = 1 / (+1)
S
- 1
x 1.
X
= 17 (1 + x) −7 (1-x), I.E.,
and in like manner
ART. 27.]
75
INTRODUCTION TO THE CALCULUS.
Putting 1 + x¨
dx (1-
xy, we have 1 + x² = x² y² and
y) = xydy, whence
d x
d y
+y`
dl
and
√1 + x²
1 — y²
d x
XIII.)
= 1 (x + √1 + x²), and also
√1 + x
d x
XIV.)
= 1 (x + √x² - 1).
1
= √ 1 + x²
d x
ART. 27. In order to find the integral of tang.-' x =
we have only to change
integrate each member.
I.)
• d x
1 + x
1
1 + x²
1
1 + x²
into a series, by division, and then
We obtain thus
· 1 — x² - x² -- x² + x²
-
and
= ƒ ã x − ƒ x a x + ƒ x à x −— f 'dx+….., consequently
π
4
T
π
6
=
tang.~¹1 = 1 − } + { → ¦ + }−
X:9
tang.'xx- +
X5
x²
…etc., E.G.,
3
5
7
…., and the half circumference
3
•
...],
√ √
189
4 (1 − } + ¦ − ¦ + ↓ − ……. ), or
= tang.~* √} = √} [1 − } · } + } (} )² − 4 (4)³ + ···],
whence 76 (1 − 1 + 15 - T8 + ...) = 3,1415926....
√1
45
In the same manner we obtain from
1
x
5
(1 − x²)¯ = 1 + x² + § x² + 18 2" + ...
∞²)¯} ¦ æ*
d z = = fax + 1 ƒ x² d x + } ƒ x* d x + v « fx d x + . . ., L.E..
S
1
II.)
sin.¯¹ x = x +
6
1 x³ 1.3x5 1.3.5 x
+
2.3 2.4.5
1
+
+
2.4.6.7
5
E.G., = sin. 11 = (1 + 24 +880 + 7188 + ...),
π
{
6
3
640
...,
!
76
[ART. 28.
INTRODUCTION TO THE CALCULUS.
1,04167
0,00469
π = 3.
=
3,1416 . .
0.00070
0,00012
1
3
When we put sin. x = A。 + A₁x + А₂x² + А¸Ñ³ + A4 x¹ +
etc., we obtain by successive differentiation
d (sin. x)
d x
d (cos. x)
d x
= cos. x = A₁ + 2 A¸x + 3 A3 x² + 4 Â 4x³ + ...
d (sin. x)
d x
d (cos. x)
d x
Now for c
sin.x = 2 A₂ + 2.3 A₂ x + 3. 4 Á₁x² + …….
3
cos. x = 2.3. A3 + 2.3.4. ¸¤ + ...
sin. x 2.3.4. A₁ + ...
O we have sin. x = 0, and cos. x = 1, therefore
we obtain from the first series A, 0, from the second A₁ = cos. 0
1
2.3'
from
1, from the third A₂ = = 0, from the fourth A3
the fifth A₁ = 0, etc. If we substitute these values in the supposed
series, we have the series of the sine
III.) sin. x =
+
X
X³
X5
1 1.2.3 1.2.3.4.5
1.2.3.4.5.6.
7
In the same way we obtain
x²
X¹
IV.)
cos. x = 1 -
+
+
1.
2
1.2.3.4
1.2.3.4.5.6
X³
2 ენ
17 x²
V.)
tang. x = x +
+
+
+
and ·
3
3.5 3.5.7.3
1
20
X³
2 x
VI.)
cotang.x =
etc.
X 3
3.5.3
3.5.7.9
(See Ingenieur, page 159.)
ART. 28. When we integrate the differential formula d (u v)
= u d v + v du, of Art. 8, we obtain the expression u v = u d v
+ S v du, and the following formula for integration:
S v du = u vud v, or
f
S p (x) d ƒ (x) = ¢ (x) ƒ (x) − Sƒ (x) d ø (x).
f
This is known as the integration by parts.
This rule is always employed if the integral
v du
ƒ ø (x) d ƒ (x) is not known, and if, on the contrary, S u d v=
ART. 28.]
INTRODUCTION TO THE CALCULUS.
ƒ ƒ (x) do x is. E.G. By means of this formula we can refer the
integration of the formula,
d y = √1 + x². d x
to another known integral. We must substitute
$ (x) = √1 + x², whence do (x)
x d x
=
√1 + x²
and f(x) = x, whence d f (x) = d x, then we have,
√1 + x² d x = x √1 + x² - √
1
S
x² d x
J
√1
1 +2.2-
x
√
but
x²
1 + x²
1
√1 + x²
√1 + x²
√1 + x²
√1 + x²
√1+229
whence it follows that
√
I
√1 + x² d x = x √ 1 + x²
√1
=
√
·SI+
√1 + x² d x +
S
√ √ Azz
d x
or
+ x²
2
S NI +
√ √1 + x² d x = x √1 + 2ª +
and consequently,
x²
√
Svar
d x
11 + x²²
d x
I.)
S
x²
[ √1 + x d x = 4 x √1 + 2 + 1 Svizz
/
+
x²
II.)
In like manner,
S
x²
½ [x √1 + x² + 7 ( x + √ 1 + x³)]·
d x
√ √ 1 − x² d x = 4 x √ I − ∞² + 1 S
x ²
√1 − x²
= { [x √1 − x² + sin.-¹ x], and
III.)
√ √ x − 1 d x = 4 x √ x − 1
V x² −
We have also
I S
d x
√x² - 1
−
¿ [x √ x² − 1 − 1 ( x + √x² − 1)].
L'(sin. x)' dx=fsin. x sin. x dx=-fsin. x d (cos.x)=-sin. x cos. x
+fcos. x d (sin. x)
sin. x cos. x +S (cos. x)² d x
= — sin. x cos. x +
[1 − (sin. x)'] d x,
whence it follows that
78
[ART. 29.
INTRODUCTION TO THE CALCULUS.
1
V.)
S (cos. x) d x =
2
VI.)
VII.)
IV.)
2 S (sin. x) d x = dx sin. x cos. x, and
I
1 1
ƒ (sin. x)' d x = (x − sin. x cos. x) = (x − sin. 2 x).
In like manner
1
(x + sin. x cos. x) = (x + ¦ sin. 2 x), and
1
Isin. x cos. x d x = sin. 2 x d (2x)=cos. 2 x,
1 S
Ï (tang. x) d x = tang. xx, and
VIII.)
/ (cotg. a) d x = −
x)²
(cotg. x + x).
IX.)
S
X.)
S
XI.)
fix. d
-S
Finally we have
ƒ x sin. x d x=−x cos. x + ƒ cos. x d x=
Sxed x = x d (e*) = x e* - Je* dx = (x − 1) e",
l x. d x = xl x
x cos. x+sin. x,
х
S xa x
d
x (1 x
1), and
X
XII.) Sxix. dx = 21
l
1
x
−
S
ƒ x d x = (1 x − )
x²
1))
2
2 x
ART. 29. If we wish to find the quadrature of a curve, A P B,
FIG. 35.
P
Fig. 35, 1.E., to determine or express by
a function of the abscissas o this curve
the area of the surface A B C, which
is enclosed by the curve A P B and
its co-ordinates AC and B C, we im-
agine this surface divided by an in-
finite number of ordinates M P, N Q,
etc., into elementary strips, like M N
P Q, with the constant width d x, and
the variable length M P = y..
y. Since
we can put the area of such an element of the surface

MN
C
d F
MP + NQ).
2
. M N = (y + i dy) d x = y d x
we will find the area of the entire surface by integrating the differ-
ential Y
d
X, and we have
F = Syd x;
E.G., for the parabola whose parameter is p we have y = px, and,
therefore, its surface
x V p. x
v
Z x
r = √ √ pxd x = √p f x d x = p = } x √ p z = } z y
F S
ART. 29.]
79
INTRODUCTION TO THE CALCULUS.
The surface of the parabola A B C is therefore two-thirds of
the rectangle A CBD which encloses it.
This formula holds good also for oblique co-ordinates inclined
at an angle X A Y = a, E.G., for the surface A B C, Fig. 36, we
have when we substitute instead of B C = y the normal distance
BN = y sin. a
F = sin. a
Syd x,
E.G., for the parabola when the axis of abscissas AX is a diameter,
and the axis of ordinates AY is tangent to the curve, we have
y²
p x
= P₁ X
P₁ x =
sin.2 a
(See "Ingenieur," page 177.)
and Fxy sin. a,
I.E., the surface A B C parallelogram A B C D.
FIG. 36.
Y
FIG. 37.


B
1
D
B
B
P
A
X
A
MN
C
N
C₁
For a surface B C C₁ B₁
=
F, between the abscissa AC₁
and AC c, Fig. 37, we obtain, according to Art. 17,
S
ว 1
√^¹
F = y d x
a²
E.G., for y =
X
I.E.,
F = √"
a² d x
(C1)
F = o² 1 (&
a²
X
= a² (l c₁ — 1 c),
The equation corresponds to the curve P Q, Fig. 38, dis-
X
cussed in Art. 3, and if we have A Mc and A N = c₁, the area
of the surface M N Q P is
80
[ART. 30.
INTRODUCTION TO THE CALCULUS.
FIG. 38.
Y
4
2
F = a² 1 (G
1
P
1
2
A
1 M
2 N 3
4
2
-X
If we suppose, for simplicity,
that a = c =
c = 1, and c₁ = x, we
obtain

F= 1x;
hence the surfaces (1 M P 1),
(1 N Q 1), etc., are the Naperian
logarithms of the abscissas AM,
A N, etc. The curve itself is the
so-called equilateral hyperbola in
which the two semi-axes a and b
are equal; hence the angle formed
=
by the asymptotes with the axes is a 45°; and the right lines
A X and A Y, which approach nearer and nearer the curve with-
out ever attaining it, are its asymptotes. In consequence of the
relation between the abscissas and the area of the surfaces, the
Naperian logarithms are often styled hyperbolic logarithms.
B
FIG. 39.
ART. 30. We can put every integral y d x = S ¢ (x) d x
equal to the area of a surface F, and if the inte-
gration cannot be effected by means of one of the
known rules, we can find it, at least approximately,
by calculating the area of the corresponding
surface by means of a well-known geometrical
device.

P
R
Y
2
A
N
M
If a surface ABPQN, Fig. 39, is deter-
mined by the base A Na, and by three equi-
distant ordinates A B Yo, M P = Y1, N Q
= 2, we have the area of the trapezoid
A BQ N = F₁ = (Yo + Y2)
and that of the segment B P Q S B, if we consider B P Q
to be a parabola
Yo
2
F, = ¦ P S. B R = } (M P — MS). A N = 2 (y, — % +) .r.
2
Hence the entire surface is
•
ART. 30.]
81
INTRODUCTION TO THE CALCULUS.
F = F, ÷ F₁ = [ 1 (Yo + Y2) + 3(Yı
[į
2
Yo + Y³)]x
2
X
6
= [¿ (Yo + Y₂) + 3 y₁] x = (Yo + 4 Y₁ + Y₂) · ☎·
If we introduce in the equation a mean ordinate
F= xy, we obtain
and put
Y
y=
Yo + 4 Y₁ + Y2
6
In order to find the area of a surface, lying above a given base
M N =x, and determined by an uneven number of ordinates
Yo, Y1, Y2, Y3 ... Yn, by which it is divided into an even number
of equally wide strips, we have only to make repeated application
of this rule. The width of a strip is and the area of the first
FIG. 40.
X
n'
pair of strips is
Yo + 4 Y₁ + Y2 2 x

A
Ꭹ
M
6
B
of the second pair
N
Y½ + 4 Ys + Y ₁ 2 x
6
N
of the third pair,
'N
Y + + 4 Y5 + Y6 Q x
و
etc.;
ጎ
6
and the area of the first six strips, or of the first three pair, fur
which n =
6, is
F= (Yo + 4 y₁ + 2 Y₂ + 4 Y3 + 2 Y1 + 4 Ys + Y6)
X
Yo ) 3 2 16
3.6
X
Y3
[Yo + Ye + 4 (Y1 + Ya + Ys) + 2 (Y₂ + Y+)] 18;
it is easy to perceive that the area of a surface divided in four pair
of strips is
F= [Yo + Ys + 4 (Y1 + Ys + Ys + Y1) + 2 (Y2 + Ys + Yo) ] 38
X
3.8'
and in general, for a surface divided in n strips, we have
こど
​F= [Yo+ Y₂+ 4 (Y₁ + Ys + ... + Yπ-1) + 2 (Y: + Y + + ... + Yn−2) ] 37 π
n
82
[ART. 30.
INTRODUCTION TO THE CALCULUS.
and the mean altitude of such a surface is
Yo + Yn + 4 (Y₁ + Y3 + ... + Yn-1) + 2 (Y½ + Ys + ... + Yn−2)
3 n
in which n must be an even number.
This formula, well known under the name of Simpson's Rule
(see "Ingenieur," page 190), can be employed for the determina-
1
tion of an integral y d x = ƒ`` (x) d x, if we divide x = a
—–
C
into an even number n of equal parts, and calculate the ordinates
Yo = (c), y₁ = &
¢ Yı $/c+
(c + 2), y₂ = 4
Y 2
2x
+
n
3
Y₂ = $(e +
³ (c + 3 x )
:)....
n
... up to y₁ = (x),
Уп
$
and then substitute these values in the formula
ƒ` y d x = ƒˆ' 4 ( x ) d x
[Yo + Yn+4 (Yı + Y8 + .. + Yn−1) + 2 (Y½ + Y4 + … + Yn−2) ]
C
3 n
S
22
x
1 X
E.G.,
when we assume n = 6 or
da gives, since here c₁-c=2—1=1 and y=4 (x) =
1
x²
X
N
C1 C
6
1
=
Yo
11
1
1,0000, yı
7
§ = 0,8571, Y2
6
1100130
== 0,7500,
1
1
6
Y3 =
§=0,6666, Y4=10
= 0,6000, s
=0,5454, and y=0,5000,
11
6
therefore
Yз Y5
3
Yo+y= 1,5000, y₁ + уs + уs = 2,0692, and y₂+ y₁ = 1,3500,
and we have the required integral
y½ Ys
d x
12,4768
=(1,5000+4. 2,0692+2. 1,3500). T'E
=0,69315.
18
I
a
From Art. 22, III, we have
1
So
² d x
= 7271 = 0,693147.
X
We see that the results of the two methods agree very well.
ART. 31.]
دن
INTRODUCTION TO THE CALCULUS.
CO
£3
FIG. 41.
S T
ART. 31. Further on, another rule will be given which can be
employed for an uneven number
of strips. If we treat a very flat
segment AM B, Fig 41, as a seg-
ment of a parabola, we have from
Art. 29 the area of the same,

M
A
B
CDE
FAB. MD,
or, if A T and B T are the tan-
gents at the ends A and B, and therefore C T = 2 C M, we have
A B.TE
2
F =
height, and therefore A C. CS3 AC tang. SA C.
The angle SA CSBC is TAC+TAS = T B C –
TB S; putting the small angles T A S and T B S, equal to each
other, we obtain for the same
of the isosceles triangle A S B of the same
TAS=TBS:
TBC-TA C
2
and
ТВС -ТАС
SAC TAC+
2
TAC+TB C
2
δτε
2
when we denote the tangential angles TA Cand TB C by d and ɛ.
Now since ACB C
AB the chords, we have
A
F = ↓ s² tang. (§ + €).
2
This formula can be employed for the portion of surface
MAB N, Fig. 42, whose tangential
angles T A D = a and T B E = 3
are given; putting the angle formed
by the chord B A D = A BK } – 0,
E
we have
T
FIG. 42.

G
S
E B
F
M
0
N
or, since a
S*
F
12
TAB=&= TAD BAD
TBA
= @
☛ and
¤ — A B E – TBE
G 6, whence
δ + επ a B,
and the segment over A B
F = ↓ s² tary. (°
- ß is small,
}
a
2
tang. (a — B) = 33 ( tang. a —
(
121+ tang. a tang. B
84
[ART. 31.
INTRODUCTION TO THE CALCULUS.
!
or since a and ẞ differ but little from each other, and therefore we
can substitute in tang. a tang. ẞ instead of a and ẞ the mean value
σ, we have
F= 128².
tang. atang. B
1 + tang. σ²
1/2 s² cos.² o (tang. a
tang. B),
and substituting for s cos. o the base M N = x,
ха
F =
12 (tang. a-tang.ẞ),
therefore the area of the entire portion of surface MA B N, when
y, and y, designate its ordinates MA and N B, is
x
F₁ = (Yo + y₁) — + (tang, a — tang. B)
2
x²
12
= y₁ and
If another portion of the surface N B C O adjoins the first and
has a base N 0 = x, and the ordinates BN and CO
Y2, and the tangential angles SBF ẞ and S C Gy, we have
for the area of the same
2.
X
2
F₂ = (Y₁ + Y₂) + (tang. ẞ — tang. y)
ß
x²
12'
and therefore for the whole surface, since-tang. ẞ cancels +tang. B,
x²
F= F₁ + F₂ = (½ Yo + Y₁ + 1 y2)x+(tang. a tang. v) 12.
For a surface composed of strips of like width we have, when a
is the tangential angle at the commencement and 8 at the end,
F = (} Yo + Y₁ + Y₂+ ¦ Y3) x + (tang. a tang. 8)
G
x²
12'
and in general for a portion of surface, determined by the abscissas
22
a 22
3 x
n'
N
N
x, and by the ordinates yo y₁, Y2... Yn, and by the
tangential angles a, and a, of the ends,
F = ( { Yo + Y₁ + Y2 + ... + Y₂-1 +
Yn−1 ½ Yn)
An Integral
+(lang.a-tang. a)
tang. a.)
(2)
ƒˆˆ y d x = ƒˆ' 4 ( x ) d x
f.” (x)
f
812
= (3 Yo + Y₁ + Y₂ + . . . + Yn−1 + ½ Y») 2/2
N
1
1
ΙΣ
+ I2 (tang. a -
tang. a₁) (2)
ART. 32.]
85
INTRODUCTION TO THE CALCULUS.
N
+ (c
can be found by putting x = cc, calculating the values
Yo = (c), y₁ = ¢
$ Yı
Ф
X
Y2
x
+ 2/2 ), y 2 = 4 (c + 2 2),
Y3
4 (c + 3.2) ..., 3₂
x
Yn
4 (c + 2) = ¢ (cs),
d y
as well as tang. a =
d x
Y = p(x) =
stituting them in the equation.
E.G., for
Yo
1
с
=
d x
=
(c) and tang.a,= 4 (c,), and sub-
Lod we have, if we take n = 6, since
1
x = c₁ — c = 2 − 1 and y = 4 (x) = 1,
1, y₁ =
also, since
tang. a =
x
Led r
d (x−¹)
6
6
6
6
;, Y2 = 8, Y3 = {, Y₁= 10, у5=11 and y=1;
1
1
1 + //
1
d y
d x
1
d x
11
1 and tang. B =
(1)²=
4.
1, and therefore
x29
6
= (1 + 9 + § + 9 + 1% + √1 + 1 ) · }+(−1 + 1). 12·36
4,1692
6
6
10
3
1
2. 72.38 0,69487
0,00173 0,69314.
(Compare the example of the last article.)
ART. 32. To rectify a curve, or from its equation y = f(x) be-
tween the co-ordinates A M = x and M P = y, Fig. 43, to deduce
an equation between the arc A P -s and one or other of the
co-ordinates, we determine the differential of the arc A P of the
curve, and then we seek its integral. If x be increased by a quan-
tity M N P R= d x, y is increased by R Q dy, and s by
FIG. 43.
P
R
T
X
A
M N
the element P Q = d s, and
according to the Theorem
of Pythagoras we have

I.E.,
P Q² = P R² + Q R²,
d s²= d x² + d y',
d s = √ d x² + d y²,
hence the arc of the curve
itself is
s =
S
= √ √
d x² + d y².
86
[ART. 32.
INTRODUCTION TO THE CALCULUS.
E.G., for Neil's parabola (see Art. 9, Fig. 17), whose equation is
a y=x³, we have 2 a y d y
3 x² d x, whence
3x² d x
9 x* d x²
9 x d x²
dy=
and d y
2 a y
4 a* y²
4 a
9
and
d s²
a) d x²,
da, hence
=S√1
S=
and
4 a
1 +
taf
9
=(1+
9 x
(1 + 2
4 a
22)
4 a
d x =
a
x
40 / (1 + 2)²d (2)
√(1+
9
4 a
4 a
4 a
us d u =
3 us = £7 α V
8
27
a√(1 +
9
11+
9 x
3
Sus
4 al
In order to find the necessary constant, we make s begin with c
y, and we obtain
27
0 = 8
a √1³ + Con., or Con.
8
27
α
and
8 = 2, a [√ (1 + 2 2 )
- 1}
8
27
9 x 3
4 a
E.G., for the piece A P₁ whose abscissa a = a, we have
3
s = 2; a [ √ (18)³ – 1] = 1,736 a.
8
27
Introducing the tangential angle QPR
43) we have
I.E.,
PTM a (Fig.
=
Q R = P Q. sin. Q P R and PR PQ cos. Q P R,
d Y
=
d s sin. a and d x = ds cos. a,
and besides, tang. a =
d
y
(see Art. 6),
d x
d x
d y
also,
sin. a =
and cos. a =
;
and finally,
d s
d s
$ =
SVI
√1 + tang.² a . d x =
If the equation between any two of the quantities x, y, s and a
is given, we can find the equation between any two others.
S
sin. a
dy = √ cos. a
S
d x
S
If, E.G., cos. a =
we have
√ c² + s²
s d s
d x = ds cos. a =
and
√c² + 8"
=S-
s d s
2 s d s
√ c² + s²
Nc
Nc² + s²
ja
d u
Vu
1/ u¯& du= x²
Ju¯`
= √ c² + s² + Const., and if x and s are equal to zero at the
same time, x = √ c² + s² C.
ART. 33.
87
INTRODUCTION TO THE CALCULUS.
ART. 33. A right line perpendicular to the tangent P T, Fig.
44, is also normal to the curve at the point of tangency, for the
FIG. 44.

H
S
R
X
T A M
N
0
K
tangent gives the direction of the curve at this point.
The portion PK of the line between the point of tangency P
and the axis of abscissas is called simply the NORMAL, and the pro-
jection of the same M K on the axis of abscissas the SUBNORMAL.
We have for the latter, since the angle M P K is equal to the tan-
gential angle PTM = a,
I.E., the subnormal
MK = M P . tang. a,
dy
y tang. a =
Y T x
d
Since for the system of curves y = a", tang. a = mxm-1, it fol-
lows that the subnormal is = mx". ɔm-
272-1 = m
M x²m−1
m y²
and
X
for the common parabola, whose equation is y = p, we have the
subnormal
Ρ Ρ
Y 2, that is constant.
2 y
If to a second point Q, infinitely near the point P, we draw
another normal QC, we obtain in the point of intersection of
these two lines the centre C of a circle which can be described
through the points of tangency P and Q. It is called the circle of
curvature, and the portions C P and C Q of the normals are radii
of this circle, or, as they are styled, the radii of curvature. This
circle is the one of all those, which can be made to pass through P
and Q, which keeps closest to the element P Q of the curve, and
we can therefore assume that its arc P Q coincides with the ele-
ment PQ of the curve. It is called the osculatory circle.
Denoting the radius CP CQ by r, the arc A P of curve by
s or its element PQ by ds, and the tangential angle or are of
P TM by a, and its element SUM-STM, I.E.,-UST=-
88
[ART. 33.
INTRODUCTION TO THE CALCULUS.
PCQ by da, we have, since PQ-CP. arc of the angle PCQ,
ds=-r da, whence the radius of curvatures r ——
FIG. 45.
d s
da

H
S
R
T A
M N
0
X
K
We can generally determine a from the equation of the co-ordi-
nates by putting tang. a =
d y
d x
Now d tang. a =
d a
cos.² a
d x
and cos. a=
whence
>
d s
d x²
da cos.² a. d tang. a =
d tang. a and
d s².
d s
r =
d s
da
d s³
d x² d tang. a
d s³
d x d tang. a
and for a
cos.' a d tang. a
For a convex curve r = + = +
point of inflexion r = ∞.
For the co-ordinates A Ou and 0 Cv of the centre C of
curvature, we have
u=AM+H C=x+ C P sin. C P H, I.E., u=x+r sin. a, and
v=0 C=M P-H P=y-CP cos. O P H, I.E., v=y—r cos. a.
FIG. 46
Y
BI
D
P
A
X
C
KM A T
D₁
B
L
The continuous line formed
by the centres of curvature forms
a curve, which is called evolute
of A P, and whose course is de-
termined by the co-ordinates u
and v.

1
If the ellipse A D A, D₁, Fig.
-x 46, is laid upon the circle A B
A, B₁, its co-ordinates C M = x
and M Qy can be expressed
by means of the central angle
P CB = 4 of the circle. We have
here
ART. 33.]
89
INTRODUCTION TO THE CALCULUS.
x = C P sin. C P M = C P sin. B C P = a sin. p, and
b
y =
M Q
MP =
а
b
a
CP cos. C P M = b cos. p.
- b
From the latter we obtain dr = a cos. o do and d Y
sin. o do, and consequently for the tangential angle of the ellipse
Q TX = a
tang. a =
d y
d x
b sin.
a cos. o
b
tang. 4, and for its com-
α
a
cotg. p.
plementary angle
Q TC = a, = 180°— a,
b
a, tang. p and cotg. a₁ =
tang. a₁ =
a
Hence the subtangent of the ellipse is
MT = MQ cotg. M T Q
= y
: y cotg. a =
a y
b
cotg. ¢ = cotg.,
31
when y, designates the ordinate M P of the circle. Since the tan-
gent P T to the latter is perpendicular to the radius C P, we have
also P T M=P C' B=0, and therefore the subtangent M T of the
same is also MP cotg. MTP-y, cotg. o.
Therefore the two points of the ellipse and circle which have
the same ordinate, have one and the same subtangent.
Farther, for an elementary arc of the ellipse
d s²= d x² + d y²= (a² cos.²+b² sin.² ) d o²,
and the differential of tang. a,
b
b do
d tang. a =
d tang. O
α
a cos.2
whence it follows that the
d s³
radius of curvature of the ellipse is
(a² cos. + b² sin. p);
r = d x² d tang. a
b
a² cos.²
2
•
a cos. o
?
(a² cos.² + b² sin. p)
a b
E.G., for y = 0, I.E., for sin. p = 0, and cos. = 1, we have the
maximum radius of curvature
Q³
I'm
a b
a²
b
90
[ART 34.
INTRODUCTION TO THE CALCULUS.
and, on the contrary, for = 90°, I.E., for sin. 1 and cos.
4 = o
=0, the minimum radius of curvature
b³
rn ab
a
The first value of r corresponds to the point D, and the last to
the point A, and both are determined by the portions of the axes
CL and CK, which are cut off by the perpendiculars erected upon
the chord A, D at its ends A, and D.
ART. 34. Many functions, which occur in practice, are com-
posed of the various functions which we have already studied,
such as
Y
,
Y =e, and y = sin. x, y = cos. x, etc.;
and it is easy, with the assistance of the foregoing rules, to deter-
mine their properties, such as the position of their tangents, their
quadrature, their radius of curvature, etc., as well as to construct
the curves, as is shown by the following examples:
For the curve, whose equation is y = x²
we have
whence
Since this tangent becomes = 0 for x
d y = 2 x d x - x² d x,
tang. a = 2 x x² = x (2
x²
(1 – 3) = } ử,
- 1)
x (2 — x).
0 and x = 2, its direction
at these two points is parallel to that of the axis of abscissas.
=
Farther, d tang. a 2 dx 2 x dx 2 (1 - x) d x,
whence for
and for
=
= 0, d tang. a =
x = 2,
2, d tang. a =
+ 2 d x
2 d x,
−
and therefore the ordinate of the first point is a minimum, and that
of the second point a maximum. If we put d tang. a =
2
3
0, we ob-
tain x = 1 and y , the co-ordinates of a point of inflexion in
which the concave portion of the curve joins the convex.
Farther, for an element ds of the curve we have
d s² = d x² + d y² = d x² + x² (2− x)' d x² = [1 + x² (2 ---x)²] ‹ x²,
d
whence the radius of curvature is
r =
d s³
d x² d tang a
[1 + x² (2 − x)²]³ .
2 (1 − x)
-1
29
E.G., for x = 0 we have r =
2
1, =
, for x 1, r = −
-1
x =
for a 2, r =
= 1,
+, and for x=3, r. 101 +7,906.
-2
0
1.10+7,906.
ART. 341
91
INTRODUCTION TO THE CALCULUS.
The corresponding curve is shown in Fig. 47, in which is the
I.
B.
FIG. 47.
Y
16
B
9
4
M
P
1
W
K
3_2
1 A
2
3
I
II.
M
_Y
C
Κ
M₁
X
origin and XX, Y Y the
axes of co-ordinates. The
parabola BA B₁, which ex-
tends symmetrically upon
both sides of the axis of A Y,
represents the first part y₁=x²
of the equation, and, on the
contrary, the curve CA C₁,
which upon the right-hand
side of Y descends below
XX, and on the left-hand
side rises above it, and thus
diverges more and more from
the axis II, as it increases
its distance from IF, cor-
responds to the second part
Y₂ = 4x³.

Y2
138
In order to find for a given
abscissa, the corresponding
point of the curve y = x²
2, we have but to add alge-
braically the corresponding
ordinates of the first
first two
1
curves; E.G., since for x =
we have y = 1 and y
it follows that the correspond-
ing ordinate of the point I
is y = y₁ + y = 1 − } = };
Y Y
farther, for a 2 we have
S
§, and
Y₁
4, and y
hence the co-ordinate of the
point is y = 4 — § = !·
៖
In the same way = 3 gives
x
Y = Y₁ + Y₂ = 990;=
4, y = 16 = 4; Z=
64
−1, y = 1 + 3 = 1; x = − 2, y = 4 + $ = 30, etc., and we per-
ceive that the curve from A towards the right has the form A W
MK L, and that in the beginning it runs above the abscissa A K
3, but from that point it extends to infinity below the axis
92
[ART. 34.
INTRODUCTION TO THE CALCULUS.
JA, and that from A towards the left it forms but one branch
APQ..., which rises to infinity. From what precedes we see
that Wis a point of inflexion, and M a point of the curve where
the ordinate is a maximum. While the curve has in A and M the
direction of XX, in W it rises at an angle of 45°, for we have for
the latter tang. a = x (2x) = 1; on the contrary, the angle of
inclination at K, is tang. a = 3, consequently a is = 71° 34',
etc. The quadrature of the curve is given by the integral
F= Ĵy
ƒ y d x = ƒ (x² − 4 x°) dx = ƒ x à x − § ƒ x à x
хо
X4
3 12 3
(1
2).
d
S a
d
Hence, E.G., we have for the area of the portion of surface
A WMK above A K = 3
33
F = 2/3 (1 − 1) = 3,
and on the contrary the area of the portion of surface 3 L 4 below
the abscissa 3 4 is
43
F₁
3
33
(1 – 4) — 23 (1 − }) = 0 − 2 = −1·
−
Finally, to find the length of a portion of the curve, E.G., A W M,
we put
S
= √ √1 + x² (2 − x)² d
d x S' 4 ( x ) d x,
and employ the method of integration explained in Art. 30. Here
C1
C
c is =
=
0, and c, 2, and taking n
2, and taking n = 4 we have d x
n
2-0
ولو
4
, then substituting successively the values 0, 1, 1, and
2 for x in the function
values
$ (0)= √1=1, ¢ (1) =
(x) √ 1 + x² (2 − x), we obtain the
√1+; %=§, 6 (1)= √1+1= √2=1,414....
9
16
$ (?) = √1 + √ % = and (2)= √1=1,
2 §
6
and therefore the length of the arc A W M is
$ =
=
(† (0) + 4 ☀ (5) + 2 ☀ (1) +4 ¢ (§) +¢ (2))
($(0)
1
(1+5+2,828+5+1). = 2,471.
C1
C
3.4
ART. 35.]
93
INTRODUCTION TO THE CALCULUS.
By means of the curve y=x²
(1–319)
JC
3
we can easily determine the
course of the curve y=x 1- by extracting the square roots of
the values of the co-ordinates of the first, which give the corre-
sponding co-ordinates of the latter. But since the square root
of negative quantities are imaginary, this curve does not continue
beyond the point K to the right; and since every square root of a
positive number gives two values, equal and with opposite signs,
the new curve (II) runs in two symmetrical branches Q AMK
and Q₁ A M₁ K on both sides of the axis of abscissas.
1
ART. 35. When the quotient y =
φ
$ (x)
4(x)
and y (x) takes the indeterminate form of
of x, which always occurs when, as E.G., in y
of two functions (x)
for a certain value a
x²
a²
the numer
X a
و
ator and denominator of a fraction have a common factor x α,
we can find the real value of the same by differentiating the nu-
merator and denominator.
If x is increased by d x, and y by the corresponding element
dy, we have
y + dy
¤ (x) = 0
$ (x) + d ¢ (x)
4(x) + d 4 (x)'
and 4 (x)
but for x = a
=
0, whence
do (x)
y + d y =
d 4 (x)
but since dy is infinitely small in comparison to y, we have
Y =
$ (il²)
(x)
do (x)
Φι (2)
d p (x)
4, (x)
in which o, (x) and 4, (x) designate the differential quotients of
1
(x) and 4 (x).
Φι (*)
0
If y =
is also =
we can differentiate it anew, and put
Y₁ (x)'
O'
Φι
d p₁ (x)
Φ. (2)
y =
d p₁ (x)
Y₂ (xx)'
In the same way the indeterminate expressions y
and
94
[ART, 35.
INTRODUCTION TO THE CALCULUS.
1
0 × ∞, etc., can be treated, for ∞
whence and 0 × ∞
O'
can be put = 0:
3 x³
- 77 x²
8x + 20
0
E.G., y =
becomes for x = 2, y
5 x⁹
21 x² + 24 x 4
C
10
For this we can put
d (3 x³ — " x² - 8x+20)
Y
d (5 x³ − 21 x² + 24 x
4)
9 x²
15 x² -
14 x - 8
42 x + 24'
0
which for x = 2 gives again y
=
and we can again put
0'
y =
d (9 x²
d (15 x*
14 x
8)
18 x
- 14
9 х
17
11
42 x + 24)
30 x
42
15 x
21
9
The factor (x
and twice in the
obtain
2) is really contained twice in the numerator,
denominator. If we divide both by x 2, we
A
Y
3 x²
5 x² - 11 x + 2'
x 10
and dividing the last again by (x
2)
3x + 5
y =
5 x
1'
11
›which for x = 2 gives y
We have also for y =
9
a
Na²
X
х
when x = 0,
j d x
0'
but since
d (a
Va²
a² — x) =
d (a³
x)} =
Na²
2
X
14
1
in this case
y
Na²
2 a
X
1 x
0
further
y =
for x= 1, gives y
XC
d x
d x
but
λιχ
d l x =
and d √1 − x
X
2 V1
x²
2 √1
X
2.0
hence it follows that y =
=
= 0.
XC
1
ART. 36.]
95
INTRODUCTION TO THE CALCULUS.
Finally, y =
1 − sin. x + cos. x
·1 + sin. x + cos. x
π
gives for x =
(90°)
1-1+0
0
y =
we have therefore
1 +1 +0 0'
d (1
sin. x + cos. x)
cos. X
sin. x
y =
d ( − 1 + sin x + cos.x)
cos. x
sin. x
0 - 1
= 1.
0 - 1
ART. 36. When, for a function y = a u + ẞ v, a series of
corresponding values of the variables u, v and y has been deter-
mined by observation or measurement, we can require the values
of the constants a and ẞ which are the freest from accidental or
irregular errors of observation and measurement, and which
express most exactly the relation between the quantities u, v and
y, of which u and v are known functions of one and the same
variable, x. Of all the methods that can be employed for the
resolution of this problem, I.E., for the determination of the most
possible, or the most probably correct, values of the constants, the
method of the least squares is the most general, and rests upon the
most scientific basis.
If the results of the observations corresponding to the func-
tion y = a u + ß v are,
U19 V1, Y1
U2, V2, Y2
W3, V39 Y3
Un VR, Y n
we have the following values for the errors of observation, and for
their corresponding squares.
Z₁ = Y₁ = (a u₁ + ẞ v₁)
-
Z2
Z₂ = Y₂
Y2
(a uz + B v₂)
2
Z3=Y3
a
-
(α Us + B v3)
Yn − (α îla + B v₂)
(a
Za = Y n
96
[ART. 36.
INTRODUCTION TO THE CALCULUS.
z,² = y₁² — 2 a
1
1
u₁ у ₁ − 2 ẞ v₁ Y₁ + a² u₁² + 2 a ẞ u₁ v₁
2
1
+ ß² v₁²
2
2
Z2
z₂² = y₂² - 2
123²=y3² - 2
Z3 Y3
α
u ₂ Y2 - 2 ẞ v ₂ y z + a² u₂+2 a ẞuş v½
+ ẞ² v₂
1
2
ß²
α
z
2
2
2
2
Uз Yз-2 ẞ v3 уз+a² u² + 2 a ẞ us vs + B² vz
3
Zn² = у„² — 2 α u» Yn − 2 ß vn Yn + a² Un² + 2 а ß U₂ v n + B² V n
Y n
a
Employing the sign of summation
quantities of the same kind, y + y₂² +
•
2
+ Vn Yn
V₁ Y 1 + Vş Y x + V3 Y3 +
sum of the squares of the errors
Σ
Σ (x²) = Σ (y²) - 2 a (u y)
+ 2 a ẞ Σ (u v)
−
-
2
a
n
2
to denote the sum of
2
2
y² + . . . + yn² = Σ (y²),
Σ (v y), etc., we have for the
2 B (v y) + a² Σ (u²)
β Σ
+ B² Σ (v²).
In this equation, besides the sum of the squares of the errors
Σ (2), which is to be considered as the dependent variable, only
a and ẞ are unknown. The method of the smallest squares
requires us to choose such values for a and ẞ as shall cause Σ (z²)
to be a minimum; and therefore we must differentiate the
function Σ (z²), which we have obtained, once in reference to a
and once in reference to ẞ, and put each differential quotient
of Σ (z²) thus obtained by itself equal to zero. In this way we
obtain the following equations of condition for a and ß,
- Σ (u y) + a Σ (u³) + BE (u v) = 0,
-Σ (vy) + BΣ (v²) + a Σ (u v) = 0,
and resolving these we have
Σ (v²) Σ (u y) -
a
B =
Σ (u') Σ (v²) —
Σ (u²) (vy)
Σ
£
Σ (u²) Σ (v²) —
These formulas give
Σ
Σ (u v) (v y)
Σ (u v) Σ (u v)'
and
Σ
(u v) Σ (u y)
(See Ingenieur, page 77.)
Σ (u, v) Σ (u v)
for a function y = a + ẞ v, since here
Σ
11, and 2 (u v) = Σ (v), Σ (u y) = 2 (y), and Σ (u) = 1 + 1
+ 1 + ... = n, I.E., the number of equations or observations,
Σ (v²) Σ (y) — Σ (v) Σ (v y)
a
nΣ (v²) - Σ (v) Σ (v)
ηΣ
nΣ (vy) - Σ (v) Σ (y)
B
n Σ (v') — Σ (v) Σ (v) *
For the still simpler fanction y = ẞ v, in which a = 0, we have
В
Σ (v y)
(v)'
ART. 36.]
97
INTRODUCTION TO THE CALCULUS.
and, finally, for the most simple case y = a, where we have to de-
termine the most probable value of a single quantity,
a =
Σ (y)
N
that is the arithmetical mean of all the values found by measure-
ment or by observation.
EXAMPLE.-In order to discover the law of a uniformly accelerated mo-
tion, L.E., the initial velocity c and the acceleration p, we have measured
the different times t₁, të, t, etc., and the corresponding spaces 81, 82, 88,
etc., described, and have found the following results,
Times
•
Spaces.
I
3
сл
5
7
IO Sec.

5
20
38
58/1/1
IOI feet.
p t
2
is the fundamental law of this motion, we are re-
Now if 8 c t +
quired to determine the constants c and p.
mulas u = t, and v =ť³, and also a = c, ß
Putting in the foregoing for.
p
2
the calculation of c and p the following formulas:
C =
and y = 8, we obtain for
Σ (*) Σ (8 t) - Σ (*) Σ (8 t*)
t²)
Σ (t²) Σ (t¹)
Σ (t³) Σ (t³)
P
22
Σ (t²) Σ (8 t²)
Σ (1) Σ (8 t)
Σ (t²) Σ (t') — Σ (t³) Σ (t³)
from which the following calculations can be made,
and
->
t
ť
t³
ť
♡
s t
s t²
I
I
I
I
5
5
5
350
9
27
81
20
бо
180
25
125
625
38
190
950
7
49
343
2401
58.5
409.5
2866.5
ΙΟ
100
1000
10000
ΙΟΙ
ΙΟΙΟ
ΙΟΙΟΟ

Sum
184
1496
13108
222.5
1674.5
14101.5
Σ Σ | Σ
·
=Σ (t³)| =Σ (t³)| =Σ (t) = (s) = (st) = (st).
|
98
[ART. 37.
INTRODUCTION TO THE CALCULUS.
from which we obtain
13108. 1674,5
C =
P
184. 13108
1496. 14101,5
1496. 1496
85340
17386
4,908 feet, and
89624
=
1496. 1496
0,5155 feet.
173860
184. 14101,5 1496. 1674,5
184. 13108
W
Whence the formula for the observed movement is
4,908 t + 0,5155 tỉ,
and from this formula we have
For the times.
о
I
3
5
7
IO Sec.
For the spaces
о
5.43
FIG. 48.
B
N
P
19.36 37.43 59.62 100.63 feet.

A
M
Χ
t
= 0 1
3
5
7
10
If we consider the times
(t) as abscissas, and lay off
the calculated as well as
the observed spaces (8) as
ordinates, we can draw a
curve through the extrem-
ities of the calculated ordi-
nates, which will pass be-
tween the points M, N, O, P,
Q, determined by the ob-
served co-ordinates, so that
the sum of the squares of the
deviation of the curve from
these points shall be as small
as possible.
ART. 37. If we have no formula for the successive values of a
FIG. 49.
P₂
P
N
Po
A
Mo
2
MI
M
M2
C
quantity y, or for its dependence
upon another quantity a, and we
wish to determine its value for a
given value of x, determined by
experiment, or taken from a table,
we employ the so-called method
of interpolation, of which only
the most important part will be
given here.
If the abscissas A M = Xor
A M₁ = a, and A M, Fig.
49, and the corresponding ordi-
nates M, P, Yo, M, P₁ = Y₁,
M₁ P₁y are given, we can

Y u
=
19
ART. 37.]
99
INTRODUCTION TO THE CALCULUS.
express the ordinate MP=y, corresponding to the new abscissa A M
=x, by the formula y=a+ẞx+y x², provided three given points Po
P₁, P2, lie nearly in a straight line or in a slightly curved arc. If we
change the origin of co-ordinates from A to M, the generality of
the expression will not be affected, and we obtain for x = 0 simply
y = a, and consequently the constant member a =
Yo. Substi-
tuting in the supposed equation, in the first place x, and y₁, and
then in the second place x and y, we obtain the two following
equations of condition,
Y2
Y₁ - Yo =
Y₂- Yo =
B x₁
+ y x², and
2
1
B x
+
y x2, hence
B
(Y₁ — Yo) x₂² —
2
(Y₂ — Yɔ) x,²
2
and
(w
Y =
2
X₁ X²² — X2 X₁²
(Y₁ — Yo) X₂
from which we have
y=Yo+
2
2
X1
2
(Y2 - Yo) X1
2
Xj X 2
X 2 X 1
—
—
(Y› — Yo) x; ² — (Y:— Yo) 2º¹ ") x + ((Y₁ — Y₁) x 2 − ( Y₂ — Yo) x 1 ) x².
X ₁ X2² — X2 X₁ª
2
2
X 1 X 2 X2″ X 1
If the ordinate y, lies midway between y, and y, we have r
221, and therefore more simply
Yo
У
= Yo
(3 y₁ — 4 y₁ + Y :)
Yo
2 ½ + Y:):
x²
x +
2 x2
2X1
If but two pair of co-ordinates x, y, and ₁, y₁ are given, we
must regard the limiting line P, P, as a straight line, and conse-
quently put
Y
Yo + B x
and
Y1
Yo + B X1,
whence we have
Y₁- Yo
B
and
X1
y = Yo +
(1/₁ = Yo) x.
༡༨.༧)
When it is required to interpolate by construction between
three ordinates yo, y, y a fourth ordinate y, we draw, through the
extremities P., P1, P2 of these ordinates a circle, and take y
= to
the ordinate of the same. The centre C of the circle is determined
in the usual way by joining the points P, P, P. by straight lines
and erecting perpendiculars at the middle points of the chords.
The point of intersection C of the perpendiculars is the required
centre.
1
If the distances of the middle point P, from the two others P.
1
100
[ART. 38.
INTRODUCTION TO THE CALCULUS.
and P2, are s, and s₂, and the distance P, K of the point P from the
chord P, P,
a = P; P, P₂ =
8
=
2
h, we have for the angle at the periphery
the angle at the centre P, CP,
h
sin. a =
So
and consequently the radius of curvature CP CP, = C P₁ =
C P₂ is
0
S2
So S2
r =
2 sin. a
2 h
0
1
=
consequently we find the centre C of the circle passing through the
points Po, P₁, Pe, by describing from P, or P, or P, with a radius
equal to the value of r, calculated by means of this formula, an arc
whose intersection with the perpendicular to the chord P, P, erected
at its centre D is the required point.
2
2
ART. 38. The mean of all the ordinates upon the line M, M, is
the altitude of a rectangle M M, N, N, with the same base M. M½,
and having the same area as the surface M, M, P, P, P., and can
therefore easily be determined from this surface. According to
Art. 29 we have
2
F = √ ” y d x = [ ˜ˆ ˆ (y. + ß x + y x²) dx
B x2²
3
Yo Xz +
Y X2
+
2
= Yc X 2 +
+
(
(y,
1
3
2
xi
Yo) x;² — (Y₂ — Yo) x;²) xz
2
X1 X22 — X2 X₁²
X32
Yo) X2
2
X12 X2
(Y₁ - Y₁) xz²
2
6 x₁ (X2 X₁)
= (1.
+
= (Y₁ + Y₁)
X 2 + ((Y₁ —
2
(Y2
X22X1
2
Yo) X1) X ₂
xr
3
2
1
(Y₂ — Yo) (3 x₁ — 2 X₂) ) xr
2
6 (x-x1)
Yo) X 2 — (Y½ — Yo)
Yo) X,
2₁)
6 X₁ (X2 X₁)
and consequently the mean ordinate is
X2²,
F
Ym =
X2
(Yo + Y₂)
2
+
((y₁ —
(Y₁ — Yo) X 2 — (Y 2
(Y₂ — Yo) X₁
X2.
6 x₁ (X2 2₁)
Y2
Yo
X2
If
were =
Yı
Yo
we would have simply
the boundary would be a right line, and
Yo Y 2
F = (₁ + 1 ) =
2
X2
(Y. + Y2)
and
Ут
2
ART. 38.]
101
INTRODUCTION TO THE CALCULUS.
If also x2 = 2 x₁, that is, if y, is equidistant between y, and
Y 2, we have
Po
У
X2
6
F = (y, + 4 y₁ + y) **² (see Art. 30), and y
21
FIG. 50.
P₂ Q3
P3
¡R 2
22
Y/2
13
M3
Mo
NM N₂ M2N3
3
m
Yo + 4 Y₁ + Y 2
6
0
3
If a surface M. M½ P3 P., Fig.
50, is determined by four co-or-
dinates M. P₁ = Yo, M₁ P₁ = y'₁₂
0
P.

3
P3
M₂ P₂ = y, M, P₁ = y, which
are equidistant from one an-
other, we can determine approx-
imately the area of the same in
the following simple manner:
0
Let us denote by the base
M, M3, by Z, Z1 Z3, three ordinates
intercalated between y, and y,
and equidistant from each other,
we can then put approximatively the surface
2
X3
4
; but
0
3
Z3
21 + 22 + 23 2 z₁ +222 + 2 Z3 2 Z₁ + Z2
Y₁ = 2₁ + } (2₂ — 2₁) =
M¸ M, P¸ P¸ = F = (↓ y. + 21 + 22 + 23 + 1 Y3) 22;
2 Z3 + Ziz
and
6
6
2 Z3+ Zą
as well as y₂ =
3
6
22₁ + Zą
3
whence it follows that
Z1 + Z2 + Z3
3
Y₁ + y 2
and
2
X3
F= [{ Yo + 3 (Y₁ + Y₂) + ¦ Y³]
4
X3
[y. + 3 (y₁ + y2) + Y³]
and also
8'
Yo + 3 (Y₁ + Y₂) + Ys
Ут
8
m
While the former formula for y is employed when the surface.
is divided into an even number of strips, the latter is employed
when the number of these divisions is uneven.
201
Hence we can write approximately
acı
S.“ 'y d x = ƒ“ 4 (x) d x = [y. + 3 (y, + y₂) + Y.]
if
8
102
[ART. £8.
INTRODUCTION TO THE CALCULUS.
2 c
Y.
Y。 = $(c), Y₁ =
Φ
four known values of У
=
(x). E.G., for
(30+ c₁), Y₂ = 4 (c + 2c) and y₁ = ¢ (c,) arc
Y2
3
x
Y3
1
X
Seda (see example, Art.
1
30) we have c = 1, c, 2 and p (x) =
=
x
Yo
111
y₁ = = 1, y₁ =
3
2 + 2
3
3, Y 2
whence it follows that
and y =
1, and that
1+ 4
the approximate value of this integral is
x d
d
8
ƒª ¹ x = [1 + 3 (¦ + §) + ']·↓
1 x
111
=
=
160
0,694.
PART FIRST.
GENERAL PRINCIPLES OF MECHANICS.
FIRST SECTION.
PHORONOMICS OR THE PURELY MATHEMATICAL
THEORY OF MOTION.
CHAPTER I.
SIMPLE MOTION.
§ 1. Rest and Motion.-Everybody occupies a certain posi-
tion in space, and a body is said to be at rest, (Fr. repos, Ger. Ruhe),
when it does not change that position, and, on the contrary, a body
is said to be in motion, (Fr. mouvement, Ger. Bewegung), when it
passes continually from one position to another.
The rest and motion of a body are either absolute or relative,
according as its position is referred to a point which is itself at rest
or in motion.
On the earth there is no rest, for all bodies upon it participate
in its motion about its axis and around the sun.
If we suppose
-the earth at rest, all the terrestrial bodies which do not change
their position in regard to the earth are at rest.
§ 2 Kinds of Motion.-The uninterrupted succession of po-
sitions which a body occupies in its motion forms a space, that is
called the path or trajectory (Fr. Chemin, trajectoire, Ger. Weg) of
the moving body. The path of a point is a line. The path of a
geometrical body is, it is true, a figure, but we generally under-
stand by it the path of a certain point of the moving body, as, E.G.,
its centre. Motion is rectilinear (Fr. rectiligne, Ger. geradlinig)
106
[§ 3-5.
GENERAL PRINCIPLES OF MECHANICS.
when the path is a right line, and curvilinear (Fr. curviligne, Ger.
krummlinig) when the path of the moving body is a curved line.
§ 3. In reference to time (Fr. temps, Ger. Zeit) motion is either
uniform or variable. Motion is uniform (Fr. uniforme, G. gleich-
förmig) when equal spaces are passed through in equal arbitrary
portions of time. It is variable (Fr. varié, Ger. ungleichförmig)
when this equality does not exist. When the spaces described in
equal times become greater and greater as the time during which
the body is in motion increases, the variable motion is said to be
accelerated (Fr. accéléré, Ger. beschleunigt); but if they decrease
more and more with the increase of time, this motion is said to be
retarded (Fr. retardé, Ger. verzögert). Periodic (Fr. périodique, Ger.
periodisch) motion differs from uniform motion in this, that equal
spaces are described only within certain finite spaces of time, which
are called periods. The best example of uniform motion is given
by the apparent revolution of the fixed stars, or by the motion of
the hands of a clock. Examples of variable motion are furnished
by falling bodies, by bodies thrown upwards, by the sinking of the
surface of water in a vessel which is emptying itself, etc. The
play of the piston of a steam engine, and the oscillations of a pen-
dulum, afford good examples of periodic motion.
§ 4. Uniform Motion.-Velocity (Fr. vitesse, Ger. Geschwin-
digkeit) is the rate or measure of a motion. The larger the space
that a body passes through in a given time, the greater is its mo-
tion or its velocity. In uniform motion the velocity is constant,
and in variable motion it changes at each instant. The measure
of the velocity at a given moment of time is the space that this
body either really describes, or which it would describe, if at that
instant the motion became uniform or the velocity remained con-
stant. We generally call this measure simply the velocity.
§ 5. If a body in each instant of time describes the space o, and
if a second of time is made up of n (very many) such instants,
then the space described within a second is the velocity, or rather
the measure of the velocity, and it is
c = n. 0.
During a time t (seconds) n. t instants elapse, and in each in-
6, 7.]
107
SIMPLE MOTION.
stant the body passes through the space o, and therefore the total
space, (Fr. l'espace, Ger. Weg), which corresponds to the time t, is
s = n. t. o = n . o . t, I.E.
I.) s = ct.
In uniform motion the space (s) is a product of the velocity (c)
and the time (t).
Inversely II.) c =
S
S
III.) t =
C
EXAMPLE.-1. A locomotive advancing with a velocity of 30 feet passes
in two hours 120 minutes 7200 seconds, over the space 8 30. 7200
216000 feet.
2. If we require 43 minutes = 270 seconds to raise a bucket out of a
1200
270
pit, which is 1200 feet deep, we have its mean velocity (c)
44
4,444... feet.
40
9
3. A horse advancing with a velocity of 6 feet requires, to pass over five
miles, or 26400 feet, the time
26400
6
4400 seconds, or 1 hour 13
minutes and 20 seconds.
§ 6. If we compare two different uniform motions, we obtain
the following result:
As the spaces are sct and s₁ = c₁ t₁ their ratio is
S
c t
$1
C₁ ti
S
C
If we put t = t₁ we have
S
if we take c = c, we obtain
$1
C1
81
t
Ꮳ
and finally, if s = s, it follows that
t₁
t
C1
t
The spaces described in the same time in different uniform mo-
tions are to cach other as the velocities; the spaces described with
equal velocities are to each other as the times; and the velocities cor-
responding to equal spaces are inversely as the times.
§ 7. Uniformly Variable Motion.-A motion is uniformly
variable, (Fr. uniformément varié, Ger. gleichförmig verändert),
when the increase or diminution of the velocity within equal, ar-
bitrarily small, portions of time is always the same. It is either
uniformly accelerated (Fr. uniformément accéleré, Ger. gleichför-
108
[§ 8, 9.
GENERAL PRINCIPLES OF MECHANICS.
mig beschleunigt) or uniformly retarded (Fr. uniformément retardé,
Ger. gleichförmig verzögert). In the first case a gradual augmen-
tation, and in the second a gradual diminution of velocity takes
place.
A body falling in vacuo is uniformly accelerated, and a body
projected vertically upwards would be uniformly retarded, if the
air exerted no influence upon it.
§ 8. The amount of the change in the velocity of a body is
called the acceleration (Fr. accélération, Ger. Beschleunigung and
Acceleration). It is either positive (acceleration) or negative (re-
tardation), the former when there is an increase, and the latter.
when there is a diminution of velocity. In uniformly variable mo-
tion the acceleration is constant. We can therefore measure it
by the increase or decrease of velocity which takes place in a
second. For any other motion, the acceleration is the increase or
decrease of velocity, which a body would undergo if, from the instant
for which we wish to give the acceleration, the acceleration became
constant, and the motion was changed to a uniformly varied one.
This measure is generally called simply the acceleration.
§ 9. If the velocity of an uniformly accelerated motion in a very
small (infinitely small) instant of time is increased by a quantity
K, and if the second of time is composed of n (an infinite number
of) such instants, the increase of velocity in a second, or the so-
called acceleration, is
p = n K,
and the increase after t seconds is = nt.
n t . k = n k . t = p t.
If the initial velocity (at the moment from which we begin to
count t) is c, we have for the final velocity, I.E., for the velocity
at the end of the time t,
v = c + pt.
For a motion starting from rest c is 0, whence v = pt; and
when the motion is uniformly retarded, in which case the accelera-
tion (p) is negative, we have
v = c
pt.
EXAMPLE.-1. The acceleration of a body falling freely in vacuo is
32,20 feet. It acquires therefore after 3 seconds the velocity v =
32,20 . 3 = 96,60 feet.
pt
2. A ball rolling down an inclined plane has in the beginning a velocity
§ 10.]
109
SIMPLE MOTION.
of 25 feet, and the acceleration is 5 feet per second. Its velocity after 21 sec-
onds is therefore v = 25 + 5. 2,5 37,5 fect; 1.Ê., if from the last moment
it moved forward uniformly, it would pass over 37,5 feet in every second.
3. A locomotive moving with a velocity of 30 feet loses, in consequence
of the action of the brake, 3,5 feet of its velocity every second; its accelera-
tion is therefore - 3,5 feet and its velocity after 6 seconds is v = 30 3,5.6
= 30 -- 21
- 219 feet.
=
M
§ 10. Uniformly Accelerated Motion. Within an infinitely
small instant of time we can consider the velocity of every
motion as constant, and put the space passed through in this
instant
σ = v. T,
and we obtain the space passed through in the finite time t by
summing these small spaces. But the time in which all these
small spaces were described is one and the same 7, and we can put
their sum equal to the product of this instant of time and the sum
of the velocities corresponding to the different equal instants.
For uniformly accelerated motion the sum (0 + v) of the ve-
locities in the first and last instant is just as great as the sum
p + + (v − p 7) of those in the second and last but one instants,
and equal to the sum 2 p 7 + (v – 2 p ) of those in the third and
last but two instants, etc., and this sum is in general equal to v;
T
Τ
the sum of all these velocities is therefore equal to
(r.) the pro-
duct of the final velocity and half the number of the elements
of the time, and the space described is equal to the product
(v.3. 7) of the final velocity v and half the number of the elements
(†)
of the time and one of these elements. Now the magnitude (7)
of an element of the time multiplied by their number gives the
whole time t, whence the space described in the time t with an
v t
uniformly accelerated motion is s =
2
The space described with uniformly accelerated motion is the
same as that described with uniform motion when the velocity of
the latter is half the final velocity of the former.
EXAMPLE.-1. If a body in uniformly varied motion has acquired in 10
seconds a velocity v = 26 feet, the space described in the same time is
26 . 10
130 feet.
2
m
+
110
[§ 11, 12.
GENERAL PRINCIPLES OF MECHANICS.
2. A wagon whose motion is uniformly accelerated and which describes
25 feet in 21 seconds, possesses at the end of that time the velocity
V
2.25
2,25
50.4
9
22,22 . . . feet.
§ 11. The two fundamental formulas of uniformly accelerated
motion
I.) v = p t and
II.) s =
v t
2'
which show that the velocity is a product of the acceleration and
the time, and that the space is the product of half the terminal ve-
locity and the time, furnish two other equations, when we eliminate
in the first place v and in the second t. By this operation
we obtain
III.) s =
pt²
2
and
v²
IV.) s =
2 p
Hence, in uniformly accelerated motion, the space described is
equal to the product of half the acceleration and the square of the
time, and also to the square of the terminal velocity divided by dou-
ble the acceleration.
From these four principal formulas we deduce by inversion,
and by the elimination of one or other of the quantities contained
in them, eight other formulas, which are collected together in a table
in the "Ingenieur," page 325.
EXAMPLE.-1. A body moving with the acceleration 15,625 feet, describes
15,625. (1,5)2
2
in 1,5 seconds the space
9
15,625
=
•
8
17,578 feet.
2. A body, which acquires a velocity v = 16,5 in consequence of an
acceleration p 4,5 feet, has described in so doing the space 8 =
(16,5)2
2. 4,5
=30,25 feet.
§ 12. On comparing two different uniformly accelerated mo-
tions, we arrive at the following conclusions.
The velocities are v = p t and v
the contrary, are s =
Putting t
V
P₁ tr²
Pr tj. The spaces, on
whence we have
p ť²
2
and si
2
S
and
p t²
P₁ t₁
S1
Piti
V ₁ tr
ว
Ρι
V1
t we obtain :
v t
v² Pi
2
v₁ p
§ 13.]
111
SIMPLE MOTION.
S
S1
ย
υι
p
Ρι
; the times being equal, the ratio of the spaces de-
scribed is equal to that of the final velocities or of the accel-
erations.
If we put p₁ = p we have
V
t
S
t²
v²
and
U1
な
​2
2
t₁
$1
The acceleration being the same, I.E., when we have the same
uniformly accelerated motion, the final velocities are to each other
as the times, the spaces described as the squares of the times, and
also as the squares of the final velocities.
Farther, if we take v₁ = vit gives
p
20
tr
S
t
and
; for the
P1
t
&1
t₁
same final velocities the accelerations are to each other inversely,
and the spaces directly as the times.
Finally, for s₁ = s we have
P t,² 2,2
P1
t³
2,2; for equal spaces de-
scribed the accelerations are to each other inversely as the squares
of the times and directly as the squares of the velocities.
§ 13. For a uniformly accelerated motion with the initial veloc-
ity c we have from § 9
I.) v = c + pt,
and since the space ct belongs to the constant velocity c, and the
p t
space
to the acceleration p
2
p
t²
II.) s = c t + 2
Eliminating p from the two equations, we obtain
or eliminating t, we find
&
III.) 8 =
c + v
2
t,
IV.) s =
2 p
EXAMPLE.—1. A body moving with the initial velocity c = 3 feet and
with the acceleration p 5 feet describes in 7 seconds the space
72
s = 3.7 + 5 .
= 21 + 122,5 = 143,5 feet.
2
2. Another body, which
in 3 minutes = 180 seconds changes its ve-
locity from 21 feet to 7 feet, describes during this time the space
2,5 + 7,5
. 180 900 feet.
2
112
[S 14.
GENERAL PRINCIPLES OF MECHANICS.
§ 14. Uniformly Retarded Motion.-For uniformly retarded
motion with the initial velocity c we have the following formulas,
which are deduced from those of the foregoing paragraph by mak-
ing p negative.
I.) v=c - pt,
II.) s = c t -
p ť³
ર
c + v
III.) s =
t,
2
C²
- v²
2 p
IV.) s =
While in uniformly accelerated motion the velocity increases
without limit, in uniformly retarded motion the velocity decreases
up to a certain time, when it is 0, and afterwards it becomes
negative, I.E., the motion continues in the opposite direction.
If we put v = 0 in the first formula, we obtain p t
C
Ρ
t = c, whence
the time in which the velocity becomes 0 is t
substituting this value of t in the second equation, we obtain the
space described by the body during this time, s =
C
c²
2 P
If the time is greater than the space is smaller than
2 c
p
p
2 P
and if the time is the space becomes = 0, the body having re-
turned to its point of departure; finally, if the time is greater than
2 c
s is negative, I.E., the body is on the opposite side of the point
of departure.
p
EXAMPLE.-A body which is rolled up an inclined plane with an in-
itial velocity of 40 feet, and which suffers a retardation of 8 feet per sec-
40
40°
ond, rises only during 5 seconds and reaches a height of
100
8
2.8
feet, after which it rolls back and arrives after 10 seconds with a velocity
of 40 feet at the point from whence it started, and after 12 seconds is al-
40.12
ready 40. 124. 122 or (40.2 + 4. 2²) = 96 feet below its point of de-
parture, if the plane continues beneath it.
& 15, 16.]
113
SIMPLE MOTION.
§ 15. The Free Fall of Bodies.-The free or vertical fall of
bodies in vacuo (Fr. mouvement vertical des corps pesants, Ger.
der freie oder senkrechte Fall der Körper) furnishes the most im-
portant example of uniformly accelerated motion. The acceleration
of this motion produced by gravity (Fr. gravité, Ger. Schwer-
kraft) is designated by g, and its mean value is
9,81 meters.
30,20 Paris feet.
32,20 English feet.
31,03 Vienna feet.
311 31,25 Prussian feet.
32,7 Bavarian or meter feet.
If any of these values of g be substituted in the formulas v=gt,
v³
g t
S
and s =
=
2
2 g
v2 g s, all possible questions in relation
to the free fall of bodies can be answered.
For the metrical system of measures we have
v = 9,81 . t = 4,429 √s,
s = 4,905 ť² = 0,0510 v²,
t =
0,1019 v² = 0,4515 √s;
and for English measures
V
32,2 t =
8,025 vs,
s = 16,1 ť²
=
0,0155 v²,
0,249 Vs.
t = 0,031 v =
EXAMPLE.—1.) A body attains when it falls unhindered in 4 seconds a
velocity v = 32,2.4
128,8 feet, and describes in this time the space 8 =
16,1 . 42 257,6 feet. 2.) A body which has fallen from the height s
9 feet, has the velocity v = 8,025 1/9 = 24,075. 3.) A body projected ver-
tically upwards with a velocity of 10 feet rises to the height s = 0,0155.
10² =
1,55 feet, in the time
or nearly of a second.
t =
0,031 . 10 = 0,31,
§ 16. The following Table shows how the motion takes place as
the time elapses,
114
[$ 17.
GENERAL PRINCIPLES OF MECHANICS.
Time in
seconds S
о
I
2
3
4 5
6
7
8
9
IO
Velocity.
Space
Difference
I
2
2
O
Ig
29
39
49
59 6g
I
g
2
g
4
g
2
g
5
162
25
369
2
2
2
2
Iog
100!
2
2
g
7°
I I
13 15 17:
19:
2
2
2
2
2
2
2
2
2
79 8g 99
49% 64% 812
The last horizontal column of this table gives the spaces de-
scribed by a body falling freely in each single second. We see that
these spaces are to each other as the uneven numbers 1, 3, 5, 7, etc.,
while the times and the velocities are to each other as the regular
series of numbers 1, 2, 3, 4, 5, etc., and the distances fallen through
as their squares 1, 4, 9, 16, etc. Whence, E.G., the velocity after 6
seconds is = 6 g = 193,2 feet, I.E., the body, if from this moment
it continued to move uniformly as on a horizontal plane which of-
fered no resistance, would describe in every second the space 6 g
193,2 feet. It does not really describe this space in the following
or seventh second, but from the last column we see that it de-
scribes exactly 132 13. 16,1 209,3 feet, and in the eighth
second 15 92 — 15. 16,1 = 241,5 feet.
REMARK.-Older German writers designate the space 16,1 feet, de-
scribed by a body falling freely in the first second, by g, and call it also the
acceleration of gravity. They employ for the free fall of bodies the for-
mulas
v = 2 g t = 2 √
2 √ g 8,
v2
8 = 9 t²
4 g'
t =
v
29
This usage, known only in Germany, is tending gradually to disappear,
which, on account of the frequent misapprehensions and errors resulting
from it, is much to be desired.
§ 17. Free Fall with an Initial Velocity.-If the free fall
of a body takes place with an initial velocity (Fr. vitesse initiale, Ger.
§ 17.]
115
SIMPLE MOTION.
Anfangsgeschwindigkeit) c, the formulas assume the following
form:
v = c +
g t = c + 32,2 t feet = c + 9,81 t meters,
v =
√ c²
+
2 g s =
c² + 64,4 s feet = Vc² + 19,62 s meters,
g
s = c t + t² = ct + 16,1 t' feet = c t + 4,905 t2 meters,
2
v²
c²
and s =
= 0,0155 (v² - c²) feet 0,0510 (2
=
0,0510 (v² - c²) meters.
2 g
If, on the contrary, the body is projected vertically upwards, we
have
gt = c - 32,2 t feet
v = c
9,81 t meters,
v = V c²
2
9
8 = √c² — 64,4 s feet
c
19,62 s meters,
g
s = ct
t² = ct
16,1 t³ feet = c t
4,905 meters,
2
c²
3 وح
= 0,0155 (cv) feet 0,0510 (c²
=
0,0510 (c² — v²) meters.
29
and s =
If we consider a given velocity c as a velocity acquired by a free
fall, we call the space fallen through
c²
2 g
0,0155 c² feet = 0,0510 c² meters,
"the height due to the velocity" (F. hauteur due à la vitesse, Ger.
Geschwindigkeitshöhe). By the substitution of the above, several
of the foregoing formulas may be expressed more simply. If we
denote the height (.) due to the initial velocity by k, and that
འབྲུ°
due to the final velocity by h, we have for falling bodies,
h = k + s and s
k+s = h k,
and for ascending bodies,
h k s and s = k — h.
The space described in falling or ascending is therefore equal
to the difference of the heights due to the velocities.
0,3875,
EXAMPLE.-If for a uniformly varied motion the velocities are 5 feet
and 11 feet, and the heights due to the velocities are 0,0155. 52
and 0,0155. 11' 1,8755, the space described in passing from one velocity
=
to the other is 8 = 1,8755 0,3875
= €1,4880 feet.
116
[§ 18.
GENERAL PRINCIPLES OF MECHANICS.
§ 18. Vertical Ascension.-If in the formula s =
c³ — v²
29
for the vertical ascension of bodies we put the final velocity v =
0, we obtain the maximum height of ascension,
S =
C²
2 g
consequently the maximum height of ascension, corresponding to
the velocity c, is equal to the height of fall h due to the final velo-
city c, and therefore c√2 gk is not only the final velocity for
the height of free fall, but also the initial velocity for the maxi-
mum height of ascension k. Hence it follows that a body pro-
jected vertically upwards has at any point the same velocity, which
it would have, in the opposite direction, if it fell from a height
equal to the remaining height of ascension to that point, and which
it really possesses afterwards, when it reaches it upon falling back.
EXAMPLE.-A body projected vertically upwards, with a velocity of 15
feet, after ascending 2 feet meets an elastic obstruction, which throws it
back instantaneously with the same velocity with which it struck. How
great is this velocity, and how much time does the body require to ascend
and fall back again? The height due to the initial velocity 15 feet is
3,49 feet, and the height due to the velocity at the instant of collision is
h
3,49 2,00 1,49, and, consequently, the velocity itself is 8,025
√1,499,8 feet. The time necessary to ascend the entire height (3,49 feet)
would be t = 0,031 c = 0,031 . 15 0,465 seconds, while the time neces-
sary to ascend the height 1,49 is t₁
the time necessary to ascend the 2 feet is t
1
-
0,031 . 9,8
0,3038 seconds, whence
0,465 0,3038
ts
0,1612 seconds, and finally the whole time employed in ascending and fall-
=
or
3224
ing is = 2. 0,1612 0,3224 seconds. This, therefore, is but
9300'
about of the time, which would be employed by the body in rising and
falling if it met with no obstacle. This case occurs in practice in forging
red-hot iron, for we are obliged to give as many strokes of the hammer
as possible in a short space of time, on account of the gradual cooling of
the iron. If by means of an elastic spring we cause the hammer to be
thrown back, it can, under the circumstances supposed in the example,
make three times as many blows as when its rise was unimpeded.
REMARK 1.—In practical mechanics, particularly in hydraulics, we are
often obliged to convert velocity into height due to velocity, or the latter
into the former. A table, by means of which this operation can be per-
formed at once, is of the greatest service to the practical man.
Such a
one, calculated for the Prussian foot, is to be found in the "Ingenieur,"
page 326 to 329.
§ 19.]
117
SIMPLE MOTION.
REMARK 2.-The formulas deduced in the foregoing paragraphs are
strictly correct only for bodies falling freely in vacuo; they are, however,
sufficiently accurate for practical purposes, when the weight of the body is
great compared to its volume, and when the velocities are not very great.
They are, besides, employed in many other cases, as will be shown here-
after.
§ 19. Variable Motion in General.-The formula s = ct
(§ 5) for uniform motion holds good also for every variable motion,
if instead of t we substitute an element or an infinitely small in-
stant of the time 7, and instead of s the space o described in this
instant, for we can assume that during the instant the velocity c,
which we here denote by v, remains constant, and that the mo-
tion itself is uniform.
Hence, we have for every variable motion
σ
I.) σ = v 7, and v = (compare § 10).
Т
The velocity (v) for every instant is given by the quotient of the
element of the space divided by that of the time.
In like manner the formula v = pt (§ 11) for uniformly accele-
rated motion holds good also for every variable motion, if instead of
t and v we substitute the element of time and the infinitely small
increase of velocity k during that time, for the acceleration p
does not vary sensibly in an instant 7, and the motion can be re-
garded as uniformly accelerated during this instant. Consequently
we have for all motions
II.)
K
« = p т, and p = ½
T
The acceleration (p) is, therefore, equal to the element of the ve-
locity divided by the element of the time.
If we put the total duration of the motion t = n 7, and the ve-
locities in the successive instants 7 are V1, V2, 2'3 Un, the corres-
ponding elements of the space are o
· •
1 = 11 T, O₂ = V2 T, 03 = 13 T
σ„ = v″ 7, and the total space described is
n
s = (11 + 12 + 1½ . . . Vn) T
=
•
1, + 12 + ... + 1'n
1 T, I.E.,
N
I*) S
V₁ + V₂ +.
V2
+...+?
t = vt, when
N
V
V₁ + V3 + V3•••Vn denotes the mean velocity of the body while
N
describing the space s.
118
[$ 19.
GENERAL PRINCIPLES OF MECHANICS.
In like manner if c denotes the initial and v the final velocity,
and if P1, P2 · · · P, denote the accelerations in the equal successive
instants 7, we have
v − c = (P₁ + P2 + ...P„) T =
)?
P₁ + p₂+...+Pn
n T, I.E.,
II*)
V
C
(P₁ + P₂ +
•
22
· P₁) t = pt, when
c)
Pn
ጎ
p =
2
P₁+P₂+• • •+ Pn denotes the mean acceleration.
N
By combining the formulas I. and II. we obtain the following
not less important equation :
III.) v k = p o.
If, while the space s = no is described, the acceleration assumes
successively the values p₁, P2...P, the sum of the products p o is
P1,
(P₁ + P2 • • . + P„) σ =
- • •
(P₁ + P s + + P³) n
n o
N
= (²
P₁ + P₂ + ... + Pn
N
s=ps.
If the initial velocity c is transformed by successive increases
V
с
of k =
K
N
V K is
into the final velocity v, the sum of the products
CK + (C + k) K + …….. + (v −k) k + v k=[c + (c + k) + . . . + (v−k) + v] k
N K
(v + c)
(v + c) (v — c)
-
2 رح
c²
2
2
૭
and therefore we can write
v2
c²
v²
III*)
= p s, or s =
(compare IV., § 13).
2
2 p
With the aid of the foregoing formulas we can solve the most
varied problems of phoronomics and mechanics.
The time, in which the space s = n o is described with the vari-
able velocities V1, V2, ... V»,
1 1
V₁
IV.) t = σ + +
V2
၇)ရ
1
•
when we put the value (
is
1)
1
N \V
$
C
1
1
+
+..+
V₂
Vn
1
1
whose recip-
+ +
+
V n
v
n v₁
V1 V2
rocal v can be considered as the mean velocity.
EXAMPLE.-When a body moves according to the law va t², we have
v + k = a(t + 7)² = a (t² + 2 t T + T´), and к = a ↑ (2 t + 7), consequently
IC
p =
= 2 a t.
T
$ 20.]
119
SIMPLE MOTION.
The velocities of the body at the end of the times
T, 2T, 3 T... N 7 are a r², a (2 r)², a (3 т)².. α (n 7)²,
whence it follows that the space described in tn 7 seconds is
•
$ == [a T² + a (27)² + ..a (n 7)²] 7 = (1² + 2² + 3² + + n²) a +³,
but from Article 15, IV., of the Introduction to the Calculus we have
1 + 2 + 3 +..+n
hence
3
12.3
a
ат
(n =)³
=
3
a t³
3
(§ 20.) Differential and Integral Formulas of Phoronc-
mics. The general formulas of motion found in the foregoing
paragraphs assume, when the notations of the calculus are em-
ployed, I.E., when the element of time is designated by d t, the
element of space o by d s, and the element of velocity « by d v',
following form:
or ds = vdt, whence s =
I.)
v =
d s
d ť
d v
II.)
Ρ
d ť
S
= fed
v d v
or dv=pdt,
III.) v dv = p d s, or s =
S
the
ds
Svdt, and t
S
V
Sav
= pdt, whence v = pdt, and t=
P
>
and
2,2
C
7 c² = Sp a s
2
d
d v
in which e denotes the initial and v the final velocity, while the
space s is being described.
We see from the above that the difference of the squares of the
velocities is equal to twice the integral of the product of the accelera-
tion and the differential d s, or equal to the product of the mean ac-
celeration and the space described by the body in passing from the
velocity c to the velocity v.
According to the theory of maxima and minima the space is a
maximum, and the motion attains the greatest extension, when we
have
d s
= v = 0,
d t
and the velocity is a maximum or minimum when
d v
d t
p
= 0.
The foregoing are the fundamental formulas of the higher
Phoronomics and Mechanics.
EXAMPLE.-1. From the equation for the space s =
+3t+t, we
deduce by differentiation the equation for the velocity v=3+2 t, and that
120
[$ 20.
GENERAL PRINCIPLES OF MECHANICS.
for the acceleration p =
formly accelerated,
2; the latter is constant and the motion is uni-
For t
0, 1, 2, 3 . . . seconds, we have
v =
• •
3, 5, 7, 9 (Feet), and
$
2, 6, 12, 20. . . (Feet).
2. From the formula for the velocity
v = 10 + 3 t t2, we obtain by integration
$
t2 d
=f10at + Sзtat - Seat = 10
10 t + 3 ť²
3'
3 2 t.
and on the contrary by differentiation p =
P
Consequently, for 3 2 t=0, 1.E., for t=3 seconds, the acceleration is 0
and the velocity is a maximum (v 121), and for 10 + 3 t — ť² = 0, I.E.,
for
5 the velocity is = 0 and the space is a maxi-
3+7
t // + √ 10 + 1/
2
mum.
Fort
p =
0,
1,
2,
3,
4,
5,
6 seconds we have
3, 1,
1,
3,
5,
7,
9 feet,
v = 10, 12,
12,
10,
6,
0,
8 feet,
8 = 0, 11,
231, 341, 42, 455, 42 feet.
3. For the motion expressed by the formula p
ignates a constant coefficient, we have
μs, in which
μ des-
v2
C²
μ
2
² = Spa s = − u fs ds = -18",
or v² — c² — µ 8²;
2
whence
v = √c²
μs and 8 =
s 1
μ
d 8
d 8
1
d s
We have also dt
V
√c²
fl
μ 82
C
11
1 -
8
М
C
when we put
с
11
g
d
μ
с
8
μ
C
I u
√μ √1 — u²,
= u; and it follows that (see Art. 26, V., of the Introduc-
tion to the Calculus).
t ==
1
8 =
Гре
с
√μ
sin.
И
μ
sin.-1 8 √μ, and
sin. (t√μ), as well as
= c cos. (t √μ) and
μ
d
8
V =
d t
d v
c √μ sin. (t √μ).
}
с
p = d t
When the motion begins we have, for t =
0, 8 = 0, v = c and p = = 0,
and afterwards for
§ 21.]
121
SIMPLE MOTION.
t
नै
• √ μ = = /, or 1 = 2 √ μ
2
μ = π, or t =
C
g
v = 0 and p =
√, for
μ
$ =
0, v
c and p =
0, for
μ
3 π
с
ιν μ
π, or t =
v =
2
μ
ν μ
2 п
μ
t √ µ = 2 π, or t =
Vie
0 and p = c√μ, and for
0, v = c and p = 0.
The moving point has therefore a vibratory motion upon both sides of the
fixed point of beginning, to which it returns every time that it has de-
scribed, with a velocity which gradually increases from 0 to vc, the
space s = ±
C
S
t'
§ 21. Mean Velocity. The velocity c₁= which we find
when we divide the space described during a certain time, E.G.,
during the period of a periodic motion, by the time itself, differs
from the velocity v =
σ d
Т
(29)
t
for an instant or during the ele-
ment of time (dt). We call the former the mean velocity (Fr.
vitesse moyenne, Ger. mittlere Geschwindigkeit), and we can con-
sider it as the velocity that a body must have, to describe uniformly
in a certain time (t) the space (s) which it really does describe with a
variable motion in the same time. When the motion is uniformly
variable the mean velocity is equal to the half sum of the initial and
of the final velocity, for according to § 13 the space is equal to this
+") multiplied by the time (†).
c +
m 1 5
sum
In general, the mean velocity is (according to § 19) c₁ =
V₁ + V₂ +
N
FIG. 51.
0
2'n
in which ₁, Vas
2 denote the velocities corre-
•
UM ON,
sponding to equal and very small intervals of time.
EXAMPLE.—While a crank is turned uniformly in a circle U M O N,
Fig. 51, the load Q attached to it, E.G., the piston of an
air or water pump, etc., moves with a variable motion up
and down; the velocity of this load is at the highest and
lowest points U and O a minimum, and equal to zero, and
at half the height at M and Na maximum, and equal to the
velocity of the crank. Within a half revolution the mean ve-
locity is equal to the whole height of ascent, I.E., the diam-
eter UO of the circle in which the crank revolves, divided
by the time of a half revolution. If we put the radius of the
circle in which the crank revolves, C U = C 0 = r, that

M
U
N
122
[§ 22, 23.
GENERAL PRINCIPLES OF MECHANICS.
is, its diameter = 2 r, and the time equal to t, it follows that the mean
velocity.c₁ =
2 r
t
The crank in the same time describes a half circle
•
πr, and its velocity is c = and therefore the mean velocity of the load
C1
2
с
π
2 t
3,141
the crank.
Пр
t
c is 0,6366 times as great as the constant velocity c of
§ 22. Graphical Representation of the Formulas of Mo-
tion. The laws of motion which have been found in the foregoing
paragraphs can be expressed by geometrical figures, or, as we say,
graphically represented. Graphical representations, as they ren-
der the conception of the formula more easy, assist the mem-
ory, protect us from many errors, and serve also directly for
the determination of quantities which may be required, are
of the greatest use in mechanics. In uniform motion, the space
D
FIG. 52.
N
B
A
M
(s) is the product (ct) of the velocity and
the time, and in Geometry the area of a rect-
angle is equal to the product of the base by
the altitude; we can therefore represent the
space described (s) by a rectangle A B CD,
Fig. 52, whose base A B is the time t, and
whose altitude A D B C is the velocity c,
provided the time and the velocity are expressed by similar units
of length, that is, if the second and the foot are represented by
one and the same line.

§ 23. While in uniform motion the velocity (M N) at any mo-
ment (AM) is the same, in variable motion it is different for each
instant; therefore this motion can only be represented by a four-
FIG. 53.
N
P
E
D
F
K
sided figure, A B C D, Fig. 53, the base
of which A B, denotes the time (†), the
other boundaries being the three lines,
A D, B C, and C D. The first two
of these lines denote the initial and final
velocities, and the last one is determined
by the extremities (N) of the different lines
representing the velocities corresponding
to the intermediate times (M). Accord-
ing to the nature of the variable motion in question, the fourth
line CD is straight or curved, rises or sinks from its origin, and is

A
1
MOH
$ 21.]
123
SIMPLE MOTION.
concave or convex towards the base. In every case, however, the
area of this figure is equal to the space (s) described; for every sur-
face A B C D, Fig. 53, can be divided into a series of small
strips M O P N, which may be considered as rectangles, and the area
of each of which is a product of the base (MO) and the corresponding
altitude ( M N) or (0 P), and in like manner the space described
in a certain time is composed of small portions, each one of which
is a product of an element of time and the velocity of the body
during that instant. The figure also shows the difference between
the measure of the velocity and the space actually described in the
following unit of time. The rectangle M L, above the base
MH = unity (1) = v. 1 is the measure of the velocity M, and on
the contrary, the surface M K above the same base represents the
space actually described. In the same way the rectangle AF over
A I =
unity is the measure of the initial velocity A D = c, and the
surface A E that of the space actually described in the first second.
§ 24. In uniformly variable motion the increase or decrease v-c
of the velocity (= p t, § 13) is proportional to the time (t). If in
Fig. 54 and Fig. 55 we draw the line D E parallel to the base 4 B,
we cut off from the lines B C and M N, which represent the velo-
FIG. 54.
N
FIG. 55.

0

D
E
..Z
N
ΤΟ
D
E
A
B
M
A
B
M
cities, the equal portions B E and M O, which are equal to the
line A D representing the initial velocity, there remain the pieces
CE and N O, which represent the increase or decrease in velocity;
for these we have from what precedes the proportion
NO: CE DO: DE.
=
Such a proportion requires that N, as well as every point of the
line CD, shall be upon the straight line uniting C and D, or that
the line CD, which limits the velocities M N, shall be straight. Con-
sequently the space described in uniformly accelerated or retarded
motion can be represented by the area of a Trapezoid A B C D,
124
[§ 25, 26.
GENERAL PRINCIPLES OF MECHANICS.
c + v
2
whose altitude A B is the time (t) and whose two parallel bases
A D and B C are the initial and final velocity. The formula found
in § 13 s = t corresponds exactly to this figure. For uni-
formly accelerated motion the fourth side D Crises from the point
of origin, and for uniformly retarded motion this line descends
from the same point. When the uniformly accelerated motion be-
gins with a velocity equal to zero, the trapezoid becomes a trian-
gle, whose area is B C . A B i v t.
§ 25. The mean velocity of a variable motion is the quotient of
the space divided by the time; it gives, when multiplied by the
time, the space, and can be considered as the altitude A F
BE of the rectangle A B E F, Fig. 56, the base of which A Bis
equal to the time t, and the area of which is equal to that of the
four-sided figure A B C N D, which measures the space described.
The mean velocity is found by changing the four-sided figure A B
CND into an equally long rectangle A B E F. Its determina-
tion is especially important for periodic motion, which occurs in
almost all machines. The law of this motion is represented by
the serpentine line C D E F G, Fig. 57. If the right line L M,
FIG. 56.
N
F
E
L
D
D
A
M
B
FIG. 57.
E
G


M
F
A
N
O
P
B
drawn parallel to A B, cuts off the same space as the serpentine.
line, then L M is also the axis of CD EF G, and the distance A L
B M between the two parallels A B and L M is the mean ve-
locity of the periodic motion, and, on the contrary, A C, O E, BG,
etc., are the maximum, and ND and P F the minimum velocities
of a period 4 0, 0 B, etc.
§ 26. The acceleration or the continuous increase of velocity in a
second can easily be determined from the figure. In uniformly
accelerated motion it is constant, and is therefore the difference
P Q, Fig. 58 and Fig. 59, between the two velocities OP and M N,
$26.]
SIMPLE MOTION.
125
one of which corresponds to a time (M O) one second greater than
the other. If the motion is variable, but not uniformly, and the line
FIG. 58.
FIG. 59.
C

Р
N
D
A
M 0
B
D

N
P
A
MO
B
of velocity C D therefore a curve, the acceleration at every instant
is different, and consequently it is not really the difference P Q of
the velocities O P and M N = 0 Q, Figs. 60 and 61, which are
those at times differing one second MO from each other, but it
A
D
FIG. 60.
C
E
P
İR
N
MO
B
FIG. 61.

E

R
C
N
D
A
M 0
B
is the increase R Q of the velocity M N, which would take place,
if from the instant M the motion became a uniformly accelerated
one, that is if the curve N P C became a straight line N E. But
the tangent N E is the line in which a curve D N would prolong
itself, if from a certain point (N), its direction remained unchanged:
the new line of velocity coincides with the tangent, and the perpen-
dicular O R which reaches to this line is the velocity which would
have existed at the end of a second, if at the beginning of the same
the motion had become a uniformly accelerated one, and therefore
the difference R Q between this velocity and the initial one (MN)
is the acceleration for the instant which corresponds to the point
M in the time line A B. We can also of course consider the time
and the accelerations as the co-ordinates of a curve, in which case.
the velocities are represented by surfaces.
126
GENERAL PRINCIPLES OF MECHANICS. [§ 27, 28.
1
CHAPTER II.
COMPOUND MOTION.
§ 27. Composition of Motion.-The same body can possess,
at the same time, two or more motions; every (relative) motion is
composed of the motion within a certain space, and of the motion
of this space within or in relation to another space. Every point
on the earth possesses already two motions; for it revolves once
every day around the earth's axis, and with the earth once a year
around the sun. A person moving on a ship has two motions in
relation to the shore, his own motion proper and that of the ship;
the water which flows out of an opening in the side or in the bot-
tom of a vessel carried upon a wagon has two motions, that from
the vessel, and that with the vessel, etc.
Hence we distinguish simple and compound motion. The rec-
tilinear motions of which other rectilinear or curvilinear motions
are composed (Fr. composés, Ger. zusammengesetzt), or of which
we can imagine them to be composed, are simple motions (Fr. sim-
ple, Ger. einfach). How several simple motions can be united so
as to form a compound one, and how the decomposition of a com-
pound motion into several simple ones is accomplished, will be
.shown in what follows.
§ 28. If the simple motions take place in the same straight line,
'their sum or difference gives the resulting compound motion, the
former when the motions are in the same direction, and the latter
when the motions are in opposite directions. The correctness of
this proposition becomes evident, when we combine the spaces de-
scribed in the same time by virtue of the simple motions. The
spaces c₁ t and c₂ t described in the same time correspond to uni-
form motions whose velocities are c, and c, and if these motions
are in the same direction the space described in t seconds is
s = c₁ t + c₂ t = (C₁ + c₂) 't,
and consequently the resulting velocity of the compound motion is
the sum of the velocities of the simple motions. When the mo-
tions are in contrary directions, we have
$29, 30.]
127
COMPOUND MOTION.
8 = c₁ t — c₂ t = (C₁ — C₂) t,
and the resulting velocity is equal to the difference of the simple
velocities.
EXAMPLE.-1. A person, walking upon the deck of a ship with a velo-
city of 4 feet in the direction of the motion of the latter, appears to people
on shore, when the ship moves with a velocity of 6 feet, to pass by with a
velocity of 4 + 6 = 10 feet.
2. The water discharged from an opening in the side of a vessel with a
velocity of 25 feet, while it is moved simultaneously with the vessel in the
opposite direction with a velocity of 10 feet, has in reference to the other
objects which are at rest a velocity of only 25 10 15 feet.
-
§ 29. The same relations also obtain for variable motion. If
the same body has, besides the initial velocities c, and c,, the con-
stant accelerations p, and p, the corresponding spaces are c, t, c₂ t,
¿ P₁ ť², à p₂ t², and if the velocities and the accelerations have the
same directions, the total space described in virtue of the compo-
nent motions is
FIG. 62.
A
1
t
ť
8 = (c₁ + €₂) † + (P₁ + P:) ½·
9་
If we put c₁ + c₂ = c and p₁ + P₂
p, we obtain s = c t + P Q
9
whence it follows that not only the sum of the component
velocities gives the velocity of the resulting or compound
motion, but also that the sum of the accelerations of the
simple motions gives its acceleration.

B
M
EXAMPLE-A body upon the moon has imparted to it by the
moon an acceleration p₁ = 5,15 feet, and from the earth an ac-
celeration p₂ = 0,01 feet. Therefore, a body A, Fig. 62, beyond
the moon M and the earth E, falls towards the centre of the
moon with an acceleration of 5,16 feet, and a body B between M
and E with an acceleration of 5,14 feet.
§ 30. Parallelogram of Motions. If a body possesses at the
same time two motions which differ from each other in direction,
it takes a direction which lies between those of the two motions,
and if these motions are of different kinds, E.G., if one is uniform
and the other variable, the direction changes at every point, and
the motion is curvilinear.
We find the point 0, Fig. 63, which a body moving at the same
time in the direction A X and A Y, occupies at the end of a cer-
128
[៛ 31.
GENERAL PRINCIPLES OF MECHANICS.
tain time (t) by seeking the fourth corner O of the parallelogram A
MON, determined by the spaces A M =x AN
x and A N = y, de-
scribed simultaneously, and by the angle X AY which the direc-
tions of motion form with one another.
N
FIG. 63.
Y
Z
-X
We can convince ourselves of the correct-
ness of this proceeding by supposing the
spaces x and y described not simultane-
ously, but one after the other. By virtue
of one motion the body describes the space
A M = x, and by virtue of the other from
M in the direction A Y, that is on a line
M O parallel to A I, the space A Ny.
If we make MO AN, we obtain in O
the position of the body which corresponds to the two motions x
and Y, and which, according to this construction, is the fourth cor-
ner of the parallelogram. We can also imagine the space A M=
to be described in a line A X, which with all its points moves
forward in the direction A Y, and therefore carries M parallel to
AY and causes this point to describe the path MO A N = y.

M
=
§ 31. Parallelogram of Velocities.-If the two motions in
the directions A X and A Y take place uniformly with the ve-
locities c₁ and c, the spaces described in a certain time t are x =
c₁t and y
N
Y
= c, t, and their ratio
FIG. 64.
X
Y C2
is the same for all times,
C1
a peculiarity which is possessed only by
It follows
the right line A O, Fig. 64.
therefore that the direction of the com-
pound motion is always a straight
line. If we construct with the veloci-
ties A B =c₁
c₁ and A C = c, the paral-
lelogram A B C D, its fourth corner D
gives the point where the body is at
X the end of the first second, but since
the resulting motion is rectilinear, it
follows that it takes place in the direction of the diagonal of the
parallelogram constructed with the velocities. If we designate by
s the space A O really described in the time t, we have from the
similarity of the triangles AM 0 and ABD

B
M
§ 32.]
129
COMPOUND MOTION.
S
A D
X
A B'
whence it follows that this space
x. AD
c₁t. AD
= A D.t.
A B
C₁
According to the last equation the space described in the di-
agonal is proportional to the time (t), and therefore the compound
motion is itself uniform and its velocity c equal to A D.
Therefore the diagonal of a parallelogram, constructed with two
velocities and with the angle inclosed by them, gives the direction
and magnitude of the velocity, with which the resulting motion actu-
ally takes place. This parallelogram is called the parallelogram
of velocities (Fr. parallelogramme de vitesse, Ger. Parallelogram
der Geschwindigkeiten); the simple velocities are called compo-
nents (Fr. composantes, Ger. Seitengeschwindigkeiten), and the
compound velocity the resultant (Fr. resultante, Ger. die resulti-
rende or mittlere).
A
€2
§ 32. By employing trigonometrical formulas, the direction
C 1
FIG. 65.
B
α
D
X
and magnitude of the resulting veloc-
ity can be found by calculating one
of the equal triangles, E.G., A B D.
of which the parallelogram of velocities
is composed, by which we obtain the re-
sulting velocity AD c in terms of
the components A B = c, and A C
c and of the angle included between
them B A C′ = a.

and the angle B A D = 0,
velocity c₁, by the formula
tang. 4 =
We have also
For we obtain c by the formula
2
c = √ c²² + c₂² + 2 c₁ c₂ cos. a,
which the resultant makes with the
c2. sin. a
C₁ + C₂ cos. a'
C2
C₂ sin. a
sin.
=
or
с
C₁
or cotang. O
cotang. a +
c₂ sin. a
tang. ( 1 − 4 ) :
C₁ C2
C₁ + C₂
a
tang.
2
2
1
If the velocities c, and c, are equal to each other, the parallelo-
gram is a Rhombus, and in consequence of the diagonals being at
right angles to each other, we have more simply
c = 2 c, cos. 1 a and o
11 a.
130
[§ 33.
GENERAL PRINCIPLES OF MECHANICS.
If the velocities are at right angles, we have also more simply
2
C =
1
2
c+c and tang. =
C2
C1
EXAMPLE.-1. The water discharged from a vessel or from a machine
has a velocity c₁ 25 feet, while the vessel itself is 'moved with a velocity
19 feet in a direction, which forms with that of the water an angle
130°. What is the direction of the resultant or absolute velocity
of the water?
2
a"
c = √√25² + 19ª + 2 . 25 . 19 cos. 130º
625 + 361 50.19. cos. 50º.
―
= √986 950 cos. 50° = √986 — 610,7 =√375,3 = 19,37 feet
is the required resulting velocity.
Further, sin. 4 =
19 sin. 130°
19,37
= 0,9808 sin. 50° = 0,7513, hence the
angle formed by the direction of the resultant with that of the velocity c₁ is
* = 48° 42', and the angle formed by it with the direction of the motion
of the vessel is a
= 81° 18'.
2. If the foregoing velocities were at right angles to each other, we
would have cos. a=cos. 90°=0, and therefore the resulting velocity c=√986
= 31,40 feet, and also tang. = 18 = 0,76, hence the angle formed by it with
the first velocity is 37° 14'.
19
§ 33. We can also consider every velocity to be composed of
C
S
A
FIG. 66.
B
α
D
Χ
locities sought, and we have
two components, and therefore under
certain conditions can decompose it
into such components. If, for example,
the angles DA X = o, and DAY
=, Fig. 66, which the required
velocities form with the resultant
A D = c, are given, we draw through
the extremity D of the line represent-
ing c other lines parallel to the di-
rections AX and A Y: the points of
intersection B and D cut off the ve-

A B c, and A C = C2.
Trigonometry gives these velocities by the formulas
C₁
c sin. 4
sin. (+4)'
C₂ =
C2
c sin.
sin. (4 + 4)*
Generally, in the application of these formulas, the two velocities
are at right angles to each other, and
$ +4
=
90°, sin. (p + 4) = 1, whence
c₁ = c cos. & and c, = c sin. p.
§ 34.]
131
COMPOUND MOTION.
We can also determine, when one component (c) and its angle
of direction (4) are given, the magnitude and direction of the
other. Finally, if the three velocities c, c, and c, are given, we can
determine their angles of direction by the same method that we
employ to find the angles of a triangle, when three sides are given.
EXAMPLE.—If the velocity c = 10 feet is to be decomposed into two
components whose directions form with that of c the angles = 65° and
y = 70º, we have
€1
10 sin. 70'
sin. 135°
9,397
sin. 450
=13,29 feet and c₂:
10 sin. 65°
sin. 135°
9,063
0,7071
=12,81 feet.
§ 34. Composition and Decomposition of Velocities.-
By repeated use of the parallelogram of velocities, any number of
velocities can be combined so as to give a single resultant.
The construction of the parallelogram A B D C (Fig. 67) gives the
resultant A D of c, and c, the construction of the parallelogram
ADFE gives the resultant of A D and A E = c, and from the
construction of the parallelogram A F H G we obtain the result-
ant A H = = c of A Fand A G = C₁, or that of c₁, C2, C3 and C4.
The most simple manner of resolving this problem is by the
construction of a polygon A B D FH, whose sides A B, B D, D F'
and FH are parallel and equal to the given velocities C₁, C, C and
c4, and whose last side is always equal to the resulting velocity.
FIG. 67.
H
FIG. 68.

H
F

F
E
G
E
C
C
D
A
C1
B
A
C1
B
In case the velocities do not lie in the same plane, the re-
sultant can also be found by repeated application of the paral-
lelogram of velocities. The resultant A F= c (Fig. 68) of three
velocities A B = c₁, A C = c, and A E
C₂ and A E = c, not in the same
plane, is the diagonal of a parallelopipedon whose sides are equal
132
[Ş 35, 36.
GENERAL PRINCIPLES OF MECHANICS.
to the velocities. We often employ for this reason the term paral-
lelopipedon of velocities.
§ 35. Composition of Accelerations.-By the composition
of two uniformly accelerated motions, beginning with a velocity
0, we obtain also a uniformly accelerated motion in a straight line.
If we designate the accelerations of the motions in the directions
A X and ▲ Y (Fig. 69) by p, and p, the spaces described during the
time t are
A
FIG. 69.
C
N
AM = x =
P₁ t²
2
and

A N = y = P₂ t²
2'
and their ratio is
X
Y
P₁ t²
P₂ t²
P1
P₂
A
X
B
M
which is entirely independent of the
time, therefore the path AO is a
straight line. If we make A B = P1,
and B D = A CP2, we obtain a parallelogram A B D C, and
we have
A O
AD
AM
= { p₁ t²
A B
Pi
=t, whence AO AD.ť
=
{
According to this equation the space AO of the compound motion
is proportional to the square of the time; the motion itself is there-
fore uniformly accelerated, and its acceleration is the diagonal A D
of the parallelogram constructed with the two simple accelera-
tions.
We see, therefore, that we can combine several accelerations so
as to form a single one, or decompose a single one into several
others by means of the parallelogram of accelerations (Fr. parallél-
ogramme des accélérations, Ger. Parallelogram der Accelerationen)
according to exactly the same rules as we perform the composition
and decomposition of velocities by means of the parallelogram of
velocities.
§ 36. Composition of Velocities and Accelerations.
By the combination of a uniform motion with a uniformly ac-
celerated one we obtain, when the directions of the two motions
do not coincide, a motion which is completely irregular. If during
a certain time t, by virtue of the velocity c, the space
A N = y = c t
$ 36.]
133
COMPOUND MOTION.
is described in the direction A Y, Fig. 70, and if during the same
time, by virtue of a constant acceleration, the space
A M =
X
p ť²
2
is described in the direction A X at right angles to the former,
then the body will be in the corner O of the parallelogram con-
structed with y = c t and x
p t²
2
By the aid of these formulas,
it is true, we can find the position of the body for any given time,
but these positions do not lie in the same straight line; for if we
substitute the value of t
Y taken from the first equation, in the
C
second we obtain the equation of the path
X
p yo
2 c²*
According to this formula the space (x) described in one direction
varies, not as the space, but as the square (y²) of the space described
FIG. 70.
A
1
2
3
11
14
M
9
X
•
A
P
FIG. 71.
C


D
B
E
in the other direction, and the path of the body is therefore not a
straight line, but a certain curve known in Geometry as the parab-
ola (Fr. parabole, Ger. Parabel).
REMARK.- Let A B C, Fig. 71, be a cone with a circular base A E BF,
and D E F a section of the same parallel to the side B C and at right an-
gles to the section A B C, and let OP N Q be a second section parallel to
the base and therefore circular. Further, let E F be the line of intersec-
tion between the base and the first section, and finally, let us suppose the
parallel diameters A B and P Q to be drawn in the triangular section A B C
and the axis D G in the section DE F. Then for the half chord MN
= MO we have the equation M N P M. MQ; but MQ =GB and for
134
[§ 37.
GENERAL PRINCIPLES OF MECHANICS.
P M we have the proportion PM: DM = A G : DG, whence
MN2B G.
DM. AG
D G
But we have also G E²=BG. A G; whence, dividing the first equation
by the second,
D M MN2
D G G E²
The portions cut off from the axis (abscissas) are as the squares of the cor-
responding perpendiculars (Ordinates). This law coincides exactly with
the law of motion just found; the motion takes place then in a curved
line D N E, which is one of the conic sections. For the construction, po-
sition of the tangent, and other properties of the parabola, see the Inge-
nieur, page 175, etc.
§ 37. Parabolic Motion.-In order thoroughly to under-
stand the motion produced by the combination of velocity and
acceleration, we must be able to give for any time (t) the direction,
velocity, and the space described. The velocity parallel to A Y is
constant and = c, and that parallel to A X is variable and =
T
FIG. 72.
A
N
M
ዎ
PV
R
X
ང་
Y
sequently, for the angle POR
makes with the direction (axis)
have the following formula
tang.
pt;
if we construct with these ve-
locities O Q = c and 0 P = pt
the parallelogram O P R Q,
Fig. 72, we obtain in the di-
agonal OR the mean velocity,
or that with which the body in
O describes the parabolic path
A O U. This velocity itself is
v = √ c² + ( p t)².
O R gives also the tangent
or the direction in which the
body moves for an instant; con-
XTO, which the same
A X of the second motion, we

=
σε
OP
C
pt
Finally, to obtain the space described or the arc of the curve
A 0 = s, we can employ the formula σ = v T (§ 19), by the aid of
which we can calculate the small portions which we can consider
as elements. The calculus also gives a complicated formula for the
computation of an arc of a parabola.
38.]
135
COMPOUND MOTION.
FIG. 73.
N
Y
1
=
§ 38. We have previously supposed that the primitive directions
of motion were at right angles to each other, and we must now
consider the case, when the direction of the acceleration makes any
arbitrary angle with that of the velocity. If the velocity of
the body in the direction A Y₁ (Fig. 73) is c, and if, in the
direction A X, which forms
an angle X, A Y, a with
the former, the acceleration
is p, A is no longer the ver-
tex, and A X₁ no longer
the axis, but only the di-
rection of the axis of the
parabola. The vertex of the
parabola is situated at a
point whose co-ordinates, in
reference to the point of be-
B
M
X
at
T

A
F
a
Ev
D
b
منا
ginning of the motion, are CB = a and B Ab, of which the former
lies in the axis of the parabola and the latter is at right angles to
it. The velocity A D = c is composed of the two components
A F = c sin. a and A E= c cos. a. The first of these is constant,
and the latter is variable, and always equal to the variable velocity
pt, provided that the body requires the time t to pass from the
vertex C to the real point of beginning.
Hence we have
C cos. α = p t, whence t =
c. cos. a
and therefore
p
1) C B = a =
p t
2
c² cos.² a
and
2p
c² sin, a cos. a
c² sin. 2 a
2) BA= b = c sin. a. t =
ม
2 p
If we have determined by these distances the vertex C of the
parabola, starting from this point we can, for any given time, de-
termine the position O of the body. Besides, if we put CM x and
MO=y, the general formula
=
py²
X
2 c² sin." a'
or y = c sin. a
c sin. a √ 2 x
holds good.
Ρ
REMARK.—One of the most important applications of the theory of par-
abolic motion, just discussed, is to the motion of projectiles. A body pro-
jected in an inclined direction either upward or downward would describe,
in virtue of its initial velocity c and of the acceleration of gravity (g
= 32.2
feet), an arc of a parabola, if the resistance of the air were done away with,
136
[$ 39.
GENERAL PRINCIPLES OF MECHANICS.
or if its motion took place in vacuo. If the velocity of projection is not very
great and if the body is very heavy compared with its volume, the diver-
gence of the body from a parabolic path is small enough to be neglected.
The most perfect parabolic trajectories are those described by jets of water
issuing from vessels, fire-engines, etc. Bodies shot from guns, etc., E.G.,
musket balls, describe, in consequence of the great resistance of the air,
paths which differ very sensibly from a parabola.
A
§ 39. Motion of Projectiles. - A body projected in the di-
X
ર
T
I
FIG. 74.
M
Ο
α
D
N
B
tude du jet, Ger. Wurfweite).
rection A Y at an angle
of elevation YAD= a,
Fig. 74, ascends to a cer-
tain height B C, which
is called the height of
projection (Fr. hauteur
du jet, Ger. Wurfhöhe),
and it reaches the hori-
zontal plane from which
it started in A, at a dis-
tance A D from it, which
is called the range of
projection (Fr. ampli-

From the velocity c, the acceleration g and the angle of eleva-
tion, we obtain, according to § 38, when we replace p by g and
a° by 90° + aº, or cos. a by sin. a, etc.
the height of projection C B = a =
:
c³ sin.² a
and
2 g
c² sin. 2 a
half the range of projection AB = b =
2 g
From the last formula we see that the range of projection is a
maximum for sin. 2 a = 1, or 2 a = 90°, that is for a = 45º. A
body projected at an angle of elevation of 45° attains the greatest
range of projection.
We have also
a
I f²
2
2 c² cos." a
a
and for a point O in the path of the projectile for which C M = x
and M 0 = y,
g y²
X
2 c² cos.² a'
or when its position is given
N 0 = y₁, since in that case
by the co-ordinates A Nx, and
§ 39.]
137
COMPOUND MOTION.
x = CM BC-NO=ay, and
=
y = M 0 = A B
whence
A N = b
X1, we have
α
g (b − x₁)²
2
-
Yi
2 c² cos.² a
g (b− x₁)²
Y₁ = a
or since a
2 c² cos.² a
I b²
2 c² cos.² a
2
Y₁ = x₁ tang. a
gxr2
2
1
2 c² cos.² a
Substituting in the equation y₁ = x₁ tang. a
cos.² a'
2
g x₁2
xi
2 c² cos.² a'
for
the value 1+ tang. a, and resolving the same in reference.
to tang. a, we obtain the following expression for the angle of eleva-
tion (a), required to reach a point given by the co-ordinates a
1
I Y₁ c² = g² x₁2, then we have
and yu
C²
2
2
C
tang, a =
qX,
9 x
- (1
1 +
2 c² Y₁
2
g x1
If
(ها))
= 1 +
2 c² Yi
g xi
or c¹
2 g
N
V
c²
tang. a =
g xi
c = 9 (Y₁ + √x₁² + y₁”) and
Smaller values of c make tang. a imaginary, and larger values of c
give two values for tang. a; in the first case the point cannot be
attained, and in the second case it would be attained either in the
rise or in the fall of the projectile.
EXAMPLE.-1. A jet of water rises with a velocity of 20 feet at an angle
of 66°. The height due to the velocity is h =0,0155. 20²
0,0155. 20² = 6,2 feet, and
the jet ascends to a height a = h sin.² a = 6,2. (sin. 66°)²= 5,17 feet, the
range of the jet is 2 b = 2 . 6,2 sin. 132° — 2 . 6,2 sin. 48 —–
2.6,2 = 9,21 feet. The
time, which each particle of water requires to describe the entire arc A CD
2 c sin, a 2. 20 sin. 66°
32,2
of the parabola, is t =
corresponding to the horizontal distance AN
= 1,14 seconds. The height
X1
3 feet is
Y₁ = 3. tang. 66°
= 6,738 — 2,189
+
32.2.9
2. 400. cos. (66°)³
4,549 feet.
0,36225
6,738
0,16543
2. A jet of water discharged from a horizontal tube has, for a height 13
feet, a range of 51 feet; how great is its velocity?
138
[§ 40.
GENERAL PRINCIPLES OF MECHANICS.
From the formula x=
g y²
y²
2 c² 4h'
y²
we deduce h
substitute x = 1,75 and y =
= in which we must
4 x'
5,25¹
5,25, and thus we obtain h
4. 1,75
feet and the corresponding velocity c= 15,92 feet.
= €3,937
§ 40. Jets of Water.-The peculiarities of the motion of jets
of water are explained and shown in what follows. From what
precedes we have
y = x tang. a
Y₁ = x₁ tang. a、
g x² [1 + (tang. a)2]
2 c²
gx2 [1 + (tang. a,)2]
2 c²
and
for the equations of the parabolas formed by the paths of two as-
cending jets of water whose velocities c are the same, and whose
angles of elevation a and a₁ are different. If we put x₁ = x and
subtract these equations from one another, we obtain
g x²
y — y₁ = x (tang. a
tang. a)
2
c²
[(tang. a)² -
(tang. a,)2]
= x (tang. a
tang. a,) (1 –
g x
2 c²
(tang. a + tang. a,)).
If we assume that the two streams have nearly the same angle
of elevation and require the two parabolas to have a point in com-
mon, we must put y, y and consequently we have
x (tang. a
tang. a,) (1
g x
2 c*
(tang. a + tang. a,)) = 0, or
I x
2 c²
(tang, a + tang. a₁) = 1,
or, since we can put a₁ = a we have simply
Substituting this value in the equation
g x tang. a
c²
=
1, whence tang. a =
c²
I x
y = x tang. a
g x2
2 c²
[1 + (tang. a)³],
we obtain the equation
c²
9х3
Y
g
2 c²
(1 +
c"
I x²
2
2
g²x² 2g 2 c²
of the curve D P SPD, Fig. 75, which passes through the neigh-
boring points, in which every two parabolas starting from the same
point A at different angles cut each other, and which, therefore,
touches or envelops the whole system of parabolas A CD, A OR,
etc.

40.]
COMPOUND MOTION.
139
c²
The height to which a vertical jet of water rises is A S=
2g'
and the range of projection of a jet A CD rising at an angle of
D
H
PO
FIG. 75
0
O P
H
R
D
B
NM
MN B
K
c² sin. 2 a
45° is A D = 2.
= 2.
=
2 AS.
2 g
2g
If we transfer the origin of co-ordinates from A to S, re-
placing the co-ordinates A N=x and N P = y by the co-ordinates
SU u and UP v, we have
and the equation
C²
y= AS-SU=
u and x = A N = UP = v,
2g
c²
g x2
y =
29
-
2 c²
is thus transformed into
I v²
2 c²
u=
or va
=
U.
2 c²
g
This equation is that of the common parabola whose parameter
is p =
2 c²
g
= 4 A S, and therefore the envelope D P SPD of all

140
GENERAL PRINCIPLES OF MECHANICS.
[$40.
the jets of water rising from the point A is a common parabola,
whose vertex is S and whose axis is S A.
D
H
FIG. 76.
H
R
BNM
D
MN
B
16
A bunch of jets rising from A in all directions would be envel-
oped by the paraboloid generated by the revolution of the envelope
DPSPD around A S. If t is the time in which a body rising in
a parabola describes the arc 4 0, Fig. 76, the co-ordinates of which
are A Mx and MO=y, we have
x= ct cos. a and y = c t sin. a
gt
2
"
whence
X
COS. a=
ct
and sin. a =
y + 1 g t
ct
Substituting these values for cos. a and sin. a in the well-known
trigonometrical formula (cos. a) + (sin. a) = 1, we obtain the fol-
lowing formula
x²
(ct)²
(y + gt²)²
+
(ct)²
= 1, or 2²+(y+gt)² = c² 12.
If from a point 4, Fig. 76, bodies be projected at the same mo-
ment and in the same vertical plane at different angles of eleva-
§ 41.]
141
COMPOUND MOTION.
tion, the positions that they occupy after the lapse of a certain
time (†) are determined by the last equation, which is that of a circle
whose radius is r = ct and whose centre is situated vertically below
A at a distance a = gt, and which can therefore be written in
the following form,
x² + (y + a)² = r².
The circumference of this circle would therefore be reached at the
same moment by all the elementary jets A CD, A O P, AL S……..
rising at the same moment from the point A.
If in the formula t₁
X
c cos. a
we substitute a 45°, and x =
C²
૦૨
4 B =
2 g
we obtain t₁ =
с
C
2
g cos. 45°
g
4, hence the time re-
quired to describe the whole arc of the parabola A C D is t =
с
2 t₁
g
√2, and the radius of the circle D L D, which is reached
simultaneously by the different elements of the water, is
c²
C³
C²
KD = r = ct
V/2
1/8 2,828
= 2,828. A S, and
g
2 g
2g
the distance of the centre K from A is
C³
c²
AK = α =
Ξα 1 g t
2
2 A S.
g
2 g
If we divide D K in 4, and A K in 16 equal parts, we can, since
r is proportional to t and a to t, from the points of division 1, 4, 9
in AK, describe other circles with the radii D K, D K, and
3D K, which cut off the parabolic arcs described in the same time,
E.G., the circle described from 1 with 1 a = DK, cuts off in the
points a, a,,...., the parabolic paths A a, 4 a,...., described simul-
taneously, and the circle described from 4 with 4 BDK cuts off
in the points ß, ẞ..... the parabolic arcs A ẞ, A B₁, etc., which are
also simultaneously described.
A
4
If these circles be revolved about the vertical axis K L, they de-
scribe spherical surfaces which bound the parabolic paths described
simultaneously, when the jets are projected all around A at all
angles of elevation.
§ 41. Curvilinear Motion in General.-By the combination
of several velocities and several constant accelerations, we obtain
also a parabolic motion, for not only the velocities but also the ac-
celerations can be combined so as to form a single resultant; the
142
[§ 42.
GENERAL PRINCIPLES OF MECHANICS.
problem is then the same as if there were one velocity and one
acceleration, I.E., as if there were but one uniform and one uni-
formly accelerated motion.
If the accelerations are variable, they can be combined so as to
give a resultant, as well as if they were constant, for we can con-
sider them as constant during an infinitely small period of time (7),
and the motion as uniformly accelerated during this time. The
resulting acceleration is, it is true, like its components themselves,
variable. If we combine this resulting acceleration with the given
velocity, we obtain the small parabolic arc, in which the motion
takes place during this instant. If we determine also for the follow-
ing instant the velocity and the acceleration, we obtain another por-
tion of an arc belonging to another parabola, and proceeding in the
same manner, we obtain approximately the entire curve of the path.
42 We can consider every small arc of a curve as an arc of a
circle. The circle to which this are belongs is called the circle of
curvature or osculatory circle (Fr. cercle osculateur, Ger. Krüm-
mungskreis), and its radius is the radius of curvature (Fr. rayon de
courbure, Ger. Krümmungshalbmesser). The path of a body in
motion can be considered as composed of such ares of circles, and
B
A
M
R
FIG. 77.
a
P
X
D
Pi
E
V
we can therefore deduce a
formula for its radii. Let
AM (Fig. 77)
vts
2

be a very small space de-
scribed in the direction A X
with uniformly accelerated
motion, A N=?= тavery
small space described uni-
formly and O the fourth cor-
ner of the parallelogram con-
Y structed with a and g, that
is, the position that the body
starting from A occupies at
the end of the instant (7).
Let us draw A C'perpen-
dicular to A I, and let us
see from what point C in this line an arc of a circle can be de-
scribed through A and O. In consequence of the smallness of 4 0
we can consider not only CA, but also C O P as perpendicular to
•
§ 43.]
143
COMPOUND MOTION.
A Y, so that in the triangle N O P the angle N P O can be
treated as a right angle. The resolution of this triangle gives
0 P = 0 N sin. O N P = A M sin. X A Y = p
and the tangent
A P = AN÷NP=vT+
2
sin. a,
Cos. a =
2
fo+
PT
cos. a
2
a) -
و آ
10
can be put
P T
= VT, for
2
cos. a can be neglected in the presence of
=
v, in consequence of the infinitely small factor. Now, from the
properties of the circle we know that AP PO. (PO+2 CO),
or since P O can be neglected in the presence of 2 CO, AP PO
2 CO; whence it follows that the radius of curvature is
CA = C O = r =
A P
2 PO
V² T²
V3
p ↑² sin. a p sin. a
=
In order to determine by construction the radius of curvature,
we lay off upon the normal to the original direction of the motion
AF the normal acceleration, I.E., its normal component p sin. a.
= AD, and join the extremity E of the velocity 4 Ev to D by
the right line D E, then we erect upon D E a perpendicular E C;
the point of its intersection with the first normal is the centre of
the osculatory circle of the point 4.
By inverting the last formula we obtain the normal accelera-
tion n = p sin, a
=
2,3
7
; from which we see that it increases di-
rectly as the square of the velocity, and inversely as the radius of
curvature, or directly as the greatness of the curvature.
EXAMPLE.—The radius of curvature of the parabolic trajectory pro-
duced by the acceleration of gravity is r =
C²
0,031
and for the vertex
sin. a'
=
1, it becomes 7 =
20 feet we obtain r
12,4 feet; the
of this curve where a 90°, and therefore sin. a
0,031 c feet. For a velocity e
farther the body is distant from the vertex the smaller a becomes, and con-
sequently the greater is the radius of curvature.
§ 43. If the point 4 has described the elementary space A 0 =
o, its velocity has changed; for the initial velocity in the direc-
tion A is now combined with the velocity pr acquired in the di-
rection AX, and consequently from the parallelogram of velocities
we have for the velocity
ບ
T
T
v² + 2 v p T cos. a + p² - = v² + p ↑ (2 v cos. a + p T),
but p ↑ vanishes in the presence of 2 v cos. a, and we have
144
[$ 43.
GENERAL PRINCIPLES OF MECHANICS.
Т
v₁² = v² + 2 p v т cos. α.
But v T is the elementary space A N = A0 = o, and p cos. a is the
tangential acceleration, I.E., the component & of the acceleration p
in the direction of the tangent or of the motion, whence we have
ບ
2
v2
= k σ.
Here o cos. a is the projection AR = 5, of the space upon the
direction of the acceleration, and consequently we have
2
v²
2
= p $1.
As the motion progresses v, changes successively into v2, v3.
v„, and the projections of the elementary spaces are increased by
the quantities S2, S3.... Sn, therefore we have
2
2
૭
Š3
2
2
2
23
V ₂
Un
-1
= p {2,
2
P Š3, •
2
=P En
3
n
ບູ
2
v3
and by addition
= p (§1 + §2 + . . . §n) = p x,
in which a denotes the total projection of the acceleration upon
A X. We can also put
vn²
2
2 وخ
P₁ + P₂+...+P₂
Pn
X,
N
when the acceleration is variable and assumes successively the val-
ues P1, P2
•
•
Pn•
We see from the above that the variation of the velocity does
not in the least depend upon the form or length of the path de-
scribed, but only on its projection x upon the direction of the ac-
celeration. For this reason all the jets of water, Fig. 76, have one
and the same velocity on reaching the same horizontal plane H H.
If c is the initial velocity or velocity of efflux, v the velocity at H H,
and b the height of the line H H above A, we have
v² — c²
2
v = √ c²
No
2
g b, whence
2 gb.
If at a certain point of the motion we have a = 90°, the tan-
gential acceleration k p cos. a becomes 0, and the normal ac-
celeration n = p sin. a is equal to the mean acceleration p. In this
case the variation of the squares of the velocities while the element
o of the space is being described, is v, v 0, and we have v₁ =
v; and if the motion continues in a curve, the direction of the ac-
2 —
1
§ 44.]
145
COMPOUND MOTION.
celeration changing in such a manner as always to remain normal
to the direction of the motion (I.E., if there is no tangential accel-
eration), — ¿9ª
¿
0, or v₁ = v remains constant while the point is
describing any finite space, and the final velocity is equal to the
initial velocity c.
18
The normal acceleration, for which the velocity remains constant,
p =
C²²
an example of which is afforded by motion in a circle, for then the ra-
dius of curvature CA CO = C D = r is constant. Inversely
a constant acceleration, which always acts
at right angles to the direction in which
the body is moving, causes uniform mo-
tion in a circle.
FIG. 78.
A
NP

M
O
Y
EXAMPLE.-A body, revolving in a circle 5
feet in diameter in such a manner as to make each
2 πr
revolution in 5 seconds, has a velocity c=
D
X
2 π.5
5
2. π
-
6,283 feet, and a normal ac-
celeration p=
(6,283)*
5
7,896 feet, I.E., in every
second it would be diverted from the straight line a distance p=1. 7,896
= 3,948 feet.
(§ 44.) Curvilinear Motion in General.-If a point P, Fig.
79, moves in two directions A X and A Fat the same time, we
Y
O
2
1
FIG. 79.
P₂
k
W
W
+
can consider the spaces de-
scribed AK LP = &
and A L - K P = y as the
co-ordinates of the curve
AP W formed by the path,
and if d t is the element
of time, in which the body
describes the elementary
spaces PR
d x and R Q
=dy, we have (from § 20)
the velocity along the ab-
scissa

X
T
A
K M
N
X
d x
1) u =
d ť
and that along the ordi-
nate
146
[§ 44.
GENERAL PRINCIPLES OF MECHANICS
dy
2) v =
d ť
and therefore the resulting tangential velocity, or that along the
curve, when the directions A X and A Y of the motions are at
right angles to each other,
d
3)
w = √ w² + v² = √ (a c )² + ( 2 )² = √ √ d x² + d y² _ å s
V´u²
d
t.
1
d ť
ď ť
in which formula ds denotes the element P Q of the curve which,
according to Art. 32 of the Introduction to the Calculus, is equal to
√d x² + d y².
The acceleration along the abscissa is, according to § 20,
FIG 80.
V
W
Y
d u
4) p
d ť

and that along the ordi-
1,
오
​•
P₁.
k
nate
W
d v
5) I
d t
22
α
X
T
A
K M
N
X
For the tangential an-
gle P TX
=
QPR = a,
formed by the direction of
motion Pw with the direc-
tion of the abscissas, we
have,
V
dy
dx'
tang. a = U
C
and also
V
dy
sin. a =
and
พ
A s
И
d x
cos. a =
го
d s
The accelerations p and q can be decomposed into the following
components in the directions of the tangent P T and of the nor-
mal P N,
P1
p cos. a and p₁ = p sin. a,
21
q sin. a and q₂
I2 q cos. a.
Consequently the tangential acceleration is
k = P₁ + J₁ = p cos. a + q sin. a
d u
W
d v
V u d u + v ď v
+
d t
W
d t
W
w d t
and the normal acceleration is
$ 44.]
147
COMPOUND MOTION.
n = X a
q2
p sin. a
q cos. a
di
V
d v
И
v d u
u d v
d t W
d t w
v²
w d t
But by differentiating u+vw we obtain
u d u + v d v = wd w,
and therefore we have more simply for the tangential acceleration
wd w
d w
6)
k
w d t
d t
V
From tang. a =
W
•
we obtain d tang. a =
u d v — v du
(Introduction to the Calculus, Art. 8) and the radius of the curva-
ture CPCQ of the elementary arc P Q (according to Art. 33
of the Introduction to the Calculus) is
d s³
r
d x d tang. a
whence it follows that
u² d s² d §³
d s
'd
v du―u dv= u³d tung. a=
r d x² 2 d t²
до
d
§ (1³)² = 20² d
22
s
and that the normal acceleration is simply
17)
'N
w² d s
r w d t
W
ds
w²
d t
k ds =
d w
d t
d s
·
d s =
d w = w d w;
dt
Finally we have
from which we obtain (as in § 20),
8)
W²
c²
2
= fkds,
when we suppose that while describing the space s the velocity
changes from c to w. Therefore, in curvilinear motion half the dif-
·ference of the squares of the velocities is equal to the product of the
mean acceleration (h) and the space s.
In like manner
p d x + q d y = u du + v d v = w d w, and therefore
w²
9)
2
c²
=S(pdx + qd y) = Spd x + Sq dy, and
10) Sk ds = Spdx + Sqdy, or
k ds = pdx + q dy.
The product of the tangential acceleration and the element of the
curve is equal to the sum of the products of the accelerations along
the co-ordinates and the corresponding elements of co-ordinates.
148
[$ 44.
GENERAL PRINCIPLES OF MECHANICS.
12 t,
EXAMPLE.-A body moves on one axis A X with the velocity u =
and on the other A Y with the velocity v = 4 ť² 9; required the other
conditions of the resulting motion. The corresponding accelerations along
the co-ordinates are
d u
d v
p =
12, and q
8 t,
d t
d t
f
and the co-ordinates, or spaces described along the axes, are
x= Sudt = 12 t d t = 6 t2, and
y =
v
= fodt = f (4 r — 9) d t =
t²
4
3
9 t,
=
in which equations the spaces count from the time t = 0. The tangential
velocity, or that along the curve, is
Nu²
r = √ u² + v²
√′144 ť² + (4 ť² — 9)² = = √ 16 t* + 72 ť² + 81 = 4 ť² + 9,
consequently the tangential acceleration is
k
d ro
d t
8t the acceleration q along the ordinate.
We have also for the space described along the curve
8=
Su
4
= S w d t = f (4 € + 9) d t = 13
3
t³ + 9t.
When the direction of the motion is given by the formula,
tang. a =
V
U
4 t² 9
z x 9
12 t
we have
d tang. a =
2 √6x
4 to +9
dt,
12 t2
and therefore the radius of curvature of the trajectory is
d 83
(4 t² + 9)³. 12 ta
r = d x² d tang. a
144 ť² (4 t² +9)
or, r
202
12.
(4 t² + 9)²
12
Consequently the normal acceleration, which produces a constant
change of direction of the motion of the body, is
n =
202
p
12, or constant.
The equation of the curve of the trajectory of the body is found by sub-
in the foregoing equation, and it is
X
stituting t =
6
4
3
√(3)
2
9
X
6
6
9
- 9) √
X
6
9
The ordinate y is a (negative) maximum for v = 0, 1.E., for t²:
ort =
4
3
9 27
and x = 6. ť² — 6 .
and then
2
4
2”
4
9
3
3
Y
9.
9
3
4
2
2
27
3
and on the contrary, it is =
0, for t² =
or t
√3, and x =
4
2
81
2
D
$ 45.]
149
COMPOUND MOTION.
The curve which forms the path of the body runs at first below the axis
27
of abscissas, and after the time t
V
it cuts it at a point whose
4
abscissa is x =
81
2'
and from that time it remains above the axis.
The following table contains a collection of the corresponding values
of t, u, v, w, x, y, tang. a, r and s, from which the curve A B C D E, Fig. 81,
is constructed.
[
Y
FIG. 81.

18
27
D
AO 6
13,5
24
40,15
9
18
B
6
36
18V3
15
E
49 1/3
36
96-X

t
гл
V
W
X
Y
tang. a
go
S
27
о
-9
9
O
O
∞
O
4
I
12
- 5
13
6
23
5
169
31
3
12
I 2
3
27
I }}\
18
18
9
27
18
2
22
2
24
7
25
7
86
24
625
3
24
I 2
3
∞
81
√3 18√3
18
36
2
3
3
675
63
3
36
27 45
54
+9
4
4
148
55
1875
364
4
48
3
48
4 i 3
HI MOM! +
108
1/3
27 1
3
2
55 75 96 +
§ 45. Relative Motion.-If two bodies are moving simul-
taneously, a continual change in their relative positions, distances
apart, etc., takes place, the value of which may be determined for
any instant by the aid of what precedes. Let A, Fig. 82, be the
point where one and B that where the other motion begins; the first
150
[$ 46.
GENERAL PRINCIPLES OF MECHANICS.
body passes in a given time (t) in the direction A X to the position
M, and the other body in the same time in the direction B Yto
the point N. Now if we draw M N, this line will give us the rela-
Q
FIG. 82.
N
B
Ac
M
--X
tive position and distance from
each other of the bodies A and
B at the end of this time. Draw-
ing A O parallel to M N, and
making AOM N, the line
A O will also give the relative
position of the bodies A and B.
If we now draw O N, we obtain

a parallelogram, in which O N is = A M. If, finally, we make B Q
equal and parallel to N O and draw O Q, we obtain a new parallel-
ogram B N O Q, in which the one side B N is the absolute space
(y) described by the second body, the other side BQ is the space
(x) described by the other body in the opposite direction, and the
fourth corner is the relative position of the second body, that is,
in reference to the position of the first body, which we consider
to be fixed. Hence we can determine the relative position O of a
moving body (B) by giving to this body besides its motion (B N)
another, equal to but in the opposite direction from that A M of
the body (A), to which its position is referred, and then by com-
bining in the ordinary way, as, E.G., by the aid of a parallelogram,
these two motions.
§ 46. If the motions of the bodies A and B are uniform, we
can substitute for A M and B N the velocities c and c,, that is the
spaces described in one second. In this way we obtain the rela-
tive velocity of one body when we give to it besides its own abso-
lute velocity, that of the body to which we refer the first velocity,
but in the opposite direction.
P
A
FIG. 83.
B
M
K
NOR
Y
α
C
I
The same relation holds good for
the accelerations. If, E.G., a body
A, Fig. 83, moves uniformly in the
direction with the velocity c,
and a body B moves in the direc-
tion BF, which makes an angle a
with BX, with an initial velocity
= 0 and with the constant accele-
ration p, we can also suppose that
A stands still and that B possesses,
}

§ 46.]
151
COMPOUND MOTION.
besides the acceleration p, also the velocity (c) in the direc-
tion BX, parallel to A X; the body will then describe the parabolic
path B 0 P.
The spaces described in the time t in the directions BY
ct, the first of which can be
and BX, are B N =
D t²
2
and B M =
decomposed into the components NR
p t²
2
cos, a and B R =
p t
2
sin. a, which are parallel and at right angles to A X.
Now if A C =a and C B = b are the original co-ordinates of
the co-
the point B in reference to A, and A K = x and K 0
ordinates of the same after the time t, we have, since A K = A C
-ON- N R and K O
NR 0 = C B — BR,
x = a
c t
p t²
2
cos, a and y = b
p
t²
2
sin. a,
and consequently the corresponding relative velocities
W
C
pt cos. a and v = pt sin. a.
From the abscissa x we determine the time by the formula
t =
2 (a
p cos. a
2
x)
C
+
。 cos. a)
p cos. a
С
p cos. a
and, on the contrary, from the ordinate y by the formula
2 (b y)
t =
1
p sin. a
If the body B moves in the line A X towards A, we have b = 0
and also a = 0, and therefore
t
2 (a x)
P
+
(
с
p
putting x = 0, we obtain for the time, when two bodies will meet,
t = √
2 a
+
ม
2
C √2 ap + c² C
p
p
If, on the contrary, the body B moves in the line AI ahead
of the body A, then a = 180°, and the distance of the former from
the latter body is x = a
c t +
V t²
2
and, inversely, the time, at
the end of which the bodies are at a distance x from each other, is
t
± √
2 (a
x)
P
+
C
P
The corresponding velocity u = c + p t is 0, and the dis-
=
C²
с
tance 2 is a minimum for t
and its value is x = a
p
2 P
152
[$ 46.
GENERAL PRINCIPLES OF MECHANICS.
For every other value of x we have two values for the time, one
of which is greater and the other less than
с
p
REMARK.-The foregoing theory of relative motion is often applied, not
only in celestial mechanics, but also in the mechanics of machines. Let us
consider the following case.
1
2
2
2
A body 4, Fig. 84, moves in the direction A X with the velocity c₁, and
should be met by another body B which has the velocity c₂; what direction
must we give the latter? If we draw A B, lay off from B, c, in the op-
posite direction and complete with c₁ and c₂ a parallelogram B c₁ c c₂,
whose diagonal c coincides with A B, we obtain in the direction B c₂ = C₂
of its side, not only the direction B Yin which the body B must move,
but also in the point of intersection C of the two
directions AX and B Y, the point where the two
bodies will meet. If a is the angle B A X formed
by A X, and 3 the angle A BY formed by BY with
A B, we have
FIG. 84.
D

C
C2
B
-X
sin. B
sin. a
C1
C2
1
This formula is applicable to the aberration of the
light of the stars which is caused by the compo-
sition of the velocity c, of the earth A around the
sun with the velocity c₂ of the light of the star B.
is about 19 miles, and c, about 192,000
miles, consequently
Here c
C1
2
A
Y
sin. B
C1
C2
sin. a =
19 sin. a
192000
sin. a
10105'
FIG. 85.
Y
hence the aberration or the angle A B C = 3, formed by the apparent di-
rection A B of the star (which can be supposed to be infinitely distant) with
the true direction B C or AD, is ẞ=20" sin. a; and for a=90°, that is, for a
star, which is vertically above the path of the earth (in the so-called pole of
the ecliptic), we have ẞ 20". In consequence of this divergence we al-
ways see a star 20" in the direction of the
motion of the earth behind its true posi-
tion, and consequently a star in the
neighborhood of the pole of the ecliptic
describes apparently in the course of a
year a small circle of 20" radius around
its true position. For stars in the plane
of the earth's path this apparent motion
takes place in a straight line, and for
the other stars in an apparent ellipse.

A
D
G
F
B
E
X
EXAMPLE.-A locomotive moves from
A upon the railroad track A X, Fig. 85,
46.1
153
COMPOUND MOTION.
і
with 35 feet velocity, and another at the same time from B with 20 feet
velocity upon the track B Y, which forms an angle B D X = 56° with the
first. Now if the initial distances are A C = 30000 feet, and C B = 24000
feet, how great is the distance AO after a quarter of an hour? From the
absolute velocity B E = c₁ 20 feet of the second train, the inverse velo-
city B F = c = 35 feet of the first, and the included angle E B F = a =
180° – B D C = 180° — 56° = 124°, we obtain the relative velocity of the
second train
1
BG= = √ c² + c₁² + 2 c c₁ cos, a =
= √1225 + 400
2
1
1
√35 + 20²
2.35.20. cos. 56°
1400 cos. 56° = 1/1625 782,9 = √/842,1 = 29,02 feet.
For the angle & B F = o, included between the direction of the rela-
tive motion and the direction of the first motion, we have
c, sin. 56°
sin. Ф
29,02
20. 0,8290
29,02
; log sin. 4=0,75690-1, whence -34°,50'.
900
The relative space described in 15 minutes=900 seconds is B 0=29,02 .
26118 feet, the distance A B is = √(30000)² + (24000)² = 38419
feet, the value of the angle B A C =
A BF, whose tangent is
is y = 38° 40′, and therefore the angle
24000
0,8,
30000
A B0 = ☀ + 1/1 34° 50' + 38° 40' 73° 30′,
and the distance apart of the two trains after 15 minutes is
AO
VAB² + B 0² 2 AB. BO cos. A BO
= √384193 + 261182
2.38419. 26118 cos. 73° 30′
= √1588190000 = 39850 feet.
SECOND SECTION.
MECHANICS, OR THE PHYSICAL SCIENCE OF
MOTION IN GENERAL.
1
CHAPTER I.
FUNDAMENTAL PRINCIPLES AND LAWS OF MECHANICS.
§ 47. Mechanics.-Mechanics (Fr. mécanique, Ger. Mechanik)
is the science which treats of the laws of the motion of material
bodies. It is an application to the bodies of the exterior world of
that part of Phoronomics or Cinematics which deals with the mo-
tions of geometrical bodies without considering the cause. Me-
chanics is a part of Natural Philosophy (Fr. physique générale,
Ger. Naturlehre) or of the science of the laws, in accordance with
which the changes in the material world take place, viz., that part
of it, which treats of the changes in the material world arising from
measurable motions.
§ 48. Force.-Force (Fr. force, Ger. Kraft) is the cause of the
motion, or of the change in the motion of material bodies. Every
change in motion, E.G., every change of velocity, must be regarded
as the effect of a force. For this reason we attribute to a body
falling freely a force, which we call gravity; for the velocity of the
body changes continually. But, on the other hand, we cannot.
infer from the fact that a body is at rest or moving uniformly that
it is free from the action of any force; for forces may balance
each other without causing any visible effect. Gravity, which
causes a body to fall, acts as strongly upon it when it lies upon a
table, but its effect is here destroyed by the resistance of the table
or other support.
$ 49, 50.] PRINCIPLES AND LAWS OF MECHANICS.
155
§ 49. Equilibrium.-A body is in equilibrium (Fr. équilibre,
Ger. Gleichgewicht), or the forces acting on a body hold each other
in equilibrium, or balance each other, when they counterbalance
or neutralize each other without leaving any resulting action, or
without causing any motion or change of motion. E.G. When a
body is suspended by a string, gravity is in equilibrium with the
cohesion of the string. The equilibrium of several forces is de-
stroyed and motion produced when one of the forces is removed or
neutralized in any way. Thus a steel spring, which is bent by a
weight, begins to move as soon as the weight is removed, for then
the force of the spring, which is called its elasticity, comes into
action.
Statics (Fr. statique, Ger. Statik) is that part of mechanics which
treats of the laws of equilibrium. Dynamics (Fr. dynamique, Ger.
Dynamik), on the contrary, treats of forces as producers of motion.
§ 50. Classification of the Forces.-According to their
action, we can divide forces into motive forces (Fr. forces motrices
puissance, Ger. bewegende Kräfte), and resistances (Fr. résistances,
Ger. Widerstände). The former produce, or can produce, motion,
the latter can only prevent or diminish it. Gravity, the elasticity of
a steel spring, etc., belong to the moving forces, friction, resistance
of bodies, ctc., to the resistances; for although they can hinder or
diminish motion or neutralize moving forces, they are in no way
capable of producing motion. The moving forces are either accel-
crating (Fr. accéleratrices, Ger. beschleunigende) or retarding (Fr.
retardatrices, Ger. verzögernde). The former cause a positive, the
latter a negative, acceleration, producing in the first case an accel-
erated, and in the second a retarded motion. The resistances are
always retarding forces, but all retarding forces are not necessarily
resistances. When a body is projected vertically upward, gravity
acts as a retarding force, but gravity is not on this account a re-
sistance, for when the body falls it becomes an accelerating force.
We distinguish also uniform (Fr. constantes, Ger. beständige, con-
stante) and variable forces (Fr. variable, Ger. veranderliche). While
uniform forces act always in the same way, and therefore in the
equal instants of time produce the same effect, I.E., the same in-
crease or decrease of velocity, the effects of variable forces are
different at different times; hence the former forces produce uni-
formly variable motions, and the latter variably accelerated or
retarded motions.
156
GENERAL PRINCIPLES OF MECHANICS. [§ 51, 52, 53.
§ 51. Pressure.-Pressure (Fr. pression, Ger. Druck), and
traction (Fr. traction, Ger. Zug), are the first effects of force upon
a material body. In consequence of the action of a force bodies are
either compressed or extended, or, in general, a change of form is
caused.
The pressure or traction, produced by gravity acting vertically
downwards and to which the support of a heavy body or the string,
to which it is suspended, is subjected, is called the weight (Fr. poids,
Ger. Gewicht) of the body.
Pressure and traction, and also weight, are quantities of a pe-
culiar kind, and can be compared only with themselves; but since
they are effects of force they may be employed as measures of the
latter.
The most simple and therefore the most common way of
measuring forces is by means of weights.
§ 52. Equality of Forces.-Two weights, two pressures, two
tractions, or the two forces corresponding to them are equal, when
we can replace one by the other without producing a different
action. When, E.G., a steel spring is bent in exactly the same man-
ner by a weight G suspended to it as by another weight G, hung
upon it in exactly the same manner, the two weights, and therefore
the forces of gravity of the two bodies are equal. If in the
same way a loaded scale (Fr. balance, Ger. Waage) is made to bal-
ance as well by the weight G as by another G₁, with which we have
replaced G, then these weights are equal, although the arms of the
balance may be unequal, and the other weight be greater or less.
A pressure or weight (force) is 2, 3, 4, etc., or in general n
times as great as another pressure, etc., when it produces the same
effect as 2, 3, 4....n pressures of the second kind acting together.
If a scale loaded in any arbitrary manner is caused to balance by
the weight (G) as well as by 2, 3, 4, etc., equal weights (G,), then is
the weight (G) 2, 3, 4, etc. times as great as the weight (G₁).
§ 53. Matter.-Matter (Fr. Matière, Ger. Materie) is that, by
which the bodies of the exterior world (which in contradistinction
to geometrical bodies are called material bodies) act upon our
senses. Mass (Fr. masse, Ger. Masse) is the quantity of matter
which makes up a body.
Bodies of equal volume (Fr. volume, Ger. Volumen) or of equal
geometrical contents generally have different weights. Therefore
$ 54, 55.]
157
PRINCIPLES AND LAWS OF MECHANICS.
we can not determine from the volume of a body its weight; it is
necessary for that purpose to know the weight of the unit of
volume, E.G., of a cubic foot, cubic meter, etc.
§ 54. Unit of Weight.-The measurement of weights or
forces consists in comparing them to some given unchangeable
weight, which is assumed as the unit. We can, it is true, choose this
unit of weight or force arbitrarily, but practically it is advan-
tageous to choose for this purpose the weight of a certain volume
of some body, which is universally distributed. This volume is
generally one of the common units of space. One of the units of
weight is the gram, which is determined by the weight of a cubic
centimetre of pure water at its maximum density (at a temperature
of about 4° C.). The old Prussian pound is also a unit referred
to the weight of water. A Prussian cubic foot of distilled water
weighs at 15° R. in vacuo 66 Prussian pounds. Now a Prussian
foot is 139,13 Paris lines 0,3137946 meter; whence it follows
that a Prussian pound 467,711 grams. The Prussian new or
custom-house pound weighs exactly kilogramm. The English
pound is determined by the weight of a cubic foot of water at a
temperature of 39°, 1 F. The pound is equal to 453,5926 grams.
A cubic foot of water weighs 62,425 lbs.
1
§ 55. Inertia (Fr. inertie, Ger. Trägheit) is that property of
matter, in virtue of which matter cannot move of itself nor change
the motion, that has been imparted to it. Every material body re-
mains at rest as long as no force is applied to it, and if it has been
put in motion continues to move uniformly in a straight line, as
long as it is free from the action of any force. If, therefore,
changes in the state of motion of a material body occur, if a body
changes the direction of its motion, or if its velocity becomes
greater or less, this result must not be attributed to the body as a
certain quantity of matter, but to some exterior cause, I.E., to a
force.
Since, whenever there is a change in the state of motion of a
body, a force is developed, we can in this sense count inertia as one
of the forces. If a moving body could be removed from the influ-
ence of all the forces which act upon it, it would move forward
uniformly for ever; but such a uniform motion is nowhere to be
found, since it is impossible for us to remove a body from the in-
fluence of every force. If a mass moves upon a horizontal table
158
[$ 56, 57.
GENERAL PRINCIPLES OF MECHANICS.
the action of gravity is counterbalanced by the table, and therefore
does not act directly upon the body, but in consequence of the
pressure of the body on the table a resistance is developed, which
will be treated hereafter under the name of friction. This resist-
ance continually diminishes the velocity of the moving body, and
the body therefore assumes a uniformly retarded motion and finally
comes to rest. The air also opposes a resistance to its motion, and
even if the friction of the body could be completely put aside, a
continual decrease of velocity would be caused by the former.
But we find that the loss of velocity becomes less and less, and that
the motion approximates more and more to a uniform one, the more
we diminish the number and magnitude of these resistances, and we
can therefore conclude, that if all moving forces and resistances
were removed, a perfectly uniform motion would ensue.
§ 56. Measure of Forces.-The force (P) which accelerates an
inert mass (M) is proportional to the acceleration (p) and to the
mass (M) itself. When the masses are the same, it increases with
the infinitely small increments of velocity produced in the infin-
itely small spaces of time, and when the velocities are equal it in-
creases in the same ratio as the masses themselves. In order to
produce an m fold acceleration of the same mass, or of equal masses,
we require an m fold force, and an n fold mass requires an n fold
force to produce the same acceleration.
Since we have not as yet adopted a measure for the masses, we
can assume
P = Mp,
or that the force is equal to the product of mass and the accelera-
tion, and at the same time we can substitute instead of the force
its effect, I.E., the pressure produced by it.
The correctness of this general law of motion can be proved by
direct experiment, when we, E.G., drive along upon a horizontal
table by means of bent steel springs similar or different movable
masses; but the important proof lies in this, that all the results
and rules for compound motion, deduced from the law, correspond
exactly with our observations and with natural phenomena.
§ 57. Mass.- All bodies at the same point on the earth fall in
vacuo equally quickly, namely, with the constant acceleration
M
32,2 feet (§ 15). If the mass of a body is
g
9.81 meter
§ 58.]
159
PRINCIPLES AND LAWS OF MECHANICS.
and the weight which measures the force of gravity = G, we have
from the last formula
G = M g,
I.E., the weight of a body is a product of its mass and the acceleration
of gravity, and inversely
G
M =
g
I.E., the mass of a body is the weight of the same divided by the accel-
eration of gravity, or the mass is that weight which a body would
have if the acceleration of gravity were = 1, E.G., a meter, a foot,
etc. For that point upon or in the neighborhood of the earth or of
any other celestial body, where the bodies fall with a velocity (at the
end of the first second) of 1 meter instead of 9,81 meters, the mass,
or rather the measure of the same, is given directly by the weight
of the body.
According as the acceleration of gravity is expressed in meters
or feet we have for the masses
G
M =
0,1019 G, or
9,81
G
M =
0,031 G.
32,2
Hence the mass of a body, whose weight is 20 pounds, is
M = 0,031 × 20 0,62 pounds, and inversely the weight of a
mass of 20 pounds is G = 32,2 × 20 = 644 pounds.
§ 58. If we suppose the acceleration (g) of gravity to be con-
stant, it follows that the mass of a body is exactly proportional to its
weight, and that, when the masses of two bodies are M and M₁ and
their weights G and G₁, we have
M
G
M
Gi
Therefore, the weight of a body can be employed as a measure
of its mass, so that the greater the mass a body is the greater is its
weight.
However the acceleration of gravity is variable, becoming
greater as we approach the poles and diminishing as we approach
the equator; it is a maximum at the poles and a minimum at the
equator. It also decreases when a body is elevated above the level
160
[$ 59.
GENERAL PRINCIPLES OF MECHANICS.
of the sea.
Now since a mass, so long as we take nothing from it
nor add anything to it, is a constant quantity and remains the
same for all points on the earth, and even on the moon, it follows
that the weight of a body must be variable and depend upon the
position of the body, and that in general it must be proportional
G
༡
G₁
J1
to the acceleration of gravity, or that must be
The same steel spring would therefore be differently deflected
by the same weight at different points on the earth-at the
equator and on high mountains the least, and at the poles at the
level of the sea the most.
§ 59. Heaviness (Fr. densité, Ger. Dichtigkeit) is the in-
tensity with which matter fills space. The heavier a body is, the
more matter is contained in the space it occupies. The natural
measure of the heaviness is that quantity of matter (the mass)
which fills the unity of volume; but since matter can only be
measured by weight, the weight of a unit of volume, E.G., of a
cubic meter or of a cubic foot of another matter, must be employed
as the measure of its heaviness. Hence, the heaviness of water
at 39°.1 F. is 62,425 pounds, and that of cast iron at 32° F.
is 452 pounds, I.E., a cubic foot of water weighs 62,425, and
a cubic foot of cast iron 452. In ordinary calculations we assume
that of water to be 62 pounds. From the volume of a body
and its heaviness y we have its weight G = Ty.
The product of the volume and the heaviness is the weight.
The heaviness of a body is uniform (Fr. homogène, uniforme,
Ger. gleichförmig) or variable, (Fr. variable, hétérogène, Ger.
ungleichförmig), according as equal portions of the volume
have equal or different weights, E.G., the heaviness of the simple
metals is uniform, since equal parts of them, however small, weigh
the same.
Granite, on the contrary, is a body of variable heaviness,
since it is composed of parts of different density.
EXAMPLE.-1. If the heaviness of lead is 719 pounds, then 3,2 cubic feet
of lead weigh G
2278,4 lbs. If the weight of a cubic foot of bar
iron be 180 pounds, the volume of a piece, whose weight is 205 pounds, is
Τ
G 205
480
0,4271 cubic feet
0,4271 x 1728 = 738 cubic inches.
Note.--In German and French the word "density" is employed to express
the weight of a cubic foot, a cubic meter, etc., of any material. In English,
unfortunately, it is employed as a synonym of specific gravity.-TR.
$ 60.]
161
PRINCIPLES AND LAWS OF MECHANICS.
If 10,4 cubic feet of hemlock, thoroughly saturated with water, weighs
577, then its heaviness is
G
Y
V
577
10,4
= 55,5 pounds.
$60. Specific Gravity.-Specific weight, or specific gravity,
(Fr. poids spécifique, Ger. specifisches or eigenthümliches Gewicht)
is the ratio of the heaviness of one body to that of another body,
generally water, which is assumed as the unit. But the heaviness
is equal to the weight of the unit of volume; therefore the specific.
gravity is also the ratio of the weight of one body to that of
another, E.G., water, of equal volume.
In order to distinguish the specific gravity or specific weight
from the weight of a body of a given volume, the latter is called the
absolute weight (Fr. poids absolu, Ger. absolutes Gewicht).
If y is the heaviness of the matter (water), to which the others
are referred, and y, the heaviness of any matter whose specific
gravity is denoted by ɛ, we have the following formula:
Y1
ε
and Y'₁ = ε Y,
Y
therefore the heaviness of any matter is equal to the specific gravity
of the same multiplied by the heaviness of water.
The absolute weight G of a mass of whose volume is V, and
whose specific gravity is e, is:
G = Vy₁ = Vεy.
EXAMPLE.-1. The heaviness of pure silver is 655 pounds, and that of
water 62,425 pounds; consequently the specific gravity of the former (in
relation to water) is
655
62,425
10,50, I.E., every mass of silver is 101
times as heavy as a mass of water that occupies the same space. 2) If we
take 13,598 for the specific gravity of mercury, and the heaviness of water as
62,425, then we have for the heaviness of mercury,
13,598.62,425 848,86 pounds.
A mass of 35 cubic inches of the same weighs, since 1,728 cubic inches are
a cubic foot,
G 848,86
848,86 . 35
1728
17,19 pounds.
REMARK.-The use of the French weights and measures possesses the
advantage that we can perform the multiplication by ɛ and y by simply
changing the position of the decimal point, for a cubic centimeter weighs
a gram, and a cubic meter a million grams, or 1,000 kilograms. The
heaviness of mercury is therefore, when we employ the French measure,
Y1 13,598. 1000 = 13598 kilograms; that is, a cubic meter of mercury
weighs 13598 kilograms.
162
[§ 61, 62.
GENERAL PRINCIPLES OF MECHANICS.
§ 61. The following table contains the specific gravities of those
substances, which are met with the oftenest in practical mechanics.
A complete table of specific gravities is to be found in the
Ingenieur, page 310.
Mean specific gravity of
the wood of deciduous
Sandstone
Brick.
1,90 to 2,70
1,40 to 2,22
trees, dry
0,659
Masonry with mortar made
saturated with water
1,110
of lime and quarry stone:
Mean specific gravity of
Fresh.
2.46
the wood of evergreen
Dry
2,40
trees, dry
0,453
saturated with water
0,839*
Masonry with mortar made
of lime and sandstone:
Mercury.
13,56
Fresh.
2,12
Lead
11,33
Copper, cast and dense
Dry
2,05
8,75
66 hammered.
8,97
Brickwork with
mortar
Brass
made of lime:
8,55
Iron, cast, white
Fresh
7,50
1,55 to 1,70
66 grey
7,10
Dry.
1,47 to 1,59
((
(( medium.
7,06
Earth, clayey, stamped:
in rods.
7,60
Fresh.
2,06
Zinc, cast
(6 rolled
•
7,05
Dry
1,93
7,54
Garden earth:
Granite
Gneiss
Limestone
2,50 to 3,05
Fresh.
= 2,05
2,39 to 2,71
Dry
1,63
2,40 to 2,86
Dry poor earth
1,34
§ 62. State of Aggregation.-Bodies present themselves to
us in three different states, depending upon the manner in which
their parts are held together. This is called their state of aggrega-
tion. They are either solid (Fr. solides, Ger. fest) or fluid (Fr.
fluides, Ger. flüssig), and the latter are either liquid (Fr. liquides,
Ger. tropfbar flüssig) or gaseous ((Fr. gazeux, aériformes, Ger. elas-
tisch flüssig). Solid bodies are those, whose parts are held together
so firmly, that a certain force is necessary to change their forms or
to produce a separation of their parts. Fluids are bodies, the
position of whose parts in reference to each other is changed by the
smallest force. Elastic fluids, the representative of which is the
air, are distinguished from liquids, the representative of which is
*See the absorption of water by wood, polytechnische Mittheilingen,
Vol. II, 1845.
§ 63, 64.]
163
PRINCIPLES AND LAWS OF MECHANICS.
water, by the fact that they tend continually to expand more and
more, which tendency is not possessed by water, etc.
While every solid body possesses a peculiar form of its own and
a definite volume, liquids have only a determined volume, but no
peculiar form. Gases or aeriform fluids possess neither one nor the
other.
§ 63. Classification of the Forces.-Forces are very differ-
ent in their nature; we give here only the most important ones:
1) Gravity, by virtue of which all bodies tend to approach the
centre of the earth.
2) The Force of Inertia, which manifests itself when a change
in the velocity or in the direction of the moving body
takes place.
3) The Muscular Force of living beings, or the force produced
by means of the muscles of men and animals.
4) The Elastic Force, or that of springs, which bodies exhibit
when a change of form or of volume occurs.
5) The Force of Heat, by virtue of which bodies expand and
contract, when a change of temperature takes place.
6) The Force of Cohesion, or the force by which the parts of a
body hold together, and with which they resist separa-
tion.
7) The Force of Adhesion, or the force with which bodies
brought into close contact attract each other.
8) The Magnetic Force, or the attractive and repulsive force of
the magnet.
Then we have the electric and the electro-magnetic forces, etc.
The resistances due to friction, rigidity, resistance of bodies,
etc., are due principally to the force of cohesion, which, like the
clasticity, etc., is due to the so-called molecular force, or the force
with which the molecules, or the smallest parts of a body, act upon
one another.
§ 64. Forces, how Determined.--For every force, we must
distinguish :
1) The point of application (Fr. point d'application: Ger. An-
griffspunkt), the point of the body to which the force is
directly applied.
2) The direction of the force (Fr. direction, Ger. Richtung), the
right line, in which a force moves the point of applica-
164
[$ 65, 66.
GENERAL PRINCIPLES OF MECHANICS.
tion, or tends to move it or hinder its motion. The direc-
tion of a force has, like every direction of motion, two
senses. It can take place from left to right, or from right
to left, from above downwards, or from below upwards.
One is considered as positive, and the other as negative.
As we read and write from left to right, and from above
downwards, it is natural to consider these motions as
positive, and the opposite motions as negative.
3) The absolute magnitude or intensity (Fr. grandeur absolue,
intensité, Ger. absolute Grösse) of the force, which we
have seen is measured by weights, E.G. pounds, kilograms,
etc.
Forces are graphically represented by straight lines, whose
direction and length indicate the direction and magnitude of the
forces, and one of whose extremities can be considered as the point
of application of the forces.
§ 65. Action and Reaction.--The first effect produced by a
force upon a body is an extension or compression, combined with
a change of form or of volume, which commences at the point of
application, and from there gradually spreads itself farther and
farther into the body. By this inward change in the body the
elasticity inherent in it comes into action and sets itself in equi-
librium with the force, and is, therefore, equal to it, but acts in the
opposite direction. Hence, action and reaction are equal and oppo-
site. This law is true, not only for the effects of forces acting
by contact, but also for those acting by attraction and repulsion,
among which the magnetic forces, and also that of gravity, must
be counted. A bar of iron attracts a magnet exactly as much as it
is attracted itself by the magnet. The force, with which the moon
is attracted towards the earth (by gravity), is equal to the force
with which the moon reacts upon the earth.
The force with which a weight presses upon its support is
returned by the latter in the opposite direction. The force, with
which a workman pulls, pushes, etc., a machine, reacts upon the
workman, and tends to move him in the opposite direction. When
one body impinges upon another, the first presses upon the second
exactly as much, as the second does upon the first.
§ 66. Division of Mechanics.-General mechanics are di-
vided into two principal divisions, according to the state of aggre-
gation of the bodies:
§ 67.]
165
MECHANICS OF A MATERIAL POINT.
1) Into the mechanics of solid or rigid bodies (Fr. mécanique
des corps solides, Ger. Mechanik der festen oder starren
Körper).
2) Into the mechanics of fluids (Fr. mécanique des fluides,
Ger. Mechanik der flüssigen Körper). The latter can
again be divided:
a) Into the mechanics of water and other liquids or hydraulics
(Fr. hydraulique, Ger. Hydraulik, Hydromechanik); and
b) Into the mechanics of air and other aëriform bodies (Fr. mé-
canique des fluides aëriformes, Ger. Mechanik der luft-
förmigen Körper).
If we take into consideration the division of mechanics into
statics and dynamics, we can again divide it into:
1) Statics of rigid bodies.
2) Dynamics of rigid bodies.
3) Statics of water, etc., or hydrostatics.
4) Dynamics of water, etc., or hydrodynamics.
5) Statics of air (of gases and vapor) or aërostatics.
6) Dynamics of air (of gases and vapors) or aërodynamics or
pneumatics.
CHAPTER II.
MECHANICS OF A MATERIAL POINT.
§ 67. A material point (Fr. point matérial, Ger. materieller
Punkt) is a material body whose dimensions in all directions are in-
finitely small compared with the space described by it. In order to
simplify the discussion, we will now consider the motion and equili-
brium of a material point alone. A (finite) body is a continuous
union of an infinite number of material points or molecules. If
the different points or elements of a body move in exactly the same
manner, I.E., with same velocity in parallel straight lines, the
theory of the motion of material point is applicable to the whole
body; for in this case we can suppose that equal portions of the
mass fre impelled by equal portions of the force.
166
[$ 68, 69.
GENERAL PRINCIPLES OF MECHANICS.
§ 68. Simple Constant Force.-If p is the acceleration with
which a mass Mis impelled by a force P, we have from § 56
P
P = Mp, or inversely the acceleration p = M®
G
Putting the mass M =
g'
G denoting the weight of the body
and g the acceleration of gravity, we obtain the force
and the acceleration
1) P = 2 G,
2) p =
g
P
9.
GI.
We find then the force (P) which moves a body with the accel-
eration (p) by multiplying the weight (G) of the body by the
ratio of its acceleration to that of gravity.
()
Inversely we obtain the acceleration (p), with which a force (P)
will move a mass M, by multiplying the acceleration (g) of gravity
by the ratio () of the force to the weight of the body.
EXAMPLE.-Let us imagine a body placed upon a very smooth horizon-
tal table, which opposes no resistance to its motion, but which counteracts
the effect of gravity. If this body be subjected to the action of a horizon-
tal force, the body yields and moves forward in the direction of the force.
If the weight of the body is G = 50 pounds and the force which acts
uninterruptedly upon it is P = 10 pounds, it will assume a uniformly accel-
P
erated motion, the acceleration of which is p
feet. If, on the contrary, the acceleration produced in a body weighing
10
G
g
32,2 = 6,44
50
9
.42 =
42 pounds by a force Pis
p 9 feet, then the force is P
0,031.378 = 11,7 pounds.
P
G
9
32,20
§ 69. If the force acting upon a body is constant, a uniformly
variable motion is the result, and it is uniformly accelerated, when
the direction of the force coincides with the original direction of
motion, and uniformly retarded, when the force acts in the opposite
direction. If we substitute in the formulas of § 13 and § 14, in-
P P
g, we obtain the following:
stead of p, its value
M
G
§ 69.]
167
MECHANICS OF A MATERIAL POINT.
I. For uniformly accelerated motion:
P
P
P
G
1) v = c + 2 gt = c + 32,2 t feet = c + 9,81
t metres,
G
G
2) s = ct +
P g t²
G 2
ct + 16,1
P
G
P
t² feet = ct + 4,905
t2 metres.
G
II. For uniformly retarded motion:
P
P
P
1) v = c
g t = c
=
32,2
=
t feet — c — 9,81
t metres.
G
G
G
Pg t
P
P
2) s = ct
G 2
G
G
= ct - 16,1 t² feet ct - 4,905 t2 metres.
By means of the above formulas all questions, which can arise
in reference to the rectilinear motions produced by a constant force,
can be answered.
EXAMPLE.—1) A wagon weighing 2,000 pounds moves upon a horizon-
tal road, which opposes no resistance to it, with a velocity of 4 feet, and
is impelled during 15 seconds by a constant force of twenty-five pounds;
with what velocity will it proceed after the action of this force? The
P
G
required velocity is v = c + 32,2 t, but here we have c = 4, P
4, P = 25,
25
15, whence v = 4 + 32,2.
•
15 = 4 + 6,037
2000
G 2,000 and t = 15, whence v
10,037 feet. 2) Under the same circumstances a wagon weighing 5,500
pounds, which in the three previous minutes had described uniformly 950
feet, was impelled during 30 seconds by a constant force, so that after-
wards it described 1650 feet uniformly in three minutes. What was this
950
3.60
force? The initial velocity is c =
1650
3.60
3,889.G
gt
locity is =
force P
5,277 feet, and the final ve-
P
9,166 feet, whence
g t = v c = 3,889, and the
0,031. 3,889.
5500
30
0,120559.
550
3
22,10 pounds.
3) A sled weighing 1500 pounds and sliding on a horizontal support with
a velocity of 15 feet loses, in consequence of the friction, in 25 seconds, the
whole of its velocity. What is the amount of the friction? The motion is
here uniformly retarded and the final velocity is v = 0, hence c = 32,2.
Pt
G'
and P
0,031
A c
t
=
0,031
1500.15
25
0,031 . 900
27,9 pounds,
which is the friction in question. 4) Another sled, weighing 1200 pounds
and moving with an initial velocity of 12 feet, is obliged to overcome a
168
[$ 70.
GENERAL PRINCIPLES OF MECHANICS.
friction of 45 pounds when in motion. What is its velocity after 8 seconds,
and what is the space described?
The final velocity is
v = 12. 32,2
45.8
1200
= 12
9,662,34 feet,
and the space described is
$
•
8: = 57,36 feet.
= (³ + 2) t = ( 12 + 2,84)
2
§ 70. Mechanical Effect.*-Mechanical effect or work done
(Fr. travail mecanique, Ger. Leistung or Arbeit der Kraft) is that
effect which a force accomplishes in overcoming a resistance, as,
E.G., gravity, friction, inertia, etc. Work is done when we elevate
a weight, when a greater velocity is communicated to a body, when
the forms of bodies are changed, when they are divided, etc. The
work done depends not only upon the force, but also on the space
during which it is in action, or during which it overcomes a re-
sistance. If we raise a body slowly enough to be able to disregard
the inertia, the work done is proportional to its weight and to the
height which it is lifted for 1) the effect is the same if a body of
the m (3) fold weight is lifted a certain height, or if m (3) bodies
of the weight (G) are lifted the same height; it is m times as
great as that necessary to raise the simple weight the same height;
and in like manner 2) the work done is the same, if one and the
same weight be raised the n (5) fold height (n h) or if it is raised
n (5) times to the simple height, and in general n (5) times so great,
as when it is raised to the simple height. In like manner, the
work done by a weight sinking slowly is proportional to the weight
and to the distance it sinks. This proportion is, however, true for
every other kind of work done; in order to make a saw cut of
twice the length and of the same depth as another we are obliged
to separate twice as many particles, and the work done is therefore
double; the double length requires the force to describe double the
distance, and consequently the work is proportional to the space
described. In like manner the work done by a run of millstones
increases evidently with the number of grains of a certain kind
of corn which it grinds to a certain fineness. This quantity is,
however, under the same circumstances proportional to the number
* Energy is the capacity of a body to perform work. Energy is said to be
stored when this capacity is increased, and to be restored when it is diminished.
The unit of energy is the same as that of work.TR.
§ 71, 72.]
169
MECHANICS OF A MATERIAL POINT.
of revolutions, or rather to the space described by the upper mill-
stone while this quantity of corn is being ground. The work
done increases, therefore, directly with the space described.
§ 71. As the work done by a force depends upon the inten-
sity of the force and the space described by it, we can assume as
the unit of work or dynamical unit (Fr. unité dynamique, Ger.
Einheit der mechanischen Arbeit oder Leistung) the work done
in overcoming a resistance, whose intensity is the unit of weight
(pound, kilogram) over a space equal to the unit of length (foot,
metre), and we can also put this measure equal to the product
of the force or resistance into the space described by it in its
direction while overcoming the resistance.
If we put the amount of the resistance itself = P and the
space described by the force, or rather by its point of application,
while overcoming its, then the work done in overcoming this
resistance is
A Ps units of work.
In order better to define the units of work (which we can style
simply dynam) the units of both factors P and s are generally
given, and instead of units of work we say kilogram-meters and
pound-feet, or inversely meterkilograms, foot-pounds, etc., accord-
ing as the weight and the space are expressed in kilograms and
meters, or in pounds and feet. For simplicity we write instead of
meterkilogram, mk or km; and instead of foot pound, lb. ft.,
or ft. lb.
15
12
EXAMPLE.-1. In order to raise a stamp weighing 210 pounds, 15 inches
high, the work to be done is A 210.
262,5 ft. lbs.
2. By a me-
chanical effect of 1500 foot pounds a sled, which when moving must over-
come a friction of 75 pounds, will be drawn forward a distance
$
P
A 1500
75
20 feet.
§ 72. Not only when the force is invariable, or the resistance is
constant, but also when the resistance varies while the force is
overcoming it, can the work done be expressed by the product of the
force and the space described, provided we assume for the value of
the force the mean value of the continuous succession of forces. The
relation between the time, velocity and space is therefore the same
here; for we can regard the latter as the product of the time and
170
[$ 73.
GENERAL PRINCIPLES OF MECHANICS.
of the mean of the velocities. We can also employ here the same
graphical representations. The work done can be regarded as the
area of a rectangle A B C D, Fig. 86, whose base A B is the space
(s) described and whose height is either the constant force Por
the mean value of the different forces. In general, however, the
work done can be represented by the area of a figure A B CND,
Fig. 87, the base of which is the space s described, and the height
D
FIG. 86.
N
FIG. 87.

N

F
E
D
A
B
M
A
M
B
of which above each point of the base is equal to the force corre-
sponding to that point of the path. If we transform the figure
ABCND in a rectangle ABEF with the same base and the
same area, its altitude A F = B E gives the mean value of the
force.
§ 73. Arithmetic and Geometry give several different methods
for finding the mean value of a continuous succession of quanti-
ties, the most important of which are to be found in the Ingenieur.
The method known as Simpson's Rule is, however, the one most.
generally employed in practice, because in many cases it unites
great simplicity with a high degree of
accuracy.
FIG. 88.
H K
F
D
C
A
E G
J B
In
every case it is necessary to divide
the space A B = = s (Fig. 88), in n (as
many as possible) equal parts, such as
A EEG GJ, etc., and to deter-
mine the forces EF P₁, GH = PJK
=P, etc., at the ends of these portions
of the path. If we put the initial force.
AD P, and the final one B C
P₂
+

0
we have the mean force P (į P。 + P₁ + P₂ + P¿ +
P₁-1 + 1 P₁) : n, and consequently its work
S
Ps (P + P₁ + P₂ + ... + P₂-1 + P₂)
=
2
Pn−1 ¦
22
§ 74.1
171
MECHANICS OF A MATERIAL POINT.
If the number of parts (n) be even, I.E. . 2, 4, 6, 8, etc., Simp-
son's Rule gives more exactly the mean force
P = (P¸ + 4 P₁ + 2 P₂ + 4 P₂ + ...
whence the corresponding work done is
P s = (P。 + 4 P₁ + 2 P₂ + 4 P3 +
1
2
If n is an uneven number, we can put
+ 4 P-1 + Pn): 3n,
P„)
+4
+ 4 Pr−1 + P„)
S
3 n
P 3 = [3 (P, + 3 P₁ + 3 P½ + P3) + ¦ (P¸ + 4 P¸ + 2 Ps
S
+ . . . . + 4 P₂-1 + P₂)] (See Art. 38 of the Introduction
Pn−1
to the Calculus.)
N
EXAMPLE.—In order to determine the work done by a horse, in drawing
a wagon along a road, we employ a dynamometer (or force measurer, one
side of which is attached to the wagon and the other to the horse, and we
observe from time to time the intensity of the force. If the initial force is
P = 110 pounds, that after moving 25 feet 122 pounds, that after 50 feet 127
pounds, that after 75 feet 120 pounds, and that at the end of the whole dis-
tance, 100 feet, 114 pounds, we have for the mean value of the force ac-
cording to the first formula
Р 1
= ( (§. 110 + 122 + 127 + 120 + 1 × 114): 4
and for the mechanical effect
Ps 120,25 × 100
=
=
120,25 pounds,
12025 foot-pounds.
According to the second formula we have
P⇒(110+4. 122 +2. 127 +4 . 120 + 114) : (3 . 4)
and the mechanical effect
་
1446
12
=120,5 pounds,
Ps = 120,5 . 100 = 12050 foot-pounds.
Ꮲ
§ 74. Principle of the Vis Viva or Living Forces.-If、
ぴ
​the formula s =
c²
or p s =
v² -- c²
found in § 14, we substi-
2 P
2
P
tute for p its value
g, we obtain the mechanical effect A
P:
G
G,
2 g
= (³*~,^)6, or designating the heights due to the velocities
2 g
C²
and
by h and k,
2g
Ps (hk) G.
-
This equation, so important in practical mechanics, means that
the mechanical effect (Ps), which a mass absorbs when its velocity
changes from a lesser to a greater, or that which it gives out, when
its velocity is forced to change from a greater to a less, is always
172
[$ 75.
GENERAL PRINCIPLES OF MECHANICS.
equal to the product of the weight of the mass into the difference of
the heights due to the different velocities
v2
g
2 g
EXAMPLE.-1. In order to impart, upon a perfectly smooth railroad, a
velocity of 30 feet to a wagon weighing 4000 pounds, the work to be done
212
is Ps
G= 0.0155 v² G 0,0155 × 900 × 4000 = 55800 pounds, and this
29
wagon will perform the same amount of work if a resistance ne opposed to
it, so as to cause it gradually to come to rest. 2. Another wagon, weighing
6000 pounds and moving with a velocity of 15 feet, acquires in consequence
of the action of a force a velocity of 24 feet; how much mechanical effect
is stored by the wagon, or how much work is performed by the force?
h =
=
The heights due to the velocities 15 and 24 feet are k
22
2 g
c³
2 g
= 3,487 and
8,928 feet. Consequently the work done P s = (h k) G
=
(8,928 3,487) × 6000 = 5,441 × 6000 32646 foot-pounds.
―
If the space described is known the force can be found, and if the
force is known the space can be found. Let us suppose, E.G., in the last
case, that the space described by the wagon, while the velocity changes from
11 to 24 feet, is but 100 feet, we have then the force P (h
326,46 pounds. If, however, the force was 2000 pounds,
32646
100
-
h:)
G
S
G
the space would be 8 = (h k)
P
32646
2000
= 16,323 feet. 3. If a sled
weighing 500 pounds, and moving with a velocity of 16 feet, loses in con-
sequence of the friction the whole of its velocity while describing 100 feet,
the resistance of the friction is
h x G
P
0,0155 × 16² ×
8
500
100
0,0155 × 256 × 519,84 pounds.
§ 75. The formula for the work done, found in the preceding
paragraph,
Ps=
(220) G
C.
=
(12 k) G,
g
holds good not only when the forces are constant, but also when
they are variable, if we substitute (according to § 73) instead of P
the mean value of the force; for according to III*), in § 19, we
have, in general, for every continuous motion
2² — c³
= p s
2
in which p =
P₁ + P₂ +
+P" denotes the mean acceleration
N
$75.]
173
MECHANICS OF A MATERIAL POINT.
with which the space s is described, and we have also
1
P₁ + P₂ +
Pr
p
2
n M
·
•
whence
,
v²
P₁ + P₂ + .
•
+ Pr
M
P)'s and
2
M
P s = (³°³
2,2
c²
M =
G = (h -
k) G,
2
2 g
P₁ + ... + Pn
in which P =
denotes the mean of all the forces
N
s 2s 3s
n s
measured after the spaces
are described.
ጎ ጎ N
n
The force P can also be calculated by means of one of the
formulas of § 73, when the number n of the parts is not assumed
to be very great.
We are very often required to calculate the change of velocity
that a given mass M undergoes, when a given amount of me-
chanical effect Ps is imparted to it. The principal equation
which we have found is then to be employed in the following form
h = k +
Ps
G
or v=
√c² + 29
Ps
G
If we have calculated by means of this formula the velocities
v, ... v, which correspond to the spaces.
212
calculate by means of the formula
s 2s 3s
•
S, we can
ጎ N N
t
S 1
N
1
1
1
+ +
+...+
V2 V3
Un
the time in which the space s is described.
=
2 P s
v² — c²
Ps
the principal
i (v + c) (v c)
In the form G = Mg
formula we have found serves to determine the mass M, which in
consequence of the mechanical effect P s imparted to it will un-
dergo a change of velocity v - c.
When the motion of a body is continuous and the final velocity
v is equal to the initial one c, then the work done becomes = 0,
I.E, the accelerated part of the motion absorbs exactly as much
work as the retarding portion gives out.
EXAMPLE.—If a wagon weighing 2500 pounds, moving without fric-
tion with an initial velocity of 10 feet, has imparted to it a mechanical
effect of 8000 foot-pounds, what is its final velocity ?
174
[§ 76.
GENERAL PRINCIPLES OF MECHANICS.
Here
10+ 64,4.
8000
2500
√ 100 + 206,08 = 17,49 feet,
G
9
REMARK. We call, without attaching any particular idea to the term,
the product of the mass M: into the square of the velocity (v2), that is
M v³, the vis viva (Fr. force vive, Ger. lebendige Kraft) of the moving mass,
and we can therefore put the mechanical effect, which a mass which is
moved absorbs, equal to the half of its vis viva. If an inert mass passes
from a velocity c to another v, then the work gained or lost is equal to the
half difference of the vis viva at the beginning and of that at the end of
the change of velocity. This law of the mechanical effect bodies produce
by virtue of their inertia is called the principle of vis viva (Fr. principe
des forces vives, Ger. Princip der lebendigen Kräfte).
2
§ 76. Composition of Forces.-If two forces P, and P, act
upon the same body 1) in the same or 2) in opposite direc-
tions, then their effect is the same as when a single force equal to
1) the sum or 2) the difference of these forces acted upon the body;
for these forces impart to the mass the accelerations
P₁
P
M
and pr
2
P₂
M
whence, according to § 28, the resulting acceleration is
p = p₁ = p₂ =
P2
P₁ ± P₂
M
and consequently the corresponding force is
P = Mp P₁± P2.
We call the force P derived from the two forces and capable of
producing the same effect (equipollent) their resultant (Fr. résult-
ante, Ger. Resultirende), and its constituents P, and P, its com-
ponents (Fr. composantes, Ger. Componenten).
1
2
EXAMPLE.-1) A body lying upon the flat of the hand presses with its
absolute weight on it only so long as the hand is at rest, or is moved with
the body uniformly up or down; but if we lift the hand with an accelerated
motion, it experiences a heavier pressure; and if, on the contrary, we allow
it to sink with an accelerated motion, then the pressure becomes less than
the weight, and even 0 when the hand is lowered with an acceleration
equal to that of gravity. If the pressure on the hand is P, then the body
falls with the force G
P only, if its mass is M
Ꮐ
G
; if we put the ac-
g
P, and
p
therefore the pressure P = G–
G
y
(1.-
2)
G. If, on the contrary,
celeration with which the hand descends = p we have G P =
g
§ 76.]
175
MECHANICS OF A MATERIAL POINT.
we raise the body upon the hand with an acceleration p, then the accelera-
tion Ρ is opposite to the acceleration g, and the pressure becomes P =(1
Ρ
+ G. According as we lower or raise a body with an acceleration of
the hand is (1 -
20 feet, the pressure upon the hand is
20
32,2
:)
=
G (1 0,62) G =
1,62 times the same
0,38 times the weight of the body, or 1 + 0,62
weight. 2) If with the flat of the hand I throw a body weighing 3
pounds 14 feet vertically upward, by urging it on continuously during the
first two feet, then the work done is P s =
= G h = 3.14 = 42 pounds, and
the pressure of the body on the hand is P
the body when at rest presses with a weight of three pounds upon the
hand, and, on the contrary, during the act of throwing it, it reacts with a
force of 21 pounds upon the hand.
42
2
= 21 pounds. Hence
3) What load Q can a piston movable in a cylinder A A C C, Fig. 89,
raise to the height D K = 8 = 6 feet, if during the first half of its course
FIG. 89.
H
C
P₁₂
P
G
P.
F
M
Р B
E
B
V
In
the air which flows in from a very large res-
ervoir acts upon it with a force of 6000
pounds, and if during the second half of its
course this air enclosed in the cylinder ex-
pands according to the law of Mariotte, while
the exterior air acts with a constant pressure
of 2000 pounds in the opposite direction.
Since the air shut in the cylinder at the end
of the second half of the course of the piston
has expanded to double its volume, the
pressure of the same upon the piston at the
end of the course is only . P = 3000 pounds.
The air inclosed in the cylinder, when the
piston has traveled 3 feet, presses with a force of 6000 pounds upon it, on
the contrary at the end of four feet with a force of §. 6000 = 4500 pounds,
at the end of 5 feet with .6000 3600 pounds, and at the end of the
entire course with a force of .6000 3000 pounds. Hence the mean
} [6000 + 3 (4500 + 3600) + 3000]

P
A
force during the expansion
23300
= 4162 pounds, and consequently the mean force during the whole
8
of the course of the piston is
6000 + 4162
2
5081 pounds. If we sub-
tract the constant opposing force of 2000 pounds from this, it follows that
the weight to be raised by the piston is
Q = 5081 2000 3081 pounds.
The motive force for the first half of the course is then P (8 +
2,000) = 6000 - 5081 919 pounds, and consequently the acceleration
of the motion is p =
(
P− (Q + 2000)
919
g
Q
3081
. 32,2 9,6 feet, and
176
[$ 76.
GENERAL PRINCIPLES OF MECHANICS.
the velocity at the end of the first half of the course of the piston 8₁
3 feet is v=
√ 2 ps
= √ 6.9,6 = √57,6
which this space is described by the piston is t₁
=
1
1
02
7,589 feet, and the time in
281
6
7,589
0,790
seconds. The distance, which has been traveled by the piston when the
force and the load balance each other, that is, when the motive force and
consequently the acceleration is 0, and the velocity of the piston is a
maximum, is
x =
le
P
Q + 2000
6,543
500)
8
6000. 3
3,543 feet.
2
5081
When the distance
= 3,2715 feet has been described, the force act-
2
6000.3
3,2715
M
ing on the inside piston is
force is = 5502
5081421 pounds, and the mean value of the same
5502, and consequently the motive
919+ 4.421 +0
while the piston passes from 3 to 3,543 feet is
434
6.
434
434. 32,2
3081
pounds. The corresponding mean acceleration is = 3081
4,535 feet, and consequently the maximum velocity of the piston at the
end of the space x = 81 + $ 2
8 = 3,543 feet is
vm
√ v² + 2 p s 2
=
7,907 feet.
✓ 62,525
√ 57,6 + 2 × 4,535 × 0,543
The time required to describe the space 82 = 0,543 can be put
8
2
t2
+
2
V
1) = 0,2715 (7,589 + 907) = 0,070
€
If the piston has described the space 5,5 the motive force is
seconds.
18000
5,500
1808 pounds, and if the piston is midway between this point
5081
and the point of maximum velocity, this force is then =
18000
4,5215
5081
1808 × 32,2
1100 pounds, and the corresponding accelerations are—— 3081
=
18,89 feet, and =
1100 × 32,2
3081
11,49 feet.
The mean acceleration while the piston describes the portion of the
space 5,500 - 3,543
1,957 feet is consequently
0+4 × 11,49 +18,89
6
is
10,81 feet, and therefore the velocity acquired at the end of this space
√62,525 2 × 10,81 × 1,957 = √20,215 = 4,496 feet. On the contrary,
during the first half of the last portion of the course, the mean acceleration is
5,745 feet, and therefore the velocity at the end of the
0 + 11,49
2
= √ 62,525 2 × 5,745 × 0,9785
√51,282
space 4,5215 feet v₁
7,161 feet, and we have for the time required to describe the space s
8
§ 77.]
177
MECHANICS OF A MATERIAL POINT.
1
4
1,957, tz
+
+
6
心​仉
​V
1
1
V2
4
1
0,326 (7,90
+
+
,907
7,161
4,496) = 0,326
× 0,9075
==
0,296 seconds. Finally, we can put the time during which the
2
284 1
last portion 84
0,5 of the whole course is described t 4
D2 4,496
0,2224 seconds, and the time required by the piston to describe its entire
course t=t₁+tq + t z + t₁ = 0,790 + 0,070 +0,296 +0,2224 = 1,378 seconds.
1
4
77. Parallelogram of Forces. If a mass (a material
point) M, Fig. 90, is acted upon by two forces, P, and P, whose
direction, M X and M Y, form an angle X M Y = a with each
other, the forces cause in these directions the accelerations
P1
P₁
M
1
and p2
P2
M'
and by combining them, a resulting acceleration (§ 35) in the
direction M Z, which is determined by
the diagonal of a parallelogram con-
structed with p₁, P., and a, is obtained;
this resulting acceleration is
FIG. 90.
P₁
Pi

M
X
P₂
P
Y
P
Z
2
2
p = √ p₁² + p²² + 2 pr p₂ cos. α,
and we have for the angle 4, which its
direction makes with the
MX of the acceleration p₁
sin.
direction
P₂ sin, a
P
Substituting in these two formulas the given values of p, and p,
we obtain
P
p = √(ii)² + (²)
M
sin. =
• = (fi)
Po sin. a
M P
P
9
cos, a and
M
and multiplying the first equation by M, we have
Mp = √ P² + P²² + 2 P、 P₂ cos. a,
1
or since Mp is the force P corresponding to the acceleration p, we
1) P = √ P,² + P¸² + 2 P、 P₂ cos. a,
find
2) sin.
P, sin. a
P
The resultant or mean force is determined in magnitude and di-
rection from the component forces in exactly the same manner, as the
resulting acceleration is determined from the component accelerations.
If we represent the forces by right lines, making the ratio of
12
178
[$ 77.
GENERAL PRINCIPLES OF MECHANICS.
their length the same as that of the weights (E.G. pounds) to
each other, the resultant can then be represented by the di-
agonal of the parallelogram whose sides are formed by the compo-
nent forces, and one angle of which is equal to the angle formed by
the component forces with each other. The parallelogram thus
constructed with the component forces, the diagonal of which rep-
resents the resultant, is called the parallelogram of forces.
1
EXAMPLE.-If a body, Fig. 91, weighing 150 pounds and resting on a
perfectly smooth table, is acted on by two forces P₁ = 30 pounds, and
P₂ = 24 pounds, which form with each other an angle P₁ M P2
2
a = 105",
in what direction and with what acceleration will the motion take place?
Since cos. a=cos. 105° =
the resultant
FIG. 91.
X
P√302 + 242
= √900 + 576
=
cos. 75, we have

2 × 24 x 30 x cos. 75°
1440 cos. 75"
√1476 - 372,7 = 33,22 pounds;
and the corresponding acceleration
P
M
Z
y
Ρ
P Pg
M = G
33,22 × 32,2
=
7,13 feet.
Ꮐ
150
The direction of the motion forms an angle o
with the direction of the first force, which is de-
termined by the following formula
P
sin. &=
24
33,22
sin. 105"=0,7224 sin. 75° -0,6978;
and is 44° 15′.
Φ
REMARK.—The resultant (P) depends (according to the formula just
found) upon the components alone, and not upon the mass (M) of the
body upon which the forces act. For this reason we find in many works
upon mechanics the correctness of the parallelogram of forces demonstrated
without reference to the mass, but with the assumption of some one of the
fundamental laws of statics. Such pure statical demonstrations are
numerous. In each of the following works we find a different one:
"Eytelwein's Handbuch der Statik fester Körper;" "Gerstner's Hand-
buch der Mechanik;" "Kayser's Handbuch der Statik;" "Mobius' Lehr-
buch der Statik;""Rühlman's Technische Mechanik." The demonstration
in Gerstner's Mechanik" is based upon the theory of the lever; it is really
very simple, and is to be found in old, and also in later works, E.G., in those
of Kästner, Monge, Whewell, etc. Kayser's demonstration is that of Poisson
in elementary shape. Mobius' discussion of it is based upon a particular
theory of couples (des couples) introduced by Poisson (Elements de
Statique). A peculiar demonstration is given by Duchayla in the Corre-
spondence sur l'école polytechnique No. 4, which is reproduced by Brix in
เ
$78.]
179
MECHANICS OF A MATERIAL POINT.
his Lehrbuch der Statik fester Körper. It is also given in many other
works, E.G., in Moseley's Mechanical Principles, etc. The demonstration
of the parallelogram of forces given by Navier in his " Leçons de Mécan-
ique" (German by Meier, 1851) is also to be found in Rühlmann's “Grund-
züge der Mechanik," Leipzig, 1860. A theory of this parallelogram,
founded on the laws of motion, is to be found in Newton's "Principia."
It is also employed in many later works, I.E., by Venturoli, Poncelet, Burg,
etc. See Elementi di Mecanica e d'Idraulica di Venturoli," "Mecanique
industrielle par Poncelet," "Compendium der populären Mechanik and
Machinenlehre von Burg." A new demonstration by Mobius is to be found
in the Berichten der Gesellshaft der Wissenshaften zu Leipzig (1850), an-
other by Ettingshausen in the papers of the Academy of Vienna (1851), and
a third, by Schlömich in his "Zeitschrift für Mathematik and Physik"
(1857).
§ 78. Decomposition of Forces.--With the aid of the paral-
lelogram of forces we can not only combine two or more forces so
as to find a single resultant, but also decompose a given force.
under given circumstances, into two or more forces. If the angles
and, which the components MP, P, and MP, P., Fig. 91,
make with the given force M P = P are given, then the compo-
nents are determined by the following formulas
P₁
P sin.
sin. (Þ + 4)³
P₂
P sin.
sin. (+4)
¢ ¥
If the components are at right angles, then + 4 = 90° and sin.
(Ø + 4)
=
1, and we have
P₁
P cos.and P₂ = P sin. ọ.
and if, finally, and ☀ are equal, we have
P
P₁ = P₁ =
P. sin. O
sin. 20
2 cos. O
EXAMPLE.-1) How heavily will a table A B, Fig. 92, be pressed by a
body M whose weight is G = 70 pounds, and which acted on by a force
FIG. 92.
P
Mi
P
A
B
P = 50 pounds, which is inclined to the horizon
at an angle P M P₁
component is
1
40°? The horizontal

P₁ P cos. † = 50 cos. 40° = 38,30 pounds,
and the vertical component
P₂ P sin. ¢ = 50 sin. 40° = 32,14 pounds.
The latter tends to raise the body from the table,
and consequently the pressure on the table is
· P₂ = 70 32,14 = 37,86 pounds.
G — P₂ =
2) If a body M, Fig. 91, weighing 110 pounds,
is moved upon a horizontal support by two forces.
so that in the first second it describes a distance
of 6,5 feet in a direction, which forms with the two directions of the forces
180
[$ 79.
GENERAL PRINCIPLES OF MECHANICS.
the angles o
acceleration is
case p
2.6,5
P
52º and y = 77°, the forces can be found as follows: The
double the space described in the first second, or in this
= 13 feet, and the resultant is
P Q
0,031. 13. 110 44,33 pounds.
Hence one of the components is
9
P sin. 770
sin. (520 + 770)
44,33 sin. 77°
sin. 51°
55,58 pounds,
and the other is
P₂
44,33 sin. 52°
sin. 51°
= 44,95 pounds.
§ 79. Composition of Forces in a Plane.--In order to find
the resultant P of a number of component forces P1, P2, P3, etc.
(Fig. 93), we can pursue exactly the same method that we em-
ployed in the composition of velocities. We can, by employing
repeatedly the parallelogram of forces, combine the forces two by
two so as to form one, until but one is left. The force P, and P₂
give, E.G., by means of the parallelogram M P, Q P, the resultant
Q; and if we combine this with P, we obtain, by means
of the parallelogram M Q R P3
the resultant M R = R, and the
latter, combined with P, gives,
by means of the diagonal M P
M Q
=
FIG. 93.
P₁
M
P₁
3

P, the resultant of all four
forces P1, P2, P, and P4. It is
not necessary, when combining
these forces, to complete the par-
allelograms and to find their
diagonals. We have but to con-
struct a polygon M P₁ Q R P
by drawing its sides M Pi, P, Q,
QR, R P, equal and parallel to the given components P₁, P2, P3,
P₁. The last side M P, which closes the parallelogram, is the re-
sultant required, or rather the measure of the same.
R
Р
1
1
REMARK.-The solution of mechanical problems by construction is very
useful. Although the results are not as accurate as those obtained by cal-
culation, yet they are of great value as checks against gross errors, and can
therefore always be employed as proofs of calculations. In Fig. 93 we
have drawn the forces as meeting each other and forming the given angles
P₁ MP₂ 72° 30', P₂ M P₂ = 33° 20', and P3 M P₁ 92° 40'; and
their length is such, that a pound is represented by a line or of a
1
2
4
§ 80.]
181
MECHANICS OF A MATERIAL POINT.
4
11,5 pounds, P₂
2
=
Prussian inch. The forces P₁
10,8 pounds, P3
8,5 pounds, and P₁ = 12,2, are therefore expressed by sides 11,5 lines.
10,8 lines, 8,5 lines, and 12,2 lines long. A careful construction of the
polygon of forces gives the value of the resultant P 14,6 pounds, and
the angle formed by the direction MP with the direction MP, of the first
force a = 861°.
-
1
§ 80. We can determine the resultant P more simply by de-
composing each of the components P1, P2, P3, etc., into two com-
ponents Q, and R, Q, and R, Q and R, etc., in the direction of
the rectangular axes IX and Y F, Fig. 94, by then adding alge
braically the forces which lie in the same axis, and by seeking the
intensity and direction of the resultant of the two forces which
have been thus obtained, and whose directions are at right angles
to each other. If the angles P, M X, P, MX, P, M X, etc., P₁,
P2, P, etc., form with the axis of X are = α, ɑ ɑз, etc., we
have the components Q, P, cos. a₁, R₁ = P; sin. a,; Q₂ = P₂
cos. α, R₂ = = P₂ sin. a, etc.; whence it follows from the equation.
Q = Q1 + Q₂ + Q3 +
1) Q
1
1
• •
that
P₁ cos. a₁ + P₂ cos. a₂ + P3 cos. α3 +
1
2
and also from R = R₁ + R₂+ R3..
2) R
=
that
P₁ sin. a₁ + P₂ sin. a₂ + P3 sin. az +
1
2
3
We find the value of the resultant of the two components Q and R,
just obtained, by the aid of the formula
3) P = √ Q² + R²,
and that of the angle P M X = a, formed by its direction with the
axis II, by means of the formula
P
R
4) tang. a =
Q
FIG. 94.
P
Y
R
R3
R2
R
M Q3 Q4
X
Q 1 Q z
R
4
Y
PA
X
In adding algebraically the forces
we must pay particular attention to
their signs; for if they are different
for two different forces, I.E. if these
forces act in opposite directions
from the point of application, then
this addition becomes an arithmeti-
cal subtraction. The angle a is
acute as long as R and Q are posi-
tive; it is between 90º--180", when
Qis negative and R positive; it is
between 180°-270°, when Q and R
are both negative, and is finally be-

tween 270°-360°, when R alone is negative.
182
[$ 81.
GENERAL PRINCIPLES OF MECHANICS.
3
EXAMPLE.-What is the direction and intensity of the resultant of
the forces P₁ = 30 pounds, P₂ = 70 pounds, aud P 50 pounds, whose
directions lie in the same plane and form the angles P, MP₂ = 56º and
P₂ MP3 104° with each other? If we lay the axis X X, Fig. 94, in the
direction of the first force, we obtain a = 0º, a₂ = 56º, and
56º, and a¸ =
2
104º
160º
160°; hence
1
1) Q = 30. cos. 0° + 70. cos. 56° + 50 cos. 160°
= 22,16 pounds,
1
3
= 56° +
30 + 39,14 - 46,98
2) R = 30. sin. 0° + 70. sin. 56° + 50. sin. 160º 0 + 58,03 + 17,10
= 75,13 pounds, and
3) tang. a =
75,13
22,16
3,3903,
and therefore the angle formed by the resultant with the positive portion
of the axis MX is a 73° 34′, and the resultant itself is P =√√ Q² + R²
R
Q
cos. a
sin. a
75,13
sin. 73° 34′ 0,9591
75,13
=
78,33 pounds.
§ 81. Forces in Space. If the direction of the forces do not
lie in the same plane, we pass a plane through the point of appli-
cation and decompose the forces into two others, one of which lies.
in the plane, and the other at right angles to it. The components
thus obtained, which lie in the plane, are combined according to
the rule given in the last paragraph, so as to give a single result-
ant, and those at right angles to the plane give, by simple addition,
another resultant. From these two components, which are at right
angles to each other, we find the resultant according to the well-
known rule (§ 77).
=
3
3
This method of proceeding is graphically represented in Fig.
95. MP, P₁, M P₂ = P, MP, P, are the simple forces,
A B is the plane (plane of projection) and Z Z is the axis at right
angles to it. From the decomposition of the forces P, P, etc., we
obtain the forces S, S2, etc., in the plane, and the forces N, N, etc.,
along the normal Z Z. The former are again decomposed into the
components Q1, Q2, etc., R1, R, etc., which, by addition, give the
resultants Q and R, from which, as components, we determine the
resultant S, which, combined with the sum of all the normal forces
N, N, etc., gives the required resultant P.
If we put the angles of inclination of the directions of the
forces to the plane equal to B, B, etc., we obtain for the forces in
the plane S₁ = P₁ cos. ß₁, S₂ = P₂ cos. ß2, etc., and for the normal
forces N₁ = P, sin. B, N, P, sin. B, etc. Designating the angles
which the projections of the directions of the forces in the plane
N₂
=
2
2
§ 81.]
183
MECHANICS OF A MATERIAL POINT
A B form with the axis X X by a₁, a₂, etc., that is, putting S₁ MX
= a₁, S₂ MX = a, etc., we obtain the following three forces, which
FIG. 95.

Z
PR
A
S
Ri
R₂
P
N
N₁
1
N2
M
Q1
S
S
X
Z
Ps
Y
R3
S3
B
Q
1) Q
orm the edges of a rectangular parallelopipedon (parallelopipe-
don of forces):
S₁cos. a₁ + S₂ cos. a₂
·
+ . or
1
P₁ cos. ß₁ cos. a、
1
+
2
P₂ cos. ß₂ cos. a₂ + ...,
2) R
P、 cos. ẞ, sin. a、
+ P₂ cos. ß. sin, a
•
and
3) N
1
P₁ sin. B₁ + P₂ sin. B₂ +
2
From these three forces we obtain the final resultant
4) P = √ Q² + R² + N²,
and its angle P M S
ẞ of inclination to the plane of pro、
jection by the aid of the formula
N
5) tang. B S
N
√ Q² + R²
MX
Finally, the angle S MY= a, which the projection of the re-
sultant in the plane A B forms with first axis XX, is given by
the formula
6) tang. a =
flos
R
If 2,, λ„, etc., are the angles formed by the forces P₁, P. with the
axis MX, 1, ..., the angles formed by them with the axis
MY and ₁, ½, etc., the angles formed by them with the axis M Z,
We have also
184
GENERAL PRINCIPLES OF MECHANICS.
[§ 81.
1*) Q
2*) R
=
P₁ cos. λ₁ +
1
P₁ cos. µ₁ +
3*) N = P₁ cos. v, +
2
P₂ cos. λ₂ +
P₂ cos. µ₂ + and
P₂ cos. v½ +
2
2
•
The value of the resultant is given by the formula
4*) P = √ Q² + R² + N²,
and the direction of the same by the formulas
5*) cos. λ
Q
P'
cos. μ =
R
P'
CCS. V =
N
p'
in which 2, μ and
MX, MY, M Z.
v denote the angles formed by P with the axes
We have also cos. λ= cos. a cos. ß, cos. µ = sin. a cos. ß, and
v = 90°
ẞ, or cos. v =
FIG. 96.
B
I.
A
P
II.
G
N
NY
R₂
R
2
A
S
2
sin. B.
EXAMPLE.-In order to raise vertically a
weight G, Fig. 96, I and II, by means of a
rope passing over a fixed pully, three work-
men pull at the end of the rope A with the
forces Pi 50 pounds, P₂ = 100 pounds and
40 pounds; the directions of these
forces are inclined at an angle of 60° to the
horizon, and form the horizontal angles
S₁ A S₂ S₂ A S3

P 3
1
2
1
2
90° with each other.
2
135° and S, A S₁
3
What is the inten-
sity and direction of the resultant which we
can put equal to the weight G, and how
great could this weight be made, if the forces
had the same direction? ·
The vertical components of the forces are
N₁=P₁ sin. ß₁=50 sin. 60°=43,30 pounds,
N₂=P₂ sin. ẞ2=100 sin. 60°=86,60 pounds
1
2
and N3 Ps sin. ẞ3 40 sin. 60° = 34,64
3
pounds; consequently, the vertical force is
N = N₁ + N₂ + N3 164,54 pounds.
1
2
The horizontal components are
81 = P. cos. B
50 cos. 60° = 25 pounds,
S₂ = P₂ cos. ẞ ₂ = 100 cos. 60° - 50 pounds
and S₂=P3 cos. ß3 = 40 cos. 60°—= 20 pounds.
If we pass an axis XX in the direction
of the force S₁, we have for the component
forces in this direction

Q
=
Q1 + Q2 + Q3 = S₁ cos. α₁ + S₂ cos. a₂+
S3 cos. az
20 cos. 270°
25-35,355
1
1
25 cos. 0° + 50 cos. 135° +
25.1-50. 0,7071-20.0=
10,355 pounds, and for the
component in the direction Y Y
$82.]
MECHANICS OF A MATERIAL POINT.
185
R
=
1
R₁ + R₂+ R 3
2
1
1
2
α
2
3
25 sin. 0° +
S₁ sin. a₁ + S₂ sin. a₂ + S₂ sin. a3 =
50 sin. 135° + 20 sin. 270° = 50 . 0,7071
50. 0,7071-2015,355 pounds, and for the
horizontal resultant
S = √ Q² + R² = √10,355 + 15,355² = 18,520 pounds.
The angle a, formed by this resultant with the axis XX, is determined
by the formula
tang. a=
R
Q
15,355
10,355
1,4828, whence a = 180°
180°
56° :
56° = 124°.
The final resultant is
PVN²+ S²
√ 164,54² + 18,5202 165,58 pounds.
=
The angle of inclination of this force to the horizon is determined by
the formula
N 164,54
tang. ẞ =
S
18,520
8,8848, whence we have ẞ = 83° 35′.
If all the forces acted in the same direction, the resultant would be
50 + 100 + 40 = 190 pounds, or 190 — 165,58
165,58 = 24,42 pounds greater
than the one just found.
§ 82. Principle of Virtual Velocities. From the fore-
going rules for the composition of forces, two others can be
deduced, which are of great importance in their practical appli-
cations. Let M, Fig. 97, be a ma-
P
P
FIG. 97.
Y
L
P₂
R
R₂
R
1
2
1
MR
terial point, MR = P₁ and MP
=P, the forces acting upon it,
and M P = P the resultant of the
forces P, and P.
If we pass
through two axes MX and
M Yat right angles to each other,
and decompose the forces P, and
P2, as well as their resultant P,
into their components in the di-
rection of these axes, I.E., P, into Q₁,
and R1, P in Q, and R. and Pinto
Qand R, we obtain the forces in the
direction of one axis Q1, Q, and Q, and those in the direction of the
other R, R₁ and R₂, and we have Q Q1 + Q₂ and R = R₁ + R₂.
If from any point O in the axis MX we let fall the perpendiculars
O L1, O L. and O L upon the directions of the forces P1, P, and
P, we obtain the right-angled triangles M O L, M O L, and MO L,
which are similar to the triangles formed by the three forces, viz.,

X-
Q₂
M
▲ M O L, ∞ ≤
Δ ΜΟΙΦ Δ
▲ MOLS ▲
=
M P₁ Q1,
Μ Ρ
M P₂ Q29
M P Q.
186
[$ 83.
GENERAL PRINCIPLES OF MECHANICS.
In consequence of this similarity we have
Q2
P.2
M L2 Q
and
MO P
ML
1
M Q₁
I.E.,
MP, P₁
Q₁
ML₁
Ꮇ Ꭴ
ΜΟ ; substituting these values of Q1, Q, and
ذن
Q in the formula Q = Q1 + Q2, we obtain
P.ML = P₁. M L + P₂. M L.
In like manner we have
R₁
P
O L₁ Ra
MO' P
1
O L₂
R O L
and
MO P
Μ Ο'
whence
P. OL = P₁. O L₁ + P₂. O L
The formulas hold good, when P is the resultant of three or
more forces P₁, P2, P3, etc., since we have, in general,
Q
Q1 + Q₂ + Q3 +
R = R₁ + R2 + R3 +
We can, therefore, put, in general,
1) P.ML = P₁ . M L₁ + P₂ . M L₂+ P₂. M L3 + . . .,
1
2) P.OL= P₁. O L₁ + P₂. O L₂ + P3. O Ls + . . .
L3
The resultant P of the forces P1, P2, P., etc., must correspond
to both these equations, and they can therefore be employed to de-
termine P.
The first of these two formulas can also be employed for a sys-
tem of forces in space, N, Q, R, Fig. 95, since here we have also
N N₁ + N₂ + ѳ +
2
1
1
or
3
P cos. v = P₁ cos. v₁ + P₂ coș. v½ + P3 cos. v; + . . ., and also
P. MO cos. v=P₁. M O cos. v₁ + P₂ M O cos. v₂+ P3 M O cos. v3 +..
1
§ 83. If the point of application M, Fig. 98 and Fig. 99, moves
to O, or if we imagine the point of application moved forward
FIG. 98.
L
M
FIG. 99.


PR
M
L
through the space M Ox, we call the projection ML s of
this space x upon the direction of the force M P the space described
by the force P, and the product Ps of the force by the space is the
$84.]
187
MECHANICS OF A MATERIAL POINT.
work done by the force. If we substitute these quantities in the
equation (1) of the last paragraph we obtain
Ps = P₁ s₁ + P₂ §2 + P 3 $3 +
2
hence the work done by the resultant is equal to the sum of the work
done by the component forces.
In adding the mechanical effects we must, as in adding the
forces, pay attention to the signs of the same. If one of the forces
Q1, Q, of the foregoing paragraph, acts in the opposite direction to
the others, then it must be introduced as negative quantity; this
force, as for example, Q3 in Fig. 94, § 80, is, however, a component
of a force P which, under the circumstances supposed in the fore-
going paragraph, opposes the motion M L of its point of applica-
tion; we are, therefore, obliged to treat the force P, Fig. 99, which
acts in opposition to the motion M L, as negative, if we consider
the force P, Fig. 98, which acts in the direction of the motion M L,
to be positive.
If the forces are variable, either in magnitude or in direction,
then the formula
Ps P₁ s₁ + P₂ 82 + P3 83 + ...
=
1
2
S3
is correct only for an infinitely small space s, 81, 82, etc.
We call the infinitely small spacesσ1, 2, 3, etc., described by the
forces corresponding to the infinitely small space described by the
material point, the virtual velocities (Fr. vitesses virtuelles, Ger.
virtuelle Geschwindigkeiten) of the same, and the law correspond-
ing to the formula Po P, o₁ + P₂ σ + P₂ o, is known as the
principle of virtual velocities.
=
1
3
§ 84. Transmission of Mechanical Effect.-According to
the principle of vis viva for a rectilinear motion the work (P 8)
done by a force (P), when the velocity c of a mass M is changed
into a velocity v, is
༡
Ps = (1² 2 ) M
Now if P is the resultant of the forces P1, P, etc., which act
on the mass M, and if the spaces described by them are s₁, s, etc.,
while the mass M describes the space s, we have, from the forego-
ing paragraph,
Ps= P₁ §₁ + P₂ §; +
1
from which we deduce the following general formula,
P₁ 8₁ + P½ 8½ + . . . =
§1
2
= (³¹² 5 °² ) M ;
!
188
[$ 84.
GENERAL PRINCIPLES OF MECHANICS.
therefore the sum of the work done by the single forces is equal to
half the increase of the vis viva of the mass.
If the velocity during the motion is constant, I.E., if v = c and
the motion itself is uniform, we have v" - c²= 0, and therefore
there is neither gain nor loss of vis viva, whence
P₁ 8, + P₂ S₂ + P3 83 + ... = 0;
1
2
S3
and the sum of the mechanical effects of the single forces is null.
If, on the contrary, the sum of the mechanical effects is null,
then the forces do not change the motion of the body in the given
direction; if the body has no motion in the given direction, it will
not have any imparted to it in this direction by the action of the
forces; if it had before a certain velocity in a given direction, it will
retain the same.
If the forces are variable, the variable velocity v can, after a cer-
tain time, become the initial. This phenomena occurs in all peri-
odic motions, which are very common in machinery. But v = c
M = 0, and therefore the gain or
gives the work done. (0² = 0)
2
loss of mechanical effect during a period of the motion is = 0.
Example.—A wagon, Fig. 100, weighing G 5000 pounds is moved
forward on a horizontal road by a force P₁ = 660 pounds, inclined at an an-
P
A
by the force P₁
=
FIG. 100,
I
B
gle a 24° to the horizon,
and is obliged to overcome
a horizontal resistance P₂
= 450 produced by the fric-
tion, what work must the
force P₁ do, in order to
change the initial velocity
of 2 feet of the wagon into
a velocity of 5 feet?
If we put the space de-
scribed by the wagon M O
=s, we have the work done

P₁. M L = P₁ s cos. a = 660 . 8 cos. 24° =
1
1
602,94 . 8,
and the work done by the force P₂ acting as a resistance is
= (— P₂). 8
450. 8,
2
consequently the work done by the motive force is
P 3 = P₁s cos. a
M
P2 8 cos. 0 =
(602,94 — 450) 8 =
1
450) 8 = 152,94 8 foot pounds.
The mass, however, absorbed during the change of velocity the me-
chanical effect
§ 85.]
189
MECHANICS OF A MATERIAL POINT.
(0² = 0")
G
2 g
52-22
5000=0,0155. (25-4). 5000-1627,5 foot-pounds:
2 g
putting the two effects equal to each other we obtain 152,94 . 8 = 1627,5,
whence the space described by the wagon is
MO= 8
1627,5
152,94
10,64 feet,
and finally the mechanical effect of the force P₁ is
1
P₁ s cos. a = 602,94. 10,64
10,64 = 6415 foot-pounds.
1
§ 85. Curvilinear Motion.-If we suppose the spaces (σ, 1,
etc.,) infinitely small, we can apply the foregoing formulas to cur-
vilinear motion. Let MO S, Fig. 101, be the trajectory of the
M O
K
FIG. 101.
P
N
N
R
S
Pn
Ka
material point, and M P = P
the resultant of all the forces act-
ing upon it. If we decompose
this force into two others, the
one of which M K = K is tan-
gent and the other MN = N
normal to the curve, we call the
former the tangential and the
latter the normal force.
While the material point de-

scribes the element MOσ of its curved path M O S, and its
velocity changes from c to v₁, the mass M absorbs the mechanical
effect (*) M, during the same time the tangential force K
2
=
performs the work Ko, and the normal force the work N. 0 0,
and consequently we have
120°)
Κ σ
Ko= (0² = 0) M.
(^
?
If, while the point describes the space M O S = s = n o, the
tangential velocity changes from c to v, and at the same time
the tangential force assumes successively the values K₁, K.,.. K„
then
(K₁ + K₂ + . . + K₂) σ
and the work done is
=
K₁ + K₂ + . . + K„
N
A = K s = ('¹² ≈ c²) M, when K =
2
K₁ + k + . . + K₂
ጎ
denotes the mean value of the variable tangential force.
M.
If we put the projection of the elementary space M 0 = σ upon
190
ES 85.
GENERAL PRINCIPLES OF MECHANICS.
the direction ML of the force, we have also P = Ko; if,
therefore, while the point describes the space MOS s no
the resultant P assumes successively the values P1, P2... Pn, the
projections of the elementary spaces are successively 51, 52... En
and we have also
Pn §n
Pn Šn
= =
P₁₁ + P₂ §2 + .. + P₂ n = (K₁ + K₂ + .. + K₁) σ,
and therefore
2
A = P、 §; + P₂ §; + . . + P„ 5,₁ = (³¹² — c² )
C
2) M.
Šn
2
When the direction of the force P remains constant, the pro-
jections,.. of the portions o. . . of the space or that of
S σ
the whole space s = n o form a straight line
MR
M R = a = % + t
है,
• •
If we put x = m §, we can also write
と
​་
*
A
(P₁ + P₂ + .. + Pm) § = (P₁ + P₂ + ... + P„)
P₁ + P₂ +
•
+ Pm
m
OC
m
P2,
of the forces, which
of the projections of the
where P denotes the mean
correspond to the equal portions =
path on the direction of the force.
We have, therefore, also
Px =
M
M = (h−k) G,
"in which k denotes the height due to the initial velocity c and h
that due to the final velocity v, and G the weight M g of the
moving body.
Therefore, in curvilinear motion, the entire work done is equal
to the product of the weight of the body moved and the difference
of the heights due to the velocities.
REMARK.-The formulas, thus obtained by the combination of the prin-
ciple of vis viva with that of virtual velocities, are particularly appli-
cable to the cases of bodies, which are compelled to describe a given path,
either because there is a support placed under them, or because they are
suspended by a string, etc. If such a body is impelled by gravity alone,
then the work performed by its weight & in descending a distance, whose
vertical projection is 8, is G 8, whence
G8
(hk) G, L.E. 8h-k.
$ 85.]
191
MECHANICS OF A MATERIAL POINT.
Whatever may be the path on which a body descends from one hori-
zontal plane A B, Fig. 102, to another horizontal one CD, the difference
A E3
FIG. 102.
E,
E
El
EA

B
G
G
-G-
F3
F2
F
D
1)
of the heights due to the velocities is always equal to the vertical
height of descent. Bodies, which begin to describe the paths E F, E₁ F
E2 F2, etc., with equal velocities (c), arrive at the end of these paths with
the same velocity, although they require different times to acquire it.
If, for example, the initial velocity is c = 10 feet, and the vertical height
of fall = $=
20 feet, or h = 8 + k = 20 + 0,0155 . 10² = 21,55 feet, we have
for the final velocity
.
v = √2 g h = 8,025 √21,55
37,24 feet,
whatever may be the straight or curved line in which the descent takes
place.
THIRD SECTION.
STATICS OF RIGID BODIES.
CHAPTER I.
GENERAL PRINCIPLES OF THE STATICS OF RIGID BODIES.
§ 86. Transference of the Point of Application.-Al-
though the form of every rigid body is changed by the forces which
act upon it, that is, it is compressed, extended, bent, etc., yet in
many cases we can consider the body as perfectly rigid, not only
because this change of form or displacement of its parts is often
very small, but also because it takes place during a very short
space of time. For the sake of simplicity we will therefore con-
sider, when nothing to the contrary is stated, a rigid body to be a
system of points rigidly united to each other.
A force P, Fig. 103, which acts upon a rigid body at a point A,
FIG. 103.
P
P
A
AT
Χ
M
FIG. 104.
A 2
A 3
X
P
P
M
transmits itself unchanged
in its own direction XX
-x through the whole body,

-X
and an equal opposite force
P, will balance it, when its
1
point of application A, lies
in the direction FX
The distance of these
points of application A

and A, from each other has no influence upon the state of equi-
librium; the two opposite forces balance each other, whatever the
distance may be, if the points are rigidly connected.
We can
€ 87, 88.]
193
STATICS OF RIGID BODIES.
therefore assert, that the action of a force P, (Fig. 104) remains
the same, no matter in what point A, A, A3, etc., of its direction
it may be applied or act upon the body M.
§ 87. If two forces P, and P2, Fig. 105, acting in the same
plane are applied at different points A, and A. to a body, their
Qi
FIG. 105.
P
B
_P
AP
A 2
2
action upon the body is the same as if
the point Cat which the two directions
intersect were the common point of ap-
plication C of these forces; for, accord-
ing to the law just laid down, both
points of application can be transferred
to C without producing any change in
the action of the forces. If, therefore,
we make
C Q₁ = A, P₁ = P, and

1
C Q₂ = A, P₁ = P₁₂
and complete the parallelogram CQ
QQ, its diagonal will give us the result-
ant C Q = P of C Q, and C Q, and also of the forces P, and P.
The point of application of this resultant can be any other point A
in the direction of the diagonal.
-
If at a point B on the diagonal we apply a force B P P
equal and opposite to the resultant A P = P, the forces P₁, P, and
– P will balance each other.
§ 88. Statical Moment.-If from any point 0, Fig. 106, in
the plane of the forces we let fall the perpendiculars 0 L₁, O L
and L upon the directions of the component forces P, and P₂
and of the resultant P, we have, according to § 82,
P.OL P₁. O L₁ + P₂. O L
=
and, therefore, from the perpendiculars or distances O L, and 0 L,
of the components we can find that of the resultant by putting
OL=
P. O LP. O L
P
While the intensity and direction of the resultant is found by
means of the parallelogram of forces, the position L of the point
of application is obtained by means of the last formula.
13
194
[$ 88.
GENERAL PRINCIPLES OF MECHANICS.
If the directions of the forces, when sufficiently prolonged, form
a, the value of the resultant is
an angle P, CP,
1)
A₁
FIG. 106.
P
a
2
P = √P;² + P,¸² + 2 P₁ P₂ cos. a.
PA2
If the direction of the resultant
forms an angle PCP, a, with the
direction of the component P₁, We
have

2) sin. a,
P₂ sin. a
=
2
P
If, finally, the distances from any
point to the directions C P₁ and C' P₂
of the given forces are O L,
1
a
a, and
O L₂
a, then the distance OL
from this point to the direction C P
of the resultant is
L1
P₁ α₁ + P₂ α₂
2
3)
a =
P
By the aid of the last distance a we can determine the position
of the resultant without reference to any auxiliary point C by de-
scribing from O with the radius a a circle, and by drawing a tan-
gent L P to it, the direction of which is given by the angle a,.
1
EXAMPLE.—A body is acted upon by the forces P₁ ⇒ 20 pounds and
P₂ = 34 pounds, whose directions form an angle P₁ C P½ = a = 70° with
each other, and their distances from a certain point are 0 L₁ α 4
feet and O L₂ = A q
1 foot; what is the intensity, direction and posi-
tion of the resultant? The value of the resultant is
P = √ 20² + 34² + 2 .20 . 34 cos. 70°
√ 2021,15 = 44,96 feet;
1
400+ 1156 + 1360.0,34202
and its direction is determined by the angle a₁, whose sine is
sin. a₁ =
1
34. sin. 70°
44,96
hence log sin. a 1
=
0,85163 — 1,
and the angle formed by the direction of the resultant with that of the
force P₁ is a₁
α =
1
45° 17'. The position or line of application of the result-
ant is finally determined by its distance O L from 0, which is
a
20.4 +34.1
44,96
114
= = 2,536 feet.
44,96
80, 90]
195
STATICS OF RIGID BODIES.
FIG. 107.
P
L₁
= ay
§ 89. We call the normal distances () L₁ -a, and O L₂
of the directions of the forces from an arbitrary point 0, Fig. 107,
the arms of the lever, or simply the arms
(Fr. bras du levier, Ger. Hebelarme) of the
forces, because they form an important ele-
ment in the theory of the lever, which will
be discussed hereafter. The product P a of
the force and the arm of the lever is called
the statical moment of the force (Fr. moment
L
22
21
des forces, Ger. statisches or Kraftmoment). Since P a P, a,
+ P₂ a, the statical moment of the resultant is equal to the sum
of the statical moments of the two components.
2
2
In adding the moments, we must pay attention to the positive
and negative signs. If the forces P, and P, act in the same direc-
tion around 0, as in Fig. 107, if, E.G., the direction of the forces
coincide with the direction of motion of the bands of a watch, they
and their moments are said to have the same sign, and if one of
them is taken as positive, the other must also be considered as
positive. If, on the contrary, the two forces, as in Fig. 108, act in
FIG. 108.
FIG. 109.


Pr
L₂
a
L
A
A
มา
P₁
A2
A
PV
BY
opposite directions around the point O, they and their statical mo-
ments are said to be opposite to each other, and when one is
assumed to be positive, the other must be taken as negative.
9
In the combination represented in Fig. 109 we have P a =
P₁ a₁ - P. a, since P is opposite to the force P₁, or its moment
P₂ a, is negative, while in the combination in Fig. 106 P a =
P₁ a₁ + P₂ α,
§ 90. Composition of Forces in the Same Plane.--If
three forces P₁, P, P., Fig. 110, are applied to a body at three
different points A, A, A, in the same plane, we first combine two
(P₁, P.) of these forces so as to obtain a resultant CQ
Q, and
then combine the latter with the third force (P3) according to the
196
[$ 91.
GENERAL PRINCIPLES OF MECHANICS.
=
P3
same rule, constructing with DR, CQ and D RA, Ps the
parallelogram DR, RR. The diagonal D R is the required re-
sultant P of P1, P2, and P3. It is easy to see how we must pro-
ceed, when a fourth force P. is added.
4
Here the intensity and direction of the resultant is found in ex-
actly the same manner as when the forces are applied at the same
FIG. 110.
R
P2
L3
Q2
R
P
Q1
Q₁ A
C
A 3
R2
P3
L1
K
point (see § 80); the rules
given in § 80 can be employed
to calculate the first two ele-
ments of the resultant, but
the third element, viz., the
position of the resultant or
its line of application, must
be determined by means of
the formula for the statical
moments. If O L₁ = α19
O L₂ = α2, O L3 = αz and
O L a are the arms of
the three component forces
P1, P2, P3 and of their re-

sultant P in reference to an arbitrary point 0, and if Q is the re-
sultant of P₁ and P, and O K its arm, we have
1
1
Pa = Q.OK + P3 a3 and Q. OK P₁ a₁ + P₂ α2.
Combining these two equations, we obtain
Pa = P₁ a₁ + P₂ α² + P3 α3,
and in like manner when there are several forces
Pa = P₁ α₁ + P₂ α² + P3 α3 +
2
L.E., the (statical) moment of the resultant is always equal to the alge-
braical sum of the (statical) moments of the components.
2
2
3
§ 91. If P1, P2, P3, etc., Fig. 111, are the individual forces of a
system, a, a, a, etc., the angles P, D, X, P, D. X, P₂ D. X, etc.,
formed by the directions of these forces with any arbitrary axis
XX and a, a, a, etc., their arms 0 L₁, O L, O L, etc., in refer-
ence to the point of intersection O of the two axes
we have, according to §§ 80 and 90,
1) the component parallel to the axis XX
Q
= P₁ cos. a₁ + P₂ cos. a₂ +
2) the component parallel to the axis Y Y
R = P₁ sin, a₁ + P₂ sin. a₂ + ...
1
and Y Y,
§ 91.]
197
STATICS OF RIGID BODIES.
3) the resultant of the whole system
P = √ Q² + R²,
4) the angle a formed by the resultant with the axis for which
R
tang. a =
Q
5) and the arm of the resultant or the radius of the circle to
which the direction of the resultant is tangent
a =
P₁ a₁ + P₂ α +...
2
P₁ + P₂+
FIG. 111.
Y

P
A
L
X
D
D₂
R
P (P)
L1
Y
A 3
13
D3
P
If b, b₁, b₂, etc., denote the distances O D, O D1, O D2, etc., cut
off from the axis XI, we have
a = b sin. a, a₁ = b₁ sin. a1, a2 = b₂ sin. a,, etc.,
and therefore also
1
P₁ b₁ sin. a₁ + P₂ b₂ sin. a, +
R₁ b₁ + R₂ b₂ + ...
b =
P sin. a
R
If we replace the resultant (P) by an equal opposite force (- P),
the forces P1, P2, P3... (- P) will balance each other.
If x1, x2
• •
and y₁, y. . . . denote the co-ordinates of the points
of application A, A, ... of the given forces P1, P2..., the mo-
ments of the components of the latter are R, 21, Rx... and Q1 y,
and the moment of the resultant is
Q x Y 2
Pa =
and its arm is
(R₁ X1 + R2 X2 + . . .) − ( Q1 Y₁ + Q2 Y2 + . . .),
Qi
198
[§ 91.
GENERAL PRINCIPLES OF MECHANICS.
(R₁ x₁ + R₂ 2₁₂ + ...)
a =
(Q₁ Y₁ + Q: Y₂ + ...)
√ (R₁ + R₂ + . . .)" + ( Q1 + Q₂ + . . .)"
2
EXAMPLE.The forces P₁ = 40 pounds, P₂ = 30 pounds, P3
pounds, Fig. 112, form with the axis X X the angles a
3
2
1
=
60³, ɑ½
= 70
80°,
az 140°, and the distances between the points of intersection D₁, D2, D3
of the directions of the forces with the axis are D₁ D₂ 4 feet, and D2 D3
5 feet. Required the elements of the resultant. The sum of the com-
ponents parallel to the axis X X is
Q
=
40 cos. 60° + 30 cos. (— 80°) + 70 cos. 142°
= 40 cos. 60° + 30 cos. 80°
=
20+ 5,209 - 55,161 =
The sum of those parallel to the axis
R
=
70 cos. 38°
29,952 pounds.
Fis
40 sin. 60° + 30 sin. (— 80°) + 70 sin. 142°
40 sin. 60° — 30 sin. 80° + 70 sin. 38°
34,641 - 29,544 + 43,096 :
= 48,193.
FIG. 112.

Y
P
P
X
D₁
L₂ L
P₂
D,
2
D
Hence it follows that the resultant
=X
D3
P = √ Q² + R² = √/29,952² + 48,193²
1/29,952 + 48,1932 = 1/3219,68
1/3219,68 = 56,742 pounds.
The angle a formed by the latter with the axis is determined by the
formula
tang. a =
R
Q
48,193
29,952
- 1,6090, from which we obtain
a = 180°
58° 8' 121° 52'.
=
If we transfer the origin of the co-ordinates to D3, we have the
arın of the force
O
1
P₁ sin. a, b₁ + P₂ sin. ag b₂ +
a1 1
2
1
R₁ b₁ + R₂ b₂ +
2
+..
P
P
164,049
56,742
= 2,891 feet,
0 L = a =
34,641. (4 + 5) — 29,544. 5+ 0
56,742
§ 92.]
199
STATICS OF RIGID BODIES.
and, on the contrary, the distance cut off on the axis XX
O D = b =
164,049
48,193
=
3,404 feet.
§ 92. Parallel Forces.-If the forces P1, P2, P3, etc., Fig. 113,
of a rigid system of forces are parallel, their arms O L1, O L9,
O La, etc., coincide with each other; if through the origin O we
draw an arbitrary line XX, the directions of the forces will cut off
from it the portions O D₁, O D, O D, etc., which are proportional
to the arms O L1, O L2, O L, etc., for we have ▲ O D, L, ∞ A
O D₂ L ∞ O D3 L, etc. Designating the angle D, OL, D, OL,
etc., by a, the arms () L₁, O L, etc., by a, a, etc., and the distances
cut off Ò D₁, O D, etc., by b₁, b, etc., we have
α₁ = b₁ cos. ɑ, ɑ? b₂ cos. a, etc.
Finally, substituting these values in the formula
we obtain
Pa = P₁ a₁ + P₂ α₂+....
1
Pb cos. a= P₁ b₁ cos. a + P₂ b₂ cos. a +.
or, omitting the common factor cos. a, we have
X
Г.
LI
D₁ D₂
P b = P₁ b₁ + P₂ b₂ + ..
FIG. 113.
P₂
2
12 13
P3
D3
=
In every system of parallel
forces we can substitute for the
arms the distances O D₁, O D2
etc., cut off from any oblique line
by the directions of the forces.
Since the intensity and direction
of the resultant of a system of
forces with different points of
application is the same as that

of a system of forces applied in one point, the resultant of the sys-
tem of parallel forces has the same direction as the components,
and is equal to their algebraical sum; hence we have
1)
2)
3)
P = P₁ + P₂ + P3+... and
α =
b =
P₁ α₁ + P₂ α + · ··, or
P₁ + P₂ +
2
P,
•
P₁ b₁ + P₂ b + ..
P+ P+...
200
[§ 93.
GENERAL PRINCIPLES OF MECHANICS.
2
EXAMPLE.-The directions of the three forces P₁ = 12 pounds, P₂
32 pounds and P3 25 pounds cut a straight line in the points D₁, D₂
and D3, Fig. 113, whose distances from each other are D, D₂
21 inches,
and D₂ D3
30 inches; required the resultant. The intensity of this
force is
1
P = 12 — 32 + 25 5 pounds,
1
and the distance D₁ D of its point of application D in the axis X X from
the point D₁ is
12.0 32.21 + 25. (21 + 30)
5
0 - 672 + 1275
=
5
120,6 inches.
§ 93. Couples.-The resultant of two equal and opposite
forces P, and — P₁ is
and its arm is
1
P = P₁ + (− P₁) = P₁ - P₁ = 0,
=
1
P₁ a₁ + P₂ a
а
=∞(infinitely great).
0
FIG. 115.
FIG. 114.
-P

P₁
Pa

I18
La
Ma
D₁
DB
P
M₁
-P₂
Li
No finite force acting at a finite distance can balance a couple,
but two such couples can balance each other. Let P, and P₁
and -P, and P, Fig. 115, be two such couples, and O L, a₁, O M₁
= 0 L, L, M₁ = a₁ b₁, O L = a, and O M, O L, L, M
= α 2
A
M
2
2
1
1
=
b, their arms measured from a certain point O, then, when
equilibrium exists, we have
P₁ a₁ - P₁ (a,
- b₁) - (az
P₂ α₂ + P₂ (α₂ — b₂)
1
P₁ (α₁ — b₁)
1
P₁ b₁ = P₂ b
2
= 0, I.E.
Two such couples balance each other when the product of one
force by its distance from the opposite one is the same for both
couples.
A pair of equal opposite forces is called simply a couple (Fr.
couple, Ger. Kräftepaar), and the product of one of its forces by
their normal distance apart is called the moment of the couple.
§ 93.]
201
STATICS OF RIGID BODIES.
From the foregoing we see that two couples acting in opposite
directions balance each other, when their moments are equal.
That this rule is correct can be proved in the following manner.
If we transfer the points of application of the forces P1, P, and
- P₁, P, of the couples (P₁, - P₁) and (P2, - P2), Fig. 116,
to the points of intersection A and B of their lines of application,
we can combine P, and P₂
as well as P₁ and
B
-R
-P₂
FIG. 116.
P₂
R
D
A
L₂
1
P₂

by means of the parallelo-
gram of forces and obtain
the resultants. If the di-
rections of these resultants
lie in the prolongation of
the line A B, then these
forces, and consequently the
corresponding couples (P1
P₁), and (P2, - P.), bal-
-
ance each other. If equilibrium exists, the triangle A B C formed
by A B and by the directions of the forces P and P, must be
similar to the triangles R A P, and B R P₁, and consequently we
have the proportion
C B
CA
P₁
P₂
=
or the equation P,. CA P. C B.
But the perpendiculars A L₁ = b, and B L₂ = b. to the di-
rections of the couples are proportional to the hypothenuses CA
and B of the similar triangles AC L, and B C L, and we can
therefore put
P₁ b₁ = P. b..
1
The moments of two couples which balance each other are con-
sequently equal to each other.
If in the formula (§ 91) for the arm a of the resultant
a =
P₁ α₁ + Pɔ ɑ₂ +
P
we substitute P = 0, while the sum of the statical moments has a
finite value, we obtain a∞, a proof that in this case there can
be no other resultant than a couple.
If the forces of a system shall balance each other, it is necessary
not only that the resultant PQ+R of the components Q
and R, but also that its moment
P α = P₁ α₁ + P, α, + . . . shall be = 0.
202
[$ 94.
GENERAL PRINCIPLES OF MECHANICS.
EXAMPLE.—If one couple consists of the forces P₁
་་
1
P
P₂ 2
25 pounds and
25 pounds and the other of the forces P₂ 18 pounds and
18 pounds, and if the normal distance between the first couple
is b 3 feet, then to produce equilibrium it is necessary that the normal
distance or arm of the second couple shall be
b₂
25.3
18
4 feet.
§ 94. Composition and Decomposition of Couples.-The
composition and decomposition of couples acting in the same plane
is accomplished by a mere algebraical addition, and is therefore
much simpler than the composition and decomposition of single
forces. Since two opposite couples balance each other, when their
moments are equal, the action of two couples is the same and the
couples are said to be equivalent, when the moment of one couple
P.
A
FIG. 117.
P₂
D
-
is equal to that of the other. If, therefore,
the two couples (P₁, P₁) and (P, P),
Fig. 117, are to be combined, we can replace
the one (P, P.) by another which has
the same arm A Bb, as the former couple.
(P 19
P₁), and can then add the forces thus
obtained to the others, and thus obtain a
single couple. If b, is the arm CD of the
one couple and (Q,
we have Q bi
P₂ b₂

P₁
1
Q
resulting couple is
2
b₁
2
- Q) the reduced couple,
P₂ by, and consequently
2
hence one component of the
Pab
P₁ + Q = P₁ + b₁
and the required moment of the resulting couple is
(P₁ + Q) b₁ = P₁ b₁ + P₂ bạ
1
In same manner the resultant of three couples may be found.
If P₁ b₁, P. b, and P3 bs be the moments of these couples, we
can put
P, b.
3
Qb, and P, b₁ = R b₁, or
Paba
3
P₂ b₂
Q
and R =
V₁
from which we obtain the resultant
(P₁ + Q + R) b₁ = P₁ b₁ + P₂ b₂ + P3 b₂.
2
V3.
In combining these couples to obtain a single resultant we
must pay attention to the signs, since the moments of the couples
§ 95.]
203
STATICS OF RIGID BODIES.
tending to turn the body in one direction are positive, and the mo-
ments of those tending to turn it in the other are negative. We can
now adopt the following principle for indicating the direction of
rotation of a couple. Let us assume arbitrarily a centre of rotation
between the lines of application of the forces of a couple; then if
the couple tends to turn in the direction of the hands of a watch,
the couple is to be considered as positive, and if in the other
direction, as negative.
The foregoing rule for the composition of couples is also appli-
M
FIG. 118.
P₁+ P₂
N
M
P(P+P₂)
N
cable, when the forces act in parallel
planes. If the parallel couples (P1,

2
P) and (P, P), Fig. 118, in
the parallel planes M M and N N
have equal moments P, b, and
P₂ by and act in opposite directions
to each other, they will also balance
each other; for they give rise to two
resultants P, + P, and (P₁ +
P), which balance each other, as
they are applied in the same point
E, which is determined by the equa-
tions
EA.REC. P, E B. P₁ = ED. P, and
P₁b, P₂ b, I.E. A B. P₁ = CD. P, whence
=
2
1
EA: EB: A B = E C: ED: CD;
hence this point coincides with the point of intersection of the two
transverse lines A Cand B D.
Since the couple (P2, - P₂) balances every other couple acting
in a parallel plane with an equal and opposite moment, it follows
that every couple can be replaced by another which has the same
moment, and which acts in a plane parallel to that of the first.
If, therefore, several couples whose planes of action are parallel
are applied to a body, they can be replaced by a single couple whose
moment is the algebraical sum of their moments, and whose plane,
which in other respects is arbitrary, is parallel to the planes of the
given system.
§ 95. If two couples (P₁, - P₁) and (P, — P₂) act in two differ-
ent planes E ME, and FNF, Fig. 119, whose line of intersection is
204
[§ 95.
GENERAL PRINCIPLES OF MECHANICS.
the straight line A B, and which form with each other a given
angle
P
Ꭱ
E A FE, B F₁
FIG. 119.
N
F
P₂
F
M
P₂
B
_R
P
E₁
1
а
we can, after having reduced them
to the same arm A B, combine
them by means of the parallelo-
gram of forces. We obtain thus
from P₁ and P, the resultant R,
and from - P, and - P₂ the result-
ant R. These two resultants
being equal and opposite, form
another couple, whose plane is
given by the direction of R and
R.

2
The resultant R can be found
according to § 77 by means of the
formulas
R = √ P²² + P¸² + 2 P、 P₂ cos. a and
2
sin. B
P₂ sin, a
R
Pa
in which ẞ denotes the angle E ARE, BR formed by the
direction of the resultant with that of the component P₁. If the
arm is A B = c, and if we put the moment P₁ c
moment P, c = Q b or P₁
Pa and the
Q b
and P₂
we obtain
с
C
R =
(Pa)
Q b
Pa Q b
+
+ 2
Cos. a,
C
C
C
P) and
or the moment of the resultant of the couples (P,
(Q,
Q)
Rc = √(Pa)² + (Q b)² + 2 P a. Qb. cos. a,
and in like manner for the angle formed by its plane with that of
the first couple (P, P) we have
Q b
sin. B
sin. a.
Rc
We can therefore combine and decompose couples acting in the
different planes in exactly the same manner as forces applied at the
same point, by substituting instead of the latter the moments of
the former, and instead of the angles, which the directions of the
former make with each other, those formed by their planes of action.
§ 96.]
205
STATICS OF RIGID BODIES.
The referring back of the theory of couples to the principle of
the decomposition of simple forces can be greatly simplified by in-
troducing the axis of rotation instead of the plane of rotation of
the couple. We understand by the axis of rotation or axis of a
couple, any perpendicular to its plane. Since every couple can be
arbitrarily displaced in its plane without changing its action upon
the body, we can pass the axis of the couple through any given
point.
FIG. 120.
Since the plane and the axis of a couple are at right angles to
each other, the axes AX, A Y
and A Z, Fig. 120, form the same
angles with each other as the
planes A EK, A FK and AG K
themselves. If one of the couples
is the resultant of the other two,
we see from what precedes, that
the diagonal of the parallelogram
constructed with the moments Pa
and Qb will give the moment
of the resultant; if therefore we

Qb
Pa
Re
X
G
F
E
lay off upon the axes A X and Y the moments Pa and Q b,
and then complete the parallelogram, we obtain in its diagonal not
only the axis AZ of the resulting couple, but also its moment Rc.
We see, therefore, that couples are combined and decomposed in ex-
actly the same way as simple forces, provided we substitute for the
directions of the forces the axes of the couples and the moments
of the latter for the forces themselves. All the rules for the com-
position and decomposition of forces given in § 76 and § 77, etc.,
are in this sense applicable to the composition and decomposition
of couples.
§ 96. Centre of Parallel Forces.-If the parallel forces lie
in different planes, their composition must be effected in the fol-
lowing manner. Prolonging the straight line A, A., Fig. 121,
which joins the points of application of two parallel forces P₁ and
P₁, until it meets the plane which contains the axes MA and M Y,
which are at right angles to each other, and taking the point of
intersection K as the origin, we have for the point of application
A of the resultant P₁ + P, of these forces
1
2
(P₁ + P₂) . K A = P₁ . K A₁ + P¸ . K A¸.
206
[$ 96.
GENERAL PRINCIPLES OF MECHANICS.
1
2
Now since B, B, and B, are the projections of the points of ap-
plication A, A, and A, upon the plane X Y, we have
:
A B: A, B₁: A, B₂ = K A K A₁: K A
and therefore also
29
1
(P₁ + P₂) . A B = P₁ . A₁ B₁ + P½. A½ B2.
FIG. 121.
Ꮓ
1
2
2
?
If we designate the normal distances A, B₁, A½ B2, A3 В„,,etc.,
of the points of application
from the plane X X by Z1, Z29
Z3, etc., and the normal dis-
tance of the point of applica-
tion A from this plane by z,
we have for two forces

D
A 2
M
A
1
B
B₁
Ba
K
X
P
Consequently we have in general
1
2
(P₁ + P₂) z = P₁ %1 + P₂ %2 ;
and for three forces, since (P₁
+ P₂) can be considered as
one force with the moment
P₁ Z₁ + P₂ Z2,
2
(P₁ + P₂ + P3) %
2
= P₁ %₁ + P₂ %2 + P3 Z3, etc.
2
(P₁ + P₂+ P3 + ...) z = P₁ z₁ + P₂ Zą + P323...,
1
2
and therefore
1)
2 =
2
P₁ z₁ + P₂ Z₂ + ...
2
P₁ + P₂ +
1
2
If, in like manner, we denote the distances A Cand A D of the
point of application A of the resultant from the planes X Z and
Y Z by y and x, and the distances of the points of application A1,
A,... from the same planes by Y1, Y2 and x1, x2 ..., we obtain
•
P₁ Y₁ +
P₂ Y₂ +
2)
Y
and
P₁ +
2
P₂ +
1
P₁ x₁ +
P₂ X₂ + ...
3)
x =
2
P₁ + P₂+...
1
2
The distances, x, y and z, from three fixed planes, as, E.G., from
the floor and two sides of a room, determine completely the point A ;
for it is the eighth corner of the parallelopipedon constructed with
x, y and z; hence there is but one point of application of the re-
sultant of such a system of forces.
Since the three formulas for x, y and z do not contain the angles
formed by the forces with the fixed planes, the point of application
is not dependent upon them or upon the direction of the forces;
§ 97.]
207
STATICS OF RIGID BODIES.
the whole system can therefore be turned about this point without
its ceasing to be the point of application, as long as the forces re-
main parallel.
In a system of parallel forces we call the product of a force by
the distance of its point of application from a plane or line the
moment of this force in reference to the plane or line, and it is also
customary to call the point of application of the resultant the cen-
tre of parallel forces (Fr. centre des forces parallèles, Ger. Mittel-
punkt des ganzen Systems). We obtain the distance of the centre
of a system of parallel forces from any plane or line (the latter, when
the forces are in the same plane) by dividing the sum of the stati-
cal moments by the sum of the forces themselves.

EXAMPLE. If the forces are P
and their distances or the co-
ordinates of their points
of application are
сл
5
7
10
4 pounds.
In
1
૫
2
0
9 feet.
Yn
p
2
4
5
3
Zn
8
3
7 10 (C
Pn X
5
14
036 foot pounds.
we will have the moments
Pn Yn
10
28
50 12
(C
Pr 2n 40
21 70 40
Now the sum of the forces is 19
19 7 12 pounds, and therefore
the distances of the centre of parallel forces from the three co-ordinate
planes are
536 14 27
9
X
2,25 feet,
12
12
4
10 + 50 + 12 28
44
11
У
12
12
3,66 feet, and
3
40+70 + 40 - 21
129
43
2
12
10,75 feet.
12
4
§ 97. Forces in Space.-If we wish to combine a system of
forces directed in different directions, we pass a plane through
them and transfer all their points of application to this plane, and
then decompose each force into two components, one perpendicular
to and the other in the plane. If B, B... are the angles formed
by the directions of the forces with the plane, the components nor-
mal to the plane are P, sin. B1, P2 sin. B.... and those in the plane.
are P₁ cos. ß₁, P. cos. ß, etc. The resultant of the latter can be ob-
tained as indicated in § 91, and that of the former as indicated in
1
208
[$ 97.
GENERAL PRINCIPLES OF MECHANICS.
the last paragraph. Generally the directions of the two resultants
do not cut each other at all, and the composition of the forces so
as to form a single resultant is not possible. If, however, the re-
sultant of the parallel forces passes through a point K, Fig. 122,
in the direction A B of the resultant P of the forces lying in the
plane (that of the paper), a composition is possible. Putting the
ordinates of the points of application K of the first resultant O C
DK = u and O D = C K = v, the arm of the other O L
and the angle BA O formed by the latter with the axis XX,= a,
then the condition for the possibility of the composition is
u sin. a + v cos. α = a.
a
If this equation is not satisfied, if, E.G., the resultant of the nor-
mal forces passes through K₁, it is not possible to refer the whole.
system of forces to a single resultant, but they can be replaced by
FIG. 122.
Y
FIG. 123.


B
L
K
D
K.
D.
1
X
N
P
a resultant R, Fig. 123, and a couple (P,
P) by decomposing
P and R,
the resultant N of the parallel forces into the forces
one of which is equal, parallel and opposite to the resultant P of
the forces in the plane.
We can accomplish directly this referring of a system of forces
to a single force and to a couple by imagining a system of couples,
whose positive components are exactly equal in amount and direc-
tion to the given forçes, to be applied to the body at any arbi-
trary point. These couples naturally do not change the state of
equilibrium, for being applied at the same point they counteract
themselves. On the contrary, the positive. components can be
combined according to known rules (§ 81) so as to give one result-
ant, while the negative components form with the given forces
couples, whose resultant (according to § 95) is a single couple.
After these operations have been performed, we have only one force
and one couple.
§ 98.]
209
STATICS OF RIGID BODIES.
3
§ 98. Principle of Virtual Velocities.-If a system of forces
P1, P2, P3, Fig. 124, which act in a plane, have a motion of trans-
lation, that is, if all the points of application A1, A2, A, describe
equal parallel spaces A, B, A, B, A, B, then (according to
the meaning of § 81) the work done by the resultant is equal to
FIG. 124.
3

L2
X
B1
LI
B3
PK
B
B
L
[13
Q3.
ใน
A
X
P3
sum is
= 0. If the pro-
A, B, etc., upon the di-
the sum of the work done by the components, and consequently,
when the forces balance each other, this
jections of the common space A, B₁
rections of the forces are A₁ L1, A2 L2, etc., 81, 82, etc., the work
done by the resultant is
P s = P₁ s₁ + P₂ 82 +
2
•
This law is a consequence of one of the formulas in § 91. Ac-
cording to it, the component Q of the resultant parallel to the axis
X X is equal to the sum
Q1 + Q₂ + Q3 +
of the components of the forces P1, P2, etc., which are parallel to
it. Now from the similarity of the triangles A, B, L, and A, P₁ Qı
we know that
1
Q₁
A₁ L₁
S1
P₁
1
A₁ B₁
A B'
and therefore we have
Q₁
=
P₁ s₁
A B'
1 1
Q₂
P₂ S 2
A B'
Ps
etc. and Q
A B
Hence, instead of
we can put
Q = Q₁ + Q₂ + ...
P s = P₁ s₁ + P₂ §2 +
მა
14
210
[§ 99, 100.
GENERAL PRINCIPLES OF MECHANICS.
§ 99. Equilibrium in a Rotary Motion.-If a system of
forces P1, P2, etc., Fig. 125, acting in the same plane, is caused to
C2
A2
L2
FIG. 125.
A,
B₁
·P₁
turn a very small distance
about a point O, the principle
of virtual velocities announced
in § 83 and § 98 is applicable
here also, as can be demon-
strated in the following man-
ner. According to § 89 the mo-
ment of the resultant P.OL
= Pa is equal to the sum
of the moments of the com-
ponents, or
Pa = P₁ α₁ + P₂ α₂ +

1
2
The space A, B₁, corresponding to a rotation through a small
Bº
angle A₁ 0 B₁ = ß° or a small are ß
=
·
180°
π, is situated at right
angles to the radius 0 A₁, therefore the triangle A, B, C, formed
by letting fall the perpendicular B, C, upon the direction of the
force, is similar to the triangle OA, L, formed by the arm O L₁ = a₁₂
and we have
O L₁
O A₁
1
A₁ C₁
A₁ B₁
1
If we put the virtual velocity A, C₁ = σ, and the arc A, B,
= 0 A₁. ẞ, we obtain
1
στ
α₁ =
Ο Αγ. στ
02
and in like manner a₂ = etc.
O A₁.ẞ β'
Substituting these values of a₁, a½, etc., in the above equation,
we obtain
Ρο Ρισι P₂ 02
2
+
B
B
B
+ . . ., etc.,
or since ẞ is a common divisor,
Po
P₁ σ₁ + P₂ σ½ +
1 1
as we found in § 83.
2 2
Therefore, for a small rotation, the work (P a) done by the re-
sultant is equal to the sum of the work done by the components.
§ 100. The principle of virtual velocities holds good for any
arbitrarily great rotation, when, instead of the virtual velocities
of the points of application, we substitute the projections
§ 101.]
211
STATICS OF RIGID BODIES.
L₁ C₁, L₂ C₂, Fig. 126, of the spaces described by the ends L, Ly
2
Lo
FIG. 126.

C₂ Br
1
D₁
P₁
P₂
etc., of the perpendiculars; for multiplying the well-known equa-
tion for the statical moment
Pa
2
P₁ α₁ + P₂ α2 +
by sin. ẞ and substituting in the new equation
P a sin. ß = P₁ a₁ sin. ẞ + P₂ a₂ sin. B,
instead of a sin. ß, a, sin. ß ... the spaces
we obtain
O B, sin. L, O B₁ =
D₁ B₁ = L₁ C₁ = 8₁,
O B₂ sin. L. O B₁ =
D₂ B₂ = La C₂ = 8, etc.,
B, 0 B. L, C,
P s = P₁ s₁ + P₂ S₂ +
2
This principle remains correct for finite rotations, when the di-
rections of the forces revolve with the system, or when the point
of application or end of the perpendicular changes continually so
that the arms ( L₁ = 0 B₁, etc., remain constant; for from
Pa = P₁ α₁ + P₂ ɑ2 +
by multiplying it by ẞ we obtain
P a ß =
1
P₁ a₁ ß + P½ α₂ ß + . . ., I.E.,
Ps = P₁ s₁ + P₂ 82 +
1
2
when s₁ s, etc., denote the arcs L₁ B₁, L, B, etc., described by
the points of application L₁, L, etc.
§ 101. A Small Displacement Referred to a Rotation.—
Every small motion or displacement of a body in a plane can be
considered as a small rotation about a movable centre as we will
now proceed to show. Let A and B, Fig. 127, two points of the
body (surface or line), be subjected to a small displacement, in con-
sequence of which they now occupy the positions 4, and B₁, A, B₁
being = A B. If we erect at these points perpendiculars to the
paths A A₁, and B B₁, they will cut each other at a point C, about
which we can imagine the spaces A A, and B B₁, considered as
arcs of circles, to be described. But since A B = A₁ Â₁, A C
1
1
"
212
[$ 102.
GENERAL PRINCIPLES OF MECHANICS.
4, C and B C B, C, the two triangles A B C and A, B, C
B₁
are similar; the angle B, C A, is therefore equal to the angle
BCA, and the angle of rotation A CA, equal to the angle
of rotation BC B₁. If we make A, D, A D we obtain, since the
angles D, A, C and D A C and the sides CA, and CA are equal
1
to each other, two equal, similar triangles CA, D, and CA D,
in which CD, CD and A, C D₁ = LA CD.
= ≤
quently,
E
FIG. 127.
B₁
1
Conse-
A CA, is also = LDC D₁, and when the displace-
placement of the line A B is small,
every other point D of it will de-
scribe an arc of a circle. Finally, if E
is a point lying without the line A B
but rigidly connected with it, the small
space E E, described by it can also
be regarded as a small are of a circle,
whose centre is at C; for if we make
the angle E, A, B, EA B and the
distance A, E, = AE, we obtain
again two equal and similar trian-
gles A, C E, and A CE, whose sides

1
1
CE, and CE and whose angles
A, CE, and A CE are equal to each other, and the same thing
can be proved for every other point rigidly connected with A B.
We can, therefore, consider any small motion of a surface or of a
solid body rigidly connected with AB as a small rotation about
a centre, which is determined by the point of intersection C of the
perpendiculars to the spaces A A, and B B, described by two
points of the body.
§ 102. Generality of the Principle of Virtual Velocities.
-According to a foregoing paragraph (99) the mechanical effect
of the resultant is equal to the mechanical effect of its components
for a small revolution of the system, and according to the last
paragraph (101) any small motion can be considered as a revolu-
tion; the principle of virtual velocities is therefore applicable to
any small motion of a body or of a system of forces.
If, therefore, a system of forces is in equilibrium, I.E., if the re-
sultant is null, then after a small arbitrary motion the sum of the
mechanical effects must be equal to 0. If, on the contrary, for a
small motion of the body the sum of all the mechanical effects is
equal to zero, it does not necessarily follow that the system is in
§ 103, 104.]
213
CENTRE OF GRAVITY.
equilibrium, for then this sum must be = 0 for all possible small
motions. Since the formula expressing the principle of virtual
velocities fulfils but one of the conditions of equilibrium, in order
that equilibrium shall exist it is necessary that this formula shall
be true for as many independent motions as there are conditions,
E.G., for a system of forces in a plane for three independent
motions.
CHAPTER II.
THE THEORY OF THE CENTRE OF GRAVITY.
§ 103. Centre of Gravity.-The weights of the different
parts of a heavy body form a system of parallel forces, whose re-
sultant is the weight of the whole body and whose centre can be
determined by the three formulas of paragraph 96. We call this
centre of the forces of gravity of a body or system of bodies the
centre of gravity (Fr. centre de gravité, Ger. Schwerpunkt), and
also the centre of the mass of the body or system of bodies. If a
body be caused to rotate about its centre of gravity, that point will
never cease to be the centre of gravity, for if we suppose the fixed
planes, to which the points of application of the single weights are
referred, to rotate with the body, during this rotation the position
of the directions of the forces in regard to these planes change, and
on the contrary the distances of the points of application from
these planes remain constant. Therefore the centre of gravity is
that point at which the weight of a body acts as a force vertically
downwards, and at which it must be supported in order to keep
the body at rest.
§ 104. Line and Plane of Gravity.—Every straight line,
which contains the centre of gravity, is called a line of gravity, and
every plane passing through the centre of gravity a plane of gravity.
The centre of gravity is determined by the intersection of two lines
of gravity, or by that of a line of gravity and a plane of gravity, or
by the point where three planes of gravity cut each other.
Since the point of application of a force can be transferred arbi-
trarily in the direction of the force without affecting the action of
the latter, a body is in equilibrium whenever any point of the ver-
tical line passing through the centre of gravity is held fast.
214
[§ 105.
GENERAL PRINCIPLES OF MECHANICS.
If a body M, Fig. 128, be suspended at the end of a string C A,
we obtain in the prolongation A B of this string a line of gravity, and
M
B
FIG. 128,
D
E
M
B
if it be suspended in another
way we find a second line of
gravity DE. The point of inter-
section S of the two lines is the
centre of gravity of the whole
body. If we suspend a body
by means of an axis, or if we
balance it upon a sharp edge
(knife edge), the vertical plane
passing through the axis or
knife edge is a plane of gravity.
Empirical determinations
of the centre of gravity, such


as we have just given, are seldom applicable; we generally employ
some of the geometrical methods, given in the following pages, to
determine with accuracy the centre of gravity. In many bodies,
such as rings, etc., the centre of gravity is without the body. If
such a body is to be suspended by its centre of gravity, it is neces-
sary to fasten to it a second body in such a manner that the cen-
tres of gravity of the two bodies shall coincide.
§ 105. Determination of the Centre of Gravity.-Let x₁,
X, X3, etc., be the distances of the parts of a heavy body from one
co-ordinate plane, y₁, Ye, Y, etc., those from the second, and 1, Z2,
zą, etc., those from the third, and let P1, P2, P3, etc., be the weights
of these parts, we have, from § 96, for the distances of the centre
of gravity of the body from the three planes
x =
y
=
2 =
1
P₁ x₁ + P₂ X₂ + P3 X3 + ...
P₁ + P₂ + P3 +
2
P₁ Y₁ + P₂ Y½ + P3Y3 + ...
2
P₁+ P₂+ P3+...
P₁ z₁ + P₂ Zą + P3 %3 +...
P₁ + P₂ + P3
1
and
If we denote the volume of these parts of the body by V1, V29
V₁, etc., and the weight of their units of volume by Y1, Y, Y, etc.,
we can write
32
x =
2
V₁ Y₁ X₁ + V₂ Y 2 X 2 + V3 Y 3 X 3 + ...
V₁Y₁ + V₂Y₂ + V₂ Y3 + ...
etc.
§ 106.]
215
CENTRE OF GRAVITY.
If the body is homogeneous, I.E., if y is the same for all the
parts, we have
X
1
( V₁ X₁ + V₂ Xq + . . .) Y
2
(V₁ + V₂ + ...) Y
2
or, cancelling the common factor y,
1
V₁ x₁ + V₂ X₂+...
1) x =
V₁ + V₂ +
2) y =
V₁ Y ₁ + V₂ Y₂ +
1 1
V₁ + V₂ +
V₁ z₁ + V₂ Z2 + ..
3) z =
V₁ + V₂ + ...
and
Consequently we can substitute for the weights of the different
parts their volumes, and the determination of the centre of gravity
becomes a question of pure geometry.
When one or two of the dimensions of a body are very small
compared with the others, E.G., in the case of sheet-iron, wire, etc.,
we can regard them as planes or lines, and determine their centres
of gravity by means of the last three formulas, substituting instead
of the volumes V₁, V2, etc., the surfaces F1, F2, etc., or the lengths
71, 72, etc.
§ 106. In regular spaces the centre of gravity coincides with
their centre, E.G., in the case of the cube, sphere, equilateral trian-
gle, circle, etc. Symmetrical spaces have their centre of gravity in
the axis or plane of symmetry. A body A D F H, Fig. 129, is di-
D
FIG. 129.
H
E
B
F
vided by the plane of symmetry A B CD
into two halves, which differ only in their
position in regard to the plane, and the
conditions are therefore the same on both
sides of the plane; the moments are con-
sequently the same on both sides, and
the centre of gravity is to be found in
this plane.
Since the axis of symmetry E F di-
vides the plane surface A B F C D, Fig.
130, into two parts, one of which is the
reflected image of the other, the conditions are the same on each

216
[§ 107.
GENERAL PRINCIPLES OF MECHANICS.
side; consequently the moments on both sides are the same, and
the centre of gravity of the whole surface lies in this line.
Finally, the axis of symmetry K L of a body A B G H, Fig.
131, is also a line of gravity of it; for it is formed by the intersec-
FIG. 130.
D E A
FIG. 131.
F


F
B
D
F
S*
C
H
tion of two planes of symmetry A B C D and E F G H
For this reason the centre of gravity of a cylinder, of a cone and
of a solid of rotation, formed by the revolution of a surface, or by
being turned upon a lathe, is to be found in the axis of the body.
§ 107. Centre of Gravity of Lines.-The centre of grav-
ity of a straight line is at its centre.
The centre of gravity of the arc of a circle A M B = b, Fig. 132,
is to be found in the radius drawn to the middle M of the arc; for
this radius is an axis of symmetry of the arc. In order to deter-
mine the distance C Sy of the centre of gravity S from the cen-
X
A
FIG. 132.
M
B
C
N
X
tre of the circle, we divide
the arc into a very great
number of parts and deter-
mine their statical moment
in reference to an axis XX,
which passes through the
centre C and is parallel to
the chord A B= s. If P Q
is a part of the arc and P N

its distance from XX, its statical moment is PQ. P N.
Drawing the radius P C M Cr and the projection QR of P Q
parallel to A B, we obtain two similar triangles P Q R and CPN,
for which we have
§ 108.]
217
CENTRE OF GRAVITY.
PQ: Q R = CP: PN,
whence we obtain for the statical moment of an element of the arc
P Q. P N = QR. CPQ R. r.
But in the statical moments of all the other elements of the arc
r is a common factor, and the sum of all the projections Q R of the
elements of the arc is equal to the chord, which is the projection of
the entire arc; consequently the moment the arc is the chord s
multiplied by the radius r. Putting this moment equal to the arc
b multiplied by the distance y, or b y = s r, we obtain
У S
Sr
or y
go
The distance of the centre of gravity from the centre is to the ra-
dius as the chord is to the arc.
If the angle subtended by the arc b is = ß° and the arc cor-
responding to the radius 1 = ß =
B°
π, we have b = ßr and
180°
B
s = 2r sin.
2'
and consequently
2 sin. § ß.r
Y
в
B
For a semicircle ß = π and sin.
1, whence
2
2
77
Y
r = 0,6366 ... r, approximatively =
r.
π
11
§ 108. In order to find the centre of gravity of a polygon or
FIG. 133.
Y
D
L2
K
༦༠
S₂
B
2
H
A
X
H₁ K₁S₁
H
2
combination of lines A B C D, Fig.
133, we first obtain the distances of
the centres H, K, L of the lines
AB=1, B C = 1, CD=13, etc.,
from the two axes and 0 Y,
viz., HH, y₁, HH₂ = 2₁, KK₁ =
Y₂, K K₂ = x², etc. The distances
of the centre of gravity from these

axes are
OS₁ = SS₂ = x=
O S₁ = S S₁ = y =
hiển + lọt
1₁ + 12 + ...
l₁ Y₁ + l? Y ¿ + ..
b₁ + b + ...
218
[S 109.
GENERAL PRINCIPLES OF MECHANICS.
E.G., the distance of the centre of gravity S of a wire A B C, Fig.
134, bent in the shape of a triangle from the base A B is
H
ah + 1 b h
NS=
=Y
y =
a + b + c
FIG. 134.
D K
IS
F
A
G
N M
E
B
h
a + b
a + b + c ' Q2'
when the sides opposite the angles
A, B, C are denoted by a, b, c
and the altitude CG by h.
If we join the middles H, K, M
of the sides of the triangle and
inscribe a circle in the triangle
thus obtained, its centre will co-
incide with the centre of gravity
S; for the distance of this point
from one of the sides HK is

h
a + b
h
S D = ND – N S
2
a + b + c
2
ch
2 (a + b + c)
▲ A B C
a + b + c
"
or constant, and therefore
= the dis-
tances SE and SF from the other sides.
§ 109. Centre of Gravity of Plane Figures.-The centre
of gravity of a parallelogram A B CD, Fig. 135, is situated at the
A
FIG. 135.
D
K
L
B
C
point of intersection S of its diagonals;
for all strips KL, formed by drawing
lines parallel to one of the diagonals
B D, are divided by the other diagonal
A Cinto two equal parts; each of the
diagonals is therefore a line of gravity.
In a triangle A B C, Fig. 136, every
line CD drawn from an angle to the
centre D of the opposite side A B is a line of gravity; for it bisects
every element K L of the triangle formed by drawing lines paral-
lel to A B. If from a second angle A we draw a second line of
gravity to the middle E of the opposite side B C, the point of in-
tersection S of the two lines of gravity gives the centre of gravity
of the whole triangle.

=
Since B D BA and B E B C, D E is parallel to AC
and equal to A C, the triangle D E S is similar to the triangle
CA S and CS = 2 S D. Adding S D, we obtain CS + S D,
§ 110.]
219
CENTRE OF GRAVITY.
I.E. CD = 3 SD and inversely SD = CD. The centre of
gravity S is at a distance equal to
base and at a distance equal to
3
CD from the middle D of the
CD from the angle C. If we
draw the perpendiculars CH and S N to the base, we have also
FIG. 136.
M
K
E
Մ
FIG. 137.


S
OB
D
A
II ND
B
X
-X
Α1
C₁ S₁D₁
B₁
NS
CH; the centre of gravity S is at a distance from the
base of the triangle equal to one third of the altitude.
The distance of the centre of gravity of a triangle A B C, Fig.
137, from an axis Xis S S₁ = D D₁ + 1 (C C, D D,), but
D D₁ = (A A, + B B₁), and consequently we have
Ꭰ
y = SS₁ = C C₁₂ + · (A A, + B B₁)
} ¦ ¦
•
A A₁ + B B₁ + C C₁
1
3
I.E., the arithmetical mean of the distances of the angles from XX.
Since the distance of the centre of gravity of three equal weights,
applied at the corners of a triangle, is determined in the same way,
the centre of gravity of a plane triangle coincides with the centre
of gravity of these three weights.
§ 110. The determination of the centre of gravity of a trape-
zoid A B C D, Fig. 138, can be made in the following manner.
The right line M N, which joins the centres of the two bases A B
and CD, is a line of gravity of the trapezoid; for if we draw a great
number of lines parallel to the bases, the figure will be divided into
a number of small strips whose centres or centres of gravity lie
upon the line M N. In order to determine completely the centre
of gravity S, we have only to find its distance SH from the
base A B.
Let the bases A B and C D be denoted by b, and b, and the al-
titude or normal distance between the latter by h. Now if we
draw D E parallel to the side B C, we obtain a parallelogram
220
[§ 110.
GENERAL PRINCIPLES OF MECHANICS.
BCDE, whose area is b₂ h and the distance of whose centre of
h
gravity S, from A B is
(b₁ — b₂) h
and a triangle A D E, whose area is
2'
and the distance of whose centre of gravity from A B
૭
is
h
اين
K
D
N
FIG. 138.

N
S
$1
L
F
A O
HME
B
The statical moment of the trapezoid in reference to A B
is therefore
h
Fy = b₂ h
(b₁
=
+
2
2
b₂) h h
3
h"
(b₁ + 2 b₂)
6'
h
but the area of the trapezoid is F = (b₁ + b₂) 2'
consequently the normal distance of the
centre of gravity from
the base is
HS = y
6
} (b₁ + 2 b.) h²
1
b₁ + 2 b ₂ h
У
½ (b₁ + b₂) h
b₁ + b₂ '3
b₁ + b₂
The distance of this point from the middle line K L
of the trapezoid is
h
US = 1/2 -
HS=
2
3 (b₁ + b₁) — 2 (b₁ + 2 b₂) h
b₁ + b y
1
b₁ - b₂ h
6'
I.E., Y1
b₁ + b₂ ° 6°
In order to find the centre of gravity by construction, we have
only to prolong the two bases, make the prolongation C G = b₁ and
the prolongation AF b, and join the extremities Fand G thus
obtained by a straight line; the point of intersection S with the line
M N is the required centre of gravity; for from HS=
2 b₁ + b₂ MN
2
1
b₁+2b₂ h
b₁ + b₁
3
t follows that
b₁ + 2 b, MN
2
MS
and NS
or
b₁ + b₂
3
MS
b₁ + 2 b
2
i b₁ + b ₂
MF
NS 2 b₁ + b₂
b₁ + ½ b₂
CGNC
NG
bj + b 3
MA+AF
§ 111, 112.]
221
CENTRE OF GRAVITY
which, in consequence of the similarity of the triangles M S Fand
NS G, is perfectly true.
If we denote by a the projection A O of the side A D upon
A B, the distance of the centre of gravity from the corner A is
determined by the formula
A H = x =
2
b₁² + b₁ b₂ + b²² + a (b₁ + 2 b.)
2
3 (b₁ + b₂)
§ 111. In order to find the centre of gravity of any other four-
FIG. 139.
D
sided figure A B C D, Fig. 139, we can
divide it by means of the diagonal A C
into two triangles, and then determine
their centres of gravity S, and S. by
means of the foregoing rules; thus we
obtain a line of gravity S₁ S2. If we
again divide the figure by the diagonal
BD into two other triangles, and de-
termine their centres of gravity, we
obtain a second line of gravity, whose
intersection with S S gives the centre of gravity of the whole
figure.

E
S
M
F
S₂
B
We can proceed more simply by bisecting the diagonal A Cat
M and laying off the longer portion B E of the other diagonal
upon the shorter portion, so as to have D F = BE. We then
draw FM and divide this line into three equal parts; the centre
of gravity is at the first point of division S from M as can be
proved in the following manner. We have M S 1 M D and
M SM B; consequently S, S, is parallel to B D, but SS,
multiplied by ▲ A CD = S S. multiplied by ▲ A CB or S S₁ .
DE SS. BE, whence S S₁: S S = BE: D E. But we have
=
BED F and D E B F, consequently also S S S S =
DF: BF. Hence the right line M F cuts the line of gravity
S, S, at the centre of gravity S of the whole figure.
§ 112. If we are required to find the centre of gravity S of a
polygon A B C D E, Fig. 140, we divide it into triangles and find the
statical moments in reference to two rectangular axes
1
and Y F.
X2, O B₂
= 2:29
If the co-ordinates 0 A₁ = x₁, O A2 = Y1, 0 B,
y, etc., of the corners are given, the statical moments of the tri-
angles A B 0, B C O, C D O, etc., can be determined very simply
in the following manner. The area of the triangle A B O is, ac-
cording to the remark which follows, D₁ = (X ₁ Y₂ — X2 Y₁),
1 2
—
222
[§ 112.
GENERAL PRINCIPLES OF MECHANICS.
= 2
that of the following triangle B C O is D₂ = 1 (X2 Y3 — X3 Y2),
etc., the distance of the centre of gravity of AB O from Y Y is,
according to § 109,
X₁ + X2 + 0 X1 + X2
21
U₁ =
3
3
and that from X X is
v₁
Y₁ + Y 2
3
those of the centre of grav-
(2
=
X2 + X3
3
and v₂
ity of the triangle B C O are
Multiplying these distances by the areas of the triangles we ob-
tain the statical moments of the latter, and substituting the values
thus found in the formulas
Y2 + Y 3
etc.
3
1 1
W =
D₁ u₁ + D₂ u₂ +
D₁ + D₂+..
D₁ v₁ + Dş v₂ +...
and v₁ =
2
we obtain the distances u = 0 S, and v
=
D₁ + D₂ +
0 S, of the required
centre of gravity S from the axes Y Y and X X.
If we divide in two ways a polygon of n sides by means of a di-
agonal into a triangle and a polygon of (n 1) sides, and then
join the centre of the former with that of the latter, we obtain in
this way two lines of gravity, whose intersection gives the centre.
of gravity. By repeated application of this operation, we can find
by construction the centre of gravity of any polygon.
EXAMPLE.-A pentagon A B C D E, Fig. 140, is given by the co-ordi-
FIG. 140.

A
X-
A₁
Ei
B
S
B2
Cal
B. S
A₂
S₂
Ꭰ
D
E
E2
_Y
1
nates of its corners A, B, C, etc., and the co-ordinates of its centre of
gravity are required.
§ 113.]
223
CENTRE OF GRAVITY.

The triple co-ordi-
Co-ordinates
given.
nate of the centre
of gravity.
The sextuple statical
moment.
Double area of the triangles.
४
У
3 Un
3 Vn
6 Dn un
Un
6 Dn V
Dan
2 I
7.15 +21.16=441
15
16.9+12.15=324
9
12. 12+18.9=306
Total,
1984
24
7
- 16
- 12
18
II
12 18.11+24.12=486 | +42
24.21-7. II=427
31
32
13237
13664
9
36-3969
15876
28
6
-9072
1944
+
6
21
1836
-6426
I
20412
486
22444
24572
The distance of the centre of gravity from the axis Y Y is therefore
1 22444
S S2
= U =
= 3,771
3' 1984
and from X X it is
S S₁ =
1 24572
3 1984
= 4,128.
REMARK.—If C A.
= x 19
C B₁ = x2, CA 2
1
1
y₁ and C B₂ = y₂ are the
co-ordinates of two corners of a triangle A B C, Fig. 141, the third corner
C of which coincides with the origin of co-ordinates, its area is
FIG. 141.
A
B
X
Α1
B₁
A 2
Bo
D= trapezoid A B B₁ A¸ +triangle
1 1
CBB, triangle CA A₁

1
= (V ₁ + Y²) (x, − x₂) + 2 2 3 2 –
2
(X1
X1 Y1 X1 Y 2-X2 Y1
2
2
2
The area of this triangle is there-
fore the difference between those
of two other triangles CB, A, and
CA, B, and one co-ordinate of
2
1
one point is the base of one trian-
gle and the other co-ordinate is the altitude of the second triangle. In
like manner one co-ordinate of the second point is the altitude of the first
triangle and the other co-ordinate is the base of the second triangle.
113. The Centre of Gravity of a Sector, A C B, Fig.
142, coincides with centre of gravity S of the arc A, B₁, which has
the same central angle as the former and whose radius C A, is two
thirds of that CA of the sector; for the latter can be divided by an
224
[S 114.
GENERAL PRINCIPLES OF MECHANICS.
infinite number of radii into small triangles, whose centres of gravity
FIG. 142.
M
AK
B
M
D
B,
+-
are situated at a distance from the
centre C equal to two thirds of ra-
dius; the continuous succession of
these centres forms the arc A, M₁ B₁.
The centre of gravity S of the sector
lies, therefore, upon the radius which
bisects this surface and at the distance

CS = y =
chord 2
arc 3
CA
4 sin. 1 B
3 B
r
from the centre, when r denotes the radius of sector and ẞ the
arc which measures its central angle A C B.
For the semicircle ß = π, sin. § ß = sin. 90º = 1, whence
У
4
3 п
0,4244 r, or approximatively
For a quadrant we have
14
1.
33
4 √
Y
r =
3 . T
4 1/2
3 π
r = 0,6002 r,
and for a sextant
4
124
Σ
У
r =
r = 0,6366 r.
3
π
π
FIG. 143.
§ 114. The Centre of Gravity of the Segment of a Circle,
A B M, Fig. 143, is found by putting
its moment equal to the difference
of the moments of the sector A CBM
and of the triangle AC B. If r is
the radius C A, s the chord A B and
A the area of the segment A B M, we
have the moment of the sector
M

* S₂
C
the moment of triangle
B
triangle multiplied by C S
= sector multiplied by C S₁
r.arc chord 2 1
7= $ 12,
2
arc 3
3
s
$
4
4
$ 72
مودی
3
12'
and consequently the moment of the segment A
$ 115.]
225
CENTRE OF GRAVITY.
A.
1
3
3
Hence the required distance is y =
A . C S = A y = {} s r² − ( ³ ³ - 12)
12'
12 A
1
For a semicircle s = 2 r and A =
π r², and therefore
2
8 p³
4 r
Y
π jo²
3 π'
12.
2
as we have already found.
FIG. 144.
M
E
D
B
In the same way the centre of grav-
ity S of a section of a ring A B D E,
Fig. 144, can be found; for it is the
difference of two sectors ACB and
DCE. If the radii are C A = r₁ and
CE r₂ and the chords A B = $1
and DE =S2, we have the statical

2
1
moment of the sectors
and 82 12,
3
3
and consequently that of the portion of the ring
M =
S₁ ri
3
Sa 12
or since
S2
81
r
r.
1
M =
91
r2
3
The area of the piece of the ring is F=
j
2
Bri
Br₂
2
2
2
1
3
2
in which ẞ denotes the arc which measures the central angle
A CB; hence the centre of gravity Sof the section of the ring is
determined by the formula.
M
C S
= y
F
4 sin. ß ri
3
グ
​3
B
= b and r₁ + r² 2
1
ri
1
3
9
J'
2
r2
2
r
7.
3
2 r.
2
3
3
$1
r B
2 /r₁
chord
2
7.
arc
(1
2 r, when r
2,
グ
​1
sin. B
B
1 12
+ 14 (-/-)²)
130°,
EXAMPLE.-If the radius of the extrados of an arch is r1 5 feet, and
that of the intrados is re 3 feet, and if the central angle is pº
2
the distance of the centre of gravity of the front surface of the arch from
its centre is
y
4 sin. 65° 59 3,53
3 arc.130° 52 3,52
3,430 feet.
4. 0,9063 125 42,875
3. 2,2689 25 12,25
3.6252 . 82,125
6,8067 12,75
#
15
226
[$ 115.
GENERAL PRINCIPLES OF MECHANICS.
(§ 115.) Determination of the Centre of Gravity by the
Aid of the Calculus.-The determination of the centre of gravity
by means of the calculus is accomplished in the following man-
FIG. 145.
Y
P
O
M
L
A
K
N
X
F from the axis A Y,
ner. Let A N P, Fig. 145, be the given
surface, A N = x its abscissa and N P = y
its ordinate. The area of an element
of the surface is
d Fy dx (see Introduction to the
Calculus, Art. 29) and its moment in ref-
erence to the axis of ordinates A Fis
OM.dFAN.dFxy dx;
if we put the distance L S A K of the
centre of gravity S of the whole surface
=u, we have

Fu= √xy d x,
and consequently 1) u =
S x y d x
S x y d x
L y d x
Since the centre or centre
F
of gravity M of the element N M P
is at the distance N M y from the axis A X, the moment of
d Fin reference to this axis AX is
NM.dF y d F = 1 y' d x ;
11
y²
putting the distance K SAL of the centre of gravity S of the
whole surface F from the axis A X,
v, we have
Fv = Sy² da, and therefore
2) v =
1
z Sy² d x
F
172
Ly'd x
Syd x
E.G., for the parabola, whose equation is y = px or y = √p. x,
we have
SV
W =
Npsx
Np S x d x
S x d x
S Np. x d x
Np S xs d x
S x d x
Who ToiRo
Z xt
= x,
Z x}
091100
or LSA KA N, and, on the contrary,
v = 1/1/2
==
or
09:00
S p x d x
fxdx
P x
√ p s ad x = 1 √p / n d x = 1 v p
S
p x = 3 y,
KS=ALNP.
x²
= x²
§ 116.]
227
CENTRE OF GRAVITY.
§ 116. The Centre of Gravity of Curved Surfaces.-The
centre of gravity of the curved surface (envelope) of a cylinder
A B CD, Fig. 146, lies in the middle S of
D
FIG. 146.
S
the axis M N of this body; for all the ring-
N C shaped elements of the envelope of the cyl-
inder, obtained by cutting the body parallel
to its base, have their centres and centres of
gravity upon this axis; the centres of grav-
ity form then a homogeneous heavy line.
For the same reason the centre of gravity
of the envelope of a prism lies in the middle
of the line, which unites the centres of gravity of its bases.

A
M
B
The centre of gravity S of the envelope of a right cone A B C,
Fig. 147, lies in the axis of the cone one-third of its length from
the base, or two-thirds from the apex; for this curved surface can
be divided into an infinite number of infinitely small triangles by
means of straight lines (called sides of the cone). The centre of
gravity of all these triangles form a circle HK, which is situated
at a distance equal to two-thirds of the axis from the apex C, and
whose centre or centre of gravity S lies in the axis C M.
FIG. 147.
C
FIG. 148.


E
D
K
.N
H
S
M
F
H
K
S
ΟΙ
C
B
M
The centre of gravity of a zone A B D E, Fig. 148, of a sphere,
and also that of spherical shell, lies in the middle § of its height
M N; for, according to the teachings of geometry, the zone has
the same area as the envelope F G H K of a cylinder, whose height
is equal to that M N of the zone and whose radius is equal to that
CO of the sphere, and this holds good even in the ring-shaped ele-
ments obtained by passing an infinite number of planes parallel to
the base through the zone; hence the centre of gravity of the zone
and of the envelope of the cylinder coincide.
REMARK.-The centre of gravity of the envelope of an oblique cone or
228
[§ 117.
GENERAL PRINCIPLES OF MECHANICS.
pyramid is to be found, it is true, at a distance from the base equal to one-
third of the altitude, but not in the right line joining the apex to the
centre of gravity of the periphery of the base, since by cutting the en-
velope parallel to the latter we divide it into rings of different thicknesses
on different sides.
FIG. 149.
§ 117. Centre of Gravity of Bodies.-The centre of gravity
S of a prism A K, Fig. 149, is the centre of the line uniting the
centres of gravity M and N of the two bases
AD and GK; for by passing planes parallel
to the base through the body we divide it
into similar slices, whose centres lie in M N,
and whose continuous succession form the
homogeneous heavy line M N.
G
A
B
L

K
D
For the same reason the centre of gravity
of a cylinder is to be found in the middle of
its axis.
The centre of gravity of pyramid A D F, Fig. 150, lies in the
straight line M Fjoining the apex F with the centre of gravity M
of the base; for all slices such as N O P Q R have, in consequence
of their similarity to the base A B C D E, their centre of gravity
upon this line.
FIG. 150.
F
FIG. 151.

D

A
N
P
B
C
D
N
MG
H
E
B
If the body is a triangular pyramid, like A B C D, Fig. 151, we
can consider each of the four corners as the apex and the opposite
side as the base. The centre of gravity is therefore determined by
the intersection of the two straight lines drawn from the corners
D and A to the centres of gravity M and N of the opposite surfaces
A B C and B C D.
If the right lines E A and E D are also given, we have (accord-
§ 118.]
229
CENTRE OF GRAVITY.
FIG. 152.
D
3
ing to § 109) EMEA and E NE D. MN is therefore
parallel to A D and AD, and the triangle M N S is similar
to the triangle D A S. In conse-
quence of this similarity we have
also MSDS or DS=3 MS
and M D = MS + SD = 4 M S,
or inversely MS MD. The
distance of the centre of gravity
of a triangular pyramid from its
base along the line joining the
centre of gravity M of the base to
the apex D of the pyramid is equal
to one-fourth of this line.

M
B
*N
E
If the altitudes D H and S G
are given and if we draw the line
H M, we obtain the similar triangles D H M and S G M, in which,
as we have just seen, S G DH. We can therefore assert that
the distance of the centre of gravity of a triangular pyramid from
its base is one-fourth and from its apex three-fourths of its altitude.
Finally, since every pyramid and every cone is composed of tri-
angular pyramids of the same height, the centre of gravity of every
pyramid and of every cone lies at a distance from the base equal to
one-fourth of the altitude and at a distance from the apex equal to
three-fourths of the altitude.
We determine the centre of gravity of a pyramid or of a cone
by passing a plane, at a distance from the base equal to one-fourth
the altitude, through the body parallel to its base and by finding
the centre of gravity of this section or the point where a line
drawn from the centre of gravity of the base to the apex will cut it.
§ 118. If we know the distances A A, B B₁, etc., of the four
corners of a triangular pyramid A B C D, Fig. 153, from a plane
H K, the distance S S₁ of its centre of gravity S from the plane is
their mean value
S S₁
A A₁ + B B₁ + C C₁ + D D₁
4
which can be proved in the following manner. The distance of
the centre of gravity M of the base A B C from this plane is (§ 109)
M M₁ =
A A₁ + B B₁ + CC,
Co
3
230
[$ 118.
GENERAL PRINCIPLES OF MECHANICS.
and the distance of the centre of gravity S of the pyramid is
S S₁ = M M₁ + ¦ (D D, − M M₁),
4
FIG. 153.

A
B₁
B
K
D,
in which D D₁ is the distance of the apex. Combining the last two
equations, we obtain
SS₁ = y = 3 M M₁ + D D₁ =
¦
A A₁ + B B₁ + C C₁ + D D₁
4
The distance of the centre of gravity of four equal weights
placed at the corners of the triangular pyramid is also equal to the
arithmetical mean
Y
A A₁ + B B₁ + C C₁ + D D₁
4
DD₁.
consequently the centre of gravity of the pyramid coincides with
that of these weights.
Y
A
FIG. 154.
1. I
Z
Ο
B
B₁
REMARK.-The determination
of the volume of a triangular pyra-
mid from the co-ordinates of its
corners is very simple. If we pass
through the apex 0 of such a
pyramid A B C 0, Fig. 154, three
co-ordinate planes X Y, X Z, Y Z,
and denote the distances of the
corners A, B, C from these planes
by 21, 22, 23; Y 1, Y 2, Y3 and α1, X2, X 8,
we have the volume of the pyramid

1
11
2
V = ± { [≈₁Y 2 * 3 + X 2 Y 3 1 +
X3 Y 122 ( X 1 Y 3 Z 2 + X g Y 1 2 3
+ X3 Y 2 21)],
2
§ 119.]
231
CENTRE OF GRAVITY.
which is found by considering the pyramid as the aggregate of four ob-
liquely truncated prisms.
The distances of the centre of gravity of this pyramid from the three
co-ordinate planes YZ, XZ and X Y are
X1 + X 2 + Xxz Y 1 + Y 2 + Y 3
21 + 22 + 23
X
4
1
Y
and z =
4
4
§ 119. The centre of gravity S of any polyhedron, such as
A B C D O, Fig. 155, can be found by calculating the statical
FIG. 155.
Di
A
D₂
B₁
Z

X
moments and volumes of the triangular pyramids, such as A B C O,
BCD O, into which it can be decomposed.
If the distances of the corners A, B, C, etc., from the co-ordinate
planes FZ, XZ and X Y, passing through the common apex O of
all the pyramids, are x1, x2, x, etc., y1, Y2, Yз, etc., and Z1, Z2, Z3, etc.,
we have the volumes of the various pyramids.
1
6
—
V₁ = ± f (X1 Y½ Z3 + X 2 Y 3 Z 1 + X3 Y 1 Zą — X₁ Yз Zą — X₂ Y1 Z3 — X3 Ye Z1),
Z2
2
V₂ = ± ↓ ( X 2 Y 3 Z 4 + X3 Y 4 Z2 + X Y 2 Z3 — X2 Ys Zz — Xş Yş Z↓ — X₁ Y3 Z2),
etc., and the distances of their centres of gravity from the co-ordi-
nate planes are
X₁ + X 2 + X3
U r =
V₁ =
4
Y₁ + Y ½ + Y3
4
Z₁ + 22 + 23
4
X2 + X3 + X₁
Из
Uz =
"
༧༠
V₂ =
4
etc.
From these values we calculate the distances u, v, w of the
centre of gravity S of the whole body by means of the formulas
Y½ + Y3 + Y +
4
22 + 28 + 2:4
འU2
4
232
[§ 119.
GENERAL PRINCIPLES OF MECHANICS.
V₁ u₁ + √₂ U₂ +
u =
V
V₁ + V₂ +
V₁ v₁ + √2 V₂ +...
V₁ + V₂ +
9
and
V; w₁ + V₂ w₂ +
•
W 20 =
1
V₁ + V₂ +
EXAMPLE.-A body A B C D O, Fig. 155, bounded by six triangles, is
determined by the following values of its co-ordinates, and we wish to find
the co-ordinates of the centre of gravity.

Given Co-
ordinates.
The sextuple volume of the
triangular pyramids
ABCO and B C D 0.
Quadruple
Co-ordi-
nates of the
Centres of
Gravity.
8
Y
2
4
Un
4
V n
και ξ
Twenty-four fold
Statical Moments.
24
Vn
Un
24
Vn V n
24
Vn Wn
2023 41
20.29.28
20.40.30
6 = 23.30.12
23.28.45
31072 77 92 99 2392544 2858624 3076128)
45 29 30
41.45.40
41.12.29
12 40 28
=
38.35 20
45.35.28
6 ½ 29.20.12
30.38.40
(45.40.20
29.28.38 17204 95 104 781634380 1789216 1341912
30.12.35
Total 48276
4026924 4647840 4418040
From the results of the above calculation we deduce the distances of
the centre of gravity S of the whole body from the planes Y Z, X Z and X Y,
1
U
4
4026924
48276
20,853,
1
4647840
=24,069, and,
4
48276
1 4418040
20
22,879.
4
48276
REMARK.-We can also determine the centre of gravity of a polyhedron
by dividing it in two ways by means of a plane into two pieces and by
joining the centres of gravity of each two pieces; the intersection of the two
lines gives the required centre of gravity. Since both lines are lines of
gravity, the intersection must be the centre of gravity of the whole body.
If the body has a great number of corners, this process becomes very long,
in consequence of the number of times this division must be repeated.
The five-cornered body in Fig. 155, which must be divided in two ways
into two triangular pyramids, has its centre of gravity at the intersection
of the lines joining the centres of gravity of each two of these pyramids.
§ 120.]
233
CENTRE OF GRAVITY.
A
FIG. 156.
F
§ 120. The centre of gravity of a truncated pyramid or frus-
tum of a pyramid A D Q N, Fig. 156,
lies in the line G M joining the centres of
gravity of the two (parallel) bases. In or-
der to determine the distance of this point.
from one of the bases we must calculate
the volumes and moments of the complete
pyramid A D F and of the portion NQF,
which has been cut away.
If the areas
of the bases A D and NQ are = G₁ and
Ga, and if the perpendicular distance be-
tween them=h, the height x of the por-
tion of the pyramid, which is wanting, is
determined by the formula

N
K
S
BG S
M P
C D
2
G₁
G2
(h + x)²
x²
whence
218
h
G₁
h N G z
+ 1 =
or x =
G₂
NG₁
NG₂
h N G
h + x =
NG₁
and
The moment of the whole pyramid in reference to its base is
NGⓇ
G₁ (h + x)
3
h + x
4
1
h³ G₂ 2
12 (NG,
NG.)"'
and that of the part of pyramid, that is wanting, is
1
+
2
h² G₂²
h² V G
3 NG₁ NG₂ 12 ⋅ (NG₁ — NG.)
G₂ x
3
1
h +
1 )
hence the moment of the truncated pyramid is
2?
hª
12(VG, NG₂)°
•
[G,²
4 (VG, G₂
G₂³) — G₂³] =
2
h² (G₁² — 4 G₂ VG₁ G₂ + 3 G:)
12 (G₁ - 2 VG₁ G₂ + G₂)
h²
12
. (G₁ + 2 NG, G₂ + 3 G.).
1
Now the contents of the truncated pyramid are
V = G₁
(G₁ + NG, G₂ + G₂)
(G₂ ) } }
h
3
and therefore the distance of the centre of gravity S from the
base is
234
[§ 121.
GENERAL PRINCIPLES OF MECHANICS.
G₁ + 2 VG, G₂ + 3 G₂ h
Y
G₁ + VG, G₂ + G ₂
4
The distance S. S of this point from the plane K L, passing
through the middle of the body parallel to its base and dividing its
height into two equal parts, is
h
Y1
2
y =
[2 (G₁ + VG, G₂ + G₂)~(G₁ + 2 VG₁ G₂+ 3 G₂)] h
G₁ + NG₁ G 2 + G₂
G₁
G2
G₁ + √ G₁ G₂ + G₂
If the radii of the
h
4
4
bases of a frustum of cone are r, and r½, or
1
G₁ =πr₁² and G₂ =
2
π r²², we have
y =
Y1
=
r₁² + 2 r₁ r₂ + 3 r₂²
2
2
1 2
r²² + r1 r2 + r₂²
r r₂³
I
√2
r₁₂² + r₁ r₂ + r₂²
h
and
4
2
h
2
4
1
EXAMPLE. The centre of gravity of a truncated cone whose altitude
is h 20 inches and whose radii are r = 12 inches and r₁ 8 inches lies,
as is always the case, in the line joining the centres of the bases, and at a
distance
1222.12.8 +3.82.
12² + 12.8 + 8²
5.528
304
2640
304
8,684 inches
y = 20.
from the greater base.
§ 121. An obelisk, I.E., a body A CO Q, Fig. 157, bounded
by two dissimilar rectangular bases and by four trapezoids, can
E
FIG. 157.
P
B
H
R
be decomposed into a parallelopipedon
AFR P, into two triangular prisms
EHRQ and G KRO and into a
four-sided pyramid HK R. By the aid
of the moments of these component
parts we can find the centre of gravity
of the whole body.
It is easy to see that the right line
C joining the middle of one base to that
of the other is a line of gravity of the
body; we have, therefore, but the distance of the centre of gravity
from one of the bases to determine. Let us denote the length
B C and the width A B of one base by, and b₁, and the length
QR and the width P Q of the other base by l, and b, and the
height of the body or the distance of the bases apart by h. The

§ 121.]
235
CENTRE OF GRAVITY.
contents of the parallelopipedon are then --- b, l, h, and its moment
h
is ba la h.
2
j b₂ la h².
The contents of the two triangular
2
prisms are
h
([b₁ − b₂] l2 + [lı — l2] b₂)
and their moments are
2'
h h
=
=
2
and finally the contents of the pyramid are
( [b, −b₂] l, + [l, − ls] b₂) 252 · 23
2' 3'
h
=
(b₁ — b₂) (l, —
12) 233
3'
h h
=
(b₁ — b.) (1 — 12)
34
and its moment is
From the above we deduce the volume of the whole body
h
6
V = ( 6 b₂ l₂ + 3 b₁ l½ + 3 l₁ b₂ − 6 b¸ l, + 2 b₁ l₁ + 2 b¸ l2 − 2 b¸ l½ — 2 b₂ l₁) .
l, l, —
= (2 b l + ? b, b
+ b b + ab)
h
6'
its moment
Vy = (6 b₂ l₂+ 2 b₁ l₂ + 2 1, b₂ − 4 b₂ l₂ + b₁ l₁ + b₂ la — b₁ l₂ — l₁ b₂) ·
= (3 b₂ l₂ + b₁ l₁ + b₁ l₂ + b₂ 1)
h²
12'
and the distance of its centre of gravity S from the base b₁ ↳₁
h2
12
We can also put (see the "Planimetrie und Stereometrie" of
C. Koppe)
y =
b, l + 3 B, C + bi lọ + b h h
2 b₁ l₁ + 2 b la + b₁ la + b¸ l¸° 2°
т
b₁ + b² h + b h +
2
2
b₁ b. l l
2
h
2 3*
cross section
h
Y1
Y
2
3 (b, + b₂) (l
b₁ lr — bs l
+ ?2) + (b, − b.) (l
h
•
1.)
The distance y, of the centre of gravity from the
through the middle is determined by the formula
REMARK.-This formula is also applicable to bodies with elliptical
bases. If the semi-axes of one base are a, and b, and those of the other
a₂ and b₂, the volume of such a body is
2
√ =
Th
6
1
1
(2 α, b₁ + 2 a₂ b ₂ + α ₁ b ₂ + ag b₁),
1 1
2 2
1 2
and the distance of its centre of gravity from the base
y =
α1
1 1
1 2
2
1 1
a, b, is
h
1
g
b ₁ + 3 α 2 b 2 + α ₁ b₂ + α ₂ b.
2 a₁ b₁ + 2 α z b₂ + a₁ b g + a
a
236
[$ 122
GENERAL PRINCIPLES OF MECHANICS.
EXAMPLE.—If the embankment A CO Q, Fig. 158, for a dam is 20 feet
high, 250 feet long and 40 wide at the bottom, and 400 feet long and 15
A
D
0
FIG. 158.
R
B

feet wide on top, what is the distance of its centre of gravity from its base?
400, and h = 20, and consequently
Here b =
1
40, 7,
=
1
250, b₂
2
= 15, l₂
the distance is
20
40.250 + 3. 15. 400 + 40. 400 + 15.250
2.40.250 + 2. 15. 400 + 40. 400 + 15. 250 2
Y
4775
5175
10 =
1910
207
9,227 feet.
§ 122. If the circular sector A CD, Fig. 159, is revolved about
its radius CD, a spherical sector AC B is generated, the centre
FIG. 159.
D
A
M
D1
A
s
Mil
B₁
B
of gravity of which can be determined in
the following manner. We can consider
this body as the aggregate of an infinite.
number of infinitely thin pyramids, whose
common apex is the centre C and whose
bases form the spherical zone AD B. The
centres of gravity of each of these pyramids
are situated at a distance equal to 3 of the
radius CD of the sphere from its centre.
C, and they form a second spherical zone
A, D, B₁, whose radius CD, CD.
3
The centre of gravity of this curved surface is also that of the
spherical sector; for the weights of the elementary pyramids are
equally distributed over this surface, which is therefore every-
where equally heavy.

4
If we put the radius C A C' Dr and the altitude D M of
the exterior zone = h, we have for the interior zone CD,
3 r
and M, D₁ = 3 h, and consequently (§ 116) S D₁ = M, D₁ = jh,
and the distance of the centre of gravity of the spherical sector
from the centre Cis
1
2
CS CD,
$ ( − 1 ) .
h
Ꭰ
S D₁ = 3 r - 3 h = 3
=
For a hemisphere r = h, and therefore the distance of its centre
of gravity S from the centre Cis
$123, 124.]
237
CENTRE OF GRAVITY.
CS
3
→
4 2
02.00
r.
§ 123. We obtain the centre of gravity S of a spherical seg-
FIG. 160.
Ꭰ
S
A
M
E! L
C
B
F
ment A B D, Fig. 160, by putting
the moment of the segment equal
to that of the spherical sector
AD B C less that of the cone
A B C. Denoting again the radius
CD of the sphere by r and the
altitude D M by h, we have the
moment of the sector
= 3 π r³ h. 3 (2 r − h) = ¦ π r² h (2 r — h),
and that of the cone

}
= πh (2 r - h). (r- h). (r− h) =
§
hence the moment of the segment is
1 π h (2 r — h) (r — h)';
Vy = { πh (2 r − h) (r² — [r — h]')
h]²) = {
= {
The contents of the segment are
V = { π h² (3 r — h),
and consequently the required distance is
C S = Y
4
1 π h² (2 r — h)²
· π h³ (3 r h)
·
π h² (2 r — h)*.
3 r
(2 r — h)²
h
If we put again hr, the segment becomes a hemisphere, and,
as before, we have CS z r.
This formula is also true for the segment A, D B, of a spheroid
generated by the revolution of the arc D A, of an ellipse about its'
major axis CD r; for if we cut the two segments by means of
planes parallel to the base A B into thin slices, the ratio of the
corresponding slices is constant and
MA₁2
b-
2
MA²
CE
С E³ ༡
when
We must multiply
b denotes the smaller semi-axis of the ellipse.
not only the volume, but also the moment of the spherical segment
b²
by to obtain the volume and moment of the segment of the
22
spheroid, and therefore the quotient C S =
moment
volume
is not changed.
(2 r
S
})²
y
3
4
in which r de-
3 r
—---
h
In general we have CS
notes that semi-axis about which the ellipse is revolved, when gen-
erating the spheroid.
§ 124. Application of Simpson's Rule.-In order to find
the centre of gravity of an irregular body A B C D, Fig. 161, we
238
[§ 124.
GENERAL PRINCIPLES OF MECHANICS.
divide it, by means of planes equally distant from each other, into
thin slices and determine the area of the cross sections thus ob-
tained and their moments in reference to the first parallel plane
FIG. 161.
D
F!
F!
F₁
A
F. M
B
A B, which serves as base, and we then
combine the latter by means of Simpson's
rule.
If the areas of the cross-sections are
Fo, F1, F2, F3, F, and the total height or
distance M N between the two parallel
planes farthest apart h, we have, ac-
cording to Simpson's rule, the volume of
the body

=
h
V = (F₂+4 F +2 F +4 F + F₁) 12*
Multiplying in this formula each surface by its distance from
its base we obtain the moment of the body, viz.,
h
h
4' 12'
Vy = (0. F +1.4 F +2.2 F +3.4 F +4 F)
and dividing the last equation by the first we obtain the required
distance of the centre of gravity S
Y
M Sy
(0. F +1.4 F +2.2 F +3.4 F +4 F) h
F +4 F +2 F +4 F + F
If the number of slices = 6, we have
4
0. F +1.4 F +2.2 F +3.4 F +4.2 F +5.4 F +6 F h
F +4 F + 2 F₂ + 4 F₂ + 2 F₁ + 4 F¿ + F。
2
3
6
It is easy to see how this formula varies, when the number of
slices is changed. The rule, however, requires, that the number of
slices shall be an even one, or the number of surfaces an uneven one.
In many cases we need determine but one distance, as a line of
gravity is also known. Solids of rotation formed upon the turn-
ing lathe are very common examples of such bodies. Their axis
of rotation is a line of gravity.
A
This formula is also applicable to the determination of the
centre of gravity of a surface, in which case the
cross sections Fo, F1, F2, etc., become lines.
FIG. 162.

D
L
S
EXAMPLE 1. For the parabolic conoid A B C, Fig.
162, formed by the revolution of a portion A B M of
a parabola about its axis A M, we obtain, when we make
Mbut one section DNE through the middle, the following.
Let the altitude A M=h, the radius B M =
ANN M
and consequently the radius D N
h
2
C =r√. The area of the section through A is F。 = 0,
0
§ 125.]
239
CENTRE OF GRAVITY.
π jo?
that through N F₁
πD N²
1
and that through M, F₂ = r².
π
2
2
Hence it follows that the volume of this body is
h
V = ½/ (0 + 4 F, + F₂)
6
and that its moment is
h2
V y = 12
h
=
6
= ~ ( 2 π r²
+
(1 . 2 π r² + 2 π p²)
π r²) = } π r² h = j F₂ h,
Π
= } π r² h²
§ F₂ h².
2
Consequently the distance of the centre of gravity S from the vertex is
A
A S = Y
1 F₂ h²
2
h.
1 F₂ h
2
FIG. 163.
FIG. 164.
M

B
A
B

D
S
M
Ꭰ
C
E
C
EXAMPLE 2. The mean half widths
1 inch, r₁
=
1
are ro
1.1 inches, 72
= 0,4 inches, and its height M N
1
of the vessel A B C D, Fig. 164,
0,9 inches, 73 = 0,7 inches, and r
2,5 inches; required the centre of
4
4
gravity of the space within it. The cross sections are Fo
1 T,
F₁ 1,21, F 0,81 π, F3 = 0,49 π and F₁ = 0,16 π, and therefore
the distance of its centre of gravity from the horizontal plane A Bis
0.1 1.4. 1,21 + 2.2.0,81 +3.4. 0,49 +4.0.16.π 2,5
1 π + 4 . 1,21 π + 2.0,81 π + 4. 0,49 π + 0,16 . π
14,60 2,5 36,50
9,58 4 38,32
MS
0,9502 inches.
The vacant space in the vessel is √ = 9,58 π .
4
2,5
12
6,270 cubic inches.
(§ 125) Determination of the Centre of Gravity of Sur-
faces and Solids of Rotation.-The centre of gravity of curved
A
FIG. 165.
P
M
P
QL
surfaces and of bodies with curved sur-
faces can be determined generally by the
aid of the calculus. In practice, solids
and surfaces of rotation occur most fre-
quently, and we will therefore here treat
only of the determination of the centre
of gravity of these forms. If the plane
curve A P, Fig. 165, revolves about its
axis AC, it describes a so-called surface
of rotation A P P₁; and if the surface
A P M bounded by the curve A P and

240
GENERAL PRINCIPLES OF MECHANICS.
[$ 125.
its co-ordinates A M and M P is revolved about the same axis a
solid of rotation bounded by a circular surface P M P, and by a
surface of rotation A P P, is produced.
If we denote the abscissa A M by x, the corresponding ordinate
by y and the corresponding arc 4 P by s, and also the element
M N = P R of the abscissa by d x, the element QR of the ordi-
nate by dy and the element P Q of the curve by d s, we have the
area of the belt-shaped element P Q Q₁ P, generated by the revo-
lution of ds, when we put the surface of rotation A P P、
=
1
d 0 2 π. P M. P Q = 2π y d s,
0,
and, on the contrary, the contents of the element of the solid of ro-
tation AP P₁ V, limited by this element of the surface, are
Ꮩ
d Vπ P M². MN πy' d x.
= y²
Since the distance of both elements from a plane passing
through A at right angles to the axis A Cis equal to the abscissa
2, the moment of d O is
and that of d Vis
x d 0 = 2 π x y d s,
ad V =
= π x у² d x.
Now since
0 = £ 2 ñ y ds = 2πfyds and
V = ƒ πy d x =
S
S
πf y² dx,
and since according to the above formulas the moment of O is
T S
£ 2 π x y ds = 2 πf x y d s,
and that of Vis
Sπ x y² d x = π / x y² d x,
пху
it follows, that the distance A Sy of the centre of gravity S from
the origin A is
1) for surfaces of rotation
2 π S x y d s
U
Q
2π Jyds
L x y d s
Syd s'
W =
and, on the contrary,
2) for solids of rotation,
E.G., for a spherical zone whose
2
radius C Q
π Ï x y² d x
1 x y d x
π Ï y² d x
Sy' d x '
= r we have, since
P Q
α Q
d s
J
PR Q N
I.E.
or y ds = r dx,
d x
Y
$126.]
241
CENTRE OF GRAVITY.
Lxrd x
ƒ x d x
} x²
A Su
= 1 x = 1 A M.
frd x
jdx
X
(Compare § 116.)
For a segment of a sphere, on the contrary, we have, since we
can put y' 2 r x - x²,
=
S
ASU =
(2rx-x²) x d x
J (2 rx-x²) d x
S2 r x d x - S x³ d x
j 2 rx dx
S x d x
{ rx³ — } x*
2
4
( r —
¦ x) x
-
8 r 3 x x
rx²
r -
3
j x
3 r
X 4'
and consequently
CS = r - U =
4/00
(2
2 r
x)²
(Compare § 123.)
3 r
X
§ 126. Properties of Guldinus.—An interesting and often
very useful application of the theory of the centre of gravity is
the properties of Guldinus (Fr. méthode centrobarique, Ger. die
Guldinische Regel). According to these the contents of a solid
of rotation (or the area of a surface of rotation) is equal to the
product of the generating surface (or generating line) and the
space described by its centre of gravity while generating the body
(or surface). The correctness of this rule can be proved as follows:
If a plane surface A B D, Fig. 166, is revolved about an axis
XX, every element F1, F2, etc. of it describes a ring; if the dis-
tances of these elements F, F, etc. from the axis of rotation
XX are F₁ K₁, F, K, etc. r₁, r., etc.,
and if the angle of rotation is FK F
= SCS₁ = a" or the arc corresponding
to the radius 1, a, the arc-shaped
paths described by the elements are
r₁a, r₂ a, etc. The spaces described by
B₁ the elements F, F, etc., can be re-
garded as curved prisms whose alti-
tudes are r, a, r, a, etc., their contents
are therefore F₁r, a, Fr, a, etc., and
consequently the volume of the whole body A B D D, B, A, is
FIG. 166.
Ꭰ
B
X
A₁
F
1
2

1 1
=
V = F₁ r₁ a + F₂r, a + .... (F₁ r'₁ + F₁ r₂ + ...) a.
1
16
242
[$ 126.
GENERAL PRINCIPLES OF MECHANICS.
If y
C'S is the distance of the centre of gravity S of the gen-
erating surface from the axis of rotation, we have
(F₁ + F₂ + ...) y
·) y = F₁ r₁ + F₂ r₂ +
and consequently the volume of the whole body
V = (F₁ + F₂ + ...) y a.
But F₁ + F₂ + ... is the area of the surface F, and Y a is the arc
S S₁ =
= w described by the centre of gravity; hence it follows
that V = Fw, which is what was to be proved.
This formula is also applicable to the case of the rotation of a
line, since the latter can be considered as a surface of infinitely
small width. In this instance we have F = l w, I.E. the surface
of rotation is the product of the generating line (1) and the space
(w) described by its centre of gravity.
EXAMPLE 1. If the semi-axes of the elliptical cross section A B E D,
Fig. 167, of a half ring are C A a and C B = b, and if the distance C M
of its centre C from the axis XX=r, the elliptical generating surface
will be F = π a b, and the space described by its centre of gravity (C)
will be w = πr.
Hence the volume of this half ring is V = π² a br;
1
2 V 2 m² abr.
Ÿ, and
that of the whole ring is ₁
If the dimensions are a = 5 inches, b
5 inches, 6 = 3 inches and r = 6 inches, the
volume of one-quarter of the ring is
½ . π² . 5 . 3 . 6 = 9,8696 . 5 . 9 = 444,132 cubic inches.
FIG. 167.
X
FIG. 168.
X


DE B
1
B
B
CD
M
D
D 1
E
X
X
B₁
EXAMPLE 2. The volume of a ring with the semi-circular cross section
A B D, Fig. 168, is, when CA = C B = a denotes the radius of this cross
section and M C = r that of the hollow space,
π a²
√ =
2 π 7 +
4 a
3 π
= π a² (ñ r + § a).
EXAMPLE 3. If the segment of a circle A D B, Fig. 169, revolves about
the diameter E F parallel to its chord A B, it describes a sphere A D₁B
with a cylindrical hole A B B₁ A₁in it. If A is the area of the segment
§ 126.]
CENTRE OF GRAVITY.
1
243
and & the length of its chord A B = A₁ B₁, we have (§ 114) for the distance
of its centre of gravity S from the centre C
في شكلك
83
CS =y 12 A'
and consequently the volume of the sphere with the cylindrical hole is
83
π 83
V = 2 π у A = 2 π
12
6
For a complete sphere we have the chord or height of the hole equal to
the diameter d of the sphere, and consequently its volume
as we know.
V
=
πα
6
EXAMPLE 4. We are required to find the area of the surface and the
contents of the cupola A D B, Fig. 170, of a cloistered arch, when the half
D
S
FIG. 169.
E
A
A1
B
BL
F
FIG. 170.
D


A
B
SMC
D
1
width M A = M B = a and the altitude M D = h are given. From the
two given dimensions we obtain the radius C A CD of the generating
circle
a² + h²
2 a
The central angle A C D = a is given by the formula
h
sin, a ==
The centre of gravity S of an arc D A D₁ = 2 A D is determined by
the distances
CS
↑ .
chord M D
arc A D
r sin. a
and CM = r cos. a;
a
consequently the distance of the centre of gravity S from the axis M D is
r sin. a
MS
T cos. a =r
a
(sin
sin. a
cos. a
a),
and the space described by the centre of gravity in describing the surface
ADB is
'sin. a
20 = 2 πη
008. a
α
a).
244
[§ 127.
GENERAL PRINCIPLES OF MECHANICS.
1
The generatrix D A D₁ is 2 r a, consequently its half is A D
the surface of rotation A D B generated by the latter is
=ra, and
sin. a
0 = ra. 2 π r
cos. a
α
)
2 π p² (sin. a
a cos, a).
a
sin. a =
3'
0 = π p² √ √ 3
Very often we have a° = 60°, or
П
hence the required area is
3
The distance of the centre of gravity of the segment D A D₁ = A
(a — ½ sin. 2 a) from the centre Cis
√3 and cos, a =
مرجة
(√3 − ) = 2,1515. º.
r².
= A = p²
(2. M D)³
12 A
2 p³ sin.³ a
3
A
and, therefore, its distance from the axis is
MS CS-CM
=
2 p³ sin.³ a
r cos. a,
3
A
and the space described by this centre of gravity in one revolution around
MD is
20
2πη
A
ㅠ
​2 пра
(ff r² sin,³ a
A cos. a)
[ f sin.³ a - (a — sin. 2 a) cos. a].
A
The volume of the body generated by the revolution of the segment
DAD, is found by multiplying this space by A, and the volume of the
cupola by dividing the last product by two. The latter volume is
V = π r³ { } sin.³ a
(a — 1 sin. 2 a) cos. a]
E.G., if a° =
60°, we have
П
α
sin. a =1 √3, sin. 2 a =
√3, cos. a =
, and therefore
3'
V = προς
($ √8 - 5 )
= 0.3956, r³.
§ 127. The properties of Guldinus are also applicable to bodies
formed by the motion of the centre of gravity of the generating
surface along any curve, as long as the surface remains at right-
angles to the curve; for every curve can be regarded as composed
of an infinite number of infinitely small arcs of circles. The vol-
ume of the body is here also equal to the product of the generating
surface and of the space described by its centre of gravity. The
properties can also be made use of, when the generating surface in
moving forwards is always at right angles to the projection of the
path of its centre of gravity upon any plane. In this case the
generating surface is to be multiplied not by the space described,
but by its projection.
128.]
245
CENTRE OF GRAVITY.
Hence, for example, the volume of one turn of the thread
B
FIG. 171.
H
E
D
K
C
A H K, Fig 171, of a screw is de-
termined by the product of its cross
section A B D E by the circum-
ference of the circle, whose radius
is the distance M S of the centre
of gravity S of the surface 4 BDE
from the axis C M of the screw.
In many cases we can combine
the use of the properties of Guldi-
nus with that of Simpson's rule.
E.G., to find the contents of the
curved embankment A, D, B₁ D₂ A,
Fig. 172, we need only know the central angles S, CS, 2 S, CS,
= 2 S, CS, B, the cross sections A, D, F, A, D, F, A₂ D₂
= =

Di
Do
B
E
A
FIG. 172.
D2

X
В 2
S
F, and the distances C S
= ro, C S₁ = r₁ and C S₂ = r₂ of the
centres of gravity So, S, and S₂ of these cross sections from the cen-
tral axis C X. The volume V of the body is determined by the
formula
V
=
r₂
B (Fo ro + 4 F₁ r₁ + F ₂
F₂ r₂)
β
0,01745 3º
6
βο
F₂ T₂)
Bº π (For。 + 4 F₁ r₁ + F₂ re
180°
6
·F。 r。 + 4 F₁ r₁ + F₂ r₂
(Foro
+ 4
1
(1
T₁ + P₂ r²).
6
If the radii r。, r, and r, are equal to each other, or if they differ
but little, we can put r。
2
1*2
and therefore
F +4 F + F
V – 0,01745 ߺ 2°
6
§ 128. The following is another application of the theory of
the centre of gravity, which is closely allied to the foregoing.
246
[§ 128.
GENERAL PRINCIPLES OF MECHANICS.
We can assume that every obliquely truncated prismatic body
AB KL, Fig. 173, is composed of infinitely thin prisms, such as
F₁ G₁. If G₁, G, etc., are the bases and
h₁ h₂, etc., the altitudes of these prismatic
elements, we have the contents
FIG. 173.
L

F
S
A
G, H, ов
1
K
G
G₁ h, G₂ ha, etc.
and consequently the volume of the
whole obliquely truncated prism
V = G₁ h₁ + G₂ h₂ +....
Now an element F of the oblique.
section K L is to the element G, of the
base A B = G as the whole oblique sur-
face Fis to the base G; hence we have
Ꮐ
G₁ == F₁, G₂ == F₂, etc., and
F
F
G
√ =
(F₁ h₁ + F₂ h₂ + ...).
F
Finally, since F₁ h₂ + F₂ h₂ + ….. is the moment Fh of the
whole oblique section, we can put
V
G
F
Fh Gh,
I.E., the volume of an obliquely truncated prism is equal to the volume
of a complete prism, which stands on the same base and whose alti-
tude is equal to the distance SO of the centre of gravity S of the
oblique section from the base.
The distance of the centre of gravity of the oblique section of a
right triangular prism, which is truncated obliquely, from the
base is
h
=
h₁ + h₂ + h₂
3
and consequently the volume of this prism is
V = G h = G
(h₁ + h₂ + h3)
3
§ 129.] EQUILIBRIUM OF BODIES RIGIDLY FASTENED.
247
CHAPTER III.
EQUILIBRIUM OF BODIES RIGIDLY FASTENED AND SUPPORTED.
§ 129. Method of Fastening.-The propositions relative to
the equilibrium of rigid systems of forces, demonstrated in the first
chapter of this section, are applicable to solid bodies subjected to
the action of forces, when we consider the weight of the body as a
force applied at the centre of gravity and acting vertically down-
wards.
Bodies, which are held in equilibrium by forces, are capable of
moving freely, I.E., they can obey the influence of the forces, or
they are in one or more points rigidly fastened, or they are sup-
ported by other bodies.
If a point C, Fig. 174, of a solid body is rigidly fastened, any
FIG. 174.

P
C
M
D
other point P of the body, when put in motion, will describe a path,
which lies upon the surface of a sphere, whose centre is the fixed
point C and whose radius is the distance CP of the other point
from C. If, on the contrary, we fasten a body in two points C
and D, the paths described by all other points in consequence of
any possible motion would be circles; for the path of each point is
the intersection O P Q of two spherical surfaces described from
the two fixed points.
The planes of these circles are parallel to each other and per-
pendicular to the straight line joining the two fixed points. The
points upon the latter line remain immovable; the body, therefore,
248
[§ 130.
GENERAL PRINCIPLES OF MECHANICS.
revolves around this line CD, which is called, for this reason, the
axis of rotation or revclution of the body.
The planes perpendicular to this axis, and in which the different
points revolve, are called the planes of rotation or revolution of the
body. We obtain the radius M P of the circle O P Q by letting
fall a perpendicular upon the axis of revolution CD. The greater
this perpendicular is, the greater is the circle, in which the point
revolves.
If three points of a body, not in the same straight line, are firmly
fastened, then the body does not move in any direction, since
the three spherical surfaces, in which the body must move, cut each
other only in a point.
130. Equilibrium of Supported Bodies.-Every force pass-
ing through the fixed point of a body, E.G., through the centre of a
ball and socket joint, is counteracted by the support of the body,
and has, therefore, no influence upon the state of equilibrium of
the body. In like manner, if a body is supported in two points or
bearings, every force whose direction cuts the axis passing through
these fixed points is counteracted by the supports, without pro-
ducing any other effect on the body. A couple would also be
counteracted by the supports of a body, if the plane of the couple.
contains the axis of revolution passing through these points, or is
parallel to the same. Every other couple (P, P), Fig. 175,
produces, on the contrary, a revolution of the body A CB about
the axis of revolution C, if it is not balanced by another couple
(see § 95 and § 97). If the couple retains its direction during the
rotation, its lever arm and consequently its moment is variable, and
both become = 0, when the body occupies a certain position. If a
B
-P
FIG. 175.
B
D
P
body A C B, Fig. 175, is rigidly fast-
ened at C, and if the direction of the
force forms the angle B A P = a with
the line AB passing through the
two points of application, a rotation
A CA₁ = ß = 180° a is necessary
to annul the moment of the couple
(P,- P); the same is also true of a
body rigidly fastened in an axis and
acted upon by a couple, whose plane
is perpendicular to this axis.
If a body A B, Fig. 176, rigidly fastened at C, is acted on by a

§ 131.)
249
EQUILIBRIUM OF BODIES RIGIDLY FASTENED.
FIG. 176.
P
A₁
force P, whose direction does not pass through C, we can, by the
addition of two opposite forces P and P, decompose this force
into a couple (P, – · P) and a force + P, applied in Cand coun-
teracted by the point of support. The rela-
tions are the same, when the axis of a body is
rigidly fastened and a force acts upon it in a
plane of revolution. Here, however, the force
+ P is divided between the two points of sup-
port. If a is the distance CA of the point of
application A of the force from the axis C and a
the angle A CA,, formed by the line CA with
the direction of the force, we have the moment
of the couple (P, P), which tends to turn the
body, M Pa sin. a. If the direction of the
force P remains unchanged during the rotation,
M changes with a and is a maximum for a = 90°
and for a = 0° or 180° it is
= = 0. The work done by the force
P or by the couple (P, P) during the rotation of the body is
A = P . K A₁ = Pa (1 cos. a).

-P
K/+P
=
131. Stability of a Suspended Body.-If the force acting
upon a body, supported at one point or in a line, consists only of
its weight, the conditions of equilibrium require, that the centre of
gravity shall be supported, I.E., that the vertical line of gravity
shall pass through the point of support.
If the centre of gravity coincides with the point of support, we
have a case of indifferent equilibrium (Fr. équilibre indifférent, Ger.
indifferentes Gleichgewicht); for the body remains in equilibrium,
FIG. 177.
B
FIG. 178.


P
B
G
P
G
B₁
N
no matter how we may turn it. If, on the contrary, the body is
250
[§ 132.
GENERAL PRINCIPLES OF MECHANICS.
rigidly fastened or supported at a point C, lying above the centre
of gravity S, the body is in stable equilibrium (Fr. stable, Ger. sich-
eres or stabiles); for, if we bring the body into another position, one
of the components N of the weight S causes the body to return to
its original position, and the other component P is counteracted by
the fixed point C. If finally the body A B, Fig. 178, is fastened
at a point C, which lies below the centre of gravity, the body
is in unstable equilibrium (Fr. éq. instable, Ger. unsicheres or
labiles Gleichgewicht); for if we move the centre of gravity out of
the vertical line passing through C, the weight G is resolved into
two components, one N of which, instead of tending to bring the
body back to its original position, moves it more and more from it,
until the centre of gravity comes vertically below the point of
support.
The circumstances are the same, when a body is supported in
two points or in an axis; it is either in indifferent, stable or unstable
equilibrium as the centre of gravity coincides with, or is vertically
below or above the point of support. If a body is supported at a
point or in a horizontal axis, the moment with which the body seeks
to return to its position of stable equilibrium is M = G a sin. a,
in which formula G denotes the weight, a the distance CS, of the
centre of gravity S, from the axis C' and a the angle of revolution
SCS. The work done is A = G a (1 cos. a).
§ 132. Pressure upon the Points of Support of a Body.
-When a body C A D, Fig. 179, supported in two points C and
FIG. 179.

Z
P
+P
$2
-Y
E
A
D
-N
N
S1
+N
X
N
Y
P
D, is acted upon by a system of forces, in order to determine the
conditions of its equilibrium we refer (according to § 97) the
§
251
139.]
EQUILIBRIUM OF BODIES RIGIDLY FASTENED.
whole system to two forces, the direction of one of which is parallel
to the axis, while that of the other lies in a plane normal to this
line. Let E N = N, Fig. 180, be the force parallel to the axis XX
passing through the points of support C and D and AP = P the
other force, whose direction lies in a plane Y Z Y perpendicular to
XY. We can resolve the first force into a force + N, tending to
displace the axis in its own direction, and a couple (N,N),
which is transmitted to the points of support in the shape of an-
other couple (N₁, — N₁), the components of which are
d
N₁
N and N₁ =
d
N,
d denoting the distance O E of the parallel force N from the axis
CD and the distance CD of the two points of support from
each other.
FIG. 180.
Z

Р.
S
+P
P₂
-Y
E
-X
D
-N
I
N
S
+N
X
N
In like manner we decompose the force P into a force + P and
a couple (P, P), and the former again into its components P₁
and P, the first applied in Cand the second in D. Designating
the distances C O and D O of the points of application O from the
two points of support and D by 1, and l, we have
1
P.
7
P and P₁ = ½½ P,
and it is now easy, by employing the parallelogram of forces, to find
the resultant S, of the forces N, and P₁ at C, and also the resultant
S, of the forces N, and P, at D.
If we put the angle 10 (+ P) formed by the plane N O F
with the direction of the force P or + P =a, we have also the
252
IS 133.
GENERAL PRINCIPLES OF MECHANICS
1
2
angle N, CP₁ = a and Ñ, D P₂ = 180° a, and consequently the
N₁ C'
resulting pressures in C and D are
and
S₁ √ N₁² + P₁² + 2 N, P, cos. a
1
1
£₁ = √ N² + P-2 N, P, cos, a.
S
P₂
If, finally, a denotes the perpendicular O L to the direction of
the force, the moment of the couple (P, P), which tends to turn.
the body, is M = P a. If the body is in a state of equilibrium, a
must naturally be 0, and therefore P must pass through the
axis CD.
EXAMPLE.—Let the entire system of forces acting on a body rigidly
supported in the axis X X be reduced to the normal force P = 36 pounds,
and the parallel force N = 20 pounds; let the distance of the latter force
from the axis be OE d 1 feet, and the distance CD between the
two points of support be l = 4 feet; required the pressure upon the axis
or on the fixed points C and D supposing that the direction of the force P
forms an angle a = 65° with the plane X Y, and that its point of applica-
tion O is at a distance CO 71 1 foot from the point C.
N
1
The force N = 20 produces in the axis in its own direction a thrust
= 20 pounds and also the forces
which are counteracted by the supports C and D.
N₁
1
d
Z
N
1,5
4
20 = 7,5 pounds and
N₁
1
7,5 pounds,
The force P gives rise to
the forces
P =
4.
4
1
P = 1.36 9 pounds.
P₁
36 = 27 pounds and P₂ = '1
P2
Combining the latter with the former force, we obtain the resultants
S₁ = √ 7,5² + 27" + 2.7,5. 27. cos. 65° √56,25 + 729 + 171,160
✓ 956,410
S2
√7,5 + 92
✓ 80,196
30,926 pounds, and
2.7,5.9. cos. 65° = √ 56,25 + 81 57,054
8,955 pounds.
§ 133. If a body CBD, Fig. 181, firmly supported in two
points C and D, is acted upon by a single force R, whose direction
forms an angle PARẞ with the plane of rotation Y O Z, we
can decompose this force into the components
A P = P = R cos. ẞ and
AN N = R sin. B,
the first of which acts in the plane of rotation and the second
parallel to the axis, and we can treat these forces in exactly the
same manner as the resultants P and N of the system of forces in
§ 133.]
253
EQUILIBRIUM OF BODIES RIGIDLY FASTENED
the last paragraph. Here the force which the axis must counter-
act in its own direction is N = R sin. ẞ, and the components of
FIG. 181.

Z
+P
P
R
S
-Y
S₂
R
-X
-N
B
+N
-N₁
C
Y
the couple (N₁, -- N₁), which act in Cand D in opposite directions
and at right angles to C' D, are
d
d
d
N₁
N =
R sin. ẞ and
N
77
R sin. ẞ,
7 denoting the distance C D of the two points of support C and D
from each other and d the distance O A of the point of application
A of the force R from the point O on the axis.
In like manner the force acting in O at right angles to CD is
+ P = R cos. ẞ and its components in Care
P₁
P
R cos. ẞ, and in D
7.
P₁ = }} P
R cos. B,
ī
l, and l, again denoting the distances CO and D O of the points C
and D from the plane of rotation Y Z Y.
Substituting the values of N, P₁, and P, in the formulas
√ N₁₂² + P²² + 2 N, P, cos. a
S₁
S₂
√ N²² + P¿ − 2 N, P₂ cos. a
for the normal pressures in C and D, in which we designate by a
the angle YA P formed by the component P with the plane
A CD, we obtain
254
[§ 134.
GENERAL PRINCIPLES OF MECHANICS.
S₁
11
R
√ (d sin. B)² + (l½ cos. ß)² + 2 d l, sin. ß cos. ß cos. a
Ꭱ
S₂
√ (d sın. B)² + (l, cos. ẞ)² — 2 d l, sin. ẞ cos. ß cos. a
The moment of the remaining couple (P, P) is
P.OB
FIG. 182.
C
-
Pa Rd sin. a cos. B.
These formulas are applicable to the discussion of the stability
of a body O A, Fig. 182, revolving about an inclined axis CD. R
is here the weight G of the body,
d the distance O S O S₁ of its
centre of gravity from the axis of
rotation, a the angle S O S₁ = 0 S, L,
which the centre of gravity has de-
scribed in turning from its position
of equilibrium S in the plane Y SY
perpendicular to CD, and ẞ the angle
G S P formed by the plane of revo-
lution with the vertical line, or that
formed by the axis of revolution CD
with the horizontal line D H.

D
-Y
N
K
P
H
A
The work done, when the body is
brought back by its weight to its position of equilibrium and
S₁ to S, is
A = G. KS cos. ß = G d cos. ß (1 - cos. a).
§ 134. Equilibrium of Forces around an Axis.--The re-
suitant P is produced by all the component forces, whose directions
lie in one or more planes normal to the axis. But in this case
(according to § 89) the statical moment Pa is equal to the sum
P₁ α₁ + P₂ α + of the statical moments of the components,
and, when the forces are in equilibrium, the arm a is 0; for this
force then passes through the axis itself, and consequently this sum
2
• •
P₁ α, + P₂ ɑ? + = 0;
2
I.E., a body rigidly supported in an axis is in equilibrium, and
therefore remains without turning, when the sum of the statical
moments of all the forces in relation to this axis is
0, or when
the sum of the moments of the forces acting in one direction of
§ 135.]
255
EQUILIBRIUM OF BODIES RIGIDLY FASTENED.
rotation is equal to the sum of the moments of those acting in the
other.
By the aid of the last formula any element of a balanced sys-
tem of forces, such as a force or an arm, can be found, and any
force of rotation reduced from one arm to another.
If we wish to produce a state of equilibrium in a body movable
about its axis, and whose moment of rotation is P a, we have only
to apply a force of rotation Q or a couple, the moment of which
Q b P a, the difference in the two cases being that by the addi-
tion of the couple (2, Q) the pressure on the axis is not changed,
while by that of a force Q a force + Q is added to the pressure on
the axis. If the force Q or its lever arm b is given, we can calcu-
late either
b
Pa
Q
or Q
Pa
b
In the latter case we call Q the force P reduced from the arm a
to the arm b, and we can thus reduce the given force of rotation P
to any arbitrary arm, or we can replace or balance it by another
force acting with any arbitrary arm.
We can also, by means of the formula
1
P₁ α₁ + P₂ α2 +
Q
b
reduce a whole system of forces to one and the same arm.
1
EXAMPLE. The forces P₁ 50 pounds and P₂ 35 pounds act
on a body movable about an axis with the arms a₁ = 11 feet and a₂ =
23 feet; required the force P which must act with an arm ag 4 feet, in
order to produce equilibrium or to prevent motion about the axis.
have
3
We
50. 1,25
P 3
87,5
4
35. 2,5 + 4 P3
62,5
=
0, and
= 6,25 pounds.
§ 135. The Lever.-A body movable about a fixed axis and
acted on by forces is called a lever (Fr. levier, Ger. Hebel). If we
imagine it imponderable, we have a mathematical lever; but if not,
it is a material lever.
We generally assume the forces of a lever to act in a plane at
right angles to the axis and substitute for the axis a fixed point
called the fulcrum (Fr. point d'appui, Ger. Ruhe, Dreh, or Stütz-
punkt). The perpendiculars let fall from this point upon the di-
rection of the forces are called (§ 89) the arms of the lever. If the
directions of the forces of a lever are parallel, the arms of the lever
256
[§ 136.
GENERAL PRINCIPLES OF MECHANICS.
form a single right line, and the lever is then called a straight
lever (Fr. levier droit, Ger. geradliniger or gerader Hebel). The
straight lever acted on by two forces only is one or two armed, ac-
cording as the points of application of the forces lie upon the same
or upon opposite sides of the fulcrum. We distinguish also levers
of the first, second and third sort, calling the two-armed lever a
lever of the first sort, the one-armed lever a lever of the second
sort or of the third sort, according as the force (load), which acts
vertically downwards, or that (power), which acts vertically up-
wards, is nearest the fulcrum.
§ 136. The theory of the equilibrium of the lever has been
completely demonstrated in what precedes, and we have only to
make special applications of it.
For the two-armed lever A C B, Fig. 183, when the arm CA
of the force P is denoted by a and that CB of the other force Q,
which is generally called the load, by b, we have, according to the
general theory PaQb, L.E. the moment of the force is equal to
FIG. 183.
FIG. 184,

Αθ
B
Р

P
R
AC
B
R
the moment of the load, or also P : Q = b : a, I.E. the force is to the
load as the arm of the latter is to the arm of the former. The
pressure on the fulcrum is R = P + Q.
For the one-armed lever A B C, Fig. 184 and BA C, Fig. 185,
the relations between force (P) and load (Q) are the same, but the
direction of the power is opposite to that of the load, and therefore
the pressure on the fulcrum is equal to the difference of the two; in
the first case we have
R = Q – P, and in the second R = P — Q.
§ 136.]
257
EQUILIBRIUM OF BODIES RIGIDLY FASTENED.
If in the bent lever A CB the arms are CN a and CO
=b, Fig. 186, we have again P : Q = b : a, but in this case the
FIG. 185.
P
FIG. 186.
D


A
B
R
N
B
P
P₁
21
1
R
pressure R on the fulcrum is the diagonal R of the parallelogram
C P₁ R Q₁, constructed with the force P, the load Q and with the
angle P₁ C Q₁ = PDQ a formed by their directions with each
other.
=
=
If G is the weight of the lever and CE e, Fig. 187, the dis-
tance of the fulcrum from the vertical line S G passing through
the centre of gravity S of the lever, we must put Pa Ge Q b,
and we must employ the plus sign of G, when the centre of gravity
lies on the same side as the force P, and the minus sign, when
upon that of the load Q.
The theory of the lever is often applicable to tools and ma-
FIG. 187.
FIG. 188.

--
1
E
A
B
N
S
E
Р
G
B

DL
P
chinery. The knee lever A B C D, Fig. 188, which is sometimes
cited as a peculiar sort of lever, is simply a bent lever. The arm,
which is movable around an axis C, is acted upon by a force at its
17
258
[S 136.
GENERAL PRINCIPLES OF MECHANICS.
end A, and acts by means of a rod B D, (which forms with the arm
an acute angle A B D C B E = a) upon the load, which is ap-
plied at D. If a denotes the length of the arm CA and b the
length of the arm C B, we have the lever arm of Q
CE =b sin. a, whence
Pa = Q b sin. a, or
b
P =
Q sin. a, and inversely
a
α
P
Q
b sin. a
This lever is employed for pressing together materials. The
a
pressure increases directly with P and and inversely as sin. a. By
ō'
diminishing the angle a this force Q can be arbitrarily increased.
EXAMPLE—1) If the end A of a crowbar A CB, Fig. 189, be pressed
down with a force P of 60 pounds, and if the arm CA of the power is 12
A
FIG. 189.
P
P₁
A
FIG. 190.
S C
VQ
B
P₂
times as great as the arm C B
of the load, then the latter, or
rather the force Q developed in
B, is 12 times as great as P, and
we have
Q = 12.60 = 720 pounds.
2) If a load Q, Fig. 190, hang-
ing from a bar, be carried by
two workmen, one of whom
takes hold at A and the other
at B, we can determine how
much weight each has to sus-
tain. Let the load be Q = 120
pounds, the weight of the
rod be G 12 pounds, the
distance A B of the two work-
men from each other be 6
feet, the distance of the load
from one of them B be B C
23 feet and the distance of the
centre of gravity of the bar S
from the same point be B S
31 feet. If we regard B as the
fulcrum, the force P, at A must
balance the load Q and G, and
therefore we have


1
$137.] EQUILIBRIUM OF BODIES RIGIDLY FASTENED.
259
P₁.BAQ.BC+ G. BS, L.E.,
1
6 P₁
2,5 . 120 + 3,5 . 12 = 300 + 42 =
342,
1
and therefore
342
P₁
= 57 pounds.
6
If, on the contrary, A be regarded as the fulcrum, we can put
Q. A C + G. A S, or in numbers
450,
3,5 . 120 + 2,5 . 12 = 420 + 30 =
P₂ · A B =
6 P2
=
and the force exerted of the second workman is
P &
450
6
75 pounds.
The sum of the forces, which act upwards, is therefore correctly
P₁ + P₂ 57 + 75 = 132 pounds,
or as great as the sum of those acting downwards
Q + G = 120 + 12 = 132 pounds.
3) The load upon a bent lever A CB, Fig. 191, weighing 150 pounds,
acts vertically downwards and is Q = 650 pounds, and its arm C B = 4
feet, and, on the contrary, the arm of the force
P, CA
6 feet and that of the weight CE 1 foot:
required the force P necessary to produce equili-
brium and the pressure R on the bearings. We have
CA.P C B. Q + CE. G, I.E.,
FIG. 191.

P←
A
S
G
B
E
6 P = 4. 650 + 1. 150 = 2750,
and consequently
P
2750
6
= 458 pounds.
The pressure on the bearings is composed of the
vertical force Q + G = 650 + 150 = 800 pounds,
and of the horizontal force P = 4583 pounds, and
consequently we have
R = √ (Q + G)² + P²
= √ (800)² + (458})²
= √ 850070 = 922 pounds.
§ 137. More than two forces P and Q may act on a lever; it
also is not necessary that these forces act upon the lever in one and
the same plane of rotation. If Q1, Q2, Q3 are the loads on a lever
A CB, Fig. 192, and b₁, b, by their lever arms CB, CB, C B3
while the power acts with the lever arm CA = a, we have
P, a = Q₁ b₁ + Qq ba + Qs b 3 ;
3
and if the lever is straight, the pressure on the fulcrum is
R = P + Q₁ + Q² + Qs.
If the several forces of a lever act in different planes of rotation
260
[§ 137.
GENERAL PRINCIPLES OF MECHANICS.
upon the lever A C D B, B2, Fig. 193, the formula for the moment
Pa = Q₁ b₁ + Q₂ b₂ + ... does not therefore change, but a differ-
ent distribution of the total pressure R = P + Q₁ + Q2 + Qs
FIG. 192.
2
1
FIG. 193.

Р
C B₁ Bg
B3
m
R
Q₂
Q.
De be
B₁
141

upon the axis takes place between the two points of support or
bearings and D. If we denote by the length of the axis
CD of the lever or the distance of the fulcrums from each other
and by l。, 11, 12, ... the distances C O, C O₁, C O₂ of the planes of
revolution from the fulcrum C, the pressures R, and R, on the
bearings at D and Care determined by the following formulas
Plo + Q₁ l₁ + Q₂ 12 + ...
R₂ =
R₁
7
and
P (ll) + Q₁ (1 − (1) + Q₂ (1 — 7₂)
Q2 12)
1
R - R₂
If the forces acting upon a bent lever are not parallel, the ex-
pression Pa = Q₁ b₁ + Q₂ ba+... remains unchanged, but the
Plo Qh Qala2
pressures in the axis reduced to the fulcrum, E.G.,
' 7' 7
act
in different directions and cannot, therefore, be combined by simple
addition, but, on the contrary, we must combine them in the same
manner as several forces applied to a point and acting in the same
plane (see §§ 79 and 80).
1
2
2 2
=
1
EXAMPLE. The lever represented in Fig. 193 supports the loads Q₁
300 pounds and Q2 480, acting at the distances C 0₁ = 1₁ = 12 inches
and C 02
24 inches from the bearing with the arms 0₁ B₁
1
16 inches and 0₂ B₂ = b₂ 10 inches; required the force P, which,
acting with the arm 0 A = a = 60 inches, is necessary to produce equili-
brium, and the pressure on the bearings at C and D, under the assumption,
that the force acts at a distance CO
18 inches from the journal C.
and that the length of the entire axis is C D
The force required is
Q ₁ b₁ + Q2 b 2
1
a
300.16 +480.10
60
0
732 inches.
30.16 + 480
80 80 160
6
P =
pounds, and the pressures on the bearings are
Co
261
138.] EQUILIBRIUM OF BODIES RIGIDLY FASTENED.
R₂
1
160.18 + 300.12 + 480.24
32
=562,5 pounds and
=
562,5
= 300 + 480 + 160
―
R₁ R - R₂
= R
377,5 pounds.
REMARK.—The action of gravity on the lever can be employed with
advantage to determine the centre of gravity S and the weight G of a
B
G
FIG. 194.
P
A
1
1
body A B, Fig. 194. We support the body
first at a point C and then at a point C, at a
distance CC₁ = d from the former, and each
time we bring the body into equilibrium by a
force acting at the distances CA a and
С₁ A = a₁ = a
If the force necessary

1
а
1
in the first case be
=
d.
P and in the second case
P₁, and if the weight of the body be G and
the distance of its centre of gravity S from A be A B = x, we have
Pa = G (x − a) and P₁ α₁ G (x — a₁), whence
G
(P P₁) a α1 and
Pa - P₁ a₁
Ра - Ра1
Pa
a
a1
1
=
§ 138. Pressure of Bodies upon one another.-The law
deduced from experiment and announced in § 65: Action and
reaction are equal to each other," is the basis of the whole mechan-
ics of machines, and we must here explain at greater length its
meaning. If two bodies M and M, Fig. 195, act upon each other
P
-X
M
N₁
FIG. 195.
S
P
Y
with the forces P and P₁, the directions
of which do not coincide with that of the
common normal to the two surfaces
of contact, a decomposition of the forces.
always occurs; only that force N or N₁,
whose direction is that of the normal, is
transmitted from one body to the other,
the other component force S or S₁, on
the contrary, remains in the body and
must be counteracted by some other force
or obstacle, when the bodies are to be
held in equilibrium. But according to
the principle announced, the two normal.
components N and N must be exactly
equal. If the direction of the force P
forms an angle N A P = a with the normal A X and an angle
SAP ẞ with the direction of the other component S, we
have (see § 78).

S,
A
N
X
262
[$ 139.
GENERAL PRINCIPLES OF MECHANICS.
N =
P sin. B
S
sin. (a + B)'
P sin. a
sin. (a + B)*
Designating in like manner N, A, P, by a, and S, A, P, by B₁
we have also
N₁ =
P sin. B₁
sin. (a, + B₁)
and S₁
P₁ sin. a,
sin. (a₁ + B₁)
and, finally, since N = N₁
P sin. ẞ
P₁ sin. B₁
sin. (a + B)
sin. (a, + B₁)*
EXAMPLE.-How are the forces decomposed, when a body M₁, Fig. 196,
FIG. 196.
-X
S
P
M
D
S₁
M
E
X
P
held fast by an impediment D E, is pressed
upon by another body M, movable about
its axis C, with a force P 250 pounds?
The angles formed by the directions are
the following:

PAN= a = 35°
PAS B = 48°
P₁ A₁ N₁
1
a1
= 65°
P₁ A₁ S₁ = ẞ₁ = 50°.
1
1
The normal pressure between the two
bodies is determined by the first formula
and is
N
= N₁
1
P sin. B
sin. (a + ß)
250 sin. 48°
sin. 83°
187,18 pounds;
from the second we have the pressure on the axis or bearing C
S
P sin. a
sin. (a + p)
250 sin. 35°
=
144,47 pounds;
sin. 830
and, finally, by combining the third and fourth formula we obtain the
component which presses against the impediment DE
1
S₁
N₁ sin. a₁
sin. B₁
187,18 sin. 65"
sin. 500
=221,46 pounds.
§ 139. In consequence of the equality of action and reaction,
the equilibrium of a supported body is not changed, when, instead
of the support, we substitute a force, which counteracts the pressure
or tension transmitted to the support, and which is, therefore,
equal in magnitude and opposite in direction to it. After having
introduced this force, any body supported or partially retained
may be considered as entirely free, and consequently its state of
equilibrium can be treated in the same manner as that of a free
body or of a rigid system of forces.
140]
263
EQUILIBRIUM OF BODIES RIGIDLY FASTENED.
If, E.G., a hody M, Fig. 197, is movable around its axis C, the
force N is transmitted to a second body M₁, the force S' is counter-
acted by the axis C and we can assume, that the body is entirely
free and that besides P two other forces N and S act upon it.
If the body M₁ presses upon M with the force N, and against the
fixed plane D E with the force S₁, the equilibrium would not be
disturbed, if instead of these impediments we should substitute two
opposite forces - N₁ and S, and combine the same with the
forces (E.G. with P₁), which act upon the body. In a state of
equilibrium the resultant of the forces in the one as well as that
-S
Ꮲ
FIG. 197.
N
-S
FIG. 198.

-S₂

Ο
P
N
N and
in the other body must be null, and therefore the resultant of
S must be counteracted by P and the resultant of
S₁ by P₁.
N₁ and
Since the forces N and N₁, with which the two bodies act upon
each other, are in equilibrium, the forces P, S, P, and - S
must be in equilibrium, when the combination of the two bodies.
(M, M₁) is in equilibrium. The forces N, N, are called the interior
and the forces P, S, P, and S, the exterior or extraneous
— P₁
forces of the combination of bodies or of the system of forces, and
we can therefore assert that not only the interior forces are in equi-
librium, but that the exterior forces are so also, when, as is repre-
-sented in Fig. 198, we suppose the forces applied in any point 0.
§ 140. Stability.-When a body supported upon a horizontal
plane is acted on by no other force than that of gravity, it has no
tendency to move forwards; for its weight, acting vertically down-
wards, is completely counteracted by this plane, but a rotation of
264
[$ 140.
GENERAL PRINCIPLES OF MECHANICS.
the body may be produced. If the body A D B F, Fig. 199, rests
with the point D on the horizontal plane H R, it will remain at
FIG. 199.
K
A
S.
G
L
B
H
E
R
rest as long as its centre of gravity Sis
supported, I.E., as long as it lies in the
vertical line (vertical line of gravity),
passing through the point of support D.
But if a body is supported in two points
upon the horizontal surface of another
body, the conditions of equilibrium
require, that the vertical line of gravity
shall pass through the line joining the
two points of support. If, finally, a body
rests upon three or more points on a horizontal plane, equilibrium
exists, when the vertical line of gravity passes through the triangle
or polygon formed by joining these points by straight lines.
We must also distinguish for supported bodies, stable and un-

FIG. 200.
E
B
stable equilibrium. The weight G of a
body A B, Fig. 200, draws the centre
of gravity S of the same downwards;
if there is no obstacle to the action
of this force, it produces a rotation of
the body, which continues until the
centre of gravity has assumed its lowest
position and the body has assumed a
state of equilibrium. We can assert
that the equilibrium is stable, when the
centre of gravity occupies its lowest position (Fig. 201), that it is
unstable, when it occupies its highest position (Fig. 202), and that

H
FIG. 201.
C
C
R
FIG. 202.
B
FIG. 203.



S
A
B
D
H
R
E
H
H
S
finally the equilibrium is indifferent, when the centre of gravity re-
mains at the same height, no matter what may be the position of
the body (Fig. 203).
§ 141.]
265
EQUILIBRIUM OF BODIES RIGIDLY FASTENED.
FIG. 204.
K
E
EXAMPLES-1) The homogeneous body A D B F, Fig. 204, composed
of a hemisphere and a cylinder, rests upon a horizontal plane H R. Re-
quired the height S F h of the cylindri-
cal portion in order that this body shall be
in equilibrium. Any radius of a sphere is
perpendicular to the tangent plane corre-
sponding to it, but the horizontal plane is
such a plane, and consequently the radius
SD must be perpendicular to it and contain
the centre of gravity. The axis FSL
passing through the centre of the sphere is
also a line of gravity; the centre S, as inter-
section of the two lines of gravity, is therefore the centre of gravity of the
body. If we put the radius of the sphere and of the cylinder SA
SB = SL r, and the altitude of the cylinder S F = BE = h, we have
for the volume of the hemisphere V₁ = ³, and for the volume of the
cylinder V = r²h, for the distance of the centre of gravity of the sphere

G
L
B
H
S₁, S S₁
S2, S S2
=
2
R
1
r and for that of the centre of gravity of the cylinder
h. In order that the centre of gravity of the whole body fall
in 8 we must make the moment of the hemisphere r³. r equal to the
moment of the cylinder 2 h.h, whence we have
h²
r² or h r √3 = 0,7071 r.
If the body is not homogeneous, but on the contrary the hemispherical
portion has the specific gravity e, and the cylindrical portion the specific
gravity, then the moments of these portions are ². εr and
π r² h ɛ, . 1 h, and consequently by equating them we have
E
1
2 & 2
h²
= ε₁ r³, or h = "V
0,7071
282
П дой 1
1
- . r.
£2
2) The pressure, which each of three legs A, B, C, Fig. 205, of an arbi-
FIG. 205.
trarily loaded table has to bear, can be
determined in the following manner.
Let S be the centre of gravity of the
loaded table, and SE, C D perpendicu-
lars upon A B. Designating the weight
of the entire table by G and the pres-
sure in by R, we can treat A B as an
axis and put the moment of R = the mo-
ment of G, I.E., R. C D = G.SE,
from which we obtain

R =
SE
Ꮯ Ꭰ
AABS
G =
G;
•
▲ A B C
and in like manner for the pressure in B, we have
▲ ACS
G, and for that in 4
Q:
AACB
A B CS
P
G.
A ABC
266
[§ 141, 142.
GENERAL PRINCIPLES OF MECHANICS.
§ 141. Let us now investigate more fully the case of a body
resting with one base upon a horizontal plane. Such a body pos-
sesses stability or is in stable equilibrium, when its centre of gravity
is supported, I.E. when the vertical line passing through its centre
of gravity passes also through its base, since in this case the rota-
tion, which the weight of the body tends to produce, is prevented
by the resistance of the body. If the vertical line passes through
the periphery of the base, the body is in unstable equilibrium; and
if it passes outside of the base, the body is not in equilibrium, but
will rotate around one of the sides of the periphery of its base and
be overturned. The triangular prism A B C, Fig. 206, is conse-
quently in stable equilibrium, since the vertical line S G passes
through a point Nof its base B C. The parallelopipedon A B CD,
Fig. 207, is in unstable equilibrium, because the vertical line S G
passes through one of the edges D of the base CD. Finally, the
cylinder A B CD, Fig. 208, is without stability; for S G does not
pass through its base CD.
FIG. 206.
Λ
FIG. 207.
A
B


B
D
G
G
Stability (Fr. stabilité, Ger. Stabilität or Standfähigkeit) is the
A
FIG. 208.
B
NI D
capacity of a body to maintain by
its weight alone its position and
to resist any cause of rotation. If
we wish to select a measure for the
stability of a body, it is necessary
to distinguish the case of simply
moving the body from that of
actually overturning it. Let us
first consider the former case
alone.
§ 142. Formulas for Stability-A force P whose direction
is not vertical tends not only to overturn, but also to push forward
the body A B C D, Fig. 209. Let us suppose that there is an

$143.]
267
EQUILIBRIUM OF BODIES RIGIDLY FASTENED.
=
obstacle to its pushing or pulling the body forwards, and let us
consider only the rotation around an edge C. If from this edge
we let fall a perpendicular CE a upon the direction of the
force and another perpendicular
CN = e upon the vertical line of
gravity SG of the body, we have
then a bent lever E C N, to which
FIG. 209.

A
B
S
Մ
D
N
G
F
Р
the formula Pa Ge or P
=
С
a
Ꮐ
is applicable. If, therefore, the ex-
terior force P is slightly greater than
Ge
a
, the body begins to turn around
Cand thus loses its stability. Its stability is therefore dependent
upon the product (G e) of the weight of the body and the smallest
distance of a side of the periphery of the base from the vertical line
passing through the centre of gravity, and Ge can therefore be
considered as a measure of stability, and we will henceforth call it
simply the stability. Hence we see that the stability increases
equally with the weight G and with the distance e, and conse-
quently we can conclude that under the same circumstances a wall,
etc., whose weight is two or three tons, does not possess any more
stability than one, whose weight is one ton and in which the dis-
tance or arm of the lever c is two or three fold.
§ 143. 1) The weight of a parallelopipedon A B C D, Fig. 210,
whose length is 1, whose breadth is A B = CDb and whose
height is AD BC h, is G Vybhly, and its stability
= =
1
St = G. D N = G . ! C' D =
Gb
2
= { b² h l y,
ý denoting the heaviness of the material of the parallelopipedon.
FIG. 210.
A
B
DI
G
E
F
FIG. 211.
A
B


Gi
G
2) The stabilities of a body B D E, Fig. 211, composed of two
268
[§ 143.
GENERAL PRINCIPLES OF MECHANICS.
parallelopipedons, in reference to the two edges of the base Cand
F, are different from each other. If the heights are B C and E F
hand h, and the widths CD and D F = b and b₁, we have the
weights G and G, of the two portions
arms in reference to Care CN
b,
=
bhly and b, h, ly; the
band C' O = b + 1 b₁, and
those in reference to Fare b₁ + b and b₁, and the stability is,
first, for a rotation around C
1
St = 1 Gb + G₁ (b + √ b₁), = ( ! b* h + b b₁ h₁ + 1 b₁₂² h₁) ly,
and, secondly, for a rotation about F
1
St₁ = G (b₁ + b) + 1 G₁ b₁ = ( ); b,² h₁ + b b₁ h + 1 b² h) ly.
FIG. 212.
L
A
B
1
The latter stability is St, St = (h― h) b b, ly greater than
the former. If we wish to increase the stability of a wall AC by
offsets D E, we must put them upon the side of the wall, towards
which the force of rotation (wind, water, pressure of earth, etc.)
acts. The stability of a wall A B C E, Fig. 212, which is battered
on one side, is determined as follows. Let
the length of the wall be 7, the width on
top A Bb, the height B Ch and the
batter = n, I.E. when the height A K =
1 foot the batter K L =n, or for a height
h feet, nh. The weight of the parallel-
opipedon AC is Gbhly, that of the
triangular prism A D E G₁ = in h.
hly; the arms for a rotation about E are
ENED + b b = n h + b and E O
= { E D = 3 n h. Hence the stability is
G₁ n h = (! b² + n hb + nº h²) h ly.
A parallelopipedical wall of the same volume is b + in h wide,
and its stability is

E
G1
G
St = G (n h + 1 b) + 3
1
St₁ = ! (b + & nh)' h l y =
the stability is therefore St
1
1
(! b² + 1 n
h b + { n² h² ) hly ;
5
St, = (b + √z n h). i n h² l y
x
smaller than that of a battering wall.
The stability of a wall with a batter on the other side is
I
St₂ = (b² + n h b + n² h²). jhly,
3
and consequently smaller than St by an amount
St St₂ =
1
(b + { n h) . i n h³ ly,
3
2
but greater by an amount St, St₁ = 's n² 737y than the sta
bility of a parallelopipedical wall of the same volume.
A –
EXAMPLE. What is the stability per running foot of a stone wall 10
feet high, 14 feet wide on top and with a batter of † of a foot on its back?
The density of this wall can be put (§ 61) = 2,4, consequently its heaviness
§
269
144.]
EQUILIBRIUM OF BODIES RIGIDLY FASTENED.
is y =
=
=
62,4. 2,4 149,76 pounds; but we have 7 1, h =
and n = 0,2, and consequently the required stability is
St =
=
10, b = 1,25
[½ . (1,25)² + 0,2. 1,25 . 10 + (0,2). 10-] 10. 1. 149,76
(0,78125 +2,5+1,3333) 1497,6
4,6146. 1497,6 = 6911 foot-pounds.
If the same quantity of materials is used, under the same circumstances
the stability of a parallelopipedical wall would be
Sti
St 2
[½. (1,25)² + . 0,2. 1,25. 10 +
(0,2)². 102]. 149,76. 10
= (0,78125 + 1,25 + 0,5) 1497,6 = 2,531. 1497,6 = 3790 foot-pounds.
The stability of the same wall with a batter on its front would be
=
[½ (1,25)² + ½ . 0,2 . 1,25 . 10 + † (0.2)² . 10²] 149,76 . 10
(0,78125 +1,25+0,666) 1497,62,6979. 1497,6 4040 foot-pounds.
REMARK.-We see from the above that we economize material by bat-
tering the wall, by furnishing it with counterforts or offsets, by building
it on plinths, etc. This subject will be treated more in detail in the second
volume, where the pressure of earth, arches, bridges, etc., will be con-
sidered.
§ 144. Dynamical Stability.-We must distinguish from
the measure of stability given in the last paragraph another meas-
ure of the stability of a body, in which we bring into consideration
the mechanical effect necessary to overturn the body. The work
done is equal to the product of the force and the space; the force
in a heavy body is its weight, and the space is the vertical pro-
jection of the space described by the centre of gravity, and, con-
sequently, in the latter sense the product G s can be employed as
the measure of the stability of a body, when s is the vertical height,
which the centre of gravity of the body must rise, in order to bring
the body from its state of stable into one of unstable equilibrium.
Let C be the axis of rotation and S the centre of gravity of a
A
D
FIG. 213.
B
S
1
N
B
body A B CD, Fig. 213, whose dy-
namical stability is to be deter-
mined. If we cause the body to
rotate, so that its centre of gravity
S comes to S₁, I.E. vertically above
C, the body is in unstable equili-
brium; for if it is caused to revolve
a little more, it will tumble over.
If we draw the horizontal line
SN, it will cut off the height
N S₁ = 8, which the centre of gravity
has ascended, by the aid of which we obtain the dynamical sta-
bility G s. If now we have CS C S₁ = r, C M NS = e
and the altitude C N = M S = a, we obtain

270
[S 144.
GENERAL PRINCÍPLES OF MECHANICS.
S₁ N = s = r a =
Na + e²
α,
and the stability in the second sense is
St = G (√ a² + es
a).
a gives, for a
(
= 0, s = e, for a =
√ n² + 1 − n) e, ap-
0.414 e, for a = n e, s =
The factor s = √ a² + e²
s = e ( 12
1)
proximatively = (n
and for a = ∞, 8 =
1
+
n) e
2n
e
∞
е
e = thus
2n'
for a = 10 e, s =
е
20
= 0; this stability, therefore, becomes greater
and greater as the centre of gravity becomes lower and lower, and
it approaches more and more to zero as the centre of gravity is
elevated more and more above the base. Sleds, wagons, ships etc.
should therefore be loaded in such a manner, that the centre of
gravity shall lie not only as low as possible, but also as near as
possible above the centre of the base.
If the body is a prism with a symmetrical trapezoidal section,
such as is represented in Fig. 213, and if the dimensions are the
following length, height M Oh, lower breadth C D = b₁,
upper breadth A Bb, we have
b₁ + 2 b h
1
b₁ + b₂ '3
MS = a =
C M = e =
11 619
C S = r =
whence
2
+
(§ 110) and
(b₁ + 2 b₂ h
b₁ + b x 3/
2
and the dynamical stability or the mechanical effect necessary to
overturn this body is

+ 2 b₂
h
2
b₁ + 2 by h
= G [√ (2₂ )² + (b + + b²
St G
=
3/
b₁ + b₂ 3
2
EXAMPLE.-What is the stability of, or what is the mechanical effect
necessary to overturn, the granite obelisk A B C D,
FIG. 214.
A
Fig. 214, when its height is h
and breadth 7, 14 and b
length and breadth 72 4 feet
volume of this body is
30 feet, its upper length

1 foot and its lower
1
2
and b₂ = 3 feet? The
V = (2 b b + 20 lg tôi lg +
Ե, Ն4 b + b b
2 2
1
= (2. 3. 1 + 2 . 4 . 3 + 1 4 +
40,25 . 5 = 201,25 cubic feet.
h
l
})
2 6
30
. 7) 20
3. 62,4
= 187.2
If a cubic foot of granite weighs y
pounds, we have for the total weight of the body
G
201,25. 187,2 = 37674.
The height of its centre of gravity above the base is
§ 145.]
271
EQUILIBRIUM OF BODIES RIGIDLY FASTENED.
2
1
2
b₂ l2 + 3 b₁
1 1
l₁ + b₂ l₁ + b₁ l½
h
a =
2
4+음​. 글
​27,75.15
3,0
40,25
40,25
2 b 2 b 2 + 2 b₁ l₁ + b 2 l ₁ + b ₁ l 2
1 1
4. 3 + 3.3.1 + 1 . 4 + 3 .
10,342 feet.
Supposing a rotation around the longer edge of the base, we have the
horizontal distance of the centre of gravity from this edge, e —
•
1/2 · 1/1/0
axis is
7
& b 2
feet, and therefore the distance of the centre of gravity from the
C S = r = √æ te²
√a² + e² = √(1,75)² + (10,342)* √110,002
hence the height that centre of gravity must be lifted is
s = r a = 10,489 10,342 0,147 feet,
and the work to be done or the stability
S t = G s = 37674 . 0,147
=
5538 foot-pounds.
10,489;
§ 145. Work Done in Moving a Heavy Body.-In order
to find the mechanical effect, which is necessary to change the
position of a heavy body by causing a rotation, we must pursue the
same course as in calculating its dynamical stability. If we cause
a heavy body A C, Fig. 215, to rotate about a horizontal axis to
such an extent, that the inclination M C S = a of the line of
gravity C S = r becomes M C S₁
= a19
the centre of gravity S will describe the
vertical space H S, M, S₁ - MS=8₁
=r (sin. a, sin. a), and therefore if we
designate by G the weight of the body,
the mechanical effect required is
D
FIG. 215.

ΤΩ
$1
B
H
Μ Μ,
horizon, we have
=
A₁ = Gs, Gr (sin a, — sin. a).
=
If the axis of rotation is not horizon-
tal, but inclined at an angle ẞ to the
s₁ = r cos. ẞ (sin. a,
$1
-- =
sin. a) and
A, Gs, Gr cos. B (sin. a, sin. a). (Compare § 133.)
If in addition the body is moved in such a manner as not to
change its position in relation to the direction of gravity, and if its
centre of gravity and all its parts describe one and the same space,
the vertical projection of which is 8, then the moving of the
body will require, in addition to the above mechanical effect, an
amount of work A, G s, and consequently the total work done
will be
A = Â₁ + ½
=
G [r cos. ẞ (sin. a,
sin. a) + 8.] ·
The space described by the body in a horizontal direction does
272
[§ 146
GENERAL PRINCIPLES OF MECHANICS.
not enter into the question, when we suppose the motion to be slow,
in which case the work of inertia can be put equal to zero.
If a body A C, Fig. 216, lying upon a horizontal plane B C' is to
be placed upright upon another plane C, D, we have ẞ 0°, or
FIG. 216.
A2
E
B₂
2
B₁
D2
D
S₁
IN
H
F
D
E
K
B
cos. ẞ1; and if a and e
denote the horizontal and
vertical co-ordinates of the
centre of gravity of the body,
when it is in an upright
position, the radius C S =
r = √ a² + e², and the height
E₁ S₁ = a = r sin. a₁. If a
is the angle of inclination
BCS formed by the side
B C'of the body with the line
of gravity CS, we have the
original height of the centre
of gravity above the surface
on which the body rests

KS = CS sin. B C'S = r sin. a =
Va² + e² sin. a,
and consequently the height, which the centre of gravity is raised,
while the body is being placed upright is
HS, 8, E, S, E, Ha Na² + e sin. a.
=
—
2
If now s₂ is the vertical distance of the plane C, D, above the
first plane B C, we have for the entire work done in placing the
body upon C, D₂
A = G (a Va' + e sin. a + 8₂).
This determination of the work necessary to move the body is
perfectly correct only, when the centre of gravity is raised by a con-
tinuous movement from S to S. If, on the contrary, the body is
first placed upright and then raised, the mechanical effect neces-
sary is
A = G (F 0 + 8₂) = G (C' O − K S+ s₂) = G [ Va² + e² (1 − sin. a) + s.] ;
for the work G. ON which the body performs, when the centre of
gravity sinks from 0 to 8, is lost.
§ 146. Stability of a Body on an Inclined Plane.-A body
A C, Fig. 217, resting upon an inclined plane (Fr. plan incliné,
Ger. schiefe Ebene), can assume two motions; it can slide down
$ 146.] EQUILIBRIUM OF BODIES RIGIDLY FASTENED.
273
the inclined plane, or it can overturn by a revolution around one
of the edges of its base. If the body is left to itself the weight G is
decomposed into a force Nat right angles to and a force P parallel to
the base; the first is counteracted entirely by the inclined plane,
the latter, however, moves the body down the plane. If we put
the angle of inclination of the plane to the horizon = a, we have
also the angle G S N = a, and
consequently the normal pressure
N = G cos, a and
H
FIG. 217.
B
F

A
P
D
G
R
the sliding force
PG sin. a.
If the vertical line of gravity
S G passes through the base CD,
as is shown in Fig. 217, the sliding
motion alone can take place; but
if the line of gravity, as in Fig. 218, passes without the base, the
body will be overturned and is without stability.
The stability of a body A Cupon an inclined plane F H, Fig.
FIG. 218.
B
FIG. 219.
B


F
A
F
E,O
N
M
D
II
R
G
H
R
219, is different from that of a body upon a horizontal plane H R.
If D M = e and Ms a are the rectangular co-ordinates of the
centre of gravity S, we have for the arm of the stability
=
DE DO M Ne cos. a a sin. a,
while, on the contrary, it is = e, when the body stands upon a hori-
zontal plane. Since e> e cos. a a sin. a, the stability in refer-
ence to the lower edge D is always smaller upon the inclined plane,
and become null, when e cos. a = a sin. a, I.E. when tang. a =
e
a
If, then, a body, whose stability is Ge when standing upon a hori-
zontal piane, is placed upon an inclined plane, whose angle of incli-
e
nation corresponds to the expression tang. a =
it loses its sta-
a
18
274
[§ 147
GENERAL PRINCIPLES OF MECHANICS.
=
bility. On the other hand, a body can acquire stabilitý upon an
inclined plane, although wanting it when placed upon a horizontal
one. For a rotation about the upper edge C the arm is CE, CO,
+ M N = e₁ cos. a + a sin. a, while for the same position on a
horizontal plane it is C M = e. If, however, e, is negative, the
body possesses no stability as long as it rests upon a horizontal
plane; but if placed upon an inclined plane, the angle of inclina-
tion a of which is such that we have tang. a>, the body acquires
a position of stable equilibrium. If, in addition to the force of
gravity, another force P acts upon the body A B C D, Fig. 209, it
retains its stability, if the direction of the resultant N of the weight
G of the body and of the force P passes through the base C D
of the body.
a
and
EXAMPLE.—In the obelisk in the example of paragraph 144, e =
10,342 feet; consequently it will lose its stability, when placed upon
an inclined plane, for whose angle of inclination we have
tang. a =
7000
=
0,16922,
17
4. 10,342 41368
and whose angle of inclination is therefore
a 9° 36'.
§ 147. Theory of the Inclined Plane.-Since the inclined
FIG. 220.
E
P
B
K
A
F
H
G
P
G
plane counteracts only the pressure
perpendicular to it, the force P, ne-
cessary to retain the body, which is
prevented from turning over, on the
inclined plane, is determined by the
consideration, that the resultant N,
Fig. 220, of P and G must be per-
pendicular to the inclined plane. Ac-
R cording to the theory of the parallel-
ogram of forces, we have

sin. P NO
sin. PON
but the angle P N 0 = angle G O N = F H R = a, and the
angle P ON POK+ KONẞ+90°, when we denote
the angle P E F = P O K formed by the direction of the force
with the inclined plane by ẞ; hence we have
P
G
sin, a
sin. (B + 90)'
Р
sin. a
I.E.
G
cos. B'
$148.]
275
EQUILIBRIUM OF BODIES RIGIDLY FASTENED.
P
and the force, which holds the body on the inclined plane, is
For the normal pressure we have
G sin. a
cos. B
N
sin. O G N
G
sin. O N G
or, since the angle O G N
90°
(a + B) and O NG PON
=
= 90 + ß,
N
sin. [90º
(a + B)]
cos. (a + B)
G
sin. (90° +B)
cos. B
N
=
G cos. (a + B)
cos. B
and the normal pressure against the inclined plane is
If a + B is > 90° or ẞ > 90° a, N becomes negative, and
FIG. 221.
P
F
R
H
then, as is represented in Fig. 221, the inclined
plane HF must be placed above the body O, to
which the force P is applied. If the force P is
parallel to the inclined plane, ẞ becomes = 0 and
cos. B = 1, and we have
PG sin. a and N = G cos. a.
If the force P acts vertically a + ẞ is
and we have
cos. ß = sin. a, cos. (a + ß) 0,

90º,
PG and N = 0. In this case the inclined
plane has no influence upon the body.
Finally, if the force is horizontal, ẞ becomes
a and cos. ẞ
=cos. a, and we have
P =
G sin. a
cos. a
G cos. O
G
= G tang, a and N
cos. a
cos. a
EXAMPLE. In order to retain a body weighing 500 pounds upon a
plane inclined to the horizon at an angle of 50°, a force is employed, whose
direction forms an angle of 75° with the horizon: required the intensity of
the force and the pressure of the body upon the inclined plane. The
force is
P =
500 sin, 50°
cos. (75" — 50°)
and the pressure upon the plane is
N =
500 cos. 75°
cos. 25°
500 sin. 50°
cos. 25°
=
422,6 pounds.
142,8 pounds.
§ 148. The Principle of Virtual Velocities.—If we com-
bine the principle of the equality of action and reaction, explained
276
[§ 148.
GENERAL PRINCIPLES OF MECHANICS.
2
in § 138, with the principle of virtual velocities (§ 83 and § 98), we
obtain the following law. If two bodies M, and M₂ hold each other
in equilibrium, then, for a finite rectilinear or for an infinitely small
curvilinear motion of the point A of pressure or contact, not only
the sum of the mechanical effects of the forces of each separate
PK
P
N₂
FIG. 222.
D
En
C
D.
E₁
body, but also the sum of the me-
chanical effects of the exterior
forces acting upon the two bodies
(taken together) is equal to zero.
If P, and S, are the forces in

1
one body and P, and S₂ those
in the other, when the point of
contact is moved from A to B,
the spaces described by these
forces are A D₁, A E₁, A D, and
A E, and according to the law
announced above we have
2
2
P₁ . A D₁ + S₁ . A E₁ + P₂ A D½ + S½ . A E₂ = 0,
or without reference to the direction
+
P, AD, S. AE,
The correctness of this
P₂. A D₂ + S₂ · A E₂.
2
2
2
law can be demonstrated as follows.
and N, are equal, their mechanical
Since the normal forces N,
effects N₁ . A С and N. A C'must also be equal to each other, the
only difference being, that one of the forces is positive and the
other negative. But according to what we have already seen, the
mechanical effect of the resultant N. A C'is equal to the sum
of those P₁ A D, + S. AE, of its components, and in like man-
ner N, A C' – P₂. A D½ + S½ . A E; consequently we have
H
2
1
1
2
P₁. A D₁ + S₁. A E₁ = P₂. A D₂ + S₂. A E
FIG. 223.
E,
D
F
P
IB
2
2
This more general application of the principle of virtual
velocities is of great importance in
researches in statics, the determina-
tion of formulas for equilibrium be-
ing much simplified by it. If, E.G.,
we move a body A upon an inclined
plane, FH, Fig. 223, a distance A B,
the space described by its weight G
ACAB sin. A B C

A
'G
N
R
is
A B sin. FH R = A B sin a,
and, on the contrary, the space de-
149.]
277
EQUILIBRIUM OF BODIES RIGIDLY FASTENED.
scribed by the force P is = A D = A B cos. B A D = A B cos. ß, and
finally that described by the normal force N is = 0; but the work
done by N is equal to the work done by G plus the work done by
P, and we can therefore put
N.0 G.AC+ P.AD,
consequently the force, which holds the body upon an inclined
plane, is
P =
A C
AD
G
G sin. a
cos. ß'
a result, which agrees perfectly with that obtained in the foregoing
paragraph.
H
FIG. 224.
N
Hi
G
P
F
Ꭱ
D
B
R₁
On the contrary, to find the
normal force N, we must move
the inclined plane H F, Fig. 224,
an arbitrary distance A B at
right angles to the direction
of the force P, determine the
space described by the exterior
forces and then put the me-
chanical effect of the weight G
and of the force P equal to the
mechanical effect of the pressure
Nupon the inclined plane.
The space described by Nis
A D — A B cos. B A D = A B cos. ß;

that described by G is
A C = A B cos. B A C A B cos. (a + ß),
=
and that described by the force P is = 0, hence the mechanical
effect is
N. AD G. AC+ P. 0,
and
=
G.AC
N=
= G
cos. (a + B)
AD
cos. B
as we found in the foregoing paragraph.
§ 149. Theory of the Wedge.-We can now deduce very
simply the theory of the wedge. The wedge (Fr. coin, Ger. Keil) is a
movable inclined plane formed by a three-sided prism F H G,
278
[$ 149.
GENERAL PRINCIPLES OF MECHANICS.
Fig. 225. The force K P = P acts generally at right angles to
the back F G of the wedge and balances another force or weight
FIG. 225.
L
B
D
HT
H
E
R

F
R₂
K
P
A Q = Q, which presses against a side FH of the wedge. If the
angle, which measures the sharpness of the wedge, is F H G = a
and the angle formed by the direction K P or AD of the force
with the side G H is GEK = BAD — d, and, finally, if the
angle L A H formed by the direction of the load Q with the side
FH is B, the spaces described, when the wedge is moved from
the position F H G to the position F, H, G₁, are found in the fol-
lowing manner. The space described by the wedge is
ABFF = H H₁,
that described by the force is
ADAB cos. B A D
A B cos. 8,
and that described by the rod A L or by the load Q is
A B sin. A B C
A C
sin. A CB
A B sin. a
sin. HAC
A B sin. a
sin. B
On the contrary, the space described by the reaction R of the
base E G as well as that described by the reaction corresponding
to the pressure against the guides of the rod is 0.
Now putting the sum of the mechanical effects of the exterior
forces P, Q, R and R₁ = 0, we have
1
P.AD-Q. ACR.0+ R,. 0 = 0,
from which we obtain the equation of condition
Q. A C
P =
AD
Q. A B sin. a
A B cos. § sin. ß
Q sin. a
sin. ẞ cos. S ď
If the direction KE of the force passes through the edge H of
a
the wedge and bisects the angle FH G, we have =, and therefore
S
2'
$ 149.] EQUILIBRIUM OF BODIES RIGIDLY FASTENED
279
2 Q sin.
P =
Q sin. a
a
sin. B
sin. ẞ cos.
2
a
2
If the direction of the force is parallel to the base or side G H,
we have 80, and consequently
P =
Q sin. a
sin. B'
and if the direction of the load is also perpendicular to the side
FH, we have ẞ 90°, and consequently
ß
=
P= Q sin. a.
EXAMPLE. The sharpness F H G = a of a wedge is 25º, the direction
of the force is parallel to the base, and therefore d is = 0, and the load acts
at right angles to the side F H, 1.E., ẞ is = 90': required the relations of
the force and load to each other; in this case we have
P
P = Q sin. a or
sin. 25° = 0,4226.
Q
If the load is Q = 130 pounds, the force is
P130. 0,4226
=
54,938 pounds.
In order to move the load or rod a foot, the wedge must describe the
space
A C
1
A B
= 2,3662 feet.
sin. a 0,4226
REMARK 1. The relation between the force P and the load Q of the
wedge F G H, Fig. 226, can be determined by the application of the
KP
H
FIG. 226.
S
R
N
K
parallelogram of forces in the
following manner.
=
The load

upon the rod 4 Q Qis de-
composed into a component
AN = N perpendicular to the
side FH and into a component
AS = S perpendicular to the
axis of the rod. While S is
counteracted by the guides of
the rod, A N N is transmit-
ted to the wedge and combines
there as 4, N, with the force
1
APP of the wedge to form a resultant AR = R, whose
direction must be perpendicular to the base G H of the wedge, in which
case it will be transmitted completely to the support of the wedge. The
parallelogram of forces A, PR N₁ gives
1
280
GENERAL PRINCIPLES OF MECHANICS.
[$ 150.
P
N₁
1 1
sin. a
1
sin. PA, R
cos. s
sin. R A, N sin. FH G
sin. A₁ R N₁
1
and from the parallelogram of forces A N Q S we have
N
sin. N Q A
sin. QAS
1
Q
sin. A N Q
sin. LA H
sin. B'
but since N₁ is =
1
N, we obtain
by multiplying these proportions together,
P
N P
N Q
Q
sin. a
sin. ẞ cos. S'
or
P =
Q sin. a
sin. ß cos. S'
as was found in the large text of this paragraph.
REMARK 2. The theory of the lever, inclined plane and wedge will
be discussed at length in the fifth chapter, when the influence of friction
will also be taken into consideration.
CHAPTER IV.
EQUILIBRIUM IN FUNICULAR MACHINES.
§ 150. Funicular Machines. We have previously considered
the solid bodies to be perfectly rigid or stiff bodies (Fr. corps
rigides; Ger. starre or steife Körper); I.E., as bodies, whose vol-
ume and form are unchanged by the action of exterior forces upon
them. Very often in the practical application of mechanics the
supposition, that bodies are perfectly rigid, is not permissible, and
it becomes necessary, therefore, to consider these bodies in two
other states. These states are those of perfect flexibility and
of perfect elasticity, and consequently we distinguish flexible
bodies (Fr. corps flexible; Ger. biegsame Körper) and elastic
bodies (Fr. corps élastiques; Ger. elastische Körper). Flexible
bodies counteract without change of form forces in one direction
only and follow perfectly those acting in other directions; elastic
bodies, on the contrary, yield to a certain extent to every force
acting upon them.
A rigid body A B, Fig. 227, I, counteracts completely the force
$ 151.]
281
EQUILIBRIUM IN FUNICULAR MACHINES.
P, a flexible body A B, Fig. 227, II, follows the direction of the
force P, which acts upon it, in such a manner, that its axis assumes
FIG. 227.
I
11

TAX
P
P
P
the direction of the force, and an elastic body A B, Fig. 227, III,
resists the force P to a certain extent only, so that its axis under-
goes a certain deflection. Cords, ropes, straps and in a certain
sense chains are representatives of flexible bodies, although they do
not possess perfect flexibility. These bodies will be the subject of
the present chapter; elastic bodies, or rather the elasticity of rigid
bodies, will be treated of in the fourth section.
We understand by a funicular machine (Fr. machine funicu-
laire; Ger. Seilmaschine) a cord or a combination of cords (the
word cord being employed in a general sense), which is stretched
by forces, and we will occupy ourselves in this chapter with the
theory of the equilibrium of this machine. The point of the
funiculaire machine to which a force is applied, and where, conse-
quently, the cord forms an angle or undergoes a change of direc-
tion is called a knot (Fr. noeud; Ger. Knoten). The same is either
fixed (Fr. fixe; Ger. fest) or movable (Fr. coulant; Ger. beweg-
lich). Tension (Fr. tension; Ger. Spannung) is the force propa-
gated in the direction of its axis by a stretched cord. The ten-
sions at the ends of a straight cord or piece of cord are equal and
opposite (§ 86). A straight cord cannot propagate any other force
but the tension acting in the direction of its axis; for if it did, it
would bend and would no longer be straight.
§ 151. Equilibrium in a Knot.-Equilibrium exists in a
funicular machine, when each of its knots is in equilibrium. Con-
sequently we must begin with the study of the conditions of equi-
librium in a single knot.
Equilibrium exists in a knot K formed by a piece of cord
282
[§ 151.
GENERAL PRINCIPLES OF MECHANICS.
A K B, Fig. 228, when the resultant KS = S of the two tensions
= S₁ and KS, =S, is equal and opposite to the
of the cord K S
R
FIG. 228..
α
A
S
B
force P applied at the knot; for the
tensions of the cord S, and S pro-
duce the same effect in the knot
K as two forces equal to them and
acting in the same direction as
they do, and the three forces are in
equilibrium, when one of them is
equal and opposite to the resultant
of the other two (§ 87). In like.
manner the resultant R of the
force P and of one of the tensions .
S is equal and opposite to the
second tension S, etc. We can
profit by this equality to determine two conditions, E.G., the ten-
sion and direction of one of the ropes. If, E.G., the force P, the
tension S, and the angle formed by them

AKP = 180°
AKS = 180°
are given, we have for the other tension
S₂ √ P² + S₁² 2 P S₁ cos. a
- a
and for its direction or for the angle B KSB formed by it
with K S
sin. B
S₁ sin. a
S₂
EXAMPLE. If the rope A KB, Fig. 228, is fastened at its end B and
stretched at its end A by a weight G = 135 pounds and at its centre
K by a force P = 109 pounds, whose direction is upwards at an angle of
25° to the horizon, what will be the direction of the tension in the
piece of cord K B?
The intensity of the required tension is
S₂
= √ 1092 + 1352
= √11881 + 18225
For the angle ẞ we have
2. 109. 135 cos. (90"
25")
29430. cos. 65° = ✓17668,3 132,92 pounds.
1
sin. B
S₁ sin. a
S2
135. sin. 65°
132,92
log sin. ß = 0,86401 — 1,
whence ß = 67° 0', and the inclination of the piece of cord to the horizon is
во
25° = 67° 0' 25° 0'
A p
42° 0'.
A 152.]
283
EQUILIBRIUM IN FUNICULAR MACHINES.
§ 152. If a cord A K B, Fig. 229, forms a fixed knot K in con-
sequence of one portion of the cord B K lying upon a firm sup-
FIG. 229.
B
K
α
_S₁
८२
port M, while the other portion of
the cord is stretched by a force KS
S, whose direction forms a certain
angle S K S₁ = a with the direction
of the first portion of the cord, we
have the tension in the portion K B
of the cord

K S
S₁
= S cos. a,
while the second component K N = N = S sin. a is counteracted
by the support M.
We have also
S₁ = S √ 1 − (sin. a)³,
and therefore, when the angle of divergence is small,
or inversely
S, = (1 − 1 (sin. a)') § =
S (1 − 2) s,
S
–
2
α
-
S
+
(1 + 2) S.
α
1
2
If a cord is laid upon a prismatical body, and its directions thus
changed successively an amount measured by the angles a, a, Az,
-S3
FIG. 230.
S3
K:
αz K₂
Ch.2
B
S
K⋅
¡S₂
S₂ cos. α3
0.1
n
S₁
the foregoing decomposition
of the force is repeated, so
that in the knot the ten-
sion Sis changed into S₁ =
S cos. a₁, and in the knot K,
the tension S, into
S₂ = S₁ cos, a₂ = $_cos, a cos. Ag,
and in the knot As the ten-
sion S into
S cos. a, cos. a¿ cos. ɑz.

If the angles a₁, a, a, are equal to each other and a, we have
S₁ = S (cos. a)
S3
SS (cos, a)".
If the prism M becomes a cylinder, a is infinitely small and n
infinitely great, and consequently
α
8. (1) 8 (1) s
Sn
2
S
2
or if we denote the total angle of divergence n a by ẞ, we have
284
[$ 153.
GENERAL PRINCIPLES OF MECHANICS.
Sn
S₁ = (1 - 2, 3³)
α β
S', I.E.
2
αβ
‚§. = ‚§, because a and consequently is infinitely small compared
S,
with 1.
If, therefore, a cord is laid upon a smooth body so as to cover a
portion of the periphery of its cross section, its tension is not
changed thereby; and when a state of equilibrium exists the ten-
sion at both ends of the cord are equal to each other.
§ 153. If the knot A is movable, if, E.G., the force P is applied
by means of a ring to the cord A K B, Fig. 231, which is passed
through it, the resultant S of the tensions S, and S, of the cord is
equal and opposite to the force P applied to the ring; besides the
tensions of the cord are equal to each other. This equality is a
consequence of § 152, but it can also be proved in the following
manner. If we pull the rope a certain distance through the ring,
one of the tensions S, describes the space s, the other tension S, the
space
s, and the force P the space 0. If, therefore, we assume
perfect flexibility, the work done is
P.0 S.s S. 8, I.E. S, 8 S8 or SS.
=
1
=
The equality of the angles 4 K S and B K S, formed by the
direction of the resultant S with the directions of the rope, is also a
consequence of this equality of the tensions. Putting this angle
a the resolution of the rhomb K S S S gives
S = P = 2 S₁ cos. a, and inversely
P
S₁ = S₂
=
2 cos. a
FIG. 231.
P
K
M
Si
A
S2
B
Q
FIG. 232.
S
B


M
K
D
P
S
If A and B, Fig. 232, are fixed points of a cord A K B of a
:
§ 153.
285
EQUILIBRIUM IN FUNICULAR MACHINES.
given length (2 a) with a movable knot K, we can find the posi-
tions of this knot by constructing an ellipse, whose foci are at A and
B and whose major axis is equal to the length of the rope 2 a,
and by drawing a tangent to this curve perpendicular to the given
direction of the force. The point of tangency thus found is the
position of the knot; for the normal K S to the ellipse forms equal
angles with the radii vectores K A and K B, exactly as the result-
ant S does with the tensions S, and S, of the cord.
If we draw A D parallel to the direction of the given force,
make B D equal to the given length of the cord, divide A D in
two equal parts at M and erect the perpendicular M K, we obtain
the position K of the knot without constructing an ellipse; for the
angle AK M = angle DK M and A K D K, and consequently
the angle A KS angle B K S and A K + K B = D K +
KBDB.
FIG. 233.
C
S
=
EXAMPLE.-Between the points A and B, Fig. 233, a cord 9 feet long is
stretched by a weight G = 170 pounds, hung upon it by means of a ring.
The horizontal distance of the two points
from each other is A C 63 feet and the
vertical distance of the same C B 2 feet:
required the position of the knot as well as
the tensions and directions of the two por-
tions of the cord. From the length A D
9 feet as hypothenuse and the horizontal
distance A C = 63 feet, we obtain the ver-
tical line

B
Si
K
M
G
D
CD = √ 92
= √ 38,75
6,52 = √ 81
√81
6,225 feet,
42,25
angle BD K
and from this the base of the isosceles tri-
BD = C D — C B = 6,225 2 = 4,225 feet.
On account of the similarity of the triangles DKM and DA C, we have
=
DK BK=
DM
.DA
Ꭰ Ꮯ
4,225.9
2.6,225
3,054 feet,
whence
AK = 9
3,054 = 5,946 feet.
Hence for the angle a formed by the two portions of the cord with the ver-
tical line we have
cos. a
BM
B K
2,1125
3,054
and finally the tension in the cord is
0,6917, whence a = 46° 14′,
G
S₁
So
170
2 cos. a 2.0,6917
122,9 pounds.
286
[§ 154.
GENERAL PRINCIPLES OF MECHANICS.
§ 154. Equilibrium of a Funicular Polygon.-The con-
ditions of equilibrium of a funicular polygon, I.E. of a stretched
P₂
S₁
1
I.
FIG. 234.
S2
B
K
S
P
P3
P2. S2
Ps
11.
Si
Ps
P
-Sa
B
plication of this tension from
2
P₁
cord acted upon in different
points by forces, are the same
as those of the equilibrium
of forces applied at the same
point. Let A KB, Fig. 234, I,
be a cord stretched by the
forces P1, P2, P3, P, P6; P₁
and P, being applied in A,
P3 in K and P, and P, in B.
Let us denote the tension of
the portion of the cord A K
by S, and that of the portion
BK by S2, then we have S
as the resultant of the two
forces P, and P, applied in A.
Transferring the point of ap-

A to K, we have S as resultant of
S and P or of P1, P. and P. Transferring the point of applica-
tion of the force S₂ from K to B, we have S, as the resultant of P、
and P.; now, since S, is the resultant of P1, P2, and P3, this system
of forces is in equilibrium; we can therefore assert, that if certain
forces P1, P2, P3, etc., of a funicular polygon are in equilibrium,
FIG. 235.
S₁
V
A
S2
V
2
V₁ + V
K 2
K3
KA
Vn
KI
H2
V8
$3
II 8
G1
G ₂
Vis
Hm
C
༦ཐ
B
they will also hold
each other in equi-
librium, when they
are applied without
change of direction
S or intensity to a sin-

EL"
gle point, E.G. to C
(II). If the rope.
A K₁ K………. B, Fig.
235, is stretched in
the knots K₁, K₂
19
etc., by the weights
G₁, G, etc., and if
its extremities are
held fast by the ver-
tical forces V, and
V, and by the hori-
§ 155.]
287
EQUILIBRIUM IN FUNICULAR MACHINES.
كم
zontal forces H, and H, the sum of the vertical forces is
V₁ + V₂ − (G₁ + G₂ + G3 + ...),
1
and the sum of the horizontal forces is H₁ H. The conditions
of equilibrium require both these sums to be = 0, and therefore
we have
1) V₁ +
2) H₁ =
V₁ = G₁ + G₂ + G3 + ... and
Vn
H, I.E.
the sum of the vertical forces or tensions at the extremities of the
ropes of a funicular polygon stretched by weights is equal to the sum
of weights hung upon it, and the horizontal tension at one extremity
is equal and opposite to that at the other.
If we prolong the directions of the tensions S, and S. at the
extremities A and B, until they cut each other in C, and if we
transfer the point of application of these tensions to this point, we
obtain a single force P=V, + V; for the horizontal forces H, and H
balance each other. Since this force balances the sum G₁ + G₂ +
G+... of the weights attached to it, the point of application or
centre of gravity of these weights must be in the direction of this
force, I.E. in the vertical line passing through C.
:
§ 155. From the tension S, of the first portion A K, of the
S₁
V
B1
S 2
H₂
FIG. 235.
V
11
A
V₁+VD
V
B
K₁
V
3
K
Sn
rope and from the
angle of inclination
S₁ A H₁ = α, we ob-
tain the vertical ten-

1
sion Fi S₁ sin. a₁
and the horizontal
tension H₁ = S₁ cos. a₂.
If we transfer the
H, point of application
of these forces from
A to K, we have, in
addition to them,
the weight G₁, which
acts vertically down-
wards, and the verti-
cal tension in the
following portion
S₁ sin. a, G₁, while the
horizontal tension HH, H remains unchanged. The two
latter forces, when combined, give the axial tension of the second
portion of the rope
$ 3
H8
E3
G₁
G e
Gs
GA
H
11
H1
C
K, K, of the rope is V. V₁
=
G₁
=
288
[§ 155.
GENERAL PRINCIPLES OF MECHANICS.
S₂ = √ V¸² + H³,
2
and its inclination a, is determined by the formula
V₂
S₁ sin. a₁
G₁
tang. a₂ =
I.E.
,
H
S₁ cos. a₁
G₁
tang. a₂ = tang. a,
a₁
H
Transferring the point of application of
V, and H, from K, to
2
2
K2, we have, by the addition of the weight G₂, a new vertical force
29
V₂ = V₂ — G₂ = V₁ − (G₁ + G₂) = S, sin. a, —
3
(G₁ + G₂),
which is that of the third portion of the rope, while the horizontal
force H H remains unchanged. The total tension in this third
portion of the cord is
2
S₁ = √V3 + H²,
S3
and its angle of inclination a, is determined by the formula
√ 3
Si sin. a1
(G₁+ G₂), L.E.
tang. a₂ =
H
S₁ cos. a₁
1
G₁ + G
H
tang. a₂ =
a3 tang. a₁
For the angle of inclination of the fourth portion of the cord
we have
G₁ + G₂+ G3
2
4
tang. a, tang. a,
=
etc.
H
If
G₁ + G₂+ G3
H
2
3
becomes > tang. a, or G₁ + G₂ + G3 > V₁,
2
then tang. a, and consequently a, becomes negative, and the cor-
responding side K, K, of the polygon is no longer directed down-
ward, but upward. The conditions are the same for any point, for
which G₁+ G + G + ... is > V₁.
A
FIG. 236.
C
B
The tensions S1, S2, S3, etc., as well as the angles of inclination
a1, a2, ag, etc., of the different portions of the rope can easily be
represented geometrically. If we make the horizontal line CA =
CB, Fig. 236, the horizontal tension
Hand the vertical line CK, the vertical
tension V, at the point of suspension 1,
the hypothenuse A K₁ will give the total
tension S₁ of the first portion of the rope,
and the angle CA K, the inclination of
the same to the horizon. If, now, we
lay off upon C K₁the weights G1, G2, G3,
etc., as the divisions K, K2, K, K, etc.,
and draw the transverse lines A K, A K„,

K3
K
Ke
Ks
K₁
1
§ 156.]
289
EQUILIBRIUM IN FUNICULAR MACHINES.
the latter will indicate the tensions of the different succeeding
portions of the cord, and the angles CA K2, C A K, etc., the
angles of inclination a,, a, etc., of these portions.
§ 156. From the investigations in the foregoing paragraph we
can deduce the following law for the equilibrium of a cord stretched
by weights:
1) The horizontal tension is in all parts of the cord one and the
same, viz.:
= Sn cos. αn•
H = S₁ cos. α,
a₁ =
2) The vertical tension in any portion is equal to the vertical
tension of the cord at the end above it minus the sum of the weights
suspended above it, or
V₁ = V₁ (G₁ + G₂+... G'm-1).
M
M
−
This law can be expressed more generally thus: The vertical
tension in any point is equal to the tension in any other lower or
higher point plus or minus the sum of the weights suspended be-
tween them.
If we know besides the weights the angle a, and the horizontal
tension H, we obtain the vertical tension at the extremity A by
means of the formula
V₁ = H. tang. a₁,
and that at the extremity B is
√ n = (G₁ + G₂ + ... + G₂)
- VI.
If, on the contrary, the two angles of inclination a, and a, at
the two points of suspension A and B are known, the horizontal
and vertical tensions are determined in the following manner;
we have
and therefore
V
tang. an
V₁
tang. a
V₁ tang. an
V
n
tang. a,
But since V₁ + Vn G₁ + G₂ + ... I.E.,
a,
(tang, a₁ + tang. a₁) V₁ = G₁ + G₂ • • •
tang. a₁
we have
(G₁ + G₂ + ...) tang. a₁
V₁
(G₁ + G₂ +
sin. a, cos. an
)
tang. a, + tang. an
sin. (a₁ +a)
and
2
(G₁ + G₂ + ...) tang. a.
an
sin, an cos, a,
(G₁ + G₂ + . .
)
tang. a, +tang. a
sin. (a₁ + a,)'
and consequently
19
290
[§ 156.
GENERAL PRINCIPLES OF MECHANICS.
H = V₁ cotg. α₁ = V₁ cotg. an
1
If the two ends of the cord
V₁ = V₁ =
G₁ + G₂ +
as the other end B.
2
(G₁ + G₂ + ...)
cos. a¡ cos. ɑn
sin. (a,+ a₂)
have the same inclination, we have
+ Gn
; then one end A carries as much
These formulas are applicable to any pair of points or knots of
the funicular polygon, when we substitute instead of G, + G + ...
the sum of the weights, etc., suspended to the cord between the
two points. The vertical tensions of a cord, on which a weight G
is hung and the angles of inclination of which are a„ and am +1, are
sin. am cos. Am + 1
V₁ = Gm
sin. (am + am+
and
m + 1
= Gm
)
cos. a
sin. am + 1
sơn la tat
in
G
m
1 + cotg. am tang am t
Gm
1 + tang am cotg an t
'm 1
ап
'm
ก
These laws are applicable to any funicular polygon stretched by
parallel forces, when we substitute instead of the vertical the direc
tion of the forces.
1
2 3
EXAMPLE.-The funicular polygon A K₁ K₂ K, B, Fig. 237, is stretched
by three weights G₁ = 20, G₂
H₁
3
Vi
FIG. 237.
A
K₁
1
K3
G₁
K
-2
4
3
G3
B
=
30 and G3
The angles of inclination a, and a
by the formulas
S4
HA
= 16 pounds as well as by
the horizontal force H₁ = 25
pounds; required the axial ten-
sions, supposing the extremities
A and B to have the same angle
of inclination. The vertical ten-
sions at the ends are equal and are
G₁ + G₂ + G
1

V = V =
1
20+30 +16
2
2
2
3
= 33 pounds.
The vertical tension of the
second portion of the cord is
2
1
G₁
= 33
G s
20
pounds; that of the third is,
33 - 16
=
13
16 = 17 pounds.
of these extremities are determined
Ga
√3 V Gз (or G₁ + G₂ —
V₁)
ɑ4
V
33
1
tang. a₁ = tang. a4
1,32;
H
25
tnose of the second and third portions
by the formulas
G₁
20
1
tang. a₂ = tang. a¸
=
1,32
=
= 0,52 and
H
25
G
16
8
tang, az = tang. as
1,32
0,68;
Η
25
§ 157.]
291
EQUILIBRIUM IN FUNICULAR MACHINES.
whence we have
α. = = α4
1
52° 51′, a₂
= 27° 28′, az
= 27° 28', a3 = 34° 13'.
Finally the axial tensions are
2
√ V₁² + H² = √ 33² + 25² =
√ 33² + 25º = √ 1714 = 41,40 pounds,
S₁
S₁
2
S₂
√√₂² + H² = √ 13ª + 25ª
=
√ 794 =
18,18 pounds and
$3
√V₂² + H² =•√ 17² + 25² = 30,23 pounds.
3
§ 157. The Parabola as Catenary.-Let us suppose, that
the cord A CB, Fig. 238, is stretched by the weights G1, G2, G3,
A
N
G
FIG. 238.
X
M
N
T
B
etc., hung at equal horizontal
distances from each other.
Let
us denote the horizontal dis-
tance A M between the point of
suspension 4 and the lowest
point C by b and the vertical
distance CM by a; let us also
put the similarly placed co-ordi-
nates of a point O of the funicu-
lar polygon O Ny and CN
= x. If the vertical tension in

V,
A is = V, that in 0 is consequently. F, and therefore we
have for the angle of inclination to the horizon N O TROQ
of the portion of the cord O Q
tang. o
Y V
b ' H
in which I designates the horizontal tension.
Y
b
From this we obtain Q R = OR. tang. = OR.
V
H'
which is the difference of height of two neighboring corners
of the funicular polygon. If we put y successively = 0 R, 2 0 R,
30 R, etc., the latter formula gives the difference of height of
the first, second, third, etc., corners, counting from the lowest
point upwards; if now we add all these values, whose number we
can suppose to bem, we obtain the height CN of the point 0 (
above the lowest point C. Here we have
(OR + 2 OR + 3 OR + ... + m. OR)
V OR
x = CN
H
Ъ
V OR2
b
(1 + 2 + 3 + ... + m)
V m (m + 1) OP
H
Ъ
in accordance with the rule for summing an arithmetical series.
H
1.2
292
[§ 157.
GENERAL PRINCIPLES OF MECHANICS.
Finally, putting O R
y
>
we obtain
m
V
m (m + 1)
y³
X
H
2 m²
Ъ
b'
or substituting for the value of the tangent of the angle of inclina-
tion a of the end A of the rope tang. a =
X
V
H
m (m + 1) y³ tang. a
2 m² b
If the number of the weights is very great, we can put m + 1
=m, and consequently
x=
V y²
y²
H 26 26
tang. a.
For x = a, y =
b, and consequently we have
V
b
a
H
2
b tang. a
2
X
or more simply
y²
a b29
which is the equation of a parabola.
If, therefore, an imponderable string is stretched by an infinite
number of equal weights applied at equal horizontal distances from
each other, the funicular polygon becomes a parabola.
For the angle of inclination & we have
Y
tang. $
b
2 α
h
а
= 2 y
·
b2
2 y
tang. a =
2 a
b
•
X
y³
2 x
and
Y
The subtangent for the point O is
NTON tang. = y
2 x
2 x = 2 C N.
Y
If the chains and rods of a chain bridge A B D F, Fig. 239, were
B
FIG. 239.

M
E
A
without weight or very light in proportion to that of the loaded
bridge D E F, the latter weights alone would have to be considered,
and the chain A CB would form a parabola.
§ 158.]
293
EQUILIBRIUM IN FUNICULAR MACHINES.
CM
EXAMPLE.—The entire load of the chain bridge in Fig. 239 is G = 2 V
320000 pounds, the span is A B = 2b 150 feet, the height of the arc
15 feet; required the tension and other conditions of the
chain. The inclination of the chain to the horizon is determined by the
formula
tang. a =
2 a 30 2
b 75 5
0,4, whence a = 21° 48′.
The vertical tension in each point of suspension is
V weight 160000 pounds,
=
the horizontal tension is
=
1
H=Vcotg. a = 160000 .
=400000 pounds,
0,4
and the total tension at one end is
2
S = VV + H* = VV1 + cotg? a = 160000 . V1+
= 160000 V
29
4
80000 √29
430813 pounds.
(64)
§ 158. The Catenary.-If a perfectly flexible and inextensible
cord, or a chain composed of short links, is stretched by its own
weight, the axis of the same will form a curved line, which has re-
ceived the name of the catenary curve (Fr. chainette, Gr. Ketten-
linie). The strings, ropes, ribbons, chains, etc., which we meet
with in practice, are imperfectly elastic and extensible, and conse-
quently form curves, which only approach the catenary, but which
can generally be treated as such. From what precedes we know,
that the horizontal tension in the catenary is equal at all points,
while, on the contrary, the vertical tension in one point is equal to
the vertical tension in the point of attachment above it minus the
weight of the portion of the chain between this point and the point
of suspension. Since the vertical tension at the vertex, where the
catenary is horizontal, is = 0, or since the vertical tension at the
N
FIG. 240.
SI-G1
M
н
H
Ꭹ
S
H
N
Z X
C
B
point of suspension is equal to
the weight of the chain from
the point of attachment to the
vertex, the vertical tension in
any point is equal to the weight
of the portion of the chain or
cord below it.
If equal portions of the chain
are equally heavy, the curve
produced is the common cate-
nary, which is the only one we

294
[§ 159.
GENERAL PRINCIPLES OF MECHANICS.
will discuss here. If a portion of the chain or cord one foot long
weighs y, and if the arc corresponding to the co-ordinates C M = a
and MA = b, Fig. 240, is 4 O Cl, we have for the weight of
the portion A O C of the chain Gly.
If, on the contrary, the length of the arc corresponding to the
co-ordinates C N = x and N 0 = y is = s, we have for the weight
of this arc Vsy. Putting, finally, the length of a similar piece
of chain, whose weight is equal to the horizontal tension H, = c,
we have H = c y, and we have for the angles of inclination a and
ø in the points A and O
tang. a = tang. SA H
G
ly
Z
and
H
CY
C
V
SY
S
tang. & = tang. N O T =
H
CY
C
§ 159. If we make the horizontal line CH, Fig. 241, equal to
the length c of the portion of the chain measuring the horizontal
tension and C G equal to the length 7 of arc of the chain on one
side, in accordance with § 155, the hypothenuse G H gives the
intensity and direction of the tension of the cord at the point of
suspension A; for
A
P
६
G H
FIG. 241.
tang. C H G =
NCG +
G²
C G
CH
and
с
CH2 = √1+ c², or
S = √ G² + H =
M
√1+ c². y GH. y.
T² =
l²
If we divide CG into equal parts and draw from H to the
points of division 1, 2, 3, etc., straight
lines, the latter give the intensity and di-
rection of the tensions obtained by dividing
the length of the arc of the chain A Cinto
as many equal parts. For example, the line
HK gives the magnitude and direction of
the tension or tangent at the point of di-
1 vision (P) of the arc A P C, since at this

б
Y
d
α
H
C
82
a
C
2
K3
4
G
point the vertical tension= CK. 2, while
the horizontal tension is constant and
c. y, and therefore for this point we have
CK. Y
Ꮶ CK
tang.
CY
CH
as is really shown by the figure.
This peculiarity of the catenary can be
made use of to construct mechanically, approximatively correctly.
§ 160.]
295
EQUILIBRIUM IN FUNICULAR MACHINES.
this curve.
After having divided the given length of the catenary
to be constructed in very many equal parts and laid off the line
CH = c, which measures the horizontal tension, we draw the
transverse lines H 1, H 2, H 3, etc., and lay off on C H a division
C' 1 of the arc of the curve as Ca, pass through the point of division
(a) thus obtained a parallel to the transverse line H1 and cut off
again from it a part a b = = C 1. In like manner we draw through
the point (b) thus obtained a parallel to the transverse line H2 and
cut off from it bc C1 equal to a division of the arc.
We now
draw through the new point (c) a parallel to H 3 and make c d
equal to a division of the arc and continue in this way, until we
have obtained the polygon C'a b c def. We now construct another
polygon Caß y d ɛ o by drawing C a parallel to H 1, a ẞ to H2,
By to H 3, etc., and by making Ca aẞ By, etc., C1 =
= ав
=
= =
12 = 23, etc. If, finally, we pass through the centre of the lines
a a, b B, cy... fo a curve, we obtain approximatively the catenary
required.
For practical purposes we can often obtain accurately enough
a catenary corresponding to given conditions, E.G. to a given width
and height of the arc or to a given width and length of arc, etc.,
by hanging a chain with small links against a vertical wall.
§ 160. Approximate Equation of the Catenary.-In
many cases, and particularly in its application to architecture and
machinery, the horizontal tension of the catenary is very great
compared to its vertical one, and therefore the height of the arc is
small, compared with its width. Under this assumption, an equa-
tion for this curve can be found in the following manner:
Let s denote the length, a the abscissa C N and y the ordinate
R
FIG. 242.
M
N
U
B
•
NO of a very low arc C 0,
Fig. 242. We can, according
to the remark upon page 298,
put approximatively

S
= [1
1 +
2
( 1 )']
C
T
Y,
and therefore the vertical tension in a point O of a low arc of a
catenary is
V
= [ 1 + 3 ( 2 ) ] 3 Y
У
296
[§ 160.
GENERAL PRINCIPLES OF MECHANICS.
and the tangent of the tangential angle T O N = is
S
2
X
tang. = = [ 1 + 23 ( 1 ) ² ] /~
12/
с
Y
)
If we divide the ordinate y into m equal parts, we find the
portion R Q = N U of the abscissa X corresponding to such a
division OR by putting
Ꭱ
RQ= OR tang.
=
•
OR. 2 [ 1 + 23 ( 1 ) ]
门
​Since x is very small compared to y, we have approximatively
C
R Q = OR. 2. Substituting now 0 R =
the values y 2y 3y
m' m
,
772
Y
กาว
and successively for y
etc., we obtain one after the other the differ-
ent portions of x, the sum of which is
y²
X
c m²
2
(1 + 2 + 3 + ... + m)
y'
m (m + 1)
=
c m²
2
(§ 157)
y²
2 c
the latter equation is that of the parabola.
If we proceed more accurately and substitute in the formula
X
2
QR = σ R. 2 [1 + 23 ( 3 ) ],
y²
Y
just found, we obtain
y²
6
C²²
1
instead of x, the value
2c
Y
OR
C
:( 3
y
y 2 y 3y
etc., and
QR = OR. (1 + 1 ) = 0 (x + 2 + 2)
C
Putting y again successively equal to
instead of O R, 2
m' m
6
m
У we obtain successively the different portions of
m
x, and consequently their sum
X=
Y Y
= [ 12/24 (1 + 2 + 3 + . . . + m) +
c m L M
1
6 c²
(%) (10
(1ª + 2ª +3³ + ... + m³) ].
When the number of members is very great, the sum of the cardi-
m³
nal numbers 1 + 2 + 3 ... + mis
=
and the sum of their cubes
2
m¹
is
4
(see "Ingenieur," page 88). Hence we have
x
x = 2 (1
y²
+
1,
y
2 6 c² 4
2), LE
I.E.
?
=
yo y³
1) == 2 + 1
c 24 c³
2 c
[1 + 15. (2)].
12
the equation of very powerfully stretched catenary.
§ 160.]
297
EQUILIBRIUM IN FUNICULAR MACHINES.
By inversion we obtain
y¹
y2
= 2cx
=2cx
12 ca
2
4 c² x²
12 c²
x²
= 2 c x
whence
3'
x²
2) y = √ 2 cx-
3'
or approximatively,
y = √ 2 c x
(
X
1 -
12 c
The measure of the horizontal tension is given by the formula
y² y' 4 x²
y²
2
y'
C =
+
+
2x
2 x . 12 c²
2 x
Х
24x
4
yt
I.E.
y²
3) c = +
2 x 6*
The tangential angle is determined by the formula
Y
2
tang = [1 + ()']-
Þ
2 x
Y
C
1 +
Y
x
[
2
3
X
3 y
X
].
1 + 1} ( 3 ) ]
3 Y
y[
y²
1 +
[1
2 x
[ 1 +
1 X
3 y
2
2 Ꮖ
3
1
Y
81
3 Y
I.E.
The formula for the rectification of the curve is
4) tang. 4 =
2
s
5) • = y [1 +
3
2
y
()'] =
2
1 Y
6
2
[(1 + } (!)'].
EXAMPLE-1) The length of the catenary for a width of arc 2 b
feet and for a height of arc a =
21 feet is
2 2,
3 8
27 = 28[1 + (†)'] = 16. [1+ (25)']
21
2/3/2
a 2
= 16+
16 + 16. 0,065
=
17,04 feet;
16
and the length of the portion of the chain, which measures the horizontal
tension, is
b2
α 64 5
C =
+
2 a 6
+ =
5 12
12,8 + 0,417 = 13,217 feet;
the tangent of the angle of inclination at the point of suspension is
2 a
3 ª [ 1 + 1 ( 7 ) ] = [(1 + 1-(-8)'] =
5
8
tang. a =
whence the angle itself is a = 32° 50′.
5.1,03255
8
=0,6453...,
2) If a chain is 10 feet long and the width of span is 93 feet, the height
of arc is
3 (1091)
- b) b = √ 2
2
3
α =
(1 — b) b :
√
√1,7812 = 1,335 feet,
9/13/
3
19
57
=
16
32
201
298
[§ 160.
GENERAL PRINCIPLES OF MECHANICS.
and the measure of the horizontal tension is
b2
α
4.752 1,335
C =
+
+
2 a
6 2 . 1,335 6
8,673 feet.
3) If a string 30 feet long and weighing 8 pounds is stretched as nearly
horizontal as possible by a force of 20 pounds, the vertical tension is
V = } & = 4 pounds, and the horizontal force
H = √ S² — V²
√/20² — 42 √384 = 19,596 pounds,
the tangent of the angle of inclination at the point of suspension is
V
tang. =
4
H 19,596
0,20412,
and the angle itself is 11° 32'; the measure of the horizontal tension is
the width of the span is
2
26=21[1 - 1 · (-/-)²] =
· (4)'] = 80. [1 -
H
8
30
C =
= H:
H=73,485 feet,
Y
30
30
[1 - 31.
6
15 2
73,48
= 30—0,208=29,792 ft.,
a =
√ 38 (70) = 1
3 29,792 . 0,208
(l b)
✓ 29,792 . 0,078
1,524 feet.
2.2
6
and the height of the arc
2
2
REMARK 1.-We find from the radius CA
C B = C D = r and the
ordinate 4 M = y of an arc of a circle A B, Fig. 243, the ordinate
A NBN of half the arc A D = BD, by putting
Y 1
2
A B² = A M² + B M² = A M² + (C B
CM)³
A M² + ( C B — √ C A² — A M²)² — 2 C A² – 2 C A √˜˜˜Â²
= B—
CA²
2 2 r2 - 2 r √r² — y².
2
A M²,
A
I.E. 4 Y1
FIG. 243.
Hence we have

p2
Y 1
=
r V x2
2
—
yo
or approximatively, if y is
small compared with r,

y 2
4
Y1
p
2 r
873
M
N
D
y2
4
y 2
Y
y 2
1 +
1 +
4 72
2
8 r2
B
By repeated application of this formula we find the
ordinate of a quarter of the arc
Y₁
1
2
y 2
y 2
3 = (1+3)=(1+8) (1+4)
1 ½ (1 + 2 1 2 ) = 1
/ (1 + 3/2 ) ( 2
Y 2
2
and that of an eighth of the arc
ปูน
r
y2
2
Y s
1 ½ (1 + 2; 2)
=
Y
1/
8
(1 + 2/2 ) ( 1 + ±· 3,7%) (1 + (4)³ 3,7%)
y 2
8 r2
8
1 + [1 + †
8p2
+ (1)³) 1/2-2).
8 r2
•
Since the ordinates of very small arcs can be put equal to the arcs
themselves, we obtain for the arc A B approximatively
§ 161.]
299
EQUILIBRIUM IN FUNICULAR MACHINES.
y2
8 = 8 . Y₂ = y (1 + [1 + 3 + (1)²] 31).
3
1
= y (1 + [1 + 1 + (1)² + (1)³ +
is =
But 1++ (1)² + (1)³ +
(1)² + (1)³ + ... is
8 r2
1
•
•
]
or more accurately
y2
Love).
8r2
(see Ingenieur, page 82),
and therefore
8 =
(1
y2
1 +
6 r2
):
y;
or substituting instead of r the abscissa B M = ≈ by putting 2 r x = y²,
we obtain
= [1 + ÷ (†) ' ] 3
=
2/
This formula is not only applicable to the arc of a circle, but also to all
low arcs of curves.
REMARK 2. If we compare the equation
FIG. 244.
x
Y V 2 c x
3'
υ
K
E
b
found above, with the equation of the ellipse

b
Y
V 2 a x
X2
a
y
C
X
B
(see Ingenieur, page 169), we find
b2
a
b2
a2
= c and =, and consequently
a = 3 c and b = a √1 = c v3.
The curve formed by a powerfully stretched string can therefore be
considered as the arc A CB, Fig. 244, of an ellipse, the major axis of which
a = 3 c and the minor axis is K D = K E = b = c v3
0,577 a.
is K Ca =
a √ 3
=
(§ 161.) Equation of the Catenary.-The complete equa-
tion of the catenary can be found in the following manner by the
aid of the calculus. According to § 158, we have for the angle of
R
P
Y
S
FIG. 245.
M
T
N
Z X
B
=
suspension TON o. Fig. 245,
formed by the tangent O T to a
point O of the catenary A CB with
the horizontal co-ordinate O N
when the are CO is denoted by s
and the horizontal tension by H

CY,
tang. =
p C
But is also equal to the angle
O PR formed by the element of
300
[$ 162.
GENERAL PRINCIPLES OF MECHANICS.
the arc O P = ds with the element P R
ds with the element PR=dy of the ordinate
O N = y, and hence
OR
tang. O PR=
PR
d x
dy
in which O R is considered as an element d x of the abscissa C N
= x.
x. From the above it follows, that
d y
2.
d x
S d y²
C²
or
C
dx²
S
= d s²
dx²
d x², whence
c²
S
82.
But d s² is = d x² + d y², or d y²
d s²
dx²
な
​Clearing the equation of fractions and transposing, we obtain
dx² (s² + c²) = sd s2, or d x =
s d s
s² + c²°
Putting s² + c² = u, we have
2 s ds = du and d x
i du
=
u du.
udu.
U s
By integration we obtain (according to Article 18 of the In-
troduction to the Calculus)
X
ບ
- du:
+ Const. = √u + Const.
12
s² + c² + Const.
Finally, since for x = 0, s is also = 0, we have 0
0, we have 0 = √ c² + Const.,
I.E. Const. =
1) x =
S =
C =
c and
√ s² + c²
√ s² + c² - c, or inversely
√ (x + c)³ — c² :
s²
2 x
x²
√ 2 c x + x², and
EXAMPLE.—If a chain A C B, 10 feet long and weighing 30 pounds, is
suspended in such a manner that the height of the arc is C M = 4 feet, we
have
C
= 33808
$2
2 x
3 pounds,
x2
52 42
8
},
H = c y = 3. f
33 pounds.
and consequently the horizontal tension
(§ 162) As in the last paragraph by eliminating d y we obtained
an equation between the arcs and the abscissa x, in like manner
by eliminating d x we can deduce an equation between the arc s
and the ordinate y. For this purpose we substitute in the equation
§ 162.]
301
EQUILIBRIUM IN FUNICULAR MACHINES.
*
d y²
c²
d x²
—
ç³, d x² = d s² — d y³
2
and obtain the equation
>
c²
d y²
c d s
d s² — d y², or d y² (s² + c²) = c² d s², whence
dy
s² + c²°
Dividing the numerator and denominator by c and putting
s
=v, we obtain ·
C
d y
cd (2)
√1+()
c d v
√ 1 + v²
and the formula XIII, in Article 26 of the Introduction to the
Calculus, gives us the corresponding integral
y = c S
d v
√1 + v²
2) y = cl
c l (v + √ 1 + v³), I.E.
+ √ s² + c²
C
Substituting in this formula s = √2cx+x², we obtain the
proper equation for the co-ordinates of the common catenary
'c + x + √ 2 c x + x²
y = c 1 (c +
3) y =
cl
C
P?
or
4) y = c1 ( + 2) = ²² = 2² 1 (0 ± 2)
2 x
(8 +
X
Finally, by inverting 2 and 3, we obtain
5) s =
=
ec
and
e
6) x = [(+)-1]
X
e
C,
e denoting the base 2,71828 ... of the Naperian system of loga-
rithms (see Article 19 of the Introduction to the Calculus).
EXAMPLE.—The two corresponding co-ordinates of a point of the cate-
nary are x = 2 and y
3; required the horizontal tension c of this curve.
Approximatively, according to No. 3 of paragraph 160, we have
C =
y2
+
x
2 x
6
11
9
2
+
2,58.
4
6
302
[§ 162.
GENERAL PRINCIPLES OF MECHANICS.
But according to No. 3 of this paragraph (162), we have exactly
c + x + √ 2 c x + x²
y = c l
I.E.
C
3 cl
√ 40 + 4)
'c + 2 + √ 4 c + 4
C
Substituting for c, 2,58, we find the error
f
ƒ = 3
2,58 7
= 3
4,58 + 2√3,58
2,58
3,035 = – 0,035.
= 3 — 2,58 l
8,3642
2,58
If, however, we assume c = 2,53, we find the error
f
ƒ₁ = 3 — 2,53 1 (4,53
4,53 + 2 √ 3,53
2,53
= 8 - 2,58 1 (8,2,976)
1
2,53
= 3 3,002
0,002.
In order to find the true value of c, we put according to a well known
rule (see Ingenieur, page 76)
C 2,58 f
C 2,53 f₁ 0,002
0,035
17,5;
41,69 and
whence it follows that 16,5. c = 17,5. 2,53 - 2,58
c =
41,69
16,5
2,527 feet.
REMARK.—We can express very simply 8, x and y for the common cate-
nary in terms of the angle of suspension ; for from what precedes we have
8 = c tang. $ =
c sin. Ф
cos.
c (1
cos. )
x = c(√ 1 + tang.” ¢ — 1)
and
cos. o
y = cl (tang. p + √ 1 + tang." ()
c l
(
1 + sin.
cos.
By means of these formulas we can easily calculate the lengths of the
arcs and of the co-ordinates for different angles of suspension, and a useful
table, such as is given in the Ingenieur, page 353, may be thus prepared.
For this purpose we need adopt as base but a single catenary, and in this
case the best one is that, in which the measure of the horizontal tension is
= 1; to obtain s, x and y for another catenary corresponding to the hori-
zontal tension c, we have but to multiply the values of s, x and y given in
ย
we would have the com-
the table by c. If tang. & were not
› mon parabola, for which
8
but to
с
8 =
sin.
+ I tang.
cos." o
П+
2
C
c (sin.
x
tang. p
2 cos.
(sin. 6)
and
c sin.
y = c tang. O
cos.
§ 163.]
303
EQUILIBRIUM IN FUNICULAR MACHINES.
§ 163. Equilibrium of the Pulley.-Ropes, belts, etc., are
the ordinary means employed to transmit forces to the pulley and
the wheel and axle. We will here discuss only the most general part
of the theory of these two apparatuses, so far as it can be done with-
out taking into consideration the friction and the rigidity of cordage.
A pulley (Fr. poulie; Ger. Rolle) is a circular disc or sheave
A B C, Figs. 246 and 247, movable about an axis and around
FIG. 246.
FIG. 247.


T
P
A
R
Q
D
B
P
A
B
D
whose circumference a string is laid, the extremities of which are
pulled by the forces P and Q. The block (Fr. chape; Ger. Gehäuse
or Lager) of a fixed pulley (Fr. p. fixe; Ger. feste R.), in which the
axles or journals rest, is immovable. That of a movable pulley
(Fr. p. mobile; Ger. lose R.) on the contrary is free to move.
When a pulley is in equilibrium, the forces P and Q at the ex-
tremities of the cord are equal to each other; for every pulley is a
lever with equal arms, which we obtain by letting fall from the
axis the perpendiculars CA and C B upon the directions D P
and DQ of the forces or cords. It is also evident, that during any
rotation about the forces P and Q describe equal spaces r 3.
when r denotes the radius C A = CB and 3° the angle of rotation.
and from this we can conclude, that P and Q are equal. The forces
P and Q give rise to a resultant CR = R, which is counteracted
by the journal or axle and is dependent upon the angle A D B = a
formed by the directions of the cords, it is given by construction
as the diagonal of the rhomb CP, R Q, constructed with P and a;
its value is
R = 2 P cos.
α
201
304
[S 164.
GENERAL PRINCIPLES OF MECHANICS.
§ 164. The weight to be raised or the resistance Q to be overcome
in a fixed pulley, Fig. 246, acts exactly in the same manner as the
force P, and the force is therefore equal to the resistance, and
the use of this pulley produces no other effect than a change of
direction.
On the contrary, in a movable pulley, Fig. 247, the weight R
acts on the hook-shaped end of the bearings of the axle, while one
end of the rope is made fast to some immovable object; here the
force is
P
R
a
2 cos.
2
Designating the chord A M B corresponding to the arc covered
by the string by a and the radius C A C B, as before, by r, we
have
=
2 AM = 2 C A cos. C A M = 2 C A cos. A D M = 2 r cos.
and therefore
a
2
r
1
P
r
and
a
α
R
a
2 cos.
2
Hence, in a movable pulley, the force is to the load as the
radius of the pulley is to the chord of the arc covered by the string.
If a = 2 r, I.E. if the string covers a semicircle, Fig. 248, the
P
FIG. 248.
force is a minimum and is P
R; if a = r or
if 60° of the pulley is covered by the string, we
have P = R. The smaller a becomes, the greater
is P; I.E., when the arc covered by the cord is
infinitely small, the force P is infinitely great.
The relation is inverted, when we consider the
spaces described; if s is the space described by
P, while R describes the space h, we have Ps
Rh, whence

A
B
S
a
R
h γ
The movable pulley is a means of changing
the force, and is used to gain power; by means
of it we can, E.G., raise a given load with a smaller
force; but in the same ratio as the force is in-
creased the space described is diminished.
§ 165.]
305
EQUILIBRIUM IN FUNICULAR MACHINES.
REMARK.-The combinations of pulleys, such as block and tackle, etc.,
as well as the influence of friction and of the rigidity of cordage upon the
state of equilibrium of pulleys, will be treated in the third volume.
FIG. 249.
R₁
-P
V
§ 165. Wheel and Axle.—The wheel and axle (Fr. roue sur
l'arbre, Ger. Radwelle) is a rigid combination AB FE, Fig. 249,
of two pulleys or wheels mov-
able about a common axis.
The smaller of these wheels
is called the axle (Fr. arbre,
Ger. Welle), and the larger
the wheel (Fr. roue, Ger. Rad).
The rounded ends E and F,
upon which the apparatus
rests, are called the journals
(Fr. tourillons, Ger. Zapfen).
The axis of revolution of a
wheel and axle is either hori-
zontal, vertical or inclined.
We will now discuss only
the wheel and axle, movable
around a horizontal axis. We

H
B
P
S
F
D
H₂
V/2
Ra
P
E
will also suppose, that the forces P and Q or the force P and the
weight Q act at the ends of perfectly flexible ropes, which are
wound around the circumferences of the wheel and of the axle.
The questions to be answered are, what is the relation between the
force P and the weight Q, and what is the pressure upon the bear-
ings at E and F?
If at the point C, where the plane of rotation of the force P
cuts the axis E F, we imagine two equal opposite forces CP = P
and CPP to be acting in a direction parallel to that of the
force of rotation P, we obtain by the combination of these three
forces a force C P = P, which acts upon the axis, and a couple
— =
(P, P), whose moment is P. CA Pa, when a designates
=
the arm of the force A P Por the radius CA of the wheel.
Now if we imagine the two forces DQ = Q and D Q = Q to be
applied at the point D, where the plane of revolution of the weight
Q cuts the axis E F, we obtain also a force D Q = Q acting upon
the axis and a couple (Q, — Q), whose moment is =
-Q), whose moment is Q. D B = Qb,
when designates the arm of the weight Q applied in B or the
!
20
306
[§ 166.
GENERAL PRINCIPLES OF MECHANICS.
radius D B of the axle. Since the axial forces C P = P and
D Q = Q are counteracted by the bearings, and consequently
can have no influence upon the revolution of the machine, it is
necessary, in order to have a state of equilibrium, that the two
couples, which act in parallel planes, shall have equal moments
(compare § 94), or that
Pa=
Q b, or
P
b
Q
a
In every wheel and axle which is in equilibrium, whatever may
be its length, the moment P a of the power is, as in the lever, equal to
the moment Qb of the load, or the ratio of the power to the load is
equal to that of the arm of the load to the arm of the power.
If more than two forces act upon the wheel and axle, the sum
of moments of the forces tending to turn it in one direction is
naturally equal to the sum of those tending to turn it the other.
2
=
§ 166. The axial forces C P P and D Q
Q can be
decomposed into the vertical forces CP, = P, and D Q₁ = Q, and
into the horizontal forces C P. P, and D Q₂ = 2; the first two
forces combined with the weight of the machine G, which acts at
the centre of gravity S of the machine, give the total vertical
pressure on the bearings, which is
V₁ + V₁ = P₁ + Q₁ + G,
2
while the horizontal forces P, and Q. produce the lateral pressures
H₁ and H, on the bearings. If a is the angle of inclination P CP,
of the direction of the force P to the horizon and ẞ that Q D Q.
of the load, we have
P₁ = P sin. a and P₂
1
Q₁
=Q sin. ẞ and Q2
P cos. a, as well as
Q cos. B.
If now 7 is the total length of the axis E F, d the distance CE,
e the distance D E and c the distance S E of the points of the axis
C, D and S from one extremity E of the axis, we have, according to
the theory of the lever:
1) When we consider E as fulcrum of the lever E F, which is
acted on by the forces P₁, Q₁ and G,
V₁. EF P₁. ECQ.ED+G.ES, I.E.
1
V₂ l = P₁d + Q₁e + G 8,
2
§ 166.
307
EQUILIBRIUM IN FUNICULAR MACHINES.
whence we obtain the vertical pressure
V₂
P₁d + Q₁e + G s
}
and
>
2) considering F as the fulcrum of the supposed lever,
V₁. FE P₁. FC+ Q₁. FDG. FS, I.E.
1
=
V₁l = P₁ (l - d) + Q₁ (l—e) + G (1 − s),
whence we deduce the vertical pressure
V₁
P₁ (l - d) + Q₁ (1 − e) + G (18)
(?
7
F
FIG. 249.
R₁

A
I
H
B
P
E
D
H₂
P
I
Rv2
The horizontal pressures H₁ and H, are found, as follows, from
the horizontal forces P. and Q.
1) Considering E as the fulcrum of the lever E F acted on by
the forces P₂ and Q, we obtain
2
2
H₁. EF
P₂. EC-Q.. ED, I.E.
H₁l P₂d - Q₂ e,
whence we obtain the horizontal pressure
H₂ =
P₂ d Q₂ e
1
2) Considering Fas the fulcrum, we have
H₁.FEP,.FC-Q..F D, I.E.
H₁l P₂ (l - d) Q₂ (le),
from which we deduce the horizontal pressure
P₂ (l - d) — Q. (? —
(1 e)
-
H₁ =
7
308
[$ 166.
GENERAL PRINCIPLES OF MECHANICS.
By the application of the parallelogram of forces, we obtain the
total pressures R, and R, upon the bearings E and F, and they are
R₁ = √ V₁² + H," and R₂ = √ √½² + H².
1
2
Finally, if d, and d, are the angles R, E H, and R, FH, formed
by these pressures with the horizon, we have
tang. 8,
V₁
H₁
V₁
and tang. d₂
H₂
4
=
EXAMPLE.-The weight Q, suspended to a wheel and axle, acts verti-
eally and weighs 365 pounds; the radius of the wheel is a = 12 feet; the
radius of the axle is b = foot; the weight of the wheel and axle together
is 200 pounds; the distance of its centre of gravity from the journal E is
1 feet; the centre of the wheel is at a distance d from this journal
E, and the vertical plane, in which the weight acts, is e = 2 feet distant
from the same point, while the whole length of the axis is EF = l = 4
feet; now if the force necessary to produce equilibrium acts downwards at
an angle of inclination to the horizon of a = 50°, how great must it be and
what are the pressures upon the bearings? Here we have Q
90°, and consequently Q₁
unknown, and a is = 50°, whence P
is = P cɔs. a =
=
=
=
365, ß
Q cos. B 0, P is
0,7660. P and P₂
2
Q sin. B
Q
and Q2
P sin. a =
1
0,6428. P, but a is
12 =
and b
=
å, whence
b
P
Q = .365
156,4 pounds, P₁
=
1
a
Since 7 = 4,
4, d ž, e
2 and 8 = , we have l d
=
13,
- e e = 2
=
119,8 and P₂ = 100,5 pounds.
and l 8
1) On the bearing F the vertical pressure is
V₂
4
119,8. 3 + 365 . 2 + 200. §
4
and the horizontal pressure is
=280,0 pounds,
H₂
100,5. 2
4
0.2
=
18,8 pounds,
R₂ √ √₂² + H₂²
and consequently the resulting pressure is
2
2
and its inclination to the horizon is determined by the formula
280,0
tang. $2
log tang. S₂ = 1,17300, from which we obtain d
18,8'
2) For the bearing at E
1/2802 + 18,82
=
280,6 pounds,
86° 9' 5".
2
119.8.18+ 365. 2 + 200. §
=404,8 pounds and
V₁
4
日​々
​100,5 . 13
4
0
81,7 pounds,
and consequently the resulting pressure is
Ꭱ
R₁ = √ V₁2 + H₁2
1
²
²
=
1/404,82 +81,72 = 413,0 pounds,
§ 167.]
309
RESISTANCE OF FRICTION, ETC.
and for its inclination d₁ to the horizon we have
1
tang. d₁
1
404,8
81,7
1
log tang. 8₁ = 0,69502 or d
78° 35'.
1
We see that these results are correct, for we have
V₁ + √ ₂ 280 + 404,8
1
2
H₁ + H₂ = 81,7 + 18,8
P₁ + Q₁ + G, and
1
=
684,8
100,5 =
P₂ + Q2.
2
CHAPTER V.
THE RESISTANCE OF FRICTION AND THE RIGIDITY OF CORDAGE.
§ 167. Resistance of Friction.-Heretofore we have sup-
posed (§ 138) that two bodies could act upon one another only by
forces perpendicular to their common plane of contact. If these
bodies were perfectly rigid and their surfaces of contact mathemat-
ical planes, I.E. unbroken by the smallest hills or hollows, this law
would also be confirmed by experiment; but since every material
body possesses a certain degree of elasticity or even of softness, and
since the surface of all bodies, even the most highly polished, con-
tains small hills and valleys and in consequence of the porosity of
matter does not form a perfectly continuous plane, when two bodies
press upon each other their points of contact penetrate, pro-
ducing an adhesion of the parts, which can only be overcome by a
particular force, whose direction is that of the plane of contact.
This adhesion of bodies in contact, produced by their mutual pene-
tration and grasping of each other, is what is called friction (Fr.
frottement, Ger. Reibung). Friction presents itself in the motion
of a body as a passive force or resistance, since it can only hinder
or prevent motion, but can never produce or aid it. In investiga-
tions in mechanics it can be considered as a force acting in opposi-
tion to every motion, whose direction lies in the plane of contact
of the two bodies. Whatever the direction may be in which we
move a body resting upon a horizontal or inclined plane, the fric-
tion will always act in the opposite direction to that of the motion,
E.G., when we slide the body down an inclined plane, it will appear
as motion up the same. If a system of forces is in equilibrium, the
smallest additional force produces motion as long as the friction
does not come into play; but when friction is called into existence
a greater addition of force, the amount of which depends upon the
friction, is necessary to disturb the equilibrium.
310
[$ 168, 169.
GENERAL PRINCIPLES OF MECHANICS.
§ 168. In overcoming the friction, the parts which come in
contact are compressed, the projecting parts bent over, or perhaps
torn away, broken off, etc. The friction is therefore dependent not
only upon the roughness or smoothness of the surfaces, but also
upon the nature of the material of which the bodies are composed.
The harder metals generally cause less friction than the softer
ones. We cannot establish á priori any general rules for the de-
pendence of friction upon the natural properties of bodies; it is in
fact necessary to make experiments upon friction with different
materials, in order to be able to determine the friction cxisting
between bodies under other circumstances. The unguents (Fr. les
enduits; Ger. die Schmieren) have a great influence upon the
friction and upon the wearing away of bodies in contact. The
pores of the bodies are filled and the other roughnesses diminished
by the fluid or half fluid unguents, such as oil, tallow, fat, soaps,
etc., and the mutual penetration of the bodies much diminished;
for this reason they diminish very considerably the friction.
But we must not confound friction with adhesion, I.E., with
that union of two bodies which takes place when the bodies come
in contact in very many points without the existence of any pres-
sure between them. The adhesion increases with the surface of
contact and is independent of the pressure, while for friction the
reverse is true. When the pressures are small, the adhesion appears
to be very great compared with the friction, but if the pressures
are great, it becomes but a very small portion of the friction and
can generally be neglected. Unguents generally increase the adhe-
sions, since they produce a greater number of points of contact.
§ 169. Kinds of Friction-We distinguish two kinds of
friction, viz., sliding and rolling friction. The sliding friction
(Fr. frottement de glissement; Ger. gleitende Reibung) is that
resistance of friction produced, when a body slides, I.E., moves so
that all its points describe parallel lines. Rolling friction (Fr. f. de
roulement; Ger. rollende or wälzende Reibung) on the contrary,
is that resistance developed, when a body rolls, I.E., when every
point of the body at the same time progresses and revolves and
when the point of contact describes the same space upon the
moving body as upon the immovable one. A body M, Fig. 250,
supported on the plane II R, slides, for example, upon the plane
and must overcome sliding friction, when all points such as A, B, C,
etc., describe the parallel trajectories A A, B B, C C, etc., and
therefore the same points of the moving body come in contact with
§ 170.]
311
RESISTANCE OF FRICTION, ETC.
different ones of the support. The body M, Fig. 251, rolls upon
the plane H R and must therefore overcome rolling friction, when
FIG. 250.
H
C
M
M
B
FIG. 251.


C
B A
B₁
H
R
R
the points A, B, etc., of its surface move in such a manner, that
the space A E B₁ = the space A D B A, D, B, and also that
space A Eis
the space A D and the space B₁ E B₁ D₁, etc.
A particular kind of friction is the friction of axles or journals
which is produced, when a cylindrical axle, journal or gudgeon
revolves in its bearing. We distinguish two kinds of axles, hori-
zontal and vertical. The horizontal axle, journal or gudgeon
(Fr. tourillon; Ger. liegende Zapfen) moves in such a manner that
different points of the gudgeon, etc., come successively in contact
with the same point of the support. The vertical axle or pivot
(Fr. pivot; Ger. stehende Zapfen) presses with its circular base
upon the step, on which the different points of it revolve in con-
centric circles.
Particular kinds of friction are produced, when a body oscillates
upon an edge, as, E.G., a balance, or when a vibrating body is sup-
ported upon a point, as, E.G., the needle of a compass.
Friction can also be divided into immediate (Fr. immédiat ;
Ger. unmittelbare) and mediate (Fr. médiat; Ger. mittelbare). In
the first case the bodies are in immediate contact; in the latter,
on the contrary, they are separated by unguents, as, E.G., a thin
layer of oil.
We distinguish also the friction of repose or quiescence (Fr. f. de
répos; Ger. R. der Ruhe), which must be overcome when a body
at rest is put in motion, from the friction of motion (Fr. f. de
mouvement; Ger. R. der Bewegung), which resists the continuance
of a motion.
§ 170. Laws of Frictions.-1. The friction is proportional
to the normal pressure between the rubbing bodies. If we press
a body twice as much against its support as before, the friction
becomes double. A triple pressure gives a triple friction, etc.
If this law varies slightly for small pressures, we must ascribe the
variations to the proportionally greater influence of the adhesion.
312
[S 171.
GENERAL PRINCIPLES OF MECHANICS.
2. The friction is independent of the rubbing surfaces or sur-
faces of contact. The greater the rubbing surfaces the greater is,
it is true, the number of the rubbing parts, but the pressure upon
each part is so much the smaller, and consequently the resistance
of friction upon it is less. The sum of the frictions of all the parts
is therefore the same for a large and for a small surface, when the
pressure and other circumstances are the same. If the surfaces of
the sides of a parallelopipedical brick are of the same nature, the
force necessary to move the brick on a horizontal plane is the same
whether it lies on the smallest, medium, or greatest surface. When
the surfaces are very great and the pressures very small, this rule
appears to be subject to exceptions on account of the effect of the
adhesion.
3. The friction of quiescence is generally greater than that of
motion, but the latter is independent of the velocity; it is the
same for high and low velocities.
4. The friction of greased surfaces (mediate friction) is gene-
rally smaller than that of ungreased surfaces (immediate friction)
and depends less upon the rubbing bodies themselves than upon
the unguent.
5. The friction on axles is less than the ordinary friction of
sliding. The rolling friction between smooth surfaces is in most
cases so small, that we need scarcely take it into account in com-
parison with the friction of sliding.
REMARK.-The foregoing rules are strictly true only, when the pressure
upon the unit of surface of the bearings is a medium one, and when the
velocity of the circumference of the journal does not exceed certain limits.
This medium pressure is from 250 to 500 pounds per square inch, and the
mean velocity of the circumference should be 2 to 10 inches. When the
pressure is much smaller, the adhesion forms a very sensible portion of the
resistance which then becomes dependent upon the magnitude of the rub-
bing surfaces, and where the pressure and velocity are very great a large
quantity of heat is developed, which volatilizes the unguents, thus causing
the journals to cut very quickly. When, as in the case of turbines, rail-
road cars, etc., we cannot avoid these
great velocities, we must counteract this
heating of the axle by increasing the rub-
Fbing surfaces, 1.E., by increasing the length
and thickness of the axles.
D
H
A
FIG. 252.
B

N
R
§ 171. Co-efficient of Friction.
-From the first law of the foregoing
paragraph we can deduce the fol-
lowing. If in the first place a body
§ 171.]
313
RESISTANCE OF FRICTION, ETC.
A C, Fig. 252, presses with a force N against its support, and if
to move it along, I.E., to overcome its friction, we require the
force F, and if in the second place, when pressing with the force
N₁ a force F is necessary to transfer it from a state of rest into
one of motion, we will have, according to the foregoing paragraph,
1
F N
F N
F 1
whence F =
N.
N₁
If by experiment we have found for a certain pressure N, the
corresponding friction F, we see from the above, that if the rub-
bing bodies and other circumstances are the same, the friction F
corresponding to another pressure N can be found by multiplying
this pressure by the ratio
F
() between the values F, and N, cor-
responding to the first observation.
This ratio of the friction to the pressure or the friction for a
pressure 1, E.G. pound, is called the coefficient of friction
(Fr. coëfficient du frottement; Ger. Reibungscoefficient) and will
in future be designated by p. Hence we can put in general
F = Φ Ν.
The coefficient of friction is different for different materials
and for different conditions of the same material and must there-
fore be determined by experiments undertaken for that purpose.
If the body AC is pulled along a distance s upon its support, the
work to be performed is Fs. The mechanical effect o N s ab-
sorbed by the friction is equal to the product of the coefficient of
friction, the normal pressure and the space described. If the sup-
port is also movable, we must understand by s = 81 8, the relative
space described by the body, and Fs Ns is the work done by
the friction between the two bodies. The body that moves the
most quickly must perform, while describing the space s₁, the me-
chanical effect o N s, and the body which moves slower gains in
consequence of the friction while describing the space s, the me-
chanical effect o N s₂; the loss of mechanical effect caused by the
friction between the two bodies is
81
=
4 N 81 - 4 N 8₂ = N (8, 8) = N 8.
Φ
&
EXAMPLES-1. If for a pressure of 260 pounds the friction is 91 pounds,
the corresponding coefficient of friction is o = 260 2 = 0,35.
91
2. In order to pull forward a sled weighing 500 pounds on a horizontal
and very smooth snow-covered road, when the coefficient of friction is
0,04, a force F = 0,04 . 500 = 20 pounds is necessary.
3. If the coefficient of friction of a sled loaded with 500 pounds and
314
[§ 172.
GENERAL PRINCIPLES OF MECHANICS.
pulled over a paved road is 0,45, the mechanical effect required to move
the sled 480 feet is ☀ N 8 – 0,45 . 500. 480 — 108000 foot-pounds.
FIG. 253.
§ 172. The Angle of Friction or of Repose and the
Cone of Friction.-If a body
A C, Fig. 253, lies upon an in-
clined plane F H, whose angle of
inclination is F H R = a, we can
decompose its weight into the nor-
mal pressure N
G cos. a, and
B
F

A
S
K
H
into the force S
G sin. a paral-
R
lel to the plane.
The first force
causes the friction F
o G cos. a,
which resists every motion upon
the plane; consequently the force necessary to push the body up
the plane is
P = F + S =
¢ G cos. a + G sin. a
(sin. a +
cos. a) G,
and the force necessary to push
it down the same is
sin. a) G.
P₁ = F — S = ($ cos. a
The latter force becomes = 0, 1.E. the body holds itself upon
the inclined plane by its friction when sin. a = cos. a, I.E. when
tang. a = 4. As long as the inclined plane has an angle of incli-
nation, whose tangent is less than o, so long will the body remain
at rest upon the inclined plane; but if the tangent of the angle of
inclination is a little greater than 4, the body will slide down the in-
clined plane. We call this angle, I.E. the one whose tangent is equal
to the coefficient of friction, the angle of friction or of repose or of
resistance (Fr. angle du frottement, Ger. Reibungs-or Ruhewinkel).
Hence we obtain the coefficient of friction (for the friction of qui-
escence) by observing the angle of friction p and putting tang. p.
In consequence of the friction, the surface FH, Fig. 254, of a
body counteracts not only the normal pressure N of another body
F
N
FIG. 254.
D
A
E
A B, but also any oblique pressure P when
the angle N B P = a formed by its direc-
tion with the normal to the surface does
not excced the angle of friction; since the
force P gives rise to the normal pressure
BN = P cos. a, and to the lateral or
tangential pressure BSSP sin. a
and since the normal pressure P cos, a pro-
duces the friction & P cos. a, which opposes

كل
315
173.]
RESISTANCE OF FRICTION, ETC.
every movement in the plane F H, S can produce no motion as
long as we have
o P cos. a > P sin. a or o cos. a > sin. a, I.E.
tang. a < por a < p.
If we cause the angle of friction C B D = p to revolve about
the normal C B, it describes a cone, which we call the cone of fric-
tion or of resistance (Fr. cone de fr., Ger. Reibungskegel). The
cone of friction embraces the directions of all the forces, which are
completely counteracted by the inclined plane.
EXAMPLE.-In order to draw a full bucket weighing 200 pounds up a
wooden plane inclined to the horizon at an angle of 50°, the coefficient of
friction being p = 0,48, we would require a force
་
P = († cos, a + sin. a) G
(0,48 cos. 50° + sin. 50°). 200
(0,308 +'0,766). 200 = 215 pounds.
In order to let it down or to prevent its sliding down, we would have need
of a force
P₁
=
( cos. a
sin. a) G
(0,766 — 0,308). 200
(sin. 50°
0,48 cos. 50°). 200
91,5 pounds.
§ 173. Experiments on Friction.-Experiments on friction
have been made by many persons; those, which were most ex-
tended and upon the largest scale, are the experiments of Coulomb
and Morin. Both these experimenters employed, for the determina-
tion of the coefficient of friction of sliding, a sled movable upon a
horizontal surface and dragged along by a rope passing over a fixed
pulley, to the end of which a weight was attached, as is shown in
Fig. 255, in which 4 B is the surface, CD the sled, E the pulley,
and F the weight. In order to obtain the coefficients of frictions
for different substances, not only the runners of the sled, but also
the surface upon which it slid, were covered with the smoothest
possible plates of the material to be experimented on, such as wood,
iron, etc. The coefficients of friction
E
FIG. 255.
C
of rest were given by the
weight necessary to bring
the sled from a state of
rest into motion, and the
coefficients of friction of
motion were determined
by aid of the time required
by the sled to describe a
certain space 8. If G is
the weight of the sled and

P the weight necessary to move the same, we have the friction
316
[§ 173
GENERAL PRINCIPLES OF MECHANICS.
G, the moving force = P
P - ф G and the mass M
PG,
9
whence, according to § 68, the acceleration of the uniformly acceler-
ated motion engendered is
and inversely the coefficient of friction is
P
& G
p
P + G 9,
P
Ф
G
P + G Ρ
G g
28
But we have also (§ 11) spt', whence p = and
t²
P
PG 2 s
Ø
G
G g to
If we allow the sled to slide down an inclined plane, the moving
force is G (sin. a
consequently the acceleration is
cos. a), and the accelerated mass is =
G
g
p =
2 s
t³
G (sin. a
o cos. a)
g (sin. a
G
o cos. a)
g
or
2 s
g t
=
friction is
sin. a cos. a, and consequently the coefficient of sliding
28
Ф
p = tang. a
g t² cos. a
If h denotes the altitude, 7 the length and a the base of the
inclined plane, we have also =
h
28 1
I a t²°
a
In order to determine the coefficient of friction for the friction
of axles or journals, they employed a fixed pulley A CB, Fig. 256,
around which a rope was wound, to which the weights P and Q
were suspended; from the sum of the weights P + Q we have
the pressure R upon the axle, and from their difference P
Q the
force at the periphery of the pulley, which is held in equilibrium by
the friction F = (P + Q) on the surface of the axle.
If now
CA
a= the radius of the axle and CD=r the radius of the
journal, we have, since the statical moments are equal,
(P − Q) a = F r = ¢ (P + Q) r,
and consequently the coefficient of friction of rest
P
Q
α
Ф
P + Q
2³
and, on the contrary, when the weight P falls and Q rises in the
time t a distance s, the coefficient of friction of motion is
$173.]
317
RESISTANCE OF FRICTION, ETC.
P
Q
2 s\ a
P + Q
g
9 t
t²
r
The engineer Hirn employed in his (the latest) experiments
upon friction of journals the apparatus represented in Fig. 257,
FIG. 256.
FIG. 257.


A
B
A
R
Р
R
P
B
which he called a friction balance (Fr. balance de frottement, Ger.
Reibungswage). Here is an axle, which is kept in constant
rotation, as, E.G., by a water-wheel, D is the bearing, and AD B
is a lever of equal arms, which produces the pressure between
the journal and its bearing by means of the weights P and Q. The
pressure on the axle R= P+Q produces the friction
F = 4 R = $ (P + Q)
between the journal and its bearing. With this force the revolving
shaft seeks to turn the bearing and the lever A D B, which is attached
to it, in the direction of the arrow; and therefore, in order to keep
the whole in equilibrium, we must make the weight P on one side
A so much greater than the weight Q on the other, that P Q
will balance the friction. But the friction F acts with the arm
C D = r = the radius of the bearing and the difference of the
weights PQ with the arm C4 = a, which is equal to the hori-
CA
zontal distance between the axis C of the shaft and the vertical
line through the point of suspension A, and therefore we have
Fr=&R r = ¢ (P + Q) r = (P — Q) a,
φ
and the coefficient of friction required
$
P
Q
a
P + Q
REMARK.—Before Coulomb, Amontons, Camus, Bülffinger, Muschen-
brock, Ferguson, Vince and others had studied the subject of friction and
made experiments upon it. The results of all these researches have, however,
little practical value; for the experiments were made upon too small a scale.
The same objection applies to those of Ximenes, which were made about
318
[§ 174.
GENERAL PRINCIPLES OF MECHANICS.
the same time as those of Coulomb. The results of Ximenes are to be found
in the work "Teoria e Pratica delle resistenze de' solidi ne' loro attriti,
Pisa, 1782." Coulomb's experiments are described in detail in the work:
"Théorie des machines simples, etc., par Coulomb. Nouv. édit., 1821."
An abstract from it is to be found in the prize essay of Metternich, "Vom
Widerstande der Reibung, Frankfurt und Mainz, 1789." The later experi-
ments on friction were made by Rennie and Morin. Rennie employed in
his experiments in some cases a sled, which slid upon a horizontal surface,
and in others an inclined plane, down which he caused the bodies to slide,
and from the angle of inclination determined the amount of the friction.
Rennie's experiments were made with most of the substances, which we
meet with in practice, such as ice, cloth, leather, wood, stone and the
metals; they also give important data in relation to the manner in which
bodies wear, but the apparatus and the manner of conducting these experi-
ments do not allow us to hope for as great accuracy as Morin scems to have
attained in his experiments. A German translation of Rennie's Experiments
is to be found in the 17th volume (1832) of the Wiener Jahrbücher des
K. K. Polytechnischen Institutes, and also in the 34th volume (1829) of
Dingler's Polytechnisches Journal. The most extensive experiments and
those, which probably give the most accurate results, are those made by
Morin, although it cannot be denied that they leave certain points doubtful
and uncertain, and that here and there there are points, upon which more
information could be desired. This is not the place to describe the method
and apparatus employed in these experiments; we can only refer to Morin's
writings: "Nouvelles Expériences sur le frottement," etc. A capital discus-
sion of the subject "friction," and a rather full description of almost all the
experiments upon it, Morin's included, is given by Brix in the transactions
of the Society for the Advancement of Industry in Prussia, 16th and 17th
Jahrgang-Berlin, 1837 and 1838. Later experiments on mediate friction,
with particular reference to the different unguents, made by M. C. Ad.
Hirn, are described in the "Bulletin de la société industrielle de Mulhouse,
Nos. 128 and 129, 1855," under the title of "Etudes sur les principaux
phénomènes que présentent les frottements médiats, etc.;” an abstract of it
is to be found in the "Polytechnisches Centralblatt, 1855. Lieferung, 10.”
The latest researches upon friction by Bochet are described under the title,
"Nouv. Recherches expérimentales sur le frottement de glissement, par M.
Bochet," in the Annales des Mines, Cinq. Série, Tome XIX., Paris, 1861.
Prof. Rühlmann gives some information in regard to the experiments with
Waltjen's friction balance in the "Polytechnisches Centralblatt, 1861.
Heft 10."
§ 174. Friction Tables.-The following tables contain a con-
densed summary of the coefficients of friction of the substances,
most generally employed in practice.
€ 174.]
319
RESISTANCE OF FRICTION, ETC.
TABLE I.
COEFFICIENTS OF FRICTION OF REST.
Name of the rubbing bodies.
Dry.
Condition of the surfaces and nature of the unguents.
Moistened with water,
With olive oil.
Hog's lard.
Tallow.
Minimum value. 0,30 0,65
Dry soap.
Polished and greasy.
Greasy and moistened.
0,14 0,22 0,30
Wood upon
Mean
0,50 0,68
0,21 0,19 0,36 0,35
wood.
•
Maximum
0,70 0,71
0,25 0,44 0,40
Minimum value. 0,15
0,11
Metal
upon
Mean
66
0,18
0,12 0,10 0,11
0,15
metal..
Maximum
0,24
0,16
Wood on metal.
0,60 0,65 0,10 0,12 0,12
•
0,10
Hemp in ropes,
plaits, etc., on
wood
•
Thick sole leath-
er as packing
on wood or
cast iron ..
Black leather
straps over
drums
Mini'm value. 0,50
Mean
(6
0,63 0,87
Max'm
((
0,80
On edge
Flat
0,43 0,62 0,12
0,62 0,80 0,13
0,27
Made of wood. 0,47
metal 0,54
0,28 0,38
Stone or brick
upon stone or
Mini'm value. 0,67
brick, well pol-
Max'm
0,75
ished
Stone upon wrought Min. val. 0,42
iron..
Max. (( 0,49
Pearwood upon stone
•
0,64
320
[§ 175.
GENERAL PRINCIPLES OF MECHANICS.
TABLE II.
COEFFICIENTS OF FRICTION OF MOTION.
Name of the rubbing bodies.
Condition of the surfaces and nature of the unguents.
Dry.
With water.
Olive oil.
Hog's lard.
Tallow.
Hog's fat and plumbago.
Pure wagon grease.
Dry soap.
Greasy.
Min. value. 0,20
0,06
Wood
upon
wood .
Mean ((
Max.
0,36 0,25
0,48
0,06
0,07 0,07
0,14 0,08
0,15 0,12
0,07 0,08
0,16 0,15
Min. value. 0,15
Metal
upon
metal.
Mean แ
0,18 0,31
Max. (6 0,24
0,06 0,07 0,07 0,06 0,12
0,07 0,09 0,09 0,08 0,15 0,20 0,13
0,11
0,08 0,11 0,11 0,09 0,17
Wood
Min. value. 0,20
0,05 0,07 0,06
0,17
0,10
upon
metal.
Mean
•
Max. (( 0,62
0,42 0,24 0,06 0,07 0,08 0,08 0,10 0,20 0,14
Hemp in ropes, ( On wood. 0,45 0,33|
0,16
0,08 0,08 0,10
་
etc
On iron
•
0,15
0,19
Sole leather flat ( Raw
·
•
0,54 0,36 0,16
0,20
Pounded. 0,30
Greasy.
0,25
Dry
0,34 0,31 0,14
0,14
Greasy
•
0,24
upon wood or
metal.
The same on
edge for pis-
ton packing.
REMARK.-More complete tables of the coefficients of friction are to be
found in the "Ingenieur," page 403, etc. The coefficients of friction of
loose granular masses will be given in the second volume, when the theory
of the pressure of earth is treated.
§ 175. The Latest Experiments on Friction.-From the
experiments of Bochet upon sliding friction, we find, that the
results obtained by the older experimenters Coulomb and Morin
must undergo some important modifications. The former experi-
§ 175.]
321
RESISTANCE OF FRICTION, ETC.
ments were made with railroad wagons weighing from 6 to 10 tons,
which were caused to slide on a horizontal railroad either upon
their wheels, which were made fast, or upon a kind of shoe (patin).
The shocs were fastened to the frame of the wagon before, between
and behind the wheels, and in the different series of experiments
they were covered with soles of different materials, such as wood,
leather, iron, etc., on which a pressure of 2, 4, 6, 10 and 15 kilograms
per square centimetre could be produced. The wagon, thus
transformed into a sled, was moved by a locomotive attached in
front by means of a spring dynameter, which gave the pull or force,
which balanced the sliding friction. In order to prevent, as much
as possible, the resistance of the air, the wagon, which preceded the
sled, had a greater cross-section than the latter.
The correctness of the formula FN, according to which
the friction F is proportional to the pressure, is proved anew by
these experiments; but it was found, that the co-efficient of fric-
tion was dependent not only upon the nature and state of the rub-
bing surfaces, but also upon other circumstances, viz.: the velocity
of the sliding body and the specific pressure, I.E., the pressure per
unit of surface. Bochet puts
$
K Y
1 + a v
+ Y';
in which v denotes the velocity of sliding, & the value of o for infi-
nitely slow and y the value o for a very rapid motion. According
to this formula the coefficient decreases gradually from to y as
the velocity increases. The mean value of the coefficient a is
= 0,3, when v is expressed in meters, and on the contrary = 0,091,
when v is given in feet. Hence we can assume the co-efficient of
friction to be constant only, when the velocities vary from 0 to at
most 1 foot and when the other circumstances remain the same.
The co-efficients x and y are different for different materials and
depend upon the degree of smoothness of the rubbing surfaces,
upon the unguents, upon the specific pressure etc.
k
The co-efficient of friction attains its maximum value for
wood, particularly soft wood, leather and gutta-percha sliding upon
dry and ungreased iron rails. Here we have = 0,40 to 0,70. The
mean value for soft wood is к = 0,60 and for hard wood = 0,55.
The value κ is also very different for the friction of iron upon
iron. ɛ
If the surfaces are not polished we have = 0,25 to 0,60;
and, on the contrary, for polished surfaces we have = 0,12 to
K
21
322
[§ 175.
GENERAL PRINCIPLES OF MECHANICS.
0,40. The friction of iron upon iron is not diminished by sprink-
ling it with water, but the friction of wood, leather and gutta-
percha is considerably diminished by wetting the rail. When the
surfaces are oiled, sinks to from 0,05 to 0,20.
The co-efficient y is always smaller than «. When the velocities
are great, the surfaces smooth, the unguent properly applied and
the specific pressure a medium one, y has nearly the same value for
all substances.
The friction of rest is greater only in those cases where wood
or leather slide upon wet or greased rails, and then it is twice as great.
According to these experiments, we have.
1. for dry soft wood, when the pressure is at least 10 kilo-
grams per square centimeter or 142 pounds per square inch,
0,30
1 + 0,3 v
+ 0,30;
2. for dry hard wood under the same pressure
0,30
&
+ 0,25 ·
1 + 0,3 v
3. for half polished iron, dry or wet, under a pressure of more
than 300 kilograms per square centimeter or 4267 pounds
per square inch,
0,15
1 + 0,3 v
+ 0,15;
4. for the same either dry, under a pressure of at least 100
kilograms per square centimeter or polished and greased
under specific pressure of at least 20 kilograms, and also
for resinous wood with water as unguent under the same
pressure,
Φ
0,175
1 + 0,3 v
+ 0,075 ;
5. for wood properly polished and rubbed with fatty water or
fat under a pressure of at least 20 kilograms per square
centimeter (284 pounds per square inch),
0,10
1 + 0,3 v
+ 0,06.
If v is given in feet, we must substitute in the denominator
0,091 v instead of 0,3 v.
REMARK.-It is very desirable that these experiments, made on so large
a scale and giving results which differ so much from those already known,
should be repeated.
§ 176.]
323
RESISTANCE TO FRICTION, ETC.
FIG. 258.
P
§ 176. Inclined Plane.-One of the most important applica-
tions of the theory of sliding friction is to the determination of the
conditions of equilibrium of a body A C upon an inclined plane
FH, Fig. 258. If, as in § 146,
F HR = a is the angle of incli-
nation of the inclined plane and
POS, ẞ the angle formed by
the direction of the force P with
the inclined plane, we have the
normal force due to the weight G
G cos. α,

H
S,
N₁
F
B
A
0
G
R
=
No
the force which tends to move
the body down the plane = S=
G sin. a, the force N₁, with which
the force P seeks to raise the
body from the plane, = P sin. ẞ and the force S, with which it
draws the body up the plane = P cos. B. The resulting normal
force is
N N. - N₁ = G cos. a — P sin. ß,
and consequently the friction is
F= 4 (G cos. a
P sin. B).
If we wish to find the force necessary to draw the body up the
plane, the friction must be overcome, and therefore we have
S₁
=
P sin. B).
ß
S+ F, I.E. P cos. ẞ G sin. a + 4 (G cos. a
But if the force necessary to prevent the body from sliding down
the plane is required, as the friction assists the force, we will have
S₁ + F = S, I.E. P cos. ß + $ (G cos. a — P sin. B) = G sin. a.
From these equations we obtain in the first case
P =
sin. a +
cos. B +
cos, a
sin. B
G, and in the second case,
sin. a
o cos. a
P =
G.
cos. B $ sin. B
If we introduce the angle of friction or of repose p by putting
sin.p
ф
tang. p =
we obtain
cos. p
P:
sin. a cos. p ± cos. a sin. P. G,
cos. ẞ cos. p± sin. ẞ sin. p
324
[§ 176.
GENERAL PRINCIPLES OF MECHANICS.
or according to a well-known trigonometrical formula
P =
sin. (a ± p)
G;
cos. (B = p)
the upper signs are for the case, when motion is to be produced, and
the lower ones, when motion is to be prevented.
As long as we have
-
P>
sin. (a — p)
G and <
cos. (B + p)
sin. (a + p)
cos. (ẞ — p)
G,
the body will move neither up nor down.
If a is <p, the force necessary to push the body down the
plane is
P =
sin. (p
a)
G.
cos. (p + B)
The latter formula can be found by the simple application of
the parallelogram of forces O P Q G, Fig. 259. Since a body
H
E
A
FIG. 259.
B
P
S
F
R
counteracts any force from another body,
when the angle of divergence of the di
rection of the force from that of the normal
to the surface is equal to the angle of
friction p (§ 172), a state of equilibrium
will exist in the foregoing case, when the
resultant OQ Q of the forces P and G
forms an angle N O Q = p with the nor-
mal. If, in the general formula

P
G
sin. G O Q
sin. P O Q
we substitute G O Q = GO N + NOQ=
a + p and POQ=POS+S O Q = B +
90°
P, we obtain
P
sin. (a + p)
sin. (a + p)
G
sin. (B − p + 90°)
cos. (B − p)*
1
If the force P, is to prevent the body from sliding down the
inclined plane, the resultant Q, falls on the lower side of the normal
ON, and the angle of friction p enters in the calculation with a
negative sign, and consequently we have
P
sin. (a p)
G
cos. (B + p)'
$176.]
325
RESISTANCE TO FRICTION, ETC.
If the body lies upon a horizontal plane, a is = 0, and the force
necessary to move it forward becomes
P =
ф G
cos.ẞ+osin. B
G sin. p
cos. (ẞ — p)
the inclined plane, I.E., in the
ẞ 0, and therefore
If the force acts parallel to
direction of its slope, we have
P =
(sin. a ±o cos. a) G
sin. (a ±P). G. (Compare § 172.)
cos. p
If, finally, the force acts horizontally, we have
B =
P
a, cos. ẞ cos. a and sin. B
=
=
sin. a ±o cos, a
cos. a F & sin. a
sin. a, and consequently
tang. a ±0
G =
G, I.E.
1
& tang. a
P = tang. (a ± p) G, which is
resolution of the parallelogram O P Q G.
also given by the direct
Further, the force necessary to push the body up the plane
becomes a minimum, when the denominator cos. (ẞB − p) becomes a
maximum, that is, when it is = 1, or when ẞ — p is = 0, I.E. when
B
= p. When the angle formed by the direction of the force with
that of the inclined plane is equal to the angle of friction, this
force is a minimum and is P = sin. (a + p). G.
EXAMPLE.-What is the pressure along the axis of a wooden prop
A E, Fig. 260, which prevents the mass of rock A B C D, weighing G
5000 pounds, from sliding down an inclined plane (the floor of a mine),
when the inclination of the prop to the horizon is 35°, that of the inclined
Q is
Here
plane C D, 50° and when the coefficient of friction
we have
G = 5000, a = 50°, ß = 35°
and the formula gives
P =
=
0,75 ?
50° = 15° and
0,75,
cos. a
sin. 50° — 0.75 cos. 50°
5000
•
cos. 15° +
0,75 sin. 15°
0,482
1420
5000 =
1,160
sin, a
cos. B ó sin. B
Р
E
0,766
0,966 +0,194
FIG. 260.
B
A
G
1224 pounds.
If the prop was horizontal, we would have
50° and tang. p = 0,75, or p = 36° 52′,
from which we obtain
B

PG tang. (a p) = 5000 tang. (50° — 36° 52')
=5000 tang. 13° 8'=5000. 0,2333=1166 pounds.
In order to push the same mass of rock by
means of a horizontal force up the floor, when
the other circumstances are the same, a force
P = G tang. (a + p) 5000 tang. 86° 52′
= 5000. 18,2676 = 91338 pounds would
be necessary.
326
[§ 177.
GENERAL PRINCIPLES OF MECHANICS.
177. The normal pressure, with which a body A C presses
upon the inclined plane F H, Fig. 261, while being pushed up it, is
G sin. O P Q
N
=
Q cos. p
cos. p =
sin. P O Q
G sin. (90°
sin. (B + 90º − p)
a
B)
cos. p
G cos. (a + B) cos. p
cos. (ẞ − p)
and, on the contrary, when we prevent its sliding down, we have
G cos. (a + B) cos. p
=
N₁ Q₁ cos. Q₁ O N₁ =
Q₁ cos. p =
cos. (B + p).
If the direction of the force is parallel to the direction of the
plane, we have ß = 0 and N G cos. a, and when its direction is
horizontal, we have ẞ
a and
H
E
N
G cos. p
cos. (a ± p)*
FIG. 261.
P
S
F
P
B
R
FIG. 262.

P

R
H
B
N
Ꭰ
The normal pressure becomes null, when cos. (a + B) = 0 or
a + B = 90°, and becomes negative, when a + ẞ is > 90" or 3 is
> 90°
α. In the latter case the inclined plane is not under but
over the body, as is represented in Fig. 262. Here again the two
extreme cases of equilibrium exist when the resultant Q or Q1,
which is transmitted to the inclined plane F H, diverges from the
normal either above or below it at an angle, which is that of the
friction NO Q NO Q₁ = P.
==
1
In the foregoing development of the formulas for the equili-
brium of a body upon an inclined plane it is supposed, that the
resultant Q can be completely transmitted from the body AC to
the support F H R, which forms the inclined plane; this is only
§ 177.]
327
RESISTANCE OF FRICTION, ETC.
possible (according to § 146), when the direction of this force passes
PN
Hi
H
N
FIG. 263.
E
N₁
K
then the moment, with which
right about C, is Qe Pa
'R
through the supporting surface
CD of the body A C. Other-
wise the body A C, Fig. 263, has
a tendency to revolve or overturn
about the outer edge C, and this
tendency increases with the dis-
tance C K
=e of this edge from
the direction OQ of the result-
ant Q.
If a denotes the distance CL
of the direction O P of the force
and b the distance CE of the
vertical line of gravity O G of
the body from the outer edge C,
the body seeks to turn from left to
G b.

P
If P a were = Gb or
G
b
a
the resultant Q would pass
through the edge C and would be counteracted by the inclined
plane; if P a were < G b, the body would have a tendency to turn
from right to left, which turning would be prevented by its im-
penetrability.
If, on the contrary, P a is > Gb the body must receive a second
support or be guided by a second inclined plane F, H. If this
second inclined plane counteracts in A the force N and the fric-
tion N caused by it, the inclined plane F, H, will react upon the
body in A with the opposite forces N and N, which pre-
vent the turning of the body about C, and the sum of the moments
of these forces must be equal to the moment of rotation of the
force Q, I.E. N l + ¢ N d =
M
Q
e =
Pa
G b, or
Gb,
=
1) N (l + d) Pa
o
7 and a designating the distances C D and C B of the edge A from C
in the directions parallel and at right angles to the inclined plane.
If, further, N, is the pressure of the body upon the inclined
plane FH at Cand o N, the friction caused by it, we can put
2) P cos. B G sin. a + ¢ (N + N₁) and
3) P sin. B = G cos. a + N
N₁.
Eliminating N, from the last two equations we obtain the equa-
tion of condition.
328
[S 178.
GENERAL PRINCIPLES OF MECHANICS.
........
P (cos. ẞ + sin. ß)
G (sin. a +
cos. a) + 2 ¢ N,
Pa
G b
and substituting the value N =
from equation (1) we
l +
& d
have the equation
2
(Pa Gb)
P (cos. B +
sin. ẞ)
=
b + p d
or
P
2
(+
φα
-
l + & d
G ( +$d (sin. a + 4 cos. a) — ¢ b),
= G
2
from which we obtain finally
G (sin. a + cos. a) +
(cos. B + 4 sin. B) — 4 a)
P
(l + ф d) (sin, a + ф c08. a) - яфъ
cos. 2 b
(l + ø d) (cos. ẞ + & sin. B)
G
2фа
(1 + $ d) sin. (a + p) −
(l + p d) cos. (1 − p)
2 p b cos. p
G.
•
2¢ a cos. p
If N is =
0, we have Pa
G band
sin. (a + p)
b
whence
cos. (B − p)
a
α'
sin. (a + p)
P
G,
cos. (B − p)
as we found before.
$178. The Theory of the Equilibrium of Supported
Bodies referred to the Equilibrium of Free Bodies.-In
investigating the conditions of equilibrium of a body, taking into
consideration the friction, we will accomplish more surely our
object, if we imagine the body entirely free and suppose, that every
body, with which it comes in contact, acts upon it with two forces,
viz.: with one force N, which proceeds from it and is normal to the
surface of contact, and with another force o N, which opposes the
supposed motion of the point of contact on this surface and which
is caused by the friction between the two bodies. In this way
we obtain a rigid system of forces, whose state of equilibrium can
easily be determined according to the rules given in § 90, as is
shown in the following special case.
A prismatical bar A B, Fig. 264, is so placed, that its lower end
rests upon a horizontal floor CH and that its upper end leans
against the vertical wall CV; at what inclination BAC a
does it lose its equilibrium? We can here express the reactions
of the floor upon the body by a vertical force R and by the fric-
tion R, which acts horizontally, and, on the contrary, the reaction
§ 179.]
329
RESISTANCE OF FRICTION, ETC.
of the wall by a horizontal force N and by a friction o N acting
upwards. Hence, if G is the weight of the rod acting at its centre
of gravity S, we have here a system of ver-
tical forces G, R, o N and a system of
horizontal ones N and o R.
R
FIG. 264.

N
N
D.
B
When these forces are in equilibrium,
we have
1) GR o N,
+
N.AD + N. A C
2) & R
N and
3) G. A E
&
But the arm A E is
E
HI
AS cos. ɑ =
A B cos. a, the arm A D – A B sin. a
and the arm A CAB cos. a, hence the
third equation becomes simply
G cos. a = N (sin. a + o cos. a).
Combining the first two equations, we obtain
G
R + $³ R =(
G
R =
and N
1 + 0²
(1 + p²) R, whence
Go
1 + ب
Substituting this value of N in the equation (3), we have
į G cos. a =
фG
1 + س
(sin. a + ¢ cos. a), or
1 + س
2 φ
tang. a + $,
and the tangent of the required angle of inclination is
1 + φ
2 φ
1
1 — tang.³ p
tang. a =
20
20
2 tang. p
cos.² p
sin.² p
cos. 2 p
=cotg. 2 p
2 sin. p cos. p
sin. 2 p
2 p); therefore
= tang. (90°
L. BAC a 90°
= =
2 p and ABC B = 2 p.
▲ =
§ 179. Theory of the Wedge.-Friction has also a great
influence upon the conditions of equilibrium of the wedge (see
§ 149). Let us suppose, that its cross section forms an isosceles
triangle A B 8, Fig. 265, the acute angle of which 4 SB = a,
that the force acts in the centre M of the back of the wedge A B
330
[$ 179.
GENERAL PRINCIPLES OF MECHANICS.
and at right angles to it and that the body C H K presses with a
certain force N against the surface of the wedge B S, while the
FIG. 265
C
B,
B
M
A
wedge reposes with its
surface AS upon a
horizontal plane. The
body CH K is also in-
closed in two guides.
G and K, which com-
pel it, when the wedge
is pushed forward upon
the horizontal plane,
to rise with the load Q
in the direction E C

perpendicular to the surface B S of the wedge.
Since the direction of the force P forms equal angles with the
two surfaces AS and BS of the wedge, the normal pressures N, N,
and consequently the frictions o N, o N caused by them, are equal
to each other, and the forces P, N, N, o N and N must hold
each other in equilibrium. If we decompose each of the last four
forces into two components, one parallel and the other perpendicu-
lar to the direction of the force P, the sum of the forces having the
same direction as P must, of course, be in equilibrium with P.
But the directions of the forces N, N form, with the direction M S
of the force P, an angle 90 and those of the forces o N, ¢ N
a
α
2'
an angle, and therefore the components of N, N in the direction
α
α
MS are N sin. and N sin. and those of o N and o N are o N
COS. and o N cos.
a
2'
2'
and consequently we can put
а
2*
a
P = 2 N sin. 5 + 2 N ccs.
2 N
2
N (sin. 235
a
o
+ cos.
08. ).
).
а
2
In consequence of the
o
friction o N between the surface BS of
the wedge and the base of the body CIIK, this body is pressed
with an opposite force
Φ. Φ
N against the guide G H, which causes
p, N, which resists the upward move-
a friction F₁. N =
ment of the body CH K; hence we have
NFQor N (1,) =
= Q and
N
Q
1 - ФФ,
179.]
331
RESISTANCE OF FRICTION, ETC.
Substituting this value for N in the above equation, we obtain
the force necessary to raise the weight Q
2
2 Q
P
1
Φ Φι
(sin. +
a
ф
cos.), approximatively
α
+ & cos.
2
02
a
= 2
+ & cos.
2
+ 4 4, sin. 2),
= 2 Q (1 + 4 4₁) (sin.
2 Q (sin.
a
2
or putting the coefficient of friction
Φ Φι
along the guides equal to
that along the surfaces AS and B S of the wedge, we obtain
P =
2 Q
1
(sin.
a
+ & cos.
a
a
2), approximatively
cos. 3).
+ p³) sin. 2 + cos.
= 2 Q ((1 + 4²) sin.
FIG. 266.
M B
F
2
When a wedge A B C, Fig. 266, is used
for splitting or compressing bodies, the force.
upon the back A B corresponding to the
normal pressure Q against the sides A C
and B C is

necessary to move the load
a
P = 2 Q (sin. 2
+ & cos.
EXAMPLE.-Let the load on the wedge repre-
sented in Fig. 265 be Q 650, the sharpness of
the wedge 25 and the coefficient of friction
0,36; required the mechanical effect
foot along its guides.
φ
01
The force is
2.650
P =
(sin. 121° + 0,36 cos. 121°)
1.
(0,36)2
1300
(0,2164 + 0,36 . 0,9763)
1 -0,1296
1300
737,27
(0,2164 +0,3515)
0,8704
0,8704
$48,2 pounds.
The space described by the load is E E,
$1
foot, and that de
scribed by the force is
e EE,
a
B L = 8 — B B₁ cos.
1
1
cos.
sin. a
n
0,25
sin. 121°
2 sin.
0,25
0,2164
1,155 feet,
and consequently the mechanical effect necessary is
Ps= 848,2. 1,155 979,6 foot-pounds.
1 = 1.
1
If we neglected the friction, the work done would be Ps = Q s
650 325; consequently the friction nearly triples the mechanical effect
necessary to raise Q.
332
[§ 180.
GENERAL PRINCIPLES OF MECHANICS.
§ 180. In the same way we can find the force P required, when
a wedge A B C, Fig. 267, raises a load Q vertically upwards, while
moving forward itself upon a horizontal plane H O. Let the
normal pressure between the wedge A B C and the block D, which
is pressed vertically downwards by the load Q, be = N, the normal
pressure of the wedge upon the support H O be = R and the normal
pressure of the block against the
guide E E be = S. Then P must bal-
ance the forces R, 4, R, N and
N, and Q the forces S, 4. S, N
FIG. 267.

N
D
เ
ss
N
E
E
ΦΝ
B
and N.
If a is the angle of inclination
A B C of the surface A B of the
wedge, we can decompose N into the
vertical force N cos. a and the hori-
zontal force N sin. a, and o N into
the vertical force o N sin. a and the
horizontal force o N cos. a, and there-
fore we can put
4, R+ N
P
-N
AR
1) P
H
ዎ ይ
0
2) R = N cos. a
3) Q = N cos. a−
4) S = N sin. a +
From the first two equations we obtain
P
=
sin. a + 4 N cos. a,
o N sin. a,
N sin. a—4, S and
o N cos. a.
[(1 $ 4₁) sin. a + (p + $₁) cos. a] N,
and from the last two
Q
=
[(1 Ф Ф2) cos. a
(+42) sin. a] N;
and dividing the first by the second, we have
P
(1
Q (1
$ P₁) sin. a + (p + $₁) cos. a
2) cos. a
If 4 = 4; = 92, we have, since &
1
2
(Ø + $₂) sin. a
tang. p and
2 ф
=
tang. 2 p,
1
P
sin. a + cos. a tang. 2 p
tang. a + tang. 2 p
sin. a tang. 2 p
Q COS. a
=
If we disregard the friction upon the points of support, we can
put, and
= 0, and consequently
1- tang. a tang. 2 p
tang. (a + 2 p).
P
Q
sin. a + cos. a
cos. a Ф sin. a
tang. a + p
tang. (a+p). (Comp. § 176.)
1-otang. a
§ 181.]
333
RESISTANCE OF FRICTION, ETC.
When the load Q acts at right angles to the surface of the
wedge, the equations (3) and (4) must be replaced by the following
whence Q = (1 − ø 4½)
Q
N-₂ S and
S = & N,
N, or inversely,
Q
N
and
1
Φ Φ.
P
(1
When is
$1
P2, it becomes
P
Q
Þ Þ₁) sin. a + (☀ + $₁) cos. a
1 $ 2
= sin. a + cos. a . tang. 2 p.
The formula P = Q tang. (a + 2 p) is applicable to the deter-
mination of the conditions of equilibrium, when two bodies Mand N
FIG. 268.
I
II
N
M
B
1/2
R
M
A
P
are fastened together by
means of a key A B, Fig.
268, I. and II. The force
P applied to the back of the
wedge causes the tension,
with which the two bodies
are drawn against one an-
other,


Q
= P cotg. (a + 3p).
On the contrary, the
force, with which we must
press upon the bottom B
of the key in order to loosen
it, I.E. to drive it back in the direction B A, is, since a is neg-
ative here,
P₁ = Q tang. (2 pa),
or substituting the former value of Q, we have
tang. (2 pa)
P₁
P
tang. (2p+ a)
In order to prevent the wedge from jumping back of itself, a
must <2 p.
§ 181. Coefficients of Friction of Axles.-For axles the
friction of motion alone is important, and for this reason only the
results of experiments upon it are given.
334
[§ 181.
GENERAL PRINCIPLES OF MECHANICS.
TABLE III.
COEFFICIENTS OF FRICTION OF AXLES, ACCORDING TO MORIN.


Name of the rubbing bodies.
Bell metal upon bell metal.
Condition of the surfaces and nature of the unguents.
Oil, Tallow,
or Lard.
(6
cast iron..
•
Wro't iron
66
bell metal. 0,251 0,189
"(
(6
cast iron...
Cast iron
((
66
(C
Wro't iron (6
Cast iron
Lign❜m vitæ "
،،
bell metal.. 0,194 0,161
lig. vitæ... 0,188
((
66
0,185
((
cast iron..
lig. vitæ...
*
0,097
0,049
0,075 0,054 0,090 0,111
0,075 0,054
Greasy.
0,137 0,079 0,075 0,054
0,137
0,075 0,054 0,065
0,125
0,166
0,100 0,092
0,109 0,140
0,116
0,153
0,070
From this table the following practically important conclusions
can be drawn for axles, journals or gudgeons of wrought or cast
iron running in bearings of cast iron or bell-metal (brass), greased
with oil, tallow or lard, the coefficient of friction
is, when the lubrication is well sustained, = 0,054,
and with ordinary lubrication, = 0,070 to 0,080.
The values found by Coulomb differ in some respects from the
above.
REMARK.-By his experiments upon mediate friction, by means of the
friction balance, Hirn obtained several results, which differ somewhat from
those already known. The axle employed by him, consisting of a hollow
cast-iron drum 0,23 metres in diameter, and 0,22 metres long, was lubri-
cated upon the outer surface by dipping it in oil and kept cool by causing
water to pass through its interior. The bronze bearing (8 of copper and 1
of tin) was pressed upon it by means of a lever 14 metre long aud weigh-
ing 50 kilogr. while the axle made 50 to 100 revolutions per minute. It
is easy to see, that in the experiments made with this apparatus the fluidity
and adhesion of the oil employed as unguent must have played an import-
ant part, since not only the velocity of revolution, but also the rubbing
surface was very great compared to the pressure. The velocity at the cir-
§ 182.]
335
RESISTANCE OF FRICTION, ETC.
cumference of the drum, since its circumference was 72 centimetres and
since it revolved to 10 times in a second, was 60 to 120 centimetres, or 24
to 48 inches, while in machines it is generally but from 2 to 6 inches. The
horizontal section of the axle was 22. 23 = 506 square centimetres, and
consequently the pressure on each square centimetre of this section was
50
only 0,1 kilogram, I.E. 6,45. 0,220 = 1,42 pounds upon a square inch,
506
while this pressure in ordinary machines is generally more than one hun-
dred pounds. Hirn's experiments were consequently made under condi-
tions different from those generally met with in very large and powerful
machinery, and under which the other experiments, such as, E.G., those of
Morin, were tried, and therefore the variation in the results obtained is
perfectly explicable. The principal results of Hirn's experiments are the
following.
The mediate friction is dependent not only upon the pressure and the na-
ture and character of the rubbing surfaces and of unguent, but also upon
the velocity and upon the temperature of the rubbing surfaces and of the
surroundings, as well as upon the magnitude of these surfaces. The fric-
tion is directly proportional to the velocity, when the temperature is con-
stant; and if the temperature is disregarded, it increases with the square
root of the velocity. From other experiments Hirn concludes, that the
mediate friction is also proportional to the square root of the rubbing sur-
faces as well as to the square root of the pressure. In regard to the par-
ticular influence of the temperature, the following formula was given by
these experiments:
F=
Fo
1,04920
in which t denotes the temperature of the rubbing surface, F, the friction
at 0°, and F that at t degrees of temperature.
One of the principal results of these experiments was the determination
of the mechanical equivalent of heat. This subject will be treated more at
length, when we discuss the theory of heat.
§ 182. Work Done by the Friction of Axles.—If we
know the pressure R between the axle and its bearing, and if the
radius r of the axle, Fig. 269, is given, we can easily calculate the
work done by the friction on the axle during each revolution. The
friction is F o R, the space described is the circumference 2 π î
of the axle, and consequently the mechanical effect lost by the
friction is A = 4 R . 2 π r = 2 π p R r. If the axle makes u
revolutions per minute, the mechanical effect expended in each
second is
T
L = 2 π p R r .
ገራ
60
πυφ Rr
=
0,105 . u ₫ Rr.
30
336
[$ 182.
GENERAL PRINCIPLES OF MECHANICS.
The work done by the friction increases, therefore, with the
pressure on the axle, with the radius of the axle and with the
number of revolutions. We have therefore the following practical
rule, not to increase unnecessarily the pressure on the axles in
rotating machines, to make them as small as possible without en-
dangering their solidity and durability and not to allow them to
make too many revolutions in a minute, at least, when the other
circumstances do not require it.
FIG. 269.
A
B
R
FIG. 270.


Å
B
Er
H
H₁
N
Ni
R
By the use of friction-wheels instead of plumber-blocks, the
work done by the friction is diminished. In Fig. 270 A B is a
shaft, whose journal C E E, rests upon the circumferences E H
and E, H₁ of the wheels (friction-wheels), which revolve around D
and D, and lie close behind one another. The given pressure R
upon the shaft gives rise to the pressure
1
N = N₁
R
2 cos.
α
2
Here a denotes the angle D C D, included between the lines join-
ing the centres, which are also lines of pressure. In consequence
of the rolling friction between the axle Cand the circumference of
the wheels, the latter revolve with this axle, and the frictions o N
and N, are produced on the bearings D and D,, the sum of which
& R
is F = (N + N₁) =
If the radius DE D₁ E₁ be de-
$
a
COS.
noted by a, and the radius of the axle by r₁, we obtain the force,
which must be exerted at the circumference of the wheels or at
that of the axle C to overcome the friction, and it is
§ 182.]
337
RESISTANCE OF FRICTION, ETC.
r₁
21 ФР
F₁
F
ar
a₁
cos.
2
while, on the contrary, it is R, when the axle lies directly on a
step.
FIG. 271.
If we neglect the weight of the friction-
wheels, the work done when these wheels are

A
KCL B
employed is =
G
H
T1
α
a₁ cos. 2
times as great as
when the shaft revolves in a plumber-block.
If we oppose a single friction-wheel G H,
Fig. 271, to the pressure R of the axles and
if we counteract the lateral forces, which in
other respects can be neglected, by the fixed
cheeks K and L, a becomes = 0, cos.
1 and the above ratio
ψ
71
R
a
EXAMPLE.-A water-wheel weighs 30000 pounds, the radius of its cir-
cumference a is 16 feet and that of its gudgeon is r = 5 inches; how much
force is required at the circumference of the wheel to overcome the friction
or to maintain the wheel in uniform motion, when running empty, and how
great is the corresponding expenditure of mechanical effect, when it makes
5 revolutions per minute? We can here assume a coefficient of friction
0,075, and consequently the friction is R 0,075 . 30000 = 2250
pounds. Since the radius of the wheel is
16.12
5
192
5
38,4 times as
great as that of the gudgeon or the arm of the friction, the friction re-
duced to the circumference of the wheel is
=
$ R 2250
38,4 38,4
58,59 pounds.
2.5. π
12
2,618 feet; and conse-
The circumference of the gudgeon is
quently the space described by the friction in a second is
2,618 . 5
60
0,2182 feet,
and the work done by the friction during one second is
L =
0,2182 . ø R = 0,2182 . 2250 = 491 foot-pounds.
71 =
α1
If the gudgeon of this wheel is placed on friction wheels, whose radii
are but 5 times as great as the radius of the gudgeon, that is, if 3.
the force necessary at the circumference of the wheel to overcome the fric-
22
338
[§ 183.
GENERAL PRINCIPLES OF MECHANICS.
=
tion would be only. 58,59 11,72 pounds and the mechanical effect
expended but 491 98,2 pounds. But in this case the support would be
much less safe.
FIG. 272.
§ 183. Friction on a Partially Worn Bearing.-The fric-
tion of an axle A CB, Fig. 272, upon a bearing, which is partially
worn, so that the shaft is supported in a single point A, is smaller
than that of a new axle, which touches all points of its bearing.
If no rotation takes place, the axle presses
upon the point B, through which the direction
of the resulting pressure R passes; if the shaft
begins to rotate in the direction A B, the axle
rises in consequence of the friction on its
bearing, until the force S tending to move it
down balances the friction F. The result-
ant R is decomposed into a normal force N
and a tangential one S, N is transmitted to
the plumber block and produces the friction FN, which acts
tangentially, S, however, puts itself in equilibrium with F, and
we have, therefore, S N. According to the theorem of Pytha-
goras, we have R² = N² + S2, whence

R
N
R2
R² = (1 + $³) N°,
or inversely the normal pressure
R
фR
N =
and the friction F
√ 1 + 0²
√1 +0²
tang.p
or introducing the angle of friction p or =
F
R tang. P
√ 1 + tang.²p
R tang. ę cos. p = R sin. p.
Consequently, when the shaft begins to turn, the point of pres-
sure B moves in its bearing in the opposite direction through an
angle A C B = the angle of friction p.
The moment F. CA = Fr of the friction on the axle is
naturally equal to the moment Rr sin. p of the pressure R upon
the bed, both being referred to the axis of revolution C. If there
were no motion, we would have
F = R
Φ = R tang. p =
the friction after the motion begins is
R sin. p
cos. p
cos. p times as great as
before. Generally = tang. p is scarcely and cos. p > 0,995,
1
TO
zo we can, therefore,
1
200
so that the difference is scarcely 1000
5
in ordinary cases neglect the effect of the motion.
§ 184.]
339
RESISTANCE TO FRICTION, ETC.
FIG. 273.
If the wheel A B revolves with a nave, Fig.
273, about a fixed axle A C, the friction is the
same as if the axle moved in an ordinary plumber-
block, but when the nave is worn the arm of the
friction is not the radius of the shaft, but that of
the opening in the nave.

FIG. 274.
§ 184. Friction on a Triangular Bearing.-If we lay the
axle in a prismatical bearing, we have more pressure on the bearing,
and consequently more friction than, when the bearing is circular.
If the bearing A D B, Fig. 274, is tri-
angular, the axle is supported at two
points A and B and at both of them
friction must be overcome. The result-
ing pressure R is decomposed into two
components Q and Q₁, each of which is
again decomposed into a normal stress
Nor N, and into a tangential one, which
is equal to the friction FN and

N
D
R
FN. According to the foregoing paragraph, we can put
Q sin. p and Qi sin. p, consequently the total
these frictions
friction is F + F
The forces
=
(Q + Q₁) sin. p.
and Q, are found, by the resolution of a parallelo-
gram of forces formed of Q and Q₁, with the aid of the resultant R,
of the angle of friction p and of the angle 1 CB = 2 a, corespond-
ing to the arc A B included between the two points of contact;
now we have
ACDC A 0 = a
QORA C D
p and
Q₁ OR = B C D + C BO= a + p and therefore
Q o Qi
= ɑ p + a + p = 2 a.
By employing the formula of § 78, we obtain
Q₁
=
sin. (a p)
sin. 2 a
R and Q
sin. (a + P), R;
sin. 2 a
R sin. p
sin. 2 a
whence the required friction is
(sin. [a — p] + sin. [a + p])
F + F₁ = (Q + Q₁) sin. p =
·
But from trigonometry we know, that sin. (a — p) + sin. (a + p)
= 2 sin. a cos. p, and that sin. 2 a 2 sin. a cos. a, and we can
therefore put
2 sin. a R sin. p cos. p
R sin. 2p
F + F =
2 sin. a cos. a
2 cos. a
340
[§ 185.
GENERAL PRINCIPLES OF MECHANICS.
which, owing to the smallness of p, we can make
a triangular bearing is used, the friction becomes
R sin. p
When
cos. a
1
times greater
cos. a
than when a circular one is employed. If, E.G., A D B is 60º,
60° = 120° and ACD = α = 60°, we have
A CB is 180°
1
cos. 60°
times =
twice as much friction as for a circular bearing.
§ 185. Friction of a New Bearing. By the aid of the latter
formula we can find the friction on a new circular bearing, when
the axle is supported at all points. Let A D B, Fig. 275, be such
FIG. 275.
R
B
a bearing. Let us divide the arc A D B along
which the bearing and axle are in contact into
very many parts, such as A N, N O, etc., whose
projections upon the chord A B are equal, and
let us suppose that each one of these parts
transmits from the axle to the bearing equal
R
portions of the whole pressure R. Here n

n
denotes the number of these parts. According
to the foregoing paragraph, the friction of two
parts N O and N, O, opposite to each other is
R sin. 2 p
n
cos. N CD
But cos. N CD is also = cos. O N P
NP
N O'
NP represent-
•
ing the projection of the part N O on A B, and therefore
NP
chord A B
N
consequently the friction corresponding to these two parts N O and
N, O, is
R sin. 2p n. NO
п
· chord
p
R sin. 2P. NO.
chord
In order to find the friction for the entire arc A D B, we have
only to substitute instead of N O the arc A D
A D B; for the
sum of all the frictions is equal to
R sin. 2 p
chord
the sum of all the
parts of the arc; consequently the friction on a new bearing is
arc A D
F = R sin. 2 P · chord A B’
or putting the angle at the centre A C B corresponding to the arc
contained in the bearing = 2 a° and the chord A4 B2 A C sin. a,
we have
§ 186.]
341
RESISTANCE OF FRICTION, ETC.
F=
R sin. 2p
2
α
or approximatively,
sin. a
assuming 2 p = 2 sin. p,
a
F = R sin. P · sin. a
Hence the initial friction increases with the depth, that the axle
is sunk in its bearing, E.G., if the bearing includes the semi-circum-
ference of the axle, we have a = π and sin. a = 1, and therefore
π
π
F = . R sin. p is = 1,57 times as great as it is when a bearing
2
2
has been worn. If the axle does not lie deep in its bearing, or if a is
small, we can put sin. a = a
lows that F = (1 +
small.
a³
2
6
3
α
α
-
6
(1 − 2),
α
whence it fol-
6
R sin. p or R sin. p, when a is very
(§ 186.) Poncelet's Theorem.-The pressure R on the bear-
ings is generally given as the resultant of two forces P and Q,
which act at right angles to each other, and it is consequently
= √′ P² + Q². So far as we need it for the determination of the
friction
F = & R = √ P² + Q²,
$
we can content ourselves with an approximate value of √ P² + Q³,
partly because an exact value of the coefficient can never be
given, as it depends upon so many accidental circumstances,
partly, also, because the producto R is generally but a small
fraction of the other forces, which act on the machine, E.G., the
lever, pulley, wheel and axle, etc., which is supported by the bear-
ings. The formula for calculating the approximate value of
√ P² + Q² is known as Poncelet's theorem, and its truth can be
demonstrated in the following manner. We have
√ P² + Q² = P√ 1 + (2)* = P √1 + x²,
in which x=
Q
2, and if Q is the smaller force, x is a simple frac-
p'
tion. Now let us put V1 + x² = µ + v x, and let us determine
the coefficients μ and v corresponding to certain conditions. The
relative error is
y =
√ 1 + x² μ
√1 + x²
V X
- 1 -
μ + να
√1 + x²
342
[$ 180.
GENERAL PRINCIPLES OF MECHANICS.
This equation corresponds to the curve O S P, Fig. 276, whose
ordinate, when the abscissa x = 0, is A 0 = y=1-μ, and, when
O
the abscissa A B = 1,
1, is y
=1
μ + ν
1/2
The curve also cuts the
axis of abscissas in two points K and N and at S lies at its greatest
distance CS from this axis. If
Ο
K
FIG. 276.
P
N
B
we put y = 0 or

√ 1 + x²
μ + να,
and solve the equation in relation
to x, we obtain
S
X
μννμ
μe v = √ μ² + 2² – 1
1 22
2
the values of which are the abscissas A K and A N of the points
K and N, where the curve cuts the axis, and also those values for
which the error is 0. In order to find the abscissa A C of the
maximum negative error C S, we must put the differential ratio
(µ + v x) (1 + x²)—} x v (1 + x², }
dy
d c
1 + x²
(see Article 13 of the Introduction to the Calculus).
This condition is fulfilled by putting
(µ + v x) (1 + x²)− x = v (1 + x²) or
(μ
(µ + v x) x = v (1 + x²), I.E. X =
บ
ν
0
According to this formula, the abscissa A C - gives the greatest
negative ordinate.
ре
บ
μ +ν.
и
CS=1
μ² + 2²
2,2
2
√ 1 +
√μ² + 2²
- 1)
=
22
1).
In order to have neither a great positive nor a great negative
error, let us put the three ordinates A 0 = 1 − µ, B P = 1
2
1 -
μ + ν
and CS Vu+ v² 1 equal to each other, and deter-
√2
mine from them the coefficients μ and v. We have
√ 2
μ + ν
μl =
2
and consequently
μ =
, I.E., v = ( √ 2 − 1) μ
2
1) μ = 0,414 u and
µ² + v², I.E., 2 = µ (1 + √ 1 + 0,414²)
§ 187.]
343
RESISTANCE OF FRICTION, ETC.
2
1 + √ 1,1714
0,96 and v = 0,414. 0,96
0,40.
We can, therefore, put V1 + x² = 0,96 + 0,40 . x, and in like
manner the resultant
R 0,96 P+ 0,40 Q,
=
and we know that in this case the greatest error we can make is
1 μ 1 0,96 = 0,04 = four per cent. of the true
± y
value.
This formula supposes, that we know, which of the two forces
is the greater; if this is unknown to us, we assume
√1 + 20² = μ (1 + x)
and obtain in that way
y = 1
μ (1 + x)
√ 1 + x²
In this case not only x = 0, but also x = ∞ gives an error
V
1
μ. If we put x =
= 1, we have the greatest negative error
μ
1-μ= μ 17- 1, or μ =
√ 2
1 + √2
(2-1)
Putting these errors equal to each other, we obtain
2
μ
- (µ √ 2 − 1),
2
1
0,828.
2,414
1,207
In case we do not know, which of the forces is the greatest, we
can write
R =
0,83 (P + Q),
then the greatest error we can make is ± y = 1 − 0,83 = 17 per
cent.
of the true value.
If, finally, we know that x is not over 0,2, we do best to neglect
x altogether and to put VP + Qª P; if, however, x is over
0,2, it is better to make
√ P² + Q² = 0,888 P + 0,490. Q.
In both cases the maximum error is about 2 per cent.*
§ 187. The Lever.-The theory of friction just given is appli-
cable to the material lever, to the wheel and axle and to other
machines. Let us now take up the subject of the lever, discussing
at once the most general case, that of the bent lever A C B,
* Polytechnische Mittheilungen, Vol. I.
344
[$ 187.
GENERAL PRINCIPLES OF MECHANICS.
Fig. 277. Let us denote, as formerly (§ 136), the arm of the lever
FIG. 277.
B
P
K
C
CA of the power P by a, the
lever arm C B of the load Q by b
and the radius of axle by r, and
let us put the weight of the lever
G, the arm CE of the same
=s and the angles A P K and
B QK formed by the directions
of the forces with the horizon

=a and B. The power P produces the vertical pressure P sin. a
and the load Q the vertical pressure Q sin. B, and the total vertical
pressure is VG+ P sin. a + Q sin. B. The force P produces
also the horizontal pressure P cos. a and the load an opposite
pressure cos. B, and the resulting horizontal pressure is H
P cos. a
R
Q cos. B, and the total pressure on the axle is
μl V + v H µ (G+P sin. a + Q sin. B) + v (P cos. a — Q cos. ß)
in which, however, the second part v (P cos. a - Q cos. B) is never
to be taken as negative, and, therefore, when Q cos. B is > P cos. a
the sign must be changed, or rather P cos. a must be subtracted
from Q cos. B. In order to find the value of the force correspond-
ing to a state of unstable equilibrium so that for the smallest addi-
tion of force motion will take place, we put the statical moment
of the power equal to the statical moment of the load plus or minus
the moment of the weight of the machine (§ 136) and plus the
moment of the friction; thus we have
P a = Q b ± G s + o R r
Q b ± G s
+
¢ (µ V + v H ) r, whence
Qb ± G s
+
¢ [µ (G + Q sin. B) = v Q cos. B] r
P =
a
•
―
μ or sin. a F v or cos. a
If P and Q act vertically, we have simply R= P+Q+ G
and therefore Pa = Q b ± G s + $ (P + Q + G) r. If the lever
is one armed, P and Q act in opposite directions to each other and
Ris P Q+ G and therefore the friction is less. But R
must always enter into the calculation with a positive sign, for the
friction R only resists motion and never produces it. We see
from this, that a single armed lever is mechanically more perfect
than a double armed one.
EXAMPLE.—If the arms of the bent lever represented in Fig. 277 are
a = 6 feet, b = 4 feet, s = foot and r = 1½ inches, if the angles of incli-
nation are a = 70º, ß = 50º, and if the load is Q = 5600 pounds and the
weight of the lever G is 900 pounds, the force necessary to produce
$188.]
345
RESISTANCE OF FRICTION, ETC.
unstable equilibrium is determined as follows. The friction being disre-
garded, we have Pa + G s = Qb and therefore
Q b
-
G 8
P =
If we put μ =
=
a
0,96 and v =
5600. 4 900.
6
= 3658 pounds.
0,40, we obtain
µ ( A + Q sin. ß) = 0,96 (900 + 5600 sin. 50º)
= 4982 pounds,
v Q cos. ß = 0,40 . 5600 cos. 50° = 1440 pounds,
µ sin. a = 0,96 . sin. 70° = 0,902 and
v cos. a = 0,40 . cos. 70° = 0,137.
It is easy to see, that P cos. a is here smaller than Q cos. 3; for since F is
approximatively 3658 pounds, we have P cos. a = 1251 pounds, while, on
the contrary, Q cos. B is 3600 pounds; therefore we must employ in this
case for vQ cos. ẞ and for v or cos. a the lower sign and put
P=
5600.4 900. + ér (4982 + 1440)
6 q.r (0,902 — 0,137)
·
Assuming the coefficient of friction o 0,075, we obtain
3
24
or = 0,075. =
0,009375 and 6422 o r = 60
and the force required
P =
6
-
0,0717
5,9928
= 3673 pounds.
22400 450 + 60 22010
Here the vertical pressure, when we substitute the force P = 3658 pounds
determined without reference to the friction, is
་
V = 3658 sin. 70° + 5600 sin. 50' + 900 = 3437 + 4290 + 900
= 8627 pounds.
and, on the contrary, the horizontal pressure is
2349 pounds.
H=5600 cos. 50 - 3658 cos. 70 3600 1251 =
Here His > 0,2 V, and therefore we have more correctly
R = 0,888. H + 0.490 V = 0,888. 8627 +0,490. 2349
consequently the moment of the friction is
= q r R = 0,009375 . 8811
and finally the force
=
82,6 foot-pounds;
22400
P-
450 + 82,6
6
3672 pounds,
8811, and
which value differs very little, it is true, from the one obtained above.
§ 188. Friction of a Pivot.—If in a wheel and axle there is
a pressure in the direction of the axis, which is always the case,
when the axle is vertical, in consequence of the weight of the
machine, friction is produced upon the base of one of the journals.
Since there is pressure at all points of the base between the pivot
and the step (or footstep), this friction approaches nearer to the
ordinary friction of sliding, than to what we have previously con-
sidered as axle friction, and we must therefore employ in this case
the coefficients of friction given in Table II. (page 320). In order
346
[§ 188.
GENERAL PRINCIPLES OF MECHANICS.
FIG. 278.
to find the work done by this friction, we must know the mean
space described by the base A B, Fig. 278, of such a pivot. We
assume that the pressure R is equally distributed over the whole
surface, that is, we suppose that the friction upon equal portions
of the base is equally great. If we divide the base by means of the
radii C D, C E, etc., in equal sections or triangles, such as D C E,
these correspond not only to equal frictions, but
also to equal moments, and we need therefore only
find the moment of the friction of one of these
triangles. The frictions on such a triangle can be
considered as parallel forces, since they all act
tangentially, I.E., at right angles to the radius CD;
and since the centre of gravity of a body or of a
surface is nothing else than the point of application
of the resultant of the parallel forces, which are
equally distributed over the body or surface, we can
consider the centre of gravity S of this sector or
triangle D CE as the point of application of the
resultant of all the frictions upon it. If the pressure on this sector
is and radius CD CE r, it follows (according to § 113),
that the statical moment of the friction of this sector is

A
D
E
R
R
B
B
& R
=CS.
& R
グ
​N
N
and finally that the statical moment of entire friction of the pivot is
FIG. 279.
M = n. 3 r ? R
N
= 3 Rr.
Sometimes the rubbing surface is a ring A B E D, Fig. 279.
If the radii of the same are CA r, and C'′ D =
r2, we have here to determine the centre of gravity
S of a portion of a ring. Hence, according to
§ 114, the arm is

B
DE
7.
CS = 3
r.
R
A
B
and therefore the moment of the friction is
3
M = 1R ("=")
2
If we introduce the mean radius
11 + 12
2
and the breadth of the ring r、 r₂ = b, we obtain also for the
moment of the friction
§ 189.]
347
RESISTANCE OF FRICTION, ETC.
M =
Ø
& R
R (r
b²
+
12 r
The mechanical effect of the friction is, in the first case,
A = 2 π . R r
o
3
π Rr, and, in the second case,
A = πφ Ρ
R (
(==)
= 2 π φ R
2
2
2
R (r
+
127).
From the above data it is easy to calculate the friction upon a
journal composed of one or more collars, when a vertical shaft is
borne by it. It is also easy to see, that, in order to diminish the
loss of mechanical effect, the pivots should be made as small as
the
possible, and that, when the other circumstances are the same,
friction is greater on a ring than on a full circle.
EXAMPLE.-A turbine, weighing 1800 pounds, makes 100 revolutions
per minute, and the diameter of the base of the pivot is 1 inch; how much
mechanical effect is consumed in a second by the friction of this pivot ?
Assuming the coefficient of friction o 0,100, we obtain
o R = 0,100. 1800 = 180 pounds,
the space described in a revolution is
π r = . 3,14.4 0,1745 feet,
and therefore the work done in one revolution is
= 180 . 0,1745 = 31,41 foot-pounds.
60
But this machine makes in a second 100
the required loss of mechanical effect is
revolutions, and therefore
314,1
6
52,3 foot-pounds.
§ 189. Friction on Conical Pivots.-If the end of the axle
A B D, Fig. 280, is conical, the friction is greater than when the
FIG. 280.
A
B
D
N
N
pivot is flat, for the axial pressure R is
decomposed into the normal forces N, N,
etc., which produce friction

and whose
sum is greater than R alone. If half the
angle of convergence A D C BDC= a,
we have
2 N
R
sin. a
1 and therefore the friction of this conical
R
pivot is
F
= 0
R
sin, a
If we denote the radius C A = C B of the axle at the place of
entrance in the step by r₁, we have, in accordance with what pre-
cedes, the statical moment,
348
[$ 189.
GENERAL PRINCIPLES OF MECHANICS.
6305
Rr₁
sin. a
ФР
M
sin. a
or, since
ri
CA
sin. a
sin. a
M =
3 o R a.
the side D A of the cone a, we have
If we allow the axle to penetrate a very short distance into the
step, the friction is less than for a flat pivot, and for this reason
we can employ conical pivots with advantage. If, E.G.,
a =
r1
sin. a
p
or 71 =
2'
½ r sin. a,
the conical pivot, whose radius is r₁, occasions only half as much
loss of mechanical effect as the flat pivot, whose radius is r.
If the pivot forms a truncated cone, Fig. 281, friction is pro-
duced on the conical surface and on the flat base, and we have for
the statical moment of the friction
M
дод
+
sin. a
Ф R
૧૦
23
202
when r denotes the radius C A at the point, where the pivot enters
the step, r, the radius of the base and a half the angle of conver-
gence. In consequence of the great lateral pressure N the step
becomes soon so worn that finally only the pressure on the base
EF remains and the moment of the friction becomes M 3 o R r₁.
FIG. 281.
FIG. 282.
FIG. 283.



A
B
N
A
B
YR
Vertical shafts or pivots are very often rounded off as in Figs.
282 and 283. Although by this rounding the friction is not in
any way diminished, yet a diminution of the moment of the fric-
tion can be produced by diminishing the penetration of the pivot.
into the step. If we suppose the rounded surface to be spherical,
we obtain with the aid of the calculus, for a hemispherical step the
moment of friction
M
φπ
2
Rr;
and for a step forming a low segment approximatively
M
[1 +0,3
(4)] RT
Rr
Ꭱ 1,
$ 190.j
349
RESISTANCE OF FRICTION, ETC.
in which formular denotes the radius of the sphere M A M B
and r, the radius of the step C'A= C B.
REMARK.-The pressure R upon the centre A D B, Fig. 284, of the
FIG. 284.
X
B
spindle of a turning-lathe is perpendicular to
the direction of the axis DX and is decom-
posed into a normal pressure N and a lateral
pressure S parallel to the axis. Retaining the
same notation, that we employed above for
conical pivots, we have

N =
Ꭱ
cos. a
and SR tang a.
The moment of the friction caused by Nis
N
R
1
M = 9 N. 3 r,
1
Ꭱ Ꭲ .
cos, a'
or since r =
CADA sin. A D C = a sin. a, when a denotes the length
CD of the portion of the centre which is buried, we have M
tang. a.
& & R a
The lateral force S is entirely or partly counteracted by an opposite
force S, on the other centre.
EXAMPLE. If the weight of the shaft and other parts of a whim gin is
R = 6000 pounds, the radius of its conical pivot is = r = 1 inch and the
angle of convergence 2 a of the latter is = 90°, the statical moment
of the friction is
M
&.9
•
Rr
sin. a
.0,1.
6000 1 100
sin. 45° 12 3
47,1 foot-pounds.
If the shaft in hoisting a bucket out of a mine makes u = 24 revolu-
tions, the mechanical effect consumed by the friction of the pivot during
this time is
A
2 π U. & $
Rr
sin. a
= 2.24. 47,1
7103 foot-pounds.
$ 190. The so-called Anti-friction Pivots. Supposing
that the axial pressure on a pivot A B B A, Fig. 285, is propor-
tional to the surface of the cross-
section, we can put the vertical
FIG. 285.
X

D
N
R
T
N₁
pressure per square inch R₁
R
G⁹
R being the total pressure and G
the area of the vertical projection
ADD A of the whole rubbing
surface A BBA. If now a is the
angle of inclination CTO of the
element 0 of the surface to the
axis C T of the pivot, the normal
pressure on each square inch
350
[$ 190.
GENERAL PRINCIPLES OF MECHANICS.
R₁
of the bearing, will be N₁ =
will be
and the corresponding friction
sin. a
R₁
& R
F₁ = & N₁ =
= 0
sin. a
G sin. a'
and if y denotes the distance or radius of friction M O, the moment
of this friction is
R
y
F₁ y = 0
G
sin. a'
or, since
Y
tangent O T,
sin. a
F₁ Y =ф
R
от.
G
In order to obtain a regular wearing away of the axle and of its
step, the moment F, y must be the same for all positions, and con-
sequently the tangent O T must have the same value for all points
of the generating curve A O B of the axle, and therefore the mo-
ment of the friction on the whole pivot is when O T:
M = F₁y. G = Ra.
o
a
The curve A O B, whose tangent O T, measured from the
point of tangency to the axis C X, is constant, is a tractrix or trac-
tory, and is generated by drawing a heavy point 4, Fig. 286, over a
FIG. 286.
ü
1
2
B
D
3
=
= ll,
horizontal plane by means of a string,
whose end moves along a straight line
CX. This string forms the constant
tangent lines A Ca1=ẞ 2 y 3,
etc. a. In order to construct this
curve, we draw CA a perpendicular
to the axis C X and take in CA, a
near to A, and lay off a 1 a, take ß
in a 1, near to a and lay off ß 2
here again take y near to ẞ and lay off
y 3 = a, etc., and we then draw a curve
tangent to the sides A a, a ẞ, By, yd...,
etc. This method gives the tractory
the more accurately the smaller the
sides A a, a ß, ß Y, Y S . . ., etc., are.
Schiele calls this curve the anti-friction
curve. (See the Practical Mechanics'
Journal, June number, 1849, translated in the Polytechnisches
Centralblatt, Jahrgang, 1849.)

4
IX
If, as is represented in Fig. 285, we make the anti-friction curve
190.]
351
RESISTANCE OF FRICTION, ETC.
end at the circumference of the shaft the maximum radius of
friction CA r is at the same time the constant tangent a, and
therefore the moment of the friction M = ø R r is independent of
the length of the pivot. When the rubbing surface is flat and of
the same radius, the moment of friction is M₁ Rr, that is,
one third smaller, and it decreases still more in time; for the exte-
rior portions are more worn than the interior ones, and thus the
surface of friction becomes less.
1
The plugs and chambers of cocks are sometimes made in the
form of the anti-friction curve; for in this case the conditions are
the same as in a pivot.
REMARK.—When the pressure R on the pivot is so distributed that the
amount of the wearing, measured in the direction of the pressure, is equal
in all points of the circumference of the pivot, we have
N3
N₂ Y 3
1
N₁ Y1
sin. a 1
2
N, 2
sin. a q
sin. a 3
1
a₁ = a2 = az
a; N₁ Y₁
2 2
1
N₂ Y₂ = N 3 Y3
•
3
•
•
and for conical pivots, where
If 01, 02, 03 ... denote the surfaces,
N₁, N2, N3 act, we have
•
upon which the normal pressures
R = N₁ 0₁ sin. a₁ + N₂ 0₂ sin. a, + N3 03 sin, az +
1
or for conical pivots R (N, 0, + N, O,+Ng Ong + . . .) sin. a.
The portions of the surface can be considered as rings of the same
h
height whose widths are
n'
quently we have
and whose radii are y₁, Y2, Y3, conse-
h
n sin, a'
h
h
1
0₁ = 2 π Y 1
n sin. a
Y 2 01, 0 3
Y 1
N₁ 0₁ = N½ 02
=
h
03 =
2 пуз
3
etc.
n sin, a'
1
2πY 2
n sin, a'
=
Y³ 0₁, etc., and also
3
Y1
and R = n. N₁ 0₁ sin. a.
Ng 03..., and R.
Therefore, under the above assumption, the normal pressure on the
equally high rings of the circumference of the pivot are equal.
R
Inversely we have N₁ 0₁
hence the moment of the friction
n sin, a
on the pivot is
1
1 1
& R
1
1
& N₁ O₁ (Y₁ + Y
•
•
½ + + Yn)
Yn) =
n sin. a
(Y₁ + Y½ + . . . + }»).
M = (N₁ 0₁₁ + Ng Og Y2 + N3 03 Y3 +...)
1
2 2
If we have a truncated conical pivot, whose radii are ", and r we must
put y₁ + y + . . . + Yn
2
$ R (r₁ + T2)
2 sin, a
=
1
21
n ("1 + re̱) from which it follows that M
2
For a complete conical pivot, whose radius is r = 0, we have M
352
IS 191.
GENERAL PRINCIPLES OF MECHANICS.
Ꭱ
• R1, while in a foregoing paragraph (§ 189) we found M
2 sin, a
Ꭱ Ꭲ Ꭺ
1
፪ ቀ
sin. c
See the article by Mr. Reye upon the Theory of Friction of Axles in Vol.
6 of the Civilingenieur, as well as the article upon the same subject by
Director Grashof in the 5th volume of the Journal of the Association of
German Ingenieurs.
§ 191. Friction on Points and Knife-Edges.-In order to
diminish as much as possible the friction of the axles of rotating
bodies, they are often supported on sharp points, knife-edges, etc.
If the bodies employed were perfectly solid and inelastic, no loss of
mechanical effect in consequence of the friction would take place
by this method, since the space described by the friction is immeas-
urably small; but since every body possesses a certain degree of
elasticity, upon placing it upon the point or knife-edge, a slight
penetration takes place and a surface of friction is produced, upon
which the friction describes a certain space, which, although small,
occasions a loss of mechanical effect. When the rotation or vibra-
tion of a body supported in this way has continued some time, such
surfaces of friction are arcs developed by the wearing away of the
point or knife-edge, and the friction is then to be treated as we have
previously done. This mode of support is therefore only employed
in instruments such as compasses, balances, etc., where it is impor-
tant to diminish the friction and where the motion is not constant.
Coulomb made experiments upon the friction of a body sup-
ported by a hard steel point and 'movable around it. According
to these experiments, the friction increases somewhat faster than
the pressure, and changes with the degree of sharpness of the
supporting point. It is a minimum for a surface of garnet, greater
for a surface of agate, greater for a surface of rock crystal, still
greater for a surface of glass, and the greatest for a steel surface.
For very small pressures, as, E.G., in the magnetic needle, the point
can be sharpened to an angle of convergence of 10° to 20°. If,
however, the pressure is great, we must employ a much larger
angle of convergence (30° to 45°). The friction is less, when a body
lies with a plane surface upon a point than when the point plays in
a conical or spherical hollow. The circumstances are the same for
a knife-edge such as that of a balance. Balances, which are to be
heavily loaded, have knife-edges with an angle of convergence of
90°. When the balance is light, an angle of 30° is sufficient.
If we assume that the needle AB, Fig. 287. has pressed down
the point F C G an amount D C E, the height of which C M = h,
and the radius of which D M =r, and if we suppose the volume
§ 192.]
353
RESISTANCE OF FRICTION, ETC.
the measure of the fric-
1
3
½ π r²h to be proportional to the pressure R,
tion can be found in the following manner. If we put
If we put ¦ π r² h
R, in which is a coefficient given by experiment, and substi-
tute the angle of convergence D C E = 2 a or h = r cotg. a, we
obtain for the radius of the base
1' =
V
• R r = 0
π
μ
3 µ R tang. a
>
and
i
3
π
π
3 μ. VR tang. a.
R*
3
3 µ R* tang. a
From this we see that we can assume, that the friction on. a pivot
increases with the cube root of the fourth power of the pressure
and with the cube root of the tangent of half angle of convergence.
FIG. 287.
F
G
FIG. 288.

C₁

G
The amount of friction of a beam A B, Fig. 288, oscillating on
a knife-edge C C₁, can be found in like manner. If a is the half
angle of convergence DC M, 1 the length C C, of the edge and R
the pressure, we have
(R tang. a)3
& R r = μ
FIG. 289.
K
F
A
7
§ 192. Friction of Rolling.-The theory of rolling friction
is as yet by no means established upon a firm basis. We know,
that the friction increases with the pressure, and that it is greater,
when the radius of the roller is small than when it is large; but
we cannot yet give the exact algebraical relation of the friction to
the pressure and to the radius of the rolling body. Coulomb made
a few experiments with rollers of
lignum-vitæ and elm from 2 to 10
inches thick, which were rolled
upon supports of oak by winding
a thin string around the roller and
attaching to the ends of it the un-
equal weights P and Q, Fig. 289.
According to the results of these
experiments, the rolling friction is
directly proportional to the pressure
and inversely to the radius of the

PE
B
F
R
354
[$ 192.
GENERAL PRINCIPLES OF MECHANICS.
rollers, so that the force necessary to overcome the rolling friction
R
can be expressed by the formula F = f., R denoting the press-
p
ure, r the radius of the roller and f the coefficient of friction to be
determined by experiment. If r is given in English inches, we
have, according to these experiments,
For rollers of lignum-vitæ, f = 0,0189
For rollers of elm,
f
= 0,0320.
The author found for cast-iron wheels 20 inches in diameter,
rolling on cast-iron rails,
f = 0,0183, and Sectionsrath Rittinger
ƒ = 0,0193.
f
According to Pambour, we have for iron railroad wheels about
39,4 inches in diameter
0,0196 to 0,0216.
f
R
The formula F = ƒ
r
supposes that the force F, which over-
comes the friction, acts with a lever-arm H C = H L =
= H L = r equal to
the radius of the roller, and that it describes the same space as the
latter. If, however, it acts on a lever arm H K = 2 r,
2 r, the space
described by it is double that described by the roller on the sup-
port, and the friction is therefore
F₁ = √ F = ƒ
R
far
The conditions of equilibrium of rolling friction can be found
in the following manner. In consequence of the pressure Q of the
roller A C B upon the base A O, Fig. 290, the latter is compressed ;
the roller rests, therefore, not upon its lowest point 4, but upon
the point which lies a little in front of it. Transferring the
points of application A and B of the forces Q and F, of which the
latter F is the force necessary to overcome the friction, to their
FIG. 290.
D
K
B
N
point of intersection D, and constructing
with Q and F the parallelogram of forces,
we obtain in its diagonal D R the force
R, with which the roller presses upon its
support in O, and it is therefore necessary
that the moments of the forces of the bent
lever A O N shall be equal to each other.
If we put the distance O N of the point
of support from the direction of the
force = a, and the distance O M of the
same point from the vertical line of grav-

§ 192.)
355
RESISTANCE OF FRICTION, ETC.
ity of the body=f, we have
from which we obtain the required equation
Fa = Qf,
= £2.
f
F=
a
The arm ƒ is a quantity to be determined by experiment and is
so small, that we can substitute instead of a the distance of the
lowest point A from the direction of the force F, as well as instead
of Q the total pressure
Hence we have F
R.
f
I R, and consequently, when the force
a
acts horizontally and through the centre C, a = r or
f
F
R,
r
and on the contrary, when this force acts tangentially at the high-
est point K of the roller,
f
F
R.
2 r
The so-called coefficient of friction ƒ of rolling friction is there-
fore no nameless quantity, but a line, and must therefore be ex-
pressed in the same unit of measure as a.
If a body A S B is placed upon two rollers C and D, Fig. 291,
and moved forward, the force P required to move the body is very
©MAA
FIG. 291.
C₁
R
small, as we have only two rolling
frictions to overcome, viz., one
between A B and the rollers and
the other between the rollers and
-P the surface H K. The space de-
scribed progressively by the roll-
ers is but one-half that described
by the load R, so that new rollers
must be continually pushed under
it in front, for the points of con-
tact A and B between the rollers and the body A B move exactly
as much backward, in consequence of the rolling, as the axes of
the rollers move forward. If the roller A H has turned an are
A O, it has also moved forward the space A 4, equal to this arc,
O has come in contact with O,, and the new point of contact 0,
has gone backward behind the former one (1) a distance 4 O
A O. If we designate the coefficients of friction on H K and A B
by f and f₁, we have for the force necessary to move the body forward

fi)
R
P = (ƒ + ƒ₁) ₂+++
2 r
356
[$ 193.
GENERAL PRINCIPLES OF MECHANICS.
REMARK.-The extensive experiments of Morin upon the resistance of
wagons on roads confirm this law, according to which this resistance in-
creases directly as the pressure and inversely as the thickness of the rollers.
Another French engineer, Dupuit, on the contrary, infers from his experi-
ments, that rolling friction increases directly as the pressure and inversely
as the square root of the radius of the rollers. The newer experiments of
Poirée and Sauvage by means of railroad wagons, also lead to the conclu-
sion, that rolling friction increases inversely as the square root of the radius
of the wheel. See Comptes rendues de la société des ingenieurs civils à
Paris, 5 et 6 année. Particular theoretical views upon the subject of roll-
ing friction are to be found in Von Gerstner's Mechanics, Vol. I, § 537, and
in Brix's treatise on friction, Art. 6. This subject will be treated with
more detail in the Third Part, under the head of transportation on roads
and railroads.
§ 193. Friction of Cords.-We have now to study the fric-
D
ď
C
A
B
FIG. 292.
P
18
tion of flexible bodies. If a perfectly
flexible cord stretched by a force Q is
laid over the edge C of a rigid body
A B E, Fig. 292, and is thus compelled
to deviate from its original direction an
angle D C B = a°, a pressure R is pro-
duced at this edge, which gives rise to a
friction F, in consequence of which a
force P, which is either greater or less
than 2, is necessary to produce unstable.
equilibrium. The pressure is (§ 77)
R = √ P² + Q-2 P Q cos. a, and consequently the friction
F = ¢ √ P² + Q² 2 P Q cos. a.
If now we substitute PF+Q and P' approximatively
Q² + 2 Q F, we obtain

F = & V Q² + 2 Q F + Q
2 Q* cos. a
2 F Q cos. a
a
ང་
√2 (1
cos. a) (Q² + Q F)
=
2 Q sin. « √ Q² + Q F',
α
for which we can write 2 p sin. (Q +
F), when we take into
account only the first two members of the square root. Hence we
have
a
FFsin. + 2 Q Q sin.
2
and consequently the friction required is
a
§ 193.]
357
RESISTANCE OF FRICTION, ETC.
a
2 Q sin. 2
F
a'
1
o sin.
2
for which we can generally write accurately enough
α
F = 2 Q sin.
фQ :(1
1 + 4 sin. 2), and very often
a
F = 2 & Q sin. 2
2'
when the angle of deviation a is very small. Hence, in order to
draw the rope over the edge C, we need a force
·a
2 sin.
2
P = Q + F =
1 +
e
a
1 — o sin.
and, on the contrary, the force necessary to prevent the weight Q
from sinking is
a
2 sin.
2
P₁ = Q: 1+
a
1 – o sin.
we can put approximatively
P =
P =
P₁ =
P₁
1
α
a
[1 + 2 4 sin. 2 (1
sin. / (1 + $ sin. 22, or more simply
ф
1 + 2 4 sin. Q and
1 + 2 ¢ sin.
Q
1 + 2 p sin.
α
2
α
2
2
Q
1 + 4 sin. 2)
or
a
sin. 2) Q.
(1
1-
2 sin.
20
If the cord passes over several edges, the forces P and P, at the
other end of the cord can be calculated by repeated application of
these formulas. Let us consider the simple case, where the cord
A B C, Fig. 293, is laid upon a body with n edges, and where the
deviation at each edge is the same and equal to a. The tension of
the first portion of the cord is
1 + 2 4 sin. Q,
ф
Q₁
(1
when that at the end is
Q; that of the second is
358
[§ 193.
GENERAL PRINCIPLES OF MECHANICS.
a
Q2
1 + 2 p sin.
Qi
1 + 2 4 sin. 2)² 2,
a
that of the third is
Q3 = (1 + 2 4 sin.
Qs
a
2) 2: = (1 + 2 4 sin. 2)* 2,
and in general the tension at the other end is.
P =
(1 +
a
1 + 2 p sin.
ф
2)*
2
when it is required to produce motion in the direction of the force
P. Interchanging P and Q, we obtain the force necessary to pre-
vent motion in the direction of the force Q and it is
A
α
B
Q
P₁
a
1 + 2 sin.
FIG. 293.
The friction in the first case is
B
A
FIG. 294.
L

| D

P
4
n
α
-
F = P − Q = [(1 + 2
sin. 2)" – 1] 0,
and in the second
F
Q -
1] P,
a
P₁ = [(1 + 2 4 sin. 2)" —
Φ -
∙n
= [1 − (1 + 2 ø sin. 2) * ] Q.
The same formulas are also applicable to the case of a body
composed of links, as, E.G., a chain A B E, Fig. 294, which is
passed round a cylindrical body, when n is the number of links
lying upon the body. If the length of one joint of the chain is
and the distance CA of the axis A of a link from the centre
§ 194.]
359
RESISTANCE OF FRICTION, ETC.
C of the arc, which is covered, r, we have for the angle of devia-
tion D B L = ACB = a, sin.
a
2
2 r
EXAMPLE.-How great is the friction on the circumference of a wheel
4 feet high, covered with twenty links of a chain, each five inches long
and 1 inch thick, when one of the ends is fastened and the other subjected
to a strain of 50 pounds? Here we have
P₁
a
5
= 50 pounds, n = 20, sin. 2 48 + 1
5
•
49
now if we substitute for the mean value, 0,35, we obtain the friction, with
which the chain opposes the revolution of the wheel
20
49
F
[(1 .
+ 2. 0,35. 15) - 1].
50
•
490
= [(1 + 35)" - 1].
20
50
20
= [(15)² - 1].
14
50 =
2,974 . 50 = 149 pounds.
§ 194. If a stretched cord 4 B, Fig. 295, lies upon a fixed
cylindrically rounded body A CB, the friction can also be found
by the rule given in the foregoing paragraph.
Here the angle of deviation is E D B = a°
angle at the centre 4 C B of the arc 4 B of the
cord; if we divide the same in n equal parts and
regard the arc A B as consisting of n straight
FIG. 295.

E
P
B
D
A
ао
ก
lines, we obtain n edges with the deviation
and therefore the equation between the power
and the load is as in the foregoing paragraph
a
n
P = (1 +
1 + 2 p sin.
ф
2 n
a
a
On account of the smallness of the arc
sin.
N
can be re-
2 n
α
placed by
and we can put
2 n
P
= (1
(1 +
φα
+
N
Da) Q.
Developing according to the binomial theorem, we obtain
=
P = (1 + 1
n
φα
ፃ
+
n (n − 1) (ø a)² + n ( n − 1) (n − 2) ( a )² + ...) Q,
1.2
(φ
N³.
1.2.3
3
no
or, since n is very great and we can put n − 1 = ǹ − 2 = n − 3 ... —N,
360
[§ 194.
GENERAL PRINCIPLES OF MECHANICS.
P =
(
1 + ø a +
$
113 (a) +1.3. (a)³
a)²
1.2
1
1. 2. 3 ⋅ ( a)³ + ...) Q.
(Ø
x²
X³
But 1 + x +
-+
+
1.2 1.2.3
= e, e being the base
2,71828 of the Naperian system of logarithms (see Introduction to
the Calculus), and we can therefore write.
P =
= e
ex
α
Q or Q = Pe¯
- φα
, and inversely
1
P
2,3026
a
Q
Ф
(log P
log Q).
7
If the arc of the cord is not given in parts of π, but in degrees,
αυ
then we must substitute a = 180°-, and if finally it is expressed
by the number u of coils of the rope, we must put a = 2 π u.
еф
The formula P = e
, фа
. Q shows, that the friction of a cord
FPQ on a fixed cylinder does not depend at all upon the
diameter of the same, but upon the number of coils of the cord,
and also that it can easily be increased to almost infinity. If we
put =
, we have
for
(6
(6
(6
17
coils, P = 1,69 Q
1
(6
2
P = 2,85 Q
P = 8,12 Q
P = 65,94 Q
4348,56 Q.
4
P =
= (1 + 2
(REMARK.)—From the equation P
lows that
+ 2 p sin.
Q
2) in § 198, it fol-
a
P - Q = 2 4 sin.
Φ
Q,
2
or substituting instead of a the element d a of the arc and instead of P — Q,
the corresponding increase d P of the variable tension P of the cord and
putting QP, we obtain
d P =
20
d a
2
P, 0
or
whence by integration we obtain
d P
P
l P = ¢ a + Con.
In the beginning a is = 0 and P
or inversely
1 Q = 0 + Con. and l P
o da,
=
Q, and therefore we have
P
I Q
Q
P
ea, or Pea
Q.
Q
§ 195.]
361
RESISTANCE OF FRICTION, ETC.
EXAMPLE.-In order to let down a shaft a very great but indivisible
FIG. 296.
weight P = 1200 pounds, we wind the
rope, to which this weight is attached,
13 times around a firmly fastened log
A B, Fig. 296, and we hold the other
end of the rope in the hand. What force
must be exerted at this end of the rope,
when we wish the weight to descend
slowly and uniformly? If we put here
0,3, we obtain for this force
Рефа = 1200 . е 0.3. 냥 ​2ㅠ
​
e¯¯
B
D
E
Q
33
= 1200. e
33
l Q = 7 1200
π = 7,0901
2,5918
40
4,4983,
or
log Q
=
1,9536, whence
Q 89,9 pounds.
§ 195. Rigidity of Chains.-If ropes or bodies composed of
links, etc., are laid on a pulley or a cylinder movable about its
axis, the friction of cords and chains considered in the last para-
graph ceases, because the circumference of the wheel and the cord
have the same velocity, and hence force is only necessary to bend
the rope as it lays itself upon the pulley, and sometimes to
straighten it as it is unrolled from the pulley.
If it is a chain, which winds itself around a drum, the resistance
during the rolling and unrolling consists of the friction of the bolts.
FIG. 297.
against the links, since the
K
former are turned through a
certain angle in their bear-
ings. If 4 B, Fig. 297, is a
link of the chain and B G
the following one, if C is the
axis of rotation of the pulley,
upon which the chain, stretch-
ed by the weight Q, winds.
and if finally CM and CN
are perpendiculars let fall
upon the major axis of the
links A B and B G, then
MCN a is the angle

362
[§ 195.
GENERAL PRINCIPLES OF MECHANICS.
turned through by the pulley, while a new link lays itself upon it,
and KB G = 180°
ABG is the angle described by the link
B G with its bolt B D upon the link A B during the same time.
If BDB Er, is the radius of the bearing of the bolt, the
point D of the pressure or friction describes an arc D E = r, a,
while a link lays itself upon the roller, and the work done by the
friction at the point D is,
Φι
P₁ Q. r'₁ a. Supposing the force P,
necessary to overcome this friction to act in the direction of the
greater axis B G, we have the space described by it in the same
times CN multiplied by the arc of the angle M CN=CN. a,
and therefore the work done P,. C N. a, equating the two
mechanical effects, we have P₁. C' N. a = 4₁ . Q r, a, and the force
required is
=
ཡ
=
P₁ = &₁ Q 2/1,
а
1
a denoting C N the radius of the drum plus half the thickness of
the chain.
If we neglect the friction, the force necessary to turn the
pulley would be
P = Q,
but when we take into account the friction caused by the winding
of the chain upon the pulley, we have
P = Q + P₁ =
(1 + $1
Φι
124) 2.
a
Q.
If the chain unwinds from the drum, the resistance is the same;
if, therefore, as on a fixed pulley, the rope is wound upon one side.
and unwound upon the other, the required force is
P
1+
(1 + $
1
2)2, or approximatively = (1 + 2 ¢,
а
77) 2.
If, finally, the pressure on the axle is = R and the radius of the
axle = r, the force necessary to overcome all the resistances is
P
(1
1
1 + 2
201
;) 2 + ¢
R.
a
a
EXAMPLE.-How great is the force P at the end of a chain passing
FIG. 298.
A
OB
round a roller AC B, Fig. 298, when the weight
acting vertically is Q 110 pounds, the weight
of the roller and chain is 50 pounds, the radius a
of the roller, measured to the middle of the chain,
is a
7 inches, the radius of the axle Cis
inch and that of the bolts of the chain is
inch? If we put ☀ = 0,075 and ₁ = 0,15, we obtain,
&
according to the last formula, the force

1
§ of an
g of an
P
5
P=
Q M
=(1+2. 0,15. ). 110+0,075.
8.7
(110 + 50+P),
كل
363
196.]
RESISTANCE OF FRICTION, ETC.
=
or assuming in the right-hand member P approximatively 110
P = 1,016 . 110 +0,0067 . 270 111,76 + 1,81 = 113,6 pounds.
§ 196. Rigidity of Cordage.-If a rope is passed over a pulley
or winds itself upon a shaft, its rigidity (Fr. roideur, Ger. Steifig-
keit) comes into play as a resistance to its motion. The resistance
is not only dependent upon the material, of which the rope is made,
but also upon the manner, in which it is put together, and upon the
thickness of the rope; it can consequently be determined by experi-
ment alone.
The principal experiments for this object are those made by
Coulomb and those made more recently by the author himself.
While Coulomb employed only small hemp ropes from to at most.
14 inches in thickness and made them wind upon rollers of 1 to at
most 6 inches in diameter, the author employed hemp ropes 2
inches thick and wire ropes from to 1 inch thick and passed
them over rollers from 2 to 6 feet in diameter. Coulomb's experi-
FIG. 299.
B
ments were made in two different ways. In
one case, like Amonton, he employed the
apparatus represented in Fig. 299, where A B
is a roller around which two ropes are wound,
the tension being produced by a weight Q
and the rolling down of this roller by a weight
P, which pulls upon this roller by means of a
thin string. In the other case he laid the
ropes around a cylinder rolling upon a hori-
zontal surface and, after having subtracted the
rolling friction, calculated the resistance of the
rigidity from the difference of the weights,
which were suspended to the two ends of the
rope and which produced a slow rolling motion.

AP
Q
According to the experiments of Coulomb, the resistance of the
rigidity increases tolerably regularly with the amount of the ten-
sion of the rope; but there is also a constant member K, as might
have been expected; for a certain force is necessary to bend an un-
stretched rope. It was also shown, that this resistance was inversely
proportional to the radius of the roller; that for a roller of twice.
the diameter it is only one-half, for one of three times the diam-
eter, one-third, etc. Finally, the relation between the thickness
and rigidity of a rope can only be determined approximatively from
these experiments, as we might have supposed; for this rigidity de-
364
[R 197
GENERAL PRINCIPLES OF MECHANICS.
rends upon the nature of the material of the ropes and upon the
size of the fibres and strands. When a rope is new, the rigidity is pro-
portional, approximatively, to d, and when it is old, to d, d
denoting the diameter of the rope. The assumption by some
authors that it varies with the first power, and that of others that
it varies with the square of the thickness of the rope, are therefore
only approximatively true.
$197. Prony's Formula for the Rigidity of Hemp
Ropes. According to the last paragraph, the rigidity of hemp
ropes can be expressed by the following formula:
d"
S = (K + v Q),
a
in which d denotes the thickness of the rope, a the radius of the
pulley measured to the axis of the rope, Q the tension of the rope,
which passes round the pulley, and n, K and v empirical con-
stants. Prony found from Coulomb's experiments for new ropes
d 1,7
S
(2,45 + 0,053 Q),
a
and for old ones
d ¹, +
S₁
(2,45 + 0,053 Q),
a
in which formulas a and d are expressed in lines and Q and S in
pounds. These formulas are, however, based upon Paris measures ;
for English measures they become, when expressed in inches and
pounds,
S
d 1,7
(14,39 0,289 Q)
a
S
(6,96 + 0,14 Q).
a
d 1,4
Since even these complicated formulas do not agree as well as
could be wished with the results of experiment, we can, as long as
we do not take into account the later experiments, write with
Eytelwein
$ = 1.
d²
Q
a
d² Q
3604a
In this formula a must be expressed in English feet and din
English lines, but Q and S may be expressed in any arbitrary sys-
tem of weights. If we employ the metrical system of measures,
we have
$ 197.]
365
RESISTANCE OF FRICTION, ETC.
S
8 = 18,6. d² Q
a
The results given by this formula are not sufficiently accurate, ex-
cept when the tension upon the rope, as is generally the case in
practice, is very great.
The rigidity of tarred ropes was found to be about one-sixth
greater than that of untarred ones, and wet ropes were found to be
about one-twelfth more rigid than dry ones.
EXAMPLE.—If the tension upon a new rope 9 lines thick, which passes
round a pulley 5 inches diameter, is 350 pounds, the rigidity, according to
Prony, is
S =
✯ (§)¹² (14,39 + 0,289 . 350) = 0,613˚. 46,216 28,33 pounds,
and according to Eytelwein
S
92.350
3604.5
37,75 pounds.
24
If the tension were but Q 150 pounds, we would have, according to
Prony,
S = 0,613 . 23,1
=
= 14,16,
and according to Eytelwein
81. 150
S =
=
16,2.
3604 . 5
In this case the formulas give results, which coincide better with each
other. We see from the above example, how uncertain these formulas are.
REMARK.-Tables for facilitating the calculation of the resistance due
to the rigidity of cordage will be found in the Ingenieur, page 365. Ac-
cording to Morin (see his Leçons de Mécanique Pratique), we have, when
n denotes the number of strands in the rope and a the radius of the pulley
in centimetres, for untarred ropes
d = √0,1338 n centimetres and
S
n
(0,0297 + 0,0245 n + 0,0363 Q) kilograms
2 a
d²
a
and for tarred ropes
(0,1110 + 0,6843 ď² + 0,1357 Q) kilograms,
d = √ 0,186 n centimetres and
=8
n
(0,14575 + 0,0346 n + 0,0418 Q) kilograms
2 a
d'
α
(0,3918 + 0,5001 ♂² + 0,1124 Q) kilograms.
366
[§ 198.
GENERAL PRINCIPLES OF MECHANICS.
If, however, d and a are expressed in inches, and S and Q in pounds,
we can put for untarred ropes
d2
S
(0,621 + 24,70 d² + 0,3445 Q),
a
and for tarred ones
Q
α
d²
(2,193 + 18,06 d² + 0,2889 Q).
inch, a = ½ inches and
If, E.G., for an untarred rope we have d =
350 pounds, then
(0,621 + 24,70. + 0,3445.350)
S=
9 2
16 5
9
16
9
(0,621 + 13,893 + 120,575)
=
40
30,4 pounds,
while in this case (last example) Prony's formula gave S 28,33 pounds.
§ 198. Experiments Upon the Rigidity of Thick Ropes.-
The author, in his experiments upon the rigidity of cordage, made
use of the apparatus represented in Fig. 300. The sheave or roller
B
A
།
FIG. 300.
D
M
E
R
B D E, over which the rope to be
tested is passed, was, together with a
pair of iron wheels CL M, fastened
upon a shaft or axle C, and these
wheels ran upon two horizontal rails
H R. To one end F of the rope a
weight G was attached, and to the
other end A a cross K, upon which
weights were hung until the wheels.
and pulley began to roll forward
slowly. In order to be as independ-
ent as possible of errors arising from
imperfections in the apparatus, addi-
tional weights were afterwards added
at F until a rolling motion in the
opposite direction was produced. The
arithmetical mean of the weights.
added gave, when the rolling fric-
tion was deducted, the rigidity of
the rope. The coefficient of rolling
friction to be used was determined in the same way, except that a
thin string, whose rigidity could be neglected, was employed instead
of a rope.
The mean value of this coefficient was given in § 192.

K
F
€ 199.]
367
RESISTANCE TO FRICTION, ETC.
The resistance due to the rigidity is, according to the author's
views, due less to the rigidity proper than to the friction of the
different wires or strands upon each other; for in passing over the
pulley, they naturally change their relative positions. When a
wire rope passes round a fixed pulley, the first part of this resist-
ance is wanting, as the rope, in consequence of its elasticity, gives
out, when it straightens itself, as much mechanical effect as was em-
ployed in bending it around the pulley. Hence the rigidity of the
rope in this case consists solely of the friction of the wires upon
one another, a conclusion which is confirmed by the author's ex-
periments; for he found the resistance to be forty per cent. less,
when the ropes were freshly oiled or tarred than when they were
dry. The conditions are different in the case of hemp ropes, for
they do not possess, especially after long use, any elasticity, and
the strands and fibres require force not only to bend them, but also
to straighten them.
§ 199. New Formulas for the Resistance Due to the
Rigidity of Cordage.-Since the rigidity of a rope depends not
only upon its thickness, but also upon the amount of bending it is
subjected to, and also upon the manner in which it is put together,
the author considers, that these conditions can be very well ex-
pressed by the formula
K + v Q
S =
l
the constants K and v must be determined specially for each kind
of rope. The experiments of the author also showed, that for wire
ropes we should put simply K instead of
K
or
a
S = K +
v Q
a
1. For tarred hemp ropes 1,6 inches thick passing round sheaves
from 4 to 6 feet in diameter, he found
Q
S = 1,5 + 0,00565 kilograms,
a
when the radius a is expressed in metres, or
Q
S = 3,31 + 0,222 pounds,
when a is expressed in inches.
a
2. For a new untarred hemp rope inch thick, upon a pulley
21 inches in diameter, he found
368
[$ 200.
GENERAL PRINCIPLES OF MECHANICS.
S = 0,086 + 0,00164
୧
kilograms
a
Q
0,1896 + 0,06457
pounds.
(l
3. A wire rope 8 lines in diameter, formed of 16 wires, each 11
lines thick, and weighing 0,68 pound per running foot, was passed
around pulleys from 4 to 6 feet in diameter, and gave
S = 0,49 + 0,00238
Q
a
Q
kilograms = 1,08 + 0,0937 pounds.
A
4. For a freshly-tarred wire rope, with a hemp centre in each
strand and in the rope, which was 7 lines in diameter, was com-
posed of 4.4 = 16 wires, each 1 lines thick, and weighed 0,67 .
pound per running foot, he found, with a pulley 21 inches in
diameter.
8
= 0,57 +0,0006942 kilograms
A
=
Q
1,26 + 0,0272 pounds.
a
REMARK.-A detailed description of the author's experiments is to be
found in the Zeitschrift für Ingenieurwesen (dem Ingenieur), by Borne-
mann, Brückmann and Röting, Vol. I, Freiberg, 1848. The hemp ropes
of 1 were formerly employed in Freiberg for hoisting from the shafts by
means of a water-wheel and drum (Ger. Wassergöpel), but of late they
have been replaced by the wire ropes of 3 and 4. Both of these kinds of
ropes can support with sextuple security a load of 30 cwt. It was shown
by the above experiments that, when the load was the same, the resistance
due to the rigidity of wire ropes was less than that due to the rigidity of
hemp ones. If we assume the tension of the rope to be Q = 2000, and the
radius of the sheave to be a = 40 inches, we have for hemp ropes
S = 3,31 + 0,222 2000
14,41 pounds,
S = 1.08 + 0,0937 2000
=
40
5,8 pounds.
and, on the contrary, for wire ropes
§ 200. Theory of the Fixed Pulley.-Let us now apply
the principles just enunciated to the theory of the fixed pulley.
FIG. 301.
FIG. 302.


Er
D
E
A
P
P
Let A (B, Fig. 301 or Fig. 302, be the pulley, and let a be its
§ 200.]
369
RESISTANCE TO FRICTION, ETC
radius C A = C B, r the radius of its axle, G its weight, d the
thickness of the rope, Q the weight suspended to one end of the
latter, S the resistance due to the rigidity, F the friction upon the
axle, reduced to the circumference, and PQ+F+ S the force
at the other end of the rope. The rigidity of the rope is shown by
the fact that the rope does not immediately assume the curvature
of the pulley as it is wound upon the sheave, nor straighten itself
immediately, when it is unwound. On the contrary, it approaches
the sheave in an arc, the curvature of which constantly increases,
and leaves in an arc, the curvature of which constantly diminishes.
The difference between the elastic wire ropes and the unelastic
hemp ones is that the former leave the sheave somewhat sooner
and the latter somewhat later; hence the arm CD of the force in
the first case (Fig. 301) is somewhat greater, and in the second case
(Fig. 302) somewhat less than the radius C A = a of the sheave.
If we neglect the friction upon the axle and put P = (Q + S),
we have
(Q + S). C D = Q. CE,
and consequently the rigidity of the rope is
CE
S=
(C BED
C D
2 = (c
C D
(0 - 1) 2
and the ratio of the arms is
CE
S
1 +
CD
Q'
the value of which can easily be calculated by substituting one of
the values of S.
We can also determine this force P Q + S + F without
employing the ratio of the arms of the lever by substituting in
that formula either with Prony for thin hemp ropes
d"
S=
a
(K + v' Q),
or with the author for wire or thick hemp ropes.
v Q
S = K +
α
and the friction upon the axles reduced to the circumference of
the pulley is
F = 4 ~ (Q + G + P), or approximatively,
Ф
a
F = 4 = (2 Q + G).
a
24
370
[$ 200.
GENERAL PRINCIPLES OF MECHANICS.
Hence, in the first case, we have
dn
P = Q +
(K + v Q) + $ = (2 Q + G)
α
α
and in the second
v Q
P = Q + K +
a
+ 9 = (2 Q + G).
α
In the case of the wheel and axle a reduction of the force from
the circumference of the axle to that of the wheel is necessary.
EXAMPLE.-If a wire rope 8 lines in diameter passes over a pulley
5 feet high, whose axles are 3 inches in diameter, and if the tension upon
the rope is 1200 pounds, we have the required force, when the coefficient
of friction is p = 0,075 and the weight of the pulley 1500 pounds
P 1200 + 1,08 + 0,0937 . 180º + 0,075.8% (2400 + 1500)
= 1200 + 1,08 + 3,748 + 14,62 =1219 pounds;
hence 1929
=
=
1,6 per cent. of the force is lost in consequence of the rope's
passing round the pulley.
If instead of a wire rope we employed a hemp one 1,6 inches thick, we
would have
P =
1200+ 3,31 + 0,222. 1800 + 14,62
and the loss of force would be
= 1227
27
P - Q
2,25 per cent.
12
FOURTH SECTION.
THE APPLICATION OF STATICS TO THE ELAS-
TICITY AND STRENGTH OF BODIES.
CHAPTER I.
ELASTICITY AND STRENGTH OF EXTENSION, COMPRESSION
AND SHEARING.
occurs.
§201. Elasticity. The molecules or parts of a solid or rigid
body are held together by a certain force, called cohesion (Fr. cohé-
sion; Ger. Cohäsion), which must be overcome, when the body
changes its form and size, or if it is divided. The first effect, which
forces produce upon a body, is a variation in the relative position.
of its parts, in consequence of which a change of form and volume
If the forces acting upon a body exceed certain limits, a
separation of the parts takes place and perhaps a division of the
whole body into pieces. The capability of a body to resume its
original form, after the force which caused its change of shape has
been removed, is called in the most general sense of the word its
elasticity (Fr. élasticité; Ger. Elasticität). The elasticity of every
body has certain limits. If the change of form and volume exceeds
a certain amount, the body remains of the same form after such a
change, although the forces which have produced the variation
have ceased to act. The limit of elasticity is very different for
different bodies. The bodies, which permit a great change of
volume before their limit of elasticity is reached, are called perfectly
elastic; those, whose limit of elasticity is reached when they have
undergone a very slight change of form, are called inelastic,
372
[§ 202.
GENERAL PRINCIPLES OF MECHANICS.
although no such bodies really exist. It is an important rule in
architecture and in the construction of machinery, not to load the
materials employed to such an extent that the change of form
produced shall reach, much less exceed, the limit of elasticity.
§ 202. Elasticity and Strength.-Different bodies present
different phenomena, when they are changed in their form beyond
the limit of elasticity. If a body is brittle (Fr. cassant; Ger. spröde),
it flies in pieces, when its form is changed beyond its limit of elas-
ticity; if, however, it is ductile or malleable (Fr. ductile; Ger. ge-
schmeidig), as, E.G., many metals, we can cause considerable
changes in its form beyond its limit of elasticity, without causing
a separation of its parts. Some bodies are hard (Fr. dur; Ger. hart),
others soft (Fr. mou; Ger. weich); while the former oppose great
resistance to a separation of their parts, the latter permit it with-
out much difficulty.
We understand by elasticity, in the more restricted sense of the
word, the resistance with which a body opposes a change of its
form, and by strength (Fr. résistance, Ger. Festigkeit) the resistance
with which a body opposes division. In what follows, both sub-
jects will be treated. According to the manner in which the extra-
neous forces act upon bodies, we can divide elasticity and strength
into
I. Simple and
II. Combined;
and the former again into
1) Absolute or the elasticity and strength of extension,
2) Reacting, or the elasticity and strength of compression,
3) Relative, or the elasticity and strength of flexure,
4) The elasticity and strength of sheering and
5) The elasticity and strength of torsion or twisting.
If two extraneous forces P and P act by extension (Fr.
traction, Ger. Zug) in the direction of the axis of a body A B, Fig.
303, the latter resists the
extension and tearing by
means of its absolute elas-
P
FIG. 303.
-P
B
ticity and strength or its elasticity and strength of extension (Fr.
élasticité et résistance de traction, Ger. Zug oder absolute Elasticität
202] ELASTICITY AND STRENGTH OF EXTENSION, ETC. 373
FIG. 304.
Togg
und Festigkeit); if, on the contrary, two forces P and P press
the body together in the direction
of the axis of the body A B, Fig.
304, so that the latter is compressed
A
B
and finally crushed, the elasticity and strength of compression or the
reacting elasticity and strength (Fr. élasticité et résistance de com-
pression, Ger. Druck or rückwirkende Elasticität und Festigkeit)
must be overcome. If, farther, three forces P, Q, R, which balance
each other, are applied at three different points A, B, C, in the
axis of the body A B, Fig. 305, and act at right angles to the same,
this body would be bent or perhaps broken, and it is the relative
elasticity and strength, or the elasticity and strength of flexure (Fr.
élasticité et résistance de flexion, Ger. Biegungs oder relative Elas-
ticität und Festigkeit), that must be overcome, in order to bend
or break it. If, in the latter case, the points of application A and
Clie close together, as is represented in Fig. 305, a distortion is
FIG. 305.
FIG. 306.


A
Ꭰ
B
P
-
—
produced in the cross section D D, between the two points A and
C; if the force P is great enough, the body is divided into two
parts, and in this case the elasticity and strength of sheering (Fr.
élasticité et résistance par glissement cisaillement ou tranchant,
Ger. Elasticität und Festigkeit des Abschierens) is overcome. If
two couples (P, P), (Q, Q), which balance each other, act upon
a body CA, Fig. 306, in such a manner that their planes are at
right angles to the axis of the body, a twisting of the body is pro-
duced, which may become a wrenching, and here the elasticity and
strength of torsion (Fr. élasticité et résistance de torsion, Ger. Dreh-
ungs-elasticität und Festigkeit) is to be overcome.
If several of the forces here enumerated act at the same time
upon a body, the combined elasticity and strength or a combination
of two or more of the simple elasticities and strengths comes into
play.
374
[§ 203.
GENERAL PRINCIPLES OF MECHANICS.
§ 203. Extension and Compression.-The most simple
case of elasticity and strength is presented by the extension and
compression of prismatic bodies, when they are acted upon by
forces whose directions coincide with the axis of these bodies. It is
FIG. 307.
D
C1
P
FIG. 308.
P
D
of course not necessary
that both should be
motive forces. The ac-
tion is the same, when
the body is firmly sus-
pended or supported at
one end and at the
other end subjected to
a pull or to a thrust.
We can obtain an ex-
ample of this case ei-
ther by suspending to
a prism ABC' D, Fig.


307, which hangs vertically, a weight P, or by loading with a weight
P a prism A B C D, Fig. 308, which is supported at the bottom.
In the first case, the body is extended a certain amount C' C
D D₁ = 2, and in the second case, it undergoes a similar compres-
sion; if, therefore, the initial length of the body is A D = B C =
7, it becomes, in the first case,
Ꭰ
A D₁ = B C₁ = AD + D D₁ = 1 + 2,
Ꭰ
and in the second case,
С
--
A D₁ = B C₁ = A D D D₁ = 1 — λ.
The extension or compression λ increases with the pull or thrust
P, and is a function of the same. This function or algebraical
relation between P and λ cannot be determined à priori; it is
dependent upon the physical properties of the body, and is different
for different materials. If we regard P and 2 as the co-ordinates of a
curve and construct this curve with the corresponding values of P
and 2 determined by experiment, we obtain by this means not only
a graphic representation of the law, according to which bodies are
extended and compressed by extraneous forces, but also a means of
determining the peculiarities of this law.
If we lay off from 4 on the positive side of the axis II,
Fig. 309, the tensions or tensile forces, which act upon a body, as
abscissas A B, A M, etc., and at their ends the corresponding
§ 203.] ELASTICITY AND STRENGTH OF EXTENSION, ETC. 375
extensions as ordinates B D, M O, etc., parallel to FY, we obtain
a curve A DO W, which represents the law of the extension of
FIG. 309.

V
N
C
Ա.
M, B₁ A
Ꭰ
Q
-X
Q₁
DL
B
M
C₁
R
N₁
G₁
W
R
X
this body; and if, on the contrary, we cut off on the negative side
of the axis XX from A the pressures or thrusts as abscissas A B₁,
A M₁, etc, and at the extremity of the same lay off the correspond-
ing compressions as ordinates B, D₁, M, O, etc, we obtain a curve
A D, O, W₁, by which the law of compression of the body is graph-
ically represented. According to the results of many experiments,
these two curves pass without interruption into one another, have
consequently at A a common tangent G A G₁, and are therefore
properly only branches of the same curved line WODA D, O, W₁.
Although the curve as a whole differs considerably from a right.
line, yet in the neigborhood of the origin of co-ordinates 4 it
nearly coincides with the tangent G A G₁, and since for this line
the ordinates are proportional to the abscissas, we can also assume
that the small extensions and compressions produced by the pulls or
thrusts A B, A B₁, etc., are proportional to these forces (Hooks'
Law).
The total extension M O, produced by the pull 4 M, consists
of two parts, viz.: the permanent extension or set MQ, which
remains in the body, when the stress has ceased to act, and the
elastic extension Q 0, which vanishes with the pull. It is the same
for compression. The totul compression M, O, is the sum M, Q₁ +
376
[$ 204.
GENERAL PRINCIPLES OF MECHANICS.
Q. O, of the permanent compression or set M, Q, and of the elastic
one Q₁ O₁. When the forces are small, the permanent change is so
Qi
very small compared with the total one, that it can be regarded as
not existing, and consequently the total extensions and compres-
sions can be treated as the elastic ones. If the force exceeds a cer-
tain limit A B (A B₁), the so-called limit of elasticity, if, E.G., it
becomes A M (A M₁), the permanent change of length or set forms
a considerable portion of the total extension MO or of the total
compression M, O₁. If the pull or thrust reaches a certain value
A U or A U₁, the extensions U R, U W and the compressions U, R,
and U₁₁ attain the limit at which the cohesive force of the body
is no longer able to balance the pull or thrust, and consequently a
tearing asunder or a crushing of the body takes place.
1
1
If a body has been subjected to a force, which has not extended
or compressed it beyond the limit of elasticity, the body will not
assume any further set, when subjected to another pull or thrust,
which does not reach the limit of elasticity.
§ 204. Fundamental Laws of Elasticity. Modulus of
Elasticity. The lengthening or extension of a prismatical body,
produced by a force P, is proportional, in the first place, to the
length of the body, since we can assume that equally long por-
tions are equally extended, and it is inversely proportional to the
1
6
B
FIG. 310.
F
G
P
2
=
cross-section F of the body, since we can sup-
pose the entire stretching force to be equally dis-
tributed over the entire cross-section of the body.
If, therefore, a body A B, Fig. 310, whose length
is unity and whose cross-section unity, is
extended an amount o by a stress P, the exten-
sion produced in another body F G of the same
material, whose length is = 7 and whose cross-
section is = F, by the same stress is

=
λ
στ
F
The extension oσ is of course dependent upon
the pull P alone and is different for different
materials; but according to what precedes
(§ 203) we can assume that for small pulls, which do not exceed
the limits of elasticity, the extension is proportional to the cor-
σ
responding stress, or that the quotient is a constant quantity.
P
§ 204.] ELASTICITY AND STRENGTH OF EXTENSION, ETC. 377
}
Now if A B, Fig. 311, represents the tension P of a prism,
whose length is = unity and whose cross-section
FIG. 311.
Y
=
unity, within

W
R
N
U
M, B₁ A
D
-X-
X
D
B
M
Q₁
C₁
R.
N₁
G₁
W₁
Y
the limits of elasticity and BD the corresponding extension σ,
and if we denote the angle GA U = D A B of the tangent to the
curve of extension at 4 by a, we have also
BD σ
A B
P:
tang. a =
p. and therefore
P tang. a, whence we obtain
1)
σ =
λ =
Pl tang, a
F
The quantity tang, a is dependent upon the physical proper-
ties of the body and can be determined by experiment only. If
we assume 7 1, F
1 and P = 1, we obtain tang. a = λ, and
this quantity tang. a, to be determined by experiment, is the exten-
sion which is produced in a prism, whose length is unity and whose
cross-section is unity, by the tensile force unity (see Combes: Traité
de Pexploitation des mines, tome I.). If in the formula (1) we
assume F = 1 and λ = 7, we obtain the expression
1
1 = P tang, a, or
tang, a
cotang. a P:
=
1
hence
tang. a
is that force, which would stretch a prism, whose cross-
section is one square inch (1), its own length, were that possible with-
out surpassing the limit of elasticity.
378
[§ 204.
GENERAL PRINCIPLES OF MECHANICS.
1
This hypothetical empirical quantity
tang. a
cotg. a is called
the modulus of elasticity (Fr. coefficient d'élasticité; Ger. Elastici-
tätsmodul) of the body or material and will hereafter be designated
by the letter E.
According to this we have
ΡΙ
2) 2
FE'
or the relative extension, I.E., its ratio to the entire length of the
body
λ
3)
P
FE
Inversely the force corresponding to the extension à is
λ
4) P = F E.
The same formulas obtain also for the compression 2, caused by
a thrust P, and the modulus of elasticity E = cotang. a is the same
as for extension as long as the limit of elasticity is not sur-
passed, although in this case it denotes that force, which would
compress a prism of the cross-section unity its whole length, or to
an infinitely thin plate, provided that this were possible without
exceeding the limits of elasticity.
REMARK 1.-We can also put the modulus of elasticity E equal to the
weight of a prism of the same material as the body, upon which E acts, and
of the same cross-section unity. If a is the length of this body and Y the
heaviness or the weight of one cubic inch of the same material, we have
E
γ
E a y, and therefore inversely a =
Tredgold (after Young) used this length as the measure of the elasticity
(see T. Tredgold on the strength of cast iron and other metals). If E is,
E.G., 30000000 pounds for cast steel and y 0,3 pounds, we have
a
30000000
0,3
= 100000000 inches,
I.E., a steel rod 100000000 inches long, would extend a steel bar of the same
cross-section its whole length, if the law of extension given above were true
for all limits.
REMARK 2.-During the extension or compression of a body a change
of cross-section takes place, which, according to Wertheim (see Comptes
rendues, T. 26), amounts to of the longitudinal extension or compression.
If l is the initial length, F the initial cross-section and the initial volume
Fl of the body, 1, and F₁ being the length and cross-section during the
action of the force P, we have the corresponding volume
1
F l + F (l₁ — 1) — (F' — F₁) l, or
V₁ = F₁ l₁ = —
1
1
·
V₁ — V = F (1, 1) — (F - F₁) 1,
— l,
1
-
§ 205.] ELASTICITY AND STRENGTH OF EXTENSION, ETC. 379
and the relative change of volume is
1
V₁V
V
1-1 F-F
F
1
But we know that
whence it follows that
F
―
F
1
F
(5)
V
V
1
V
= + (²² ²).
1
I.E., the increase in volume is one-third the increase in length.
C
According to the theory of Poisson.
Ꮴ
1
14
(1=-=-=).
EXAMPLE—1) If the modulus of elasticity of brass wire is 14000000
pounds, what force is necessary to stretch a wire 10 feet long and 2 lines
thick one line? Here we have
λ
7 = 10 . 12 = 120 inches, λ =
π d²
inch and consequently Į
T+10:
but F =
4
= 0,7854 (12)²
0,0218 square inches, hence the force re-
quired is
P =
T. 0,0218. 14000000 = 212 pounds.
40
2) If the modulus of elasticity of iron wire is 31000000 pounds, and an
iron surveyor's chain 66 feet long and 0,2 inch thick is submitted to a pull
of 150 pounds, the increase in length is
2. =
150
0.7854. (0.2)
66.12
31000000
0,122 inches = 1,464 lines.
§ 205. Proof Load, Proof Strength, Ultimate Strength.—
The force A B, Fig. 312, which stretches a prismatical body, whose
V
FIG. 312.

W
G
R
N
C
U₁
X
M₁ B₁ A
D
B
M
U
C₁
N₁
R
G₁₂
W
צי
380
[$ 205.
GENERAL PRINCIPLES OF MECHANICS.
cross-section is unity, to the limit of elasticity, is called the modulus
f proof strength of extension, and will in future be designated by
7, while the thrust necessary to compress the same to its limit of
clasticity is called the modulus of proof strength of compression, and
will hereafter be designated by T.
From the moduli of proof strength T and T, with the aid of
the modulus of elasticity E, the extension σ and the compression o
at the limit of elasticity can easily be found; for we have
σ T
1 E
and 61
T₁
E
If is the cross section of a prismatical body, whose moduli of
proof strength are Tand T, we have their proof strength or proof
load
1)
S for a pull,
P = FT
and for a thrust, P, FT.
In constructions the bodies should never be loaded beyond their
limit of elasticity, and the loads should therefore never surpass the
proof strength of the cross-section of the prismatical bodies em-
ployed. Cross-sections must therefore be determined by the follow-
ing formulas:
૭
P
F
and
T
P₁
F
Ti
On account of the accidental overloading and concussions, to
which buildings and machines may be subjected, and also on ac-
count of the changes, which the bodies undergo in the course of
time, owing to the action of air, water, etc., we render these con-
structions safer by substituting in the foregoing formula, instead
of the proof load, only one-half or one-third of the same, I.E. by
making the cross-section two or three times as great as those given
directly by the formula. In order to have an mfold security, we
instead of T
must substitute in the formulas F
P
T'
P₁
or F=
T
T T
ጎ
m
or T₁, the working or safe loads or
The force AU, Fig. 313, necessary to tear apart a prismatical-
body, whose cross-section is unity, is called its modulus of rupture
or of ultimate strength of extension, and is denoted by the letter K;
and in like manner we call the force A U, which crushes a body,
whose cross-section is unity, the modulus of rupture or of ultimate
7
§ 205.] ELASTICITY AND STRENGTH OF EXTENSION, ETC. 381
strength of compression, and we denote it by K₁. If the cross-sec-
tion of the prismatical body is F, we have
FIG. 313.

W
G
R
N
U
M₁ B₁ A
D
Q
-X-
X
D
B M
U
R
JZ
C₁
N₁
W
3) { P=FK for the force, which will tear the body, and
P, FK, for the force, which will crush it.
The cross-section of bodies is often determined from the modu-
lus of rupture by substituting in the formulas
P
F =
and
K
4)
P₁
F₁
K₁
instead of K the working load of rupture, I.E. a small part
K K
ጎ
or
n
E.G., a fourth, sixth, tenth, etc., of the numbers determined by ex-
periment. We call n a factor of safety. If the proof strength of
all substances were the same fraction of the ultimate strength, that
A B
is, if the ratios
T
K
A B₁
T
and
were fixed constants, the
A U
K₁
A U
determination of the cross-section by means of the moduli of proof
strength would give the same result as that by means of the work-
ing load of rupture; but since this ratio is different for different
bodies, the determinations by the aid of the moduli of proof
strengths T and T₁, or rather by means of the working or safe loads
T T
M
and are generally more correct and proper, and the deter-
M
382
[§ 206.
GENERAL PRINCIPLES OF MECHANICS.
mination by the working or safe loads of rupture
K K₁
and
d. S
N
N
is only
to be employed, when the modulus of proof strength is unknown.
If the cross-section of a body is a circle, whose diameter is d, we
πα
have
F, whence P =
4
π d²
4
T= 0,7854 ď Tand
4 F
d =
1
=
1,128 √ F = 1,128
√
P
ㅠ
​Τ
EXAMPLE 1.—What weight can a hanging column of fir support, if it is
5 inches wide and 4 inches thick? Assuming the modulus of proof
5.4 20 square
strength to be 3000 pounds, the cross-section being F
inches, we have P = F T =
F T = 20. 3000 60000 pounds as the proof load
of this column. If, however, we assume the modulus of rupture to be
K = 10000 pounds, and we desire a quadruple security, we have P=FK
= 20. 10000 = 50000 pounds. In order to be secure for a great length of
time, we take but a tenth part of K, and obtain thus P = 20. 1000 =
20000 pounds.
EXAMPLE 2.—A round wrought-iron rod is to be turned so as to bear a
weight of 4500 pounds; what should be its diameter? Here T is 18700
pounds, whence d = 1,128
4500
18700
1,128 1/45
=
187
0,553 inches. The
modulus of rupture of average wrought-iron is = 58000 pounds; if, how-
ever, we wish five-fold security, we take K = 11600 pounds, and we have
d = 1,128
4500
11600
= 1,128
45
116
= 0,7025 inches.
§ 206. Modulus of Resilience and Fragility.-When we
stretch a prismatical body by a force, which gradually increases from
0 to PAM = N 0, Fig. 314, and by this means lengthen it
from 0 to 2 = M O = A N, a certain amount of work is done,
which is determined by the product of the space or total extension
A N and the mean value of the pull. which increases gradually from
0 to P N O. This product can be expressed by the surface
A NO, whose abscissa is the extension A Nλ and whose ordi-
nate is the pulling stress N 0 = A M = P. If the extension does
not exceed the limit of elasticity, the surface A NO can be con-
sidered as a right-angle triangle, whose base and altitude are 2 and
P, and the work done, corresponding to it, is
L = ¦ & P.
If we substitute in it
λ = σl and P = F T,
206.] ELASTICITY AND STRENGTH OF EXTENSION, ETC. 3S3
we obtain the work to be done in stretching it to the limit of elas-
ticity o
L= {ol. FTo T. Fl= A V,
FIG. 314.

Y
V
W
G
N
C
U
M₁ B₁ A
D
Q
-X-
D
B
M
U
Q₁
G
N₁
R.
G₁
W.
in which denotes the volume Fl of the body and A a number,
given by experiment, which is called modulus of resilience for
extension and is determined by the expression
T2
A =
{ A C' . C D = { o T
1
½ σ² E.
E
In like manner the work necessary to compress it to the limit
of elasticity is
in which
L₁ = VA
A₁ = A C₁. C₁ D₁ = ¦ σ₁ T₁ = {
¦
T³
E
denotes the modulus of resilience for compression at the limit of
elasticity.
Similar formulas can be employed for the work done in tearing
or crushing prismatical bodies; for the first case we have
and for the second,
B = the surface A
tearing; and B,
for crushing.
L = V B,
Ꮮ
L₁ = V B₁₂
UW denoting the modulus of fragility for
the surface A U, W₁, the modulus of fragility
034
[$ 207.
GENERAL PRINCIPLES OF MECHANICS.
We see from the foregoing that the mechanical effect necessary
to stretch or compress a prismatical body to the limit of elasticity,
as well as that, which is necessary to produce a tearing or crushing
of the same, is not at all dependent upon the different dimensions,
but only upon the volume V of the body; that, E. G., for two prisms
of the same material the expenditure of mechanical effect in pro-
ducing rupture is the same, when one is twice as long as the other
and the cross-section of the former but one-half that of the latter.
EXAMPLE.—if the modulus of elasticity of wrought iron is E
pounds and the extension of the same at the limit of elasticity o
T
the modulus of proof strength is, since σ = E'
T = σE=
28000000
1500
and consequently the modulus of resilience for extension is
28000000
1
1500'
18700, (approximatively)
A 10 T
σ
T2
2 E
1 0² E
18700
2.1500
= 6,23 pounds.
Hence, in order to stretch a prismatical body of wrought iron to the limit
of elasticity, the mechanical effect
L = AV = 6,23 V is necessary.
If, E.G., the volume of this body were V = 20 cubic inches, the me-
chanical effect would be L =
10,38 foot-pounds.
6,23 . 20 = 124,6 inch-pounds
=
124,6
12
(§ 207.) Extension of a Body by its Own Weight.-
If a prismatical body A B, Fig. 315, has a considerable length 1,
it undergoes, in consequence of its weight, a notable extension,
which can be determined in the following manner. Let F denote
the cross-section of the body, y its heaviness or the weight of a cu-
bic inch of the matter composing it and x the variable length of a
portion of it; the tension in an element M N is produced
by the weight of the part of the body B M lying below
it, and consequently [according to § 204, (2)] the cor-
responding extension of the length M N = 8x of this
element is
FIG. 315.
N
M
ZE
B
P
Y F x
d λ =
FE
d x
Y
E
x d x.
By integration we obtain the extension of the entire
piece B M
λ
Y
E
z Sx dx =
Y x²
2 E'
and consequently that of the entire body A B is
$207.] ELASTICITY AND STRENGTH OF EXTENSION, ETC.
385
y l²
Y FI
1 G
λ
1,
2 E
2 FE
FE
in which Gy Fl denotes the weight of the whole body.
If this weight was not equally distributed in the body, but
applied at its end B, the extension would be
Ꮐ l
λι
= 2 λ.
FE
1
The extension λ = λ, of a body in consequence of its own
weight, is but one half as great as that produced by the same weight
at the end of the body.
The same law obtains of course for the compression λ produced
in a body by its own weight.
If in either case a pull or thrust P acts upon the body, we have
the extension or compression produced
î
ΡΙ
FE
± 1/10
G l
FE
( P ± G) l
FE
in which the upper sign is to be employed, when the force P acts
in the same direction as the weight G, and the lower one, when it
acts in the opposite direction. In the latter case, the extension is
of course smaller than when P is the only tensile or compressive
force.
The total extension or compression is 0, when
Ꮐ GP, or Gy Fl 2 P, or
2 P
Y F
=
The force P, acting at the end of the body, extends it equally
while, on the contrary, the
λ
P
in all parts, viz., in the ratio
7
FE'
d λ
weight G stretches or compresses it in the variable ratio
Y X
d x
Ε
The ratio of the total extension at any point, at the distance x from
the point of application of the force P, is
2
λι λ αλ P
土
​dx
1
±y x
F
Ε
If the force P acts in the same direction as G, the maximum
ratio of extension or compression is for x = 1, and it is then
λι
1 = (-/F
P
+ λ
川
​1
E
25
P+ G
FE
386
[$ 207.
GENERAL PRINCIPLES OF MECHANICS
and, on the contrary, the minimum is for x = 0, I.E., at the point
λα P
of application of P, and it is
FE'
If P and G act in opposite directions, we must distinguish the
P
P
cases, in which 7 <
and in which 7 >
FY
> FY
In the first case
the ratio of extension or compression = ( y )
λι
P
x
is a
E
P
and a minimum and
EF'
maximum for x = 0 and
P
1
( − r ) /
Y 1) for
maximum
E
P
EF
7. In the latter case there is a positive.
0, and a negative maximum (yl
for x =
x = 0, and
for x = l, and, on the contrary, for x =
= zero.
P
FY
P、 1
)
FE
the function becomes
In order that the body shall be extended or compressed to the
limit of elasticity only, the maximum of the ratio of extension or
1
T
±y should be at most = σ = or more
E
E
compression (±yx
simply the maximum of
(
have the same direction, this
±y x) = T. But, when P and G
maximum is
P
P + y Fl
P+ G
+ y l =
F
yl
F
F
P + y F l
=T, or P F (T — y l),
F
and therefore we must put
hence the required cross-section is
P
F
Τ
If, on the contrary, the forces P and G act in opposite directions,
P
F
P
), and
we have two maxima, one = and the other = (y1-1),
F
therefore the corresponding cross-section is equal to the greater of
the values
P
P
F=
and F
T
T.
γι
If in the formulas we substitute K instead of T, we obtain the
conditions of tearing and crushing, that is, in the first case,
PF(Ky 7), and in the second either
PFK or P= F(yl K).
·
A
§ 208.] ELASTICITY AND STRENGTH OF EXTENSION, ETC.
387
For P0 we have either
T
yl- T= 0 and 7 =
or
γ
K
Y
y l − K = 0 and ? = ;
the first formula. being applicable to the case, when the body is ex-
tended or compressed to the limits of elasticity, and the second to
the case, when a tearing or crushing of the body takes place.
REMARK.-The energy stored by a body, which is extended or com-
pressed by its own weight, can be calculated in the following manner. The
element M N, Fig. 316, whose length is dx, is gradually stretched by the
weight y Fx of the portion of the body B M an amount, which
y x d x
increases gradually from 0 to d 2
and the work done
E
in accomplishing it is
FIG. 316.
N
M
ZZ
B
1 y F x. 8 2 =
y² Fx²
E
d x.
Integrating this expression, we obtain the expression for the
quantity of work done in extending all the elements of the rod
from В to M,
E
and that done in extending the entire rod
L = 4 ⋅ 2 F S x d x = ↓
y² Fx³
x²
1.
3 E
L = 1.
12 F13
SE
22 F2 12 1
3 FE
•
G2 i
3 FE
3 G 7,
in which (according to § 207) 2
the rod.
Ꮐ l
FE
denotes the total extension of
EXAMPLE. If a lead wire, whose modulus of rupture is K = 3100 and
the weight of a cubic inch of which is = 0,412 pounds, is suspended verti-
cally, it will break by its own weight, when its length is
7
K 3100
γ 0,412
— 7524 inches = 627 feet.
If the modulus of proof strength is T 670, it is stretched to the limit
of elasticity, when its length is
T 670
Y 0,412
1626 inches = 135,5 feet,
and if its modulus of elasticity is E = 1000000 pounds, we have for the
corresponding extension
T
λ= Ն,
670
E 1 1000000
•
135,5 = 0,090785 feet = 1,0894 inches.
§ 208. Bodies of Uniform Strength.-If the pull or thrust
P upon a vertical prismatical body is sensibly augmented by its
weight G, we must of course put
P+ GF Tor PFT-G= F (T — l y),
"
388
[§ 208.
GENERAL PRINCIPLES OF MECHANICS.
and determine the cross-section of this body by means of the for-
mula (compare § 207)
FIG. 317.
P
F
T-ly
If this body, as, E.G., A B, Fig. 317, is composed of prismatical
parts, we can save material by giving to each of these parts a cross-
section calculated by means of this formula. If the
length of these portions of the body are la, la, la, etc., and
if the load P is gradually increased by the weights Fly,
Fa la Y, Fa la Y, etc., of the portions to P1, P2, P3, etc., the
F4 required cross-section of the first portion is
A

F5
FA
E3
F2
F
B
P
F
P
Th₁Y
that of the second should be
P₁
1
F₂ =
T-ly
FT
T - l₂ Y
that of the third
P₂
2
F, T
2
etc.
F3
T - ls Y
Τ
T-by
Y
If the length of all the parts is the
we have more simply
P
P
T
-
same, or l₁ = l₂ = l,, etc., = 1,
F₁ = r = 17 = F(1~13)
T-ly
TT-1 ly
し
​FT
F₂ =
T-ly
PT
(T — 1 y)²
P
T
T\T
ī
F2
P
3
T
F3
F₁ = 1² Ty = = (x = 17), etc,
T \T
or in general for the cross-section of the nth portion
F
P
T
P. = " (1 = 17)".
TT ly
If the cross-section of all the pieces are to be the same, that
cross-section should be
P
F=
T-nly
P
所
​T
77).
T \T — n l
While in this case the volume of the whole body would be
V = n Fl
=
n Pl
T-nly
in the former case, where every piece has its own proper cross-sec-
tion, the volume is determined by the geometrical series.
5
§ 208.] ELASTICITY AND STRENGTH OF EXTENSION, ETC.
389
V₁ = (F₁ + F₂ + ... + F) l
ΡΙ
1
1 +
T — l
ly
T
T-ly T
T
+
+ ... +
(2)
T
T ly
But the sum of the geometrical series in the parenthesis is (see
Ingenieur, page 82)
T
T
= [(x-1,)"'— ¹]: (x ——1, − 1);
whence it follows, that
P
T
n
TY
(Fn —
Y
V₁ = = [( 27 ) - 1] = (E. - F) T,
То
Y
T-ly
and that the weight of the whole body is
G = (FF)
(F₂ — F₁) T.
If the length of the parts is very small, and, on the contrary,
their number ʼn very great, and if we denote the total length n l by
a, we have, reasoning as in § 194,
αγ
αγ
a Y
―
(T — 1 y)" = (T — ²/?')" = T" (1 − ″ 7)"
n
=
-
“
=T" €
n T
The T'
in which e 2,71828 is the base of the Naperian system of loga-
=
rithms, and therefore we have
F
P
=
T
n
P
P
a Y
a Y
e
=
Te-a7
T
T
F.&T,
T\T — 1
P
in which F.
T
denotes the area of the first cross-section.
We have also approximatively
αγ
Τ
a
F = [1 + 7 + 1 ( 77 ) ]
P
2/2
T
and, on the contrary,
a
T
T
F = [1 + 7 + (7)]
α
T
The volume of the body, composed of very many small por-
tions, is found in the manner shown above to be
P
T
#
P
- 2 ет
= — [ ( 7 — — 1 ;)" − 1
approximatively
Ра
Y T -
Pa [1 + 1 (27)
Τ
] = — ( · 7 −1),
ay
+
T
T
)']
390
[$ 208.
GENERAL PRINCIPLES OF MECHANICS.
while on the contrary, the volume of the body with a constant
cross-section is approximatively
V =
Pa
Tay
Pa
T
[1 +
αγ
α
2
+
Τ
T
The formulas
F₁ =
Р
αγ
P
eT and V₁ =
T
T
γ
(-1)
FIG. 318. FIG. 319.
P
En
A
hold good, of course, for every body, such as A B, Fig. 318, and
A B, Fig. 319, in which there is a constant variation of the cross-
section. In order to find the cross-section F
for any position M and the volume of the body
cut off at the same point, we have only to sub-
stitute in this formula for a the distance B M
of the given position from the point of applica-
tion B of the tensile or compressive force. The
bodies thus determined have at every point a
cross-section corresponding to the load they
support, and are therefore called bodies of uni-
form strength (Fr. solides d'égale résistance,
Ger. Körper von gleichem Widerstande). These
bodies have (the other circumstances being the
same) the smallest volume, require therefore the least quantity of
material and are for this reason generally the cheapest and most
advantageous that we can employ. If we compare such a body
with a prismatical one, we find from the above approximate formu-
las, that the economy of volume is

Fo
P
B
A
F
Parlay 5
2
Pa² Y
T2
αγ
V - V. = 2 [ 1 7 7 + ( 7 ) ] = 272 (1+57)
n
Τ 2 T
αγ
6 T
3 T
REMARK.-Since the relative extension and compression of a body of
T
Ε
E'
uniform strength is everywhere the same, viz., σ = its total extension is
T
E
λ= σα= a, while for a prismatical body it is only
2=
(P + ½ G) a
FE
P+ G T
P + G E
a.
EXAMPLE. What must be the cross-section of a wrought-iron pump
rod, whose length is 1000 feet, when, in addition to its own weight, it must
support a load P = 75000 pounds ? If instead of the modulus of proof
strength T = 18600 we employ for safety a working load
and put the weight of a cubic inch of wrought-iron
T
2
= 9300 pounds
Y =
7,70 . 62,425
12. 12. 12
= 0,2782 pounds,
$209.] ELASTICITY AND STRENGTH OF EXTENSION, ETC. 391
the required cross-section is
P
75000
75000
F=
= 12,58 square inches,
T αγ
9300 -
12000.0,2782
5962
and the weight of the rod is
G = F. ay = 12,58. 12000. 0,2782
42000 pounds.
If we could give this rod the form of a body of uniform strength, we
would have for the smallest cross-section
P
75000
F
= 8,06 square inches,
Τ
9300
and for the greatest
F
8,06. eº,2782.1,29 = 8,06 €º,3689
=
8,06 . 1,432 =
11,54 square inches,
G₁ =Vny (FF) T (11,548,06) 9300
Gn
and the weight of the rod would be
=
If the modulus of elasticity of wrought iron is E
the extension of the rod in the latter case would be
T
λ =
a
E
93
feet =
7,97 inches,
280
140
18600. 1000 186
28000000
and, on the contrary, in the first case it is
=
32364 pounds.
28000000 pounds,
P + G
P+ G
λ=
75000 + 21000
75000 + 42000
7,97
96000
117000
•
7,976,54 inches.
§ 209. Experiments upon Extension and Compression.
-In order to study thoroughly the laws of the elasticity of any
substance, it is necessary not only to submit prismatical bodies of
this substance (which should be made as long as possible) to
extension or compression by weights, which are gradually increased
in amount until rupture is produced, but also to observe the exact
extension or compression produced by each weight. If we place
the bodies to be experimented upon in a vertical position, the
weights can be hung or laid upon them, and they then give
directly the pull or thrust to which the body is subjected. In
order to avoid experimenting with too great weights, we generally
prefer to let the weights act upon the body by means of a lever
with unequal arms; the weights are always hung upon the long
arm (a), and the body is acted upon by the shorter arm (b). Mul-
tiplying the weight G by the ratio of the arms, we find the corre-
a
sponding pull or thrust P =
00
G.
The so-called hydraulic press
392
[$ 209.
GENERAL PRINCIPLES OF MECHANICS.
can also be employed with advantage instead of weights to produce
very great tensile or compressive forces. In order to observe the
amount of the extension or compression, a fine line is drawn upon
the bar to be experimented with near each of its ends, or a pair of
pointers, with verniers attached, are fastened to it at those points,
and in order to determine not only the elastic, but also the perma-
nent extension or set, we measure the distance between these lines
or pointers not only before and during the application of the
weights, but also after they have been removed, and it is generally
preferable to allow several minutes or even hours to elapse between
the application or removal of the weights and the measurement;
for when the forces are very great the extension and compression
do not assume the true value in a moment, but only after a certain
time. This distance is measured either with a bar compass or
directly by means of a division on the rod itself. The so-called
cathometer is also employed for this purpose; it consists essentially
of a vertical staff and of a spirit-level, which is capable of sliding
up and down the former (see Ingenieur, page 234). In order to
observe the compression on long rods, we must enclose them in
tube-shaped guides; they must also be well greased from time to
time, so that they can slide without resistance in their guides.
FIG. 320.
E
A
B
If we wish to determine the modulus of ultimate strength of a
body, we can employ shorter pieces for the experiments. In
experimenting upon rupture by
cxtension we employ bodies with
large heads A and B, Fig. 320,
through which holes are bored
exactly in the axis. In the
middle of each hole a circular

D
knife-edge is made, so that the body shall be pulled exactly in the
line of the axis by means of the bolt CD and the
clevis FE, which is applied to its ends.

FIG. 321.
بتا
E
F
In experimenting upon rupture by crushing,
the two bases of the body (A, Fig. 321) are
made parallel, it is then brought between two
cylinders B and C, whose bases are ground flat;
while the rounded head of one of the cylinders
is acted on by the compressive force, the other
is supported by the large bed-plate D, and both
slide in the interior of cylinder E F. The
pressure P upon the head H of the cylinder is
§ 210.] ELASTICITY AND STRENGTH OF EXTENSION, ETC. 393
produced either by a hydraulic press or by a one-armed lever
L O, such as is partially represented in the figure.
While the rupture of a body by tearing occurs in the smallest
cross-section, and the body is therefore divided in two parts only,
the rupture by crushing takes place generally in inclined surfaces,
and the body is divided into several pieces. Prismatical bodies are
divided, in the first place, into two pyramids, whose bases are those
of the body and whose apexes are at its centre, and in the second
place, into other pyramidical bodies, whose bases form the sides of
the body and whose apexes are also situated at its centre. Bodies,
whose structure in different directions is different, of course do not
act thus; E.G., a piece of wood would be compressed by a force
acting in the direction of the fibres, in such manner, that at its
smallest cross-section the fibres would be bent out in a spherical
form.
§ 210. Experiments upon Extension.-We are indebted
to Gerstner for the first thorough experiments upon the extension
and elasticity of iron wire. He employed in his experiments iron
wire from 0,2 to 0,8 lines in diameter and made use of the lever
apparatus represented in Fig. 322 with the pointer CD 15 feet
FIG. 322.

F
D
B
D
A
long, the counter-balance G and the sliding weight Q. The wire
E F, which was about 4 feet long, was firmly fastened at one end E
and the other was wound round a pin F, which was turned by the
394
[§ 210.
GENERAL PRINCIPLES OF MECHANICS.
endless screw S, so that the wire could be subjected to any desired
strain. The extension of the wire was shown by the pointer D
upon a rod A B in 54 times its natural size. The knife-edge C of
the lever, the pin F, around which the upper end of the wire is
wound, and the endless screw S, which turns the pin, are all repre-
sented on a larger scale in Fig. 323.
FIG. 323.
S
Gerstner proves by his ex-
periments, that every extension.
is the sum of two extensions, one
of which (the elastic extension)
disappears, when the weight is
removed, and the other (the per-
manent extension, or set) remains,
so that the extension 2 is not ex-

aetly proportional to P within the limits of elasticity, and that it
is more proper to replace the formula
λ
P FE [S 204 (4)]
by the following series
λ
λ
P = [(1 + a + 9 ( 4 ) ] FB,
즐 ​B(주​)],
N
De
in which a and ẞ are numbers determined by experiment.
Quite extensive experiments upon the elasticity and strength
of wrought iron and iron wife were afterwards made by Lagerhjelm
and by Brix. Both experimenters employed in their researches a
bent lever A C B, Fig. 324, the longer arm C B of which was de-
pressed by the weights G, which were laid upon a scale-pan W, and
FIG. 324.

A Dd
E S
BE
Vañan
W
§ 210.] ELASTICITY AND STRENGTH OF EXTENSION, ETC. 395
thus the iron bar or wire D E, which was fastened to the shorter
arm CA, was stretched to any desired extent. In the apparatus
CA
C B
1
201
used by Brix, the ratio of the arms of the lever was
and one end D of the wire was attached to the arm CA with
clamps, hooks and bolts, and the other end was fastened in the
same way to a screw S, which was turned by means of a train of
wheels by a crank K. The increase in length was given by two
verniers, which were screwed fast to the ends of the wire and
moved along two scales divided into quarter lines. When the wire
had been firmly fastened in the clamps, the scale-pan was gradually
loaded with heavy weights, and in each experiment the wire was
stretched by turning the crank K until the lever was lifted from
its support and the tension of the wire balanced the weight G.
The experiments were made with wire 1 to 1 lines thick and
gave for the average value of the modulus of rupture of unannealed
wire K = 98000 pounds, and, on the contrary, after annealing,
K 64500 pounds. The average modulus of elasticity, on the
contrary, for annealed and unannealed wire was found to be
E = 29000000 pounds; it was also found, that the limit of elas-
ticity was reached, when the strain was 0,5 K for unannealed and
0,6 K for annealed wire.
3
When the tensions were greater, the extension became perma-
nent, and the total extension of unannealed wire at the instant of
rupture was
λ
ī
λ
}
= 0,0034, and that of annealed wire = 0,0885,
or 26 times as much. In the apparatus used by Lagerhjelm the
tension on the wire was produced by a hydraulic press, the piston
rod of which was attached to the end of the iron bar.
Lagerhjelm employed in his experiments iron rods 36 inches
long, inch thick, the cross-sections of which were circular and
square. According to his experiments, the average modulus of
elasticity for Swedish wrought iron is
E = 46000000 pounds;
the modulus of rupture or of ultimate strength is
and the modulus of proof strength
1
K
=
=
E 92000 pounds;
500
1
T = o . E =
. 46000000 = 28750 pounds.
1600
396
[§ 210.
GENERAL PRINCIPLES OF MECHANICS.
Wertheim, in his experiments upon the elasticity and cohesion
of the metals, allowed the wire to hang freely, and fastened to the
end of the same a weight-box, which was supported upon the floor
by means of feet, which could be raised or lowered by turning a
screw. In order to stretch the wire by means of the weights
placed in the box, the foot-screws were turned until the box swung
freely. A cathometer was employed to determine the extension of
the wire.
The experiments were performed at very different tempera-
tures, and with wire made of various metals, such as iron, steel,
brass, tin, lead, zinc, silver, etc. The principal results of these ex-
periments will be found in the table given in § 212.
The apparatus, with which Fairbairn performed his experiments,
consists essentially of a strong wrought-iron lever or balance-beam
A CD, Fig. 325, whose fulcrum D is firmly retained by a strong
bolt F, which can be raised or lowered by means of a nut. Two
FIG. 325.

H
S
P
U
E
K
K
A
Z
R
B
Ն
M
NE
V
V
iron pillars give the necessary resistance to the bed-plate H H,
through which F passes. The piece of iron L M to be experi-
mented upon is suspended by means of a chain to the support KK,
which reposes upon the two columns T T and is connected by a
bolt and clevis to the stirrup C of the lever A CD. To the longer
§ 211.] ELASTICITY AND STRENGTH OF EXTENSION, ETC. 397
arm of the latter there is suspended not only a constant weight G,
but also a scale-board for the reception of smaller weights; the
bolt I serves to support the lever, and the latter is raised by means
of a rope ( P, which passes over a pulley and is wound upon the
shaft W of a windlass U Y Z. After the weights had been laid on,
the arm E of the lever was allowed to sink gradually by turning
the crank U, until the piece of iron to be tested was finally sub-
jected to the tension produced by N and G.
REMARK.-Gerstner's experiments upon the elasticity of iron wire, etc.,
are discussed in Gerstner's Mechanics, Vol. I. For the experiments of
Lagerhjelm, see Pfaff's translation of the treatise: Researches for the pur-
pose of determining the density, homogeneity, elasticity, malleability, and
strength of bar iron, etc., by Lagerhjelm (Nürnberg, 1829), and the informa-
tion in regard to the experiments of Brix is to be found in the treatise on
the cohesion and elasticity of some of the iron wires employed in the con-
struction of suspension bridges (Berlin, 1837).
The experiments of Wertheim upon the elasticity and cohesion of the
metals, etc., as well as of glass and wood, are discussed in "Poggendorf's
Annalen der Physik und Chemie," Ergänzungsband II, 1845. In the
latter experiments the modulus of elasticity of the bodies named was de-
termined not only by experiments upon extension, but also by experiments
upon flexion and vibration. For Fairbairn's experiments on the strength
of materials, his "Useful Information for Engineers" can be consulted.
§ 211. Iron and Wood.--The most complete set of experi-
ments upon the elasticity and strength of cast and wrought iron
are those more recently made by Hodgkinson. By these we have
for the first time acquired a complete knowledge of the laws of ex-
tension and compression for these materials, which are of such
great importance in their practical applications. Although, accord-
ing to these experiments, iron produced in different ways has
different degrees of elasticity and strength, yet it is possible to
express the behavior of this body in regard to extension and com-
pression by means of curves.
The average modulus of elasticity of cast iron (Fr. fonte, Ger.
Gusseisen) is, according to these experiments, for extension as well
as for compression
E = 1000000 kilograms, when the cross-section is one centime-
ter, and consequently
E 14,22. 1000000 = 14220000 pounds when the cross-section is
=
one inch.
The extension at the limit of elasticity is
λ
1
σ =
7
1500
398
[S 211.
GENERAL PRINCIPLES OF MECHANICS.
This extension corresponds to the modulus of proof strength
T
1000000
1500
= 667 kilograms, or
14220000
T=
= 9480 pounds.
1500
1
The compression at the limit of elasticity, on the contrary, is
1
01
750'
and therefore the modulus of proof strength is
T
=
1000000
750
=1383 kilograms
14220000
750
= 18960 pounds.
The modulus of rupture for tearing was found by these experi-
ments to be
K 1300 kilograms
and, on the contrary, that for crushing
=
18486 pounds,
K₁ = 7200 kilograms 102400 pounds.
1
The resistance of cast iron to crushing is, therefore, 5 times as
great as that to tearing.
For wrought iron (Fr. fer; Ger. Schmiedeisen) we have for
extension as well as compression
E2000000 kilograms. =28440000 pounds,
λ
and the limit of elasticity is reached, when σ =
the modulus of proof strength is
T
2000000
1500
= 1333 kilograms
1
whence
1500'
18960 pounds.
Finally the modulus of rupture or of ultimate strength of
wrought iron was found to be for tearing
K = 4000 kilograms = 56880 pounds,
and for crushing
K₁ = 3000 kilograms 42660 pounds.
1
3
=
The modulus of elasticity of wrought iron is therefore about
double that of cast iron, and while the modulus of rupture by tearing
of cast iron is but about that of wrought iron, the modulus of rup-
ture by crushing of cast iron is nearly 2 times as great as that of
wrought iron. The relations of the elasticity and strength of cast
and wrought iron are graphically represented in Fig. 326. From
the origin A on the right-hand side of the axis of abscissas X X
the tensile forces, given in thousand pounds per square inch, are
laid off and on the left-hand side the compressive forces, while the
§ 211.] ELASTICITY AND STRENGTH OF EXTENSION, ETC. 399
upper half of the axis of ordinates Y Y represents the correspond-
ing extensions, and the lower half the compressions. It will at
once strike the eye, that the curve of cast iron has a great develop-
ment on the side of compression and that of wrought iron on the
side of extension; and we also remark, that the curves form
approximatively straight lines near the origin A.
FIG. 326.
Wrought Iron

Thousandths 10TY
9+
8+
7
Wood
6+
5
4
3
40 30 20 10
2313
Cast Iron
a
Thousand pounds
90 80 70 60 50
-X
+
d
Wrought Iron
Cast Iron
10
20 30 40
X
50 60
1
&
Thousand pounds
2
3
4
Wood
5
+6
-7
+8
-9
Y 10 Thousandths
As next to iron wood (Fr. bois; Ger. Holz) is most generally
employed in construction, the relations of the elasticity of fir,
beach and oak wood are graphically represented in the figure by
a curve. The average modulus of elasticity of these kinds of
wood is
E = 110000 kilograms =1564200 pounds.
1
The limit of elasticity is reached, when σ = of the length, and
600
the corresponding modulus of proof strength is
110000
T
600
= 180 kilograms = 2607 pounds.
Finally, the modulus of rupture for tearing is
K = 650 kilograms =9243 pounds,
400
[§ 211.
GENERAL PRINCIPLES OF MECHANICS.
and, on the contrary, for crushing
K = 450 kilograms 6399 pounds.
=
The ratio 156: 1422: 2844 approximatively = 1:9: 19 of the
moduli of elasticity of wood, cast and wrought iron to each other
is expressed in the figure by the subtangents ab, ac and ad.
FIG. 327.
Thousandths 10TY
Thousand pounds
80 70
90
X
60
50
40
30
20
10
Cast Iron
d
Wrought Iron
b
9+
8+
7-
Wood
6+
5
4
со
Wrought Iron

2212
Cast Iron
a
d
X
10
1
20
Thousand pounds
30 40 50
60
а
2
3
4
Wood
5
·7
8
9
-Y-10 Thousandths
oT
The modulus of resilience Aσ T for the limit of elasticity
is expressed by the triangles A a b, A α, c, and A a, d₁, the bases
a
of which are the small ratios of extension σ = A α =
1
and
600
1
σ = A a₁ =
(approximatively).
1500
From the above, we have for wood
1
A = σ T = 1
=
•
180 0,15 kilogram centimeters
600
1
2607 = 2,17 inch-pounds,
600
for cast iron
A
•
1
1500
667 0,222 kilogram centimeters 3,16 inch-
pounds, and for wrought iron
=
8 212.] ELASTICITY AND STRENGTH OF EXTENSION, ETC. 401
A = 1/1/
1333
1500=0,444 kilogram centimeters = 6,32 inch-pounds.
Properly, a complete series of experiments is necessary to deter-
mine the modulus of fragility for tearing or crushing; for this
modulus is found by the quadrature (see Art. 29, Introduction to
the Calculus) of the complete branches of the curve on either side,
and this is especially necessary for the extension of wrought iron and
for the compression of cast iron, since the curves corresponding to
the changes in these bodies differ considerably from right lines.
The extension and compression of wood at the instant of rupture
by tearing or crushing is so little known, that we are unable to
give with any degree of certainty its moduli of fragility. If we
treat the corresponding curve as a right line, we obtain the modu-
lus of resilience for tearing
B =
K2
14
E
6502
110000
= 1,91 kilogram centimeters =
27,2 inch-
pounds, and, on the contrary, the modulus of fragility for crushing is
K₁2
B=1
E
450❜
110000
= 0,92 kilogram centimetres 13,07 inch-lbs.
When cast iron is ruptured by tearing, assuming the extension
to be σ₁ =
01
0,0016 and the mean value of the force to be 560 kilo-
grams, the modulus of fragility is
=
=
B = 0,0016. 650 1,04 kilogram-centimetres 14,8 inch-lbs.
When cast iron is ruptured by crushing, the maximum exten-
sion can be assumed to be σ, 0,008 and the mean crushing force
to be 3600 kilograms; hence the corresponding modulus of
fragility is
=
1
=
B₁ = 0,008. 3600 29 kilogram-centimetres 411 inch-lbs.
We can assume as the mean value of o, for the rupture of
wrought iron by tearing, 0,008 and for the mean value of the
force 3000 kilograms; hence the corresponding modulus of fra-
gility is
B = 0,008. 3000 = 24 kilogram-centimetres = 341 inch-lbs.
On the contrary, for the rupture of wrought iron by crushing,
we must assume σ = 0,0018 and the mean force to be
kilograms; whence the corresponding modulus of fragility is
B = 0,0018. 1300 2,34 kilogram-centimetres = 33,3 inch-lbs.
—
= 1300
§ 212. Numbers Determined by Experiment. In the
following tables I and II the mean values of the moduli of elas-
20
402
[§ 212.
GENERAL PRINCIPLES OF MECHANICS.
ticity, of proof strength and of ultimate strength of the materials
generally employed in constructions are given. The first table is
for tensile and the second for compressive forces.
The value of the relative extension σ = for the limit of elas-
λ
7
ticity given in the second column of the tables expresses also the
T
ratio of the values of Tand E given in the third and fourth
E
1
columns. In practice the bodies are only loaded with T, E.G.,
M
Tto T, or the cross-section is determined by substituting in
the formula
F=
P
K?
1
instead of K, for metals the modulus of safe load
-
K = K, for
N
20
ΤΟ
1
wood and stone = 1
o K, and for masonry but
trary, for ropes we can employ K to K.
safety.
K. On the con-
We call n a factor of
The lower numbers in the parenthesis give the values in
kilograms, assuming a cross-section of 1 centimetre square; the
upper numbers express the values in pounds referred to a cross-
section of one square inch.
REMARK.-The moduli given in these tables are for unannealed metals.
For annealed metals (Fr. metaux cuits, Ger. ausgeglüte Metalle) the modu-
lus of elasticity is generally the same as for unannealed metals, while the
modulus of rupture by tearing of annealed metals is generally from 30 to
40 per cent. less than that of unannealed ones. Tempered and annealed steel
(Fr. acier trempé et recuit, Ger. gehärteter und angelassener Stahl) has the
same modulus of elasticity as untempered steel, but its modulus of proof
strength is 20 to 30 per cent. greater than that of untempered steel. When
it is not otherwise stated, the moduli for metals were determined with
wire, which had on the outside a harder crust (caused by the drawing)
than hammered or cast metal rods. For some materials, E.G. wood, iron,
and stone, the moduli of elasticity, of proof strength and of ultimate
strength vary so much that in particular cases a value differing 25 per cent.
(more or less) from those here given may be found.
& 212.]
403
ELASTICITY AND STRENGTH OF EXTENSION,
ETC.
TABLE L
MODULI OF ELASTICITY AND STRENGTH FOR EXTENSION.
Name of the material.
جانا
Extension

σ =
at the limit of
Elasticity.
Modulus
of Elasticity E.
I
Cast iron...
0,000667
1500
I
Wro't iron in rods.
0,000667
1500
I
in wire...
= 0,001000
I000
I
in sheets...
German steel, tem-
pered and annealed 835
= 0,000800
1250
I
= 0,001198
I
Fine cast steel...
= 0,002222
450
I
Hammered copper
= 0,000250
4000
I
Sheet copper..
= 0,000274
3650
I
Copper wire...
Zinc, melted…. . . .
Brass...
= 0,001000
1000
I
= 0,000241
4150
I
0,000758
1320
I
Brass wire...
0,001350
742
I
=
Bronze, gun metal.. 1590
0,000629
I
= 0,00210
Lead..
477
I
I 000000
= 0,000667
Lead wire....
1500
14 220000 9480 3,16
I 000000 667 0,222
28 000000 18700 6,23
1 970000 1313 0,44
31 000000 31000 15,5
2 190000 2190 1,10
(26 000000 20800 8,32
1 830000 1475| 1,18
29 000000 34730 20,8
2050000 2460 1,48
41 500000 92200|102,4
2920000 6490| 7,20
15 640000 3910 0,49
VI 100000 275 0,034
(15 640000 4285 0,59
18500
1300
58200
4090
88300)
6210)
46800
3290)
116500
8190)!
145500)
10230
33800 !
2380)
30400!
2140 j
60300!
4240 j
7500
5261
17700
1242
51960
3654)
36400)
2560)
1850
130
3100
47 0,016
220
I 100000 301 0,041
I 720000 1720 8,60
I 210000 1210 0,605
13.500000 3250 0,392
950000 229 0,029
9 100000 6890 2,61
640000 485 0,184
14 000000 18900 12,76
987000 1330 0,90
9 800000 6160 1,94
690000 434 0,136
711000 1490 1,56
50000 105 0,110
70000
667 0,22
404
[§ 212.
GENERAL PRINCIPLES OF MECHANICS.
MODULI OF ELASTICITY AND STRENGTH FOR EXTENSION—Continued.
Extension
σ
Modulus
Name of the material.
1
of Elasticity E.
at the limit of Elasticity.
I
Tin...
= 0,00111I
900
I
Silver...
660
= 0,001515
I
Gold...
0,001667
600
5 700000 6300 3,50
400000 440 0,24
10 400000 15800 12,00
730000 1100 0,83
II 400000 19000 15,8
800000 1300 1,09
5000
350
41200)
2900
I
Platina..
0,001667
600
22 800000 38000 31,7
11600000 2700 2,25
38400)
2700
48300
3400
Aluminum...
Glass...
WOOD: beach, oak,
pine, spruce, fir,
in the direction
of the fibres....
9 6000000
28900
675000
2030
Io 000000
၁ဝဝ
3530
700000
248)
I
600
I 560000 2600 2,17
9200
= 0,001667
I 10000
180 0,15
650
The same kinds of
wood in the di-
rection of the
radii to
the
yearly rings.....
185000
13000
570
40
The same kinds of
wood parallel to
114000
the yearly rings.
8000
Light hemp rope..
Strong hemp rope.
Wire rope...
Chain cable.....
Leather straps (cow
leather).....
Sheet iron (riveted
with one row of
rivets)...
1
640)
45
8700!
610
6830)
1 480)
47000
3300
ܐ
51900
3650)
ނ
10400
731
| |
1
4100
290
37000
2600
§ 212.] ELASTICITY AND STRENGTH OF EXTENSION, ETC.
405
TABLE II.
THE MODULI OF ELASTICITY AND STRENGTH FOR COMPRESSION.

Name of the
material.
Compression
σ
λ
at the limit of elasticity.
Modulus of elasticity E.
Modulus of proof strength
Τ = σ Ε.
Modulus of resilience
Α = + σ Τ.
Modulus of ultimate strength K.
I
Cast iron
= 0,001333
750
S14000000 18700 12,44 104000)
990000 1320 0,88
7310)
Wrought"
I
28000000 18700
6,23
31000
0,000667
1500
I
Copper
= 0,000250
1970000 1320 0,44
S15640000 3910 0,49
2200
58300
4000
I 100000
275 0,039
4100
Brass
S 10400
731
Lead..
7250
510
Wood in
the direc-
tion of the
fibre..
Basalt.
·
Gneiss and
granite.
Limestone.
Sandstone.
·
Brick ..
Mortar..
(6800)
480
28000
1970
8300
585
5200
365 S
4150
292)
830
59

526
37
EXAMPLE 1. What should be the cross-section of a wrought-iron rod
1500 feet long, which is subjected to a pull of 60000 pounds?
T
Neglecting the weight of the rod and allowing a strain of
9350
2
60000
9350
pounds per square inch, we obtain the required cross-section F =
= 6,42 square inches. Taking into account the weight of the rod, the
weight of a cubic inch of iron being y = 0,280 pounds, we have
406
[§ 213
GENERAL PRINCIPLES OF MECHANICS.
60000
F=
▬▬
9350 1500.12. 0,280
60000
9350 5040
6000
431
=13,92 square inches.
The weight of the rod is G
Fly = 5040 . 13,92 = 70157 pounds, and
the extension of the same by the pull P = 60000 pounds and by the weight
G = 70157 pounds is
(P + 3 G) l
λ
FE
95078.18000
13,92.28000000
142617
32480
=
4,39 inches.
EXAMPLE 2. How thick must the foundation walls of a building 60 feet
long and 40 feet wide on the outside, and weighing 35000000 pounds, be
made when we employ good cut pieces of gneiss? If we make the thick-
ness of the wall equal to z, we can put the mean length of the wall
60
≈ and the mean breadth.= 40 x, and therefore the mean periphery
2.(60 x + 40 x)
= 200
4 x, and consequently the base of the
whole masonry is (200
4 x) x square feet 144 (200 4 x) x = 576
The modulus of rupture of gueiss for crushing
(50
x) x square inches.
20
is 8300 pounds. If, therefore, we assume a coefficient of security of or
a factor of safety of 20 for the wall, we can put the allowable pressure
upon a square inch
whence
8300
-
415 pounds; hence we have
20
415.576 (50 x) x = 35000000,
50 x
146,4,
and finally the required thickness of the wall
X
146,4 + x²
50
2,928 +
8,57
50
3,10 feet.
§ 213. Strength of Shearing.-The strength of shearing (Fr.
résistance par glissement ou cisaillement, Ger. Schubfestigkeit or
Widerstand des Abdrückens oder Abscheerens), which comes into
play when the surface of separation coincides with the direction
of the force, can be treated in the same manner as the strength of
extension. We have here to consider the action of three parallel
forces P. Q, and R, Fig. 328, when the points of application A and
C of two of the forces lie so near each other, that bending is not.
possible, and therefore a separation of the body in two parts takes
FIG. 328.
FIG. 329.


A
Ꭰ
B
AR
迅
​P
place between A and C in a surface D D at right angles to the axis
of the body. The strength of shearing, like that of tearing and
§ 213.] ELASTICITY AND STRENGTH OF EXTENSION, ETC.
407
crushing, is proportional to the section of the body, or rather to the
area F of the surface of separation, and in the case of wrought iron
is approximately equal to that for tearing, so that the modulus of
rupture K for tearing can also be employed as the modulus of rup-
ture for shearing, and consequently we can put the force necessary
to produce rupture by shearing, when the cross-section is F,
PFK. In general we have P= FK, K, denoting the ultimate
strength of shearing per unit of surface determined by experiment.
λ
The formula P = FE = 6 FE for tensile and compressive
forces within the limit of elasticity can also be employed for the
}
shearing force P, Fig. 329, but here o denotes the ratio =
CA
C B
of the displacement C A to the distance C B of the directions A P
and E F of the two forces from each other.
The following Table III. contains the modulus of elasticity (C)
and that of rupture or ultimate strength (K) for all bodies, for
which they are known at present, and they correspond to the
formulas PFC and P, FK, for the elasticity and strength
of shearing.
TABLE III.
MODULI OF THE ELASTICITY AND ULTIMATE STRENGTH OF
SHEARING
Names of the Bodies.
Modulus of Elasticity C.
Modulus of Ultimate
Strength K,.
Cast Iron
{
3840,000
200000^
32300
2270
Wrought Iron
{
9000000
50000
630000
3500
Fine Cast Steel
14220000
1000000
92400
6500 S
Copper
Brass.
Wood of deciduous Trees.
Wood of evergreen Trees
5260000
370000
569000
40000
616000
C is generally taken E and K, K.
3 =
}
683
48 €
22901
161
{
6260000
440000

{
{
43300
408
[§ 213
GENERAL PRINCIPLES OF MECHANICS.
FIG. 330.
B
D
E
A
N
N
C
NON
0
0
0
D
E
B
D
C
FIG. 331.
B
EA
C
=
2
The most important application of the formula P F K, is
to the determination of the thickness d of bolts and rivets, with
which plates and other flat bodies are fastened together. There
are two modes in which bodies
may be fastened together in this
way; either the plates 4 B and
CD to be joined together are
laid upon one another, as in Fig.
330, and then fastened together
by the bolts or rivets N N and
0 0, or, as is represented in Fig.
331, the plates are butted to-
gether and covered with splicing
pieces D D and E E, and they
are then fastened together by
means of the rivets N N and O 0,
which pass through both the plate and the splicing pieces. In the
first method of joining the plates the tensile stress passes from one
plate to the other through the intervention of a couple, which
causes both of the plates to undergo in addition to the stretching
also a bending, and consequently their safe or working load is
diminished. The second method, where no such couple is called
into action and where, consequently, no bending takes place, is for
this reason to be preferred. Since the plates and splicing pieces,
which are thus joined, press upon each other with no inconsidera-
ble force, the strength of the joint is considerably augmented by the
friction arising from this pressure. For greater safety we disregard
this action in determining the thickness of the rivets. On the other
hand, the working load of the plate is diminished by the holes
made for the rivets or bolts, and we must therefore take care that
it is not exceeded by the working load of the rivets. If d is the
thickness of the rivets and v their number, in the case of the joint
in two plates represented in Fig. 331, we have for the working
load of the rivets
π ď² K₂
4 N

P = v
Now, if b is the width and s the thickness of the pieces to be joined
and v, the number of the rivets in one row, the cross-section of the
plate submitted to the force P is
F = (b − v,
v₁ d) s, and therefore we have P = (b − v, d) s
K
N
K denoting the modulus of rupture of sheet iron; equating these
two values, we obtain
§ 214.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 409
V π ď²
4
В2
4 (b
ν
π d²
v₁ d) s K, or
v₁ d) s K
K
When the holes in the plates are punched, the strength of
shearing must be overcome, but in this case the surface is not
plane, but cylindrical. If s is the thickness of the plate and d the
diameter of the hole in it, we have the area of the surface of
separation
F = πds,
and consequently the force necessary to punch the hole is
P = F K₂
π d s K₂.
(Compare in the "Civil Ingenieur," Vol. I, 1854, the article “John
Jones' experiments on the force necessary to punch sheet-iron," by
C. Borneman).
EXAMPLE-1) An iron rivet 13 inch thick can resist with safety, if we as-
sume K. = .50000 8300 pounds, a force
P = K₂
π d²
2
=
8300
4
•
9.2075 π
4
14670 pounds,
and the force necessary to punch the hole through the sheet-iron, which is
inch thick, is
Р₁ = ñ d 8 . K₂
ds. =
·
3 1
2
50000 = 37500
117810 pounds.
2) If two pieces of sheet-iron are to be joined together by a row of
rivets, and if we denote the thickness of the plate by s and its width for
each rivet by b, we have
(b − d) s =
ñ d²
4
whence
πα
E.G., for d =
& and 8 =
b = d +
inch
= d
4 8
a (1 + 1d);
πα
4
3 п
b
1 -
+
= 5 inches.
4
CHAPTER II.
ELASTICITY AND STRENGTH OF FLEXURE OR BENDING.
§ 214. Flexure.-The most simple case of flexure is that of a
body A B C, Fig. 332, acted upon by a force AP P, whose di-
rection is normal to its axis A B, while the body at the same time
is retained at two points B and C. Let and 7, be the distances
7
410
[§ 214.
GENERAL PRINCIPLES OF MECHANICS.
CA and C B of the points of application A and B from the cen-
tral fulcrum or point of application C, then the force at B is
Q
=
and consequently the resultant is
ΡΙ
R = P + Q = (1 + 2 ) P.
P
FIG. 332.
R
CAM
B
Р
FIG. 333.


If we wish to prevent one portion of the body from bending,
we must insert between the two points of support an infinite num-
ber of others, or the body must be fastened or solidly walled in
along B C, as is represented in Fig. 333, and we have then to study
only the flexure of the free portion AC of the body. Let us sup-
pose the body to be a prism, and let us assume, that it is composed
of long parallel fibres placed above and alongside of one another
and that, when the body is bent, they neither lose their parallelism
nor slide upon one another.
By this flexure those fibres, which are on the convex side of
the body, are extended, and those on the concave side are com-
pressed, while a certain mean layer undergoes neither extension
nor compression. This is called the neutral surface of a deflected
beam (Fr. couche des fibres invariables, Ger. neutrale Axenschicht).
The extension and compression of the various fibres above and
below this layer are proportional to their distance from it. The ex-
tension of the fibres on one side and the compression of those on
the other increase gradually, so that the fibres most distant from
this surface on the one side undergo the maximum extension, and
those on the other the maximum compression. A portion of the
body A K B, Fig. 334, bounded before the flexure by the cross-
sections K L and N O, assumes, in consequence of the flexure, the
form K L 0, N₁, by which the cross-section N O becomes N₁ O₁,
§ 214.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 411
that is, it ceases to be parallel to K L and assumes a position per-
K
FIG. 334.
-P
NN,
Q
H
R
W
Α
P
P
P
1
pendicular to the neu-
tral surface R S The
length K N of the up-
permost fibre becomes,
in consequence, K N₁,
and that of L O the
lowest fibre becomes
LO,. The increase in
length of the former is
therefore NN, and the
decrease of the latter is
O O₁, while the fibre
RS in the neutral sur-
face retains its primi-
tive length unaltered.
The intermediate fibres,
such as T U, V IV, etc.,
are increased or dimin-
ished in length becom-
ing TU₁, V W₁, etc., and
the amount U U₁, W W₁,
etc., of the increase or
decrease is determined
by the proportions

U U₁
SU
NN₁
SN'
M
W W₁
S W
etc.
0 0,
SO'
1
Let us assume the length of the fibre
RS = K N − L 0 = unity (1),
and let us denote the extension or compression of the fibres, which
are situated at the distance unity (1) from the neutral surface, by o,
then we have for a fibre, which is situated at a distance S U or
SW z from this surface, the extension or compression
U U₁ or W W O %.
If the body is but little bent, so that the limit of clasticity is
nowhere surpassed, we can put the strain on the different fibres
proportional to their extensions, etc., and we can consequently as-
sume, that these strains are proportional to their distance from the
neutral surface, as is represented in the figure by the arrows.
412
[$ 215.
GENERAL PRINCIPLES OF MECHANICS.
If the cross-section of a fibre is unity, we have in general the
tension upon it = 0 % E; and if the cross-section of the fibre = F,
the tensile or compressive strain is expressed by the formula
S = oz FE =
= o E. F %,
and its moment in reference to the point S upon the axis is
2.0 z FE = o z² F E = 0 E . F z².
M = 2
§ 215. Moment of Flexure.-The tensile and compressive
strains in the cross-section N, O, balance the bending force Pat
the end of the body A b. We can therefore apply to these
forces the well-known laws of equilibrium. If we imagine that
there are in action at S two other forces + P and P, which
are not only equal but also parallel in direction to the given force
P, we obtain
1) A couple (P,
around S, and
P), which produces the flexure or bending
2) A simple shearing force & P= P, which tends to cut off the
portion A S of the body in the direction S P or A P. The latter
force can be decomposed into two components P, and P, whose
directions lie in the plane of the cross-section N, O, and in the neu-
tral axis S R. If a is the angle formed by the cross-section N, O,
with the direction A P of the bending force, we have
P、
P₂
P cos, a and
P sin. a.
In ordinary cases in practice the flexure of the body and also a is
so small, that we can put sin. a = 0 and cos. a= 1, and consequently
we can neglect the component P, which tends to tear off the por-
tion A Sat N, O,, and, on the contrary, we can put the force P₁,
which tends to rupture by shearing the piece A Sin N, O,, equal
to the bending stress P.
If F denote the arca of the cross-section N, O, and A, the modu-
lus of rupture for shearing, the shearing force is determined by
the product F K2.
If we are considering a long prismatical body, P is generally
so small a portion of FK, that rupture by shearing can scarcely
occur, and for this reason it will be considered in particular cases
only. (See the following chapter.)
Since one couple (P, P) can be balanced only by another
couple, it follows, that the tensile strains on one side form with the
compressive strains on the other another couple (2, — Q), and
that the moments of the two couples must be equal. If F, F2, F3
etc., are elements or infinitely small portions of the entire surface
§ 215.]
ELASTICITY AND STRENGTH OF FLEXURE, ETC. 413
F of the cross-section N ( = N, O,, and if the distance of these
portions from the neutral surface or S be denoted by Z1, Z2, Z3, etc.,
the strains in these elements are
o E. F, 21, 0 E. F. Za, o E . F3 Z3, etc.,
and their moments
2
o E. Fz, E. F. z, o E. F, z, etc.
o z²,
?
Since these forces form a couple (Q, Q), their sum
o E (F, z, + Fo Z₂ + F 3 Zz +...), and consequently
F₁ z₁ + F₂ Z2 + F 3 Z3 + . must be = 0.
FIG. 334 a.
K
NN
H
R
P
V
-P
W
P
P
1
M
But this sum can
only be 0, when the

point S of the axis co-
incides with the centre
of gravity of the sur-
face F F + F +
F+...; consequently
the neutral axis of a bent
body passes through the
centre of gravity S of
its cross-section F. The
moment of the couple
(Q - Q),
o E (Fz² + F₂ z₂²
2
+ F3 23² + + . . .).
should now be
now be put
equal to the moment
of the couple (P, — P).
If we denote the dis-
tance SH of the cen-
tre of gravity S from
the direction AP of
the bending force by x,
we have the moment
of the latter couple =
P x, and therefore
Pxo E (F₁ z2
+ Foz₂² + ...).
Finally, we have for the radius of curvature MR MS of
the neutral surface the proportion
MR SU
RS
UU
414
GENERAL PRINCIPLES OF MECHANICS.
[§ 216.
=
or, substituting MR r, RS 1, SU 1 and U U₁ = 0,
=
r
1
1
σ
1
whence the moment of force is
Consequently rσ = 1 or σ =
P x =
E
The radius of curvature at
r =
1
E
P x
2
2
2
(F₁ z₁² + F₂ z₂² + ...).
2
S is therefore
2
(F₁ z₁² + F₂ z2² + . . .).
The expression F₁z² + F₂ z2² + ... is dependent only upon
the form and size of the cross-section, and can therefore be deter-
mined by the rules of geometry. We will hereafter denote it by
W and we will call the quantity corresponding to it the measure
of the moment of flexure, and W E the moment of flexure itself
(Fr. moment de flexion; Ger. Biegungs-moment).*
From the above, we have for the radius of curvature
1
r =
WE
Pa
"
and we can assert that the radius of curvature of the neutral axis
of a deflected body is directly proportional to the measure W of the
moment of flexure and the modulus of elasticity E, and, on the con-
trary, inversely proportional to the moment P x of the force.
The curvature itself, being inversely proportional to the radius.
of curvature, increases with the moment Px of the force, and
decreases, when the moment of flexure WE increases.
§ 216. Elastic Curve.-If we have determined the moments
of flexure WE for the cross-sections of the bodies, which generally
occur in practice, we can determine by means of these values the
curvature and from it the form of the neutral axis or of the so-
called elastic curve. The equation
P x r = WE or r
WE
P x
indicates, that in the case a prismatical body the product of the
radius of curvature and the moment of the stress is constant for
all parts of the elastic curve A B, Fig. 335, and that consequently
r becomes greater or less as the arm x of the force is diminished or
increased, or as the distance of the point S considered from the
end A of the neutral axis is less or greater. At A we have x = 0,
and consequently the radius of curvature is infinitely great; at the
fixed point B, on the contrary, x is a maximum, and the radius of
curvature is therefore a minimum; hence the radius of curvature
-TR.
* Moment of flexure is also used for the bending moment Px.-
§ 216.]
ELASTICITY AND STRENGTH OF FLEXURE, ETC. 415
increases by degrees from a certain finite value to infinity, when
we proceed from the fixed point B to the end A.
If we divide a portion A S of the elastic curve, the length of
which iss, into equal parts, and erect at the end A and at the
points of division S1, S2, S3, etc., perpendiculars to the curve, they
will intersect each other at the centres Mo, M₁, M, of the osculatory
circles, and the portions cut off M, A = M, S₁, M, S₁ = M, S,
U
FIG. 335.
B
H
H3
T
AS 3
江
​H₂
H₁
Р
K
obtain d₁
2
ន
N ri
"
K
$2
s
N 12
-X,
M
-X
M3
AM 2
S
M1
$3
etc.
N 73
Mo
M, S₂ = M₂ S3, etc., are
the required radii of
curvature r₁, T2, T3 of
the elastic curve. (See
Introduction to the
Calculus, Art. 33.) If
n is the number of di-
visions of this line, we
have the length of a di-

S
vision = ; and if we
N
denote the length of
the arc (for the radius
1) of the angles of
curvature A M₂ S₁ =
8°, S, M, S = 8º,
S₁ M₂ S₁ = 83°, etc., by
S3
$1, 62, 63, etc., we can
S
put d₁ r₁ = √₂ r₂ =
N
1
2
dra, etc., whence we
If we suppose the elastic line to be but slightly curved, we can
substitute for the divisions of the arc their projections upon the
axis of abscissas A X perpendicular to the direction of the force,
I.E. we can put A K,
K, K, etc., so that the
arms of the force in reference to the points S1, S2, S3, etc., are
H₁ S = K₁ K₂
H₁ S₁ = 1/2
H, S, H, S₁ + S₁ L₂ = 2
=
es
H3 S₂ = H₂ S₂ + S₂ L3 = 3, etc.,
S3
416
[§ 216.
GENERAL PRINCIPLES OF MECHANICS.
and consequently the corresponding moments of the force or the
values for Px are · Ps 2 Ps 3 Ps
n
N
N
etc.
Substituting successively these values for P x in the formula
WE
P x
for the radius of curvature, we obtain the following series
of values for the radii of curvature
r₁ = n
WE
Ps
r₂ =
n WE
2 P s
n WE
r3
etc.;
3 Ps
hence the corresponding angles which measure the curvature are
S
$1
η γι
S
P s²
n² WE'
P s²
S
So
2.
2
n ro
P $2
n² WE
9
бз
= 3
etc.
n r3
n² WE
?
Summing these angles, we obtain for the angle of curvature
A OS = o' of the entire arc A S 8 = x
Ф= d₁ + d₂ + d +
+ бл
P s²
(1 + 2 + 3 +
+ n)
n² WE'
n³
or, since we know that 1 + 2 + 3 +
+ n =
we have
2
Φ
2
P s²
n² WE
P s3
2 WE'
FIG. 336.
U
H
H3
T
H₂
H₁
A
K
P
2
K
S3
-X₁
K
for which we can write,
according to the above
supposition,

P x²
M
-X
M
3
M₂
2
M1
$
2 WE
This arc or angle
(since the angle be-
tween two lines is equal
to that between their
normals) is equal to the
angle STU included
between the tangents
AT and S T to the
two points 4 and S
or to the angle, which
expresses the differ
ence between the in-
clination of the curve
to the axis in A and in
Mo S. If we pass from the
§ 217 ELASTICITY AND STRENGTH OF FLEXURE, ETC. 417
undetermined point S to the fixed point B, we must substitute
instead of s the entire length of A S B, or approximately the
projection of the same upon the axis of abscissas, and under
the supposition that the curve at B is perpendicular to the direc-
tion of the stress or parallel to the axis of abscissas, the angle o
becomes
P 13
A D B = ẞ =
2 WE'
and, on the contrary, the angle of inclination or tangential angle
TSH = STX becomes
P ľ P s'
2 W E 2 WE
P (t² 8²)
2 WE
P(1²x²)
a = В-ф
2 WE
If the curve at the fixed point B is not perpendicular to the
direction of the force, but inclined at a small angle a, to the axis,
we will have
B = a₁ +
P ľ
2 WE'
and therefore
P (1³ — x²)
2 WE
a = a₁ +
§ 217. Equation of the Elastic Curve. By the aid of the
latter formula we can now deduce the equation of the elastic curve.
The ordinate of the curve K S y is composed of an infinite
number (n) of parts, such as K, S1, L2 S2, L3 S3, etc., which are
found by multiplying an element of the arc
A S₁ = S₁ S₂ = S S3, etc. =
S
N
by the sine of the corresponding tangential angle
S₁ A K₁, S, S₁ L2, S3 S₂ L3, etc.
Hence we have
KS = K₁ S₁ + L2 S2 + L3 S3+..., or
S
N
3
y (sin. S, A K+ sin. S₂ S₁ L₂ + sin. S3 S₂ L₂+...).
Substituting the abscissa A K = 2 instead of the arc A S
and for the sines the arcs calculated from the formula
=
a =
P (l² — x²)
2 W E
x 2 x 3 x
>
etc., we obtain
n
N N
and introducing instead of successively
X
1
(²)' + ♪ - (³)² + ...
(32) *
N
X
P
y =
n 2 WE
[~~ (1)²+
2 x
+ ľ² -
N X
+ 1
N
)]
Now we have + P² + ... + l² = n l² and
27
418
[$ 217.
GENERAL PRINCIPLES OF MECHANICS.
() + (~~)' + C)²++ ("~~~)"
(+)+(2)+
(na)
= + 2
(1 + 2 +3 +...+n)
(see Ingenieur, page 88), whence
XxX
P
n³
X
N
3
(27)
X
12 WE [ " " - ()]..
n n
y =
P x (l²
y =
2 WE
3
or
which is the required equation of the elastic curve, when we suppose
that the curvature is not very great.
If we substitute in this equation x = 7, we obtain instead of y
the height of the arc or the deflection
BC = a =
P 1³
3 WE
.While the tangential angle a increases with the force and with
the square of the length, the deflection increases with the force and
with the cube of the length.
The work done in bending the body is determined, since the
force
P =
3 W E a
7³
increases gradually with the space described and its mean value is
1 P
P =
છ
WE a
73
by the expresssion
L = ¦ Pa =
WE a²
P² 1s
colat
16
73
WE
If a girder A B A, Fig. 337, whose length is 1, is supported at
both ends and acted on in the centre B by a force P, the ends are
FIG. 337.
P
K
C
1/2
P

HUBANE FARK
P
bent exactly in the same way as in the case just treated, but in
this case we must substitute for the force acting at A, ¦ P
3218.3 ELASTICITY AND STRENGTH OF FLEXURE, ETC.
119
4
and for the length of the arc A B = ¦ A A = 1. Consequently
the equation for the co-ordinates AK and K Sy is
=x
で
​P x ( l²x²)
1
Px (31² - 4 x)
3
Y
4 WE
48 WE
so that for x = A C =
the deflection is
2
y = B C = a₁ =
P P
48 W E
PP
1
16
3 W E'
I.E., one sixteenth of the deflection of a girder (Fig. 333) loaded at
one end with an equal weight.
If in the first case the elastic curve A B, Fig. 336, is inclined at
a small angle a, to the axis at the fixed point B, we must add to
the former expression for y the vertical projection of the portion x
of the tangent, I.E., a₁ x, so that we have for the ordinate
- 3
3 = (a + P (P = 1 ²²));
Y
and for the deflection
α
(a,
+
2 WE
Pr
3 WE
1.
X
༼ ཀ ཏ ཝིཏི
J
(§ 218.) More General Equation of the Elastic Curve.-
A more exact equation of the curve A S B, Fig. 338, formed by
the neutral axis of a deflected beam, can be deduced in the follow-
ing manner by the aid of the calculus.
If we substitute in the general equa-
tion of § 216, WE Pxr the value
of radius of curvature (from Art. 33 of
the Introduction to Calculus),
FIG. 338.
T
A
NO
P
R
M
WE =
B
d s³ 3
λ= dx² d (tang. a)
and in the latter, according to Art. 32,
ds=11+ (tang. a). d x,
we obtain
Px dx [1 + (tang. a)²]}
d tang. a.
When the girder is but moderately deflected, the angle a formed
by the tangent with the axis of abscissas is but small, and we can
therefore write
[1 + (tang. a)'] = 1 + (tang. a)',
and consequently
3
4.
P
LP

420
[§ 218.
GENERAL PRINCIPLES OF MECHANICS.
WE =
d (tang. a)
or inversely
Pxdx
d tang. a
WE
Px [1 + 3 (tang. a)²] d x
[1 - 3 (tang. a)³] d (tang. a).
1+ 3 (tang. a)³
α
From the latter we obtain
S.P
Px dx
WE
d
- Sa (tang. a) + S (tang. a)' d (tang. a),
3
or, according to Art. 18 of the Introduction to the Calculus,
P x²
2 WE
tang. a + ½ (tang. a)³ + Con.
b of the
But at the vertex B the curve is parallel to the axis of abscissas
and a = 0; substituting, therefore, the projection CA
elastic line on the axis of abscissas, we obtain
P b²
2 WE
tang. 0 + } (tang. 0) + Con. = 0 + Con.
Subtracting from this the former equation, we have
P (b²
2 WE
x²)
tang. a ½ (tang. a)³,
or inversely, for the tangential angle S T N = α,
tang. a =
2
P (b²x²)
2 WE
P (b²x²)
+ 1 (tang. a)³
+ 1/1/1
2 WE
P (b²
x²)
I.E.,
1) tang. a =
2 WE
(1 +
P³ (b² — x³)³
8 W³ E³
P² (b³
-
8 W' E'
But tang. a =
dy hence we have
d x²
dy
(1 +
P² (b²
8 W2 E²
ོད་ད
P (b²x²) d x
and
,
P
p2
2 WE
S
$
y = ₂ W g ( S (b − r ) d x + 5 w g⋅f (8 − 2) d x)
+
2 WE
P
(b²
8 W + E = —
z W B [ S v d x − S x d x
2 WE
:༡༠ x
Praz-f3vrdz + f3 v rdz - frdz)]
Sva: b' x' x
x²
8 W2 E2
P
b c
X
x
x³
P2
b°
= ₂ W B [ " + − + 8 W x ( x − d'
2 WE
3
WE'
b²
x
xx³
~ + 3y² ² - 2)]
5
+ Con.
§ 218.]
421
ELASTICITY AND STRENGTH OF FLEXURE, ETC.
Since for x = 0, y
= 0, we have also Con. = 0, and
Ꮲ Ꮖ
2) y
y =
2 WE [b
x²
P2
b²
+
3
At the vertex x = b
8 W2 E2
and y is the deflection C B = a, and
(b° – b' x² + § b² x'
x²
7
therefore
a
P
2 WE
(3 b³ +
P2
8 WE'
•
16
35
•
18 · 6'),
3) a =
P² b+
35 W2 E*
[1 + į (tang. a)²] d x we
P b³
3 WE
(1 + 38
From ds = 1 + (tang. a)². d x =
√
P (b² - x²)
obtain, by substituting tang. a = 2 WE
P² (b²
a
s
8 = (1 + ↓
P (b)) d x
=
P2
W² E²
Sax + 8 W ² E = [ S ( b' d x − 2 b² x² d x + x^d x
dx)]
= x +
P
8 WE
X
2 b² xs
3
+
5
4) s =
[1 +
P2
8 W² E²
(b¹ — § b³ x² + 2)] x.
5
I.E, the length of the arc
If we assume x = b, we have the total length of the girder
5)
1 = (1 +
P² b
15
W³
Er)0
³) b = (1 + } .
a²
b.
(1-
P³ l
15 W E2
73) 1,
Inversely we have
6) b=
and therefore
I.E.,
1 +
PP
WE
P² b*
15 W³ E'
a = 3 W (1
P 13
3 WE
ΡΙ
-
(1
-
P² 1
3
15W 6-) (1+..), or
W² E*
3 P² l
15 W² E*
35 WE
=) (1 + 38. 17).
35
WE
NE
36a
7) a = 1; (1 - 2. Pr) = (-2 (1-3-14")
3 WE
4
35
WE
3.11-2
Neglecting the members containing the higher powers of
P
we obtain, as in the last paragraph,
WE'
422
[§ 219.
GENERAL PRINCIPLES OF MECHANICS.
tang. a =
for
P (l' — x²)
2 WE
x = 0, tang. a =
P x
and y
P12
and for x = b = l, y
2 WE'
2 WE (-2), therefore,
l, y = a =
P 13
3 WE
§ 219. Flexure Produced by two Parallel Forces.—If a
girder A A, B, Fig. 339, I. and II., fixed at one end, is bent by two
FIG. 339.
I.
N₁ T
A₁
T
IS
A
PY
II.
1
B
Α
B
L
H
H
G1
forces P and P₁, whose points of
application A and 4, are at a dis-
tance from each other, while the
point of application A, of the force
P, is at a distance A, B = 1, from
the fixed point B, the moment of
flexure at a point S of the portion
A A, is

M = P x,
and, on the contrary, that of a
point S, in the portion A, B is
M₁ = P(1 + x₁) + P₁ x₁,

1
in which
and a, denote the ab-
scissas A and A, A.
In order to obtain a clear idea
of the manner in which these moments vary, we can lay off, as in
II., their different values for the different points as ordinates, E.G.,
M = y = K L, M₁ = y₁ = K₁ L₁, and join their extremities L, L,
etc., by a line A LH L, G₁, which will limit the values of M and
M₁ for the whole length of the beam.
If the girder were subjected to the force P alone, the line
bounding all the values of M or y = P x would be the straight
line AG, the ordinate of the extremity G of which is BG=
P.AB= P(+). By the addition of the force P, the por-
tion H G of this right line is replaced by the right line H G₁, whose
extremities H and G, are determined by the co-ordinates A Á₁ = 1
and A, H = P 1, and also AB=1+ 1, and B G
BG +
G G₁ = P(l + l₁) + P₁ l₁.
(1 1,)
Á,
1
If the force P is negative, the moment My = P x of a point
Kupon A A₁ = 7 remains unchanged, while, on the contrary, that
of a point K, upon A, B becomes M, y, = P (l + x₁) — P₁ x₁,
and the moment of flexure at the fixed point B is = P(+1₁) —
19
$219.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 423
P₁₁, and it is positive or negative as P (l + ₁) is greater or less
than P₁₁. In both cases the moment of flexure decreases grad-
ually from A,, remaining in the first case, Fig. 340, positive, and,
1
FIG. 340.
I.
P₁
FIG. 341.
P


A
PY
S
K
I.
T
K₁
1
A
K
IST
II.
PY
II.


1
K
Ki
B
K
G₁
A
B
ΤΟ
1
L
H
H
H₁
U H₁
on the contrary, in the second case, Fig. 341, becoming = 0 for a
point O at a distance A, 0 = x₁=
Pl
P₁ − P
from A₁, for greater
values it takes the negative sign, and at the fixed point B it is
= − [P₁ l₁ − P (1 + 4)].
In the first case the right line H G, Fig. 340, II., which repre-
sents the moment of flexure at a point K, between A and B, passes
below the base line A B and ends at a point G₁, whose ordinate is
B G₁ = P (1 + 1) - P, . In the second case, on the contrary,
the right line H G₁, Fig. 341, II., rises from the point O above A B,
and the ordinates become K₁ L₁
Y₁ =
[P₁ x₁ − P (l + x₁)]
and B G₁ = α₁ = − [P₁ l, − P (l + h)].
= Y ₁
Since the radius of curvature r =
WE
M
x1
of the girder is inversely
and consequently the curvature itself is directly proportional to the
moment of flexure M, the graphic representations in II. of figures
339, 340 and 341 furnish us also a representation of the variation
of the curvature of the girder. In the case represented in Fig. 339,
where the forces P and P, acting upon the girder have the same
direction, the curvature increases gradually in going from A to B,
but if P and P, have opposite directions, it decreases again grad-
ually as we recede from 4.
424
[§ 219.
GENERAL PRINCIPLES OF MECHANICS.
If, as in Fig. 340, P₁ l, < P (l + 7), the beam is bent in one
direction only; but if P, l, > P (+1), there is no flexure at the
point and also at a point O, Fig. 343, where a point of inflection
is formed (see Art. 14, Introduction to the Calculus), and from O
FIG. 342.
I.
P₁
FIG. 343.
Ꭱ


PY
SK
IS
C
II.
I.
T
1
K₁
Ο
A
K
G
B
PY
II.


G1
B
1
K
Ki
B
K
G₁
A
L₁
H
H
U
HI
Ι
towards B the curvature of the girder gradually increases in the
opposite direction. If in the second case, Fig. 342, the forces P and
P₁ are equal, for a point K, between A, and B,
M = P (1 + x₁) — P x₁ = Pl
is constant, and the curvatures of that portion A, B of the girder
is the same everywhere, I.E., the elastic curve is a circle.
The radius of curvature of the portion A A, is determined in
all three cases by the well-known formula
1° =
WE
>
Px
and that of the portion A, B, in the first case by the formula
r₁ =
WE
P (l + x₁) + P₁ xi
and, on the contrary, in the second and third cases by the formula
Ti P (l + x₁) – P₁ x₁
WE
Pixi
P, î; becomes =
WE
ΡΙ
or con-
When, in the second case, P、
stant, and in the third case, where P, > P(+), for the point.
O, whose abscissa x₁ =
Pl
P₁ − P’
we have r₁ = ∞ (infinitely great),
§ 220.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 425
and, on the contrary, for the point A,,r=
WE
Ρι
and for the point B,
WE
P₁ l₁ − P (l + 1₁)'
A
l,
1
According as Pl is greater or less than P₁ - P (1 + 1) etc.,
I. E., P ≥ r₁, in the latter case we have rr, or the curvature at
A, greater or less than that at B.
§ 220. The Elastic Curve for Two Forces.-The equa-
tions of the elastic curve, formed by the axis of a girder subjected
to the action of two forces P and P₁, can easily be deduced from
the formulas found in paragraphs 216 and 217.
A
Pr
T
FIG. 344.
I.
N₁ T₁
A
S
K
C
=
B
Α
L
H
G
G1
If a denote the angle of incli-
nation of the elastic line at 4,, we
have first for the portion of the
curve A A₁, Fig. 344, I, the arc
measuring the inclination of the
same at S

1) a = a +
P (T² - x²)
2 WE
and the ordinate KS corresponding
to the abscissa A K = x
2) y = a₁ x +
1
(compare § 217).
1
P x († — 3 x²)
(F

2 WE
By putting x = 0 in (1), we deter-
mine the angle of inclination in A
Pr
a = α1 2 WE'
and, on the contrary, by putting 7 in (2), we obtain the ordi-
nate at A,
›
A₁ C = α = a₁ ? +
P ↑
3 IF E
x)
a
1
For a point in the second portion of the girder A, B the mo-
ment of flexure P (l + ¿₁) + P₁ ¿'; = P ! + (P + P₁) ?, is com-
posed of the two parts P 7 and (P + P₁) ≈₁, one of which, being
constant, bends this portion of the beam in an arc of a circle,
ΠΕ
ΡΙ
whose radius is r = and whose angle of inclination at a point
= ↳₁
7'1
S situated a distance 4, S,, from A and B S from
B is measured by the arc
7,
B₁
*
P 1 (l, — x';)
WE
426
[§ 220.
GENERAL PRINCIPLES OF MECHANICS.
The inclination at S of this portion of the girder, due to the
FIG. 344 a.
I.
N₁ Ti
B
AL
T
IS
K
PY
II.
T
B
flexure produced by the moment
(P+ P₁) x₁, is measured by the arc

B₁₂
2
2
(P + P₁) (l² — x₁²).
2 WE
and consequently the total inclina-
tion at the same point is
3) B B₁ + ẞ₂ =
+
2
Pl(lx₁)
WE
(P + P₁) (l,² — x₁²)
2 WE

The deflection of B S₁, due to the
curvature in a circle measured by ß₁,
is according to the well-known for-
mula for the circle
Α
L
H
H
G
G1
BS2
(l₁ — x₁)² Pl
2
N₁ S₁ =
2 r
2 WE
hence that of the entire piece B A, is
and the height of the point S₁ above A₁ is
B C₁
Pll2
2 WE'
2
K₁ S₁ = B C₁ - N₁ S₁ =
P1 [l² — (1₁ — x₁)*]
2 WE
According to what precedes (§ 217) the
(P + P₁)x, (l,² — { x²)
B₂
2 W E
2
1
(P + P₁) (l² - 2,2)
2 WE
4) K₁ S₁ = y₁ =
Pl (2 l, x, x₁₂²)
2 WE
deflection K, S₁ =
corresponds to the angle of curvature
2
and the total deflection is therefore
2
Pl (2 l, x, − x,²) + (P + P₁) x, (1,² — } x, ²)
=
2 WE
Substituting in (3) x, 0, we obtain the angle of inclination ß,
which we had assumed as given, and its value is
2 Pll + (P + P₁) l²
2 WE
2
Now if we substitute in (4) x, = l₁, we obtain by this means the
deflection
BC₁ = a₁ =
2
3 Pl² + 2 (P + P₁) ↳³
6 WE
Finally, the total deflection of the whole girder is
221.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 427
BD = a + a₁ = a₁ l +
5
= a₁l +
= a₁l +
2
P 13 3 Pl² + 2 (P + P₁) ↳³
3 WE
+
6 WE
P l (2 l² + 3 1,²) + 2 (P ÷ P₁) l‚³
6 WE
2
P (213 +3 71,² + 2 1,2) + 2 P₁ 1,3
6 WE
1
3
If the beam A B is not horizontal at B, but inclined at a cer-
tain angle ẞ, we must add in (3) ẞ, to ẞ, and in (4) to y₁, ß, x,.
0
0
P₁ instead
If the force P₁, acts in an opposite direction to P, we must sub-
stitute in the fundamental formulas (3) and (4) P
of P + P₁.
§ 221. Girders Supported at One End. The formulas
of the foregoing paragraph are applicable to numerous cases in
P
T
FIG. 345.
C
B
C
P₁
practice. If, for exam-
ple, a girder A B, Fig.
345, is at one end im-
bedded in a wall and
at the other merely
supported, the question
arises, what is the bend-
ing force at 4, or what
force has the support at
A to bear, when the
beam is loaded with a

weight P₁, suspended at an intermediate point 4,?
P is here negative, ß. = 0 and, since A and B are at the same
level, the sum of the deflections C A,
1 +
=a C₁
a and C, B = = a₁, is
} PU² + } (P – P₁) }, ³
3
WE
I.E.
(ar
+
or since a₁
Pl²
3 WE
Pll, + ¦ ( P
WE
P)
we have
"
0,
= 0,
1
!
2
P l² l₁ + 1 (P − P₁) l² l + ¦ P Fª + ¦ P I l¦² + ¦ (P — P₁) 4º = 0.
3
From this it follows that
P =
(37 +27) 7,2
P₁
*
1* + 3 (1² 1 + 7 4₁³) + 4,³ 2 '
E.G., for 77, that is, when P, is applied in the middle of the
girder, we have
5
P =
P₁.
16
Hence the moment of flexure at A, is
1
5
ΡΙ
=
P₁l,
16
and, on the contrary, that at B is
428
[§ 222.
GENERAL PRINCIPLES OF MECHANICS.
3
6
P₁l, 2 Pl=
P₁l
P₁l,
16
or greater than that at A.
If 7 7, and the points A and B are not situated upon the
same level, if, for example, A lies a distance a, higher than B, we
must put a + A₁ = A₂•
But in this case
a
(3 P- P₁) 12
α1
2 WE
PB3
a = a₁l +
α1
3 WE
[3 P + 2 (P
hence we have
6 WE
(11 P - 3 P.) 13 and
6 WE
P₁)] 1³
(16 P – 5 P₁) ♫ª³
(5 P − 2 P₁) ™³
6 WE
= α 2
and consequently
6 WE
6 W E α,
5
+
P₁
P =
16 7³
16
If the moments at A, and B should be equal and opposite,
we must put
Pl= P₁l - 2 Pl,
or
3 P = P₁, I.E. P
P₁
=
3
in which case we must make
A₁₂ =
P 13
6 WE
P. 19
で
​18 W E
If, therefore, the end of the girder lies 0,0555
P₁ 1
WE
で
​higher than
P₁l
B, the moment of flexure in A and B is ±
when A and B are at the same height.
or smaller than
3
With the aid of the values found for P we can calculate the
radii of curvature, the tangential angles, etc., of the portions A A,
and A, B of the curve.
1
§ 222. Flexure of a Girder supported at both Ends.—
Another case, to which the formulas of the last paragraph are
applicable, is that of a girder A B, Fig. 346, supported at both ends
P
T
FIG. 346.
T
Q
A and B and acted
upon by a force P,,
whose point of ap-
plication A, is at a
distance / from one of
the points of support
A, and at a distance
1, from the other.

§ 222.] ELASTICITY AND STRENGTH OF FLEXURE, ETC.
429
1.E.
Here the moment
P. BA = the moment P₁. B A₁,
P (l + l₁) = P₁₁,
h)
and consequently the pressure on the point of support A is
P =
P₁₁
1 + 1
and, on the contrary, the pressure on the point of support B is
ΡΙ
Q
P₁ − P =
l + li
Since A and B are situated in a horizontal plane, we have
a + a₁ 0,
and the angle ẞ is not here = 0, but is a negative quantity C B T,
to be determined.
We have here
and also
Prl + (P − P₁) 1 l,'
¦
WE
P ľ³
a =
- Bl+
+
3 WE'
2
↓ P ! l² + ¦ (P – P₁) 1,3
ат
Bl₁ +
3
WE
and therefore their sum
or
P
B (l + l₁)
(2 7³ + 6 7³ 1, + 6 7 7² + 2 7½³)
6 WE
+
P₁
6 WE
7
(3 11² + 2 4³) = 0,
6 3 (1 + 1) WE=P (21+61² 7, + 6 7 7,² + 2 1,³) — P₁ (3 11,² + 21,³)
7
1
= [27° + 67² 7₁ + 677,² + 24³ — (377, + 27,³) (1 + 4)] P,
1,
from which we deduce the angle of inclination at B
P1 (27² +371 + b²)
P₁ll (2 7² + 3 7 1₁ + b²)
B
6 (11) WE
6 (1+1) WE
and that at A
ɑ =
P₁ 1 l, (1² + 3 7 1₁ + 2 1º)
6 (1 + 1)² WE
If, for example, P, is suspended in the middle, we have
and therefore
1
1₁ = 1 and P = Q
Pr
B
2 WE
=
P₁
2
で
​1
(compare § 216).
P₁ P
4 WF E
With the aid of the angle 6, thus determined, all the relations
430
[$ 223.
GENERAL PRINCIPLES OF MECHANICS.
of the flexure of the girder can be determined by the formulas.
found in what precedes.
The maximum value of the moment of flexure is for the point
of application A₁, and it is
M = Pl= Q b₁
1
P₁ll
2
P₁
[(+)
7 + b₁ 1 + 1₁
=
2
2
and it is a maximum for ll, I.E., when the weight is hung in
the middle, its value is then
P₁ (l + 1)
M
4
=
¦ P₁ l.
1
$223. A uniformly loaded Girder.-If the load is uniformly
distributed over the girder A B, Fig. 347, and if the unit of length
bears a weight =q, or the whole
girder, whose length is 1, bears the
load Q = 7 q, and a portion of the
girder A S s the load q s, we must
substitute, instead of the moments
FIG. 347.

Q
**66+
T
DARDANGAN:
A
L
K
զտ
1
п
PS,
2
N
± 7 ( 2 )
ጎ
3
Ps,
—-
N
Ps, etc., the moments
a ( 2 )
el
N
2
3 s
N
:)
etc. :
) etc..
for the centres of gravity of the loads q (*), 4(3), 4 (3)
N
9
s
2 8 3 8
S
2 $
lie in the middle of
etc., and their arms are
>
n'
ጎ
N
N
N
3 s
14
ሰፊ
etc. In this way we find the angles of curvature of the ele-
ments of the arc
3
q
б
d₁ = .
12
n³ WE'
22.9
12
14
3
N WE'
32.9 83
n³ WE'
etc.,
and therefore the angle of curvature of A S = s is
I s³
3
9 8
Ó
11t
(1² + 2² + 3² +
+ N³)
n³ WE
3
2 n³ WE' 3
9
9x8
6 WE'
approximatively
6 WE'
If x = 1, we have the tangential angle T A C = UTB of the
end A
B
=
q 73
6 WE 6 WE'
and therefore for a point S, whose abscissa is A K = x,
В-ф
q
a = ẞ − 4 = 6 W Ē (lº — x³).
—
WE
$223.1 ELASTICITY AND STRENGTH OF FLEXURE, ETC. 431
From the latter measure for the angle we find for an element
of the ordinate
X
X
a =
m
m
q
6 W E
(1³ — x³);
3 2x
3
substituting instead of a successively (); ( 3 ); (3);
M
obtain the required equation for the ordinate KS = y,
m' 6 WE
m
•
3
X
y =
VE [m で
​[ m 1
= "
( — ) ' . (1² + 2
23
+ . . . + m²³) ]
X
M
9
z [m r
กะ
( 2 ) 1], LE
I.E.
4
, we
Y
x
6 WE
73
y = 1 / ( " - 2)
6 WE
4
Assuming again x = 7, we obtain the deflection
?
a
3 ľ³ =
6 WE
q 7+
8 IV E
Q F
SWE
Q P
3 WE'
I.E., g of what it would be, if the load acted at the end of the girder.
The ordinate of the middle of the girder is
q r
12 WE
(P
73
7.3
32
31 q l¹ +
12.32 WE'
hence the distance of this point below the horizontal line passing
through B is
Y₂ = a
Y₁ =
17 q l¹
12.32 W E
and therefore the mechanical effect corresponding to the deflection
a or to the sinking (y.) of the centre of gravity of the load Q = 1 q,
when Qis gradually applied, is
L
=
12
2
1 Q Y₂ = 1 q 7 y₂ =
l
17 9° 7°
24.32 . IIE
17 Q² 7³
24. 32. IV E
If the girder is acted upon simultaneously by a uniformly dis-
tributed load Q and a force P at the end, we have the deflection
a =
P 13
+
Q F
3 WE 8 WE
3
(5+ 옹​)
WE'
If the girder A B A, Fig. 348, is supported at both ends and
carries not only the weight P applied at its centre, but also the
ભારતમાંરા
(P+Q)
*q
FIG. 348.
P
1/2 (P+Q)
load Q = 7q uniformly dis-
tributed over its length, we
find the deflection C' B = a by
substituting in the expression

a =
P
3
+
옹​)
73
C) TE
for the case represented in
Fig. 347, instead of P the
432
[§ 224.
GENERAL PRINCIPLES OF MECHANICS.
pressure or reaction
load
Q
P + Q
2
at the extremity A, instead of Q the
?, which is equally distributed upon one-half B A, and
2'
instead of half the length of the girder B A =
In this manner we obtain
1 A A = 1 l
1.
α
P + Q
6
73
%)
16/ 8 WE
(P+Q)
73
48 WE
Q 19
48 WE'
If P = 0, we have a =
that is, when the entire
load is uniformly distributed upon a beam, supported at both ends,
the deflection is but § of what it would be, if the load was sus-
pended at the centre of the girder.
The weight G of the beam has exactly the same influence upon
the deflection as a load Q, which is equally distributed, and there-
fore enters in exactly the same manner into the calculation.
FIG. 349.
A
F
D
K S
N
H
NF
B
N
1
1
§ 224. Reduction of the Moment of Flexure-If we
know the moment of flexure W E of a body A B C D, Fig. 349,
in reference to an axis N, N, without the
centre of gravity, we can easily find this
moment in reference to another axis N N,
passing through the centre of gravity and
parallel to the first. If the distance H H,
d, and
KK, between the two axes is
if the distances of the elements of the sur-
faces F, F, etc., from the neutral axis
N N are = %1, Z, etc., we have their dis-
tances from the axis N, N₁, = d + %1, d + Zy;
etc., and the moment of flexure is

N
K₁
H₁
-N₁
W₁ E = [F(d + %₁)² +
=
2,
= [F₁ (d² + 2 d z₁
F₂ (d + 2)² + ...] E
1
+ z₁²) + F₂ (d² + 2 d z + %2² ) + ...] E
z₂ zy²)
= [² (F, + F₂ + ...) + 2 d (F₁ z₁ + F₂ Z₂ + ...)
But
2
2
2
+ (F₁ z,² + Fq%2² + ...)] E.
F + E + ...
F
being the sum of all the elements is the cross-section F of the
entire body, and
F₁ z₁ + F₂ z₂+...
being the sum of the statical moments in relation to an axis pass-
ing through the centre of gravity is 0, and
2
(F, z,² + F, z₂² + ...) E
§ 225.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 433
is the moment of flexure WE in relation to the neutral axis N N;
consequently we have
and inversely
W,
W₁ E = (W + Fd2) E, or
W₁ = W + Fd²
1
W = W₁ - Fd2.
Therefore, the measure W of the moment of flexure in reference to the
neutral axis is equal to the measure W, of the moment of flexure in
reference to a second parallel axis minus the product of the cross-
section F and the square (d²) of the distance between these axes.
From this we see that, under any circumstances, the moment
of flexure in relation to the neutral axis is always the smallest.
The moment of flexure of many bodies in reference to some par-
ticular axis can often be found very easily, and we can employ it
to determine, by the aid of the formula just found, the moment in
reference to the neutral axis.
§ 225. Let CK = ≈ and C L
FIG. 350.
Y
A
N
B
F
-U-
L
-X-
K
M
-V
_Y
x² + y²
=
U
X
2)²
y, Fig. 350, be the coördinates
of a point F, referred to a sys-
tem of rectangular co-ordinates
XX, Y, and let u
and C N - v be the co-ordinates
of the same point, referred to an-
other system of rectangular co-
ordinates U U, V V, and, finally
let CF- =r be the distance of the
point Ffrom the common origin
C of the two systems of co-ordi-
nates; according to the theorem
of Pythagoras we have
u° + ?y° r², and also

Fx² + Fy²
Fu² + Fv²= Fr².
If in this equation, instead of F, we substitute successively the
elements F, F, F, etc., of the entire cross-section, and in like
manner, instead of x, y, u and r, the corresponding co-ordinates.
X1, X2, X3, etc., Y1, Y2, Y3, etc., U1, U2, Uç, etc., and v1, v2, v3, etc., we obtain
by addition the following formulas
2
2
F₁ x² + F₂ x²² + ... + F₁yi² + F₂ y₂² + ...
1
2
2
= F₁ u₁² + Fq uz² + ... + F₁ v₁² + F₁ v²² +...
U
= F₁ r²² + F‚½ r²² + · · ·,
and if we denote
28
434
[$ 225.
GENERAL PRINCIPLES OF MECHANICS.
6
2
F₁ x²² + F₂ x²² + ... by Σ (Fx²)
2
F₁ y₁² + F₂ y₂²+... by Σ (Fy³)
1
2
2 2
2
F₁ u²² + F₂ Û²² + ... by Σ (Fu²)
2
u₂²
2
F₁ v²² + F₂ v²² + ... by Σ (Fv²) and
1
2
2
F₁ r²² + F₂ r²² + ... by Σ (Fr³),
we have
1
2
Σ (Fx²) + Σ (Fy²) = Σ (Fu²) + Σ (F v²) = Σ (Fr²).
Therefore the sum of the measures of the moment of flexure, in
reference to the two axes X X and Y Y of one system of axes, is
equal to the sum of the measures of the moments of flexure, in refer-
ence to the two axes of another system of axes, and equal to the
measure of the moment of flexure, in reference to the origin, I.E.
equal to the sum of the products of the elements of the cross-section
and the square of the distances from the axis C.
If the cross-section A C C₁, Fig. 351, of a deflected body is a
symmetrical figure, and if the axis XX at right angles to the
FIG. 351.
Y
C
F
B
U
F2
N
M
H
S
A
K
F₂
-U
Fi
B₁
plane of flexure is an
axis of symmetry of
the figure, there will
be still another rela-
tion between the mo-
ments of flexure of the
body. Let S K = 20
and K F = y be the
X co-ordinates of an el-
ement of the surface
F in reference to the
system of axes XX

and
I, and let Ę N
v be the distance
of the same element
_Y
from the axis UU,
which forms an angle X SU = a with the first axis XX, we
have then
v = MF, MN MF - KL
and therefore
= KF, cos. KĘ M
S K sin. K S L = y cos. a
x sin. a,
v² = x² (sin. a)² + y² (cos. a)
-- 2 x y sin. a cos. a,
2
F₁x
F, v² = (sin. a) F x² + (cos. a) Fy sin. 2 a F, xy, and
v²) Σ
Σ (Fv') = (sin. a) E (F) + (cos. a)' (Fy') sin. 2 a E (Fx y).
§ 226.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 435
In consequence of the symmetry of the figure, every element
F, F... corresponds to another opposite element F, F..., for
which y, and consequently the entire product, is negative; hence
the sum of the corresponding products for two such elements, and
also the whole sum
and therefore we have
( F xy) = 0,
Σ (F' v²) = (sin. a)' Σ (Fx²) + (cos. a) Σ (Fy'), or
W = (sin. a) W₁ + (cos. a)? W₂,
2
in which W denotes the measure of the moment of flexure in refer-
ence to any axis U U, W, that in reference to the axis of symme-
try XX and W, that in reference to the axis Y Y at right angles
to the axis of symmetry, provided that the axes U U and Yas
well as the axis of symmetry X X pass through the centre of
gravity S of the figure.
By the aid of foregoing formulas we can often find, from the
known moments of flexure of a body in reference to a certain axis,
its moment of flexure in reference to another axis.
§ 226. Moment of Flexure of a Strip.-In order to find
the moment of flexure of a known cross-section AB, Fig. 352, I,
of a body in reference to an axis XX, let us imagine the cross-
section divided by lines perpendicular to into small strips
and every such strip as CA to be divided again into rectangular
elements F, F, F, etc. If 21, Z2, Z3, etc. are the distances (CF) of
these elements from the axis II, we have the measure of the
moment of such a strip
F₁ z₁² + F₂ z₂² + F3 23² +
F₁ Z₁ · Z₁ + F₂ Z₂. Z₂ + F 3 23. 23 +
Now if we lay off in Fig. 352,
F
FIG. 352.
I.
II.
A
F
-X-
C
-X
B
II, A B at right angles to and
equal to 4, and join B and
C by a straight line, it cuts
off from the perpendiculars to
CA, erected at the distances
(CF) Z1, Z2, Z3, etc., pieces
of the same length (FG)
Z1, Z2, Z3, etc., and F, z₁, Fa zo,
etc., can be regarded as the
volumes of prisms, and F, z₁ . z₁,
F₂ z2. z, etc., as their statical
moments with reference to the

436
[$ 227.
GENERAL PRINCIPLES OF MECHANICS.
1
axis C. The prisms F₁ z₁, Fa za, etc., however, form together a tri-
angular prism, whose base is A B C, and whose height is the
width of the strip A C(I); the sum of the above statical moments
is therefore equal to the moment of the prism A B C in reference
to the axis XX If we put the height C A = z and the width of
the prism=b, we have the volume of such a triangular prism
= 1b z²,
and since the distance of the centre of gravity from Cisz (see
§ 109), we have the statical moment of the above prisms, and con-
sequently the measure of the moment of flexure of the strip C A
W 1 b z². 3 z = 3 b z³.
In order to find the moment of flexure of the entire cross-sec-
tion A D, we have only to add together the moments of flexure of
the strips, such as CA, into which the entire surface is decomposed
by the perpendiculars to the axis X X.
The most simple case is that of a rectangular cross-section
A B CD, Fig. 353. The strips into which the surface is divided
N
FIG. 353.
A
TA*-
D
N
are here all of the same size and form to-
gether but a single strip, whose width A D
= b is that of the entire rectangle.. If the
height A B of this rectangle is h, we
have for the height of a strip

2 = 1 h ;
consequently the measure of the moment
of flexure of half of this surface is
B
C
1 b
6 (1)
bh
9
24
;
finally, the measure of the moment of the entire rectangle is
W = 2
b h³
24
3
b h³
3
12
§ 227. Moment of Flexure of a Girder, whose Form is
that of a Parallelopipedon.-From the foregoing we see that
b hs
12
E
the moment of flexure of a parallelopipedical girder W E
increases with the width and with the cube of the height of the girder.
Substituting this value for W E in the first formula
a =
P 13
3 WE
of § 217.
we obtain the deflection of a girder, whose cross-section is rectangu-
lar, and which is fixed at one end,
§ 228.] ELASTICITY AND STRENGTH OF FLEXURE, ETC.
437
P 13
a = 4. 3
b h³ E'
Substituting it in the second formula of the same paragraph
a =
1 Pl³
48 WE'
we have for a beam supported at both ends
a
Inversely, from the deflection
modulus of elasticity
and in the second
P 13
3
4b h³ E'
a we obtain in the first case the
4 P 13
39
a b h ³⁹
E =
P 13
E
4 a bh³
EXAMPLE-1) A wooden girder 10 feet 120 inches long, 8 inches
wide and 10 inches high is supported at both ends and carries a uniformly
distributed load of Q = 10000 pounds; how much will it be bent?
The deflection is
α =
Q t³
4 b h E
TO
· 10000. 1203
8. 10. E
50000. 12³
32.8 E
1350000
4. E
Substituting E = 1560000, we have a =
135
4. 156
= 0,216 inches.
2) If a parallelopipedical cast-iron rod, supported at both ends, is 2
inches wide and an inch thick, and is deflected 1 of an inch by a weight
P 18 pounds placed upon it at its centre, the distance of the supports
from each other being 5 feet, the modulus of elasticity is
P 13
18.60³
E
4 a b h³ 4. 1. 2. (1)
18.60³
1
72.216000=15552000 pounds.
§ 228. Hollow, Double-Webbed cr Tubular Girders.—
The moment of flexure of a hollow parallelopipedical girder
D
FIG. 354.
N
N
A
B
A B C D, Fig. 354, is determined by subtract-
ing from the moment of the whole cross-sec-
tion the moment of the hollow portion. If
A Bb and B C = h are the exterior and
A, B, b, and B, C, h, the interior width
=
and height, we have the measures of the mo-
ments of flexure of the surfaces AC and A, C,

b h³ b, 1,3
and
12
12
and consequently by subtraction the measure of
the moment of flexure of the tubular girder
b hs — b₁ hr³
W =
12
438
[$ 228.
GENERAL PRINCIPLES OF MECHANICS.
N
FIG. 355.
D
The moment of flexure of the single-webbed girder A B C D,
Fig. 355, is determined in exactly the same man-
ner. If A B = b and B = h are the exterior
height and width, and if A B A, B, b, and
B₁ C₁ h, are the sum of the widths and the
IN height of the two cavities, we have by subtrac
tion

A
B
FIG. 356.
D
73
b h³ — b₁ hr³
W =
12
The moment of flexure of the body 4 B CD, Fig. 356, the
cross-section of which is a cross, is found in a
similar manner. If A B = b and B C = h are
the height and width of the central portion, and
A₁ B₁ A B = b₁ is the sum of the widths,
and B, C, h, the height of the lateral por-
tions, we obtain by addition the measure of the
moment of flexure

N-
D
AL
A
B
Bi
-N
W
(
3
b h² + b, ha
12
In the same manner we can determine the moments of flexure
of many bodies which occur in practice. Thus for a body A, B, CD,
Fig. 357, with a T-shaped cross-section, whose dimensions are
N
D
D₁
FIG. 357.
C₁
T
A B
A B —
C D = b,
1
A, B₁ = A A, + B B₁ = b₁,
AD BC h and
A D₁ = B C₁ = BC-CC₁ = h₁,

1
N the measure of the moment of flexure in
reference to the lower edge A, B, is = mo-
ment of the rectangle ABCD minus moment
of the rectangles A, D, and B, C'₁, I.E.,
b (2 h)³ b₁ (2 h,)³
B
N₁
-N,
A₁ Bi
W₁ =
18
12
11.
12
3
bh" — b, h,"
www.
3
These moments are found by assuming each of these rectangles to
be the half of rectangles twice as high; for these the axis N, N, is
the neutral axis.
Now the surface A, C, D = F = b h — b₁ h₁, and its statical
moment is
1
h
h₁
F.e₁ = bh.
b₁ hi
= 1 (b h² — b, h₂²);
2
2
consequently the lever arm is
§ 229.] ELASTICITY AND STRENGTH OF FLEXURE, ETC.
439
MS = e₁
the product
F. e,² =
2
{
2
b h² - b₁ h₁²
2 (b h — b₁ h)
1
(b h² - b, h,²): (bh — b, h₁)
and the measure of the moment of flexure of the body in reference
to the neutral axis N N, passing through the centre of gravity S, is
3
2
W = W₁ F.e
2
b h³ — b₁ hr³
3
3
— { (b h² — b, h²)²: (bh — b, h₁)
3
4 (b h³ — b, h³) (b h − b, h,) — 3 (b h² — b₁ h,²)²
22
12 (bh b₁ h₁)
(b h² — b, h,²)² — 4 b h b, h, (h — h,)²
12 (bh — b, h,)
It is also easy to perceive, that the high webbed and flanged
girders have, for the same quantity of material, a greater moment
of flexure than the wide and massive ones. Since this moment
increases with the surface (F) and with the square (z) of the dis-
tance from the neutral axis, the same fibre is better able to resist
the bending the farther it is removed from the neutral axis. If,
for example, the height of a massive parallelopipedical girder is
double the width b, the measure of moment of flexure is either
Η
b. (2b)³
12
3 b¹, or =
2 b. b3
12
= { b',
the first formula obtaining, when we place its greater dimension
2 b vertical, and the latter, when it is placed horizontal; in the
first case the moment of flexure is four times as great as in the
second. If, again, we replace the solid girder, whose cross-section
is bh by a double webbed one, in which the hollow is equal to the
massive part of the cross-section b, h, bh, or if b, h, b h = bh,
—
I.E., b₁ h₁ = 2 b h, or b₁ b √ 2 and h₁ h
1
the moment of flexure for the latter girder is
3
b₁ h³ - b h³ b √ 2 (h √ 2)³ — b h³
12
12
I.E., three times as great as for the first one.
-
2, the measure of
·
= √³b h²
§ 229. Triangular Girders.-The measure of the moment of
flexure of a body with a triangular cross-section A B C, Fig. 358,
can be found, in accordance with what has been stated in the last
paragraphs, in the following manner.
The measure of the moment of flexure for the prism with a rec-
tangular cross-section A B C D is, when we retain the notations
of the next to the last paragraph, =
b ha
12
and consequently that of
440
[$ 229.
GENERAL PRINCIPLES OF MECHANICS.
its half with the triangular cross-section A B C in reference to the
central line N, N, is
-Z-
-NT
-N
FIG. 358.
Y
Y
D
Z
-N,
I₁ = {}
b hs
12
bh
24'

But the line of gravity N N of the
triangle is at a distance A B = { h
from the central line or line of gravity
N N of the rectangle, and, therefore,
according to § 224, the measure of the
moment in reference to N N is
1
N
S
-Zī
-Z₁
B
bh³
b h³
II = II₁
F=
24
12
Y-Y
bhs
bh³
1
36
3
12
The measure of the moment of flexure W of a girder with a
triangular cross-section is but one-third of the measure of the mo-
ment of flexure of a parallelopipedical one, the cross-section of
which has the same base and altitude. But since the latter girder
has but double the volume of the former, it follows, that for equal
dimensions the moment of flexure of a triangular girder is but
that of a rectangular one.
For the axis Z, Z, passing through the base B C, the measure
of this moment is
IT₂ = IT +
(1)². F
b h ³
b hs b h s
F =
+
36
18
12'
and for the axis Z Z, passing through the edge A,
I'₁ = IT +
(2)
Ch)² 5 h
b
2
b h s 4 b h ³
+
36
b h³
18
4
These formulas do not require the cross-section to be a right-
angled triangle. They hold good for any other triangle A B C,
Fig. 359, whose base B C is at right angles to the bending force
FIG. 359.
I.
A
II.
D
B
ՍՈ
N
S
B
D
C
A
A B C, so that we have for this triangle
P; for it can be de-
composed into two
right-angled trian-
gles AD B and A CD
N whose bases B D = b₁
and D C = b₂ form
together the base B C
=b of the triangle

§ 230.]
441
ELASTICITY AND STRENGTH OF FLEXURE, ETC.
(
b h³
W = 3's b₁ h² + 3'z b₂ h² = 3'5 (b₁ + b₂) h³ =
1
1
36
36
36
It is also of no importance whether the base B C lies
below the axis, I.E., whether it is placed as in I or II.
ment of flexure in both cases is
above or
The mo-
b h³
3
WE =
E,
36
when the modulus of elasticity for extension is the same as that for
compression. The same formulas can also be employed, when the
cross-section is a rhomb A B CD, Fig. 360, with the horizontal
diagonal B D. If B D = b is the width and A Ch the height,
we have for the body with this cross-section
T = 2.
b
is (1) '=
12 2
b h³
48
3
1 bh³
4 12'
I.E., one quarter of the measure of the moment of a girder with a
rectangular cross-section of the same height and width. From this
it follows, that for a double trapezoid A BE D, Fig. 361, the height
of which is A C' — B D = 1, the exterior width A B = C D = b
and the interior width E F = V₁9
b₁)
48
(3 b + b₁) h³
48
FIG. 362.
Y
bh³
II™
12
FIG. 360.
FIG. 361.

E
A F


A
B
E
A
IS
D
*D
U
X
D
K
Χ
C
D
II
_Y
B
§ 230. Polygonal Girders.-The foregoing theory can be
applied to a body with a regular polygonal cross-section A CE,
Fig. 362, whose neutral axis is at the same time an axis of
symmetry. Since such a polygon can be resolved into triangles.
having a common vertex S, the determination of its moment
442
[§ 230.
GENERAL PRINCIPLES OF MECHANICS.
consists essentially in the calculation of the moment of flexure of one
of those triangles A S B. If we denote the side A B B C = CD
of the polygon or the base of one of the triangles composing it by s
and the altitude S K of the
same by h, we have the measure
of its moment of flexure in ref-
FIG. 362 a.

E
erence to the axis XX
X
Ꭰ
A
U
K
X
L
B
-Y
h s³
h s³
114
12
48
on the contrary, this moment
in reference to a second axis
shs
4
Y Y is =
and conse-
quently the sum of these two
moments is
´s h³
3
h s³
sh
S?
+
h² +
4
48
4
12
This sum holds good (according to § 225) for every other trian
gle, and therefore, for a polygon of n
n s h
sides, we have
F
h²
(12
1 );
W₁ + W₁ = " + " (~² + 1 ) = ( + £2)
when its area n
sh
2
4
12
2, is denoted by F
2
If we designate the angle ASX by a, the measure of the
moment in reference to the axis ASL is
= W₁ (sin. a)² + W₂ (cos. a)';
1
2
but the latter is also equal to the measure of the moment W, in
reference to ASD or X X, and therefore we have
or
I.E.
2
W₁ = W', (sin. a) + W (cos. a)²,
W₁ [1 - (sin. a)'] = W. (cos. a)³,
2
W₁ (cos. a) W (cos. a)", and consequently
W₁
W 2.
For an axis UU, forming an arbitrary angle X SU
the axis XX of symmetry, the measure of the moment is
W = W₁ sin.² + W₂ cos.' W₁ (sin.³ & + cos.² p)
Now if we substitute in the above equation
F
W₁ + W₁ = 2 (2² + 12 ),
W,
TT :29
١
١١
W =
φ
with
= W₁.
1.
we obtain for any arbitrary axis of a regular polygon the measure
of the moment of flexure
$231.] ELASTICITY AND STRENGTH OF FLEXURE, ETC.
443
W = W₁ = w₂ = =
F
4
(h²
h³ +
12/
or, putting the radius of the polygon S A S Br and there-
=
s²
fore h² = r²
4'
F
w =
W
4
+ (~² - 8²).
6
§ 231. Cylindrical or Elliptical Girders.-For the circle,
considered as the polygon of an infinite number of infinitely small
sides, s = 0, and therefore the measure of the moment of flexure
of a cylinder is
W =
F
4
π pr
4
0,7854 r¹.
For a hollow cylinder or tube, whose exterior radius is r, and
whose interior one is r, we have by subtraction
2
1
π (r₁² — r₂²) (r,² + r₂²)
2
F(r
F (r₂² + r₂²)
4
4
4
π (r₁₁ — r.")
W =
4
Fr²
2
1 +
2
2 r
}
2
=
グ
​in which Fπ (r² — r²) denotes the area of the ring-shaped
r₁ + r₂
cross-section, r =
2
the mean radius and b = r
7 the
thickness of the wall of the tube. The horizontal diameter divides
the entire circle D E, Fig. 363, into two
semicircles ADB and A E B, and the
measure of the moment for such a
semicircle in reference to the diameter
A B is
FIG. 363.
D

N
-N
A
C
π
Пров
II₁
4
centre C of the circle is CS
But the distance of the centre of
gravity S of the semicircle from the
4 r
(see § 113), and therefore the
3 п
measure of the moment for the parallel axis N Nis
W = W₁ F. CSW — F.
-
3
( + - )²
8
= T
π jot
= 0,1098. rª,
8
Ο π
while, on the contrary, for the semicircle, whose diameter is vertical,
444
[§ 231.
GENERAL PRINCIPLES OF MECHANICS.
W =
π pot
8
0,3927 r¹.
In reference to an axis N N, which forms an angle N S X = a
with the axis of symmetry C D, Fig. 364, the measure of the
moment of the semicircle is.
-N-
π pot
8
W =
sin. a² + π jor
8
8
9 П
2
=) cos.
cos.² a
=
2
(0,3927 sin. a + 0,1098 cos.' a) r¹.
FIG. 364.
Մ
FIG. 365.
A


D
E₁
'D
A
-N
-N- -B
B
N
Bi
C
B 1
Y
B
From the formula
π pr
W =
4
A
for the measure of the moment of flexure of the full circle, that of
an ellipse A B A B, Fig. 365, is easily deduced. In consequence
of the relation of the ellipse to the circle given in Art. 12 of the
Introduction to the Calculus, when A B, A B, represents a circle.
whose radius C A is equal to the major semi-axis a of the ellipse,
and when the other semi-axis B of the ellipse is represented by
DE
b, we have the ratio of the width D E of an element of the
DE
ellipse to that D, E, of a similarly placed and equally high element
of the circle
BB
B₁ B₁
CB
b
CB
a
But since the moment of flexure of such a strip increases with the
simple width, the moment of a strip D E of the ellipse is to that
of the corresponding strip of the circle as b is to a, and conse-
quently the measure of the moment of flexure of a body with an
elliptical cross-section is equal times that of a body with a circu-
lar cross-section, I.E.
b
b
Η
па
παι
a
4
4
$232.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 445
If this body contains also an elliptical hollow, the semi-axes of
which are a, and b₁, we have for this body
A
E
W =
3
π (a³ b — a₁³ b₁)
4
If a body with a rectangular cross-section has an elliptical hol-
low around its axis, or, as is represented in Fig.
366, has an elliptical cavity on the side, we have
the measure of its moment of flexure
FIG. 366.
B

E
W =
b h³
12
3
παιδ
4
D
A
B
cb and h denoting the length A B and the height
A A BB of the rectangular cross-section
DABBA, and, on the contrary, a, and b, the
semi-axes CE and C F of the semi-elliptical hol-
low D FE.
§ 232. The measure of the moment of flexure of a cylinder
or a segment of a cylinder may be determined very simply in the
following manner. We divide the quadrant A D O of the segment,
of the cylinder A O B N, Fig. 367, into n equal parts, pass
N
Fig. 367.
A
B
K
E
F
through the points of division vertical
planes, such as D E, F G, etc. and de-
termine the moment of flexure for each
one of the slices D E F G, which we
consider to be right parallelopipedons.
The sum of the moments of these
slices gives the moment of flexure of the
semi-cylinder A O .B, and by doubling
this moment we obtain the moment of
flexure of the entire cvlinder. If de-
notes the radius CA CO of the cir-

r
cular cross-section 4 0 B N, a division D G of the arc
1
π I'
N
2
π I' and in consequence of the similarity of the triangles
2 n
ད
D G H and CDK, we have for the thickness K L of the slice of
the cylinder DEFG 2 DGLK
=
KD
KL = G H
D G =
CD
KD
CD 2 n
π ↑
K D.
•
•
2 n
Now according to the formula of § 226, the measure of the moment
of flexure of the slice D E F G is
K L. (2 KD)
12
S
π
T
Κ' D'
K D*.
12 2 n
3 n
446
[§ 232.
GENERAL PRINCIPLES OF MECHANICS.
If we put the variable angle A CD, which determines the dis-
tance of the slice from the vertical diameter,
ordinate or half-height of the slice, D Kr cos. p, and therefore
, we obtain the
the last measure of the moment of flexure can be put
π pr
3 n
(cos. $)*
8
-, as (cos. )'=
3 + 4 cos. 24+ cos. 4
8
π p 3 + 4 cos. 2 + cos. 4
3 n
(see the "Ingenieur," page 157). In order to find the measure of
the moment of flexure for the semi-cylinder, we must substitute in
the factor 3 + 4 cos. 2 + cos. 4 o, for o successively the values
then add the results found, and
π
π
π
π
1.
2. 3.
2 n 2 n
to n
•
2 n'
2 n
finally multiply by the common factor
24 n
Now the number 3
added n times to itself gives 3 n, the sum of the cosines from 0 to π
is 0, since the cosines in the second quadrant to π are equal
Π
2
π
and opposite to the cosines in the first quadrant 0 to, and the sum
2'
3
of the cosines in the third quadrant π to
π cancel those in the
2
3
2
fourth quadrant π to 2 π; therefore the measure of the moment of
flexure of the semi-cylinder is
IV
π pr
2
24 n
and that of the entire cylinder is
•
π pot
3 n =
8
π pl
W =
0,7854 ‚¹, or
4
πα
W =
0,04909 d*,
64
=
d 2 r denoting the diameter of the cylinder.
(REMARK.)--If we employ the formulas of the Calculus, d & denotes an
element of the arc 6, and the element D G
I'π
=r89; hence the meas-
2 n
ure of the moment of the element D E F G of the surface is
2 r¹ do /3 + 4 cos. 2 ¢ + cos. 4
||
2 do. p¹
200
12
12
3
(cos. $)¹
3
(3 + 4 cos. 2 ¢ + cos. 4 p) d p
=
8
46)
(3 do + 4 cos. 24 do + cos. 44 do)
12
[3 dø + 2 cos. 2 † d (2 p) + † cos. 4 o d (4 ø) ],
$238.] ELASTICITY AND STRENGTH OF FLEXURE, ETC.
447
and consequently that of the portion A B E D of the cylinder is
W =
W =
до
12
12
§ 26, I.).
fa 0)),
(3 fa 4+2 S cos
cos. 2 4 d (2 q) + † ƒ cos. 4 o d (4 9
1
I.E..
(3 ¢ + 2 sin. 2 4 + 1 sin. 4 ø). (See Introduction to the Calculus,
π
Substituting &
sin. 2 é $ = sin.
sin. π = 0, and sin. 4 ¢ = sin. 2 π = 0,
2'
and doubling the result obtained, we have the measure of the moment of
flexure of the entire cylinder
дов
W =
3 π
12 2
2
π pr
4
W =
For the segment D O E, on the contrary, we have
π pt
8
(3 ¢ + 2 sin. 2 p + 1 sin. 4 ọ)
доя
12
= [
29
8
12
)]
24
=
―
sin. 4 ]
48°
208
12 sin. 2 + 1 sin. 4 o`
[6 (π — 2 ø) 8 sin. 2 &
4
By simple subtraction we obtain, by means of the latter formula, the
measure of the moment W of a board D E F G of a finite thickness K L.
(§ 233.) Beams with Curvilinear Cross-sections.-The
measure of the moment of flexure I' of bodies with regular curvi-
linear cross-sections is determined most surely by the aid of the
calculus. For this purpose we decompose such a surface ANP,
Fig. 368, by ordinates into its elements, and we determine the
I
FIG. 368.
P
M
L
Ꭺ
K
N
X
moments of such an element in reference to
the axis of abscissas A and also in refer-
ence to the axis of ordinates A Y
If x is the abscissa A N and y the ordi-
nate N P, we have the area of an element
d F = y d x
(see Introduction to the Calculus, Art. 29)
and therefore the measure of the moment
of flexure in reference to the axis A F

1
d W₁ = ¦ y² . d F = y d x
' .
3
(see § 226), and, on the contrary, that in reference to the axis A Y
d W₁ = x² y d x,
X
since all points of the element are at the same distance x from A Y.
By integration we obtain for the whole surface A N P F
448
[§ 233.
GENERAL PRINCIPLES OF MECHANICS.
=
W₁ = } Sy³ d x
3
and
W₁ = Sx² y d x.
If we have determined (according to § 115) the centre of gravity
of the surface A N P and its co-ordinates A K = u and K S = v,
we find the measures of the moments of flexure in reference to the
axes passing through the centre of gravity and parallel to the co-
ordinate axes by putting
and
W₁
= fydz-vF
1 Sy
y³
x v²
W₁ = √ x² y d x − u² F.
S
E.G., for a parabolic surface A NP, whose equation is y' = px,
we have (according to
Art. 29 of the Introduction to the Calculus)
xy, and (according to § 115)
x and v = Y,
F=
hence
2
v2 F
F=
F y = ( 3 ) ' y · 3 ·
2
3
x Y
x: y³
32
and
3
w
2
6
u² F =
Fx²
=
x²
x Y
x" y.
3
25
y
Since also from y
px, it follows, that x =
and d x =
P
Q y dy
we have
p
1 Sy a x = 3
1
2
Q y d y
2
2 y³
2
y³
y'd y
y³ x
p
3 pc
15 P
15
1 2
1
x Y. Y²
Fy
3
5
and
y¹
1
W₁ = = Fy
y° – (3)
Sx y d x = Sy
Finally we obtain
(3) Fy = ( − ¿4) Fy
2y'dy
2
2 y
y'd y
x³ Y
ก
p³
7 p
3 2
7. zxy. x² =
73
3
F²x².
7
9
64!
19
Fy and
320
W 2
3
ry
3
12
Fx²
Fx²
F' x².
175
17
€ 234.]
ELASTICITY AND STRENGTH OF FLEXURE, ETC. 449
-X-D
FIG. 369.
S
Y
A
B
-Y
-X
For a symmetrical parabolic surface
AD B, Fig. 369, whose cord ABs and
whose altitude CD = h, we can put the
measure of the moment in reference to the
axis of symmetry X X

W
1 F
Fs2
s³ h
20
30
while, on the contrary, that in reference
to the axis Y Yat right angles to it re-
mains
W
12
175
Fh2
8
.175
h³ s.
§ 234. Curvilinear Cross-sections. - If we are required to
calculate the moment of flexure of a body, whose cross-section
forms a compound or irregular figure, we must either divide this
cross-section into parts, for which the measure II' is already known,
or we must decompose the same by vertical lines, calculate the
measures of the moment of flexure of these strips (according to
§ 226), and, finally, add these values together, in doing which we
can employ with advantage Simpson's or Cotes' rule.
Y
B
If, E.G., A B E C, Fig 370, is such a figure or such a portion of
Z
Z
FIG. 370.
2
~
♡
3
E
A C
D
X
1 х
·
W₁
3' 12
the cross-section of a body and if its mo-
ment of flexure in reference to the axis
A X is to be determined, we calculate first
the measure for the portion of surface
ABG D and then the measure I, for the
part CED; subtracting the latter from
the former, we obtain the required moment
IT = IT'₁ II.

= x,
If the base A D of the first part
and the altitudes of the same at equal dis-
tances from each other are zo, Z1, Z9, Z3, Z4, we
have the corresponding measure of the mo-
ment, according to Simpson's rule,
3
(2,3 + 4 %,³ + 2 %,³ + ± 23² + %₁³).
นา
z
If, on the contrary, the width CD of the piece C D E to be
subtracted be, and the altitudes of the same are Yo Y1、 Ye, Y3,
we have, according to Cotes' rule (see Introduction to the Calculus,
1
Art. 38),
W₂
3
3
(y,³ + 3 yı³ + 3 y₂³ + ys³).
3 8
29
450
[$ 235.
GENERAL PRINCIPLES OF MECHANICS.
If AX does not pass through the centre of gravity S of the
entire surface, we must reduce it by the well-known rule (§ 224) to
the axis passing through S. In the same manner other parts of
the cross-section, which lie below or alongside of A Y, may
be treated. The centre of gravity S can be determined either
according to § 124, or empirically by cutting a pattern of the
section out of thin sheet iron or paper and laying it upon a sharp
knife-edge. If we determine in this way two lines of gravity, their
point of intersection gives the centre of gravity.
EXAMPLE.—A B G E C, in Fig. 370, is a portion of the cross-section
of an iron rail, which can be considered as the difference of two surfaces
ABGD and C E D. If the width of the first is inches and that of the
second 1 inch, and if the heights of the first are
20 2,85; 21
and those of the second
Уо
2,82; 2₂ = 2,74; 23 = 2,60; and 2₁ = 2,30,
0,20; y₁ = 1,50; Y½ 1,80 and y 2,15,
1
2
3
4
we have for the measure of the moment of flexure of the first portion
W₁
1
1
3
1
4
1
12:
[2,85³ + 2,30³ + 4. (2,823 + 2,60³) + 2. 2,74³]
(23,149 + 12,167 + 4. 40,002 + 2. 20,571).
•
27
1
•
236,47
27
8,7584,
and, on the contrary, that of the second portion
W 2
1
[0,20³ + 2,15³ + 3 (1,50³ + 1,80³)]
1
1 .
8
•
24
(0,0080 + 9,9384 + 27,6210)
=
37,5674
24
= 1,5653,
3
1
consequently, the required measure for the entire surface ABGE C is
W = W₁ W₂ = 8,7584 — 1,5653 7,1931.
1
2
REMARK.-We can also put
2
2
(1.0². 12. 2
W. ( ) (1 . 0² 9, + 4 . 1² 3, + 2 . 2° . Yg + 4. 8° /s + 1.4² . §₂)
12 4
ལྟུ
192
1
(4 Y₁ + 8 Y ₂ + 36 y¸ + 16 y±),
3
•
8
when y。, Y₁, Y₂, Yз, Y4 denote the widths measured at the distances
وں
3
£ 2 1 2 3 2, † 2, 4 z from A X.
4
2,
§ 235. Strength of Flexure.-If we know the moment of
flexure of a body A K′ O B, Fig. 371, fixed at one end B and at the
other end A subjected to a force P, we can find the strain in every
one of its cross-sections N O. If S denotes the strains per square
inch at a distance S Ne from the neutral axis S, the strains at
21
Zu
the distances Z1, Z2,
>
are S₁
S, S₂
S, and their mo-
e
e
§ 235.]
451
ELASTICITY AND STRENGTH OF FLEXURE, ETC.
ments for the cross-sections F, F,...., are
S
M₁ = F₁ S₁ z₁ = F₁ z₁².
2
1
and consequently the sum
M = M₁ + M₂ + ..
=
FIG. 371.
-P
K
NN,
H
T
R
V
S
5, M₁₂ = F₁₂ S₁ %₂ = F₁ z₂ =, etc.,
e
2
z2
e
of the strains in the cross-section N O is
2
S W S
(F; z₁² + F₂ z₂² + ...) e
W
e
Now if is the dis-
X
tance SH of the cross-
section NO from the
point of application 1
of the force P, we have
alsó M Px, and
consequently
1) P x = "/e

IT S
or
Pre WS
=
and the strain in the
body at the distance e
P
from the neutral axis is
1
Me
Pxe
2) S =
Π
I'
P
M
The latter increases
with a, and is therefore
a maximum for a = 1,
I.E., at the fixed point.
B. In like manner it
increases with e, and is
therefore a maximum
for the point most dis-
tant from the neutral
axis.
If the body is no-
where to be stretched
beyond the limit of elasticity, the maximum strain S should at
most be equal to the modulus proof strength T, and consequently
we must put
S=T
Ple
W
W T
Pl=
С
or
from which we obtain the proof strength of the girder 4 K OB
452
[$ 235.
GENERAL PRINCIPLES OF MECHANICS.
W T
P =
In like manner we have for the ultimate strength or force
necessary to break the body at B
W K
P₁ =
le
in which we must substitute for K the modulus of ultimate
strength determined by experiment upon rupture. The funda-
found in § 215, can be obtained
mental formula P x =
directly as follows.
σ
WE
r
If we denote by the extension NN, produced by the strain S,
we have So E, and substituting in the proportion
=
NN₁
SN
Ꭱ Ꮪ
MR'
N N₁ = 0, S N = e,
R S = 1, and M R
r, the radius of curva-
σ
1
e
ture, we have
or σ = ; hence it follows, that
e
j
e
S
E
S
E or
γ
e
and therefore also
WE
W S
P x =
γ
e
P² t³
If in the formula L
18
6
(§ 217) for the work done in
WE
T W
e
bending the body A K B we substitute the moment Pl
and the modulus of proof-strength To E, we obtain
L = 18
T² W²
e²
WE
σ² E
W L
3 e**
But (according to § 206) o² E is the modulus of resilience 1;
therefore the work done in bending a body to the limit of elasticity
is
L = A.
W Z
3 e²
Ifb is the greatest width of the body, we can imagine the whole
cross-section F of the body to be divided in n equally wide strips,
b
whose width is and whose altitudes are z₁, Z2, Z3
;
• •
and we can put
N
F
b
(%1 + %2 + Z3 + . . .) and
N
§ 236.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 453
b
3
W =
(21³ + 20³ + 23³ ...),
12 n
and therefore also
Z
3
3
21
+ 2½³ + 23³ + ..
Fl
W Z
12*
21 + 22 + 23 +
My
My µy,
We can make z₁ = µ, e, z₂ = µ₂ C, Z3 = µ3 e, μ₁, μo, μz denoting
numbers dependent upon the form of the cross-section, and there-
fore we have
II" Z
e²
3
3
μι + μου + μ3 +
3
Fl
12'
м2 + м2 + Мз +
and consequently the mechanical effect.
But
μι με
L =
3
3
μlos
3
A [µ³ + μ₂³ + flz³ +
3 My + Mz + μz +
3
2
-)
Fl
12°
M₁² + μ₂ + M3 is a coefficient y, dependent upon the form
Mr + Miz + flz
1
of the body alone, and F1 = V is the volume of the body; hence
the work done L= AV is not dependent upon the indi-
vidual dimensions, but only upon the form of the cross-section and
the volume of the body, which is bent. When the bodies are of the
same nature and of similar cross-sections, the work done is propor-
tional to the volume of the body.
For the work done in producing rupture we must put
I Z
L₁
= B.
3 es
B denoting the modulus of fragility.
§ 236. Formulas for the Strength of Bodies.-For a paral-
lelopipedical girder 4 C B, Fig. 372, the length of which is 7, the
width b and the height h, we have
FIG. 372.
h, and, according to § 226,
e =

b h³
Π
W
; hence
12
B
b h²
6
the proof
P
strength of the girder is P
b h T
7 6
T
and its moment is P7-bh². -
6
From this it follows, that the mechanical effect necessary to bend
the girder to the limit of elasticity is
L =
AW 7 A b h² 2 l
3 e e
3
6 h
У
= ¦ A b h l = ¦ A V.
454
[§ 236.
GENERAL PRINCIPLES OF MECHANICS.
If the girder is hollow, and if its cross-section is shaped as is
represented in Fig. 373 and Fig. 374, we have
3
W
b h s
e
V₁ hy
12. h
3
b h b₁ h
6 h
b h³ — b, hr³
3
P
T
6 h l
whence
b and h being the exterior and b, and h, the interior width and
IN
FIG. 373.
FIG. 374.
FIG. 375.
FIG. 376.
E A F


A
B
Ꭰ
DI


N
N
D
B
A
B
A
B
H
height of the cross-section. For a body with a rhombic cross-sec-
tion, such as Fig. 375, we have
W
e
b h³
48.h
b h²
3
P
b h²
and from this
24
T
1
•
24
4
b h²
6
T
I.E. as great as for a parallelopipedical girder of the same height
A C h and width B D = b. For a girder, whose cross-section is
a double trapezoid, such as is represented in Fig. 376, we have
W
e
(3 b + b₁) h³
48. ½ h
1
(3 b + b₁) h²
24
;
hence the moment of the proof strength is
P l =
(3 b + b₁) h² T
4
6
b denoting the upper and b, the central width and h the height of
the cross-section.
For a girder with a regular 2n sided base, such as A D F, Fig.
377, I and II, we have, if r denotes the exterior radius CA, s the
length of the side A B, h the interior radius C L and F the entire
area of the cross-section,
W =
¹ (rº − ¿ s°) = 1 (w² + iz 8º) =
F
4
F
4
F' (r² + 2 h²)
12
§ 236.]
ELASTICITY AND STRENGTH OF FLEXURE, ETC. 455
If the neutral axis N 0, as in Fig. 377, I, passes through the
middle of the opposite sides,c = r; and if, as in Fig. 377, II, it
passes through the opposite corners,
e = h = √ r² — (½ 8)².
FIG. 377.
I.
A
IL
III.

A
L B
B
F
B
D
G
N
F
D
F
E
E
E
D
Pl=
12 r
F (r² + 2 h²)
P₁ l
Hence it follows, that in the first case
F' (r² + 2 h²)
T, and, on the contrary, in the second
T, while in both cases
12 h
F
=
i n s h
= n h v p²
V
hº
j n s V p²
p² — (¦ §)³.
P
↑
The ratio
of the proof strengths is
P
h
If the number n of the sides of a polygon is uneven, as in Fig.
377, III, we must substitute e = r, and therefore we must employ
the first formula only; provided always that the direction of the
force coincides with that of the axis of symmetry.
For a square cross-section we have s = 2 h
= 2 h = r √2, F - 8³;
and the moment of the proof load
§³
༡༠
ΡΙ
T
Τ
T= 0,333 ³ T,
612
3
and, on the contrary,
73 1 2
P₁l=
Ꭲ
6
3
2 h
s = r =
F
3
53
For a hexagonal cross-section we have
F
T = 0,471 r³ T.
343
2,598 s', and therefore
Pl
=
sᏜ Ꭲ
16
5 1 3
16
7³ T = 0,541 r³ T, and
P₁l = { s³ T = r T 0,625 ³ T.
§ =
For a regular octagonal cross-section we have
456
[§ 236.
GENERAL PRINCIPLES OF MECHANICS.
s = r
√ 2 — √ 2, h
r
√ 2 + √ 2 and
2
2√2
F = 4 s h = 2 √ 2 . r²
=
s²; hence
2 √ 2
p³ T
23 Т = 0,638 р³ T,
µ³
Pl
4 (2√2 + 1)
3 √ 20 + 14 √ 2
2 √2 +
1
8.3 T =
6
and
4 (2 1 2 + 1)
2√2+1
P₁l =
s³ T =
p³ T = 0,691 µ³ T.
3
√ 17 + 12 √ 2
3 √2 + √2
For a massive cylinder, whose radius is r, we have
W
π pt
Проз
and therefore
"
e
4 r
4
π
ΡΙ
r³ T = 0,785 r³ T =
Fr. T, and
4
A
π 73 7
L
•
3
4 j
12
12
But if the cylinder is hollow, we have, on the contrary,
= √½ A. π r² 1 =
Α.πρι 1.
√½ A V.
1
π (r₁* — r₂¹)
1 +
(2)
Fr
ΡΙ
T=
T (compare § 231),
4
b
2
7'1
1 +
2 r
r, denoting the exterior, r, the interior and r =
radius, Fπ (r,² —
=
FIG. 378.
A
the mean
2
r²) the annular cross-section of the cylinder
and b = "1 r₂ its width.
FIG. 379.
A
B
E
E,
D
E
C
N
-B
C
D
B
A
Pl=
E
D
For a girder, whose
cross-section is elliptical,
as is represented in Fig.
378, when the direction
of the semi-axis C A = a
is that of the force, and
that of the semi-axis C B
=b coincides with the
neutral axis, we have


π a² b T = ¦ F a T.
4
111
Finally, for a parallelopipedical girder hollowed out on each
side in the shape of a semi-ellipse, as is represented in Fig. 379,
we have
1 ½ b h ³
Pl=
ΡΙ
4
апы, а, 3
T
I h
b h³ - 3 π b₁ a³
6h
3
T
§ 237.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 457
and, on the contrary, if the cross-sections of the hollows are para-
bolas,
3
ΡΙ
12
1 ½ b h s
8
15
b₁ a i
3
5 b h 32 b, a,³
Τ
↑ =
T
! h
30 h
b denoting the exterior width, h the exterior height, b, the depth
of the hollow and a, the height of the same.
§ 237. Difference in the Moduli of Proof Strength.
The formula P
FIG. 380.
W T
e l
for the proof load of a girder fixed at one
end A, Fig. 380, holds good only, when the extension σ and the
compression σ of the body are equal
to each other at the limit of elas-
ticity; for under those circumstances
only can the modulus of proof
strength for extension

Ρ
C
T= 6 E
be equal to that of compression
T₁ = o, E.
1
For wrought iron this assumption seems to be nearly correct, and
for wood approximately so, but these relations are entirely different
in the case of cast iron; the latter has not only a much greater
modulus of ultimate strength for crushing than for tearing, but
also the compression o, at the limit of elasticity, which can, how-
ever, be given only approximatively, is about twice as great as the
extension σ, and consequently the modulus of proof strength T₁
for compression is twice as great as the modulus of proof strength
T for extension.
In order to find the proof strength of cast iron or of any other
body, for which there is a perceptible difference between σ and o,
or between T and T, we must first see which of the quotients
T
T
e
T
е
e1
and is the lesser,
formula
and substitute that instead of
P =
ΠΤ
e l'
in the
The other half of the beam, corresponding to the greater ratio
T T₁
e
or is of course not stretched to the limit of elasticity. In
order to reduce this cross-section and consequently that of the
whole body to a minimum and thus to economize as much mate-
rial as possible, it is necessary, that both the halves of the girder
shall be strained to the limit of elasticity. Therefore we must give
the beam such a form and such a position that we will have
458
[§ 237.
GENERAL PRINCIPLES OF MECHANICS.
T
e
T
e1
с T
σ
or
C1 T
στ
I.E., that the ratio of the greatest distances e and e, of the fibres on
the two sides from the neutral axis shall be equal to the ratio of
the moduli of proof strength T and T, for compression and ex-
tension.
T
σ
If, then, for cast iron we have
T
= 2 (see § 211), we
01
21
C
must so fashion the cross-section of a cast iron girder that shall
be as near as possible = 2.
A triangular girder must be so placed,
that the half with a triangular cross-section shall be compressed,
and that with the trapezoidal cross-section shall be stretched. If
we place one of the sides of the prism horizontal or at right angles
21 2
while in the opposite position, we
1'
to the force, we have
C1
1
have
e
2
e
We can also give cast-iron girders, whose cross-section approach
the shape of a T (as is represented in Fig. 381), such dimensions
C₁
that the ratio
с
FIG. 381.
B
C
B
D
G
1
1
1
t
I
A H M H
shall be equal to 2.
Let the entire height of the beam be A B
h, the width of the upper flange be B B =
2 BC=b, the height of the hollow on the side be
A D = h₁ = µ4h,
the width of the same be
2 D G = b₁ = v', b,
the height of the lower flange be
HL = h₂ = μ₁ h

fly
and its projection on both sides be
2 L N = b = 1'. b,
then the distance of the centre of gravity s of the whole surface
from the lower edge His
2
1 b h² — b₁ h,² + b ₂ h²
2 bhb, h₁ + by h₂
MS = c
h/1
2
2
2' +
7'
2
1
M₁ 2 + μg Vg
If we substitute.
2 and + C₂ h, we
(see § 105 and § 109).
have e
1
hand,
h, and therefore the equation of condition
$237.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 459
223 h
0915
h 1
2
2
2 1
μl₁ 2½ + μl; V z
µ.) =
1.
which, when transformed, becomes.
μr 2' (4 — 3 μr) — µ₂ v₂ (4 — 3 μ.)
My
H V
2
By the aid of this formula, when three of the ratios ₁, 1, 2 and 1,
of the dimensions are given, we can calculate the fourth. If we
make μ, 0, we have the cross-section represented in Fig. 382, the
moment of flexure of which has already been determined (§ 228),
and for which we have u, v, (4 3 μM) 1.
e
REMARK.—Moll and Reuleaux (see their work, "Die Festigkeit der
Materialen," Brunswick, 1853) recommend for the determination of the
most advantageous cross-section the use of a balance, the beam of which
forms a table. Patterns of the cross-section, cut out of sheet-iron, are
placed upon it in such a manner that the neutral axis, determined by the
shall lie exactly above the centre of rotation of the beam.
If the pattern has the most advantageous form, the beam will balance; if
it does not, we must cause it to do so by cutting away portions from the
side of the body, until the beam balances, when the pattern occupies the
above position.
ratio
1
σ
01
EXAMPLE 1.-If the cross-section of a cast-iron beam has the form of
Fig. 381, and if the ratios of the heights are
μ₁ =
h₁
h
1
7
812
7
1
8
8'
we have for the ratios of the width the condition
(
21
8
1
"'1
29 v
64.
2
= 1, I.E.
77 21
If the lower flange is omitted, then r
64
1
1'1
b
77
and the thickness of the web proper is b
If, on the contrary, we make "2
N
FIG. 382.
D
D₁
ت
0, and we have
0,831,
b₁
= 0,169 b.
1
we have
6
(77 — 39)
=
64, and
1
1
l'a
0,887
consequently v 0.887 and r
1
0,148. For h = 8 inches and b
= 7 inches, h 1 inch, b
1
51 inclres, 7, is

1
5 inches and by
inch; so that the thickness of the upper and
lower flange is 1 inch, and that of the vertical
web but inch.
=
N
EXAMPLE 2.—For a girder with a T-shaped
cross-section, Fig. 382, we have found (§ 228)
B
N
A₁ B₁
N₁
W
=
2
(b h² — b₁ h, 2)² - 4 bb, hh, (h-h₁)²
1
1
12 (bh b₁ h₁)
1
460
[$ 238.
GENERAL PRINCIPLES OF MECHANICS.
in which we must put
1 bh² - b, h, 2
1 1
e1
e ₁ = 2 b h
b₁ h₁ ;
1 1
hence, if one end is fixed and the other loaded, we have
Pl=
(b h² — b¸ h¸²)² — 4 b b¸ h h¸ (h — h₁)² T₁
1
1
1
1
b h² — b₁ h₁²
2
6
1
If we put h₁
h and b₁
1
= v₁ b, we obtain
Pl=
(1 — 14, ² v',
1
) ³ —
ս
1 1
4 u, ³½ (1 — µ4, )² b h²
bhe T₁
1
2
| 21
6
Pl=
5
14
and therefore if the beam is cast-iron and we substitute μ₁ =
(14)² - 3 (4)² -b h
If, E.G., h is = 10 and b
8 inches, and consequently
욱 ​and v
1
Ꭲ .
1
6
13 bh2
70 6
T..
1
.
h₁
1
욕​. 10
6,0
we have
13 8.100
520
Pl=
•
T₁
70
6
1 21
T..
4.
7.8 = 7 and b — b
1
= 1 inch,
84, h — h₁
=
13 inches,
1
If we substitute T₁ 18700 pounds, we have for the moment of the
proof strength, which, for the sake of safety, we should put = 150000
Pl=
520
21
•
18700 = 463048 pounds.
If this beam is 100 inches long, its safe load at the free end is
P
150000
100
1500 pounds.
If the girder is supported at both ends and carries the load in the middle,
we have
P = 4. 1500 = 6000 pounds.
While in the first case the flange must be placed on top, in the latter it
must be put at the bottom.
§ 238. Difference in the Moduli of Ultimate Strength.
If we determine the moduli of elasticity and of proof strength
by means of experiments on bending, making use of the formulas
E
Pl r
И
and T =
Ple
W'
the values found for E and T generally agree very well with those
given by direct experiments on extension and compression, when the
formulas
E
Pl
2 F
and T
P
F
are employed.
But this relation is entirely different for the modulus of ulti-
$238.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 461
mate strength. Since we cannot consider the modulus of elasticity
E to be constant beyond the limits of elasticity (for it decreases,
when the extension or compression increases), and since the mod-
ulus of elasticity for extension is no longer equal to that for
compression, the strains in the superposed fibres are no longer
proportional to their distances from the neutral axis, and conse-
quently that axis no longer passes through the centre of gravity;
the values of e and e, differ in that case essentially from what they
are, when the limit of elasticity is not surpassed.
If W denotes the measure of the moment of flexure for the
stretched half of the girder and E the mean modulus of elasticity
of the same, and if W, denotes this measure for the compressed
portion and E, the mean modulus of elasticity, we have for the
moment of the bending force, when the bending becomes excessive,
WE + W, E,
Pl =
2°
and if we put, at least approximately,
K
and
E J E₁
Kand
K, denoting the moduli of ultimate strength for tearing and
crushing, the moment of the force necessary to break it is
K (WE + W₁ E₁)
1
Pl either
or =
E e
K₁ (WE W₁ E₁)
E₁ e₁
If we again denote the statical moment of the cross-section of the
stretched portion of the body in reference to the neutral axis by M
and that of the cross-section of the compressed portion of the body
in reference to the same axis by M₁, we have the force on one side
ME
M, E,
and on the other
and since the two forces must
form a couple, M E = M₁ E₁. This equation serves to determine
the neutral axis by means of the distances e and e₁.
For a girder with a rectangular cross-section we have
M =
b e²
2
be,2
and M₁
2
and therefore
From this we obtain
E e² = E₁ e₁².
e₁ = e
VE
Ei
462
[§ 238.
GENERAL PRINCIPLES CF MECHANICS.
Substituting this value in the equation e + e, h, we have
h VE
e =
and e
NE + VE
h NE
NE + VE
The measures of the moments of flexure are in this case
W =
be³
3
and W₁
be,"
3
3
and consequently we have
b
bh (EE, VE,
EE, VE
Pl=
(Ee³ + E₁ e,³)
3r
3 r
(NE + NE,)³
3
b h³
E E
3 r (NE + NE₁)
29
and therefore the moment necessary to produce rupture is
NE₁
bh²
3
Ꮶ .
•
NE + VE
K. b h³
3
E E
Pl either
3 Ee
(VE+VE₁)²
b h²
VE
or
K
Ꮶ .
3
NE + VE
For EE, we have, of course,
ΡΙ
b h²
6
2
Ꮶ .
For wood and wrought iron, E is really about - E₁, and there-
fore we can write approximately
bh²
Pl= Ꮶ,
6
in which we must substitute for K the smaller value of the modulus
of ultimate strength. For cast iron, E, is much greater than E,
3
bhs
and therefore P 1 approaches the value K, K being the modu-
3
lus of rupture for extension. For wood we must substitute the
mean value of the modulus of ultimate strength for crushing,
K480 kilograms 6800 pounds, which value agrees very well
with the results of the experiments of Eytelwein, Gerstner, etc.
In like manner, for a wrought iron girder we must substitute
instead of the modulus of ultimate strength for crushing K
2200 kilograms = 31000 pounds. While under the same circum-
stances wood and wrought iron break by crushing, cast iron breaks
by tearing. If for the latter were about A₁, we would have
to substitute for cast iron girders, in the above formulas, the
modulus of ultimate strength of tearing, I.E., K 1300 kilograms
239] ELASTICITY AND STRENGTH OF FLEXURE, ETC.
463
=18500 pounds; but, according to the results of many experi-
ments, we must put
=
K 3200 kilograms 45500 pounds,
I.E., about the mean value of the modulus of ultimate strength for
tearing and of that for crushing.
This great difference is caused not only by the difference of the
moduli of elasticity E and E₁, but also by the granular texture of
the cast iron, which precludes the supposition that the beam is
composed of a bundle of rods.
Many different circumstances influence the elasticity, the
proof strength and the ultimate strength of a body, so that nota-
ble differences occur in the results of experiment.
The wood, for example, near the heart and root of the tree is
stronger than the sap wood and that near the top, and wood will
resist a greater force, when the latter acts parallel to the yearly
rings than when it acts at right angles to them; finally, the soil
and position of the place where the wood grew, the state of
humidity, the age, etc. influence the strength of wood. Finally,
the deflection of a body, which has been loaded very long, is always
a little greater than that produced, when the weight is first laid on.
§ 239. Experiments upon Flexure and Rupture.-Experi-
ments upon elasticity and strength were made by Eytelwein and
Gerstner with the apparatus represented in Fig. 383. A B and
A B are two trestles, upon which two iron bed-plates Cand Care
fastened, and DD is the body to be experimented upon, which is
F
M
D
GAR
N
FIG. 383.
H

P
M
N
MPEN DEN RE
B
E
CIST HEROSHER CONTR
E
B
464
[$ 239.
GENERAL PRINCIPLES OF MECHANICS.
placed upon them. The weight P, which is to bend the body, is
placed on a scale board E E, which is suspended to a stirrup M N,
whose upper end is rounded and rests upon the centre M of the
girder. In order to find the deflection produced by the weight,
Eytelwein employed two horizontal strings FF and G G and a
scale M H, placed upon the middle of the girder. Gerstner, on
the contrary, employed a long sensitive one-armed lever, which
rested upon the beam near its fulcrum and whose end indicated on
a vertical scale the deflection of M in 15 times its real size.
Lagerhjelm employed a pointer, which was moved by means of a
string passing over a pulley, and which showed the deflection of
the beam magnified upon a graduated circular dial. Others, as,
E.G., Morin, made use of a cathometer to determine the deflection.
The object observed was a point fastened in the centre of the girder.
In the English experiments a long wedge was used to measure this
deflection; it was inserted between the centre of the beam and a
fixed support. In order that the accuracy of the measurement
may not be affected by the yielding of the supports of the girder,
it should rest during the experiments either upon stone founda-
tions (Morin), or a long ruler should be placed a certain distance
above the girder and fastened at its ends to the ends of the latter,
but in such a manner that it cannot bend with the beam, and in
each experiment the distance between the lower edge of the ruler and
the centre of the deflected girder should be measured (Fairbairn).
The manner in which Stephenson, etc., determined the deflec-
tion and strength of tubular sheet iron girders, is shown with the
principal details in Fig. 384. The tube 4 B is 75 feet long (the
front portion being omitted in the figure), is supported at both
ends, as, E.G., in C, upon blocks of wood and its centre rests upon
a beam DD, which is carried by two screw-jacks. An iron arm,
the end F of which only can be seen in the figure, passes through
the middle of the tubular girder near the bottom, and from this
two stirrups G, G hang, to which the scale-board H II to receive.
the weight P is suspended. Before the experiment and during
the laying on of the weights, the entire load was supported by the
beam DD; when the screw-jacks were lowered DD sank and
placed itself upon the supports E, E, while the centre of the tube
A F, loaded with P, remained free and could assume a defloc-
tion corresponding to the force P. This deflection was measured
by means of a wedge.
In order to avoid the use of very large weights in experiment-
§ 239.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 465
ing upon large girders, they are generally made to act upon the
latter by means of a lever with unequal arms. With the same
object in view, Hodgkinson caused the force of the lever to be
FIG. 384

-W
ཕམམ་---་་་་་
applied not to the centre of a girder supported at both ends, but to
one end of a girder, which was supported in the middle and the
other end of which was fastened by a bolt to the foundation.
The results of experiments, made under very different circum-
stances and with very different kinds of materials, particularly of
wood and iron, have shown the theory laid down in the foregoing
pages to be correct in all important particulars. In regard to the
rupture of parallelopipedical girders it was proved, that those of
wood and wrought iron, under the same circumstances, gave way
by crushing, and that in the case of cast iron the rupture began
either by the exterior fibres being torn apart or by a wedge break-
ing out at the most compressed part (in the middle).
We can satisfy ourselves of the truth of the hypothesis. made
in § 214, in regard to the behaviour of the fibres of a body, sub-
jected to flexure, by making saw cuts upon the compressed side of
466
[$ 240.
GENERAL PRINCIPLES OF MECHANICS.
parallelopipedical wooden rods and then filling them up with pieces
of wood, by drawing a series of lines upon the side of a beam at
right angles to its longitudinal axis, and finally by fastening two
thin rods to the beam, one along the extended and the other along
the compressed side.
§ 240. Moduli of Proof and Ultimate Strength.-In the
following table the moduli of elasticity, of proof strength and of
ultimate strength or of rupture, as determined by experiments
upon bending and breaking are given. The first differ but little
from those determined by the experiments on extension and com-
pression; but, for the reasons given above (§ 238), this is not
true of the modulus of ultimate strength. The upper of the
two quantities in a parenthesis {} gives the value in English meas-
ures (pounds per square inch) and the lower one the same in
French measures (kilograms per square centimeter).
TABLE
OF THE MODULI OF ELASTICITY, OF PROOF STRENGTH AND OF
ULTIMATE STRENGTH OR OF RUPTURE OF DIFFERENT BODIES
IN RELATION TO BENDING AND BREAKING.
Names of the Bodies.
Modulus of Elasticity
E.
Modulus of
Proof
Strength T.
Modulus of Rup-
ture or of Ultimate
Strength A (K",).
Wood of deciduous Trees
S 1280000
90000
3100
220
9240
650)
Wood of evergreen Trees
2130000
4300
12800
150000
300
900
17000000
10670
Cast Iron
45500
1200000
750
3200
Wrought Iron.
28400000
17000
2000000
1200
32700
2300
Limestone and Sandstone
Clayslate.
350
1760
}
5000!
124
In order to determine from the value in the foregoing table the
load, which a girder can carry securely, we must introduce a factor
$ 240.] ELASTICITY AND STRENGTH OF FLEXURE, ETC 467
of safety and substitute in the formulas for the proof strength
already found for wood
either instead of T, Tor instead of K, K,
for cast iron
3
1
10
either instead of T, Tor instead of K, K,
and for wrought iron
Tor instead of K, K.
Consequently we can hereafter put for wood
either instead of T,
T = 73 kilograms
for cast iron
and for wrought iron
=
1000 pounds,
T510 kilograms 7000 pounds
=
=
T = 660 kilograms 9000 pounds.
We cannot employ these values in calculating the dimensions
of shafts and other parts of machines; for, on account of their
constant motion and of the wearing away of the parts, a greater
factor of safety must be introduced, which requires us to assume a
smaller value for T.
If we substitute these values in the formulas
Pl = b h
T
6
T
T
and Plπ p³
= πα
π d³
4
32
for parallelopipedical and for cylindrical girders, we obtain the fol-
lowing practical formulas :
For wood
= đ³
Pl=167 bh² = 785 r³ 98 d' inch-pounds.
For cast iron
№
Pl=1167 bh² 5500 687 inch-pounds.
p³
And for wrought iron the greatest value
203
Pl 1500 b h 7070 884 d' inch-pounds.
=
If with Morin, and in accordance with the practice in England,
we put for cast iron
K K
instead of T,
to = 750 kilograms,
4
5
K
instead of T
600 kilograms,
5
and for wrought iron
we obtain for cast iron
=
Pl=1778 b h = 8376r 1047 d' inch-pounds,
and for wrought iron the smaller value.
P l = 1422 b h² = 6700 7³ = 838 d³ inch-pounds.
If the load Q is not applied at the end of the beam, but is
468
[$ 240.
GENERAL PRINCIPLES OF MECHANICS.
equally distributed over the same, the arm of the load is no longer
l, but
2'
must put
and consequently, the moment being but half as great, we
??
2
W T
W T
or Q l = 2.
;
e
e
If the girder is supported at both ends (Fig. 337) and the load
P acts midway between the two points of support, whose distance
from each other is = 1, the force at each end is =
P
2
, its arm is
and its moment
2
ΡΙ
4
W T
W T
and P l = 4
e
е
Therefore, under the same circumstances, the girder bears twice
as great a load in the second and four times as great a one in the
third as in the first case.
If, finally, a girder uniformly loaded along its whole length is
supported at both ends, it is in the first place bent upwards by a
and in the second place downwards by a
Q
force whose arm is
2
2
2'
force whose point of application is the centre of gravity of one
}
of the halves of the load, whose lever arm is therefore and whose
4
moment is
27. Consequently the moment with which one end of
ρι
8
the girder is bent upwards is
ρι ρι
ρι
4
8
8
W T
bence we have Q7 = 8
The proof load of the girder is in
e
this case 8 times as great as in the first one.
For a parallelopipedical girder we have in the first case
T
Pl=b h²
in the second
"
6
T
Q l = Q b h²
in the third
6
T
Pl=4bh²
and in the fourth
6
T
Q l = 8b h
2
6'
b denoting the width and h the height of the rectangular cross-section.
$241.]
469
ELASTICITY AND STRENGTH OF FLEXURE, ETC.
EXAMPLE-1) What load can a girder of fir carry at its middle, when
its width is b = 7 and its height h 9 inches, and when the point of ap-
plication of the load is 10 feet distant from the supports? Here we have
≥ 7 = 10. 12 = 120 inches, and therefore, according to the above formula,
Pl4.
4 . 167 b h² = 4 . 167 . 7 . 81,
and the required working load is
4676.81
P =
240
58,45. 27
1578 pounds.
2) A cylindrical stick of wood, firmly imbedded at one end in masonry,
is required to bear a weight Q 10000, uniformly distributed over its
whole length l = 5 feet; what should be its diameter? We have here
про т
ρι 2
and consequently by inversion
4
3 Q7 3 10000. 60
р
V
1570
1570
2.785. r³,
3
√382 = 7,26 inches,
and the required diameter is = 2 r = 14,52 inches.
§ 241. Relative Deflection.-The bending of the moving
parts of machines, such as the shafts, axles, etc., has often a very
FIG. 385.
U
B
-X₂
H
H3
Ꭲ
H₂
H₁
A
-X
K
K
P
D
M
Ms
AM₂
M
1
Mo
bad effect upon their
working, either by giv-
ing rise to vibrations
and concussions, or by
preventing the different
parts of the machine
from engaging perfect-
ly. We are therefore
in certain cases re-
quired to determine the
cross-sections of these
parts of machines, not
with reference to the
modulus of proof
strength, but to the
deflection, by assum-
ing the deflection to
be a very small definite
portion of the entire
length of the body or
part of the machine.

We have already found (§ 217) the deflection for a prismatic
body A S B, Fig. 385, fixed at one end B and loaded at the other
A, to be
470
[§ 241.
GENERAL PRINCIPLES OF MECHANICS.
B C = a
P 1³
3 WE'
and we can put its ratio to the length A B, which is given
a
0 =
7
P 12
3 WE'
whence, by inversion,
P ľ² = 30 W E.
Hence we have for a parallelopipedical girder
b h³
3
O b h³ E
3
P ľ² = 30
E =
12
4
and for a cylindrical one
π poz
3
P ľ² = 30
E =
π O p¹ E.
4
4
α
Generally a relative deflection ℗
1
00
50 is admissible, and
we can put
1
3 πT
1) Pr² =
b h³ E =
p¹ E.
2000
2000
If we substitute for wood the modulus of elasticity E = 1600000,
we obtain
Pl2 =
= 800 b h³ 7540 2¹.
For cast iron we have E = 15000000 pounds, and therefore
P 12
7500 b h³ 70700 r',
and for wrought iron E 22000000 pounds and
3
P ľ² = 11000 b h³ = 103700 r¹.
On the contrary, when the deflection reaches the limit of elas-
ticity, we have (§ 235)
W 3
W TI
2) Pl=
or Pr² =
Pl
e
е
and, therefore, equating the two values of Pl, we obtain
W TI
e
= 3 0 W E,
and consequently the ratio of the length of the beam to the maxi-
mum distance e, when both the deflection and strain reach at the
same time their limit values
and T, is
30
}
30 E
e
T
σ
hence for parallelopipedical bodies
1
h
10/09
A
§ 241.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 471
and for cylindrical ones.
Z
30 Z
or
T
J
d
o denoting the extension or compression at the limit of elasticity
corresponding to the strain T.
30
If < we obtain from the first formula the greater value
σ
30
e
Z
for P and if, on the contrary,
> the second formula gives
с
σ
the greater value for the moment of the force. Therefore for a
given moment of force (P 7) the greater dimensions for the cross-
section are given in the first case, where the length of the body is
less than 7
e, by the formula
(30) e, by
W T
= P !
e
and in the second case, where 7 >
σ
(20)
30
e, by the formula
30 WE = Pl.
If we substitute in the ratio
7
30
1
for the limit, 0 =
σ
500'
3
0,006
we have for all materials
С 500 σ
, and, therefore, for
σ
wood, for which σ =
1 7
600' e
ticularly for a prismatical beam of this material
= 0,006 . 600 =
3,6, and more par-
and
} 18
d
.10
1,8.
1
If we assume for cast and wrought iron σ =
we obtain for
1500'
these substances
7
3. 1500
= 9 and therefore
e
500
7
9
or
= 4,5.
h
d
The formula
b h³
3
3 π r* E
Pr² =
E:
2000
2000
is of course applicable to the normal case above, I.E., when the body
is loaded at one end and fixed at the other. For a load equally
distributed we must substitute (according to § 223), instead of
P, 3 Q. If the body is supported at both ends and the load is sus-
472
[§ 242.
GENERAL PRINCIPLES OF MECHANICS.
pended in the middle, we have, instead of P,
and therefore
P
2
7
and, instead of 1,2,
P ľ² = 8.
b h³
2000
3 π r* E
E = 8.
2000
If the girder is supported in the same manner and the load
uniformly distributed, we must substitute for P,
5 Q
8
1
5001
, if its
Example—1) What load placed upon the centre of a wooden beam,
produce a relative deflection 0 =
9 inches and the distance between the sup-
supported at both ends, will
width is b 7, its height h
ports is 7 = 20 feet? Here we have
P8.
800 b h³
7
6400.7.93
(20.12)²
7.92 = 567 pounds,
1578 pounds.
while in the foregoing paragraph, under the assumption that the beam
should be bent to the limit of elasticity, we found P
2) How high and wide must we make a cast iron girder (the ratio of
its dimensions being
sustain a load Q
h
b
4), which, when supported at both ends, will
4000 pounds, uniformly distributed over its length,
which is 8 feet? Under the latter supposition, we have
// Q 12
8.7500 b h³,
7 +
1.E.,
8
consequently
§. 4000. 82. 12² = 8. 7500 or h¹ = 4¹. 6,
4
h
4 √6
=
1,565.4
6,26 inches and
h
b
1,565 inches.
4
According to the formulas of the foregoing paragraph, we would have
whence the required height is
ρι
8. 1167 b h2, or 4000 . 8 . 12 = 8. 1167
•
4'
h = 4
4 1
3 3000
1187
4. 1,37
=
5,48 inches,
and the required width
h
1,37 inches.
§ 242. Moments of Proof Load. From the formula
T
Pl = b h²
6
for the moment of the proof load of a parallelopipedical girder we
perceive that this moment increases with the width b and with the
square of the height h, that the proof load itself
§ 242.] ELASTICITY AND STRENGTH OF FLEXURE, ETC.
473
P =
b h² T
7 6
is inversely proportioned to the length (1) and that the height has a
much greater influence than the width upon the sclidity of such a
girder. A girder, whose width is double that of another, will bear
but twice as great a load as the latter, or as much as two such
girders placed side by side. A girder, whose height is double that
of another, bears, on the contrary, (2) = 4 times as much as the
latter, when their widths are the same. For this reason we make
the height of parallelopipedical girders greater than their width,
or we place them on edge, or in such a position, that the smaller
dimension shall be perpendicular to the direction of force P and
that the greater dimension shall be parallel to it.
Since b h expresses the cross-section F of the beam, we have also
T
Pl = F h
;
6
hence the moments of the proof load of bodies of equal cross-section,
mass or weight are proportional to their height. If, for example,
b and h are the width and height of one body and and 3 h those
of another body or F =
b
3
b
3 h = b h the area of both their cross-
sections, the bodies have the same weight, when the other circum-
stances are the same, but the latter bears three times as great a
load as the former.
If b h, the cross-section of the beam is a square, and we can
diminish the moment of proof load by placing the diagonal in a
vertical position. In this case, IV, as we know from § 230, remains
unchanged and is =
b h³
b
12
12'
while e becomes equal to the semi-
Therefore we have
Pl=
12 b √ !
T
T = b³
6
T
= 0,707 b³
6'
diagonal b√2 b.
Q =
b+
while, if it were laid on one of its sides, we would have P 1 = b³
See § 236.
T
6
The equations for parallelopipedical girders are analogous to
those for girders with an elliptical cross-section. We have in the
latter case (according to § 231) W =
по 19:3
4
and ea, the semi-
axis a being supposed parallel and the semi-axis b perpendicular to
A
474
[$ 240.
GENERAL PRINCIPLES OF MECHANICS.
the direction of the force or, as is generally the case, horizontal.
Here we have for such a girder
Pl
π b a²
4
T
T=
Fa
the area of the elliptical cross-section being Fπ a b. The mo-
ment of the proof load of this beam increases, therefore, with the
area and with the height a of the cross-section.
If b
= a = r, we have a cylindrical girder, whose radius is r,
and the equation becomes
p³
Pl=
"T² T = Fr
4
T
4
The moment of proof load of this body increases, therefore, with
the product of the area of the cross-section and its radius.
If the cross-sections or weights are equal, the ratio of the mo-
ment of proof load of a body with an elliptical cross-section to that
of one with a circular cross-section is Therefore, we should
a
グ
​always prefer the elliptical to the cylindrical girder.
This holds good for all other forms of cross-section; the regu-
lar form (the square, the regular hexagon, the circle, etc.) has
always, for the same area, a smaller moment of proof load than a
form of greater height and less width.
Regular forms of cross-section should, therefore, be employed
only for shafts and other bodies, revolving about their longitudinal
axis, in which case during the rotation a continual change in the
position of the dimension of the cross-section takes place, I.E., after
one-quarter of a rotation the height becomes the width and the
width the height.
§ 243. Cross-section of Wooden Girders.—If a cylindri-
cal girder has the same cross-section F = π jo² b² as a parallelo-
pipedical beam, whose height and width is = b, we have the ratio
Ъ
NπT
=
1,77245,
go
and, on the contrary, the ratio between the moments of proof load
M and M, (M) is in the first place, when the latter body is laid
upon one of its sides,
M
j
b
3 r
M₁
4 6 26
3
= 1,5. 0,5642 0,8462,
2 νπ
and in the second place, when its diagonal is placed in a vertical
position,
§ 243.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 475
M
r
b √2
3
3. 0,3989
M2
4
12
√2 π
1,1967.
The moment of proof load of the cylinder (with circular base)
is in the first place smaller and in the second place greater than
that of a parallelopipedon with a square base.
Since wooden parallelopipedical girders are cut or sawed from
the round trunks of trees, the question arises, what must be the
ratio of the dimensions of the cross-section of such a beam, in order
that it shall have the greatest moment of working load?
Let A B D E, Fig. 386, be the cross-section of the trunk of the
tree, A D d its diameter and
FIG. 386.
A
B
E
Μ
C
N
=
the breadth and
A B D E = b

A EBD = h
the height of the beam; then we have
b² + h²
d³
d², or
- b²,
and the moment of proof load is
T
T
Pl=
b h
b (ď² — b²).
6
6
The problem now is to make
b (d² — b²)
as great as possible. If we put, instead of b, b ± x, x being very
small, we obtain for the last expression
(b ± x) ‹ — (b±x)' = b d — b³± (d — 3 b) x 3 b x²,
when a is neglected. Now the difference of the two expressions is
a³
y = = (ď² — 3 b') x + 3 b x².
In order that the first value shall always be greater than the
second, the difference
y = = (d² · - 3 b²) x + 3 b x²
must be positive, whether we increase or diminish 6 by r. But
this is only possible when d² - 3 6 0; for this difference is then
² =
= 3 ba² or positive, while, on the contrary, when dª 36 has a
real positive or negative value, 3b 2² can be neglected, and the sign
of the difference (d 3 b) a varies with that of x. Therefore,
‡
w
putting d² - 3 b = 0, we obtain the required width
b = d V, and the corresponding height
v
h = √ d² — b² = d √ √ ;
the ratio of the height to the width is
,
476
[$ 243.
GENERAL PRINCIPLES OF MECHANICS.
h
√2
1,414 or about 3.
Ъ
11
A
FIG. 387.
M
B
We should, therefore, cut the trunk of the tree in such a man-
ner as to produce a beam, whose height is to its
width as 7 is to 5. In order to find the cross-
section corresponding to the greatest strength,
we divide the diameter A D, Fig. 387, into three
equal parts, erect in the points of division M and
N the perpendiculars M B and N E and join
the points B and E, where they cut the circum-
ference, with the extremities A and D by straight
lines. ABDE is the cross-section of greatest resistance; for we
have

E
N
D
=
AM: A B A B: AD and AN: AEAE: AD,
and consequently
A B = b
=bVA M. AD
WAM. = 1
√ } d. d = d √ } and
AE= h = √AN. ADN d. d = dv, or
v ş
=
3
h
√2
which is what was required.
b
1 '
REMARK 1. The moment of proof load of the trunk of the tree is
π Τ
Pl=
4
703
>
and that of the beam of greatest resistance, cut from the same, is
Τ
8 T
T
6
Pl= d √ I . & d²
d³ =
√243
✓ 243
1
8 4
√ 243 π
1 0,65
=
0,35,
and consequently the beam loses by being cut
1.E. 35 per cent. of its proof strength. In order to reduce this loss, the
beam is often made imperfectly four-sided, I.E. with the corners wanting.
The moment of the proof load of a beam with a square cross-section, hewed
from a tree of the same size is
T
d2
Pl
d
•
6
2
since the width is
height
=d
d √½ = 0,707 d; the loss is
8
4
8
1
1
1 0,60
=
6.2 √2 π
0, 40,
3 π √2
1.E. 40 per cent.
(REMARK 2.) In order to cut from a trunk of a tree a parallelopipedical
beam, whose moment of flexure is a minimum, or for which 0 = is
small as possible (compare § 241), we must have
a
is as
§ 244.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 477
b h³
W = or b h³ = h³ √√d² — h², or (b h³)2 = h® (d² — h³)
12
= d2 ho h8
as great as possible. The first differential coefficient of the latter expres-
sion in reference to b is 6 d2 h5 - 8h7, which is equal to zero for h²
d²
I.E. for
=
♦ d?,
h = d √ &
૭
d √3
2
and
d
b
= Vd2
Vd² — h²
2
For these values the moment of flexure of the beam is a minimum (see
Introduction to the Calculus, Art. 13).
h
√3
Here we have
1
1,7321, or about 1, while above we found for
h
the maximum of the moment of proof load 굼​.
b
This condition corresponds to the construction in Fig. 387, when we
make A M = D N = † A D.
§ 244. Hollow and Webbed Girders.-We have, accord-
ing to § 228, for a hollow parallelopipedical beam
W =
b h³ — b₁ hr³
12
3
and therefore the moment of proof load is
W T
W T
ΡΙ
'b h³
パー
​b₁ h₁³ T
е
} h
h
If we put
b₁
μ and
h
=v, we obtain
6
h
Ъ
b h³ - b₁ h
3
= b h²
h
b h² (1
3
µ³ v),
and, since the cross-section of the body is
F = b h b₁ h₁ = b h (1 μ v).
Τ
— µ
1
με ν
Pl=
•
1
μυ
Fh.
6
1 με ν
1
Since
μν + μν
3
μν
1 +
1
μυ
1
-
μ. 2
(1 – μ) μ ν
1
increases with v, we obtain the maximum value of P 1 for v
and it is
2
1) P l = [(1 + ( — — " ") μ ] Fh Z = (1 + µ + µ²) Fh
1
T
6
If, on the contrary, we put µ = v, we obtain
μυ
1,
T
6'
T
2) P l = (1 + µ²) F h
6'
478
[S 244.
GENERAL PRINCIPLES OF MECHANICS.
1
In both cases we must make u as great as possible, and there-
fore nearly 1. If we wish the proof strength of the girder to be
a maximum, we must make the webs as thin as possible. Hence
we have for µ = 1 in the first case
T
T
Pl 3 Fh
ΡΙ =
= Fh
and in the second case
6
2'
T
6
, and, on the contrary, forµ = 0,
Pl2Fh = Fh
P l = F h Z
T
6
In all three cases the proof load of the girder, when the cross-
section (F) or the weight is the same, increases with the height
(h); but in the first case, where the girder consists of two flanges,
it is a maximum; and in the second case, where it forms a paral-
lelopipedical tube, it has a mean value; and in the third case,
where it is composed of one or two webs, a minimum one.
If, for example, a massive girder, whose dimensions are b, and
,, has the same cross-section or weight as the supposed tubular
girder, we have
F = b₁ h₁ = b h
b₁ h₁, I.E. 2 b, h, bhor
=
b, h₁
bh
= μν= !.
1
h₁
If we assume
we have µ = v =
√, and therefore the
b
h'
3
ratio of the proof loads of the two beams is
P (1 µ³ v) h
√2 = 3√2 = 3 4 = 2,12;
1
P₁
1
1 μν 么
​the tubular girder is therefore capable of carrying more than double
the load that an cqually heavy massive girder can, whose form is
that of the hollow of the first girder.
The same relations also obtain for I-shaped girders, since (see
§ 228) the measure of the moment of flexure W is the same for
both. These formulas can also be employed for bodies with more
than two webs, as, E.G., bodies with the cross-section represented in
Fig. 388, in which case b denotes the width of the
FIG. 388.
upper and lower rib, h the entire height A D = B C,
b₁ the sum of the widths and h, the height of the
hollow spaces M, N, O, P.

A
D
B
1
The formulas for a pipe or hollow cylinder are
analogous to those for a parallelopipedical beam. If r
is the exterior and r, pr the interior radius, the
moment of proof load of this body is
=
244.]
479
ELASTICITY AND STRENGTH OF FLEXURE, ETC.
r₁₁) T
4
π
(208
Pl=
r
T
= (1 + µ¿²) Fr.
4°
T
=
(1 — µ³) π p³
4
= ( = 1) Fr
T
4
This value increases as μµ = approaches unity, and therefore
r 1
g
as the wall of the pipe becomes thinner. If we put µ = 1, we ob-
tain the corresponding maximum moment of proof load
T
T
ΡΙ = 2 Fr
44
= Fr
2*
If we compare the proof load of this tube with that of a massive
iron cylinder, whose radius r₁ = µ r = r √!!, we have then for the
latter
1
P₁l = Fr₂
T
4
T
= μ Fr
and
4
P
1 + fe
P₁
3
= (1 + !) √2 = 2 √2 = 2,12,
1/2
μ
exactly what we found under the same suppositions for parallelo-
pipedical girders.
We can see from the general equation
WT
С
2
2
.
2
(F₁ z,² + F₂ z²² + · ·) T = (F₁ µ²² + F₂ µ₂² + …. ) e T,
е
..
Pl=
that the moment of proof load of a body increases as the distances
Z₁ = µ₁ C, Z₂ = μ, e, etc., of the portions F, F, etc., of the cross-sec-
tion from the neutral axis become greater. But since this distance
can at most be e, those girders will have the greatest moment
of proof load, the different portions of whose cross-section are at
one and the same distance (the maximum one) from the neutral
axis. Such a beam consists of two flanges. Since the webs which
unite the two flanges cannot satisfy the conditions of maximum
moment of proof load, it is impossible to attain this maximum, and
we must therefore content ourselves with increasing the proof
strength of the girder by hollowing it out, by thinning it in the
neighborhood of the neutral axis, or by adding flanges at the
greatest possible distance from the same axis.
The thickness, which the web must possess in order to resist the
shearing strain, will be determined in the following chapter.
REMARK.—Under the supposition that the proof strength increases and
decreases with the ultimate strength, the English engineers increase the
size of that portion of cast-iron girders, which is subject to compression;
for that material resists compression best. On the contrary, they increase
the dimensions of the compressed side of girders of wrought iron, as the
480
[$ 245.
GENERAL PRINCIPLES OF MECHANICS.
latter resists extension best. If the girders are to be supported at both
ends, their form must depend upon the substance of which they are made.
If the beam is of cast iron, we make the bottom flange larger than the
other; if of wrought iron, the upper flange, or the upper part of the
girder is constructed of two flanges, united by vertical webs, as is repre-
sented in Fig. 388. The forms T and T, discussed in a previous paragraplı
(§ 237), are employed for cast iron.
EXAMPLE.—An oak girder 9 inches wide and 11 inches high, which has
up to the present time shown sufficient strength, is to be replaced by a
cast-iron girder, whose exterior width is 5 inches and whose height is 10
inches; how thick should it be made? If we put the double thickness of
the metal =x, the width of the hollow is 5 2, and its height is
= 10
x, and consequently we have for the hollow girder
b, hg3 = 5.103 (5 — x) (10 — x)³ = 2500 x
2 2
1
If the moment of proof load of the massive wooden beam is Pl
b ₁ h ₁
3
-450 x2 + 35 x³ — x',
1
hence the moment of proof load is Pl
7000
6.10
(2500 x 450 x² + 35 x³ —x¹).
1000
6
B
35 x³
x*)
1089000, or
x1
1556.
=
1556
2500
= 0,62, for which, how-
9 . 11² = † . 1089000, we must put
700. (2500 x 450 x
2500 x 450 x2 + 35 x³
In the first place, a is approximatively
ever, a = 0,65 should be put.
=
190,12, 35 x³ = 9,61, x¹
X
From this we obtain 450 a² = 450 . 0,4225
0,18, and finally
1556 + 190,12 — 7,56 + 0,18 1738,7
2500
2500
and consequently the required thickness of metal is
x
0,3475 inches.
2
0,695 inches,
§ 245. Excentric Load-If the force which acts upon a
FIG. 389.
P
1
A
ID
P
H
D
E
N
B
P2
B
girder supported at both ends
A and B, Fig. 389, is not applied
at the centre, but at some inter-
mediate point, situated at the
distances DA = l, and D B =
7 from the points of support, the
proof load is greater than when
the force is applied in the mid-
dle. Let us denote the forces,
with which the points of support
A and B react, by P, and P,
and the entire length of the gir
der A B = ↳₁ + lą by l. Now,
if we put the moment of P, in


§ 245.]
ELASTICITY AND STRENGTH OF FLEXURE, ETC.
481
reference to the point of support B equal to that of P in reference
to the same point and in like manner the moment of P, in refer-
ence to A equal to that of P or P₁ l = P l, and P½ l
P l₁, we
obtain the reactions at the points of support
い
​P₁ = '}; P and P, = {}; P,
=
and consequently their moments in reference to the points of
application
P₁ l₁ = P₂ l₂ =
2
Pl₁ l₂
For any other point E, whose distance B E from the point of
support B is = x, we have this moment
P₂. B E =
Pl, x
7
smaller than that just found, and consequently at B we have the
greatest deflection, and therefore we must determine the proof load
in reference to this point alone, for which we have
P l₁ l½
IT
e
し
​If we substitute 1,
x and l
2
201
+x, we obtain the
moment of the force
P
Pl, la
(x) (x + x)
P
(
222)
hence the proof load is
W T
I W T
P
7, 1,
e
( - x²) e
and therefore greater or less as x is greater or less. For x =
=
}
2'
I.E., for 1, 0, in which case P is transferred to the point of sup-
port 4, we have
P=
I W T
0.0
= ∞09
and on the contrary for x = 0, I.E. if the force P is applied at the
centre, the proof load is a minimum and is
P = 4
WT
le
as we know already from § 240. A prismatical girder supported
at both ends will sustain the smallest load, when the latter is ap-
plied at the centre, and more and more as the weight approaches
the points of support.
If we lay off as ordinates the moments of the force, which are
482
[$ 245.
GENERAL PRINCIPLES OF MECHANICS.
inversely proportional to the radius of curvature and directly to the
curvature itself, as ordinates at the different points upon the girder,
we obtain a clear representation of the variation of the deflection
at the different points upon the girder.
If, in the case just discussed, the moment of the force
Pl, l₂
in
し
​D is represented by the ordinate D L and if from its extremity L
the right lines L A and L B be drawn to the extremities of the
abscissas D A = l, and D B = l, these lines will limit the differ-
ent ordinates (as for example E N) representing the measures of
the deflection for the different portions of the body; for since
EN DL
EB D B'
it follows that
EN=
E B
DB
х
Pl₁ l
Pl₁₂ x
DL =
as we had previously found.
FIG. 390.
E
E
1
G
R
])
K
F
II
B
P₂
2
Another case which
often occurs in practice.
is, when the weight is
equally distributed over
a portion EF = c of
the entire length l of
the girder A B, Fig.
390. Let us again de-
note the distances of
the middle D of this
weight from the points
of support A and B by
l and l, and the reac-
tion of the abutments
by P₁ and P, then we
have again


1
L
P₁ = 4; Q = 4
P₁ = } Q =
0
c q
b c q
If Q were not distributed, but if, on the
plied at D, the moment for D would be
and
contrary, the force was ap-
Q4, and, representing
l
$ 245.]
ELASTICITY AND STRENGTH OF FLEXURE, ETC. 489
the same by an ordinate D L, the moment for the other points of
A B will be cut off by the right lines L A and L B. But, since
for the points within E F the forces P, and P, act in opposition
to the weight placed upon it, the ordinates between E G and FH
will be diminished. For the centre D of the loaded portion E F
the moment of half the weight
Q
C
M L
2 4
must be subtracted, and there remains, therefore, of the ordinate
DL
Q
2 4 22
1
only the portion
1, 1, C
8
D M = D L - M L = Q ( ¹ ? — § ).
p
For another point N, whose abscissa is A N, the moment is, on
the contrary,
NE
2
P..NA-NE. q. = P₁ x
( x − 1 + 1/2 c )² q
2
( x − l₁ + b c ) ³ q
and if P, is represented by the ordinate N R and
2
by the portion SR, the ordinate N S will give the total moment
( x − l. + ½ c) ¨ q
P₁ x
2
This is of course very different for different values of r, I.E. for dif-
ferent points, but is a maximum for x − 1 + ¦ c
{ =
its value is
P
P₁
1
and then
q
P*
29
=
P.
(j + h − 4 c)
P₁ -호이
​(1
-
Q
P(+hic) - P
P₁
C
c la
= P(1, − 1 + 1/4) = P, 1,
2 27
2 զ
4) = 244 (1-5).
Hence we must put the proof load of this girder
42
C
Q 1, 4, (1 - 27 ) =
7
W T
e
21
EXAMPLE.—What weight will a hollow parallelopipedical girder, made
of inch thick sheet iron, support, if its exterior height is 16 inches and its
18
exterior width is 4 inches, when it is loaded uniformly along 5 feet of its
length, the middle of the loaded portion being 8 and 4 feet distant from
the points of support? Here we have
b h³ — b₁ h₁³ 4. 16³
h
1
3.159
391,2
16
and
484
[§ 246.
GENERAL PRINCIPLES OF MECHANICS.
4 12½ (1 - 217) -
с
•
l
2
3
48 (1-34)
5
32. 19
76
24
24
3'
and the weight required is therefore
3 T
Q
391,2.
76 6
195,6
76
•
9000 =23160 pounds.
REMARK.-If the weight Q is not uniformly distributed over E F, but
if half is applied at the extremity E and half at the extremity F, the line
G M His then a right line, and the maximum moment is the ordinate
GE, for which
Q l 2
Z
WT
е
2
1, denoting the greater distance D A and l₂ the smaller distance D B of
the middle D from the two extremities A and B.
§ 246. Girders Fixed at Both Ends.-If a beam A B,
Fig. 391, is loaded in the centre C and fixed at both ends, it will be
P
FIG. 391.
- 1 P

Data Bu
R
A
M
1
નોન
B
D
E
C
P
1
VP

L
A
M:
B
D
C
E
N
H
K
curved upwards at the centre, and at the points of support A and
B downwards, and there will be formed at the centres D and E of
the semi-girders CA and C B points of inflection, where there is
no curvature or where the radius of curvature is infinitely great.
One-half of the weight P is supported by AD and the other half
by B E, and we can therefore assume that both the quarters A D
and B E of the beam are bent downwards at their ends D and E by
1
P
2
and that, on the contrary, the half D E of the girder is bent
upwards at its ends D and E by (
P
The arm of each of these
246.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 485
A B
1
forces A D = CD, etc., is =
ment is
4
4
; consequently their mo-
Pl
ΡΙ
and therefore
2 4
8
ΡΙ
W T
;
hence we can put the proof load
8
e
8 II T
4 Π΄ Τ
P =
= 2.
le
le
Such a girder will bear twice as great a load as when it is
simply supported at both ends.
ΡΙ
CL =
and
8
If we make the ordinates A H = B K = C L
draw the right lines H L and K L, they will cut off ordinates
(M N) for every other point (M) upon the beam proportional to
the moments of the force and to the deflection.
If in the formula, which we have found, we substitute the modu-
lus of rupture K instead of the modulus of proof strength T, we
obtain, of course, the force necessary to break the beam, which is
P
8 WK
le
Since the curvature is the same in A, B and C, the rupture will
take place at the same time in A, B and C.
=
If the position of the girder is the same and the load Q1q
is uniformly distributed, the girder assumes, it is true, two curva-
tures upwards and two downwards, but the points of inflection
H
20
A
D
FIG. 392.
- R
- R

B-
E
q 4
૧ ૧
R
L

A
D
E
B
C
F
K
D and E, Fig. 392, do not lie at the centres of the semi-girders;
for the deflecting forces R, R of the portions A D and B E are
486
GENERAL PRINCIPLES OF MECHANICS.
་་
[S 246.
aided by the weight upon the latter, and, on the contrary, the
action of the bending forces R, R of the central piece D is
diminished by this load. Let us put the length A D = B E い​っ
​the length C D = CE = 1, and the total length of the beam 7 =
2 (l + l), and let us denote the weight upon A D or BE by
Q₁ =ql, and that upon D E by Q2
= 2 R 2 g 12.
12. Now,
Now, since
A D is bent downwards by R and Q, we have, according to § 216
and § 223, the angle of inclination to the horizon E D T = DET
=a at the point of inflection D
CD
a
R 1,2 Q₁ 7, 2
+
1
2 WE 6 WE'
ང་
and since C D is bent upwards by (R) and downwards by Q,
we have for the same position D also
R 122
α =
2 W E
Q₂ 12
6 WE'
Equating the two values of a,
we obtain the relation
3 R (12 — 1,²) = Q₁ 4² + ·Q: 12, or
3 q l (1² — 1,2) = q (l² + 1"), I.E.,
1
3
3
3
31 [1-(-1)=4(-1).
C
12
Resolving this equation, we obtain
L
12
l
1₁ = 1/2 √ 4 and 1₁ = (1
?
—
2
and, therefore, the moment of force in relation to the middle C is
M = R l₂
2
R l R l
2
9 72
2
24
and that in reference to the extremity A or B is
Q1
24'
ינ.י.
A
D
FIG. 393.
- R
- R

B
E
MAMMONIU
༥ ༥
R
R

L
A
D
E
B
C
H
K
$247.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 487
M₁ = Rh₁ +
Ꭱ
Q₁₁
2
=gh l₂ +
2
9 7, 2
= q
9 h
“
2
1322
972
= 11² (1
8
で
​q t² (1
8
-
√ ! ) (1 + √ ! )
ρι ρι
The proof load of this beam is therefore
2
12
24
W T
3
8 WT
= 12.
le
2
le
I.E., 3 times as great as in the former case, where the weight acted
at the centre C
If we lay off
οι
12
as ordinate in A and B and also
as ordi-
24
nate in C, making A H = B K
Q 7
12
and CL
Q I
we ob-
24
tain three points H, K and L of the curve HD LEK, which
represents the variation of the deflection of the girder.
EXAMPLE.-How high can grain be piled in a grain house, when the
floor rests on beams 25 feet long, 10 inches wide and 12 inches high, if the
distance between two beams is 3 feet and if a cubic foot of corn weighs
46,7 pounds? If we employ the last formula Q7 = 12. 167 . b h², we
must put
b = 10, h = 12, l
Q
= 25. 12 = 300, and consequently
12. 167. 10. 144
=9619 pounds.
300
Now a parallelopipedical mass of grain 25 feet long, 3 feet wide and
z feet high weighs 25. 3. x. 46,7 pounds; if we substitute this value for
Q, we obtain the required height of the mass
X
9619
75 . 46,7
2,75 feet.
§ 247. Beams Dissimilarly Supported.-If a beam 4 B C,
Fig. 394, is fixed at one end A and supported at the other B and if
the load acts in the middle between A and B, we have, according
to § 221, the reaction of the support B
5
P₁
P;
16
and therefore the moment of the force in reference to C
Pl
5
CL=
Pl,
2
32
and, on the contrary, that in reference to A is
488
[§ 247.
GENERAL PRINCIPLES OF MECHANICS.
5
3
6
A H = P
P₁l=
= Pl
Pl=
Pl,
2
16/
16
32
FIG. 394.
AR TORMIDAD 2870
MA
Pradini ma
B
or greater, and consequent-
ly we can put the proof

load
16 W T
P
3
le
DI
N
YP
LAMARUAR MUAN
A
B
DMC
For an intermediate point
M, at a distance C M = x
from the centre C, this mo-
ment is
M N = P₁ (; + x)
P x = P₁

2
- (P – P₁) x.
H
P₁l
5
5
If we assume =
P - P₁
16 - 5 2
22
P₁
O
l, we obtain
that point, for which the moment is equal to zero and the radius
of curvature infinitely great. The variation of this moment and
the deflection of the girder are represented by the ordinates of the
right lines H L and L B, passing through the extremities of A H
6
Pl and of C L
32
5
32
Pl.
If, finally, a girder A B, Fig. 395, supported in the same man-
APAMTI FO
FIG. 395.
M
A
D:
N C E
TORMIND ONLA
A
D
S
L
K
R CE
B
AP₁
1
ner as the last, is uniformly
loaded, as we have previous-
ly generally supposed, with
a certain weight q upon the
running foot of the girder,
we can determine the reac-
tion P₁ at the support В in
the following manner. If
the length of the beam is 1,
the entire load is Q l g
and the moment of the force
in reference to a point M,
at a distance B M = x from
the point of support B, is


H
$248.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 489
RS = P₁x
Ꮪ
q x²
2
and consequently the angle of inclination
a
P₁ (1² — x³)
2 WE
q (l³ — x³)
6 WE
and (according to § 217 and § 223) the corresponding deflection is
y = M N =
P₁ (l² x − 1 x³)
2 WE
q (1³ x — ¦ x¹)
6 WE
4
But since A lies on the same level with B, the ordinate in A,
I. E. for x = 7, is y
0, and we must put
=
3 P₁. 3 l = q. 31,
from which we obtain the reaction at B
Р
P₁ = & ql = 3 Q.
If we substitute this value for P, in the expression for the mo-
inent, we obtain
RS =
‹ Q x
Չ
q x²
Q X
2
2
(
1 − x); and therefore for x = = 1
For x B D =
3 l it is a maximum
AH=
37 this moment is = 0, and for a = B E =
q r²
8
Q 1
8
EK=
q
9 a l²
9
Q7.
128
128
9
Since
οι 16
8 128
ρι> Q1, the moment A H in reference
128
to the fixed point A is greater than the moment K E in reference
to the middle E of B D, and the proof load corresponding to the
οι
8
moment must therefore be determined, I.E. we must put
Q = 8
W T
le
in which case we assume, of course, that the modulus of proof
strength for extension is the same as that for compression.
This proof load is 8. 13 3 times as great as it would be if
the weight were concentrated in the middle.
§ 248. Girders Loaded at Intermediate Points.—If a
girder A B, Fig. 396, loaded at both ends with equal weights P, P,
490
[$ 248.
GENERAL PRINCIPLES OF MECHANICS.
FIG. 396.
P
-P
S
A
IC
M
L
B
is supported at two points C
and D, which are at the same
distance A C = B D
BD = l, from
the ends, the reaction of each
of these points of support is
equal to the force P, and for a
point Mupon CD the moment
of flexure
CL = DOM N
P(x1)- P x₁ = — Pl₁


A
C
B
M
D
=
(x₁
1
-
is constant, and the form of
neutral axis of CD is therefore a circle, while, on the contrary, for
a point U upon A C this moment UV = P x is variable and
smaller than Pl.
WE
The radius of curvature of the middle piece C D is = r = Pl
and the angle of inclination of the axis of the beam in C and D is
7
Pl
2 r 2 WE'
7 denoting the length of this
consequently a₁ =
middle piece. From this we obtain the deflection
as well as the deflection of CA
(47) 2
73
MS = a =
2 r
8 r
PP L₁
8 WE
Pl³
a₁ = a₁ l₁ +
3 WE
Ριζ
2 WE
+
Pl³
3 WE
3
Pl² / l
WE 2
2
(
+
13)
W T
1
с
The moment of proof load for this girder is Pl,
If the same beam A B is uniformly loaded, as is shown in Fig.
FIG. 397.
A
UC U
M
DE
397, with 7 per running
foot, under certain circum-
stances the moment of
flexure for some points is
B positive, and for others
negative, and therefore at
two points U and V it is
equal to zero.


R
M
B
C
V D
ΤΩ
For a point upon A C
and B D this moment is
1
qa², and, on the con-
trary, for a point between
Cand the middle M, or between D and M, since the value of the
reaction at C and D is Q (7+) q, it is RS= y=!
§ 249.] ELASTICITY AND STRENGTH OF FLEXURE, ETC.
491 .
2
(2x + 7)² q − ( 7 + 7₁) x q = √(x² - 1 x + 1) q, and therefore
=0 for x²
2
1 x =
l₁², I.E. for
2
CU = x =
2
√ (')* — 1,² and for
7
C V = x =
2
+ √()
ぴっ
​2
7
1,
which of course requires that ↳, = <, I.E. C A < C' M. Under
any other circumstances the moment of flexure remains always
positive, as is shown in Fig. 398.
FIG. 398.
The moment of flexure is a maxi-
A
:1,
N
AZ
C
M
D
7
mum or minimum for x = and
2

is
2
MN=
2
B
9,
Q
B
while the moment of flexure in C'
1
and D is C L = DO = ! q l²·
If, therefore, in the first case,
Fig. 397, (13) — 4° > 4,² or (3)">

2
2 l,², I.E. 7 > ↳ V8, we have M N
we must put the moment of proof
2
[(3)² - "']
> CL, and since q =
load equal to
2
7 + 2 1'
们
​Q
W T
while, on the contrary, we have
2 (1 + 24)
e
W T
when 71, 18.
e
Q4
2(+27)
§ 249. Girders not Uniformly Loaded.-If a beam A B,
Fig. 399, is not uniformly loaded, but in such a manner that the
load on the running foot increases
towards the extremities of the girder
regularly with the distance from its
centre, the statical relations will be as
follows.
FIG. 399.
D
N
H
K
E

B
B
If l = A B = 2 CA=20B is
the length of the beam, measured be-
tween the points of support A and B,
q the weight of the load per unit of
surface of the cross-section and ρ
the
angle of inclination A CD = B C E

492
[§ 250.
GENERAL PRINCIPLES OF MECHANICS.
of the planes CD and CE, which bound the load, we have the
weight of the prism A C D B CE of the load, sustained by one
point of support,
Q
C = ↓ A C. AD. q = ! ( ) tang. p . q = ¿ q lª tang. p,
2
and consequently the moment of this force in reference to a point
N, at a distance A N = x from A, is
Q
Уг
Y₁ =
2
x= q
1 x tang. p.
2
(AD + NL) A
AN,
The weight of the heavy prism above A N=x is q (
and the centre of gravity of the same is at a distance N O
RAD + N L AN
AD+NL
3
from N, and consequently the moment of
this prism in reference to N is
A N
Y:
Y₂ = q(2 AD + NL)
6
[1
= q ltang. p +
(1 - x) tang. p] €
6
2 x2
tang. p (x),
6
q tang.p
24
and the entire moment of flexure for the girder at N is
x T = y = y₁ — Y₂ =
gr tang.p
24
(3 F — 6 1 x + 4 x²) = ? [ (6)ˆ
(3 F x − 6 7 x² + 4 2³)
•] tang. P,
7
if we put CN = &;
æ z or measure the abscissa x, from C
2
9 で
​This is a maximum for 7 = and equal to
48
7
tang. p. and
I f
48
Q1
tang. p. I.E.,
12
Tr==
q l x
2
9...
q x
(7
2
27
{[(3)² - ~·]
QI
IT
e
the moment of proof load of this girder is
while for an uniformly loaded beam the moment of flexure is
NT
ΠΤ
e
hence the moment of proof load is
$250. Girders Sustaining Two Loads.-If a girder A B,
Fig. 400. supported at both ends is loaded at a point C. which is at
the distances CA = 1, and CB, from the points of support
§ 250.1
ELASTICITY AND STRENGTH OF FLEXURE, ETC. 493
A and B, with a weight P and in addition carries a uniformly dis-
tributed load Q = q, the reaction of points of support and B
A
are R₁
la P
Q
+ and Ra
2
l, P Q
し
​+
2
and the moment of
flexure at a point N, situated at a distance A N from the
point of support A, is
(R − 2 2 ) x = 1/
NV = y = R₁ x
Ꭱ
Q x²
2
FIG. 400.
R₁
R2
R₁
2
2 R₁
2
q
一​小
​X.
Rg
FIG. 401.


A
C
B
Q₁₂
N
D C
B
A
K
U
IC
B
K
B
This moment is a maximum for
Չ
2 R,
q
R
2
x = x, L.E., for x
R₁
R₁
9
(+ρ2)
y = DU = ? ( ) = ; = 1, (P
29 2 զ
?
and is then
1
2 Q
(
P+
7
2)
It is here assumed, that CA > C B, I.E., l, > l, and x < h.
If x = 1, the maximum of the moment of flexure is at C' (Fig. 401),
and consequently
y = C K = R₁ l₁₂-
q h
7, 19
P+
2
Q h
2
Q4 2
27
(P + 2 ) 4 4
Q
2
7,
If we substitute
R₁₁
P
lo
X
P
Q
q
and the moment of proof load of the girder, when
( 1½ ² + 2
h₁, we obtain
2 ! Q
24₁
1₁ — 1
2 72
2 l2
P
7,
-
<
2 l
is
·494
[§ 251
GENERAL PRINCIPLES OF MECHANICS.
P la
Q
Z
W T
+
, and, on the contrary, when
2
2 Q
e
P
it is
Q
219
(P + 2)² 4 - WT
e
These formulas are specially applicable to cases, where the
weight G of the beam is taken into consideration; here G must
be substituted for Q.
§ 251. Cross-section of Rupture.-In all the cases, which
we have previously treated, we have assumed the body A B,
A GA
FIG. 402.
Fig. 402, to be prismatical and, there-
fore, the moment of flexure W E to
be constant, hence we could conclude
from the fundamental formula

Pxr WE,
=
that the radius of curvature
P
WE
P x
was inversely, or the curvature itself directly, proportional to the
moment (Pa) of the force P acting upon the body and that con-
sequently the curvature becomes a maximum or a minimum at the
same time that P x does. If, therefore, the force P is constant,
or if it increases with r (as, E.G., in the case represented in Fig. 403,
where Q
A
L
ន
FIG. 403.
K
T
qx), the curvature in-
creases or diminishes with x and be-
comes with it a maximum and mini-
mum. When, on the contrary, the
cross-section F of the body is differ-
ent in different points, then II =
(F) is also variable, the radius of
curvature is proportional to the quo
W
and the curvature itself to
Px

tient
.the expression
Pr
If we are required to find the points of great-
W
est and least curvature, we have only to determine those, for which
Px
is a maximum and a minimum.
W
§ 251.]
ELASTICITY AND STRENGTH OF FLEXURE, ETC. 495
In like manner, according to the formula
Pxe
S
W
of § 235, the strain S in a body is proportional to the expression
Pxe
W
and becomes a maximum or a minimum simultaneously
with it.
W
e
If the body is prismatical, is constant, and the maximum
strain S is proportional to the moment Pr of the force only. If
II
C
the cross-section of the body varies, is a variable quantity, and
q x uni-
this strain is dependent upon this quotient also. In the first case
the strain becomes a maximum with P x, E.G., when the beam is
acted upon at one point by a force P and by a load Q
formly distributed over a distance x, for x = = 7; in the second case
this maximum cannot be determined unless we know how the
cross-section varies. In order to find the point of maximum strain,
it is necessary to determine by algebra the maximum of the expres-
Pre
sion
In any case the part of the body where this maximum
W
strain occurs is also that point at which, if the load is sufficient, the
strain S first becomes equal to T and also to K, and, consequently,
where the limit of elasticity will first be attained or where rupture
will take place. This cross-section of the body corresponding to
is therefore called the section of rup-
the maximum value of (Pre) is
ture (Fr. section de rupture, Ger. Brechungsquerschnitt) or also
the dangerous (weak) section.
If the body has a rectangular cross-section, with the variable
width u and the variable height, we have
IF
е
U 292
69
and the section of rupture is determined by the maximum of
or by the minimum of
U v2
Px
P x
U 2,2
For a body with an elliptical cross-section, whose variable semi-
axes are u and v, we have
W
е
π U v²
4
496
[§ 252.
GENERAL PRINCIPLES OF MECHANICS.
and we must therefore again determine the minimum value of
li vs
Px
when we wish to know the weakest point in the body.
When the weight is constant, P can be left out of consideration,
U v²
20
and we have to determine only the minimum of If, on the
contrary, the weight Qq x is uniformly distributed upon the
girder, we must determine the minimum of in order to find the
section of rupture.
u v²
x²
§ 252. If a body A C D F, Fig. 404, forms a truncated wedge
or a horizontal prism with a trapezoidal base A E B F, whose con-
stant width is B C D E = b, and if the force P acts at the ex-
tremity D F of the same, we
have to find only the mini-
D
FIG. 404.

I
G
K
F
L
P
mum of
v²
X
in order to deter-
mine the section of rupture.
Putting the height D G =
EF of the end h and the
height K U of the truncated
portion HK U= c, and as-
suming, as previously, that
the section of rupture L M N is at a distance UV- =x from the
extremity D E F, we obtain the height of this section
X
M L
V = h +
h = ½ (1 + 2),
C
and we have therefore but to determine the minimum of the ex-
pression
h
h²
? = (1 + J = C + + 2)
J
Ꮖ
C
or, since h and c are determined, only that of
2
C
1
X
+
X.
c²
; but if
C
If we assume re, the latter expression becomes
z
we make x a little (.) greater or less than c, we obtain
1
1
X
C ± 4
с
c (1
1
1 ±
1
C
1
X X1
= (1 + ) and
C
1千
​C
2
§ 252.] ELASTICITY AND STRENGTH OF FLEXURE, ETC. 497
x
c ± x₁
1
c²
C
+
X₁
C²,
29
consequently
1
х
+
२ ।
"- |
X c²
C
C³
2
Hence x = c gives the minimum
с
or in any case greater than
required, I.E. the section of rupture L M N is at a distance from
the end D E F equal to the height K U = c or to the distance of
the truncated edge H K from the same end D E F in the other
direction.
The height of this section of rupture is
h
v = h + c = 2 h,
C
and consequently the proof load is
b (2 h)²
P =
T
4 b h² T
C
6
C
6
For a parallelopipedical girder, which has the same length = c
the same width b and equal volume V = b h l, the height is
h,
h + 2 h
2
3 h,
bh T
9b h² T
P =
and consequently the proof load is
C 6 4 C 6'
19%
and such a girder bears, therefore, but as much as the wedged-
shape body just treated. If the body is a truncated pyramid, the
edges A E, B D, etc., when sufficiently prolonged, cut each other
in a point, and if we designate the height of the truncated portion
by c, we have
MN= u = b 1 +
=0(1
0 (1+1)
2) and L M = v = h (1
h¦ +
and therefore the minimum of
X
or of
U vs
b h²
X
X
(1 + 2 ) °
C
1
3 x
x²
+
+
X
C²
C³
must be determined, in order to find the section of rupture. By
the differential calculus we obtain
x = = c,
32
498
[$ 253.
GENERAL PRINCIPLES OF MECHANICS.
and we can casily satisfy ourselves that this value is correct by first
substituting x = cx, and then x = 21. In both cases
we obtain a greater value than
2
+
3
1
15
+
which is the value
C
2 c
4 c
4 c
the expression
1
3 x
x²
+
+
X
c²
3
assumes for x = 14
1/2 c.
C
The distance of the section of rupture from the end D F is then
equal to half the height c of the portion of the pyramid, which is
cut off. The dimensions of this surface are
3 b and v = 3 h,
and, consequently, the required proof load of the beam is
u = b (1 + 1)
+)
Р
3 b (3 h): T
27 bh² T
1 1 c 6
4 C 6
For a body, the form of which is a truncated cone, we have,
when the radius of extremity is r and the height of the truncated
portion is c, the radius of the section of rupture r₁ =r, and
therefore
27
P =
4
π på
T
C
4
§ 253. Bodies of Uniform Strength.-If a body is so bent,
that the maximum strain S upon the extended and compressed
side of the neutral axis is at all points the same, we have a body of
the strongest form, or of uniform strength (Fr. corps d'égale résist-
ance, Ger. Körper von gleichem Widerstande). By a certain force
such a body is strained to the limit of elasticity in all its cross-
section at the same time, and has, therefore, in each part a
cross-section corresponding to its proof strength; it requires,
therefore, when the other circumstances are the same, a smaller
quantity of material than any other body of the same strength.
Therefore, for the sake of economy and to avoid unnecessary
weight, such forms are to be preferred in construction.
Since the greatest strain in a cross-section is determined by
the expression
S
Pxe
W
(see § 251),
body of uniform strength requires that
P x e
W
shall be constant for
all cross-sections of the body.
$ 253.]
ELASTICITY AND STRENGTH OF FLEXURE, ETC. 499
If the force P is constant and applied at the end of the body,
we have only to make
ех
W
or
W e x
constant, and when the force Q =q is uniformly distributed
upon the girder.
e x²
W
or
W
e x²²
must be constant. For a girder with a rectangular cross-section (sec
§ 251), whose dimensions are u and v, we must make in the first
U v2
case
,
and in the second
X
U v²
X*
constant.
If at another place at the distance from the extremity the
width is b and the height h, we must have consequently in the
U v²
first case
X
b k²
7
}
and in the second
U 2,2
x²
b h²
で
​For the constant width u = b, we have in the first case
2 وج
X
h²
I.E..
Since the equation
v2
X
hs
2
2,2
h²
7
X
or
h
Ꮖ
1
Į
is that of a parabola (see § 35, Re-
mark), the longitudinal profile AB E, Fig. 405, of such a body
FIG. 405.
FIG. 406.

D
F
E
L
P
1311

MUAMMING
P
has the form of a parabola, whose vertex E coincides with the ex-
tremity or point of application of the load P.
If a beam A B, Fig. 406, whose width is constant, is supported
at both ends and sustains the load P in the middle, or if the beam
500
[§ 253.
GENERAL PRINCIPLES OF MECHANICS.
FIG. 407.
A2P
A B, Fig. 407, is supported in the middle and is acted upon at its
ends A and B by two forces, which balance each other, its eleva-
tion must have the form of two para-
bolas united in the middle. As ex-
amples of the latter case, we may
mention working beams, balance
beams, etc. As the beam is weak-
ened by the eyes, made for the shafts
A, B and C, lateral or central ribs
are added to it.

E
D
P
P
B
b
X し
​If the height v = h is constant,
we have
or
b
X
ין
and the width is proportional to the distance from the end; the
horizontal projection of the beam A CE, Fig. 408, is a triangle
B C D and the entire girder is a wedge, the vertical edge of which
coincides with the direction of the force.
D
FIG. 408.
B
E
FIG. 409.


B
N
P
Instead of the parabolic girders, Fig. 405, we generally make
use of girders, Fig. 409, with plane surfaces. In order to econo-
mize as much material as possible the girder is made in the mid-
dle M of the same height MO= h = h√, as the parabolic
girder would have been, and the limiting plane surface C' D is made
tangent to the corresponding parabolic surface. We have
0
BC 3 AM
MO 2 AM
3. and
A D A M
MO 2 AM
14
and consequently, if we denote the greater height B C by h, and
the lesser one A D by h₂, we obtain
hm
h₁ = 3 h₁ = 3 h v
!
=
1,0607 h and
h₂ = ! hm = { h N√ = 0,3536 h,
2
§ 254.]
ELASTICITY AND STRENGTH OF FLEXURE, ETC. 501
for which we must determine the height B Nh by means of
T'
6
the well-known formula P l = b h²
The volume of such a girder, whose faces are planes, is
b l (h₁ + h₂)
2
= 0,7071 b / h, while that of the parabolic girder of
equal strength is blh 0,667 b 1 h, L.E., 5,7 per cent. smaller.
3 =
D
M
=
FIG. 410.
C
YP
D₁
M₁
In like manner we can
construct the girder A NA,,
Fig. 410, which is supported
at its extremities A and A₁,
of two portions, bounded by
plane surfaces, which have a
common height B C = h₁
1,0607 h at the point of ap-
plication of the load, and at the extremities the altitude

A D = A₁ D₁ = ho
1
= 0,3536 h.
Here the altitude B N = h must be determined by the formula
Phlo b h² T
Z
6
§ 254. If the body A B D, Fig. 411, is to be made with all its
cross-sections L M N, A B C, etc., similar, we must put
FIG. 411.
ย
and therefore
D
N
P
L
h
b

•
и гез тез
2
b h²
b² x
B
I.E.,
b³
X
or
U
ጎ
b
で
​The width and height are therefore
proportional to the cube root of corres-
ponding arms of the lever. When the
distance from the end becomes eight-fold, the height and width
are only doubled.
We can replace this body by a truncated pyramid A C E G,
Fig. 412, at the middle of whose length the height is MO = h₂ =
hm
V.h= 0,7937 h and the width M N = b₁ = √! . b = 0,7937 b
bm
and the strength of this body is exactly the same as that of the body
2:
just discussed. For the tangential angle of the curve
h
יך
or
h
で
​Vī
x, we have, according to Art. 10 of the Introduction.
502
[$ 254.
GENERAL PRINCIPLES OF MECHANICS.
to the Calculus, tang. a =
h
3
h
x=3=
therefore it follows,
3 Vi
3
3 ³√ T x²
that for
3
X
し
​= {, ↓ 1 tang. a = ¿ h √ ( )² = ↓ h √4 = h V↓ = 0,2646 h,
and in like manner we have for the curve
3
v
6
U
x
b
V
b
"'
tang. ẞ
and
3 √7x2
b
1 l tang. B
3
+
½
1 tang. a = VI. h
bm
+
1 l
From this we can calculate the dimensions of the base A B C
A B = h₁ = h
B C = b₁ =
4 V. h = 1,0583 h and
1 tang. B = V. b = 1,0583 b,
7473
and those of the smaller base E F G
FIG. 412.
FIG. 413.
B
C
N
E
B
G
D
A
A
=
=
F G = h₂ = hm −1 tang. a VI. h 0,5291 h and
EF = b₂ = b - I =
½ l tang. BV. b = 0,5291 b.
2
in
We must of course put Pl
b h² T
6
If we make the cross-section of the body of uniform strength
circular, we have for the variable radius the equation
U = V = 2 =
3
X
1
and if we replace this body by a truncated cone A B E, Fig. 413,
its radii must be
MO = rm
0,7937 r, C A = r₁
r₁ = 1,0583 r and
D E = r₂ = 0,5291 r,
2
and the radius r of the base of the solid of uniform strength must
be calculated according to the formula
π p3
Pl=
T.
4
•


$254.]
ELASTICITY AND STRENGTH OF FLEXURE, ETC. 503
If the girder is uniformly loaded and its width is constant, I.E.
if u = b, we have
22
x²
or
h²
V
72 2
х
h
ין
and its form must be that of a wedge, whose elevation is a trian-
gle A B D, Fig. 414.
FIG. 414.
FIG. 415.


MIMENTAL
SNSA!
B
1
MRUMS
emmown
CINAR
D
A
FIBROM
9:
16:
B
If the height is constant, we have
૧૫
x2
;
hence the horizontal
b
72
section of the girder is a surface limited by the two inverted arcs
of a parabola B D and CD, as is shown in Fig. 415.
If we again make the cross-sections similar, we have
می میر رستم
U
b3 h³ 19
で
​and the vertical and horizontal profiles are cubic parabolas, the
cubes of the ordinates of which are proportional to the squares of
the abscissas.
If a body A E B, Fig. 416, supported at both ends, is uni-
FIG. 416.
F
E
Amit
TIPAPPROA
B
DIAN
UN MARSTOM
M
D
FUTUREYIT!
G
PRIESTOR
formly loaded with the weight q
per running foot or upon its whole
length A B = 7 with Q ql, we
have the moment of the force at
a point O, situated at the distance
A0=x from one of the sup-
ports A,

Q
2. x − q x
X
2
· 3 = }} (1 x − x²),
AWAM LATRIO
M
D1
and, on the contrary, at the cen-
tre C
Q 7
Q 1
Q?
q t
2
2
2 4
8
8
Assuming the width b of the body to be constant, we have
504
[§ 255.
GENERAL PRINCIPLES OF MECHANICS.
T
Չ
b v².
6
92 (2
х
x²) and
T
b h².
q
6
8
h denoting the height C E of the body at the centre, and by divi-
sion we obtain
v2
1 x =
x²
or
h²
=(A)
2
(1 x − x²).
If hl, v² would be
1
=
xx, and therefore the longitu-
dinal profile would be the circle A D, B, described with the
radius ¦ 1; but since 1 x — æª must be multiplied by (+)* in order
=
to obtain the square v² of the height M O NO at any point, the
circle becomes an ellipse A D B or A E B, whose semi-axes are
С A = α₁
F
l and CD = C E
h.
We can replace this body by a girder A A B D B, Fig. 417,
FIG. 417.
G
with plane surfaces, whose
height at the distance A M
7 from the points of sup-
port B and B is M 0 = hm

B
h
147
16
12
187² = 1 √3. h.
D
The angle of inclination a of
the surface BD to the axis A Cis given by the equation
tang. a =
h
3 7 X
1 l '
2 h
4
| l
2 h
h
V lx - x²
7
13
√ 3 12
W3
√3.
7
consequently we have
tang. a = √3. h and the height of the
4
body in the middle
Z
C D = MO +
tang. a =
3 √3 . h = 1,1548 h,
4
and, on the contrary, the height at the ends is
7
=
AB MO -
tang. a =
} √3. h =
0,5774 h.
4
(§ 255.) The deflection of a body of uniform strength is, of
course, under the same circumstances, greater than that of a pris-
matical girder. For the case, where the beam is fixed at one end
§ 255.]
505
ELASTICITY AND STRENGTH OF FLEXURE, ETC.
and subjected to a stress P at the other, the deflection is found as
follows. The well-known proportion
r =
E e
Τ
E
gives us the formula
e
T
in which the radius of curvature is a function of the dis-
tance e.
If we know the dependence of e and x upon each other,
we obtain an equation between r and x, from which we can deduce
(in the way explained in § 218) the equation of the co-ordinates of
the elastic curve. If we assume the deflection to be small, we can
again put the length of arc s equal to the abscissa r, and conse-
quently equate the differentials ds and dx; hence we can, as be-
fore, assume
d x
da
r =
From this we obtain
E
d x =
e da,
T
and, by integration, the tangential angle
α
T
If d x
Ee
e
For a girder with a rectangular cross-section ev, and
therefore
a =
-2 T Sax.
E
If the width is constant or u = b, we have
2,2
h²
7(see § 253), and therefore
X
v = h
h V and
a
E
T. 17 J
h
or, since for x = 1, a = 0 and consequently
2 T √ √x dx = -
2 T NT
•
2 √x + Cons.,
V
Eh
h
2 T V
Con. =
E h .217.
7,
4 T VI
α
E
h (√l - √x).
d y
If we put a =
we obtain
d x
4 T V T
dy
E h (Vi√x) d x,
and, therefore, the required equation of the co-ordinates is
4Ꭲ Ꮙ Ꮣ
y E T ( x V T − 2 x √x) = 4
T VI
E
h i -
π ( N l − 3 N x) x
506
[§ 256.
GENERAL PRINCIPLES OF MECHANICS.
For x = 1, y becomes a; the deflection is then
Tľ²
4
α =
137
E h
T
6 Pl
But Pl = b h².
or T =
and, therefore, the deflection.
6
bh2
is given by the formula
8 P 13
4 P 1³
a =
2.
Eb hs
3
Ebh39
I.E, it is twice as great as in the case of a parallelopipedical girder,
whose height is h and whose width is b (compare § 227).
If the force acts at the middle of a girder, supported at both
1
P
ends, we have only to substitute
for P, and for 1, and we obtain
2
8 P l³
α =
To⋅ Eb h
16
I.E., it is 16 times smaller than when the force acts at the end.
For a body of uniform strength with a triangular base, as is
represented in Fig. 408, the variable width is u =
X
b, and
u h³
Prx =
E
12
b h³ x
127
E;
hence the radius of curvature r =
b h³ E
12l' P
is constant, the curve
formed is a circle, and the corresponding deflection is
T
а
2 1
6 P 19
bh
b h E
4 P B
3
⋅
b h³ E'
I.E., times as great as for a parallelopipedical girder.
§ 256. Deflection of Metal Springs.-The most common
examples of bodies of uniform strength, as well as of those which
bend in a circle, are steel or other metal springs. The springs, of
which the spring dynamometers are made, are of the finest steel and
are from to 1 meter long, from 4 to 5 centimeters wide and in
the middle from 8 to 21 millemeters thick. They form bodies of
uniform strength, and their longitudinal profile is composed of two
parabolas united in the middle (see § 253). In order to increase
the action, the spring dynamometer is made of two such parabolic
springs A A and B B, Fig. 418, which are united at their ends A
$ 256.]
ELASTICITY AND STRENGTH OF FLEXURE, ETC. 507
by means of the links A B, A B (see Morin's Leçons de Mécanique
FIG. 418.
C
2
B
D
A
B
Pratique, Résistance des Matériaux,
No. 198). These dynamometers
measure the force P, which is ap-
plied to the hook D in the middle
of one of the springs, by the space
described by the point Z, which is
of course equal to the sum of the
deflections of the two springs. But
from what precedes we know that
8 P P

and consequently we have here
a =
16
bh E
s = 2a =
P B
bh E"
and, therefore, the force
P = ( b h E ) s
13
S,
corresponding to the space s described by the pointer.
0.0211, 7 = 1,0 meter,
In experimenting with such an instrument, whose springs were
of the following dimensions: b 0,05, h
the space described by the pointer was s
9,7 millemeter, when
the load was P 1000 kilograms; the coefficient of this dynam-
eter was therefore
b h³ E
13
s
P 1000
9,7
103,09,
and for other cases we must put
P = 103,09 s kilograms,
when s is given in millimeters, or when the scale is divided into
millimeters.
If, instead of parabolic springs, we employ triangular ones of
uniform strength, we have
S
6 P 1³
= α =
3
and, therefore,
2
P =
TO
b ñ ³ E'
b h³ E
(BRE)
S,
I.E., one-third greater than for a dynamometer with parabolic
springs.
Wagon springs should unite great flexibility with great strength,
while, on the contrary, it is not necessary to know the exact relation
between P and s. For this reason, these springs are often formed
of a number of simple springs laid upon one another.
508
[$ 256.
GENERAL PRINCIPLES OF MECHANICS.
If the compound spring is composed of n simple parallelopiped-
ical springs, placed upon one another, we have, when the width is
b. the thickness 7 and the length 7, the deflection corresponding to
the force P at the end A of the entire spring a =
proof load
4 P 1³
n E b h³
P = n
b h² T
7 6
and therefore also
T 12
Τι
·
a
and the
or
E h し ​3 E h
If the entire spring A CD, Fig. 419, consists of n simple tri-
angular springs, we have
6 Pl³
а
while P = n
n E b h 39
bh² T
6
remains unchanged, and therefore
Tで ​α Τι
a
or
E h
El
A
of the flexibility in-
Therefore, in both cases the measure
7
h
creases with the ratios and and is the same as for a simple
T
E
spring of n times the width (n b).
C
A
FIG. 419.
FIG. 420.
D
В
+ P
+P
+P
+P


P
I
A
_PV
D
B
G
H
In order to economize material, we superpose springs of differ-
ent lengths and construct them of such a shape, that by the action.
of the force P at the end A of the entire spring they are bent in
arcs of circles of nearly or exactly the same radius. The force P
bends the lowest triangular piece A A of the the entire spring
A B H, Fig. 420, whose length in the arc of a circle, whose
radius is r = N
b h³ E
127' P'
7
n'
and in order that the remaining paral-
lelopipedical portion shall be bent in like manner, it is necessary
256.] ELASTICITY AND STRENGTH OF FLEXURE, ETC.
509
that the same shall exert a pressure at A upon the succeeding
spring, which shall be equal to the force P; for the moment of
Pl
of a couple
flexure of this spring is then equal to the moment ጎ
?
(P, – P) whose arm is The relations of the flexure of the first
N
spring repeat themselves in the second, which is
7
shorter than
N
it; it is bent in a circle whose radius r =
n b h³
127
E
when its end
•
P'
A, A, is triangular and the other portion is parallelopipedical, and
if it presses on the third spring with a force P. This is also the
case for the third spring 4, G D, etc., up to the last piece, which has
no parallelopipedical portion, and which, by the action of the force
P, is bent in a circle of the above radius r. The entire deflection of
で
​this compound spring is a =
2 r
6 P 13
n E b h ³⁹
and the proof load is
P = n
b h² T
16
hence
T 12
a
Tl
a
E hori
Eh
The relations of the flexure are here exactly the same as for a
spring composed of single triangular springs; it can also easily
be proved, that both sets of springs require the same amount of
material.
It is not, however, necessary to make the ends of the springs
exactly triangular; we can employ any other form of equal curva-
ture, E.G., we can make them of the constant width b and then at
the distance x from the end A the height must be
3
y = h v n x
Such a double spring is represented in Fig. 421. Here the
FIG. 421.
OP

VP
A
B
B
DD
P
total proof load is 2 P; the length must not, however, be meas-
ured from the middle, but from the ends BD, BD of the fastening.
510
[§ 257.
GENERAL PRINCIPLES OF MECHANICS.
REMARK. The reader can consult upon the subject of wagon springs:
F. Reuleux: Die Construction und Berechnung der für den Maschinenbau
wichtigsten Federarten. Winterthur, 1857; also Redtenbacher: die
Gesetze des Locomotivenbaues, Mannheim 1855, and Philips: Mémoire sur
les ressorts en acier, etc., in the Annales des Mines, Tome I., 1852.
CHAPTER III.
THE ACTION OF THE SHEARING ELASTICITY IN THE BENDING
AND TWISTING OF BODIES.
§ 257. The Shearing Force Parallel to the Neutral
Axis.-In a body, which is subjected only to a tensile or com-
pressive force, the bases A B and C D of an element A B C D of
-P
FIG. 422.

P
the body, Fig. 422, are only acted upon by the two opposite forces
P and P, which balance each other, while the sides A B and
VP
FIG. 423.
A.P
S1
E
ន
PV
N
CD remain free from the ac-
tion of extraneous forces; for
the neighboring elements of
the body are subjected to the
same axial strain as the sup-
posed element A B C D itself.
But the case is different when
the body is bent; for on one
side A B of the element
ABCD a strain is pro-
duced which is opposite in di-
rection to that upon the other
side CD of the element, and
in consequence of the cohesion

§ 257.]
ACTION OF THE SHEARING ELASTICITY, ETC.
511
in A B and CD, the element A B C D is subjected to the action
of a couple. This couple is a maximum for an element which lies.
in the neutral axis; for the element is here subjected on the side
A B to an extension, and on the side CD to a compression.
If S is the strain upon a fibre at the distance e from the neu-
tral axis, when the cross-section = 1, the strains upon the portions
F, F, F... of the entire cross-section, which are situated at the
distances Z1, Z2, Z3 ... from the neutral axis, are
F3 Z3
F₁ Z1
F₂ Z2
2
S,
S.
е
e
e
S, etc.,
and the total strain in the cross-section F₁ + F₂+ F3...
..is
S
Q
e
(F₁ z₁ + F₂ z 2 + ...) =
S
Σ (Fz).
e
Now if F + F + ... is the part of the cross-section on one
side of the neutral axis, Q is the total strain on that side of the
neutral axis. The strain on the other side is, according to the
theory of the centre of gravity (compare § 215), equal in intensity
to it, but opposite in direction.
Besides we have, according to § 235, S =
Pre S
P x
or
W
e
W'
whence also Q
P x
W
(F₁ %₁ + F₂ %9 + . . .).
2
%₂
In a cross-section, which is at a distance A B = x, from the first
one, the strain is
Q₁
P(xx)
W
(F₁ z₁ + F₂ ž₂ + . . .),
and therefore the total force with which the piece A B E tends to
slide upon A B is
Q - Q₁ =
Px₁
TV
(F₁ z₁ + F₂ Zg+...).
%1
Now if b, is the width of the cross-section at the neutral axis,
the shearing force along the unit of surface in this axis is
X.
Q - Q₁
P
0
b。 x 1 b. IV
(F₁ 21 + F₂ 22 + ...)
=
Ρ Σ (F)
b. W
If, therefore, the girder is not to be ruptured by a sliding along
the neutral axis, we must put = the modulus of ultimate
strength, and in order that it shall be as secure against rupture
by shearing as against breaking across, it is necessary that I shall
be at most equal to the modulus of proof strength T, I.E. that
#
512
[š 257.
GENERAL PRINCIPLES OF MECHANICS.
P
T=
b. W
(Fz), or P =
bo W T
Σ (Fz)'
and
P
bo
Σ (Fz).
TW
Σ (F z) is also = F₁ 8₁ = F₂ S2, when F, and F, denote the
areas of the portions of the entire cross-section F F₁ + F₂, lying
on the opposite sides of the neutral axis, and s, and s, the distances
of the centres of gravity of the two portions from that axis.
For a rectangular girder, whose cross-section Fb h, we have
b h h
Σ (Fz) = F 8₁ =
2 4
b h²
8
b h³
W =
and b₁ = b, whence
"
12
P
P
=
3 b h Tand b。 = b
=
32
Th
π ď²
For a cylindrical girder, whose cross-section is F =
we
4
the centre,
Σ (F z)
F₁ s₁ =
S1
πα 2
d =
83 π
d³
παι
W =
and bo
=
d, whence
64
π d³
3 π
P
T=
64.
d³
16
12
P
d = 4 1
Р
T
have, since the centre of gravity is situated at a distance.
12, and, according to § 232,
2
d from
3 п
3 пТ
d² T, and
= 1,303 √/P
π a² b
In like manner for an elliptical girder, since W = 4
α = a² b and b, = 2 b, we have P
пав 2
F₁ s₁ =
or b
2 Π
4 P
P
= 0,4244
a Ţ
2 b, we have P = ¾ ñ a b T,
3 παΤ
Finally, for a tubular parallelopipedical girder, whose cross-
section is F = b h − b, h, (Fig. 354, § 228), we have
3
b h² — b, h₂²
2
F₁ s₁ =
81
W =
8
b h² - b₁ hr
12
3
and b₁ = b
b,,
hence
P =
OSIŁO
bh - b, h,²
3
(b — b,) (b h³ — b, h‚³) T
2
The shearing force X diminishes as the distance of the surface,
in which it exists, from the neutral axis increases, and becomes
finally null at the surface of the body, where the distance from the
neutral axis is a maximum. The intensity of the shearing force
§ 258.]
ACTION OF THE SHEARING ELASTICITY, ETC.
513
X at a given distance 0 B = h, from the neutral axis of the body
PΣ (Fz)
M N, Fig. 424, is also given by the formula X =
FIG. 424.
1P
ZA
S
B
S
M
Z
+ PV
N
found
b. IV
above, if instead of Σ (Fz)
we substitute the sums of the
products F₁ z₁, F₂ Z2 on
one side of A B C D, and in-
stead of the width b, of the

0
•
surface at the given distance
h. The sums of the products
Fn Zn Fn + 1 Zn + 1 for the other
side is, however, equal to the
sum of the products F₁ z₁,
since the products
F₂ Z2
2
of the elements, situated on
the opposite sides of the neutral axis within the distanceh,
balance each other.
E.G. if the cross-section of a girder is rectangular, we have for
the point situated midway between the neutral axis and the limit
ing surfaces, I.E., at the distance
h
from the neutral axis
4
b h
Σ (F2) = F₁ 8₁ =
⋅ 3 h = 3 bh',
4
32
and, therefore, the shearing force is
33
P. 3 b h²
P
X =
b h³
b h
b.
12
P
=
لاين
b h
while at the neutral axis its value is X
§ 258. The Shearing Force in the Plane of the Cross
section.—As the tensile and compressive forces of the ends of ar
element A B C D, Fig. 424, are in equilibrium, so also the shearing
forces in this element, which form two couples, balance each other.
Now if § is the length A B and the height B C of the element,
we have the shearing forces along A B and C D,
and the moment of the couple, formed by them, §
and the shearing forces along B C and D A are
and the moment of the couple formed by the latter is
Z; now if equilibrium exists, we must have 5 X
that X = Z.
X and
Ꮧ,
X. 5 =
§ X,
§ 5 X,
Z and
Z,
Z. § =
§
5 Z, I.E.,
33
514
[$ 258.
GENERAL PRINCIPLES OF MECHANICS.
The formula X =
PΣ (Fz)
b W
is, therefore, also applicable to the
determination of the shearing force Z along the entire cross-section.
It is, E.G., in a girder with a rectangular cross-section, for an ele-
ment in the neutral axis =
P
14338
and for one at a distance ị h
bh'
4
from the neutral axis 9
P
06 etc.
b h
The sum of the shearing forces along the entire cross-section,
must of course be equal to the force P, or, if several forces act at
right angles to the axis of the beam, equal to the sum Σ (P) of
these forces. This can be proved as follows: if we divide the
maximum distance e of the elements of the surface from the neutral
axis into n equal parts, we can imagine the cross-section upon the
corresponding side of the neutral axis to be composed of the strips
h
h
b b₂ b39
n
axis are
n
h
n
>
etc., whose moments in reference to the neutral
h
2
b3
b. (4), 2 6. (4), 38, (A), etc.,
b₂
and the sum of the latter is
h
2
= ( 1 2 )² (1 b₁ + 2 b₂ + 3 b₂ + 4 b, + …..).
n
4
In reference to the axis, which is at a distance
tral axis, the sum of these moments is
h
2
b½
= ( 1 2 ) ( 2 b₂ + 3 b₂ + 4 b₁ + ...),
N
in reference to the axis at the distance 2
2
h
it is
n'
( 12 ) (3 6, + 4 b; + · · · ),
N
h
from the neu-
-
п
and therefore the sum of all these sums to the distance e is
= (32) i
[b₁ + (2 + 2) b₂ + (3 + 3 + 3) b¸ + ...]
( 12 ) (1² .
2
b₁ + 2³ . b₂ + 3ª . b; + . . . + n³ bn).
N
It follows that the sum of all the shearing forces along cross-
section on one side of the neutral axis is
$ 259.
ACTION OF THE SHEARING ELASTICITY, ETC.
515
R₁ =
Ph
X, 6, (h
b₂
h
Xs bs
h.
b₁ ( 1 ) + X, b. ( 14 ) + x; b₂ ( 12 ) +
N
times the sum last found
W n
Ph
W
N
( 12 )
N
(1². b₁ + 2². b₂ + 3². b₂ + ... + n². b₂).
But the measure of the moment of flexure for this half of the
cross-section is
h
h
W₁ = 2 ( F x ) = 1 [b, (1)
3
= (1) ˚ (1² . b,
22%
+ 2² . b₂ +
+ b₂
(3, 4)
2 h
3 h\²
+b3
+
N
...]
3² . b₂ + ... + n² . b₂),
whence it follows, that the required shearing force along this sur-
face is
R₁ =
P W₁
W
In like manner we find for the half of the cross-section, situated
PW.
on the other side of the neutral axis, the shearing force R₂ = Π
and finally it follows that the shearing strain for the entire cross-
P (W₁+ W₂)
section is R
P, since the measure of the mo-
W
=
ment of flexure of the entire cross-section is equal to the sum
W₁+W of measures of the moments of flexure of the two por-
tions of it.
I
2
§ 259. Maximum and Minimum Strain. If the strains
in any section are known, the strain in any given cross-section
can be found by employing the ordinary methods for the com-
position and decomposition of forces. In order to find the
FIG. 425.
A
B
strains in an element 1 C, Fig. 425, of
the surface, whose plane forms the varia-
ble angle B A C = with the longitu-
dinal axis of the body, we decompose the
tensions in the projections A B and B C
of this element of the surface into two
components, one of which acts in the
plane of AC and the other at right-angles
to it, and we then combine the compo-
nents in A C, so as to form a single
shearing force, and the components, acting
in a direction at right-angles to A C, so as to form a single tensile
or compressive force. If the width of the elements A B, B C and
AC of the surfaces is unity, we can put the shearing force along

D
516
[§ 259.
GENERAL PRINCIPLES OF MECHANICS.
A B,
– A B. X and decompose it into its components A B. X
cos. 4 and A B. X sin. 4, and in like manner we can put the
shearing force along B C, BC. ZBC. X and decompose
it into its components
BC. X sin.
and B C. X cos. 4.
The components of the tensile force B C . Q = B C .
Sz
e
whose
direction is perpendicular to B C, on the contrary, are B C. Q cos. Y
and B C. Q sin. 4, and it follows that the entire shearing strain
along A Creferred to the unit of surface is
4
U (AB. X cos. - BC. X sin. + BC. Q cos. 4): A C,
and that the tensile strain at right-angles to AC is for the unit
of surface
V = (A B. X sin. + BC. X cos. + BC. Q cos. ): A C.
But
= cos. and
2
U =
A B
BC
= sin. p, whence it follows also that
A C
A C
U = X (cos. 4)² - X (sin. 4)² + Q sin. 4 cos. 4 and
U = 2 X sin. & cos. 4+ Q (sin. 4), qr, since
(cos. 4)² — (sin. 4)² =cos. 2 and 2 sin. 4 cos.
X cos. 2 4 + 1 Q sin. 2 4 = 24
X cos. 2 4
= sin. 24,
Sz
+
sin. 2 4 and
2 e
Sz
V = X sin. 2 4 + Q (sin. 4)²
X sin. 24+
(1
2 e
cos. 2 4).
The strains in the surfaces A D and C D, which together with
the surfaces A B and C D fully limit the element A B C D, give,
of course, equal and opposite shearing and tensile forces. On the
contrary, for a similar element of the body upon the compressed
side Q is negative, and therefore
U = X cos. 2 4
1 Q sin. 2 4 = X cos. 2 &
V=X sin. 24 - Q(1-
Sz
sin. 2 y and
2 e
Ꮪ
cos. 2)=X sin. 24-
(1
cos. 24).
2 e
In order now to find the values of the angle of inclination 4,
for which the shearing force U and the normal one V assume their
maximum or minimum values, we substitute for 4,2 4 + µ, µ de-
noting a very small increment, and require that by it the corres-
ponding values of U and V shall not be changed. For U =
X cos. 2 4 + Q sin. 24, we obtain thus a second value
U₁ = X cos. (2 4 + µ) + ½ Q sin. (2 4 + µ)
= X (cos. 24 cos. µ
+ cos. 2 4 sin. μ), or,
sin. 24 sin. µ) + ¦ ¦ (sin. 2 4 cos. µ
since we can put cos. μl 1,
€ 259.]
517
ACTION OF THE SHEARING ELASTICITY, ETC.
U₁ = X cos. 2 + 1 Q sin. 2 4 - (X sin. 24 - Q cos. 2 ) sin. p.
4
Now if we put U₁ = U, we must have X sin. 2 4 — ¦ ¦ cos. 2 4 = 0
and therefore
cos. 24, I.E.,
2
Q
sin. 24
2 X
Q
Sz
tang. 24 =
2 X
2 X e
From this it follows also that
Q
Sz
sin. 24
and
√ Q² + 4X²
2 X e
√ (S z)² + (2 X e)²
2 X e
cos. 2 & =
√ Q² + 4 X²
V (S z)² + (2 X e)”
and that, finally, the required maximum value of the shearing force
Sz
U is
2 X² + ¦ Q²
Um
²
√ Q² + 4 X ²
√ ( ! Q)² + X * = √ √ ( 2 ) *
1
+ X?
In the neutral axis Q is
2 e
=
0, and therefore UX and tang.
= 0, I.E. 2 ¥ = 0 and 180°, or & 0 and 90°. For the most
remote fibres, on the contrary, X is = 0 and z =e; therefore
24
Q
Um
2
S
2
and tang. 2 4 = ∞, or 2 ¥ = 90° and = 45.
In passing from the neutral axis to the outmost fibre, the
angles of inclination for the maximum strain change gradually
from 0 and 90 degrees to 45 degrees, and the maximum strain
S
varies from X, to 5
2*
In order to be certain that this strain shall not become greater
than the axial strain S, which is calculated by the aid of the for-
Pxe
and is equal to the modulus of proof strength T,
W
mula S
we must make X, at most — S, or rather
PΣ (Fz)
b。 W
<
Pxe
W W ?
I.E.
,
Σ (Fz)
b.
x e.
Q
√ ₁
1
=
If, then, in the formula V = X sin. 2 4 + (1
we put instead of
2
and again make cos. µ =
X (sin. 24 cos. µ + cos. 2 4 sin.µ) +
+ sin. 2 4 sin. µ)
cos. 24)
1, we obtain
Q
92 (1
cos. 24 cos. µl
X sin. 2 4
Q
+ ? (1
cos. 24)
+
(x
Q
cos. 2 4 +
2
sin. 2 4) sin. µ,
2
518
[$ 259.
GENERAL PRINCIPLES OF MECHANICS,
and in order that shall cause V to become a maximum or a min-
Q
imum, V, must be = V or X cos. 2 4 +
sin. 24
= 0, I.E.
2
tang. 24
2X
Q
2 Xe
as well
as
Sz
2 X
Q
2 X 2
V₂
√ Q² + 4 X ²
+
sin. 2 4 = = √ Q² + 4 X ²
The corresponding minimum of Vis
√ Q² + 48²)
and cos. 24
= ±
士 ​√ Q² + 4 X"
Q
2
(1 –
S z
2 e
Q
?
2
S
+ X²:
e:
2 X²
4 X ²
Q
4
2
Q
2
+ I'
and, on the contrary, its maximum is
Q
+
V₁ = √² + + + + 2 (1 + √ Q² + 1 x ) = ? + 1 ( ! ) = r²
ก
Sz
+
2 e
2
e
We must require the maximum to be at most cqual to the
modulus of proof strength Tor
Sz
2 e
+ √(√ ²)² + X² < T.
In the neutral axis Q is = 0, and therefore tang. 2 4
=
n
∞
I, on
ون
or 24 270° and 135 or 45 degrees, and √₁ =
= 4
the contrary, V, + X. In the most distant fibre, on the con-
trary, Fis = 0 and Q S, and therefore tang. 2 y = 0 or 2 y
0 or 180° and y = 0 or 90°, and V₁ = 0, on the contrary,
F= S. In ordinary girders the maximum strain increases
=
A
F
gradually from X =
P ≤ ( F z)
b W
to S
Pre
W
as we pass from the
neutral axis to the outmost fibre.
bha
For a parallelopipedical girder we have Σ (F z)
B
=
8
b h³
h
. b.
b₁ = b and e =
and therefore the limit values are X = 3.
12
2'
P
6 P x
P ( 1 − 2 ) ( )
+2
(2)
and $ =
S
; but in general we have
b h
b h*
2 W
6 P
b h³
3
-
"[(')' — 2'] and
Sz
12 P x z
and therefore
e
b h³
§ 260.] ACTION OF THE SHEARING ELASTICITY, ETC.
519

6 x z
V
x
6 P
2
Y = "PE² + √ ( P ) + (CD)` [(')' − ·]
Vm
bh³
6 P
b h³
3
b h²
3
(x 2)² + (('¹²) — 2²
for example, for z=1h,
6 ½³ [x x + √ ( x 2 )² +
る​だ
​[x + √ x² + (3)* h²], and for x = 0,
3 P
V...
2 b h²
9 P
V
etc.
m
8bh'
If such a girder A B, Fig. 426, is fixed at one end B, the di-
rections of the maximum and minimum normal forces V and V
D
A
D
FIG. 426.
n
can be represented by two systems
of lines, which cut the neutral axis
at an angle of 45°, and the outer
fibre and each other at an angle of
90°. The curves, which are concave
downwards, correspond to the tensile
forces, and those which are concave
upwards to the compressive forces.
The steeper end of any curve cor-
responds to the minimum and the flatter end, on the contrary, to
the maximum forces. At the ends D and D, both these strains
become equal to zero, while for the ends C and C, their values are
the greatest.
§ 260. Influence of the Strength of Shearing upon the
Proof Load of a Girder.-The capability of a girder to support
a certain load requires not only that the strain S =

outermost fibre, but also that the shearing force X
Pxe
W
in the
PE (Fz)
b. W
in
the neutral axis shall not exceed the modulus of proof strength T.
In the last chapter we have repeatedly given the moments which,
in ordinary cases, we must substitute for P z in the expression for
S; we have, therefore, only to give the values, which we must sub-
stitute for the force P in the expression for X.
If the girder is fixed at one end and acted on by a force P
at the other end, P can be directly employed in the formula
P (F2)
I₁ =
If the beam supports, in addition, a uniformly
I.
b. IT
distributed load, whose intensity upon the unit of length is q, we
must substitute for P in this expression P+ qz and P + q b
520
[§ 260.
GENERAL PRINCIPLES OF MECHANICS.
when we wish to determine the maximum value of X. If, on the
contrary, the girder is supported at both ends and sustains at the
distances l, and l₂ = 1 — l, from the points of support a load P,
we must substitute for one portion of the beam P, and for the
other P instead of P in the formula for X, in order to find the
shearing force in the neutral axis. If, on the contrary, this girder
sustains an equally distributed load q l, each of the points of sup-
port bears
and the shearing force of the whole cross-section at
any point at the distance x from the points of support is P = q
X
q r
2,
(x). The latter is = 0 in the middle, where x =
2'
becomes
greater and greater towards the end, and at the point of support
is P q
2
If a girder, supported at both ends, sustains a load, which is
equally distributed over a part c of its total length, while the other
portionc is not loaded, the point of support of the first por-
с
q c
tion bears a part q c (1-2) of the total load y e and that of the
second portion a load c
q c²
and the vertical shearing force at the
2 ľ
distance x from the first point of support is.
P = q c (1
C
2 ) − q x = q(c-
C³
21
- x).
q c²
and this value
21'
The value of the latter becomes for x = c,
remains the same for any distances x > c.
exactly one-half of the girder, I.E. if c
P = q
2
2'
If the load covers
1
=
we have
(32 - x)
0
or for x =
P =
2'
q l
8
8
FIG. 427.
R
qc
R
If, finally, the girder A B,
Fig. 427, bears a load p l equal-
ly distributed over its entire
length and a load q c equally
distributed over the length A C
= c, the reactions of the points
of support are

O
R₁
2
pl + q (c -
(c
2 ²)
– 17) and R,
2
pl
Չ
I c²
+
21'
$ 260.] ACTION OF THE SHEARING ELASTICITY, ETC.
521
whence it follows, that the vertical shearing force at the distance
A 0 = x from the point of support A is
P
ρι
C
p² + q ( c − c ) − ( p + q) x,
2
-
2
for the latter expression becomes p
x=
any distances x > c it is
(½ - c) ·
q c²
and for
pl
p
2
21'
+ 9 c²
21
C³
+ p x.
q² in C' is = 0
21
p l
2
q c²
+
21
p (1 − x)
=
The vertical shearing force P = p
G-
( - c) -
2 p
p
for c² +
lc =
ľ, I.E., for
q
q
C =
=(
p
+1
9
(
p
+
q! 9
1.
If, in general, at a point of the girder the shearing force is
P = R − q x, we have for the
moment of flexure
q x²
M = R x
2
2
q x (27
R
q
- x).
2 R
R
This, however, for x =
x, I.E., for x =
is a maxi-
Չ
q
mum, in which case P becomes = 0; the moment of flexure of a
girder becomes a maximum for the same point at which the verti-
cal shearing force is 0, and in the foregoing case c gives that
length of the load q c, for which the moment
[2]
+ q
q ( c − 2 ²)] c
(p + q) c²
2
(p + q) c²
2
becomes a maximum, and it is then
These formulas are applicable to girders for bridges, where q c
denotes the intensity of the moving load.
P
The shearing force X
b. 11
(Fz)
W
must be specially consid-
ered in the case of bodies of uniform strength, the cross-section of
which, according to what we have seen above (§ 253), might in
some parts be infinitely small. For example, for the parabolic
girder in Fig. 406, we have ₁ = T = 3.
necessary cross-section at each end is F,
b ho
P
and therefore, the
P
b, h, = 3
in which
T'
T denotes the modulus of proof strength for shearing.
522
[§ 261.
GENERAL PRINCIPLES OF MECHANICS.
§ 261. Influence of the Elasticity of Shearing upon the
Form of the Elastic Curve.-We have yet to determine what
influence the elasticity of shearing has upon the form of the elastic
curve or upon the form of the neutral axis of a loaded girder A B,
Fig. 428. According to the formula PFC, in which C de-
notes the modulus of the elasticity of
shearing and F the cross-section of the
bcam, the inclination the beam A, B pro-
Ꮧ.
duced by the shearing force is ‹ =
Ai
A
P
FIG. 428.

B
C
and, therefore, the corresponding deflec-
tion of the end A, of the girder, whose
length A, B = 7, is
Xol
0
ΡΙΣ (F 2)
A。 А₁ = α₁
=il=
C
b, IC
=a,
To this must be added the deflection A, A = a, produced by
the flexure of the beam, and which, according to § 217, is a2 =
P 13
3 WE; the total deflection of the girder is therefore
b h³
19 ,
Pl
B C = A, A = a =
= α = a₁ + a₂ =
P)
aş
For a parallelopipedical girder b
, consequently
α
4 P 19
b h³ E
3
(Fz)
(= (0) + 378)
W b C
b, Σ (F z)
+ 8 = ( ) ² )
3
=
E
b h²
8
and W=
[1 +
E
or, assuming
= 3,
Ꮯ
a
4 P 19
b hs E
[1
1 + 1/
(7)]
4 P 13
E.G., for l
10 h, we have a = 1,01125 .
if then the
•
bh E
3
girder is ten times as long as thick, the deflection at the loaded
end, due to the shearing force, is so small compared with that due
to the flexure of the girder, that in most cases we can neglect it.
In order to determine the modulus of elasticity of a girder A B,
we load it first with a small weight P at the greatest distance l, and
afterwards with a large weight P, at a smaller distance 7, from the
point of support B, and we observe the corresponding deflections
a and a, of the length 7 of the girder. Now we have
262.]
523
ACTION OF THE SHEARING ELASTICITY, ETC.
a
α =
ΡΙΣ (F 2)
b。 W c
P 13
+
and
3 WE
P₁ lΣ (Fz)
1
P₁ 1,3
2
P₁ l² (l — l).
a₁ =
α1
+
+
b. W C
3 W E
2 WE
In order to eliminate C, divide
the first equation by P and the
second by P, and subtract the equations obtained from one
another.
1
Thus we obtain
3
2
a
α,
1
P P₁
WE 3
:(
13
1₁₂² (1 — 7₁₂)`
2
1
で
​1 1,2
WE 3
2
+ 꽁​),
6
and therefore the modulus of elasticity for tensile and compressive
PP
73
forces is
E
(a P₁a, P) W 3
1 7, 2
2
\3
+
6
With the aid of this expression and the formula for a, we
determine the modulus of elasticity for shearing by the formula
C =
ΡΙ
b.
3 Σ (Fz) E
3 WE a P 1³°
§ 262. Elasticity of Tersion.-In order to investigate the
theory of the twisting or torsion of a body (see § 202), we can again
begin with the case of a body H C D L, Fig. 429, fixed at one end,
FIG. 429.
A
H
-P
NSI
P
H
D
but, in order to avoid any
complex change of form, we
must assume that the free
end is acted upon by a couple
(P,-P) whose plane AHB
coincides with the plane of
rotation of the axis C D.
Let us imagine the body to
be composed of long fibres,
such as H K, which, in
consequence of the torsion.
assume the form of a helix,
by which H K comes into
the position L K and the

whole base is turned through an angle HCL=a. If the portions.
H₁ K₁, H. K, etc., of the fibres, whose lengths are unity and whose
cross-sections are F, F, etc., undergo a lateral displacement through
the distance H, L₁ = 6₁, H₂ L₂ = σ, etc., we can put, when the modu-
lus of elasticity for shearing is C, the corresponding shearing forces
S₁ = o, F₁ C, S, σ, F, C, etc. Now if the corresponding angle
=
524
$ 263.
GENERAL PRINCIPLES OF MECHANICS.
=
=
of torsion is H, O L, H, O L, and if the distances of these
fibres from the axis CD of the body are O H₁ = 21, 0 H₂ = Z2, we
have σ₁
= Ф CF, 71,
S₂
•
$ 21902 =0%2 .; hence the strains are S
ФС F2 72 and their moments are
•
2
S₁ z₁ = 4 C F₁ z₁², Są Z₂ = C F₂ z2...
%1
4
2
2
All the forces S₁, S₂ . . . of a cross-section H₁ O L, must in any
case balance the couple (P, — P); if then a is the lever arm A B
of this couple or P a its moment, we can put
2
Pa S₁ z₁ + S₂ %g + ... = C F₁ z² + G F₂ z2² + ...
• $
1
=
2
= C (F₁₂ z²² + F₂ 23² + ...).
$
1
2
Now if we designate the geometrical measure F₁ z₁² + F₂ z₂² + ……..
of the moment of torsion by W', we have Pa
ФСИ.
But the angle of torsion for the entire length C D = 1 of the
body is a = ol, therefore we can put
a C W
1) Pa =
or Pal = a CW,
し
​and the angle of torsion
Pal
2) a =
C W
As we have done previously (§ 215), we can call W C the
moment of torsion, and consequently W the measure of the moment
of torsion, and we can then assert, that the moment of the force P a
increases directly as the angle of torsion and inversely as the length
of the body.
The work done in producing a torsion equal to the angle a is
P
L =
2
•
а α =
a² W C
27
2 W C
P² a² l
;
for the space described by the force P, which causes it, is a a.
These formulas hold good for prismatical bodies alone, for bodies.
with other forms we must substitute in them instead of the ratio
11
a mean value of it.
W
2
§ 263. Moment of Torsion or Twisting Moment.-The
measure W F₁₁² + F₂ Z22 + ... of the moment of torsion can
easily be calculated, according to the rule explained in § 225, from
§ 263.]
525
ACTION OF THE SHEARING ELASTICITY, ETC.
the measure of the moment of flexure for the same cross-section.
If, for example, W, is the measure of the moment of flexure of a
1
FIG. 430.
V
A
N
-U
-X
K
M
-V
Y
B
X
surface A B D, Fig. 430, re-
ferred to an axis XX and W
the same in reference to an axis
Y Y at right angles to the first,
we have for the measure of the
moment of torsion in reference
to the intersection of the two axes
W = W₁ + W.
For a shaft with a square cross-
section A B D E, Fig. 431, we
have, when b denotes the length
of the side A B=D E, according
to § 226, the measure of the mo-
ment of flexure in reference to each axis XX and Y Y

b hs
b₁
W₁
W 2
12
12'
and consequently the measure of the moment of torsion is
b₁ b₁
W = W₁ + W₂ = 2
12
6'
and the moment of the force
Pa =
a W C
Z
a b C
=
0,1667
6 7
a C f₁
?
For a shaft with a rectangular cross-section (b h) we would
have, on the contrary,
a b h (b² + h²)
a b h (b² + h²) C
Pa=
C = 0,0833
127
¿
FIG. 431.
P
FIG. 432.

H

P
X
K
.P
K
H
P
N
For a cylindrical shaft with circular cross-section AB, Fig.
432, whose radius is CA = =r, the measure of the moment of
flexure in reference to an axis FX or YF is (according to § 231)
526
[§ 263.
GENERAL PRINCIPLES OF MECHANICS.
П зов
W₁ = W₂
4
and therefore the measure of the moment of torsion in reference to
the point in that axis is
W = 2 W₁
П зов
2
Now if the twisting couple (P,
P) acts with an arm H K
= a, or each of its components with an arm C H = C K
α
2'
8163
we have
Pa=
a W C
1
апр с
27
=
1,5708
а зов с
7
If the shaft is hollow and its radii are r, and r₂, we have the fol-
lowing formula:
απ
a π (r₁₁ — r₂¹) C
2
(r₁ — r₂') C
Pa =
= 1,5708 a
21
2
The torsion of a shaft A B M, Fig. 432, is generally produced
by two couples (P, P), (Q, —
P), (Q, — Q), which balance each other,
and therefore, instead of 1, we must substitute not the entire length
of the shaft, but the distance between the planes in which the two
couples act; it makes no difference, however, whether we make the
moment of torsion equal to the moment of the couple (P, P) or
to that of the couple (Q, Q). If we denote the arm H K of the
couple (P, P) by a, and the arm N O of the other couple
(Q, — Q) by b, we have
Pa = Q b
W C
7
The foregoing theory gives us for bodies limited by plane sur-
'face moments of torsion, which vary somewhat from the exact
truth; for we suppose, in calculating them, that the bases of the
prism subjected to the torsion remain plane surfaces, while, in re-
ality, they become warped. According to the researches of Saint
Venant, Werthheim, etc. (see the "Comptes rendus des séances de
l'académie des sciences à Paris," T. 24 and T. 27, as well as "l'In-
génieur," Nos. 1 and 2, 1858; in German in the "Civilingenieur,"
4 Vol., 1858), we have for a square shaft
Pa = 0,841
a b₁ C
61
a b¹ C
0,1402
in which b denotes the length of the side of the square cross-section.
For bodies, the dimensions of whose cross-sections differ very
§ 263.] ACTION OF THE SHEARING ELASTICITY, ETC.
527
much from each other, these variations are greater; E.G., for a pris-
matical body with a rectangular cross-section, whose width is b and
whose height is h, we have
h b³ b h (b² + h³)
b h³
W =
W₁ + W₂
+
12
12
a W C
Pa=
a b¹ C
"
67
a b h (b² + h³) C
12 /
12
and therefore
Now if this formula requires a correction, when hb, in which
case P a =
it is natural to expect that when 6 differs ma-
terially from h, in which case the surface of the sides will become
more warped, it will no longer be sufficiently accurate. In fact,
taking into consideration the warping of the surfaces, we find by
means of the calculus
Pa=
a h³ b³ C
3 (b² + h²) l'
and according to the later experiments of Werthheim, the mean
value of the required coefficient of correction is
quently we must put
0,903; conse-
a hs b⁹ C
P a = 0,903
0,301
3 (b² + h²) l
a h³ bs C
(b³ + h³) l®
Ifb is very small compared to h, we have
Pa = 0,301
a h b³ C
}
If the angle of torsion is given in degrees, putting
0,017453 aº, we obtain
180°
1) for prismatic girders or shafts with a circular cross-section,
the diameter of which is d = 2 r
ai π pr
Pal=
2
=
C =
r¹
απα
C
32
a° π² p¹
180º.2
aº π² d¹
C
C
180° 32
r¹
0,001714 a° d' C;
1,571 ar C = 0,0982 a d' C = 0,02742 aº pt C
2) for prismatic girders, axles or shafts with a square cross-section,
the length of whose side is b, when we neglect the coefficient of
correction,
a b C
Pal=
= 0,1667 a b' C
6
a° ñ b₁ C
1080°
= 0,00291 a° b* C.
528
[§ 264
GENERAL PRINCIPLES OF MECHANICS.
Inversely we have
Pal
Pal
Pal
a = 0,637
= 10,18
6-
and
pt C
ď¹ C
b* C
a° = 36,4
Pal
Pal
Pal
= 583
= 344
4
p¹ C
d C
b¹ Cr
The values for C must be taken from Table III. in § 213.
Hence we have, E.G.,
1
1) For cast iron, C = 2840000, whence
Pal = 77900 aº pª
4867 a° d* = 8264 a° b' and
Pal
¤° = 0,00001281°
pr
Pal
0,0001211°
Pal
0,0002053°
d'
b₁
= =
2) For wrought iron, C 9000000, whence
P a l = 246780 a° r¹ = 15426 a° d' = 26190 a° b' and
Pal
a° = 0,00000404°
=0,0000648°
Pal
d4
= 0,0000382°
Pal
b₁
-
3) For wood, C= 590000,
Pal
161800 aº p¹
Pal
1011 a° d' = 1712 aº b¹ and
a° 0,0000617°
= :
0,000988°
pt
Pal
d¹
Pal
0,000583°
EXAMPLE-1) What moment of torsion can a square wrought-iron shaft
10 feet long and 5 inches thick withstand, without suffering the angle of
torsion to become more than of a degree? Here, according to this table,
we have
625
P a = 26190 · 1 · 10. 12
=
34102 inch-pounds 2842 foot-pounds.
2) What is the amount of torsion sustained by a hollow cast-iron shaft,
= 6 inches and
whose length is 100 inches and whose radii are r,
r 2
= 4 inches, when the moment of the force is Pa = 10000 foot-pounds?
Here
consequently
4
aº (r₁ + — r₂¹)
Pa 77900
7
Pal
10000. 12. 100
aº
77900 (rı r₂+)
77900 (62 + 42) (62 — 42)
1
ទ
120000
779.52. 20
1500
10127
degrees 8,887 minutes = 8 minutes 53 seconds.
§ 264. Resistance to Rupture by Torsion.-If in a prism
CK L, Fig. 433, twisted by a couple (P, P) the shearing force
per unit of surface at a certain distance e from the axis CD is S,
§ 264.]
529
ACTION OF THE SHEARING ELASTICITY,
ETC.
the shearing force at any other distance z, is =
21
S, and its mo-
е
A
H
P
P
FIG. 433.
Ho
S₂
D
NS₁
K
ment is = S, and for a
e

cross-section F, it is
F₁ z₁2
Z
e
S
S = Fizi²;
e
in like manner the moments
of the shearing forces of other
cross-sections F, F...
which are at the distances
Z2, Z3 from the axis C D,
•
S
2
are F₂ z₂²,
e
S
2
F'3, Z3², etc.;
е
hence the total moment of tor-
sion of the body is
e
S
Pa =
F₁ x² + — F₂ 22² +
S
S
F3 23² +
e
e
S
e
S W
1) P a =
W
Pa
е
S
е
(F₁ z1² + F₂ 22² + . . .), I.E.
or Pa e = SW, and
Substituting for S the modulus of proof strength T for shearing,
and for e the greatest distance of the elements of the cross-section
from the neutral axis, we obtain in the formula
=
2) Pae TW an equation for determining the dimensions
of the cross-section, which the body must have if it is not to be
strained at any point beyond the limit of elasticity. If, instead of
the modulus of proof strength T, we substitute the modulus of
rupture K for shearing, we obtain the moment P, a, which will
break the body by wrenching; it is
KW
3) P₁ a = e
we
For a massive cylindrical shaft, whose diameter d= 2 r, we
bave
W
п зов
π 73
and therefore
ė
2 r
2
π p³ T
π d³ T
Pa =
2
0,1963 d³ T, and also
π µ³ K
16
π ď³ K
P₁a:
= 0,1963 ď K.
2
16
34
530
[§ 264.
GENERAL PRINCIPLES OF MECHANICS.
If the shaft is hollow and the diameters are d₁ = 2 r, and d₂ =
2 r, in which case
Pa =
in which F
W
π (r₁₁ — r₂¹)
we have, on the contrary,
e
2 Ti
r₂')
π (d₁¹
T=
T =
d2) F (d₂² + d₂²)
16 di
4 d₁
2
da)
π (r₁ˆ
2 ri
π (dr²
4
T,
denotes the cross-section of the body.
For a prismatical body with a square cross-section, the length
of whose side is b, we have
v
and e = 1 b √ 2 = b √ 1, whence
b₁
W =
6
W
bs
b
b³ T
and P a =
0,2357 b³ T
e
3√2
3 √ 2
If in the fundamental formula P a =
o C W of § 262 we substi-
tute o
σ
e
tang. s
in which e denotes the distance of the most
e
remote fibre from the axis of rotation CD and d the angle H K L,
which this fibre has been turned from its original position by the
torsion, we obtain
Pae
Pae
CW tang. d; but we have also
SW, hence
in which
S = C tang. 8, and therefore
T = C tang. 8, or tang. d =
T
C
denotes the angle of displacement, when the strain
has reached the limit of elasticity.
The mechanical effect, which is required
through an angle a, is, according to § 262, L =
to twist the shaft
P² a² l
2 W C'
and there-
S W
S² W I
fore if we substitute Pa =
e
which S denotes the maximum
we can put L
strain.
in
C
2 e²
At the limit of elasticity S T; hence it follows that the me-
chanical effect necessary to twist the body to the limit of its elas-
ticity is
L
T2
C
•
WI
2 €**
e
§ 264.]
531
ACTION OF THE SHEARING ELASTICITY, ETC.
πρ
For a prismatic body with a circular cross-section W =
and er, whence
2
T¹2
π p² l
Ta
L
V₁
2 C'
2
4 C
and, on the contrary, when the cross-section is a square
b*
b2
W =
and e²
and therefore
6
2'
T2
T2
T2
L
2
V.
2 C
3 b2
6 C
6 C
T3
o CT
στ
Now
is the modulus of resilience for the
2 C
20
2
limit of elasticity; hence we have for the cylinder LA V, and
for the parallelopipedon L} A V.
The work done in both cases is proportional to the volume of
the body alone (compare § 206 and § 235).
1
3
We can also put for the mechanical effect necessary to rupture
of the body by wrenching LBV and BV, in which B
denotes the modulus of fragility for wrenching.
If we assume with General Morin for all substances
T
C
tang. & = 0,000667
or the angle of displacement d = 2 min. 18 sec., we obtain for
cast iron
T = 200000. 0,000667 = 134 kilo. = 1906 lbs.,
therefore, when we employ the French measures
Pa 26,3 d³ 31,6 6' kilogr. centimeters,
= =
and, on the contrary, when we employ the English measures
Pa = 374 d³ = 449 b' inch-pounds.
Under the same conditions we have for wrought iron
T = 630000. 0,000667 = 420 kilo. = 5974 lbs.,
and therefore
or
Pa = 82,4 d³ 99,2 6 kilogram centimeters,
Pa = 1173 d² = 1408 6 inch-pounds.
Likewise under the same conditions we have as a mean for
wood
F = 41650. 0,000667
=
27,8 kilogr.
= 395 lbs.,
whence
or
Pa = 5,46 d = 6,55 b³ kilogr. centimeters,
P a = 77,5 d³ = 93,1 b³ inch-pounds.
532
[§ 265.
GENERAL PRINCIPLES OF MECHANICS.
The coefficients of these formulas are correct only for bodies
at rest or for shafts, which turn slowly and smoothly; for common
shafts we give double security, I.E., we make the coefficients but
half as great. When their motion is very quick and accompanied
by concussions, we are obliged to make the coefficient but one-
eighth of those given above.
EXAMPLE-1) The cast iron shaft of a turbine wheel exerts at the cir-
cumference of the cog-wheel upon it, which is 6 inches in diameter, a
pressure of 4000 pounds. Required the thickness of the shaft. Here the
moment of the force is P a = 4000. 6 = 24000 inch-pounds, and conse-
374
quently the diameter of the wheel, when we put Pa = d", is
2
d =
24000
187
=
5,04 inches.
If the distance from the cog-wheel to the water-wheel is 7 = 48 inches,
we have, according to the foregoing paragraph, the angle of torsion
=
0,0002053°
24000 . 48
5,044
0,367° = 22'.
2) A force P = 600 lbs. acts with a lever arm a = 15 feet = 180 inches
upon a square fir shaft, while the load Q acts with an arm of 2 feet at´a
distance = 6 feet 72 inches in the direction of the axis; how thick
should the shaft be made and what is the angle of torsion?
In order to have quadruple safety, we must put
Pa = 600. 180 = 108000 =
93,1 b³
hence the width of the side is
b =
3 4. 108000
93,1
and the angle of torsion is
=OD
0,000583
108000 . 72
(16,68)¹
16,68 inches,
= 0,0586 degrees 33 minutes.
=
CHAPTER IV.
OF THE PROOF STRENGTH OF LONG COLUMNS OR THE RESIST-
ANCE TO CRUSHING BY BENDING OR BREAKING ACROSS.
§ 265. Proof Strength of a Long Pillar Fixed at One
End.-If a prismatic body A B (I), Fig. 434, is fastened at one end
§ 265.]
533
PROOF STRENGTH OF LONG COLUMNS, ETC.
B and acted upon at the other by a force P, whose direction is that
of the longitudinal axis of the pillar, the relations of the flexure,
FIG. 434.
I
II

A
L
B
T
4 YP
A
M
K
under these circumstances, are very different from what they are
where the force acts, as we have seen in § 214, etc., at right angles
to this axis. The neutral axis A
The neutral axis A B (II) assumes in this case
another form; for the lever arm of the force P is represented by
the ordinate M 0 = y and not by the abscissa A M = x, and its
moment is not P, but P y; consequently the radius of curva-
ture KO= r is determined by the expression
r =
WE
Py
while, according to § 215, for a bending force acting at right
angles to the axis we must put
WE
Px
At the point B, where the pillar is fastened, y becomes the de-
flection B C = a, the radius of curvature r =
WE
is a minimum
Pa
and the curvature itself a maximum. On the contrary, at the point
of application A, where y = 0, the radius of curvature is infinite
and the curvature itself null.
If we denote by the arc, which measures the angle 0 K 0₁of
curvature of the element 0 0₁ = σ of the curve, we have r =
O O
o
σ
Ꮄ
and therefore Pyo = WE 8; and if ẞ° is the angle of inclina-
tion 0 0, N of the same to the axis A C, we can put the element
NO of the ordinate = v = σ B, and therefore
P y v = WE ß §, and in like manner
Ρ Σ
Ε
PE (y v) = WEΣ (38).
}
534
[§ 265.
GENERAL PRINCIPLES OF MECHANICS.
In order to find the sum Σ (y v) for the arc A O, let us substi-
tute for y, v, 2 v, 3 v . . . n v in the above equation. Thus we
obtain Σ (y v)
= V (y) = v (v + 2 v + 3 v + + n v)
n² v
n² v²
2
2
>
or since n v = M 0 = y,
Σ (yv) =
y² and P Σ (y v) = { P y².
2
•
= v
In like manner, to find Σ (8 d), we substitute for ẞ successively
B, B+ 8, B + 2 d... ß + n d, and complete the summation as
follows:
Σ (38) = 5Σ (3) = 8 (ẞ + B + 8 + B + 28 + ... + B + n 8)
(β δ) δ (β) (ß ß ß ß
= 8 [nẞ + (1 + 2
+ 3 + ... + n) s]
= 8 (n ß + ¹² ³)
=ns B
2
n ô (ß +
(8 + 220).
If the angle of inclination at A, = a, we can put ß + n 8
and therefore
Σ (3 8) = (a — B) (15 .
whence
—
a,
α
В
+
WEΣ (38) =
ΠΕΣ (β
Py'
For the end B, y
2
WE (a ẞ"), and finally
WE (a³ - B²).
=
a and B 0, and therefore
Pa² = WE a² and
—
P (a² y²) = WEB²,
from this we obtain the tangential angle
~ ¹³) = į ( a − B) (a + B) = ! (a² — B³),
1) BV
'P (a² — y)²
WE
From ẞ and the element N 0 = v of the ordinate we obtain
the element of the abscissa
บ
NO=
§
= v z
B
WE
P (a² - y³)
P
§ V
WE
Va²
บ
υ
WE
or
√a²
y²
Р
>
να
yo
If with the hypothenuse C B = a of the right-angled triangle
FIG. 435.
B
0
=
BCD, Fig. 435, whose altitude is B D = y
and whose base is CD Nay, we de-
scribe an arc A B, we have for the element
BO the proportion

=
BO
BN
C B
CD'
ψ
a
I.E.
υ
Na²
a² — yº
A
D
whence
§ 265.] PROOF STRENGTH OF LONG COLUMNS, ETC.
535
να
υ
2
ป
a
and
un
y
P
必
​as well as
WE
a
P
Σ
WE
× (5) = 1/
Σ (ψ).
a
But () is the sum of all the elements of the abscissa and is
= x, and Σ (4) is the sum of all the elements of the arc A B and
is equal to the arc A B itself; therefore we have also
x v
P
WE
arc A B
= sin.-1
У
α
α
The abscissa of the elastic curve A B, Fig. 434, II, is therefore
and its ordinate is
2) x = 1
WE
P
•
sin.-1
y
a
P
WE
3) y = a sin. ( x √
(a
=
If x = A B A C 1, the length of the column, we have
y = the deflection B C = a; therefore
a = a sin.
whence
Р
WP
4) P
WE
P=
π
2'
P
E),
I.E., sin.
sin. (1 V
(IN WPE)
Р
1,
W E
from which we obtain the bending force
WE
( 77 )' w
Since this formula does not contain the deflection a, we can
assume that the force P, determined by it, is capable of holding the
body in equilibrium, however much the body may be bent. This
peculiar circumstance is owing to the fact that the increase of
the flexure is accompanied not only by an increase of resistance, but
also by an increase of the lever arm a, and consequently of the
moment P a of the force.
The force necessary to rupture the pillar by breaking it across,
is therefore
P = = ( 577 ) WE
=
2,4674
WE
で
​REMARK.-If we substitute in the formula y = a sin. ( x 1
2
(5)
P
P=
WE
W E, we obtain the following equation of the elastic curve for this
case of the action of a force
536
[§ 266.
GENERAL PRINCIPLES OF MECHANICS.
Substituting in this
we obtain..
y = a sin.
(717).

x = 0
2
·
27 37 47
57 6l, etc.,
0
a
0
•
a
0 a
0, etc.
If, then, a column, whose length is l, is increased any amount in length,
2
π
a force P
2
1
WE will bend it in the shape of the serpentine line
A B А‚ В¸ ø . . ., Fig. 436, which is composed of a number of similar arcs
ø
A Band is cut by the axis A X at the distances A A₁, A A2,
and at the distances A C, A C₁, A C₂, the curve is
a, C₁ B₁ a, C B₂
FIG. 436.
A
B
C
A2
Be
Ca
at its maximum distances C B
from this same axis.

= a
§ 266. Parallelopipedical and Cylindrical
Columns. For a parallelopipedical column, the
greater dimension of whose cross-section is b and the
smaller one is h, we have W
3
b h³
12
(see § 226), and con-
sequently the force necessary to rupture the same
B₁ by breaking it across is
P
=
( 577 ) 20 21 2
Π b h³ E
b h³ E
0,2056
The resistance of a parallelopipedon to breaking
across is directly proportional to the width b and to the
cube (h) of the thickness or smaller dimension h of its
cross-section and inversely proportional to the square
(1²) of the length.
For a cylindrical pillar, whose radius is r or whose diameter is d,
π pot
πα
(see § 231), consequently we have
W =
π pr
4
64
π r¹ E
тя
d' E
r¹ E
P
E
=
1,9381.
4
256
で
​72
d' E
16 で
​= 0,1211
Therefore the (reacting) strength of a cylindrical column, by
which it resists bending or breaking across, is directly proportional
to the fourth power of its diameter and inversely proportional to the
square of the length.
For a hollow column, whose radii are r and r₁, and whose diam-
eters are d and d₁ = µ d, we have
§ 266.] PROOF STRENGTH OF LONG COLUMNS, ETC.
537
π³ (p❝ — r₁') E
π³ (d' — d₁') E
-
P
16
12
256
で
​ポ
​256
d' E
12
0,1211 (1 μ¹)
d* F
12
(1 — µ³)
If the column A B A, Fig. 437, is not fixed at the lower end
A, but only stands upon it, it will bend in a symmetrical curve,
each half B A and B A, having the form of the axis of a column
fixed at one end (Fig. 434). The above formula can be applied
1
directly to this case by substituting instead of 1; of course denotes
the total length of the pillar. The proof load is therefore four
times as great as in the first case, and it is
P
(7)*
( 7 ) WE
π² b h³
12 F
E =
πT³ d¹
64 3
E.
This case of flexure occurs when, as is represented in Fig. 437,
FIG. 437
I
III
II
A.
A
P
P
P
C
B
A₁
FIG. 438.
JII
11


B
B
B
A
P
M-C
B
B
3.
1
I. and III., the ends of the pillar are rounded or when they are
movable around bolts. An example of the latter case is the con-
necting rod of a steam engine.
1
If a pillar is fixed at both ends, as is represented by B A B₁,
Fig. 438, I. and III., its axis will be bent in a curve B A C A, B₁,
Fig. 438, II., with two points of inflection A and A₁, and in which
the normal case of curvature is repeated four times, substituting,
7
therefore, in the formula for the normal case instead of 1, we ob-
4'
538
[§ 266.
GENERAL PRINCIPLES OF MECHANICS.
tain the proof load of such a pillar fixed at both ends
P = 1
( 27 )* w
π 2
WE
π² b h³
3 12
3
По а
E
E.
16 72
According to Hodgkinson's experiments, the proof load is only
twelve times as great as in the normal case, while according to the
above formula it would be sixteen times as great.
The principal example of this case of flexure is that of the
piston rod of steam engines, etc.
If, finally, a column A O B, Fig. 439, is fixed at one end B and
at the other prevented from sliding sideways,
the proof load P is eight times as great as in
the normal case, or
FIG. 439.
I

II
2
A
P = 8 8 (17) Ⓡ
WE =
6
P
B
B
2b hs E T³ d' E'
32 で
​The force which is necessary to crush a
column, whose cross-section is F and whose
modulus of rupture is A, is given, according to
§ 205, by the simple formula P = FK.
If we put this force equal to the force
P =
WE
necessary to produce rupture by breaking across
in the normal case, we obtain the equation
E
K'
1
Fr
IV
(1)² F, or 7 V
F π E
W 2 K
For a cylindrical pillar, whose thickness is d, in which case
F
16
it follows that
W
d29
1
d
π
E
1
8
K
0,3927 V
E
K'
For cast iron E- 17000000 and K =
17000000 and K = 104500, hence
VE
7
K
= √ 162,68 = 12,8 and
= 5.
d
For wrought iron E = 28400000 and K =
31000, hence
E
K
/
916 = 30,3 and
12.
d
Finally for wood we have as a mean
E
1664000 and K
6770, hence
§ 267.]
539
PROOF STRENGTH OF LONG COLUMNS, ETC.
E
K
√ 246 =
15,7 and
= 6.
1
If a column is free at both ends, the values of
d
great as those found above.
are twice as
When the ratio of the length to the thickness is that just given,
the resistance to breaking across is equal to that of crushing, and
it is only when the pillars are longer than this, that the resistance
to breaking across exceeds the resistance to crushing. In this case
the dimensions of the cross-section are to be calculated by the
above formula.
EXAMPLE-1) The working load of a cylindrical pine column 12 feet
is
long and 11 inches thick, assuming 10 as a factor of safety,
P
3d E
64 7 10
0,4845 (11) *
12
•
166400 = 80620. 0,7061 = 56900.
་
2) How thick must such a column of cast iron be made, when its length
is to be 20 feet and the load 10000 pounds? Here, if we put instead of E,
E
1700000, we have
10
64 Pl
d =
3
T
-3.1700000
2402
82,34375
640000 . 2402
31. 1700000
240
9,074
5,14 inches.
According to the formula for the strength of crushing
d =
V
4 P
π K'
K
10
or, substituting = 10400 pounds in the calculation, we have
4. 10000
400
50
d 1
π. 10400
1,106 inches.
介 ​104
13 T
•
If the length of the pillar does not exceed 10. 1,106
required thickness would then be but 1,106 inches.
=
11,06 inches, the
(§ 267.) Bodies of Uniform Resistance to Breaking
Across. If a pillar A B, Fig. 440, fixed at one end, is so shaped,
that in all its cross-section the strain is the same, a solid of uni-
form resistance is formed, which requires the minimum amount of
material for its construction (see § 208 and § 253). The cross-
section of such a body is certainly a maximum at the fixed end B,
and it decreases gradually towards the end 4. The law of this
decrease is found as follows: denoting again by x and y the co-
ordinates of a point 0 in the axis of the column, by a the tangen-
540
[§ 267.
GENERAL PRINCIPLES OF MECHANICS.
tial angle M A O for this point, by W the measure of the moment
of flexure, by z the radius 0 0, of the column at this point and by
S the strain at the surface A O, B₁, which is there-
fore that at the point 0, of the cross-section through
O, we have
FIG. 440.
A

PV
M
S =
Mz
W
Py z
(see § 235) and
W
WE
M =
= P y =
Py
WE
d tang, a
d x
j
(see § 218), whence
B B₁
d tang. a
S
=
E z
d x
S d y
or, since tang. a =
Ez tang. a d tang. a.
But, since for a circular cross-section
W π 23
Z
4 Py
S
Py
W
π Z
πT S
п
3 π S
d y
d (z³)
=
4 P
whence
dy
d x
Z
4
y,
and we have
3 π S²
z² d z,
4 P
4 P
or S z³ P
П
4
z² d z and Sd y
3 π S2
z d z =
4 PE
tang. a d tang. a.
By integration we obtain
z² = Const.
S'2
17 PE* =
tang. a,
2
and, if we denote the radius of the cross-section at B by r, we have
Pla
St2
π (p² — z²). = tang.² a, since a = 0; hence
PE
3 п
tang. a SV
=
√ p² — z².
4 PE
Putting tang, a =
dy
S
z² dz
درم است
π
we obtain
dx
P
d x
3 π E
z² d z
√ p²
z² and
4 P
d x
d x =
√ 3
3 ПЕ
z² d z
= r² V
4 P
3 π E
4 P
u² d u
་
Z
when
за
is denoted by u.
§ 267.]
541
PROOF STRENGTH OF LONG COLUMNS, ETC.
But
u²
1
1
чег
1
+
1
ге?
2
√1 − u² +
√1
u²
√1 — u²
√1 -
u²
and therefore
S
u² d u
√1 -
U2
Sv
√1 — u². du +
S
d u
#1 U²
d u
= − ¦ u √1 — u² + ½
1
u²
{ u √ 1 − u² + 1
u² + ¦ sin.~' U.
1
(See the Introduction to the Calculus, Art. 27 and 26.)
Hence we have
3 π Ε
X =
16 P
[2²
2
rº sin.¹ %; — z √r³² — z' ] · .
For x = l, z = r, the radius of cross-section of the base, for
2
which sin.
z v p²
1 =
P =
π
=r,
Π
-1 1 =
and
— sin.-¹ 1
༡༠༠
z² = 0. Therefore it follows that
(77)
2
3 π E
16 P
3 π 2*
16
and that the proof load is
E =
م حلم
(177)
π jot
E
4
that is, three-fourths of the proof load of a cylindrical pillar, whose
radius is r (compare § 265). Consequently the radius of the base
of a column of uniform strength is = V 1,075 times the
radius of a column of the same length, whose proof strength is the
same.
Comparing the abscissa x with the total length of the column,
we obtain
X
π
Ї
2
[ [ sin.¹ r
Z
11
(3)']
2
に
​times the area of the
2 z
7'
segment of a circle, whose radius = 1 and whose chord
2 x
πι
If, then, we regard as the area of the segment of a circle,
пе
can determine, by means of a table of segments (see the Ingenieur,
page 152), the corresponding angle at the centre, and from it we
can calculate for a given abscissa x the corresponding radius of the
پینے می
542
[$ 268.
GENERAL PRINCIPLES OF MÉCHANICS.
cross-section z = r sin.
$
2 x
1
; E.G., for x =
1 1,
2
πι
π
and we find from the table of segments
radius of the cross-section of the pillar is
= 0,3183,
= 93° 49′; hence the
z = r sin. 46° 50′ = 0,729 r.
To resist rupture by crushing, the radius of the cross-section.
P
V
>
π T
and this radius must
of the pillar at the top must be r.
always be employed for all points, where the formula for breaking
across gives smaller values for z.
If the pillar stands with its base unretained, as is represented
in Fig. 437, the calculation must be made in the same manner for
one-half () of it. The maximum radius r is, of course, that of
the cross-section in the middle, and it corresponds to the formula
π Ÿ³ E
(Z)²
4
P = }} ( Z )' ( 7 ) * T
§ 268. Hodgkinson's Experiments.-The recent experi-
ments of Mr. Hodgkinson upon the resistance of columns to
breaking across (see Barlow's report in the "Philosophical Trans-
actions," 1840) confirm, at least approximatively, the correctness
of the formulas deduced in the foregoing pages. According to this
experimenter the formula
2
2
2
π π d' E
64
(77)
b E
12
= (~~) w E = ( 7 )² = 1
P =
for prismatical columns with circular or square cross-sections is
correct for wood when we introduce a particular value for E; but,
on the contrary, it can be employed for wrought iron only when
we substitute for d' the power d,, and for cast iron it is suffi-
ciently correct when d' and l' are replaced by the powers d, and '".
The chief results of Hodgkinson's experiments upon prismatic
pillars with circular and square cross-sections are given in the fol-
lowing table. The coefficients given in it refer to the case when
the pillars are cut off at both ends at right angles to their longitu-
dinal axis and repose upon these bases. When the ends are rounded
so that these extremities of the columns are not prevented from
assuming any inclination, these coefficients are nearly three times
as small. If, on the contrary, the column is fixed at one end and
the other capable of turning, the coefficient is but half as great as
in the first case. If, finally, one end of the pillar is fixed and the
§ 268.]
543
PROOF STRENGTH OF LONG COLUMNS, ETC.
other capable of being turned and of sliding, the proof load is but
one-tenth of that of the first case, where both ends are fixed.
TABLE OF THE FORCES NECESSARY TO RUPTURE COLUMNS BY
BREAKING THEM ACROSS.

Name of the prismatic pillars.
Breaking stress.
English measure, French measure, Prussian measure,
tons.
kilograms.
new pounds.
Cast-iron pillars with circu-
d 3,53
d 3,55
3,55
lar cross-section.
Wrought-iron pillars with
44,16
10900
94700
71,7
71,7
d 9,55
d 3,50
d 3,55
circular cross-section
133,75
46140
284400

12
12
Square pillars of dry Dantzic
oak
b+
b₁
b*
10,95
2480
12
2357°F
b₁
b+
Square pillars of dry fir
7,81
1770
16840
で
​7.2
で
​In the column for English measured and b are given in inches,
7 in feet, and P in tons of 2240 pounds. In that for the French
measures, on the contrary, d and b are given in centimetres, 7 in
decimetres, and P in kilograms, and in the last column d and bare
expressed in inches, 7 in feet, and P in new pounds.
Mr. Hodgkinson also found that cast-iron pillars, with round
ends, were sooner crushed than broken across, when < 15 d, and
when the ends were flat as long as I was < 30 d. Dry wood possesses
double as much strength as timber just felled. When employing
this formula for calculating the working load of columns, we employ
a coefficient of security of to, or a factor of safety of from 4 to 12.
Hence, with sextuple security, we can put for cast-iron pillars,
when d and I are given in inches,
ΤΣ
P =
44,16
6
121,7
•
₫3,55 44,16
71,7
6
73,55
68,3
= 502,688
21,7
€ 3,55
71,7
tons,
and d = (P)
0,0173 (P l¹‚7) 0,2817 inches.
For wrought-iron pillars we have, when we adopt the same
coefficient of security,
(3,55
P = 3210
tons and
d = (Pt)
0,01028 (P ľ²) 0,2817 inches.
544
[§ 269.
GENERAL PRINCIPLES OF MECHANICS.
For pillars of oak wood, employing a coefficient of security of
ro, we have
b
= 157,68 2/ = 267,69 tons,
P =
= 157,68 (
(²)²°F =
d'
12
b
0,2822 (P 7²); and d = 0,2472 (P P²); inches.
Finally, for pillars of fir wood, we have
P =
112,46 (2)
. F
F = 112,46
12
d¹
190,92
b = 0,307 (P l²) and d = 0,269 (Pl²)}.
EXAMPLE. For a cylindrical fir post, 11 inches thick and 12. 12
inches long, fixed at both ends, the proof load is
P
2
121
Р
190,92 (144
)=
= 134,802 tons.
= 144
If the ends of such a pillar are capable of moving freely, the proof load
1
} P = 44,934 tons, while according to the theoretical formula we
have P₁
-
56900 lbs. 25,402 tons. (See Example 1 of § 266.)
§ 269. More Simple Determination of the Proof Load
of Columns.-The foregoing formulas for the bending and
breaking across of pillars are calculated upon the assumption that
the force P is applied exactly at the end A of the longitudinal axis
of the pillar. Now since m practice this is scarcely ever perfectly
true, and since the action of the force ceases to be central as soon
as the pillar bends, it is advisable, in determining the proof load
of a beam, to take into consideration from the beginning the
eccentricity of the point of application of the force. Assuming
that the point of application D of the force P is at a distance
DA = c from the end 4 of the axis AB, Fig. 441, of the column
and that the deflection B C = a of the pillar is small,
compared with e, we can consider the elastic curve
formed by the axis of the pillar to be a circle, whose
radius is
FIG. 441.

D
P
T
=
But now
2 å
P (a + c) r = WE. whence
P (a + c) l²
a
P V c
2 WE a, as well as
and
2WE-PE
a + c =
2 W Ec
2 E PP
If F denotes the cross-section of the pillar and e half its thick-
ness, measured in the plane A B D, the uniform strain produced
in each cross-section by the force P is
§ 269.] PROOF STRENGTH OF LONG COLUMNS, ETC.
545:
S₁
P
F"
and the strain produced at the exterior surface by the moment
P (a + c) of the force is
S2
P (a + c) e
W
2 P Ece
2 WE PP
and consequently the maximum strain in the pillar is
2 EFce
PP
P
2 P Ece
P
S = S₁ + S₁ =
+
F 2 WE – P ľ²
F'
(1
1 +
2 WE-PE).
2 E Fce
Putting $ = to the modulus proof strength T, we have
S
P (2 WE
(1
P(1 +
2 WE-PU
Pr+ 2 EFce)
=
Now if P is small compared with
= FT, or
(2 W E – P F²) F T.
(W+ Fc e), we can put
FT
Fee FT
+
2 WEFT
P
or
2 E (W + Fc e) + F Tr²
1 +
Р
W 2 WE
FT
P
Φ + ψ
da
in which and are empirical numbers.
The civil engineer Love (see "Mémoire sur la Résistance du fer
et de la fonte, etc.," Paris, 1852) deduced from the experiments of
Hodgkinson the values = 0,45 and 4 0,00337; hence we have
P = XFT
FT
1,45 + 0,00337 (~1) *
from which the following table for the coefficient
1
x =
1,45 + 0,00337
(3)
has been calculated.

7.
ΙΟ
20 30 40
50
60
70 80 90
100
x=0,559 0,357 0,223 0,146 0,101 0,0735 0,0556 0,0435 0,0347 0,0285
These values of x must be multiplied by the modulus of proof
strength T for compression, when the modulus of proof strength
for long pillars is to be determined for a given ratio of length.
General Morin gives, after Rondelet, the following table, which
35
546
[§ 269.
GENERAL PRINCIPLES OF MECHANICS.
furnishes too great values for x, when the pillars are of medium
length.
I
d
12
24
36
48
60
72
七​=
I
LOKO
5C
6
100
-12
3
18
6
1
24
EXAMPLE—1) What load can a pine post bear, whose length is 15 feet
and whose thickness is 12 inches? According to the table upon page 404,
the modulus of proof strength for a short pillar is T = 2600; but since the
ratio of the length to the thickness is
ī
d
15, we have
1
1
x =
=
1,45 + 0,00337. 152
0,453,
2,208
T=
0,453 . 2600 =
whence we obtain the modulus of proof strength x T
1178 pounds; hence the proof strength of the pillar is
P1178
π d²
4
1178. 0,7854. 144 133000 pounds.
If we employ a factor of safety 3, we can put
P =
133000
3
=44300 pounds.
2) How thick must a hollow cylindrical pillar of cast iron, 25 feet
long, be made, when it stands vertical and is required to support a
load P 100000 pounds? Assuming the diameter d, of the hollow part
to be three-fifths of the exterior diameter (d) of the pillar, we can substi-
tute in the theoretical formula
π-B
P =
24
d =
П3
4
d'
•
16
72
d.
4
1
E (§ 226),
4 Pl²
d¹
16
.3
0,0544 π³ E
[1
(8)'] = 0,0544 d', whence we obtain
Substituting in this expression P = 100000, l² = (25. 12)² = 90000,
31, and, instead of E,
E
10
14220000
10
1422000,
we obtain the required thickness of the pillar

d =
4
V
=V
400000.90
0,0544 . 31. 1422
187500
0,0527. 237
6000000
1,6864. 237
11,07 inches.
If we make d = 11,25 inches, we obtain d₁ = 0,6. 11,25 = 6,75 inches,
1
$ 270.]
547
COMBINED ELASTICITY AND STRENGTH.
According to our last formula we have, when we assume
25
d = 1
= 25,
for the required cross-section of the pillar
F = [1,45 + 0,00337
+ 0,00387 (2)²]
21 P
T
3,556. 100000
T
355600
T
and putting, according to § 212,
18700
T
= 6200 pounds,
3
we obtain
F
355600
6200
57,35, and therefore, since
F
4
the required exterior diameter of the pillar
— (d² — d¸³) = [1 − (})²]
πα
4
0,16 π α,
d
F
0,16
57,35
10,68 inches.
0,16 π
Assuming & = 11 inches, we obtain
d₁ = 0,6 d
0,6. 11 = 6,6 inches.
CHAPTER V.
COMBINED ELASTICITY AND STRENGTH.
§ 270. Combined Elasticity and Strength-A body is
often acted upon at the same time by two forces, E.G. a tensile and
a bending one, etc., by which a double change of form is produced,
as, E. G., an extension and a bending. We call the force with which
a body resists this two-fold change of form its combined elasticity
and strength, and we will proceed to investigate the most important
cases of this kind.
Properly speaking, the case (§ 214) of the bending of a body
A K B 0, Fig. 442, is really one of combined strength; for the
force A P = P, which acts at the end 4 of the body, can be re-
solved into a couple (P, P) and a force SP P. The former,
which alone we have previously considered, tends to bend the por-
tion A S of the body, and the latter tends to tear this piece from
548
[§ 270.
GENERAL PRINCIPLES OF MECHANICS.
the remaining portion S B. The latter force can be resolved into
K
H
R
P
FIG. 442.
L
NN,
P₂T
-P
W
Ꮲ
P
1
M
two components
= P cos. a
and
P₁

P₂ = P sin. a
(§ 215), one of which
acts at right angles to
the direction of the fibres
and the other in the di-
rection of the axis of the
fibres. The latter com-
ponent combines with
the strain in the fibres
produced by the bend-
ing and increases the ex-
tensions upon the side of
the tensile strains and
decreases the compres-
sion upon the other side.
The magnitude of the
extension of each fibre
RS = KN,
etc., whose length
1,
by the tensile force P
sin. a is (§ 204)
σ
P sin. a
FE
>
F denoting the cross-section NO of the body.
2
If at this distance from the line N, O,, Fig. 443, which deter-
mines the ends of the fibres, that have been extended by the bend-
ing, we draw a line N, O, parallel to it, it will form the boundary
of the fibres which have been submitted to both causes of change
of length, and it will cut the original limit in a point S, which
corresponds to the fibre, that is unchanged in length, and conse-
quently gives the new or true position of the neutral axis. The
distance S S e of this neutral axis from the original one, which
corresponds to the moment of flexure, is determined by the pro-
portion
270.]
549
COMBINED ELASTICITY AND STRENGTH.
FIG. 443.
N1
NN2
K
P2
H
R
Α
S' S₂
SN
e1
I.E.,
S S₁
N N
е

e
whence e₁ =
σ
B
But we have also
hence
σ
1
(§ 235),
е
Pr sin. a
02
01
P P₁
P
l₁ = r o₁ =
FE
The radius of curvature r₁ of
the neutral axis determined in
this more accurate manner is
greater by the quantity (e,) than
that of the neutral axis previously considered; hence we have
r₁ = r + c = r (1 + 0) = r (1 -
P sin. a
FE).
The angle a, which the variable cross-section N, O, or N, 0,
forms with the direction of the force P, is equal to the tangential
angle a (found in § 216); hence, as this angle is small, we can put
P(x²)
or, since
sin. a = a =
2 WE
WE
r=
(§ 215),
Px
T²
x²
r sin. a = r a =
from which we obtain
2 x
e₁ =
P (1²
2 F E x
x²)
Hence for the point B, where the beam is fixed and for which
0, and for the point A at the other end, where
X =
= 1, we have e
0, e₁ =
x = 0, e,
Pl
0
=∞; on the contrary, for x =
P (l²x²)
2 FE
We
have e₁ = e; consequently the neutral axis coincides at B with
the original one, and in passing from B to A it separates more and
more from it, until, finally, it reaches the concave side of the body,
and, if prolonged beyond the body, at the end 4 it is at an infinite
distance from the other axis.
The maximum extension produced by the flexure is
•
6 Ξ
Pex
WE'
550
[$ 270.
GENERAL PRINCIPLES OF MECHANICS.
and that produced by the tensile force P sin. a is
σ₁ =
hence the total extension is
P sin. a
FE
N N N N₁ + N₁ N₁ =
,
P ех
E W
sin. a
+
F
T
we can put
E'
and, if the latter has reached the limit of elasticity
e x
P
+
W
F
sin, a) =
T
and the proof load is
W T
W T
P =
W
P (1²
x²)
e x +
sin. a
e x +
F
2 FE
For a moderate deflection, which is all these girders are gene-
rally exposed to, this value is a minimum for x = l, and it is
as we have already found.
A
U
FIG. 444.
S
A1
V+
V༡
I.
II.
B
Wi
F
B
U
W
VP
-P₁ III.
B
P =
W T
e l
1
REMARK.--If the girder, as, E.G., A ¸
A, B,
Fig. 444, I., II., III., is acted upon by two
forces, two or even three displacements of
the neutral axis from the centre of gravity
may take place. If the two forces act in
the same direction as represented in Fig.
444, I., this displacement on one side of the
cross-section A is determined by the
formula

1
Pr sin, a
FE
and, on the contrary, on the other side by
the formula
(P + P₁) r sin, a
FE
At the point of application A, this dis-
placement changes from
A
U
Pr sin, a
to
A, V₁
1
1
FE
P+P
019
A
V₂
P
2
P
when we pass from one side to the other,
on the contrary, at the fixed point B, where a = 0, we have e
2
= 0.
§ 271.]
551
COMBINED ELASTICITY AND STRENGTH.
If the two forces act in opposite directions and the moment
P,
P₁. A₁ B = P₁₂
1
of the negative force is greater than the moment
P . A B = P (l₁ + l)
2
2
1
of the positive one, in which case the girder is bent in two opposite
directions, which meet in a point of inflection F, the neutral axis consists
of three branches U V₁, V₂ W, and W, B (Fig. 444, II.), which are not
continuous, and the normals at the point of inflection F is an asymptote
to the last two of these curves; for here r = ∞ and consequently
Pr sin, a
FE
If, although the forces act in opposite directions, we have P (l + l₁) > P₁₁
as represented in Fig. 444, III., the displacement of the neutral axis upon
one side of A₁
is
1
A₁ V₁ = l
Pr sin. a
FE
>
and that upon the other is
1
ez =
(P — P₁)r sin a
FE
1
and at the cross-section through 4, there is a break in the two branches
U V₁ and V₂ B of the neutral axis, the value of which is
1
2
V₁ V 2
1
P, r sin, a
FE
§ 271. Eccentric Pull and Thrust.-If a column A B, Fig.
445 and 446, acted upon by a tensile or compressive force, whose
direction, although parallel to, is not that of the longitudinal axis
of the body, the combined elasticity and strength will come into
play. This eccentric force can, as we know, be replaced by a force
FIG. 445.
B
C
-P
P P
+P
FIG. 446.
+P P
1-P
P in the direction of the axis,
and a couple (P, P), whose
lever arm c is the distance CA
of the point of application of the
force P from the axis of the body,
and whose moment is therefore
Pc. The force A P = P in
the line of the axis produces in
all the fibres the constant strain


=
13
S₁
P
F'
m
in which F denotes the
cross-section of the body;.the
552
[§ 271.
GENERAL PRINCIPLES OF MECHANICS.
couple, on the contrary, bends the body in a curve, whose radius
is determined by the well-known formula (§ 215) P x r = W E,
in which we must substitute for the moment of the force the
WE
Pc
moment Pc of the couple. Consequently r =
is constant,
when W or the cross-section Fis constant, and therefore the curve
formed by the axis of the body is an arc of a circle.
Ife is the maximum distance of the fibres from the neutral axis
passing through the cross-section of the body, we have the maxi-
mum strain produced in the body by the couple
Pce
S2
W?
and hence the total strain is
P
Pce
S = S₁ + S₂
+
F
W'
consequently, when we put this
equal to the modulus of proof
strength T, or assume that the most remote fibre is strained to the
limit of elasticity, we obtain
T=
P Pee
+
F W
(1 + Fce) 7.
P
W
Hence the proof load of the pillar is
FT
P =
Fce
1 +
W
E.G., for one with a rectangular cross-section, the dimensions of
which are b and h,
P =
1 +
FT
6 c'
h
and for one with a circular cross-section, whose radius is r,
P
FT
4 c
1 +
From this we see that the strength of a body is tried much
more severely by an eccentric pull or thrust than by an equal one
acting in the direction of the longitudinal axis of the body.
If the column is prevented from bending by a support upon
the
side, as, E.G., BA C, Fig. 447, represents, P remains of course
= FT.
If the force acts at the periphery of a parallelopipedical pillar
h
A B, Fig. 448, and at the distance c =
from the axis, we have
2
§ 272.]
C
COMBINED ELASTICITY AND STRENGTH.
553
P =
FT
1 + 3
= FT;
and the proof load is but one-fourth of what it would be if the
weight were applied in the prolongation of the axis of the body
(Fig. 449).
FIG. 447.
FIG. 448. FIG. 449.
B
B
B
For a cylindrical pillar, with
a force acting at the circum-
= r, and
ference, we have c
consequently

FT
P
1+ 4
5
= FT
C
A
N
P
P
P
I.E., but one-fifth what it would
be if its point of application was
in the axis of the body.
These formulas can be applied
to rupture by extension, com-
pression and breaking across; it
is only necessary for each species
of separation to substitute a different coefficient of ultimate
strength, or put
FK
F
P
1 +
Fee
Η
1
K
Fce
WK
1
in which A, denotes the modulus of rupture by compression (or
extension) and A, that for breaking across.
§ 272. Oblique Pull or Thrust.-The theory of combined
elasticity and strength is particularly applicable to the case, where
the direction of the force P forms an acute angle RAP = 8 with
the axis of the beam A B, Fig. 450. One of the two components.
R
FIG. 450.
P
N
B
S2
RP cos. d acts as a tensile force
and the other P sin. & as a bending
one upon the body, and the strain

S
P cos. d
F
produced in the whole cross-section
by the first component combines with
the strain
P sin. & . l e
We
produced by the moment Plsin. d of the second component in
the outside fibres, and causes the strain
554
[§ 272.
GENERAL PRINCIPLES OF MECHANICS.
T=S=S₁ + S₂ =
P cos. d
F
Plesin. S
+
W
or more simply
T = P (COS.
cos. dle sin. d
+
Hence the required proof load is
W
FT
P =
Fle
cos. $ +
sin. S
И
P
F
Τ
(cos.
Fle
cos. 8 +
5 6
W
Ꮃ .
sin. 8).
아
​or, inversely, the required cross-section is
Or, if we substitute a modulus of proof strength T, for bending
different from that (T) for extension we have
Fle
F = P(cos, 8 +
T W T
For a parallelopipedical girder we have
sin. 8).
Fe 6
W
h'
and consequently
F
= P (
cos. S
6 7
+
T
h T
sin. 5),
and for a cylindrical one
Fe 4
whence
W
F = P (COS, +
'cos. ɗ
4 l
T
r T
sin. ô).
The same formula holds good for the case represented in Fig.
451, in which the first component R produces compression in the
FIG. 451.
A
C....
B
N
bine the strain produced by R
S₁ =
girder. If here again & denotes
the angle, which the direction
of the force P makes with the
axis of the girder, the values of
the components are
R = P cos. d and
NP sin. 8.
In order to find the proof
load of the girder, we must com-

P cos. 8
F
$ 272.]
555
COMBINED ELASTICITY AND STRENGTH.
with the greatest strain
Plesin. d
S₂ =
W
produced by the bending, and then we must substitute in the
formula
T = P (
P(
cos. d le sin. &
+
F
W
8)
or
P
F=
Τ
(cos. 8+
Fle
W
sin. 8)
이
​just found, instead of T, not the modulus of proof strength for ex-
tension, but that for compression.
In both the cases treated above the displacement of the neutral
layer of fibres from the centre of gravity is
01
S₁
G₁ =
e
O 2
S₂
W cotg. d
Fex
б
which, E.G., for parallelopipedical beams, becomes
&r=
1 cotg. Ô
6 x
It is also easy to see that by the combination of the maximum
extension or compression with the extension or compression of the
fibres, which is equally distributed over the entire cross-section of
the body, there is produced an extension or compression
0₁ ± 0₂ =
S₁ ± S₂ P cos. d
E F
E
+ le sin. o.)..
I'
If we introduce the modulus of proof strength T and for the
T
sake of security employ for wood and iron only we obtain
1) for wood in both cases
P =
780 F
67
h
3'
780 F
47
sin. d
cos. 8 + sin. d cos. d +
2) for cast iron, in the first case (Fig. 450)
3640 F
3640 F
P =
67
47
cos. d +
sin. S
cos. d +
sin. ɗ
h
7°
and in the second case (Fig. 451)
9360 F
9360 F
P =
cos. d +
6 7
47
sin. S
cos. d +
sin. d
h
2
556
[§ 273.
GENERAL PRINCIPLES OF MECHANICS.
§ 273. The case just treated occurs often in practice. If, E.G.,
a weight P is hung from a girder A B, Fig. 452, which is inclined
to the horizon, we have, when the angle of inclination of the direc-
tion of the axis is P A R d, the tensile force R P cos. d and
the bending force N P sin. d, and therefore
=
FIG. 452.
B
FT
P
=
b l
cos. & +
sin. 8
h
FIG. 453.


C
A
A
D
R
N
P
P
BLA
If, as is represented in Fig. 453, not only the direction of the
stress P is inclined to the axis of the body, but also its point of
application lies without it, in calculating the proof load we must
consider the point of application transported to D in the pro-
longation of the axis A B of the girder, I.E. we must substitute in
place of the length B 7 the length BDBA+AD= 1 +
C in which the horizontal distance CA is denoted by e, and
the angle CD A, formed by the axis of the girder with the verti-
cal, is represented by d.
sin. S
=
In like manner, for the pillar 4 B, Fig. 454, which is inclined
at an angle & to the vertical, we have the proof load
P
FT
6 7
cos. d + sin. &
h
FT
47
cos. d +
sin. s
in which we must substitute the modulus of proof strength for
compression, while in the former case we should employ that for
extension.
If a loaded girder A A. Fig. 455, is not freely supported, but
wedged between two walls, a decomposition of the forces takes
place into components producing compression and into compo-
nents producing a flexure. If the terminal surfaces A, A of this
§ 273.]
557
COMBINED ELASTICITY AND STRENGTH.
beam form an angle d with its cross-section, and if a force P acts
in the middle B of the girder, the reactions of the walls upon the
ends of the girder are Q and Q, and these forces are inclined at an
FIG. 454.
S
FIG. 455.

-P
P
N
P
angled to the horizon and give a resultant CP
balances the force P.
P = 2 Q cos. ACP 2 Q sin. 8,
Hence
or inversely
Q
P
=
DEN JUBE
- P,
which
2 sin. S
The reactions of the walls can be decomposed into a compres-
sive force in the direction of the axis of the girder
R = Q cos. 5 =
P cos. S
2 sin. 8
P cotg. &
and into a force
P
N =
Q sin. s
2'
which is perpendicular to the latter and produces a bending; con-
sequently we have
I.E.
R
N. le
T = S
=
S₁ + S.
S+
+
F
W
T
P cotg. 8. Ple
2 F
+
4 W
and the proof load of the girder is
2 FT
P
cotg. Ô +}
Fle
Π
The condition of affairs is the same, when an inclined prop A B,
Fig. 456, carries a load which has been dumped upon it. But here
Q can be resolved into a force Q, at right angles to the axis of the
558
[§ 273.
GENERAL PRINCIPLES OF MECHANICS
prop and into a force N, at right angles to the side (in miners'
language, the floor). Neglecting, for greater safety, the friction of
Q
1
FIG. 456.
B
the loose masses of stone upon
the floor and denoting the angle
formed by the terminal surfaces.
of the prop with its cross-section
by d, and the inclination of the
floor B C to the horizon by ß, we
obtain Qi
Q sin. ẞ and

2 FT
1
Fle
cotg. Ô t W
(see § 240), and therefore
Q
(coty.
8 + 1
2 FT
1.) sin. B.
W
:)
EXAMPLE-1) What must be the dimensions of the cross-section of the
inclined girder A B, Fig. 452, which is made of pine and is 9 feet long and
whose direction forms an angle of 60' with the horizon, when it bears at
the extremity A a weight P 6000 pounds? The formula
=
Р
FT
67
cos. +
б
sin. S
gives, when we substitute P
h
6000 pounds, T
b
780, d
90° - 60°
30° and Z 9.12
F = b h =
6000
780
108 inches, and assume h
5 h² =
712
h² = 10,77 (0,86€
0,866 +
Approximatively, we have
cos. 30° +
(cos
648. 0,500
10.500)
h
6.108
20) = 9,33 +
sin. 30°
3489
h
30°),
I.E.
h = √3489
more accurately
15,17,
h = √3489 + 9,33 . 15,17
and consequently
15 h 10,98 inches.
√3631 =
15,37 inches,
2) At what distance from each other must two 12 inches thick collars
A B of a so-called overhand stoping A B C, Fig. 456, be laid, when the
gob is piled 60 feet high upon it in a vein 4 feet thick, dipping at 70°, if
we assume that the weight of the gob is 65 pounds per cubic foot? De-
noting the required distance by a, we have the weight upon each collar
274.]
559
COMB NED ELASTICITY AND STRENGTH.
Q
= 4 . 60 . 65 x = 15600 x, and consequently the pressure upon each
collar is
Q₁
3 =
Q sin. 70º = 15600 x sin. 70° = 15600. 0,9397 x =
15600 . 0,9397 x = 14559 x ìbs.
If the ends A A of the collar form an angle of 70° with the axis, or if
20°, we have
2 FT
2.113,1.780
176436
14659 x =
cotg. 20° +
27
d
2,747 +
2.48
12
10,747 '
and therefore
x =
176436
10,747 . 14659
1,12 feet =
13,44 inches.
The required distance between the two collars is therefore
x
d = 1,44 inches.
(§ 274.) Flexure of Girders Subjected to a Tensile
Force.—The normal proof load P of a girder 4 B, Fig. 457, is dimin-
ished by the application of a small force in the direction of the axis
only when the girder is short. If, on the contrary, the length of the
FIG. 457.
girder and the tensile
force exceed certain
limits, the moment of

B
Q
T
the latter acts in the
C
A
K
P
opposite direction to
the moment of the
bending stress, thus di-
minishing the deflec-
tion of the body and increasing its proof load.
If we put again the co-ordinates of the elastic curve A & B,
Fig. 457, formed by the axis of the girder, 4 K -x x and K 8=Y,
we have the moment of the forces in reference to a point S in the
axis
P x Qy,
we can therefore write (according to § 215)
substituting
(Px - Qy) r = WE,
d x
da'
in which a denotes the tangential angle S T K, and denoting, in
p
Q
order to simplify the expression, V by p, and
by q, we
WE
ΠΕ
da =
↑
obtain the equation
d x (P x - Qy) d x
WE
= − (p³ x − qªy) d x.
560
GENERAL PRINCIPLES OF MECHANICS.
[$ 274.
Now making
p² x
1) y
q
FIG. 458.
Q
T
A
S
K
p³
P
2) a
d y
d x
q²
(m ε² ² + n ε¹³),
B
in which m and n de-
note constants, to be
determined, and ɛ the
base of the Naperian
system of logarithms
(see Introduction to the
Calculus, Art. 19), we
obtain

P² — (m e¹² — n e¹¹) q.
ε
and since the differential of the last equation, viz.,
(M εI * + N ε¯¹¤) q³ d x,
d a =
when substituted in
formula
n
2
equation 1), gives the above fundamental
q²
da = ( y − ¹²² ) îº d x = − ( p° x — q° y) d x,
x 2
q²
the correctness of the above expression for y is proved.
x
Since for = 0 we have y = 0, we obtain by substituting these
values in 1) the following equation
=
1,
0 = 0 − (m ε° + n ɛº), I.E.,
m + n = 0,
and since for xl, a = 0, we obtain by substituting these values.
in 2) the equation
p²
0
(m. ε??
N. EI
q
and substituting the value n = m taken from the foregoing
equation, we have
whence
p*
0:
0 =
2
m q (ε²² + ε?'),
m =
N
and the moment of the forces is
103
P x
Q y
Q m (ε 7 x
Р
ε
Ε
EU
+ ε
Q
•
-9
The latter is certainly a maximum for the fixed point B, the
co-ordinates of which are x = A C
then its value is
I and y = B Ç' = a, and
§ 275.]
561
COMBINED ELASTICITY AND STRENGTH.
Pl - Q a
P
६१
q
+ε9
If q is a proper fraction, that is, if the girder is short and the
force in the direction of the axis is small, we can put
ET /
= 1 + q l +
q² p
and also
2
3
q³ 1³
6
!
E
= 1 − q l +
q² t²
3
2
Q° 13
6
+.
hence we have the moment of the forces
= P l (1 + ¦ q² l²) (1 — § q² 7)
Pl (1 + i q² l²)
P l - Q a
Q a =
1 + q² ²
PI(1-
= P l (1 − ¦ q² l') = P l
Q P
E).
3 WE
If, on the contrary, the force Q is so great that q l becomes at
least = 2, we can then neglect
ε
1
EŸ 19
when it occurs with e'', and therefore we can put
हीरा
हा +हजा
E?
ह??
=
1,
so that the moment of the forces becomes simply
P l − Q a
P
Q a = = P P◊
WE
Q
(§ 275.) Proof Load of a Girder Subjected to a Ten-
sile Force. By the aid of the moments of the forces P and Q,
found in the foregoing paragraph, we can determine by the method,
which we have so often employed, the proof load of the girder.
The force produces a tension per unit of surface
Q
S
F
in the direction of the axis of the body, and the moment P – Q a
of the two forces P and Q produces a tension in the fibres at the
maximum distance e from the neutral axis, which is
S2
hence the total tension is
(Pl-Qa) e
W
S = S₁ + S₂ =
Q
+
F
(Pl — Q a) e
W
36
562
[§ 275.
GENERAL PRINCIPLES OF MECHANICS.
When the latter reaches the limit of elasticity, S =
can put
Q
T
+
F
(Pl-Qa) e
W
T, and we
If the modulus of proof strength 7, for compression is different
from that 7' for extension, we have
T
Q
+
F
(Pl-Qa) e
W
in which e denotes the maximum distance of the compressed fibres
from the neutral axis. In both cases we must substitute
P
P l - Q a
Q a =
(
q
ε
ε + ε
so that the required proof load of the body becomes either
or
P =
P =
+ ε
(+) )
E9 1
ह?
+ ε
q
(1-2) WT1
FT
e
Ꮃ
Q WT. q
For a small tensile force Q we can put
:) (1
+
FT)
Pl− Q a = P l
7 (1
Q 12
3 WE
e
so that, when we take into consideration the extension only, we
have
P =
(1
(FT-Q) W
Q 12
3 WE Fle
Q
Ꭲ
(1 + 27 ) (1-2) WT
3 W E
FT le
Without the tensile force Q the proof load of the body would be
W T
P₁
le
hence we have the ratio
P
Q
P₁ = (1+977) (1 - p).
3 WE
FT
from which it is easy to see, that the proof load is increased or
diminished by Q, as FT
Q
is greater or less than
Q 12
I.E., as
3 WE'
3 W
T
is greater or less than
Fl
E
When the tensile force is great, in which case we can put
Pl− Qa = P√ WE
Q
§ 276.]
563
COMBINED ELASTICITY AND STRENGTH.
we have the proof load
P
(1 -
Q
FT
Q W T
E
е
This expression becomes a maximum with the expression
√ Q³
N Q
By differentiating the latter and putting the differ-
FT
ential equation obtained equal to zero, we obtain
This maximum value is
FT
Q
3
FWT T
e
P: 3333 3 E
and the ratio of the latter to the proof load P, of a girder, which
is not subjected to a tensile force, is
P
P₁
IN
FT
3 WE
¤ F
3 W
1
For a parallelopipedical beam, whose height is h and whose
b h³
width is b, we have F = b h, W =
and e =
h, whence
12
P 47 T
4 7
1
No.
1
P₁ 3 h
E
3 h
If the beam is of wood,
T
1
σ =
E
600'
and therefore
P
1 Į
1
0,0544
P₁
600 h
h'
7
E.G., for
=
30, P = 1,632 P₁;
h
the girder carries nearly two-thirds more than when it is not sub-
jected to a tensile force.
For
7
10000
h 544
than 18,4, P, is smaller than P, and the proof load P of the beam
is diminished by the stress Q.
= 18,4, P₁ = P, and for values of smaller
h
§ 276. Torsion Combined with a Tensile or Com-
pressive Force.-If a column A B, Fig. 459, is acted upon at
the same time by a force Q, whose direction is that of its axis, and
564
[$ 276.
GENERAL PRINCIPLES OF MECHANICS.
by a couple (P, — P), which tends to twist it, both the elasticity
of torsion and that of extension (or compression) come into play.
The result of the combination of these two elasticities may be in-
vestigated as follows: If the strain per unit of surface produced
Q
F
by the force Q is S₁ = and that produced by the moment of
torsion at the distance e from the longitudinal axis of the body is
Pae
W
S₂
we can assume, that a parallelopipedical element
FIG. 459.

Η
A
FIG. 460.

S₁
Z
B
A
B
S₂
ลง
D
K
X
-2 -S1
A B C D, Fig. 460, of the body, is acted upon by the normal forces
A B. S₁ and CD. S upon A B and CD and by the couple
(AB.S, CD. S) along A B and CD and by the opposite
couple (B C. Z, A D. Z) along Band AD. If the diagonal
plane forms an angle with the axis of the body or with the
direction of the strain S, the components of the forces S₁, S, and
Z upon one side of A Care
A B . §, sin. 4, A B . §, cos. 4, and B C. Z sin. 4,
and consequently the total normal force upon A Cis
AC.SAB. S, sin. + AB. S, cos. + BC. Z sin. 4,
or, since the moment of (B C.Z, A D. Z) is equal to the mo-
4
ment of (A B. So, CD. S), L.E.
AB.B C.Z BC. A B. S, or Z
$29
A C.SAB. S, sin. 4+ (A B cos. + B C sin. 4) S2,
so that, finally, the normal strain upon the unit of surface of
A C is
§ 276.]
565
COMBINED ELASTICITY AND STRENGTH.
A B
S
S₁ sin. 4 +
A C
(4
A B
cos. p +
A C
B C
A C
sin. 4) S₂.
)
A B
BC
But
= sin. and
=cos., whence
A C
A C
4
S = S₁ (sin. 4²) + 2 S₂ sin. & cos. 4 = S₁ (sin. 4)² + S₂ sin. 2 4
S₁
5. (²
cos. 2 &
2
༧༧)
+ S₂ sin. 24 (compare § 259).
This equation gives a maximum value for S, when tang. 2 =
2 S₂
2 S2
S₁
or sin. 24
and cos. 24=
S₁
2
√ S₁² + (2 S₂)²
√ S₁² + (28₂)2
and this maximum value is
S₁
S₁
2 822
Sm=
1 +
+
√ S₁² + (2 S₂)²
√S₁² + (2 S₂)²
călo
+
+ S. 2.
2
Substituting the above values for S, and S, in this equation,
we obtain the required maximum strain
0
SIT 2 F + √ ( 2 )² + (P a c)².
ITL
+1
2 F
W
Now, since the body should resist with safety the actions of
these forces P and Q, we must put Sm to the modulus of proof
strength Tor
Q
2 F
+
Рае
√ ( 2 ) + (Pac)² = T,
W
from which we obtain the equation of condition
= T² _ QT
(Pac)² =
W
F
The allowable moment of torsion is therefore
1) Pa =
W
e
√ T2
Ꭲ
Q T
F
,
and the allowable force in the direction of the axis is
2) Q = FT
FP a e
T
W
In order to find the dimensions of the cross-section correspond-
ing to the forces P and Q, we put
W
√
Ρα
Q T
T²
F
when the force producing torsior is the greater, and, on the
contrary,
566
[$ 276.
GENERAL PRINCIPLES OF MECHANICS.
2
F
T-1
+ (Pae)
b h
F= bh, W = (b² + h²)
12
when that in the direction of the axis is the greater.
For a parallelopipedical column, whose dimensions are b and h,
we have
and è = 3 √b² + h², consequently
W b h
Pa
Pa
√b² + h² =
e
6
T³
2
Q T
bh
T
(1 -
Q
b h T
)
and
F-bh
36
Q
Pa
T
(b² + h²) T
b k
(Ba)
b
6 P a
√ b² + h⋅ b h T
:)'T
If we know the ratio v =
of the dimensions, we can calculate
the dimensions themselves by means of this formula.
For a pillar with a square base bh, and therefore
h² √2
6
h = b
Pa
T
PC (1-9)
3
h² T
3
,
/GV) Pa (1 - 0)
12
Τ
3
Pa
T
T
Tand
T
w² = 2 [ 1 − 1 ( PM) J
h =
= b
√ / [1 − 1 ( P ) T
= T
h³ T
For a cylindrical pillar or shaft we have
π p3
2
li
π 2,4
F
Ε πρ
= π p², I
W =
2
and e = r,
whence
Pa
Q T
and r =
2 Pa
π p²
Q
1 /2 P a
Τ π тра
ΠΤ
and r = √ [1-(P)T
π T
(
1
π r²² T
"
Q
as well as
Q
2 Pa
про =
T2
T
$277.]
567
COMBINED ELASTICITY AND STRENGTH.
FIG. 461.
2
If the force Q in the direction of the axis is a compressive one,
the formulas found above still hold good; for
not only the direction of the force S (Fig.
461) is opposite, but also the forces S, and Z
can be assumed to act in the opposite direc-
tion, when we wish to obtain the maximum
resultant S.
A
T =
-Se
S2
B

400 pounds, is
3
r = V
2 P a
z
EXAMPLE.-If a vertical wooden shaft weigh-
ing 10000 pounds is subjected to a moment of tor-
sion Pa = 72000, the required radius, assuming
Q
1
=
π T
π p² T
3/0,6366.72000
400
10000
400 π p²
/0,6366. 180 (1
(1 — 7,958) -
Approximatively, we have
γ
√114,6
4,85, whence
7,958
7,958
0,3383, and
23,52
(1 -
7,958)-
1
=
1,071,
✓ 0,6617
so that the required radius is, more accurately,
r = 4,85 . 1,071 = 5,194 inches,
and consequently the diameter of the shaft is
d 10,39 inches.
§ 277. Flexure and Torsion Combined.-Cases often oc、
cur where a girder or shaft is acted upon at the same time by a
bending force and a twisting couple. Horizontal shafts are gel-
erally submitted to both of these actions. In order to investigate
II
FIG. 462.
P
++
K
the relations of the combined action of
these two forces, let us imagine a pris-
matic body A B CD, Fig. 462, fixed at
one end B D, to be acted upon at the
other end by a bending force Q and
at the same time by a twisting couple
(P, P). If is the length AC of
the shaft, II, the measure of the mo-
ment of flexure and e, the maximum
distance of an element of the cross-sec-

1
568
[§ 277.
GENERAL PRINCIPLES OF MECHANICS.
譬
​tion from the neutral axis, we have the maximum strain produced
in the direction of the axis by the force Q
S₁
Q l₁ e₁
(compare § 235).
W₁
If, on the contrary, a denotes the lever arm H K of the couple
(P, P), W the measure of the moment of torsion and e the
greatest distance of any element of the cross-section from the axis
CD of the body, we can put the maximum shearing strain pro-
duced by the couple
S₂ =
Pae
W
240
Now here, as we can easily understand, the strain S
W₁
takes the place of the absolute strain S
Q
F
of the foregoing par-
agraph, and therefore we can put for the maximum strain in the
whole body A B C D, Fig. 462,
S₁
Sm
+1
2
Si)² + S², or
Q₁he₁
T=
2 W
+
+
e₁
2
√ (Q+ h ; )² + (Pa®);
2 W
from which we obtain the equation of condition
2
(P#c) = r•
T2
Q Le T
W₁
The allowable moment of torsion is therefore
W
1) Pa =
e
and the bending force is
2) Q =
Le II 2
W₁
T
W
e
√ T2
Ql₁e₁ T
>
W√ T²
2
Qhe T W T
W
V1.
e
1
Qhe
W₁ T'
W
(Pae)'];
>
from which we obtain either
Pa
or
W₁
W₁
Q h₁₂
1
e₁
T
T
(Pae)
For a square shaft
W
h³ VQ
W
h³
and
whence
e
6
e
6'
§ 277.]
569
COMBINED ELASTICITY AND STRENGTH.
h³ =
h
as well as
h³
=
6 √ Pa
T
3
1
6
(1 –
Ра
و
6 Q 4
h³ T )
624
Q
and
· Vars Pa (1994) +
T
-
だ​T
6 P
1604 [1-(a)] and
3
T
6 Q4
T
h = √ © 24 [1-(-
h³ T
6 P a
3
h³ T
-)'T'; .
I'₁
1
and
2
e
Проз
4
; hence we can put
while, on the contrary, for a cylindrical shaft,
W
e
Прод
as well as
913
r =
2 Pa
π T
(1 - 404) +
3 2 Pa
π T
Q
T
(1.
пров
and
4 Q 4₁ ==
Ql
трот
2 P
~=4941-Gand
r =
4 Q h
T
1
про т
2Pa
# 2 [¹² - (PD)] +
T
Very often it is not a couple, but a force P, acting eccentrically
to the axis, which produces the torsion in the body B CD, Fig. 463.
FIG. 463.
P
Since such a force can be decomposed into an
equal central force CP+P and into a
couple (P, P), whose lever arm is the dis-
tance CA between the axis CD of the body
and the line of application of the force P. we
have here a case of combined strength, al-
though there is no other force Q for the
twisting produced by the couple (P. - P),
combines with the bending produced by the
axial force + P. The above formulas can
be employed directly for determining the
thickness of such a body, when we substitute in them P Q 4.

+P
·
If, in addition to the eccentric force P, there is another Q,
whose moment is Q, we must substitute instead of P 1, P l + Q h₂.
570
[§ 278.
GENERAL PRINCIPLES OF MECHANICS.
§ 278. Bending Forces in Different Planes.-If a girder
or shaft B C, Fig. 464, is acted upon by two bending forces Q, and
Q, whose directions C, Q,and C Qu
FIG. 464.
C₁
C₂
R₁
B
29
although at right angles to the axis C, B
of the body, are not parallel to each
other, the portion C, B of the body will
be bent by two couples (21, 2₁) and
(Q2 Q2), the resultant of which must
be found, when we wish to determine
the nature and magnitude of the bend-
ing. If and l, denote the arms of the
forces Q, and Q, in reference to the fixed
point B, Q, and Q l are their mo-

2
ments, and if a is the angle formed by the
directions of the forces, when passing
through the same point, we have, according to § 95, the moment
of the resulting couple
Rc
2
√ (Q₁ 4)² + (Q½ 42)² + 2 (Q1 4) (Q₂ la) cos. a,
and for the angle B, which the plane of this couple makes with that
of the couple (21,
(R,
Q1),
sin. B
Qy la
Re
R c
In order to find the intensity and the plane of this couple
R), we can reduce the force Q, from C₂ to C₁, combine the
reduced force Q
Qe la
1,
by means of the parallelogram of forces
with the force Q, and thus determine the resultant R; the pro-
duct R₁ l₁ = Rc is the value of the moment of the resulting couple
and the angle Q, C, R is the angle ß, which the plane of this couple
forms with that of the couple (Q1, - Q1). This plane is of course.
that in which the body is bent, and by the aid of the moment R, l,
Rc, just found, we obtain the maximum strain in the body
S
Ree
W
or, putting this equal to the modulus of proof strength T, we have
TW
√ (Q₁ 4₁)² + (Q2 42)² + 2 (Q1 41) (Q₂ ↳½) cos. a.
с
If a twisting couple (P, P), whose moment is P a, also acts
upon this body 4 B, the maximum strain becomes.
ST=
R. c e
2 W₁
+
√( cc.)²+ (Pac);
2 W
W
§ 278.]
571
COMBINED ELASTICITY AND STRENGTH.
in which W, denotes the measure of the moment of flexure, W that
of torsion, e, the greatest distance of any element of the body from
the neutral axis and e that of any element from the longitudinal
axis of the body at D.
From the above we obtain
2
(Pae)² =
= T²
Rce T
W
= T² — [(Q, 4,)² + (Q? ↳½)² + 2 ( Q₁ 4) (Q) cos. a]
& T
11
By the aid of the formulas of the foregoing paragraph the
required dimensions of the cross-section of the body can be found
by substituting in them instead of Q7 the sum Q₁ ↳₁ + Q₂ l..
If only one bending force Q, acts upon the body and if at the
same time it is acted upon by a single twisting force P instead of
a couple (P. P), this force P can be resolved into a twisting
couple (P,P) and a force P acting upon the axis, so that
instead of Q2 l, we must substitute in the latter formula P 1.
FINAL REMARK.-Although there is no portion of mechanics which has
been the subject of so many experiments as the elasticity and strength of
bodies, yet much remains to be investigated and many points are still
uncertain. Experiments upon this subject have been made by Ardant,
Banks, Barlow, Bevan, Brix, Busson, Burg, Duleau, Ebbels, Eytelwein,
Finchan, Gerstner, Girard, Gauthey, Fairbairn and Hodgkinson, Lagerjhelm,
Musschenbrock, Morveau, Navier, Rennie, Rondelet, Tredgold. Wertheim,
etc. The older experiments are discussed at length in Eytelwein's “ Hand-
buch der Statik fester Körper," Vol. II., and also in Gerstner's "Handbuch
der Mechanik," Vol. I. A copious treatise on this subject by v. Burg is
given in the 19th and 20th volumes of the Jahrbücher des Polytechn.
Instituts zu Wien. Theories which differ somewhat from those given in
this work are also to be found in this treatise. The experiments of Brix
and Lagerjhelm have already been mentioned (page 304). New and very
varied experiments upon the reacting strength of different kinds of stone
by Brix are reported in the 320 year (1853) of the transactions of the
“Verein zur Beförderung des Gewerbefleiszes in Preussen." A simple
theory of flexure by Brix is to be found in the treatise "Elementare Berech-
nung des Widerstandes prismatischer Körper gegen die Biegung," which is
printed separately from the transactions of the Preussischen Gewerbeve-
reins. Wertheim's latest experiments upon elasticity have already been
mentioned (page 396). An abstract of Hodgkinson's experiments is to
be found in Moseley's "Mechanical Principles of Engineering and Archi-
tecture." Hodgkinson's principal work, the title of which is "Experimen-
tal Researches on the strength and other properties of cast iron, etc.," was
published by John Weale in 1846. A French translation of it by Pirel
572
[§ 278.
GENERAL PRINCIPLES OF MECHANICS.
appeared in Tome IX., 1855, of the "Annales des Ponts et Chaussées,” and
an abstract of it by Couche in Tome XX., 1855, of the "Annales des
Mines." Tredgold has published a treatise upon the strength of cast iron
and other metals. The following works are also recommended for study.
Poncelet's " “Introduction à la Mécanique Industrielle," Part I., Navier's
Résumé des Leons sur l'application de la Mécanique, Part I., translated
into German by Westphal under the title "Mechanik der Bankunst," to
which work Poncelet has made some additions in his theory of the resist-
ance of rigid bodies (see his Manual of Applied Mechanics, Vol. II., trans-
lated into German by Schnuse). We would also recommend particularly
the "
Resistance des Materiaux " (Leçons de Mécanique Pratique), by A.
Morin, which has been much used in preparing this work. We may men-
tion further the "Theorie der Holz-und Eisenconstructionen mit besonderer
Rücksicht auf das Bauwesen," by George Rebhan, Vienna, 1856, the work
of Moll and Reuleaux (already quoted in page 469) upon “die Festigeit
der Materialien," a "Memoire sur la Resistance du Fer et de la Fonte, par
G. H. Love, Paris, 1852," as well as Tate's work upon the strength of mate-
rials as applied to tubular bridges, etc. The theory of combined elas-
ticity and strength was first treated by the author in "der Zeitschrift für
das gesammte Ingenieurwesen (dem Ingenieur), by Bornemann, etc., Vol. I.
In the first volume of the new series of this magazine (Civilingenieur,
1854) the graphic representation of the relative strength is treated by Mr.
Bornemann, and the results of the experiments made by Bornemann and
by Lemarle are also given.
The theory of elasticity and strength will be treated of again when we
discuss the theory of oscillation and of impact.
Mr. Fairbairn's Useful Information for Engineers, I. and II. Series, gives
the results of many experiments upon the strength of wrought iron of dif-
ferent forms, as well as upon stone, glass, etc. From a theoretical point
of view, we can particularly recommend, "Legons sur la theorie mathe-
matique de l'élasticité des corps solides," par Lamé, "A Manual of Applied
Mechanics," by W. J. Rankine, the "Cours de Mécanique appliquée," I.
Partic, by Bresse, and the "Théorie de la résistance et de la flexion plane
des solides," par Belanger. The treatise of Laissle and Schüblen, "Ueber
den Bau der Brückenträger," is a fair exponent of the state of science upon
this question, when it was written, and is therefore to be recommended.
Rühlmann's "Grundzüge der Mechanik," 3. Auflage (1860), contains also
a treatise upon the resistance of materials worth reading.
The "Civilingenieur" and the "Zeitschrift des deutschen Ingenieur-
vereins" contain several valuable treatises upon the theory of elasticity
and strength, particularly those by Grashof, Schwedler, Winkler, etc., as
well as several good translations from the French and English of Barlow,
Bouniceau, Fairbairn, Love, etc. The results of many experiments by Fair-
bairn, Karmarsch, Schönemann, Völkers, etc., are also given in these journals.
FIFTH SECTION.
DYNAMICS OF RIGID BODIES.
CHAPTER I.
THEORY OF THE MOMENT OF INERTIA.
§ 279. Kinds of Motion. The motion of a rigid body is
either one of translation, or of rotation, or a combination of the two.
In the motion of translation (Fr. mouvement de translation; Ger.
fortschreitende or progressive Bewegung) the spaces described
K
FIG. 465.
A
D F
E
simultaneously by the different parts of the
body are parallel and equal to each other;
in the motion of rotation (Fr. mouvement
de rotation; Ger. drehende or rotirende
Bewegung), on the contrary, the parts of
the body describe concentric arcs of circles.
about a certain line, called the axis of rota-
tion (Fr. axe de rotation; Ger. Umdre
hungsaxe). Every compound motion can
be considered as a motion of rotation around
a movable axis. The latter is either varia-
ble or constant. The piston DE and the
piston-rod B F of a pump or steam engine,
Fig. 465, have a motion of translation, and
the crank A C has a motion of rotation.
The connecting rod A B has a compound
motion; for one of its extremities B has a
motion of translation, while the other A
has a motion of rotation. The axis of rota-
tion of a cylinder, which is rolling, is con-

574
[$ 280
GENERAL PRINCIPLES OF MECHANICS.
stant, while that of the connecting rod A B is variable; for its
position is determined by the intersection M of the perpendicular
BK to direction CB of the axis of the piston-rod and of the pro-
longation of the crank C4 (see § 101).
§ 280. Rectilinear Motion.-The laws of motion of a mate-
rial point, discussed in § 82 and § 98, are directly applicable to a
rectilincar motion of translation. The elements of the mass M₁,
M2, M3, etc., of a body, moving with the acceleration p, resist the
motion, by virtue of their inertia, with the forces M, p, M. p, M3 p,
etc. (§ 54), and since the motions of all these elements take place
in parallel lines, the directions of these forces are also parallel; the
resultant of all these forces due to the inertia is equal to the sum
P ÷
1
M₁p M₂p + Map + ... = (M₁ + M₂ + M₂ + ...) p = Mp,
when M denotes the mass of the whole body, and the point of ap-
plication of the resultant coincides with the centre of gravity. In
order to set in motion a body, whose mass is M and whose weight
is G Mg and which in other respects is free to move, we re-
quire a force
G p
g
P = Mp =
whose direction must pass through the centre of gravity S of the
body.
If, in consequence of the action of the force P, the velocity e is
changed to the velocity v while the space s is described, the energy
stored by the mass is (§ 72)
•
Ps
じ
​M = ("² = ~ ) G
(h
k) G.
2
2 J
If
EXAMPLE.-The motion of the piston and piston-rod of a pump, steam-
engine, blowing-machine, etc., is variable; at the beginning and end of its
stroke the velocity is = 0, and near the middle of it it is a maximum.
the weight of the piston and piston-rod G, and if the maximum velocity
at the middle of its stroke = 2, the energy stored by them in the first half
of the stroke and restored in the second half is
=
v"
L
G'.
29
If a 800 pounds and v = 5 feet, we have
L = 0,0155 .53.800
310 foot-pounds.
Now if half the stroke of the piston is 8 = 4 feet, we have the mean
force, which is necessary to produce the acceleration of the piston in the
first half of the stroke and which the piston exerts in the second half, when
it is retarded,
§ 281.J
575
THEORY OF THE MOMENT OF INERTIA.
L
212
310
P:
G =
S 2 g s
77 pounds.
4
P
FIG. 466.
§ 281. Mction of Rotation.-If the motive force P of a
body A B, Fig. 466, does not pass through its centre of gravity S,
the body turns around that point, and at the
same time moves forward exactly as if the force
acted directly at the point S, as can be shown in
the following manner. Let us let fall from the
centre of gravity S a perpendicular S A upon
the direction of the force and continue it in the
other direction until the prolongation S B is
equal to the perpendicular SA, and let us sup-
pose that two forces +P and
P, which

P₁
12
P
S
↓ P,
balance each other and are parallel to P, are applied at B. The
force+P combines with half the force P acting in A and gives
rise to the resultant
PPP = P
applied at the centre of gravity, while, on the contrary, the force
- P forms with the other half (P) of the force P applied in .1
a couple; hence the force P, applied eccentrically, is equivalent to
a force P₁ = P, which is applied at the centre of gravity, and which
moves this point and with it the body, and to a couple ( P,
P), which causes the body to turn around its centre of gravity S
without producing a pressure upon it. The statical moment of
this couple is
SA
= ¦ P . § Ã ÷ ! P. SB = P. § Ã = P a,
SA +
or equal to the statical moment of the force P applied in A in
reference to the centre of gravity S; the resulting rotation would
therefore be the same if the centre of gravity S were fixed and P
alone were acting.
D
N
FIG. 467.
-P
B
D,
N
IK
E
Р
E₁
If a body AB, Fig. 467, is compelled,
by means of guides D E, D, E, to assume
a motion of translation, the eccentric force
APP produces the came effect upen
the motion of the body as an equal force
acting at the centre of gravity, and the
couple (¦ P, –F) is counteracted by
the guides If a is the eccentricity S4
of the force P, or the distance of its diree-
tion from the centre of gravity S of the
body, and if b denotes the distance HK

576
[§ 282.
GENERAL PRINCIPLES OF MECHANICS.
between the perpendiculars to the guides at the diagonally opposite.
points F and G and (N,- N) the couple, with which the body
acts on the guides, we have, by equating the moment of the
couples (P, - į P) and ( N, — N),
Nb Pa, and therefore
a
N = 1/1
P.
b
If, finally, the body 4 B, Fig. 468, is prevented from moving
FIG. 468.
P
P
P₁
forward by the fixed axis C, the eccentric force
APP produces the same effect upon the
rotation of the body about this axis Cas a
couple (P,P) with the arm 2 CA =
2 ( B = 2 a, or with the moment ¦ P. 2 a
Pa; for the remaining central force CP,
P is counteracted by the bearings of the
axis (compare § 130).

P₁
§ 282. Moment of Inertia.-During the rotation of a body
A B, Fig. 469, about a fixed axis C, all points M₁, M, etc., of it de-
P
FIG. 469.
B₂
M
N
2
scribe equal angles at the centre M, C' N,
=M, CN, etc., °, which, when the
radii C D₁ C' D, etc., one (1) are
equal, correspond to the same arc
D, E, D, E, etc.,
= = &

=
Ø°
πT.
180°
Since the velocity is determined by
the quotient of the clement of the space
and the corresponding element of the
time, the angular velocity (Fr. vitesse an-
gulaire, Ger. Winkelgeschwindigkeit), I.E. the velocity of those
points of the body which are situated at a distance equal to the
unit of length (E.G. a foot) from the axis of rotation, is therefore
one and the same for the whole body, and its value is
()
&
T
and in like manner the angular acceleration, or the acceleration of
the rotating body at the distance = unity from the axis of rota-
tion, is the same for the whole body, and its value is
•
K
W
T
§ 282.]
577
THEORY OF THE MOMENT OF INERTIA.
denoting the increase of angular velocity in the element of
time T.
1
In order to find the spaces s1, S2, etc., the velocities 1, 2, etc.,
and the accelerations P1, P2, etc., of the points M1, M2, etc., of the
C M₂ = 129
body, which are situated at the distances C M₁
r₁, C M₂
etc., from the axis of rotation C, we must multiply the angular
space 4, the angular velocity w, and the angular acceleration p by
r₁, r₂, etc.; thus we obtain
8₁ = $r₁, 82, etc.,
1,
V2
V₁ = w r 1, V₂ = wr, etc., and
P₁ = k r₁, P₂ = k r½, etc.
q
If the whole mass M of the body is composed of the parts M₁,
M2, etc., which are at distances equal to the radii r₁, r, etc., from
the axis of rotation C, the forces with which these elements of the
mass resist the rotation are
P₁ = M₁ p₁ = к M₁ r₁, P₂ = M. pak Mar, etc.,
1
and their moments are
k M,
2
P₁ r₁ = k M₁ r², Par₂ = K M. r, etc.,
1
and the moment necessary to cause the body to rotate with the
angular acceleration « is
Pa = & M₁ r₁², + k M¸ r²² +
K
2
= K (M₁ r₁² + M₂ r²² + M3 r3² + ...).
1
In like manner (according to § 84) the energy stored by the
elements M₁, M, etc., while they acquire the velocities v₁, v', etc., is
2
A₁ = ↓ M₁ v²² = { w² M, r₁²,
A,
1
v,
2
2
A₂ = ↓ M₂ v₂² = w² M₂ r², etc.,
• }
2
and therefore the work done in communicating to the whole body
the angular velocity w is
A = Á₁ + Á₂ +
= ! w² (M₁ r₁² + M₂ r2² + M₂ r'3' + ...).
3
1
The force of and the energy stored by a body in rotation de-
pends principally upon the sum of the products M, ri+ Mara² +
M3 r3³ + ... of the different elements M1, M2, etc., of the mass and
of the squares of the distances r₁, r, etc., from the axis of revolu-
tion. This sum is called the moment of inertia (Fr. moment d'in-
ertie, Ger. Trägheits-, Drehungs- or Massenmoment), and we will
hereafter denote it by Mor W. Hence the moment of the force,
by which the mass M = M₁ + M₂ + whose moment of
inertia is
W = M r² = M₁ r²² + M₂ r² + ..
37
578
[§ 283.
GENERAL PRINCIPLES OF MECHANICS.
has imparted to it the angular acceleration, is
1) Pa = к M ² = k W,
and, on the contrary, the work done in putting the mass M in ro-
tation with the angular velocity w is
2) P s = { w² M r² = { w² W.
If the initial angular velocity of the mass was e, the work done
in increasing it to w is
Ps = ! w² W - √ ε² W = ↓ (w² — ε³) W.
1
We can also determine from the work done and the initial ve-
locity ɛ the final velocity w; it is
W = ε² +
2 Ps
W
EXAMPLE. If the body A B, Fig. 469, movable about a fixed axis C
and in the beginning at rest, possesses a moment of inertia of 50 foot-
pounds, and if it is set in rotation, by means of a rope passing round a
pulley, by a force P 20 pounds, which describes the spaces = 5 feet,
the angular velocity produced is
W =
/2 P8
2 P 8
W
2.20.5
50
√4 =
2 feet,
I.E., every point at the distance of a foot from the axis of rotation de-
scribes, after this work has been done, 2 feet in each second.
one revolution is
The time of
2 π
t
3,1416 seconds,
W
and the number of revolutions in a minute is
U
60
t
60
3,1416
19,1..
If the angular velocity w =
= 2 feet, just fourd, is transformed into a ve-
foot, the work performed by the body is
locity & =
P₁ 81
[2² — (3)²]. 50 (4-). 25
=
ft.25 85,93 foot-pounds,
E.G., it has lifted a weight of 10 pounds 8,593 feet high.
1
2
1
2
§ 283. Reduction of the Mass.-If the angular velocities of
two masses M₁ and M, are the same, if, E.G., they belong to the
same rotating body, their living forces are to each other as their
moments of incrtia W, M₁r and W, M. r, and if the latter
are equal, both masses have the same living force. Two masses
have, then, equal influence upon the state of motion of a rotating
body, and one can be replaced by the other, without causing a
change in that state, when their moments of inertia M₁_r,² and
M₂ r are equal, or when the masses themselves are to each other
inversely as the square of their distances from the axis of rotation.
2
$238.]
579
THEORY OF THE MOMENT OF INERTIA.
With the aid of the formula M₁ r₁²
2
My r² we can reduce a mass
r₂
from one distance to another, I.E. we can find a mass M, which at
the distance r, has the same influence on the state of motion of the
rotating body as the given mass M, at the distance, and this
mass is
M₂
M₁ ri
1'2
2
1
١١١
r 2
I.E., the mass reduced to the distance r, is equal to the moment of
inertia of the mass divided by the square of that distance.
Two weights Q and Q, fixed upon a disc 1 CB, Fig. 470, at
FIG. 470.
X
X
P
the distances C B = b and C' B, = a from
the axis of rotation II, have the same
influence upon the movement of the disc
in consequence of their inertia, when Q, a
= Q b² or Q₁ = If, therefore, a force

Q b
a™
P, whose arm is ('A
C B₁
= a, causes
a body, whose weight is Q and whose dis-
tance from the axis of rotation is C B b,
to rotate, we must reduce the latter to the
arm a of the force P and put instead of Q,
Q₁
Q b
and the mass moved by P is
a²
M
(P + 27²): 9,
Q
a
consequently the acceleration of the weight P is
Force
P
p
•
9
Mass
b³
Pa
P a² + Q b²
⋅ 9
9,
P + Q
a²
and the angular acceleration is
p
Pa
K
•
a
Pa² + Q b*
g.
EXAMPLE. If the weight of the rotating mass is Q
distance from the axis of rotation is b
moving force is P = 24 pounds and its
accelerated by Pis
M
=
360 pounds, its
2,5 feet, the weight acting as
arm is a = 1,5 feet, the mass
= [P + (21.5; )² 2 ] : g = 0,031 (24 +
= 31,74 pounds,
Q
and the acceleration of the weight is
25
9
360)
= 0,031. 1024
24
Ρ
=
31,74
0,756 feet,
580
[$ 284
GENERAL PRINCIPLES OF MECHANICS.
on the contrary, that of the mass Q is
b
5
5. 0,756
1 =
I
a
•
10
3
p
3
1,26 feet,
and the angular acceleration is
K
Ρ
a
0,504.
After four seconds the angular velocity is
W = 0,504 . 4 = 2,016 feet,
and the corresponding space described is
1 w t =
2,016. 4
2
4,032 feet,
hence the angle of rotation is
4,032
·
π
180° = 1,2834. 180° — 231° 1′
and the space described by the weight Pis
FIG. 471.
$ =
p t²
2
0,756 . 42
2
= 6,048 feet.
§ 284. Reduction of the Moments of Inertia.-If the
moment of inertia of a body or of a system of bodies in reference
to an axis passing through the centre of gravity S of the body is
known, the moment of inertia in reference
to any other axis, parallel to the former, can
easily be determined. Let S, Fig. 471, be
the first axis of rotation, which passes through
the centre of gravity, and D the other axis
of rotation, for which the moment of inertia
is to be determined; let SD d be the dis-
tance between the two axes and S N₁

D
S
Xr
and N, M₁ = y, the rectangular co-ordinates of an element M, of
the mass of the whole body. The moment of inertia of this ele-
ment in reference to D will be
= M₁ . D M‚² = M₁ (D N²² + N₁ M,³) = M₁ [(d + x;)² + yi²]
and in reference to S
2
2
= M₁. S M₁ = M, (S N² + N₁ M,³) = M, (x² + y₁²),
and, therefore, the difference of these moments is
= M₁ (ď² + 2 d x + x² + y²) — M, (x² + y²)= M, d² + 2 M, dx.
1
For another element of the mass it is
for a third it is
2
M₁₂ d² + 2 M» d x»,
= M₁ d² + 2 M₂ d x3
3
and, therefore, the moment of all the elements together is
= (M₁ + M. + Ms + . . .) d² + 2 d (M₁ x₁ + M₂ X2 + M3 X3 + ...).
3
§ 285.]
581
THEORY OF THE MOMENT OF INERTIA.
1
But M₁ + M₂ + ... is the sum M of all the masses and M₁₁ +
M½ x² + M3 x3 is the sum Ma of the statical moments; hence it
follows that the difference between the moment of inertia I, of
the whole body in reference to the axis D and its moment of inertia
Win reference to Sis
2
M đ²
M ď² + 2 d M x.
1
W₁ – W =
But since the sum of the statical moments of all the elements
upon one side of every plane passing through the centre of gravity
is equal to that of the moment of those on the other, the alge-
braical sum of all the moments is = 0, and we have M x = 0, and
consequently
I.E
M
W₁
W W = M đ,
W₁ W + M ďª².
1
The moment of inertia of a body in reference to an eccentric axis
is equal to the moment of inertia in reference to a parallel axis
passing through the centre of gravity plus the product of the mass
of the body by the square of the distance of the two axes from each
other.
We see from this that of all the moments of inertia in reference
to a set of parallel axes that one is the least, whose axis is a line
of gravity of the body.
§ 285. Radius of Gyration.-It is very important to deter-
mine the moment of inertia for various geometrical bodies; for the
values thus deduced are frequently employed in the different calcu-
lations in mechanics. If the bodies, as we will hereafter suppose,
are homogeneous, the different portions M,, M, etc., of the mass, are
proportional to the corresponding portions V₁, V, etc., of the vol-
ume, and the measure of the moment of inertia, or as it is generally
called, the moment of inertia, can be replaced by the sum of the
products of the portions of the volume and the square of their
distances from the axis of rotation. In this sense we can also
determine the moment of inertia of lines and surfaces.
If we
imagine the entire mass of a body concentrated in one point, we
can determine the distance of the same from the axis, if we sup-
pose that the moment of inertia of the mass, which is thus concen-
trated, is the same as it was, when distributed through the whole
space. This is called the radius of gyration (Fr. rayon d'inertie,
Ger. Drehungs- or Trägheitshalbmesser). If W is the moment of
inertia, M the mass and k the radius of gyration, we have
Mk² = W, and therefore
2
k = √ W
M
582
[$ 286.
GENERAL PRINCIPLES OF MECHANICS.
We must also remember that this radius does not give a definite
point, but only a circle, in whose circumference the mass can be
distributed arbitrarily.
If in the formula 11, W + M ď² we substitute W- Mk3
and I'₁ = J k₁”, we obtain
1
k₁₁ = k² + d,
I.E., the square of the radius of gyration in relation to any axis is
equal to the square of the radius of gyration in reference to the line
of gravity parallel to that axis plus the square of the distance of
the two axes from each other.
FIG. 472.
X
§ 286. Moment of Inertia of a Rod.—The moment of inertia
of a rod AB, Fig. 472, which revolves about an axis II passing
through its middle §, is determined in the fol-
lowing manner. Let the cross-section of the
rod be and half its length be, and the
= F
angle, which its axis makes with the axis of
B rotation, I.E. 1 SI, be = a. Let us divide the
half length of the rod into a parts, the contents
ΕΙ
of each of which are ; the distances of the

N
different portions of it from the centre Sare
7 27 37
etc., hence their distances from the
A
n' n N
Z
ས་
axis of IX, such as M N, are
sin. a,
N
27
31
sin. a,
ጎ
-X
):
N
sin. a, etc., and the squares of the
latter are (sin. a) (sin. a), 9 (sin. a) etc.
N
4 (2
N
F!
N
getc.
Multiplying these squares by the contents of an element
N
and adding the products thus obtained, we obtain the moment of
inertia of the rod
F1
sin. a
P! [([sin. a) + 4-(1 sin. 9) + 9 (9) '+...]
N
T
N
亿
​4. a)²
FP sin. a
223
(1² + 2² + 3² +
·· + n²),
but since 1² + 2² + 3² + . . . + n³
n³
=
3'
§ 287.]
583
THEORY OF THE MOMENT OF INERTIA.
we have
W =
FP sin.² a
3
2
Now since Fl is the volume of the half rod, which we treat as
the mass of the body, we have
W = } M l' sin.² a.
The distance of one end of the rod from the axis XX is
ACBD = a = 1 sin. a,
and, therefore, we have more simply
W = } M a²,
which formula applies to the entire rod, when we understand by
M the mass of the whole rod.
The moment of inertia of a mass A, at the end 4 of the rod is
M₁ a²; if, therefore, we make M₁ M, M has the same moment
of inertia as the rod. Hence, so far as the moment of inertia is
concerned, it makes no difference whether the mass is equally
distributed along the rod, or whether one-third of it is concentrated
at the end 4. If we put = M², we obtain l ja, and,
therefore, the radius of gyration of the rod is
A
FIG. 473.
D
X
X-
B
ka v
N s
= 0,5773.a.
If the rod is at right angles to the axis
of rotation a = 1, and consequently

W = ! MP,
19
3
If, finally, the rod does not lie in the
same plane as the axis of rotation, if the
shortest distance between the axis of rota-
tion and the axis of the rod is
=
Ꭰ
SS C C D D₁ = d,
and if the normal distances A C= B D of
the ends A and B of the rod from the axis
CD, passing through the centre of gravity
S of the rod and parallel to C, D, is a, we
have (according to § 284) the moment of
inertia of the rod
W₁ = W + ¦ Ma² = M (d² + { a²).
}
§ 287. Rectangle and Parallelopipedon -The moments
of inertia of plane surfaces are found in exactly the same way as
their moments of flexure I F₁ zi² + F, ≈2+... We can, con-
584
[§ 287.
GENERAL PRINCIPLES OF MECHANICS.
sequently, employ here the values of W, found in the last section
for various surfaces, as their moments of inertia W.
For the rectangle A B C D, Fig. 474, the moment of inertia in
Y
FIG. 474.
Z
B
F
X-
G
D
-Y
-2
-X
reference to the axis XX, which runs
parallel to one side and through the
middle S of the figure, is, according to
§ 226,

W
b h³
12
b denoting the width A B C D paral-
lel to the axis of rotation and h the
length A D = B C of the surface.
But the area of this surface can be re-
garded as the mass M, and therefore
we have
W
M h²
12
M
3
1.E. equal to that of one-third of this mass concentrated at the dis-
h
tance SFS G
from the axis of rotation. •
2
If this rectangle turns upon an axis Z Z, which is at right
angles to its plane and which at the same time passes through the
middle S of the figure, we have, according to § 225,
IV
་
M h²
+
M b² M (h² + b²)
M
12
12
12
3
14 [()² + ()"]
M
d
( 3 ) 3
3 2
d designating the diagonal A C = B D of the rectangle. We can
imagine here also one-third of the whole mass to be concentrated
at one of the corners A, B ...
FIG. 475.
F
B
-X.
Since a regular parallelopipedon B E F, Fig. 475, can be decom-
posed by parallel planes into equal
rectangular slices, this formula is
applicable, when the axis of rota-
tion passes through the centres of
two opposite surfaces. It follows
also that the moment of inertia of
the parallelopipedon is equal to the
moment of inertia of one-third of

D
E
X
its mass applied at one of the corners A.
$288.]
585
THEORY OF THE MOMENT OF INERTIA.
FIG. 476.
§ 288. Prism and Cylinder.—By the aid of the formula for
the moment of inertia of a parallelopipedon, we can also calculate
that of a triangular prism. The diagonal plane A D F divides the
parallelopipedon into two equal triangu-
lar prisms, whose bases A B D, Fig. 476,
are right-angled triangles. The moment
of inertia for a rotation about an axis
XX, passing through the middles Cand
A of the hypothenuses, is = √½ M d².
Now if we employ the rule given in
§ 284, we obtain the moment of inertia
H

-Y、 F
-X
K
B
Y
X
D
in reference to an axis y passing through the centres of gravity
S and S
I.E.
J
M. CS² = M (12
d2
· (CB)')
W = √½ Md M.CS M
12
M
= ~[ 12/2² - (~) ²]
M[
W = js M đ,
18
and it follows also that the moment of inertia in reference to the
edge B His
W₁ = W + M. S B² = 1'8 M d² + M (d)² = 3 M d²
= M d²,
d denoting the hypothenuse A D of the triangular base.
For a prism A D F E, Fig. 477, whose bases are isosceles tri-
angles, the moment of inertia in reference to an axis IX, joining
the centres of gravity of the bases, is
FIG. 477.
D
Y
B
F
E
X
Y
6
= M d d denoting the side A D =
AE of one of the bases; for this surface
can be divided by the perpendicular A B
into two right-angled triangles. Now if
the altitude A B of the isosceles triangles,
which form the bases, is h, we have the
moment of inertia of this prism in refer-
ence to the axis passing through the

centres of gravity of the bases
W = & M dr
M
(4)² = M (j ď² — ¿ ¿³)
= } M ({ d® — ! h²),
586
[S 288.
GENERAL PRINCIPLES OF MECHANICS.
and, finally, the moment of inertia in reference to the edge, passing
through the points A and F of the bases, is
d³
h²
1
W₁ = W + M (3 h)²
4 h
= M6
+
9
9
= {M ( d² + h²).
FIG. 478.
K
BD
= C B
By the aid of the latter formula, we can calculate the moment
of inertia of a regular right prism A D F A. Fig. 478, which re-
volves about its geometrical axis. Let CA
be the radius of base or of one
of the triangles composing the base, the al-
titude CN of one of these triangles ACB,
and M the mass of the whole prism, then, ac-
cording to the last formula, when we substi-
tute r for d, we have

-X-
H
F
E
X
1
= ¦ M (??²
2
+/
78).
The regular prism becomes a cylinder, when he becomes equal
to r, and the moment of inertia of the cylinder in reference to its
geometrical axis is
W = ↓ M (12² + r²) = ↓ Mr².
11
The moment of inertia of a cylinder is equal to the moment of
inertia of half the mass of the cylinder concentrated upon its cir-
cumference, or equal to the moment of inertia of the whole mass at
the distance
-X
!
k = r √ = 0,7071. r.
If the cylinder A B D E, Fig. 479, is hollow, we must subtract
FIG. 479.
E
A
D
B
X
the moment of inertia of the hollow space
from that of the solid cylinder. Let 7
denote the length, r the radius C A of
the exterior and r, that C G of the interior
cylinder, then we have, according to the
above formula, for the moment of inertia
of the hollow cylinder
W = ↓ (M, r₁' — M₂ r²) = π (vì² . rî² — r²². rë²) ? = 4 π (r." — r:') l

1
2
1
2
2
1
(1
½ r₂2²
2
1
1
= { π (r₁² — r₂²) (r²² + r₂²) 7 = ↓ M (r²² + r;²);
-
2
for the volume of the body, which may also be considered as its
mass, is (r,² r²) 7. If r denotes the mean diameter
and b the width r,r, of the annular surface, we have
1
r₁ + r₂
2
2
W
M
M\r₂ +
(r₂
b
4
$ 289]
587
THEORY OF THE MOMENT OF INERTIA.
FIG. 480.
C
X
§ 289. Cone and Pyramid.-With the aid of the formula
for the moment of inertia of a cylinder we can
calculate those of a right cone and of a pyramid.
Let A CB, Fig. 480, be a cone turning upon its
geometrical axis and let r = D A D B be the
radius of its base and h = CD its altitude, which
coincides with the axis. If by passing planes
through it, parallel to the base and at equal dis-
tances from each other, we divide it into n slices,
we obtain n discs, whose radii are

F
E
D
Α
B
π
·O'
(
N.
j
N
and whose common height is
7
1
2
3
N
N
N
N
h
; the volumes of
N
etc,
these slices are
·
h
N
π
C)
N
h
N
π
3
C)
h
n
and consequently their moments of inertia are
cone
π
() · —
h
2n
(2)
C). 4C) 2
N
etc.
n'
The sum of these values gives the moment of inertia of the entire
W =
π pt h
2 n'
I.E., since 1ª + 2ª + 3* + . . . + 12ª =
(1ª + 2ª + 3* + ... + m¹),
า and the mass of the cone is
5
π p² h
M
CO
3
π pt h
W =
3
π p³ h
3
M r³.
10
10
3
10
A
FIG. 481.
X
-X
B
E
In like manner we have under the same cir-
cumstances for a right pyramid A C E, Fig. 481,
whose base is a rectangle,

W = } Md²,
in which formula d denotes the half DA of the
diagonal of the base.
We obtain, by subtracting one moment of
inertia from another, the moment of inertia of a
frustum of a cone (A B E F, Fig. 480) in refer-
ence to its geometrical axis FI
If we denote the radii DA and 0 Fby r, and r,
and the altitudes CD and CO by h, and h₂, we have
588
GENERAL PRINCIPLES OF MECHANICS.
[$ 290.
π
W
10
or, since the mass is
π
4
(r₁' h₂ — r₂* h₂) =
π h₁
10 Υ1
(1'‚º — r½³),
2
(r,³ — r‚³),
M =
2
3
1 (r₁² h₁ — r₂² h₂)
2
π h₁
3 11
γ.
2
5
W = Ax (-?)
M
グ
​3
グ
​§ 290. Sphere.-In the same manner the moment of inertia
of a sphere, revolving upon one of its diameters DE 2 r, is
determined. Let us divide the hemisphere A D B, Fig. 482, by
FIG. 482.
X
G
H
K
Α
C
B
-X
E
planes parallel to its base A CB, into n
equally thick slices, such as G K H, etc.,
and let us determine their moments.
square of the radius G of one of these
slices is
G K² = C G² – CK² = r² — CK²,
and, therefore, its moment of inertia is

= ! π . 2 (x²
П
π ľ
2 n
(204
ጎረ
CK)
(7 — 2 p². CK² + CK*).
Substituting successively for C K,
The
r 2 r 3r
n r
>
etc., to
and
N
ጎ
N.
adding the results, we obtain the moment of inertia of the hemisphere
W
I.E.,
put
[n
π r
n. p¹ — 2 p²
2 n
22.
π r
2n [nr
W =
2
N²
2
2
(1*+2*+...+n
C)² (1² + 2 +
+ 2 ² + . . . + m² ) + ( - ) ( 1 + 2 +….. + 22') ]
Проб
2
1.3
3
+
C
4
22.
4 π 2.5
(1
1 +
15
Now since the contents of a hemisphere are M
and if we consider
still holds good.
W = .2 . q₁² = }} M r²,
% π p.²
π r³, we can
as the mass of the whole sphere, the formula
The radius of gyration is
k = r 信
​0,6324 . r;
two-fifths of the mass of the sphere, at a distance equal to the
radius of the sphere from the axis of rotation, has the same moment
of inertia as the entire sphere. The formula
W = Mr²
3
holds good also for any spheroid whose equatorial radius is = r.
(See § 123.)
i
i
3 291.]
589
THEORY OF THE MOMENT OF INERTIA.
If the sphere revolves about another axis at the distance d from
the centre, we must put the moment of inertia
W = M (T² + { r²).
%
§ 291. Cylinder and Cone.-The moment of inertia of a
circle ABD E, Fig. 483, in reference to an axis passing through
its centre Cand at right angles to the plane of the circle, since all
points are at a distance (Ar from the axis, is
M r²,
and consequently that in reference to a diameter For FY
(compare § 231) is
W₁ = ¦ W = ! Mr².
On the contrary, the moment of inertia of a circular disc
A B D E, Fig. 483, which revolves about its diameter B E, is
found to be, like the moment of flexure of a cylinder,
πλα
4
M 12
4
consequently the radius of gyration of this surface is
ん
​h =
v
= r √ √ = { r,
1.E., half the radius of the circle.
FIG. 483.
Y
A
FIG. 484.


A
B
-X
G
E
F
I
X-
C
-X
B
E
D
D
Y
its diameter F G, which
Let be the half height
From this we can calculate the moment of inertia of a cylinder
A B D E, Fig. 484, which revolves around
passes through its centre of gravity S.
A Fandr the radius C A = C B of the cylinder, then the volume
of one half of it is π² 7, and if we pass through it planes
parallel to the base and at equal distances from each other, we
decompose this body into n equal parts, each of which is = N
N
and the first of which is at a distance the second at a distance
31
N.
27
N
the third at a distance etc., from the centre of gravity S. By
means of the formula in § 284, we obtain the moments of inertia
of these discs or slices
590
[§ 292.
GENERAL PRINCIPLES OF MECHANICS.
π l
Z
² ² [ { " + ( )']); " " [1 ² + (²²)']
ጎ
Про2 г
п
4
ጎ
3 \
N
²² [1² + (2/4)'], etc.,
whose sum is the moment of inertia
π p² l r n p²
?
W =
[
+
N
4
( . )' (1' + 2² + 3² + . . . + n²) ]
72
12.3
= π p² l
3
N 3
= M (~² + 1)
4
of half the cylinder. This formula holds good for the whole cylinder.
when M denotes its mass.
The moment of inertia of a right prism A B D, Fig. 485, in
reference to a transverse axis I passing through the centre of
gravity is determined in a similar way. Let & be the radius of
gyration of the base A B of the prism in reference to an axis N N,
passing through the centre of gravity C of the base and parallel
to YY, and let 7 denote the half length or height C S = D S of
the prism; we have the required moment of inertia in reference
to the axis XX
W = M (k² + } ľ²).
FIG. 485.
FIG. 486.
A

-IT
D

N
X
X
X
A
B
In like manner we find for the right cone A B D, Fig. 486,
whose axis of rotation passes through its centre of gravity at right
angles to its geometrical axis (D,
Η 3. Mr²² +
= 30
4
$292. Segments.-The moment of inertia of a paraboloid of
revolution B A D, Fig. 487, which revolves around its axis of
revolution AC, is determined in a similar manner to that of a
sphere. If the radius of the base is CB
altitude CA h, and if we divide the body
}
height we have their contents
n'
C D = a, and the
into slices of the
§ 292.]
591
THEORY OF THE MOMENT OF INERTIA.
h
N
π
1
h
2
h
•
a²,
N
N
N
N
3
π a², etc.,
•
N
for the squares of the radii are as the altitudes or distances from
the vertex A. From this we obtain the moments of inertia of the
successive disc-shaped elements of the body, which are
a² h π
4 a* h
π
9 a*
>
N
2
n²
27
h π
Π
n' 2 ' n² n 2 N"
etc.,
and consequently the moment of inertia of the whole paraboloid is
π at h
W =
(1² + 2² + 3² +
+ n²)
3
2 n²
παh
2 n³ 3
223
παι
6
π a² h
a²
3
= 1 Ma²;
2
3
for the volume of this body is M
FIG. 487.
B
π а² h
2
This formula may be applied to a low
segment of a sphere.
If the altitude of such a segment is
not very small compared with a, we have
for the moment of inertia of one of its

A
-X
X
slices
π h
2 n
D
π h
G
a¹ =
π h
2 n
h³ (2 r — h)²
(4 p² h²
2 n
•
in which r denotes the radius of the sphere.
4 r h³ + h*),
Now if we substitute for h successively the values
h 2 h 3 h
ጎረ N N
we obtain the moment of inertia of the segment of the sphere
h
N
ޑ
(分​)
N
π h
W
4 p²
2 n
(
h
N ³
4 r
N
3
4
(4): 7 + ( ): 7]
5
π/23
(20 72
30
The volume or the
and therefore
15r h + 3 h³).
mass of the segment of the sphere is
— π h¨ († — } h),
M
W = π h³ ( r — } h).
3
2 h ( r
— jt h + só ·
2
90
グ
​18
3 M h
T
(r — f h + só ·
ha
90
13
, etc.,
generally it is sufficiently correct to put
W = 3 M h (r — † h) = } M (a² + } h²).
TR
This formula is applicable to the bob of a pendulum.
592
[$ 293.
GENERAL PRINCIPLES OF MECHANICS.
FIG. 488.
Y
§ 293. Parabola and Ellipse.-For the surface A B D,
Fig. 488, of a parabola, if, instead of the surface F, we substitute
the mass M or change F into M, and if we
denote the chord A B by s and the height
of the arc CD by h, we have (according to
§ 233) the moment of inertia in reference
to the geometrical axis of this surface
M s²
20

-X-D
S
A
B
-Y
-X
W₁
and that in reference to the axis Y Y,
passing through the centre of gravity S at
right angles to XX, is
W₂ = 12; Mh².
2
T75
Hence the moment of inertia in reference
to an axis, passing through S at right angles to the surface of the
parabola, is
W = W₁ + W₂
=
M
12
175
( 3 ) + 1'4; 4') = { } [ ( ; )² + {} ~ ' ]
M
12
35
们
​20
For such an axis, passing through the vertex D of the parabola,
the moment is, since D S 3 h (§ 115),
W = (h)²
W + M (3 h)²
=
18
M
+ ¹²² h²
2
and, on the contrary, the moment in reference to an axis passing
through the centre C of the chord is
M
S
W₁ = W + M (h)² = } ~ [ ( ) ² + ; π ² ]
h
This formula is also applicable to a prism whose bases are para-
bolas, E.G. a working-beam, which consists of two such prisms
oscillating about an axis passing through their middle C.
The moment of inertia of an ellipse A B A B, Fig. 489, whose
semi-axes are С A = a and C B b. in
reference to the axis B B, is (according
to § 231)
A
FIG. 489.

Di..
D
E
-N-
B
C
παι
4
Ma
4.
and that in reference to the axis A 4 is
A
IT'
павз
4
M b²
4
hence the moment of inertia in reference
§ 294.]
593
THEORY OF THE MOMENT OF INERTIA.
to an axis, passing through the centre Cat right angles to the plane
of the figure, is
W = W₁ + W₂ = ¦ M (a² + b²).
(§ 294.) Surfaces and Solids of Revolution.-The mo-
ments of inertia of surfaces and solids of revolution can be determined
with the aid of the Calculus by means of the following formulas.
1) If a zone or belt P Q Q₁ P₁, Fig. 490, whose radius is M P
y and whose width is PQds, is
caused to revolve around its geometrical
axis AC, we have (according to § 125) its
FIG. 490.
P
R
M
Z
PL Q L
C
area
1

d
а 0 = 2 π у ds,
y
and its moment of inertia is
3
у² d 0 = 2 π у³ d s ;
hence the moment of inertia of the whole
surface of revolution AP P, in reference
to its axis A Cis
3
W = 2 = fy ds.
π
π
2) For a slice P Q Q P, whose volume is dV y dx, the
moment of inertia in reference to the axis AC is (according to
$288)
d V. y³
2
πy'd x
2
and consequently the moment of inertia for the whole solid of rev-
olution AP P, is
W = &fy' d x.
If A P is an arc of a circle, in which case the surface generated
by its revolution is a spherical cup or zone, we have
y³
= 2 rx x² and y ds = r d x,
and consequently the moment of inertia of this zone is
(2 r fx d x − fx d x )
W = 2 = f ( 2 r x x²º) r d x = 2 = r ( 2 r ƒ x d x
= 2 πr (r2 = 3)
x³
or, if we substitute h for the altitude A M = 2, we have
h
W = 2 x r k (r
=
Mh (r
since the area or mass of the zone is M = 2 πr h.
38
594
IS 294
GENERAL PRINCIPLES OF MECHANICS.
For the entire surface of the sphere h = 2 r, and therefore
W = ¦ M r².
3
1
If, on the contrary, A P is the arc of an ellipse, and conse-
quently the solid of revolution A P P, generated by the rotation
of the plane surface A P M a segment of an ellipsoid of revolution,
we will have
y²
1/2 (2 a x − xº),
a²
and therefore its moment of inertia in reference to the axis AC is
π
W = ¦ · = f ( 2 a x − x²)' d x
2 a'
пъ
7, a' f (4 a² x² − 4 a x² + x') d x
2 a¹
пъх
3
2 a
( § a² x³ − a x' +
это
5
E.G. for the entire ellipsoid, in which case x =
4
W = b'a = } } π a b². b² =
185
•
for the contents of this body are expressed by
(compare § 123).
1
2 a,
z M b² ;
b²;
で
​323 π a³
παι
a²
3) If the belt P Q Q, P₁ revolves about an axis passing through
A at right angles to its geometrical axis AC, we have (see § 284
and § 291) its moment of inertia
1
= d 0 (x² + y²) = 2 π (x² + y²) y d s,
½
and, therefore, the moment of inertia of the whole zone A P P₁ is
W
π
S (2x² + y²) y d s.
1
4) If the entire disc P Q Q, P₁ revolves around this same axis
passing through A, its moment of inertia is
d V (x² + ¦ y²) = π у² (x² + ¦ y²) d x,
4
and, therefore, that of the entire body AP P₁ is
W =
ㅠ
​2
f(x² + ¦ y³) y² d x.
For a paraboloid of revolution (see § 292), we have, when we
denote its altitude A M by h and the radius of its base M P by a,
y²
2
a²
X
h
and consequently the moment of inertia in reference to the axis of
ordinates passing through A is
πα
x
x³
h
S
w = "c² (x + 1 Z ) x d x = " " (1 x + 41°)"),
πα
Τ
h
h
$ 295.]
595
THEORY OF THE MOMENT OF INERTIA.
or, when we substitute x = h,
W = { π a² h (h² + } a²) = ! M (h² + { a²),
since the volume of this body isah (comp. § 124).
Hence we have the moment of inertia of this body in reference
to an axis, passing through the centre of gravity S at right angles
to A C
W₁ = ! M (h² + a) — (3)² M h² = } M (a² + } }²).
3
}
3
§ 295. Accelerated Rotation of a Wheel and Axle.—
The most frequent applications of the theory of the moment of
inertia are to machines and instruments; for rotary motions
around a fixed axis are very common in them. Since throughout
this work we shall meet with very many applications of this theory,
we shall treat here but a few simple cases.
B
FIG. 491.
A
X
If two weights P and Q act by means of two perfectly flexible
strings upon the wheel and arle A C D B, Fig. 491, if their arms are
CA a and DB = b and if the jour-
nals are so small that the friction can
be neglected, the machine is in equi-
librium, when the statical moments
P. CA, and Q.D B, are equal to
each other, or when Pa = Qb. If,
on the contrary, the moment of the
weight P is greater than that of Q.
or Pa > Qb, P will fall and Q will
rise; on the contrary, if P a < Q b;
P will rise and Q will fall. Let us therefore seek the relations of
the motions in this case, taking, E.G., Pa > Q b. The force,
which acts with the arm b and corresponds to the weight Q, pro-
duces a force whose arm is a and which acts in opposition to
the force corresponding to the weight P, so that the motive force

E
Q b
a
in action at A is P
P
Q
g
The mass reduced from the arm b to
Q b
Ü
the arm a, is
I a³
Q b²
M = (P +
(P + b²): g
Q
:g,
α
hence the mass moved by the force P
Q b
is
α
or, if the moment of inertia of the wheel and axle is W =
G k
and
g
596
[$ 295.
GENERAL PRINCIPLES OF MECHANICS.
therefore the mass of the same reduced to 4 is =
G k²
we have
M = (P
more accurately
P+
I a
Q b² Gk2
+
a²
a²
: g = (Pa² + Q b² + G h²): ga².
α.
Hence the acceleration of P or of the circumference of the
wheel is
p =
motive force
mass
Pa
Q b
Q b
P
a
P a² + Q b² + G k² · 9 a²
P a² + Q b² + G kg a;
hence the acceleration
ference of the axle is
9
b
a
k²
of the rising weight Q or of the circum-
P a − Q b
p
P a² + Q b² + G k²
•
g b.
The tension of the cord, to which P is attached, is
S = P _ P p
-
g
and that of the cord, to which Q is attached, is
= P(1
-
(1 − 1)) (see § 76),
Q X
S₁
=
= Q +
g
· 2 (1 + 1),
on the bearings is
S + S₁ = P + Q
PP
g
Q q
(Pa- Qb)²
+ = P + Q
g
Pa² + Q b² + G k"
and, therefore, the pressure
The pressure on the bearing of a wheel and axle, when in rota-
tion, is consequently less than when it is standing still.
From the accelerations p and q the other relations of the mo-
tion can be found; after t seconds the velocity of P is
pt
v = p
and that of Q is
v₁ = q
q t ;
FIG. 492.
-X
B
F
A
X
E
P
the space described by P is

8 = ½ p t²
and that by Q,
8₁ = √ q t².
EXAMPLE.- Let the weight upon the
wheel, Fig. 492, be P-60 pounds and that
on the axle, Q 160 pounds; let the arm
of the former be CA a = 20 inches
and that of the latter DB b = 6 inches,
and let the axle be composed of a massive cylinder, weighing 10 pounds,
F
§ 295.]
597
THEORY OF THE MOMENT OF INERTIA.
and the wheel of two rings, one weighing 40 pounds and the other 12
pounds, and of four arms, weighing together 15 pounds; finally, let the
radii of the large ring A E be = 20 and 19 inches and those of the smaller
one F G be 8 and 6 inches. Required the conditions of motion of this
machine. The motive force at the circumference of the wheel is
P
b
a
Q
6
60
ΤΟ
. 160 60-
—
48 = 12 pounds,
and the momeut of inertia of the machine, when we disregard the masses
of the ropes and journals, is equal to the moment of inertia of the axle,
which is
W be
2
10.62
2
180,
plus the moment of the smaller ring, which is
2
(~1
2
R₁ (r₁² + 12²)
plus the moment of the larger ring, which is
2
R₂ (13² + 742)
2
12. (82 +62)
600,
2
40.(20² + 192)
2
15220,
15. (192 +19.8+82)
= 2885;
3
plus the moment of the arms, which is, approximately,
A (r 3
2'
1
4
3 (71)
A (1
2
3 ▲ (~₁² + ~ 1 7 4 + ²±²
hence, by addition, we obtain
3
G 1:2 180 + 600 + 15220 + 2885 = 18885,
or, taking the foot as the unit of measure,
The whole mass, reduced to the radius of the wheel, is
18885
144
=
131,14.
M = (P+
Q b² + G k2
A 2
6
:g=
09] =
60 + 160
(990) * +
18885
+
202
町
​']:"
g
= (60
18885
400
. 0,031
60 + 160.0,09 +
= (60 + 14,4 + 47,21) . 0,031
=
121,61 . 0,031 = 3,76991 pounds.
Hence we have the acceleration of the weight P, or that of the circum-
ference of the wheel,
P-
b
Q
α
Ρ
Pa² + Qb² + G k²
12
3,76991
3,183 feet;
ee
a²
and, on the contrary, that of Qis
9
α
Ρ ਭੰਨ . 8, 183
=
0,955 feet;
the tension on the rope to which P is hung is
3,183
-(1-32,20
60 =
and that of the rope supporting Q is
(1-2)..
P =
1
s₁ = (
1 +
2). 8
(1 — 0,099) . 60 = 54,06 pounds,
Q = (1 + 0,955 . 0,031). 160 = 1,03. 160 = 164,8 pounds;
consequently the pressure on the bearings is S + S₁ = 54,06 + 164,8
218,86 lbs., or, if we include the weight of the machine, it is 218,86
598
[§ 296.
GENERAL PRINCIPLES OF MECHANICS.
+ 77
295,86 pounds. At the end of 10 seconds P has attained the ve-
locity v = pt = 3,183. 10 = 31,83 feet, and has described the space s =
v t
b
2
31,83.5 = 159,2 feet, and Q has been raised up s₁
s=0,3. 159,2
a
=
47,76 feet.
§ 296. The weight P, which imparts to the weight Q the ac-
celeration
q
P a b — Q b²
P a² + Q b² + G k²
9,
can be replaced by another P, without changing the acceleration
of Q, when the arm of the latter is a,, in which case we have
P₁a, - Qb
2
P₁ai + Qb² + Gh"
If we designate the quantity
Ct
car
Pa
Q b
Pa² + Q b² + G k²°
P a² + Q b² + G k²
Pa- Qb
Q b (b + c) + G k²
and the required arm of the lever
P₁
Q k²
a₁ = 1 o ± √ ( ) _ ? b (b + c) G lô
P₁
by c, we obtain
We find, also, by the differential calculus that the greatest ac-
celeration is imparted to Q by P, when the arm of the latter cor-
responds to the equation Pa — 2 Q a b = Q² + G k², or when
a =
b Q
Р
+1
2
(( 0 )² + Q b² + G & ²
(^^)
Р
The foregoing formulas become very complicated, when we take
into consideration the friction of the journals and the rigidity of
the ropes. If we denote the resistance due to both of these, reduced
to a radius r, by F, we must substitute, instead of the motive force
Q b + Fr
Ъ
P Q, the expression P
a
acceleration of Q
q
and
a
(Pa-Fr)b Q b²
Pa² + Q b² + G k²
a
•
· I
g
and then we have the
"
2
F
b²
Q b + P r + √ √ Q b + P r ) + Q u + G F
P
P
P
EXAMPLE-1) If the weights P = 30 pounds and Q = 80 pounds act
with the arms a = 2 feet and b =
moment of inertia of this machine is G
rising weight Q will be
foot upon a wheel and axle, and if the
60, the acceleration of the
$297.]
599
THEORY OF THE MOMENT OF INERTIA.
20.2.1
80. (1)²
•
g
30 - 20
120 + 20 + 60
32,2
32,2 =
20
Չ
2
30. 2² + 80. (1)² + 60
=
1,61 feet.
Now if we wish to produce the same acceleration with a weight P₁
45 pounds, the arm of P, must be
1
а 1
с
(
80 . 1 ( 1 + c) + 60
45
200
but
C =
10,
60 40
32
hence
а 1
as
5 ± 1 25
5 ± 1.11,358 = 5 ± 3,786
3
=
8,786 or 1,214 feet.
2) The acceleration of Q is a maximum when the arm of the force or
radius of the wheel is
1.80
a
+
30
1
4012
+
20 + 60
4
16 24
+
+
30
3
9
9
4 + √40
3
3,4415 feet,
and this maximum acceleration is then
1=
(3
30.1,7207 20
36)
g=
31,621
435,32
g = 2,339 feet.
30. (3,4415) + 80
3) If the moment of the friction and of the rigidity of the ropes be
Fr=
8, we must substitute, instead of Q b, Qb + F r = 40 + 8 = 48,
whence it follows that
48
a
30
+ 1
48
30
8
3
√/ ( 18 )² + 8 = 1,6 + √5,227
= 3,886 feet,
and that the corresponding maximum acceleration is
30. 1,943 8.20
30. (3,886)² + 80
•
J =
34,29
533
32,2 2,07 feet.
§ 297. Atwood's Machine.-The formulas for the wheel
and axle found in § 295 are applicable to the simple fixed pulley;
for if we put b = a, the wheel and axle becomes a fixed pulley. Re-
taining the same notations that we employed in the foregoing
paragraphs, we have the acceleration with which P sinks and Q
rises
(P Q) a²
p = q (P + Q) a² + G k² • I,
or, taking the friction into consideration,
(P
p = q
Q) a² Far
(P + Q) a* + G k²
· g.
In order to diminish the friction, the axle C of the pulley A B,
Fig. 493, is placed upon the friction-wheels D E F and D, E, F,.
Now if the moment of inertia of these wheels is G₁k and their
2
600
[$ 297.
GENERAL PRINCIPLES OF MECHANICS.
radius is D E = D, E, = a,, we have, when F designates the fric-
tion reduced to the circumference of the axle C,
p = q
(P − Q) a² Far
(P + Q) a² + G k² + G₁
2
g;
k² p²
a₂
2
reduced to their
G₁ h
for the moment of inertia of these friction rollers,
circumference or that of the axle of the wheel, is
Inversely we have the acceleration of gravity
A
9 =
FIG. 493.
(P + Q) a² + G k² + G₁
B
H
(P · Q) a² Far
2
2
•
P.
a₁
2
When the difference PQ, of the two weights is small, the
acceleration p is small and the motion is
consequently very slow; hence the resist-
ance opposed to the weights by the air
is unimportant, and the acceleration of
gravity can be determined with a certain
degree of accuracy by means of such an
apparatus, while the determination of it by
observations upon a body falling freely is
impossible. Experiments of this kind were
first made by an Englishman named At-
wood (see Atwood's treatise on Rectilinear
and Rotary Motion), and for this reason
the apparatus is known as Atwood's Ma-
chine. The scale HK, along which the
weight P falls, serves to measure the
distance fallen through. From the spaces.
fallen through and the corresponding time
t we obtain

L
•
P
GA TE
N
N
2.8
p
ť
山
​K
but if during the fall we remove the motive
force by causing the weight L L, which is
made in the shape of a ring and is equal to the force, to be caught
by the fixed ring N N₁, the remainder of the space s,, through which
the weight P falls, will be described uniformly, and the velocity,
which is determined by the time t, (which can be observed by
means of a good watch), is
V
$1
t₁
§ 298.]
601
THEORY OF THE MOMENT OF INERTIA.
and the acceleration is
2
S
$1
P
t
t ti
1
If we make t₁ = t1, we obtain directly by the experiment.
p = $1. Substituting this value of p in the above-mentioned
formula, we obtain the acceleration g of gravity.
$298. Accelerated Motion of a System of Pulleys or
Tackle. The accelerations of the weights P and Q, which are
supported by a system composed of a fixed pulley A B, and a loose
pulley E G, Fig. 494, are found in the following
Let the weight of the pulleys A B and
E G be G and G₁, their moments of inertia Gh²
and G₁ h₁², their radii C' A = a and D E = a, and
their masses reduced to the circumference M
FIG. 494.
H
Α
B
manner.

G k²
G₁
and M₁
ki
k₁2
g
a²
2
2.
If the weight P sinks
9
E
G
P
a certain distance s, Q+ G, rises s (§ 164), the
work done is therefore P s − ( Q + G₁) Now if
2
in sinking the weight P has acquired the velocity v,
2
then the velocity is communicated to Q + G₁, the velocity of the
pulley A B at the circumference is v and the pulley E G acquires
at its circumference the velocity; for in rolling motion the mo-
The
で
​tions of translation and of rotation are equal to each other.
sum of the living forces, corresponding to the masses and velocities, is
P
¢° 」
g
Q + G₁
g
(3)
G k²
G₁ k
2
+
7:² +
I a²
2
gas
(3)
putting the half of it equal to the work done, we obtain the equation
G₁)
P
2
(Q + G¹)) & = (F
=(p
Q + G₁
Gle
P +
+
+
G₁ h₁)
4
a
2 g
Hence the velocity corresponding to the space s, described by P, is
P +
2 g s ( p
Q + G₁
1
+
Q + G₁
₁)
G, k
+
Gh
a²
602
IS 298.
GENERAL PRINCIPLES OF MECHANICS.
1,2
For the acceleration p we have p s
and therefore
2'
P
Q + G₁
2
p =
g
P+
Q + G₁
4
Gk2
G, ki
2
+
+
a²
4 ai
2
G₁ is pi
P
The acceleration of Q +
tion of G, is also the same.
unites the two pulleys, is
2, and the rotary accelera-
2'
The tension on the rope B E, which
G k³ \ P
a² I g
S = P
P+
2
P
Gh) is expended in producing the accel-
g
for the force (
(P
P+
a²
eration of P and G; the tension on the rope G H, which is
fastened at one end, is, on the contrary,
S S
G₁k P
2
2
ai
2
for the pulley E G is set in rotation by the difference SS, of the
tensions on the rope.
EXAMPLE. The weights P 40 pounds and Q = 66 pounds hang
upon the system of pulleys or tackle represented in Fig. 494, and each of the
pulleys weighs 6 pounds; required the acceleration of each of the weights.
The motive force is
P -
Q + G₁
2
1 = 40
66 +6
2
= 4 pounds.
The masses of these pulleys, reduced to their circumferences, are
G 2
G, I
h 2 G 6
3
1 1
(§ 288),
I a²
ga1
2g
2 J
and the total mass is
= (1
F +
Q + a ₁
4
G k²
1
1
G₁k 1
2
247
+
+
a²
4 a
2
: g = (40 + 42 + 3 +
3) : g
4 g
1
p =
4
247
•
hence the acceleration of the sinking weight is
and that of the rising weight is
The tension of the rope B E is
16.g
247
16. 32,2
247
515,2
247
2,086 feet,
Ρ
P1
1,043 feet
2
S
P - (P +
(P + 3 ) 23/
G p
2
— 40 — 43 .
2,086
32, 2
40 - 2,785
3'7,215 pounds.
and that of the rope G His
G
S₁ = S
37,215 - 3.
2
29
1.043
32,2
37,118 pounds.
$299.]
C03
THEORY OF THE MOMENT OF INERTIA.
§ 299. The motion is more complicated, when the pulley E G,
Fig. 495, hangs only upon a cord wound around it. Let us sup-
pose that P sinks with the acceleration p, and that Q
rises with the acceleration q, then the acceleration of
B the motion at the circumference of the loose pulley is
q₁ = p q (§ 45).
Ρ
FIG. 495.

A
Now if we put the tension of the cord E, S, we
obtain
P
G
E
-
P − S = (P
P +
G k² \ P
aⓇ g
and
8 − ( Q + G₁) = ( Q + G₁) 2;
S
g
for, according to § 281, we can assume, that S acts at the centre of
gravity D of E G.
Finally we have
G₁ kr qi
S
2
રી I'
since we can assume that the centre of gravity D is fixed and that
the pulley is put in rotation by S
The last three formulas give the accelerations
P
S
S
p =
I, I
Gk
(Q + G₁)
Q + G₁
G₁)) 9
Sa
2
g and qı
g;
G₁ ki
P+
a™
substituting all three in the equation q₁ = p- 7, we obtain
Sa,
P
S
G₁ ki
9
G k²
I
8 − (Q + G₁)
S
Q + G ₁
g,
P+
a™
whence it follows that the tension of the rope is
2 P a² + G k²
S
aj
1
+
2
Gh
Q
´¯6) (P a² + G k)² + a².
- G
From this value of S we find by the application of the above formula
the accelerations of the weights P and Q.
If we neglect the mass G of the fixed pulley and put Q
we obtain simply
S =
2 Pa. G, k,"
2
2
2 PG, k
P (a,² + k,²) a² + Ga² k²² G₁ h‚² + P (αi² + ki³)
0,
If the end of the cord 4 E, instead of passing over the pulley,
is fixed, we have the acceleration p = 0, and therefore 9,
and the tension
1
9:
604
[$ 299.
GENERAL PRINCIPLES OF MECHANICS.
(Q + G₁) G₁ ki
2
1
2
( Q + G₁) a₁² + G₁ ki
29
S
for Q
= 0, we have
G₁k, 2
S
a₁² + ki
2
If the rolling body G, is a massive cylinder, we have
2
2
= 1 G₁,
a₁
and the tension in the first case is
2 P G₁
S
3 P + Gi
and in the second
G₁₁
3
If in the first case the weight P must rise, we have p negative
and S> P, I.E.,
2
2
2 PG, k> PG, k+ P (a,² + k₁²),
or simply
G₁
P
a₂
> 1 +
in order that G, shall sink it is necessary that S < G₁, or that
G₁
> 1
P
a²
1
2'
EXAMPLE.-If the rope G H of the system of pulleys in the example of
§ 298, Fig. 494, suddenly breaks, the rope B will be, for an instant at least,
stretched by a force
2 P+
G k²
a²
1
5976
1147
Մ
S
2
a
1
+
2
1
1
83.72
Q + G,
25.43 + 72
= 5,210 pounds.
Hence the acceleration of the sinking weight P is
2.40 + 3
;) (e
P+
G Z²
a²
+1
(}} + 7½) (40 + 3) + 1
Р
S
P
GE²
P +
a²
40 - 5,210)
5,210). 32,2 =
34,79
32,2
=
26,05,
40 + 3
43
and that of the sinking pulley is
9
(?
1
Q + G ₁
Q + G ₁
1
72
=
(73 - 5.210)
66,79
32,2 =
32,2
=
29,87 feet,
72
and the acceleration of rotation of this pulley is
2
21
G₁
•
k
2
g
5,210
3
•
32,2
=
55,92 feet.
1 1
$ 300.]
605
THEORY OF THE MOMENT OF INERTIA.
§ 300. Rolling Motion of a Body on a Horizontal
Plane.—If a round body A C D, Fig. 496, is pushed forward with
A
C
D
FIG. 496.
S
E
a certain initial velocity
c upon the horizontal
path D E, it will, in
consequence of the fric-
tion upon this path, as-
sume a motion of rota-
tion, the velocity of

which will gradually increase; its acceleration p is determined by
the formula
Ρ
Force
Mass
2
& G a² φα
M k²
9,
k2
in which
denotes the coefficient of friction, G = Mg the weight,
G the friction, Mh the moment of inertia and a the radius CD
of rotation of the body. The velocity of rotation at the distance
CD from the axis c, engendered by this acceleration in the
time t, is
a²
v = pt:
$
12
For I t.
On the contrary, the forward motion of the body suffers a re-
tardation q, which is determined by the formula
q
Resistance
Mass
фG
=
& J,
M
hence the velocity of this motion after t seconds is
0
v₁ = c
q t = c — og t.
Now if we put v₁ = v, or
a²
&gt = cogt,
we obtain the time after which the velocity of rotation becomes
equal to that of translation and the rolling of the body begins.
This time is
t
1
C
k2
C
7:2
a² + k² & g
At the end of this time the common velocity is
and the space described by the centre
a²
a² c
G₁ =
og t
a² + h²?
of the body is
$ =
= (c + G₁)
2 a + k² c
t =
2
だ
​C
(2 a² + h²) h²²
a² + k² 2´ a² + h² ˆ ọ g
(a² + h²)* *20.
606
[$ 301.
GENERAL PRINCIPLES OF MECHANICS.
If the coefficient of rolling friction was 0, the body A C
=
would roll on forever with the constant velocity c₁
a c
a² + k²
upon
the horizontal plane without coming to rest; but since the rolling
friction fG constantly opposes this motion (see § 192), the body,
a
after describing a certain space $1, will come to rest. At the end
of this friction has consumed the whole
of this space the work
f G sr
a
of the energy
G c₁²
2 g
2
G k²
2
+
a²
2 g
a²
a² + k² \ G c₁₂²
2g
2
stored by the mass of the
body, and therefore we can put
+)
2
a²
2 g
f G s₁
'a² + k² \ G c
ņ
hence the space
2
a³
a² + k²
2
f a
2 g
ƒ (a² + k²) 2 g
is described in the time
2 S1
a² + h²
C₁ ас
t₁
C1
f a
g
fg
k2
h2
For a rolling ball
3, and for a cylinder
= (see § 290).
a²
a²
c²
C
In the latter case t
183
C₁ 3 c, s =
and s
''ૐ
O g
249
a
c²
f 2 g
CHAPTER II.
THE CENTRIFUGAL FORCE OF RIGID BODIES..
§ 301. The Normal Force.-The force of inertia manifests
itself not only when the velocity of a moving body changes, but also
when there is a change in the direction of the motion; for a body,
§ 301.]
THE CENTRIFUGAL FORCE OF RIGID BODIES.
C07
by virtue of its inertia, moves uniformly and in a straight line (see
§ 55). The action of inertia, when the direction changes continu-
aliy, I.E. when the motion of a body takes place in a curved line,
and particularly in a circle, will be the subject discussed in this
chapter.
If a material point moves in a curved line, it is at every point
subjected to an acceleration, which causes it to deviate from its
former direction. This acceleration has already been treated of in
phoronomics under the name of the normal acceleration. Let the
radius of curvature of the path of the moving body ber and its
velocity v, then the normal acceleration is
v2
p
(§ i³).
グ
​=
Now if the mass of the point 1, the acceleration corres-
ponds to a force
M v²
P = Mp
グ
​which we must consider as the original cause of the continued
change of the direction or motion of the point. If the point is
acted upon by no other (tangential) force than the normal one, its
velocity will be constant and c, and therefore the normal force
ture.
FIG. 497.
P =
=
=
Mc²
is dependent only upon the curvature or radius of curvature, I.E.
smaller for a smaller curvature or for a greater radius of curvature,
and greater for a greater curvature or for a smaller radius of curva-
When the radius of curvature is doubled, the normal force
is but one-half as great as before. If a material point M, Fig. 497,
is obliged to pass over a horizontal
plane in a curved line ABD FH,
if we neglect the friction, the point
will have in all points the same ve-
locity and the pressure against the
side wall in every position will be
Kequal
equal to the normal force. While
the point describes the arc 4 B

C
Më
this pressure is
; while
CA
it describes BD it is =
Mc²
EB
Mc²
for the arc D F it is
and
GD
*
608
[$ 302.
GENERAL PRINCIPLES OF MECHANICS.
Mc²
for the arc FH,
CA, E B, G D and K F denoting the
KF
radii of curvature of the portions A B, B D, D F and FH of the
path.
§ 302. Centripetal and Centrifugal Forces.-If a material
point or body moves in a circle, the normal force acts radially
inwards, and for this reason it is called the centripetal force (Fr.
force centripède, Ger. Centripetal- or Annäherungskraft), and the
force in the opposite direction, L.E. radially outwards, with which
the body through its inertia resists the former force, has received.
the name of the centrifugal force (Fr. force centrifuge, Ger. Centrif-
ugal-, Flich- or Schwungkraft). The centripetal force is the one
which acts upon the body inwards, and the centrifugal force is the
resistance of the body, which acts in the opposite direction. In the
revolution of the planets around the sun, the attraction of the sun
is the centripetal force; if the moving body is compelled to describe
a circle by a guide, such as is represented in Fig. 497, the guide.
acts by its resistance as the centripetal force and opposes the centrif
ugal force of the body. If, finally, the revolving body is connected
by means of a string or rod with the centre of rotation, then it is
the elasticity of the rod, which puts itself in equilibrium with the
centrifugal force of the body and acts as the centripetal force.
If G is the weight, and therefore M
G
g
the mass of the re-
volving body, r the radius of the circle, in which the revolution
takes place, and v the velocity of revolution, we have, according to
the last paragraph, for the centrifugal force
or
M v²
P
2'
G v²
gr
شماوج
G
= 2.
2 g
v2
P : G = 2 .
: 1',
2 g
I.E., the centrifugal force is to the weight of the body as double the
height due to the velocity is to the radius of rotation.
If the motion is uniform, which is always the case when no
other force (tangential force) besides the centripetal force acts
upon the body, we can then express velocity = e in terms of the
and the
duration t of a revolution by putting c =
space
time
2 πη
t
$ 302.]
609
THE CENTRIFUGAL FORCE OF RIGID BODIES.
expression for the centrifugal force becomes
יך ה
P =
t
M
4 π²
472
Mr =
•
t²
g
I to
Gr.
1
g
Since 439,4784, and in feet = 0,031, we have, in a more
convenient form for calculation, the value of the centrifugal force
P
39,4784
t'
Mr =
1,2238
Gr pounds.
t
The number u of revolutions per minute is often given, in which
case, substituting for t,
60
W
we have
u² =
u²
u² Mr = 0,010966 u Mr 0,0003399 u Gr pounds.
P
39,4784
3600
We have also P
2 п
Since
t
Gr
4,0243 = 0,001118 u Gr kilograms.
ť
is the angular velocity w, we can also write
P = w². Mr.
Hence it follows that for equal times of revolution, I.E. for the
same number of revolutions in a given time or for the same angular
velocities, the centrifugal force increases as the product of the mass
and the radius of gyration; and if the other circumstances are the
same, it is inversely proportional to the square of the time of revolu-
tion, or directly proportional to the square of the number of
revolutions and to the square of the angular velocity.
EXAMPLE-1) If a body, weighing 50 pounds, describes a circle of 3 feet
radius 400 times in a minute, the centrifugal force is P = 0,0003399 .
4002 .50 .3 3,399 . 16. 50 . 3 339,9. 24 8158 pounds.
If this body is connected with the axis by a hemp rope, the modulus
of ultimate strength of which is (§ 212) 7000 lbs., we should put 8158
7000. F, and therefore the cross-section of rope should be F =
1,165 square inches, and its diameter should be
8158
7000
d =
/4 F
1
0,5642 . √4.660 — 0,5642. 2,159
-
1,22 inches.
π
In order to have triple security, we must make d 1,22 v3 =
1,22 . 1,732 = 2,11 inches.
2) From the radius of the earth r = 204 million feet, and the time of
39
610
[$ 203.
GENERAL PRINCIPLES OF MECHANICS.
revolution or length of day t 24 hours 24.60.60 86400 seconds,
we obtain for the centrifugal force of body upon the earth at the equator
20750000 G 2539
86400"
1
P = 1,2238.
G
•
8642
G,
290
24
but if the day were 17 times as short, or
17
1b. 24' 42", this force would
be 17 289 times as great, and the centrifugal force would be nearly
equal to the weight G of the body. At the equator, in that case, the cen-
trifugal force would be equal to the force of gravity, and the body would
neither fall nor rise.
3) The centrifugal force arising from the revolution of the moon around
the earth is counteracted by the attraction of the latter. If G is the weight
of the moon and r is its distance from the earth, and t the time of revolu-
tion around the latter, the centrifugal force of this body is
1,2238.
Gr
t²
Now let a be the radius of the earth, and let us assume that the force
of gravity at different distances from its centre is inversely proportional to
the nth power of this distance; we have the weight of the moon or the
attraction of the earth
28
and putting both forces equal to each other
= G
(笑​)
>
(9)=
= 1,2238.
t2
a 1
But
グ
​r = 1251 million feet, t = 27 days 7 hours 43 minutes
60'
39342 minutes = 39342.60 = 2360520 seconds, whence
72
(65) * -
1,2238.1251
393,42.36
1
3600
hence n = 2, 1.E. the attraction of the earth (or gravity) is inversely pro-
portional to the square of the distance from its centre.
$303. Mechanical Effect of the Centrifugal Force.-
If the path (A B, Fig, 498, in which the body M moves, is not
P
FIG. 498.
M
N
B
at rest, but turning upon an axis C, it
imparts to the body a centrifugal force
P, by virtue of which it either gives out
or absorbs a certain amount of mechanical
effect. The former occurs when, in mov-
ing in its path, it departs from, and the
latter when it approaches the axis of rota-
tion C Let M be the mass of the body,
w the constant angular velocity with which
the path, E.G. a top (Fr. sabot, Ger. Krei-
sel), turns around its axis C, and let z de-

$303.] THE CENTRIFUGAL FORCE OF RIGID BODIES.
611
note the variable distance (M of the body, which is moving in
the path CAB; we have the centrifugal force of the body
P = w² M z,
and the work done by this force, while the body describes an cle-
ment MO of its path and the radius C M is increased by an
amount N 0 =
O 5, is
PC = w² Mz. S.
5.
Let us imagine the radius z to be composed of n parts, each = 5
then if we put z = n 5 and assume that the body begins to move
at the centre of rotation (', we obtain the work done by the cen-
trifugal force of the body, while the body is describing the space
CAM, during which time the distance of the body is gradually
increasing from 0 to z. By substituting successively in the last
equation, instead of z, the values 5, 25, 3,... n, and then adding
the values thus found, we obtain this mechanical effect
A = w² MS (5+25+35+ ... + n 5) = w² _M 5° (1 + 2 + 3 + ... + n),
or, since 1 + 2 + 3 + +n, when the number of members is
n²
great,
we can write
2'
n°
A = w² M 52
I
w² M zº.
2
Now the velocity of rotation of the top at the distance CM = z
from its axis is
2' = W Z.
hence we can write more simply
A z M v²
G.
2 g
when we substitute, instead of the mass of the body, the weight
G = Mg.
If the body begins its motion, not at C, but at any other point
A without the axis of rotation, and at a distance (A = z₁ from
, where the velocity of rotation is
12
2
2'₁ = W Z19
the work M z done by the centrifugal force while the body is
passing from C to A must be omitted, and we have the work done
by the centrifugal force while the body passes from A to M
A = ↓ w° M z² — ! w² M z,² = { w° M (z° — z₁°)
{
= 4 M (v² — x') = ('² 3,,"")
2 g
G.
If a body moves in a rigid path or groove, which revolves about
a fixed axis, the vis viva of this body is increased or diminished by
612
[§ 304.
GENERAL PRINCIPLES OF MECHANICS.
the product of the mass (M) and the difference of the heights due
to the velocities of revolution (
and -) at the two ends A
v2
v,2
g 2g
and M of the path. The increase takes place when the motion is
from within outward, and the decrease when the motion is from
without inward.
FIG. 499.
2
§ 304. If a body begins its path A M B upon a top A B C,
Fig. 499, at A with a relative velocity c₁,
and leaves the top at B with the relative
velocity c, and if the velocities of rotation
of the top in A and B are v₁ and v2, the
energy stored by the body in describing
the path A M B, supposing no other force
to act upon it, is

M
B
vi
2
C₂
V₂
vi
A
G
G,
2 g
2 g
C 2
and therefore
or
c₁² + 1:2²
2
c² c²² = v₂² — v₁²,
and consequently the velocity of exit is
Co
2
2
=
+212 v₁² = √ci² + w² (r,² — ri“),
denoting the angular velocity of the top and r, and r, the dis-
tances CA and C B of the points (A and B) of entrance and exit
from the axis of rotation C.
The relative velocity of exit c, is determined in like manner,
when the body enters at B upon the top with the relative velocity
c and moves upon it from without inwards. It is then
C₁ =
2
2
2
√ c₂ — (v₂² — v₁²) = √c₂ — w² (r² — r‚²).
C1 √c²
Since the body in describing the path A M B has, besides its
relative velocity (c) in the path, also the velocity of rotation of
the path, it must be introduced at 4 with an absolute velocity
A w, w₁, which is determined in intensity and direction by the
diagonal of the parallelogram constructed with e, and v₁, and the
body leaves at B with an absolute velocity B w₂ = w., determined
by the diagonal of the parallelogram B c, wa , constructed with
the relative velocities c₂ and v₂.
The energy restored, or stored, by the body in describing the
path AMB on the top, which has been gained or lost by the
top, is
§ 304.] THE CENTRIFUGAL FORCE OF RIGID BODIES.
613
A = ±
W2
2
2
· (10, 10") 07.
29
G.
If a body should transmit all its energy
20
2
2g
G to the top, while
describing the path A M B, the absolute velocity of exit must be
w₂0, and c, must be not only equal to v, but also exactly oppo-
site to it; the path must therefore be tangent to the circumference
at B.
W2
EXAMPLE.—If the interior radius of the top, represented in Fig. 499, is
CA = r
= 1 foot and the exterior one C B =
1 feet and if it
revolves 100 times per minute, the angular velocity is
1
Пи
W
30
3,1416.
10
3
10,472 feet,
and consequently the velocity at the interior circumference is
V v₁ = wr₁ = 10,472 feet, and at the exterior one
1
02
= wr. = 10,472 . 1,5 = 15,708 feet.
2
1
1
Now if we cause a body, whose velocity is w₁ 25, to enter the top at
A, in such a direction that the angle , A, formed by its absolute mo-
tion with the direction of revolution is a = 30', we have for the relative
velocity c₁, with which the body begins its motion on the top,
1
1
and therefore
c₁2
2
2
1
C v ₁ ² + w ₁² — 2 v₁ w₁ cos, a =
v1 101
109,66 — 453,45 + 625,00
=
281,21,
C 16,77 feet.
If the body is to enter without impact, we must have for the angle
v₁ A c₁ ẞ formed by the path with the inner circumference of the top
1
1
sin. B
20
1
or
sin. a
C1
25 sản. 30
sin. ẞ =
16,77
whence ß = 48° 12' .
For the relative velocity of exit c, we have
2
2
2
1
2
2
1
C₂² = C₁ ² + v₂² — v₁² = 281,21 + 109,66 [(§)² — 1ª] = 418,28,
and consequently
C2 20,45 feet.
And, on the contrary, for the absolute velocity of exit w, when the canal
or groove A M B forms with the exterior circumference an angle d
2
or v₂ B cz
102
2
=
2
2
W₂² = C₂² + Vg³·
and consequently
го
2012
2 g
160°, we have
2 C 2 v2 cos. & = 418,28 246,74 — 603,72
+
61,30,
202 7,80 feet.
ខ
9,69 feet, and
20 2*
2 g
0,0155. 61,31
=
0,95 feet,
Finally, the heights due to the velocities are
202
0,0155 .625 =
and the amount of mechanical effect imparted to the top by a body, whose
weight is G, while passing over the top, is
614
[s 305.
GENERAL PRINCIPLES OF MECHANICS.
(9,69 0,95) G = 8,74 G,
2
го
ՂՐ
1
2
A
*) & = (9,60 – 0,95)
2 g
or, if its weight G = 10 pounds,
A 8,74 . 10 = 87,4 foot-pounds.
REMARK.—The foregoing theory of the motion of a body on a top is
directly applicable to turbine wheels.
§ 305. Centrifugal Force of Masses of Finite Dimen-
sions -The formulas for the centrifugal force found in the fore-
going paragraphs are not directly applicable to an aggregate of
masses or to a mass of finite extent; for we do not know what
radius r of gyration must be substituted in the calculation. To
R
K
FIG. 500.
Z
Q
determine this radius, the following
method may be adopted. Let CZ,
Fig. 500, be the axis of rotation and
CX and Y two rectangular co-ordi-
nate axes and let M be an element of
the mass and M K = », M L = y and
M N = z its distances from the co-cr-
dinate planes Y Z, Y Z and I I
Since the contrifugal force P acts in the
direction of the radius, we can transfer
its point of application to its point of
intersection with the axis of rotation.
If we decompose this force into two components in the directions
of the axes CX and CF, we obtain O Q = Q and OR = R, for
which we have

Y
X
O Q: 0 P = 0 L: 0 M and OR: 0 P = 0 K: 0M,
whence
M
t.
Y
Q
P and R
P,
Τ
r designating the distance O M of the element of the mass from
the axis of rotation. If we proceed in the same way with all the
elements of the mass, we obtain two systems of parallel forces, one
in the plane X Z and the other in the plane FZ, and each of
which acts at right angles to the axis CZ. Employing the indices
1, 2, 3, etc., to distinguish the various clements of the mass, I.E.
putting them M₁, M, M, etc., and their distances = 1, X2, X
etc., we have the resultant of one system of forces
Q = Q1 + Q₂ + Qs + ..
P₁ x²,
+
Pude +
ro
P 3X3
13
+
= w₂. (M{ x₁ + My xy + . . .),
$ 305.]
615
THE CENTRIFUGAL FORCE OF RIGID BODIES.
and that of the other
R
Y
R
R₁ + R₂ +
R3
R2
R₂
and
26 =
FIG. 501.
2
じ
​02
V
01
C
Q,
Q8
= w². (M₁ y₁ + M₂ Y½ + .).
Q
X
2
If, finally, we put the dis-
tance CO₁, C' O, etc., of the
elements of the mass from
the plane of Y Y = 21, 2,
X
etc., we obtain for the points
of application U and V of
these resultants the ordi-
nates C U u and C' V = v
by means of the formulas
(Q₁ + Q: + ...) u
Q₁ Z₁ + Q₂ ZQ +

1
and (R₁+ R₂ +
R₁ z₁ +
R₂ z₂ +
•
Q₁ Z1 + Q2 Z2 +
Q₁ + Q₂ +
1 1
M₁ x₁ z₁ +
M, x, z, +
M₁ x₁ +
M₂ X 2 +
•
.) v =
whence
v =
R₁ Z₁ + R₂ Z2 +
1
R₁ + R₂ +
M₁ y₁ z₁ + Ms Y Z + . . .
M₁ y₁ + M» Y½ +
Hence we see that generally the centrifugal forces of a system
of masses or of finite bodies can be referred to two forces, which
cannot be combined so as to give but a single resultant when u
and v are unequal.
EXAMPLE.-Let the masses of a system be
1
2
3
M₁ = 10 pounds, M, 15 pounds, M,
and their distances
18 pounds, M, 12 pounds,
x1
=0 inches, 2
3
4 inches, z = 2 inches,
4
6 inches,
Y
3
1
Y 2
1
(C
Y 3
5
(C
3
CC
4
21
2
(6
22
3
((
23
3
((
24
then the resultants of the centrifugal forces are
Q w² . (10 . 0 + 15 . 4 + 18 . 2 + 12 . 6)
R = w². (10.3 + 15. 1 + 18.5+ 12 . 3)
and consequently their distances from the origin Care
10.0.2 + 15.4.3 + 18.2.3 + 12.6.0
10.0 + 15. 4 + 18. 2 + 12 . 6
=
168 . w² and
171. w²,
น
288
168 7
12
1,714 inches,
and
10.3.2 + 15.1.3 + 18.5.3 + 12.3.0
V
10.3 + 15.1 + 18.5 + 12.3
375 125
171 57
2,193 inches.
The difference of these values of u and shows that the centrifugal
forces cannot be replaced by a single force.
616
[S 306.
GENERAL PRINCIPLES OF MECHANICS.
Y
§ 306. If the elements of the mass lie in a plane of rotation,
RA
L
P2
FIG. 502.
Z
M₂
P = √ Q² + R² =
Now if C K
centre of gravity
we have
P
K
I.E. in a plane X CY, Fig. 502,
which is at right angles to the
axis of rotation, as M₁, M... ., do,
their centrifugal forces will give
a single resultant; for their di-
rections cut each other at one
X point C of the axis C Z. If we
retain the notations of the last
paragraph, we obtain the re-
sulting centrifugal force in this

case
w² √ [(M, x₁+M₂ x₂+...)² + (M₁ y₁ + M₂ y₂+ ...)'].
= x and C L = y are the co-ordinates of the
of the system of masses M = M₁ + M₂ +
M₁ x₁ + M½ x ½ + = M x
2
M₁ Y₁ + M₂ Y₂ +
•
= My,
whence it follows that the centrifugal force is
M²x²
P = w² √ M² x² + M² y² = w² M √x² + y² = w² Mr,
in which r = √x² + y² designates the distance C S of the centre
of gravity from the axis of rotation CZ.
For the angle P C X = a, formed by this force with the axis
CX, we have
R
tang. a =
My
Mx
Q
Y
,
X
consequently, the direction of the
centrifugal force passes through
the centre of gravity of the system, and that force is precisely the
same as it would be if all the elements of the mass were concentrated
at the centre of gravity.
FIG. 503.
Z
For a disc A B at right angles to the axis of rotation Z Z,
Fig. 503, the centrifugal force is also =
w² M r, if M denotes its mass and r the dis-
tance C S of its centre of gravity from the
axis. If the centres of gravity of the ele-
ments of the mass of a body lie in a plane of
rotation, or if this plane is a plane of symme-
try of the body A D F F, Fig. 504, the cen-
trifugal forces of the elements of the mass of
the body can be combined so as to give a
single resultant acting at the centre of gravity of the body, and

-%
X
Р
§ 306.]
617
THE CENTRIFUGAL FORCE OF RIGID BODIES.
this resultant corresponds to the distance of this point S from the
axis of rotation and can therefore be determined by the formula
P = w² Mr.
Ꮓ
FIG. 504.
F
FIG. 505.
Z


A
-X
A
E
P X
D
E
Y
-Z
F1
-Z
In order to find the centrifugal force of a body A B D E,
Fig. 505, let us divide it into disc-shaped elements by planes per-
pendicular to the axis Z Z, and then find their centres of gravity
S1, S2, etc.; we can then determine by the aid of the latter the cen-
trifugal forces, by decomposing these into their components in the
directions of the axes CX and C' I and by combining the compo-
nents in the plane Z CX, we obtain the resultant Q, and by com-
bining those in the plane Z CY, we obtain their resultant R.
If the centre of gravity of all the discs lie in a line parallel to
the axis of rotation, we have r = x₁ = x², etc., and y Y₁ = Ya, etc.,
and therefore r = r₁ = r., etc., whence it follows that the centrif
ugal force of the whole body is
P = w² (M₁r + Mɔr + . . .) = w² Mr
and that the distance of the point of application from the plane
XY is
z =
1
(M₁ za
2₁ + M₂ Z2 + . . .) r
M₁ z₁ + Mɔ za +
M₁ + M₂ +
= 2.
(M₁ + M₂ + ...) r
2
From these equations we see that the centrifugal force of a body,
which can be divided into discs, whose centres of gravity lie in a
line parallel to the axis of rotation, is equal to the centrifugal force
of the mass of the body concentrated at its centre of gravity, and
the point of application of this force is at the centre of gravity.
Hence we can find in this manner the centrifugal forces of all
symmetrical bodies (see § 106), whose axis of symmetry is parallel
to their axis of rotation, and also that of all solids of revolution,
whose geometrical axis is parallel to the axis of rotation. If the
axis of rotation and the geometrical axis coincide the resulting
centrifugal force is 0.
=
618
[$ 307.
GENERAL PRINCIPLES OF MECHANICS.
EXAMPLE.-The dimensions, heaviness and strength of a mill-stone
A B D E, Fig. 506, are given; required the angular velocity when the
stone is torn apart by the centrifugal force. Putting the radius of the
millstone = r r₁, the radius of its eye
FIG. 506.
F
A
B
-P
P
D
E
L
11
= r₂, its height A E
its heaviness
= HL །
γ and the modu-
lus of ultimate strength K, we
have the force necessary to tear the
stone apart in a diametral plane
P = 2 (r₁ re)

2) I K
the weight of the stone
G
2
= ñ (1¸² — 12²) ly,
ན་
and the radius of rotation for each
half of the stone, I.E. the distance
of its centre of gravity from the
-2
axis of revolution (see § 114),
4
r
1
2
r
3 π
2
2
T
з
1
2
At the moment of tearing apart the centrifugal force of one-half the stone
is equal to the breaking load of the stone, and we have
I.E.,
Gr
w². 131
2 (r1
2
r₂) I K
l z
w². & (r
3
1
P 2)
= 2 (71
- r₂) l K.
2)
g
W
1/3 g (r₂
3
1
3 g K
3
1
1
(r₁" + r₁ 7 g + 1 2 ² ) 7
2
= 4 inches, K
4 inches, K = 750 pounds and
Cancelling 27 on both sides of the equation,.we have
/3 g (r,r₂) K
Now if ri = 2 feet = 24 inches, r
24 inches, r₂
1
the specific gravity of the stone 2,5, or the weight of a cubic inch of it
•
62,4 2,5
Y
1728
the tearing begins,
0,09028 pounds, we have the angular velocity, when
3. 12. 32,2 . 750
W =
V
688 . 0,09028
3375 . 16,1
43. 0,09028
118,3 inches.
If the number of revolutions in a minute
u, we have w
2π u
60
and
30 ω
30 . 118,3
inversely u =
or in this case,
= 11293.
}
1
に
​2
Generally the number of revolutions of such a stone is 120 or about nine
times less. For a fly-wheel we can put r₁ + ~ ₁ ²² 2 + re²
p 3 r², denoting
the radius of the middle of the ring, and consequently we have
1
1
2
r
9 K
ω
r² y
or v = @ r = V
K
}
§ 307. If all the parts M, M of a system of masses, Fig. 507,
or the centres of gravity of the elements of a body are in a plane
307]
€19
THE CENTRIFUGAL FORCE OF RIGID BODIES.
passing through the axis of rotation, the centrifugal forces form a
system of parallel forces and can be referred to a single force. Let
Y
Z
01
02
FIG. 507.
M1
F
M2
M3
03
13
E
X
the distances of the elements of the
mass from the axis of rotation ZZ be
O, M₁ = O₂
r₁, O. M₂ = r», etc.,
then the centrifugal forces are
P₁ = w² M₁ r₁, P₁ = w² M. r, etc.,

1
and their resultant is
P₁
2
= w² (M₁ r₁ + M, r₂ + . . .)
= w² M r,
r denoting the distance of the centre
of gravity of the whole mass M from
the axis of rotation. The distance
of the centre of gravity from the axis.
of rotation must be considered here as the radius of rotation. In
order to find the point of application O of the resulting centrifugal
force P, we substitute the distance of the elements of the mass
from the normal plane, viz., C' O₁ = 21, C′ 0₂ = z», etc., in the formula
2
2
CO = z =
M₁ r₁ z₁ +
M₁ r₁ +
Mş rạ za + ...
M₂ va
+
• ·
By the aid of the formula P w Mr the centrifugal forces
of solids of revolution and of other geometrical bodies can be deter-
mined, when the axis of these bodies is in the same plane as the axis
of revolution.
For a rod A C, Fig. 508, whose length is 4 C = 7 and whose
angle of inclination A CZ to the axis
of rotation is = a, we have
FIG. 508.
Z
A
L
K
C
Ꭾ
r = K S = ! 1 sin a,
and consequently the centrifugal force
P = w². MI sin. a;

!
but in order to find the point of appli-
cation of this force, we must substi-
tute in the expression
ľ
00 30
for the moment
M
N
M
w².
x sin, a, x cos. a
N
M
= w².
า
x² sin. a cos. a
of the rod successively, instead of the ele-
620
[$ 307.
GENERAL PRINCIPLES OF MECHANICS.
ments
7 27 3 1
•
N N N
,
etc., and add the expressions thus obtained to-
gether. In this manner we find
M
で
​P u = w² sin. a cos. a (1² + 2² + 3² + ... + n²)
N
N
¹ w² M 1² sin. a cos. a,
3
hence the arm CL
u =
0, 0 or
2
l
} w² M l² sin. a cos. a : ¦ w² M 1 sin, a = 3 l cos. a,
and the distance of the point O from the end C of the rod, which
lies on the axis, is
( 0 = 3 l
ૐ 7.
If the rod A B, Fig. 509, does not reach the axis, we have
P
I w² Fl² sin. a w² Fl sin. a
½ w² F sin. a (li
2
1.2),
and the moment
P u = 16)² F sin. a cos. a (1³
3
1.³) ;
for the mass of CA (= cross-section multiplied by the length) is
Fl and the mass of C B,
Fly.
It follows, therefore, that the distance of the point of applica-
tion from the point of intersection C with the axis is
3
7,³ - 13
7.) 2
CO =
or C 0 = 1 +
3
2
12 /
7 denoting the distance C' S of the centre of gravity and ₁ — l, the
length of the rod.
L
K
Z FIG. 509.
A
FIG. 510.

2
A
F B

K
B
E
D
P
C
X
Y
סדי
P
X
Y
C Bo SoOo AO
This formula holds good also for a rectangular plate A B D E,
Fig. 510, which is divided into two similar rectangles by the axial
plane CO Z, and whose plane is at right angles to this axial plane;
for the points of application of the centrifugal forces of the slices,
§ 308.] THE CENTRIFUGAL FORCE OF RIGID BODIES.
621
obtained by passing planes through it perpendicular to C Z, are
in the medial line F G. Now if the distances CF and C G of the
two bases A B and F E from the origin Care , and le, we have
here also
CO
3.
1,3 13
7,2 122
(4 — 1½)²
1 +
12 7
In like manner the centrifugal force of a right cone A B D,
with a circular base, Fig. 511, which turns about an axis C D
D
LK
FIG. 511.
2
L
P
K
B
K
α
passing through its apex, is found by
substituting in the formula P = 2 Mr
グ
​for the distance K S of the centre of
gravity S of this body from CZ. If h
denote the altitude KD of the cone, and
a the angle B CZ formed by the base
of the cone with the axis of rotation,
we will have
KSDS cos. D SK
–
3 h cos. a,
and consequently the required centrifu-
gal force is
P = w² M 3 h cos. a.
The point of application O of this
force is determined by the co-ordinates
DL = LO
u and L 0 = v, for which we
find with the aid of the Calculus, under
the supposition that the axis of rotation C Z does not pass through
the cone, the following expression
and
2°
И =
3 h sin. a [1 − ( —¿)']
V =
-
2 h
4 h cos. a [1 + (r tang. a)'].
r denoting the radius = KB of the base.
§ 308. If all the different parts of the body lie neither in a
plane normal to the axis of revolution, nor in one containing that
axis, the resulting centrifugal forces
Q = w² (M₁ x₁ + M₂ a + ...) and Rw (M₁ y₁ + M₂ y₂ + ...)
will not give a single force, but it is possible to replace them by a
force
P = √ Q² + R* = w² M 1,
Q
applied at the centre of gravity, and by a couple composed of Q
and R. If we apply at the centre of gravity four forces + Q and
as well as R and R, which balance each other, the positive
forces will give the resultant
+
$

622
[S 308.
GENERAL PRINCIPLES OF MECHANICS.
P = √ Q² + R²,
while the negative ones
Q and
R, together with the centrifu-
gal forces applied at U and V (see Fig. 501) form the couples.
(Q, - Q) and (R,
a single couple.
W
R), which can be combined so as to form
In order to understand better this referring of the centrifugal
Z
FIG. 512.
B
R
E
R
P
Q1
R
R1
X
A
E1
S1
Οι
B1
-2
forces of a revolving body to
one force and one couple,
let us consider the following
simple case. The rod A B,
Fig. 512, which revolves
about the axis Z Z, is paral-
lel to the plane Y Z and its
end A reposes upon the axis
C X. Let us denote the
length A B of the rod by 1,
its weight by G, the angle
A B B₁, formed by the rod
with the axis of rotation, by
a and its distance CA from
the plane Z, which is also
its shortest distance from
the axis Z Z by a.

M
Now if
E is an element of the rod,
N
and y A E, the horizontal projection of its distance A E from
the end 4, we have the components of the centrifugal force P, of
this element
M
M
M
M
Q₁
= w².
CA = w².
a and R₁ = w². A E₁ = w². Y,
n
22
N
N
and their moments in reference to the plane X CF of the base,
since the distance of the element from this plane X is
E, E = A E, cotg, a — y cotg. a, are
Q₁ 2₁ = w². CA. E₁ E = w²
M
n
M
R₁ %₁ = w².
Ꭱ 2
•
g? . cotg, a.
N
M
a y cotg. a and
n
The resultant of all the components parallel to Y Z is
M
Q = Q1 + Q2 +
... = n. w².
n.w².
a = w². Ma,
N
§ 308.] THE CENTRIFUGAL FORCE OF RIGID BODIES.
623
Q u =
1 sin. a
and its moment is
Q₁ Z1 + Q2 Z2 + .
w².
M
a cotg. a (y + 2 + …),
N
2 7 sin. a
3 l sin. a
or, since y₁
, Y2
Y3 =
,etc., and cotg. a.
n
n
N
sin. a - cos. a, we have
M
7
M
て ​n2
Qu = w². a cos. a..
(1 + 2 + 3 + . . . + n) = w².
a cos. a
n
N
N
2
N
w². Mal cos. a.
The distance of the point of application of this component from
the plane X Y of the base is
1 w² M a l cos. a
i i cos. a,
S₁ S = u
w² M a
I.E., this point coincides with the centre of gravity of the rod.
The resultant of the components parallel to Y Z is
M
= w³. (Y₁ + Y₂ + ...)
2
N
w² Ml sin. a, and its moment is
cotg, a (i + 3 + .
R = R₁ + R₂ +
M 1 sin. a
N
= w²
N
N
2
M
R v
= w?
22
M
= 2
N
M
1º
= 32
w2
n
M
P
= w²
2
N
N
cotg. a
222
.)
(1 sin. a)²
(2 l sin. a)*
+
n²
η
+ -...)
(sin. a) cotg. a (1+4+9+..+n)
sin, a cos. a .
I w³ M l² sin. a cos. a.
723
3
Hence the distance of the point of application O of this force
from the plane X Fis
0₁ 0 = v
§ 6² Ml² sin, a cos. a
s w² Ml sin, a
ql cos. a,
I.E. this point lies at a distance (3 1) l cos. a = l cos, a verti-
cally above the centre of gravity, or, in general, SO of the
length of the rod A B.
1
From the two components Q = w* M_a_and_ R = ½ w³ M
sin. a, it follows that the final resultant, which acts at the centre
of gravity of the rod, is
P Ꮙ Ꭱ
Q² + R² = w² M √ a² + ¦ lª² sin. a²,
that the couple is (R, - R), and that its moment is
624
GENERAL PRINCIPLES OF MECHANICS. [§ 309, 310.
R.SO
i w² M l sin. a . } l cos, a
1 w² M l² sin. a cos. a =
1 w² M l² sin. 2 a.
12
24
§ 309. Free Axes.-The centrifugal forces of a body revolv-
ing uniformly upon its axis generally exert a pressure upon the
axis, yet it is possible for these forces to balance each other, in
which case the axis is subjected to no pressure from them. As ex-
amples of this case we may mention solids of revolution turning
around their axis of symmetry, or geometrical axis, the wheel and
axle, water wheels, etc. If a body in this condition is acted upon
by no other forces, it will remain forever in revolution, although
the axis is not fixed. This axis of rotation is therefore called a
free axis (Fr. axe libre, Ger. freie Axe). From what precedes, we
know the conditions, which are necessary when an axis of rotation
becomes a free axis. It is necessary that not only the two re-
sultants Q and R of the forces parallel to the co-ordinate planes
AZ and Y Z, but also that the sums of the moments of each of
the two systems of forces shall be = 0, whence it follows that
1) M₁ x₁ +
2) M₁ y₁ +
3) M₁ x₁ %,
M½ x2 +
Mạ½ Y½ +
0,
•
0,
+
2
M₂ X» Z9 +
M₂ Y½ %2
+
= 0 and
0.
•
1
4) M₁ y₁ %₁ +
The first two conditions require the free axis to pass through
the centre of gravity of the body or system of masses. The two
latter, however, give the elements required for determining the po-
sition of this axis. It can also be proved that every body or system
of masses has at least three free axes, and that these axes are at
right angles to cach other and intersect each other at the centre of
gravity of the system.
:
The higher mechanics distinguishes from the free axes other
axes, which may intersect each other at any point of the system and
which are called principal axes (Fr. axes principaux, Ger. Haupt-
axen). It is also proved that the moment of inertia of a body in
reference to one of the principal axes is a maximum, and in rela-
tion to the second it is a minimum, and in relation to the third it
has a mean value, and that for a point which lies in the free axes
the principal axes are parallel to the free axes, I.E. to the principal
axes passing through the centre of gravity.
§ 310. Free Axes of a System of Masses in a Plane.--
If the parts of a mass are in a plane, E.G., if they form a thin plate
§ 310.]
THE CENTRIFUGAL FORCE OF RIGID BODIES.
625
or plane figure, then the straight line, passing through the centre
of gravity of the entire mass at right angle to that plane, is
a free axis of the mass; for in this case the mass has no radius
of rotation, and therefore the only possible centrifugal force is = 0.
In order to find the other two free axes, we employ the following
method. Let S, Fig. 513, be the centre of gravity of a mass and
let UU and V V be two co-ordinate
axes in the plane of the mass and let us
determine the elements of the mass by
means of co-ordinates parallel to these
axes, E.G. the element M, by the co-or-
dinates M₁ N = u₁ and M₁ 0 = 2'1•
U
X
FIG. 513.
V
Y

N
M1
F
K
R
S
-Y
U
X
1
=2. Now
if XX is one free axis and Y an axis
at right-angles to the same and if the
angle US X, which the free axis makes
with the axis of co-ordinates S U and
which is to be determined, o, then
putting for the co-ordinates of the elements of the mass in refer-
ence to XX and Y Y, ≈1, 2 . . . and y₁, Y
the mass M,
we obtain
M₁ K = x, and M, L = y₁,
1,
1
E.G. for those of
x₁ =M₁K=SR+RL=SO ccs. + 0 M₁ sin. p=u, cos. $+v'1 sim. O,
y₁ = M₁ L=
1
OR+OF
u₁ sin. & + v', cos. Ò,
and therefore the product.
1
1
SO sin. + OM, cos.
& O
X₁ Y₁ = (u, cos. $ + v, sin. 4) . (— u, sin. 9 + v', cos. $)
(ur²
(U
2
or, since sin. & cos. P
X₁ Y 1
v₁³) sin. & cos. & + U₁ v₁ (cos. $² sin. '),
2
2
sin. 2 4 and cos. O³ sin. p³ = cos. 2 4,
§ (u,³ — v‚³) sin. 2 4 + u₁ v₁ cos. 2 0,
and therefore the moment of the element M, is
M
2
M₁ x₁ y₁ == (u₁² — v‚²) sin. 2 ¢ + M₁ u₁ vi cos. 2 4,
1 1
1
and in like manner the moment of the element M is
M₂ Xş Y₂ = -
?
M 2
2
(U₂² a,) sin. 2 p + M, 2t, U, c0s. 2 p, etc.,
and the sum of the moments of all the elements or the moment of
the entire mass itself is
M₁ x₁ Y₁ + M₂ X Y ₂ + ... = sin. 2 [(M, u₁² + M₂ u₂² + ...)
1 1 1
1 ¢
- (M₁ v₁² + M₂ v² + ...)] + cos. 2 † (M¡ u₁ v₁ + Mş Uş Vg + ...).
– v;
1
40
626
[$ 311.
GENERAL PRINCIPLES OF MECHANICS.
In order that I shall be a free axis, this moment must be
0; we must therefore
2
put
½ sin. 2¢ [(M, u₁² + M₂ U ₂² + ...) —
u,
2
2
2
u + ...) — (M, v₁² + M₂ v₂² + …..)]
cos. 2 p (M¸ u₁ V½ + M½ U₂ V2 + ...) = 0,
from this we obtain the equation of condition.
sin. 2 o
tang. 2 p
=
2 (M₁ u, v, + M₂ Uq V2 + ...)
2
cos. 2 p (M₁ u₁² + M₂ u₂² + ...)
1
2
2
2
1
(M, v₁² + M₂ v₂² + ...)
2
Double the moment of the centrifugal force
Difference of the moments of inertia.
This formula gives two values for 2 4, which differ from each
other 180°, or two values of differing 90° from each other; this
angle therefore determines not only the free axis XX, but also
the free axis Y Y perpendicular to it.
§ 311. The free axes of many surfaces and bodies can be given
without any calculation. In a symmetrical figure, E.G., the axis of
symmetry is a free axis, the perpendicular at the centre of gravity
is the second, and the axis at right-angles to the surface of the
figure the third free axis. For a solid of revolution A B, Fig. 514,
the axis of rotation Z Z is one free axis and in like manner every
normal XX, Y F... to this line and passing through the centre.
of gravity is another. For a sphere every diameter is a free axis, and
for a right parallelopipedon A B D, Fig. 515, bounded by 6 rectan-
FIG. 514.
Z
FIG. 515.
Z


Y
Y
A
-X
A
X
X
X
B
Y
B
Y
Ꮓ
Z
gles they are the three axes XX, Y Y and Z Z, passing through
the centre of gravity perpendicular to the sides B D, A B and A D,
and parallel to the edges.
Let us now determine the three axes for a rhomboid A B C D,
Fig. 516. We begin by passing two rectangular co-ordinate axes
UU and VV through the centre of gravity, so that one is paral-
:
§ 311.]
THE CENTRIFUGAL FORCE OF RIGID BODIES.
627
lel to the side A B of the rhomboid, and by decomposing the rhom-
boid by parallel lines in 2 n equal strips, such as FG. Now if one
side A B = 2 a and the other A D = 2b and the acute angle A D C
between two sides = a, we have the length of the strip E G,
situated at a distance SE
from U U,
FIG. 516.
Y
A
FA
-U
-X.
D
B
X
U
= K G + E K = a + 2 cotg. a,
and that of the other part E F

= α a cotg. a,
b
and since sin. a is the width of
Pag
N
both, we have the areas of these
strips
b sin. a
(a + x cotg. a) and
N
b sin. a
N
(a – a cotg. a) ;
and consequently the measures of the centrifugal forces of the two
portions in reference to the axis V Vare
b sin. a
(a + x cotg. a) . (a + cotg. a)
=
ጎ
b sin. a
2 n
(a + c cotg. a)
³
and
b sin. a
2 n
(a – a cotg. a),
and their moments in reference to the axis U U are
b sin. a
b sin. a
(a + x cotg. a)' x and
(a
. cotg. a)? 2.
α
2 n
2 n
Since the two forces act in opposition to each other in reference
to V V, by combining their moments we obtain the difference
2
fx sin, a
2
[(a + æ cotg. a) — (a — a cotg. a)
a b x² cos. a.
N
2 n
If we substitute in this formula successively
3 b sin. a
N
b sin, a 2 b sin. a
N
n
,etc., and add the results, we obtain the measure of the
moment of the centrifugal force of one-half the parallelogram
2 ab
b² sin² a
22³
cos. a .
(1² + 2² +32 + ... + n°)=2 a b³ sin. a cos. a
•+n²)=2
N
༡༡?
3 n°
3 a b³ sin² a cos. α,
and for the whole parallelogram we have
628
[§ 311.
GENERAL PRINCIPLES OF MECHANICS.
M₁ u, v₁ + M₂ u₂ v½ +
4
a b³ sin.² a cos. a.
The moment of inertia of one strip F G in reference to Vis
b sin. a | (a + a cotg. a)
N
( a + 2cotg.
3
2
2 b sin. a
( +3 a ? cotg a )
3 n
Substituting for a successively
3
+
(a
а. Ъ
N
xcotg. a)
3
a)²)
sin a (a + 3 x cotg. a).
b sin. a 2 b sin. a 3 b sin. a
N
N,
N
etc.,
and summing the resulting values, we obtain the moment of inertia
of one-half the rhomboid, which is
=3ab sin. a (a + b² cos.' a),
and for the whole rhomboid it is
a b sin. a (a² + b² cos.² a).
In reference to the axis of rotation U U the moment of inertia
of the parallelogram is
= 4 a b sin. a
b² sin, a
3
= a b³ sin.³ a (see § 287),
and the required difference of the moments is given by the equation
2
2
2
(M₁ u₁² + M» u²² + .) — (M, v,² + M₂ v2² + + ...)
•
4 a b sin. a (a² + b² cos.² a)
3 a b sin. a [a² + b² (cos.² a
3 a b sin. a (a² + b² cos. 2 a).
3
§ a b³ sin.³ a
sin. a)]
Finally, we have for the angle US X, which the free axis
XX makes with the co-ordinate axis U U or with the side A B,
according to § 310,
tang. 2 p =
2
2 (M₁ u₁ v₁ + M 2 U2 V2 +
+ ...)
(M₁ u₂² + M₂ u2² +
)
2. ‡ a b³ sin.² a cos. a
4
3
2
(M₁ v₁² + M₂ v2² +
3 a b sin. a (a² + b² cos. 2 a)
For the rhombus ɑ = b, and
b² sin. 2 a
a² + b² cos. 2 a
tang.2p=
sin. 2 a
1+ cos. 2 a
2 sin, a cos. a
2 sin.a cos.a
=tang.a,
1 + cos." a
sin* a
2 cos.² a
2
a
or
20 = a and o
201
Since this angle gives the direction of the diagonal, it follows
that the diagonals are free axes of the rhombus.
§ 312.J
629
THE CENTRIFUGAL FORCE OF RIGID BODIES.
EXAMPLE. The sides of the rhomboid A B C D, Fig. 516, are A B
2 a
16 inches, and B C 2 b 10 inches, and the angle A B C — a =
60°; what are the directions of the free axes?
Here we have
tang. 2 $ =
52 sin. 120°
25 sin. 60'
82 + 52 cos. 120° 64 — 25 cos. 60º
25. 0,86603
64-25. 0,5
= 0,42040 = tang. 22° 48′ or tang. 202° 48′;
hence it follows that the angles of inclination of the first two free axes to
the side A B are o 11° 24′ and 101° 24'. The third free axis is perpen-
dicular to the plane of the parallelogram. These angles determine the free
axes of a right parallelopipedon with a rhomboidal base.
§ 312. Action upon the Axis of Rotation.-If a material
point M, Fig. 517, revolves with a variable motion around a fixed
-X-
P
FIG. 517.
-Y
-P
K
A
N\P
P
L
MI
Y
N2 N
X
axis C, the latter must coun-
teract röt only the centrifu-
gal force, but also the force
of inertia of this point. While
the centrifugal force acts ra-
dially outwards, the force of
inertia acts tangentially either
in the opposite or in the same.
direction as the movement of
rotation, according as the ac-
celeration of this motion is
positive or negative (Retard-
ation). We can therefore as-
sume that the centrifugal force

MN= CN= N acts directly upon the axis C, and that the force
of inertia M P P is composed of a couple (P, P) and an
axial force, P, and consequently the entire force, acting upon the
axis, CRR is represented by the diagonal of a right-angled
parallelogram formed of N and P. If r is the distance CM of
the mass M from the axis of rotation C, the angular velocity and
*`the angular acceleration, we have, according to § 302 and § 282,
N = w² M r
and
P = к Mr,
and therefore the required resultant is
R = √ Nº + P² = √w* + k² . M r,
630
[§ 312.
GENERAL PRINCIPLES OF MECHANICS.
and for the angle R C N = 4, made by this force with the
direction CM of the centrifugal force, we have
tang. p =
P
N
P
N
K
W2.
Since in consequence of the acceleration , w is variable, the
centrifugal force N and the resultant R are variable.
In order to combine the centrifugal forces and the forces of
inertia of the masses M1, M2, etc., we decompose each of these forces
into two components parallel to the directions of two axes XX and
YY, then if we combine them by algebraical addition, so as to
obtain two forces acting in the direction of each axis, we have only
to determine the resultant of these two forces. If x and y are the
co-ordinates CK and CL of the material point M in reference to
the co-ordinate axes XX and FF, we have the two components
of the centrifugal force N
X
X
N₁
N = w² Mx and
Y
N
N = w² My,
and, on the contrary, those of the force of inertia
Y
P = k M y and
P₁
X
P₂
P = к Mx,
K
and therefore the entire force in the axis XX is
Q
=
N₁ + P₁ = w² M x + к My,
and that in the axis Y Y is
R
=
N½ − P₂ = w² M K M x.
Y
If we have a system of points or masses M₁, M, etc., which are
revolving about a fixed axis C, Fig. 518, and if the co-ordinates of
these points in reference to the axis XX are
C K
= x1, C K₂ = X2, etc.,
2
and those in reference to the axis Y Y are
the entire force in the direction of the first axis is
C L₁ = Y₁, C L» = y., etc.,
Q
1 1
w² M₁ x²₁ + к M₁ y₁ + w² M½ x ½ + к M ₂ Y½
+
• •
I.E.
2
Q = w² (M₁ x₁ + Mg Xg + ...) + K (M₁ Y₁ + M₂ Y2 + ...),
1
2
and that in the direction of the other axis is
R = w² (M₁ y₁ + M₂ Y₂ + ...) — к (M, x₁ + M₂ X₂ + ...).
2
X2
1
Y½
2 2
§ 312.] THE CENTRIFUGAL FORCE OF RIGID BODIES.
631
•
Now if we denote the entire mass M₁ + M½ + ... by M and
the co-ordinates of its centre of gravity in reference to the axes
XX and Y Y by x and y, we have (see § 305)
Y
P1
FIG. 518.

2
-X
K2
K1
Q
X
R
N₁
N2
M1
L₁
L2
M2
Y
M₁ x₁ + M₂ x2 +
Mx
M₁ y₁ + M₂ Y½ +
My,
and therefore, more simply,
Q = w² Mx + к My and
R = w² M Y K M x.
From Q and R we obtain the resultant
S = √ Q² + R²,
and for the angle X CS
p
of its direction
R
tang. =
Q
Since Mx and My are the statical moments of the centre of
gravity, it follows that in determining the pressure S upon the axis
of a system of masses, situated in one and the same plane of revo-
lution, we can consider the whole mass to be concentrated at the
centre of gravity; and since the distance of the centre of gravity
of the system of masses from the axis of rotation is
we have also
r = √x² + y²,
S = √ [(w² Mx + k My)² + (w² My
= M √ [w¹ (2² + y²) + k² (x² + y²)]
к M x)³]
= M√w* + k² √x² + y² = √ w¹ + k² . Mr
REMARK.—If a triangle A B C, Fig. 519, revolves about its corner O,
and if the other corners A and B are determined by the co-ordinates
632
[§ 312.
GENERAL PRINCIPLES OF MECHANICS.
(x1, y₁) and (x2, y2), we have, according to § 112, the co-ordinates of its
centre of gravity S
x1 + x 2
C S₁
= x
3
1
Y ₁ + Y 2
C S₂
= y
3
and the mass, if we measure it by its super-
ficial area, is
M
X 1 Y 2 — X 2 Y 1
2
Its moment of inertia in reference to the axis
of rotation C can be determined by the for-
FIG. 519.
P
and

C
B₁$1
A1-X
R
S2
S
A2
B2
B
Y
M
2
6
mula
3
3
W =
M (1
6 x1
Y 1
92
น
+
X2
Y 1
Y 2
2
(X₁² + X1 X 2 + X₂² + Y 1² + Y1 Y2 + Y2²).
This formula is also applicable to a right prism, whose base is the tri-
angle A B C.
EXAMPLE.-A right prism with the triangular base A B C is caused to
revolve around its edge C by a force which acts uninterruptedly, so that
at the end of the time t = 1 it has made u = revolutions; required not
only the moment of this couple, but also the action of this motion upon
the axis C. Let the base of this body be determined by the co-ordinates
0,5; x2 0,4, Y2
X1
1,5, Y1
=
1,0 feet,
and let its length or height be l = 2 feet, and its heaviness y = 30 pounds.
From these data we calculate, first, the area of the base
1
F
X, Y 2
X2 Y1
2
1,5 1,0 0,4. 0,5
.
2
1,3
2
0,65 square feet,
and the mass of the whole body
πιγ
M=
0,031 . 0,65 . 2. 30 1,209 pounds.
=
g
Now
P
1
X₁² + X 1 X 2 + x2
2,25 + 0,60 + 0,16
Y ₁ ² + Y ₁ Y 2 + Y ₂ ²
2
0,25 + 0,50 + 1,00
=
1
hence the moment of inertia of the body is
2
3,01 and
1,75,
M
W = (3,01 + 1,75)
4,76
•
6
1,209
6
=
0,95914.
In consequence of the constant action of the couple, the movement of
rotation is uniformly accelerated, and consequently the angular velocity of
the body at the end of the time t = 1 second is (see § 10)
W
28
t
2.2 πα
t
and the mechanical effect required is
2.2.5 π
31,416 feet,
૭
§ 312.] THE CENTRIFUGAL FORCE OF RIGID BODIES.
633
1
=
A = { w² W = (31,416). 0,95914 473,3 foot-pounds.
The angular acceleration is
W
31,416
K
t
1
=
31,416 feet,
and therefore the moment of the couple
Pa = kW = 31,416. 0,95914
30,13 foot-pounds.
The distances of the centre of gravity S of the base from the co-ordi-
nate axes XX and Y Y are
X
Xx1 + x 2
3
1,5 +0,4
3
0,6333 and
1
Y ₁ + Y 2
0,5 + 1,0
У
0,5000,
3
= 3
consequently the distance of the centre of gravity from the axis is
CS r =
√x² + y² = 0,6511.
Besides we have
ان
w¹ = 31,416¹
974090 and
K²
31,4162
987,
whence
√975077 = 987,46,
and the pressure upon the aris increases during the accelerated rotation from
to
-
P = k Mr 31,416. 1,209 . 0,6511
24,73 pounds
RV
Vale + k². Mr
K
987,46. 1,209. 0,6511
777,33 pounds.
If after one second of time the couple ceases to act, the motion of rota-
tion of the body becomes uniform, and the pressure upon the axis from
that moment consists only of the centrifugal force, which is
N = w² Mr = 986,96 . 0,7872 = 776,94 pounds.
The pressure upon the axis, which increases gradually from 24,73 to
777,33 pounds, is in the beginning at right-angles to the central line of
gravity CS, but approaches more and more this line as the velocity
increases, so that at the end of the time t 1 second, it makes but an
angle with that line, and this angle is determined by the expression
tang.
P
N
= 1° 49'.
24,73
776,94
0,03183,
for which
If the couple ceases to act, the direction of the
axial force N 776,94 pounds, coincides of course with the central line of
gravity CS and revolves with this line in a circle. If instead of the couple
a single force P acts with the arm a upon the body, another pressure equal
to this force P must be added to the pressure on the axis.
$
634
[§ 313.
GENERAL PRINCIPLES OF MECHANICS.
I
§ 313. Centre of Percussion.-If the different portions
M₁, M2, etc., Fig. 520, of a system of revolving masses are not in
R
Y
N
FIG. 520.
X2
K2
Ꮓ
B
Y2
S2
R2
L2
V
M2
F
U
0,
Q1
K₁
R1
P
Y₁
A
Mi
X1
C
H
X
one and the same plane, the
directions of the forces
Q₁
= w² M₁ x₁ + K M₁ Y₁,

1 1
19
M₂ y₂, etc.,
Q₂ X₂
₂ = w² M½ x ½ + k M₂
2
2 2
no longer coincide with the
co-ordinate axis XX, but
lie in the co-ordinate plane
X Z, and those of the forces
R₁ = w² M₁ yı K M₁ X₁,
= w² M₂ Y½
R2
2
1
1
K M, x2, etc.,
no longer lie in the axis
YY, but in the co-ordinate
plane Z. The system of
forces Q1, Q2, etc., and R1, R2,
etc., give, according to § 305,
the resultants
Q = Q1 + Q₂ + ... and
R R₁ + R₂ +
Now since the lines of ap-
plication UQ and VR do
not generally lie in the same
plane, but cut the axis C Z
of rotation at different points U and V, it is impossible to obtain a
single resultant by combining them, but we can refer them to a
single force and a couple. The components are, of course, as above,
Q = w² (M₁ x₁ + M₂ X 2 + ...) + к (M₁ y₁ + M₂ Y₂ +...)
= w² M x + к My
and
1
2
R = w² (M; y₁ + M » Y½ + . . .) − k (M½ x₁ + M½ x½ + ...)
1
w² M y + « M x,
M denoting the entire mass M₁ + M₂ + ... and x and y the dis-
tances of its centre of gravity S from the co-ordinate planes Y Z
and X Z.
Now if we put the distances of the masses M₁, M., etc., from the
plane of rotation X Y, which is perpendicular to the axis of rota-
tion CZ, equal to z₁, z, etc., we obtain, as in § 305, the distances
of the points of application U and V of the forces Q and R from
the origin C.
§ 313.]
635
THE CENTRIFUGAL FORCE OF RIGID BODIES.
Q₁ z₁ + Q₂ %₂ +...
Q2Z₂
U
and
v =
Q₁%₁
Q₁ + Q₂ +
w² (M, x, z₁ + Mn X₂ Zn + . . .) + к (M₁ y₁ z₁ + M₂ Y₂ %₂ + ...)
w³ (M, Xx₁ + M½ X½ + . . .) + k (M₂ Y₁ + M₂ Y₂ + . . .)
+ ...) + giả + Mg +
k
R₁ Z₁ + Rg Zq +...
R₁ + R₂ +
w² (M½ Y, 2, + M» Y2 Z2 + . . .) — k ( M; X; z; + M₂ Xɔ Z½ + • • •)
1
w™ ( M₁ Y₁ + M ₂ Y ½ + . . .) — K (M; X₁ + M½ X9 + . . .)
1
If the axis CZ is retained at two points A and B (the pillow
blocks), which are at the distance C A = l, and C B = l, from
the origin of co-ordinates, the force Q is decomposed into two com-
ponents
U
Q
X₁
1
( 1 − 1 ) 2 and X₁ = ( ~ — 4) &
1,
X.
and the force R into the components
x₁ = (1/2 = 1)
R and I, = ( 2 — 4) R.
7. —
Now the pressure upon the bearing A is
2
S₁ = √ X₁² + YI”,
and that upon the bearing B is
S₂ √X² + X₂²³·
If the acceleration of the rotation is produced not by a couple,
whose moment is Pa, but by a force P, whose arm is a, a third
pressure equal to the force P is added to the two axial forces Q
and R. If we cause this force P to act, at the distance F 0 = a
from the axis of rotation, parallel to the axis CF and perpendicu-
lar to the plane X Z, and if we assume that its line of application
is at a distance CF HO=b from the co-ordinate plane X Y,
H0
the force R only will be increased by an amount P, and the portion
of it I, at the bearing 4 will be increased by
Р
F₁ = (=)) P
and the part F, at the bearing B by
x = (₂ = 1)
Y₁
P.
636
[§ 313.
GENERAL PRINCIPLES OF MECHANICS.
If
M₁ x₁ + M½ ï½ +
0,
M₁ Y₁ + M₂ Y ½ + ... = 0,
2
M₁ x₁ z₁ + M₂ xXą z₂+... = 0 and
Μ
2
M₁ Y₁ Z₁ + M½ Y₂ Z2 +
1
2 2
0,
CZ is a free axis, and not only the forces Q and R, but also
their moments Q u and R v become = 0; and we can, therefore,
conclude that when a system of masses rotates about a free axis
not only the centrifugal forces, but also the moments of inertia.
balance each other (compare § 309).
Let us assume that the system of masses is at rest, I.E., w = = 0,
or let us neglect the action of the centrifugal force upon the axis
of rotation, then we have more simply for the pressures in the axes
y = k (M₁ y₁ + M₂ y₂ +
Q
= * M
Y
2
•
.) and
R = ~ KM x = к (M¸ X₁ + M¸¤¸ + . . .), and also
Qu := K
(M, Y, Z ₁ +
M₂ Y z Zz + ...) and
R v =
к (M, X, z₁
k 2,
+ M₂ Xą z₂+...).
2
When the plane of XZ is plane of symmetry and consequently
↑ pœne of gravity,
M₁ Y₁ + M₂ Y2 + ... = 0
and
2
M₁ Y₁ z₁ + M₂ Y½ Z₂+= 0,
and, therefore,
FIG. 521.
Z
Y2Z2

B
X2
Q = 0
S2
and also
0
Q2
K2
L2
V
M3
Q u = 0.
Now if we require that
the force of rotation
R2
F
U
Q1
K₁
R1
P
L1
A
Y₁
C
Μι
X1
KW
P =
α
shall be counteracted by the
force of inertia R, so that
there shall be no action upon
the axis of rotation, we must
have
P+R = 0
S1
H
X
and
Pb+ Rv = 0,
I.E.,
Y
:
§ 313.]
637
THE CENTRIFUGAL FORCE OF RIGID BODIES.
and
K W
a
k W b
a
and consequently
k (M₂ x, + M½ x ½ + . . .) = 0
2
k ( M, x, z₁ + M₂ X. %» + . . .) 0,
W
a
Mx
M₁ r²² + M₂ r² +
M₁ x₁ + M₂ x2 +
Moment of inertia
Statical moment
and
1
M₁ ≈₁ z₁ + M₂ xɔ za +
.
b
a
IF
•
Moment of the centrifugal force
Statical moment.
M₁ x₁ z₁ + Mş x y z₂ + . . .
M₁ ŵ₁ + M₂ xX2 +
These co-ordinates determine a point 0, which is called the
centre of percussion (Fr. centre de percussion; Ger. Mittelpunkt
des Stosses); for every force of impact P, whose direction passes
through this point and is at right angles to the plane of symmetry
AZ of the body passing through the axis of rotation or fixed axis
C Z, will be completely balanced, when the collision takes place,
by the inertia of the mass, without producing any action upon the
axis of the body.
EXAMPLE-1) The moment of inertia of a straight line or rod C E,
Fig. 522, of uniform thickness throughout, which at one end C' meets the
axis CZ at a given angle Z CE, when is its mass.
and the distance D E of its other end from the axis
of rotation, is
FIG. 522.
Z

D
D
F
5
01
M₁
W = M k² 3 M r² (see § 286),
and, on the contrary, the statical moment is
M x = 1 Mr,
and finally the moment of the centrifugal force, since,
if h denotes the projection C D of the length CE of the
rod on the axis of rotation CZ, we have
or
is
M₁ 11
1
с 01
21
h
1
1
X 1
h
h
M¸ x¸², M. x, ², = ², M, x¸², etc.,
2
1
1
n
M₁ x₁ ²₁ + M₂ X 2 Z 2 +
1
h
(M₁ x₁² + M₂
+ M₂ ¤q² + . . .)
} Mr² = } Mhr.
1 1
X 2
Therefore, the co-ordinates of the centre of percussion of this rod are
determined by the formulas
638
[§ 313.
GENERAL PRINCIPLES OF MECHANICS.
Moment of inertia
FOα =
Statical moment
! M r²
&
1. M r
cojto
r
and
C F = b
Moment of centrifugal force
Statical moment
Mh r
2 h,
1 M r
and this centre is situated at of the length CE of the rod from the end
C and of the same from the end E.
2) The moment of inertia of a surface A B C, Fig. 523, whose form is a
right-angled triangle, which turns around its base C A,
is, when we denote the mass by M and its base and
perpendicular C A and C B by h and r,
FIG. 523.
Z

h p³ hr p2
T =
1 M r² (see § 229),
12
2
6
A
B
and its statical moment, since the centre of gravity S
F
2
is at a distance
from the axis CA, is
K
L
Mr
Mx
3
consequently the distance of the centre of percussion (
of this surface from this axis is
FO
6
a
} M p²
z M r
2 r.
h
For an element KL of the triangle, whose shape is that of a strip,
whose length is a and whose width is N
and which is situated at a dis-
tance C K = z from the apex C, the moment of the centrifugal force is
Mx z =
N.
x . } x z,
or, since
2
11
7'
or r
= 1/2
2,
h'
h
M x z
h
22.
(C)²
23.
(1), 2 (1), ³ () .……..
ก
(
Substituting for z successively the values 1
3
and adding the values thus obtained for Mrz, we have the total moment
of the centrifugal forces
(~7) (1³ + 2² + 8° + . . . + 22³)
h
1
M₁ x1 ≈₁ + M₂ t₂ ²º +
•
2
N
1.
3
مرات
(
N
4
N
(4)²
3
rh
r h
1 Mrh,
and, therefore, the distance of the centre of percussion O from the corner
C is
C F = b =
1 M r h
§ M r
& h.
§ 314.]
639
THE ACTION OF GRAVITY, ETC.
CHAPTER III.
OF THE ACTION OF GRAVITY UPON THE MOTION OF BODIES
IN PRESCRIBED PATHS.
§ 314. Sliding upon an Inclined Plane.-A heavy body can
be hindered in many ways from falling freely. We will, however,
consider but two cases here, viz., the case of a body supported by
an inclined plane and the case of a body movable around a hori-
zontal axis. In both cases the paths of the bodies are contained in
a vertical plane. If a body lies upon an inclined plane, its weight
is decomposed into two components, one of which is normal to the
plane and is counteracted by it, and the other is parallel to the
plane and acts upon the body as a motive force. Let G be the
weight of the body A B C D, Fig. 524, and a angle of inclination of
FIG. 524.
B
F
A
P
H
G
R
the inclined plane FHR to the
horizon, according to § 146 tỷ
normal force is

NG cos. a,
and the motive force is
PG sin. a.
The motion of the body can
be either a sliding or a rolling
one. Let us consider the former
case first. In this case all the
parts of the body participate equally in its motion, and have there-
fore a common acceleration p, determined by the well-known formula
p =
force
mass'
hence
G sin. a
g
g sin. a;
P
M
G
p: g = sin. a : 1,
1.E., the acceleration of a body upon an inclined plane is to the accel-
eration of gravity as the sine of the angle of inclination of the plane
is to unity. But on account of the friction this formula is seldom
sufficiently accurate. It is. therefore, very often necessary in prac-
tice to take the friction into consideration.
If a body moves upon a curved surface the acceleration is
640
[$ 315.
GENERAL PRINCIPLES OF MECHANICS.
variable, and is in every point equal to the acceleration correspond-
ing to the plane, which is tangent to the curved surface at that
point.
§ 315. If a body slides down an inclined plane without fric-
tion and its initial velocity is 0, then, according to § 11, the
final velocity after t seconds is
v = g sin. a . t
t =
32,2 sin. a . t feet = 9,81 sin. a t meters,
and the space described is
•
16,1 sin. a. t° feet = 4,905 sin. a . t meters.
$ = § g sin. a ť² =
When a body falls freely v₁
therefore put
gt and s₁
gť, and we can
V: Vi
v: v₁ = S: S₁ =
sin. a : 1,
I.E., the final velocity and the space described by a body sliding upon
the inclined plane are to the velocity and the space described by a
body falling freely as the sine of the angle of inclination of the plane
is to unity.
H
FIG. 525.
In the right-angled triangle F G H, Fig. 525, whose hypothenuse
F G is vertical, the base is F H = F G sin. F G H =
F G sin. F H R = F G sin. a, when a denotes the
inclination of the base to the horizon, and therefore
FH: FG = sin, a :
: 1;
F

R
the body, therefore, describes the vertical hypothenuse
F G and the inclined base F H in the same time.
Hence the space described by a body upon an inclined
plane in the time, in which, if falling freely, it would
describe a given space, can be found by construction.
G
Since all the angles FH, G, FH, G, etc., inscribed in a semi-
circle FH, G, Fig. 526, are right angles, the semicircle subtended
H2
H₁
H
3
FIG. 526.
R
R
R2
G
153
K2
by F G will cut off from all inclined
planes beginning at F the distances
F H₁, F H, etc., described simultane-
ously with the diameter. For this rea-
son we say that the chords or diameter
of a circle are described simultaneously
or isochronously. This is true not only
when the chords, as, E.G., F H₁, FH,
etc., begin at the highest point F, but
also when the chords, as, E.G., K, G, Kọ G,
etc., end at its lowest point G; for we

§ 316.]
641
THE ACTION OF GRAVITY, ETC.
can draw through F the chords FK, FK, etc., which have the
same length and position as the chords G H, G H, etc.
§ 316. From the equation
219
$
2p
v2
2g. sin. a
for the space described,
we obtain
༧ཐ
s sin, a =
and inversely,
2 g'
FIG. 527.
F
v = √ 2 g s sin. a.
Now s sin. a is the height FR (Fig. 527) of the inclined plane
or the vertical projection h of the space FH = s. If, therefore,
several bodies, whose initial velocities are = 0, descend inclined
planes FH, FH, etc., of different inclina-
tions, but of the same height, their final
velocity will be the same and equal to that
acquired by a body falling freely through
the distance FR (compare § 43 and § 84).
From the equation s = ¦ g sin. a
ig sin. a. t² we

H
H
R
obtain the formula for the time
2 s
1
t =
1
g sin. a
sin. a
2 s sin. a
9
1
2 h
sin. a
g
If a body falls freely through the height FR h, the time is
t₁ = V
2 h
9
whence
t : t₁ = 1 : sin. a = s : h = F H : F R.
The time required by a body to descend an inclined plane is to the
time of falling freely through the height of this plane as the length
of the plane is to its height.
EXAMPLE-1) The top F of an inclined plane F H, Fig. 528, is given,
and we are required to determine the other extremity H, which is situated
in such a position upon a line A B that a body descending the plane will
reach this line in the shortest time. If through ♬ we draw the horizontal
line F G until it cuts A B, and make G H = G_F, we obtain in Ħ the
point required, and in F H the plane of the quickest descent; for if we
pass through Fand H a circle, to which the lines FH and G H are tan-
41
642
[§ 317.
GENERAL PRINCIPLES OF MECHANICS.
gents, the chords F K₁, F Kg, etc., described simultaneously, are shorter
FIG. 528.
A
G
H
H
H₂
K2
=
F
C
D
B
1
than the lines F H₁, FH₂, etc., drawn from F to
the line A B; consequently the time required to
descend this chord is less than that required to
descend these lines, and the inclined plane F H,
which coincides with that chord, is the plane of
quickest descent.
2) Required the inclination of the inclined
plane F H, Fig. 527, which a body will descend
in the same time as it will fall freely through the
height FR and move with the acquired velocity
upon a horizontal plane to H. The time required
to fall through the vertical distance FR = h is

2 h
g
v = √ 2 gh.
√2gh.
,
and the velocity acquired is
If no velocity is lost in passing from the vertical to the horizontal mo-
tion, which is the case when the corner R is rounded off, the space R H
hcotg. a will be described uniformly and in the time
h cotg. a
tz
ว
h cotg. a
√ 2 g h
2 h
coty, a
1
g
The time in which a body will descend the inclined plane is
t
1
1
2 h
9
sin. a
Now if we put t = t₁ + t₂, we obtain the equation of condition
1
1 + 1 cotg. a or
sin. a
tang. a
sin. a
= tang. a + 1.
1
•
Resolving this equation, we obtain tang. a = In the corresponding
inclined plane the height is to the base is to the length as 3 is to 4 is to
5, and the angle of inclination is a = 36° 52' 11".
3) The time in which a body will
base is a, is
t
28
g sin. a
slide down an inclined plane, whose
2 α
g sin, a cos. a
4 α
g
sin. 2 a
this is a minimum when sin. 2 a is a maximum, I.E. = 1; then 2 a° — 90
or a° 45°. Water flows quickest down roofs whose pitch is 45°.
§ 317. If the initial velocity of a body upon an inclined plane
is c, we must employ the formula found in § 13 and § 14; hence,
when a body ascends an inclined plane, we have the velocity
v = c g sin. a. t,
and the space described
§ 318.]
643
THE ACTION OF GRAVITY, ETC.
8 = ct 1 g sin. a. t,
and for a body descending the inclined plane we must put
v = c + g sin. a t and s = c t + 1 g sin. a. t.
•
In both cases, however, the following formula
S =
v2
2 g sin. a'
c²
v²
c²
2,2
C³
or s sin. a = h =
29
2 g
2 g
is applicable.
The vertical projection (h) of the space (s) described upon the in-
clined plane is always equal to the difference of the heights due to the
velocities.
FIG. 529.
F
When two inclined planes F G Q and G H R, Fig. 529, meet in
a rounded edge, a body descending the plane will experience no
impact in passing from one to the other;
hence, if we have such a combination of
planes, there will be no loss of velocity,
and the following rule will be applicable
Q to the case of a body descending these
planes: height of fall equal to height due
to velocity. We can easily understand that
when a body ascends or descends a series of such planes or a curved
line or surface, its motion will take place according to the same law.

H
G
R
EXAMPLE--1) A body ascends, with an initial velocity of 21 feet, an
inclined plane, the inclination of which is 22°. What is its velocity and
what is the space described after 13 seconds?
The velocity is
v = 21 32,2 sin. 22° . 1,5 = 21
2,91 feet,
and the space is
c + v
2
21 + 2,91
2
32,2 . 0,3746. 1,5
21 — 18,09
23,91.3
4
17,93 feet.
2) How high will a body, whose initial velocity is 36 feet, rise upon a
plane inclined at 48° to the horizon? The vertical height is
h
v2
?
2 J
0,0155. v³ = 0,0155. 36º = 20,088 feet,
and therefore the entire space described upon the inclined plane is
h
20,088
8
sin. a sin. 48°
27,031 feet,
and the time required to describe it is
2.8 2.27,031 27,031
t =
V
36
1,5 seconds.
18
§ 318. Sliding upon an Inclined Plane when the Fric-
tion is taken into Consideration.-The sliding friction has
644
[§ 318.
GENERAL PRINCIPLES OF MECHANICS.
great influence upon the ascent or descent of a body upon an in-
clined plane. From the weight G of the body and from the angle
of inclination a we obtain the normal pressure
NG cos. ɑ,
and consequently the friction
F = Φ Ν & G cos. a.
1
G sin. a, with which
If we subtract the latter from the force P₁
the gravity pulls it down the plane, there remains the motive force
P = G sin. a & G cos. a,
and we have for acceleration of a body moving down the inclined
plane
force
Ρ
(?
G sin, a-
o G cos. a
G
a)
J =
(sin. a -
o cos. a) g.
mass
For a body ascending an inclined plane the motive force is reg-
ative and G sin. a + ¢. G cos. a, and the acceleration p is also
negative and = (sin. a + cos. a) g.
B
p
If two bodies placed upon two different inclined planes F G and
C
H R
FIG. 530.
A
descends and draws up
P = G sin, a
G (sin. a
and the mass moved
De-
FH, Fig. 530, are united by a perfectly
flexible cord, which passes over a pulley
C', it is possible that one of the bodies
will descend and raise the other.
noting the weight of these bodies by G
and G₁, and the angles of inclination
of the inclined planes, upon which they
rest, by a and a,, and assuming that G
G₁, we obtain the motive force
G₁ sin. a、 & G cos. a & G₁ cos. a₁
G (sin. a, + cos. a₁),

G
o
cos. a)
G+ G₁
M -
g
$
and therefore the acceleration with which G descends and G₁
ascends is
G (sin. a o cos. a)
p
G₁ (sin. a, + o cos. a₁)
G + G₁
Since the friction, which is a resistance, cannot produce mo-
tion, we must have, if G descends and G, ascends,
cos. a) > G₁ (sin. a,
G (sin. a
+ & cos. a₁), or
G
sin. a, + cos. a,
ф
G
sin. (a, + p)
I.E.
>
G₁
sin. a Ф cos. a
G₁
sin. (a — p)
§ 318.]
645
THE ACTION OF GRAVITY, ETC.
If, on the contrary, G, descends and G ascends, we must have
G₁
sin. a +
cos. a
or
G
sin. a1
cos. a₁
G
sin. a₁
cos. a₁
I.E.
G
<
sin. (a,p)
G₁
sin. a
+
cos. a
G₁
sin. (a + p)
G
As long as the ratio
is within the limits
G₁
sin. a, +
cos. a
and
sin. a
• cos. a
sin. a
sin. a +
-
cos. a
or
>
cos. a
sin. (a
sin. (a,+p)
P)
sin. (a, - p)
and
sin. (a + p)
the friction will prevent any motion.
EXAMPLE-1) A sled slides down an inclined plane covered with snow,
150 feet long and inclined at an angle of 20 degrees, and on arriving at the
bottom it slides forward upon a horizontal plane until the friction brings
it to rest. If the coefficient of friction between the snow and the sled is
0,03, what space will the sled describe upon the horizontal plane (the
resistance of the air being neglected) ?
The acceleration of the sled is
p =
(sin. a
(0,3420
o cos. a) g = (sin. 20° -0,03. cos. 20°). 32,2
0,03. 0,9397). 32,2 = 0,3138 . 32,2 = 10,104 feet,
and therefore its velocity on arriving at the bottom of the inclined plane is
v = √ 2 p s = √ 2. 10,104 . 150 =
√ 2. 10,104 . 150 = √3031,2 = 55,06 feet.
Upon the horizontal plane the acceleration is
Pi
0,03. 32,2
and therefore the space described is
81 =
v2 3031,2
29 1,932
0,966 feet,
1569 feet.
The time required to slide down the inclined plane is
28 300
t =
V
that required to slide along on the horizontal plane is
H
K
FIG. 531.
F
G
R
5,45 seconds;
55,06
t₁
2 81 3138
= 57 seconds,
ว 55,06

and therefore the duration of the entire journey is
t + t₁
62,45 seconds = 1 minute 2,45 seconds.
2) A bucket K, Fig. 531, which, when filled, weighs
250 pounds, is drawn up a plane, 70 feet long and in-
clined at an angle of 50°, by a weight G = 260; what
time will be required when the coefficient of the fric-
tion of the bucket upon the floor is 0,36?
646
GENERAL PRINCIPLES OF MECHANICS.
[$ 319.
The motive force is
G
260
==
260
(sin. 50° + 0,36 cos. 50°). 250
(sin. a + & cos. a) K
0,9974. 250 = 10,6 pounds,
and therefore the acceleration is
Ρ
10,6
250 +- 260
10,6
0,0208 feet;
510
the time of the motion is
28
t
1
Ρ
140
0,0208
=
√6731 =
82,04 sec. = 1 min. 22 sec.,
28 140
v
=
1,70 feet.
t
82
and the final velocity
§ 319. Rolling Motion upon an Inclined Plane.-When
a wagon runs down an inclined plane, it is the friction on the axle
which offers the principal resistance to the acceleration. If G is
the weight of the wagon, r the radius of the axle and a that of the
wheel, we have
or
N
a
φγ
G cos. α,
α
and therefore the acceleration
p
=
(sin. a - pr
cos. a g.
a) g.
a
If a round body A B, as, E.G., a cylinder or a sphere, etc., rolls
down an inclined plane F H, Fig. 532, we have at the same time a
H
FIG. 532.
B
K
F
R
motion of translation and of rotation. As
the acceleration of translation p is generally
equal to that of rotation (§ 169), if we put
the moment of inertia of the rotating body
= Gk and the radius C A of rotation = ɑ,
we obtain for the force A K A, with
which the roller (in consequence of the mu-
tual penetration of its surface and that of

the inclined plane) is set in rotation,
K = P
Gh²
ga²*
But the force K opposes the force G sin. a, which tends to
cause the body to slide down the plane, and therefore the motive
force for the motion of translation is
P = G sin. a
K,
and its acceleration is
G sin, a
K
p
•
G
$319.]
647
THE ACTION OF GRAVITY, ETC.
Eliminating K from the two equations, we obtain
G k²
G P
= G q sin. a
•
·P,
a²
and consequently the required acceleration
g sin. a
p
h
1 +
a²
For a homogeneous cylinder h²
a² (§ 288), and therefore
g sin. a
p
1 + 1/1/
3 g sin. a,
but for a sphere k² =
a² (§ 290), and therefore
p
g sin. a
1 + 2/
5 g sin. a;
5
the acceleration of a rolling cylinder is but and that of a rolling
sphere is but as great as that of a body sliding without friction.
The force which produces the rotation is
7
K
g sin. a
h:2
Gh²
G k² sin. a
g a²
a² + h
1 +
α
As long as this force is less than the sliding friction o G cos. a,
so long will the body descend the plane with a perfect rolling
motion. But if
K> & G cos. a, I.E., if tang. a > ø (1 +
tang, a > ø (1 + %)
だ
​the friction is no longer sufficient to impart a velocity of rotation
equal to that of translation; the acceleration of translation
becomes, as in the case of sliding friction,
G sin, a
& G cos. a
p =
G
• g =
(sin. a
o cos. a) g,
and that of rotation
o G cos. a
a²
P₁
•
G k² : a³
9 = 0
α
k²
g cos. a.
If the weight of a wagon is G, the radius of its wheels a and
their moment of inertia G k², we will have
G sin, a
G cos. a
K
a
K = p
I a³
and p =
•
g,
G
648
[§ 320.
GENERAL PRINCIPLES OF MECHANICS.
I.E.,
2°
g (sin. a
cos. a)
a
p
G₁ k₁2
1 +
Ga
EXAMPLE-1) A wagon, which, when loaded, weighs 3600 pounds and
whose wheels are 4 feet high and have a moment of inertia of 2000 foot-
pounds, rolls down a plane whose inclination is 12°; required the accelera-
tion, when the coefficient of friction upon the axles is
thickness of the axles is 2 r = 3 inches.
Here we have
G 1
G₁k
Ga
1
2
10
0,15 and the
2000
3600 .22
5
0,139 and
0,15.
36
a
1
4.4
0,0094,
and therefore the required acceleration is
32,2 (sin. 12° —0,0094. cos. 12°)
32,2. (0,2079 — 0,0094. 0,978)
P
32,2.0,1987
1,139
1 + 0,139
5,617 feet.
1,139
2) With what acceleration will a massive roller roll down a plane whose
angle of inclination is a = 40° ?
If the coefficient of sliding friction of the roller upon the plane is
= 0,24, we have
&
=
α
(14
+ 22) = 0,24 (1 + 2) = 0,72.
Now tang. 40° 0,839, and tang. a is therefore greater than
ø (1 -
(1 + 22).
and the acceleration of the rolling motion is smaller than that of the mo-
tion of translation.
The latter is
Ρ (sin, a & cos, a) g = (0,648 0,24. 0,7660). 32,2 = 0,459. 32,2
14,78 feet, and the former is
P₁ = 0,24 . 2 . 32,20 cos. 40° = 15,456. 0,776
=
11,99 feet.
§ 320. The Circular Pendulum.-A body suspended from
a horizontal axis is in equilibrium as long as its centre of gravity
is vertically under this axis; but if we move the centre of gravity
out of the vertical plane containing the axis and abandon the body
to itself, it assumes an oscillating or vibrating motion (Fr. oscilla-
tion, Ger. Schwingende Bewegung), I.E., a reciprocating motion in
a circle. A body oscillating about a horizontal axis is called a
pendulum (Fr. pendule, Ger. Pendel or Kreispendel). If the
oscillating body is a material point, and if it is connected with the
axis of rotation by a line without weight, we have a simple or
theoretical pendulum (Fr. p. simple, Ger. einfaches or mathema-*
§ 321.]
649
THE ACTION OF GRAVITY, ETC.
tisches P.); but if the pendulum consists of a body or of several
bodies of finite dimensions, it is called a compound pendulum (Fr.
pendule composé, Ger. zuzammengeseztes, physisches or materielles
Pendel). Such a pendulum can be considered as a rigid combina-
tion of a number of simple pendulums, oscillating around a
common axis. The simple pendulum has no real existence, but it
is of great use in discussing the theory of the compound pendu-
If the
lum, which can be deduced from that of the simple one.
pendulum, which is suspended in C, Fig. 533, is moved from its
vertical position CM to the position CA and left to itself, by
virtue of its weight it will return towards C M with an accelerated
FIG. 533.
C
A
M
D
B
motion, and it will arrive at the point M
with a velocity, the height due to which is
equal to D M. In consequence of this
velocity it describes upon the other side
the arc M B MA, and rises to the
height D M. It falls back again from B
to M and A and continues to move back-
wards and forwards in the arc A B.
If we
could do away with the friction on the
axis and the resistance of the air, this
oscillating motion of the pendulum would continue forever; but
since these resistances can never be entirely removed, the arc in
which the oscillation takes place will gradually decrease until the
pendulum comes to rest.
The motion of the pendulum from A to B is called an oscilla-
tion (Fr. oscillation, Ger. Schwung or Pendelschlag), the arc A B, the
amplitude (Fr. amplitude, Ger. Swingungsbogen), and the angle
measured by half the amplitude is called the angle of displacement.
The time in which the pendulum makes an oscillation is called the
time, duration, or period of an oscillation (Fr. durée d'une oscilla-
tion, Ger. Schwingungszeit or Schwingungsdauer).

§ 321. Theory of the Simple Pendulum.-In consequence
of the frequent use of the pendulura in common life, viz. for clocks,
it is important to know the duration of an oscillation; its demon-
stration is therefore one of the most important problems in
Mechanics. To solve this problem, let us put the length of the
pendulum A C = M Cr, Fig. 534, and the height of rise and
fall during an oscillation M D = h. Assuming that the pendulum
650
[§ 321.
GENERAL PRINCIPLES OF MECHANICS.
has fallen from A to G, and making the vertical height D H of fall
corresponding to this motion, we have the velocity acquired
at G
A
FIG. 534.
C
D
L
II
Ki
R
ΔΙ
v = √ 2 g x,
and the element of time, during which
the element of its path G K is described,

G K
V
GK
19
√2gx
If we describe from the middle O of
with the radius O M = O D
MD
the semicircle M ND, we can cut from
the latter an elementary arc N P, which
will have the same altitude P Q
KL = RH as GK, and whose relation
to the latter can be very simply ex-
pressed. In consequence of the sim-
ilarity of the triangles G K L and C G H we have
GK
K L
C G
GH'
and in consequence of the similarity of the triangles N P Q and
ONH
dividing the first of
bering that KL
of the arcs
NP
P Q
ON
NH'
these proportions by the second and remem-
P Q, we obtain the ratio of the above elements
G K CG. NH
NP GH. ON'
From a well-known property of the circle we have
GH² = MH (2 CMM II) and N H = M H.D H;
whence it follows that
GK
NP
CG.NDH
r V x
ON. √2 CM - M II
j h √ Q r
N
(π x)
and the time required to describe an element of the path is
r V x
T
3 h v z r
g
h
1
NP
2 r
(h-x) 12g x h √ 2 g [2 r — (h — x)]
NP
h
X
2 r
NP
222.]
651
THE ACTION OF GRAVITY, ETC.
Generally in practice the angle of displacement is small, and
X
then
h
and
2 r² 2 r
2. х
2 ↑
are such small quantities, that we can
neglect them and their higher powers and put
NP
T = 1
g
h
The duration of a semi-oscillation or the time within which the
pendulum describes the arc A M is equal to the sum of all the
elements of the time corresponding to the elements G K or N P.
1
Now since
h
g
1
to
1
h
g
is a constant factor, we can put the sum equal
times the sum of all the elements forming the semi-
circle D N M, I.E.,
14
1
times the semicircle
(끝​), Or
or
g
1
T
π h
t₁
h
g
2
2
The same time is required by the pendulum for its ascent; for
the velocities are the same but opposite in direction, hence the
duration of a complete oscillation is double the latter, or
t = 2 t₁ = = π |
g
(§ 322) More Exact Formula for the Duration of an
Oscillation of the Circular Pendulum.-In order to determine
the duration of an oscillation with greater precision. as is some-
times necessary, when angles of displacement are large, we can
transform the equation
(^_
1
h
X
1
2 r
into the series
h
X
1 + 1/
2 r
+ 8 . ( ² 5 7 2 )² + ...,
2 r
and then we have the time in which an element of the path is
described
h
X
X*
T
= 1 + 1/1/2
+ 3/
2 r
2 r
+...]
NP
g
h
652
[§ 322.
GENERAL PRINCIPLES OF MECHANICS.
A
Putting the central angle D O Np, or the are
FIG. 535.
DNDO. =
Φ
ho
2

we obtain the height
MH = h
x = M O - H 0 =
712
h
h
h
D
+ cos. = (1 + cos. o)
2
2
or, since
(1 + cos. $)²
ト
​il
[
and therefore the element of time
10
II
R
T
K
M
1.
[1 + ½ . (1 + cos. ¤)
h
+ § (1 + cos. 4)³ (17,. )² +
= 1 + 2 cos. & + (cos. ø)
4 r
3 + 2 cos. + & cos. 2 4,
1 + 1 (1 +.cos. p)
h
4 r
h
4 r
...]
NP
I
h
1 + 2 cos. +
+ 3 (3 + 2 cos. &
+ 1 008. 24) (4) + . . ]
cos.
= [1 + & 4
h
4γ
9
√
+...] g
+ √'s ("; )' +
16
+ (1³18 + . . .)
= ([1
1 + 1
h
4 r
9
4
h
h
+ 1/6
グ
​4
NP
h
h
3
1 + cos. 2 p
2
cos.
... + (1 + 2 (14.) + ..) 08. 4
4 r
• +...]
cos. 2 +
1
NP
h
2
+...]
NP
+
h
+
4 ľ
)²+...]
}
NP cos. $ + (1/2 + ...) (4.) NP cos. 2)√
h
3
h
g
NP
Now the sum of all the elements N P is = the arc DN P
& h
2
h
NP cos. is NQ and the sum of all the NQ is the
2 N P cos. 20
ordinate N H
sin. p and also the sum of all the
2
h
h
1 + 1/2
4 ľ
+18 (
9
(1)
+
is sin. 2 4, therefore the time required to describe the arc A G is
16
2
(1)² +...] sin.o
グ
​h
-...] + [
$
+
4 r
4
h
+ (18 + ….
4 r
sin. 20
2
•
1
§ 322.]
653
THE ACTION OF GRAVITY, ETC.
The time required to describe the arc A M is, since we have
here, sin. 4 = sin. π and sin. 2 p = sin. 2 π = 0,
h
h
4 = [(1 + 1 ) + 8 · 3 - ( - ) + ...] . ! √ √ ]}
ħ
2
3 . 33
4 r
g
=
[1 + (3)²
h
1.3
h
+
+
2 r
2.
...]
1
2 g
As the velocity decreases in the same manner, when the pen-
dulum ascends on the other side, as it increased during the descent,
the time required for describing the entire arc or the duration of
the complete oscillation is
t = 26₁ = [1 + (1)
h
2 r
2
1.3
ts
= 37,
+ (3)" (-)"
h
)
(
h 3
)² + ··· ] - √ ²²
....].
g
1.3.5
+ ( 3
\2.4.6/
-
π
1
If the pendulum oscillates in a semicircle, we have h = r, and
consequently the duration of an oscillation is
t =
(1 + 1 + +
9
225
+..
...).
πν 1,180 π 4
g
256 18432
In the most cases in practice the amplitude of the oscillations
is much less than a semicircle, and the formula
t = (1 + 3/4) = √//;
8 r
π
g
If the angle of displacement be denoted by a, we have cos. a=
is sufficiently accurate.
グ ​h
h
h
= 1
or =1
cos. a, and therefore
?'
h
1
cos. a
Sr
2
= 4 (sin. 2);
from the latter formula we can determine the correction to be
applied for any given amplitude. If, for example, this angle is
a = 15º, we have
h
1
8 r
= (sin. 15)² = 0,00426.
2
and, on the contrary, for a = 5°
h
0,00017;
8 r
for this last amplitude the duration of an oscillation is
t = 1,00047 . πT W
654
[$ 323.
GENERAL PRINCIPLES OF MECHANICS.
Consequently if the amplitude is less than 5°, we can put with
sufficient accuracy the duration of an oscillation
π
t = π √
g
NT
Ng
=
0,554 Vr.
§ 323. Length of the Pendulum.-Since in the formula
t = π |
2'
9
the angle of displacement does not appear, it follows that the
duration of small oscillations of a pendulum does not depend upon
this angle, and that pendulums of the same lengths, when their
amplitudes, although different, are small, oscillate isochronally or
have the same duration of oscillation. A pendulum, when its am-
plitude is 4 degrees, make an oscillation in (almost) the same time
as when it is 1 degree.
If we compare the duration t of an oscillation with the time t₁
of the free fall, we find the following relation. The time required
by a body to fall freely a distance r is
hence
g
t₁ = √/ 2 r
√2.
1
恒​;
t: t₁ = π: √2;
1
the duration of an oscillation of a pendulum is to the time required
by a body to fall freely a distance equal to the length of the pen-
dulum as the number is to the square root of 2. The time re-
quired to fall the distance 2 r is
t₁
π
2.2 r
g
γ
= 2 V
g
therefore the duration of an oscillation is to the time required to fall
a height equal to twice the length of the pendulum as π is to 2.
If we put the durations of the oscillations of two pendulums,
whose lengths are r and r, equal to t and t₁, we obtain
t: t₁ = NT NT.
:
When the acceleration of gravity is the same, the durations of the
oscillations are proportional to the square roots of the lengths of the
pendulums. Now if n is the number of oscillations made by one
pendulum in a certain time, as, E.G., in a minute, and n, the num-
ber made in the same time by another pendulum, we have
1
1
√r: Nr₁ =
•
N
N1
§ 324.]
655
THE ACTION OF GRAVITY, ETC.
and inversely
:
n: n₁ = NT, NT,
1
I.E. the number of oscillations is inversely proportional to the square
root of the length of the pendulum. A pendulum four times as
long as another makes but one-half as many oscillations in the
same time.
A pendulum is called a second pendulum (Fr. pendule à seconde,
Ger. Secundenpendel), when the duration of its oscillation is a
second. Substituting in the formula t = π 1
g
t
1, we obtain
g
the length of the second pendulum r = ; for English system
of measures
π
r 3,26255 feet 39,1506 inches,
and for the metrical system
=
r = 0,9938 metres.
By inverting the formula tπ √
√
,
we obtain g
(7)²
r, by
means of which we can deduce from the length r of the pendulum
and the duration of its oscillation the acceleration g of gravity.
We
e can determine the value of g more simply and more accurately
in this manner than with Atwood's machine.
REMARK. By observations upon the pendulum, the decrease of the force
of gravity, as we proceed from the equator to the poles, has been proved,
and its intensity determined. This diminution is caused by the centrifugal
force arising from the daily revolution of the earth upon its axis, and also
by the increase of the radius of the earth from the poles to the equator.
The centrifugal force diminishes the action of gravity at the equator ¿¿ of
its value (§ 302), while at the poles the action of the centrifugal force is null.
By observation upon the pendulum we can determine the acceleration of
gravity at the place of observation. This acceleration, when ẞ denotes the
latitude of the place, is
g =
9,8056 (1 — 0,00259 cos. 2 ß) metres;
therefore at the equator, where 3 = 0 and cos. 2 3
O and cos. 23 = 1, we have,
g = 9,8056 (1 — 0,00259) 9,780 metres,
and at the poles, where ß 90°, cos. 2 µ = cos. 180°
ġ = 9,8056. 1,00259 9,831 metres.
Upon mountains g is smaller than at the level of the sea.
1,
1
290
§ 324. Cycloid.-We can put a body in oscillation or cause it
to assume a reciprocating motion in an infinite number of ways
Any body moving in such a manner is called a pendulum. We
distinguish several kinds of pendulums, as, for example, the circu-
tar pendulum, which we have just discussed, the cycloidal pendulum,
where the body, by virtue of its weight, swings backwards and for-
656
IS 325.
GENERAL PRINCIPLES OF MECHANICS.
wards in a cycloid, and the torsion pendulum or torsion balance,
where a body oscillates in consequence of the torsion of a string or
wire, etc. We will here discuss only the cycloidal pendulum.
The cycloid (Fr. cycloide, Ger. Cycloide) A P, D, Fig. 536, is a
1
B
FIG. 536.
B1 D
C1
Q₁
P
P₁
0
M
·R··
A
1
1
12
A1
1
curve described by a
point A of a circle
A P B, which rolls
upon a straight line.
B D. If this gene-
rating circle rolls for-
ward the distance
B B₁ C C₁ and
comes into the posi-
tion A₁ B₁, it turns
through the arc A P

= A, P₁ = B B₁ = P P₁, and the ordinate M P, corresponding
to any abscissa A M is ordinate M P of the circle plus the arc
A P, which the circle has turned. In this rolling the generating
circle turns always upon its point of tangency to the base line B D;
if it is in A, B₁, it turns about B₁, and thus describes the element
P₁ Q, of an arc of the cycloid; consequently the chord B, P, gives
the direction of the normal and the chord A, P, that of the tangent
P₁ Tat the point P, of the cycloid. The prolongation P Q of the
chord AP reaching to the ordinate O Q, is equal to the element
P₁ Q₁ of the cycloid; since the space P R due to the motion of ro-
tation is equal to that R Q due to the motion of translation, P Q is
the base of an isosceles triangle, and is equal to twice the line P N,
which is cut off by the perpendicular R N; PN is finally the dif
ference of the two neighboring chords AR and AP, and conse-
quently the element P Q of the cycloid is equal to twice the
difference (A R AP) of the chords. Since the successive èle-
ments of the cycloid compose the arc 4 P, and the sum of the
differences of the chords the entire chord A P, we have the length
of the arc 4 P, of the cycloid equal to twice the chord A P of the
generating circle. The diameter of the circle is the chord corre-
sponding to the semi-cycloid, and the length of the semi-cycloid is
therefore twice the diameter (21 B) of the generating circle.
$325 Cycloidal Pendulum.-From the properties of the
cycloid, found in the foregoing paragraph, we can easily deduce the
theory of the cycloidal pendulum, or the formula for the duration
of an oscillation of a body vibrating in the arc of a cycloid. Let
325.]
657
THE ACTION OF GRAVITY, ETC.
AK M, Fig. 537, be half the arc of the cycloid, in which a body
oscillates, and ME the generating circle, whose radius is CE =
S
A
FIG. 537.
E

D
LF
H
G
R
K
P
M
B
CM = r. If the body has described the arc A G or fallen from
the height D H = x (compare § 321), it has attained the velocity
v = 1/2 gx, with which it describes the element G K of the arc in
the time
G K
GK
V
√ 2 g x
In consequence of the similarity of the triangles G L K and FHM,
we have
G K
FM
K L
MH
or, since FM² = MH. ME,
GK
KL
WM H.ME
M H
WM E
W MH
and in consequence of the similarity of the triangles N P Q and
Ο Ν Η
NP
ON
P Q
NH
or, since N H² = MH.D H,
NP
P Q
ON
WMH.DH
Now K L =
P Q, hence by division we have
G K
NP
WME VMH.DH VME. DH
WM H
ON
or, since O N, half the height fallen through,
DH = x,
G K
NP
√2rx
j h
2 1/2 r x
h
O N
h
ME = 2 r and
2'
42
658
[§ 325.
GENERAL PRINCIPLES OF MECHANICS,
If we substitute G K
2 √ 2 r x
h
NP in the formula
T=
G K
√ 2 g x
-"
we obtain
2 V 2 r x
2
T
•
NP =
1
N P.
•
√2gx.h
h
g
The time required to fall from A to M is the sum of all the
values of 7, obtained by substituting for N P all the divisions of
the semicircle D N M, or
2
h
1
times the semicircle D N M
g
(r).
Hence we have the time required to describe the arc A M
π 2
t₁
h.
1
π
2
h
g
g
and since the time for ascending the arc M B is equal to it, we have
for the time required to describe the whole are AM B
/
t = 2 t₁ = 2 π
= π V
g
4 r
g
Since this quantity is entirely independent of the length of the
arc, it follows that the times of the oscillations for all arcs of the
same cycloid are mathematically exactly equal, or that the cycloidal
pendulum is perfectly isochronal. If we compare this formula with
the formula for the duration of the oscillations of a circular pen-
dulum, we find that the durations are the same for both pendulums,
when the length of the circular pendulum is four times the radius
of the generating circle of the cycloid.
REMARK.-In order to make a body suspended by a flexible cord oscil-
iate in a cycloid and thereby to form a cycloidal pendulum, we must hang
B
FIG. 538.
C
E
F
A
H
K
P
N
D
01
the same between two arcs C O and C 01,
Fig. 538, of a cycloid, so that during
each oscillation the cord will unwind
from one and wind upon the other arc.
It can easily be shown that, when the
cord C O P wraps and unwraps, the end
P describes a cycloid equal to the given
one, but in an inverted position. The
length of the semi-cycloid is CO A
CD = 2 A B and the arc O A is

the
straight line O P, which has been un-
wound; but the arc 0 A = twice the
chord A F 2 GO, and therefore
326.]
659
ETC.
THE ACTION OF GRAVITY,
PGGO
G 0 = A F and HN = A E. Describing upon D H = A B a
semicircle D KH and drawing the ordinate NP, we have KH = P G
and, therefore, also
PK = GH
GH A H
arc B F = = arc D K,
AGAH-FO =arc A F B – arc A F =
and, finally, N P is the ordinate N K of the circle plus the correspond-
ing arc DK; NP is therefore the ordinate of a cycloid D P A corre-
sponding to the generating circle D K H.
"Jahrbü-
Also Prechtl's
Upon the application of cycloidal pendulums to clocks, see
cher des polytechn. Institutes in Wien," Vol. 20, Art. II.
technologische Encyclopädie, Bd. 19.
(§ 326.) The Curve of Quickest Descent.-It can be proved
by the Calculus that the cycloid, besides the property of isochronism
or tautochronism, possesses also that of brachystonism, I.E. it is the
line in which a body descends from one given point to another in
the shortest time.
We can prove this (as Jacob Bernoulli did) in the following
manner.
FIG. 539.
N
M
S
Let the relative position of two points A and B, Fig. 539, be
given by the vertical distance A C =a and the horizontal one
B C = b, and that of a horizontal
line D E by the vertical distance
AD = h; required the point A, in
which a body falling from A to B
must intersect the line D E in order
to reach B in the shortest time. If
the body arrives at A with the ve-
locity, the velocity at A is

A
E
K
B
L
D
C
h:
21
212 + 2 g ;
and supposing that 4, K and B are
infinitely near each other, or that a, b and h are very small com-
pared to , we can assume that A K is described uniformly with
the velocity v and K B uniformly with the velocity ₁, or that the
time, in which A K B is described, is
A K
K B
t =
+
༡!
V₁
Denoting D K by z, we have
A K = √h² + 2² and K B = √(a − h)² + (b − 2)²,
and therefore
660
[§ 326.
GENERAL PRINCIPLES OF MECHANICS ·
Wh² + 23
√(a
—
h)² + (b z)²
t
+
υ
V₁
This quantity will be a minimum, when we make its first dif-
ferential coefficient
d t
d z
2
b - z
v₁ √ (a − h)² + (b
= 0.
2)²
But
א
Z
Wh² + z²
K D
KA
= cos. A K D = cos. &
and
b
ช
V (a
h)² + (b
%) 2
BL
BK
=cos. KBL = cos. $1,
1
4 and 4, denoting the inclination of the paths A K and K B to the
horizon; hence we have for the equation of condition
cos.
V
cos. 01
U1
=
Putting the heights due to the velocities v and v₁, MA y and
NK = y₁, or
v = √2gy and v₁ = √2 g y₁,
our equation becomes
cos.
Vy
cos. 1
Nyi
and if we apply this formula to the case of a curved line S A K B,
it follows that for every point of this curve the quotient cos.
be a constant quantity, such as
1
√2 r
must
Ny
This property corresponds to a cycloid S G M, Fig. 540; for
we have for an element G K of this curve
6
A
FIG. 540.
E

D
B
L FNO
H
G
R
K
P
M
§ 327.]
661
THE ACTION OF GRAVITY, ETC.
GL
FH
WM H. EH
E H
Y
cos. =
G K
FM
WM H. EM
EM
and therefore
cos. o
1
=
CE
√2r
r denoting the radius C M C E of the generating circle EF M.
An arc SG of a cycloid is therefore the arc in which a body
descends in the shortest. time from one point S to another point G.
FIG. 541.
Λ
$327. The Compound or Material Pendulum.-In order
to determine the duration of an oscillation of a compound pendulum
or of any body A B, Fig. 541, oscillating about a horizontal axis C,
we must first find the centre of oscillation (Fr. centre
d'oscillation, Ger. Mittelpunkt des Schwunges or
Schwingungspunkt), L.E., that point A of the body
which, if it oscillates alone around C or forms a
simple pendulum, has the same duration of oscilla-
tion as the entire body. We can easily perceive
that there are several such points in a body, but we
generally understand by it only that one, which
lies in the same perpendicular to the horizontal

F
B
R--......
H
axis as the centre of gravity does.
From the variable angle of displacement K CF
the acceleration of the isolated point A, which is
=g sin. o;
Ø
we obtain
for we can imagine that it slides down a plane, whose inclination is
K H R = K C F = 9. If M is the moment of inertia of the
entire body or system of bodies A B, Ms its statical moment, I.E.
the product of the mass and the distance C S = s of its centre of
gravity from the axis of oscillation C, and r the distance CA of
the centre of oscillation from the axis of rotation or the length of
the simple pendulum, which vibrates isochronally with the material
pendulum AB, we have the mass reduced to K
Mk²
2.2
and therefore the rotary force reduced to this point is
S
M g sin. ;
consequently the acceleration is
662
[§ 327.
GENERAL PRINCIPLES OF MECHANICS.
force
S
Ρ
Mg sin. :
mass
7'
Mk2
2.2
Ms r
•
Mk2
g sin. p.
In order that the duration of an oscillation of this pendulum
shall be the same as that of the simple pendulum, it must have in
every position the same acceleration as the other; hence
Msr
Mk²
•
g sin. p
= g
g sin. p.
This equation gives
Mk2
Ms
moment of inertia
statical moment
We find, then, that the distance of the centre of oscillation from
the point about which the rotation takes place, or the length of the
simple pendulum having the same duration of oscillation as the com-
pound pendulum, is equal to the moment of inertia of the compound
pendulum divided by its statical moment or the moment of its weight.
N
g
we ob-
Substituting this value of r in the formula t = π |
tain for the duration of an oscillation of a compound pendulum
Mk²
t = π π 1
= π
Mg s
or more accurately
t = π
+(1.
h
+
8 r
g s
g
S'
By inversion we obtain from the duration of an oscillation of a
suspended body its moment of inertia by putting
2
Mk2
(-4):
Mgs or k² =
(4)²
2
I s.
REMARK-1) In order to determine the moment of inertia M k² of a
body from the duration of one of its oscillations, it is necessary to know its
statical moment Mg 8 =
G s.
The latter is found by drawing the body
A C, Fig. 542, out of its position of equilibrium by means of a rope A BD,
which passes over a pulley and to which a weight P is suspended. The
perpendicular C N, let fall from the axis Cupon the direction of the rope
A B, is the arm a of the weight P, and Pa is equal to the moment G. CH
of the weight G, which acts vertically at the centre of gravity S. Denoting
by a the angle VC S C SH, which the body is raised by the weight P,
we have
and therefore
CH
CS sin. a = s sin. a,
G s sin. a = Pa,
from which we deduce the required statical moment
A s =
Pa
sin, a
D
§ 327.]
663
THE ACTION OF GRAVITY, ETC.
2) A very simple and useful pendulum A D F, Fig. 543, may be made
of a ball of lead A about 1 inch in diameter, suspended by a silk thread,
FIG. 542.
H
M
G
D
N
Р
FIG. 543.
G

F
F
D

This
whose upper end is fastened into a ferrule D by a clamping screw.
ferrule has upon its end a screw, which passes through the arm E F
and is made fast by a nut G, when the arm has been screwed into a
door-frame or some other solid support. If the length is C A = 0,2485
or nearly meter, then this pendulum will beat half-seconds for almost
an hour, although the arcs in which it oscillates will continually decrease.
EXAMPLE-1) If the point of suspension of a prismatical rod 4 B,
Fig. 544, is at a distance CA 1 from one end A and C B = lz
from the other B, its moment of inertia, when F denotes its cross-sec-
tion, is (§ 286)
and its statical moment is
1
ME² = } F (1³ + l₂³),
M 3 = } F (l¸² — 1½³) ;
hence the length of the simple pendulum, which oscillates isochronally,
is
FIG. 544.
B
?=
Mk3
M s
3
1³ + 3 d²
1
6 d
7 denoting the sum l₁+ l₂ and d the difference l₁l. If this
1
2
rod should beat half-seconds, we must make
g
r = 1
1. 39,159,79 inches,
π
and if the rod is 12 inches long we must put
144 + 3 d²
9,79
or de
19,58 d =
48,
ва
hence
19,58
√191,3764
19,58
13,83
d
23 inches;
2
20
A
from which we obtain
664
[§ 328.
GENERAL PRINCIPLES OF MECHANICS.
l + d
21
6+176
=
2
776 and 12
l – d
2
6 - 116
7
416.
2) If G is the weight and 7 the length of the rod of a pendulum with
a spheroidal bob A B, Fig. 545, and if K is the weight and r, the diam-
eter M A = M B of the latter, we will have
FIG. 545.
M
B
A
} Gl² + K [(l + r₁)² + } r₁²]
1
į G l + K (l + ~ ₁)
1
1

If the wire weighs 0,05 pounds and the ball 1,5 pounds, and
if the length of the wire is 1 foot and the radius of the ball 1,15
inches, we have the distance of the centre of oscillation of this
pendulum from the axis of rotation
·
0,05.12 + K (13,15+. 1,15)
1.0,05 12 + 1,5. 13,15
262,577
20,025
=
•
13,112 inches.
If we neglect the wire, r =
260,177
19,725
2,4 + 260,177
0,3 + 19,725
13,190 inches, and if we assume
The
the mass of the ball to be concentrated at its centre r = 13,15 inches.
duration of an oscillation of this pendulum is
t = π 1
= 0,554
13,112
12
0,554 √1,0926
0,5791 seconds.
§ 328. Reciprocity of the Point of Suspension and the
Centre of Oscillation.-The point of suspension and the centre
of oscillation are reciprocal (Fr. réciproque; Ger. wechselseitig),
I.E. one can be changed for the other, or the pendulum can be sus-
pended at the centre of oscillation without changing the duration
of the oscillation. This can be proved, by the aid of what was
said in § 284, in the following manner. Let W be the moment of
inertia of the compound pendulum A B, Fig. 546, referred to an
axis of rotation passing through its centre of grav-
ity S, for an axis of rotation passing through C,
which is at a distance CS s from the centre of
gravity S, we have
FIG. 546.

W₁ = W + M s²,
and therefore the distance of the centre of oscilla-
tion from the axis of rotation Cis
F
B
R.......
H
W + M s²
W
+ s.
Ms
Ms
Ms
Denoting the distance K S = r
s of the centre
of oscillation K from the centre of gravity by s₁, we obtain the
W
M'
equation s s₁ = in which s and s, present themselves in the
§ 329.]
665
THE ACTION OF GRAVITY, ETC.
A
same manner, and therefore can be changed for one another. This
formula is consequently applicable not only to the case, where s
expresses the distance of centre of rotation and s, that of the cen-
tre of oscillation from the centre of gravity, but also to the case,
where s expresses the distance of the centre of oscillation and s
that of the centre of rotation from the centre of gravity. There-
fore C becomes the centre of oscillation, when K becomes the point
of suspension. We employ this property in the revers-
FIG. 547. able pendulum 4 B, Fig. 547, first suggested by Bohnen-
berger and afterwards employed by Kater. It is provided
with two knife-edge axes C and A, which are so placed,
that the duration of an oscillation remains the same,
whether the pendulum is suspended from one axis or the
other. In order to avoid changing the position of the
axes in reference to each other, two sliding weights are
p applied to it, the smaller of which can be moved by a
small screw. If by sliding the weights we have brought
them to such a position, that the duration of an oscilla-
tion is the same, whether the pendulum be suspended in
Cor K, we obtain in the distance the length r
of the simple pendulum, which vibrates isochronally with
the reversable pendulum, and the duration of the oscilla
tion is given by the formula
B
Q
FIG. 548.
t = π
r
g
§ 329. Rocking Pendulum.-The rocking of a body with a
cylindrical base can be compared to the oscillation of a pendulum.
This rocking, like every other rolling motion, is composed of a mo-
tion of translation and one of rotation, but we can consider it as a
rotation about a variable axis. This axis of rotation is the point
of support, where the rocking body A B C, Fig. 548, rests upon
the horizontal support HR. Let
the radius C D = C P of the cylin-
drical base ADB be = r and the
distance CS of the centre of gravity
S of the whole body from the centre
Cof this base bes, then we have
for the distance SP = y of the cen-
tre of gravity from the centre of rota-
tion, corresponding to the angle
SCP = 0,
C

N
S
A
H
P
B
R
666
[$ 329.
GENERAL PRINCIPLES OF MECHANICS.
2
y² = r² + s²
p² + §³ — 2 r s cos. O
=
(r− s)² + 4 rs (sin.
· (sin. 2).
If we denote the moment of inertia of the whole body in reference
to the centre of gravity S by M h, we obtain the moment of inertia
in reference to the point of support P
W = M (k² + y²) = M [k² + (r
M [ k² + (v − s)² + 4 r s
for which for small angles we can put M [+ (r
(sin. $)'],
s)² + r s p³] or
even M [k² + (r $):]. Now since the moment of the force =
Mgs sin. p, we have the angular
G.SN = M g. C S sin. O
acceleration for a rotation around P
moment of force
Mgs sin. p
g s sin. p
K
moment of inertia
M[k² + (r − s)*]
h² + (v — 8)²°
g sin. o
For the simple pendulum it is =
, when r, denotes its length.
グン
​If they should oscillate isochronally, we must have
gs sin. p
g sin. p
7² + (1-8)²
だ​+(
h² + (2 8) 2
I.E., 7'1 =
21
$
The duration of an oscillation of the rocking body is, therefore,
FIG. 549.
A
21
1
t Π
g
7½² + (r
Jy s
2

This theory is applicable to a pendulum AB, Fig. 549,
with a rounded axis of rotation CM, when we substitute
for r the radius of curvature CM of this axis. If instead
of the rounded axis a knife-edge axis D is used, the dura-
tion of an oscillation would be
S12
t₁
π
k² + D S¹²
g. DS
だ ​+ (S
x)²
π
J
g (8x)
when the distance CD of the knife-edge D from the cen-
tre of the rounded axis is denoted by x.
The two pen-
dulums will have the same duration of oscillation, when
k² + ( s − x)²
π½² + (~ 8)²
2
1;2
7² + 2²
S
X
S
or
S
X =
2r;
X
S
k2
k²
k² x
putting approximatively
+
and neglecting r², we
CA
s
X
S
S
obtain
2 r s³
X =
REMARK.-The conical pendulum will be discussed in the third part,
in the article upon the "Governor."
In the appendix to this volume the subject of oscillation is treated at
length.
330.]
667
THE THEORY OF IMPACT.
CHAPTER IV.
THE THEORY OF IMPACT.
§ 330. Impact in General.-On account of the impenetra-
bility of matter, two bodies cannot occupy the same space at the
same time. If two bodies come together in such a way that one
seeks to force itself into the space occupied by the other, a recipro-
cal action between them takes place, which causes a change in the
conditions of motion of these bodies. This reciprocal action is
what is called impact or collision (Fr. choc, Ger. Stoss).
The conditions of impact depend, in the first place, upon the
law of the equality of action and reaction (§ 65); during the im-
pact one body presses exactly as much upon the other as the other
does upon it in the opposite direction. The straight line, normal
to the surfaces, in which the two bodies touch each other, and
passing through the point of tangency, is the direction of the
force of impact. If the centre of gravity of the two bodies is upon
this line, the impact is said to be central; if not, it is said to be
eccentric. When the bodies and B, Fig. 550, collide, the impact
FIG. 550.
D
A
B
N
N.
E
FIG. 551.
D


B
N
E
is central; for their centres of gravity S, and S. lie in the normal
NN to the tangent plane. In the case represented in Fig. 551 the
impact of 4 is central and that of B eccentric; for S lies in and
S without the normal line or line of impact N N.
When we consider the direction of motion, we distinguish direct
impact (Fr. choc direct, Ger. gerader Stoss) and oblique impact (Fr.
choc oblique, Ger. shiefer Stoss). In direct impact the line of im-
663
[$ 331.
GENERAL PRINCIPLES OF MECHANICS.
pact coincides with the direction of motion; in oblique impact the
two directions diverge from each other. If the two bodies 4 and.
N
FIG. 552.
D C₁ C₂
E
B
32
S
tially or entirely retained.
N
B, Fig. 552, move in the directions
S₁ C, and S. Ca, which diverge from
the line of impact N N, the impact
which takes place is oblique, while,
on the contrary, it would have been
direct if the directions of motion had
coincided with N N.
We distinguish, also, the impact
of free bodies from that of those par-

§ 331. The time during which motion is imparted to a body or
a change in its motion is produced is, it is true, very small, but by
no means infinitely so; it depends not only upon the force of im-
pact, but also upon the mass, velocity and elasticity of the colliding
bodies. We can assume this time to consist of two parts. In the
first period the bodies compress each other, and in the second they
expand again, either totally or partially. The elasticity of the
body, which is brought into action by the compression, puts itself
into equilibrium with the inertia, and thus changes the condition
of motion of the body. If during the compression the limit of
elasticity is not surpassed, the body returns to exactly its former
shape, and it is said to be perfectly elastic; but if the body, after
the impact, only partially resumes its original form, we say it is
imperfectly elastic; and if, finally, the body retains the shape it as-
sumed under the maximum of compression or possesses no ten-
dency to re-expand, we say that the body is inelastic. This classi-
fication of impact is correct within certain limits only; for it is
possible that the same body will act as an elastic one when the im-
pact is slight, and as an inelastic one when the impact is violent.
Strictly speaking, perfectly elastic and perfectly inelastic bodies
have no existence; but we will hereafter consider elastic bodies to
be those which apparently resume their original form, and inelastic
bodies to be those which undergo a considerable change of form in
consequence of the impact.
In practical mechanics the bodies, such as wood, iron, etc.,
which are subjected to impact, are very often regarded as inelastic,
because they either possess but little elasticity or lose the greater
part of their elasticity in consequence of the repetition of the im-
§ 332.]
THE THEORY OF IMPACT.
669
pact. It is very important in constructing machinery, etc., to avoid
impacts as much as possible. If this cannot be done, we should
diminish their intensity or change them into elastic ones; for they
give rise to jars or concussions and cause the machinery to wear
very fast, and in consequence a portion of the energy of the ma-
chine is consumed.
§ 332. Central Impact. - Let us first investigate the laws of
the direct central impact of bodies moving freely. Let us suppose
the duration of the impact composed of the equal elements 7, and
the pressure between the bodies during the first element of time to
be P₁, during the second to be = P, during the third to be
P, etc. Now if the mass of the
body A, Fig. 553, M,, we have the
corresponding accelerations.
=
FIG. 553.
D

P:
NP
P₁
N
P1
P
Mi
M₁
M
M
P
P3 =
etc.
M
But, according to § 19, the vari-
ation in velocity corresponding to p
and to an element of the time is
K
k = pt;
hence the elementary increments and diminutions of velocity in
the foregoing case are
K₁ =
P₁ T
M₁
P₂ T
Рот
Kg =
K3
etc.,
M₁
>
M₁
>
and the increase or decrease in velocity of the mass M, after a cer-
tain time is
1
K₁ + K₂ + K3 +
=
= (P₁ + P₂ + P3 + …..)
Mi
and the corresponding variation in velocity of the body B, whose
mass is M, is
=
(P+ P+P + . . . .)
T
M
The pressure acts in the following or impinging body in oppo-
sition to the velocity e, producing a diminution of velocity, and
after a certain time the velocity, which the body still possesses, is
670
[$ 332.
GENERAL PRINCIPLES OF MECHANICS.
Т
v₁ = c − (P₁ + P₂ + ...) 11.
2
Mi
The pressure acts upon the body B, which is in advance and which
is impinged upon, in the direction of motion, its velocity c₂ is
increased and becomes
Vg = Cg + (P₁ + P₂ + P₂ + ...)
2
3
T
M
2
Eliminating from the two equations (P, + P₂ + P3 + ...) T,
we have the general formula
I. M₁ (c₁
-
v₁) == M₂ (v₂
— c₂), or
M₁ v₁ + M₂ v₂ = M₁ c₁ + M₂ C2.
1
The product of the mass of a body and its velocity is called its
momentum (Fr. quantité de mouvement; Ger. Bewegungsmoment),
and we can consequently assert that at every instant of the impact
the sum of the momentums (M₁ v₁ + M½ v₂) of the two bodies is the
same as before the impact took place.
At the instant of greatest compression, the two bodies have the
same velocity v, hence if we substitute this value v for v, and v, in
the formula just found, we obtain
M₁ v + M₂ v = M₁ c₁ + M¿ C.,
2
from which we deduce the velocity of the bodies at the moment of
greatest compression
v =
M₁ c₁ + M₂ C2
M₁ + M₂
If the bodies A and B are inelastic, I.E. if after compression
they have no tendency to expand, all imparting or changing of
motion ceases, when the bodies have been subjected to the maxi-
mum compression, and they then move on with the common
velocity
t
v =
M₁ c₁ + M₂ ca
M₁ + M₂
EXAMPLE-1) If an inelastic body B weighing 30 pounds is moving
with a velocity of 3 feet and is impinged upon by another inelastic body
A weighing 50 pounds and moving with a velocity of 7 feet, the two move
on after the collision with a velocity
50.7 + 30 . 3
350 + 90 44
11
5 feet.
50 + 30
80
8
2
2) In order to cause a body weighing 120 pounds to change its velocity
+
333.]
671
THE THEORY OF IMPACT.
from c =
11 feet to v = 2 feet, we let a body weighing 50 pounds strike
it; what velocity must the latter have? Here we have
1
c₁ = 0 +
M
(v — c₂) M₂
Ꮇ .
2
2 +
(2 — 1,5). 120
50
6
2 +
3,2 feet.
5
333. Elastic Impact.-If the colliding bodies are perfectly
elastic, they expand gradually during the second period of the im-
pact after having been compressed in the first one, and when they
have finally assumed their original form, they continue their mo-
tion with different velocities. Since the work done in compressing
an elastic body is equal to the energy restored by the body, when
it expands again, no loss of vis viva is caused by the impact of
elastic bodies. Hence we have for the vis viva the following equa-
tion
2
II. M₁ v² + M₂ v₂ = M₁ c₁₂+ M₂ ca², or
v.
2
v':
M₁ (c₁² v¸²) = M₂ (v₂² — c₂²²).
From equations I. and II. the velocities, and 2 of the bodies
after the impact can be found. First by division we have
I.E.,
2
2
V₂ Ca
V2
C₂
C₁ + V₁ = V2 + C, or V2
V₁ = C₁ = Ca;
substituting the value
2₂ = G₁ + V₁
Ca,
deduced from the last equation, in equation I., we have
whence
M₁ v₁ + M₂ v₁ + M₂ (c₁
(M, + M.) v₁ = (M₁ +
and
— c₂) = M₁ c₁ + Ma ce, or
-
M₂) c₁ − 2 M₂ (c₁ — c.),
V₁ = C₁
2 M₂ (C₁— C₂)
M₁ + M₂
2 M₂ (c - C₂)
2 M₁ (c₁ — c₂)
1
1'2 = C₁ — C₂ + C₁
= C₂+
M₁ + M₂
M₁ + M
Hence if the bodies are inelastic, the loss of velocity of one
body is
M₁ C₁ + M₂ C₂
v = C₁
1
M₁ + M₂
M. (erce)
M₁ + M₂
and when they are elastic, it is double that amount, or
672
[§ 334.
GENERAL PRINCIPLES OF MECHANICS.
:
2 M₂ (c₁ - c₂)
C - 2₁ =
'
M₁ + M₂
and while for inelastic bodies we have the gain in velocity of the
other body
V- Ca
M₁ C₁ + M Ca
M₁ + M₂
C₂ =
C2
M₁ (c₁- C₂)
M₁ + M₂
for elastic bodies it is
2 M₁ (C₁ - C₂)
Vo C 2
M₁ + M₂
or double as much.
EXAMPLE.-Two perfectly elastic balls, one weighing 10 pounds and
the other 16 pounds, collide with the velocities 12 and 6 feet. What are
their velocities after the impact? Here M₁
C2
10, C1
6 feet, and the loss of velocity of the first body is
2.16 (12 + 6)
1
10 + 16
2
12, M₂ = 16 and
2.16. 18
26
22,154 feet.
and the increase of the velocity of the other is
v2 C₂
2.10.18
26
=13,846 feet.
?₁ = 12-22,154
1
13,846
=
The first body, therefore, rebounds after the collision with the velocity
- 10,154 feet, and the other with the velocity v2
- 6 +
7,846 feet. The vis viva of these bodies after the impact is
= M₁ v + M₂ v₂² 10. 10,154² + 16 . 7,846° 1031 + 985 2016 or
the same as that before impact M₁ c₁ + M₂ c₂ = 10. 12² + 16 . 6² = 1440 +
576 = 2016.
1 1
2
2
1 1
2
If the bodies were inelastic, the first body would lose but
ቂ.
2
C 1
1
2
6,928
11,077 feet of its velocity and the other would gain
feet; the velocity of the first body after the impact would be 12 - 11,077
0,923 feet, and that of the second 6 + 6,923 = 0,923; a loss of me-
chanical effect
[2016 — (10 + 16) 0,923²] : 2 g = (2016
however, takes place.
22,2) . 0,0155 = 30,9 foot-pounds,
$334. Particular Cases. The formulas found in the fore-
going paragraph for the final velocities of impact are of course
applicable, when one of the bodies is at rest, or when the two
bodies move in opposite directions and towards each other, or
when the mass of one of the bodies is infinitely great compared
to that of the other, etc. If the mass M, is at rest, we have c = 0
and therefore for inelastic bodies
§ 334.]
673
THE THEORY OF IMPACT.
V
1
M₁ c₁
M₁ + M₂
and for elastic ones
V₁ = C₁
2 M₂ C₁
M₁ + M₂
M₁ - M
1
M₁ + M₂
C₁, and
2 M₁ C₁
2 M₁
1½ = 0 +
C1.
M₁ + M₂
2
M₁ + M₂
If the bodies move towards each other, c, is negative, and there-
fore for inelastic bodies
V
M₁ c₁
M₂ C₂
M₁ + M₂
2 M₂ (c₁ + C₂)
and for elastic ones
2 M₁ (C₁ + C₂)
V₁ = C₁
and v₂ =
v2
C₂ +
M₁ + M₂
1
M₁ + M₂
2
If in this case the momenta of the bodies are equal, or M, c,
M₂ c₂, when the bodies are inelastic, v = 0, I.E., the bodies bring
each other to rest, but if they are elastic,
12 =
2 (M, c₁ + M₁ c₁)
M₁ + M₂
= C₁
2 c₁ =
C1
—
C₁, and
2 (M¿ c₂ + M₁ c₂)
M₁ + M, 2
C₂ + 2 C₂ = + C₂;
the bodies after the impact proceed in the opposite direction with
the same velocity they originally had.
If, on the contrary, the
masses are equal, we have for inelastic bodies
رح
v =
C1
2
Cz
and for elastic ones
V1
c. and v₂ = C1,
va
I.E., each body returns with the same velocity that the other body
had before the impact. If the bodies move in the same direction,
and if the one in advance is infinitely great, we have for inelastic
bodies
and for elastic ones
Maca
M
V₁ = C₁ 2 (C₁ — C₂) = 2 ca
1
=
C29
C1, V2 = C₂ + 0 = c₂ ;
the velocity of the infinitely great body is not changed by the
impact. If the infinitely great body is at rest, or if c₂ = 0, we have
for inelastic bodies
v = 0,
and for elastic ones
Vi
C1, V2 = 0;
here the infinitely great body remains at rest; but in the first case
43
674
[§ 335.
GENERAL PRINCIPLES OF MECHANICS.
the impinging body loses its velocity completely, and in the second
case it is transformed into an equal opposite one.
EXAMPLE-1) With what velocity must a body weighing 8 pounds
strike a body weighing 25 pounds in order to communicate to the latter a
velocity of 2 feet? If the bodies are inelastic, we must put
1
v =
1
M₁ C1
M₁ + M₂
I.E. 2
8.01
8 + 25'
whence we obtain c₁ =
33
=
4
81 feet, which is the required velocity; if
they were elastic, we would have
V 2
v₂ =
2 M₁ c₁
M₁ + M₂'
whence c₁ =
3341 feet.
2) If a ball M₁,Fig. 554, strikes with the velocity c, the mass M₂
FIG. 554.
M
M₂
M3
MA
S₁-
2
2
= n M₁,
11
which is at rest, if the second mass
strikes a third M₂ = n M, = n² M,
3
with the velocity imparted to it by
the impact, and if this third mass
strikes a fourth M
n M3
n³ M₁, etc., we have, when these

masses are perfectly elastic, the velocities
4
2
"₂ =
1
2 M₁
M₁ + n M₁
2
2 M2
પે
2
C1
1 + n
C1, vg
•
1
2
M₂ + n M,
1 + n
2
2
2
3
C1, V
=
4
+ n
1 + n
If, for example, the weight of each mass is one-half that of the pre-
ceding one, we have the ratio of the geometrical series formed by the
masses
n =
hence
v2
C1, V3
(4)² C1, V4
(*) C1
·
•
V10 =
(31) 9 C 1
13,32 . •
$335. Loss of Energy.-When two inelastic bodies collide,
a loss of vis viva always takes place, and therefore they do not
possess so much energy after the impact as before.
pact the vis viva of the masses M, and M2, which
velocities c₁ and C29 is
M₁ c₁² + M₂ ca²,
2
but after the impact they move with the velocity
υ
M₁ c₁ + M₂ ca
M₁ + M₂
2
and
Before the im-
move with the
1
355.]
675
THE THEORY OF IMPACT.
their vis viva is
M₁ v² + M₂ v²;
2
by subtraction we obtain the loss of vis viva caused by the impact
2
K = M₁ (c,²
v²) + M₂ (c₂²³
= M; (c₁ + v') (C₁
— v) — M₂ (c₂
+
v) (v
c₂), but
1
M₁ M₂ (c₁
c₂)
M₁ (c, — v)
— v) =
M₂ (v — c₂) ·
-
M₁ + M₂
whence
M, M (cc)
(c₁ — c.)² M₁ M₂
2
K = (c₁ + v − C2 — v)
=
1
M₁ + M₂
M₁ + M₂
If the weights of the bodies are G, and G, or if
G₁
M₁ =
and M₂
g
G2
g
we have the loss of energy or the work done
1
1
1
+
M₁ M₂
We call
(C₁ — C₂)²
(1
G, G₂
A
2 g
1
G₁ + G₂
the harmonic mean between G, and G, and we
G₁ G₂
G₁ + G₂
can assert that the loss of energy, caused by the impact of two inelastic
bodies and expended in changing their form, is equal to the product
of harmonic mean of the two masses and the height due to the differ-
ence of their velocities.
If one of the masses M is at rest, we have the loss of
mechanical effect
1
A
2 g
G₁ G₂
G₁ + G?
and if the moving mass M, is very great, compared to the mass at
rest, G, disappears before G, and the formula becomes
А
We can also put
2
2
Go
•
2 g
K = M₁ (c v²) + M₂ (c₂² - 212)
−
= M₁ (e-2 c, v + v² +2 c, v−2 v²) + M₂ (c₂2-2 c; v + v² + 2 c; v−2 v₂³)
= M₁ (c₁ — v)² + 2 M₁ v (c, v) + M, (c₂ — v)² + 2 My v (c, — v)
— M, (c, — v)² + M₂ (C₂ — v)²;
for M₁ (c,v) = M₂ (vca).
From this we see that the vis viva lost by the inelastic impact is
676
[$ 336.
GENERAL PRINCIPLES OF MECHANICS.
equal to the sum of the products of the masses and the squares of
their gain or loss of velocity.
EXAMPLE-1) If in a machine 16 impacts per minute take place be-
tween the masses
1
1000
1200
M₁
lbs. and M₂
lbs.,
g
9
2 feet, the loss of energy, in con-
=
5
1. 9. 0,0155 000
20,29 foot-lbs.
2200
whose velocities are c = 5 feet and c₂
sequence of these impacts, is
6
0
(5 — 2)² 1000 . 1200
A =
per second.
2 ÿ
1
2
2) If two trains of cars, weighing 120000 and 160000 pounds, come into
collision upon a railroad when their velocities are c₁ = 20 and c₂ = 15
feet, a loss of mechanical effect, which is expended in destroying the loco-
motives and cars, ensues; its value is
20 + 15\² 120000. 160000
(20 15)
29
280000
352. 0,0155.
1920000
28
1302000 foot-lbs.
§ 336. Hardness.-If we know the modulus of elasticity of
the colliding bodies, we can find also the compressive force and the
amount of compression. Let the cross-section of the bodies A and
B, Fig. 555, be F, and F, their length
1 and 2, and their moduli of elasti-
city be E, and E. If they impinge
upon one another, the compressions.
produced are, according to § 204,
FIG. 555.
D

B
_N__P
N
M
M
λι
Pl₁
FE
Pl₂
and λ =
F. E
E
and their ratio is
2,
λε
F, E, 1
FE, 12
F, E₁
If, for the sake of simplicity, we denote
by H₁ and
2
F₂ E,
la
by H2, we obtain
and
P
P
λι
=
and 2,
H₁
H₂
श
2₁
H
2
2.22
H
Calling, with Whewell (see the Mechanics of Engineering,
FE
§ 207), the quantity
the hardness (Fr. dureté raideur, Ger.
1
§ 336.]
677
THE THEORY OF IMPACT.
Härte) of a body, it follows that the depth of compression is in-
versely proportional to the hardness.
G
If the mass M
impinges with the velocity c upon an im-
g
movable or infinitely great mass, all its vis viva is expended in com-
pressing the latter body, whence, according to § 206,
¦ P o =
1 Po
M c²
2 2 g
c²
G.
But the space o is equal to the sum of the compressions λ, and
P
P
A, and we have 2,
and λ₂ =
whence
H₁
H
1
1
5 = 21
λι + λο
= P
+
or inversely
H H₂
H₁ H
)
H₁ + H₂
P,
H₁ H₂
P
σ.
H₁ + H
Substituting this value of P in the above equation, we obtain the
equation of condition
H₁ H
c²
0² =
G,
12
•
H₁ + H₂
2 g
or
σ = c v
HH G
H₁ + H₂
H₁ H
g
by the aid of which the values P, 2, and 4, can be calculated.
EXAMPLE.—If with a sledge, that weighs 50 pounds and is 6 inches long
and the area of whose face is 4 square inches, we strike a lead plate one
inch thick, and the area of whose cross-section is 2 square inches, with a
velocity of 50 feet, the effect can be discussed as follows. Assuming E,
29000000 as the modulus of elasticity of iron and E
lead, we find the hardness of the two bodies to be
=
F, E. 4. 29000000
1
1
700000 as that of
H₁
= 19333333 and
2
H₂
F₂ E.
lo
6
2.700000
1
= 1400000.
Substituting these values in the formula
H₁ + H. G
1
σ = c p
1
H₁ H₂
g
= 4. 6. 0,29 = 7 pounds, or
and putting the weight of the sledge = 4. 6. 0,29
G
7.0,031
0,217,
we have for the space described by the sledge in compressing the lead
20733333 . 0,217
σ = 50
50 √ 19333333. 1400000
=501
0.44991
2706666
=0,0204 inches=0,245 lines.
زا
*
678
[$ 337.
GENERAL PRINCIPLES OF MECHANICS.
Hence the pressure is
1
P
1
H₁ H₂
H₁ + H
2
19333333. 1400000
. 0,020426632 pounds;
20733333
the compression of the hammer is
P
21
H₁
1
26632
19333333
=
0,0014 inches 0,016 lines,
and that of the lead
λε
P
H₂
26632
1400000
0,019 inches = 0,228 lines.
§ 337. Elastic-inelastic Impact.—If two masses M, and M₂
are moving with the velocities c, and c, in the same direction, their
common velocity at the moment of maximum compression is, ac-
cording to § 332,
وح
1
2
and the work done during the compression, according to § 335, is
A
(c₁ — C₂)" M, M₂
2
M₁ c₁ + M₂ C₂
M₁ + M₂
(Ci
(2) 2
G₁ G₂
1
M₁+ M
2g
G₁ + G₁₂
H₁ H
112
H₁ + H₂
but this mechanical effect can be put
= Po = P (λ, + 2₂)
¦
↓
whence we obtain for the sum of the compressions of the two
masses
σ = (C₁ — C₂) V
*
H₁ + H₂
H,
G₁ G
g (G₁ + G₂) H₁ H
from which the compressive force P and the compressions λ, and
2, of the two masses can be found.
If the bodies are inelastic, they remain compressed after the
impact; but if one only is inelastic, the other resumes its original.
form in a second period, and the work done in expanding produces
another change of velocity. If, for example, the mass M
G₁
g
1
is elastic, the work done in the second period of the impact is
p²
1 H₁ H
2
1 PM₁ = • H₁
Ρ λ
12
2 H₁ \H₁ + H₂
2
C₂)² G₁ Ga
H
2 g
G₁ + G₂' H₁ + H₂
We have, therefore, when the velocities after the impact are ", and
v2, the formulas
§ 337.]
679
THE THEORY OF IMPACT
H₂
M₁ + M₂' H₁ + H₂
M₁ M,
1
M₁ v₁ +
M₂ v₂ = M₁ c₁ +
M, c, and
M₁ M
M₁ v₁² + M₂ v₂² = M₁ c*
2
+
2
M₂ C₂² + (C₁ -
C₂)².
M₁ M₂
2
2
=M₁ c₁² + M₂ c2 — (c₁—c₂)² ·
1
M₁ + M₂
I.E.
2
.
H₁₂
+ (C₁—C₂)² · M₁ + M₂' H₁ + H
.
M, M.,
1
2
H₁
M₁ v‚² + M¸ v‚² = M₁ c₁² + M₂ c₂² — (c₁ — c₂)²⋅ M₁ + M₂' H₁ + H;
'₁
If we put the loss of velocity c₁ = 2, we have the gain in
velocity
M₁ a
1
V2
C2
M₂'
and the last equation assumes the following form:
x
1
x (2 e, − x) − x ( 2
c₂
x (2 c, + 111, 7) — (c₁ — c.)²
2
or
M₁ + M₂
M₂
M.
H
M₁ + M₂* H₁ + H₂
M₂
H
2² − 2 (c, − c₂) x + (c₁ − c₂)² · M₁ + M₂ * H¸ + H¸
=
0,
= 0.
M₂
Multiplying by
M₁
M₁ + M₂
and remembering that
H₁
1
HI₁ + II.
H₂
H₁ + H?
we obtain the quadratic equation
M
x² - 2 (c₁
C₂)
1
M₁ + M₂
2
M.
= (c₁ − cs)" (M,²±² M.) H₁ + H;
+
cs)" (M,
M
M₂
M.)
M₁ + M₂
x + (c₁
C
H₂
or
(x – (er -
C₂)
M
M₁ + M.
2
= (c₁ — c₂)² (M, + M₂
H
•
H+H
by resolving which we obtain the loss of velocity x of the first body
H
M₁
C1
V₁ = (C
C₂)
M₁ + M. (1 + V
H₁ + H.
11.).
and the gain of velocity of the other
M₁
M₁ + M₂
+
V 2 C₂ = (c₁ C₂) (1 + √
H
H₁+ H
EXAMPLE.—If we assume that in the example of the foregoing para-
graph the iron sledge is perfectly elastic and that the lead plate is perfectly
inelastic, we obtain the loss of velocity of the hammer, which weighs 7
pounds and falls with the velocity of 50 feet, since we must put c = 0 and
M₂ = ∞,
2
680
[S 338.
GENERAL PRINCIPLES OF MECHANICS.
C1
1 (1
H₂
1 –
1 + 1
1
H₁ + H₂
50 (1 +
1400000
+1
20733333
2
= 50 (1 + 0,26) = 63 feet;
hence the velocity of the sledge after the blow is
2
1
63 = 50
C1
-
63
13 feet.
The velocity of the lead plate, which is retained, of course remains 0.
§ 338. Imperfectly Elastic Impact. If the colliding
bodies are imperfectly elastic, they expand only partially in the second
period of the impact and the mechanical effect expended in pro-
ducing the compression in the first period is not entirely restored
in the second period. If 2, and 2, again denote the amount of
compression and P the pressure (called also the force of distorsion),
we have the mechanical effects expended during the compression
¦ P λ, and ¦ P 22, and if during the expansion but the µth
part or more generally during the expansion of the first body
but the pth and during that of the other but the path part of the
mechanical effect is restored, the entire loss of mechanical effect is
A = ↓ P [(1 − µ₁) λ, + (1 − µs) λ],
—
Р
P
or, putting λ₁ =
and 22
H₁
H
1
A =
Į P² [
μι 1 μ 2
+
H
H 2
The force with which the bodies react in the second period is
called the force of restitution..
But according to the foregoing paragraph we have.
P
H₁ H σ
and σ = (C₁
H₁ + H.
C₂) 1
1
M M
M₁ + M₂
1
H₁ + H₂
H, H₂
hence the required loss of mechanical effect is
(c₁
C₂)²
2
M₁ M
A =
2
MM
(c₁
C₂) 2
M₁ M
2
M₁ + M₂
1
fr H₂ + µ₂ H
2
H₁ + H₂
To find the velocities v, and v, after the impact, we employ the
equations
M₁ v₁ + M¸ v₂ = M₁ c₁ + M₂ c, and
2
H₁ H
H₁ + H₂
(
1
μ.
+
H₁
H₂
(1 -
2
1
2
v²²
2
M₁ v₁² + M₂ v₂ = M₁ c² + M₂ c₂²
2
(c₁ — C₂)²
2
•
M₁ M2
(1 − µ₁) H₂ + (1 µ₂) H₁
M₁ + M₂
H₁ + H₂
which we must combine and resolve. In exactly the same manner
as in the last paragraph the loss of velocity of the first body is found
to be
§ 338.]
THE THEORY OF IMPACT.
681
จ
C₁ = V₁ = (C₁ — C₂)
M₁ + M
(1
1 + V
+ √ M₂ H₁ + M₁ H
H₁ + H₁
1
and the gain in velocity of the body, which is in advance,
M₁
V₂ — C₂ = (C₁ — C₂)
-
M₁ + M.
(1 + V
μ ₂ H₁ + μ₁ H
H₁ + H
These two formulas include also the laws of perfectly elastic
and of inelastic impact. If we substitute in them µ, = µ₂ =
μ₂ = 1, we
obtain the formula already found for perfectly elastic bodies, and
if we assume μ₁ = μl₂ =
µ,
0, we obtain the formulas for inelastic im-
pact, etc. If both bodies are equally elastic, or μ₁ =μ, we have
more simply
M₂
C₁ — v'₁ = (c₁
C₂)
(1 + √ µ)
μ
M₁ + M₂
and
M₁
V2 — C q = (C₂ — C₂)
(1 + √ μ).
M₁ + M»
If the mass M, is at rest and infinitely great, it follows that
c₁ − v₁ = c, (1 + √ µ), I.E.,
v₁ =
c₁ √ μ, or inversely
μ =
If we cause a mass M, to fall from a height h upon a rigidly
supported mass M, and if it bounces back to a height h₁, we can
determine the coefficient of imperfect elasticity of the body by the
formula
h
μ
h'
Newton found in this way for ivory,
for glass
µ = (3)² =
=
§ ÷ 0,79,
μ =
and for cork, steel and
6
=
64
SI
( }¦ § ) ² 0,9375° = 0,879,
wool
(§) = 0,555 = 0,309.
We assume, in this case, that the falling body is a sphere and
that the body upon which it falls is flat.
General Morin by causing cannon balls, weighing from 6 to 20
kilograms, to fall upon masses of clay, wood and cast-iron, which
were suspended from a spring balance or spring dynamometer
found that for clay and wood is nearly = 0, and that, on the
contrary, for cast-iron it is nearly 1, 1.E. that the impact of
=
682
[§ 339.
GENERAL PRINCIPLES OF MECHANICS.
bodies of former substances can be considered as inelastic and that
of those of the latter substances as perfectly elastic (see A. Morin,
Notions fondamentales de Mécanique, Art. 67-70).
EXAMPLE.—What will be the velocities of two steel plates after impact, if
before the impact their velocities were c₁ = 10 and c₁ = 6 feet, and if one
weighs 30 and the other 40 pounds? Here we have
40
~₁ = (10 + 6)
(10 + 6). 48 (1 + 5)
1
hence the required velocities are
and
1
16.4.1
V1 C1 14,22 10
=
14,22
v2
C2 + 10,66
1
16.8
9
14,22 feet,
4,22 feet
FIG. 556.
A
B
N
E2
G₁ C1
E
6 + 10,66 — 4,66 feet.
§ 339. Oblique Impact. If the directions of motion S, C
and SC of the two bodies 1 and B, Fig. 556, diverge from the
normal N N to the tangent plane,
an oblique impact takes place. The
theory of oblique impact can be re-
ferred to that of direct impact by
G₂ decomposing the velocities S, C₁ = c;
and S, Cc, into their components.
in the direction of the normal and
IN tangent; the components in the di-
rection of the normal produce a
direct impact, and are, therefore, changed exactly as in the case of
direct impact, while the velocities parallel to the tangent plane
cause no impact, and, therefore, remain unchanged. If we combine
the normal velocity of any body, obtained according to the rules
for direct impact, with the tangential velocity, which has remained
unchanged, the resultant is the velocity of the body after the im-
pact. Putting the angles formed by the directions of motion with
the normal equal to a, and a,, or C, S] N = a, and C₂ S, N = α, we
obtain for the normal velocities S, E, and S. E, the values c, cos. a
and c, cos. ɑ, and, on the contrary, for the tangential velocities S₁ F₁
and S, F, the values c, sin. a, and c, sın. aɔ.
The normal velocities are changed by the collision, the first one
becoming

Me
V₁ = C₁ cos. a¡
(c₁ cos. α1
C2 cos. a
a)
(I + Nế
1
M₁ + M₂
and the second
M₁
V₂ = c2 cos. a₂ + (c, cos. a,
C₂ cos. α)
(1 + √μ),
M₁ + M₂
in which M, and M, denote the masses of the two bodies.
$ 40.]
683
THE THEORY OF IMPACT.
1
From , and c, sin. a, we obtain the velocity S, G, cf the first
body after the impact
2
201
=
Voi² + ci sin.“ aŋ,
and from v and c, sin. a. the velocity S. G. of the second body
W₂ =
W2
2
2
√v²² + c² sin² a2;
the angles formed by the directions of the velocities with the
normal are given by the formulas
tang. ß₁ =
C1 SIN. A1
でも
​cɔ sin. a₂
and tang. B.
V₂
B, denoting the angle G, S, N and 3, the angle G, S, N.
1
EXAMPLE-1) Two balls, weighing 30 and 50 pounds, strike each other
with the velocities c₁ = 20 and co 25 feet, whose directions form the
angles a
21° 35′ and a = 65° 20′ with the direction of the normal to
the tangent plane; in what direction and with what velocity will these
bodies move after the impact? The constant components are
c₁ sin. a.
C₂ sin. ɑ2
1
20. sin. 21° 35′ 7,357 feet and
25. sin. 65° 20′ = 22,719 feet,
and the variable ones are
C1 cos. α1
a
20. cos. 21° 35'
C₂ cos. αg
25. cos. 65° 20′ =
18,598 feet and
10,433 feet.
If the bodies are inelastic, we have
velocities after the impact are
21
Ve
50
0, and therefore the normal
18,598 — (18,598 — 10,433). 58 18,598 — 5,103
=
80
10,433 + 8,165. § 10,433 + 3,062
Hence the resulting velocities are
201
=
20%
√13,495 +7,357" =
v13,495² + 22,7192 =
—
13,495 feet and
13,495 feet.
√236.24 = 15,37 feet and
√698,27 = 26.42 feet;
and their directions are determined by the formulas
tang. B₁
1
tang. B₂
7,357
13,495'
22,719
13,495'
log. tang. B₁
0,73653 — 1, 31
28° 36′ and
log. tang. Be
=
0,22622, B₂
= 59° 17'
§ 340. Impact against an Infinitely Great Mass.-If the
mass A, Fig. 557, strikes against another mass, which is infinitely
great, or against an immovable object B B, or if c = 0 and
M₂ =∞, we have
>
V₁ = C₁ cos. a
c₁ cos. a₁ (1 + V´µ)
M₁ (1 + Vμ)
c₁ cos. a₁ vμ and
= 0 + 0 = 0,
∞
V₂ = 0 + C₁ cos. a¡
684
[$ 340.
GENERAL PRINCIPLES OF MECHANICS.
if in addition µ = 0, we have ₁ = 0, but if µ = 1, ?',
G
FIG. 557.
F
C
1
=
C₁ cos. α19
I E., when the impact is inelastic, the normal force is complete-
ly annihilated, but, on the contrary,
when it is perfectly elastic, the normal
force is changed into an equal opposite
one. The angle formed by the di-
rection of motion after the impact
with the normal is determined by the
equation

B
_N
N
_E
E
c₁ sin, a
B
tang. B₁
221
c₁ sin. a,
c₁ cos. a₁ Vμ
νμ
X
tang. a, V
μ
for inelastic bodies
tang. B₁
tang. as
0
=∞; I.E. B₁
90°;
and for elastic ones
tang. B₁
tang. a,, I.E. ß,
a1.
1
After an inelastic body has impinged upon an inelastic obstacle,
it moves on with the velocity c, sin. a, in the direction of the tan-
gent plane. When an elastic body has impinged upon an elastic
obstacle, it moves on with its velocity unchanged in the direction
S G, which lies in the same plane as the normal N N and the
original direction X S, and makes with the normal the same angle
G SN that the direction of motion before the impact made with
it. The angle X SN, formed by the direction of motion before
the impact with the normal or perpendicular, is called the angle of
incidence (Fr. angle d'incidence; Ger. Einfalls-winkel), and the angle
GSN, formed by the direction of motion after the impact with
the same, is called the angle of reflexion (Fr. angle de reflexion; Ger.
Austritts- or Reflexionswinkel); we can therefore assert that when
the impact is perfectly elastic, the angles of incidence and of reflexion
he in the same plane as the normal and are equal to cach other.
When the impact is imperfectly elastic, the ratio 4 of the
tangents of these angles is equal to the ratio of the velocity pro-
duced by the expansion to the velocity lost by the compression.
By the aid of this law we can easily find the direction in which
§ 341.]
685
THE THEORY OF IMPACT.
a body A, Fig. 558, must strike against an immovable obstacle
B B, when we wish it to take a given direction SI' after the im-
Y
-N
1
0
Y
FIG. 558.
R
B
B
X
N
imperfectly elastic, we must
is the required direction, for
pact. If the impact is elastic, we let fall
from a point Y of the given direction
a perpendicular Y O upon the normal
N N and prolong it until the pro-
longation, is equal to the per-
pendicular itself; S Y is then the
direction in question; for, accord-
ing to the construction, the angle
NSY₁ = NSY. If the impact is
make O F, V. OF; then Y, S

1
tang, a
tang. B₁
Ο Ι,
OY
If we let fall the perpendicular Y R upon the line SR parallel
to the tangent plane and make the prolongation RX=1
1
RY,
fl
SX will be, as we can easily see, the required direction of incidence.
REMARK. The principal application of the theory of oblique impact.
is to the game of billiards. See "Théorie Mathematique des effets du jeu de
billard, par Coriolis." According to Coriolis, when a billiard ball strikes
the cushion the ratio of the velocity of recoil to the velocity of impact is
0,5 to 0,6 or μ is 0,5² = 0,25 to 0,6ª 0,36. By the aid of these values
the direction, in which a ball A must strike the cushion B B when it is
to be thrown back towards a point Y, can be determined. We let fall
from the perpendicular YR to the line of gravity parallel to the cushion,
prolong the same a distance RX
1
11
10 to 10 of its length, and
draw the line ₁ X; the point of intersection D is the point towards which
the ball must be driven, when we wish it to rebound towards Y. The twist
of the ball causes this relation to vary somewhat.
§ 341. Friction of Impact.-When oblique impact occurs,
the pressure between the colliding bodies gives rise to friction, in
consequence of which the components in the direction of the tan-
gent plane are caused to vary. The friction F of impact is deter-
mined in the same way as that of pressure. If P denote the
pressure of impact and the coefficient of friction, then F P.
It differs from the friction of pressure in this only, that, like the
impact itself, it acts but for an instant. The changes in velocity
o
686
[§ 341.
GENERAL PRINCIPLES OF MECHANICS.
produced by it are not, however, immeasurably small; for the
pressure P during impact (and therefore the portion P of it) is
generally very great. Denoting the impinging mass by M and the
normal acceleration produced by the force of impact P by p, we
have
P = = Mp and F =
o Mp,
and also the retardation or negative acceleration of the friction
during the impact
F
= & P,
M
I.E. O times that of the normal force. Now the duration of the ac-
tion is the same for both forces; therefore the change of velocity pro-
duced by the friction is times the change of the normal velocity
produced by the impact.
If a mass M falls vertically upon a horizontal sled, and if the
velocity c of this mass is entirely lost by the collision, the retarda-
tion of the motion of the sled, whose mass is M₁, is
F
& Mp
M + M₁ M + M²
and consequently the loss of velocity is
وة
& M
M+ M₁
C.
Morin has proved the correctness of this formula by experiment
(see his Notions fondamentales de Mécanique).
If a body strikes an immovable mass B B at an angle a, Fig.
559, the change in the normal velocity is, according to the last
paragraph,
GI
1
FIG. 559.
F2:
B
N
E1
α
E
F
N
w = c cos. a (1 + õ);
hence the variation produced in the
tangential velocity is
$ w = $ c (1 + Vµ) cos. a.
After the impact the component csin.a
becomes

=
c sin. a
oc (1 + √μ) cos. a
фе
[sin. a cos. a (1 + √μ)] c;
$
for perfectly elastic bodies it is
=
(sin. a
2 & cos. a) c,
and, on the contrary, for inelastic bodies it is
(sin. a o cos. a) c.
§ 341.]
687
THE THEORY OF IMPACT.
The friction very often causes the bodies to turn around their
centres of gravity, or if, before the impact, a motion of rotation ex-
ists, it is changed. If the moment of inertia of a round body A in
reference to its centre of gravity S′ is = M k³, and if its radius S C
= a, we have the mass of the body reduced to the point of tan-
gency C
M k²
a²
and therefore the acceleration of the rotation produced by the fric-
tion Fis
P₁
F
Mk: a²
& Mp
M k²: a²
a²
φρ.
and the corresponding change of velocity is
a²
W1
Φ
·
W =
k2
a²
ha
(1 + Vµ) c cos, a.
a²
a²
For a cylinder
h²
=2, and for a sphere =, therefore, it fol-
h
lows that the changes of velocity of rotation of these round bodies,
produced by impact against a plane, are
w₁ = 2 ¢ (1 + √μ) cos. a and w =
§ (1 + Vμ) cos. a.
EXAMPLE.-If a billiard ball strikes the cushion with a velocity of 15
feet, in such a manner that the angle of incidence a = 45°, what will be
the conditions of motion after the impact? Putting for vu its mean value
0,55, we have the normal component of the velocity after the impact
√μ. c cos. a= 0,55. 15. cos. 45° — —
•
V
8,25.11
5,833 feet,
and assuming, with Coriolis, & = 0,20, we obtain the component of the ve
locity parallel to the cushion, which is
= c sin, a 9 (1 + õ) e cos, a = (1-0,20. 1,55). 10,607 = 0,69. 10,607
= 7,319 feet,
and consequently for the angle of reflection we have
tang. B
7,319
5,833
= 1,2548 or 3 = 51° 27′;
hence the velocity after impact is
5,833
9,360 feet.
cos. 51° 27/
The ball also acquires the velocity of rotation
688
[$ 342.
GENERAL PRINCIPLES OF MECHANICS.
§. 1,55. 10,607
=
8,220 feet
about its vertical line of gravity.
Since the ball does not slide, but rolls upon the billiard table, we must
assume that, besides its velocity e =- 15 feet of translation, it has an equal
velocity of rotation, and that this can also be resolved into the components
c cos, a 10,607 and c sin, a = 10,607.
The first component corresponds to a rotation about an axis parallel to the
axis of the cushion, and becomes
c cos. a
M
10,607 - 8,220
§ 4 (1 + √µ) c cos, a =
www
2,387 feet;
the other component e sin. a = 10,607 feet corresponds to a rotation about
au axis normal to the cushion and remains unchanged.
§ 342. Impact of Revolving Bodies.-If two bodies A and
B, Fig. 560, revolving around the fixed axes G and A, impinge upon
FIG. 560.
N
血
​H
P
E/B
P
masses reduced to the
to the line of impact
2
2
one another, changes of velocity take
place, which can be determined from
the moments of inertia M, k, and
My k of these bodies in reference to
their fixed axes by the aid of the
formulas found in the preceding para-
graphs. If the perpendiculars GH
and KL, let fall from the axes of ro-
tation upon the line of impact, be
denoted by a, and a,, we will have the
extremities II and L of these perpendicular
M₁ ki M. h.2

N
1
a,"
and
aa
substituting these values
for M, and M, in the formula for central impact, we obtain the vari-
ations of velocity of the points H and L (§ 338).
= ((1 − (2) M₁ k₁² a² + M¸ k¸² a¦*
M₂ k²² : α²²
2
C₁ = V₁ =
("
(2)
M₁ k: a + M₂ ha
2
(1 + √ µ)
:
M₂ k₂ aj
2
2
(1 + 4) and
212
· (1 − e)
M₁ k: a
(1 + 1 µe)
1
M₁ k‚² a²² + M, k; aj
(1 + √ μl),
the velocities of these points before the
= (~₁ = c₂)
in which e, and e, denote
impact.
»
M₁k": a + M. ka
2
M₁ br² az
2
To introduce the angular velocities, let us denote the angular
velocities before the impact by ε, and ɛ, and those after the impact
§ 342.]
689
THE THEORY OF IMPACT.
V1
by w, and w, thus we obtain c₁ = a₁ ɛ₁, c₂ = A» Ɛs, V'] = A₂ w, and
v₂ = ɑ, w₂, and the loss of velocity of the impinging body is
V2
E1
w₁ = α₁ (α¸ ɛ₁
and the gain in velocity of the impinged body is
Mak
M₁ k‚“ a²² + M, ki α,
2
2
(1 + √ μ),
Aɔ Ɛą)
2
M₁ k₁2
M₁ k₁² a² + M₂ k₂ a₁³
2
2
(1 + √ µ).
2 2
2
M₂ k₂2
2
2
M₁ k₁² a₂² + M₂k₂² a‚³
2
2
2
Wy Ɛ₂ = α¿ (α, ε1
The angular velocities after impact are
ω, = €1
M
- α₂ Ɛ2) (1 + √ µ)
and
M₁ k₂ 2
W₂ = Ɛ2 + A₂ (α; €1
α ₂ ε¸) (1 + √ µ)
M₁
1
ki a
2
2
k₁² α²² + M₂ k² aj
བྱ་
If both bodies are perfectly elastic, we
have μ = 1, or
1 + √ μ = 2,
and if they are inelastic, µ
0, or
1 + 1
μl = 1.
In the latter case the loss of vis viva occasioned by the impact is
=
2
M₁ k. M₂ ks
E. M₁ k₁² a₂² + M, k," a,**
ɑş Ɛ»)².
EXAMPLE. The moment of inertia of the shaft A G, Fig. 561, in refer-
FIG. 561.
R
ence to its axis of rota-
tion, Gis

2
M₁ k¸² 40000 : g,
1
and that of the tilt ham-
mer BK in reference to
its axis Kis
= 150000 g,
the arm G C of the shaft
is two feet and that K C
of the hammer is 6 feet, and the angular velocity of the shaft at the mo-
ment it impinges upon the hammer is 1,05 feet; how great is the velocity
after the impact and how much mechanical effect is lost by each blow, sup-
posing both bodies to be completely inelastic? The required angular
velocity of the shaft is
4. 1,05. 150000
40000. 36 + 150000 . 4
1
= 1,05
0,741 feet,
and that of the hammer is
60
= 105
105 (1 -
(1 − 204)
=
1,05. 0,706
2.6. 1,05. 4
also
ω1
204
G C
K C
0,741.
0,247 feet
44
690
[§ 343.
GENERAL PRINCIPLES OF MECHANICS.
I.E., three times as small as that of the shaft. The loss of mechanical effect
for each impact is
(2. 1,052)
A
2 g
40000. 150000
40000. 36 + 150000. 4
0,0155 (2,1)²
•
600000
144 + 60
0,0155. 4,41
150000
51
10253,25
51
FIG. 562.
=201,05 foot-pounds.
§ 343. Impact of an Oscillating Body.-If a body A,
Fig. 562, which has a motion of translation and is
unretained, impinges upon a body B CK, movable
around an axis K, we can find the velocities after
impact by substituting in the formulas of the pre-
ceding paragraph instead of a, ε, and a, w, the ve-
locities of translation c, and v, and instead of
N M, k
the mass M, of the first body; the other no-
tations remain unchanged. The velocity of the

B
ar
2
first mass after the impact is therefore
and the angular velocity of the second
1
2
M₂ ka
2
V₁ = c₁ — (c₁ — α₂ ɛ2) (1 + √ µ)
•
M₁ a₂+ M₂ k"
is
M
W₂ = E2 + A₂ (C1 − α₂ ɛ2) (1
+
√
µì) .
2
• M₁ a² + M₂ ka
2
If the mass M, is at rest, or if ɛ
=
0, we have
V₁ C1 c₂ (I' + √ µ)
M2
M₂ ka
2
M₁ a₂² + M₂ ka
and
M₁ ag
2
w₂ = c₁ (1 + √ μ) • M₁ a² + M₂ k₂²
1
ky
If M₁ is at rest, I.E., if the oscillating mass impinges upon it,
we have c₁ = 0, and hence
1
v₁ = α₂ εy (1 + √ µ¹) • M, a₂² + M₂ k₂"
M₂ k₂²
2
2
k²
2
and
W2
(1 − (1 + √ µ)
M₁ a₂
2
2
M₁ a² + My k
砂
​2
a, at which
The velocity, which is imparted to a mass at rest by another
by a blow, depends not only upon the velocity of the blow and the
masses of the bodies, but also upon the distance KL
the direction of the impact is situated from the axis
which is capable of rotation. If the free body impinges upon the
oscillating body, the angular velocity of the other becomes
of the body
§ 343.]
691
THE THEORY OF IMPACT.
M₁₂
c₁ (1 + √ μ)
2
M₁ a₂² + M₂, k,"
29
and if the oscillating body strikes against the free one, the latter
acquires the velocity
2
both velocities increase, therefore, when
as
Ma kaz
2
21
= ε₁₂ (1 + √ µe)
2
M₁ a₂+ M₂ k
2)
1
or
2
M₂ h²
2
M₁ as +
a²
h:2
increases, or M₁ a₂ +
M₂
decreases.
2
Cla
M₁ a² + M₂ k
Substituting for a, a± x, x being very small, we obtain for
the value of last expression
My h
2
M₂ ka²
だ​。
M₁ (a ± x) +
M₁ a ±. M₁₂ x +
a + x
or, since the powers of 2 are very small,
a
(1
1 +
F
Ma ka
= M₁ a +
a
± (1,
M₂ lis²)
≈ +
+ .
218
+-
aⓇ
F...),
Now if a is to correspond to the minimum value of M₁ a₂+
2
the member ± (M, M, ₂²)
+
2
a²
Mək
as
x must disappear; for its sign is
different, when a is increased a quantity (r) from what it becomes,
when a is decreased by a quantity (− x); hence we must have
(M₁ - Mak₂²)
FIG. 563.
λ = 0, I.E.,

N
Mo his²
ёг
M₁, and consequently
a
M₂ k²
M
M
ka
M
N
B
Now if one body strike against the other at
this distance (a), the latter assumes its maximum
velocity, which is
1) w₂ = (1 + √ √ π¹) ; 1;
μ) 1
/M
(1 + 1 µl)
2 h M
2 a
M
= (1 + √ π¹) ², α
a
M₁
2
when the oscillating body is impinged upon; and
2) v₁ = h, ε, (1 + √ μ) |
↓ µ)
when the free body receives the blow.
692
[§ 343.
GENERAL PRINCIPLES OF MECHANICS.
The extremity L of the distance or lever arm a =
M
K₂ V
M
2
which corresponds to the maximum velocity, or the point, where
the latter line intersects the line of impact, is sometimes, though
incorrectly, called the centre of percussion; a more correct term
would be the point of percussion.
,
We should be careful not to confound it with the centre of per-
cussion (§ 313), whose distance from the axis of rotation is ex-
pressed by the equation
2
α =
Maka
M₂ s
K: 2
S
in which s denotes the distance of the centre of gravity of the mass
M, from the axis of rotation. If the direction N N of the impact
of the masses M, and M, passes through the centre of percussion,
the reaction upon the axis of rotation becomes = 0.
In order, for example, to prevent a hammer from jarring, L.E.
reacting upon the hand, which holds it, or upon the axis, about
which it turns, it is necessary that the direction of the blow shall
pass through the centre of percussion.
If a suspended body A B is struck by a mass M, with force P
at the point of percussion, or at a distance a = k√ M,
axis K, the reaction upon the axis is
P₁ = P+R PK M₂ s (see § 313).
M₁
from the
=
K M₂ k₂²
ん​。
2
Since P
, we have the angular acceleration к =
a
Pa
M₂ k
2
and K M₂s =
₁ = P(1 –
P₁
My s a
M, k³
2
Ms a
P; hence the required reaction is
s a
M, k,) = P(1 − 84) = P (1
-
2
S
M
-
ky
M
EXAMPLE-1) The centre of percussion of a prismatical rod C A, Fig.
564, which revolves about one of its ends, is
at a distance
FIG. 564.

A
3
2
at the distance C O
α O = a =
½ r
= } CA
from the axis. Now if we grasp the rod at
one end and strike with the point 0, which is
CA, upon an obstacle, we will feel no recoil.
The point of percussion, on the contrary, is at a distance r
C, and if the mass of the body struck M₁
7'
√3
Me
from
3 M,
1
= M.
M2, we have this distance
0,5774 r. The rod CA must therefore strike a mass at rest at this
344.]
693
THE THEORY OF IMPACT.
distance from C, when we wish to communicate the greatest possible ve-
locity to the latter.
2) The distance of the centre of percussion 0 of a parallelopipedon
B D E, Fig. 565, from an axis XX, which is parallel to four of its sides
and is at a distance SA = s from the centre of gravity, and about which
the body rotates, is
-X
B
FIG. 565.
E
+ √ d²
a
8

d denoting the semi-diagonal CD of the sides,
through which the axis X X passes (§ 287). If the
force of impact passed through the point of per-
cussion, we would have
M.
M
2
a = k₂ V
√ (s² + { d²)
M₁
M₁
1
1
D
and the reaction upon the axis would be
P₁ = P(1 − 1)
= P(1 -
2
2
Co
M₂
2
1
+ z d² M 1
§ 344. Ballistic Pendulum.-The principles discussed in the
preceding paragraphs are applicable to the theory of the ballistic
pendulum of Robins (Fr. pendule ballistique; Ger. ballistische Pen-
del). It consists of a large mass M, Fig. 566, which is capable of
B
FIG. 566,
E
D
turning around a horizon-
tal axis C. It is set in os-
cillation by means of a
cannon-ball, which is shot
against it, and serves to
determine its velocity. In
order to render the im-·
pact as inelastic as possi-
ble, upon the side where
the ball strikes, a large
cavity is made, which from
time to time is filled with
fresh wood or clay, etc.
The ball remains, there-
fore, after every shot,
sticking in this mass, and
oscillates together with
the whole body. In order
to determine the velocity

694
[S 344.
GENERAL PRINCIPLES OF MECHANICS.
of the hall, it is necessary to know the angle of displacement; to
determine this angle, a graduated arc B D is placed under it, along
which a pointer, placed directly below the centre of gravity of the
pendulum, moves.
According to the foregoing paragraph, the angular velocity of
the ballistic pendulum, after the impact of the ball, is
W
1
M₁ α c₁
2
M₁ as² + M₂ k'
2
M, denoting the mass of the ball, M h the moment of inertia
of the pendulum, c, the velocity of the ball and a, the arm C G of
the impact or the distance of the line of impact N N from the axis
of rotation. If the distance CM of the centre of oscillation M of
the entire mass, including the ball, from the axis of rotation C', I.E.
if the length of the simple pendulum, oscillating isochronously
with the ballistic one, 7, and the angle of displacement ECD
=a, the height ascended by a pendulum oscillating isochronously
will be
r cos.a = r (1
cos. a)
h = C M — C H = r
hence the velocity at the lowest point of its path is
V
v = √2 g h = 2 √ gr sin.
and the corresponding angular velocity
V
W
g
21 sin.
•
a.
2
α
2'
equating these values of the angular velocity, we have
C1
=
2
M₁ a₂² + M₂ k¸² /g
M₁ az
sin.
2 1
α
2*
(sin.");
2 r (sin. 3)
Now, according to the theory of the simple pendulum,
moment of inertia M₁ a² + M₂ k₂²
r =
statical moment
(M + M.) 8
s denoting the distance CS of the centre of gravity from the axis
of rotation; hence it follows that
2
M₁ as² + M₂ ha² = (M₁ + M) s r and
= 2
(M₁
M₁ + M₁ $
M.)
V gr. sin.
a
2
M₁
If the pendulum makes n oscillations per minute, the duration
of an oscillation is
グ
​πν
TV" - 60"
g
J
and therefore g
60" . g
дро
N
N π
hence the required velocity of the ball is
§ 345.]
695
THE THEORY OF IMPACT.
M₁ + M½ 120 g s
M₁
η πα
α
sin.
2
EXAMPLE.—If a ballistic pendulum weighing 3000 pounds is set in os-
cillation by a 6-pound ball shot at it, and the angle of displacement is 15°,
if the distance s of the centre of gravity from the axis = 5 feet and the
distance of the direction of the shot from this axis is = 5,5 feet, and,
finally, if the number of oscillations per minute is n = 40, the velocity of
the ball, according to the above formula, is
3006 120. 32,2.5
C
sin. 749
•
6 40. 3,1416. 5,5
501.3864. sin. 7° 30′
44. 3,1416
= 1828 feet.
§ 345. Eccentric Impact.-Let us now examine a simple
case of eccentric impact, where the two masses are perfectly free.
If two bodies A and B E, Fig. 567, strike each other in such a
FIG. 567.
B
F
S
D
E
-N
manner that the direction N N of the impact
passes through the centre of gravity S₁ of one
body, and beyond the centre of gravity S of
the other, the impact will be central for one
body and eccentric for the other. The action
of this eccentric impact can be found accord-
ing to the theorem of § 281, if we assume, in
the first place, that the second body is free
and that the direction of impact passes
through its centre of gravity S, and, in the
second place, that this body is held fast at
the centre of gravity, and that the force of impact acts as a rotating
force. Now if e, is the initial velocity of A, c that of the centre of
gravity of B E, and if the two velocities become , and v, we have,
as in § 332, M₁ v₁ + M v = M₁ c₁ + M c. If, further, & is the
initial angular velocity of the body B E, in turning about the axis
passing through its centre of gravity perpendicular to the plane
N N S, and if, in consequence of the impact, this becomes w, de-
noting the moment of inertia of this body in reference to S by
Mk, and the eccentricity or the distance SA of the centre of
gravity S from the line of impact by s, we have

M₁ v₁ +
Mk²
s W = M₁ c₁ +
v₁ = +
Mk²
S E.
If the bodies are inelastic, both points of tangency have the
same velocity after impact, then vs w. Determining from
the foregoing equations v and w in terms of r,, and substituting the
values thus obtained in the last equation, we obtain
V₁ =
M₁ (c₁ — v₁)
M
+c+
M₁ 8² (c₁ — v₁)
Mk2
+ S &,
696
[$ 346.
GENERAL PRINCIPLES OF MECHANICS.
from which we determine the loss of velocity of the first body
C1 V₁ =
Mk² (
s ε)
(M₁ + M) k² + M₁ s²
the gain in velocity of translation of the second
رح
C
M₁ k² (c, C s ε)
(M₁ + M) k² + M₁ s²
and its gain in angular velocity
W
ε
1
- SE
ε)
M₁s (C1 с
(M₁ + M) k² + M₁ s²*
When the impact is a perfectly elastic one, these values are
doubled, and when it is imperfectly elastic, they are (1) times
as great.
EXAMPLE.-If an iron ball A, weighing 65 pounds, strikes with a ve-
locity of 36 feet the parallelopipedon B E, Fig. 567, which is at rest and
is made of spruce, if this body is 5 feet long, 3 feet wide and 2 feet thick,
and if the direction of impact N N is at a distance S K 13 feet from
the centre of gravity S, we obtain the following values for the velocities after
the impact. If the specific gravity of spruce is 0,45, the weight of the
parallelopipedon is = 5. 3. 2. 62,4 . 0,45 pounds 842,4 pounds. The
square of the semi-diagonal of side B D F parallel to the direction of the
impact is
whence (according to § 287),
and
(§)² + (3)² = 7,25,
k2 . 7,25
=
2,416,
g Mk
3
842,4 . 2,416 = 2035,2,
g (M₁ + M) k³ 907,4. 2,416 = 2192,3;
hence the velocity of the ball after the impact is
=
01
= 36
(1
Mk² c
1
(M₁ + M) k² + M₁ $2
2035.2
2391,4
= 36
(1
2035,2
2192,3 + 65 1,752,
36. 0,149 =
5,364 feet,
1
and that of the centre of gravity of the body struck is
M₁ k² c₁
1
(M₁ + M) k² + M₁ s²
1
157,08 .36
2,364 feet;
2391,4
1
Ms c C1
113,75 . 36
W
= 1,712 feet.
(M, + M ) k² + M₁ 82
2391,4
and finally the angular velocity is
1
§ 346. Uses of the Force of Impact. The weight of a body
is a force which depends upon its mass alone and increases uni-
formly with it; the force of impact, on the contrary, increases not.
only with the mass, but also with the velocity and with the hard-
ness of the colliding bodies (see § 336 and § 338), and it can therefore
be increased at will. Impact is consequently an excellent method of
§ 346.]
697
THE THEORY OF IMPACT.
obtaining great forces with small masses or weights, and it is very
often made use of for breaking or stamping rock, cutting or com-
pressing metals, driving nails, piles, etc. On the other hand, im-
pact occasions not only a loss of mechanical effect, but also causes
the different portions of the machine to wear rapidly or even to
break, and the durability of the structure or machine is seriously
affected by it. For this reason it is necessary to make the dimen-
sions of those parts of the machine larger than when the latter are
subjected to extension, compression, weight, etc., without impact.
FIG. 568.
If a rigid body A B, Fig. 568, strikes upon an unlimited mass
C D C of soft matter, it compresses the latter with a certain force,
whose mean value P is determined by means
of the depth of the impression K L = 8.
when we put the work done P s during
the compression equal to the energy of the
mass of the striking body. If M be the
A

P
C
K
C
B
D
mass, or G
g
M the weight, of this body
A B and v the velocity with which it strikes
upon CDC, we will have
212
{ M v² = G
2 g
and the required force with which the soft matter will be com-
pressed is
Р
==
M²
S
2,2
2g8 S
G.
Dividing this force by the cross-section of the body F, we obtain
the force with which each unit of surface of the soft material is
compressed and which such a unit can bear without giving way,
P
P F
2.2 G
2g F's
For safety we only load such a mass with a small portion of p,
for example with one-tenth part (?
(2%).
The body M acquires its velocity v by being allowed to fall freely
from a height h
2g
།
22
If we substitute this height, instead of
in the foregoing formula, we have
2 g
P =
G h
3
Gh
or for the unit of surface p
Fs
698
[§ 347.
GENERAL PRINCIPLES OF MECHANICS.
The force or resistance P, with which soft or loose granular
masses oppose the penetration of a rigid body A B, is generally
variable and increases with the depths of the penetration. In
many cases we can assume it to increase directly with 8, I.E., that
it is null at the beginning and double at the end what it is in the
middle. Now since the value of P, deduced from the above
formula, is the mean value, the resistance or proof load P, of soft
materials is twice as great as the value P obtained by the formula,
I.E.,
P₁ = 2 P
2 P =
2 Gh
S
EXAMPLE.-If a commander AB, Fig. 568, whose weight G 120 lbs.
falls upon a mass of earth from a height h = 4 feet, and if the latter is
compressed an inch by the last blow, a surface of this material equal to
the cross-section of the stamper will support a weight
P
Gh
8
120.4
=23040 pounds.
1
4 ड
4
Now if the cross-section F of the commander is square feet, the force
per square foot supported by this mass of earth would be
p
P
F
23040
1,25
18432 pounds;
instead of which, for the sake of safety, we should take but √5 P 1843,2
pounds.
FIG. 569.
10
§ 347. Pile-driving.—If we drive piles
such as 1 B, Fig. 569, into earth or any
other soft material (D C, we increase its
resistance much more than we would by
simply stamping it. Such piles (Fr. pieux ;
Ger. Pfäble) are from 10 to 30 feet long, 8
to 20 inches thick, and are provided with
an iron shoe B. The body M, the so-called
ram (Fr. mouton; Ger. Rammklotz, Ranım-
bär or Hoyer), which is allowed to fall from
3 to 30 feet upon the top of the pile, is gen-
erally made of cast iron, more rarely of oak,
and weighs from 5 to 20 hundred weights.
If the ram falls the vertical distance h, the
velocity with which it strikes the pile is
C = 12 gh,
and if its weight = G and that of the pile

$347.]
699
THE THEORY OF IMPACT.
= G,, we have, when we suppose that both bodies are inelastic, the
velocity of the same at the end of the impact (see § 332)
V =
G c
G + Gi
hence the corresponding height due to the velocity is
شرح
G
C c²
2 g
= (646) · 22 = (6 +6;)
G
h.
G+ G
G +
•
g
Now if the pile sink during the last blow a distance s, the re-
sistance of the earth and the load which the pile can support is
v2
P = (G+ G₁)
2gs
G2
h
G+ G
or more correctly, since the weight G+ G, of the pile and ram act
in opposition to the resistance of the earth,
72
h Ᏻ
G + G₁
P
s
+ (G + G₁).
In most cases G+ G, is so small, compared to P, that we can
neglect the latter part of the formula.
If the weight G, of the pile is much smaller than the weight G
of the ram, we can write
and simply
V
G c
G+ G₁
= C
h
G.
s
P
The foregoing theory suffices in practice, when the resistance P
is moderate and, consequently, the depths of the impression is
not very small; for in that case the compression of the pile, etc.,
can be neglected. If, on the contrary, the resistance P is very
great and, consequently, the depths of the impression very small,
the compression o of the pile can no longer be regarded as null, and
must therefore be introduced into the calculation.
The pile of course does not begin to sink until the force of
impact has become equal to the resistance P of the earth. Now
FE
if H =
and H
FE
7,
denote the hardness of the ram and
that of the pile (in the sense of § 336), the sum of the compressions
of the two bodies, when the force of impact is P, is
P P
+
H H
C
( + 1 ) ¹
H H
P.
and the mechanical effect expended in producing this com-
pression is
Ρο
L
2
(+)
P
2
700
[$ 347.
GENERAL PRINCIPLES OF MECHANICS.
Now if this first impact of the two bodies causes the velocity c of
G
the ram to become v, its mass M =
performs the work
g
M
2
L = ! M e² — ↓ M v² = (c² — v¹²) ¹² = (~ ~,, ″ ) & ;
—
G
2 g
hence we can put
1
1
P2
G
+
2 g
H
H 2
from which we obtain
v²
2g
2g
-6
1
+
H
H
:)
1) P
G
2 G
consequently the velocity of the ram, when the pile begins to pen-
etrate the earth, is
v =
C²
2
g
G
1
1 p"
H) 2 G
We infer from this that a pile (and also a bolt or nail in a wall)
will begin to enter the resisting obstacle when
G>
2 g
H II
1 P2
+
2'
or when the weight of the ram and its velocity have the proper re-
lation to the resistance of the earth. During the penetration of
the pile the force of impact and, consequently, the compression of
the pile, etc., diminish as long as the velocity of the ram exceeds.
that of the pile; when both attain a common velocity, and the
force of impact becomes a maximum, the bodies begin to expand
again. During this expansion not only the velocity of the ram,
but also that of the pile becomes gradually = 0; the pressure be-
tween the two bodies becomes again P, and consequently at the
moment, when the pile ceases to penetrate, the whole energy
G of the ram is consumed by the work
C²
29
G
1
+
II
:)
p²
2
expended in compressing, and by the work
Ps
done in driving the pile to the depth s.
Hence we have
c²
G
G h
2 g
( /1/1 + 1 )
II
)?
p
+ P 8,
2
§ 347.]
701
THE THEORY OF IMPACT.
and therefore the load which corresponds to the depth of penetra-
tion s is
P =
し
​H H₁
H+ H₁
-) (
2
H+ H
II II
c²
G + s²
2g
1
If the compression
+
11) 1/2
:)
P2
is considerably smaller than
the space s described by the pile, we can write simply
c G
Gh
P =
, or, more accurately,
2g s
S
Gh
P =
s +
G
1
Gh
+
28
II H₁ 2
Comparing the work done in driving in the pile
Gh
Ps
1 +
G
1
1 P
+
H H₁/ 2 s
with the work done Gh in raising the ram, we see that the former
approaches the latter more and more as
FE
smaller or as the hardness H:
ت)
1
1
+
H
H
:)
P
becomes
28
FE
}
of the ram and that I
of the pile become greater, I.E. the greater the cross-sections F and
Fand the moduli of elasticity E and E, of these bodies are and the
smaller the lengths are.
The action of the weights of the two bodies can be entirely neg-
lected, since they generally form but a small portion of the resist-
ance P. We can also neglect the energy, which the bodies possess
in consequence of their elasticity (although the latter is imperfect)
after the pile has come to rest; for the body, which is thrown back
by their expansion, is generally, upon falling again, incapable of
overcoming P and setting the pile in motion. For safety's sake,
the pile, which has been driven in, is loaded with only part of
the resistance P, just found, or perhaps with even less. According
to some late experiments made by Major John Sanders, U. S. A.,
at Fort Delaware (communicated by letter) we can put, approxi-
matively, the resistance
Gh
P =
3 s
1
Tō
EXAMPLE.-A pile, whose cross-section is 1 foot = 144 square inches,
whose length is 25 feet 25.12 300 inches, and whose weight is 1200
pounds, is driven by the last tally of ten blows of a ram, weighing 2000
702
[§ 348.
GENERAL PRINCIPLES OF MECHANICS.
pounds and falling 6 feet = 72 inches, 2 inches deeper, what is the resist-
ance of the earth? If we neglect the inconsiderable compression of the
cast iron and put (according to § 212) the modulus of elasticity of wood
1.560000, we obtain
E₁
1
1
300
1
= 0 +
2 F₁ E₁
1
2. 144. 1,560000
1
Now since Gh 2000.72
1
1497600*
+
H H
144000 inch-pounds and the depth of the pen-
2 =
etration after one blow is s =
tion of P the following equation :
P2
1497600
0,2 inches, we obtain for the determina-
+ 0,2 P = = 144000 or P² + 299520 P = 215654400000.
Resolving this equation, we obtain
P: 149760 + √238082457600 = 338177 pounds.
According to Sanders' formula
Gh
P =
3 s
144000
0,6
240000,
while the old formula, on the contrary, gives
P
G² h
G
Gh 2000 144000
.720000
3200 0,2
(G + G 2) 8 G+ G, 8
450000 pounds.
T
From P
338177 pounds we obtain
1
P2
( 2 + 1 )
76365 inch-pounds,
2
II
H
and therefore the height from which a ram weighing 2000 pounds must
fall in order to move the pile is
!
h =
1
H H
+
1 P3 76365
2 G 2000
=
38,2 inches.
§ 348. Absolute Strength of Impact.-By the aid of the
moduli of resilience and fragility (see § 206) we can
FIG. 570. easily calculate the conditions under which a prismatical
G
body A B, Fig. 570, will be stretched to the limit of elas-
ticity or broken by a blow in the direction of its axis. If
G be the weight and c the velocity of the impinging
body, the work done, when the prismatical body, whose
weight we will denote by G,, is struck, is

h
C³
L
•
2g
Gra
G + GR
or denoting the height due to the velocity
G₁
B
have more simply
L
G²
12 h
G + G i
c²
by h, we
2g
§ 348.]
703
THE THEORY OF IMPACT.
This mechanical effect is chiefly expended in stretching the rod.
A B, upon which the second body hangs; if, therefore, H is the
hardness, the length, F the cross-section, E the modulus of elasti-
city, P the force of impact and λ the extension of the rod produced
by it, we have
Πλ
P2
FE
L
} Hλ² =
2
12.
2
2 H
21
and consequently
FE
G² h
λ² =
λε
27
G + Gi
from which the extension 2 of the rod, caused by this impact, can
be easily calculated.
If the rod is to be extended only to the limit of elasticity, we
have, when A denotes the modulus of resilience (§ 206),
L= AV = A FI,
and therefore
A Fl=
G² h
G + G₁;
the velocity of impact c = 2 g h, which is necessary to stretch it
to the limit of elasticity, is determined by the height
G+ G₁
h =
GB
A Fl.
If we are required to find the conditions of rupture of the rod,
we must substitute, instead of the modulus of resilience 4, the
modulus of fragility B.
We see from this that the
greater is the blow it can bear.
greater the mass of the rod is, the
Hence we have the following im-
portant rule, that the mass of bodies subjected to impacts should
be made as great as possible.
1
Since G and G, fall the distance λ during the impact, it is more
correct to put
L
G² h
G + G₁
+ (G + G₁) 2,
or for the case, when the limit of elasticity is reached,
λ
in which
ī
of elasticity.
A F
G³ h
GT
G + G₂ l
+ (G + G₁) },
Τ
=σ expresses the extension corresponding to the limit.
If, finally, we wish to take into consideration the mass and
weight G₂ of the rod, we have, since its centre of gravity sinks but
112 2,0
2
704
GENERAL PRINCIPLES OF MECHANICS.
[§ 348.
A F =
G2
h
1
G + G₁ + G₂ •Ï + ( G + G₁ + { G₂) 0.
We have a similar instance of the action of impact, when a
G
moving mass M =
"
Fig. 571, puts another mass M₁ = in mo-
g
G₁
g
FIG. 571

C
tion by means of a chain or rope. If c is the velocity of M at the
moment, when the chain is stretched, v the velocity with which
both bodies move after the impact, we have again
Мс
V = M + M₁
G G
G + Gi
while, on the contrary, the work expended in stretching the chain is
1
L = √ Mc² — ↓ (M + M₁) v² =
! (M + M,) v² = (M
(μ- M+M) S
2
M M
M+M₁ 2
C²
G G₁
G + G₁
h.
•
If, therefore, this chain, etc., is to be stretched only to the limit
of elasticity, we must put
AFl=
G G₁
G + G₁
h,
F denoting the cross-section and I the length of the chain.
EXAMPLE-1) If two opposite suspension-rods of a chain bridge sup-
port a constant weight of 5000 pounds, which is increased 6000 pounds by
a passing wagon, if the modulus of resilience 4 of wrought iron is 7 inch-
pounds and if the length of the suspension-rods is 200 inches and their
cross-section 1,5 square inches, we have the dangerous height of fall
A FI (G + G₁)
h.=
G
7.2.1.5.200.11000
36000000
7.11
77
1,28 inches.
60
60
If the wagon passes over an obstacle 1,3 inches high, the suspension-rods
would already be in danger of being stretched beyond the limit of elas-
ticity.
2) If a full bucket or loaded cage in a shaft is not gradually set in mo-
ton, but if by means of the rope, which has been hanging loosely, it is sud-
denly brought to a certain velocity by the revolving drun, the rope will
often be stretched beyond the limit of elasticity, and sometimes even
$ 349.]
705
THE THEORY OF IMPACT.
broken. If the mass of the drum and shaft, reduced to the circumference
of the former, is M =
is G1
G
100000
"
the weight of the full bucket or cage
9
2000 pounds, and the weight of the rope = 400 pounds, then if
the weight of a cubic inch of rope is =
0,3 pounds, its volume will be
1
Fl=
G
2
400 4000
0,3 3
cubic inches,
and, finally, if the modulus of fragility of this rope is = 350 pounds, we
have the height due to the velocity, which will break the rope,
h = B Fl.
=238 inches 19,83 feet,
G+ G.
1 = 350.
G G ₁
4000 100000+ 2000 1400000
3 100000. 2000
102
3
200000
and, therefore, the velocity of the rope at the beginning of the strain is
C = √2gh = ✓ 64,4. 19,83 = 35,74 feet.
§ 349. Relative Strength of Impact.-The foregoing the-
ory is also applicable to the case of a prismatic body B B, Fig.
572, supported at both ends and exposed to the blows of a body
A, which falls from the height A C h upon its middle C. Let
Ꮐ
g
= M be the mass of the falling body and M₁ that of the body
B B, reduced to its middle C, then the energy of the bodies after
the impact is
c²
M²
c²
M
L
2
M + M₁
1
2 g⋅ M + M₁'
M 9
M
Gh.
M + M₁
FIG. 572.
A
B
B
D
FIG. 573.
B
N
C
B
0
D
The mass M, of the beam B B can be determined in the follow-
ing manner. Let & be the weight, half the length B D, Fig.
573, of this beam, a the abscissa B N and y the corresponding
ordinate NO of the curve, formed by BB at the moment of
greatest flexure, and, finally, let a denote the maximum deflection
CD of this curve. If we imagine B C to be divided into n (an infi-
nite number of) parts, the weight of an element of the rod will
•


45
706
[$ 349.
GENERAL PRINCIPLES OF MECHANICS.
be
ጎ
G₁
and therefore the mass of an element of the rod, reduced
from N to D, is
But, according to § 217,
N
&
G₁
n g
C D
(~9)=
G₁ y²
nga²
Px
72
Y
2 WE
(~ — 23 *°),
or
3
P² 1º
P²x²
y²
で ​- 3 l²x² +
4 W2 E2
9 and a² =
9 W2 E"
whence it follows that the element of the mass of the rod is
X
9 G₁ x² 14
3 ľ²x² +
9
4 ng
で
​7
2 3
n' n
12
Now if instead of a we substitute successively
n l
N
and add, etc., the values thus obtained, we obtain the mass of the
rod B B, reduced to its middle C,
M₁
9 G₁
4 q l®
12
で
​7+
•
3
74
5
27
G₁
17
35
g
If we substitute this value, we can put the work done by the
impact
L
=
M
M+ M₁
G² h
Gh
=
G+
17 Gi
35
and obtain the condition of bending to the limit of elasticity (see
§ 235),
WI
G² h
A.
3 e²
G + 1/1/2/1/13
11 G
35
If the beam is a parallelopipedon, we have
1 A V₁
=
G² h
17
35
G + } } G {
and therefore
A V₁ (G + 1
h
17
35
G₁)
G₁
9 G²
or putting V₁ =
Y
h
A G₁ (G + 3 Z
9 y G³
35
17 G)
If we substitute B instead of A, the expression becomes
h
B G₁ (G + { Z G₁)
9 Y G²
17
35
and gives the height, from which the weight G must fall in order
to break the parallelopipedical rod.
"
350.]
707
THE THEORY OF IMPACT.
EXAMPLE.—From what a height must an iron weight G 200 pounds
fall, in order to break by striking it in the middle a cast iron plate 36
inches long, 12 inches wide and 3 inches thick, which is supported at both
ends?
The modulus of fragility
B = 14,8 inch-pounds
(see § 211), and the volume of the plate is
V = b h l 12.3.36
1
1296 cubic inches,
and, since a cubic inch of cast iron weighs y = 0,259 pounds, its weight is
G₁ 1296. 0,259 = 335,7 pounds;
1
the required height is
h =
14,8. 335,7 (200 + 17. 356,4)
9. 0,275.40000
183 inches.
§ 350. Mechanical Effect of the Strength of Torsion.-
We can also investigate the action of impact in twisting shafts.
According to § 262 the mechanical effect which is required to pro-
duce a torsion a in a shaft, whose length is 7 and the measure of
whose moment of flexure is W, is
L
Ραα
2
a². W C
27
Pa² l
2 W C''
we can also put
S² WI
L
(see § 264),
2 C e²
e denoting the distance of the most remote fibre from the neutral
axis and S the strain in that fibre.
Τ
20
If we substitute for S the modulus of proof strength T, and for
στ
2
the modulus of resilience 4, we obtain the work to be
performed in stretching the remotest fibre to the limit of elasticity
Wi
L = A .
and the mechanical effect necessary to rupture the shaft by wrench-
ing, when we substitute for the modulus of resilience A the modu-
lus of fragility B; its value is
Πι
L₁ = B.
e²·
Пров
For a cylindrical shaft W =
and er, hence
2
A
L
π r² 1 =
2
A
2
I and L₁
B
2
B
π r² 1 =
2
when
r² 7 denotes the volume of this shaft.
For a shaft with a square cross-section, the length of whose side
is b, we have
708
[$ 350.
GENERAL PRINCIPLES OF MECHANICS.
b+
W =
and e = b √
6
ولا
A
A
L =
b² l =
3
3
and consequently
If a revolving wheel and axle, whose mass reduced to the point
B
V and L₁
V.
3
G
G₁
of impact is M =
impinges upon a mass M₁ = which is at
"
g
g
rest, with the velocity c, both will move on after the impact with
the velocity
Mc
V v = M + M₁
G c
G + G₁
(see § 334), and
G G
c²
consequently the mechanical effect
L = G+ G₁ 2 g'
which is expended in twisting the axle and bending the arms of the
wheel, is lost (see § 335).
But L is also the sum of the mechanical effects expended in
producing the torsion of the axle and in bending the arms of the
wheel, etc., I.E.,
L
= A .
W l
Ꭺ .
е
W₁
W, l
+ A₁
2
e₁
when A, denotes the modulus of resilience, W, the measure of mo-
ment of flexure and e, the distance of the exterior fibre from the
neutral axis (see § 235); we can therefore put
A WI
2
A₁ W₁₁
1
+
3 e²
G G₁
G+ G₁ 20
c²
W z
T
If the shaft is cylindrical, we have
and if it is four-
e²
2
WI
Ꮴ
sided, we have
3
,
when V denotes its volume; and for the
W₁l
2
3 ei 9
>
four-sided arm we have
of the arm.
V₁
1 where V, denotes the volume
Hence for a cylindrical shaft we have
A
2
14
A₁
√ + V₁
9
G G₁ C³
G + G₁ 2 g
A
A₁
G G₁ c²
1
Ꮴ +
V₁
1
3
9
G+ G₁ 2 g
and, on the contrary, for a four-sided shaft
§ 350.]
709
THE THEORY OF IMPACT.
1
The volumes V and V, have a certain relation to each other,
which can be expressed as follows. The moment of flexure of the
arms is equal to the moment of torsion of the shaft.
Hence
1)
WT
е
W₁ T
e₁
or
π cl³
b³ T
T=
16
3√2
b₁ h₁² T₁
6
}
T and T, denoting the moduli of proof strength for torsion and
bending and d the diameter of a round, and b the length of the
sides of a four-sided shaft, while h, is the thickness and b, the sum
of the widths of all the arms of the wheel.
But we have also V
π
ď²
4
1 =
= b² and V₁ ==
b₁ h₁₁, and
therefore
π ď² l A
8
+
b₁ h₁ hj Aj
9
G G₁
and
G + G、 2 g
2)
b, h, l, A,
G G₁
c²
ļ b² 7 A +
9
G + G₁ 2 g
b₁
Now if the ratio v =
of the dimensions is given, we can cal-
h₁
culate the thickness d or h of the shaft or the thickness h, and the
width b₁ of the arms by means of equations (1) and (2). We must
introduce into this calculation
1
1) for cast iron
T2
A
=
3,16 and A
2 C
19062
2. 2840000
0,640 inch-lbs.,
2) for wrought iron
T¹2
=
A, 6,23 and A =
20
59742
2. 9000000
1,983 inch-lbs.,
3) and for wood, the mean value
T¹³
Α
1
A₁ = 2,17 and A =
20
395*
2.590000
0,132 inch-lbs.
EXAMPLE.-Let the mass of the wheel, etc., of a tilt-hammer, reduced to
the point of application of the cam, be M-
200000
pounds, and the mass of
g
the hammer reduced to the same point be M
25000
pounds, and let the
g
710
[$ 350.
GENERAL PRINCIPLES OF MECHANICS.
distance from the wheel to the ring, in which the cams are set, be l = 15
feet 225 inches, and the length of the arms of the wheel be li 10 feet
120 inches. Now if the hammer, every time it is lifted, is struck with a
velocity of 2 feet, how thick must the shaft and the arms of the wheel be
made in order to sustain this impact without being damaged? If the shaft
and arms are of wood, we have
395
πα
16
1000
1
b₁ h₁²
6
2
and if the number of arms is n = 16, we can put
1
b₁ =v.nh
whence we obtain
0,707. 16 h,
=
11,3. h₁,
3
16000. 11,3
2,9. h₁.
6.395 π
·
But
d = hi
П
F100
A l = 0,132 . 225
π
11,66,
8
† 4, 7, §. 2,17. 120 28,9,
1
and also
G G
C?
G+ G₁
2g
12. 0,0155.4.
200000. 25000
200000 + 25000
= 16533 inch-pounds ;
hence we have the equation of condition
5000000
0,744.
225
(2,9). 11,66 h₁² + 11,3. 28,9 ¸² 16533, I.E.,
1
98,1 h¸² + 326,6 h¸¡
2
1
1
hence the required thickness of the arm
hi
2
16533,
the width of the arm
h₁
✓
16533
6,24 inches
424,7
1
b₁ = 0,707 h1
4,41 inches,
d = 2,9 h₁
=
18,1 inches.
and the thickness of the shaft
1
For the sake of security we make the dimensions considerably larger.
REMARK.-It is only of late years that much attention has been paid to
the strength of impact. We find something in regard to it in Tredgold's
work on the strength of cast iron, in Poncelet's "Introduction à la
Mécanique Industrielle," and in Rühlmann's “Grundzüge der Mechanik
und Geostatik." The discussion in the latter work is based principally
upon Hodgkinson's experiments on the resistance of prismatic bodies to
impact, upon which subject an article by Bornemann is to be found in the
"Zeitschrift für das gesammte Ingenieurwesen" (the Ingenieur).
The experiments of Hodgkinson agree essentially with the foregoing
theory of the strength of impact; they apply particularly to relative
3 350.1
711
THE THEORY OF IMPACT.
1
strength, and were made in the following manner: large weights swinging
like pendulums were caused to strike against rods supported at both
G2 h
ends. The formula L =
which we found by assuming that the
G + 1 G ₁ '
impact was perfectly inelastic, was verified completely; the mechanical
effect L was found not to depend upon the nature of the colliding bodies.
Equally heavy bodies of different materials (cast iron, cast steel, bell metal,
lead) produced, when they fell from the same height, equal deflections of
the same rod (of cast iron or cast steel); the deflections were almost ex-
actly the same as those given by the theory for a perfectly elastic rod.
FINAL REMARK.-For the study of the Mechanics of rigid bodies, be-
sides the older works of Euler, Poisson, Poinsot, Poncelet, Navier and
Coriolis, and those of Whewell, Mosely, Eytelwein and Gerstner, the follow-
ing are recommended :
Duhamel, Cours de Mécanique, Paris, chez Mallet-Bachelier, 1854;
Sohnke, Analytische Theorie der Statik und Dynamik, Halle, 1854; Broch's
Lehrbuch der Mechanik, Berlin, 1854; Morin, Leçons de Mécanique pra-
tique, Delaunay, Traité de Mécanique rationelle, Paris, 1856; Rankine, A
Manual of Applied Mechanics, second edition, London, 1861-a valuable
work, too little prized in England. A translation of a new Monograph
upon impact, by Poinsot, has lately appeared in the third year of Schlö-
mich's Zeitschrift für Mathematik und Physik.
SIXTH SECTION.
STATICS OF FLUIDS.
CHAPTER I.
OF THE EQUILIBRIUM AND PRESSURE OF WATER IN VESSELS
§ 351. Fluids.-We consider fluids to be bodies composed of
material points, whose coherence is so slight that the smallest force
suffices to separate them from each other (§ 62). Many bodies
which are met with in nature, such as air, water, etc., possess
this distinguishing property of fluids in an eminent degree, while
others, on the contrary, such as oil, tallow, softened clay, etc., pos-
sess a less degree of fluidity. The former are called perfectly, and
the latter imperfectly fluid, or viscous bodies. Certain bodies, as,
E.G., dough, lie midway between the solids and the fluids.
Perfectly fluid bodies, of which only we will treat in the discus-
sion which is to follow, are at the same time perfectly elastic, I.E.
they can be compressed by extraneous forces, and when these forces
are removed, they reassume the primitive volume. But the amount
of change of volume corresponding to a certain pressure is very dif-
ferent for different fluids; while in liquids this change is quite un-
important, in gaseous or aeriform fluids it is very great, and they
are therefore called clastic or compressible fluids. On account of
the slight degree of compressibility of liquids, they are treated in
most of the researches in hydrostatics (§ 66) as incompressible or
inelastic fluids. As water is the most generally diffused of all
liquids and is the most generally employed in practical life, we
regard it as the representative of all these fluids, and in the re-
searches in the mechanics of liquids we speak only of water, with
$ 352.]
EQUILIBRIUM AND PRESSURE OF WATER, ETC. 713
the tacit understanding that the mechanical relations of other
liquids are the same.
For the same reason, in the mechanics of elastic fluids we speak
only of common atmospheric air.
REMARK.-A column of water, whose cross-section is one square inch,
is compressed by a weight of 14,7 pounds, corresponding to the weight of
the atmosphere, about 0,00005 or one fifty millionth of its volume, while a
column of air under the same pressure occupies but one-half of its primi-
tive volume. See Aimé "Ueber die Zusammendrückung der Flüssigkeiten"
in Poggendorff's Annalen, Ergänzungsband (to Vol. 72), 1848. According
FE (§ 204), we have, when P = 14,7 pounds, F
to the formula P =
1 square inch and
تر
ī
1
the modulus of elasticity of water
2
5
ī 100000 20000'
E=
Pl
F2
—
14,7. 20000 = 294000 pounds.
$352. Principle of Equal Pressure.-The characteristic
property of fluids, by which they are principally distinguished from
solid bodies and which forms the basis of the theory of the equili-
brium of fluids, is the capacity of transmitting the pressure exerted
upon a portion of their surface unchanged in all directions. In solid
bodies the pressure is transmitted only in its own direction (§ 86);
if, on the contrary, water is subjected to pressure on one side, the
same pressure is exerted throughout all the mass of fluid and can
consequently be observed at all parts of the surface. In order to
convince ourselves of the correctness of this law, we can employ
F
P
FIG. 574.
P
B
CG
D
P
an apparatus filled with water, like
the one whose horizontal cross-sec-
tion is represented in Fig. 574. The
tubes A E, B F, etc., which are of
the same size and at the same dis-
tance above the base, are closed by
pistons, which are easily movable
and which fit the tubes perfectly;
the water will then press upon each
of them, by virtue of its weight, ex-
actly as much as upon the others.
Let us for the present disregard
this pressure and regard the water as imponderable. If we exert
against one of the pistons A a certain pressure P, the water will

714
LS 352.
GENERAL PRINCIPLES OF MECHANICS.
transmit the same pressure to the other pistons B, C, D, and to
preserve the equilibrium or to prevent these pistons from moving
backwards, an equal opposite pressure P (Fig. 575) must be exerted
against each of the other pistons. We are therefore authorized to
assume that the pressure P exerted upon a portion 4 of the surface
produces a strain which is propagated not only in the straight
line AC, but also in every other direction B F, D H, etc., upon any
equally large portions C, B, D of the surface.
FIG. 575.
P
FIG. 576.


E
D
B
E
B C
D.
P P P
t
P
If the axes of the pipes B F, C G, etc., Fig. 576, are parallel to
each other, the forces acting on the pistons can be combined so as
to give a single resultant; if n is the number of the equally large
pistons, the total pressure upon them will be
P₁ = n P;
in the case represented in the figure
P₁ = 3 P.
1
Now the aggregate arca F, of the surfaces B, C, D, upon which
the pressures are exerted, is also = n times the area F of one
of the pistons; n is therefore not only =
P₁
p'
F
but also
or in
general
P₁
F
F
and P₁ =
P.
p
F
F
Now if we cause the tubes B, C, D to approach each other,
until they form, as in Fig. 577, a single one, and if we close the
latter by a single piston, F, becomes a single surface and P, is the
pressure exerted upon it; hence we have the general law the
pressures cxorted by a fluid upon the different parts of the walls of
the vessel are proportional to the areas of those parts.
:
€ 353.]
EQUILIBRIUM AND PRESSURE OF WATER, ETC. 715
This law corresponds also to the principle of virtual veloci-
ties. If the piston A D = F, Fig. 578, moves a distance 1 A
inwards, it presses
FIG. 577.
P
や
​FIG. 578.
B B.
A A,
PE
D D₁
1
E E/
P
1
S
the prism of water
F's out of its tube,
and the piston BE
= F moves out-
wards the distance
BB, s, and leaves
behind it the pris-
matical space F, 81.

=
1
Now as we have
assumed that water can be neither expanded nor compressed, its
volume must remain the same after the pistons have been moved,
or the increase Fs must be equal to the decrease F₁ s₁. But the
equation F₁s, Fs gives
F
S
F
S
and by combining this proportion with the proportion
we obtain
P₁
P
S
$1
hence the mechanical effect P₁ s₁ = Ps (see § 83).
P₁
F
P
F
EXAMPLE. If the diameter of the piston A D is 13 inches and that of
the piston B E is 10 inches, and if the pressure exerted by the former upon
the water is 36 pounds, that exerted upon the latter piston is
1
P₁ = P
पुरा
F
103
1,52
36 =
400
9
•
36 = 1600 pounds.
If the first piston moves 6 inches, the second moves but
81
F
F
1
9.6
400
22070
=
0,135 inches.
REMARK.—In the following pages we will meet with many applications
of this law, E.G., to the hydraulic press, water pressure engines, pumps, etc.
§ 353. Pressure in the Water. The pressure exerted by
FIG. 579.
E
B
P
D
H
the particles of water against each other
must be estimated in exactly the same
manner as the pressure of the water against
the wall of the vessel. The pressures upon
both sides of any surface E C G, which di-
vides the water in a vessel B G H. Fig. 579,
into two parts, when equilibrium exists,
are equal. Now as a rigid body counter-

716
[$ 353.
GENERAL PRINCIPLES OF MECHANICS.
FIG. 580.
E
P
H
acts all forces whose directions are at right angles to its surface,
the conditions of equilibrium will not be disturbed, when one-half
E G H of the liquid becomes rigid, or if its limiting surface
becomes a wall of the vessel. If the fluid
half E B G in one portion CD = F₁ of
the imaginary surface of separation E CG
exerts a pressure P, upon the rigid half
EG II, the latter counteracts this pres-
sure completely and will react with an equal
opposite pressure (— P₁) upon C' D = F₁.
Since the conditions of equilibrium will
not be changed, when this mass of water
EGH becomes fluid again, the latter will react with an equal
pressure (— P) upon the mass of water E B G; hence the pres-
sure of the water upon both sides of a surface C D F is also de-
termined by the proportion

1
-P
B
1
·P
D
P₁
P
F
יF
when all the water is pressed in a surface A B = F by a force P.
Hence the pressure upon any given surface F in any arbitrary
position is
FIG. 581.
F
P₁
P.
F
The law of the transmission of pressure in water, expressed by
the last proportion, is only applicable when we consider water as
an imponderable fluid, and it must therefore be modified, when it
is required to determine in addition the pressure arising from the
weight of the water. If we imagine a part of the water in a vessel
CD E, Fig. 581, to become rigid and to have the form of an infi-
nitely thin horizontal prism 4A B,
it is easy to see, that the pressures
of the water, that remains fluid,
upon the sides of the rigid part
balances the weight G of the prism
and that the horizontal pressures
upon the vertical bases A and B
of this part counteract each other.
These pressures (P and P) must
therefore be equal and opposite to

G
D
K
G
L
E
each other. Since the state of equilibrium is not changed, when
A B again becomes fluid, it follows that the pressures of the
$ 353.]
EQUILIBRIUM AND PRESSURE OF WATER, ETC. 717
water against the vertical elements A and B of the surface, which
are situated in one and the same horizontal plane, must be equal
to each other, and since the pressure upon an element does not
change, when its inclination or direction changes, it follows that
the water in a horizontal layer, as, E.G., G H, K L, etc., exerts the
same pressure in all directions and in all positions.
FIG: 582.
If we imagine a vertical prism A B, whose cross-section is infi-
nitely small, to become rigid in the mass of water CH K, Fig. 582,
we can conclude from the conditions of its equi-
librium with the remaining liquid that the
pressures exerted by the latter upon the vertical
sides of the prism balance each other and that
the weight G of the latter body is in equilibrium
with the excess P₁ P of the pressure P, upon
lower base B above the pressure P upon the
upper base 4. Hence P₁
P G, I.E. the

C
D
P
G
H
B
K
AP
L
1
1
pressure P, of the water upon any elementary
surface B is equal to its pressure P upon an ele-
ment 1, of equal size and situated above it, plus
the weight G of a column of water A B, whose
base is one or other elementary surface and
whose height is the vertical distance between the
two elements. According to what precedes this
rule is not only applicable to two elements,
situated vertically above one another, but can also be employed
for determining the pressure upon the walls of the vessel; for the
two pressures P and P, are transmitted unchanged in the hori-
zontal planes G H and K L. Hence the pressure Rupon an ele-
mentary surface B, K or L of the horizontal plane AL is equal to
the pressure P upon an equally great element 1, G or H in a
higher horizontal plane plus the weight of the column of water.
whose base is this element F and whose height is the distance
A B =
h of the horizontal planes G H and K L from one another.
If y is the heaviness of water, this weight is
G = Fhy, and therefore P,₁ = P + G = P + Fhy.
1
If the areas of the elements of surface are unequal; if, E.G., the
area of the upper one (in GH) is F and that of the lower one
(in KL) is F, the pressure upon the latter is
P₁ = }} (P + Fh y) =
F
F
P+Fhy.
F
F
By means of the same formula the pressure P upon an element
718
[§ 354.
GENERAL PRINCIPLES OF MECHANICS.
F in the horizontal plane G H can be determined, when the exterior
pressure P, upon an element of the surface ( D = F, which is at
a distance above or below G II, is known.
P
It is
F
F
P. ± Fhy.
Since the pressures upon equal elements in a horizontal plane
are equal to each other, it follows that the foregoing formula is
applicable to horizontal surfaces of finite dimension, as, E.G., where
the water serves to transmit the force P,
which acts upon a horizontal piston F,
Fig. 583, to another horizontal piston F
This formula
Р
FIG. 583.

A
T
ок
B
D
P
P₁ = 1/1
F
F
F
P + Eh y = E (
+17)
F
gives directly the pressure P, upon this
surface, when I denotes the vertical height.
CD between the surfaces of the two pistons.
P
P₁
and
F
If we denote the pressures F
upon the units of surface by p and p,, we
have more simply
Pr p + h y
EXAMPLE.--If the diameters of the two pistons F and F of a hydrosta-
tic press AC B, Fig. 583, are d
-
1
21 inches and d₁ = 9 inches, and if they
are situated at the distance CD h 60 inches above one another, and
if the larger piston is to exert a pressure R 1600 pounds, we have the
force which must be applied to the smaller piston
F
P
P₁ - Fhy
F
(1)² R
\
d²
π
4
hy
(
π
1600
25
18
4
60.62,5
1728
123,46 - 10,66
=
112,8 pounds.
§ 354. Surface of Water.-In consequence of the action of
gravity upon water, all the elements of it tend to descend, and
really do so when they are not prevented. In order to keep a quan-
tity of water together, it is necessary to confine it in a vessel. The
water in a vessel A B C Fig. 584, can only be in equilibrium when
the free surface HR is at right angles to the direction of gravity, or
§ 354.] EQUILIBRIUM AND PRESSURE OF WATER, ETC. 719
horizontal; for so long as this surface is curved or inclined to the
horizon there will be elements of the water, such as E, which, be-
ing situated above the others, will, in consequence of their great
A
FIG. 584.
H---
G
mobility and their weight, slide down those
below them as upon an inclined plane. Since,
when the distances are very great, the direc-
Rtions of gravity cannot be considered as paral-
lel lines, the free surface or the surface of the
water in a very large vessel, E.G. in a large sea,
will not, under these circumstances, form a
plane surface, but a portion of the surface of a
sphere. If another force acts, in addition to gravity, upon the ele-
ments of the water, then, when equilibrium exists, the free surface
of the water is at right angles to the resultant of this force and that
of gravity.
If a vessel A B C, Fig. 585, is moved forward with the constant
acceleration p, the free surface of the water forms an inclined plane

A
FIG. 585.
D
P
H
F
P
RG
B
D F; for in this case every element E
of this surface is drawn vertically down-
wards by its weight G and in a horizon-
P
tal direction by its inertia P = G, the
g
two forces giving rise to a resultant R,
whose direction forms, with that of
gravity, a constant angle R E G = a.
This angle is at the same time the angle D F H formed by the
surface of the water (which is at right angles to the resultant) with
the horizon. It is determined by the equation

P
p
tang. a = G
g
FIG. 586.
A
X
C
If, on the contrary, a vessel A B C, Fig. 586, is caused to re-
volve uniformly about its vertical axis II,
the surface of the revolving water forms a
hollow 4 O C, whose cross-section through
the axis is a parabola. If o is the angular
velocity of the vessel and of the water in it,
G the weight of an element E of the water,
and y its distance ME from the vertical axis,
we have the centrifugal force of this ele-
ment

-X B
G
R
720
[$ 354.
GENERAL PRINCIPLES OF MECHANICS.
G Y
g
F = w² (§ 302),
and therefore for the angle REGTEM, formed by the
resultant with the vertical or by the tangent E T to the profile of
the water with the horizontal line M E,
tang.
F
w² y
G
g
From this formula we see that the tangent of the angle, formed
by the tangent line with the ordinate, is proportional to the ordi-
nate. Since this is one of the properties of the common parabola
(see § 157), the vertical cross-section A O C of the surface of the
water is a parabola, whose axis coincides with the axis of rotation
X X.
If the velocity of rotation of the water in the vessel A B D, Fig
c² G
588, were constant and = c, we would have F
and there
gy
D
FIG. 597.
A
X
C

FIG. 589.
F
0

ΟΙ
R
G
K
-X B
T
FIG. 588.

B
A
H
D
F
R
B
G
R
fore tang. =
; hence the subtangent of the curve, formed by
gy
C²
the cross-section A E B of the water, MT = m
or constant.
g
According to Article 20 of the Introduction to the Calculus, the
equation of such a curve is
$355.] EQUILIBRIUM AND PRESSURE OF WATER, ETC. 721
y = r e = r @ ‹‹²,
r denoting the ordinate of the beginning 1.
If we cause a vessel A B H, Fig. 589, to move uniformly in a
vertical circle around a horizontal axis (, the surface of the water
will assume a cylindrical form, with a circular cross-section D E H.
If we prolong the direction of the resultant R of the weight G and
of the centrifugal force F of an element E until it cuts the vertical
line CA, passing through the centre of rotation, we obtain the two
similar triangles E CO and E F R, for which we have
CO FR
EU
G
;
EF F
but if we put the radius of gyration E C y and retain the last
wo Gy whence it follows that the line
w² GY
notations, we have F =
g
g
30
00
g
W3
π U
2936
W U²
feet =
894,6
U²
meters,
u denoting the number of revolutions per minute. Since this value
of CO is the same for all the elements of the water, it follows that
the resultants of all the elements of the water forming the cross-
section D E II are directed towards O, and that the cross-section,
which is at right-angles to all these directions, is the are of a circle
described from 0. Hence the surfaces of the water in the buckets
of an overshot water-wheel are always cylindrical ones, described
from the same horizontal axis.
H
H
A
FIG. 590.
B
R.
R1
R2
§ 355. Pressure upon the Bottom.-The pressure in a
vessel A B C D, Fig. 590, is a minimum immediately below the
surface, increases with the depth, and is
a maximum at the bottom. This, al-
though a consequence of § 353, can also
be proved as follows. Let us suppose
that the area of the surface II, R, of the
water is F, and that a pressure P, is ex-
erted uniformly upon, E.G. by the at-
mosphere lying above it or by a piston,
and let us imagine the entire mass of
water to be divided by very many hori-
zontal planes, such as H, R₁, HI, R, etc., into equally thick layers.
If F, is the area of the first layer H, R₁, λ its thickness, and y the
heaviness of water, we have the weight of the first layer G, F, λy,
and that portion of the pressure in H, R, produced by the pressure

♡
K
D
C
R 3
46
722
[$ 3:5
GENERAL PRINCIPLES OF MECHANICS.
P. upon the surface of the water II, R, according to the principles
enunciated in § 352, is
P. F
F
Adding both these pressures, we obtain the pressure in the horizon-
tal section II, R₁
P. F,
P₁
+ Ελγ.
F
Dividing by F₁, we obtain the equation
F
P₁
Р
P.
+ λγ,
Fo
P.
0
or, since
and denote the pressures p, and p, in II, R. and
F
P
F
1
H₁ R, referred to the unit of surface, we have
î
P1 Po + λ y.
2
The pressure in the following horizontal layer H. R, is deter-
mined exactly in the same manner as the pressure in the layer
H₁ R₁, but we must not forget that the initial pressure upon an
element of the surface is in this case p, Py, while in the
first case it was Po. Hence the pressure in the horizontal layer
H₂ R is
A
P₂ = p₁ + λ y = Po + λ y + λ y = p₁ + 2 2 y,
FIG. 591.
B
in like manner the pressure in the third
layer H R is

= 2₂+32);
Ro
H
40-
R1
in the fourth
R2
= Po + 4 λ y',
Rs
H
and in the nth
3
K
D
C
we can therefore put the
the nth horizontal layer
= 1 + ηλγ.
But n 2 is the depth O K = h of this
nth layer below the surface of the water;
pressure upon each unit of surface in
p = Po + hy (compare § 353).
We call the depth h of one element of surface below the level
of the water its head or height of water (Fr. charge d'eau; Ger.
Druckhöhe), and we find the pressure of the water upon any unit
of surface by adding to the pressure applied from without the
weight of a column of water, whose base is unity and whose height
is the head of water. When a surface is horizontal, as E.G. the
bottom CD (Fig. 591), the head of water h is the same for all
positions, and if its area is F, the pressure of the water upon it is
€355.
EQUILIBRIUM AND PRESSURE OF WATER, ETC. 723
P = (p。 + h y) F = Fp. + Fhy = P, + Fhy,
or, if we neglect the external pressure, P
FhY. The pressure
of the water upon a horizontal surface is therefore equal to the weight
of the column of water Fh above it.
This pressure of the water upon a horizontal surface, E.G. upon
the horizontal bottom or upon a horizontal portion of the wall of a
vessel, is entirely independent of the form of the vessel; whether
the vessel 4 C, Fig. 592, is prismatic as in a, or wider above than
below as in b, or wider below than above as in e, or inclined as in
d, or with spherical walls as in e, etc., the pressure upon the bottom
is always equal to the weight of a column of water, whose base is
the bottom of the vessel and whose height is its depth below the
level of the water. Since the pressure of water is transmitted in
all directions, this law is also applicable when the surface, as E.G.
B C, in Fig. 593, is pressed from below upwards. Each unit of
surface of the layer of water BK, touching B C, is subjected to the
pressure of a column of water, whose height 18 H BRK = 1,
and the pressure against the surface C B is = Fhy, F denoting
the area of that surface.
FIG. 592.
A
B
A
a
D
C
D
C
A
B
D
D
d
FIG. 593.

A
H
R
B
K

FIG. 594.
D
A
B
A B
B
E
A
C
D
C
-H
R
('
D
E

Hence it follows that the water in the communicating tubes
ABC and DE F, Fig. 594, will stand at the same height, when
in equilibrium, or that the surfaces 1 B and E F will be in the
same horizontal plane. In order to preserve the equilibrium it is
necessary that the layer of water HR shall be pressed downwards
by the column of water E R above it as much as it is pressed up-
wards by the mass of water below it. Since in both cases the
724
[§ 356.
GENERAL PRINCIPLES OF MECHANICS.
surface pressed upon is the same, the head of water must be the
same, and the level of the water at 4 B must be at the same height
above HR as that at E F
§ 356. Lateral Pressure. The formula just found for the
pressure of water against a horizontal surface, is not directly appli-
cable to a plane surface inclined to the horizon; for in this case the
head of water is different at different points.
The pressure p = hy upon every unit of surface within the
horizontal layer at the depth h below the surface of the water acts
in all directions (§ 352), and, consequently, at right angles to the
walls of the vessel, by which (§ 138) it is entirely counteracted.
Now if F, is the area of an element of the side A B C, Fig. 595,
and h, its head of water F, H, we have the pressure perpendicular
to it
FIG. 595.
A
·R.
B
Y
S
P₁
H1
M
F
P₁ = F₁.h₁y;
1
if F is the surface of a second ele-
ment and h, its head of water, we have
the normal pressure upon it

P₁ = F₂ b₂ Y;
and in like manner for a third ele-
ment
P = F3 h3 y, etc.
These normal pressures form a
system of parallel forces, whose result-
ant P is the sum of these pressures,
I.E.,
1
P = (F,h,+ F₂ h₂ + ...) y.
Fh +...)y.
But F, h₁ + F, h₂ + ... is the sum of the statical moments of
F, F, etc., in reference to the surface A O B of the water and
Fh, when F' denotes the area of the whole surface and h the
depth SO of its centre of gravity S below the surface of the water;
hence the entire normal pressure against the plane surface is
PFhy.
If we understand by the head of water of a surface the depth of
its centre of gravity below the surface of the water, the following
rule will be generally applicable, viz.: the pressure of water against
a plane surface is equal to the weight of a column of water, whose
base is the surface and whose height is its head of water.
We must here observe that this pressure does not depend upon
the quantity of water above or in front of the surface pressed, thus,
$357.]
725
EQUILIBRIUM AND PRESSURE OF WATER, ETC.
E.G., if the other circumstances are the same, a wall A B C D, Fig.
596, has to resist the same pressure whether it dams up the water of a
small trough A CEF or that of a large dam A C G H or that of a
lake. From the width A B C D = b_and_the height A D
BC a of the rectangular wall we obtain the surface of the same
Fab and the head of
FIG. 596.
a
H
F
A
water SO
and, there-

2'
G
E
B
fore, the pressure of the wa-
ter against it is
a
P = a b y = { a² by
2
The pressure increases
therefore with the width and with the square of the height of the
surface pressed upon.
EXAMPLE.—If the water in front of a sluice gate, made of oak, 4 feet
wide, 5 feet high and 23 inches thick, stands 33 feet high, how great a force
is required to lift it?
The volume of this gate
4.5. 5 25 cubic feet.
21
Assuming the heaviness of oak, saturated with water, to be according to
§ 61, 62,5 . 1,11 = 69,375 pounds, the weight of this gate is
G 25.69,375 25. 11,5625 289,06 pounds.
The pressure of the water against the gate and the pressure of the lat-
ter against its guides is
-
Р ¿ (3)². 4. 62,5 49. 31,25 = 1531,25 pounds;
putting the coefficient of friction for wet wood (§ 174) &
the friction of the gate upon its guides
F
P = 0,68. 1531,25 = 1041,25.
0,68, we have
Adding to the latter the weight of the gate, we have the force necessary to
draw it up
1041,25 + 289,06 = 1330,31 pounds.
357. Centre of Pressure of Water.-The resultant P
Fhy of all the elementary pressures Fly, F. ha y, etc., has, like
the resultant of any other system of parallel forces, a definite point
of application, which is called the centre of pressure. By retain-
ing or supporting this point the whole pressure of the water
upon a surface will be held in equilibrium. The statical moments
of the elementary pressures F, h, y, F, họ y, etc., in reference to the
plane of the surface of the water A B 0, Fig. 595, are
F, h, y. h₁ = F, hy, F. h. y. h₂ = F. ho y, etc..
and the statical moment of the entire pressure of the water in
reference to this plane is
726
[$ 357.
GENERAL PRINCIPLES OF MECHANICS.
2
(F, h₂² + F, h₂² + . . .) Y.
Denoting the distance K M of the centre M of this pressure from
the surface of the water by z, we have the moment of the pressure
of the water
P z = (F₁ h₁ + F₂ h₂ + . . .) z Y,
+ Fk+) 2,
and by putting these moments equal to each other we obtain the
distance of this centre M below the surface of the water
1) z =
F₁ h‚² + F₂ h₂² +
F₁ h₁ + F₂ hy +
2
2
or =
F₁ h₁² + F, h²² +
Fh
2
when, as above, F denotes the area of the entire surface and h the
depth of the centre of gravity below the surface of the water.
In order to determine completely this point of pressure, we
must find its distance from another line or plane. If we put the
distances F, G, F G, etc., of the elements F, F, etc., of the sur-
face from the line A C, which determines the angle of inclination
of the plane,
= Y1, Y2, etc., we have the
moments of the elementary pressures
in reference to this line
A
E
IB
B
FIG. 597.
Y

N
O
S
K
M
H₁
F
= F, h, y₁ Y, F₂ h₂ ya Y, etc.,
and the moment of the entire surface
2 2
= (Eh, y₁ + F₂ h₂ Y + ...) Y';
denoting the distance M N of this
centre M from that line by ', we
have also this moment
= (Fh, + F, hq + ...) v y.
Equating these two moments, we
obtain the second ordinate
P₁
X
F₁ h₁ y ₁ +
F ₂ hy
No Yo
Y +
2) v =
or
F₁ h₁ + F₂ h₂ +
F₁ h₁ y₁ + F₂ h₂ Y 2 + ···
Fh
If a denote the angle of inclination of the plane A B C to the
horizon, x1, x2, etc., the distances E, F, E, F, etc., of the elements
F, F, etc., and u is the distance L M of the centre of pressure M
from the line of intersection A B of the plane with the surface of
the water, we have h₁ = x, sin. a, x, sin. a, etc., and also z = U
sin. a; substituting these values of z and v in the expression, we
obtain
F₁ x² +
2
+2
F₂ x²² + ...
Moment of inertia
U
F₁ x₁ +
F₂ Xq + ...
!
V =
1
F₁ x₁ Y₁ +
x
F₁ X₁ +
2
x2
F₂ X 2 Y 2 +
2
F ₂ X q +...
Statical moment
and
Moment of the centrifugal force
Statical moment
§ 358.]
EQUILIBRIUM AND PRESSURE OF WATER, ETC. 727
We find then the distances u and v of the centre of pressure
from the horizontal axis and from the axis AX, formed by
the line of dip, when we divide by the statical moment of the sur-
face with reference to the first axis, in the first place, the moment of
inertia in reference to the same axis, and, in the second, the mo-
ment of the centrifugal force of the same in reference to both axes.
The first distance is also that of the centre of oscillation from the
line of intersection with the surface of the water. Besides it is
easy to perceive that the centre of pressure of water coincides per-
fectly with the centre of percussion, determined in § 313, when the
line of intersection of the surface with the surface of the
water is regarded as the axis of rotation.
§ 358. Pressure of Water against Rectangles and Tri-
angles.—If the surface pressed upon is a rectangle A C, Fig. 598,
with a horizontal base line CD, the centre M of pressure is found
in the line of dip KL, which bisects the base line, and it is at a
distance equal to two-thirds of this line from the side A B, which
lies in the surface of water. If the rectangle, as in Fig. 599, does
FIG. 598.
FIG. 599.
FIG. 600.


OK
A
K
B

H
K
K
¡M
D
L
C
SM
A
B
M
"
D
B
not reach the surface of the water, then, if the distance KL of the
lower line CD from the surface of the water, and that K O
of the upper one A B, l2, we have the distance KM of the cen-
tre of pressure from the surface of the water
૧૨
3
1,³ — 72 ³
3
The distance AM of the centre of pressure M of a right-angle
triangle A B C, Fig. 600, whose base A B lies in the surface of the
water, from A B (Example § 313) is
¦ F. P
U
F.l
117,
when 7 denotes the altitude B C of the triangle.
The distance of this point M from the other side B C is, since
this point lies in the line CO, which bisects the triangle and
728
[§ 358.
GENERAL PRINCIPLES OF MECHANICS.
runs from the apex C' to the middle of the base, N M
b denoting the base A B.
1 b,
If the apex C is situated at the surface of the water, as in Fig.
601, and if the base A B is below the apex, we have
K M = u
1 F 12
37 and
Fl
b
NM v =
=
3 b.
2
If the whole triangle A B C, Fig. 602, is immersed in the water,
FIG. 602.
FIG. 601.
K
D
H
R
A
N
A
B
C
18 F (4
− 1)² + F (1s
W =
F (b + 4 = 4)
3
and the base A B is at a dis-
tance A H = l, and the apex
Cat the distance C H 7,
from surface H R, we deter-
mine the distance MK of the
centre of pressure M below the
surface of the water HR by
means of the formula


+
تن
1
18
T'S (4
-
9
1½)² + § (2 l + %₁)²
2
7₂² + 2 71 72 + 3 l½
1
3 ( 2 72 + 1 )
2 (1₁ +242)
The centre of pressure of other plane figures can be determined
in the same manner.
EXAMPLE.—What force P must we employ to raise a circular clack-
FIG. 603.
P
R
B
A
F = π p²
πα
4
valve A B, Fig. 603, which is movable
about a horizontal axis D? Let the

length of this valve be
=
14 feet, its di-
ameter A B be = 14 feet, and the distance
of its centre of gravity S from the axis D
be D S = 0,75 feet, and its weight be
G = 35 pounds; further, let the distance
DH of the axis of rotation D from the
surface of the water, measured in the
plane of the valve, be 1 foot and the
angle of inclination of this plane to the
horizon be a = 68°.
The surface upon which the pressure
is exerted is
25
0,7854. = 1,2272 square feet,
16
$ 359.]
729
EQUILIBRIUM AND PRESSURE OF WATER, ETC.
and the head of water or depth of its centre C below the water level is
O C = h h = H C. sin, a = (HD + D C ) sin. a = (H D + DB + B C') sin. a
= (1 + 0,25 + 0,625) sin. 68° = 1,875 . 0,9272 — 1,7385 feet,
and, therefore, the pressure of the water upon the surface A B = Fis
Q = F h
Fhy
1,2272. 1,7385. 62,5 = 133,34;
the arm b of this force with reference to the axis of rotation D is the dis-
tance D M of the centre of pressure M from it, hence
b = HM
HD.
1
1,875 +
4. 1,875
But we have
5
H M = HC +
4 H C
whence
LO
2
= 1,9271 feet,
b = 1,9271 1,0000 = 0,9271 feet,
and the required statical moment of the pressure is
Q b
b = 133,34. 0,9271 123,62 foot-pounds.
The arm of the weight of the valve is
DK DS cos, a = 0,75 . cos. 68° = 0,75 . 0,3746
and therefore its statical moment is
35 . 0,2810 = 9,84 foot-pounds.
0,2810 feet,
By adding these moments, we obtain the entire moment necessary to
open the valve
Pa = 123,62 + 9,84
=
133,46 foot-pounds.
Now if the arm of the force, which opens the valve, is D N = a = 0,75
feet, the intensity of that force must be
P =
133,46
0,75
177,95 pounds.
§ 359. Pressure upon Both Sides of a Surface.—If a plane
surface A B, Fig. 604, is subjected upon both sides to the pressure
FIG. 604.
A
Si
B
of water, the two resultants of the pres-
sures on the two sides give rise to a
new resultant, which, as they act in
opposite directions, is obtained by sub-
tracting one from the other.
If F is the area of the portion A B
subjected to pressure on one side of the
surface, and h the depth A S of its
centre of gravity below the surface of

1
1
the water, and if F, is the area of the portion A, B, on the other
side, which is subjected to the pressure of the water, and h, the
depth A, S of its centre of gravity below the corresponding surface
of the water, the required resultant will be
PFhy Fhy (Fh-F, h) y.
If the moment of inertia of the first portion of the surface with
reference to the line, in which the plane of the surface cuts the first
730
I'S 359.
GENERAL PRINCIPLES OF MECHANICS.
surface of the water, Fh, we have the statical moment of the
pressure of the water upon one side
2
= Fly,
and if the moment of inertia of the second portion of the surface,
with reference to its line of intersection with the other surface of
the water,
Fh, we will have in like manner the statical mo-
ment of the pressure of the water on the other side, with reference
to the axis in the second surface of the water,
2
F₁ ki² Y.
Putting the difference of level A A, of the two surfaces of the
water = a, we have the increase of the latter moment, when we pass
from the axis A, to the axis A,
1
= F₁h, ay,
and consequently the statical moment of the pressure F, h, y, in
reference to the axis A in the first surface of water, is
2
2
= F, ki y + F, h₁. a. y = (F, k + F, a h₁) y.
1
Hence it follows that the statical moment of the difference of
the two resultants is
2
= (Fk² — F, ka F, h) Y,
=
and the arm of this difference or the distance of the centre of
pressure from the axis in the first surface of water is
U W =
2
F k² — F, k²² — a F, h₁
Fh-Fh
If the portions of surface which are subjected to pressure are
equal, as is represented in Fig. 605, where the whole surface A B
= F is submerged, we have more simply
PF (hh₁) Y,
Fig. 605.

H
R
and since k
k²
2
k₁² + 2 a h₁ + a' (see § 224)
R₁ and h
1 =
a, we have
2
a h₁
a h₁ + a²
U
h
1
a
B
h
h₁ + a = h.
1
In the latter case the pressure is equal to
the weight of a column of water, whose base
is the surface pressed upon and whose height is the difference of
level R H of the water on the two sides of the surfaces, and the centre
of pressure coincides with the centre of gravity S of the surface.
This law is also correct when the two surfaces of water are subjected to
equal pressure, E.G. by means of pistons or by the atmosphere; for if
the pressure upon each unit of surface = p and the height of the
Ρ
2 (§ 355), we must substi-
corresponding column of water is 7=
Y
$360.] EQUILIBRIUM AND PRESSURE OF WATER, ETC. 731
tute, instead of h, h +1, and instead of h, h₁ + 1; by subtraction.
we obtain the pressure
P = (h + 1 − [h₁ + 1]) Fy = (h — h₁) Fy.
l
For this reason we generally neglect the pressure of the air in
hydrostatical experiments.
EXAMPLE. The depth 4 B of the water in the head-bay, Fig. 606, is
7 feet, the water in the chamber of the lock rises 4 feet upon the gate, and
the width of the canal and lock-chamber is
7,5 feet; what is the resulting pressure upon
the gate of the lock?
FIG. 606.
A

S
B
Here
F = 7. 7,5 = 52,5 square feet,
F₁ = 4 . 7,5 30,0 square feet,
1
7
7= h₁
4
2 feet,
2' 1 2
h
a = 7
4 = 3 feet,
1
49
1
16
72 =
and ki
42 =
•
3
1
3
3
;
hence the required resultant is
P = (F h − F₁ h₂) y = (52,5 . ·
1
123,75.62,5 =
7734,4 pounds,
and the depth of the point of application below the surface of the water is
30.
2). 62,5
52,5
•
49
3
16
30.
- 3.60
3
гл
7
52,5.
60
2
517,5
123,75
4,182 feet.
§ 360. Pressure in a Given Direction.-In many cases we
wish to know but one part of the pressure, viz.: that exerted in a
certain direction. In order to find such a component, we decom-
pose the normal pressure M P P on the surface A B F, Fig.
607, into two components, one in the given direction X and one
at right angles to it, viz.:
FIG. 607.
Τ
-21
•
P
X
P..
M P
P, and M P.
Now if a is the angle P MX
formed by the direction of the normal
pressure with the given direction I
of the component, the components
will be
P₁ P cos, a and P
If we project the surface

P sin. a.
upon
a plane perpendicular to the given di-
rection MX, we have the area of the projection B C
732
[$ 360
GENERAL PRINCIPLES OF MECHANICS.
F₁
F. cos. A B C
or, since the angle of inclination A B C of the surface to its pro-
jection is equal to the angle P M X = a, formed by the direction
of the normal pressure and that of its component P₁,
FF cos. a, and inversely
F
cos. a =
F
F
1
P₁ = P .
F
the required component is therefore.
P₁ = F₁hy,
Y,
=
Now, since the value of the normal pressure is P Fhy, we
have
I.E., the pressure exerted by water in any direction upon a surface is
equal to the weight of a column of water, whose base is the projection
of the surface at right angles to the given direction and whose height
is the depth of the centre of gravity of the surface below the surface
of the water.
In most cases in practice we are only required to determine
the vertical or a horizontal component of the pressure of the
water against the surface. Since the projection at right angles to
the vertical direction is the horizontal projection and that at right
angles to a horizontal direction is a vertical one, we find the ver-
tical pressure of the water against a surface by treating its hori-
zontal projection as the surface subjected to pressure, and, on the
contrary, the horizontal pressure of the water in any direction by
treating the vertical projection, or elevation, of the surface at right
angles to the given direction as the surface pressed upon, and in
both cases we must regard the depth OS of the centre of gravity
S of the surface below the surface of the water as the head of water.
Hence, if we wish to determine in the case of a prismatical em-
bankment or dam A B D E, Fig. 608, the horizontal pressure of
FIG. 608.
M
v¥__
G
E
the water, we must con-
sider the longitudinal
elevation AC, and if
the vertical pressure is
to be determined, the ho-
rizontal projection BC
of the surface A B must
be considered as the sur-
face pressed upon. Put-
ting the length of the

$ 360.]
EQUILIBRIUM AND PRESSURE OF WATER, ETC. 733
dam = 1, its height A Ch and horizontal projection of the slope
B C' = a, we have the horizontal pressure of the water
h
H=lh.
γ y = ½ h² ly
thly
2
and its vertical pressure
h
V = al. y =
alhy.
2
Now if the width of the top of this dam is A E = b, the hori-
zontal projection of the other slope D F = a, and the heaviness of
the material of the dam = y₁, the weight of the dam is
G = (b
+
2
a + a ₁ ) k l y v
h Y
and the entire vertical pressure
V+G= ↓ alhy + (b
V+Galh +
a + a
2
of the dam upon its horizontal base is
hly₁ = [şa y + (b + ª + α) v. ] n l.
Y
Putting the coefficient of friction
force necessary to push the dam forward
F = 4 (V + G) = [¹ a y + (b +
a a₁
2
o, we have the friction or
a
り
​+ a ¹) v₁ ] ¢ h l .
l.
2
When the horizontal pressure pushes the embankment forward,
we must have
ya
swly = [¿va +
h²
or more simply
a
(b + " + ").]. 11,
2
Y:
h = 4 (a + (2 b + a + a) 2).
If we wish to prevent the dam from being moved, we must make
h < • (a +
(2 b + a + a,) 2¹²),
b > 4 [( / /
-
− a) 2/2 − (a + a.)]
or
γ
''1
For the sake of greater security we assume that the water has
penetrated below the base of the dam to a great extent, and for
this reason, in the worst case, we must consider that an opposite
pressure = (b + a + a₁) ? hy is acting from below upwards;
hence we must put
h < 4 [ (2 b + a + a, ) ( ~ 1 − 1) – a.].
Y
− –
EXAMPLE.-If the density of the clay composing the dam is nearly
double that of water, or
Y1
= 2 and
Y1
- 1 = 1,
Y
734
[§ 361.
GENERAL PRINCIPLES OF MECHANICS.
we can write simply
h < ¢ (2 b + a).
It has been found by experiment that a dam resists sufficiently, when
its height, top and the horizontal projections of its slopes are equal to
each other. Hence, if we substitute in the last formula
h
Ъ
ɑ, we obtain
for which reason in other cases we must put
h
1
k = [(@28 + a + a) (2) - 1)-a].
+ b
and for clay dams in particular
h = { (2 b + a), or inversely
b
3 h α
2
If the height of the dam is 20 feet and the angle of inclination of the
slope is a = 36º, the horizontal projection is
a
h cot. a = 20 . cotg. 36 20. 1,3764
and therefore the width of the top of the dam must be
b =
60 — 27,53
2
=
16,24 feet.
= 27,53 feet,
§ 361. Pressure upon Curved Surfaces.-The law of the
pressure of water in a given direction, deduced in the foregoing
paragraph, is applicable only to plane surfaces or to a single cle-
ment of a curved surface, but not to curved surfaces in general.
The normal pressures upon the different elements of curved sur-
faces can be decomposed into components parallel to a given direc-
tion and into others perpendicular to the first. The first set of
parallel components forms a system of parallel forces, whose result-
ant gives the pressure in the given direction, and the other set of
components can also be combined so as to form a single resultant,
but the two resultants are not capable of further combination,
unless their directions intersect each other (§ 97). Hence we are
generally unable to combine all the pressures upon the elements.
of curved surfaces so as to form a single resultant; there are, how-
ever, cases where it is possible.
If G, G, G, etc., are the projections and ha, ha, ha, etc., the
heads of water of the elements F, F, F, etc., of a curved surface,
the pressure of the water in the direction perpendicular to the
plane of projection is
P
P = ...),
(G₁ hq + G₂ họ + G3 h3 +
and its moment in reference to the plane of the surface of the water is
P u = (G₁ h₂² + G₂ h₂² + G ; lis² + · · ·) )'.
2
2
3
h
2
•
If we can decompose the curved surface subjected to the pressure
§ 361.]
EQUILIBRIUM AND PRESSURE OF WATER, ETC. 735
G₁
Go G3
F
F'
into elements, which have a constant ratio to their projections,
I.E., if we can put
etc. = n, we will have
F
F
G₁
G₂
etc., and therefore
N
忆
​P
(F
F₁ h₁
F, ha
+
+
+...)
F₁ h₁ + F₂ h₂ +
Fh
•
y =
I'
N
N
N
22.
F denoting the area and h the depth of the centre of gravity of the
entire surface below the level of the water. But we have
F = F₁ + F₂ + ... = n G₁ + n G₂ + ... = n (G₁ + G₂ + ...) = n G,
G denoting the area of the projection of the entire surface; hence
P =
Fh
N
Y =
G h Y,
as in the case of a plane surface, or the pressure of water in one
direction is equal to the weight of a prism of water, whose base is
the projection of the curved surface upon a plane perpendicular to
the given direction and whose height is the depth of the centre of
gravity of the curved surface below the surface of the water.
FIG. 609.
C
II
Thus, E.G., the vertical pressure against the side of a conical
vessel ACB, Fig. 609, which is filled with water, is equal to the
weight of a column of water, whose base is the
base of the cone and whose height is two-thirds
the length of the axis CM; for the horizontal
projections of the surface of a right cone, as
well as the surface itself, can be decomposed
into elementary triangles, and the centre of
gravity S of the surface of the cone is at a dis-
tance from the apex of the cone equal to two-
thirds of its height (§ 116). If r is the radius
h
of the base and h the height of the cone, we
have the pressure upon the basey and
the vertical pressure upon the sides gπrhy; now as the base
and the side are united together and the pressures are in opposite
directions, it follows that the force with which the entire vessel is
pressed downwards is

K
B
= (1 — ¦) π r²² hy = j = r² hy
3)
= the weight of the entire mass of water. If we cut the base loose
from the conical portion of the vessel it will exert a pressure upon
its support hy, and to prevent the side of the vessel from
being raised by the water we would have to exert a pressure upon
3 π r² hy.
it
736
[§ 362.
GENERAL PRINCIPLES OF MECHANICS.
FIG. 610.
H
R
REMARK.-The pressure exerted by the steam of a steam-engine or the
water of a water-pressure engine is perfectly independent of the shape of
the piston. No matter how much we may increase the
surface pressed upon by hollowing out or rounding the
piston, the force, with which the water or steam moves
the piston, remains the same and is equal to the product
of the cross-section or horizontal projection of the piston
and the pressure upon the unit of surface. If the piston
A B, Fig. 610, is funnel-shaped and if its greater radius
is CA
=r and its smaller G D G E = "₁,
= C B
pressure upon the base is
upon the conical surface is
resulting pressure is

Α
B
Р
P
D
E
G
7.
1
the
π r² p and the reaction
=
=
π (p2
1
π (r² — r₁²)p; hence the
P = ñ r² p − ñ (r² — r¸²) p = π r¸² p
2
πρ
1
1
= the cross-section of the cylinder multiplied by the pressure upon the
unit of surface.
§ 362. Horizontal and Vertical Pressure.-Whatever may
be the form of a curved surface A B, Fig. 611, the horizontal pres-
sure of the water against it is always equal to the weight of a
A1
P
B
FIG. 611.
11
R
TBn= B
column of water, whose base is
the vertical projection A, B, of
the surface at right angles to the
given direction and whose height
is the depth OS of the centre
of gravity S of this projection
below the surface of the water.
The correctness of this assump
P tion is shown directly by the
formula

1
P = (G₁h₂+ G₂ h + ...) Y,
when we remember that the heads of water h, h, etc., of the ele-
ments of the surface are also the heads of water of their projections
or that G₁ h₁ + G₂ h¸ + ... is the statical moment of the entire
projection, I.E., the product & h of the vertical projection & multi-
plied by the depth h of its centre of gravity below the surface of
the water. Hence we must again put
P = Ghy
and remember that h is the head of water of the vertical projection.
The vertical section, by which we divide a vessel and the water
contained in it into two equal or unequal parts, is at the same time
the vertical projection of both parts, the horizontal pressure upon
one part of the vessel is proportional to its vertical projection
§ 362.]
EQUILIBRIUM AND PRESSURE OF WATER, ETC. 737
multiplied by the depth of its centre of gravity below the surface
of the water; consequently the horizontal pressure upon one portion
A B of the wall of the vessel is exactly as great as the horizontal
pressure upon the opposite portion A, B,, which acts in the opposite
direction, and the two pressures balance each other. The vessel
will therefore be subjected to equal pressure in all directions by the
water contained in it.
A,
Q
FIG. 612.
R
The vertical pressure P, G₁ h, y of the water against an ele、
ment F, Fig. 612, of the wall of the vessel is, since the horizontal
projection G, of the element can be regarded as the cross-section,
and the head of water h, as the height, or G, h₁
as the volume, of a prism, equal to the weight
of a column of water H F, extending above.
the element to the plane H R of the surface
of the water. Hence the elementary surfaces,
which form a finite portion A B of the bot-
tom or wall of the vessel, support a pressure
which is equal to the weight of the columns
of water above them, I.E. to the weight of the
column of water above the entire portion.
Putting its volume equal to V, we obtain
the vertical pressure of the water

A
P
P
B
Bo
B
P = F₁ Y•
1
The vertical pressure upon another portion
A, B, of the wall of the vessel, which lies
vertically above the former and which limits the volume 4, B₁ H =
V, is
Q
= I;
1
but if the two portions are rigidly connected together, the result-
ant of the two forces, which acts vertically downwards, will be
R = (P − Q) = (V,
F₂) Y Vy
= to the weight of the column of water contained between the two
surfaces. If we apply this rule to the entire vessel, it follows that
the entire vertical pressure of the water against the vessel is equal to
the weight of the water contained in it.
If we make an opening O in the side of the vessel H B R. Fig.
613, I and II, that portion of the pressure, which corresponds to
the cross-section of this opening, is wanting and the pressure upon
the surface Fopposite to it remains unbalanced. If the opening
(as in I) is closed by a stopper K, which is prevented from yielding
by a resisting object L on the outside, an equal distribution of the
47
738
[§ 363.
GENERAL PRINCIPLES OF MECHANICS.
horizontal pressure upon the walls of the vessel no longer takes
place, but, on the contrary, the vessel is moved forward with a
force P = Fhy, which is counteracted on the opposite side by
FIG. 613.
I
H
h
II

R
H
R
Th
P
F
-प्र
B
B
the stopper. If the stopper is removed and the water allowed to
flow through the opening O, as in II, the reaction of the discharg-
ing water increases this pressure P from Fhy to P₁ = 2 F hy,.
as will be shown hereafter.
FIG. 614.
HOS R
1
EXAMPLE.—The vertical pressure P₁ upon the lower hemispherical sur-
face A D B, Fig. 614, is equal to the weight of a column of water bounded
above by the surface of the water H R and below
by the hemispherical surface. If r is the radius C A
= CD of this surface and h the height CO of the
surface of the water above the horizontal plane A B,
which limits it, the volume of the hemisphere A BD
will be V,
³, and that of the cylinder above
A B, V2 π² h; hence

A
F
D
B
1
P₁ = (V₁ + V₂) y = }} = r² + = r² h) y = (k + f r) π r² y.
The pressure, which is directed vertically upwards
upon the upper hemisphere A E B, is, on the contrary,
P₂ = (V₂ − V₁) y = (h — } r) π p² 7',
and therefore the entire vertical pressure
P = P₁
P₁ − P₂ 2 V₂ = 4 π προγ
2
3
is equal to the weight of water in the entire sphere.
The horizontal pressure upon one of the hemi-
spheres D A E and D B E, which join each other in
the vertical plane D C E, is measured by the weight of a prism, whose
base is D C E = 1 and whose height is CO = h; this pressure is
π
R = π p² hу.
$ 363. Thickness of Pipes.-The application of the theory
of the pressure of water to the determination of the thickness of
pipes, boilers, etc., is of great importance. In order that these
vessels shall sufficiently resist the pressure of the water and not be
§ 363.] EQUILIBRIUM AND PRESSURE OF WATER, ETC. 739
broken, their walls must be made of a certain thickness, which de-
pends upon the head of water and the internal diameter of the
vessel. The rupture of the pipe may be caused either by a trans-
verse or by a longitudinal tearing. The latter form of rupture is
most likely to occur, as will appear from the following discussion.
If the head of the water in a pipe = h or the pressure upon the
unit of surface of the pipe is phy, the width of the pipe J N
= 2 CM = 2 r, Fig. 615, and the cross-section of the body of
water in it F = r², the pressure, which is exerted upon the sur-
face of the end of the pipe and which must be sustained by the
cross-section of the tube, is
π
ה
P=Fp= r² hy = = r² p.
Now if the thickness of the pipe is A D = BE = e, its cross-
section is
= 2 = r e + e² = 2 = r e ( 1
1 +
2
= π (†' + e)² — π yoz
(r
Прод
and if we denote the modulus of proof strength of the material, of
which the pipe is composed, by T, the proof strength of the entire
tube in the direction of the axis is
FIG. 615.
R
M
N
B
E
S
R
P
(1
+
21
Hence we can put
(1 + 2/2)
(1 +
Ր
2 r
2 πρε Τ

2 π reТ = r² p, or
T
2 e T = rp (see § 205);
the resolution of this equation gives
the thickness
r p
e =
2(1) T
of the pipe, for which we can generally write with sufficient accu-
racy
e
r p
2 T
rhy
2 T
The mean pressure, which the water exerts upon a portion of
the wall AM B, whose length is 7 and whose central angle is
ACB = 2 aº, is, since the projection of this portion at right.
angles to the line CM passing through the centre is a rectangle,
whose area is A B . 1 = 2 r 1 sin, a
P = 2rl sin. a. p 2 r1h sin. a . y'.
740
[§ 363.
GENERAL PRINCIPLES OF MECHANICS.
This force is held in equilibrium by the forces of cohesion R,
Rin the cross-sections AD. 1 and BE. 1 el of the wall of the
pipe; it is therefore equal to the sum 2Q of the components
DQ = Q and EQ Q of the latter forces, which are parallel to
the line C M. Now if we put R = el T, we obtain
Q
= R sin. A R Q = R sin. A C M = el T sin. a,
and therefore
2 el T sin. a = 2 r l p
2 r lp sin. a, L.E. e T = r p;
hence the required thickness of the pipe is
e
r p
T
=
rhy
T
which is entirely independent of its length.
Since the first calculation gave e only
ק יך
2 T'
it follows that
to prevent a longitudinal tear we must make the wall twice as
thick as would be necessary to prevent a transverse one.
From the formula
r p
rhy
e
T
T
just found, it follows that the thickness of similar pipes must be
proportional to the width and to the head of water or pressure upon
the unit of surface. A pipe, which is three times as wide as
another and which has to bear a pressure five times as great as the
first, must be fifteen times as thick.
We must give to hollow spheres which sustain a pressure p upon
each unit of surface the thickness
r p
2 T
for here the projection of the surface pressed upon is the great
circle, and the surface of separation of the ring is 2 = r e
ה
(1+). or approximatively, when the thickness is small, = 2 = re.
:
The formulas just found, give for p = 0 also e = 0; hence
pipes, which have no internal pressure to resist, can be made infi-
nitely thin but since every pipe in consequence of its own weight
must sustain a certain pressure and also must be made of a certain
thickness to be water-tight, we must add to the value found a
certain thickness e in order to have a pipe, which under all circum-
stances will be strong enough. Hence for a cylindrical tube or
boiler we have
€ 363.] EQUILIBRIUM AND PRESSURE OF WATER, ETC.
741
e = e₁ +
rhy
Τ
T'
or more simply, if d is the interior width of the tube, p the pressure
in atmospheres, each corresponding to a column of water 34 feet
high, and μ a coefficient determined by experiment,
e = e₁ + µ p d.
It has been experimentally determined that for tubes made of
Sheet iron
Cast iron
Copper
•
Lead
Zinc.
Wood
Natural stone
Artificial stone .
e = 0,00083 p d + 0,12 inches,
e = 0,00238 p d ÷ 0,34
e = 0,00148 p d + 0,16
е
0,00507 p d + 0,21
e = 0,00242 p d ÷ 0,16
e = 0,0323_p d + 1,07
e = 0.0369 P d + 1,13
e = 0,0538 pd ÷ 1,58
(C
66
CC
EXAMPLE. — If a vertical water-pressure engine has an inlet cast-iron
pipe 10 inches wide inside, how thick must its walls be for a depth of 100,
200 and 300 feet? For a depth of 100 feet this thickness is 0.00238 .
100. 10+ 0.34 0.07 + 0.34 0.41 inches for a depth of 200 feet
014 + 0,34 = 0.48 inches; and for a depth of 300 feet = 0,21 + 0,34
0,55 inches. Cast-iron conduit pipes are generally tested to 10 atmo-
spheres, in which case we have
0,0238 d +.0,34 inches,
and for pipes of 10 inches internal diameter we must make the thickness
e = 0,24 + 0,34 0,58 inches.
=
REMARK-1) In the second part of this work the thickness of tubes ex-
posed not only to hydrostatic pressure, but also to hydraulic impact, will
be calculated.
2) In the second part the thickness of the walls of steam-boilers will be
treated. Upon the theory of the thickness of pipes, we can consult the
treatise of Geb. Regierungsrath Brix in the proceedings of the "Vereins
zur Beförderung des Gewerbefleisses, in Preussen," year 1834, and Wiehes
“Lehre von den einfachen Maschinentheilen," Vol. I, and also Rankine's
Manual of Applied Mechanics." page 289, and Scheffler's ¨ Monographien
über die Gitter- und Bogenträger, und über die Festigkeit der Gefasstinde.”
The technical relations and the testing of pipes are treated in Hagen's
"Handbuch der Wasserbaukunst," Part 1st, and also in Genier's "Essai
sur les moyens de conduire, etc., les eaux,” and in the “Traité theoretique
et pratique de la conduite et de la distribution des eaux," par Dupuit,
Paris. 1854.
742
[S 364.
GENERAL PRINCIPLES OF MECHANICS.
CHAPTER II.
EQUILIBRIUM OF WATER WITH OTHER BODIES.
§ 364. Upward Pressure, Buoyant Effort. A body im-
mersed in water is subjected to pressure upon all sides, and the
question arises, what is the magnitude, direction, and point of
application of the resultant of all these pressures? Let us imagine
this resultant composed of a vertical and two horizontal compo-
nents, and let us determine them according to the rules of § 362.
The horizontal pressure of the water against a body is equal to the
horizontal pressure against its vertical projection; but every eleva-
tion AC, Fig. 616, of a body is at the same time the projection of
the rear part 1 D C and of the fore part A B C of its surface, and
consequently the pressure Pupon the hind part of the surface of a
body is equal to the pressure P upon the fore part; and as the
directions of these pressures are opposite, their resultant is = 0.
Since this relation exists for any given horizontal direction and its
corresponding vertical projection, it follows that the resultant of
all the horizontal pressures is equal to zero, and that the body A C,
which is under water, is subjected to equal pressure in all horizontal
directions, and therefore has no tendency to move horizontally.
FIG. 616.
FIG. 617.

0
H
R
R
D

BY
D
E
P
P
C
X
In order to find the vertical pressure of the water upon a body
A B D, Fig. 617, immersed in it, let us imagine it to be decomposed
§ 365] EQUILIBRIUM OF WATER WITH OTHER BODIES.
743
into the vertical elementary prisms A B, C D, etc., and let us de-
termine the vertical pressure upon their bases A and B, C and D,
etc. Let the lengths of these columns be l₁, l, etc., the depths
H B, K D of their upper ends B, D below the surface of the water
OR be h₁, hy, etc., and their horizontal cross-sections be F, F, etc.,
then we have the vertical pressures which act from above down-
wards upon their ends B, D, etc.,
Q1, Qe, etc., F, h, y, F, h, y, etc.,
and, on the contrary, the vertical pressures which act from below
upwards against the ends 4, C, etc., are
F, (hr + 1,) y, F. (h₂ + 1) y, etc.
parallel forces we obtain the resultant
− ( Q1 + Q₂ + ...)
F,
P₁, P., etc.,
By combining these
P = P₁ + P₂ +
= F, (h, + b) Y +
= (F, 1, + F. 1₂ +
l.
in which denotes the volume of the immersed body or of the
water displaced by it. Hence the upward pressure or buoyant effort,
with which water tends to raise up a body immersed in it, is equal to
the weight of the water displaced or of a quantity of water which has
the same volume as the submerged body.
F₂ (h₂ + la) y + ... - F, h, y - F₂ h₂y-...
. . .) y = V y,
Finally, in order to determine the point of application of this
resultant, let us put the distances E F₁, E F, etc., of the elemen-
tary columns B, C D, etc., from a vertical plane O equal to
a1, a, etc., and let us determine their moments in reference to this
plane. If S is the point of application of the upward thrust, which
is called the centre of buoyancy, and E S = x its distance from
that plane, we have
and therefore
Vyx = F₁ l₁ y . a₁ + F₂ la y . a₂ +
1
F₁l, a₁ + F, 1, α +
V₁ α₁ + V₂ α₂ + ...
2
X =
I'₁ + V₂ +
F₁ h₁ + F₂ l₂ +
•
1
2
F₁, V½, etc., denoting the contents of the elementary columns. Since
(according to § 105) the centre of gravity of a body is determined
by exactly the same formula, it follows that the point of application
S of the upward thrust coincides with the centre of gravity of the
water displaced. The direction of the buoyant effort is called the
line of support; when it passes through the centre of gravity of the
body, it is called the line of rest.
$ 365. Upward Pressure, or Buoyant Effort, when the
Body is Partially Surrounded by Water.-If a body, such as
A B D, Fig. 618, is not entirely surrounded by the water A HR,
744
[§ 365.
GENERAL PRINCIPLES OF MECHANICS.
and the surface A B, whose area is F, is united to the wall of the
vessel, or if the body, where its cross-section is 1 B F, passes
through the wall of the vessel, the pressure which the water would
have exerted upon this surface A B, if the body was free or in con-
tact with the water alone, is absent.
RA
FIG. 618.
B
D
If we denote the head of water upon
A B, I.E. the depth of its centre of grav-
ity below the surface of water II R, by
h, the pressure of the water upon A B
will be P = Fhy; and if V, denotes
the volume of water displaced by A B D,
the buoyant effort of the water, or the
force, with which the body would tend
to rise if it were free, is P₁
V₁ Y.

ཡ=
However, since the pressure upon
A B is wanting, the entire action of the
water upon the body is the resultant R
of P₁ = V₁ y and P
FhY.
In order to determine this resultant, we prolong the vertical line
of gravity of the water displaced and the right line passing through
the centre M of the pressure perpendicular to A B until they meet
at the point C; then, assuming the forces P, and P to be ap-
plied at this point, we combine them by means of the parallelogram
of forces and obtain the resultant C R R.
-
1
1
If the inclination of the surface A B to the horizon as well as
the deviation of the force P from the vertical = a, the angle
formed by the directions of the forces P and — P₁ with each other
will be = M C R₁ = 180 a, and therefore the resultant, which
measures the whole effect of the pressure of the water upon the
body A B D, will be
R
= Y
√ R² + P²
2 P P₁ cos. a
1
√ Vi + (Fh)*
2 V, Fh cos. a.
1
1
According to the principle of action and reaction, the body will
react with a pressure Rupon the water.
If V is the volume
0
of the water in the vessel or V, y its weight G, the pressure, which
acts vertically downwards upon the vessel, is
Q = V₁ y + P₁ = (V。 + V₁) Y, I.E. Q = V y,
when VV + V, denotes the volume of the space occupied by
the water and the body A B D.
=
Combining this with the pressure P Fhy, we have the entire
pressure sustained by the vessel
366.] EQUILIBRIUM OF WATER WITH OTHER BODIES. 745
R₁
= √ Q² + P² 2 Q P cos. a
= Y
√ √² + (Fl)² 2 V Fh cos. a.
If the surface A B were horizontal or a = 0°, we would have
R = (V, Fh) y and R₁ = (V Fh) y.
If also V = 0, R would be
Fhy (see § 355).
§ 366. Equilibrium of Floating Bodies.-The buoyant
effort P upon a body floating or immersed in water is accompanied
by the weight G of the body, which acts in the opposite direction,
and the resultant of the two forces is
R = G Por= (ε 1) VY,
in which ɛ denotes the specific gravity of the body.
If the body is homogeneous, its centre of gravity and that of the
water displaced coincide, and this point is consequently the point.
of application of the resultant R GP; but if the body is
heterogeneous, the two centres of gravity do not coincide and the
point of application of the resultant does not coincide with either
of them. Putting the horizontal distance S H, Fig. 619, of the two
FIG. 619.
B
centres of gravity from each other = b
and the horizontal distance SA of the
required point of application A from the
centre of gravity S of the water dis-
placed, a, we have the equation
G b

whence we obtain
Ra,
G b
G b
a
R G
P
G
If the immersed body is abandoned to
the action of gravity, one of three cases
may occur. Either the specific gravity
ɛ of the body is equal to that of the water, or it is greater, or it is
less. In the first case the buoyant effort is equal to the weight, in
the second it is smaller, and in the third it is greater. While in
the first case the buoyant effort and the weight are in equilibrium,
in the second case the body will sink with the force
GFy (ε 1) I' y,
Ꮐ
and in the third case it will rise with the force
Fy G = (1 − e) Ty.
The body will continue to rise until the volume 1 of the water
displaced by the body and limited by the plane of the surface of the
7746
[$ 367
GENERAL PRINCIPLES OF MECHANICS.
water has the same weight as the entire body. The weight G
Vεy of the body A B, Fig. 620, and the buoyant effort P
FIG. 620.
Р
B
C
HT
V₁y form a couple, by which the body is turned
until the directions of these forces coincide or
until the centre of gravity of the body and the
centre of buoyancy come into the same vertical
line, or until the line of support becomes a line
of rest. From the equality of the forces P and
G we have the expression

V₁ = e V, or V
V
Ε
1'
The line passing through the centre of gravity of the floating
body and the centre of buoyancy is called the axis of floatation (Fr.
axe de flottaison; Ger. Schwimmaxe), and the section of the float-
ing body formed by the plane of the surface of the water is called
the plane of floatation (Fr. plan de flottaison; Ger. Schwimme-
bene). From what precedes we see that any plane, which divides.
the body in such a manner that the centres of gravity of the two
portions will be in a lime perpendicular to it, and that one portion
of the body will be to the whole as the specific gravity of the body
is to that of the liquid, will be a plane of floatation of the body.
§ 367. Depth of Floatation.-If we know the form and
weight of a floating body, we can calculate beforehand by the aid
of the foregoing rule the depth of immersion. If G is the weight
of the body, we can put the volume of the
water displaced
H
FIG. 621.
A E B
inaboading!
R
G

Y
if we combine this with the stereometric formula
for this volume V, we obtain the required
equation of condition.
For a prism A B C, Fig. 621, whose axis is
vertical, we have V, Fy, when F denotes the
cross-section and y the depth CD of immer-
sion; hence it follows that
Gh
G
Fy=
G
Y
and y
FY
TY
in which denotes the volume and h the length of the floating
prism.
For a pyramid A B C, Fig. 622, floating with its apex below
$367] EQUILIBRIUM OF WATER WITH OTHER BODIES.
747
the surface of the water, we have, since the contents of similar
pyramids are proportional to the cubes of their heights,
V = 2, and consequently the depth of immersion, is
V
CD = y = h V = h j
3 T
I
1
G
VY
in which denotes the volume and h the height of the pyramid.
FIG. 622.
A E B
H
R
H
C
FIG. 623.
C

EB
R
For a pyramid, A B C, Fig. 623, floating with its base under
water, we obtain, on the contrary, the distance CD = y₁ from the
apex to the surface of the water by putting
G
3
3
Ε
W³ - 3/1
h³
whence ?1 h V 1
Τι
Τ
T1 = 71
h 1 -
ΤΥ
For a sphere A B, Fig. 624, whose radius is CA = r,
V₁ —
= = y² (r — ?!?),
H
FIG. 624.
E
R
its radius is AC = B C = r, we
we have therefore, in this case, to
solve the cubic equation

y³ − 3 r y² +
3 G
0
π)
in order to find the depth of the
immersion DE
sphere.
y of the
If a cylinder A K, Fig. 625,
floats with its axis horizontal and
have, when aº denotes the central
angle 4 C B of the immersed are, for the depth of immersion D E
Y
= r (1
-
cos, a);
1
now in order to find the arc a we must put the volume of the water
748
GENERAL PRINCIPLES OF MECHANICS.
[§ 368.
=
displaced sector (22
(ma)
'r²
minus the triangle (2² sin. a), multiplied
by the length B K = 1 of the cylinder, or
FIG. 625.
R
K
E
(a — sin. a)
2
G
2
γ
2 G
a
sin. a =
I r² Y
and resolve the equation
2

by approximation with reference
to a..
EXAMPLE-1) If a wooden sphere
10 inches in diameter, which is float-
ing, is immersed 43 inches in the water, the volume of the water displaced is
1
V₁ = π (2)² (5
π . 81.7
8
while the volume of the sphere itself is
π d³
6
·
π 103
6
567. π
8
222,66 cubic inches,
523,6 cubic inches.
Therefore 523,6 cubic inches of the material of the sphere weigh as much
as 222,66 cubic inches of water, and the specific gravity of the former is
222,66
523,6
€ 0,425.
2) How deep will a wooden cylinder 10 inches in diameter sink, when
floating, if its specific gravity is = 0,425 ?
e 0,425? Here
a
sin. a π p² l. ε Y
= = 0,425 . π =
TE
1,3352.
2
I p²² V
Now the table of segments in the "Ingenieur," page 154, gives for the area
1,32766 a segment of a circle, whose central angle is a' =
a
sin. a
2
=
a
166", and for
sin. a
2
1,34487 an angle aº = 167"; we can, therefore,
put the angle at the centre, corresponding to the sector 1,3352
1,33520
→
1,32766
a"
166° +
1° 166" +
•
1,34487 1,32766
754"
1721
166° 26'.
The depth of immersion is, therefore,
y = r (1 — cos. § a) = 5 (1 cos. 83' 13′) = 5. 0,8819 = 4,41 inches.
§ 368. The most important application of the above principle
is to the determination of the depth of immersion of boats and
ships. If the boats have a regular form this depth can be calcu-
lated by geometrical formulas; but if the form is irregular, or if its
equation is unknown, or if it is composed of very many forms, the
depth of immersion must be determined by experiment.
$ 368.]
EQUILIBRIUM OF WATER WITH OTHER BODIES. 749
An example of the first case is furnished by the boat ACE G H,
represented in Fig. 626, whose sides are plane surfaces.
It con-
FIG. 626.
G

H
Q
P
X
T
R
K
W
B
F
10
N
E
M
sists of a parallelopipedon A CF and two four-sided pyramids
CEF and B G H, which form the bow and stern, and its plane
of floatation is composed of a parallelogram K L O P and of two
.trapezoids L M N O and K P Q R which limit the space, from
which the water is displaced and which can be decomposed into a
parallelopipedon K CO T, into two triangular prisms UV M N
and I'XR Q, and into two four-sided pyramids C Ț M and B X R.
Let us put the length A D B C of the central portion = 1, its
width A G = b and its height A B = h, the length of each of the
two beaks = c and the depth of immersion under water, I.E. B K
= CL = y. It follows that the immersed portion K COT of
the middle piece is
= BC.CS. CL = 1 b y.
Putting the width C U of the base of the pyramid CV M, = x
and the height of this pyramid=z, we have
X
Z Y whence
b
C
だ
​Ъ
C
X
h
Y
y;
and z =
hence the volume of this pyramid is
= { x y z =
bc y³
3 729
3
and those of the two pyramids (CVM and B X R) together are
b c y³
だ
​The cross-section of the triangular pyramid UVN is
cy
= = y z =
and the side M N = V O
2 h
=
= b
by
h
- 9 4 = b ( 1 − 2 ) ;
750
[§ 369.
GENERAL PRINCIPLES OF MECHANICS.
+
hence the contents of the two prisms V U N and X W Q together
are
= 2.
y²
c y²
h
cy.6(11) boy (13)
2)
2 h
Finally, by adding the volumes first found, we obtain that of
the water displaced
3
су
c y²
V = bly + 3
b c y³
h²
+
bc y²
h
bey' = (1 + 28 -1.08) by.
Now if the gross weight of the boat
cy²
h
3
h"
G, we must put
1. cy) bh y = G, or
に
​су
1 +
h
3 lh²
y³
3 h y²
C
•
3 h² G
У y +
= 0.
b c y
By resolving this cubic equation we obtain from the gross
weight G of the boat its depth y of floatation.
EXAMPLE—1) If the length of the middle portion of a boat is l 50
feet, the length of each terminal pyramid is c = 15 feet, the width b = 12
feet and the depth h 4 feet, the total load for an immersion of 2 feet is
[50+ 15. —. 15. ()²]. 12. 2.62,5
[50+ 7,5
G
? 4
2) If the gross weight of the
have for the depth of immersion
y³ — 12 y²
From this we obtain
Y
y =
213,33 + y³ 12 y2
160
1,25] 24. 62,5 = 84375 pounds.
above boat was 50000 pounds, we would
160 y + 213,33 = 0.
=
1,333 + 0,00625 yª — 0,075 y³,
approximatively, y 1,333 +0,00625 (1,333) — 0,075 (1,333)'
= 1,333 +0,0148 0,1333 1,215, and more exactly
1,333 +0,00625 (1,215)³ — 0,075 (1,215)² = 1,2338 feet.
REMARK.—In order to find the weight of the cargo, vessels are provided
on both sides with a scale. The divisions of the scale are generally deter-
mined empirically by finding the immersion for given loads. This subject
will be treated more at length in the third volume.
§ 369. Stability of Floating Bodies.-A body floats either
in an upright or inclined position, and with or without stability.
A body, E.G. a ship, floats in an upright position, when at least one
of the planes passing through the axis of floatation is a plane
of symmetry of the body, and in an inclined position, when the
body cannot be divided into two symmetrical parts by any plane
passing through the axis of floatation. A floating body is in stable
§ 369.] EQUILIBRIUM OF WATER WITH OTHER BODIES.
751
equilibrium, when it tends to maintain its position of equilibrium
(compare § 141), I.E. if work must be done to move it out of this
position, or if it returns to its original position of equilibrium after
having been moved from it. A body floats in unstable equilibrium,
when it passes into a new position of equilibrium as soon as it has
been moved from its original one by being shaken, by a blow, etc.
T
If a body A B C, Fig. 627, which was floating in an upright
position, is brought into an inclined one, the centre of buoyancy S
moves from the plane of symmetry and assumes a position S, in
the half of the body most immersed. The buoyant effort P = y;
which is applied at S, and the weight of the ship GP, which
is applied at C, form a couple which will always turn the body
(see § 93). No matter around what point this rotation takes place,
the point C, yielding to the weight G, will always sink, and the
point S, or another M, situated in the vertical line S, P, yielding
to the action P, will rise, and the axis or plane of symmetry E F
will be drawn downwards at C and upwards at M, and therefore
the body will right itself when M, as in Fig. 627, is above C, and,
H
FIG. 627
A
P
FIG. 628.

A

F
F
B
R
B
Di
H
R
E
P
on the contrary, it will incline itself more and more when, as is
represented in Fig. 628, M is situated below C. Hence the stability
of a floating body, such as a ship, depends upon the point M, where
the vertical line, which passes through the centre of buoyancy S,
cuts the plane of symmetry. This point is called the metacentre
(Fr. métacentre; Ger. Metacentrum). A ship or any other body
floats with stability when its metacentre lies above its centre of
gravity, and without stability when it lies below it; it is in indif
ferent equilibrium when these two points coincide.
from the
The horizontal distance CD of the metacentre
centre of gravity C of the ship is the arm of the couple, formed by
P and G = P, and its moment, which is the measure of the
752
[$ 370.
GENERAL PRINCIPLES OF MECHANICS.
=
stability, is P. CD. If we denote the distance CM by c, and
the angle SMS, through which the ship rolls or through which
its axis is turned, by p, we have for the measure of the stability of
the ship
S = P c sin. &;
C
it increases, therefore, with the weight, with the distance of the
metacentre from the centre of gravity of the ship and with the
angle of inclination.
§ 370. Determination of the Moment of Stability.-In
the last formula
S = P c sin. 4,
the stability of the ship depends principally upon the distance of
the metacentre from the centre of gravity of the ship, and it is,
therefore, important to obtain a formula for the determination of
this distance. While the ship
H
H
E
FIG. 629
A
P
R
B
R
A B E, Fig. 629, passes from its
upright to its inclined position,
the centre of buoyancy S moves to
S₁, and the wedge-shaped space
HOH passes out of the water
drawing the wedge-shaped piece
ROR into it, and the buoyant
effort on one side is diminished
by the force Q, acting at the cen-
tre of gravity F of the space
HOH, and upon the other side
it is increased by an equal force Q,
acting at the centre of gravity G
of the space ROR. Therefore
the force P applied at S, replaces

Q),
the force originally applied at S and the couple (Q, — Q), or, what
amounts to the same thing, an opposite force P, acting in S,
balances the force P applied at S together with the couple (Q,
or more simply a couple (P, P), whose points of application are
S and S₁, balances the couple (Q, Q). Now if the cross-section
HER = H, ER, of the immersed portion of the ship F and
the cross-section II O H ROR, of the space, which is drawn
out the water on one side and immersed on the other, F₁, if the
horizontal distance KL of the centres of gravity of these spaces
from each other = a and the horizontal distance M T of the centres
of gravity S and S from each other, or the horizontal projection
$ 370.]
EQUILIBRIUM OF WATER WITH OTHER BODIES. 753
SS, of the space described by S, during the rolling, s, we have,
since the couples balance each other,
F s = F, a, whence s =
МТ
S MI = sin.
F₁
a and
F
8
sin.
Fa
F sin. o'
The line C M =e, which enters as a factor into the measure
of the stability, is = CS+ SM; denoting, therefore, the distance
CS of the centre of gravity C of the ship from the centre of buoy-
ancy S by e, we obtain the measure of the stability
(F, a
S = P c sin. O & = Р P(
+ e sin. ø).
If the angle through which the ship rolls is small, the cross-
sections HO H, and R OR, can be treated as isosceles triangles.
If we denote the width HR H, R, of the ship at the surface of
the water by b, we can put
F₁
i . 1 b . 1 b p
ЪФ
b² and K L = a 2. 3
b,
2
as well as sin. = 4; hence the measure of the stability of the
ship is
S'
P (i
1
b³ &
F
+ eø ) = (1/2 + e) Po.
)=(1
РФ.
F
If the centre of gravity C of the ship coincides with the centre
of buoyancy S, we have e = 0, whence
b s
S
Ρ Φ,
12 F
and if the centre of gravity of the ship lies above the centre of
buoyancy, e, on the contrary, is negative and
S = (12)² 1 − e) Pp.
F
It also follows that the stability of a ship becomes null, when e
is negative and at the same time
bs
12 F
We see from the results obtained that a ship's stability is greater
the wider the ship is and the lower the centre of gravity is.
EXAMPLE. The measure of the stability of a parallelopipedon A D,
Fig. 630, whose width A B b, whose height A E h and whose depth
of immersion E H = y is, since F = by and e
h - y
e=
2
ร
ર
48
754
[§ 371.
GENERAL PRINCIPLES OF MECHANICS.
h
8 = P+ (12+)
РФ
by
or if the specific gravity of the material of which the parallelopipedon is
composed be put
FIG. 630.
A
h
S Po
12 & h
2
(1 − e))

From this we see that the stability ceases
H
B
when
b2
R
6 h² ε (1 − e), I.E., when
E
Ъ
v6 ɛ (1 − ɛ).
h
D
For ε =
we have
b
윤​. 글
​V
1,225.
h
If in this case the width is not at least 1,225 times the height, the paral-
lelopipedon floats in unstable equilibrium.
371. Inclined Floating.-The formula
S P
(
Fa
F
±e sin. &
이
​for the stability of a floating body can also be employed to determine
the various positions of floating bodies; for if we put = 0, we
obtain the equation of condition of the position of equilibrium, and
by resolving it we obtain the corresponding angle of inclination.
We have, therefore, to resolve the equation
F, a
F
± e sin. p = 0
in reference to .
In the case of a parallelopipedon A B D E, Fig. 631, the cross-
section F is HRDEH, R, DE by, b denoting the
width 4B HR and y the depth of immersion E HDR,
A
H
H
FIG. 631.
E
B
and the cross-section
F₁ = HO H₁ = R O R₁
1

is a right-angled triangle, whose base
OH = OR = b,
and whose altitude is
is
R
R?
D
whence
¦
1
HH₁ = RR₁ = b tang. 9,
Fb
į b² tang. p.
Now since the centre of gravity P
is at a distance
FU = ¦ HH₁ = b tang. o
{
$ 371.] EQUILIBRIUM OF WATER WITH OTHER BODIES.
755
30H
b from
from the base H R and at a distance O U
the centre 0, it follows that the horizontal distance of the centre
of gravity F from the centre O
= 0K
ON+ NK
O U cos. + F U sin. &
= b cos.
6
+ b tang.
sin. p,
3
and the arm of the lever is
a = K L = 20Kb cos. + b
& 1
sin.
Cos. Ó
Hence the equation of condition of the inclined position of
equilibrium is
4 • &
btang. (b cos. + b sin. o)
by cos. o
e sin. ☀ = 0,
or, substituting
sin.
cos. O
tang. 9,
¿)
sin. [('+' tang. ) b² — e y] = 0,
¢
which equation is satisfied by
sin. 6 = 0 and by
tang. = 1/21
12 ey
1.
2
o
The angle 0, determined by the first equation, is applicable
to the body when in an upright position, and that, given by the
second equation, to the body when floating in an inclined position.
ey
If the latter case is possible, must be > Now if h is the
ぶ
​7'½•
height and the specific gravity of the parallelopipedon, we have
y = ε h and e =
2
h — Y − (1 − ε)
h
= —
2
whence it follows that
6ε (1
tang. = √ 2 V
ε) }²
1:
ぶ
​and the equation of condition for inclined floating is
h
b
1
6ε (1
EXAMPLE 1) If the floating parallelopipedon is as high as wide, and if
its specific gravity is ‹ =
2, we have
tang. ḍ =
2 13.3 1 V3 3 − 2 = 1, whence = 45°.
2) If the height h =
, we have
0,9 of the width b and the specific gravity is again
tang. = 13.0,81
2 = 0,43 0,6557, whence o = 33° 15'.
=
756
GENERAL PRINCIPLES OF MECHANICS.
[§ 372.
§ 372. Specific Gravity. The law of the buoyant effort or
upward thrust of water can be made use of to determine the heavi-
ness or specific gravity of bodies. According to § 364 the buoyant
effort of the water is equal to the weight of liquid displaced; hence,
if we denote by V the volume of the body and by y, the heaviness
of the liquid, we have the buoyant effort P = Vy. Now if y, is
the heaviness of the material of the body, we have its weight
G = V Y2, whence the ratio of the heavinesses is
Y2
G
Yı P
I.E., the heaviness of the immersed body is to the heaviness of the
fluid as the absolute weight of the body is to the buoyant effort or
loss of weight during immersion.
Hence Ye
G
P
Yı and γι
F
G
Y2, or if y denotes the heaviness
of water, ε, the specific gravity of the fluid, and ε, that of the body,
we have, putting y₁ = &, y and y½ = ɛ½ Y,
E2
E
G
ε₁ and &₁ =
P
P
G
If we know the weight of a body and its loss of weight when
immersed in a liquid, we can find from the heaviness or specific
gravity of the liquid the heaviness and specific gravity of the ma-
terial of which the body is composed, and, inversely, from the
heaviness or specific gravity of the latter, the heaviness and specific
gravity of the former.
£1
=
=
If the liquid in which we weigh solid bodies is water, we have
ε, = 1 and y₁ = y 1000 kilograms 62,425 pounds; the former
when we employ the cubic meter and the latter when we employ
the cubic foot as unit of volume; therefore in this case the heavi-
ness of the body is
G
Y2 =
Y =
absolute weight
Р
loss of weight
multiplied by the heaviness of water,
and its specific gravity is
G
Eq
E9 =
P
absolute weight
loss of weight
In order to find the buoyant effort or loss of weight, we employ,
as we do when determining the weight G, an ordinary balance. To
the under side of one of its scale-pans is attached a small hook, from
which the body is suspended by means of a hair, wire or fine thread
before it is immersed in the water, which is contained in a vessel
placed under the dish of the scale. A scale thus arranged for
§ 372.] · EQUILIBRIUM OF WATER WITH OTHER BODIES.
757
weighing under water is generally called a hydrostatic balance (Fr.
balance hydrostatique; Ger. hydrostatische Wage).
If the body whose specific gravity is to be determined is less
dense than water, we can combine it mechanically with some other
heavy body, so that the compound mass will tend to sink in the
water. If the heavy body loses in the water a weight P, and the
compound mass P₁, the loss of weight of the lighter body is
P P₁ - P.
Now if G denotes the weight of the lighter body, we have its spe-
cific gravity
G
E 2
G
P P₁ - P
1
If we know the specific gravity ε of a mechanical combination
of two bodies, and also the specific gravities e, and ε of the compo-
nents, we can calculate from the weight G of the whole mass, by
means of the well-known principle of Archimedes, the weights G,
and G, of the components.
We have G₁ + G₂
G, and also
G
G₂
volume
+ volume
ει γ
ε
Ea Y
G₁
G₂
G
+
ε
G
volume or
εγ'
Combining the two equations, we obtain
G
1
G₁ = @(): (-4), or
E
a₁ = a(-1)=(2)
=G
EXAMPLE-1) If a piece of limestone weighing 310 grams becomes 121,5
grams lighter in water, the specific gravity of this body is
310
121,5
= 2,55.
2) In order to find the specific gravity of a piece of oak, a piece of lead
wire, which lost 10,5 grams in weight when weighed in water, was wrapped
around the piece of wood, which weighed 426,5 grams. The compound
mass was 484,5 grains lighter in the water than in the air; hence the spe-
cific gravity of the wood is
426,5
484,5 — 10,5
426,5
474
0,9.
3) An iron vessel completely filled with mercury weighed 500 pounds,
and lost, when weighed in water, 40 pounds. If the specific gravity of the
cast iron is = 7,2 and that of the mercury is
empty vessel is
13,6, the weight of the
•
758
[$ 373.
GENERAL PRINCIPLES OF MECHANICS.
1
= 500
40
(300 - 13,0): (-1,2- 13,6)
(0,1388 — 0,0735) =
500 . 0,00647
0,0653
500 (0,08 - 0,07353):
3235
=
49,5 pounds,
65,3
and the weight of the mercury contained in it is
G₂
500. (0,08 0,1388): (0,07353
450,2 pounds.
―
0,1388)
=
500 . 0,0588
0,0653
2940
6,5%
REMARK-1) We can determine the specific gravity of fluids, loose
granular masses, etc., by simply weighing them in the air; for by enclosing
them in vessels, we can obtain any desired volume of them. If the weight
of an empty bottle is = G, and when filled with water it is = G G₁, and if,
when filled with some other liquid, its weight is G₂, the specific gravity
of the latter liquid is
G
& 2
G
ε
G G
1
In order, E.G., to obtain the specific gravity of rye (in bulk), we filled a
bottle with grains of rye, and, after shaking it well, weighed it. After
subtracting the weight of the bottle, that of the rye was found to be
120,75 grams, and the weight of an equal quantity of water was 155,65;
hence the specific gravity of the rye in bulk is
120,75
155,65
0,776,
and a cubic foot of this grain weighs
0,776 . 62,5 48,5 pounds.
2) The problem, first solved by Archimedes, of determining from the
specific gravity of a composition, and those of its components, the propor-
tion of each of the components, is of but very limited application to chem-
ical combinations, alloys of metals, etc.; for in such cases a contraction
generally, and an expansion sometimes, takes place, so that the volume
of the composition is no longer equal to the sum of the volumes of the
components.
§ 373. Hydrometers, Areometers.-We generally employ
for the determination of the density of fluids areometers or hydrom-
eters (Fr. aréomètres; Ger. Aräometer, Senkwagen). These instru-
ments are hollow, symmetrical about an axis, have their centre of
gravity very low down, and give, when we float them in any liquid,
its specific gravity. They are made of glass, sheet brass, etc., and,
according to the uses they are applied to, are called hydrometers,
lactometers, salinometers, alcoholmeters, etc. There are two kinds.
of areometers, viz.: those with weights (Fr. à volume constant; Ger.
Gewichtsaräometer), and the graduated areometers (Fr. à poids
§373.] EQUILIBRIUM OF WATER WITH OTHER BODIES.
759
constant; Ger. Scalenaräometer). The first are often used to de-
termine the weight or specific gravity of solid bodies.
1) If V is the volume of the part of an areometer A B C, Fig.
632, which is under water, when the latter floats vertically im-
mersed to a point O, G the weight of the whole apparatus, P that
of the weights placed upon the dish A, when the apparatus floats
in water, whose heaviness = y, and P, their weight when the ap-
paratus floats in another liquid whose density is y₁, we will have
FIG. 632.
B
FIG. 634.
A
A
FIG. 633.
A
X
B
C
Vy PG,
= P + G,
=
Vy₁ = P₁ + G.
1


Hence the ratio of the heavinesses or
specific gravities of these liquids is
γι
P₁+ G
Y
P+G
2) Let P be the weight, which must
be placed upon the dish in order to im-
merse the areometer A B C, Fig. 633,
to a point O, and let P, be the weight,
which must be placed upon the dish A
with the body to be weighed in order
to produce the same immersion, then
we have simply

G₁ P — P₁.
But if we must increase P, by P₂ when
the body to be weighed is placed in the
lower dish C, which is under water, in
order to preserve the same depth of im-
mersion, the upward thrust is P, and
the specific gravity of the body is
G₁ P P₁
E P₂
P₂
The hydrometer with the dish sus-
pended below is employea for the de-
termination of the specific gravity of
solid bodies, such as minerals, etc., and
is called Nicholson's hydrometer.
3) If we put the weight of an areom-
eter BC with a graduated scale A B,
Fig. 634, G, and the immersed vol-
ume, when it floats on water, , we have G Vy. If the
areometer rises a distance O X = a, when immersed in another
760
[§ 373.
GENERAL PRINCIPLES OF MECHANICS.
liquid, we have, when the cross-section of the shaft is denoted by
F, the volume immersed
=V - Fx, and therefore G (V
=
—
Fx) Y1.
Dividing the two formulas by each other, we obtain the heaviness
of the liquid
T
Y₁
·Y=Y:
V - Fx
(1-
F
V
a)
Y
1-μπα
F
V.
in which μ denotes the constant quotient
If the liquid in which the areometer floats is lighter than water,
it will sink in it a distance x, and we will have
G = (V + F x) y, and therefore
γι
Y
1 + μα
·F
In order to find the coefficient μ = we increase its weight
by an amount P, E.G. by pouring mercury in the areometer at the
upper end, so that it passes to the bottom of it, rendering the ap-
paratus so much heavier that, when floating in water, a consid-
erable portion of the length of the stem, to which the scale is
applied, is immersed. Putting P Fly, I denoting the immer.
sion produced by P, we obtain
F
P
P
μ
V
VlY
Gl
EXAMPLE-1) If an areometer, weighing 65 grams, must have 13,5
grams removed from the dish in order to float at the same depth in alcohol
as it had done in water, the specific gravity of alcohol is
65 - 13,5
65
1 — 0,208
0,208 = 0,792.
2) The normal weight of a Nicholson hydrometer is 100 grams; that
is, we must place 100 grams upon the dish in order to sink the instrument
to 0, but we must take away 66,5 grams after having laid a piece of brass
which we wish to weigh upon the upper dish, and 7,85 grams had to be
added when the brass was removed to the lower dish. The absolute weight
of the brass is then 66,5 grams and its specific gravity is
66,5
7,85
8,47.
3) A graduated areometer, weighing 75 grams, rises, after 31 grams of
the substance, with which it was filled, has been removed, a distance = 6
inches = 72 lines; the coefficient μ is therefore
31
75.72
0,00574.
$374.] EQUILIBRIUM OF WATER WITH OTHER BODIES.
761
It was then refilled until its weight became again 75 grams, when it was
placed in a solution of salt; it then rose a distance of 29 lines; the specific
gravity of the liquid is therefore
-
1,2.
= 1: (1 - 0,00574. 29) 1: 0,833
REMARK.—A more extended treatment of this subject belongs to the
province of chemistry, physics and technology.
FIG. 635.
§ 374. Liquids of Different Densities.-If several liquids
of different densities exist in a vessel at the same time without
exerting any chemical action upon one another, they will place
themselves, in consequence of the ease with which their particles
are displaced, above each other in the order of their specific gravi-
ties, viz: the most dense at the bottom, the less
dense above it and the least dense on top. When
in equilibrium the limiting surfaces are hori-
zontal; for so long as the limiting surface E F
between the masses M and N, Fig. 635, is inclined
so long will there be columns of liquid, such as
G K, G, K₁, etc., of different weights above the
horizontal layer HR; hence the pressure upon
this layer cannot be the same everywhere and

A
K
K
M
E
N
F
R
B
consequently equilibrium cannot exist.
The liquids arrange themselves also in the communicating tubes
A B and CD, Fig. 636, according to their specific gravities above.
one another, but their surfaces 40 and D G do not lie in one
and the same horizontal plane.
G
D
FIG. 636
E
0
H
R
C-
B
FIG. 637.


G
D
E
A
H
R
C
B
If Fis the area of the cross-section HR of a piston, Fig. 637.
in one leg A B of two communicating tubes and h the head of
water or the height E H of the surface of the water in the second
tube CD above H R, we have the pressure upon the surface of the
piston
PFhy.
762
[§ 375, 376.
GENERAL PRINCIPLES OF MECHANICS.
If we replace the force, exerted by the piston, by a column of
liquid HA O R, Fig. 636, whose height is h, and whose heaviness
is y₁, we have
equating the two expressions we obtain
or the proportion
P = Fh, Y₁;
h₁ Y₁ = h Y,
hr Y
γι
h
Therefore the heads or the heights of the columns of liquid,
measured from the common plane of contact of two different liquids,
which are in equilibrium in two communicating tubes, are to each
other inversely as the heavinesses or specific gravities of these liquids.
Since mercury is about 13,6 times as heavy as water, a column
of mercury in communicating tubes will hold in equilibrium a
column of water 13,6 times as long.
CHAPTER III.
OF THE MOLECULAR ACTION OF WATER.
§ 375. Molecular Forces.-Although the cohesion of water
is very slight, it is not null. The molecules (Fr. molécules; Ger.
Theile or Moleküle) not only cohere together, but also adhere to
other bodies, E.G., to the sides of a vessel, so that a certain force is
necessary to destroy this union, which we call the adhesion (Fr. ad-
hérence; Ger. Adhäsion) of the water. A drop of water, which
hangs from a solid body, demonstrates the existence of the cɔhe-
sion and of the adhesion of the water. Without cohesion the
water could not form a drop and without adhesion it could not
remain hanging from the solid body; gravity is here overcome not
only by the cohesion, but also by the adhesion. The actions, arising
from the combination of the forces of cohesion and adhesion, are
called, to distinguish them from the actions of inertia, of gravity,
etc., the molecular actions. Capillarity or the raising or depressing
of the surface of water or mercury in narrow tubes or between plates,
placed close together, is an important instance of molecular action.
§ 376. Adhesion Plates.-The cohesion and adhesion of
water have been determined by means of adhesion plates. To
لكم
763
377.]
THE MOLECULAR ACTION OF WATER
accomplish this object, such a plate is suspended (instead of the
scale pan) at one end of the beam of a balance, which is brought
into equilibrium by means of weights; the vessel containing the
liquid to be examined is then caused to approach gradually, until
the surface of the liquid comes in contact with the plate. Weights
are now gradually placed upon the dish at the other end of the
beam, until the plate is torn away from the surface of the water.
The results of such experiments depend particularly upon the fact
whether the plate is moistened by the water or not. In the first
case after the contact a thin sheet of water remains hanging to the
plate; hence in tearing the latter from the water, we overcome not
the adhesion, but the cohesion of the water. Hence the force
necessary to tear different plates from the surface of the water
does not depend upon the nature of the material, of which
the plates are composed. Other liquids, on the contrary, require
different forces to be applied to the adhesion plates. Du Buat
found that the adhesion between water and tin plate was from 65
to 70 grains per square inch (old Prussian measure). This gives a
force of about 5 kilograms for a square meter, or 1,024 pounds per
square foot. Achard found values differing but little from the
above for lead, iron, copper, brass, tin and zinc. Gay Lussac ob-
tained the same results with a glass disc, and Huth with different
kinds of wooden plates..
If, on the contrary, the surface of the disc is not moistened by
the surface of the water, the results obtained are totally different;
for in this case it.is not the cohesion, but the adhesion of the water
which is overcome. It appears that the duration of contact has a
great influence upon the force necessary to tear the disc loose, E.G.,
Gay Lussac found that, with a glass plate 120 millimeters in diam-
eter, a force varying from 150 to 300 grams, according as the dura-
tion of contact was long or short, was necessary to tear it loose from
a surface of mercury.
REMARK.-In Frankenheim's “Lehre der Cohäsion" the phenomena of
cohesion, as, E.G., those presented when moistened plates are torn from the
surface of water, are called "Synaphy,” and, on the contrary, the phenomena
of adhesion, as, E.G., those occurring during the separation of unmoistened
plates from the surface of a liquid, “ Prosaphy.”
§ 377. Adhesion to the Sides of a Vessel.-If a drop
of water spreads itself out upon the surface of another body and
moistens it, the adhesion is in this case predominant; but if, on
764
[S 377.
GENERAL PRINCIPLES OF MECHANICS.
the contrary, the drop retains its spherical form upon the surface
of a solid or fluid body, the cohesion is the strongest. The com-
bined action of these two forces upon the surface of a liquid near
the walls of the vessel is particularly remarkable; the water rises.
up and forms a concave surface when the cohesion is less powerful
than the adhesion, and the wall becomes moistened in consequence;
the surface of the water, on the contrary, is curved downwards in
the neighborhood of the walls of the vessel and forms a convex
surface when the side of the vessel is not moistened or when the
cohesion is predominant.
These phenomena can be easily explained as follows.
A molecule E in the surface HR of the water (Fig. 638) is
drawn downwards in all directions by the surrounding water, and
the resultant of all these attractions is a single force A acting ver-
tically downwards; on the contrary, a molecule E at the vertical
wall B E, Fig. 639, of the vessel is acted upon by the wall with a
FIG. 638.
FIG. 639.


R
R
B
·
horizontal force P and by the water filling the quadrant BEO
with a force 4, whose direction is inclined downwards to the hori-
zon; the direction of the resultant R of these two forces is at
right angles to the surface of the water (see § 354). According as
the attractive force of the wall of the vessel is greater or less than
the horizontal component A, of the mean force of cohesion A of
P
7
J
R
FIG. 640.
the water, the resultant R will assume a di-
rection either from without inward or from
within outward. In the first case (Fig. 639)
the surface of the water at E rises along
the wall, and in the second case it descends
along the wall B E, as is represented in Fig.
640.
These relations change, when the water reaches to the brim of
the vessel; for the direction of the attractive force of the wall of
the vessel is then different. If, E.G., the surface of the water E 0,
Fig. 641, which in the beginning reached to the brim of the vessel
B CO, is caused to rise gradually by adding water, the inclination
of the force of adhesion to the horizon will gradually increase, and

§ 378.]
765
THE MOLECULAR ACTION OF WATER.
FIG. 641.
its horizontal component will, in consequence, gradually decrease,
until it becomes less than the horizontal component 1, of the force
of cohesion A. Consequently the form of the surface of the water
at E changes continually, until its con-
cavity becomes a convexity and the de-
pression below the brim of the vessel be-
comes an elevation, which must attain a
certain height before the water will flow
over the side of the vessel.

B
E
E,
0
§ 378. Tension of the Surface of the Water. Since each
molecule in the surface H R, Fig. 638, of a liquid is attracted down-
wards by the mass of liquid below it with a force 1, we can assume
that a condensation and a coherence among the molecules of the
liquid upon the surface will be the result and that a certain force
will therefore be necessary to overcome this coherence or to tear
the surface of the liquid. This coherence of the surface of a liquid
shows itself not only whenever a foreign body is dipped into it,
r
G
FIG. 642.
FIG. 643.
S
C
but also whenever the surface
of the liquid becomes curved,
as, E.G., in the neighborhood
of the wall of the vessel. If
we assume with Young that
the tension or cohesion of the
surface of a liquid is the same
in all parts of it, we can de-
duce, as Geheimer Oberbau-
rath Hagen has proved, from
that hypothesis all the laws
of capillary attraction which
coincide best with the results
of experiment.
In the neighborhood of a
plane wall D G, Figs. 642 and
643, the surface of the liquid
forms a cylindrical surface
DAH, which is convex either
upwards or downwards. If P
is the normal force upon an
element A E B = σ of this
surface, S the tension of this


766
[S 379.
GENERAL PRINCIPLES OF MECHANICS.
element and r its radius of curvature C A C B, we have, in
consequence of the similarity of the triangles E P S and A B C,
ន
ΤΩ
S
FIG. 644.
G
Р
A B σ
S
CA

and, therefore, the normal or
FIG. 645.
GNC bending force is
R
σ
P = = S.
Now if the element A E B
of the surface is at the vertical
distance OR y above or
below the surface of the water
which is free from the influ-
ence of the wall D G, and if y
denotes the heaviness of the
liquid, we have, according to
(§ 356) the well-known law of
hydrostatics, the pressure of
the water upon the element
A B = σ

o
P = σ Y Y ;
we can therefore put
σ
o y Y
S and
S
Y
y = ry
Hence the depression or elevation of an element of the surface
of a liquid in reference to the free or unaffected part of this surface
is inversely proportional to the radius of curvature.
§ 379. In the vicinity of a curved wall, E.G., of a vertical cylin-
drical surface, the surface of the water forms a surface of double
curvature and the column of water below the rectangular element
F G H K, Fig. 646, of the surface is solicited by two forces P₁ and
P, one of which is the resultant of the tensions S₁, S, in the nor-
mal plane A B E, parallel to the side F G H K; the other is
the resultant of the tensions S2, S, in the normal plane C D E,
parallel to the side G H = FK. The former plane corresponds
to the greater and the latter to the least radius of curvature; put-
ting the two radii = r₁ and r, and the length of the sides F G
and GH, and denoting the tension for a width
we have the tensions acting in the two planes.
= 01
unity by S,
§ 380.]
767
THE MOLECULAR ACTION OF WATER.
S₁ = σ, S and S, σ, S
=σ₂
S2 = 1
and the normal forces resulting from them
G
B
S
1
S2
P₁
2
S
=
P₁₂ S
01
So, 02
and
ri 1
2
P₂ = 01 S
02
80102
re
P = P₁ + P₂ = S 01 02
( —
FIG. 646,
H
E
S₁
F
D
A
and their resultant is
+
1
If here also y denote the
height of the element FG HK
of the surface (which may be
regarded as a rectangle, whose
area is σ, σ) above the low-
est or general surface of the
water, we have the force, with
which this element is drawn
K normally upwards or down-
wards by the water above or
below it,
P Y σ ₁ σ ₂ Y ;'
equating the two values for
P, we obtain

1
Yo, σ₂ y = S 0, 02
+
whence
y =
γ
S 1
r
1
+
グ
​When the wall is cylindrical the elevation (depression) of the
surface of the water above (below) the general water level is at
every point proportional to the sum of the reciprocals of the maxı-
mum and minimum radii of curvature. This formula contains
also that of the foregoing paragraph; for if the normal section
CED is a right line, we have
r₂ = ∞, whence
1
= 0 and
12
S 1
Y
Y
(§ 380.) Curve of the Surface of Water.-The curve
formed by the vertical cross-section of the surface of the water
768
[$ 380.
GENERAL PRINCIPLES OF MECHANICS.
near a plane wall, can be found, according to Hagen, in the follow-
ing manner. Let A R, Fig. 647, be the surface of the water
FIG. 647.
B
1-1
M-N
K
attracted by the vertical wall B K,
HR the general level of the water,
and let the point of intersection H of
the two surfaces be the origin of co-
ordinates. Let us put the co-ordinates
of a point 0 of the surface AOR, HM
= x and M 0 = y, the arc A 0 = 8,
the tangential angle O T M = a, and
the elements OQ, Q P and O P re-
spectively d x, dy and d s.

=
S
Since y = and, according to
ry
Article 33 of the Introduction to the
Calculus,
d s
and d y
d s sin. a, we have
d a
Sd a
S sin. a. d a
Y
or
y d s
y d y
S
y dy =
sin. a. da,
Y
+ y² = = √
S
γι
by integrating which we obtain
Since for the point R, a and y are both = 0, we have
•
sin, a d a = Con.
da
S
Cos. a.
Υ
S'
S
0 = Con.
cos. 0, whence Con.
and
γ
Y
y²
28
Y
4 S (1
(1
cos. a)
=
cos. a)
2
4 S
(sin. a),
Y
hence
S
Y
21
sin. a.
Y
For a
=
90°, we have sin. { a = sin. 45° V; hence the
maximum elevation of the water immediately against the wall is
S
h
21
√
1/28
, or inversely
Y
γ
S
h2 and
Y
1) y = h √2. sin. ½ a.
$380.]
769
THE MOLECULAR ACTION OF WATER.
Differentiating this expression, we obtain
d y
and since d y =
• d r = − h v ! .
= − h v .
= −
h V.
•
1 h 1/2 cos.a. da h√ cos.a. da,
= ¦
d x. tang. a, it follows that
cos. a
tang. a
da =
cos. ↓ a [(cos. ¦ a)
2 sin. a.
1-2 (sin.
2 sin.
1
= − h √ ⋅ (sin şa
•
a
a)'
2
sin. 1
cos. i a cos. a
sin. a
2
(sin. a)]
d a
cos. a
d a
1 a) a
d a.
But now
S sin. 1 a
da =
2 cos.a and
•
d a
S
d a
// sin» ! a
1 a
= 21 tang. 1 a
(see Introduction to the Calculus, Art. 29);
hence we have
x = − h √ 1 (l tang. a+ 2 cos.a) + Con.
=
90°, tang. ¦ a = tang. 221°
1
Now since for x = 0, a°
and cos.a = V, it follows that
Con. = h
1
√ [l ( √2 − 1) + 2 √], and
}
√2 1
R√ [2 + 2(√ √ cos. § a)}
2) x = h √ √ [i
= h [1
tang.a
√2 - 1
√2. cos. 1 a √ √ 1 ( √2 + 1) tang. ¦ a].
For a = 0 we have
and therefore
cos. ½ a = 1 and 1 tang. ¦ a =
x = + ∞∞ ;
HR is consequently the asymptote, which the section A O R of
the surface of the water continually approaches.
REMARK.-If we invert the formula (1) and put
sin. a =
34
we can calculate for every value of y, first a and then by means of (2) the
corresponding value of x.
49
770
[$ 381.
GENERAL PRINCIPLES OF MECHANICS.
The measurements made by Hagen to test this theory, show that it
agrees very well with the results of experiment. They were tried with a
dead polished brass plate upon spring water, and gave the following results.
แ
y measured in lines. 1,37 0,70 0,49 0,34 0,24 0,18 0,12 0,07 0,04 0,016
0,00 0,31 0,63 0,94 1,26 1,57 1,88 2,50 3,13 3,74
0,00 0,33 0,64 0,96 1,28 1,56 1,95 2,47 3,01 3,90
a calculated...
These values are given in Paris lines. From h = 1,37 lines we calcu-
S
late
0,94 and the minimum radius of curvature r =
Υ
0,68 lines. Plates
of boxwood, slate, and glass gave the same results.
§ 381. Parallel Plates.-The water between two plates
D E, D E, Fig. 648, which are placed near each other, rises not
only on the outside, but also between them
and the cross-section of its surface is nearly a
semi-ellipse. One semi-axis of the elliptical
D
FIG. 648.
D

A
B
F
E
E
A
R
cross-section is the half width C A
= a, the
other semi-axis C B = b is equal to the differ-
ence A F - B G = hy - h₁ of the maximum
and minimum elevations of the elliptical sur-
face A B A above the general water level.
According to the "Ingenieur," page 171, the
radius of curvature of the ellipse at A is
2
(h₂ — h₁)², and that at B is
a
,
b2
Τι
r₁ =
α
a²
a²
r2 =
;
b
(no
h₁)
hence we have, according to § 378, the elevation of the surface of
the water at A
a s
S
h₂ =
r₁Y
(h₂ — h₁)² y'
and, on the contrary, that at B
or
S (h₂ h₁) S
h₁
r₂ Y
2
αγ
Subtracting the latter equation from the former, we obtain
а
Y (h₂ — h₁) ²
S
h₂
h₁
S
α
1
h₂
2
1 = // (= 2) - 3 )
3
2
a²
2
a²
h₁
§ 381.]
771
THE MOLECULAR ACTION OF WATER.
whence
1) h₂ h₁ = a
-
-
2) h₂ =
α
1
1
S
S + a² y
1
2
Y
~ (+ a²).
>
S
3) h
α
Y
S' + a² y'
n =
hq - hr
h
a² Y
S
: a²:
S'
Y
1
S
h₂ h₁
a
Y
and, finally, the ratio
If a is very small, we can put
the elevation of the surface of the water is then inversely proportional
to the distance of the plates from each other.
We have, however, more accurately,
1 S
h₂
a
Y
(1 + 8 a² 7)
1
+ 3 a, and
a
γ
1
S
h₁
1
a
Y
(1 - 3 a² 2)
1 S
1
133 a.
αγ
By inversion we obtain
S
a²
= a h₁ +
Y
3*
These formulas agree very well with the results of observation,
α
especially when does not reach.
h₁
Hagen found, from his experiments with two parallel plane
plates in spring water, as a mean
h₁
=
1,55, h₂ = 2,09, and h = 1,38 Paris lines,
and by calculation
S
Y
1,04, h₂ 2,12, and h = 1,44 Paris lines.
More recent experiments (see Poggendorff's Annalen, Vol. 77)
gave for
a = 0,360; 0,5875; 0,7575 lines,
h₁= 2,562; 1,429; 1,068 lines, and
S
= 0,949; 0,907; 0,917 lines,
Y
772
[§ 382.
GENERAL PRINCIPLES OF MECHANICS.
I.E. as a mean value
S
0,9243 and S
0,01059 grams.
γ
(Compare the foregoing paragraph.)
§ 382. Capillary Tubes.-We can easily calculate the height
to which the surface of water will rise in narrow vertical tubes,
called capillary tubes (Fr. tubes capillaires; Ger. Haarröhrchen), by
starting from the formula
FIG. 649.
D
D
A
B
H
E
E
A
R
S
Y
γ
(+)
of § 379 as a basis and assuming that the sur-
face (the meniscus) forms a semi-spheroid
A B A, Fig. 649, whose circular base A A coin-
cides with the cross-section of the tube. If we
retain the notations of the foregoing paragraph,
I.E. if we put the radius CA of the tube
and minimum and maximum heights B G
and A F of the water in the tube above the
general level of the water H R, h, and has
we must substitute in

=
a
h₂
h₁
=
11
रारा
1
+
r₁ = a and r,
(h₂ — h₁)²
and in
,
2.
a
a²
ri
+
r.
و
1
r₁ = r₂ =
2
; thus we obtain
ལའི་
h₁
S
а
h₂
+
and
για
(h₂-h)
h₁
2S (ha h₁)
Y
a²
Subtracting the last equation from the one preceding it, we
2 (heh))
a
a²
h₁)),
2
obtain
S /1
а
h₂ — hr
+
Y
(h₂ — h₁)²
2
or
S
1 =
Y
145 (a th
1
+
(h₂
- h₁)
:),
and also
γ.
+
S
22) (Mr
h₁)²
= α.
If a is small, we can put
(h₂- h₂)
1
3
(N₂ — h₂)³
hr) ³ — (h₂ — hr)²
α
§ 382.]
773
THE MOLECULAR ACTION OF WATER.
1
a²
2/3 (h₂ — hr) ³ — — (h₂ —
3
(h₂ — h₁)² = a,
a
whence it follows that
h₂
h₁
= α ;
assuming h, h, ad and putting (h,
= + — h₁)² = a² + 2 a §,
h₁)³ = a³ + 3 a² d, we obtain
and also (h,
Y
—
3
8)
1
( 3 + 2/3 ) ( a² + 3 a² §) — — (a² + 2 a §) = a,
S
a³
a
or
Y
a³ +
Y
S
+
S
аё
2 ) . 8 a² 8 -
3 a 8 28 = 0,
3 −
whence it follows that
Yas
3 ya² + 4 §, or approximatively, d =
Hence we have
whence
2 S
1
h₁
S
h₂
Y
a
γ
1
α
+
12
S"
– - h₁
γα
= α
γα
4 S'
S
α
α
(a - 207)
(a -
α
4 S
γα
4 S
3
૨
and
а
+
Y
a
α
a²
S
1
+
γ
a
/ (1 + 2 )):
2
a
a
Y
2
Y a²
4 S*
(1 + 2 a²)²]
Y
α
门
​2.
4 S
The mean elevation in capillary tubes is inversely proportional
to the width of the tube.
We have also for the determination of S the formula
S
i a h₁ +
γ
a²
4
Observations made by Hagen with capillary tubes in spring
water gave the following results:

Width of tube a, lines 0,295 0,336 0,413 0,546 0,647 0,751 0,765
Elevation h₁,
Measure of) S
γ
tension
(6
10,08 8,506,87 5,17 4,28
4,283,723,59
grams
1,508 1,455 1,458 1,478 1,473 1,512 1,494
774
[$ 383.
GENERAL PRINCIPLES OF MECHANICS.
According to these experiments the mean values are
S
Y
=
=
1,482 and $ 0,0170 grams.
The variations in these values are due to the fact that the ten-
sion S of the surface of the water diminishes with the time, and is
much smaller in water that has been boiled, than in fresh. We
can now assume that the tension of the water in every strip 1 line
wide is S = 0,0106 to 0,0170 grams.
§383. The foregoing theory is also applicable, when the wall is
not moistened by the liquid; here, however, it is not an elevation
but a sinking of the surface which takes place, and the latter is
concave instead of convex. The vertical force P, which is due to
the difference of level B G and acts from below upwards, is balanced
by the tensions S and S of the surface A B A, Fig. 650, of the
liquid in the tube. The force of adhesion of the solid body does
not, according to the foregoing theory, come into play in this case.
FIG. 650.
D D
FIG. 651.
D D


A
A
H
R
S
B
A
E
H
R
E E
If we make the force, with which the wall of the tube attracts
to itself the column of fluid B G, Fig. 651, proportional to the
circumference of the tube, if, E.G., for a cylindrical tube we put this
force P = µ 2 π α, in which μ denotes a coefficient, we have
παh = 2 μπα,
and, therefore, the mean elevation of the water in the tube is
2 μ
h
a
μ
For two parallel plates, on the contrary, we have P 2µ7 and
P = 2 a h l y, I denoting the undetermined length of the column
of water, and, therefore,
h = "
α
I.E., half as great as in a tube, when the distance 2 a of the plates
383.]
775
THE MOLECULAR ACTION OF WATER.
from each other is equal to the diameter of the tube. This agrees
also with the results of the last paragraph.
According to Hagen's experiments the strength or tension of
the surface of liquid does not depend upon its degree of fluidity,
but it increases in intensity, the more the liquid adheres to other
bodies. According to others, particularly Brunner and Franken-
heim (see Poggendorf's Annalen, Vols. 70 and 72), the height h, to
which water rises in capillary tubes, increases and S consequently
diminishes, when the temperature of the liquid is augmented. For
alcohol S is about one-half and for mercury about eight times the
strength of the surface of water.
REMARK-1) Hagen found by measuring and weighing drops of liquid,
which tore themselves loose from the base of small cylinders, åbout the
same values as he did by his observations upon capillary plates. In like
manner the experiments with adhesion plates have furnished results, which
coincide very well with the former, when we assume that the force neces-
sary to tear the plate loose is balanced by the weight of the cylinder of
liquid raised and by the tension upon the surface of this cylinder.
2) The number of treatises upon capillary attraction is so great that we
cannot cite them all here. The greatest mathematicians, such as La Place,
Poisson, Gauss, etc., have given their attention to it. A complete account
of the older literature is to be found in Frankenheim's "Lehre von der Co-
hasion." The treatise which was specially used in preparing this chapter is
the following: "Ueber die Oberfläche der Flüssigkeiten," by Hagen, a
memoir read in the Royal Academy of Science in Berlin, in 1845. A new
physical theory of capillary attraction, by J. Mille, is contained in Vol. 45
of Poggendorff's Annalen (1838). Here also belong Boutigny's Studies
of Bodies in a Spheroidal Condition.
776
[$ 384.
GENERAL PRINCIPLES OF MECHANICS.
CHAPTER IV.
OF THE EQUILIBRIUM AND PRESSURE OF THE AIR.
FIG. 652.
§ 384. Tension of Gases.-The atmospheric air, which sur-
rounds us, as well as all other gases (Fr. gaz; Ger. gase) possess, in
consequence of the repulsion between their molecules, a tendency
to expand into a greater space. We can therefore obtain a limited
quantity of air only by enclosing it in a perfectly tight vessel. The
force with which the gases seek to expand is called their tension
(Fr. tension; Ger. Spannkraft, Elasticität or Expansivkraft). It
shows itself by the pressure exerted by the gas upon the walls of
the vessel enclosing it, and differs from the elasticity of solids or
liquids in this: it is in action, no matter what the density of the
gas may be, while the expansive force of solids and
liquids is null, when they are extended to a certain de-
gree. The pressure or tension of the air and other
gases is measured by barometers, manometers and valves.
The barometer (Fr. baromètre; Ger. Barometer) is em-
ployed principally to measure the pressure of the atmo-
sphere. The most common kind is the so-called cistern
barometer, Fig. 652; it consists of a glass tube, closed
at one end A and open at the other B, which, after be-
ing filled with mercury, is turned over and placed with
its open end under the mercury contained in the vessel
CD. After the instrument has been inverted, there
remains in the tube a column B S of mercury, which
(see § 374) is balanced by the pressure of the air upon
the surface H R. Since the space A S above the col-
umn of mercury is free from air, the column has no
pressure upon it from above, and the height of this
column, or rather that of the mercury in the same,
above the level HR of the mercury in the vessel can
be employed as a measure of the pressure of the air.
In order to measure easily and correctly this height,
D an accurately graduated scale is added, which can be
moved along the tube and which is sometimes provided with a
movable pointer S.

B
C
R
385.1
EQUILIBRIUM AND PRESSURE OF THE AIR.
777
REMARK.-It is the province of physics to give more detailed descrip-
tions of different barometers, to explain their use, etc. (See Müller's Lehr-
buch der Physik und Meteorologie, Vol. I.)
=
§ 385. Pressure of the Atmosphere.-By means of the
barometer it has been found that in places situated near the level
of the sea, when the atmosphere is in its average condition, the
pressure of the air is balanced by a column of mercury at a tem-
perature of 32° Fahr., 76 centimetres long or about 28 Paris inches
= 29 Prussian inches 29,92 English inches. Since the specific
gravity of mercury at 32° temperature is 13,6, it follows that the
pressure of the air is equal to the weight of a column of water
0,76. 13,6 10,336 metres = 31,73 Paris feet 32,84 Prussian
feet 33,91 English feet. We often measure the tension of the
air by the pressure upon the unit of surface. Since a cubic centi-
metre of mercury weighs 0,0136 kilograms, the atmospheric pres-
sure or the weight of a column of mercury 76 centimetres high, the
base of which is 1 square centimetre, is
p = 0,0136.76
1,0336 kilograms.
But a square inch is 6,451 square centimetres, and therefore the
mean pressure of the air is also measured by 1,0336 . 6,451 6,678
kilograms 14,701 pounds upon a square inch 2116,9 pounds.
upon a square foot. Assuming the exact height of the barometer
to be 28 Paris inches = 29 Prussian inches, we obtain for the
pressure of the air upon one square inch 14,103 Prussian pounds
and upon the square foot 2030 Prussian pounds.
The standard usually adopted, where the English system of
measure is used, is 14,7 pounds upon the square inch, which cor-
responds to a column of mercury about 30 (exactly 29,922) inches
and to a column of water about 34 (exactly 33,9) feet high. It is
very common in mechanics to take the pressure of the atmosphere
as the unit and to refer other tensions to it; they are then given in
pressures of the atmosphere, or simply in atmospheres. Thus a
column of mercury 30. n inches high, or a weight of 14,7. n Eng-
lish pounds, corresponds to the pressure of n atmospheres, and, in-
versely, a column of mercury h inches high to a tension 30
0.03333 h atmospheres and the tension
p
14.7
h
0,06803 p atmo-
spheres to a pressure of p pounds upon a square inch. Besides the
equation
h
29,922
ก
gives the formulas for reduction
14,7
h = 2,0355 p inches and p = 0,4913 h pounds.
778
IS 386.
GENERAL PRINCIPLES OF MECHANICS.
For a tension of h inches = p pounds the pressure upon a sur-
face of F square inches is
P = Fp = 0,4913 Fh pounds
=
Fhy =
2,0355 Fp inches.
EXAMPLE-1) If the level of the water is 250 feet above the piston of a
water-pressure engine, the pressure upon the piston is
250
7,4 atmospheres.
34
2) If the air in a blowing cylinder has a tension of 1,2 atmospheres, the
pressure upon every square inch of the same is
1,2 . 14,7
=
17,64 pounds,
and upon the piston, whose diameter is 50 inches,
π 502
4
17,64
=
34636 pounds.
502
•
. 14,7
= 28863
4
Since the atmosphere exerts an opposite pressure
lbs., the force of the piston is
FIG. 653.
P = 34636 28863 - 5773 pounds.
―
§ 386. Manometer.-In order to determine the tension of
gases or vapors which are enclosed in vessels, we employ instru-
ments, which resemble barometers and are called ma-
nometers (Fr. manomètres; Ger. Manometer). These
instruments are filled with mercury or water and are
either open or closed; in the latter case the upper part
may be free from air or filled with it. The manome-
ter with a vacuum above the column of mercury, as is
represented in Fig. 653, is like the common barometer.
In order to be able to measure with it the tension of
the air in a gasholder, a tube CE is added to it, one
end of which Copens into the gasholder and the other
end E enters above the level of the mercury H R into
the case II DR of the instrument. The space II E R
above the mercury is thus put in communication with
the gasholder; the air existing in this space assumes
the tension of the air or gas in the gasholder and
presses a column of mercury B Sinto the tube, which
balances the tension of the air that is to be meas-
ured.

H
R
C
The syphon manometer A B C, Fig. 654, which is
open at the end A, gives the excess of the tension of the
gas in a vessel above the pressure of the atmosphere;
for that tension is balanced by the combination of the pressure of
the atmosphere upon S and of the column of mercury R S. If b
$ 383.]
779
EQUILIBRIUM AND PRESSURE OF THE AIR.
is the height of the barometer and h that of the manometer, or the
distance R S between the surfaces H and S of the quicksilver in
the two legs of the manometer, the pressure of the air which is in
communication with the short leg will be expressed by the height
of the column of mercury
b₁ = b+h,
1
or by the pressure upon a square inch
p = 0,4913 (b + h) pounds,
or, ifb is the mean height of the barometer,
p = 14,7 +0,4913 h pounds.
The cistern manometer A B C D, Fig. 655, is more common
than the syphon manometer. Since in the former the air acts
upon the column of liquid through the medium of a large mass
of mercury or water, the vibrations of the air are not so quickly
FIG. 654.
A
FIG. €55.
FIG. 656.



S
R
B
H
R
M
D
C
R
A
N
communicated to the column of liquid, and consequently the meas-
urement of the column, which is less agitated, can be made more
easily and more accurately. In order to facilitate the reading of
the instrument, a float, which communicates by means of a string,
passing over a pulley, with a pointer, which is movable along the
scale, is often placed on top of the mercury in the tube.
Manometers can also be used for the purpose of measuring the
pressure of water and other liquids; in this case they are called
piezometers (Fr. piézometres; Ger. Piezometer).
By the aid of a valve D E, Fig. 656, the tension of the gas or
steam, contained in a vessel M N, can be determined, although not
with the same accuracy, by placing the sliding weight G in such a po-
sition that it balances the pressure of the steam. If C S = s is the
distance of the centre of gravity of the lever from the axis of rota-
tion C, CA = a the arm of the lever of the sliding weight and Q
the combined weight of the valve and lever, we have the statical
moment, with which the valve is pressed downwards by the weights,
780
[§ 387.
GENERAL PRINCIPLES OF MECHANICS.
Ga + Q s;
now if the pressure of the gas or steam upwards = P, the pressure
of the atmosphere downwards= P, and the arm of the lever C B
of the valve = b, we have the statical moment with which the
valve tends to open
(P – P₁) b,
equating the two moments, we obtain
-
Pb → P₁ b = Ga+Qs, and consequently,
P = P₁ +
Ga + Q s
b
If r denote the radius of the valve D E, p the interior and p₁
the exterior tension, measured by the pressure upon a square inch,
we have
π p²
p = p₁ +
P = π r² p and P₁ =
Ga + Q s
π jo² b
P1,
π r² р₁, whence
EXAMPLE-1) If the height of the mercury in an open manometer is
3,5 inches and that of the barometer 30 inches, the corresponding tension is
30 + 3,5 33,5 inches, or
h = b + h₁
+
Ρ 0,4913. h
0,4913. 33,5
=
16,46 pounds.
2) If the height of a water manometer is 21 inches and that of the
barometer is 29 inches, the corresponding tension is
h = 29 +
21
13,6
30,54 inches = 15,0 pounds.
3) If the statical moment of a safety valve, when not loaded, is 10 inch-
pounds, if the arm of the lever of the valve, measured from the valve to
the axis of rotation, is b = 4 inches and its radius is r
difference of the pressures upon the valve is
p - Pi
150 + 10
π (1,5)" . 4
160
5,66 pounds.
9 π
If the pressure of the atmosphere were p₁
of the air under the valve would be
p = 20,26 pounds.
1,5 inches, the
14,6 pounds, the tension
§ 387. Mariotte's Law.-The tension of a gas increases with
the condensation; the more we compress a certain quantity of air,
the greater the tension becomes, and the more we expand or attenu-
ate it, the less the tension becomes. The relation between the
tension and the density or volume of gases is expressed by the law
discovered by Mariotte (or Boyle) and named after him. It asserts,
that the density of one and the same quantity of air is proportional to
its tension, or, since the spaces occupied by one and the same mass
are inversely proportional to their densities, that the volumes of one
and the same mass of air are inversely proportional to their tensions.
§ 387.]
781
EQUILIBRIUM AND PRESSURE OF THE AIR.
D
FIG. 657.
If a certain quantity of air is compressed into half its original
volume, that is if its density doubled, its tension becomes twice as
great as it was in the beginning, and if, on the contrary, a certain
quantity of air is expanded to three times its original volume, its
density is diminished to one-third of what it was, and its original
tension is also diminished in the same proportion. If the space
below the piston E F of a cylinder AC, Fig. 657, is filled with
ordinary atmospheric air, which in the beginning
acts with a pressure of 14,7 pounds upon each
square inch, it will act with a pressure of 29,4
pounds, when we move the piston to E, F, and
thus compress the inclosed air into one-half its
initial volume; the pressure will become 3. 14,7
44,1 pounds, when the piston in passing to
E, F, describes two-thirds of the entire height.
If the area of the surface of the piston is one
square foot, the pressure of the atmosphere against it is 144. 14,7
2116,8 pounds; hence, if we wish to depress the piston one-half
the height of the cylinder, we must place upon it a gradually
increasing weight of 2116,8 pounds, and if we wish to depress it
two-thirds of the height of the cylinder, 2. 2116,8 = 4233,6 pounds
must gradually be added, etc.

E
Ei
E2
A
F
F.
B
We can also prove Mariotte's Law by pouring mercury into the
tube Go H, which communicates with the cylindrical air vessel
A C, Fig. 658. If we begin by cutting off a certain volume A (
of air, of the same tension as the exterior air, by
means of a quantity D E F H of mercury, and
G₂ if we then compress it by pouring in quicksilver,
until it occupies one-half, one-quarter, etc., of its
G₁ original volume, we will find that heights G, H,
D
D₁
D
E
FIG. 658.

B
C₂
H₂
H₁
H
F
1
G₂ H, etc., of the surface of the mercury in the
tube are equal to the height of the barometer b
multiplied by one, three, etc. Consequently, if
we add the height corresponding to the pressure
of the atmosphere, we find that the tension is
double, quadruple, etc., that of the original
volume.
The correctness of the law of Mariotte in regard to expansion
can easily be proved by dipping a cylindrical tube (of regular cali-
bre) A B, Fig. 659, vertically into mercury (water) and, after
properly closing the upper end 4, expanding the enclosed volume
782
[§ 387.
GENERAL PRINCIPLES OF MECHANICS.
of air A E (I) by carefully drawing up the tube so that the air
shall occupy a volume A, E, (II). The densities of the air in
I.
FIG. 659.
JA
A
CE,
EC
H
D
D
B
2
B
the spaces A E and A, E, are in-
versely proportional to the heights
A C and A, C₁, and its tensions are
directly proportional to the differ-
ences between the height b of the
barometer and the heights C' D and
R C, D, of the columns D E and D, E,
of mercury standing above the level
HR of the mercury; hence, accord-
ing to Mariotte's law,


1
A C
b ~ C₁ D₁
A₁ C₁
b - C D'
1
which can be verified by observing any given immersion of the
tube A B.
If h and h₁ or p and p, are the tensions, y and y, the corre-
sponding densities or heavinesses, and V and V, the corresponding
volumes of the same quantity of air, we have, according to the
above law,
Y V₁ h
p
or Vy
V₁y, and V₁ pi
1
Vp, whence
1
V
Pi
hr
h
P1
Y₁
Y =
y and V
V
PV.
h
Ρ
hr
P₁
By means of these formulas we can reduce the density and also
the volume of the air of one tension to those of another.
REMARK.-It is only when the pressures are very great that variations
from the law of Mariotte are observed. According to Regnault; when the
volume V of atmospheric air at one meter pressure becomes the volume.
V₁, the pressure is
0,0011054 (1)
(-1) + 0,000019881
V
p
V
V
-1)]meters,
V
so that for
5
10
15
20
V
we have
L
Ρ 4,97944 9,91622
EXAMPLE 1) If the manometer of a blowing machine marks 3 inches,
14,82484
19,71988 meters.
30 + 3
and the barometer stands at 30 inches, the density of the blast is
30
33
1,1 times as great as that of the exterior air.
30
2) If a cubic foot of air, when the barometer stands at 30,05 inches,
§ 388.]
783
EQUILIBRIUM AND PRESSURE OF THE AIR.
weighs
62,425
770
pounds, what is its weight when the barometer stands at 34
inches? Its weight is
62,425 34
•
770 30,05
42,449
462,77
= 0,09173 pounds.
FIG. 660.
H
R
3) How deep can a diving-bell (Fr. cloche à plongeur; Ger. Taucher-
glocke) A B C D, Fig. 660, be immersed in water, when the water is not to
rise in it above a certain height CH=y. In the
beginning the bell with its opening C D stands
above the level of the water H R, so that the
whole space is filled with air at a pressure
equal to that of a column of water, whose height
is b. If afterwards the bell sinks to a
depth 0 C = x and a volume W of water is
thus introduced into it, the volume of the in-
closed air, when none is pressed back through
the hose, becomes V Wand the height of the
water barometer becomes b + x

S
✔
y; hence
W
b + x Y
Ъ
V
V — W
D
whence we obtain
x = y b+
V b
V- W
Wo
? +
V
W
If the mean cross-section of the lower part of the bell
-
F, we can put
Fy and therefore
x = y +
F b
I FY
If the height of barometer = 34 feet of water, the volume of the bell V-
100 cubic feet, the mean cross-section of the lower half F
20 square
feet, and the height, to which the water is to be admitted, is y 3 feet, the
volume of this water is W= Fy
Fy = 20.3 = 60 cubic feet; hence that of
the confined air is
100
W = 40 cubic feet, and its density is
40
23
= 3 + 51 = 54 feet.
times that of the exterior air, and the corresponding depth of immersion is
60.34
40
X 3 +
§ 388. Work Done by Compressed Air.-The energy stored
by a given quantity of air when it is compressed to a certain degree,
as well as that restored by it when it expands again, can not be de-
termined at once; for the tension varies at every moment of the
expansion or compression. We must therefore seek out a particular
formula for the calculation of this quantity. Let us imagine a
certain quantity of air A F to be shut off in a cylinder A C, Fig.
661, by a piston E F, and let us calculate what mechanical effect is
784
[§ 388.
GENERAL PRINCIPLES OF MECHANICS.
= F F₁. If
necessary to move the piston a certain distance E E,
the initial tension = p and the initial height of the space in the
cylinder A Es, and if, on the contrary, the tension after the
space E E, has been described = p, and the height
E, A of the remaining volume of air s₁, we have
the proportion
FIG. 661.
D
C
E
E
P₁: p = s: 81,
=s: s₁, whence p₁ =
=

S
p.
S1
While the piston describes a very small portion
2
E, E = σ of the space, the tension p, can be re-
garded as constant, and the work done is
F
E2
F
A
B
Fps o
$1
Fp, σ =
F denoting the area of the piston.
According to the theory of logarithms,* a very small quantity
X = 7 (1 + x) = 2,3026 log. (1 + x),
7 denoting the Naperian and log. the common logarithm; conse-
quently we can put
σ
σ
Fps == Fps1 (1+0)
But now
? (1 +
σ
81
S₁
l
S
= 2,3026 F'p s log. (1+0).
$1
to
(8₁ + 0) = 1 (8, + 0) — 1 §,;
$1
hence the elementary work done is
Fps
σ
81
= Fps [l (s; + 0) — 7 8,].
Let us imagine the whole space E E, to be composed of n parts,
such as o, I.E., let us put E E, = no, we will then find the work
corresponding to all these parts by substituting in the last formula
successively, instead of s₁, the values s₁ + ¤, 8₁ + 2 0, 8, + 3 0, . . .
up to s₁ + (n 1) o, and instead of 8, + o, the values s, + 2 6,
s₁ + 3 σ, etc., up to s₁ + n σ or s, and if we add the values de-
duced, we will obtain the whole work done while the space s
is described
x²
?༠
+
+ ... (see § 194
and also the Introduction to the Calculus, Art. 19) for a very small a, we
have e= 1+x, and therefore
* According to the series e² = 1 + x + 1.2 1.2.3
7 (1 + x) = x.
$388.] EQUILIBRIUM AND PRESSURE OF THE AIR.
785
1 (8, + 0) — 7 8
7 (8₁ + 2 0) — 1 (8₁ + 0)
1 (s + 3 σ)
o)
1 (8, + 2 σ)
A = Fps
1 (8,
+
n o) — 1 [s, + (n − 1) o]
=
Fps (l s
= Fps [l (s₁
+ n o) — 1 s₁]
l
1 8 ) = Fps 1 (~~);
for the first term in each line is cancelled by the second term in
the next.
S h₁
Since
Pi
we can put the work done
$1
h
Ρ
A = F p s t (h)
= Fps
7 (2)
l
If we make the space described by the piston s
$1
S₁ = x, we
find for the mean value of the pressure on the piston, when the air
is compressed in the ratio.
h₁
Pi
h
p
A
(음​).
P = 4 = P P 1 (2)
X
X
Putting F 1 (square foot) and s = 1 (foot), we obtain the
following formula for the work done.
4 = pl (22) = 2,3026 p log. (2).
A
This formula gives the mechanical effect necessary to transform
a unit of volume (1 cubic foot) of air from a lower pressure or ten-
sion p to a higher one p₁, and in so doing to compress the air into
a volume of (2) cubic feet.
On the contrary,
A = p₁1 ( 2 ) = 2,3026 p, log. (2)
P
expresses the work done by the unit of volume of a gas which passes
from a greater tension p, to a lesser one p.
In order to compress a quantity of air, whose volume is V and
whose tension is p, into a volume V, of the tension P1
work to be done is V p¹ (√),
Ꮮ
P, the
and if, on the contrary, the volume
50
786
[$ 388.
GENERAL PRINCIPLES OF MECHANICS.
V₁ of the tension p₁ becomes a volume V, whose tension is p
V₁
1
V
P₁, the energy restored is
1
1
1
REMARK.--The mechanical effect necessary to produce moderate dif-
ferences of tension (P₁ p), or small changes of volume (V₁ - V) can be
expressed more simply by the formula
(૪
-
1
A = F(P +
2
P¹ ) (
−
P₁ (8 — F
) = P(1-2) (P+P₁)
2
= V (1
−
1
)
p
P1
1
(P + P₁ ),
2
or more accurately by the aid of Simpson's rule, when z denotes the press-
8 + 81 of the piston, by the formula
ure at the middle of the path 2
But now
w
P
z 1
A = V (1-2) (P + 4 + P₁)
6
2
૨
28
2 P1
2
8
p
1 (8 + 8 )
8 + 8₁
p
P + Pi
1
1 +
P1
8P P1
+
+ P₁
P + P1
whence it follows that
A = ↓ V ( 1 − 2) (p
8 (P₁ - p) - 2)
P1
=
V p
+
P₁ + p
P
1
P1
9
EXAMPLE-1) If a blowing machine changes per second 10 cubic feet
of air, at a pressure of 28 inches, into a blast at a pressure of 30 inches,
the work to be done in every second is
237711 . ( 15 – 7 16)
30
A = 17280. 0,4913 . 28 . l
28
237711. (2,708050
2,639057) = 237711. 0,068993
=16400,4 inch-pounds 1366,7 foot-pounds.
=
The approximate formula, given in the remark, gives for this work
30 8.2
28
A =
흉​.
. 237711 +
39618,5. 0,41387
28
58
30
1366,4 foot-pounds.
16396,9 inch-pounds
T π.82 =
2) If under the piston of a steam-engine, whose area is F = π . 8²
201 square inches, there is a quantity of steam 15 inches high and at a ten-
sion of 3 atmospheres, and if this steam, in expanding, moves the piston
forward 25 inches, the energy restored and transmitted to the piston is, if
we assume Mariotte's law to be true for the expansion of steam,
A = 201. 3. 14,70 . 15 l (15+25) = 132961,5 7 §
132961,5. 0,98083 = 130413 inch lbs. 10866 foot-lbs.,
and the mean force upon the piston is, when we neglect the friction and
he opposing pressure,
$389.] EQUILIBRIUM AND PRESSURE OF THE AIR.
787
P =
130413
25
=5217 pounds.
§ 389. Pressure in the Different Layers of Air.-The air
enclosed in a vessel has a different density and tension at different
depths; for the upper layers compress those below them, upon
which they rest; the density and tension are the same in the same
horizontal layer only, and both increase with the depth. In order
to find the law of this increase of the density from above down-
wards, or of the decrease from below upwards, we make use of a
method similar to that employed in the foregoing paragraph.
=
Let us imagine a vertical column A E, Fig. 662, whose cross-
section A B 1 and whose height A F= s. Putting the heavi-
ness of the lowest layer =y and its tension = p, and
the heaviness of the upper layer E F=y, and its
FIG. 662.

D
C
F
E,
E
--
A
B
tension = p₁, we have
Yı
Pi
Ρ
Y
If o denotes the height E E, of the layer E, F, its
weight, which is the decrease of the tension corre-
sponding to o, is
υ= 1.σ. γι =
hence by inversion we obtain
or, as in the foregoing paragraph,
O Y Pi
p
σ
p บ
Y' pi
σ
p
Y
1(1
บ
p
711+
[? (p₁ + v) − 1 p₁].
Pi
Y
If we substitute in it for p, successively p₁ + v, p₁ + 2 v, p₁ + 3 v,
etc., up to p p₁ + (n − 1) v and add the corresponding heights
of the layers of air or values of o, we obtain, exactly as in the fore-
going paragraph, the height of the entire column of air
(1 p₁)
p-1p) = 21(2),
S
= 2 α p
or also
P
S =
1 ( 1 ) = 2,302 2 log. (†),
γ
P
Y
when b and b, denote the tensions and p and p, the corresponding
heights of the barometer in A and F.
Inversely, if the height s is given, the corresponding tension
and density of the air can be calculated. We have
SY
SY
p
Y
p
p
e
,
or y₁ = y e
P₁
Y1
788
[$ 390.
GENERAL PRINCIPLES OF MECHANICS.
in which e = 2,71828 denotes the base of the Naperian system of
logarithms.
REMARK.—This formula is employed for the measurement of heights
by means of the barometer, a subject which is treated in the “Ingenieur,”
page 273. If we neglect the temperature, etc., we can write as a mean value
feet.
8 =
60346 log. (~)
1
EXAMPLE 1) If we have found the height of the barometer at the foot
of a mountain to be 339 and at the top 315 lines, the height of the moun-
tain given by these observations is
s = 60346 log. (313) = 60346. 0,031889
1924 feet.
2) For the density of the air at the top of a mountain 10000 feet high,
we have
1
log.
Y
60346
10000 0,165711, whence
71
Y
71
1,465 and
71
}
1,465
its density is therefore 68 per cent. of that of the air at its foot.
: 0,683;
§ 390. Stereometer and Volumeter.-Mariotte's law finds
a practical application in the determination of the volumes of pul-
verent and fibrous bodies, etc., by means of the so-called stereometer
and volumeter.
Ι
FIG. 663.
II
A
1) Say's Stereometer.-If the glass tube CD, which is immersed
in mercury HD R and at the same time is in communication with
the closed vessel 4 B, Fig. 663, I, is
raised up without being drawn entirely
out of the mercury (II), then, in conse-
quence of the expansion of the enclosed
air, a column CE of air enters into the
tube and a column of mercury D E will
remain behind in the tube, by the aid
of which the diminished tension of the
enclosed air balances the pressure of the
atmosphere.

A
E
B
H
IR
H
D
B
R
0
Now if V, is the volume of the space
ABC, V, the required volume of the
body K, which is placed in it, V the
volume of the column of air C E, b the
height of the barometer and h that of
the column of mercury D E, we have,
according to Mariotte's law, since the same quantity of air occupies
the volume V. V₁, when the tension is b, and the volume V
V₁ + V, when the tension is b — h,
15
§ 390.]
789
EQUILIBRIUM AND PRESSURE OF THE AIR.
V
V₁
1
b
h
;
V₁ − V₁ + V b
hence the required volume of the body is
V₁
h
(1)
h
V.
If we know the volume V, and if, when making the experi-
ment, we draw the tube so far out of the water that the length and
consequently the volume V of the column of air in the tube CD
becomes a certain definite one, and if we observe also the height b
of the barometer and that h of the column of mercury D E, we can
calculate by means of this formula the volume V, of the body K.
2) Regnault's Volumeter.-If the space A B CD, Fig. 664, which
is filled with atmospheric air and which contains also the body K,
whose volume V, is to be determined, is shut off
G
h
FIG. 664.
D
B
M
A
K
N
E
H
F
E
A
D
Bi
1
by the cock C from the exterior air, and if, by
opening the cock E, we let out so much mercury
from the tube D E that its level descends from
M to N, we can again employ (according to
Mariotte's law) the above formula

V₁ – V₁
V
Ъ
h
b
in which we denote the volume of the space
A B C D by V, that of the mercury drawn off
by and the height M N of the same by h. It
follows, exactly as in the above case, that the
volume of the body in A is
то
b
V.
h.
In order to fill the tube D E with mercury
again for the purpose of making a new measure-
ment, we put that tube D E in communication
with the reservoir of mercury G H by turning
the cock E.
3) Kopp's Volumeter.-The pressure of the
air enclosed in the space A B C D, Fig. 665, is
the same as that of the exterior air, when the
surface of the mercury D G touches the lower
opening D of the manometer D E. If by means
of a piston P we press the mercury into D G,
until it rises to a certain height and its surface reaches the point

H
790
[$ 391.
GENERAL PRINCIPLES OF MECHANICS.
S, the enclosed air will be compressed and the mercury will rise a
certain distance h in the manometer, which distance can be read off
upon the scale. If again V, is the volume A B C D of the air,
V₁ the required volume of the body placed in it and V the volume
of the mercury, which has been pressed into the air-vessel, we have
in this case
1
0
V₁ - V₁
V₁ — V₁ — V
b + h
b
and, therefore, the required volume of the body
V₁
V
(
b + h
十九
​V.
h
The constant volumes V, and V, are determined for each par-
ticular instrument by filling them with mercury and weighing the
quantity which they hold.
I
D
FIG. 666.
Π
D
§ 391. Air Pump.-(Fr. machine pneumatique; Ger. Luft-
pumpe.) If we raise the piston K, Fig. 666, of an air pump when
the stop-cock is in the position (I) and
push it down when the stop-cock is in
position (II), it acts as an exhausting or
rarefying pump; if, on the contrary, we
raise the piston when the stop-cock is in
position (II) and depress it when it is in
position (I), it acts as a compressing or
condensing pump. In the first case the
air in the receiver A is more and more
rarefied by the reciprocating motion of the
piston in the cylinder CD, and in the
latter case it is rendered more and more
dense.


H
IK
H
K
1) The Exhaust Pump.-If V is the
volume of the receiver, measured to the
cock H, V, the clearance between H and
the lowest position of the piston, and the volume described by
the piston K, which is also measured by the product F's of the
surface Fof the piston and the space s described by it, the pressure
b of the air originally contained in the receiver becomes, according
to Mariotte's law, at the end of a single stroke of the piston
1
V + V₁
V + V₁ + C
b.
Since
upon the return of the piston the clearance remains filled
with air at the pressure of the exterior air b, if the pressure of the
§ 391.]
791
EQUILIBRIUM AND PRESSURE OF THE AIR.
air in the receiver at the end of the second stroke is denoted by b,
we will have
(V + V₁ + C) b₂ = V b₁ + V₁ b
V, b
V + V₁ + C
1
2
V V₁b
1
V+ V₁ + C
+
2
b₂ = (
b +
V+ V₁ +
1
+ C
V
V V₁ b
Ꮴ
1
+ V₁ b, whence
V₁b
+
V + V₁ + C
1
1
(V + V₁ + C')²
In like manner for the tension b, at the end of the third stroke
we find
(V+V₁ + C) b₁ = Vb₂+ V₁ b, and therefore
V
b₂ = √ √ + √₁₂ + C
V V
V₁b
b3
3
b +
b.
1
V² V₁
b
+
V V, b
(V + V₁ + C¹) ³ (V + V₁ + C)²
V
3
V
V + V₁ + c = (v + V₁ + c ) s + [(v + v₁ + d)
C
+
V
+
+1
V+V₁ + C
+ 1]
V₁ b
V + V₁ + C
and from the foregoing we see that the pressure b, after n strokes,
will be
V
c)%
= ( √ + √ ₁₂+ c ) 0
V
V
d)
+ [ ( x + n +
V
n—1
C
V
. . .
V₁ b
+ (v + n + d) + + 1 ] √ + V₁ + o
V
V V₁ C
V + V₁ + C
If we denote
by p and
b₂
p" b + (1 + p + p² +
or, since the
p" 1
p 1
V₁
ར ་
1
V + V₁ + C
-1
V C®
1
by q, we will have
+ pr−1 ) q b,
sum of the geometrical series in the parenthesis is
1 - p"
(see Ingenieur, page 82), the required final
1 - p
tension is simply
ก
b₂ = [p" + ( — — 1") q] b.
For n∞, p" becomes
=
ble tension is
=
p
0, and consequently the smallest possi-
q b
br
1 - p
V₁ b
C + Vi
If we adopt the same notations as
2) The Condensing Pump.
for the exhaust pump, we have here for the tension of the air at
the end of the first single stroke
V
+ C) b;
(V + V₁) b₁ = (V + V₁ + C) b, whence b₁ = (+₁+ C
1
and for that b, at the end of the second stroke
V+V₁
792
[S 391.
GENERAL PRINCIPLES OF MECHANICS.
(V + V₁) b₂ = V b₁ +
(V₁ + C) b, whence
1
(V + V₁ + C') V b
V₁ + C
1
b₂
+
b
(V + V₁)²
V + V₁ 1
し
​V
V
b+
+ V₁
V + V₁
+ 1)
V₁ + C
1
b.
V + V₁
1
1
In like manner the tension at the end of the third stroke is
found to be
(V + V₁) b₂ = V b₂ + (V₁ + C) b, and therefore
2
V
V
b3
b +
+
V + V₁
V + V₁
V
V + V₁
or putting
V
1
V₁ + C
V + V₁
P1 and
V + V₁
= 91
+ 1]
1
V₁ + C
V
Ꮴ + ", b,
1
b₂ = [p₁² + (1 + p₁ + p₁²) q₁] b.
1
In general, we have for the tension at the end of the nth stroke
of the piston
n
b₁ = [p₁" + (1 + p₁ + pi² + ... + p₁"-¹) q₁] b, or, since
1 + p₁ + p₁² +
2
+ Pi
12
Ρι 1
1 Ρι
12
Pi
1
1
Pi
6. = [p² + ( = 12) 4,] 0.
n
For n = ∞, pi” = 0 and
1
91b
V₁ + C
b,
b.
n
1
P1
V₁ 1
This is of course the greatest tension that can be produced by
this condensing pump.
1
If the clearance V, were = 0, we would have for the exhaust
pump q = 0, whence
n
p" b = (x + c )" ;
V
+
p;
and, on the contrary, for the condensing pump p₁ = 1 and
1- p₁"
μι
n
1
P₁
=n, and consequently
b₁ = (1 + n q₁) b =
= (1 + n
(
V
4) b.
EXAMPLE.-If the volume of the receiver of an air pump is V = 1000
cubic inches and the clearance is 10 cubic inches, while the volume of the
cylinder is 300 cubic inches, the tension of the air after 20 strokes is
1) when rarifying, since
1000
Ρ
= 0,76336 and
1310
10
1
q
= 0,0076336,
1310
131
§ 392.1
793
EQUILIBRIUM AND PRESSURE OF THE AIR.
b₂ = b₂0 = (0,7633620 +
n o
on the contrary,
1
0,7633629
0,76336
•
0,0076336) b
1
(0,0045143 + 0,0321126) b = 0,076269 b ;
2) when condensing, in which case
1000
Ρι
0,99010 and
1010
310
21
0,30693,
1010
b₁ = b20
(0,990120 +
1
0,990120
•
0,30693) b
1 0,9901
598)
0,18046
81954 +
= (0,8195.
0,009901
0,30693) b = 6,414 b.
§ 392. Gay-Lussac's Law. The heat or temperature of
gases has an important influence upon their density and tension.
The more the air enclosed in a vessel is warmed, the greater its
tension becomes, and the more the temperature of a gas, contained
in a vessel closed by a piston, is raised, the more it will expand and
drive the piston before it. Gay-Lussac's experiments, repeated
more recently by Rudberg, Magnus and Regnault, have shown that
for the same density the tensions, and for the same tensions the
volume, of one and the same quantity of air increases with the
temperature. We can place this law by the side of that of Mariotte
and call it Gay-Lussac's Law. According to the latest researches
the increase of the tension of a given volume of air, when heated
from the freezing to the boiling point of water, is 0,367 times the
original tension, or if its temperature is raised that much, the vol-
ume of a given quantity of air is increased 36,7 per cent., when the
tension remains constant. If the temperature is given by the cen-
tigrade thermometer, in which the distance between the freezing
and boiling points of water is divided into 100 degrees, the expan-
sion for each degree is = 0,00367, and for the temperature tº it is
0,00367 tº, or if, on the contrary, we use Reaumur's division of
the same space into 80 degrees, we have the expansion for each de-
gree = 0,00459, or for a temperature of tº,
In England and America the Fahrenheit thermometer is gene-
rally used, in which the boiling point is 212° and the freezing
point is 32°; hence the increase for each degree is
for t° it is 0,00204 († 32).
0.00459 t.
0,00204, and
This ratio or coefficient of expansion d = 0,00367 or = 0,00204
is strictly correct for atmospheric air alone; its value for other
gases is generally smaller, and it varies slightly with the tempera-
ture for atmospheric air.
794
GENERAL PRINCIPLES OF MECHANICS.
[§ 392.
If a mass of air, originally of the volume V, is warmed from
the freezing point to t degrees without changing its tension, its
volume becomes
V = (1 + 0,00367 t) V₁ = [1 + 0,00204 (t - 32°)] Vo
and if it reaches the temperature t₁, the volume becomes
V、
V₁ = (1 + 0,00367 t₁) V =
hence the ratio of the volumes is
[1 + 0,00204 (t,
[1 + 0,00204 (t₁ — 32º)] V.;
(1 + 0,00367 t)
1 + 0,00204 (t
32°)
;
17
(1 + 0,00367 t₂)
1 + 0,00204 (t,
32°) '
T
on the contrary, the ratio of the densities or heavinesses is
Y
T
1
1 + 0,00367 t₁
1 +0,00204 (t₁ — 32°)
Y1
V
1 + 0,00367 t
1 + 0,00204 (t — 32°)'
γ
V₁
1 + & t₁
1 + 8 (t, - 32°)
I
1 + s t
1 + d (t - 32")
or generally
γι
When a change in the tension also occurs, if p, is the tension at
the freezing point, p that at the temperature t and p, that at t₁, we
have
V
=
(1 + 0,00367 t)
Po
Vo
p
V₁ = (1 + 0,00367 t,)
Po Vo
Pi
V
1 + 0,00367 t
Pand
V₁
1 + 0,00367 t₁ p
γ
1 + 0,00367 t, p
or
,
7/1
1 + 0,00367 t
Pi.
Y
1 + 0,0036% t
b
as well as
Y1
1 + 0,00367 t bi
20
b
1 + 0,00367 t Y
P1 b₁
1 + 0,00367 ti yi
When t is given in degrees of Fahrenheit's thermometer, we must
substitute in the latter formulas for 0,00367 t, 0,00204 (t — 32°).
EXAMPLE. If 800 cubic feet of air, at a tension of 15 pounds and
at a temperature of 50° Fahrenheit, are brought, by means of the blow-
ing engine and warming apparatus of an iron furnace, to a temperature
of 392° and to a tension of 19 lbs, its volume will be
1+0,00204 . (392 — 32)
V₁
•
15.800 =
1 1 + 0,00204. (50 — 32)
1,734 12000
1,0367 19
= 1056 cubic feet.
REMARK.--The formula
γ
1
1 + s t ₁
Y1
V
1
1 + s t 1 + S (t − 32)
1 + $ (t₁
32)
$393.]
795
EQUILIBRIUM AND PRESSURE OF THE AIR.
can be employed for solids and for some liquids; but for every solid we
must substitute a different coefficient of expansion, E.G.,
for cast iron, d
for glass,
་་་
Centigrade. Fahrenheit.
0,0000336 0,0000187,
s = 0,0000258 = 0,0000143,
for mercury, d = 0,0001802 = 0,0001001.
§ 393. Heaviness of the Air.-By the aid of the formula
at the end of the last paragraph, we can calculate the heaviness y
of the air for a given temperature and tension. Regnault, by his
recent weighings and measurements, found the weight of a cubic
meter of atmospheric air, at the temperature 0° of the centigrade
thermometer and at a tension corresponding to height of 0,76
meters of the barometer, to be 1,2935 kilograms. Since a cubic
foot (English) = 0,02832 cubic meters and 1 kilogram = 2,20460
pounds English, the heaviness of air under the given conditions is
=2,20460. 0,02832. 1,2935 0,08076 pounds English.
If the temperature is to centigrade, we have for the French
1,2935
measure
Y =
1 + 0,00367 t
kilograms,
and for the English system of measures and Fahrenheit's ther-
mometer
0.08076
Y 1 + 0,00204 († — 32º)*
If the tension differs from the mean tension, or if the height of the
barometer is not 0,76 meters, but b, we have
1,2935
b
Y 1 + 0,00367 t ' 0,76
1,702 . b
1 + 0,00367 t
kilograms,
or, since in England and America the height of the barometer is
generally given in inches, and since 0,76 meters = 29,92 English
inches,
Y
0,08076
Ъ
*
1 1+0,00204 (t – 32") 29,92
0,002699 b
lbs.
1 + 0,00204 († — 32º)
Very often we express the tension by the pressure p upon the
square centimeter or inch, and then we must introduce the factor
p
or by doing which we obtain
14,7*
1,2935
p
1 + 0,00367 t ' 1,0336
p
1,0336
1,2514 p
Y =
1 ± 0,00367 t
kilograms, or
0,08076
p
0,005494 p
Y
lbs.
1 + 0,00204 (t — 32) ˚ 14,7
*
1 + 0,00204 († — 32)
For the same temperature and tension, the density of steam is
about g of that of atmospheric air; hence for steam we have
796
CS 394.
GENERAL PRINC PLES OF MECHANICS.
+
0,8084
0,7821 p
Y
1 + 0,00367 t ' 1,0336
kilograms, or
1 + 0,00367 t
0,050475
Ρ
Y
0,003434 p
+0,00204 (t −32)
pounds.
1 + 0,00204 († — 32) ´ 14,7
EXAMPLE--1) What is the weight of the air contained in a cylindrical
regulator 40 feet long and 6 feet wide, when it is at a temperature of 50°
and its tension is 18 pounds? The heaviness of this air is
0,005494. 18
1,0367
0,098892
1,0367
=0,09539 pounds,
and the capacity of the reservoir is
V =
π . 32. 40 = 1131 cubic feet;
.
hence the air enclosed in it weighs
Vì
0,09539. 1131
=
107,9 pounds.
2) A steam-engine uses per minute 500 cubic feet of steam at a temper-
ature of 224,6° F. and at a tension of 39 inches = 0,4913. 39 19,161
pounds; how much water is needed to produce this steam? The heavi-
ness of the steam is
0,003434. 19,161
1 + 0,00204. 192,6
0,06580
1,393
=
0,04724 pounds;
hence the weight of 500 cubic feet of steam is
H
= 500. 0,04724 = 23,62 pounds.
Vy=
FIG. 667.
§ 394. Air Manometer.-From the results obtained in the
last paragraphs, the theory of the air or closed manom-
eter can be deduced. It is composed of a barometer
tube 4 B, Fig. 667, of regular calibre, the upper part
of which is filled with air and the lower part with
mercury, and of a cistern CER, which also contains
mercury and is put in communication with the gas or
vapor. From the heights of the columns of air and
mercury in A B, the tension can be calculated in the
following manner. The instrument is generally so
arranged that the mercury in the tube and in the
cistern are upon the same level, when the tempera-
ture of the enclosed air is t = 10° Cent. = 50° Fahr.
and the tension in the space ER is equal to the
mean height of the barometer b = 0,76 meter = 29,92
inches.

D
Ꭱ
If, when the height of the barometer is b, a column
of quicksilver rises from the cistern ER into the
tube to a height h, and if the length 4 S of the re-
maining column of air ish, the tension of the
latter is
$ 395.] EQUILIBRIUM AND PRESSURE OF THE AIR.
797
lo
h₂
2
z = (M₁ + h² ) 8₂
b,
b₁
=
h₁ + z = h₁ +
(M+m)
b.
and, therefore, the height of the barometer of the air in E R
'h₁ + h₁₂
Now if a change of temperature takes place, I.E., if the tem-
perature at the time when h, and h, were observed, was not as in
the beginning = t, but = t₁, we have for the tension of the column
of air A S
(h₁₂ + h₂) 0,
1 + 0,00204 (t, 32)
1 + 0,00204 (t 32) hz
and, therefore, the required height of barometer is
1 + 0,00204 (t,- 32) h₁ + h₂
1 + 0,00204 (t-32)
29,92 inches and t = 50° Fahr.
b.
b₁
=
h₁ +
h₂
For b=
h
1
b₁ = h₁ + 28,86 [1 +0,00204 (t, - 32)]
ha
=
hh,+h, denoting the total length of the tube, measured from
its upper end A to the surface HR of the mercury. From the
height of the barometer b inches we obtain the pressure upon each
square inch (English)
Pi
14,7
29,92
h₁ + 14,7.
28,86
29,92
[1 + 0,00204 (t, — 32)]
h
ha
h
= 0,4913 h, + 14,179[1 + 0,00204 (†, — 32)] ½ lbs.
1
h₂
Putting
1 + 8 (t₁ -
1 + 8 (t -
32)
=μ, we have
32)
(b₁— h₁) (h-
h₁)
uh b, and therefore
b₁ + h
'b₁
h
2
h₁
+ 1
2
2
+ ¹)² + (µ b — b₁) N.
By the aid of this formula we can calculate the values of the
divisions of a scale, upon which the pressure b can be read off from
the height of the manometer.
EXAMPLE.—If a closed manometer 25 inches long, at a temperature of
69,8° Fahr., shows a column of air 12 inches long, the corresponding height
of barometer is
1
P1
25 — 12 + 28,86 (1 + 0,00204 . 37,8) £5 13 + 28,86. 1,07707. §§
13 + 64,76 = 77,76 inches, and the pressure on a square inch is
0,4913. 77.76 38,20 pounds.
§ 395. Buoyant Effort or Upward Thrust of the Air.-
The law of the buoyant effort of water against a body immersed in
798
[$ 395.
GENERAL PRINCIPLES OF MECHANICS.
it, discussed in § 364, can of course be applied to bodies in the air.
If V is the volume of the body and y the heaviness of the air, in
which it is placed, the buoyant effort, according to this law, is
P = Vy; if the body has the apparent weight G (in the air), its
true weight (in vacuo) is
G₁ G+ VY.
If, further, y, is the heaviness of this body, we have also
V y₁, and therefore .
G₁
G₁
V
so that we can put
,
γι'
G₁ = G +
G₁ Y
Y₁
or G₁ (Y₁ — Y)
= G Y1, whence it follows that
Y₁
G₁
G.
If the body is weighed upon a scale by a weight G, whose
heaviness is y, the following equation
G₂ =
Y2
Y2
G
holds good; if we divide the last two equations by each other, we
obtain the ratio of the weights
G₁
G₂
11
1
Y
Y₁ Y2
Y2
•
Y2 Yi
1
γ'
Yı
or, approximatively, and generally accurately enough,
G₁
γ
γ
1 +
1+ Y
G₂
Y1
Y2
片
​1
Y1
Y 2
극​),
or also
= 1 + ε
G2
(
1
ㅎ​),
· ɛ, ε₁, and ɛ, denoting the specific gravities of the air, of the body
weighed, and of the weight itself.
ε
In many cases and are such small fractions that they can
ε
€1
E2
be neglected and the true weight can be put equal to the ap-
parent one.
REMARK.—The law of the buoyancy of the air can be employed to de-
termine the force, with which, and the height, to which an air-balloon
(Fr. aérostat; Ger. Luftballon) A B, Fig. 668, will rise. If Vis the vol-
ume of the balloon, G its total apparent weight, including the car, etc., y₁
the heaviness of the external and y, that of the enclosed air, we have the
buoyant effect
P = V Y ₁ = VY₂ + G, and therefore
§ 395.] EQUILIBRIUM AND PRESSURE OF THE AIR.
799
V (1₁Y2) = G;
the necessary volume of the balloon is
FIG. 668.
G
V
Y 1

Y 2
and the heaviness of the external air, when
the balloon attains the greatest height, is
G
Y₁ = 72 + √.
y
From this heaviness, by means of the
formula
Ρ
8
Z
Y
Ρ
({₁₂ ) = ² 1 (²),
γ
found in § 389, we can determine the great-
est height 8, to which the balloon will rise,
by substituting for y the heaviness of the
air at the point of beginning, which must
be calculated according to § 393.
EXAMPLE 1.-What is the ratio of the
true weight of dry hard wood to its appa-
rent weight, when it is weighed by means of brass weights at a tempera-
ture of 32° and when the height of the barometer is 29 inches. The den-
sity of the air is, according to § 393,
y = 0,002699 . 29 0,07827 pounds, that of the wood
71 0,453 . 62,425, and that of brass
1½ = 8,55 . 62,425 (see § 61),
2
consequently the ratio required is
1
G₁
1 +
Ꮐ
G2
0,07827
62,425
•
(0.453-8,55)
1 + 0,001254 . 2,091
1,00262.
Thus we see that one thousand pounds of wood lose about 25 pounds
in consequence of the buoyancy of the air.
EXAMPLE 2.—If the diameter of a spherical balloon is 30 feet and the
heaviness of the matter with which it is filled is ye
if the weight of the balloon with the car and load is G
0,017 pounds, and
500 pounds, the
heaviness of the air at the place, where the balloon ceases to rise, is
G
6 G
3000
71 = 72 +
Y3
V
72 +
=
ñ d³
= 0,017 +
π 30³ = 0,017 + 0,03537
= 0.05237 pounds.
Now if the density of the exterior air at the starting-point is 0,0800
pounds, we have
1
7 (4)
5237
0,4948,
and if we assume the ratio of the pressure per square foot to the heaviness of
the air, I.E.,
will rise
p
= 26210, we obtain the maximum height to which the balloon
8
Y
= 2 2 (~7)
=26210. 0,4948 = 12969 feet.
SEVENTH SECTION.
DYNAMICS OF FLUIDS.
CHAPTER I.
THE GENERAL THEORY OF THE EFFLUX OF WATER FROM
VESSELS.
§ 396. Efflux.-The theory of the efflux (Fr. écoulement;
Ger. Ausfluss) of fluids from vessels forms the first grand division
of hydrodynamics. We distinguish, in the first place, the efflux of
water and the efflux of air, and, in the second place, efflux under
constant and under variable pressure. We will begin with the
efflux of water under constant pressure. We can regard the pres-
sure of water as constant, when the same quantity of water enters
the vessel as is discharged from it, or when the quantity of water
discharged is very small, compared with the capacity of the vessel.
The principal problem to be solved is to determine the quantity of
water or the discharge (Fr. dépense; Ger. Wassermenge), which
passes through a given aperture or orifice (Fr. orifice; Ger. Oeff-
nung) under a given pressure and in a given time.
If the discharge per second = Q, we have the discharge in
t seconds, when the pressure is constant,
T = Qt.
But if we wish to find the discharge per second, we must know
the size of the orifice and the velocity of the effluent molecules of
the water.
To simplify our researches, we assume that the mole-
cules flow in parallel straight lines, and, consequently, form a pris-
§ 397.] THE EFFLUX OF WATER FROM VESSELS.
801
matic stream, vein or jet of water (Fr. veine, courant de fluide
Ger. Wasserstrahl). If F is the cross-section of the stream and v
the velocity of the water, or that of every one of its molecules, the
discharge per second forms a prism, whose base is F and whose
height is r, and, therefore, we have
and
Q
= F v units of volume
G = Fv y units of weight,
y denoting the heaviness of the effluent water or liquid.
EXAMPLE-1) If water flows through a sluice gate, the cross-section of
which is 1,7 square feet, with a velocity of 14 feet, the discharge per
second is
Q
14 . 1,7 = 23,8 cubic feet,
and the hourly discharge is
=
23,8 . 3600 = 85680 cubic feet.
2) If 264 cubic feet of water are discharged in 3 minutes and 10
seconds through an orifice, the area of which is 5 square inches, the mean
velocity of the liquid is
V
264
0 =
Ft
5
264. 144
5.190
=40 feet.
190
144
FIG. 669.
A
B
H
R
§ 397. Velocity of Efflux-Let us imagine a vessel A C,
Fig. 669, which is full of water, to be provided with an orifice F,
which is rounded upon the inside and is
very small, compared to the surface H R of
the water, and let us put the head of water
F G (Fr. charge d'eau; Ger. Druckhöhe),
which is to be regarded as constant during
the efflux, h, the velocity of efflux,
and the discharge per second Q, or its
weight Qy. The work, which this quan-
tity of water can perform while sinking
through the distance h, is = Qhy, and the
energy stored by the discharge, whose weight
is Qy, in passing from a state of rest to the

D
C
velocity v, is
=
Q y (§ 74). If no loss of mechanical effect takes
2 g
place during the passage through the orifice, the quantities of work
are equal to each other, or h Q Y =
Q Y, I.
2 g
va
h
2 g
1
51
802
[$ 397.
GENERAL PRINCIPLES OF MECHANICS.
and inversely
in meters
V
√2 gh,
h = 0,0510 v* and v = 4,429 √h,
and in feet (English),
—
h 0,0155 v² and v 8,025 vπ.
=
The velocity of the effluent water is the same as that of a body
which has fallen freely through a height which is equal to the head of
water.
The correctness of this law can also be shown by the following
experiment. If in the vessel A CF, Fig. 670, we make an orifice
FIG. 670.
A
B
L
H
R
K
directed upwards, the jet FK will rise verti-
cally and will nearly reach the level HR of
the water in the vessel, and we can assume
that it would actually reach it, if all impedi-
ments (such as the resistance of the air, the
friction upon the sides of the vessel, the dis-
turbance caused by the falling back of the
water upon itself, etc.) were removed. Since
a body which rises vertically to the height h
has an initial velocity

v = √ 2 g h,
√2 gh,
it follows that the velocity of efflux must be
V = √2
√2 g h.
For another head of water h, the velocity
of efflux is
C
v₁ = √ 2 g h₁₂
hence we have
v: v₁ =
Vi N h : N h₁ ;
the velocities of efflux are, therefore, to each other as the square roots
of their heads of water.
EXAMPLE--1) The discharge per second through an orifice whose area
is 10 square inches, under a head of water of 5 feet, is
Q
= Fv = 10. 12 √ 2 g h=120. 8,025 √5 = 963. 2,236-2153 cubic inches.
2) In order that 252 cubic inches of water shall pass in one second
through an opening of 6 square inches, the head of water must be
1
Q
2 g
2g
F
0,0155
12
252 2
0,0155
12
•
42² = 2,28 inches.
§ 398.]
803
THE EFFLUX OF WATER FROM VESSELS.
c²
2 g
responding to the height h₁ =
§ 398. Velocities of Influx and Efflux.-If the water flows
in with a certain velocity c, we must add to the mechanical effect
h Qy the energy Qy, possessed by the influent water and cor-
C²
2 g
due to the velocity; hence we
must put
22
(h + h₁) Q y =
Q Y, or h + h₁
2 g
2 g
and the velocity of efflux
v = √2 g (h + h₁) = √2 g h + c².
If the vessel is maintained constantly full, the quantity of the
influent water is equal to the discharge Q, and we can put G c =
Fv, in which G denotes the area of the cross-section HR (Fig.
669) of the water that is flowing in. Putting c =
F v²
F
G
v, we obtain
F
v²
2 g
5-()= [1-()]
(등​)
2 g
h =
2,2
whence
29
v =
ບ
√2 g h
2
11
()
F
G
According to this formula, the velocity increases with the ratio
√2 g h, when
of the cross-sections, and it is a minimum and =
the cross-section F of the orifice of discharge is very small, com-
pared with that G of the orifice of influx, and it approaches nearer
and nearer to infinity, the smaller the difference between the two
2 g h
0
00,
F
orifices becomes. If F G or = 1, we have v =
G
and also c∞o; this infinite value must be understood
thus: if a vessel 4 C, Fig. 671, is without a bottom, water
must flow in and out with an infinitely great velocity or
the stream of liquid G F will not fill the orifice of exit
G c
FIG. 671.
A
B
CD. Putting v =
we obtain
F
h
[-1]
and therefore F =
G
g
2 a h
g
DFC
1 +
c²
-

804
[$ 399.
GENERAL PRINCIPLES OF MECHANICS.
which expression shows that the cross-section F of the discharging
stream is always smaller for a finite velocity of influx than that G
through which the water flows in, and that it therefore does not fill the
orifice of efflux, when the latter is larger than
G
REMARK.-The correctness of the formula
√2 g h
2
>
1 +
2 g h
C²
√1 - (G)
which was first established by Daniel Bernoulli, was afterwards much
disputed. I have endeavored to prove in the "Allgemeinen Maschinen-
encyclopädie," by Hulsse, in the article “Efflux” (Ausfluss), how unfounded
were the representations, which were made.
EXAMPLE.—If water flows from a vessel, whose cross-section is 60 square
inches, through a circular orifice in the bottom 5 inches in diameter under
a head of water of six feet, its velocity is
8,025 √6
8,025. 2,449 19,653 19,653
(0,327)² √0,8931 0,945
2
25 π √1
1 -
4.60
FIG. 672.
20,79 feet.
$399. Velocity of Efflux, Pressure and Heaviness.-The
formulas, which we have found, hold good so long only as the pres-
sure of the air upon the surface of the water is the same as that
upon the orifice of efflux; but if these pressures differ, these formulas
must have an addition made to them. If the sur-
face H R, Fig. 672, is pressed upon by a piston K
with a force P₁, as occurs, E.G., in fire engines, we
can imagine this force to be replaced by the pres-
sure of a column of water. If h, is the height
LK of this column and y the heaviness of the
liquid, we can put
L
А.
P

B
H
R
D
P₁ = G h₁ y.
Substituting for h the head of water h + h₁
P₁
Gy'
which has been increased by h₁ =
tain for the velocity of efflux
h +
P₁
we ob-
G Y
V =
√2 g h +
(h
P₁
Gy
when we assume
F
G
to be very small.
If we denote the pressure
upon each unit of the surface G by p₁, we have more simply
§ 399.]
805
THE EFFLUX OF WATER FROM VESSELS.
P₁
= P19
G
and therefore
2' =
g h
√29 (4+1)
Finally, if we denote the pressure of the water at the orifice of
efflux by p, we can put
*
p = (b + 2) y, or
(+2)
h + Pi
Y.
whence
Y
√292
v =
Y
Hence the velocity of efflux is directly proportional to the square
root of the pressure upon the unit of surface and inversely to the
square root of the density or heaviness of the liquid. When the
pressure is the same, a liquid four times as heavy as another dis-
charges one-half as fast as the latter. Since air is 770 times
lighter than water, it would, if it were inelastic, flow out under the
same pressure 1770 273 times faster than water.
A
FIG. 673.
B
This theory is also applicable to the case where the effluent
water is subjected to the pressure of a column of another liquid.
If above the level H R of the water HER
in a vessel AC D, Fig. 673, there is still
a column of liquid H R₁, whose height
G G₁ h and whose heaviness = y₁, while
that of the water is y, we can replace
the latter by a column of water whose

H
M
H
E
·R-
R
K
F
C
Y1
height is h, without changing the pres-
γ
sure upon HR or causing the velocity v
of the water, which is passing through the
opening F, to vary. Hence if h is the
head E G of water, I.E., the height of the
surface of separation HR above the orifice F, we have the height
D
due to velocity
v2
2 g
= h +
Yi hv
γ
806
[$ 399.
GENERAL PRINCIPLES OF MECHANICS.
and therefore
Y1
v =
√2gh+
hr).
Y
Now if y₁ < y or h+h₁ < h+h₁, the jet FK, which rises
γι
γ
vertically, will not reach the level H₁ R, L₁of the surface of the
liquid.
If the surface of separation H R, Fig. 674, is not above, but a
FIG. 674.
M
AK
A
H
B
R,
1
D
H
C
R
certain distance E F = h_below the
orifice F of the vessel A D C, while the
surface H, R, of the liquid H, DR is
at the height G G, =h, above the sur-
face of separation H R, we have

v²
γι h₁ h,
Y
2 g
and therefore the velocity of the jet
Y
V = √29 1, − h).
Y
This supposes,>h, or >
h
γ
Y
γι
From this it is easy to see that the jet
FK, which is projected vertically up-
wards, can rise above the surface H, R, of
Y'1
the liquid H, DR. If G M = h₁ is
Y
the head of the liquid, reduced to that of water, M gives the level
to which the jet will nearly reach.
If the water does not discharge freely, but under water, a dimi-
nution of the velocity of efflux takes place owing to the opposite
pressure. If the orifice F of the vessel A C,
h below the
Fig. 675, is at a distance F G
upper level H R of the water and at a dis-
tance F G₁ h, below the lower level H, R₁,
we have the pressure from above downwards
p = hy,
FIG. 675.
A
HGR
B

Hi
-Ri
D
F
and the opposite pressure from below up-
wards
P₁ = h₁ y;
Y
hence the force, which produces the efflux, is
Y
p - p₁ = (h — h₁) y
§ 399.]
807
THE EFFLUX OF WATER FROM VESSELS.
and the velocity of efflux is
v =
√2 g
Ρ
P₁
Y
= √2 g (h — h₁).
When water discharges under water, we must regard the differ-
ence of level h
water.
FIG. 676.
A
-
h₁ between the surfaces of water as the head of
If the water at the orifice of efflux is pressed
upon with a force p and at the surface or ori-
fice of influx with a force p₁, we have in general

v =
√2gh+
Pi
p
Y
H
D
B
C
R
E
This case occurs when water flows from one
closed vessel A B C into another closed one
D E, Fig. 676. Here h is the height F G of
the surface of the water H R above the orifice
F, p₁ the pressure of the air in A HR and p
the pressure of the air or the steam in D E.
EXAMPLE—1) If the piston of a fire engine is 12
inches in diameter and it is pressed down in the
cylinder with a force of 3000 pounds, and if there are
no resistances in the pipes and hose, the water will
pass through the nozzle of the hose with a velocity
P.
P1
V =
√ 2g
√29
γ
2 J G Y
= 8,025
3
=
8,025 64.
81に
​π
T
3000
62,5
•
4
62,74 feet;
if the stream is directed vertically upwards, it will reach a height
h = 0,0155. v² = 61,007 feet.
2) If water flows into a space in which the air has been rarified, E.G.,
into the condenser of a steam engine, while its upper surface is pressed
upon by the atmosphere, we must employ the last formula for the velocity
of efflux, viz.,
v =
√29 (h
Ρι
P
g h +
J
If the head of water is h = 3 feet, the height of the barometer of the exte-
rior air 29 inches and that of the enclosed air 4 inches, we have
1
P₁ - P
= 29
4 25 inches
γ
2,083 feet of mercury
13,6. 2,083 = 28,33 feet of water,
808
[$ 400.
GENERAL PRINCIPLES OF MECHANICS.
hence the velocity of the water flowing into the space, which is filled with
rarefied air, is
v = 8,025 √3 + 28,33
=
8,025 √31,33
44,92 feet.
3) If the water in the feed-pipe of a steam boiler stands 12 feet above
the level of the water in the boiler and if the pressure of the steam in the
latter is 20 pounds and that of the exterior air is 15 pounds, the velocity
with which the water enters the boiler is
v = 8,025 12 +
(15
20). 144
62,5
5. 144
8,025 12
62,5
8,025 √12
11,52
= 5,56 feet.
§ 400. Hydraulic or Hydrodynamic Head.—If the water
in a vessel is in motion, it presses less against the sides of the ves-
sel than when it is at rest. We must, therefore, distinguish the
hydraulic or hydrodynamic from the hydrostatic head of water.
If p, is the pressure upon each unit of the surface of the water
H₁ R₁
G₁, Fig. 677, p the pressure at the orifice F and h the
head of water F G₁, we have the velocity of efflux
Οι
=
v = V2g (h+
Ρι
p
P1 p
γ
√1-(6)
v²
g
A + " — " = [(1 − ( ) ] 27, ;
h
Y
;
now if in another section H₁₂ R₂ G2, which is at a distance
A
H
I-L
FIG. 677.
But
,,
C
B
R₂
Pr
γ
F v
G₁
FG, h, above the orifice, the pressure is
2
= P2, we have in like manner
P₂- P
P2
h₁ + = [1 − (~~)'] %;
γ
F2
2 g

If we subtract these two equations from each
other, we obtain
F
()
དུ
h - h₁ + ²² = "² = [( C ) - ( 6 ) T
P₁ P2
Y
or, if we denote the head of water G₁ G₂ of the
layer H, R₂ = G₂ by h2, we have for the hydro-
dynamic head at H, R₂
2
P1
F2
= M + 2)² = [(G)² - ( 7 ) ] ; ;
[(罰​)-()]]
γ.
2 g
is the velocity v, of the water at the upper surface G₁,
and the velocity v of the water in the cross-section G₂, we
Fv
G₂
can, therefore, put
§ 400.]
809
THE EFFLUX OF WATER FROM VESSELS.
P2
P₁
+ h₂
Y
Y
The hydraulic head
P2
γ
h₂ - (
V₂
2
g
2
").
at any position in the vessel is equal to
Pi
the hydrostatic head
+ h₂, diminished by the difference of the
Y
heights due to the velocities of the water at this point and at the inlet
orifice. If the free surface G, of the water is very great, we can
neglect the velocity of influx and put
Pa
P₁
V 2
+ h₂
;
γ
γ
2
g
hence the hydraulic head is less than the hydrostatic head by an
amount equal to the height due to the velocity of the water. The
quicker the water moves, the less it presses upon the sides of the
pipe. For this reason pipes often burst or leak for the first time,
when the motion of the water is checked, when the pipes clog, etc.
By means of the apparatus A B C D, represented in Fig. 678,
H
KUI MICROBE-SUOMET KORIUMILES/VELO
E1
A
D
FIG. 678.
G2
B
E2
K
R
the difference between the hydraulic and
the hydrostatic head can be ocularly dem-
onstrated. If from the cross-section G,
we carry a tube ER upwards the latter
will fill with water, which will rise above
the level H R of the water, when G₂ > G₁ or
v, < v₁; for, since the pressure p₁ upon
the surface of the water is balanced by the
pressure of the air upon the mouth of the
tube, we can put the height, which meas-
ures the pressure in G2,

2
and x ish, when
x =
pa
Y
v,2
く
​2 g
2 g
=
h
(3
2.
V 1
g
2 g
If, on the contrary, the cross-sec-
tion G3 < G₁, the water flows more rapidly through G₁, and we
have for the height of column of water in the tube E₁, inserted at Gs,
y = b - ( - )
N3
g 2
which is less than ha, so that the water does not rise to the level
HR of G₁. If, finally, G, is very small and the corresponding ve-
810
[$ 401.
GENERAL PRINCIPLES OF MECHANICS.
locity very great,
FIG. 679.
A
R
H
E1
D
FIG. 680.
A
K
B
2
vš
V.
2 g
E2
K
R
z g
can be > h4, and the corresponding
hydraulic head z will be negative, I.E. the
pressure of the air on the outside will be
greater than that of the water within.
Hence, if a tube is carried downwards and
its end placed under water, a column of
water E, A will rise in it, which, together
with the pressure of the water, will bal-
ance the atmospheric pressure. If the
tube is short, the water in the vessel K,
which, in this experiment, should be col-
ored, will rise in the tube, enter the reser-
voir A B C D and flow, with the other
water, out at F
REMARK.-If the vessel A C E, Fig. 680, consists of a reservoir AC and
of a narrow vertical tube C E, the hydrodynamic head is
negative in all parts of this tube. If we do not regard the
pressure of the atmosphere p₁, the pressure of the water at
the orifice of efflux is 0; for here the entire head of water
is expended in producing the velocity = √2gh; on the
v
contrary, for a position D, E, which is at a distance G₁ G
h₁ under the water level, the hydraulic head is


EGD1
E
D
§ 401.
formula
= h₁ — h = — (h — h₁),
1
or negative; if, then, a hole were bored in this tube, no water
would escape, but, on the contrary, air would be sucked in
and discharged at F. This negative pressure is a maximum
directly under the reservoir, since h, is here a minimum.
Rectangular Lateral Orifices.-By the aid of the
Q
= F v = F√2 g h,
2
the discharge per second can be calculated only when the orifice is
horizontal, since in that case the velocity is uniform in the whole
cross-section F; but if the cross-section is inclined to the horizon,
if, E.G., the opening is in the side of the vessel, the molecules of
water at different depths flow out with different velocities, and the
discharge can no longer be regarded as a prism; hence the formula
Fv = F√2 g h cannot be applied directly. The general for-
Q
=
mula is
Q =
F, √ 2 g h + F₂ √ 2 g h₂ + Fz √2 g hs + ...
= √ 2 g (F; √ h₁ + F ₂ √ h₂ + F ₂3 √ π3 + .),
§ 401.]
811
THE EFFLUX OF WATER FROM VESSELS.
in which F1, F2, F3... denote the areas and h₁, h₂, h.... the heads
of water of the various portions of the orifice.
The simplest case is that of efflux through a notch in the side,
weir or overfall, Fig. 681. The notch D E G H in the wall, through
A
FIG. 681.
B
which the efflux takes place, is rec-
tangular; let us denote its width.
DEGH by b and its height
DH = E G by h. If we decom-
pose this surface bh, by horizontal
lines, into a great number n of hor-
izontal strips of equal width, we can
consider the velocity to be constant
for each of them. Since, if we pro-
ceed from above downwards, the
heads of water of these strips are

h
2 h 3 h
etc.,
N N
N
we have for the corresponding ve-
locities

√29
•
h
n'
2 h
N
N 2g. V
3 h
2 g.
etc.,
N
h
b h
•
we have
N
N
and since the area of each of these strips is b
the corresponding discharges
b h
h b h
2 h b h
3 h
29
•
N
ف22
2g.
2 J
etc.;
N
N
N
n
=
N
(√ 2
hence that of the whole section is
b h
h
2 J.
+
2 J
•
2 h
N
3 h
+1
V 2 g
+
•
....)
N
N
b h N 2 g h
n v n
(
√ I + √ 2 + √ 3 + . . . + √ ñ ).
Since (as is given in the Ingenieur, page 88)
√1 + √2 + √3 + ... + √n,
or
n'
+
1ª + 2ª + 3ª + ... + n³
1 + 1/
it follows that the required discharge is
11
{n} = j n √ n,
812
[S 401.
GENERAL PRINCIPLES OF MECHANICS.
√
b h № 2 g h
Q
=
n N n
z n N n = z b h N2 g h = b √2gh'.
3
If we understand by the mean velocity v that velocity, which must
exist at all points of the overfall, when the same quantity of water
passes through the whole cross-section with a uniform velocity as does
pass through with the variable velocity, we can put
Q
bh v, whence it follows that
v = 3 √ 2 g h,
I.E. the mean velocity of water flowing out through a rectangular
notch in the side of a vessel is the velocity at the sill or lower edge
of the notch.
A
If the rectangular orifice K G, Fig. 682, with the horizontal
FIG. 682.
•
C
base G H, does not reach to the level
of the water, we find the discharge
through it by regarding it as the dif-
ference between two notches in the
side D E G H and D E LK. If h₁
is the depth E G of the lower and
h₂
EL that of the upper edge, we
have for the discharges through these
notches

z b v 2 g h ³,
3
and
B
z N
Q = b √ 2 g h r
3
3 b √ 2 g h 2
3 •
hence the discharge through the rec-
tangular opening G H K L is
3
–
3 b 1 2 g h₂ = 3 b 1/2 g (h,³ — h¸³),
and the mean velocity of efflux is
8
Q
h₁
v =
3
b v 2 g
•
b (hr
12)
h₁
hos
ha
or the depth of
If h is the mean head of water E M =
h₁ + h₂
2
the centre of the orifice below the level of the water, and a the
height KH= LG h, h, of the orifice, we can put
v = √ 2 g.
1
= -
( n + 2)² - ( r — 9
) ²
a
(1)² ] 12 gh.
= [1 − b (9)'] ₁
or approximatively
$ 402.]
813
THE EFFLUX OF WATER FROM VESSELS.
EXAMPLE.-If a rectangular orifice of efflux is 3 feet wide and 11 feet
high and the lower edge is 23 feet below the level of the water, the dis-
charge is
Q
=
= . 8,025. 3 (2,75; — 1,51 )
2.
16,05 2,723 = 43,7 cubic feet.
According to the approximate formula
16,05 (4,560 — 1,837)
= 11,698
=
о
= [1 -
1
(1,135)']. 8,025 √2,125 =
(1 - 0,0036) 11,698
0,042
11,656 feet,
5
43,710 cubic feet.
and the discharge is, therefore,
Q = 3. §. 11,656
REMARK.-If the notch in the wall is inclined to the horizon at an
angle d, we must substitute for the height of the orifice
1
hạnh
he instead of
sin. б
h₁
1
h2, and therefore we must put
Q
ace
7 √2 g
sin. S
(√h,
3
√h₂³).
If the cross-section of the reservoir, from which the water is dis-
charging, is not much larger than the cross-section of the orifice, we must
take into account the velocity of approach v₁
F
1 G
v of the water and put
Q = fb √z g [ (h
1
2 h +
2 g
- (2₂
1
h₂ +
FIG. 683.
§ 402. Triangular Lateral Orifice.-Besides rectangular lat-
eral orifices, triangular and circular ones also occur in practice.
We will next discuss the discharge through a triangular orifice
DE G, Fig. 683, with a horizontal base E G and with its apex D
at the level of the water H R. If we put the base
E G = b and the height D E = h and if we divide
the latter into n equal parts and pass through these
divisions lines parallel to the base, we divide the
entire surface into small strips, whose areas are
h 2 b h 3 b h
etc., and whose heads of
H K
DR

b
N
N
N N
N
h 2 h 3 h
water are
etc.
n'
The discharges through them are
ጎ
N
b h
h 2 b h
2 h 3 b h
n2
√29
3 h
2 g
n N n."
N
Nº
129
etc.,
N
by summing these we obtain the discharge of the whole triangular
orifice
814
[§ 402.
GENERAL PRINCIPLES OF MECHANICS.
h
b h
Q =
2 g
n"
22
bh √ 2 g h
n² v n
(1 + 2 √ 2 + 3 √ 3 + . . . + n √ n)
(1; + 2; + 3§ + ... + n³),
or, since the series in the parenthesis
MIN
no + 1
+1
Q = & bh √ 2 g h =
2
12gh.
{{૭
= ni,
If the base D K of the orifice D G K lies in the surface of the
water and the apex G is at the depth h below it, we have the cor-
responding discharge, since that through the rectangle D EGK
is 3 b h v 2 gh,
z
Q = q b h √ 2 g h − 3 b h √ 2 g h = {z b h √ 2 gh.
15
The discharge through a trapezium A B C D, Fig. 684, whose
upper base A B = b₁ lies in the surface of the water, whose lower
base is C D = b, and whose height is D E = h, is found by com-
bining the discharge through a rectangle with those through two
triangles, and it is
Q = 3 b₂ h v 2 g h + † (b₁ − b₂) h √ 2 g h
༡
4
15
5 (2b+ 3b) h V2gh.
FIG. 684.
HA E
F BR
FIG. 685.
HA
BR


M
Further, the discharge through a triangle CD E, Fig. 685,
whose base is D E = b₁, whose altitude is 0 M = h, and whose
apex C'is situated at a depth O C = h below the level H R of the
water, is discharge through A B C minus that through AE
Q
2
= b h √ 2 g h − (2b+ 3 b,) h, √ 2 gh,
1 5
15
= 2√2 g [2b (ht — h,) — 3 b, h,].
15
Since the width A B = b is determined by the proportion
b: b₁ : : h: (h - h₁), it follows that
FIG. 686.
Q
H
R
2 √ 2 g . b, (2 h (h² — h‚³) — 3 h‚³)
እ)
(2h =
15
h - h₁

2 √2 g.br (2 h -5hh+3h3
‚³ + 3 h‚³).
15
h-h
Finally, we have for the discharge through a
triangle A CD, Fig. 686, whose apex lies above
I
its base,
§ 403.]
815
THE EFFLUX OF WATER FROM VESSELS.
2 1/2 g. b₁ (2 h − 5 hh, + 3 h
Q = 3 √2 g. b₁ (h} — h¸³)
15
- 3 h,i)
h — hr
1. (3 ht
h
J
h₁
2 1/2 g.b₁ (3 h — 5 h, hi + 2 h
15
1
EXAMPLE.—What is the discharge through the square orifice A B C D,
Fig. 687, whose vertical diagonal A C=1 foot, when the corner A reaches
to the level of the water? The discharge through the upper half of the
square is
Q = z b √ Q g h³ ₫ . 1. 8,025 √
and that through the lower half
1,605. 0,7071
=
1,135 cubic feet,
2 b √2 g (2 hr
h
H
FIG. 687.
A
1
1
15
h
h₁ 1
5AA, 1+3A, 1)
R

2.8,025 /2
5 (1) + 3 (1)
15
1 - 1
32,10
(2 — 1,7678 + 0,5303)
15
2,14. 0,7625 = 1,632 cubic feet,
consequently the total discharge is
FIG. 688.
FIN G
Q = 1,135 + 1,632
2,767 cubic feet.
§ 403. Circular Lateral Orifices.-The discharge for a cir-
cular aperture 1 B, Fig. 688, can only be determined by means
of approximate formulas obtained in the follow-
ing manner. Let us decompose the circular ori-
R fice by concentric circles into small rings of
equal width and let us consider each ring to be
composed of elements, which may be regarded as
parallelograms. If r is the radius, b the width
and n the number of elements of one of these
2 π r
is the length of one of these elements
N

B
rings,
and its area is
2 π r b
K =
N
Now if h is the depth C G of the centre C below the level of the
water H R and the angle A CK, which measures the distance of
the element from the highest point A of the ring, we have for
the head of water of this element
KN = CG CL
=
h
r cos. P,
and therefore the discharge through this element
2 π r b
N
√2 g (h r cos. 4).
816
GENERAL PRINCIPLES OF MECHANICS.
[§ 403.
But
√ h
r cos. o
7cos. -
= √h 1
[1
h
cos.³ ¢ + · · ·]
2
h
= √π [ 1 - 1 1/2
до
1
cos.
h
16
T'S (7)*
h
(1 + cos. 2 ø) + ...}
Υπη
2 π r b
and therefore the discharge through this element is
γ
√ 2 g h [1 − 1 . ½
cos.
1
16
n
h
T's (2)* (1 + cos. 2 p) + ...]
The discharge through the whole ring is found by substituting
in the parenthesis instead of 1, n. 1 = n, and instead of cos. ø the
sum of all the cosines of o from 4 = 0 to 4 = 2 π, and instead of
0 to 2 = 4 π.
cos. 2 the sum of all the cosines of 2 p from 2
Since the sum of all the cosines of a full circle is equal to 0, these
cosines disappear, and we have the discharge through the ring
2 π r b
N
= 2 = r b√2 g h
16
2
N
e (')'- ··· ]
... }
...]
√2 gh [1
[n
1
T'o (2)².
[1
−
16
r
r 2 r 3 r
If, instead of b, we substitute
and instead of r,
to
m
m' m
m
m r
M.
we obtain the discharge through each of the rings, which form
the entire circle, and finally the discharge through the entire circu-
lar aperture is
Q=2πr √2gh (2 (1+2+3+...+m) — 1's
2πη
m²
2 π r √ Q a h
√2 gh.
653
== r² √2 gb [1 —
π
or more exactly
h1
1
'
32
Q
= π r² √2 g h [
[1
m²
•
6
m* h²
² (1³ + 2² + 3³ +
...
+m³))
203
m"
m¹ h²
4
16
2
(†)'
-···})
2
5
3'2 (1) ' — 1021 (1) * - ··· ]
If the circle reaches to the level of the water, we have
Q = 987 π r² √2 g h 0,964 F 12 gh,
1024
when F = π ² denotes the area of the circle.
Moreover, it is easy to understand that in all cases, where the
head of water at the centre is equal to or greater than the diameter
of the orifice, we can put the value of the entire series = 1 and
Q = F√2 gh.
This rule can also be applied to other orifices and also to all
§ 404.]
THE EFFLUX OF WATER FROM VESSELS.
817
cases, where the depth of the centre of gravity of the orifice below
the level of the water is as great as the height of the aperture; we
can then regard the depth h of this point as the head of water and
put
Q = F √2 g h.
If we consider that the mean of all the cosines of the first
quadrant is =
2
π
>
2
π
and that of all those of the second quadrant is
or that the mean of the first and second quadrant
= 0, the
discharge for the upper semicircle, determined in the manner
shown above, is
Прод
2
2
√ 2 g h |
7[1-
= F V 2 g h [1 −
gh[1 -
п
32
327 (17) - 12 (7)]
32²
( 2 ) − ✯ (1)']
п
and that through the lower semicircle is
π 7.2
Q. = =
2
π
√28 [(1 +
= F √ 2 g h [1
gh[1
+
2
2 グ
​32
3 = ( ) - » (')' + · · ·]
п
3 2
3 = ( ) − s (1) ' + ···]
п
32
in which F denotes the area of the aperture.
The formulas for Q, Q. and Qe hold good also for elliptical
orifices with horizontal axes; for the discharges, when the other
circumstances are the same, are proportional to the widths of the
apertures and the width of an ellipse is proportional to the width
of an equally high circle (see Introduction to the Calculus, Art. 12).
EXAMPLE. What is the hourly discharge through a circular orifics 1
inch in diameter, when the level of the water is one line above the top of it?
Here we have
and
до
h
1
(1)²
+; hence ( (5)
3 (
2
36
=
49
0,735,
= 1 1- 0,023
=
0,977,
}
and consequently the discharge per second is
π. 12
7
Q
4
12. 8,025
1
•
144
per minute
0,0977 = 8,025. 0,977 √7=16,29 c. inches.
4
977,4 cubic inches, and per hour 33,94 cubic feet.
§ 404. Efflux from a Vessel in Motion-The velocity of
efflux changes when a vessel, originally at rest or moving un-
formly, is set in motion, or when a change in its condition of
motion takes place, since in this case every molecule of the water
acts upon those surrounding it not only by its weight, but also
by its inertia.
52
818
[$ 404.
GENERAL PRINCIPLES OF MECHANICS.
If the vessel A C, Fig. 689, is moved with an accelerated motion
vertically upwards, while the water flows through an opening Fin
FIG. 689.
K
G
E
A
D
C
F
B
the bottom, the velocity of efflux is
augmented, and if it descends with an
accelerated motion, the velocity is dimin-
ished. If the acceleration is p, every
molecule M of the water presses not
only with its weight Mg, but also with
its inertia M p, and in the first case we
must put the force of each molecule
equal to (g + p) M, and in the second
case equal to (g − p) M, or instead of
9, g±p. Hence it follows that

and that the velocity of efflux is
v2
(g±p) h,
2
V √2 (g ± p) h.
If the vessel rises with the velocity g, we have
v = √ 2.2 g h = 2 Ńgh,
No
and the velocity of efflux is 1,414 times as great as it would be if
the vessel stood still. If the vessel falls by its own weight or
with the acceleration g, v is = 0 and no water runs out. If
the vessel moves uniformly upwards or downwards, v remains
12 gh, but if its rise is retarded, v becomes 12 (gp) h, and if
its fall is retarded, v is
A
FIG. 690.
X
B
√2 (g+p) h.
If the vessel, from which the water flows,
is moved horizontally or at an acute angle
to the horizon, the surface (see § 354) be-
comes oblique to the horizon and a varia-
tion of the velocity of efflux is the result.
If a vessel 4 C, Fig. 690, is caused to
revolve about its vertical axis XI, its sur-
face will assume, according to § 354, the
shape of a parabolic funnel A O B, and at
the centre M of the bottom the head of
water M O is smaller than near the edge,
and the water will flow more slowly through
an orifice at the centre than through any
other equally large aperture in the bottom.
If h denotes the head of water M O at the centre M, the velocity

D
X
$ 404.]
819
THE EFFLUX OF WATER FROM VESSELS.
if
of efflux through an aperture at that point will be = N2gh; but
У
denotes the distance MF NP of an aperture F from the
axis XX and w the angular velocity, we have, since the subtan-
gent TN of the arc OP of the parabola is equal to twice the abscissu
O N, the corresponding elevation of the water above the centre O
ON TNP N. tang. N P T,
w² y
consequently if we substitute tang. N P T = tang. p = (see
g
§ 354) and denote the angular velocity wh of F by w, we can put
ON = x = y.
11
W³ y
g
w³ y³
2 g
22
2 g
Hence the velocity of efflux through the orifice Fis
D
F
D
v =
√2 g (k +
w²
2 g.
FIG. 691.
X
"TO MIGDAGBOKUSEMA USTALLINENS
ابلت
-X
E
B
= √2 g h + w³.
This formula holds good for a
vessel of any shape, even when it
is closed on top, like A C, Fig. 691,
in such a manner that the fun-
nel DOC cannot be completely
formed. Here also h is the depth
MO of the orifice below the vertex
O of the funnel and v the velocity
of rotation of the aperture. It will
be employed repeatedly in the dis-
cussion of reaction wheels and tur-
bines in another part of the work.
EXAMPLE-1) If the vessel A C, Fig. 689, which when filled with water
weighs 350 pounds, is drawn upwards by a weight G of 450 pounds by
means of a cord passing over a pulley, it rises with an acceleration

450 - 350
p
•
g
450 + 350
100
800
•
9 = 19,
and the velocity of efflux is
v = √2 (g.+ p) h =
Now if the head of water were h
√2. & gh
// = √ gh.
4 feet, the velocity of efflux would be
v = √9. g 3 √32,2 = 17,02 feet.
2) If the vessel A C, Fig. 691, which is filled with water, makes 100
revolutions per minute and if the orifice F is 2 feet below the level of the
water at the centre and at a distance from the axis XX,
velocity of efflux is
3 feet, the
√2 g h + w³ =
64,4.2 +
3. π . 1001?
30
= √128,8 + 100. π²
=
√128,8 + 987 = V
√1115,8= 33,4 feet.
If the vessel stands still, we have v = √128,8
= 11,35 feet.
820
[$ 405.
GENERAL PRINCIPLES OF MECHANICS.
CHAPTER II.
OF THE CONTRACTION OF THE VEIN OR JET OF WATER WHEN
ISSUING FROM AN ORIFICE IN A THIN PLATE.
§ 405. Coefficient of Velocity. The laws of efflux, deduced
in the last chapter, coincide almost exactly with the results ob-
tained in practice, so long as the head of water is not very small,
compared to the width of the aperture, if the orifice of efflux is
gradually widened inwards and joins bottom or sides without
forming an angle or edge. The experiments made with polished
metal mouth-pieces by Michelotti, Eytelwein and others, and also
by the author, have shown that the real effective discharge is from
96 to 99 per cent. of the theoretical one. The mouth-piece A D,
Fig. 692, which is represented in one-half its natural size, gave
under a pressure of 10 feet 98 per cent.,
under a pressure of 5 feet 97 per cent., and
under a pressure of 1 foot 96 per cent. of
the discharge calculated theoretically (Ex-
periments with large orifices, see Unter-
suchungen in dem Gebiete der Mechanik
und Hydraulik, Zweite Abtheil.). If the
efflux through such a mouth-piece is to be
as free from disturbance as possible, the
rounding must not be in the form of a
circle, but in that of a curve A D
B C,
A
FIG. 692.
D
C
B

the curvature of which gradually decreases from within outwards
(from A towards D). Since in this case the stream has the same
cross-section F as the orifice, we can assume that the diminution
of the discharge is caused by a loss of velocity arising from the
friction of the water upon, or its adhesion to, the inner surface of
the mouth-piece and from the viscosity of the water. Hereafter
we will call the ratio of the real or effective velocity to the theo-
retical velocity v = √2 gh the coefficient of velocity (Fr. coefficient
de vitesse; Ger. Geschwindigkeitscoefficient) and we will denote
it by . Thus the effective velocity of efflux in the simplest case is
v₁ = p*v = ¢ √2 gh,
· φυ Ф
403.] CONTRACTION OF THE VEIN OF JET OF WATER. 821
and the effective discharge is
feet)
φ
V2
Q = F v₁ = & F v = F √2 gh.
Substituting for its mean value 0,975, we obtain (in English
= 0,975. 8,025 F√h = 7,824 F Wh.
Q = 0,975. F√2 g h h =
The vis viva of a quantity Q of water, issuing with the velocity
V₁, is
Q Y
g
•
v,2, by virtue of which it can perform the mechanical
2
vi
effect Q Y⋅ 2 g
But since the weight Qy in descending from the
༡༡༠
height h
performs the work Q y . h
2 g
the loss of mechanical effect of the water during the efflux is
v2
=
Q x
it follows that
2 g
L= QY (1₁₂-22)
22
= (1 − ø³) Q Y ·
I
g
2g
= (1 -- 0,975³) QY⋅ 2 g
I.E.,
༩
0,049
or 4,9 per cent.
2 g
L =
The water, which issues from the vessel, will therefore perform
4,9 per cent. less work by virtue of its vis viva than by virtue of its
weight, when falling from the height h.
REMARK.—The author has tested the law of efflux, expressed by the
formula v = √√2 g h, under very different heads, viz., from the very great
head of 100 meters to the very small one of 0,02 meters. A well rounded
mouth-piece 1 centimeter wide gave for the heads

h= 0,02 meters. 0,50 meters 3,5 meters
0,959
0,967
0,975
17 meters
103 meters
0,994
0,994
See Civilingenieur, New Series, Vol. 5, first and second numbers.
§ 406. Coefficient of Contraction.-If the water issues from
an orifice in a thin plate (Fr. orifice en mince paroi; Ger. Mün-
dung in der dünnen Wand), and if the other circumstances are the
same, a considerable diminution in the discharge takes place. This
diminution is due to the fact that the directions of the molecules
of the water, which are passing through the orifice, converge and
produce a contracted stream or vein (Fr. veine contractée; Ger.
contrahirter Wasserstrahl). The measurements of the stream, made
S22
[§ 403.
GENERAL PRINCIPLES OF MECHANICS.
by several experimenters and more recently by the author himself,
have shown that the stream, at a distance from the orifice equal to
half its width, experiences its maximum contraction, and that its
thickness is 0,8 of the diameter of the orifice. If F is the cross-
section of the contracted vein and F that of the orifice, we have
therefore
F₁
F
F= 0,8² F = 0,64 F.
The ratio of these cross-sections is called the coefficient of
contraction (Fr. coefficient de contraction; Ger. Contractionscoeffi-
cient), and is denoted by a; from what precedes we see that its
mean value for the efflux of water through an orifice in a thin
plate is a = 0,64.
•
So long as we have no more accurate knowledge of the law of
the contraction of the stream, we can assume that the stream flow-
ing through a circular orifice A A, Fig. €93, forms a solid of rota-
tion A E E A, whose surface is generated by the revolution of the
arc A E of a circle about the axis CD of the stream. Putting the
གན་ཟརྒྱུུ་། ང་ད་ད་བ་ཅུ
FIG. 693.
H
R
•
•
J
diameter A A of the orifice
d
and the distance CD of the con-
tracted section E E from the orifice
d, we obtain the radius
MA — ME

of the generating arc A E by means
of the equation
A N² – EN (2 ME - EN)
ď²
or
4
M
M
NEDEN
d
10
(2 r — 21/14),
(21
d
10
from which we obtain
r = 1,3 d.
The velocity of efflux through orifices of this kind is about
V₁
0,97 v.
The contraction of the stream of water owes its origin to the
fact that not only the water immediately above the orifice flows
out, but also that the water all around flows in and is discharged
with it. The filaments of water begin to converge within the
vessel, as is shown in the figure, and the contraction of the stream
is caused by the prolongation of this convergence. We can con-
vince ourselves of this fact by employing a glass vessel and putting
into the water small bodies, such as saw-dust, bits of sealing-wax,
§ 407.]
823
CONTRACTION OF THE VEIN OR JET OF WATER.
etc., of nearly the same specific gravity as the water, and allowing
them to flow out with it.
§ 407. Contracted Vein of Water.-If the water flows
through triangular, quadrangular, etc., orifices in a thin plate, the
stream assumes particular forms. The most striking phenomenon
is the inversion of the stream or the change in position of its cross-
section in reference to the cross-section of the orifice, in conse-
quence of which a corner of the former cross-section comes into
the same position as the middle of one of the sides of the orifice.
Thus the cross-section of the stream, issuing from a triangular ori-
fice A B C, Fig. 694, is, at a certain distance from the latter, a
three-pointed star D E F; that from a square orifice A B C D,
Fig. 695, is a four-pointed star E F G H; that from a pentagonal
FIG. 694.
E
A
F
FIG. 695.
FIG. 696.

F

G
B H

A
B
A
C
E
G
B
K
F
D
C
E
D
D
H
L
orifice A B C D E, Fig. 696, is a five-pointed star E G H K L, etc.
The cross-sections are very different at different distances from the
orifice; they decrease for a certain distance and then increase again,
etc.; the stream consists, therefore, of ribs of variable width and
forms, as can be best observed when the pressure is very great,
bulges and nodes similar in form to the cactus plant. If the ori-
fice A B C D, Fig. 697, is rectangular, the cross-section at a small
B
FIG. 697.
E
F
distance from the aperture forms also
a star or cross, but at a greater dis-
tance it assumes more the form of an
inverted rectangle E F
Bidone observed the discharge
from various kinds of orifices; Pon-
celet and Lesbros have made the
only accurate measurements of the
stream issuing from square orifices
(see the Allgemeine Maschinenency
klopädie, article "Ausfluss"). The
last measurements have led to a small
coefficient of contraction 0,563.

824
[§ 408.
GENERAL PRINCIPLES OF MECHANICS.
Measurements of the water discharged through smaller openings
have given greater coefficients of contraction; they indicate that
the coefficients are greater for oblong rectangles than for rectangles,
which approach the square in form.
§ 408. Coefficient of Efflux. If the effective velocity of
water issuing from an opening in a thin plate was equal to the
theoretical v = √2g h, we would have for the effective discharge
= a F v = a F√2 g h,
Q
a F denoting the cross-section of the stream at the point of maxi-
mum contraction, where the molecules of water move in parallel
lines; but this is by no means true. It appears, from experiment,
that Q is smaller than a F 2 g h and that we must multiply the
theoretical discharge FV2 g h by a coefficient smaller than the co-
efficient of contraction, in order to obtain the real discharge. We
must therefore assume that, when water issues from an orifice in a
thin plate, a certain loss of velocity takes place, and consequently
a coefficient of velocity must also be introduced; hence the effec-
tive velocity of efflux is
v₁ = 4 v = √ 2 gh.
The effective discharge is
Q₁
1
1
p
F₁. v₁ = a F. v = a + Fv a o F√2 g h.
φ = φ
Let us call the ratio of the real discharge Q, to the theoretical or
hypothetical discharge Q the coefficient of efflux (Fr. coefficient de
dépense; Ger. Ausflusscoefficient) and let us denote it hereafter
by μ; then we have
and therefore
Q₁ = µ Q = µ Fv µF √2gh,
μ =
μ = α φ,
I.E. the coefficient of efflux is the product of the coefficient of velocity
and the coefficient of contraction.
Repeated observations, and particularly the measurements of
the author, have led to the conclusion that the coefficient of efflux
is not constant for all orifices in a thin plate, that it is greater
for small orifices and small velocities of efflux than for large
orifices and great velocities and that it is much greater for long,
narrow orifices than for those whose forms are regular or circular.
For square orifices, whose areas are from 1 to 9 square inches,
under a head of from 7 to 21 feet, according to the experiments of
§ 409.]
825
CONTRACTION OF THE VEIN OR JET OF WATER.
Bossut and Michelotti, the mean coefficient of efflux is µ
μ =
= .0,610;
for circular orifices from to 6 inches in diameter and under a head
of from 4 to 21 feet, it is u= 0,615 or about 8. The values, which
were obtained by Bossut and Michelotti from their observations,
differ materially from each other; but they do not appear to de-
pend upon the size of the orifice or upon the head. According to
the experiments of the author, under a head of 0,6 meters, the co-
efficient of efflux is for a circular orifice
1 centimeter in diameter
2 centimeters
3
66
4
66
66
(6
μ =
0,628
= 0,621
0,614
= 0,607.
On the contrary, under a head of 0,25 meters, with the same orifice,
1 centimeter in diameter, he found.
2 centimeters
CC
66
66
μ = 0,637
= 0,629
= 0,622
3
66
4
(C
66
(6
0,614.
We see from these results of experiment that the coefficient of
efflux increases when the size of the orifice and the head of water
diminish. If we assume as mean values µ = 0,62 and a = 0,64,
we obtain the coefficient of velocity for efflux through an orifice in
a thin plate
a
0,97,
or about the same as for efflux through mouth-pieces rounded in-
ternally.
REMARK-1) Experiments made by Buff (see Poggendorff's Annalen,
Vol. XLVI) show that the coefficients of velocity for small orifices and
small heads or velocities are considerably greater than for large or medium
orifices and velocities. An orifice of 2,084 lines in diameter gave, under a
head of 1½ inches, μ = 0,692 and, under a head of 35 inches, µ = 0,644.
On the contrary, an orifice 4,848 lines wide, under a head of 4 inches,
gave μ = 0,682 and, under a head of 29 inches, u = 0,653. The author
also obtained similar results.
·
2) For efflux under water, according to the experiments of the author,
the coefficients of velocity are nearly 13 per cent. smaller than for efflux
into the air.
§ 409. Experiments.-The coefficient of effluxu correspond-
ing to a certain mouth-piece can be determined, when we know the
discharge V, which passes through the known cross-section F of
the orifice under a head of water h in a certain time t; here we
have
826
[$ 409.
GENERAL PRINCIPLES OF MECHANICS.
and inversely
V = µ F√ 2 g h . t,
V
μ Ft. 12 g h
In order to find its two factors, viz.: the coefficient of contrac-
tion and that of velocity, it is necessary to measure either the cross-
section F a F of the stream or to determine the velocity of
efflux v₁ = 0 v = 4 12 g h by means of the range of the jet.
Neither measurement can be made with sufficient accuracy unless
the stream is thin and the cross-section is circular.
The circular cross-section F of a jet can be determined very
simply by means of the apparatus represented in Fig. 698. It is
FIG. 698.
F.
composed of a ring and four sharp-pointed set-
screws A, B, C, D, which screw in towards each
other. The screws are directed towards the
centre of the cross-section of the stream and are
turned until their points touch its surface; the
ring is then removed from the stream and the
distance between the opposite points of the
screws is measured; the mean d, of these two
distances is assumed to be the diameter of the
stream. Now if d is the diameter of the cross-section of the orifice,
we have

and therefore
a =
F
F
=();
d
>
fl
a
•
If we measure the range B C = b of a jet A B, Fig. 699, which
issues horizontally from the mouth-piece SA, which is at a certain
height AC a above the ground, we have, according to § 36, the
-'ocity of efflux
FIG. 699.

$
M
F
E
D
G
A
K
$409.1 CONTRACTION OF THE VEIN OR JET OF WATER.
827
g b²
21 =
2 a
and since v₁ = ¢ v
1
=
whence
=
4 v = p √ 2 g h, we obtain
V1
υ
b2
b
"
4 a h
2 V a h
2 V a h
α
Φ
μό
The determination of v is more certain when, instead of a and
b, we measure the horizontal and vertical co-ordinates of three
points of the parabolic axis of the stream; for the axis of the
mouth-piece may have an unknown inclination to the horizon.
The most simple method of proceeding is to stretch a horizontal
thread D F above the stream and to hang three plumb-lines from
three points D, E, and F, which are at equal distances from each
other; we then measure the distances D G, E H, and F K of the
axis of the stream from DF. If D F = x is the horizontal dis-
tance of the extreme points from each other, if the vertical dis-
tances D G, E H, and F K = 2, 21, and z, and if we take G as the
origin of co-ordinates, we have the co-ordinates for the point H
x₁ = G L = DE=DF and y₁ =LH=EH — D G = z₁ — %,
and for the point K
= x
X
=3ar
= =
=21 2,
= 22 Z.
x= GMDF and y = MK FK DG Z₂- %.
According to § 39, if a denotes the angle of inclination of the
axis of the stream at G,
J xj
2 vi cos. a
2
2
I x²²
vi cos.² a'
gx
2
2
2
2 >
2 v₁* cos.* a
1'1
2
g xs
and also
or
and
Y₁ = x₁ tang. a +
Y₂ = x2 tang. a +
2
Y1
x₁ tang. a =
Y2
x₂ tang, a =
whence, by division, we obtain, since
= 2 x1,
19
3/1
x, tang, a
Y 2
x₂ tang. a
42
1, and therefore tang, a =
4 y₁ - Y?
2
1
If in one of the foregoing formulas, instead of
we put 1+
cos.³ a
2 v² cos.³ a
tang. a, and for tang. a we substitute the last expression, we obtain
the required formula for the velocity of efflux
828
[S 410.
GENERAL PRINCIPLES OF MECHANICS.
ປີ 1
2 (Y₂
g x²
x tang. a) cos.²' a
√/ I [x* + (4 Y, — Y₁)³]
4 (y₂-2 y₁)
Hence the coefficient of velocity is
(1 + tang. a²) y x²
2 (2y, -4 y₁)
V₁
21
• =
1/
(x² + (4 Y₁ — Y2)*
Yı
V √ 2 g h
Y
2
8 h (y₂- 2yr)
EXAMPLE 1) The following measurements of an uncontracted stream,
which issued from a well-rounded orifice 1 centimeter wide, were made:
x = 2,480 meters,
z = —
0,267 0,1135 = 0,1535 meters,
1
Y₁ = 21
Y 2 22
2 = 0,669
0,669 -0,1135
and the head of water was h = 3,359 meters.
the coefficient of velocity to be
$ = √
2,48² + 0,0592
8. 3,359. 0,2485
0,5555
(6
From these data we find
6,185
26,872. 0,2485
μ
=0,963.
Since no contraction took place, a = 1 and therefore
1 and therefore up. The results
of measurements given in the remark to § 405 agree well with this value.
2) The measurements of a perfectly contracted stream, which passed
through a circular orifice in a thin plate, were, for a head of water h =
3,396 meters, the following:
x = 2,70 meters,
2 = 0,2465 0,1115 = 0,1350 meters,
0,6620 – 0,1115
1
Y₁ = 21
2
Y ₂ = 2 2
2
whence it follows that
2,70² + 0,012
&
1
8. 3,396. 0,2805
0,5505
7,2901
1
=0,978.
27,168. 0,2805
From the measurement of the discharge µ was calculated to be 0,617;
μ
hence the coefficient of contraction was a = = 0,631, which agreed very
well with the measurement of the cross-section of the stream.
§ 410. Rectangular Lateral Orifices.-The most accurate
experiments upon efflux through large lateral rectangular orifices
are those made at Metz by Poncelet and Lesbros. The width of
these apertures were 2 and in some cases 6 decimeters and their
heights were different, varying from 1 centimeter to 2 decimeters.
In order to produce a perfect contraction, the orifice was made in a
brass plate 4 millemeters thick. From the results of these experi-
ments, these savants have calculated, by interpolation, the tables,
which are given at the end of this paragraph, and which can be
employed for the measurement or calculation of discharges.
Ifb is the width of the orifice A L, Fig. 700, and if h, and h
$410.] CONTRACTION OF THE VEIN OR JET OF WATER. 829
are the heights E G and E L of the level of the water above the
lower and upper horizontal edge of the orifice, we have, according
to § 401, the discharge
Q
= {bV2g(h – họ).
(h₂
If we introduce the height of the orifice G L = a =
and the mean head of water E M = h
matively
α
a²
h₁ ha
h₁ + h₂
hṛ we have approxi-
2
Q = (1-6)) a b √2 g k
96
2
and, therefore, the effective discharge is
A
D
B
FIG. 700.
wo
-
-
•
Q₁ =μ Q = (1
C
-
a²
96 h³) μ a b v T g h .
If we put
a²
(1 - 987) A
м.
2

96 h³) μ = 14,
we have more simply
Q₁ = µ₁ a b √ Q gh,
v
and as it is more convenient to employ
this simple formula for the discharge,
the values of μ₁, and not those of μ
are given.
Since the water in the neighborhood
of the orifice is in motion, it stands
higher immediately in front of the
wall, in which the aperture is made
for this reason two tables are given,
one to be used, when the heads of water are measured at a distance
from the orifice, and the other, when they are measured directly at
the wall of orifice. We see from both these tables that, with some
exceptions, the less the height of the orifice and the head of water
is, the greater the coefficient of efflux is.
If the width of an orifice is different from those given, we
must employ these coefficients to calculate the discharge, as we
have no other experiments to base our calculations upon. That
we are not liable to great error can be seen by comparing the co-
efficients for the orifices, whose widths are 0,6 meters, with those,
whose widths are 0,2 meters, for the same head of water.
the apertures are not rectangular, we determine their mean height
and width and substitute in the calculation the coefficient corre
sponding to these dimensions. It is always better to measure the
head of water at a great distance from the orifice and to employ
If
830
[§ 410.
GENERAL PRINCIPLES OF MECHANICS.
the first table than to measure it immediately at the orifice, where
the surface of the water is curved and less tranquil than at a dis-
tance from it.
EXAMPLE-1) What is the discharge through an orifice 2 decimeters
wide and 1 decimeter high, when the surface of the water is 1 meters
above the upper edge? Here we have
h₁ + h ₂
1,6 + 1,5
b = 0,2, a = 0,1, h
1,55 meters,
2
2
=
0,1103 cubic meters.
and, therefore, the theoretical discharge is
Q = 0,1 . 0,2 √2 g √1,55 = 0,02 . 4,429 . 1,245
But Table I gives for a = 0,1 and h₂ = 1,5, µ1
real discharge is
Q
0,611. 0,1103 0,0674 cubic meters.
0,611, hence the
2) What is the discharge through a rectangular orifice in a thin plate,
whose height is 8 inches and whose width 2 inches, under a head of water
of 15 inches above the upper edge? The theoretical discharge is
Q = .. 8,025 √
1
0,8917. 1,1547 1,0296 cubic feet.
But two inches is about 0,05 meters and 15 inches about 0,4 meters,
we can therefore take the value μ₁ 0,628, corresponding to a =
h ₂
h₂
0,4, and put the required discharge
Q ₁
0,628 . 1,0296 = 0,647 cubic feet.
0,05 and
3) If the width is 0,25 meter, the height 0,15 and the head of water
0,045, we have
Q = 0,25 . 0,15 . 4,429 √0,12
0,166. 0,3464 0,0575 cubic meters;
=
the height 0,15 corresponds, for h₂
0,04, to the mean value
0,582 + 0,603
μ 1
0,5925,
2
and, for h₂
== 0,05, to
μli
0,585 + 0,605
2
0,595.
Now since h₂
0,045 is given, we substitute the new mean
0,5925 + 0,5950
2
0,594
as coefficient of efflux, and we obtain the required discharge
Q₁ = 0,594. 0,0575 0,03415 cubic meters.
1
==
REMARK.--The coefficients of velocity do not change sensibly for a rec-
tangular orifice, when we change the height into the width or vice versa,
as is demonstrated by the following experiments of Lesbros (see his "Ex-
periences Hydrauliques, Paris, 1851").
An orifice 0,60 meters wide and 0,02 meters high, under a head of water
from h
0,30 to 1,50 meters, gave
μ₁ = µ = 0,635 to 0,622,
1
μ
and, on the contrary, when it was set on edge, or when the height was 0,60
meters and the width 0,02 meters,
μ₁ = 0,610 to 0,626 and
μ 0,638 to 0,627.
=
$410.] CONTRACTION OF THE VEIN OR JET OF WATER.
831
TABLE I
The coefficients of efflux of water issuing from rectangular orifices in a thin
vertical plate, according to Poncelet and Lesbros.
(The heads of water are measured above the orifice at a point where the
water can be considered as still. The values below the asterisk (*) are de-
termined only by interpolation.)
HEIGHT OF THE ORIFICE, IN METERS.
Head of water or dis-
tance of the level of
the water above the
upper edge of the
orifice, in meters.
Width of the orifice = 0,2 meters.
Width of the orifice
= 0,6 meters.
0,20
0,10
0,05
0,03
0,02
0,01
0,20
0,02
0,000
((
(L
3
(6
(6
66
CC
(6
0,005
(6
66
((
66
0,705
((
0,010
(C
((
0,607
0,630
0,660
0,701
(C
0,644
0,015
(6
0,593 | 0,612 0,632
0,660
0,697
(4
0.644
0,020
0,572 0,596 0,615 | 0,634
0,030 0,578 0,600
0,659
0,694
66
0.643
0,620 0,638 0,659
0,688 0,593
0,642
་
0,040 0,582 0,603
0,623
0,640 | 0,658
0,683 0,595
0,642
0,050 0,585 0,605
0,625
0,640 0,658
0,679
0,597
0,641
0,060 0,587 0,607
0,627
0.640 0,657
0,676
0,599
0.€41
0,070 0,588 0,609
0,628
0,639
0,656 0.673 0,600 0.€40
0,080 0,589 0,610 0,629
0,638
0,656
0,670
0,001
0.640
0,090 0,591 0,610 0,629
0,637
0,655
0.668
0,601 0,639
0,100 0,592 0,611 0,630
0,637
0,654
0,666
0,602❘ 0,639
0,120 0,593 0,612
0,630
0,636
0,653 0,663
0,603 0.638
0,140 0,595 0,613
0,630
0,635 0,651
0,660
0,603
0.637
0,160 0,596 0,614
0,180 0,597 0,615
0,200 0,598 0,615
0,250 0,599 0,616 0,630
0,631
0,634 | 0,650
0,658
0,604 0,637
0,630
0,634 0,649
0,657
0,605 0,636
0,630
0,633 0,648 0,655
0,605 0,635
0,632
0,646 0,653
0,606 0,634
0,300
0,600 0,616 0,629
0.632
0,644 0,650
0.607 0,623
0,400
0,602❘ 0,617 0,628
0,631
0,642
0,647
0.007
0,631
0,500
0,603 0,617 0,628
0,600 0,604 0,617 0,627
0,700 0,604 0,616 0,627
0,800 0,605 0,616 0,627
0,900 0,605 0,615 0,626
1,000 0,605 0,615 0,626
1,100 0,604 0,614 0,625
1,200 0,604 0,614 0,624
1,300 0,603 0,613 0,622
0,625 0,622 0,603 0.624
1,400 0,603 0,612 0,621 0,622 0.622 0,618 0,603 0.624
1,500 0,602 0,611 0,620 0,620* 0,619* 0,615* 0.602 0,623
1,600 0,602 0,611 0,618 0,618 0,617 0.613 0,602 0.023
1,700 0,602* 0,610* 0,617 0,616 0,615 0,612 0,602 0.022
1,800 0,601 0,609 0,615* 0,615 0,614 0,612 0.602
1,900 0,601 0,608 | 0,614 0,613 0,612 0,611 0,602
2,000
0,601 0,607 0,613 0,612 0,612 0,611 0,602 0,620
3,000 0,601 0,603 0,606 | 0,608 | 0,610 0,609 0,601 0,615
0,630 0,640
0,644
C,607
0,630
0,627 | 0,630 | 0,638
0,642
0,607 0.629
0,629 0,637
0,640
0,607 0.628
0,629 | 0,636
0,637
0,606
0,628
0,628 0,634
0,635 0.606
0.627
0,628 0,633
0,632 0.605
0.626
0,627
0,631
0,629 0.604
0.626
0,626
0,628 0,626
0,604
0.625
0,624
0,€21*
0,021
Similar tables for the Prussian system of measures are to be found in
the Ingenieur, page 432.
832
[§ 410.
GENERAL PRINCIPLES OF MECHANICS.
TABLE II.
The coefficients of efflux of water issuing from rectangular orifices in a thin
vertical plate, according to Poncelet and Lesbros.
(The heads of water were measured directly at the orifice. The values
above and below the asterisks (*) are determined by interpolation only.)
Head of water or dis-
tance of the surface
of the water above
the upper edge of the
orifice, in meters.
HEIGHT OF THE ORIFICE, IN METERS.
Width of the orifice = 0,2 meters.
Width of the
orifice
0,6 meters.
0,20
0,10 0,05
0,03
0,02
0,01
0,20
0,000
0,619
0,667 0,713 0,776 0,783
0,795
0,586
0,005 0,597
0,010
0,630* 0,668* 0,725* 0,750*
0,778*
0,587
0,595 0,618 0,642
0,687 0,720
0,762
0,589
0,015 0,594 0,615
0,639
0,674 0,707
0,745
0,590
0,020
0,594* 0,614
0,638
0,668
0,668 | 0,697
0.729
0,591
0,030
0,593 0,613
0,637
0,659 0,685
|
0,708
0,592
0,040 0,593 0,612
0,636
0,654 0,678
0,695
0,594*
0,050 0,593 0,612
0,636
0,651
0,672
0,686
0,595
0,060 0,594 0,613
0,635
0,647
0,668
0,681
0,596
0,070 0,594 0,613
0,635
0,645
0,665
0,677
0,597
0,080 0,594 0,613
0,635
0,843
0,662
0,675
0,598
0,090 0,595 0,614
0,634
0,641 0,659
0,672
0,599
0,100 0,595 0,614
0,634
0,640 0,657
0,669
0,600
0,120 0,596 0,614
0,633
0,637
0,637 0,655
0,665
0,601
0,140 0,597 0,614
0,632
0,636 | 0,653
0,661
0,602
0,160 0,597 0,615
0,631
0,635 0,651
0,659
0,602
0,180 0,598 0,615
0,631
0,634 0,650
0,657
0,603
0,200 0,599 0,615
0,630
0,633 0,649
0,656
0,603
0,250 0,600 0,616
0,630
0,632 0,646
0,653
0,604
0,300 0,601 0,616 0,629
0,400 0,602 0,617 0,629
0,500 0,603 0,617 0,628
0,600 0,604 0,617
0,700 0,604 0,616 0,627
0,800 0,605 0,616
0,632
0,644
0,651
0,605
0,631 0,642
0,647
0,606
0,630 0,640
0,645
0,607
0,627
0,630 | 0,638
0,643
0,607
0,629 0,637
0,640
0,607
0,627
0,629 | 0,636
0,637
0,607
0,900 0,605 6,615
0,626
0,628 | 0,634
0,635
0,607
1,000
0,605 0,615
0,626
0,628 | 0,633
0,632
0,606
1,100 0,604 0,614
0,625
0,627 | 0,631
0,629
0,606
1,200 0,604 0,614
0,624
0,626 0,628
0,626
0,605
1,300 0,603 0,613
1,400 0,603 0,612 0,621
1,500 0,602 0,611 0,620
1,600 0,602 0,611 0,618
1,700 0,602* 0,610* 0,617
1,800 0,601 0,609 0,615
1,900 0,601 0,608 0,614
2,000 0,601 0,607 0,614
3,000 0,601 0,603 0,606
0,622
0,624 0,625
0,622
0,604
0,622❘ 0,622 0,618
0,603
0,620* 0,619*
0,615*
0,603
0,618 0,617 0,613
0,602
0,616 0,615
0,612
0,602
0,615 0,614
0,612
0,602
0,613 | 0,613
0,611
0,602
0,612 0,612
0,611
0,602
0,608 0,610
|
0,609
0,601
§ 411.]
833
CONTRACTION OF THE VEIN OR JET OF WATER.
FIG. 701.
§ 411. Overfalls.-If the water flows through an overfall, weir
or notch (Fr. déversoirs; Ger. Ueberfälle) in a thin wall, as, E.G., F B,
Fig. 701, the stream is contracted
on three sides and a diminution
of the discharge is produced. The
discharge through this orifice is
Q = 3 µ b h N Q g h.
μ

A
H
EL
D
C
B
Here the head of water E H = h
is to be measured, not at the edge,
but at least three feet from the
wall in which the notch is cut; for the surface of the water is de-
pressed immediately behind the orifice, and the depression increases
continually towards the orifice, and in the plane of the orifice its
value G R is from 0,1 to 0,25 of the head of water F R, so that the
thickness F G of the stream is but 0,9 to 0,75 of the head of water.
Many experiments have been made upon efflux of water through
notches in a thin plate, and the results, although very multifarious,
do not agree as well as could be desired. The following tables con-
tain the results of the experiments of Poncelet and Lesbros.
1. TABLE OF COEFFICIENTS OF EFFLUX FOR OVERFALLS
TWO DECIMETERS WIDE, ACCORDING TO PONCELET AND
LESBROS.

Head of water h 0,01 0,02 0,03 0,04 0,06 0,08 0,10 0,15 0,20 0,22
| |
in meters.
Coefficient
of efflux
μ ₁ = 1 μl.
1
0,424 0,417 0,412 0,407 0,401 0,397 0,395 0,393 0,390 0,385
2. TABLE OF THE COEFFICIENTS OF EFFLUX FOR OVERFALLS
SIX DECIMETERS WIDE.

60,08 |
Head of water h
in meters.
0,06 0,08 0,10 0,12 0,15 0,20 0,30 0,40 0,50 0,60
Coefficient
of efflux
• M.
0,412 0,409 0,406 0,4030,400 0,395 0,391 0,391 0,391 0,390
Hence for approximate determinations we can put µ,
0,4.
53
834
[S 412.
GENERAL PRINCIPLES OF MECHANICS.
=
=
Eytelwein found, by his experiments with overfalls of great width,
the mean value of µ, to be = μ 0,42, and Bidone µ₁ 30,62
0,41, etc.
The most extensive experiments were made by d'Au-
buisson and Castel. From these d'Aubuisson concludes that for
overfalls, whose width is not greater than that of the canal or of
the wall in which the weir is placed, we can put µ = 0,60 or 3 µ
0,40; that, on the contrary, when the overfall extends across the
whole wall or has the same width as the canal, we must take μ
0,665 or µ, = 0,444; that, finally, when the relations between the
width of the notch and that of the canal differ from the above, the
coefficient of efflux is very varied, the extremes being 0,58 and 0,66.
The experiments made in 1853 and 1854, at Hanswyk, upon over-
falls 3 to 6 meters wide under a head of 0,1 to 1,0 meters gave
ре 0,64 to 0,65 or 3 µ = 0,427 to 0,433 (see the "Zeitschrift des
Archit- und Ingen-Vereins für Hanover, 1857"). The researches
made by the author upon the efflux of water through overfalls re-
fer the variation of these coefficients of efflux to certain laws, which
will be noticed further on (§ 417).
=
EXAMPLE-1) The discharge per second of an overfall, 0,25 meters
wide under a head of water of 0,15 meters is
Q
0393. b h √2 g h
-
0,393. 4,429. 0,25 (0,15);
0,02527 cubic meters.
0,435. 0,0581
2) What must be the width of an overfall, which under a head of water
of 8 inches will discharge 6 cubic feet of water? Here we have
Q
Ъ
μ₁ √2 g h³
6
0,4.8,025 √()³
6
3,434 feet.
3,210 . 0,5443
If according to Eytelwein we take μ₁ =
0,42, we have
6
b
3,271.
3,37 . 0,5443
§ 412. Maximum and Minimum Contraction.-When wa-
ter flows through an orifice in a plane surface, the axis of the ori-
fice is at right angles to the wall of the vessel and we have a me-
dium contraction; if, however, the axis of the orifice or of the
stream forms an acute angle with the portion of the wall of the
vessel containing the aperture, the contraction is smaller, and if
the angle between this axis and the inner surface of the vessel is
obtuse, the contraction is greater. The first case is represented in
Fig. 702 and the second in Fig. 703. This difference of contrac-
tion is, of course, due to the fact that in the former case the
molecules of the water, which are flowing towards the orifices, are
§ 412.] CONTRACTION OF THE VEIN OR JET OF WATER. 835
deviated less, and in the latter case more, from their primitive di-
rection, while passing through this aperture and forming the vein.
The contraction is a minimum, I.E., null, if, by gradually con-
tracting the wall surrounding the orifice, the water is prevented
from flowing in upon the side and, on the contrary, a maximum
when the direction of the wall is opposite to that of the stream, so
that certain molecules must describe an angle of 180 degrees in
FIG. 702.
B
A
FIG. 703.
FIG. 704.
FIG. 705.
B
A
B
A
B




E
E
F
Ꭰ
C
D
D
F
C
order to reach the orifice. Both cases are represented in Figures
704 and 705. In the first case the coefficient of efflux is nearly 1,
viz.: 0,96 to 0,98, and in the second case, according to the measure-
ments of Borda, Bidone and of the author, its mean value is = 0,53.
In practice, variations of the coefficients of efflux, produced by
convergent walls, often occur, particularly in the case of sluices,
which are inclined to the horizon, as is shown in Fig. 706. Pon-
celet found for such an orifice the coefficient of efflux µ = 0,80,
when the gate was inclined at an angle of 45°, and, on the contrary,
μ is only = 0,74, when the inclination is 63 degrees, I.E., for a
A
FIG. 706.
དོ་་་་
B
H
FIG. 707.

B

G
G
F
E
D
C
slope of one-half to one.
D
For the overfall, represented in Fig. 707,
where, as in Poncelet's sluice, contraction takes place upon one
side only, the author found µ = 0,70 or µ, 3 µ = 0,467 for an
inclination of 45°, and µ μ 0,67 or µ₁ = 0,447 for an inclination
of 63 degrees.
According to M. Boileau (see his Traité de la mesure des eaux
836
[§ 413.
GENERAL PRINCIPLES OF MECHANICS.
=
courantes) we can put for an overfall, which is inclined upwards
in such a way that the horizontal projection is the vertical, or
that the angle of inclination is 714 degrees, the coefficient of efflux
= 0,973 times the coefficient of efflux for an overfall with a vertical
wall. We also find from the experiments of Boileau that, for ver-
tical overfalls placed at an angle to the direction of the stream, we
must put, when the angle is 45°, the coefficient of efflux 0,942
and, when the angle is 65°, only 0,911 times the coefficient of efflux
for the normal overfall; the whole length of the edge, over which
the water flows, being of course considered as the length of the
orifice.
EXAMPLE.-If a sluice gate, which is inclined at an angle of 50 degrees
and closes a trough 21 feet wide, is raised foot and if the surface of the
water then stands permanently 4 feet above the bottom of the trough, the
height of the orifice is
the head of water is
α
1 sin. 50°
h = 4
0,3830,
1. 0,3880 3,8085 feet,
and the coefficient of velocity is μ =
=
0,78, hence the discharge is
Q = 0,78. 2,25. 0,3830. 8,025 √3,8085 = 10,52 cubic feet.
§ 413. Scale of Contraction.-The more the direction of the
water which flows in from the sides differs from that of the stream,
the greater is the contraction of the vein.
When a stream flows through the orifice C, Fig. 708, in a plane
thin plate, the angle d, formed by its axis or direction of motion
FIG. 708.

Limin
N
with that of the molecules of water which flow in from the side, is
a right angle ("); when the orifice 4 is formed by the thin
A edge
of a tube, this angle d is two right angles (T); when we have
a conical divergent mouth-piece B, d is between and π;
when the discharge takes place through a conical convergent
§ 414.] CONTRACTION OF THE VEIN OR JET OF WATER.
837
π
mouth-piece, d is between 0 and and when a cylindrical mouth-
2'
piece E well rounded off internally is used, it is
=
0.
In order to discover the law, according to which the contraction
diminishes with the angle d, the author made a series of experi-
ments with a great number of mouth-pieces 2 centimeters wide and
under different pressures (from 1 to 10 feet); the results of these
experiments are given in the following table:

б 180° 1571° 135° 1124°
0,541 0,546 0,577 0,605
673° | 45° | 2230 1110
90°
6710
0
53° 0°
0,632 0,684 0,753 0,882 0,924 0,9490,966
This table gives, it is true, only the coefficients of efflux u corre-
sponding to different angles of deviation d; the coefficients of
contraction are from 1 to 2 per cent. greater, since a small loss of
velocity always takes place during the efflux. In order to prevent
any loss of vis viva, when the water enters the mouth-pieces D and
E, the latter are rounded off at the entrance. The friction, to be
overcome by the water in passing along the walls of the mouth-
piece, will be determined in the following chapter.
REMARK.-According to the calculations of Prof. Zeuner (see Civilin-
genieur, Vol. 2d, page 55) of the results of the above experiments, we can
put
ㅠ
​μg “½″ (1 + 0,33214 (cos. d)³ + 0,16672 (cos. §)¹)
介
​# denoting the coefficient of efflux for an orifice in a plane thin plate,
for which the maximum deviation of the elements of the water during efflux
is==90°, and us, on the contrary, denoting the coefficient of efflux for
an orifice in a conical thin plate, where the maximum deviation of the
elements of the water upon entering is d.
§ 414. Partial or Incomplete Contraction.-We have as
yet studied only the case, where the water flows in from all sides of
the opening and forms a stream contracted upon all sides; we must
now consider the case, where the water flows in from but one or
more sides to the orifice, and consequently produces a stream which
is incompletely contracted. In order to distinguish these condi-
tions of contraction from each other, we will call the case, where
the stream is contracted on all sides, complete contraction, and the
case, where the stream is contracted upon a part only of its
periphery, partial or incomplete contraction (Fr. contraction incom-
plète; Ger. unvollständige or partielle Contraction). Incomplete
contraction occurs whenever an orifice in a thin plane plate is
838
[§ 414.
GENERAL PRINCIPLES OF MECHANICS.
surrounded upon one or more sides by a plate placed in the
direction of the stream. In Fig. 709 there are represented four
orifices a, b, c, d of equal magnitude in the bottom A C of a vessel.
The contraction of the water flowing through the orifice a in the
middle of the bottom is complete, for in this case the water can
flow in from all sides; the contraction of the stream in passing
through b, c or d is incomplete, for the water in these cases can
flow in from only three, two or one side.
In like manner,
when a rectangular lateral orifice extends to the bottom of the
vessel, the contraction is incomplete; for that upon the side of the
base is wanting; if further the opening extend to the bottom and
sides of the trough, there will be contraction upon one side only.
Incomplete contraction manifests itself in two ways. First, it
gives an inclined direction to the stream; and secondly, it causes a
greater discharge.
FIG. 709.
A
B
ď
D
FIG. 710.

A
B

H
C
G
D
If, E.G., the lateral orifice F, Fig. 710, reaches to the bottom
C D, so that no contraction can take place there, the axis F G of
the stream will form an angle H F G of about 9 degrees with the
normal FH to the plane of the orifices. This deviation of the
stream becomes much greater when two adjoining sides are con-
fined. If the orifice has a border upon two opposite sides, the con-
traction at those points is thereby prevented, and this deviation of
the stream does not take place, but at a certain distance from the
orifice the stream becomes wider than it would have done, if it had
not been confined upon those sides. Although a greater discharge
is obtained when the contraction is incomplete, yet it is generally to
be avoided, since it is always accompanied by a deviation in the
direction and by a great increase in the width of the stream.
Experiments upon the efflux of water, when the contraction is
incomplete, have been made by Bidone and by the author.
§ 414.] CONTRACTION OF THE VEIN OR JET OF WATER. 839
very
Their results show that the coefficient of efflux increases
nearly with the ratio of the length of the border to the entire peri-
phery of the orifice; but it is easy to perceive that this relation is
different, when the periphery is nearly or entirely surrounded by a
border, in which case the contraction is almost or totally done away
with. If we put the ratio of the portion with a rim to the entire
periphery = n and denote by an empirical quantity, we can put,
approximatively, the ratio of the coefficient μ, of efflux for incom-
plete contraction to the coefficient μ, for complete contraction
μl n
Мо
k
= 1 + kn, and consequently μ₁ = (1 + n) μ。.
Bidone's experiments gave for small circular orifices = 0,128,
and for square ones = 0,152; those of the author gave for small
rectangular orifices к=0.134, and for larger ones (Poncelet's mouth-
pieces) 0,2 meter wide and 0,1 meter high x = 0,157 (see the Maga-
zine "der Ingenieur," vol. 2d). In practice rectangular orifices
with rims are almost the only ones employed; we will assume for
them, as a mean value, к = 0,155, and consequently put
µ₂ = (1 + 0,155 n) µ„
b
For a rectangular lateral orifice, whose height is a and whose
width is b, we have n =
when there is no contraction
upon the side b, if, E.G., this side is upon the bottom; n =
2 (a + b)'
one side a and one side b are provided with rims; and n =
, when
2a + b
2 (a + b)'
when the contraction is prevented upon the side b and upon the
two sides a, the latter case occurs, when the orifice occupies the
entire width of the reservoir and extends to the bottom.
EXAMPLE.-What is the discharge through a vertical sluice 3 feet wide
and 10 inches high, when the head of water is 13 feet above the upper
edge of the orifice and the lower edge is at the bottom of the trough, so
that there is no contraction upon that side? The theoretical discharge is
Q = 1§. 3. 8,025 √1,5 + ½ §. 8,025 √1,9166
= 27,77 cubic feet.
According to Poncelet's table for perfect contraction μ =
we have
5
3
=
N 2 (3+19)
9
18+ 5
23,
hence for the present case of incomplete contraction
(1 + 0,155.). 0,604 = 1,060. 0,604
=
0,640
end the effective discharge is
Q = 0,640 Q = 0,640. 27,77
=
17,77 cubic feet
0.504, but
840
[§ 415.
GENERAL PRINCIPLES OF MECHANICS.
§ 415. Imperfect Contraction.-The contraction of the vein
depends also upon this fact, viz.: whether the water is sensibly at
rest in front of the orifice or whether it arrives there with a certain
velocity; the faster the water approaches the orifice of efflux, the
less the stream is contracted, and consequently the greater is the
discharge. The various relations of contraction and efflux, given
and discussed in what precedes, are applicable only where the ori-
fice is in a large wall, in which case we can assume that the water
arrives at the orifice with a very small velocity; we must now
investigate the relations of contraction and efflux, when the cross-
section of the orifice is not much smaller than that of the approach-
ing water, in which case the water arrives with a velocity, which is
not negligable. In order to distinguish these two cases from each
other, let us call the contraction which occurs, when the water
above the orifice is at rest, perfect contraction and that which
occurs, when the water is in motion, imperfect contraction (Fr. con-
traction imparfaite; Ger. unvollkommene Contraction). The
contraction during efflux from the vessel A C, Fig. 711, is imper-
FIG. 711.
A
E
DD
B
fect; for the cross-section F of the orifice is not
much smaller than that G of the water approach-
ing it or the area of the wall CD, in which this
orifice is placed. If the vessel was of the form
A B C, D, and the area of the base C, D, was
much greater than that of the orifice F, the
efflux would take place with perfect contraction.
The imperfectly contracted stream is distin-
C guished from the perfectly contracted one not
only by its size, but also by the fact that it is not
so transparent and crystalline as the latter is.
If we denote the ratio of the area F of the orifice to that G of

F
G'
the wall in which it is situated, or by n, the coefficient of efflux for
perfect contraction by μ, and that for imperfect contraction by μ
we can put with great accuracy, according to the experiments and
calculations of the author,
1) for circular orifices
μ₂ = μo [1 +0,04564 (14,821" - 1)],
n
2) and for rectangular orifices
μ₂ = μ. [1 + 0,0760 (9" — 1)].*
µn
* Experiments upon the imperfect contraction of water, etc., Leipzig, 1843.
§ 415.] CONTRACTION OF THE VEIN OR JET OF WATER. 841
In order to facilitate the calculations which are required in
practice, the corrections
μn
Мо
of the coefficient of efflux in con-
Мо
sequence of the imperfect contraction have been arranged in the
following tables:
n
un
μο
n
TABLE L
The corrections of the coefficients of efflux for circular orifices.

Мо
0,05
0,05 0,10 0,15 0,20 0,25 0,30 0,35 0,40 0,45 0,50
0,007 0,014 0,023 0,034 0,045 0,059 0,075 0,092 0,112 0,134
0,55 0,60 0,65 0,70 0,75 0,80 0,85 0,90 0,95 1,00
,60
0,80
,90
Мир
Мо
0,161 0,189 0,223 0,260 0,303 0,351 0,408 0,471 0,546 0,631
Мо
TABLE II.
The corrections of the coefficients of efflux for rectangular orifices.
N.
un
In
140
0,05 0,10 0,15 0,20 0,25 0,30 0,35 0,40 0,45 0,50
25
0,009 0,019 0,030 0,042 0,056 0,071 0,088 0,107 0,128 0,152
Мо
n
0,55 0,60 0,65 0,70 0,75 0,80 0,85 0,90 0,95 1,00
0,60
,80 0,90
Мп
Мо
0,178 0,208 0,241 0,278 0,319 0,365 0,416 0,473 0,537 0,608
The upper lines in these tables contain various values of the
F
ratio of the cross-sections, and immediately below are the corre-
G
sponding additions to be made to the coefficient of efflux on account
of the imperfect contraction, E.G., for the ratio n = 0,35, I.E., for
the case, where the area of the orifice is 35 hundredths of the area
of the entire wall, in which the orifice is made, we have for a cir-
cular orifice
μl n
flo
=
0,075,
and for a rectangular one = 0,088; the coefficient of efflux for
842
[§ 416.
GENERAL PRINCIPLES OF MECHANICS
perfect contraction must be increased in the first case 75 thou-
sandths and in the second 88 thousandths, when we wish to obtain
the corresponding coefficient of efflux for imperfect contraction.
If the coefficient of efflux were = 0,615, we would have in the
first case
Мо 35
and in the second case
1,075. 0,615 0,661
0,669.
M0,35 = 1,088. 0,615
=
EXAMPLE.—What is the discharge through a rectangular lateral orifice
F, which is 14 feet wide and foot high, when it is cut in a rectangular
wall C D, Fig. 712, 2 feet wide and 1 foot high, and when the head of
water E H = h, where the water is at
rest, is 2 feet. The theoretical dis-
charge is
FIG. 712.
A
B

G
H
ID
Q = 1,25. 0,5 . 8.025 √2
5,0156 1,414 = 7,092 cubic feet,
and the coefficient of efflux for perfect
contraction is, according to Poncelet,
but the ratio of the cross-sections is
Мо
=
0,610,
F
1,25. 0,5
N
G
2.1
0,312,
and for n = 0,312 we have, according to Table II, page 841,
Mo
u₁ 0,071 +18 (0,088 — 0,071) = 0,071 +0,004
hence the coefficient of efflux for the present case is
0,075;
1,075 + 0,610
=
0,6557.
0,6557 . Q
10,312 1.075 Po
and the effective discharge is
Q s
0,6557 . 7,092 = 4,65 cubic feet.
§ 416. Efflux of Moving Water. We have heretofore
assumed that the head of water was measured in still water; we
must now discuss the case where the head of water can be meas-
ured only in water, which is approaching the orifice with a certain
velocity. If we assume the orifice to be rectangular and denote
the width by b, the head of water in reference to the two horizon-
tal edges by h, and h, and the height due to the velocity of ap-
proach c of the water by k, we have the theoretical discharge
Q = 3 b √ 2 g [(h, + k); − (h₂ + k)³].
This formula cannot be directly employed for the determination
of the discharge, since the height due to the velocity
2
C²
1
k =
2 g
2 g
(2)
§ 416.] CONTRACTION OF THE VEIN OR JET OF WATER. 843
depends also upon Q, and, if we transform it, we obtain a compli-
cated equation of a high degree; it is much simpler, therefore, to
put the effective discharge
Q₁ = μ, a b √2 gh
and to understand by μ, not a simple coefficient of efflux, but a co-
efficient depending principally upon the ratio of the cross-sections.
This case is often met with in practice, E. G., when we wish to
measure the quantity of water which passes through a ditch or
caual; for we can seldom dam up the water by means of a trans-
verse wall B C, Fig. 713, to such a height that the area F of the
A
FIG. 713.
B
·H-
·G-
orifice, through which the water
is discharged, will be but a small
fraction of the cross-section of the
stream which approaches it, and
it is only in the latter case that
the velocity of approach is very
small compared to the mean ve-
locity of efflux.
In the experiments made by the author with Poncelet's orifices
the head of water was measured 1 meter back from the plane of the
orifice, they gave

μn
Мо
=
Мо
0,641 (†) = 0,641 . n²,
F
0
n = denoting the ratio of the cross-sections, which should not
G
be much greater than 1, μ, denoting the coefficient of efflux for
perfect contraction, taken from Poncelet's table, and μ, the coeffi-
cient of efflux for the present case. Let b be the width and a the
height of the orifice, b, the width and a, the depth of the stream.
of water and h the depth of the upper edge of the orifice below the
level of the water, then we have the effective discharge
Q₁
= μl • A 6
[1 + 0,641 (
a b
a, b;
2 J
(12+
The following table is useful in abridging calculations in practice.

0,05 0,10 0,15 0,20 0,25 0,30 0,35 0,40 0,45 0,50
n
thepr.
ре
0,002 0,006 0,014 0,026 0,040 0,058 0,079 0,103 0,130 0,160
flo
EXAMPLE.—In order to find the amount of water brought by a ditch 3
feet wide, a transverse wall, containing a rectangular orifice 2 feet wide and
844
[§ 417.·
GENERAL PRINCIPLES OF MECHANICS.
1 foot high is put in it, and the water is thus raised so that, when its level
becomes constant, it is at a distance of 21 above the bottom and 12 feet
above the lower edge of the orifice. The corresponding theoretical dis-
charge is
Q: = a b √ 2 g h
4
h = 1,2.8,025 √1,25 = 16,05 . 1,11817,94 cubic feet.
As the coefficient of efflux for perfect contraction is 0,602 and the ratio
of the cross-sections is
n =
F
G
a b
1.2
a, b,
1 1
2,25.3
0,296,
we have the coefficient of efflux in the present case
µn = (1 + 0,641 . 0,296³) µ,
= 1,056. 0,602
=
0,6357,
and the effective discharge
Q+
=
17,94.0,6357
=
11,4 cubic feet.
§ 417. The contraction is also imperfect when water is dis-
charged through overfalls (like that in Fig. 714), if the cross-
FIG. 714
ATHRUBHECH
B
D
-H
--
section F of the stream pass-
ing over the sill C is a notable
fraction of the cross-section G
of the approaching water. The
overfall may extend over but
a portion or over the whole of
the canal or ditch. In the

latter case, as there is no contraction upon the sides of the orifice,
the discharge is greater than through orifices of the first kind.
The author has made experiments upon these cases of efflux and
deduced from the results obtained formulas, by means of which the
coefficient of efflux can be calculated with sufficient accuracy, when
the ratio n =
F
G
of the cross-sections is known.
Leth be the head of water E H above the sill of the overfall,
a, the total depth of water, b the width of the overfall, and b, that
of the approaching water; we have then
F
h b
n =
and
G
aj bi
1) for Poncelet's overfall
μl n
Мо
on the contrary,
1,718 (F) 4
( 77 ) * = 1,718 n²;
G
2) for an overfall occupying the whole width of the ditch or trough
n
μm
Мо
Мо
0,041 + 0,3693 n²;
t
§ 417.] CONTRACTION OF THE VEIN OR JET OF WATER.
845
hence the discharge in the first case is
Q₁ = 3 µ. . 6 [1 + 1,718
+ 1,718 (2
(hb)' ] √2 g h³,
and in the second case,
0
Q₁ = 3 µ.. 6 [1,041 + 0,3693 () ]
h
√2 gh³,
h denoting the head of water E H above the sill F of the overfall,
measured at a point about one meter back of it.
In the following tables the corrections
values of n are given.
n
fun
Мо
TABLE I.
fl n
Mo
for the simplest
Мо
Corrections of the coefficients of efflux for Poncelet's overfalls.

0,05 0,10 0,15 0,20 0,25 0,30 0,35 0,40 0,45 0,50
|0,000 0,000 0,001 0,003 0,007 0,014 0,026 0,044 0,070|0,107
TABLE II.
Corrections for overfalls extending over the entire width, or without lateral
flan
contraction.

|
n 0,00 0,05 0,10 0,15 0,20 0,25 0,30 0,35 0,40 0,45 0,50
Мо
μα
0,041 0,042 0,045 0,049 0,056 0,064 0,074 0,086 0,100 0,116 0,133
EXAMPLE.-In order to determine the amount of water carried by a
canal 5 feet wide, we place in it a transverse partition with the upper edge
beveled outwards and we allow the water to flow over this. After the
upper water had ceased to rise, the height of its surface above the bottom
of the canal was 33 feet and above the sill 1 feet; the theoretical dis-
charge was therefore
2 = §. 5. 8,025 (3)² = 49,14 cubic feet.
Q
=
h 1,5
a₁ 3,5
[1,041 + 0,3693 (4)³] . 0,577 = 1,110 . 0,577
The coefficient of efflux is in this case, since
and μ =
= 0,577,
= 0,64,
and therefore the effective discharge is
Q₁ = 0,64. Q
0,64 . Q = 0,64 . 49,14
31,45 cubic feet.
846
[§ 418.
GENERAL PRINCIPLES OF MECHANICS.
§ 418. Lesbros's Experiments.-We are indebted to Mons.
Lesbros for a great number of experiments upon the efflux of water
through rectangular orifices in a thin plate; the crifices, being
provided internally and externally with rims, afforded examples of
both partial and incomplete contraction (see his "Experiences hy-
drauliques sur les lois de l'écoulement de l'eau"). We will give
here only the principal results of his experiments with a rectangu-
lar orifice 2 decimeters wide. The orifices, which were surrounded
with borders of different kinds, are distinguished from each other
in Fig. 715 by the letters A, B, C, etc.
A
FIG. 715.
B
C
D
E
F
G
H
A denotes the ordinary mouth-piece without any rim or border
(as in § 410);
B denotes a similar mouth-piece with a vertical wall upon the
inside perpendicular to the plane of the orifice and at a distance
of 2 centimeters from one side of it;
C denotes the first mouth-piece enclosed on the inside by two
such walls;
D the orifice A, provided on the inside with two vertical walls,
which converge towards each other at an angle of 90° and cut
the plane of the orifice at an angle of 45° and at a distance of
2 centimeters from the side of it;
E the orifice A with a horizontal wall, which passes across the
reservoir and reaches exactly to the lower edge of the orifice;
F the orifice B,
G the orifice C, and
H the orifice D with a horizontal rim or wall, as in E, which
completely prevents the contraction at the lower edge of the
orifice.
§ 418.] CONTRACTION OF THE VEIN OR JET OF WATER
847
I.
TABLE OF THE COEFFICIENTS OF EFFLUX FOR FREE EFFLUX
THROUGH THE ORIFICES A, B, C, ETC.

Head of water above
the upper edge of
the orifice, measured
back from the plane
of the orifice
Meters.
0,020
0,050
0,100
0,200
0,500
1,000
1,500
2,000
3,000
0,020
0,050
0,100
0,200
0,500
1,000
1,500
2,000
3,000
Height of orifice.
Coefficient of efflux for the orifices.
A
B | C | D | E |
F G
H
Meters.
0,572 0,587
0,589 0,599
0,585 0,593 0,631 0,595 0,608 0,622
0,636
0,639
0,592 0,600 0,631 0,601 0,615 0,628
0,598 0,606 0,632 0,607 0,621 0,633 0,708 0,643
0,200 0,6030,6100,6310,6110,623 0,636 0,680 0,644
0,605 0,611 0,628 0,612 0,6240,637 0,676 0,642
0,602 0,611 0,627 0,611 0,6240,637 0,672 0,641
0,601 0,610 0,626 0,611 0,619 0,636 0,668 0,640
0,601 0,609 0,624 0,610 0,614 0,6340,665 0,638
0,678
0,616 0,627 0,647 0,631 0,664 0,663
0,625 0,630 0,646 0,632 0,667 0,669 0,690 0,677
0,630 0,633 0,645 0,633 0,669 0,674 0,688 0,677
0,631 0,635 0,642 0,633 0,670 0,676 0,687 0,675
0,050 0,628 0,634 0,637 0,632 0,668 0,676 0,682 0,671
0,625 0,628 0,635 0,627 0,666 0,672 0,680 0,670
0,619 0,622 0,634 0,621 0,665 0,670 0,678 0,670
0,613 0,616 0,634 0,615 0,664 0,670 0,674 0,669,
0,606 0,609 0,632 0,608 0,662 0,669 0,673 0,668
848
[§ 418.
GENERAL PRINCIPLES OF MECHANICS.
II.
TABLE OF THE COEFFICIENTS OF EFFLUX THROUGH THE ORI-
FICES A, B, C, ETC.,
With external shoots or uncovered canals of the same dimensions as
the orifice (Fr. canaux de fuite; Ger. äussere Ansatzgerinnen).
The shoots fitted the orifice exactly, and consequently the bev-
eling of the sides and bottom of the mouth-piece was done away
with. They were either horizontal and 3 meters long or (in the
experiments marked with *) inclined of their length, which was
but 2,5 meters.
ΤΟ
Head of water above the
upper edge of the orifice
measured back from the
plane of the orifice.
Height of orifice.
Coefficients of efflux for the orifices.
A
BC
E E* F F*
G
G*
H
0,488
0,520
0,480 0,489 0,496 0,480 0,527
0,511 0,517 0,531 0,510 0,553 0,509 0,546 0,528
0,542 0,545 0,563 0,538 0,574 0,534 0,569 0,560 0,593 0,552
0,574 0,576 0,591 0,566 0,592 0,562 0,589 0,589 0,617 0,582
0,200 0,599 0,602 0,621 0,592 0:607 0,591 0,608 0,591 0,632 0,613,
Meters.
Meters.
0,020
0,050
0,100
0,200
0,500
1,000
1,500
2,000
3,000
0,020
0,050
0,100
0,200
0,500
1,000
1,500
2,000
3,000
0,601 0,609 0,628 0,600 0,610 0,601 0,615 0,601 0,638 0,623
0,601 0,610 0,627 0,602 0,610 0,604 0,617 0,604 0,641 0,624
0,601 0,610 0,626 0,602 0,609 0,604 0,617 0,604 0,642 0,624
0,601 0,609 0,624,0,601 0,608 0,602 0,616 0,602 0,641 0,622
0,494
0,488 0,555 0,557 0,487 0.585 0,483 0,579 0,512
0,577 0,600 0,603 0,571 0,614 0,570 0,611 0,582 0,625 0,577
0,6240,625 0,628 0,605 0,632 0,609 0,628 0,621 0,639 0,616
0,631 0,633 0,637 0,617 0,645 0,623 0,643 0,637 0,649 0,629
0,050 0,625 0,630 0,635 0,626 0,652 0,630 0,650 0,647 0,6560,636
0,624 0,627 0,635 0,628 0,651 0,633 0,651 0,649 0,656 0,638
0,619 0,622 0,634 0,627 0,650 0,632 0,651 0,647 0,656 0,637
0,613 0,616 0,634'0,623 0,650 0,631 0,651 0,644 0,6560,635
0,606 0,609 0,632 0,618 0,649 0,628 0,651 0,639 0,656 0,632
0,656
!
§ 419.]
CONTRACTION OF THE VEIN OR JET OF WATER. 849
EXAMPLE.- What is the discharge through an orifice 2 decimeters
wide and 1 decimeter high, when the lower edge is 0,35 meters below the
level of the water and upon a level with the bottom of the vessel, 1) for
free efflux, and 2) for efflux through a short horizontal shoot? We have
in this case the orifice E, and the head of water above the upper edge is
0.35 0,10=0,25 meters. Table I gives, when the head is = 0,20 and
the height of orifice = 0, 20, the coefficient of efflux
=
μ 0,621, and, on
the contrary, when the height of the orifice is = 0,05 meters, μ = 0,670;
µ
hence for the first case of the problem we can put
μ
0,621 + 0,670
2
0,645.
Table II gives, on the contrary, by interpolation, for a head of water
0,25 meters above the upper edge of the orifice, the following values for μ.
0,566 + 3% (0,592 – 0,566) = 0,570, and
30
0,617 +3% (0,626
5
30
hence in the second case we can put
μ =
0,570 + 0,619
2
0,617)
=
0,619;
= 0,594.
=
0,020 square meters;
=
0,300 meters;
The cross-section of the orifice is
Fab 0,20. 0,10
the mean head of water is
h = 0,350 - 0,050
and, consequently, the theoretical discharge is
Q = F √2 g h = 0,02 √2. 9,81. 0,3 0,02 √5,886
=
=
0,02. 2,425 = 0,0485 cubic meters.
The effective discharge is in the first case
Q1 = μr Q
0,645. 0,0485 0,0313 cubic meters,
=
and, on the contrary, in the second case, I.E., when a shoot is added,
Q = 142
Q
0,594. 0,0485
=
0,0288 cubic meters.
According to the formula un = (1 +0,155 n) μ of § 414, we can put for
efflux with partial contraction µn μ 1 (1 + 0,52) μ。
=
1,052 μ, since
of the periphery of the orifice is surrounded by a border. But for
such an orifice with complete contraction we have, according to Table I,
page 831, μ = 0,616; hence
and the discharge is
μ Q
=
=
**
.
1,052 0,616 = 0,648,
0,648. 0,0485 = 0,0314 cubic meters,
I.E., a little greater than that obtained by employing Lesbros's table.
§ 419. M. Lesbros has also experimented upon efflux through
overfalls, employing the same orifices A, B, C, etc., but not allow-
ing the water to rise to the upper edge of the orifice. The principal
results of these experiments are to be found in the following tables.
3
54
850
[S 419.
GENERAL PRINCIPLES OF MECHANICS
TABLE I.
Table of the coefficients of efflux (3 µ) for free efflux through
overfalls or notches.
Coefficients of efflux for the orifices.
Head of water
above the sill.
measured where
the water is still.
A
B
C
D
E
F
G
Meter
0,015
0,421
0,020
0,417
0,450 0,450 | 0,441
0,446 | 0,444
0,395 0,371
0,305
0,437
0,402 0,379
0,318
0,030
0,412 0,437 0,435
0,430
0,410 |0,388
0,337
0,040
0,050
0,407 0,430
0,404
0,429 0,424
0,4II 0,394
0,352
0,425
0,426 |0,419
0,411
0,398
0,362
0,070 0,398 0,41
0,416 0,422 0,412
0,100 0,395 0,409 0,420
0,150 0,393 0,406 |0,42
0,409
0,402
0,375
0,405
0,408 0,405
0,382
0,403
0,407 0,407
0,383
0,200
0,390 0,402 0,424 0,403
0,379 0,396
0,250
0,396 0,422 0,401
0,300 0,371 0,390 0,418 0,398 0,403 0,406
0,405 | 0,408
0,383
0,404 0,407
|
0,381
0,378
TABLE II.
Table of the coefficients of efflux (μ) for efflux through weirs with
short shoots or open canals.
Head of wa-
ter above the
sill, measur-
Coefficients of efflux for the orifices.
ed where the
water is still.
A
B
C
D
E
F
G H
Meter.
0,015
0,375 0,388 0,400
0,020
0,196 0,368 0,383
0,395
0,208
0,208 0,201
0,1750,190
0,030
0,234 0,358
0,373
0,385
0,232 0,228
0,205
0,222
0,040 0,263 0,351 0,365
0,379
0,251 0,250
0,234 0,250
|
0,050 0,278 0,346 0,360
0,070 0,292 0,343 0,352 0,371
0,100 0,304 0,340 | 0,345 | 0,369
0,150 0,315 0,335 0,340
0,200 0,319 0,3310,338 0,366 | 0,323
0,250 0,321 0,328 0,336
0,300 0,324 0,326 0,334
0,302
0,367 0,314 0,316 0,313
0,322 0,322
0,364|0,329 |0,326 | 0,329 | 0,341
0,361 0,332 0,3290,332 0,345
A comparison of the coefficients in Table I and Table II shows
that the discharge through orifices provided with shoots is smaller
than that through those without them, and that the difference is
greater, the smaller the head of water is; we also see, by comparing
0,375
0,268
0,267
0,260 0,272
|
0,288
0,289
0,285 0 296
0,304 0,299
0,313
|
0,327
0,335
419.] CONTRACTION OF THE VEIN OR JET OF WATER. 851
the columns C and C*, E and E*, F and F'*, and G and G*in the
tables of the last paragraph, that the inclined shoot creates less dis-
turbance in the efflux than the horizontal one.
REMARK 1.— A different theory of the efflux of water is advanced by G.
Boileau in his "Traité sur la mesure des eaux courantes." According to
it the velocity of the effluent water is the same at all parts of the cross-sec-
tion and depends upon the depth of the upper limiting line of the vein at
the plane of the orifice below the level of the water in the reservoir.
Boileau employs the same formula for overfalls, in which case he must
know of course the height of the stream in the plane of the orifice. Later,
in the 12th volume of the 5th series of the Annales des Mines, 1857, M.
Clarinval has given another formula for efflux through overfalls in which no
empirical number μ appears, but instead of he substitutes the factor
a √ 1 - a
1
in which h denotes the head of water and a the thickness of
√2 (h² a²)
the stream above the sill of the overfall. See the "Civilingenieur," Vol.
5th. I consider the hypothesis upon which this formula is based to be
incorrect.
REMARK 2.-Mr. J. B. Francis gives in his work "The Lowell Hydraulic
Experiments, Boston, 1855," the following formula for efflux though a
wide overfall or weir.
Q = 3,33 (? — 0,1 n h) hi English cubic feet,
in which h denotes the head of water above the sill of the weir, 7 its length,
and n either 0 or 1 or 2, according as the contraction of the vein is pre-
vented upon both, one or none of the sides. Since for the English system
of measures
we have
√2 g
8,025,
3,33
expo
=
0,415.
8,025
The experiments, upon which this formula is based, were made with
weirs 10 feet wide and under heads of water from 0,6 to 1,6 feet. The edge
of the weir was formed of an iron plate beveled down stream, the reservoir
was 13,96 feet wide, and the sill was 4,6 feet above its bottom. See the
Civilingenieur, Vol. 2, 1856.
Bakewell's experiments upon efflux through weirs or overfalls give
results differing in some respects from the above. (See Polytech. Central
Blatt, 18th year, 1852.)
REMARK 3.—At the sluice-gate of the wheel at Remscheid, Herr Rönt-
chen found μ = 0,90 to 0,93. See Dingler's Journal, Vol. 158.
A new edition of Mr. J. B. Francis' work has been recently published by
D. Van Nostrand, New York.—[TR.]
852
[$ 420.
GENERAL PRINCIPLES OF MECHANICS.
CHAPTER III.
OF THE FLOW OF WATER THROUGH PIPES.
§ 420. Short Tubes.-If we allow the water to discharge
through a short tube, or pipe, called also an ajutage, (Fr. tuyau
additionel; Ger. kurze Ansatzröhre), the condition of affairs is
entirely different from that existing, when the water issues from
an orifice in a thin plate or from an orifice in thick wall, which is
rounded off on the outside. If the short tube is prismatic and 21
to 3 times as long as wide, the stream is uncontracted and non-
transparent and its range and consequently its velocity is smaller
than when it issues, under the same circumstances, from an orifice
in a thin plate. If, therefore, the tube K L has the same cross-
section as the orifice F, Fig. 716, and if the head of water is the
FIG. 716.

A
B
L
K
H
D
E
same for both, we obtain at R L a troubled and uncontracted or
thicker stream and at F H a clear and contracted or thinner one;
FIG. 717.
A
B
L
H
we can also see that the range E R
is smaller than the range D H.
This condition of efflux exists only
when the length of the tube is the
given one; if the tube is shorter,
E.G. as long as wide, the vein K R,
Fig. 717, does not touch the sides
of the tube, the latter has then no
influence upon the efflux, and the
stream issues from it as from an
orifice in a thin plate.
Sometimes it happens, when the length of the tube is greater,

§ 421.] THE FLOW OF WATER THROUGH PIPES.
853
1
that the stream does not fill it; this occurs when the water has
no opportunity of coming in contact with the sides of the tube;
if in this case we close for an instant the outside end of the tube
with the hand or with a board, the stream will fill the tube and we
have the so-called discharge of a filled tube (Fr. à gueule bée;
Ger. voller Ausfluss). The vein is contracted in this case also, but
the contracted portion is within the tube. We can satisfy our-
selves of this by employing glass tubes like K L, Fig. 718, and by
FIG. 718.
B
throwing small light bodies
into the water. Upon so do-
ing, we observe that near the
entrance there is a motion
of translation in the middle of
the cross-section F₁, but that,
on the contrary, at the peri-
phery of the same the water
forms an eddy. It is, however,
the capillarity or adhesion of

the water to the walls of the tube, which causes it to fill the end FL
of the tube completely. The pressure of the water discharging
from the tube is that of the atmosphere, but the contracted cross-
section F is only a times as great as that F of the tube; the
1
velocity v, at that point is therefore times as great as the velocity
a
of efflux v and the pressure of the water at F, is smaller than that
at the end of the tube, which is equal to the pressure of the atmo-
sphere. If we bore a small hole in the pipe near F, no water will
run out, but air will be sucked in and the discharge with a filled
tube ceases, when the hole is enlarged or when several of them are
made. We can also cause the water in the tube A B to rise and
flow through the tube K L by making it enter the latter at F.
The discharge with a filled tube ceases for cylindrical tubes, when
the head of water attains a certain magnitude (see § 439, Chap. IV).
421. Short Cylindrical Tubes.-Many experiments have
been made upon the efflux of water through short cylindrical tubes,
but the results obtained differ quite sensibly from each other. It
is particularly Bossut's coefficients of efflux which differ most from
those of others by their smallness (0,785). The results of the ex-
periments Michelotti with tubes 1 to 3 inches in diameter, under
a head of water varying from 3 to 20 feet, gave as a mean value
854
[§ 421.
GENERAL PRINCIPLES OF MECHANICS.
μl =
0,813. The results of the experiments of Bidone, Eytelwein
and d'Aubuisson differ but little from those of the latter. But,
according to the experiments of the author, we can adopt for short
cylindrical tubes as a mean value 0,815. Since we found this
coefficient for an orifice in a thin plate = 0,615, it follows that,
when the other circumstances are the same, 815 = 1,325 times as
much water is discharged through a short pipe as through an ori-
fice in a thin plate. These coefficients increase, when the diameter
of the tube becomes greater and decrease a little, when the head
of water or the velocity of efflux increases. According to some
experiments of the author's, made under heads varying from 0,23
to 0,6 meters, we have for tubes 3 times as long as wide

When the width is
1
μπ
?
2
3
4 centimeters.
0,843
0,832 0,821
0,810
According to this table the coefficients of efflux decrease sensi-
bly as the width of the tube increases. In like manner Buff found
with a tube 2,79 lines wide and 4,3 lines long that the coefficient
of efflux increased gradually from 0,825 to 0,855, when the head
of water decreased from 33 to 1 inches.
For the efflux of water through short parallelopipedical tubes
the author found the coefficient to be 0,819.
If the short tube KL, Fig. 719, is partially surrounded by a
border or rim in the inside of the vessel, if, E.G., one of its sides
is flush with the bottom CD of the vessel and if partial contrac-
tion is thus produced, according to the experiments of the author,
the coefficient of efflux is not sensibly increased, but the water
FIG. 719.
B
FIG. 720.


D
K
C
Ꮮ
A
C
B
§ 422.]
855
THE FLOW OF WATER THROUGH PIPES.
moves with different velocities in different parts of the cross-sec-
tion, viz., upon the side C more quickly than upon the opposite one.
If the face of the tube is not in the surface of the plate but
projects into the vessel, like E, F, G, Fig. 720, it is then called an
interior short tube. If the face of the tube is at the least 5 times as
wide as the bore of the tube, as at E, the coefficient of efflux remains
the same as if the face were in the plane of the wall, but if the
face of the tube is smaller, as at F and G, the coefficient of efflux
is smaller. According to the experiments of Bidone and of the
author, if the face is very small, it is 0,71, when the stream fills the
tube; on the contrary, it is 0,53 (compare § 113), when it does
not touch the internal surface of the tube. In the first case (F)
the stream is troubled and divergent like a broom, but in the
second (G) it is compact and crystalline.
o
§ 422. Coefficient of Resistance.-Since the stream of water
issues from a short prismatical tube without being contracted, it
follows that the coefficient of contraction of this mouth-piece a =
unity and that its coefficient of velocity = its coefficient of efflux μ.
The vis viva of a quantity of water Q, which issues with a velocity
v2
", is
Qy (see § 74). But the theoreti-
2g
and therefore the theoretical energy of
Q Y
v², and its energy is
g
2
cal velocity of efflux is
φ
1
༡,༠
φ
29
Qy. Hence the loss of energy
the water discharged is
of the quantity of water during the efflux is
2
=
(1) (1)0x
Q Y
j
が
​( 252 - 1) 29
2 g 2
g
For efflux through orifices in a thin plate, the mean value of
is 0,975; hence the loss of energy is
2 وح
Q y
[(0.975)-1] Y = 0,052
29
?)བཾ
2 g
QY;
for efflux through a short cylindrical pipe, on the contrary, p =
0,815, and the corresponding loss of energy is
Q
= [(0,315) - 1], 2 x = 0,505
2 g
212
2 g
QY,
I.E., nearly 10 times as much as for efflux through an orifice in a
thin plate. Consequently if the vis viva of the water is to be made
use of, it is better to allow it to flow through an orifice in a thin
plate than through a short prismatical tube. If, however, we
856
[§ 422.
GENERAL PRINCIPLES OF MECHANICS.
round off the edge of the tube, where it is united to the interior
surface of the vessel, so as to produce a gradual passage from the
vessel into the tube, the coefficient of efflux is increased to 0,96
and at the same time the loss of energy is reduced to 8 per cent.
For short tubes or ajutages, which are rounded off or shaped inter-
nally like the contracted vein, we have µ = 0,975, and the
= &
Φ
loss of mechanical effect is the same as it is for an orifice in a thin
plate, viz., 5 per cent.
The loss of mechanical effect (1)29 y corresponds to a
head of water
(-1),
v²
2 g
g
; we can therefore consider that the loss
of head due to the resistance to efflux is (-
$2
2
- 1)
v2
and we can
2 g
assume that, when this loss has been subtracted, the remaining por-
tion of the head is employed in producing the velocity.
φ
(~12 - 1) 2 3
2, 2
which increases with the square
g
This loss z =
of the velocity, is known as the height of resistance (Fr. hauteur
1
1,
de résistance; Ger. Widerstandshöhe) and the coefficient
by which the head of water must be multiplied in order to obtain
the height of resistance, is called the coefficient of resistance. Here-
after we will denote this coefficient, which also gives the ratio of
the height of resistance to the head of water, by or the height
of resistance itself by z =
By means of the formulas
v2
Š
2 g
1
1 and
FIG. 721.
L
B
1
√1 + 5
we can calculate from the coefficient
of velocity the coefficient of resistance,
or the latter from the former.
If the velocity of efflux v is the
same, the head of water of an orifice
K, Fig. 721, whose coefficient of resist-

ance is o, is h
v2
2 g p3
and the head
of water of the orifice L, through which
the water flows with this theoretical
K
D
C
§ 423.]
857
THE FLOW OF WATER THROUGH PIPES.
velocity, is h₁ =
22
2 g
consequently the first orifice must lie at a dis-
v²
v2
tance KL = z = h 么 ​m = (1/2 − 1)
ら
​below the second
2 g
2 g
one. This distance z is called the height of resistance. If they
have the same cross-section F and there is no contraction at either
orifice, the discharge QFv is the same for both.
EXAMPLE-1) What is the discharge under a head of water of 3 feet
through a tube 2 inches in diameter, whose coefficient of resistance is
= 0,4. Here
1
φ
=
0,845; hence
√1,4
v = 0,845 . 8,025 √3 = 11,745 feet;
2
Σ
F = (†½)² π =
-
0,02182 square feet,
0,256 cubic feet.
and consequently the required discharge is
Q = 0,02182. 11,745
=
2) If a tube 2 inches wide discharges under a head of 2 feet 10 cubic
feet of water in a minute, the coefficient of efflux or velocity is
Q
10
F√2 g h 60. 0,02182.8,025 √ 2
the coefficient of resistance
1
0,673
1
0,673,
1,05 √ 2
( )
1
-
1,208,
1,208. 0,0155
(2)² = 0,0187.
1,092 feet.
0.1309°
and the loss of head, caused by the resistance of the tube, is
23
2 = 5
2 g
=
v2
1,208.
1
FIG. 722.
§ 423. Inclined Short Tubes or Ajutages.-When the
tubes are applied to the vessel in an inclined position or when
they are cut off obliquely to the axis, the discharge is less than
when they are inserted into the vessel at
right angles or cut off at right angles to
E their axis; for in this case the direction of
the water is changed. The author's extended
experiments upon this subject have led to
the following conclusions. If d denotes the
angle L KN, formed by the axis of the tube
K L, Fig. 722, with the normal K N to the plane A B of the
orifice, and if denotes the coefficient of resistance for tubes cut
off at right angles, we have for the coefficient of resistance of in-
clined tubes

B
N
ૐ 5 + 0,303 sin. § + 0,226 sin.' §.
Assuming for the mean value 0,505, we obtain
858
[§ 424.
GENERAL PRINCIPLES OF MECHANICS.

for 8º =
О
ΙΟ
20
30
40
50 60 deg.
the coefficients of
resistance Ši
=
0,505
0,565 0,635 0,713
0,794
0,870 | 0,937
t
the coefficient of
0,815
efflux μ₁ =
0,799 0,782 | 0,764
0,747
0,731 0,719
Hence, E.G., the coefficient of resistance of a short tube, the
angle of deviation of whose axis is 20°, is 5, = 0,635 and the coeffi-
cient of efflux is
1
рез
=
1,635
0,782,
and, on the contrary, when the deviation is 35°, the former is
0,753 and the latter 0,755.
These inclined tubes are generally longer than those we have
previously considered, and they must be longer when they are to
be completely filled with water. The foregoing formula gives only
that part of the resistance due to the short tube at the inlet
orifice, that is, three times as long as the tube is wide. The resist-
ance of the remaining part of the tube will be given further on.
EXAMPLE. If the plane of the orifice A B of the discharge-pipe K L,
Fig. 723, as well as the inside slope of the dam, is inclined at an angle of 40°
K
BAS
FIG. 723.
D
L
C
to the horizon, the axis of the tube
will form an angle of 50° with
that plane; hence the coefficient
of resistance for efflux through
the entrance of this pipe is
0,870, and if the coefficient of re-
sistance for the remaining longer

portion is 0,650, we have the coefficient of resistance for the entire tube
5 = 0,870 + 0,650 — 1,520,
and therefore the coefficient of efflux is
μ
1
1
= 0,630.
√1+ 1,520 √2,520
If the head of water is 10 feet and the width of the pipe 1 foot, the
discharge is
Q = 0.630. 8,025 √10 12,56 cubic feet.
4
§ 424. Imperfect Contraction-If a short tube KL, Fig.
724, is inserted in a plane wall, whose area G is but little larger
than the cross-section F of the tube, the water will approach the
§ 424.]
859
THE FLOW OF WATER THROUGH PIPES.
mouth of the short tube with a velocity, which we cannot neglect,
and the stream which enters it is imperfectly contracted; hence
FIG. 724.
A
L
B
the velocity of efflux is greater than
when the water can be considered to
be at rest at the mouth of the tube.
F
Now if
G

n is the ratio of the cross-
section of the tube to that of the wall
and the coefficient of efflux for perfect
Мо
F
contraction, in which case we can put = 0, we have, according
G
to the experiments of the author, for the coefficient of efflux with im-
perfect contraction, when we put the ratio of the cross-sections = n,
μ n
Ho
0,102 n + 0,067 n³ + 0,046 n³, or
μ₁ = μ。 (1 +0,102 n + 0,067 n² + 0,046 n²).
If, E.G., we assume the cross-section of the tube to be one-sixth
of that of the wall, we have
μl
or putting
!
36
µ。 (1 + 0,102.1 + 0,067. ' + 0,046.16)
= μ。 (1 + 0,017 + 0,0019 + 0,0002)
Мо
= 0,815
μ₂ = 0,815. 1,019 = 0,830.
рет
μn
llo
The values
llo
= 1,019 μ¹o,
02
of the correction are given in the following
tables, which are more convenient for use.
TABLE OF THE CORRECTIONS OF THE COEFFICIENTS OF
EFFLUX, ON ACCOUNT OF IMPERFECT CONTRACTION, FOR
EFFLUX THROUGH SHORT CYLINDRICAL TUBES.
N
0,05 0,10 0,15 0,20 0,25 0,30 0,35 0,40 0,45 | 0,50 |
"}
μ
th,
μπ
n
рвать
Мо
Мо
0,006 0,013 0,020 0,027 0,035 0,043 0,052 0,060 0.070 0,080

1,90 0,95 | 1,00
0,55 0,60 0,65 0,70 0,75 0,80 0,85 0,90 0,95 1,00
0,090 0,102 0,114 0,127 0,138 0,152 0,166 0,181 0,198 0,227
860
[$ 424.
GENERAL PRINCIPLES OF MECHANICS.
When the water is discharged through short parallelopipedical
tubes, these corrections are about the same.
The principal applications of these corrections are to the efflux
of water through compound tubes, as, E.G., in the case represented
A
FIG. 725.
B
in Fig. 725, where the short tube K L
is inserted into another short tube
G K, and the latter into the vessel
A C. Here, when the water enters
the smaller from the larger tube, the
stream is imperfectly contracted, and
the coefficient of efflux is determined
by the last rule. If we put the coef-
ficient of resistance corresponding to this coefficient of efflux = 1,
the coefficient of resistance for its entrance into the larger tube
from the reservoir , the head of water = h, the velocity of

D
C
F
G
efflux v and the ratio of the cross-sections of the tube = n,
or the velocity of the water in the larger tube = n v, we have the
formula
v2
(n 11)²
v²
h
+ 5.
+ 5₁
I.E.
2 g
2g
2 g
2,2
h
=
(1 + n² 5 + 51)
and therefore
2 g
√2 g h
v=
√1 + n² 5 + 5₁
EXAMPLE.-What is the discharge from the vessel represented in Fig.
725, when the head of water is h 4 feet, the width of the narrow tube 2
inches and that of the larger one 3 inches? Here
N (})² = 4, whence 1,069. 0,815
end the corresponding coefficient of resistance
1 = 0,318; but we have
=
0,871
5 = (0,371)²-
ら
​= 0,505 and n² (= 1. 0,505
=
0,099,
whence it follows that
1 + n²² 6 + 5₁ 1 + 0,099 + 0,318
and the velocity of efflux
1,417,
8,025 . √4
√1,417
16,05
√1,417
= 13,48.
Finally, since the cross-section of the tube is F
it follows that the discharge is
Q = 13,48. 0.02182
π
--
144
0,02182 square feet,
=
0,294 cubic feet.
$425.]
861
THE FLOW OF WATER THROUGH PIPES.
K
D
FIG. 726.
A
B
$425. Conical Short Tubes or Ajutages.-The discharges
from conical mouth-pieces or short conical tubes are different from
those obtained from cylindrical or prismatic ones. They are either
conically convergent or conically divergent. In the first case the
outlet orifice is smaller than the
inlet, and in the second case the
inlet is smaller than the outlet.
The coefficients of efflux through
the former tubes are greater and
those of efflux through the latter
smaller than for cylindrical tubes.
The same conical tube discharges
more water when we make the
wider end the orifice of discharge, as in K, Fig. 726, than when
we put it in the wall of the reservoir, as is represented at L in the
same figure; but the ratio of the discharge is not as great as that
of the openings. When authors such as B. Venturi and Eytelwein
give greater coefficients of efflux for conically divergent than for
conically convergent tubes, it must be remembered that the smaller
cross-section is always considered as the orifice. The influence of
the conicalness of the tubes upon the discharge is shown by the
following experiments, made under heads of from 0,25 to 3,3
meters, with a tube A D, Fig. 727, 9 centimeters long. The width
of this tube at one end was D E 2,468,
at the other A B 3,228 centimeters,
and the angle of convergence, I.E. the angle
A O B, formed by the prolongation of the
opposite sides A E and B D of a section
through the axis of the tube, was 40° 50′.
When the water issued from the narrow opening, the coefficient of
efflux was = 0,920; but when it issued from the wider opening, it
was = 0,553. If we substitute in the calculation the narrower
orifice as cross-section, we find it 0,946. The stream, in the first
case, when the tube was conically convergent, was but little con-
tracted, dense and smooth; in the second case, where the mouth-
piece was conically divergent, the stream was very divergent and
torn and pulsated violently. Venturi and Eytelwein have experi-
mented upon efflux through conically divergent tubes. Both these
experimenters also attached to these conical tubes cylindrical and
conical mouth-pieces, shaped like the contracted vein. With a
compound mouth-piece, like the one represented in Fig. 728, the

FIG. 727.
A
E
C
B
D
862
[$ 426.
GENERAL PRINCIPLES OF MECHANICS.
1 3
16
diverging portion K L of which was 12 lines in diameter in the
narrowest place and 21 lines at the widest, and 8 inches long,
and whose angle of convergence was 5° 9', Eytelwein found μ =
1,5526, when he treated the narrow end as the orifice, and, on the
contrary, μ 0,483 when, as was proper, he treated the larger end
FIG. 728.
as the orifice. However,
K
G
L
1,5526
2,5 times as much
0,615
water is discharged through this compound mouth-
piece as through a simple orifice in a thin plate, and
1,9 times as much as through a short
1,5526
0,815
cylindrical pipe. When the velocities and the angle of divergence
are great, it is not possible to produce a complete efflux, even by at
first closing the end of the mouth-piece.
The author found with a short conically divergent mouth-
piece 4 centimeters long, whose minimum and maximum widths
were 1 and 1,54 centimeters and whose angle of divergence was
8° 4′, under a head of 0,4 meters, μ 0,738 when the internal edge
was rounded off, and µ = 0,395 when it was not.
§ 426. The most extensive experiments upon the efflux of
water through conically convergent tubes are those made by d'Au-
buisson and Castel. A great variety of tubes, which differed in
length, width and in the angle of convergence, were employed.
The most extensive were the experiments with tubes 1,55 centi-
meters wide at the orifice of efflux and 2,6 times as long, I.E., 4 cen-
timeters long; for this reason we give their results in the follow-
ing table. The head of water was always 3 meters. The discharge
was measured by a gauged vessel, but in order to determine not
only the coefficient of efflux, but also the coefficients of velocity
and contraction, the ranges of the jet corresponding to the given
heights were measured, and from them the velocities of efflux were
calculated.
The ratio
0
√2 g h
of the effective velocity v to the theoretical
one 2 g h gave the coefficient of velocity o, the ratio
√
Q
of
Fv2 g h
the effective discharge Q to the theoretical discharge F V2gh the
coefficient of efflux p, and, finally, the ratio of the two coefficients,
1.E.,
μ
determined the coefficient of contraction a.
§ 427.]
863
THE FLOW OF WATER THROUGH PIPES.
This determination is not accurate enough, when the velocities
of efflux are great; for in that case the resistance of the air is too
great.
TABLE OF THE COEFFICIENTS OF EFFLUX AND VELOCITY FOR
EFFLUX THROUGH CONICALLY CONVERGENT TUBES.
Angle of
convergence.
Coefficient of Coefficient of
efflux.
velocity.
Angle of
convergence.
Coefficient of Coefficient of
efflux.
velocity.
0,829
0,829
13° 24′
0,946
0,963
I
36'
0,866
0,867
14° 28'
0,941
0,966
3° 10'
0,895
0,894
16° 36′
0,938
0,971
4° Io'
0,912
0,910
19° 28′
0,924
0,970
5° 26'
0,924
0,919
21° 0'
0,919
0,972
7° 52'
0,930
0,932
23° 0
0,914
0,974
80
58'
0,934
0,942
29°
58'
0,895
0,975 •
10° 20'
0,938
0,951
40° 20'
0,870
0,980
12° 4'
0,942
0,955
48° 50′
0,847
0,984
According to this table, the coefficient of efflux attains its maxi-
mum value 0,946 for a tube, whose sides converge at an angle of 134°,
that, on the contrary, the coefficients of velocity increase continu-
ally with the angle of convergence. How the foregoing table is to
be employed in practice, is shown by the following example.
EXAMPLE.—What is the discharge through a short conical mouth-piece
1½ inches wide at the orifice of efflux and converging at an angle of 10°, when
the head of water is 16 feet? According to the author's experiments, a
cylindrical tube of this width gives μ = 0,810, d'Aubuisson tube, however,
gave µ = 0,829, or 0,829
0,829, or 0,829 — 0,810 0,019 more; now, according to the
table, for a tube converging at 10°, µ = 0,937; it is therefore better to put
for the given tube µ = 0,937 0,019 = 0,918; whence we obtain the
====
discharge
π
Q = 0,918.
0,825 √16
0,918 . 8,025 π
=
4.83
0,3616 cubic feet.
64
§ 427. Resistance of Friction. The longer prismatical or
cylindrical pipes are, the greater is the diminution of the discharge
through them; we must therefore assume that the walls of the
pipes by friction, adhesion or by the water's sticking to them resist
the motion of the water. As we might suppose, and in accordance
with many observations and measurements, we can assume that
864
[§ 427.
GENERAL PRINCIPLES OF MECHANICS.
this resistance of friction is entirely independent of the pressure,
that it is directly proportional to the length and inversely to the
diameter d of the pipe, L.E., it is proportional to the ratio
7
ď
It has
also been proved that this resistance is greater when the velocities.
are great and less when they are small, and that it increases, very
If we measure this
nearly, with the square of the velocity v.
resistance by a column of water, which must afterwards be sub-
tracted from the total head h, in order to obtain the height neces-
sary to produce the velocity, we can put this height, which we will
hereafter call the height of resistance of friction,
h = 5.
7
v²
d⋅ 2 g
denoting here an empirical number, which we can style the co-
efficient of friction. Hence the loss of head or of pressure in conse-
quence of the friction of the water in the pipe is greater, the greater
the ratio
16
d
of the length to the width and the greater the height
due to the velocity
section of the tube
2,2
is. From the discharge Q and the cross-
2 g
π d²
F =
4
4 Q
π d 29
we obtain the velocity
and, therefore, the height of resistance of friction
1 4
2
· · 2 · 2, (+ 2) = 5·2, ()²
h = 5.
d 2g πα
Š
1
29
Q²
b
If we wish to conduct a certain quantity Q of water through a
pipe with as little loss of head or fall as possible, we must make
the pipe as short and as wide as we can. If the width of the pipe
is double that of another, the friction in the former is () = 3
that in the latter.
5
If the cross-section of the pipe is a rectangle, whose height is a
and whose width is b, we must substitute
1
d
4
•
πα
j π d ²
periphery
14
area
2 (a + b)
a b
a + b
2 ab'
l (a + b) v²
2 a b 2g
whence we have
h = 5.
§ 428.]
865
THE FLOW OF WATER THROUGH PIPES.
By the aid of these formulas for the resistance of friction in
pipes, we can find the discharge and the velocity of efflux of the
water conveyed by a pipe of a given length and width, under a
given pressure. It is also of no consequence whether the tube KL,
Fig. 729, is horizontal or inclined upwards or downwards, so long
as we understand by the
head of water the depth
FIG. 729.
Α.
H
R
RL of the centre L of

the mouth of the pipe
below the level H 0 of,
B
the water in the reser-
voir.
If h is the head of water, h, the height of resistance for the ori-
fice of influx, and h, the height of resistance for the remaining part
of the tube, we have
and
h
V³
1 − (hr + hr) = 3g
(h₁
2 g'
v²
or h
+ h₁ + h₂.
2 g
If 5 denotes the coefficient of resistance for the orifice of influx
the coefficient of resistance of friction of the rest of the tube,
we can put
༡,
2.2
;
1
v2
h =
+5。
+ 5.
•
2g
2g
d 2 g
or
1) h = (1 -
(1 + 5 + 5
2, 2
2 g
and
√2 g h
2) v =
√ 1 + 5 + 5.
d
From the latter formula we obtain the discharge Q Fv.
=
For very long tubes 1 + 5, is very small, compared with
and we can write more simply
!
1 v²
h = 5 d⋅ 2 g'
d'
v = 1
or inversely,
d
2 gh.
ď
§ 428. The coefficient of friction, like the coefficient of efflux,
is not perfectly constant; it is greater for low velocities than for
high ones, I.E. the resistance of friction of the water in tubes does.
not increase exactly with the square, but with another power of the
55
}
866
[§ 428.
GENERAL PRINCIPLES OF MECHANICS.
velocities. Prony and Eytelwein have assumed that the head lost
by the resistance of friction increases with the simple velocity and
with the square of the same, and have established for it the formula
h = (a v + B v²) //
in which a and ẞ denote constants determined by experiment. In
order to determine these constants, these authors availed themselves
of 51 experiments made at different times by Couplet, Bossut, and
du Buat upon the flow of water through long tubes. Prony de-
duced from them
h =
(0,0000693 v + 0,0013932 v²)
ď
Eytelwein,
h
=
(0,0000894 v + 0,0011213 v²)
d'Aubuisson assumes
h =
(0,0000753 v + 0,001370 v²) meters.
d
The following formula, proposed by the author, coincides better
with the results of observation; it is
h
· (a +
B\ l v²
V v d 2 g
and is founded upon the assumption that the resistance of friction
increases at the same time with the square and with the square
root of the cube of the velocity. We have, therefore, for the coeffi-
cient of resistance
B
$ = a +
Νυ
and for the height of resistance of friction simply
7 v²
h == 5.
d2g
For the determination of the coefficient of resistance or of the
auxiliary constants a and ẞ the author availed himself of not only
the 51 experiments of Couplet, Bossut, and du Buat, employed by
Prony and Eytelwein, but also of 11 experiments made by himself
and one by a M. Gueymard, of Grenoble. The older experiments
were made with velocities of from 0,043 to 1,930 meters, but by the
experiments of the author this limit has been extended to 4,648
meters. The widths of the pipes in the older experiments were
27, 36, 54, 135, and 490 millimeters, and the newer experiments
$428.]
867
THE FLOW OF WATER THROUGH PIPES.
were made with pipes 33, 71, and 275 millimeters in diameter. By
the aid of the method of least squares, the author found from the
63 experiments
0,0094711
No
$ = 0,01439 +
or
h
(0,01439
0,0094711) 7
009471
v2
0,01439 +
meters,
No
d'2g
or for the English system of measure
h
(0,01439
+
0,017155) 7
No
v2
d' 2 g
REMARK-1) If we take into consideration some other experiments made
by Professor Zeuner with a zinc tube 2 centimeters wide, and with a ve-
locity of from 0,1356 to 0,4287 meters, we obtain
<= 0,014312 +
0,010327
võ
v being given in meters.
2) Newer experiments upon the flow of water with great and very great
velocities were made by the author in 1856 and 1858 (see the " Civilinge-
nieur," Vol. V, Nos. 1 and 3, as well as Vol. IX, No. 1). The results of
these experiments are contained in the following table:
Nature of the tubes.
Width of the
tubes (d).
Mean velocity of
Coefficient
the water in the of friction (.
tubes (v).

Narrow glass tubes
Wider glass tubes.
Narrow brass tubes
•
1,03 ctm.
8,51 meters. 0,01815
1,43
(6
10,18
0,01865
1,04
(C
8,64
0,01869
The same made shorter.
1,04
12,32
"C
0,01784
The same under very great pressure.
1,04
20,99
0,01690
Wider brass tubes
1,43
66
8,66
པ
0,01719
The same made shorter.
1,43
(C
12,40
(C
0,01736
The same under very great pressure .
1,43
21,59 (C
0,01478
Wider zinc tubes.
2,47
(2
3,19
0,01962
The same shorter
•
2,47
4,73
(6
0,01838
The same still shorter
2,47
(C 6,24
<C 0,01790
The same still shorter
2,47
CC
9,18
❤
0,01670
868
[$ 429.
GENERAL PRINCIPLES OF MECHANICS.
The values in the last column again show that the coefficient of resist-
ance for the friction of water in tubes decreases not only as the velocity
(v) increases, but also, although more slowly, as the width (d) of the pipe
becomes greater. However, for high velocities, the formula
5 = 0,01439 +
0,0094711
võ
agrees tolerably well with the numbers found by experiment, E.G., for
v 9 meters
5 =
and for v = 16 meters
0,01439 +0,00316 =
0,01755
<= 0,01439 + 0,00237
=
0,01676.
These coincide very well with the values in the last table, which corre、
spond most nearly to them.
REMARK 3.-M. de Saint-Venant found that the well-known formula
for the resistance of water in tubes agrees better with the results of experi-
ment, when we assume the height due to the friction to increase not with
22
v2 or but with v. (See his "Mémoire sur des formules nouvelles pour
2 g'
la solution des problèmes relatifs aux eaux courantes.") According to him
we must put
4 l
1
h =
d
•
0,00029557 v2 0,00118228
=
0372
0,023197 v-?
•
7 v²
d 2 g
•
The assumption of a fractional exponent for v is not at all new; Woltmann
put v instead of, v² and Eytelwein proposed vi instead of v' (see the
author's article upon Efflux [Ausfluss] in the "allgemeine Maschinenency-
clopädie" of Hulsse.
REMARK 4.--New and very extended experiments upon the motion of
water in pipes have been made by Monsieur H. Darcy (see the report to
the Academy of Sciences at Paris in the Comptes rendus, etc., Tom. 38,
1854, sur des recherches expérimentales relatives au mouvement des
eaux dans les tuyaux"). Mons. Darcy deduces from these experiments,
where the velocity is not less than 2 decimeters, the formula
46
z = (0,000507
=(0,01989
0,00000647
Z
+
r
+
0,0005078) 7 v²
d
meters;
d z g
hence the coefficient of resistance should be
5
<= 0,01989 +
0,0005078
d
This formula, however, is not sufficiently accurate for small velocities.
§ 429. To facilitate calculation the following table of the
coefficients of resistance has been arranged. We see from it that the
variation of this coefficient is not insignificant, since for a velocity
§ 430.]
869
THE FLOW OF WATER THROUGH PIPES.
of 0,1 meter it is = 0,0443, for one of 1 meter,
0,0239 and for
one of 5 meters,
=
0,0186.
TABLE OF THE COEFFICIENTS OF FRICTION OF WATER.
Meters.
v
0
1
0
∞
Decimeters.
૭
2
3
4
5
6
7
8
9
0,0443 0,0356 0,0317 0,0294 0,0278 0,0266 0,0257 0,0250 0,0244
0,0239 0,0234 0,0230 0,0227 0,0224 0,0221 0,0219 0,0217 0,0215 0,0213
2 0,0211 0,0209 0,0208 0,0206,0,0205 0,0204 0,0203 0,0202 0,0201 0,0200
3 0,0199 0,0198 0,0197 0,0196 0,0195 0,0195 0,0194 0,0193 0,0193 0,0192
4 0,0191 0,0191 0,0190 0,0190 0,0189 0,0189 0,0188 0,0188 0,0187 0,0187
We find in this table the coefficients of resistance correspond-
ing to a certain velocity by searching for the whole meters in the
vertical columns and for the tenths of a meter in the horizontal
column and then moving horizontally from the first number and
vertically from the last, until we arrive at the point where the two
motions meet. E.G. for v= 1,3 meters, 0,0227; for v = 2,8,
<= 0,0201.
For the English foot we can put
=

V 0,1 0,2 0,3 0,4 0,5 0,6
0,7 0,8 0,9
0,0686 0,0527 0,0457 0,0415 0,0387 0,0365 0,0349 0,0336 0,0325
ල
V
1 11 11
૫
2
CO
3
4
68
12
20
5 0,0315 0,0297 0,0284 0,0265 0,0243 0,0230 0,0214 0,0205 0,0193 0,0182
REMARK.-A more extensive and more convenient table is to be found
in the Ingenieur, pages 442 and 443.
§ 430. Long Pipes.-In considering the motion of water in
long pipes or combinations of pipes, the three principal questions
to be solved are the following.
1) The length 7 and the width d of the pipe and the quantity
Q of water to be conducted may be given and we may be required
to find the necessary head. In this case we must first calculate
the velocity
v =
Q
F
4 Q
πα
Q
1,2732.
d "
•
870
[S 430.
GENERAL PRINCIPLES OF MECHANICS.
and then search in one of the last tables for the value of the coef-
ficient of friction 5, corresponding to this value, and finally we
must substitute the values d, l, v, 5 and 5, (5, denoting the coeffi-
cient for the orifice of influx) in the first principal formula.
h = (1 + 5 + 5
v²
2 g
2) The length and width of the pipe and the head of water
may be given and the discharge may be required. The velocity
must be found by means of the formula
√
√ Q a h
1 + 5 + 5
d
Now as the coefficient of resistance is not perfectly constant,
but varies somewhat with u, we must first find v approximatively
in order to be able to calculate from it.
From v we determine
Q
=
π d²
4
v = 0,7854 d² v.
3) The discharge, the head of water and the length of the pipe
may be given, and we may be required to determine the necessary
width of the pipe.
( 42 )². 2/20
π
4
Since v =
4 Q
π d²
Q 1
or v² =
we have
+ 5 +
5
1/2)
(+ 2)².
or
π
2
2 g h . ( 47 ) = (1
1
+
7
5)
+ 5
or
d
dos
2 g h = (1
•
1
d'
2 g h . (42)
dⓇ (1 + 5) α + 51;
hence the width of the pipe is
5
d
=
(1 + 5) d + ζι
2gh
⋅ (4-2)*.
But since (4)*
4)= 1,6212 and 1 + 5, as a mean =
1
1,505 and for
the English system of measures
0,0155, we can put
29
d = 0,4787 √ (1,505. d + 5 1)
V
This formula can only be used to obtain approximative values;
Q²
feet.
h
§ 430.]
871
THE FLOW OF WATER THROUGH PIPES.
for not only the unknown quantity d, but also the coefficient,
which depends upon the velocity v = occurs in it.
4 Q
πα
EXAMPLE 1) What must the head of water be, when a set of pipes 150
feet long and 5 inches in diameter is required to deliver 25 cubic feet of
water per minute? Here we have
25. 122
1,2732
3,056 feet, On
60.52
and therefore we can make
0,0243; hence the head of water or total
fall of the pipes must be
150.
h =
· (1,505 + 0,0243 .
0.12). 0,0155. 3,0562
5
= (1,505 + 8,748) 0,0155. 9,339 = 1,484 feet.
2) What is the discharge through a set of pipes 48 feet long and 2
inches in diameter, under a head of 5 feet? Here
8,025 √5
17,945
1,505 + 5.
48.12
2
√1,505 + 288.5
For the present, assuming
0,020, we obtain
17,945
✓ 7,26
17,945
2,7
6,6;
but * =
6,6 gives more correctly $ =
0,0211, and therefore we have
v
17,945
√1,505 + 288 . 0,0211
17,945
6,52 feet, e
√7,582
and the discharge
Q = 0,7854 (22)² 6,52 = 0,142 cubic feet = 245,4 cubic inches.
3) What must be the diameter of a set of pipes 100 feet long, which are
to discharge one half of one cubic foot of water per second under a head
of 5 feet?
d
Here
0,4787 √ (1,505 d + 100 ¿) . † . (†)²
Assuming for the present = 0,02, we obtain
0,4787 V0,075 d + 5 5.
d = 0,4787 0,075 d +0,100, or approximatively
6
d = 0,4787 0,100
=
0,30; hence we have more accurately
=
d = 0,4787 0,0225 +0,100 0,4787 V0,1225
0,3145 feet
3,774 inches.
=
This diameter corresponds to the cross-section
F
0,7854. 0,31452
the velocity is consequently
Q
0.5
V
F 0.0777
=
0,0777 square feet;
6,435 feet,
and the coefficient of resistance = 0,212.
correct value, we obtain
d = 0,4787
Substituting the latter me
0,1285 0,318 feet = 3,82 inches.
=
872
[§ 431:
GENERAL PRINCIPLES OF MECHANICS.
REMARK 1.-Experiments made by the author with ordinary wooden
pipes 2 and 4 inches in diameter gave coefficients of resistance 1,75 times
greater than those for metal pipes, given in the tables in the foregoing par
agraph. While we have, when the velocity is 3 feet, for metal pipes (
0,0243, for wooden pipes its value is = 0,0243. 1,75. 0,042525; in example 1
we found for a metal pipe 150 feet long the head to be 1,484 feet, but for a
wooden pipe under the same circumstances it would be
h
=
=
(1,505 + 0,042525 . 360) 0,0155 . 9,339 16,81. 0,1448 = 2,43 feet.
According to D'Arcy's Experiments, the coefficient of resistance
in-
creases very considerably with the roughness of the walls of the pipe, and
if the walls are very rough it is doubled or even trebled. The author
found more recently the same result.
REMARK 2.—The temperature also has an important influence upon the
resistance of water in pipes. Experiments have been made upon this sub-
ject by Gerstner (see his "Handbuch der Mechanic," Vol. II), and more
recently by Geh. Rath Hagen (see his " Abhandlungen über den Einfluss
der Temperatur auf die Bewegung des Wassers in Röhren," Berlin, 1854).
The experiments of the latter, made, it is true, with very narrow tubes
(d 0,108 to 0,227 inches), have shown that under the same circumstances.
the velocity of the water in pipes does not decrease indefinitely with the
temperature, but that for every tube there is a certain temperature for
which this velocity is a maximum. For the experiments without this
maximum, Hagen finds the following formula:
h
= m l r−1,25 . 1,75, and
m = 0,000038941
0,0000017185 √t,
in which the temperature t is expressed in degrees of the Reaumur ther-
mometer, and the head h, the length 1, the radius of the tuber and the
velocity v in inches (Prussian).
(§ 431.) Conical Pipes.-The resistance of friction in a conical
pipe A D, Fig. 730, can be found in the following manner.
Let us
FIG. 730.
EK D
M
N
R
P
denote the semi-angle of convergence of the walls of the
pipe A C L = B C L by d, the diameter of the inlet
orifice by d₁, that of the outlet by do, the length K L
of the pipe by 1, and the velocity of efflux at D E by v.
At a distance K M: x from the outlet of the tube
the diameter of the tube is
=
NO=y=DE + 2 K Mtang. 8 d₂+ 2x tang. 8,
hence for the velocity w at that point, since

W
V
W
A
L B
2
d₂
y2,
d2
2
y²
we can put
V =
V
2 x
1 +
(1
d2
tang. s)
§ 431.]
873
THE FLOW OF WATER THROUGH PIPES.
O P = N R
For an element N O P R of the tube, whose length is
the height of resistance of the friction is
M Q
d x
cos. d
cos. ď
dh = 5
d x
y cos. 8° 2g
w²
d x
=5.
y cos. d (1 +
8 (1
2x
do
tang. 8)
2 g
d x
2,2
=5.
;
d₂ cos. d (1 +
√ (1 +
2x
da
tang. s)
2 9
h = 5.
v²
2g d₂
hence the height of resistance of friction for the whole tube is
d x
5
(1
2x
1 +
d₂
tang. 8)
8) cos. 8
But
S
S
(1
1 +
1 +
2 x
d₂
2 x
do
=
d x
tang. s)°
da
2 sin. Se
d.
cos. d
¡S (1
8 sin. 8
d x
5
8)%
tang.
sin. 8 [1
8
d.
da
1 +
(1.
cos. d
+
2 x
da
2x
de
[1 − (1 +
8 sin. 8 [1 -
d
x
tang. 8) a (22 tang. 8)
tang. 8)`', whence we obtain
27
do
tang. 8)]
or
(2/2)
do
8 sin. 8 [1 - (2;)]
1
since d, + 2 l tang. & expresses the diameter d₁ of the inlet orifice.
Consequently the required height of resistance is
h = 5 .
d₂
2 g d. 8 sin. 8 [1 - (d;)"]
212
2,2
d,
5
F
sin. 8 [1 - (1/2)). 19
2 g
5 cosec. 8 [1 −
8 [(1 − ( 1242 ) ]
da
d 2 g
If the inlet orifice is much larger than the outlet orifice, we can
4
put (d2
= 0, and consequently
5
v2
h
§ § cosec. §.
;
sin. § 2 g
29
ƒ
874
[§ 432.
GENERAL PRINCIPLES OF MECHANICS.
the resistance of friction in this case does not depend at all upon
the length of the tube.
1
б
EXAMPLE. If the angle of convergence of the outlet portion of the
nozzle A K, Fig. 731, of a fire-engine is 2 d = 5°, that of the inlet portion
A B, 2 $1
18°, the width of the outlet d₂ 7 lines, and the width of the
inlet d, 1½ inches = 18 lines, and if its whole length 4 K = 1 = 6 inches
= 72 lines, what is its coefficient of resistance? Putting the length of the
outlet portion B K 7, and that of the inlet portion A B = 12, we have
1 = 1₁ + l₂ and l₁ tang. 8 + 1½ tang. §,
or in figures
FIG. 731.
1
2
1
§
2
d₁-do
- d₂,
1
l ₁ + l ₂ = 72 and l₁ tang. 21° + 12 tang. 9° =
0,04362 l₁ + 0,15838 l₂
1
= 5,5.
51,54 and l₂
૭
2
11, or
Hence l₁
20,46 lines and the width at B,
where the conical surfaces meet each other, is
3
2
1
d₂ = d₂ + 2 l₁ tang. 8 = 7 + 2.51,54. 0,04362
Since this place is rounded off, we can put d3
hence for the outlet piece
1
11,53 lines.
= 13 lines;
[1 − (773)'] . cosec. 21°
K
[1 - (₂)]
2
sin. S
3

=
0,9159. 22,926
=
21,08,
and for the inlet portion
2
[1 − (~;)"]
3
cosec. S
1
[1
—
(18)]. cosec. 9°
TO
0,7795 、 6,392 = 4,98.
Therefore the height of resistance for the entire nozzle is
h =
$ 21,08 + 4,98. (1/2
3
(2)]
1
v2
2 g
v2
5
v2
21,5
•
g
2 J
21,08 + 4,98. (13)],
v2
= 2,75.29
if we substitute
1
2g
=
0,0155 and assume
=
0,02, we have
29
h = 0,054.207
I.E. about the height due to the velocity, which result coincides very well
with the results of experiments with such a nozzle.
§ 432. Conduit Pipes.-The outlet at the end of a system
of pipes is either under water or in the air. Both cases are repre-
sented in Figures 732 and 733. In the first case we must regard
as the head h the difference of level R C of the two surfaces of
water, and in the second case the vertical distance RO of the out-
let orifice O below the level H of the water in the reservoir. If the
§ 432.]
875
THE FLOW OF WATER THROUGH PIPES.
tube is everywhere of the same width d, the formulas found in
§ 430 can be applied directly; but if the tube is enlarged or nar-
HA
国
​FIG. 732.
R
H
h
B
N
M
L
FIG. 733.


Ꭱ
M
h
rowed at any point, we will have several different velocities in the
pipe, and therefore the resistance of friction for each portion of
the pipe must be calculated separately. Such a case is presented
by the pipes in Fig. 733, which lead to a fountain or jet d'eau, in
which case the mouth-piece O is narrower than the pipe B L M,
which conveys the water. If we put, as we generally do, the ve-
locity of efflux v, the width of the orifice O of efflux
width of the pipe d₁, we have the velocity of the water in the pipe
" = (2) '%
d, the
and if we denote by 7 the length of the pipe B L M and by, the
coefficient of friction, we have for the corresponding height of
friction
h1
51
7, v₁2
dr 2 g dr dr
1,
d
22
Š
2 g
Now if is the coefficient of friction for the inlet orifice Kand
that for the outlet orifice O, it follows that the loss of head caused
by the first is
h。 = ら​。
21
2 g
d
v³
=
ち​。
d 2 g
༠ཀྭ॰ •
h₂ =
ら
​2 g
and, on the contrary, that occasioned by passing through the
second is
hence we have the entire head
༡༣
d
h = 1, + h₂ + h + h = [1 + 5 ( 2 ) + 5 4 ( ˜† )² + 5 ] 3 7
29
and inversely the velocity of efflux
V =
2 g h
1 + (50 + 51
})() +
d, d
2 g
If we wish the jet to rise to the greatest height, the orifice or
mouthpiece must not only cause as little resistance as possible, but
also allow the water to issue from it with its fibres nearly parallel,
so that they may form, while rising, a stream which will hold to-
876
[§ 433.
GENERAL PRINCIPLES OF MECHANICS.
gether as long as possible, and consequently be less disturbed by
the air than a stream which was more or less torn when it left the
orifice. For this reason we prefer a short, cylindrical or slightly
conical mouth-piece, with the orifice of influx rounded off, to an
orifice in a thin plate or to the orifices of the form of the con-
tracted stream, although the former cause a greater loss of velocity
than the latter. The nodes and bulges, which a stream which has
passed through the latter orifices forms or tends to form, give the
air a much better chance to penetrate it than a cylindrical stream.
§ 433. Jets of Water-So long as the stream K L N, which
flows vertically downwards through a horizontal orifice K, Fig. 734,
FIG. 734.
S
O
M
K
L
remains continuous and is not broken up
by the air, its cross-section L decreases.
more and more as the distance K L = x
from the orifice increases. If c is the ve-
locity of efflux and the velocity at L, we
have

v² = 2 g x + c²,
denoting by F the cross-section of the ori-
fice of efflux and by that of the stream
at L, we have the following equation
Yvor F" c² = I² v²,
from which we deduce the equation.
Fc
I'² (c² + 2 g x) = F c', or
Y =
F2 C3
& + 2 g x
g?
for the form of the cataract of Newton (see
Newton's Principia Philosophiæ, Vol. II,
Sect. VII). If the cross-section of the
orifice is a circle, whose diameter is d,
the cross-section at L forms a circle, whose
diameter is y and for which we can put
c² d¹
N
yo
or
c²
+ 2 q x'
Y
4
d
2gx
1 +
c²
Experiments upon the internal consti-
tution of falling streams of water have
§ 433.]
877
THE FLOW OF WATER THROUGH PIPES.
been made by Savart. See Poggendorff's Annalen der Physik,
Vol. 33.
The cross-section O of a stream M S, which rises vertically
from a horizontal orifice M, increases gradually with its distance
M 0 = x from the orifice M; for here the velocity of the water
at O is
v = √ c²
2 g x, and therefore
2
Y' =
F¹² c²
c² — 2 g x ³
hence we have for the diameter of the cross-section at O
y¹
c² d⭑
c² - 2 gx
d
or y
4
2gx
1
c²
C
by h, we have sim-
2g
Denoting the height due to the velocity
ply and generally
Y₁ =
V
d
√1±
X
h
This formula becomes incorrect at its limits; according to it,
E.G. in the rising stream for x = h or at the apex S, the diameter
of the stream would be
y =
d
V1-1
d
1 - 1 0
= ∞
This, however, is not the case; for the various fibres of water,
of which the stream is composed, are not really at rest at the
highest point, but possess a small velocity radially outwards. If
FIG. 735.
Y
the stream of water
A O C, Fig. 735, is in-
clined to the horizon,
this formula

A
x
T
C
M
0
α
D
N
B
d
y =
X
1 ±
h
is still applicable, when
we substitute instead
of x the vertical projec-
tion NO of the stream
A O. If the jet flows
X
878
[§ 434
GENERAL PRINCIPLES OF MECHANICS.
out of the orifice at an angle v to the horizon, its maximum height
BC is
α =
a
c² (sin. v)²
2g
= h (sin. v)² (see § 39).
Therefore its diameter (at the vertex C) is
d
d
d
y = 4
α
V1
´1 — (sin. v)²
√ cos.v
11
h
In the descending portion CD of the stream, y becomes gradually
smaller and smaller, and when the stream reaches the horizontal
plane A D, from which it started, y becomes again d, if the air
has produced no disturbance in the motion of the stream.
=
§ 434. The height s, to which a vertical jet of water will rise,
is approximatively equal to height due to the velocity h
c²
2 g
only
when the velocity of efflux (c) is small. From the experiments
made by the author (see the experiments upon the height of rise
of jets of water with different mouth-pieces in the 5th vol. of the
Zeitschrift des Vereins deutscher Ingenieure), the following facts
concerning jets of water were ascertained.
1) The resistance of the air for small velocities of efflux, viz.,
from 5 to 20 feet, or for heights of rise of from 1 to 6 feet, is so
small that the height of rise of the jet may in this case without
c²
2 g
appreciable error be put equal to the height due to the velocity
2) If the height due to the velocity does not exceed 75 feet or
the velocity of efflux 56 feet, the ratio of the height of rise to the
height due to the velocity can be expressed by the formula
S
h
1
a + ß h + y h"
in which a, ẞ and y denote empirical coefficients to be determined
for each mouth-piece.
3) For jets, which issue from orifices in a thin plate, the con-
stant a can be put = 1; hence we can assume that the resistance
during the passage through the orifice is almost null, when the
velocities are small, and that it is measurable only when the
velocities are great. The coefficient of resistance for these orifices
is therefore not constant, but increases from zero gradually with
§ 434.]
879
THE FLOW OF WATER THROUGH PIPES.
the velocity; the value = 0,97, given in § 408, can only be con-
sidered as a mean one.
4) For the same velocity of efflux the height of rise increases
with the thickness of the stream, or with the width of the orifice;
consequently the resistance of the air is smaller for thick than for
thin streams. The height of rise increases, therefore, not only with
the head, but also with the thickness of the stream.
5) Under the same circumstance a stream, issuing from a circu-
lar orifice, rises higher than one discharged from an aperture of a
different shape (square, etc.)
6) If the velocities of efflux and the widths of the orifices are
the same, those streams which are not contracted rise higher than
those which are, not only because the former are thinner, but also
because the latter, in consequence of their contractions and expan-
sions, oppose less resistance to the penetration of the air.
If the other circumstances and relations are the same and if the
velocities of efflux are not very small, the jets issuing from short
cone-shaped and longer conical tubes or ajutages with an internal
rounding off attain the greatest height.
Mariotte concluded from his experiments upon the height of
rise of jets of water (see Meining's Translation of Mariotte's Prin-
ciples of Hydrostatics and Hydraulics) with orifices in a thin plate
4 to 6 lines in diameter and under heads of from 5 to 35 feet that
the head or height due to the velocity, necessary to produce the rise
8, must be
whence
h = s +
S3
300
Paris feet,
219
1 +
S
300
1 + 0,003333 s.
The very extensive and varied experiments of the author, made
under heads of from 3 to 70 feet, give, on the contrary, for circular
orifices in a thin plate, when their diameter was
1) 1 centimeter
h
1 + 0,0035305 h + 0,00005406 h³, and when it was
S
2) 1,41 centimeters
h
1 + 0,00237191 h + 0,00005609 h³,
S
h being given in English feet.
880
[§ 434
GENERAL PRINCIPLES OF MECHANICS.
With a conical mouth-piece A B C, Fig. 736, 15 centimeters
long and 1 centimeter wide at the outlet
C and 3 centimeters wide at the inlet
orifice A, which was well rounded off, the
following result was obtained:
FIG. 736.
I
FIG. 737.

AC
I
h
3) = 1,0453 + 0,0001137 h

S
+ 0,00007981 h³,
and, on the contrary, with the truncated
mouth-piece A B, Fig. 737, whose width
was 1,41 centimeters at the outlet B,
result was
4)
A
A
Height due to velocity h
h
S
= 1,0216 +0,0007294 h
+ 0,00003036 h².
the
By the aid of these formulas the follow-
ing table of the heights of jets has been
calculated.

10 20 30 40 50 60 70
Height of jet according to (1).. 9.61
66
((
66
CC
แ
(6
66
(6
(2).. 9,715
(3).. 9,48
(4).. 9,69
18,31 25,98 32,58 38,12 42,66 46,30
18,69 26,75 33,77 39,72 44,63 48,58
18,53 26,77 33,97 39,98 44,79 48,47
19,08 28,02 36,39 44,09 51,08 57,31
EXAMPLE.—If the pipe conducting the water to a fountain is 350 feet
long and 2 inches in diameter, and if the conical orifice is inch wide, how
high would the jet rise under a head of 40 feet, provided all the resist-
ances, except the friction, are small enough to be neglected ?
Here if we put
1
350
4
9,5, (~;)* = (4)• = rty and
2100,
d₁
ΤΣ
5₁ = 0,025, 5 = 0,5, (1
the height due to the velocity of efflux is
40
h。
v 2
h =
2g
1 + 5% + 51
a) (a)*
d
4
1 + (0,5 + 0,025.2100). s
1
40
1,207
33,14 feet,
and therefore the height to which the jet will rise in still air is
§ 435.]
881
THE FLOW OF WATER THROUGH PIPES.
h
33,14
8 =
1,0216 +0,0007294 h + 0,00003036 h² 1,0216 + 0,02417 +0,03334
33,14
1,0791
=
30,71 feet.
§ 435. Piezometer.-The head, lost by the water which is
passing through a set of pipes A B CD E, Fig. 738, in conse-
A
HA
K
FIG. 738.
M
BY
B
D
quence of contractions in
the conduit, friction, etc.,
can be measured by means
of the columns of water
maintained in the vertical
tubes BK, CM, DO which
are attached to the pipe;
when they serve for this
purpose only, they are called
piezometers (see § 386).

If v is the velocity of the water at the point B, Fig. 738, where
a piezometer is inserted, 7 the length and d the width of the por-
tion A B of the pipe, h the head of water or depth of the point B
below the level of the water, 5, the coefficient of resistance for the
entrance of the water from the reservoir into the pipe and 5 the
coefficient of friction, we have the height of the piezometer, which
measures the pressure in B,
z = h
v²
(1 + 5 + 5 4 ) 29 9
g
On the contrary, if the length of the portion B C of the pipe is
7, and the fall is h₁, we have the height of the piezometer at C
21 = h + h₁
(1+50
Z
3 وح
(1 + 5 + 5 + 5 ) 10/2
d d g
Hence the difference of the heights of the piezometer is
7,
v²
21 z = h₁
ら
​d' 2 g
and, inversely, the height of resistance of the portion B C of the
pipe is
ら
​.,
༧°
d 2 g
= h₁ + z = z₁ = fall of this portion of the pipe plus
the difference of the heights of the piezometers.
We see from this example that the piezometer can be employed
to measure the resistances, which the water has to overcome in
passing through the pipes. If any obstacle, if, E.G., a small body
sticks fast in the pipe, its presence will be shown immediately by
the sinking of the column of water in the piezometer, and the dis-
882
[§ 435.
GENERAL PRINCIPLES OF MECHANICS.
tance it sinks will indicate the amount of this resistance. The re-
sistances occasioned by regulating apparatuses, such as cocks, valves,
etc. (a subject which will be treated in the following chapter), can
also be expressed by the height of the piezometer. Thus the
piezometer at D is lower than at C not only on account of the fric-
tion of the water in the portion CD of the tube, but also on ac-
count of contraction in the pipe produced by the valve gate S. If,
when the valve-gate is completely open, the difference N O of the
heights of the piezometers =h, and if, when the gate is pushed in
a certain distance, it is h, the difference, or sinking, h₁ - h2,
hɩ — h₂,
gives the height of resistance due to the passage of the water
through the valve gate.
=
Finally, the velocity of efflux of the water can be calculated
from the height of the piezometer. If the height of the piezometer
PQz, the length of the last portion of the tube D E 1 and
the width of the same = d, we have
z = 8
2,2
d' 2 g
=
and therefore the velocity of efflux is
2gz
v =
1
7
d 2 g z
5
ら
​d
EXAMPLE.--If the height of the piezometer P Q = 2 upon the system
of pipes in Fig. 738 is & feet, if the length of the pipe D E, measured from
150 feet and if the diameter of
the piezometer to the outlet orifice, is l
the tube is 3 inches, it follows, when the coefficient of resistance = 0,025,
that the velocity of efflux is
3,5
v =
8,025
0,75
150. 12 · 0,025
8,025. 0,2415
=
1,94 feet,
and the discharge
2
2 = (315). 1,94 0,129 cubic feet.
П
Q
4
12
REMARK. The motion of water in a pipe B C D, Fig. 739, can easily
FIG. 739.

A
0.........
B
E
DE
K
h
be disturbed by air, which may be given off from the water or enter the
pipe from without. In order to prevent either case from occurring, we must
436.]
883
RESISTANCE TO THE MOTION OF WATER, ETC.
take care that the pressure at every point shall be positive, or rather that
it shall exceed the atmospheric pressure, or that there shall be a column
of water C E in every piezometer. The height of this column is
1
2 = h₁ = (1 + 50 + 5'3) 2012
1
g
h₁ denoting the head C O at C,1, the length of the portion B C of the pipe
and the velocity of the water in the tube. It is, therefore, necessary that
v²
h₁ > (1 + 50 + 5 ² 2 ) 2
h,
1
g'
that, E.G., the head of water in the receiving reservoir shall at least exceed
the height due to the velocity of the water in the pipe. Otherwise the
pipe may suck in air in an eddy.
1 + 50 + 5
d
h, h denoting the entire fall Ƒ K
We can also puth₁
>
7
ら
​1 + 50
of the pipe and l its entire length B C D.
If we wish to prevent the air from accumulating in the pipe, we may
lay the pipe in such a position that it will rise slightly in the direction in
which the water is moving. The air will then be carried along with the
water.
CHAPTER IV.
RESISTANCE TO THE MOTION OF WATER WHEN THE CONDUIT
IS SUDDENLY ENLARGED OR CONTRACTED
§ 436. Sudden Enlargement.-Changes in the cross-section
of a pipe or any other conduit produce a change of velocity. The
velocity is inversely proportional to the cross-section of the stream;
the wider the vessel is, the smaller is the velocity, and the narrower
the vessel is, the greater is the velocity of the water flowing through
it. If the cross-section of a vessel changes suddenly, as, E.G., that
of the tube AC E, Fig. 740, does, a sudden change of velocity,
FIG. 740.
D
A
B
C
F
E
G
accompanied by a loss of vis viva
and a corresponding diminution
of pressure, takes place. This
loss is calculated in exactly the
same manner as the loss of me-
chanical effect occasioned by the

impact of inelastic bodies (see § 335). Every element of the water,
884
[§ 436.
GENERAL PRINCIPLES OF MECHANICS.
which passes out of the narrower tube B D into the wider one D G,
impinges against the more slowly moving current in this pipe and
after the impact moves forward with it. Exactly the same phe-
nomena occur when solid inelastic bodies collide; these bodies
also move forward after the impact with a common velocity. Now
we have found that the loss of mechanical effect occasioned by
the impact of inelastic bodies is
L =
2
(V₁ V₂) G₁ G₂
2 g
G₁ + G ₂
,
and since in this case the impinging element G, is infinitely small
compared to the mass of water G, impinged upon, we can put
L
(v₁ — V₂)²
2 g
G
and consequently the corresponding loss of head is
h
(v₁ V2)³
2 g
Hence, by the sudden change of velocity, a loss of head is caused,
which is measured by the height due to this change of velocity.
Now if the cross-section of the one pipe A C, F₁, that of the
other pipe C E, which is united to it, F, the velocity of the water
in the first tube = v, and that in the other v, we have
F v
V₁
F₁
and therefore the loss of head in passing from one tube to the
other is
F
m = (1, − 1).
v2
2 g
and the corresponding coefficient of resistance, which was first
found by Borda, is
The head
5=(
2
= ( − 1 ).
F
2
M₁ = (₁ = 1)². 27
(7 -
g
which we have just found, cannot of course be lost without pro-
ducing any effect; we must rather assume that the mechanical
effect corresponding to it is employed in separating and communi-
cating a vibratory motion to the elements of the water, which before
formed a continuous mass, and in forming the eddies W, W.
The experiments made by the author confirm this theory. If
§ 437.]
885
RESISTANCE TO THE MOTION OF WATER, ETC.
the tube D G is to be maintained full of water it must not be very
FIG. 741.
E
short or much wider than the tube
A C. The loss is done away with
D
A
F
B
C
F
when, as in Fig. 741, the edges are
rounded off so as to cause a gradual
passage from one tube into the other.
EXAMPLE. If the diameter of one of the portions of the compound
pipe, Fig. 740, is twice that of the other, then
=
F
=
F 1
(4)²= 4, the coefficient
(4 1)² 9 and the corresponding height of resistance
22
9
If the
•
2g
of resistance
for the passage from the narrower to the wider tubes is
velocity of the water in the latter pipe is = 10 feet, it follows that the
height of resistance is = 9. 9,0155 . 10² = 13,95 feet.
§ 437. Contraction.-A sudden change of velocity also takes
place, when the water passes from a vessel A B, Fig. 742, into a
narrower pipe D G, particularly if at the place of inlet CD there
is a diaphragm with an opening, whose cross-section is smaller than
the cross-section of the pipe D G. If the area of this orifice = F, and
if a is the coefficient of contraction, we have the cross-section of the
contracted stream F = a F₁; and if, on the contrary, F is the
cross-section of the pipe and v the velocity of efflux, we find the
velocity of the water at the contracted cross-section F by means
of the formula
F
V 2
=
a F
ช
hence the loss of head in passing from F, to F or from v, to v is
h =
(V₂ v)²
2 g
F
a₁ F₁
2
− 1)
2
v²
2 g'
and the corresponding coefficient of resistance is
FIG. 742.
F
5=1
(-1).
F - 1).
A
B
D
E
FIG. 743.
Λ


E
B
If the diaphragm is absent, we have a common short pipe, Fig.
743, and then F F and
886
[§ 437..
GENERAL PRINCIPLES OF MECHANICS.
or inversely
5
= (-1);
1
a
1 + √3
Assuming a = 0,64, we obtain
1
2
< = ('¹ — 0,64)² = (√₂)² = 0,316.
5
0,64
is increased by the resistance at the entrance into the tube and
by the friction of the water in the exterior portion of the tube to
0,505 (§ 422).
From experiments made with a short tube, the inlet orifice of
which was contracted as is represented in Fig. 742, the author has
been led to the following conclusion: The coefficient of resistance
for the passage of the water through the diaphragm and into the
wider tube can be expressed by the following formula:
but we must put
for
F
F
1
0,1
F
F
¿ = ( 1, − 1);

|
0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0
a = 0,616 0,614 0,612 0,610 0,607 0,605 0,603 0,601 0,598 0,596
and consequently

19,789,612,5,256
(= 231,750,99 19,78 9,612 5,256 3,077 1,876 1,169 0,734 0,480
If, E.G., the narrow cross-section is half that of the pipe, the co
efficient of resistance is = 5,256, 1E. the passage through this
contracted orifice occasions a loss of head 51 times as great as the
height due to the velocity.
EXAMPLE.--What is the discharge through the apparatus represented
in Fig. 742, when the head is 13 feet, the diameter of the contracted circu-
lar orifice 1½, and that of the pipe C E, = 2 inches? Here we have
F
1
F
5 =
2
('1')})" = (4)² = ↓ = 0,56 and therefore a = 0,606, and
2
16
9. 0,606
16
− 1) = (165,5454)* = (1,54)
= 3,74.
§ 438.] RESISTANCE TO THE MOTION OF WATER, ETC. 887
Now if we put h =
v2
(1 + 5) we obtain the velocity of efflux
2 g'
√2gh
v =
√1 + 5
and consequently the discharge is
8,025√1,5
√4,74
= 4,51,
πα
π
Q
V
4
. 4. 12. 4,51
= 54,12. π = 170 cubic inches.
4
§ 438. Influence of Imperfect Contraction.-In the case
considered in the last paragraph, where the water comes from a
large vessel, the contraction can be considered as perfect; but if
the cross-section of the vessel, or that of the stream which arrives
at the narrow orifice, is not very great compared to the cross-sec-
tion F1, Fig. 744, of that orifice, the contraction is imperfect, and
the coefficient of resistance is consequently smaller than in the case
just considered. If the notations previously employed are retained,
we have again the height of resistance or the head lost in passing
through F
h
F
= (1,77 - 1)' ~~~
2 g
but we must substitute variable values for a, which increase with
the ratio
F
G
of the cross-section of the narrow orifice to that of the
pipe, which conducts the water to it. If a diaphragm is placed in
FIG. 744.
B
D
B
FIG. 745.
D
E


G
a pipe A G, Fig. 745, of constant diameter, the same reasoning
holds good; but the coefficient a depends upon
F
F
According to the author's experiments, we must substitute in
the formula for the coefficient of resistance
F
= (a,1, − 1).
F
888
[§ 439.
GENERAL PRINCIPLES OF MECHANICS.

for
F
F
=0,1 0,2 0,3 0,4 0,5 0,6 0,7
0,7 0,8 0,9 1,0
a1 0,624 0,632 0,643 0,659 0,681 0,712 0,755 0,813 0,892 1,000
whence it follows that

,290 0,060 0,000
5: 225,9 47,77 30,83 7,801 3,753 1,796 0,797 0,290 0,060 0,000|
If by rounding off the edges the contraction is diminished or
B
FIG. 747.
A
FIG. 746.
D
E
C
G
prevented, the loss of head be-
comes smaller, and it can be done
away with, almost entirely, by in-
troducing into the pipe a piece,
which widens gradually and is
shaped like M N, Fig. 746.

1
EXAMPLE.-What head is necessary, if the apparatus represented in Fig.
747 is required to deliver 8 cubic feet of water per
minute? Let the width of the diaphragm F₁ be
1½ inches, the width of the discharge-pipe D G, 2
inches, and the width of the vessel A 0,3 inches,
then we have
B

F
G= (13)
F
2
2
1, whence a = 0,637; now
(13)
(3)2
=
16,
FI
(
D
F
1
and the coefficient of resistance
16
−
9. 0,637
1)' = (10,267) *
5,733
= 3,207.
Hence it follows that the velocity of efflux is
4 Q
παε
4.8
19,2
6,112 feet,
60.
기
​π (1) 2
П
and, therefore, the required head is
h =
(1 + 5)
v2
2 g
=
4,207. 0,0155. 6,1122 2,43 feet.
§ 439. Relations of Pressure in Cylindrical Pipes.-By
B
FIG. 748.
D
E
F
C
G
the aid of Borda's formula we
can calculate the various rela-
tions of the pressure in a dis-
charge pipe, the diameter of
which is not constant. Let p₁
be the pressure and v, the ve-

§ 439.]
RESISTANCE TO THE MOTION OF WATER, ETC. 889
locity of the water at F, and p the pressure and v the velocity of
the same at F, then we have
Ρ
v²
(v₁ — v)²
2
P₁
vi
+
+
+
and therefore
Y
2 g
P1
p
+
29
2
v₁² + (v₁
γ
2 g'
p
(v₁ — v) v
or
Y
Y
P1
p
2g
v²
g
Y
Y
But the total head is
v2 (v₁ — v)²
h
+
2g
hence we have also
F
2,
2g
=
[1 + ( − 1 )'] 200
;
P1
p
2 (v, — v) v
h,
Y
γ
v² + (v₁ − v)³
Ρ
γ
F
2(-1) h.
F
1
1 + ( − 1 )
F₁
2
When a stream of water, whose cross-section is F, flows into the
P
free air, is to the height b of the water barometer, and there-
γ
fore the height of the piezometer at F, is
21
2₁ =
Pi
Y
=
2 (1) A
h
b
+(-1)
1 +
So long as p remains positive, the water will discharge at E G
with the cross-section F filled; if, on the contrary, p becomes
negative, the supposed condition of efflux ceases to exist and the
water flows through the exterior tube C E, as if it were not there,
with the theoretical velocity v₁ = √2 g h.
In order to have a full discharge at E G, it is necessary that
F
2 (1) 1
F
1 +
h
<
Ъ
F
(-1)
1 +
F
2
<b, or that
(− 1)'
G-
F
F
2 - 1)
If, then, the limits of the head h, given by this formula, are sur-
passed, the discharge with a full cross-section ceases.
890
[§ 439.
GENERAL PRINCIPLES OF MECHANICS.
These formulas are also applicable to the case of the pipe C E,
Fig. 742, with a diaphragm; but here we must substitute instead
F₁, a, F; hence, for efflux with a filled tube, we must have
1 +
h
2
<
F
(7 - 1)
F
F₁
(a₂ F-1)
2
If we remove the diaphragm, we have a simple short pipe C E,
Fig. 743, and then F₁ = F; hence we must put
h
b
く
​1 + ( − 1 )
a
-
2-1
α
-1)
2
If we substitute a = 0,54 or
1
α
1 =
0,5625, we obtain the
limit of discharge with filled cross-section through these pipes
h
<
b
1 + 0,3164
2.0,5625
h
,
I.E. 1,17%.
<
b
If we assume b = 34 feet, it follows that when the head is
greater than 1,17 . 34 = 39,8 feet, the efflux with a full cross-
section through a short pipe ceases.
The results of the author's experiments coincide perfectly with
the above conclusions (see the article upon the efflux of water under
great pressure in the 9th volume of the "Civilingenieur").
This limit is reached more quickly, when the water discharges
into rarefied air; for in that case b is less than 34 feet. If, E.G., the
height of the water barometer in this space was three feet, the
efflux with filled cross-section for a short pipe would cease when
the head became h = 1,17.3 = 3,51 feet.
If the water flows through a pipe A CE, Fig. 749, which is
gradually enlarged, the height of the piezometer at the inlet
portion AB is
p
2
p
? - ? - " - " - ? - [ ( ) - 1 ]
Y 2 g Y
F
h,
P1
γ
[(E)-1]
2 g
= {[() - ¹] ^
ター​[-1]
p
=
b,
Y
P1
21
= b
Y
consequently, if we put
FIG. 749.
D
E
F
F
- [( 7 )² - 1] n.
We must have, therefore,
B
C
h
1
Ъ
<
F
1
F
§ 440.]
891
RESISTANCE TO THE MOTION OF WATER, ETC.
when the efflux takes place with a filled cross-section. If we put
h
1,17 or
b
b
h
=
0,8547, we obtain the ratio of the cross-section, for
which, under a head h = 39,8 feet, the efflux with filled cross-
section ceases, viz. ·
F
√ 1 + 0,8547
F
=
1,362.
§ 440. The Relations of Pressure in Conical Pipes.-The
relations of efflux and pressure in a cylindrical pipe C E, with or
without diaphragm, undergo the following modifications, when an-
other mouth-piece or another tube E G H K, Fig. 750, is added to
FIG. 750.
A
C
G
H
-F
F₂
K
D
E
B
F
the former. Let F denote the
cross-section, v the velocity and p
the pressure of the water at the
outlet HK, F, the cross-section of
the inlet, a F₁ that of the con-
tracted stream of water, v, the ve-
v₁
locity and p, the pressure of the
water in the latter; in like man-
ner let F₂ be the cross-section of the tube, where the stream of
water again touches the wall, v, the velocity and p, the pressure
of the water at that point. Then we have

P2 p 213
+
and therefore
>
γ
Y
2 g
Pi
P2
V₂ (V₁ — Vş)
P
2
V₂ (V₁
V₂)
+
Y
Y
g
Y
2 g
g
P
v² + 1½
V1 V2
P
v2
+
+
Y
2 g
g
γ
21
2 g
2 V1 V2 + Vş
2
or, since we can put a F, v₁ = F₂ v₂ = F v, or
F v
Fv
21
2₁ =
and v₂ =
12
a F
F₂'
2
2 F2
71
p
P =² + [1-37 + ()]
Y
Y
a F
2 g
Now the head necessary to produce the required velocity of
efflux is
h =
2 g
+
(v₁ — v₁)² = [1
2 g
from which it follows that
F
2
F v2
F₂
=(1+
= [1 + (-)']
a F
g
892
[§ 440.
GENERAL PRINCIPLES OF MECHANICS.
1
γ
P1 p
Y
+
2 F3
+
قا
1 +
a F, F₂
F
(a-3)
(F)
F2
1
2
1
+
h
P
F₂
a F, F₂
F
+
h,
Υ
1
1
1
2
+
F
F
a F
F
2
1
+
P
a F, F₂
F',
F
I.E. 21 =
h,
γ
1
1
1
2
+
F2
a F₁
F₂
1
or, when the water is discharged into free air,
2
a F₁ F₂
(
1
+
F
Z₁ = b
h.
1
1
1
2
F2
-+
a F
F₂
If the efflux takes place with full cross-section, we must have,
according to what precedes,
1
1
1
+
h
F3
a F
F
<
or
b
2
1
1
a F, F,
h
F¹2
F2
1 +
b
(a
2
1
h
1
1
F² > 6 x x ) - ( x − 2)
F12
a F₁ F₂
Fb a F
By the aid of the foregoing formula the relations of the efflux
through the conical tubes A B D E, Figs. 751 and 752, can be
FIG. 751.
FIG. 752.


E
D
C
FaF
B
F
F
F
F2
F
0
B
E
D
given by substituting for F, the cross-section of the pipe, where the
stream touches the wall. If d denotes the semi-angle of diver-
gence ACB of one, or the semi-angle of convergence of the other
tube, and if we assume that the length F, F, of the eddy is equal to
the width A B = d of the orifice, we have the width of pipe, where
the stream reaches its wall,
d₂ = d₁ ± 2 d₁ tang. ♪ = (1 ± 2 tang. 8) d₁,
§ 440.]
893
RESISTANCE TO THE MOTION OF WATER, ETC.
and therefore the ratio of the cross-sections
F2
2
F
(d; )² = (1
= (12 tang. 8),
tang.d),²
in which the positive sign is to be employed for the divergent pipe
in Fig. 751 and the negative sign for the convergent one in Fig.
752, E.G. for d = 21 degrees, 2 tang. 8 = 0,0875 and
F₂
(1 ± 0,0875)² either =
1,1827 or 0,8327;
F
hence the velocity of efflux in the first case is
v =
2 g h
1 + ( − 1,1827)° (2)*
a
and, on the contrary, in the second
2 g h
1
2 g h
1 + 0,514
0,514 ( (+)
2 g h
ย
レ
​2
1 +
(5
2
0,8327)*()*
1 + 0,1308 (1)
α
The corresponding coefficient of efflux
u =
1
11 1 + 0,514
0,514 (7)
for the divergent tube is, of course, considerably smaller than the
coefficient of efflux
μ
1
1 +
0,1308 (#)
F
of the convergent tube.
If, E.G., the tubes were three times as long as wide at the inlet
orifice, we would have in the first case
(7)² = (1 +
F
μ =
= (1 + 6 tang. 8)*
tang. d)* = 1,2625* = 2,5405 and
1
√ 2,306
(1
1
= 0,659, and, on the contrary, in the second case
=
- 6 tang. 8)* 0,7375' 0,2958 and
√ 1,0387
= 0,981 (compare § 425).
If the efflux through these pipes takes place with filled cross-
section, we must have
894
[§ 441.
GENERAL PRINCIPLES OF MECHANICS.
h
b
2 F F
F
F\2
1 + (-2)
a
F
2
a F₁ F₁ - [1 + (1)² ]
or in the first case, when
F
1,5939
F 1,5939
=
2,4906 and
= 1,3477,
a F₁
0,64
F₂
1,1827
h
1 + 1,14292
2,3062
<
0,592,
b
6,7112 2,8163 3,8949
and the head h must be less than 34. 0,592 = 20,1 feet.
§ 441. Elbows.-A particular kind of impediment is opposed
to the motion of water in pipes, when the latter are bent or form
elbows. These resistances cannot be determined with safety by
theory and must, therefore, like so many of the phenomena of
efflux, be studied by experiment. If a pipe A CB, Fig. 753, forms
an elbow, the stream separates itself from the inner surface of the
second branch of the pipe in consequence of the centrifugal force ;
when this piece is short, the efflux with full cross-section ceases,
and the discharge is, therefore, smaller than from an equally long
straight pipe. If the exterior portion CB of the elbow A CB,
FIG. 753.
FIG. 754.


S
B
B
E
Fig. 754, is longer, an eddy S is formed beyond C, and, when the
tube is again filled, the velocity of efflux v is smaller. This dimi-
nution of the velocity of efflux must be treated exactly in the
same manner as the resistance produced by a contraction in the
pipe. If F is the cross-section of the tube and F, that of the con-
tracted vein, we have the coefficient of contraction of the latter
a =
F
F
and, therefore, the corresponding coefficient of resistance
§ 441.] RESISTANCE TO THE MOTION OF WATER, ETC. 895
F
2
2
< = C - ¹)' = ( − 1)'.
5
a
The coefficient of contraction a, and consequently the corre-
sponding coefficient of resistance 5, depends upon the semi-angle of
deviation & ACD BCE BCF, Fig. 753, and accord-
= = =
ing to the experiments of the author, made with a tube 3 centi-
meters in diameter, we can put
5=0,9457 sin.' d + 2,047 sin.* S.
The following table contains a series of coefficients of resistance,
calculated for different angles of deviation.

8º =
10 20 30 40 45
50
55
60 65
70
0,046 0,139 0,364 0,740 0,984 1,260 1,556 | 1,861 | 2,158 2,431
We see from this table that the vis viva of water in pipes
is considerably diminished by the elbows. If, E.G., the elbow
makes a right angle or d = 45°, we have the loss of head occa-
sioned by it
203
g
v2
h = 5. 2 9
0,984.
2 g'
or nearly as much as the height due to the velocity.
When the pipes are narrower, becomes considerably greater,
E.G., for an elbow 1 centimeter in diameter and with an angle of
deviation of 90°, was found 1,536. See the author's "Experi-
& =
mentalhydraulik.”
If to one elbow A CB, Fig. 755, another elbow is joined, as is
shown in Fig. 756, and Fig. 757, a peculiar, but at the same time
FIG. 755.
FIG. 756.
FIG. 757.
LUL
easily explicable, phenomenon of efflux is observed. The second
elbow B D E, Fig. 756, which turns the stream to the same side
as the first one A CB, produces no further contraction of the



896
[S 442.
GENERAL PRINCIPLES OF MECHANICS.
h
stream, and, therefore, for efflux with full cross-section is no
greater than for a simple elbow A CB. But if the elbow B D E,
Fig. 757, turns the stream to the opposite side, the contraction is
a double one, and the coefficient of resistance is consequently twice
as great as for a single elbow. If, finally, B D E is so joined to
A C B that D E stands at right-angles to the plane A B D, then
becomes about 1 times as great as for the single elbow A C B.
EXAMPLE.—If a system of pipes K L N, Fig. 758, 150 feet long and 5
inches in diameter, which should
discharge 25 cubic feet of water,
contains two elbows, the required
head will be
FIG. 758.

A H
R
h =
(1,505 +8,712 + 2.0,984)
2 g
N
B
M
D
12,185. 0,1448 1,76 feet.
(Compare Example in § 430.)
§ 442. Bends.—Curved pipes, when the other circumstances
are the same, cause much less resistance than elbows. They also
cause, in consequence of the centrifugal force of the water, a par-
tial contraction of the stream A B D, Fig. 759, so that, when the
bend is not terminated by a long straight pipe, the cross-section
F₁ of the stream at its outlet is smaller than that F of the pipe.
But if the bend A B D, Fig. 760, is terminated by a long straight
FIG. 759.
A
FIG. 760.


BME
p
D
F
B
E
D
pipe D E, an eddy F is formed and an efflux with filled cross-sec-
tion again takes place at the expense of the vis viva of water. If
the coefficient of contraction
of resistance of the bend.
F
a, we have for the coefficient
F
5
α
= (-1)
a
The coefficient of contraction a depends upon the ratio of the
γ
radius B M = E M = a, Fig. 759, of the pipe to its radius of cur-
§ 442.]
RESISTANCE TO THE MOTION OF WATER, ETC. 897
vature CMr, and it can be determined approximatively in the
following manner. If v is the velocity of the water upon entering
the bend and, that of the contracted vein, we have v, F₁ = v F,
F
whence v₁ = v, and, therefore, the head which measures the
pressure in BE is
F
h
V
1
2
v₁² — v²
2 g
v²
[(2)² - 1] 90 9
2 g
This height, multiplied by 1 and y, gives the pressure of the stream
of water in all directions upon the unit of surface at E
2
v²
ぴ
​p = ` y = [('-') - 1], y = [()' - 1], .
Y
2
Y.
–
g
Since the centrifugal force of the water acts upon the convex
side in opposition to the pressure p, it is possible that it may bal-
ance the latter completely. But in this case the exterior air would
enter and separate the stream entirely from the convex side, as is
shown in Figs. 759 and 760. The centrifugal force of a prism of
water, whose length is B E = 2 a and whose cross-section is 1, ia,
when the radius of curvature is C M = r,
9
v2
gr
•
2 a Y.
Now if we put p 9, we have the condition of separation of the
stream from the wall of the pipe
1
1 =
a²
4 a
r
and consequently the coefficient of contraction
α =
до
r + 4 a
hence the coefficient of resistance for efflux with a full pipe is
5
・W
/r+4 a
- 1).
ጥ
As this calculation is based upon a mean velocity and a mean
radius of curvature, it will, of course, give but an approximate
value of a and S.
From his own experiments and from the results of some obser-
vations made by Du Buat, the author has deduced the following
empirical formulas for the coefficients of resistance of water in
passing through bent pipes:
1) for bends with circular cross-sections
5 = 0,131 + 1,847 (~)*;
57
898
[$ 442.
GENERAL PRINCIPLES OF MECHANICS.
2) for bends with rectangular cross-sections
a
Š
=
0,124 + 3,104
+ 3,104 (2) *.
The following tables are calculated according to these formulas:
TABLE I.
Coefficients of the resistance due to the curvature of pipes with circular cross-
a
11
818
5
sections.

0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9
|
1,0
0,131 0,138,0,158 0,206 0,294 0,440 0,661 0,977 1,408 1,978
TABLE II.
Coefficients of the resistance due to the curvature of pipes with rectangular
cross-sections.
a
0,1 0,2 0,3 0,4 0,5 0,6 0,7
| 0,8
0,8 0,9 1,0
5= 0,1240,135 0,180 0,250 0,398 0,643 1,015 1,546 2,271 3,228
From the above tables we see that for a circular pipe, whose
radius of curvature is twice the radius of its cross-section, the coef-
ficient of resistance 0,294, and that for a pipe, whose radius of
curvature is at least 10 times the radius of the cross-section, the
coefficient = 0,131.
In order to check the contraction of the stream of water in a
bend A B D, Fig. 761, the cross-section of the pipe must be grad-
ually diminished in such a manner that the ratio of the cross-sec-
tion D H = F of the outlet orifice to that B E
F of the inlet
1
shall be a =
√5+ 1
FIG. 761.
A
FIG. 762.
A
FIG. 763.
Λ



B
E
BME
H
D
D
E
B
D
§ 442.]
RESISTANCE TO THE MOTION OF WATER, ETC. 850
If one bend B D, Fig. 762, is terminated by another, which
turns the stream further in the same direction, if, E.G., the axis of
the pipe forms a semicircle, like B D E, Fig. 763, the contraction
is not changed and a and have the same values as for the pipe in
Fig. 762, which forms but a quadrant. If, on the contrary, a bend
D E, Fig. 764, which turns the stream in the opposite direction, is
attached to the first one, an eddy Fis formed between the two and
a second contraction of the stream takes place, by which the resist-
ance (5) is nearly doubled.
A
FIG. 764.
FIG. 765.
FIG. 766.



A
B
E
E
B
B
F
S
D
D
E
D
The resistance to water flowing through bends can be dimin-
ished by enlarging the cross-section of the pipe, as in B D E, Fig.
765, or by inserting in it a thin partition, like S in B D E, Fig.
766; for in the first case the velocity v, and in the second the ratio
a
go
is smaller, and consequently the coefficient of resistance is ren-
dered less.
EXAMPLE.-If the system of pipes B L M, Fig. 767, in the second ex-
FIG. 767.
A
R
H
ample of § 430, contains 5 bends
each of 90°, and if the radius of
curvature of each is 2 inches, we
have

h
2/2
a
་
L
M
and according to the first of the
foregoing tables, the correspond-
ing coefficient of resistance =
0,294; consequently for the 5
bends 5 5 = 1,47; hence the velocity of the water issuing from the pipe,
instead of
6,52 feet, is
17,945
v
√7,582
17,945
17,945
=
€5,964 feet,
√ 7,582 + 1,47
√9,052
900
[§ 443.
GENERAL PRINCIPLES OF MECHANICS.
so that the discharge per second is now
=
Q = 0,7854. g. 5,964 0,1301 cubic feet
224,81 cubic inches.
$443. Valve-Gates, Cocks, Valves.-In order to regulate
the discharge of water from pipes and vessels, we employ various
kinds of apparatus, such as cocks, valve-gates, and valves, by means
of which we produce a contraction in the pipe, which occasions a
resistance to the passage of the water, the value of which is deter-
mined in the same manner as the losses of head in the foregoing
paragraph. As the stream of water is subjected to changes of
direction, is divided, etc., the coefficients a and can only be
determined by experiments made for that purpose. Such experi-
ments have been made by the author,* the principal results of
which are given in the following tables:
TABLE I.
The coefficients of resistance for the passage of water through valve-gates
or slide valves (Fr. tiroirs; Ger. Schieber or Schubventile) in parallelo-
pipedical pipes.

Ratio of the cross
F
1,0 0,9 0,8 0,7 0,6 0,5
sections
| 07 | 0
0,4 | 0,3 0,2 0,1
F
Coefficient of re-
0,00 0,09 0,39 0,95 2,08 4,02 8,12 17,8 44,5 193
sistance <
TABLE II.
The coefficients of resistance for the passage of water through valve-gates
or slide-valves in cylindrical pipes.

Relative height of opening 0
8 =
воро
Ratio of the cross-sections=1,000 0,948 0,856 0,740 0,609 0,466 0,315 0,159
Coefficient of resistance
0,00 0,07 0,26 0,81 2,06 5,52 17,0 97,8
* Experiments upon the efflux of water through valve-gates, cocks, clacks,
and valves, made and calculated by Julius Weisbach, or under the title "Un-
tersuchungen im Gebiete der Mechanik und Hydraulik, etc.," Leipzig, 1842.
§ 443.] RESISTANCE TO THE MOTION OF WATER, ETC. 901
TABLE III.
The coefficients of resistance for the passage of water through a cock (Fr.
robinet; Ger. Hahn) in parallelopipedical pipes.
Angle that the cock is
turned d =
Ratio of the cross-sec-
tions =
!
5° 10° 15° 20° 25° 30° 35°
40° 45° 50° 55 66
0,926 0,849 0,769 0,687 0,604 0,520 0,436 0,352 0,269 0,188 0,110 0
Coefficient
of resist -
ance =
0,05 0,31 0,88 1,84 3,45 6,15 11,2 20,7 41,0 95,3 275
275
TABLE IV.
The coefficients of resistance for the passage of water through a cock in a
cylindrical pipe.
Angle that the cock is turned d
Ratio of the cross-sections
5° 10° 15° 20° 25°
25° 30° 35°
0,926 0,850 0,772 0,692 0,613 0,535 0,458
Coefficient of resistance
0,05 0,29 0,75 1,56 3,10 5,47 9,68
Angle that the cock is turned & =
40° 45° 50° 55°
60°
65° 821°
Ratio of the cross-sections
0,385 0,315 0,250 0,190 0,137 0,091 0
Coefficient of resistance
17,3 31,2 52,6 106 206
486
∞
TABLE V.
The coefficients of resistance for the passage of water through throttle-valves
(Fr. valves; Ger. Drehklappen or Drosselventile) in parallelopipedical
pipes.

Angle that the valve is turned &=
Ratio of the cross-sections
Coefficients of resistance
5° 10° 15° 20°
20° 25° 30° 35°
=0,913 0,826 0,741 0,658 0,577 0,500 0,426
0,9180,826
0,28 0,45 0,77 1,84 2,16 3,54 5,7
902
[S 444.
GENERAL PRINCIPLES OF MECHANICS.
Angle that the valve is
turned &=
40° 45° 50°
55° 60°
| |
65°
70°
90°
Ratio of the cross-sec-
tions =
Coefficients of resistance
0,357 0,293 0,234 0,181 0,134 0,094 0,060 0

9,2715,07 24,9 42,7 77,4 158 368
8
TABLE VI
Coefficients of resistance for the passage of water through throttle-valves in
cylindrical pipes.
Angle that the valve is turned d=| 5°
Ratio of the cross-sections
10° 15° 20° 25° 30° 35°
= 0,913 0,826 0,741 0,658 0,577 0,500 0,426
0,24 0,52 0,90 1,54 2,51 3,91 6,22
Coefficient of resistance
Angle that the valve is
turned & =
Ratio of the cross-sec-
tions
Coefficient of resistance
40° 45° 50° 55° 60°
65°
70° 90°
0,357 0,293 0,234 0,181 0,134 0,094 0,060 0
10,8 18,7 32,6 58,8 118 256 751
8
§ 444. With the aid of the coefficients of resistance, given in
the above tables, we can find not only the loss of head for a certain
position of the valve-gate, cock or valve, but also the position we
must give to these apparatus in order to produce a certain velocity
of efflux or a certain resistance. Of course, such a determination
will be more accurate, the more the regulating apparatus resembles
that used in the experiments. Besides, the values given in the
above tables are not correct, when the water, after passing the con-
tracted orifice produced by the apparatus, does not fill the pipe.
again. In order that the efflux with a filled cross-section shall take
place, it is necessary, when the contraction is great, that the pipe
shall have a certain length. The cross-section of the parallelopiped-
ical pipe was 5 centimeters wide and 2 centimeters high, and the
diameter of the cylindrical pipe was 4 centimeters. With the slide-
§ 444.]
RESISTANCE TO THE MOTION OF WATER, ETC. 903
valve or valve-gate, Fig. 768, the cross-section is merely narrowed,
and its shape in one pipe is a simple rectangle F1, Fig. 769, and in
FIG. 768.
K
FIG. 769.
K
FIG. 770.
K



A
B
D
F
C
the other a crescent F, Fig. 770. When cocks are employed, as in
Fig. 771, there are two contractions and two changes of direction,
and the resistance is therefore in this case very great. The cross-
A
B
FIG. 771.
H
FIG. 772.
A
D

D
K
C
B
C
sections of the maximum contractions have very peculiar forms.
The stream is divided by the throttle-valve (or disc and pivot valve),
Fig. 772, into two parts, each of which passes through a contracted
orifice. The cross-sections of the contracted openings are rec-
tangular in parallelopipedical pipes and crescent-shaped in cylin-
drical ones. The following examples will sufficiently explain the
use of the foregoing tables.
EXAMPLE-1)` If in a system of cylindrical pipes 3 inches in diameter
and 500 feet long a valve-gate is introduced, and if it is raised of the
entire height, so as to close § of the diameter, what will be the discharge
through it under a bead of 4 feet? According to what precedes we can
put the coefficient of resistance for the entrance of the water into the pipe
0
1
= 0,505 and the coefficient of resistance of the pipe according to
Table II, § 443, 5,52, whence it follows that the velocity of efflux is
8,025 √4
D
√1,5
7
8,025 . 2
√7,025 + 500.4¢
16,05
√7,025 + 2000
1,505 + 5,52 + 5
d
If we put the coefficient of friction < = 0,025, we obtain
16,05
ニ
​√57,025
=2,125 feet.
904
[$ 445.
GENERAL PRINCIPLES OF MECHANICS.
0
But the velocity v = 2,125 feet corresponds more accurately to ¿ = 0,0265,
hence we have more correctly
16,05
√60,025
2,07 feet,
and the discharge per second is
Q
=
π
4
9. 12. 2,07 = 55,89π = 176 cubic inches.
2) A system of pipes 4 inches in diameter discharges under a head of
5 feet 10 cubic feet of water per minute; at what angle must a throttle
valve, placed in them, be turned to cause a discharge of 8 cubic feet per
minute? The initial velocity is
10.4
60. π (1)²
6
2
1,91 feet,
π
and that after turning the valve
Τσ
•
8 1,91 =
1,528 feet.
The coefficient of efflux in the first case is
v
1,91
√2 g h 8,025 √5
hence the coefficient of resistance is
0,107;
1
1
1
1=
2
με
0,1072
86,34,
and the coefficient of efflux in the second case is
ΤΟ
•
8 0,107 = 0,0856;
hence the coefficient of resistance is
1
0,08562
1 135,5,
and the coefficient of resistance of the throttle valve
<= 135,5 86,34
=
49,16.
Now Table VI, § 443, gives for the angle d = 50°, 5 = 32,6 and for the
angle d
55°, < = 58,8; we can, therefore, assume that, when the vaive
16,56
is placed at an angle of 50° +
5° = 53° 10′, the required quantity
26,2
of water will be discharged. If we take into consideration the fact that
the coefficient of friction changes from 0,0268 to 0,0283 when the velocity
decreases from 1,91 to 1,528, we have more correctly
283
ら
​5 = 135,5 86,34
268
135,5 91,2
=
44,3,
and consequently the angle that the valve must be turned
11,7
50° +
5° 52° 14'.
26,2
§ 445. Valves.—The knowledge of the resistance produced by
valves (Fr. soupapes; Ger. Ventile) is of the greatest importance.
Experiments have also been made by the author with them.
Those most commonly employed are the puppet valve and the
?
§ 445.]
RESISTANCE TO THE MOTION OF WATER, ETC
905
clack valve, which are represented in Figs. 773 and 774. In both
cases the water passes through an aperture in a ring R G, which
FIG. 773.
A
D
B
B
FIG. 774.


C
is called the seat. The puppet valve KL, Fig. 773, is provided
with a spindle, which runs in guides and which permits the valve
to move only in the direction of its axis; the clack K L, Fig. 774,
on the contrary, opens by turning like a door. We see that in
both apparatuses not only the ring, but also the valve are obstacles
to the motion of the water.
F₁
F
The ratio of the aperture in the seat of the puppet valve, with
which the experiments were made, to that of the pipe was 0,356,
and, on the contrary, the ratio of ring-shaped surface around the
open valve to the cross-section of the pipe was = 0,406, hence we
can put as a mean = 0,381. By observing the efflux for differ-
ent positions of the valve it was found that the coefficient of resist-
ance decreased with the lift of the valve, but that this decrease was
very inconsiderable, when the lift exceeded one-half the width of
the orifice. Its value for this position was = 11, and the height
of resistance or loss of head was
% = 5
ترجم
29
= 11.
2,2
2 g
v denoting the velocity of the water in the full pipe. This num-
ber can be used to find the coefficients of resistance corresponding
to other relations of cross-section. If we put in general
F
5 = (7-1)².
a F
we obtain for the case observed
F
= 0,381 and < =
F
(0.3
1
0.381 a
-1)
1) 11,
=
and therefore
1
1
α
= 0,608,
0,381 (1 +11)
4,317 . 0,381
and finally the general expression for the coefficient of resistance
906
[§ 445.
GENERAL PRINCIPLES OF MECHANICS.
A
5 = (0,608 F-1)=(1,645. F − 1).
1
If, E.G., the cross-section of the aperture is one half that of the
pipe, the coefficient of resistance becomes
= (1,645.21)² = 2,292 = 5,24.
In the experiments with clack-valves the ratio of the cross-
F
section of the aperture to that of the pipe, I.E., was 0,535.
F
The following table shows how the coefficients of resistance de-
crease as the opening increases.
TABLE OF THE COEFFICENTS OF RESISTANCE FOR
CLACK-VALVES.

Angle of opening.
15" 20" 25" 30" 35" 40" 45" 50" 55" 60" 65" 70°
Coefficient of resistance.. 90 62 42 30 20 14 9,5 6,6 4,6 3,2 2, 1,7
By the aid of this table the coefficient of resistance for clack-
valves can be calculated approximatively, when the relations of the
cross-sections are different. We must adopt the same method as
we did for puppet valves.
EXAMPLE.-A force-pump delivers every time the plunger descends in
4 seconds 5 cubic feet of water, the diameter of the column pipe in which
the puppet-valve is placed is 6 inches, the interior diameter of the valve-
ring is 3 inches, and the maximum diameter of the valve is 4 inches;
what resistance is to be overcome by the water in passing through this
valve? The ratio of the cross-sections for these apertures is
(3,5)² = (72)² = 0,84,
6
==
and the ratio of the ring-shaped contraction to the cross-section of the
tube is
=
= 1 -
-(4,5)² =
= 1 (4)² 0,44;
―
6
hence the mean ratio of the cross-sections is
F 0,34 + 0,44
2
0,39,
and the corresponding coefficient of resistance
1
F
ら
​1,645
0,39
2
- 1 ) = 3,22º
10,4.
The velocity of the water is
5
20
V
6,37 feet,
4.
π
П
•
( 1 )²
§ C
907
RESISTANCE TO THE MOTION OF WATER, ETC.
446.]
the height due to the velocity is
of resistance is
10,4 . 0,630
=
0,630 feet, and consequently the height
6,55 feet. The amount of water raised in
a second weighs . 62,5 78,125 lbs. ; the mechanical effect consumed
by the passage of the water through the valve in that time is therefore
6,55. 78,125 511,72 foot-pounds.
§ 446. Compound Vessels.-The foregoing theory of the re-
sistance due to the passage of water through contractions, is also
applicable to the discharge from compound vessels. The apparatus
A D, represented in Fig. 775, is divided by two walls, which contain
the orifices F and F, into three communicating
vessels. If the dividing walls were absent and
the edges at the passage from one vessel to the
other rounded off, we would have, as in the case
of a simple vessel, the velocity of efflux
C
FIG. 775.

1
H
B
'ج
V 2 g h
11 + 50
!
F
D
in which 7 denotes the depth of the orifice below
the level of the water and 5, the coefficient of re-
sistance for the passage of the water through the orifice F
But since when the water has passed through the orifices F, and
F₂ the cross-sections a F and a F, change suddenly into the cross-
sections G₁ and G, of the vessels CD and B C, and according to
§ 437 the resistances thus produced are
a
a F₁
2
2
Ꮐ
2 g
( G - 1)' ( ~ F)".
2
F
a F
2,2
G
2 g
F₁
G
2 g
and
v₂²
F
F
a F
2 g
F
G
2 g'
we have
2 g
(G1 - 1)² (af) 57
で
​a F
(a + s ) + 5 + 5 √ = [1 + 5
(1+ཤཱཿ॰) 》༡
2 g
2 J
F
F
a F
G
+ ( E )² + (-) ]
a F
F₁
G
whence we obtain the velocity of efflux
V
√2 g h
F
a F2
√ 1 + 5+ (1 - °C)² + (
F₁
G
a F
908
[§ 446.
GENERAL PRINCIPLES OF MECHANICS.
In the compound vessel, represented in Fig. 776, from which
the water is discharging, the same conditions exist, but perhaps
FIG. 776.
A
D
H
G
K
B
E
m = [1 + (
a
or, since the velocity v₁ =
-
we must here consider the fric-
tion of the water in the com-
municating tube CE. Let 1 be
the length, d the diameter, & the
coefficient of friction of this
tube, and v, the velocity of the
water in it, then we have for
the head lost by the water in
passing from AC to GL

1)+ sal
a F
ら
​2',
F
h₁ = [1 + ( − 1 )²
2
vi
2
a 2 g
1
a F
+ 5
F
2 J
If we subtract this height from the total head h, there remains the
head in the second vessel h h h; hence the velocity of
――
efflux is
V =
√ 2 g h 2
√1 + 50
A
√2 g h
2
a F
√ 1 + 5+ [1 + ( − 1 ) + 5 ] (C)
α
F
This determination becomes very simple when the apparatus is
FIG. 777.
་
D H
01
1
G
H
B
E
K
h₁
1
1
(
21
2 g α 1
=
like the one represented in Fig. 777;
for in this case we can assume the
cross-sections G, G1, G₂ to be infi-
nitely large, compared to the cross-
sections of the orifices F, F, F.
The first difference of level O H, or
the height of resistance for the pas-
sage through F, is
a F\2
a F
•
v2
2 g
and in like manner the second difference of level O, H₁ or the
height of resistance for the passage through F, is
in which a, a, and a,
orifices F, F, and F.
My = ( a F
)
2 2,2
F2 g
α ₂ F
denote the coefficients of contraction for the
Hence
A

§ 446.] RESISTANCE TO THE MOTION OF WATER, ETC.
909
V =
√1 +
and the discharge is
Q
1
√2 g h
F2
αρ
F
(a F )² + ( a )
a F V 2 g h
F
F
√ ₁ + (af) + ( F )
2
F
Nz g h

2
1
( 17 )² + ( 1 )² + ((_,_ F )
F₁
It is easy to see that under the same circumstances compound
vessels, or reservoirs, discharge less water than simple ones.
is
EXAMPLE.-If in the apparatus, Fig. 776, the total head or the depth
of the centre of the orifice F below the level of the water in the first vessel
6 feet, if the orifice is 8 inches wide and 4 inches high and if the
reservoirs are united by a pipe 10 feet long, 12 inches wide and 6 inches
high, what will be the discharge ?
The mean width of the trunk is
d =
4.1.0,5
2. 1,5
Z
3.10
feet, hence
d
= 15.
2
=
0,025, we obtain
7
d
0,025. 15 =
0,375,
Putting the coefficient of friction
and adding S
pipes, we have
0,505, the coefficient for the entrance into prismatical
1 +
− 1)
+
1 +0,505 + 0,375 =
1,88.
d
α
a F
Since
F₁
1
0,64.8.4
12.6
0,2845, it follows that the coefficient of resist-
=
ance for the entire pipe is
coefficient of resistance for the
velocity of efflux
1,88. 0,2845² 0,152, and if we put the
passage through F, = 0,07, we obtain the
8,025 V/6
v =
√1,07 + 0,152
8,025 √6
17,78 feet.
√ 1,222
The contracted cross-section is 0,64. 1. 134 0,32 square feet, and there-
fore the discharge is
Q = 0,32 . 17,78
=
5,69 cubic feet.
910
[S 447.
GENERAL PRINCIPLES OF MECHANICS.
1
CHAPTER V.
OF THE EFFLUX OF WATER UNDER VARIABLE PRESSURE.
§ 447. Prismatic Vessels.—If a vessel, from which water is
issuing through an orifice in the side or bottom, receives no sup-
ply of water from any other source, the level of the water will
gradually sink, and the vessel will finally become empty. Now if
the discharge into the vessel is greater or less than that
FV2gh from it, the water level will rise or sink, until the head
becomes h =
, and afterwards the head and velocity of
1
2g
(
g p
Q
F
2
efflux will remain constant. Our problem now is to determine the
dependence upon each other of the time, of the rising or sinking of
the surface of the water and, if it, occurs, of the emptying of the
vessel, when the latter has a given form and size. The most simple
case is that of efflux through an orifice in the bottom of a prismatic
vessel, which receives no supply of water. Let x be the variable
head FP, F the area of the orifice and G the cross-section of the
vessel A C, Fig. 778, then the theoretical velocity of efflux is
v = √ 2 g x,
A
FIG. 778.
B
and the theoretical velocity of the sinking surface
of the water is

F
F
وح
G
G
√ 2 gx,
D
and the effective velocity
V₁
µ F
G
√2gx.
In the beginning x = F 0 = h, and at the end of the efflux
x = 0, the initial velocity is therefore
X
and the final velocity
We see from the formula
μ F
C
G gh,
√2 g h,
C₁
0.
1
√2 (11) 9 x,
G
§ 448.] EFFLUX OF WATER UNDER VARIABLE PRESSURE. 911
that the motion of the surface of the water is uniformly retarded,
and that the retardation is p
that this velocity will be = 0
time.
I.E.
t =
t
21
p
μ
2
(HF)²
G
g; we also know (§ 14)
and that the efflux will cease after a
F\2
G
μ F
V 2 g h :
G
(
: ("F)' 9
G
μ F
1
2gh
g²
2 G Và
µ F N 2
√2g
We can also put
2 Gh
2 G h
2 V
t =
μ F N 2 g h
Q
µ
and consequently we can assume that a volume V = G h of water
will be discharged through an opening F in the bottom under a
head decreasing from h to 0 in double the time that it would, if the
head were constant and equal to h.
As the coefficient of efflux is not perfectly constant, but in-
creases when the head diminishes, we must employ a mean value
of this coefficient in our calculations.
EXAMPLE.-In what time will a parallelopipedical box, whose cross-
section is 14 square feet, empty itself through an orifice in the bottom,
which is circular and 2 inches in diameter, when the initial head is 4 feet?
Theoretically the time required would be
π
2. 14 √4
2. 14. 144. 2
t
8,025. ·(6)
8,025 π
8064
8,025 π
=
319",9 = 5 min. 19,9 sec.
46
At the end of half the duration of the efflux the head is =
1.4
(1)² h
1 foot, but the coefficient of efflux for an orifice in a thin plate,
corresponding to a head = 1 foot, is μ 0,613; hence the real duration
of the efflux is
=
319,9
0,613
521',8 = 8 minutes 41,8 seconds.
§ 448. Communicating Vessels.-Since for an initial head
h₁ the duration of efflux is
2 Ꮐ ᏉᏂ
t₁
ре
µ F V 2 g
F√2
and for an initial head h, the duration is
2 G N h
to
μ F. √2g
912
[S 448.
GENERAL PRINCIPLES OF MECHANICS.
it follows that by subtraction we obtain the time during which the
head changes from h, to h₂, or the level of the water sinks a dis-
tance h₁ h; its value is
2 G
t =
( Nhi
Wh₂),
µ F. √2 g
or, when the dimensions are expressed in feet,
G
t = 0,249 (Nπ, — √πs).
μ F
On the contrary, when the duration of the efflux is given, we
determine the distance s h₁ h that the surface of the water
sinks by means of the formula
μ
hy = (
u N 2 g ⋅ Ft),
•
S
1/2
2 G
μ
or
μ √2 g. Ft µ √ % g¸ Ft)
Ft (√
G
4 G
The same formulas are applicable to the case of a vessel C D,
Fig. 779, filled from another vessel 4 B, in which the level of the
FIG. 779.
A
C
H
B
water is constant. If the cross-section
of the communicating pipe or orifice
= F that of the vessel to be filled = G
and initial difference of level 0 0, of
the two surfaces of water h, we have,
since in this case the level of the water
in the second vessel rises with a uni-
formly retarded motion, the time re-
time in which the second surface of the

quired to fill it or the
water rises to the level HR of the first
t
2 G N h
μ F. N 2 g
and in like manner the time in which the distance
tween the surfaces of the water becomes 0, 0
which the level of the water rises a distance 0, 0,
2 G
t
(NT
√29
NW).
( 0,
= h₁ be-
hą, or during
s = hr
hr - hq,
•
EXAMPLE 1) How much does the surface of the water in the last exam-
ple (§ 447) sink in 2 minutes? Here we have
=
h. 4, t 2.60 120,
1
F
π
G 14. 144'
and if we assume also μ 0,605, it follows that
hg = (√h₂
Th₂ - μ. √2 g. 11)² = ( 2 –
0,605.8,025. π. 120) 2
•
Ft2
2
П
2. 14. 144
§ 449.] EFFLUX OF WATER UNDER VARIABLE PRESSURE.
913
= 1,546² = 2,3901 feet,
= (2
5 π 2
—
- 0,605. 8,025.
168/
8 = 4
= 4 — 2,3901
=
1,6099 feet.
and that the required distance that it sinks is
FIG. 780.
R
2) What time does the water require to rise in a pipe C D, Fig. 780, 18
inches in diameter, so as to overflow, when the
pipe communicates with a vessel A B by means
of a short pipe 1½ inches in diameter, and when
the surface of the water G is in the beginning
at a distance O H 6 feet below the constant
level of the water A and at a distance O C = 41
feet below the top C of the pipe. We must
substitute in the formula

B
Ο
Ꭰ
h₁
1
G
=
F
2.144
t =
0,81 . 8,025
t
2 G
后
​H √2 g. p (√h₁ — √ñ₂),
6, h₂ = 6 — 4,5 =
1812
=
1,5,
144 and μ =
F
0,81; thus we obtain
288. 1,2248
√1,5)
0,81 . 8,025
54,3 seconds.
§ 449. If the first vessel A B, Fig. 781, from which the water
passes into the second, receives no water, and if its cross-section G₁
HI
FIG. 781.
A
C
B
Ο
D
R
cannot be considered as infinitely great,
compared to the cross-section G of the
other vessel CD, we must modify our cal-
culation. If the variable distance G, O₁ of
the first surface of the water above the
level H R, at which the water in both ves-
sels stands after the efflux, = and the
distance GO of the other surface of the
water below the same plane = Y, the
variable head will be x + y and the corresponding velocity of
efflux will be v = √2 g (x + y), or, since the quantity of water
G₁ x = G y,

V =
√29 (1+
CG)
y.
μ F
G
√29 (1
(1 +
Ꮐ
y;
The velocity with which the surface of the water rises in the sec-
ond vessel is
VI
μ F
G
hence the corresponding retardation is
=
F 2
G
(H) (1 + 2) 3
G
9,
58
914
[§ 450.
GENERAL PRINCIPLES OF MECHANICS.
and the duration of efflux is
Ꮐ
μΡ
F2
G
2)
t = "F√29 (1 + G) y : ("/") (1 + €) 9
G
2g
2 G N y
# F √ 29 ( 1 + c )
(1
G
1/2)
G
Y
Substituting instead of x and y the initial difference of level h,
that is, putting ≈ + y = h or (1 + G)
x
y = h, we obtain
h
Y
G
1 +
G₁
and the time in which the two surfaces of water come to one
level is
t =
2 GVT
G
μ F1 +
F (1 + G) √ 2 g
2 G G₁. Nh
µF (G+ G₁) 2 g
The time during which the difference of level changes from h
to h₁ is, on the contrary,
t =
2 G G₁ (Nh - NTr₁)
µF (G+ G₁) 12 g
1
EXAMPLE.—If the cross-section G₁ of the vessel, from which the water
flows into the other, is 10 square feet and the cross-section G of the vessel
receiving the water is 4 square feet, if the initial difference of level be-
tween the two surfaces of water is 3 feet, and if the cylindrical pipe which
forms the communication is 1 inch in diameter, the time in which the two
surfaces of water will reach the same level is
2.10.4. √3
t
0,82. 8,025.
П 14
4 144
320.72. √3
0,82. 8,025.7 π
= 276 seconds.
§ 450. Notch in the Side.-If the water issues through a
notch, overfall or weir D E from a prismatic vessel A B C, Fig.
A
G
FIG. 782.
B
782, into which there is no water dis-
charged, the duration of the efflux is
found in the following manner. Let us
denote the cross-section of the vessel
by G, the width E F of the notch by b,
and the height DE by h, and let us di-
vide the whole orifice of efflux into
small strips, the length of each being

§ 450.] EFFLUX OF WATER UNDER VARIABLE PRESSURE
915
h
b and the height
ond is
N
If the head is constant, the discharge per sec-
Q = & µ b √ 2 g h³;
G h
dividing the contents
of a layer of water by the latter, we o
N
tain the corresponding duration of the efflux
T=
Gh
for which we will write
3 μ n b v 2 g
osho
3 Gh
2 μ n b v 2 g
h ³
h-.
In order to obtain the duration t of the efflux of a quantity
G (h
h₁) or to determine the time during which the level of the
water above the sill sinks from D E = h to D E₁ = h₁, let us put.
h₁ =
m
n
h, I.E. let h, consist of m parts, and let us substitute in the
last equation, instead of h-, successively
2
1 -
h
+
-
川
​n h
( m² n ) ¯`', (m + ¹ n ) ¯ ³, (m² ± 2 1 ) ., . . . (~ 4 ) ;;
n
>
N
N
N
and then add the results found. In this manner we obtain the re-
quired time
t
Gh
[(mh)
m
h
3 0 1 2² = [(" ") + (" + 1 ) + ... +
2 µ n b v g
2μην
3 Gh
2 µ n b v 2 g
3 Gh
2 µ n b √2g
h-
N
N
+...+
N
N
[m→ + (m + 1)− + . . . + n−1]
[ (1− + 2−³ + 3→→³ + . . . + n − 1)
— (1−3 + 2−3 + 3 + . . . + m−)],
or, according to the Ingenieur, page 88,
t =
3 Gh→
2 µn b √2 g
3 G n³
2 µ b V 2 g h
•
N− + 1
ጎጂ
+ 1
3 + 1
+ 1
3 G
2 (m −³ — n→³)
m) - 1 ]
µ b v Q g h
-]
3 G
1
ре
µ b N 2 g N h
V h
1)
1
If we put h₁ = 0, we obtain
and also t = ∞; hence the
√hr
3 G
μ b √ 2 g
[(
M
ጎ
h
-
time required for the water to sink to the level of the sill will be
infinitely great.
916
[§ 451.
GENERAL PRINCIPLES OF MECHANICS.
EXAMPLE.-If the water issues from a reservoir 110 feet long and 40
feet wide, through an overfall 8 inches wide, in what time will the level
of the water fall from 15 inches above the sill to 6 inches above it? Here
we have
3. 110. 40
1
1
t
μ. &.8,025
0,5
√1,25/
19800
μ . 8,025
―
19800
19800 . 0,5198
(1,4142
0,8944)
8,025 μ
8,025 μ
1282,5
seconds.
μ
If we assume as the coefficient of efflux
of discharge
t
1282,5
0,6
0,60, we have for the real time
2137,5 seconds 35 minutes 37,5 seconds.
REMARK.--For a rectangular orifice in the side we can put approxima-
μ
tively
t
2 G
μ F √ 2 g
((√Th₁ - √T₂)
a²
288
(√h₂
√h₂
in which F and G denote the cross-sections of the orifice and of the vessel,
a the height of the orifice, h, the initial head, and h, the head when the
discharge ceases.
a
If h 2
the orifice becomes a notch and the formula
2'
for overfalls must be employed.
A
FIG. 783.
G
B₁
§ 451. Wedge-Shaped and Pyramidal Vessels.—If the
vessel A B F, Fig. 783, from which the water is discharged, forms
a horizontal triangular prism, the time in
which it will empty itself is found in the
B following manner. If we divide the height
CE = h into n equal parts and pass hori-
zontal planes through the points of divi-
sion, the whole mass of water will be
divided into equally thick horizontal layers,
whose common length is AD land whose
width diminishes from the surface down-
wards. If the width D B of the upper layer =b, the width D, B₁
of another layer situated at the distance C E₁ = x above the orifice
F, which is located at the lower edge of the prism, is У = b,
h

Di
and its volume is y l.
of time is
h
b l x
•
n
N
х
But the discharge in the unit
Q = µ FV 2gx;
hence the small time, during which the water sinks a distance
h
n'
1...
b l x
b l
T=
: µ F V 2 g x
x³.
N
n μ F V 2 g
451.] EFFLUX OF WATER UNDER VARIABLE PRESSURE.
917
Finally, since the sum of all the x from x =
h\
h
to x =
nh.
is
N
N
h³,
(12)*. 2² = 3 n h+,
we have the time of discharge of the whole prism of water
b l
b l
33
3 n ht
C
i b l h
= 233
•
h+ =
9383
I.E.
nμ FV2g
μ FV 2 g
V
μ µ F V 2 g h
t
µ F c
in which V= blh denotes the total volume of the water and
c = √2gh the initial velocity. In this case it requires more
time to empty the vessel than if the velocity c were constant.
If the vessel A B F, Fig. 784, forms an upright paraboloid, we
have the ratio of the radii K M = y and
CD = b
A
FIG. 784.

D
B
Ъ
√x
Nh
F
hence the ratio of the horizontal section
G, through K to the base A D B G is
G₁ y³ X
=
and therefore
G
b²
h'
G x
G₁
;
h
the volume of the layer of water is
G₁ ·
•
h
N
G x
N
As this expression coincides exactly with that found for the trian-
gular prism, we can put here also
t =
j G h
µ F V 2 g h
or, since VG h (§ 124, Example),
T
313
μ Fc
This formula can also be employed in many other cases for the
approximate determination of the duration of efflux, E.G., for de-
termining the time required to empty a dam. It is applicable,
whenever the horizontal sections increase in the same ratio as the
distances from the bottom.
918
[$ 451.
GENERAL PRINCIPLES OF MECHANICS.
A
H
H
FIG. 785.
GER
F
B
If, finally, we have a pyramidal vessel.
A B F, Fig. 785, to deal with, then
G₁: G= x²: h2, and, therefore, G₁ =

1
R1
the volume of the layer H, R₁ is
G₁ h
n
G x²
n h
G x²
;
h²
and the time necessary to discharge the latter is
G
TE
G x²
n h
: μ F V 2 g x
xå.
•
n μ F h v 2 g
h
nh.
to x =
is
N
N
But since the sum of all the x from x =
n
()'.² = 4 n h³,
we have the time necessary to empty the entire pyramid
G
G ht
t
z
nhs = ៖
G2
n µ F h v Q
F
g
√2 g
3 G h
6
µ
F√ 2 g h
1
ре
or, putting Gh = V,
V
t =
μ F c
Since in this case the initial velocity gradually diminishes from
c to 0, the duration of the efflux is greater than if the velocity
remained invariable and equal to c.
EXAMPLE.--What time will a dam, the area of whose surface is 765000
square feet, require to empty itself, when the discharge pipe enters at the
deepest place and is 15 inches in diameter and 50 feet long, and when the
depth is 15 feet? Theoretically
V
t
F √2 g h
765000. 15
П
5
2
•
4
(+). 8,025 √15
19584000
π.8,025 √15
= 200568 seconds.
But the coefficient of resistance for the entrance of the water into the
pipe, which is cut off at an angle of 45°, is
ら
​0,505 + 0,327 (see § 423)
and the resistance of friction for the pipe is
=
0,025
•
22
= 0,025
d 2 g
=
0,832,
50
22
び
​5 2 g
hence the complete coefficient of efflux for the same is
1
μ √1 + 0,832 + 1
2
1
0,594,
√2,832
and the required duration of efflux is
t = 200568 ; 0,594
337655 seconds 93 hours 47 minutes 35 seconds.
§ 452.] EFFLUX OF WATER UNDER VARIABLE PRESSURE. 919
H
H₁
A
t
FIG. 786.
§ 452. Spherical and Obelisk Shaped Vessels.-By the
aid of the formulas, deduced in the foregoing
paragraph, we can find the duration of the
efflux from spherical, obelisk shaped, pyrami-
dal, etc., vessels.

=
حراتي
F
R
R₁
B
π r h³
uFV2gh
1) The time required to empty a segment
of a sphere A FB, Fig. 786, which is filled
with water, whose radius C A = C F = r and
whose height F G = h, is
π h³
aze
μ
F
π
15
√2 g h
(10r 3 h) h
μ F √ 2 g
or, if an entire sphere is to be emptied, in which case h2r,
16 π r² 1/2 r
15 μ F √2 g
and for a hemisphere, where hr, t=
=
14 π r² Vr
15 μ F √2 g
Here the horizontal layer H, R, G₁, corresponding to the
depth F G₁ = x, is
h
2 π r h x
π h x²
= π x (2 r
x) N
n
N
hence, if the velocity of efflux is v = √2 g x, the duration of the
discharge will be
2 π r h
π h
X3/
Т
n µ F V 2 g
n µ F v Q g
μ
Since the first part of this expression coincides with the formula
for the emptying of prismatic vessels and the second part with
that for the emptying of pyramidal vessels, if we in the first
case substitute 2 π r h for b 1 and in the second π h³ instead of G,
we obtain by the aid of the difference of times required to empty a
prismatic and a pyramidal vessel
t
2335
FIG. 787.
A
B
0
E
L
F
K
M
b l h
µ F N 2 g h
and t =
Gh
µ F N 2 g h
the time, in which a segment of a sphere
will empty itself, as was found above.
2) For a vessel A C K, Fig. 787, shaped
like an obelisk or a pontoon, we can employ
the above formulas; for we can consider it
to be composed of a parallelopipedon A EK,
of two prisms B E N and D E N and of a
pyramid C E N (compare § 121). Let b be

*
920
[§ 452.
GENERAL PRINCIPLES OF MECHANICS.
the width A D of the top, b, that K L of the bottom, 7 the length
A B of top, l, that K N of the bottom and h the height O F of
the vessel. Then we have the surface A C of the water
-
b l = b₁ l₁ + b₁ (1 − 1₁) + 1, (b − b₁) + (1 − 1₁) (b — b₁),
1
in which b₁ 7, is the base of the parallelopipedon A E K, b₁ (l — 11)
and 1, (b — b₁) the bases of the prisms B E N and DE K and
(1 — l₁) (b — b₁) that of the pyramid C' E N.
Now the time required to empty the paral-
lelopipedon is
FIG. 788.
A
0
E
L
M
F
N
B

t3
2 3, 4 Nh
t₁
μ F √ 2 g
that required for the triangular prisms is
t₂ =
23
[b, (1 − 1,) + 1, (b — b₁)] √h
μ F √2
g
and finally that required to empty the pyra-
mid is
(1 − 1) (b − b₁) √π
—
μ F √2 g
hence the time required to empty the entire vessel is
t = t₁ + t₂ + tz
= [30 b₁ l₁ +10 b,(l—7,) +10 7, (b−b₁) +6 (1—1,) (b−b,)]
= [3 b l + 8 b₁ ↳₁ + 2 (b l, + b, l)]
√h
15 μ F √2 g
2 Nh
15 μ F √2 g
b₁
b
When
the vessel is a truncated pyramid. Putting in
this case the base b1 = G and the base b₁ l₁ = G₁, we obtain
t =
(3 G + 8 G₁ + 4 NG G₁)
1
2 Nπ
15 μ F √2g
fl
It is easy to see that this formula will hold good for any trian-
gular or polygonal pyramid.
EXAMPLE.--An obelisk-shaped reservoir is 5 feet long and 3 feet wide on
top and at a depth of 4 feet, where a short pipe 1 inch in diameter and 3
inches long is inserted in it, it is 4 feet long and 2 feet wide; how long a
time will be required for the water to sink 24 feet?
empty it is, assuming µ =
0,815,
t = [8.4.2 + 3 . 5 . 3 + 2 (3. 4 + 5 . 2)]
The time required to
2 √4
П
2
15. 0,815. (1). 8,025
4
·
12
§ 453.] EFFLUX OF WATER UNDER VARIABLE PRESSURE.
921
153.4.4.144
153.
2304
12,225. 8,025 π
43 and
153.7,475 1144 sec.
15. 0,815. 8,025. π
At the level 4 21/ 14 feet above the tube l 11 + 3/28
b = b₁ + 3/3 23 feet; hence the time required to empty the vessel, when
it is filled to that height, is
t₁ = [8.4.2 + 3.35. 19 + 2 (2.35 + 4.19)]
= 603 seconds.
•
1152 v1,5
15.0,815. 8,025 π
The difference of these times gives the time (541 seconds) in which the
surface of the water will sink 2 feet from the top.
§ 453. Irregularly-shaped Vessels.-If we are required to
find the duration of efflux for an irregularly-shaped vessel HFR,
FIG. 789.

H
Go Gr
O
R
Gg
G2 G3 G4
O
Fig. 789, we must employ some method of approximation, such as
Simpson's Rule. If we divide the whole quantity of water into 4
equally thick layers and denote the heads corresponding to the
different horizontal sections Go, G1, G2, G3, G4, by ho, hy, ho, h3, h₁,
we obtain, according to Simpson's Rule, the duration of the efflux
4 G3 G₁
+
h. - hs
G
t
+
4 G₁
+
2 Ga
+
12 μ FV2 g
h¸ Wh
V h₂
V ha
If we divide it into six layers, we have
120
12.0
G
4 G
2 G2
t =
+
+
+
4 Gs
+
2 G₁+
4 G5
G6
+
18 μ FV2g
Wh
V hr
V h z
Vhs
Vhs
Vhs
V ho
h. - hs
V =
12
V
The discharge in the first case is
h. - he
18
(G₂ + 4 G₁ + 2 G₂ · 4 G₂ + G₁), and in the second
3
(G。 + 4 G, + 2 G₂ + 4 Gs + 2 G₁ + 4 G5 + Go).
0.2
If the form and size of the reservoir is not known, we can cal-
culate the discharge V by observing the heights ho, h, etc., of the
water at equal intervals of time. If t is the whole duration of the
efflux, we have for orifices in the side and bottom
μ Ft V2 g
12
(Nπ + 4 √ π₂ + 2 √ h₂ + 4 √ π3 + Vha),
and for overfalls or notches
V
μ b t
3
3
VZg{√h³ + 4 √π³ + 2 √π?³ + ± √πz³ + √πa³).
12
922
[$ 454.
GENERAL PRINCIPLES OF MECHANICS.
EXAMPLE.
In what time will the surface of the water of a dam sink 6
feet, when the discharge-pipe is a semi-cylinder 18 inches wide, 9 inches
high, and 60 feet long, and when the cross-sections of the surfaces of the
water are
for a head of 20 feet, G,
= 600000 square feet,
0
((
((
66
18,5 G
แ
495000
66
1
(6
((
17,0"
15,5
G
410000
(6
2
"L
G 3
325000
(6
(6
(C
14,0"
G &
265000
?
4
Now F
8
(3)²
9 п
32
0,8836 square feet. If we put, as in the Ex-
ample of § 451, the coefficient of resistance for the entrance of the water in
the pipe 0,832 and that of the friction, = 0,025
1,6356, we obtain the coefficient of efflux
0,025. 60. 1,091 =
d
1
1
0,537, and
μ =
μι
F √2 g
√1 + 0,832 + 1,6356
√3,4685
0,537. 0,8836. 8,025 = 3,808.
Now we have
Go
600000
G
495000
= 134170,
=
115090,
Vπ
√20
√hi
√18,5
G 2
410000
G3 325000
=
99440,
=
82550,
√he
√17
√hz
√15,5
265000
4
√14
4
6
t =
√h
12.3,808
1194440
7,616
70830; hence the duration of the efflux is
(134170 + 4.115090 + 2. 99440 + 4 . 82550 + 70830)
= 156833 seconds 43 hours 33 minutes 53 seconds.
The discharge is
√ =
6
f. (600000 + 4. 495000 + 2.410000 + 4. 325000 + 265000)
TE
•
4965000
2
=2882500 cubic feet.
§ 454. Influx and Efflux.—If, while water is flowing out of
the vessel, other water is flowing into it, the determination of the
time in which the level of the water will rise or sink a certain dis-
tance becomes much more complicated, and we are generally obliged
to content ourselves with an approximate result. If the discharge
per second into the vessel Q₁> µ F2 gh, the water will rise, and
if Q₁ < µ F √ 2 gh, it will sink. But the level of the water be-
comes constant, when the head is increased or diminished, until it
becomes equal to k
1
2 g \ µ F
(
Q1
2
1
The time 7, during which the
§ 454.] EFFLUX OF WATER UNDER VARIABLE PRESSURE.
923
variable head x is increased a small quantity §, is determined by the
equation
G₁ = Q₁T — µ F√2 g x. T,
1
and, on the contrary, the time, in which the surface of the water
sinks a distance §, is determined by the equation
G₁ = µ F√2 g x . T - Q₁ T.
&
Hence we have in the first case
T=
TE
G, E
Q₁ µ F V 2 g x
Q
G₁ &
µ F √ 2 g x
μ
and in the second
By employing Simpson's Rule, we obtain the time of discharge,
during which G, becomes successively G1, G, etc., and the head h
becomes h₁, h2,
h。 - hs
Go
4 G₁
+
+
t
12
μ
F√2 g h
Q₁
4 G3
µFV2gh-Q
1
G₁
μ
2 G₂
µ F V 2 g h ₂
F √ 2 g h₂ — Qı
+
+
μ F V 2 g hs
Q1
μ F V 2 g h s
Q₁
or if we denote
by
7, we have more simply
μ
FV29,
ho
hs
GⓇ
4 G₁
2 G₂
+
t
12 μ F 12 g
Vho
N k
+
V k
√ h ₂
N k
4 G3
Gx
+
+
√ h3
N k
Nhx
N k
If the vessel is prismatic and its cross-section is constant
and G, we have (see the author's "Experimentalhydraulik,"
=
§ 9, XII)
Vh
t
2 G
μ F √ 2 g
[ Nh
√ Tu + NT. I NT - N
(
for the time, in which the head h changes to h
Nh
Nh
Since for h₁ h:,
Wh
Nπ - Nk
0
∞,
it follows that the level of the water becomes permanent after an
infinite time has elapsed.
For a notch in the side we have the following formula
t =
2
G k (√h — √h)² (h₂ + √h, k + k)
3 Q1
[
(Why
+ √12 tang.-1
N/k)² (h + Nh k + k)
( Vì − 1) 112 k
3 k + (2 √h+ √π) (2 √π, + √k)
924
[§ 455.
GENERAL PRINCIPLES OF MECHANICS.
in which k =
Q₁
( 2 μ b √ 2
g
29
and 7 denotes the Naperian logarithm
and tang.-¹y the arc whose tangent is y.
According as k≤h or the discharge into the vessel
Q₁ Z z p b √ 2 g h³,
k, but in this case the cor-
a rising or a sinking of the water in the vessel takes place. The
state of permanency occurs, when h₁
responding time t becomes = ∞.
EXAMPLE.—In what time will the water in a parallelopipedical box
12 feet long and 6 feet wide rise 2 feet above the sill of a notch in the side
foot wide, when the discharge into it is 5 cubic feet per second? Here
we have h = 0 and consequently more simply
G k
t
3 Q₁
[z
1
h₁ + √/h₁ k + k
(√h - √k)
1
- √3 h₁
+ 12 tang.-1
2
1
√π + √h,
1
Now G =
12.6
5, h₁
2, b
1
29
11/19 μl ==== 0,6, and
5
72 feet, Q1
. 0,6 . . 8,025)² = 2,133 ;
hence the time required is
t =
=
•
72. 2,1330 4,1330 + √4,2660
3.5
[2
(1,4142 1,4605)*
10,238 [10.002144
[16,1984
√12
10,238 (7,969 - 1,776)
1
√6
1,4142 + 2,9210
√12 tang.~' (1,
tang. - ' (1852)]
4.3352
(1.4142
10,238 . 6,193 = 634 seconds.
2,9210)]
§ 455. Locks and Sluices.-We can make a useful applica-
tion of the principles just enunciated to the filling and emptying
of locks and sluices (Fr. écluses; Ger. Schleusen). We distinguish
FIG. 790.
A
H
R
F
~A
L
B
G
B
FIG. 791.
0
-H-
X
R
Ꮹ .
S
two kinds of locks, name-
ly, the single and the
double. The single lock
consists of a chamber B,
Fig. 790, which is sepa-
rated from the water in
the head bay A by the
gate H F and from that
in the tail bay C by the
gate RS. The double
lock, Fig. 791, on the
contrary, consists of two
chambers with an upper
gate KL, a middle one


HF, and a lower one RS.
$455.] EFFLUX OF WATER UNDER VARIABLE PRESSURE. 925
1) If we put the mean horizontal cross-section of the chamber
of a single lock G, the distance O 0, of the centre of the open-
ing in the upper gate below the surface HR of the water in the
head bay = h₁, its distance 0, 0, above the water in the tail bay = h₂
and the cross-section of the orifice in the gate F, we have the
time necessary to fill the lock to the middle of the orifice, during
which the head is constant,
t₁ = μ F
G h₂
√2 g h
and the time necessary to fill the remaining space, during which
there is a gradual diminution of head,
ta
2 G hr
μ F √2 g h
hence the time required to fill the whole lock is
t = t₁ + tą
(2 h₁ + h₂) G
μ F √ 2 g h i
If the orifice in the lower gate is entirely submerged, the head
decreases gradually during the emptying from 0 0, h, + h₂ to
zero, and the time of emptying the lock is, therefore,
t
Q G N h + h₂
μ F √2 g
But if, on the contrary, a part of the orifice lies above the
lower water level, we have to consider two quantities of water, one
discharged above and the other below the water. Putting the
height of the portion of the orifice above the water a,, the height.
of that below the water = a, and the width of the orifice =
obtain the duration of the efflux by means of the formula
t
2 G (h, + họ)
µ b v Z g (ar √ h + hq −
2 NM
=
A1 + α₂ Nh₁ + h₂
2
b, we
2) In the double lock (Fig. 791) the head in the upper cham-
ber which is cut off from the head bay gradually diminishes during
the efflux into the second chamber. If G is the horizontal cross-
section of the first chamber, and if the initial head 0 0₁ = h₁
is diminished to X 0, x, while the water in the lower chamber
rises to the middle of the orifice in the gate a distance 0, 0, hs,
we have the time corresponding to it
2 G
t₁
ре
F (√π — √x).
√2 g
But the discharge is
926
[$ 456
GENERAL PRINCIPLES OF MECHANICS.
G (h₁ - x) =
G₁ h₂; hence
G₁
x = h₁
h₂ and
G
2.G
t₁ =
- Nh₁
G₂ h.)
μ F √ 2 g
2NG
G
NG h₁
ле F √2 g
(VGL - VG h— G, k).
The time in which the water in the second chamber rises to a
level with that in the first, or in which the water in the two cham-
bers assumes a common level, is, according to § 449,
to
μ
and the whole time required to fill it is
2 G G V x
F (G+ G₁) √2 g
2 G₁ NG NG h₁ — G₁ h₂
µ F (G + G₁) √ 2 g
2NG
t = t₁ + t₂ =
μ F √2 g
( NG M
G
h
N hr
VG 1 –
G, h
h₂
G + G₁
EXAMPLE.-What time is necessary to empty and fill a single lock of
the following dimensions: mean length of the lock = 200 feet, mean width
= 24 feet or G 200 . 24 = 4800 square feet; distance of the centre of
= 5 feet, width
the orifice in the upper gate from both surfaces of water
2 feet, height of the orifice in the upper gate
of both orifices
feet, and height of the orifice (entirely submerged) of the lower gate
feet? Substituting in the formula
= 4
= 5
1
(2 h₁ + h₂) G
t
F √ 2 g h
G
=
4800, μ =
0,615, F = 4.21
required to fill it
5, h₂ = 5,
1
· h₁ = 5, h₂
we have for the time
.3.5.4800
t
6,15. 8,025 √5
14400
1,23. 8,025 √5
If we substitute in the formula
1
2 G √ h₁ + h₂ G =
t
М
F√2 g
4. 21 10 and √2 g
=
=
8,025,
652 sec. 10 min. 52 sec.
4800, h, + h₂ = 10, F = 5 . 21 = 12,5, we have
1
the time necessary to empty the lock
t
2.4800 √10
0,615. 12,5. 8,025
= 492 seconds 8 minutes 12 seconds.
§ 456. Apparatus for Hydraulic Experiments. By means
of the apparatus represented in Fig. 792, we can not only show by
more than 100 experiments the most important phenomena of
efflux, but also prove in figures the most important of its laws.
The apparatus consists of a discharging vessel A B C, which is
provided with three orifices F1, F2, Fa, whose centres are at dis-
§ 456.] EFFLUX OF WATER UNDER VARIABLE PRESSURE. 927
tances from the mean level of the water, which are to each other
as the squares 1, 4, 9.
To these orifices various mouth-pieces and
pipes can be applied, and in order to do this without being dis-
S
FIG. 792.
A
W
M
B
B
A

W
H
K2
G2
G₂t
L
F
F
3
K3
L3
D
E
E
turbed by the water, we close the orifice by means of a particular
kind of valve H2, H3, to which is attached a rod passing through a
stuffing box in the back of the apparatus. In the upper and wider
part A B of the apparatus two pointers Z₁ and Z, which are
directed upwards, are placed. These serve as fixed points, the one
marking the beginning and the other the end of the experiment.
The water which is discharged is caught in a vessel, which before
each experiment is placed on top the discharging reservoir, into
which its contents are emptied by opening an orifice that is gen-
erally closed by a stopper.
In order to find by the aid of this apparatus the coefficient of
effluxu for different mouth-pieces and tubes, we must observe by
means of a good stop-watch the time t, in which the water-level
sinks from one pointer to the other, or within which the head 7,
becomes ha; if, then, Fis the cross-section of the orifice and G the
928
[§ 456.
GENERAL PRINCIPLES OF MECHANICS.
area of the sinking surface of the water, we have the coefficient of
efflux (see § 448)
μl =
2 G (VÌ – V,)
and the corresponding mean head
F3.
Ft √2 g
h = (Nπ + Nπ₂) ®.
2
This apparatus is provided with a collection of mouth-pieces and
tubes, viz., square, rectangular, circular and triangular orifices in a
thin plate with or without an internal rim, short cylindrical and
conical tubes, long straight tubes of different diameters, elbows,
bends, etc., which can be inserted in the different openings F₁, F,
By means of an apparatus with the above accessories we can
show in a few hours all the phenomena and laws of efflux; with it
we can study not only the perfect and imperfect and complete and
incomplete contraction, but also the different degrees of the con-
traction of the jet, and we can make ourselves acquainted with the
resistance of friction, with that of elbows and bends, and also, by
observing jets of water and the sucking up of water, with the
positive and negative pressure of water. We will always find
results which agree pretty well, and sometimes extraordinarily
well, with the coefficients given by experiment (µ, 6, a, 5). In our
apparatus G = 0,125 square meters, the usual diameter of the
orifices and tubes is 1 centimeter, and for the lower orifice
h₁ = 0,96 meters and h,= 0,84 meters. (A detailed description of
this apparatus and of the experiments, etc., which can be made
with it, is given in the author's "Experimentalhydraulik.")
The following example shows how well observations with this
apparatus agree with the well-known experiments on a large scale.
With a short cylindrical tube placed in the lower aperture, t was
33, and with a long glass tube, for which the ratio
was found to be 56; from this we deduce in the one case
16
d
124, t
1
₁ = 0,815 and 5
1 =
0,504,
and in the other
1
Мог
0,480 and
1 = 3,332;
2
Mq
hence
51
52 - 5₁ = 3,332
0,504 2,828,
§ 456.] EFFLUX OF WATER UNDER VARIABLE PRESSURE. 929
and therefore the coefficient of friction for the tube is
5 = 1/18 (52
d
ī
(52 — 51)
=
2,828
124
=
0,0228.
Š
According to the first table in § 429, for the mean velocity v =
1,84 meters, with which the water was discharged from the tube,
0,0215; the results agree, therefore, very well. By means
of these experiments, we can satisfy ourselves that the velocity of
efflux of the water does not depend at all upon the inclination of
the tube, but upon the head of water above the orifice of discharge.
The duration of efflux is the same, no matter whether the long tube
is inserted in the lower or middle opening, provided its orifice of
discharge is at the same depth below the surface of the water in the
reservoir.
This apparatus has recently received many additions, so that we
can now make with it experiments upon the efflux of water under
constant pressure, upon the efflux of air, and also upon the pressure,
impact, and reaction of water.
66
CLOSING REMARK. --A very complete list of the works upon the subject
of efflux of water and upon the motion,of water in tubes is given in the
Allgemeine Machinenencyclopädie," Vol. I, Art. "Ausfluss." We will
mention here, among the later works, Gerstner's "Handbuch der Me-
chanik," Vol. 2, Prague, 1832; d'Aubuisson's "Traité d'Hydraulique à
l'usage des Ingénieurs," II édit. 1840; Eytelwein's "Handbuch der Me-
chanik fester Körper und der Hydraulik," 3d edition, 1842; Scheffler's
“Principien der Hydrostatik und Hydraulik," Braunschweig, 1847. The
older works of Bossut and du Buat upon hydraulics are always of value on
account of their practical treatment of the subject. "Die Experimental-
hydraulik, eine Anleitung zur Ausführung hydraulischer Versuche im
kleinen," by J. Weisbach, Freiberg, 1855, is particularly adapted for teach-
ing and for the practical study of hydraulics. Rühlmann's "Hydrome-
chanik" is also to be recommended. The more recent works of Lesbros,
Boileau, Francis, etc., have been mentioned before (§§ 378, 380 and 387).
We can also recommend Rankine's "Manual of Applied Mechanics," as
well as Bresse's "Cours de Mécanique Appliquée," II. But two parts of
the hydraulic experiments of the author have as yet appeared, and they are
1) "Experiments upon the efflux of water through valve-gates, cocks,
clacks, and valves;" and
2) "Experiments upon the incomplete contraction of water during
efflux, etc., Leipzig, 1843."
Several new treatises by the author upon hydraulics are contained in
the "Civilingenieur," the "Zeitschrift des Deutschen Ingenieurvereines,"
etc.
59
930
[§ 457.
GENERAL PRINCIPLES OF MECHANICS.
CHAPTER VI.
OF THE EFFLUX OF THE AIR AND OTHER FLUIDS FROM VESSELS
AND PIPES.
§ 457. Efflux of Mercury and Oil-The general formula
v = √2 g h (see § 397)
for the velocity v of efflux of water under a pressure, measured by
the head h, holds good (see § 399) also for other liquids, such as
quicksilver, oil, alcohol, etc., and can also be employed for the ef-
flux of air and other aeriform fluids, when the pressure is not very
great. If y denotes the heaviness of the fluid and Ρ its pressure
upon the unit of surface, we have in like manner h
therefore
Ρ
and
v = √ 29
Y
If we measure the pressure by means of a piezometer, filled with
a liquid whose density is y₁, the height of the column of liquid is
p
h₁
Yı
hence p = h₁ Y₁, and therefore
v =
129
γι
Y
Y¹ h₁ = √2 g ε₁ h₁,
Yı
Y
in which ε₁ =
the piezometer to that of the fluid which is being discharged.
denotes the ratio of the heaviness of the liquid in
This agreement of the laws of efflux for different fluids is not
confined to the velocity alone, but extends to the contraction of
the fluid vein. Streams of mercury, oil, air, etc., when passing
through an orifice in a thin plate, are contracted in almost exactly
the same manner as a stream of water. Some experiments made
by the author upon the efflux of mercury, oil and air, have shown
conclusively this agreement (see the Polytechn. Centralblatt, year
1851, page 386). These experiments gave
1) With a circular orifice in a thin plate 6,5 millimeters in di-
§ 457.]
EFFLUX OF THE AIR AND OTHER FLUIDS, ETC. 931
ameter, under heads of 91,5 millimeters and 329 millimeters, the
coefficients of efflux

For water.
Mercury.
Rape-seed oil.
μl = 0,709
0,670
0,674
From the above table it appears that the contraction of streams
of mercury and rape-seed oil is a little greater than that of a stream
of water.
2) With a short, well-rounded, conoidal mouth-piece, whose di-
ameter d was 6,6 m. m. and whose length was double the diameter
(1 = 2 d), the following values were found

Rape-seed oil.
For water.
Mercury.
At a temp. 12° C.
At a temp. 39° C.
0,989
0,430
0,665
μ == 0,942
3) A short cylindrical pipe, which was not rounded off inside,
whose diameter was d = 6,76 millimeters and which was three
times as long as wide (7 3 d), gave the following values:

Rape-seed oil.
For water.
Mercury.
At a temp. 124° C.
At a temp. 39° C.
μ = 0,885
0,900
0,363
0,604
From these experiments we find that mercury flows through
short mouth-pieces and pipes but little faster than water, and that,
on the contrary, the velocity of rape-seed oil increases visibly with
the temperature and is less than that of water. The great differ-
ence between the velocity of water and oil is due to the greater ad-
hesion of the oil to the walls of the pipe.
4) The following values of the coefficient of resistance were
obtained with a glass tube 6,64 millimeters in diameter and 86
times as long as wide (I) and with an iron tube 6,78 millimeters in
diameter and 85 times as long as wide (II).
932
[§ 458.
GENERAL PRINCIPLES OF MECHANICS.
Rape-seed oil.
For water.
Mercury.
At a temp. 12° C.
At a temp. 39° C.
I.
<= 0,0271
0,0277
39,21
2,722

II.
$ = 0,0403
0,0461
54,90
5,24
According to this last experiment the coefficient of resistance
of mercury in an iron or glass tube is a little greater, and, on the
contrary, that of rape-seed oil many times greater than that of
water. We also see from these tables that the coefficient of resist-
ance of the rape-seed oil diminishes as the temperature or degree of
fluidity increases. These experiments also show that the coefficient
of resistance for the iron tube is much greater than for the glass
tube, which is due to the greater smoothness of the latter.
§ 458. Velocity of Efflux of Air.-If we assume that the
air does not change its density during the efflux, the well-known
formula for the efflux of water from vessels can also be applied to
FIG. 793.
M
the efflux of air. If p is the pressure of the
exterior air and p, and y, the pressure and
heaviness of the air inside the vessel A B,
Fig. 793, we can put for the velocity of ef-
flux of the latter (see § 399)

A
F
B
(p₁
v =
№ 2 g
p)
Y₁
№ 2 g
Y1
P₁
(1 - p)
Pr
But (according to § 393), if p is the pressure in kilograms upon
a square centimeter of surface, y the weight of a cubic meter of air,
and
its temperature
p
γ
1 + 0,00367. T
1,2514
or, if p is referred to a surface of one square meter,
p 10000
γ
1,2514
hence it follows that
(1 + 0,00367 T) = 7991 (1 + 0,00367 T);
$459.] EFFLUX OF THE AIR AND OTHER FLUIDS, ETC.
933
P1
Yı
p
= √7991 √1 + 0,00367 7,
γ
or replacing 0,00367 by d
p 89,39 √1 + § 7, and v = 89,39 √ 2 g (1 + 8 r) (1
Y
= 396 / (1 + 8 T) (1
(1 + 8 7) (1 − 1),
or for the English system of measures
v =
161,9 √ 2 g (1 + 87) (1
= 1299 √ (1 + 87) (1
p
Pi
p
P1
being expressed in degrees of the centigrade thermometer.
一​别
​
If b is the height of the barometer and that of the manom-
eter (M), we have also
p
P1
b
b + h
or 1.
p
h
P₁ b + h
,
and consequently the velocity of the issuing air
h
v = 396 √ (1 + 8 τ)
δ
meters
b + h
h
feet,
b + h
= 1299 / (1 + 6 T)
or approximatively, when the height of the manometer is small, by
putting
1
1 +
h
h
= 1 -
2 b'
b
v = 396 (1
396 (1 –
- 2
h
2 b
h
6)(1 + 8 7) / meters
b
= 1299
(1
–
h
21/2) √ (1 +57)
(1 + d 7) — feet.
h
h
REMARK.-On account of the ordinary humidity of the atmosphere, it is
advisable in practice to take d = 0,004.
§ 459. Discharge.-If F is the cross-section of the orifice, we
have the effective discharge, measured at the pressure in the reser-
voir, p, or b+ h,
934
[$ 460.
GENERAL PRINCIPLES OF MECHANICS.
Ρι
g
γι
b
= F √2 2 g 1
b + h
Ρ
g
Q₁ = Fv = F √ 29 21 ( 1 − 2) = F √ 29Z √ 1 - B
Ρ
Y
E.G., for atmospheric air
(1-2) F√
Ρ
Q₁ = 396 F V
= 1299 F
(1 + 8T) h
b + h
(1 + ST) h
cubic meters
cubic feet.
b + h
If we reduce this quantity of air to the pressure of the exterior
air p or b, we obtain
= F √ 2gp b + h
FV
Y
b
21
Q
Q₁
Ρ
b + h
b
= FN
2 g P
Ρι
P
11 1.
Y
P
Pi
h
FV
2 g P
b + h
γ
(1 + 1)
h h
b'
h
cubic meters.
h\ h
b
E.G., for atmospheric air

Q
Q
= 396 F (1 + 8 T) (1 +
(1+)

= 1299 F√(1 + 87) (1 + 2/2)
√(1+87)
cubic feet.
EXAMPLE.-The air in a large reservoir is at a temperature of 120° C
and at a pressure corresponding to a height of the manometer of 5 inches,
while the barometer marks 29,2 inches; what will be the discharge through
an orifice 1½ inches in diameter ?
The theoretical velocity of efflux is
5
v = 1299 / (1 + 0,00367 . 120)
= 1299
34,2
1,4404. 5
34,2
=596 feet, and
the cross-section of the orifice is
π d²
F=
4
Π
4
· (1) *
256
L
0,01227 square feet;
hence the theoretical discharge, measured at the pressure in the reservoir, is
= F v
Fv596.
596 . 0,01227 = 7,313 cubic feet,
Q ₁
and, on the contrary, at the exterior pressure the volume is
b + h
Q
Q
34,2
29,2
•
7,313 8,565 cubic feet.
§ 460. Efflux according to Mariotte's Law.—If we suppose
that the air does not change its temperature during the discharge,
§ 460.] EFFLUX OF THE AIR AND OTHER FLUIDS, ETC.
935
we can assume that it expands according to the law of Mariotte
(see § 387), and therefore that the quantity of air Q in passing from
the pressure p to the pressure p, performs the work Q p l
v³
7 (2.).
If
we put this work equal to the energy Qy stored by Q y during
2 g
the efflux, we obtain the following formula
v²
Q y = 1
2 g
(+) 2p, or
v2
p
29
γ
(別
​hence the velocity of efflux is
v =
√2921 ( 2 );
(別​);
Now, as in the foregoing paragraph, for the metrical system of
measures
P 1 + 6т
γ 1,2514
hence we have here also
v = 396 √/ (1 + 8 7) 2 (24)
and
= 396 -
√ (1 + 8 7) 7 (b + h) meters,
v = 1299 √ (1 + 8 7) 1 (24)
b
1299 √ 7
(1 + 8 7) 1 (³ + ") feet,
b h
b
in which b denotes the height of the barometer in the exterior air
and h the height of the manometer for the confined air, 7 the tem-
perature of the latter in degrees centigrade and 80,00367 the
well-known coefficient of expansion of air. Now the theoretical
discharge per second is
Q = Fv = F√29 P
Y
(別​)
= l
1299 F √ (1 + 8 7) 7 (b + h)
√(1+87)
b
cubic feet,
or, when reduced to the pressure of the air in the reservoir,
2g
P
Ρ
Q1
Q
PF
√29
•
P1
Pr
Y
(別​)
b
P 'b
6 +7 F √ 2 9 2 1 (b + h)
b h
b
= 1299 F
b + h
Y
b
b
√ (1 + 87) 1 (0 + h)
b
936
[§ 461.
GENERAL PRINCIPLES OF MECHANICS.
If the excess of pressure of the air in the reservoir, or
small, we can put
b h
パナ​)
h
1 (6 + 4) = 1 (1 + 1 ) = 1 - 1()'
h
b'
is very
(see the Ingenieur, page 81), and therefore, approximatively,
Q = F√ 2g2 (1
√
Y
b
h
h
2 b'
while according to the first formula for the efflux (see § 459)
h\ h
Q = F √ 2 g 2 ( 1 + 4 ) /
Y
We see that if we assume that air in flowing out expands
according to Mariotte's law, we obtain a smaller discharge than
when we consider that the air acts exactly like water and does not
h
expand at all. This difference diminishes with
and in both cases
b'
h
for very small values of
we have
ō'
Q = F√ 29
p h
h
;
Y
b
= 1299 F√ (1
(1
+ 8 t)
+ 87) cubic feet.
b
§ 461. Work Done by the Heat. The logarithmic expres-
sion, found in § 388, for the work done during the compression or
expansion of air is correct only, when we assume that, while the
change of volume or density is taking place, the temperature of
the air does not alter; but this is correct only, when the change
takes place so slowly that the heat in the confined air has time
enough to communicate any excess to the walls of the vessel and
to the exterior air. But if the change of density takes place so
quickly that it is accompanied by a change of temperature, when
the air is compressed, the temperature is elevated and when it is
expanded, it is lowered. Under these circumstances the tension
cannot change according to the law of Mariotte alone. If p and
P₁ are the pressures, y and y, the heavinesses and 7 and 7, the tem-
peratures of the same air, we have, according to § 392, the formula
1 + δ τι γι
P1
Ρ
1 + δ τ Y
T
1
Now if during the sudden change of pressure the temperature
varies in the ratio
1 + δ τι
1 + δ τ
(2)
§ 461.]
EFFLUX OF THE AIR AND OTHER FLUIDS, ETC. 937
we can put
or
FIG. 794.
P1
1 + ɗ
Ρ
+от
(別​)
γι
δτ
γ
P
²² = (1 + + + ) = ( )'.
1+6т
If in a cylinder A C, Fig. 794, a prism of air, whose initial
height is E B = s, whose initial tension is p
and whose heaviness is y, is cut off by a piston
E F, and if, by suddenly raising the piston a
distance x, we cause the density of this mass
of air to become y and its tension to become
z, we have, according to the last formula,

Z
S
p
()² = (-a)'
X
and therefore
A
B
s
2
p.
F
E.
F
E
D
20=
ق)
s + X
In order to move the piston, whose area
we will for simplicity put equal to the unit
of surface, through an element o of its path
the work, which must be done, is
S
X
p o = p o s² ( s − x)−1.
Substituting instead of a successively 1 o, 2 o, 3 o... and put-
tings no and the height of the prism of air, when the piston
has described the space E E, E, B 81 = m σ, we have for the
work done by the piston in moving the distance E E
A₁ = p o sì [s−² + (s − σ)− + (8 − 2 o)−³ + ... + (s — m o)—']
(s
= po si
p s
of
((0) → + (2 o)− + (3 o)→ + ... + (n o)−1
{(0)7) + (207) + (30) ) +
— [(o) + (20) + (30) + ... + (mo)→]
→+ 2 + 3 + + m² +
•
(1− + 2− + 3− +
• •
+ m−})
+ 22-3
Now, according to page 88 of the Ingenieur, when m and n are infi-
nitely great numbers, we have
1→ + 2− + 3→ + ... + m→ =
M²
11
and
mt
938
[§ 461.
GENERAL PRINCIPLES OF MECHANICS.
1− + 2 + 3 + ... + n−3
2
=
;
no
hence
1
στ m$
1
4₁ = 24 (2-2)=2pa² (4)
= 2 p s [(~~)* − 1].
If by raising the piston another distance s we wish to force the
compressed mass of air A E, into a space R, where the pressure is
P();
P₁ = p
1
the work to be done will be
A 2
= Pi Si
p så
8,$
the exterior air presses upon the piston during the whole of its
course with a force p and transmits to it the mechanical effect
A3 = ps. Hence the total mechanical effect necessary to compress
the volume of air (1. s) and force it into the space R is
A = A₁ + Á — A3
*
· * -
st
= 2 p s [(') - 1 ] + " - ps3 ps[()-1]
ps
8, +
−
and consequently the work done in compressing a volume of air
from the pressure p to P1 is
A
s
* -
3
4 = 3 Vp [( - ) - 1] = 3 V p [(")' - 1] = 3 Vp (VE-1),
$1
−
while, according to Mariotte's law, we should put
Α A = Vpl (2)
and for perfectly incompressible fluids we have
P1
A = V (p₁ − p)
= Vp ·
1).
p
If, on the contrary, the quantity V, y, of air at the pressure Pi
is brought back by sudden expansion to the pressure p and the
density
or to the volume
Y = Y1
(2)
√ = i
V₁
1
(2)
the work done by air is
4 = 3 Vp [()'- 1] = 3 V, p[1 − (2)']
A
1
§ 462.] EFFLUX OF THE AIR AND OTHER FLUIDS, ETC.
939
EXAMPLE.-If a blowing engine converts per second 10 cubic feet of
air at a pressure b 28 inches of the barometer into a blast at the pressure
b + h = 30 inches, it requires, according to the formula,
P1
A = 3 V p
´? [ (21) * - 1]
since the pressure per square foot is
-
p 144. 0,4913 6 = 144 . 0,4913 .28 = 1981 pounds,
the mechanical effect
3
Α
30. 1981
28
(√30-1)
14
= 1382 foot-pounds.
The logarithmic formula (see Example 1, § 388) gives A
pounds, and that for water
1)=
15
= 59430
W − 1) =
= 5943.0,2326
1366,7 foot-
A V
= √ p ( P¹
(P₁-1)
= 19810 (11-1):
15
14
19810
14
1415 foot-pounds.
§ 462. Efflux of Air, when the Cooling is taken into
consideration.—The energy A = 3 Q, p₁[1 − (2)']
which is
restored during the sudden expansion of Q, to Q, can be put equal
to the work Q1 Y1
•
Q1Y1
g
v²
2 g
done in overcoming the inertia of the mass
of air when the latter assumes the velocity v.
From the equation
2. p₁ [1 − (2)']
22
Qi Xi
3 Q1 Pi
2
g
P1
we deduce the following formula for efflux:
3 P₁ [1 - (2)² ] ·
γι
v³
2 g
v =
√2
29.
hence we have in meters
•
3 Pi
Y₁
1
or

门
​v = 154,8 √ 2 g (1 + ô 7) [1 − (2)*]
στ) [1 − (3)']

= 685,8 √ (1 + d +)
and in English feet
V
-

= 280,4 √√ 2 g (1 + 8 7) [1 − ()]
-
-

8 7) [1 − (2.)'] feet.
-
= 2250 √ (1 + 8 7)
-
940
[$ 462.
GENERAL PRINCIPLES OF MECHANICS.
}
The tension of the issuing air is that of the exterior air p; its
heaviness is
and its temperature is
cop
Y = Y
(2)
1
T₂ = T1
(男​)
P1
б
Fv=FV 2 g
•
3 Pi
Y₁
[1 − (²)']
[¹ - (3)*]
cubic feet,
Ρι
and the theoretical discharge from an orifice, whose area is F, is

Q2
= (187)
280,4 F√ 2 g (1 + 8 7) [:
· [1
1
in which p₁, y₁ and 7, denote the pressure, heaviness and tempera-
ture of the confined air.
Reduced to the pressure in the reservoir, this discharge is
Y2
12. Q₁ = (+)² 2₂ =
= F
γι
√29.3 2 [1-()}
(2) 12
γι
and, finally, reduced to the pressure of the exterior air and to the
temperature of the air in the vessel or to the heaviness y
Y1
(11)
it is
Q
P1
Q1
Q₁ = F ( 2 ) + √ √ 2
3 pi
p
p
9
3 P₁
g
= F √ 29³
If we put
P1
b+ h
p
b
Yı
Y1
(分
​[1 − (93)']
(2) [ (2)² - 1}
* * − 1]
20
in which b denotes the height of the ba
rometer in the exterior air and b + h that of the barometer in the
confined air, we obtain

3 p₁ (b
Q = F √ √ 2 g ³ !': ('³ + ^ )
Yı
-
h
[ ("'
+
*)* − 1]
7)
(† †
( + ¹) [P
¹) * [(" + ")*.
h
- 1]
= 280,4 F√ 2 g (1 + 5
=2250 FV (1 + d T)
In most cases
h
h
b
h
6+16+
h
b

(+4) [ ( + 4) - 1] cubic feet
b
is very small, and we can put
(† + 4) = (1 + 1 )
(6)=(1+
b
b + h $
( 4 ) 1 =
b
} }
h
b
$
= 1
-
↓
193
+
( ) )
1
1/1/
h
}
b
+
1
- ()+
5
81
(†)
§ 463.] EFFLUX OF THE AIR AND OTHER FLUIDS, ETC. 941
and therefore
Q = F √ √ 29
2g
F√ 29
h
=
3 b
Pi h
²;
Y₁
21
Yı
1
•
· } [
h
1/4
[1
=
= F [1 −
&
()]
1
54
1
[1
-
+
1
1
h
h
5
128 + ☆ (1) *}
b
27
h
1
3
b
}; − 3 ()'] [¹ − · ¦ + ~ ()']
b
2
27 (1)²]
W 2g
p₁ h
Yr b
5
27

In the application of this formula to fans, blowing engines, etc.,
h
b
in which cases <,the theoretical discharge, measured at the ex-
terior pressure and the interior temperature, is simply
Q = FV 2g
P1
h
Y₁ b
=89,39 FV 2 g (1 + 8T)
89,39 FV
h
h
= 396 FV (1 + Ô T)
cubic meters
b
h
h
= 161,9 F 2 g (1 + dr)
b
1299 FV (1 + 87) cubic feet.
παε
4
EXAMPLE.--In the case treated in the Example of § 459, where b = 29,2,
h = 5 inches, = 120° and F =
7
discharge according to the last formula,
external air,
0,01227 square feet, we have the
measured at the pressure of the
5
Q
=
1299 F
1,4404
·
1299 F 0,2466
V
29,2
=
645,1 F = 645,1.0,01227
7,915 cubic feet,
while previously (§ 459) we found, according to the formula for water,
Q = 8,565 cubic feet, and according to the logarithmic formula in § 460,
we have
Q = 1299 F
34,2
1,4404
7
1299 F √0,2277
29,2
7,606 cubic feet.
= 619,9. 0,01227
§ 463. Efflux of Moving Air.-The formulas for efflux
already found are based upon the supposition that the pressure p
or the height h of the manometer is measured at a place, where the
air is at rest or moving very slowly; but if we measure p, and h₁
at a point, where the air is in motion, if, E.G., the manometer M,
is in communication with the air in a pipe C F, Fig. 795, we must
942
[§ 463.
GENERAL PRINCIPLES OF MECHANICS.
take into consideration, in determining the velocity of efflux, the
vis viva of the approaching air. If c be the velocity of the air pass-
ing the orifice of the manometer, we must put
c²
Q1 P1
2 g
+ 3 Q₁ p₁ [1 − (2)*]
–
Q1 Y1
Q₁Yr 29
g
If F denotes the cross-section of the orifice and G that of the
tube or of the stream, which passes the orifice of the manometer,
the discharge of air is Q. Y₁ = G cy₁ = FvY₂; hence
с
FY2
F
=== 7 (2)'
υ
G Y 1
2
G
v2
and
= ³ Q. p. [1 − (2)'].
*] 3 3 Q₁
3
en [1-()' (2) ']
29
and the required velocity of efflux is

3 pi
2g
Y₁
37 [1-(2)']
*
v =
F
1 -
G
(7)* (4)*
or approximatively, when p, is not much greater than p,

2 g
Pi
3p [1-(2)]
Y₁
1-.
F
1
G
(7)
(1 + 8 T) [1 .
δ
(別
​= 2250
feet.
1
()
Here, as in the case of the efflux of water, the velocity of efflux
FIG. 795.
M
A
M₁
ho
B
F
F
increases with the ratio of
G
the cross-section of the ori-
fice to that G of the pipe or
moving stream of air. We see
from this that, under the same
circumstances, the height p,
of the manometer decreases
as the diameter of the tube

diminishes, or as the velocity of the air in the pipe increases.
§ 463.] EFFLUX OF THE AIR AND OTHER FLUIDS, ETC. 943
If we denote by p, the tension in the reservoir, where the air is
at rest, we have also
v2
29-31 [1-(2)'],
(カ​)],
g
and if we eliminate v from the two expressions, we obtain
1
1
(2)
(2)
ૐ
F
1 -
()(2), approximatively
approximatively = 1 − (7)
If b denotes the height of the barometer in the free air, h that
of the manometer connected with the reservoir and F the area of
the orifice of efflux, we have, finally, the theoretical discharge,
measured when its heaviness is
Y =
= ()
Yı
by
b + h

p₁ h
h
2g
y b
2g (1+87)
Q = F
b
161,9 F
1
()
F\ 2
1
-(6)
G
h
(1 + δ τ)
b
= 1299 F
ユー​(
EXAMPLE.—The height of a quicksilver manometer, which is placed
upon a pipe 3 inches in diameter through which air is passing, is 23 inches,
while the air is discharged through a circular orifice 2 inches in diameter
at the end of the pipe: what is the velocity of discharge, assuming the
barometer in the external air to stand at 27 inches and the air in the pipe
to be at a temperature of 10° C? Here
б
√1 + dr = √1,0367
=
1,018, 1 √
Ѣ
√= 0,3015 and
F
π jo² =
3,141 : 144 = 0,02181 and
F\2
V
1 -
√ 492
49
162
46,314
49
=
0,9452;
Q
= 1299 F.
1,018. 0,3015
0,9452
421,8 F = 9,20 cubic feet.
hence the discharge is
For the corresponding tension p, in the reservoir, we have
Lit
· (2)² = -
[1 − (?)'] · [1 − (−)'] = (1 − VF) : 0,9452*
1-
www
Po
0,0287
0,8934
=
0,03212; hence
944
[§ 464
GENERAL PRINCIPLES OF MECHANICS.
3
p
=
0,90788, Po
1,103 p and b + h.
1,103 b
Po
and consequently the height of the manometer in the reservoir is
0,103 b = 0,103 . 27,5 2,83 inches.
ho
=
§ 464. Coefficients of Efflux-The phenomena of contrac-
tion, which we have studied for the efflux of water, are also met
with in the efflux of air from vessels. If the orifice of efflux is in
a thin plate, the stream of air has a smaller cross-section than the
orifice, and the effective discharge Q, is consequently smaller than
the theoretical Q, or the product Fv of the cross-section F of the
orifice and the theoretical velocity v. This diminution of the dis-
charge is owing principally, as we can observe in a stream of
smoke, to the contraction of the stream of air, and we can, there-
F₁
F
fore, as in the case of water (see § 406), call the ratio a =
the cross-section F of the stream of air to that F of the orifice
the coefficient of contraction,
of
V1
the ratio
(see § 408)
=
of the effective velocity v, to the theoretical v
V
the coefficient of velocity.
Q₁
1
F₁ v₁
and the ratio µ
ap of the effective discharge Qi
Q
Fv
to the theoretical discharge Q
the coefficient of efflux.
=
As in the case of water the coefficient of velocity for the ef-
flux of air through an orifice in a thin plate is nearly
1, and
therefore, so long as we have no measurements of the stream of air,
we must put the coefficient of efflux µ = a equal to the coefficient
of contraction a. The older experiments upon the efflux of air
through orifices in a thin plate vary very considerably from each
other. The experiments of Koch, calculated according to the
formula for water by Buff, gave for circular orifices from 3 to 6
lines in diameter, when the height of the water manometer was
from 0,2 to 6,2 feet, u 0,60 to 0,50; on the contrary, the experi-
ments of d'Aubuisson, calculated in the same way, give for circular
orifices 1 to 3 centimeters in diameter, when the height of the
water manometer is between 0,027 and 0,144 meters, µ = 0,65 to
0,64. Poncelet also found, upon calculating the experiments of
Pecqueur by the same formula, for an orifice 1 centimeter in diam-
eter, under an excess of pressure of 1 atmosphere, or of a column
µ
§ 464.] EFFLUX OF THE AIR AND OTHER FLUIDS, ETC. 945
of water 10 meters high, μ
timeters wide, µ = 0,566.
0,563, and for a similar one 1,5 cen-
The more extended experiments of the
author, calculated according to the last formula
Q = F [1 − &; (4) ] √ 2 g
gave the following results:
1
54
P₁ h
Y Z ³
1) When the diameter of the orifice d = 1 centimeter and the
ratio of the pressures was

P1
b + h
1,05 1,09
1,43
1,65
1,89 2,15
p
b
μl = 0,555 0,589 0,692 | 0,724 | 0,754 | 0,788
2) When the diameter of the orifice d 2,14 centimeters, for
=

b + h
1,05 1,09 1,36
1,67
2,01
b
М =
0,558 0,573 0,634 0,678 | 0,723
| |
3) When the diameter of the orifice d
1,725 centimeters, for

b + h
b
1,08
1,37
1,63
μ =
0,563
0,631
0,665
4) When the diameter of the orifice d= 2 centimeters, for

b + h
1,08
1,39
b
μ =
0,578
0,641
The coefficient of contraction for efflux through an orifice in a
thin plate increases sensibly with the head. But if the formula for
water is employed, there is much less variation; this formula gives
μ nearly 2, E.G. for
V
Pi
=
2; ó = 0,707 times as great as the
Pi
p
946
[$ 465.
GENERAL PRINCIPLES OF MECHANICS.
last formula. According to the first table, for d
μ =
0,754 + 0,788
2
0,707 . 0,771
==
celet found.
= 1 and
P1
= 2,
p
= 0,771; hence, according to the water formula,
0,555, which is nearly the same value as Pon-
For efflux through a circular orifice 1 centimeter in diameter,
situated in a conically convergent wall, the angle of convergence
being 100 degrees, the author found for

b+ h
b
1,31
1,66
μ =
0,752
0,793
In like manner with the same orifice in a conically divergent
wall, the angle of divergence being 100 degrees, the author obtained
for

b + h
b
1,30
1,66
μ = 0,589
0,663
§ 465. The variability of the coefficient of contraction a = µ
for the efflux of air through an orifice in a thin plate also affects,
according to the well-known formula
1
1
= n
(see § 422),
√1 + 5
1 +
( − 1)*
the coefficient of efflux for short pipes. According to the experi-
ments of Koch, cited above, we have for such tubes 3 to 4 lines
in diameter and from 4 to 6 times their diameter in length, when
the pressure is 0,3 to 6,2 feet of the water manometer, µ = 0,74 to
0,72, while, on the contrary, d'Aubuisson gives for similar tubes, 1
to 3 centimeters in diameter, 3 to 4 times as long as wide, and
under a pressure equal to 0,027 to 0,141 meters of the water ma-
nometer, μ 0,92 to 0,93; and Poncelet found for cylindrical
pipes 1 centimeter in diameter and from 2 to 10 centimeters long,
under twice the atmospheric pressure, u = 0 632 to 0,650.
=
The experiments made by the author, on the contrary, have led
to the following results:
§ 465.] EFFLUX OF THE AIR AND OTHER FLUIDS, ETC. 947
1) A short cylindrical tube or ajutage, 1 centimeter in diameter
and 3 centimeters long, gave for
b + h
b
1,05
1,10
1,30
М =
0,730
0,771
0,830
2) A similar tube, 1,414 centimeters in diameter and three times
as long as wide, gave for

b + h
b
1,41
1,69
μ =
0,813
0,822
3) A similar pipe, 2,44 centimeters wide and three times as long,
gave for
b + h
b
1,74,μ
=
0,833.
The increase of the coefficient of efflux as the pressure increases
is explained by the simultaneous increase of the coefficient of
contraction.
The short pipe (1), when its inlet orifice was slightly rounded
off, gave as a mean value for its coefficient of efflux µ = 0,927,
which is much greater than that for a similar pipe which is not
rounded off.
4) A short pipe, with its inlet orifice well rounded off, 1 centi-
meter wide and 1,6 centimeters long, gave for
b + h
1,24
1,38
1,59
1,85
2,14
b

μ =
0,979 0,986
0,965
0,971 0,978
The advantage of the formula for efflux
Q
"F V
µ F √ 2 g
P₁ b
over the others
Y. P
is shown by the fact that this coefficient approaches very nearly
(as it should do) unity.
The older formula gives of course for great pressures much
smaller values for μ..
948
[$ 465.
GENERAL PRINCIPLES OF MECHANICS.
On the contrary, the logarithmic formula (see § 460) gives much
greater values which may sometimes even exceed unity.
A short conical pipe, rounded off at the inlet orifice, gave nearly
the same values for μ, and a short conical tube, which was not
rounded off, and which was 1 centimeter in diameter and 4 centi-
meters long, and whose angle of convergence was 7° 9', gave for
b + h
b
1,08
1,27
1,65

μ =
0,910
0,922
0,964
Koch and Buff found with a similar tube, whose exterior diam-
eter was 2,72 lines and the angle of convergence of whose sides was
6º, under a head of 0,3 to 6,2 feet of the water manometer µ = 0,73
to 0,85, and according to d'Aubuisson a similar pipe, whose orifice
was 1,5 centimeters in diameter, gave under a pressure measured
by a height of from 0,027 to 0,144 meters of the water manometer,
μ = 0,94. The old or water formula was employed in the calcu-
lations.
The complete nozzle A C, Fig. 736, § 434, consisting of a
conical tube with an angle of convergence of 6°, which was 14,5
centimeters long, 1 centimeter wide at the outlet and 3,8 centi-
meters wide at the inlet, which was well rounded off, gave for

b + h
b
1,08
1,45
2,16
М =
0,932
0,960
0,984
By experiments upon the influx of air into vessels, Saint-
Venant and Wantzel found for a short mouth-piece, rounded off
internally in the form of a quarter of a circle, when the calcula-
tions were made according to the new formula, = 0,98, and for
an orifice in a thin plate, p = 0,61.
If the pressures are small, as is the case in the ordinary fan,
where <, we can substitute, according to what precedes, when
h
b
1
we employ the new formula for efflux
Q = µ F √ 2g
P₁ b
Y
•
h
μ
as a mean
h
√
(1 + 0,004
1299 µF (1 + 0,004 т) cubic feet,
b
1) for an orifice in a thin plate, µ =
0,56,
§ 466.] EFFLUX OF THE AIR AND OTHER FLUIDS, ETC.
949
2) for a short cylindrical pipe, p = 0,75,
3) for a well rounded off conical mouth-piece, µ = 0,98,
4) for a conical pipe, whose angle of convergence is about 6°,
µ = 0,92.
EXAMPLE.-If the sum of the areas of two conical tuyeres of a blowing
machine is 3 square inches, the temperature in the reservoir is 15°, the
height of the manometer in the regulator is 3 inches and the height of the
barometer in the exterior air is 29 inches, we have the effective discharge,
measured at the pressure of the exterior air,

Q
= 1299 µ F√(1 + 0,004 †)
μ
1299. 0,92
3
b
h
·
144 V (1 + 0,004. 15)
3
29
24,9 1
1,06 . 3
29
=
•
24,9 0,331
=
8,242 cubic feet.
§ 466. Coefficient of Friction of Air.—If air moves through
a long pipe C F, Fig. 796, it has, like water, a resistance of friction
M
M1
A
B
FIG. 796.

M
F
to overcome, and this resistance can be measured by the height of
a column of air, which is determined by the expression
1 v²
5.
d⋅ 2 g
in which, as in the case of water pipes, I denotes the length, d the
diameter of the pipe, v the velocity of the air, and the coefficient
of resistance of friction, to be determined by experiment.
Girard's experiments upon the movement of air in pipes gave a
coefficient of resistance = 0,0256, those of d'Aubuisson, as a mean,
= 0,0238, while according to the experiments of Buff the mean
value of 5 = 0,0375. Poncelet, on the contrary, found from the
data furnished by the experiments of Pecqueur, when the ratio of
pressure is Pi
2, μ = 0,0237.
p
fl
The experiments of the author, calculated according to the new
formula, gave the following results:
1) A brass tube, 1 centimeter wide and 2 meters long, gave for
950
[$ 467.
GENERAL PRINCIPLES OF MECHANICS.
velocities of from 25 to 150 meters gradually decreasing from
0,027260 to 0,01482.
2) A glass tube of the same length, when the velocities were
about the same, gave
0,02738 to 0,01390.
3) A brass tube, 1,41 centimeters wide and 3 meters long, gave
5=0,02578 to 0,01214.
4) and a similar glass tube, 5 = 0,02663 to 0,009408.
5) Finally, a zinc tube, 2,4 centimeters wide and 10 meters
long, gave, for velocities of from 25 to 80 meters, = 0,2303 to
0,01296.
From what precedes we may conclude that it is only when
velocities are about 25 meters or 80 feet, that the coefficient of re-
sistance can be put 0,024, and that it becomes smaller and
smaller as the velocity of the air in the pipe increases.
Approximatively we can write, when the velocity is expressed
=
0,217
No
The
0,120
in meters, Š
or when it is expressed in feet 5 =
V v
general relations of the flow of air in pipes are very similar to those
of water.
The resistance, caused by elbows and bends, is to be treated in
the same way as in the case of water.
In the author's experiments a rectangular elbow, 1 centimeter in
diameter, gave S 1,61, and a similar one, 1,41 centimeters in
diameter, gave = 1,24, and a pipe like the former, when bent in
the shape of a quarter of a circle, gave = 0,485, and one like the
latter, bent in the same way, gave Š 0,471.
5
§ 467. Motion of Air in Long Pipes.-By the aid of the
coefficient of the resistance of friction of a pipe B F, we can cal-
culate the velocity of efflux and the discharge for a given length
and width of the pipe.
2
If hy is the height of the manometer M, at the end of the pipe
CF, Fig. 797, directly behind the mouth-piece F, whose coefficient
A
M
M1
B
FIG. 797.

M 3
F
1
of resistance is
2
1, and if d denote the diameter of the
рез
§ 467.] EFFLUX OF THE AIR AND OTHER FLUIDS, ETC.
951
pipe and d, that of the orifice, whose area is therefore F
have, according to what precedes, the discharge
=
πα
2
we
4
P1
P₁
h₂
h.
2 g
(1 + d +)
Y₁ b
di
b
Q=μ₁ F₁
= 1299 µ, π
μι
cub.ft.
4
d₁
4 9
1 –
or, inversely, for the height h₂ of the manometer
P₁ h₂
Q
1/2 = [(1 − ( 2 ) ] + ( 2 ) ²
2 g
But the height of the manometer at the entrance of the pipe is
I v²
h₁
h₂+ 5
2
d 2 g
I denoting the length of the pipe between M, and M, and v the
velocity of the air in this pipe; hence we have
P₁ h₁
1
2
F
;-)² + 5
I v²
d 2 g
or,
Q
V₁ =
v₁ F?
substituting v =
(d) '%, and
P1
h₁
门
​γι
b
1
=([1]+5/2(2) 2
hence the discharge is
F
Q = R |
πα
2
= 1299
4
29
d₁
4
P₁
Y1
1
2
h₁
b
+ 5
7 / dy
42
d
d
h₁
b
(1 + d +)
4
1
1 − ( ) )² ] ² + ³ { ( )
2
5
d
(99)

cubic feet.
If, finally, the height h of the manometer M in the reservoir
A B is known, we have, when we denote the coefficient of resist-
ance for the entrance C by 5, and substitute
1
1+ 1, since at
μι
v2
the entrance into the pipe the head
is lost,
29
P₁ h
[(5% + 5 3 ) ( 1/2)*
d
+ 1 + 5₁
5]
2g
$] 1, (2)²;
F
and consequently the discharge
952
[§ 468.
GENERAL PRINCIPLES OF MECHANICS.
Q = F₁
-AV
P₁ h
2 g
Y₁ b
d₁
+ 5
+ 1 + $₁
0
h
πα
2
= 1299
4
(1 +0,04 T)
(5% + 5 4 ) ( 14 )
d
4
b
cubic feet.
+ 1 + 5₁
If the point where the air enters the pipe is a distance s below or
above the point where it is discharged from it, we must subtract
from or add to the quantity
ical sign a quantity s.
P₁ h
Y₁ b
in the numerator under the rad-
EXAMPLE.-The height of a quicksilver manometer, which is placed
upon a regulator at the head of a system of air pipes 320 feet long and 4
inches in diameter, is 3,1 inches, the height of the barometer in the free
air is 29 inches, the width of orifice in the conically convergent end of the
pipe is d 2 inches, and the temperature of the compressed air in the
regulator is 7 = 20° C.; what quantity of air is delivered through these
pipes?
Here (1 + 0,004 †)
b
3,1
1,08.
h
0,11545,
29
1
1
16
77
50
1
1
1
2
0,778,
0,752
9
9
а
= 0,024. 320. 3 = 23,04, (1) * = (~)
(金​)
1
0,0625,
16
1
1
51
1=
1
με
2
πα
0,922
0,330, and
2
П
1
3,1416
F₁
(중​)2
0,021817 square feet;
4
4
144
hence the required discharge is
Q:
= 1299 . 0,021817
0,11545
0,11545
(0,778 + 23,04) 0,0625 + 1,330
28,34 1,489 +1.330
28,34 √0,040954
5,735 cubic feet.
§ 468. Efflux when the Pressure Diminishes.-If there
is no influx of air into a reservoir, from which an uninterrupted
discharge of air is taking place through an orifice in it, the density
and tension gradually diminish, and consequently the velocity of
efflux becomes less and less. The relations of this diminution to
the time and to the discharge can be determined in the following
manner.
§ 468.] EFFLUX OF THE AIR AND OTHER FLUIDS, ETC. 953
b; then
Let the volume of the reservoir be V, the initial height of the
manometer be = h., and its height at the end of a certain time t be
h₁, and let that of the barometer in the free air be
the quantity of air originally in the reservoir, reduced to the
pressure of the exterior air, is
and at the end of the time t it is
V (b + ho)
b
V (b + h₁)
Ъ
hence the discharge in the time t, reduced to the external pressure,
is
Vi
V (b + h)
b
V (b + h₁)
b
V (ho
— h₁)
b
But we have also
P₁ X
V₁ = μ Ft
2 g
Yı b'
a denoting the mean height of the barometer during the time t of
efflux; hence
V (h。 - hr)
V (h。 - h₁)
(x)−.
t =
μ F√29
P₁b x
μ F
P1
№ 29
2 g
b
Y1
+
m—n
(0)-*
no
2 (0)—+
112
m
Y1
Now if we put h. = m σ and h₁ = n o, we have the mean value
(x)→→→
(1→ + 2→♪
(0) +
m-n (m²
+ m−³) — (1−³ + 2 + . . . + 22−3)
n
σ
σ
(
2 ( √ h
2 ( 4 h.
V
(see Ingenieur, p. 88);
(m n) o
hence the required time of efflux is
2 V ( N h − V T₁)
h - h₁
2 T
ん​。
t =
b
풍​).
µF 29
µ F V 2 g
P1 b
P1
µ F V 2 g
γι
γι
This determination is sufficiently correct only when the reser-
voir (V) is large, or when the orifice of efflux, as well as the
pressure, is small, in which case the cooling of the air in the reser-
voir is very slight.
EXAMPLE.-A cylindrical regulator 50 feet long and 5 feet in diameter is
filled with air at a pressure corresponding to the height h 10 inches of
the manometer and at a temperature of 6° C. Now if the air issues from
an orifice 1 inch in diameter into a space where the barometer stands at
27 inches, the question arises, in what time will the manometer sink to 7
954
[§ 468.
GENERAL PRINCIPLES OF MECHANICS.
inches and what will be the discharge in that time? The volume of the
regulator or boiler is
V
π
π
52.50 1250
4
4
981,75 cubic feet, and
10
7
1
1
b
= 0,09942,
27
27
P1
1299 √1 + 0,00367 . T 1299 √1,02202
129
Π
Π
F = (12) 2
0,005454 square feet.
576
4
= 1313 and
Now if we put the coefficient of efflux µ = 0,95, we have the required
duration of the efflux
t =
2.981,75. 0,09942
0,95. 0,005454 . 1313
28,69 seconds.
REMARK.—A more general theory of the efflux of air and steam will be
given in the second volume.
FINAL REMARK.-Experiments upon the efflux of air have been made
by Young, Schmidt, Lagerhjelm, Koch, d'Aubuisson, Buff, and more re-
cently by Saint Venant, Wantzel, and Pecqueur. In reference to the ex-
periments of Young and Schmidt, see Gilbert's Annalen, Vol. 22, 1801, and
Vol. 6, 1820, and Poggendorf's Annalen, Vol. 2, 1824; for those of Koch
and Buff, see the "Studien des Götting'schen Vereines bergmännischer
Freunde," Vol. 1, 1824; Vol. 3, 1833; Vol. 4, 1837; and Vol. 5, 1838;
also Poggendorf's Annalen, Vol. 27, 1836, and Vol. 40, 1837. See also
Gerstner's "Mechanik," Vol. 3, and Hülsse's "Algemeine Maschinenency-
klopädie," Article "Ausfluss." Lagerhjelm's experiments are discussed in
the Swedish work "Hydrauliska Försök af Lagerhjelm, Forselles och
Kallstenius," 1 Delen, Stockholm, 1818. The experiments of d'Aubuisson
are to be found in the "Annales des Mines," Vol. 11, 1825; Vol. 13, 1826;
Vols. 3 and 4, 1828; and also in d'Aubuisson's "Traité d'Hydraulique."
The experiments of Saint-Venant and Wantzel are to be found in the
Comptes rendus hebd. des séances de l'Académie des Sciences, à Paris,
1839." The latest French experiments are discussed by Poncelet in a
"note sur les experiences de M. Pecqueur relatives à l'écoulement de l'air
dans les tubes, etc.," which is contained in the Comptes rendus, and an
abstract of it is to be found in the Polytechnische Centralblatt, Vol. 6,
1845. From these experiments Poncelet concludes that air follows the
same laws of efflux as water. The greater number of these experiments
were made with very narrow orifices, for which reason they scarcely fulfill
the requirements of practice. Unfortunately these experiments do not
agree as well as could be wished, and the coefficients found by d'Aubuisson
differ very sensibly from those calculated from Koch's experiments. Com-
parative experiments upon the efflux and influx of air and upon the efflux
of water are given in the author's "Experimental-Hydraulik." The re-
sults of the latest experiments of the author, which were made upon a
large scale, are given in the 5th volume of the Civilingenieur.
(C
§ 469.] THE MOTION OF WATER IN CANALS AND RIVERS. 955
CHAPTER VII.
OF THE MOTION OF WATER IN CANALS AND RIVERS.
§ 469. Running Water.-The theory of the motion of water
in canals and rivers forms the second part of hydraulics. Water
flows either in a natural or in an artificial bed (Fr. lit; Ger. Bett).
In the first case the channel is a river, creek, rivulet, etc., in the
second case it is a canal, ditch, race, trough, etc. In the theory of
the motion of running water this difference is of but little im-
portance.
The bed of the stream consists of the bottom of the channel
(Fr. font du lit; Ger. Grundbett or Sohle) and of the two banks
or shores (Fr. bords; Ger. Ufer). If we pass a plane through the
stream of water at right angles to the direction, in which it is
flowing, we obtain a transverse section (Fr. section; Ger. Quer-
schnitt). The line bounding this section is the tranverse profile
which is composed of the water profile or wetted perimeter and of
the air profile. A vertical plane in the direction of the stream
gives us the longitudinal section or profile (Fr. profil; Ger. Profil)
of the latter. The slope of the stream (Fr. pente; Ger. Abhang) is
the angle formed by its surface with the horizon. The relative
slope is the fall in the unit of distance. The slope is determined
for any definite distance by the fall (Fr.
chute; Ger. Gefälle), which is the vertical
distance of one of the extremities of a cer-
tain portion of the stream above the other.
In the portion 4 D= 1, Fig. 798, B C is
A
the bottom of the channel, D H h the
fall and the angle D A H = d is the slope.
B
FIG. 799.
H

D
IC
The relative slope is
1.
sin. 8 =
で
​h
or approximatively d
=
•
REMARK.-The fall of creeks and rivers varics very much. The Elb
falls in a German mile (4 English miles) from Hohenelbe to Podiebrad
57 feet, from there to Leitmeritz 9 feet, from there to Mühlberg 2,5 feet.
Mountain streams fall from 8 to 80 feet per mile. For particulars see
"Vergleichende hydrographische Tabellen, etc., von Stranz." The fall in
canals and other artificial channels is much smaller. The relative slope is
generally less than 0,001, it is often 0,0001 and even less. More details.
upon this subject will be found in the second part.
956
[§ 470-
GENERAL PRINCIPLES OF MECHANICS.
§ 470. Different Velocities in a Cross-section. The ve-
locity of the water is far from being uniform in all points of the
same transverse section. The adhesion of the water to the bed of
the channel and the cohesion of the molecules of water cause the
particles of water nearest to the sides and bed of the channel to be
most hindered in their motion. For this reason, the velocity
decreases from the surface towards the bed of the channel and it is
a minimum at the shores and bottom. The maximum velocity in
a straight river is generally found in the middle or in that portion
of the surface, where the water is the deepest. That portion of the
river, where the water has its maximum velocity, is called the line
of current or axis of the stream and the deepest portion of the bed
is called the mid-channel.
When the stream bends, the axis of the stream is general near
the concave shore.
The mean velocity of the water in a cross-section, according to
§ 396, is
с
Q
F
Discharge per second
Area of the transverse section'
We can also determine the mean velocity from velocities C1, C2, C3,
etc., in the different portions of the transverse section and the
areas F, F, F, etc., of the latter. We have here
Q
=
F₁₂+ F₂ C₂ + F3 C3 + ...
and, therefore, also
c =
C
2
F₁ c₁ + F₂ C₂ +
F₁ + F₂+
Besides the mean velocity we introduce the mean depth of water,
I.E., that depth a, which a transverse section would have, if its
area was the same and the depth was uniform instead of being
variable and equal to a₁, ɑ2, ɑ3, etc. Here we have
α =
F
b
Area of the transverse section
Width of the transverse section'
If the widths of the elements corresponding to the depths a₁, α,
FIG. 799.
Dz
as, etc., Fig. 799, are b₁, b2, bз, etc., we
have

+ •
F ɑg
= ar b₁ + α b₂ +
and consequently
α =
a₁ b₁ + ɑ₂ b₂ +
1
b₁ + bx +
•
Finally, the mean velocity is
8471.] THE MOTION OF WATER IN CANALS AND RIVERS. 95%
c =
α₁ b₁ c₁ + ɑ₂ b₂ C₂ +
a₁ b₁ + a₂ b₂ +
and, when the widths b₁, b, etc., of the portions are the same,
C
A₁ C₁ + α ₂ C2 +
A₁ + α₂ +
•
A river or creek is in a state of permanency (Fr. permanence;
Ger. Beharrungszustande) or it has a fixed regimen, when the same
quantity of water passes through each of its cross-section in the
same time, I.E., if Qor the product Fc of the area of the cross-
section and the mean velocity is constant for the whole length of
the portion of the river under consideration. Hence we have the
simple law when the motion of the water is permanent the mean
velocities of two transverse sections are to each other inversely as the
areas of these sections.
EXAMPLE-1) In the transverse section A B C D, Fig. 799, of a canal,
we have found the widths of the divisions to be
1
the mean depths to be
3,1 feet, b₂
2
=
5,4 feet, b3
4,3 feet,
a1
= 3,0 feet
2,9 feet, c₂ 3,7 feet, c
3
= 3,2 feet.
2,5 feet, a 4,5 feet, a3
and the corresponding mean velocities to be
C₁
1
Here we can put the area of the section
F = 3,1 . 2,5 + 5,4. 4,5 + 4,3. 3,0 = 44,95 square feet
and the discharge
Q
3,1. 2,5. 2,9 + 5,4. 4,5. 3,7 + 4,3 .3,0. 3,2
from which we obtain the mean velocity
=
153,665 cubic feet,
C =
Q
F
153,665
44,95
3,419 feet.
2) If a ditch should carry 4,5 cubic feet of water with a mean velocity of
4,5
2 feet per second, we must make the area of its transverse section = 2,25
2
square feet.
3) If the same river is at one place 560 feet wide and as an average 9
feet deep, and if it moves with a mean velocity of 24 feet, the mean velocity
at another place, where it is 320 feet wide and as a mean 7,5 feet deep, is
c =
560.9
320. 7,5
•
2,254,725 feet.
§ 471. Mean Velocity.-If we divide the depth of the water
at any point into equal parts and lay off the corresponding veloci-
ties as ordinates, we obtain a scale A B, Fig 800, of the velocities
of the stream. Although it is very probable that the law of this
scale, or of the change of velocity, is expressed by a curve, as
958
[§ 471.
GENERAL PRINCIPLES OF MECHANICS.
E G. according to Gerstner, by an ellipse, etc., yet without risking
FIG. 800.
V
B
(m
a very great error we can substitute a
straight line, I.E., assume that the velocity
diminishes regularly with the depth; for
this diminution of the velocity is always
slight. According to the experiments of
Ximenes, Brünnings and Funk, the mean
velocity in a perpendicular line is
Cm = 0,915 Co,
c, denoting the maximum velocity or that of the surface of the
water. The diminution of the velocity from the surface to the
middle M is therefore

Co
and we can put the velocity
Cm
(1
0,915) c = 0,085 c.,
at the bottom, or at the foot of the
perpendicular,
Gn = Co
2. 0,085 c。
=
(1 — 0,170) c。 =
0,83 c。.
=
If the total depth is a, we have, if we assume the scale of velocity
to be represented by a straight line, for a depth A N x below
the water the velocity
v = co (c。 — c₂) 2 = (1 — 0,172) c.
-
α
Now if Co, C1, c₂ are the velocities at the surface of a profile,
whose depth is not very variable, we have the corresponding veloci-
ties at the mean depth
0,915 c., 0,915 C1, 0,915 C2,
and therefore the mean velocity in the whole transverse section
c = 0,915
Co + C₁ + C₂ +
n + 1
•
+ C n
If, finally, we assume that the velocity diminishes from the line
of current or axis of the stream towards the shores in the same
manner as towards the bottom, we can put the mean superficial
velocity
+ Cn 0,915 c.;
Co + G₁ +
n + 1
thus we obtain the mean velocity of the whole transverse section.
c = 0,915. 0,915 . c.
=
0,837 Co
I.E., 83 to 84 per cent. of the maximum velocity.
Prony deduced from the experiments of du Buat, which, how-
ever, were made in small ditches, the following formula, which is
perhaps more correct in such cases,
Cm
12,372 + co
3,153 + c
6)
7,78 + co
c.
c。 meter =
Со
c. feet.
10,34 + co
§ 472.] THE MOTION OF WATER IN CANALS AND RIVERS. 959
Hence for mean velocities of 3 feet we have
Cm = 0,81 Co.
If the flow of the water is impeded by a contraction of the
transverse section, the level of the water will be raised, and cm be-
comes still greater.
EXAMPLE.—If the velocity of the water in the axis of a river is 4 feet,
and if its depth 6 feet, we have the mean velocity in the corresponding
perpendicular
cm = 0,915. 4 = 3,66 feet,
the velocity at the bottom
the velocity 2 feet from the surface
= 0,83. 4
3,32 feet,
0,057).4
3,772 feet
*
v = (1 — 0,17.)
0,17 . ) 4 = (1
S
and, finally, the mean velocity of the entire transverse section
c = 0,837 . 4 = 3,348 feet;
on the contrary, according to Prony, we would have
C
11,78
14,34
4
23,56
7,17
3,29 feet.
REMARK.—This and the following subjects are treated at length in the
Allgemeine Maschinenencyklopädie, Article "Bewegung des Wassers."
New experiments and new views upon the same subject are to be found in
the following work: "Lahmeyer, Erfahrungsresultate uber die Bewegung
des Wassers in Flussbetten und Canälen, Braunschweig, 1845." Accord-
ing to Baumgarten's observations (see Annales des Ponts et Chaussées,
Paris, 1848, and also polytechnisches Centralblatt, No. 14, 1849) the values
given by this formula, when the velocities are great (above 1,5 meters), are
too large and we must put in such cases
Cm-
C'm
(3.153 + Co).
0
0,8 c。 meters.
co
Owing to the resistance of the air the maximum velocity of the water is
to be found a little below the surface of the water. From this point of
maximum velocity the velocity diminishes as the square of the depth; hence
the scale of velocity corresponds to a parabola. In like manner, according
to Boileau (see his Traité sur la mesure des eaux), the velocity decreases
as the square of the distance from the axis of the stream. If c, denotes the
velocity in the axis of the stream, the velocity at the horizontal distance x
from it will be
μ 20²,
in which μ denotes an empirical number, which is different for different
streams.
§ 472. Most Advantageous Transverse Profile. The
resistance, offered by the bed of the stream in consequence of the
adhesion, viscosity and friction of the water, is proportional to
the surface of contact, and consequently to the wetted perimeter
960
[§ 472.
GENERAL PRINCIPLES OF MECHANICS.
p, or to that portion of the profile which forms the bed. Now since
the number of filaments of water passed by any transverse section
increases with its area, the resistance to each filament is inversely
Ρ
proportional to the area, and consequently to the quotient of the
F
wetted perimeter divided by the area F of the entire transverse
section. In order to have the least resistance from friction, we
P
must give the profile such a shape that shall be as small as pos-
F
sible, I.E., that the wetted perimeter p shall be a minimum for a
given area, or that the area shall be a maximum for a given wetted
perimeter p. When the apparatus which conducts the water is
closed on all sides as in the case of pipes, p is the perimeter of the
entire transverse section. Now among all figures of the same
number of sides, the regular one, and among all the regular ones,
the one with the greatest number of sides has the smallest perim-
eter for a given area; hence in conduits closed on all sides the
resistance is smaller the more regular the shape of their transverse
section is, and the greater the number of sides is. Since the circle
is a regular figure of infinite number of sides, the resistance of
friction is the smallest when the transverse section is of that form.
When the aqueduct is open on top, the case is different; for the
upper surface is free, or in contact with the air alone, which, so
We
long as it is still, offers little or no resistance to the water.
must, therefore, in determining this resistance of friction, neglect.
the air profile.
FIG. 801.
In practice we employ in canals, ditches, troughs and flumes
only rectangular and trapezoidal profiles. A horizontal line E F
Fig. 801, passing through the centre M of the square A C, divides
the area and perimeter into two equal parts, and
what has been said of the square is truc for these
halves; hence, among all rectangular profiles,
the half square A E, or that which is twice as
wide as high, is the one which causes the smallest
resistance of friction.

D
In like manner, the regular hexagon A CE,
Fig. 802, is divided by a horizontal line CF into two equal trape-
zoids, each of which, like the entire hexagon, has the greatest
relative area, and consequently among all trapezoidal profiles, the
half of the regular hexagon, or the trapezoid A B C F, with the
§ 473.] THE MOTION OF WATER IN CANALS AND RIVERS. 961
angle of slope B C M = 60°, is the one which causes the least
resistance of friction.
In like manner, the half of a regular octagon A D E, Fig. 803,
the half of a regular decagon, etc., and finally the half circle A D B,
Fig. 804, are, under the proper circumstances, the most advan-
بتا
FIG. 802.
E
D
FIG. 803.
E
M
D
FIG. 804.



B
tageous profiles for canals, etc. The trapezoidal, or half hexagonal,
cross-section causes less resistance than the half square or rec-
tangle, the ratio of whose sides is 1:2; the relative perimeter of
the hexagon is smaller than that of the square. The half decagon
offers still less resistance, and with the semicircle the latter is a
minimum. The circular and square profiles are employed only
for troughs made of iron, stone, or wood. The trapezoid is em-
ployed in canals, which are dug out or walled up. Other forms
are rarely used, owing to the difficulty of constructing them.
§ 473. When canals are not walled up, but only dug in the
earth or sand, an angle of slope of 60° is too great or the relative
slope cotg. 60° = 0,57735 too small; for the banks would not be
sufficiently stable; we are therefore compelled to employ trapezoi-
dal transverse profiles, in which the inclination of the side to the
base is smaller than 60°, perhaps only 45° or even less. For a trapezoi-
dal cross-section A B C D, Fig. 805, which has the same area and
perimeter as the half square, the relative slope is, and the
angle of slope is 36° 52'. If we divide the height B E into three
equal parts, the bottom B C is equal to two of them, the parallel
top AD is equal to 10 and each side A B C D is 5 parts.
In many cases we make the relative slope = 2; in which case the
angle is 26° 34', and sometimes it exceeds even 2.
In any case the angle of slope B A E = 0, Fig. 806, or the slope
A E
=cotang. O is to be considered as a given quantity, dependent
BE
upon the nature of the ground in which the canal is excavated,
and therefore we have only to determine the dimensions of the pro-
61
962
[$ 473.
GENERAL PRINCIPLES OF MECHANICS.
file which will offer the least resistance. Putting the width B C of
FIG. 805.
E
D
FIG. 806.


A E
F D
A E
the bottom =
b, the depth B E = a and the slope
v, we
BE
have the wetted perimeter of the profile p =
A B + B C + C D = b + 2 √ a² + v² a² = b + 2 a √ 1 + v²,
and the area of the same
F = a b + va a = a (b + v a),·
or inversely
whence the ratio
F
b
να,
α
p
1
α
+
F α F
(2 √ v² + 1 · v).
Substituting instead of a, a + x, in which x is a small quan-
tity, we have
(a + x)
F
p
1
+
(2 √ v² + 1 − v)
F
a + x
a
(1
X
x²
a
+
α
a + x
+
(2
2
√ v² + 1 − v}
F
1
a
F
+ (2 √v² + 1 − v) +
2 √ √² + 1
ν
a
F
117) a
x²
x +
α
as
In order that this value, not only for a positive but also for a
negative value of x, shall be greater than the first value
1
a
a
+ (2 √ v² + 1 −− v),
F
or that
P
shall be a minimum, it is necessary that the members
F
with the factor x shall disappear or that
2 √ √² + 1 ν 1
F
0,
a²
whence the required depth of the canal is
F
a²
V
2 √ v² + 1
v² + 1 − v
or, since v = cotang. O and √v² + 1
1
sin. O'
§ 474.] THE MOTION OF WATER IN CANALS AND RIVERS. 963
F sin. 0
a²
2
cos. 0.
Hence for a given angle of slope ℗ and for a given area, the
most advantageous form for the transverse profile is determined by
the formulas
a =
√ Fsin. O
2
cos. O
F
and b = - a cotang. 0.
a
Consequently the width A D of the top is
b₁ = b + 2 v a =
F
+ a cotang. 0,
a
1
(2
—
+
and the ratio
p
b
+
F F
2 a
F sin. 0 α
cos. 0) g
F sin. O
2
α
EXAMPLE.-What dimensions should be given to the transverse profile
of a canal, when the angle of slope of its banks is to be 40° and when it is
to carry a quantity Q = 75 cubic feet of water with a mean velocity of
3 feet.
Here
F
=
Q
с
= 25 square feet, and therefore the required depth is
75
3
α =
2
25 sin. 40°
cos. 40°
51
0,64279
1,23396
3,609 feet,
3,609 cotang. 40° = 6,927 - 4,301
=
2,626 feet,
the width at the bottom is
b
25
3,609
the horizontal projection of the slope of the shore is
v a = a cotang. 0 = 3,609 cotang. 40° = 4,301,
the width on top is
the wetted perimeter is
1
= b + 2 a cotang. 0 = 6,927 + 4,301
=
11,228 feet,
p = b +
2 a
sin. O
2,626 +
7,218
sin. 40°
=
13,855 feet,
and the ratio which determines the resistance of friction is
Ρ
2
2
F α 3,609
0,5542.
We have for a transverse profile in the shape of the half of a regular
hexagon, where 0 = 60°, a = 3,80 feet, b = 4,39, 6, 8,78 and p
1 = 8,78 and p =- 13,16
feet, and therefore
p
F
13,16
25
0,526.
§ 474. Table of the Most Advantageous Transverse
Profiles. The following table gives the dimensions of the most
advantageous transverse profiles for different angles of slope and for
given transverse sections:
964
[$ 474.
GENERAL PRINCIPLES OF MECHANICS.
DIMENSIONS OF THE TRANSVERSE PROFILES.
Angle of
slope 0.
Relative
slope v.
Quotient
Depth a.
Width of bot-
tom b.
jection of slope
Horizontal pro- Width at the
772
va.
top b+ 2 va. F
VF
90°
O
0,707 VF 1,414 VF
NF
2,828
O
1,414 VF
NF
60°
0,577 0,760 VF 0,877 VF 0,439 VF 1,755
2,632
NF
VF
45°
1,000 0,740 VF 0,613 VF
0,613 VF 0,740 VF
NF
2,092
NF
2,704
NF
400
2,771
1,192 0,722 VF 0,525 VF 0,860 VF 2,246 VF
36° 52' 1,333 0,707 VF 0,471 NF 0,943 NF 2,357 NF
NF
2,828
NF
35°
1,4020,697 VF 0,439
NF 0,995
2,870
NF 2,430 VF
NF
30°
1,732 0,664 VF 0,356 VF 1,150 VF 2,656 VF
3,012
NF
26° 34' 2,000 0,636 VF 0,300 NF 1,272
3,144
NF 2,844 VF
NF

Semi-
circle
0,798 VF
2,507
1,596 VF
NF
NF
p
We see from the above table that the quotient
is a minimum
F
2,507
and =
NF
for the semicircle, that it is greater for the half
hexagon and still greater for the half square, and for the trapezoid
with its sides sloping at an angle of 36° 52', etc.
EXAMPLE.-What dimensions are to be given to a transverse profile
whose area is to be 40 feet, when the banks are to slope at an angle of 35°
According to the foregoing table
the depth is a = 0,697 √40
4,408 feet,
the lower breadth is b = 0,439 √40 = 2,777 feet,
the horizontal projection of the slope v a = 0,995 √40
the upper breadth b₁
=
15,363,
6,293 feet,
§ 475.] THE MOTION OF WATER IN CANALS AND RIVERS.
965
and the quotient is
P
2,870
F √40
= 0,4538.
§ 475. Uniform Motion. The motion of water in channels
is for a certain distance either uniform or variable; it is uniform,
when the mean velocity in all the cross-sections is constant, and, on
on the contrary, it is variable, when the mean velocity and also the
area of the cross-sections change. We will now treat of uniform
motion.
When the motion of water is uniform for a distance A D = 1,
Fig. 807, the entire fall h is employed in overcoming the friction
B
FIG. 807.
H
D
C
upon the bed, and the water flows away
with the same velocity, with which it
arrived, I.E., a height due to a velocity is
neither absorbed nor set free. If we meas-
ure this friction by the height of a column
of water, we can put the latter equal to
the fall. The height due to the resistance of friction increases

with the quotient P
A,
with 7 and with the square of the mean ve-
locity c (§ 427); hence the formula
l
1) h = 5 ² p
=
•
C²
F 2 g
holds good, in which is an empirical number, which is called the
coefficient of the resistance of friction.
By inversion we have
F
2) c C = 1
2 g h.
5.1 p
To determine the fall from the length, the transverse profile
and the velocity, or inversely, to determine the velocity from the
fall, the length and the transverse profile, it is necessary to know
the coefficient of friction . According to Eytelwein's calculation
of the 91 experiments of du Buat, Brünings, Funk and Woltmann,
<= 0,007565, and therefore.
h
= 0,007565
Ip
F
2 g
If we put g
9 =
9,809 meters or 32,2 feet, we obtain for the
metrical system
h = 0,0003856 ¹p
Fh
c' and c = 50,9
F
50,9 V
Ρ
pi
ין
966
[§ 476.
GENERAL PRINCIPLES OF MECHANICS.
and for the English system of measure
0,00011747 pc' and c =
h
c²
F
For conduit pipes
ιρ
πια
47
F
1
π d²
2
pipes is
h = 0,03026
d
2 g
92,26 // Fh
ρι
; hence the formula for
while we found more correctly (§ 428) for medium velocities in
the same
v2
h = 0,025
d 2 q
g
The friction upon river beds is, therefore, as might be expected,
greater than in metal conduit pipes.
EXAMPLE--1) How much fall must a canal, whose length is 7
feet, whose lower width is b = 3 feet, whose upper width
2600
7 feet
and whose depth is a = 3 feet, have in order to carry 40 cubic feet of
water per second? Here
p = 3 + 2 √2² + 3²
hence the required fall is
10,211, F =
(7 + 3) 3
2
-
15 and c = 48
10 = 8/8
2600. 10,211
0,305422. 10,211. 64
h =
0,00011747 .
(8) 2
1,48 feet.
15
15.9
2) What quantity of water will be delivered by a canal 5800 feet long,
when the fall is 3 feet, its depth 5 feet, its lower breadth 4 feet and its
upper breadth 12 feet? Here
Ρ
F
4 + 2 √5² + 4ª
5.8
16,806
40
0,42015;
hence the velocity is
C =
92,26
3
0,42015. 5800
92,26
92,26
V0,14005. 5800
√812,29
92,26
3,237 feet,
28,5
and the quantity delivered is
Q
= Fc
40. 3,237 = 129,48 cubic feet.
§ 476. Coefficients of Friction. The coefficient of friction,
for which we assumed in the foregoing paragraph the mean value
0,007565, is not constant for rivers, creeks, etc., but, as in the case
of pipes, increases slightly, when the velocity diminishes, and
decreases, when the velocity increases. We must therefore put
(1+
5 = 5 (1 + 2) or 5, (1
α
→)
or to some similar formula.
§ 476.] THE MOTION OF WATER IN CANALS AND RIVERS.
967
The author in the article quoted in the remark of § 471 found
from 255 experiments, most of which were made by himself, for
English measures
5 = 0,007409 (1 + 0,1920),
and for the metrical system of measures
5 = 0,007409 (1 0,05853).
+
C
We see that this formula gives for a velocity c = 83 feet the
above-quoted mean value = 0,007565. In order to facilitate cal-
culation, the following tables for the metrical system have been
prepared:
Velocity c=
Coefficient of re-
sistance = 0,0
0,1 0,2
0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 meters.
1175 0958 0885 0849 0828 0813 0803 0795 0789

Velocity c =
Coefficient of resistance
5 = 0,0
1
1,2 1,5
1,5 | 2 | 3
GO
3
4
5 meters.
0784 0777 0771 0763 0755 0752
0750
For English system of measures we can employ the following
table.

Velocity c
0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 11
2
છ
3
10
5
7 10 15 feet.
Coefficient of re- 1215 1097 1025 0978 0944 0918 0899 0883 0836 0812 0788 0769 0761 07551 07504
sistance = 0,0
These tables are directly applicable to all cases, where the velo-
city c is given and the fall is required, and when formula No. 1 of
the foregoing paragraph is employed. If the velocity c is unknown,
or if that is the required quantity, the tables can only be employed
directly when we have an approximate value of c. The simplest
way to proceed is to determine c approximatively by one of the
formulas
c = 50,9 V
Fh
pł
meters or c = 92, 26 √F
pl
feet,
then find by means of the table, and substitute the value so found
in the formula
968
[§ 476.
GENERAL PRINCIPLES OF MECHANICS.
c²
h F
or
2g
5 lp'
F
·
2 g h.
C =
From the velocity c we determine the quantity of water QF c.
If the quantity of water and the fall are given, as is often the
case in laying out canals, and it is required to determine the trans-
verse profile, we must substitute
p
M
F
(see table, § 474) and
NF
Q
C =
in the formula
F
h = 0,007565·5
and put
g
m l Q²
h = 0,007565
2 g F
m l Q²
³
from which we obtain
I.E. in meters
and in English feet
F = (0,007565
2 g h
F = 0,0431 (m1 Qº)²,
F
=
l
0,0268 ( (m² Q²)?.
h
>
From this we obtain the approximative value
Q
C = ;
F
if we take the corresponding value of from the table, we can cal-
culate more accurately
F=
(5.
m l Q²) 3
2 g h
from which we deduce more correct values for
C =
Q
F
m VF,
and p = m
as well as for a, b, etc.
EXAMPLE-1) What must be the fall of a canal 1500 feet long, whose
lower breadth is two feet, whose upper breadth is 8 feet, and whose depth
is 4 feet, when it is required to convey 70 cubic feet of water per second?
Here
p = 2 + 2 √ 4² + 3² = 12, F = 5.4 = 20, c = 금음 ​= 3,5;
hence
5 = 0,00784 and
1500. 12 3,52
86,436
h =
0,00784
= 1,34 feet.
20
29
2 g
§ 477.] THE MOTION OF WATER IN CANALS AND RIVERS. 969
2) What quantity of water is carried by a creek 40 feet wide, whose
mean depth is 4 feet, and whose wetted perimeter is 46 feet, when it falls
10 inches in 750 feet? Here we have approximatively
40.4,5.10
C =
92,26.1 40.750.12
92,26
= 6,1 feet;
√230
•
hence we can assume
<= 0,00765.
We have now more correctly
c²
Fh
2g
ζιρ
4,5.40.10
0,00765.46.750.12
1
1,7595
0,5683 and c =
6,05 feet.
The quantity of water carried is
Q
= Fc4,5. 40. 6,05 1089 cubic feet.
4,5.40.
3) It is required to excavate a ditch 3650 feet long, which, with a total
fall of one foot, shall carry 12 cubic feet of water per second. What must
be the dimensions of the transverse section when the form is a regular hex-
agon? Here m = 2,632 (see table, § 474); hence we have approximatively
0,0268 (2,632. 3650.144)? 7,66 square feet, and
F =
12
C
1,567.
7,66
Here we must take
0,0083, and, therefore,
F = (0,0083 . 2,632 .
3650.144\}
64,4
= 7,95 square feet.
From this we obtain the depth
α
0,760 VF
2,14, the lower width
b
0,877 VF
b₁ = 2. 2,47
=
4,94.
1
2,47, and the upper width
REMARK-1) According to Saint Venant, we can put accurately enough
h
=
0,000401
Ρ
F
vi} = 0,000401. 37. v
Ρ
v2
meters;
•
F
2 g
hence the coefficient of resistance is
5 = 0,000401.2 g. v- 0,007887 v―,
E.G. for v = 1 meter
= 0,007887
and for v =
1 meter
Š
0,007887 V4
0,007887. 1,134 · 0,008945.
(Compare § 428, Remark 3.)
2) A table, which abridges these calculations, is given in the Ingenieur,
pages 460 and 461.
§ 477. Variable Motion-The theory of the variable motion
of water in channels can be referred to the theory of uniform mo-
tion, when we consider the resistance of friction upon a small por-
tion of the length of the river to be constant and put the corre-
sponding head
970
[§ 477.
GENERAL PRINCIPLES OF MECHANICS.
l p
5
F 2g
292
•
We must also take into consideration the vis viva corresponding to
the change of velocity.
Let A B C D, Fig. 808, be a short portion of the channel of a
river, whose length A D = 1, and whose fall DH h; let v, be
the velocity of approach and v, that with which the water flows
away. If we apply the laws of efflux to
an element D at the surface of the
water, we have for its velocity v₁
FIG. 808.
H

D
G
B
E
K
2
2%
= h +
2 J
2g
but an element E, which is situated
below the water, has on one side, it is
true, a greater head A G E H, but it is pressed back by the
water below it with a head DE; hence the effective fall, which
produces motion, is only D II EH E D, and consequently
–
the following formula holds good for any element:
ບ
h
2
H
2 g
vo³
;
2
0
if we add the resistance of friction, we obtain
h =
༢༡,༣ — v
2 g
vo
2
+5.
lp ༧°
F 2g
0
in which p, F and v denote the mean values of the wetted perim-
eter, the transverse section and the velocity. If F, denote the area
of the upper and F, that of the lower transverse section, we can put
FF
F=
{}
2 g
v2
F
2
and QFv. F, v;, whence
0
Q
2 [(2)² - ()]=(-1) Ga
1
2g
[(別​)
V. + ?
F. + F
1
from which we obtain
1
F
Q²
G
and
F
2
2 g
g
1
+
2
F
F + F
ιρ
F.
V 2 g h
2) Q =
1
1
1
+ 5
F2 F
2
l p
F+F
(+)
1
2
1
F2
];
Q²
2 g'
1) A = [√ √ + 5 + (+1)] and
h [p
F.
F
By the aid of formula 1) we can calculate from the quantity of
water carried, the length and transverse sections of a section of river
477] THE MOTION OF WATER IN CANALS AND RIVERS. 971
or canal the corresponding fall h, and by the aid of formula 2) from
the fall, length and cross-section the quantity of water carried. In
order to obtain greater accuracy we should calculate these for sev-
eral small portions of the channel of the river and then take the
arithmetical mean of the results. If the total fall only is known,
we must substitute this value instead of h in the last formula and
instead of
1
1 1
1
2
F F2 F3 F' 29
0
in which F denotes the area of the last cross-section, and instead of
n
I p
F + F₁
1
5. (1/2 + 77,3),
F
the sum of all the similar values for the different portions of the
channel of the river.
EXAMPLE.-A creek falls 9,6 inches in 300 feet, the mean value of its
wetted perimeter is 40 feet, the area of its upper transverse section is 70
square feet, and that of its lower is 60 square feet. Required the discharge
of this brook. Here
8,025 √0,8
Q
1
300 .40 1
+ 0,007565.
+
602
702
7,178
130 602 702
.7,178
3541 cubic feet.
√0,0000731 + 0,0003365
√0,0004096
2 Q
709
The mean velocity is
5,45 feet; hence it is more cor-
F+F 130
0
1
rect to put
5
0,00768 instead of 0,007565,
and therefore we have more accurately
Q
7,178
√0,0000731 + 0,0003416
352,5.
If the same stream has at another place the same amount of water in it
and falls 11 inches in 450 feet, and if the area of its upper transverse section
is 50 and that of its lower 60 feet, the mean length of its wetted perimeter
being 36 feet, we have
8,025 √0,9167
Q
1
1
+ 0,00768.
602
502
450.36
110
+
60% 502

0,9167
305 cubic feet.
0.0001222 + 0,0007549
= 8,025 1
The mean of the values is
Q
=
3543051
2
330 cubic feet.
972
[$ 478.
GENERAL PRINCIPLES OF MECHANICS.
§ 478. In order to obtain the formula for the depth of the
water, let us put the upper depth = a, and the lower = a,, the slope
of the bed = a, and consequently the fall of the bed = l sin, a.
The fall of the stream is then
h = α o
hence we have the equation
1
Q2
g
a₁ + I sin. a;
1
F₁
F 2 9
Q²
2
a - a - ( 7 ) ₂ = [(5 x + x ( 1 + 1) 2 - sin a] 1,
F²
(-1)
Q²
F22 g
2
whence we deduce
α1
F
Š
0
Ρ
F. + F₁
1
2
+ 7 = 29
F
g
| Q²
sin. a.
0
By the aid of this formula we can determine the distance 7,
which corresponds to a given change a, a, in depth. If the in-
verse problem is to be solved, we must resort to the method of
approximation, I.E., we must calculate first the lengths, and a cor-
responding to the assumed changes a, a, and a, a, of depth,
and then we must find by a proportion the change of depth corre-
sponding to the given length 7 (see Ingenieur, Arithmetic, § 16,
V, page 76).
The formula can be simplified, when the width of the stream
is constant. In this case we can put
(F. — F₁) (F. + F) 2
F29
(αar) (α + α₁)
1
F2
Q²
2
Fi Q²
F2
F2 F
2 g
v²²
approximatively
απ
2 g
1
p
2
and in like manner
Q
(F²
F
(αo
v
F + F (} + 1 ) & = P(F: +E") 5
F₁
approximatively
F2 2 g
Fi
P
で​。
a₁) (1
(α%
1 =
p
5.
and consequently
Ρ
a b⋅ 2 g
20
2
b˚ 2 g
2
2
で
​2 J
a) v
2 g
(F. +F) F 2g
2
from which we obtain
sin. a
で
​0
sin. a
α,
a b
2 g
2
21
1
2 g
$479.] THE MOTION OF WATER IN CANALS AND RIVERS. 973
By the aid of this formula we can obtain directly, for a given
distance 1, the corresponding change (a, -a,) of depth of the stream.
EXAMPLE.--A horizontal ditch 800 feet long and 5 feet wide is required
to convey 20 cubic feet of water per second; the depth of water at the
entrance is 2 feet, what will be its depth at the end of the ditch? Let us
divide the entire length of the ditch into two equal sections and determine
by the last formula the fall for each of them.
800
Here sin. a = 0,
2
20
2.5
= 400 and b = 5; for the first section, v
2; hence (0,00812 and a。
=
2; now since p = 81, it follows that
8,5
4
0,00812 .
10
29
400 =
a
a1
2 4
2,7608
15,1
0,183 feet.
1
2 2g
2 — 0,183
20
= 8,2, 01
9,085
Now for the second section a₁ = 2
1,817 and p is about
2,201; the fall in the second section will be
0,00812 .
8,2
9,085
2,2012
2 g
0,2205
a
. 400 =
0,240 ;
2
2,2012
0,9172
1
1,817
2 g
0,183 + 0,240
=
0,423
hence the entire fall is
and the depth of water at the lower end is
= 2 2 - 0,423 1,577 feet 18,92 inches.
=
§ 479. Floods and Freshets.-When the water level in
rivers or canals changes, it is accompanied by changes in the ve-
locity and in the quantity of water carried. A rise of the water
level not only increases the cross-section, but also causes a greater
velocity and, therefore, for a double reason a greater discharge; in
like manner a fall of the water level causes both a diminution of
velocity and of cross-section, and consequently a two-fold diminu-
tion of the quantity of water. If the original depth = a and the
present one = a, and the upper width of the canal = b, we can
put the increase of the cross-section
1
b (α₁ a); hence the
cross-section, when the water level has risen a distance a,
F₁ = F + b (α1
a) and consequently
F
b (an
α)
=1+
F
F
and we can put approximatively
√3 = 1 + b (a₁ = a)
F
F
2 F
a, is
974
[§ 479.
GENERAL PRINCIPLES OF MECHANICS.
If the original wetted perimeter = p, the present one = p₁ and
the angle of slope of the banks = 0, we have
2 (αr
a)
P₁ = p +
whence
sin. O
P1
1 +
2 (α1 a)
and
P
p sin. O
1
P1
ar
α
= 1 +
or
Ρ
p sin. O'
VP
p
a₁
α
= 1-
P1
p sin. O
Now the velocity for the original depth of water is
hence
and for the present depth it is c₁ = 92,26 V
c = 92,26 1/Fh
C₁
F
p
C
F
P1
(1 +
+
Ъ
1 + (α₁
a)
2 F 2
sin. O
b (a₁ — a)) (1
and the relative change of velocity is
C1
C
1)
=
с
b
(a₁ − a) ( 2 F
2 F
1
o)
12.0).
p sin. 0).
а
α1
p sin.
O
F h
P1
On the contrary, the ratio of the quantity of water carried by
the river is
o
Q1
Q
Fc
&₁ = FC = (1 +
(1+
b
-
a)
Ъ
(a₁ = α) ) [1
2) [1 + (ar
+ (a₁ − a) ( 23 ³/F
F
b
= 1 + ( a − a) (3) F
p
v
1
sin.
0),
and the relative increase in the quantity of water is
2)
Q₁ - Q
Q
==
3 b
a) (2½ F
1
p sin. ).
1
p sin.
6)]
We can put less accurately, but in many cases, particularly for
wide canals with little slope, sufficiently so, Fab and neg-
1
lect
p sin. O'
in which case we have more simply
C
a1
а
142
and
C
a
Q
2000
a
а
According to these formulas the relative change in the velocity
is half as great and that in the quantity of water as great as the
relative change in the depth of the water.
The foregoing formulas are only applicable, when the motion
§ 479.] THE MOTION OF WATER IN CANALS AND RIVERS. 975
of the water in its channel is permanent, in which case the depth
of the water is constant, but they do not hold good when the depth of
the water is variable. The mean velocity in the same transverse sec-
tion is greater, when the water is rising, and smaller, when the
water is falling than when the depth of the water is constant;
hence in the first case more water and in the second case less water
passes through than when the motion is permanent.
EXAMPLE-1) If the depth of the water increases, the velocity is in-
creased and the quantity of water of its original value.
2) If the depth decreases 8 per cent., the velocity is diminished 4 per
cent. and the quantity of water 12 per cent.
3) By the aid of the more accurate formula
Q +
1
Q
3b
Q
(α₁ — α)
2 F
p sin. O
we can construct a water-level scale K M, Fig. 809, upon which we can
read off the quantity of water passing in a canal for any depth K L, when
we know the quantity of water for a certain mean
FIG. 809.
IM
depth. If b = 9 feet, b₁
1
3, a = 3, and 0 = 45°,

we have
A
D
(9 + 3) 3
N
F
18 square feet,
B
11,485, and
Q1
Q
Q
3.9
2.18
1
11,485. 0,70
10,707) (α1
= 0,627 (α1
- a).
a) = (0,750 – 0,123) (a,' — a)
p = 3 + 2 . 3. √2
sin. O √ = 0,707; hence
57) (11 − a)
If the quantity of water corresponding to the mean water level is Q
40 cubic feet, we have
If at
―
Q₁ = 40 + 40.0,627 (a¸ — a) = 40 + 25 (α1 — a).
1
α = 0,04 feet = 5,76 lines, Q1
11,52 lines, Q₁ = 42 cu. ft.; if a₁
Q1
41 cu. ft.; if a.
a
1
1
a
=
0,08
- 0,04 feet, Q1 = 39 cu. ft.,
5,76 lines apart, would
feet
etc., a scale, whose divisions are L M LN
give the quantity of water to a cubic foot. The accuracy of course di-
minishes as the difference of the depth of water from the mean depth in-
creases.
REMARK.-The construction of mill-races, canals for bringing water, as
well as the location of dams, weirs, etc., will be treated of at length in the
second volume.
FINAL REMARK.-The author has discussed at length the subject of the
motion of water in canals and rivers in the Allgemeine Encyklopädie,
Vol. II, Article “Bewegung des Wassers in Canälen und Flüssen," and has
given there a list of the treatises upon this subject up to 1844. Rittinger's
tabulated synopsis of the experiments upon the motion of water in canals
is contained in the "Zeitschrift des österreichischen Ingenieurvereins,"
year 1855.
976
[$ 480.
GENERAL PRINCIPLES OF MECHANICS.
CHAPTER VIII.
HYDROMETRY, OR THE THEORY OF MEASURING WATER.
H
FIG. 810.
G
Ι.
D
§ 480. Gauging.-The discharge of a running stream within
a certain time is measured either by gauged vessels, by a dis-
charging apparatus, or by hydrometers. The most simple method
is that by means of gauged vessels, but this is only applica-
ble to small quantities of water. The vessel is most frequently
composed of boards, and is therefore parallelopipedical in form, and
to increase its strength, it is generally bound with iron hoops. The
manner of calculating the exact contents of this vessel is given in
the Ingenieur, page 208. The water is brought to the vessel by a
trough E F, Fig. 810, at the end of which is placed a double clack,
by means of which the water can
be made to flow into the vessel or
alongside of it. In order to deter-
mine accurately the depth of the
body of water, we employ a scale
K L. If, before we begin the
measurement, we lower the pointer
Z until it touches the surface of
the water, which, perhaps, may
only cover the bottom, and read
off on the scale the depth of the
water, we obtain the depth Z Z, of the water to be measured by
subtracting the former reading from that given by the scale, when
the pointer Z, after the completion of the observation, is brought into
contact with the top of the water. The clack must of course be so
arranged before the experiment that water shall discharge alongside
of the vessel. If we are satisfied that the influx into the trough
has become constant, we observe the time upon a watch held in the
hand and turn the clack around so that the water will discharge into
the vessel; when the vessel is full, or partially so, we observe again
upon the watch the time and return the clack to its original posi-
tion. From the mean cross-section F and the depth Z Z₁ = s of
the body of water, we calculate the total discharge V Fs, which,

B
481.]
977
HYDROMETRY, ETC.
when divided by the duration of the influx, which is the difference
t of the two times observed upon the watch, gives the discharge per
second
Q
F's
t
REMARK.—If we wish to know at any time the discharge of a variable
stream of water, we can employ the apparatus represented in Fig. 811,
FIG. 811.
E
G
N
P
H
M
which is often met with in salt
works. Here there are two meas-
uring vessels A and B, which are
alternately filled and emptied. The
water, which is brought to them
by the pipe F, passes through a
short tube C G, which is rigidly
connected with the lever D E which
turns about C. If one of the ves-
sels, as, E.G., A, is filled, the water
flows through a small trough H
and fills the little bucket M, which
then draws the lever down and the

pipe C G comes into such a position as to carry the water into B. The
valves and P are opened by cords passing around pulleys and attached
to the lever. The opening of the valves is assisted by iron balls, which
give the last impulse to the lever when it is descending. The buckets M
and N contain small orifices, through which they empty themselves after
the lever has turned. A counter attached to the apparatus gives the num-
ber of strokes, which can be read off at any time. Other apparatuses of
the same sort, which were employed by Brown, are described in Dingler's
Polyt. Journal, Vol. 115. In reference to a new apparatus for measuring
water by Noeggerath, see Polyt. Centralblatt, 1856, No. 5. Compare also
the works of Francis, Lesbros, etc., which have been cited. See also further
on, § 506.
§ 481. Regulators of Efflux.-Very often small and medium
discharges are measured by causing them to pass through a known
orifice under a known head. From the area F of the orifice and
the head b we determine, by the aid of the coefficient of efflux, the
discharge per second
Q = µ F √2 gh.
ре
Poncelet's orifices are the best for this purpose; for their coeffi-
cients of efflux are known (§ 410) with great accuracy for different
heads, but they are only applicable, when the discharge is a medium
one. The author employs in his measurements of water four such
orifices, one 5, one 10, one 15 and one 20 centimeters high and all
62
978
[$ 481
GENERAL PRINCIPLES OF MECHANICS.
FIG. 812.
1)
20 centimeters wide. These orifices are cut out of brass plates,
which are placed upon wooden frames such as A C, Fig. 812, and
these frames can be fastened to any wall by means of four strong iron
screws. But in many cases we must employ much
larger orifices for which the coefficients of efflux have
not been determined so certainly; and very often we
can only employ overfalls or notches, which are even
less accurate. But in any case we should endeavor
to produce both perfect and complete contraction.
Hence, if the orifice is in a thick wall, we should bevel it off upon
the outside. The corrections to be applied for partial and incom-
plete contraction have been sufficiently explained in §§ 416, 417.
B
In order to measure the quantity of water in a trough, we first
put the mouth-piece in its place and then wait until the head.
becomes permanent. In order to measure the head, we can em-
ploy either the fixed scale KL with a pointer, Fig. 813, or the
movable one E F, Fig. 814. If we wish to observe the efflux directly
H
FIG. 813.
G
L
ர்
K
A
FIG. 814.


I
B
D
FIG. 815.
at the sluice gate, it is advisable to attach to the guides two brass
scales B C and D E, Fig. 815, with the pointers F and G by means
of which we are able to read off with more cer-
tainty the height of the orifice. It is always bet-
ter, when measuring water, to employ a new
sluice gate and new guides which are properly
beveled outwards.

A
E
The most simple way of measuring the water
in a trough is to place a board C F, Fig. 701,
beveled at the top, across it and to measure the
overfall which is produced. If the ditch or trough is long and
nearly horizontal, considerable time will elapse before the flow be-
comes permanent, and it is, therefore, advisable before beginning
§ 482.]
979
HYDROMETRY, ETC.
the measurement to put in another board, which will prevent for
some time the efflux of the water and thus hasten its rise to the
height necessary for a permanent flow.
FIG. 816.
In order to measure the discharge of a creek, we can construct
a dam A B, Fig. 816, of boards and
allow the water to flow through an
opening C in it, or we can employ a
simple overfall or weir (this subject
will be treated more at length in the
second volume).
REMARK.-The most simple method
of determining the head is to observe the
position of the pointer, first, when its point touches the surface of the
water, while the flow is permanent, and secondly, when it touches the sur-
face of the still water which is on a level with the top of the sill. The
difference of the two observed heights is the head of water or the height
of the water above the sill. We must be careful in observing the last
height of the pointer to pay attention to the action of the capillary attraċ-
tion, in consequence of which the level of the water may be 1,37 lines
above or below the sill, before efflux over the latter will begin (see § 380).

§ 482. We can easily measure the quantity of water carried by
a canal or trough A B, Figs. 817 and 818, in the following man-
A
FIG. 817.
B
FIG. 818.


ner: a board, the lower edge of which has been beveled, is inserted
in the trough in such a manner as to leave an opening between it
and the bottom of the latter, through which the water will pass.
This method has an advantage over that in which overfalls are
employed, viz.: the water, which is dammed back, comes to rest
better, and we can, therefore, measure the head more accurately.
When it is possible to have a free efflux, as in Fig. 817, we should
prefer it, since greater accuracy can be obtained, but when the
980
[§ 482.
GENERAL PRINCIPLES OF MECHANICS.
quantities of water are large, it is not possible to prevent the water
from rising, and we are obliged to be satisfied with an efflux
under water, such as is represented in Fig. 818. If the trough
ends but a short distance from the orifice, I.E., if it forms a shoot,
the water flows through it almost freely and we have one of the cases
of Lesbros' experiments (§ 418). If a denote the height and b the
width of the orifice, h the head measured to the middle of the ori-
fice and μ the coefficient of efflux, taken from Table II, § 418, we
have the discharge
Q = µ a b √ Q a h.
v
If, on the contrary, the trough is long, or if the water, which is
flowing away, is so obstructed that its surface becomes horizontal,
the water will pass all portions of the cross-section of the orifice
with the same velocity, which is that corresponding to a head equal
to the difference of level of the water A above and the water B
below the orifice, and we have only to substitute in the latter
formula for h the difference of level.
If the water flows into the air, or if the surface of the lower
water, as in Fig. 817, does not rise above the upper edge of the
orifice, we must substitute for an orifice with beveled or with
rounded edges
ре =
0,965,
and consequently, when the depth of the stream is a and its width b,
Q = 0,965 a b √2 gh,
or more accurately, when a, is the depth of the approaching water
and a that of the water flowing away (see § 398),
2gh
Q
0,965 a b V
1-(4)
α
2.
When the efflux takes place under water, in which case the lower
surface of the water is above the upper edge of the orifice (see Fig.
818), an eddy is formed behind the wall of the orifice, by which the
efflux is considerably interfered with. According to several experi-
ments of the author, for an orifice with a sharp edge we must put,
as a mean value,
μ = 0,462,
and, on the contrary, when the edge is rounded off in the shape of a
quadrant,
0,717.
ре
EXAMPLE.—In order to find the discharge of a trough A B, Fig. 818, a
sharp-edged board CD was placed in it and an efflux under water was
thus produced; the following observations were then made. Width of
orifice or trough 6 = 3 feet, height of orifice or distance C E of the edge C
§ 483.]
981
HYDROMETRY, ETC.
of the board above the bottom of the trough a = 6 inches, length of the
pointer Z above the orifice h₁ = 0,445 feet, length of the pointer Z, below
Hence the difference of level is
the orifice h₂
= 1,073.
1
h = h₂
- h₁
=
1,073
0,445
=
0,628 feet,
and the required discharge is
Q
0,462 . 8,025 .3 . 0,5 √h₂ — h¸
5,56 √0,628
=
4,40 cubic feet.
§ 483. If the coefficient of efflux were always the same for sim-
ilar cross-sections, the triangular or two-sided notch 4 B C, Fig. 819,
would have a great advantage over the notch with a horizontal sill;
FIG. 819.
but this assumption, as we have seen in the
case of circular apertures, is not correct for
small orifices, and only approximatively so for
large ones. Professor Thomson, of Belfast,
recommends such notches as useful for measur-
ing water. From the width A B = b and the
height CD h, we obtain the discharge

Q
8
=
μ b h
15 2
=
√2 g h (see § 402),
and if we put, with Prof. Thomson, the coefficient of efflux = 0,619,
b h
2
Q = 0,33 √2 g h = 0,132 b h³ cubic feet.
Orifices, so shaped that the discharge through them shall be
proportional to their height, are useful in measuring water. If they
are provided with a sluice-gate the height of the opening is the
measure of the discharge. Let the head above the upper edge of
such an orifice A B C D, Fig. 820, be O A h, the length of this
edge be A Bb, that of the lower edge,
CD b₁, and the height of the orifice,
AD = a.
Horizontal lines at the distance
Y-
FIG. 820.
h
α
b
B
n
N
M
E
D
X
=

from each other will divide this orifice
into strips of equal height, and the dis-
Q
charge through each of them should be
N
the same. For the upper strip, whose width
is b and for which the head is h, we have
Q
b a
12gh,
N
N
= x
and, on the contrary, for another strip at a distance OM be-
low the surface of the water, whose width M N y,
982
[$ 484
GENERAL PRINCIPLES OF MECHANICS.
Q
уа
n
√ 2 g x;
n
equating these two values of
we obtain
N
y √x = b √h, or
y
b
√ √ TH
h
X
The curve B N C, which bounds the orifice on the side, belongs
to one of the system of curves discussed in Article 9 of the Intro-
duction to the Calculus; its asymptotes are the horizontal line
O Yand the vertical one O X.
From Q, h and a we obtain
1) the upper width of the orifice b
Q
a N 2 g h
2) the width of orifice at the depth x, y = b
3) the lower width of the orifice b₁
The area of the orifice is
h
1
by
h + a
F= 2b (Nh (h + a) — h),
and consequently the mean head is
1
Z
29
α
h
;)'
h
�་
h
X
If the orifice is provided with a sliding gate, when it is raised a
distance D M = a₁, it gives an orifice of efflux M C, the discharge
through which is Q₁ =
a1
a
Q.
$ 484. Prony's Method.-As considerable time often elapses
before the flow of the water, which has been dammed back, be-
comes permanent, the following method, proposed by Prony, can
often be employed with advantage. We begin by closing the
orifice completely by means of a sluice-gate, and we wait until the
water has risen to a certain height, or as high as circumstances
will permit; we then open the gate enough to allow more water to
be discharged than is arriving, and we measure the height of the
water at equal intervals of time, which should be as small as pos-
sible; finally, we close the orifice again perfectly and observe the
time t, in which the water rises to its former height. Now during
the lapse of the time t+t, the same quantity of water has of
course arrived and been discharged; hence the quantity of water
which arrives in the time t+t, is equal to the discharge in the
§ 485.]
983
HYDROMETRY, ETC.
time t. If the heads, while the level of the water was sinking, were
h。, h₁, họ, h, and h, we have the mean velocity
√√29
v =
12
(Nπ₂ + 4 √ Th₂ + 2 √ h₂ + 4 √ h3 + √Ñ₁) (see § 453),
and if the area of the opening of the sluice is F, the discharge in
the time t is
ре
Ft √ 2 g
V
12
Vi
(Nπro + 4 √π₂ + 2 √ πg + 4 √ πs + √π4);
hence the quantity of water arriving in a second is
Q
V
= t + t₁
μ Ft V 2 g
(√h₂ + 4 √ π₂ + 2 √ πq + 4 √ πs + √πs).
12 (t + t₁)
EXAMPLE.-In order to measure the quantity of water in a brook, which
we wish to employ to turn a water-wheel, the stream was dammed up by a
wall of boards, as is represented in Fig. 816, and after opening the rec-
tangular orifice in it, we made the following observations: initial head, 2
feet; after 30", 1,8 feet; after 60", 1,55 feet; after 90″, 1,3 feet; after 120”,
1.15 feet; after 150,", 1,05 feet; and after 180", 0,9 feet; width of the ori-
fice =2 feet, height foot, time required for the water to rise to former
level 110". In the first place the mean velocity is
8,025
18
(√2 + 4 √1,8 + 2 √1,55 +4√1,3 + 2 √1,15 + 4 √1,05 + √0,9)
· 0,4458 (1,414 + 5,364 + 2,490 + 4,561 + 2,145 + 4,099 + 0,949)
= 0,4458. 21,022
But F = 2.
=
9,372 cubic feet.
obtain the required
Q
1
9,372 feet.
square foot, the theoretical discharge is, therefore,
If we assume that the coefficient of efflux
quantity of water
0,61. 180
180 + 110
•
9,372 = 3,548 cubic feet.
0,61, we
§ 485. Water-inch.-When we are required to measure small
quantities of water, we often allow it to discharge under a given
head through circular orifices in a thin plate, which are one inch
in diameter. We call the discharge through such an orifice, under
the smallest pressure, I.E. when the surface of the water is one line
above the uppermost part of the orifice, a water-inch (Fr. pouce
d'eau; Ger. Wasserzoll or Brunnenzoll). The French assume that
a water-inch (old Paris measure) corresponds to a discharge in 24
hours of 19,1953 cubic meters, or
in 1 hour, 0,7998 cubic meters, and
in 1 minute, 0,01333 cubic meters;
but the older data, given by Mariotte, Couplet, and Bossut, differ
considerably from the above. According to Hagen, the water-inch
(for Prussian measure) discharges 520 cubic feet in 24 hours, or
0,3611 cubic feet in a minute. Prony's double water modulus (or
984
[§ 485.
GENERAL PRINCIPLES OF MECHANICS.
"nouveau pouce d'eau"), which corresponds to an orifice 2 centi-
meters in diameter, under a pressure of 5 centimeters, and which
discharges 20 cubic meters in 24 hours, has not been adopted gen-
erally.
The observations can be made with more certainty when we
have a greater head; it is simpler to make this head equal to the
diameter 1 inch of the orifice. According to Bornemann and Rö-
ting, such a water-inch passes daily 642,8 cubic feet (Prussian) of
water (see the Ingenieur, page 463).
The apparatus, by which we measure the water with the aid of
the water-inch, is represented in Fig. 821. The water to be meas-
FIG. 821.
A
ured is discharged from the
pipe A into a box B, from
which it passes through
holes, made in the parti-
tion CD below the level
of the water, into the box
E; from it the water is
discharged through circu-
lar orifices F one inch in
diameter, which are cut
out of sheet iron, into the

1
+1
reservoir G. To preserve the level of the water 1 line above the
top of the orifice we must have a sufficient number of holes, a por-
tion of which are closed by stoppers. We employ for more accu-
rate determinations in addition the orifice F, which allows, of
a water-inch to pass through. When the quantity of water is very
great, we divide it into several portions and measure in this way
but one portion, as, E.G., a tenth. This division is easily accom-
plished by conducting the water into a reservoir with a certain
number of orifices on the same level and catching the water deliv-
ered from one of the orifices only in the above apparatus.
REMARK.- We can also employ cocks and other regulating apparatuses
for measuring water, when we know the coefficients of resistance corre-
sponding to every position. If h is the head, F the cross-section of the pipe
and μ the coefficient of efflux for the cock, when fully open, we have the
discharge
or inversely
µ F √ 2 g h,
Q
Q
= μ
1
and
F √ 2 g h
(7)
2 g h.
§ 486.]
985
HYDROMETRY, ETC.
Now if we put the coefficient of resistance for a certain position of the
cock, which may be taken from one of the tables given previously, = 5,
we have the corresponding discharge
2 g h
Q₁
=
F
1
+5
μ F V 2 g h
VI + μ² 5
Q
√1 + μ² 5
με
1+5
Q
1
2 g h
We can construct from the above formula a convenient table, and we
have only to glance at it when we wish to know the discharge correspond-
ing to a certain position of the cock. If, E.G., µ = 0,7 and F1
= 4 square
inches, we have
0,7. 4. 12. 8,025 √h
√1 +0,49 5
1 foot,
Q1
or, if h is constant and
269.64
Q
√1 +0,49 5
h
269,64
cubic inches,
1 + 0,49 5
Now if the cock is turned 5°, 10°, 15°, 20°, 25°, etc., the coefficients of
resistance are 0,057, 0,293, 0,758, 1,559, 3,095, and the corresponding dis-
charges are 266, 252,1, 230,2, 203, 170 cubic inches.
§ 486. In order to regulate the efflux through an orifice F,
Fig. 822, we employ either a cock or valve A, Fig. 822, which is
FIG. 822.
K
FIG. 823.


A.
B
regulated by means of a lever and a float K, so that the same quan-
tity of water is discharged through B as through F.
The discharge of water from a reservoir B D E, Fig. 823,
through a lower orifice or tube D can be regulated by means of a
wide overfall B, since a moderate change in the quantity of water,
discharged through A, will produce but a slight change in the
height of the water above the sill B; hence the augmentation of
the head of the orifice of efflux will be inconsiderable.
Let F denote the area of the orifice D, h the height of the sill
of the overfall above the middle of the orifice and h, the height of
986
[S 487.
GENERAL PRINCIPLES OF MECHANICS.
the surface of the water above the same sill. We have the dis-
charge through D
Q
= µ F√2 g (h + h),
when the coefficient of efflux is . Substituting the head h, above
the weir, which can be determined from the discharge Q₁, the width
b, and the coefficient of efflux μ, by means of the equation.
Q₁
3
= 3 μ, b₁ √2 g h₁³, or by the formula
we obtain the expression
Q
1
2
>
V
Q
= p F√ 2 g h + (3921),
from which it is easy to see that Q varies less with Q1, the greater
the value of h is and the greater the width b, of the overfall is.
The width b of the overfall can be easily increased by giving it
a curved form like B O B, Fig. 824. The discharge through the
DWB
FIG. 824.
orifice D is then almost
constant, although the
quantity of water flow-
ing in may be very va-
riable; for the height of
the water above the long
curved sill is always
small compared with the
height of this sill above
the orifice of efflux..

REMARK.-Such an apparatus for dividing the water was constructed
of sheet iron for the Wernergraben at Freiberg by Oberkunstmeister Schwam-
krug. It discharges through a rectangular orifice D, which is 5 feet long
and 1 foot high, almost invariably 40 cubic feet of water per second, while
the remaining water passes over the overfall, the sill of which lies 2 feet
above the upper edge of the orifice, and flows on in the ditch to where it
is wanted.
§ 487. Hydrometric Goblet.-We can employ to measurė
small quantities of running water the small vessel, represented in
Fig. 825, which I have called the hydrometric goblet. This instru-
ment consists of a tube B, 12 inches long and 3 inches in diameter
with a funnel-shaped mouth-piece A, and of a vessel D, 6 inches
high and 6 inches wide, which is united to B by an intermediate
§ 487.]
987
HYDROMETRY, ETC.
conical piece C. This vessel has an orifice L L in the side, in
which we can insert mouth-pieces containing different sized circu-
lar orifices in a thin plate. The instrument is held by means of
the handle H under the water S, which is being discharged, E.G.,
R
FIG. 825.
L
N
L
H
B
ΤΑ
F
P
Ο
from the pipe R, and the water thus caught
is allowed to discharge itself through the ori-
fices L L. In order to tranquilize the water
in the vessel a fine sieve or wire gauze is
placed in the reservoir D, and in order to ob-
serve the head of the water a glass tube O P,
which is placed close to a brass scale and ends
a half an inch from the bottom of the vessel,
is added to it. From the observed head, the
known cross-section of the orifice and the
corresponding coefficient of efflux, we can cal-
culate the discharge by means of the formula
Q = µ F √2 gh.
V2

=
If we prepare a small table, we can spare
ourselves this calculation and the only opera-
tion, which we are required to perform, is a
simple interpolation between the values in the
table. If d is the diameter of the orifice,
πα
F
and therefore
4
леп
4
ď² V2 g h =
4
√2 g. d³ vñ.
The discharge Q is double, when the cross-section or d is double,
or when the head is four times as great. If we so arrange the in-
strument that the maximum head shall be four times the minimum ;
if, E.G., the former is 12 and the latter 3 inches, and if we employ
series of orifices whose diameters form the geometrical series
I.E.
d, √2 d, 2 d, 2 12 d, 4 d, etc.
d, 1,414 d, 2 d, 2,828 d, 4 d, etc.,
we obtain a means of measuring all discharges from the minimum
given by the smallest orifice with the diameter d under the smallest
head, to the maximum, given by the largest orifice with the diam-
eter Vn. d under the greatest head 4 h.
988
GENERAL PRINCIPLES OF MECHANICS.
[$ 487.
If we assume for
1.
II.
III.
IV.
V.
VI.
VII.
d =
1 V9
1
√2
$
=0,12500,1768 =
0,1768 = 0,2500 = 0,3535 -0,5000
√2
0,7071
1 inch
1,0000
μl
0,690 0,675 0,660 0,647 0,635
0,627
0,620
we can calculate the following useful table.
Table of the hourly discharge in cubic feet for the following orifices.

Head in inches.
I.
II.
III.
IV.
V.
VI. VII.
3 + LO
0,85
1,66
3,25
6,37
12,51 24,70 48,85
4
0,98
1,92
3,75
7,36
14,44
28,52
56,40
5
1,10
2,14
4,19
8,23
16,15
31,89
63,06
6
1,20
2,35
4,60
9,01 17,69
34,93
69,08
་
1,30
2,54
4,96
9,73
19,10
37,73
74,61
8
со
1,39
2,72
5,31
10,41
20,42
40,33
79,77
9
1,47
2,88
5,63
11,04
21,66
42,78
84,60
10
1,55
3,03
5,93
11,65 22,84
45,09
89,18
11
1,63
3,18
6,22
12,20 23,95
47,30
93,53
12
1,70
3,32
6,50
12.74 25,01 49,40
97,69
13
1,77
3,46
6,77 13,26 26,04 51,42 101,68
The manner of using this table is shown by the following
example.
EXAMPLE.-In order to determine the quantity of water furnished by a
spring, the water from it was caught in a hydrometric goblet, and it was
found that a state of permanency occurred when the efflux took place
through the orifice V (one half inch in diameter) under a head of 10,4
inches. According to the table for h 10 inches
Q = 22,84 cubic feet per hour,
and for h = 11 inches
Q
=
23,95 cubic feet,
consequently the difference for one inch is 1,11 cubic feet, and for 0,4 inches
§ 488.]
989
HYDROMETRY, ETC.
0,4. 1,11 0,444. Hence the discharge under the head h = 10,4 inches is
Q = 22,84 + 0,444 = 23,284 cubic feet.
10
§ 488. Floating Bodies.-The discharge of large creeks,
canals and rivers can only be measured by means of hydrometers,
which indicate the velocity. The simplest of these instruments are
floating bodies (Fr. flotteurs; Ger. Schwimmer). We can use any
floating body for this purpose, but it is safer to employ bodies of
medium size and of but little less specific gravity than the water
itself. Bodies whose volumes are about of a foot are quite large
enough. Very large bodies do not easily assume the velocity of the
water, and very small bodies, particularly when they project much
above the level of the water, are easily disturbed in their motion by
accidental circumstances, such as the wind, etc. A simple piece of
wood is often employed, but it is better to cover the wood with a
light-colored paint; hollow floats, such as glass bottles, sheet-
iron balls, etc., are better; for we can fill them partially with water.
Floating balls are, however, most generally employed. They are
made of sheet brass and are from 4 to 12 inches in diameter; to
prevent their being lost sight of, they are covered with a coat of
light-colored oil paint. Such a floating ball A, Fig. 826, gives the
velocity at the surface only, and often only that in the axis of the
stream. By uniting two balls A and B, we can find also the
velocity at different depths. In this case one ball, which is to be
submerged, is filled with water, and the other contains enough to
prevent more than a small portion of it from projecting above the
FIG. 826.
FIG. 827.
B
B
level of the water.
The two balls are
united by a string,
wire or thin wire
chain. We first de-
termine by a single
ball the superficial
velocity c, and we

co
then determine the
mean velocity c of the two connected balls; now if we denote the
velocity at the depth of the second ball by c₁, we can put
C =
Ca + C₁, and, therefore, inversely, c, = 2 c — c
2
If we unite the balls successively by longer and longer pieces
of wire, we obtain in this way the velocities at greater and greater
990
[§ 489.
GENERAL PRINCIPLES OF MECHANICS.
depths. The mean velocity of a perpendicular is determined by
allowing the second ball to swim near the bottom and putting
C
Co + C₂
2
;
it is more accurate, however, to take the mean of all the observed
velocities in the perpendicular as the mean velocity.
To obtain the mean velocity in a perpendicular, a floating staff
A, B, represented in Fig. 828, is often employed, and it is very
FIG. 828.
B
B
convenient, when it is used for meas-
urements in canals and ditches, to have
it made of short pieces which can be
screwed together. The one used by
the author is composed of 15 hollow
pieces, each one decimeter long. In
order to make it float nearly perpen-
dicularly, the lower part is filled with
enough shot to prevent more than the
head from projecting above the water.
The number of pieces to be screwed together depends, of course,
upon the depth of the canal.
We observe, when using the floating staff and the connected
balls, that, when the movement of water in channels is not impeded,
the velocity at the surface is greater than that at the bottom; for
the top of the staff and the uppermost ball are always in advance.
It is only when the channel is contracted, as, E.G., by piers of
bridges, that the opposite phenomenon is observed.
REMARK. -- Generally, and particularly with large floating bodies such
as ships, etc, the velocity of the floating body is somewhat greater than
that of the water; this is owing less to the fact that the body, in floating,
: slides down an inclined plane formed by the surface of the water, than to
the fact that it does not participate, or at least only partially so, in the ir-
regular internal motion of the water, this variation is, however, so slight,
when the floating bodies are small, as to be negligible.
§ 489. Determination of the Velocity and of the Cross-
section. We find the velocity of a floating ball by observing by
means of a good watch with a second-hand or by means of a half-
second pendulum (§ 327) the time t, in which it describes the dis-
tance A B = s, Fig. 829, which has been previously measured and
staked off on the shore. The required velocity of the sphere is then
c = In order that the time t shall correspond exactly to the

C
$489.]
991
HYDROMETRY, ETC.
FIG. 829.
D
distance measured on the shore, it is necessary to put two rods C'
and D, by means of a suitable instrument, in such a position upon
the other side of the river that the
lines CA and D B shall be perpen-
dicular to A B. Placing ourselves
behind A, we note the instant the
float K, which has been placed in the
water some distance above, arrives at
the line AC, and then passing be-

EK
A
B
hind B, we observe upon the watch the instant that the float ar-
rives at the line B D; by subtracting the time of the first observa-
tion from that of the second, we obtain the time t, in which the
space s is described. In order to determine the discharge Q = Fc,
we must know, besides the mean velocity c, the area F of the cross-
section. To find this area, it is necessary to know the width and
the mean depth of the water. The depth is measured by a gradu-
ated sounding-rod A B, Fig. 830, the cross-section of which is
elongated and the foot of which is formed by board; when the
depth is great, we can make use of a sounding-chain, to the end of
which an iron plate is attached, which, when the measurement is
being made, lies upon the bottom. The width and the abscissas or
distances from the shore corresponding to the depths measured are
FIG. 830.
A
FIG. 831.


E
easily found for canals and small creeks EF G,
Fig. 831, by stretching a measuring chain A B or
laying a rod, etc., across the stream. When the
river is wide, we make use of a plane-table M,
placed at a proper distance 4 O from the cross-
section E F, Fig. 82, to be measured. If a o
upon the plane-table is the reduced distance 4 0
of the fixed points A and O from each other, and if we have placed
co in the direction A Ó, and thus made the direction a ƒ of the
width, which had been drawn, previously to putting the plane-table
in position, parallel to the line AF to be measured off, cach line.
of sight towards the points E, F, G, etc., in the transverse profile
cuts off upon the table the corresponding points e, f, g, and
992
[$ 490.
GENERAL PRINCIPLES OF MECHANICS.
a e, a f, a g, etc., are the distances A E, A F, A G, etc., upon the
reduced scale. When using the sounding-rod to measure the
depth, it is, therefore, not necessary to measure the distance of
G
E
FIG. 832.
M
the corresponding points from the
shore; for the engineer, who is at
the plane-table, can sight at the
sounding-rod, when it is placed in
the line E F.
Now if the width E F, Fig. 831,
of a transverse profile is made up
of the portions b₁, ba, ba, etc., and
if the mean depths of these por-
tions are a1, a2, as, and the mean velocities C1, C2, C3, etc., we have
the area of the cross-section

the discharge
F = a₁ b₁ + ɑ2 b₂ + ɑ3 b3 +
Q = a₁ b₁ c₁ + ɑ₂ by c₂ + az bz Cz + ...9
and finally the mean velocity
Q
a₁ b₁ c₁ +
A2 *}
ɑş b₂ C₂ + . . .
C F
a₁ b₁ +
αş b₂ +
•
EXAMPLE.-Upon a pretty straight and constant portion of a river the
following observations were made:


Feet. Feet. Feet. Feet. Feet.

At the centre of the divisions of the width 5 12
the depths were
10 00
20
15
7
3
6
11
8
4
•
the mean velocities were
1,9
2,3
2,8
2,4
2,1
The area of the cross-section is
F = 5 . 3 + 12 . 6 + 20 . 11 + 15 . 8 + 7 . 4 = 455 square feet,
the discharge is
Q
་་
15. 1,972. 2,3 + 220. 2,8 + 120. 2,4 + 28. 2,1 = 1156,9 cubic feet,
and the mean velocity is
C
1156,9
455
2,54 feet.
§ 490. Woltmann's Mill or Tachometer.-The best hy-
drometer is Wollmann's tachometer or Woltmann's Mill (Fr. Moulinet
de Woltmann; Ger. hydrometrisches Flügelrad von Woltmann),
Fig. 833. It consists of a horizontal shaft A B with from 2 to 5
€ 490.]
993
HYDROMETRY, ETC.
surfaces or vanes F, inclined to the direction of the axis; when
immersed in water and held opposite to the direction of motion, it
FIG. 833.

E
୯
B
Q
3
indicates by the number of its revolutions the velocity of the run-
ning water. To enable us to count the number of revolutions the
shaft has cut upon it a certain number of threads of an endless
screw C, which work into the teeth of a cog-wheel D, which indi-
cates, by means of a pointer and figures engraved upon the wheel,
the number of revolutions of the wheel F. As we often wish to
register a great number of revolutions the shaft of the cog-wheel
carries a pinion, which takes into another cog-wheel E, upon which
we can read off, as upon the hour-hand of a watch, multiples
(E.G., five or tenfold) of the number of revolutions of the vanes.
If, for example, both cog-wheels have 50 teeth and the pinion has
10, the second wheel will turn one tooth, while the first moves five,
or the shaft of the vane wheel makes five turns. When the pointer
of the first wheel is at 27 = 25 + 2 and that of the second at 32,
the corresponding number of revolutions of the vane-wheel is
= 32.5 + 2 = 162.
The entire instrument with a sheet iron vane is screwed to a
63
994
[S 490.
GENERAL PRINCIPLES OF MECHANICS.
pole, so that it may easily be immersed and held in the water. In
order to prevent the gearing from turning except during the time
of the observation, its shafts run in bearings placed upon a lever
GO, which is pressed down by means of a spring, so that the teeth
of the first cog-wheel do not take into the endless screw except
when the string G E is drawn upwards. The number of revolu-
tions in a given time is not exactly proportional to the velocity of
the water; hence we cannot put v = a. u, in which u is the num-
ber of revolutions, v the velocity and a an empirical number, but
we must put
or more accurately
v = v。 + a U,
0
v = v。 + a u + ß u²
or still more accurately
v = a u + √ v² + ß u³,
in which v, denotes the velocity of the water, when it ceases to
move the vanes, and a and ẞ are numbers to be determined by
experiment. The constants v., a, ẞ must be determined for each
particular instrument. By their aid a single observation gives the
velocity, but it is always safer to make at least two and then take
their mean value as the true one.
EXAMPLE.—If for a tachometer v 0,110 feet, a = 0,480 and ẞ= 0,
then v =
· 0,11 + 0,48 u, and if we have found the number of revolutions
of the fan to be 210 in 80 seconds, the corresponding velocity of the
water is
v = : 0,11 +0,48.
210
80
= 0,11 + 1,26
1,37 feet.
REMARK 1.—The constants v。, a and 3 depend principally upon the
angle of impact, I.E., upon the angle formed by the surface of the vanes
with direction of the motion of the water and also with the direction of
the axis of the wheel. If we wish to make, when the velocities are small,
pretty accurate observations, it is advisable to make the angle of impact
large, I.E., about 70°. It is also desirable to have vane-wheels of different
sizes and of different angles of impact, so that when the depth or velocity
of the water is greater or smaller we can employ one or the other.
REMARK 2.-If the tachometer had no resistance to overcome in turn-
ing, the vanes A B, Fig. 834, would describe the space O C₁ = C D
tang. C D C₁ while the water describes C D; hence, if we denote by v
the velocity of the water and by d the angle of impact O C B :
C D C ₁
1
=
§ 491.]
995
HYDROMETRY, ETC.
we have under this supposition the mean velocity of rotation of the vane-
FIG. 834.
FIG. 835.
A
X
D
B
wheel
v₁ = v tang. 8,
1
from which it is easy to see that,
when r denotes the mean radius of
the vane-wheel, the number of revo-
lutions is


И
01
2πη
v tang. 8
2 πρ
B₂
and that, consequently, it is directly
proportional to the velocity v of the
water and to the tangent of the angle
of impact and inversely to the mean radius of the vane-wheel.
REMARK 3.—In order to determine the superficial velocity of water we
also employ a small wheel made of metal, like the one represented in Fig.
835, and we allow only the lower part to be immersed in the water. The
number of revolutions is given by a train of wheels, exactly as in the
tachometer.
§ 491. In order to determine the constants or the coefficients
of a tachometer, it is necessary to hold the instrument in running
water, the velocity of which is known, and to observe the number
of revolutions. Although only as many observations as there are
constants are required, yet it is safer to make as many observa-
tions as possible, particularly with very different velocities, and
to employ the method of the least squares (see Introduction to the
Calculus, Art. 36) and thus do away with the accidental errors
of observation. The velocity of the water may be determined by a
floating sphere, or we may catch the water in a gauged vessel and
divide the quantity of water caught by the cross-section. If the
floating sphere is employed, the air must be still and the water must
move uniformly and in a straight line. The vane-wheel must be
immersed at several points along the path described by the floating
sphere, and to insure perfect accuracy, the diameter of the sphere
should be about equal to that of the vane-wheel.
The second method of determination by catching the water, in
which the mill is immersed, in a gauged vessel possesses many
advantages. For this purpose, and for adjusting hydrometers
generally, it is very desirable to have at one's disposition a
hydraulic observatory, which consists of a gauged vessel, a trough,
and a discharging vessel or reservoir. We can then give the
water any desired velocity; for we can regulate not only the
entrance of the water into the trough, but also, by inserting boards,
996
[§ 491.
GENERAL PRINCIPLES OF MECHANICS.
we can regulate at will the velocity in it. In making the observa-
tion, we have but to insert the tachometer at different parts of the
cross-section of the trough, to measure the depth of this section by
a scale, and then to gauge the quantity of water, which has passed
through in a given time (§ 480). The area of the cross-section is
obtained by multiplying the mean depth by the mean width, and
the discharge Q is calculated from the mean cross-section of the
receiving reservoir and the depth of the water, which has flowed
into it, by means of the formula
Q
G s
t
finally, from Q and F we deduce the mean velocity of the water
Q
G s
V F
Ft
The corresponding number u of revolutions of the vane-wheel
is the mean of all the revolutions observed when we inserted the
instrument in different parts of the transverse profile.
If by experiment we have determined a series V1, V2, V3, etc., of
mean velocities and the corresponding numbers of revolutions, we
obtain, by substituting them in the formula
v = v。 + a Uz
or in the more accurate one
v = a u + √ v² + ßu³,
as many equations of conditions for the constants v., a, ẞ, as we
made observations, and we can find from them the constants them-
selves either by employing the method given in Art. 36 of the In-
troduction to the Calculus, or by dividing these equations into as
many groups as there are unknown constants, and combining them
by addition into as many equations of condition as are necessary
for the determination of v, a and ß.
If we assume the passive resistances of the instrument to be
small enough to be neglected, we can put va u and determine
a by moving the instrument forward in still water and observing
the number n = ut of revolutions made in describing the space
s = vt; then
ย
v t S
a
U
ut
22
REMARK—1) If we employ the simple formula with two constants, we
can put, according to the method of least squares,
Σ (y²) Σ (x) — Σ (xy) Σ (y)
®。 = Σ (x²) Σ (y²) — [Σ (x y)]²
and a =
Σ (2) Σ (1) - Σ (!) Σ (2)
Σ (x²) Σ (y²) — [Σ (x y)]²
§ 491.]
997
HYDROMETRY, ETC.
1
U
in which x =
and y =
and the sign Σ denotes the sum of all the
v
v
values of the same kind as that which follows it, E.G.
1
1
1
Σ (2)
+
+
+
V1
20 2
V3
1
1
1
U 1
ՂԱ
u z
2
+
+
+
Σ (x y)
V
v
V2
2
03
V
3
1
1
EXAMPLE.
locities
We have observed with a small tachometer that for the ve-
0,163, 0,205, 0,298, 0,366, 0,610 meters
the number of revolutions per second were
0,600, 0,835, 1,467, 1,805, 3,142,
and we wish to determine the constants corresponding to this instrument.
By the aid of the formula given in the Remark, we obtain, since
1
1
Σ (3)
+
+
18,740,
0,163 0,205
Σ (3)
=
0,600 0,835
+
+
22,759,
0,163 0,205
2
Σ (2°) = (0,183) + (0,305) + · =82,246,
(y²)= 105,233, and
Σ
0,600
0,835
Σ (x y)
+
(0,163)2 (0,205)²
+ ... = 80,961,
105,233. 18,740 - 80,961. 22,759
82,846. 105,233 — (80,961)²
129,5
0,060 and
2162
368,3
α
2162
0,1703;
hence the formula for this instrument is
Substituting u =
u = 0,835 gives
W =
u =
1,467,
1,805,
and finally, u =
3,142,
v = 0,060 + 0,1703 u.
0,6, we obtain
v = 0,060 + 0,102 = 0,162;
v = 0,060 + 0,142 = 0,202;
v = 0,060 + 0,249 = 0,309;
v = 0,060 + 0,307 = 0,367;
v = : 0,060 + 0,535 = 0,595.
The calculated values therefore agree very well with the observed ones.
REMARK—2) We can also, according to Lapointe, insert the tachometer
in a cylindrical pipe, and thus obtain the velocity of the water flowing
through it. The counting apparatus can be placed outside of the pipe
and connected with the vane-wheel by means of a shaft. Lapointe calls
this instrument une tube jaugeur (see "Comptes rendues," T. XXV, 1848;
998
[S 492.
GENERAL PRINCIPLES OF MECHANICS.
also Polytechn. Centralblatt, 1847). Fig. 836 gives an ideal representation
G
FIG. 836.
E
F
B
R
R
H
of the tachometer in a pipe. The vane-wheel in
this case also puts a shaft D E in rotation by
means of an endless screw; the former passes out
of the pipe R R, in which the water to be
measured flows, through a stuffing-box F into
the case G H of the counting apparatus, the ar-
rangement of which may be very varied.
REMARK-3) The French have but lately be-
gun to give sufficient attention to the tachometer.
A complete treatise upon this instrument, by
Baumgarten, is to be found in the "Annales des
ponts et chaussées," T. XIV, 1847, and an abstract
of it in the "Polytechnisches Centralblatt, 1849." Baumgarten recommends
a screw-wheel and adds several remarks, which agree very well with our
experiments, made many years ago. A new tachometer, without wheels
and with a long screw, is described by Boileau in his "Traité de la mesure
des eaux courantes."

FIG. 837.
§ 492. Pitot's Tube.-The other hydrometers are more im-
perfect than the tachometer; for they are either less accurate or
more difficult to use. The simplest instrument of this kind is
Pitot's tube (Fr. la tube de Pitot; Ger. Pitot'sche Röhre). It con-
sists of a bent glass tube A B C, Fig. 837, which is held in the
water in such a manner that the lower part is
horizontal and opposite to the motion of the
water. By the impulse of the water a column
of water will be forced into the tube and held
above the level of the water, and this rise D E
is proportional to the impulse or to the velo-
city of the water which produces it; this rise
or difference of level can therefore serve to
measure the velocity of the water. If the height
DE above the exterior surface of the water
= h and the velocity of the water = v, we can put

B
D
E
h
༧༠
g
37
in which μ is an empirical number, or inversely
v = µ √2 g h, or more simply
v = 4√h.
$ 493.]
999
HYDROMETRY, ETC.
In order to find the constant , we hold the instrument in the
water where the velocity is known to be v₁; if the rise is
if the rise is = h₁, we
have the constant
V1
, which can be employed in other cases,
where the velocity is to be determined by this instrument.
In order to facilitate the reading off of the height h, the instru-
ment is composed of two tubes A B and C D, as is represented in
Fig. 838; from one of the tubes a pipe proceeds in the direction of
the stream, and from the other two pipes F and F at right-angles
E
FIG. 838.
F
SH
F
Κ
to that direction, but by means of the same cock
both tubes can be closed at once. If we draw the
instrument out of the water, we can easily read off
the difference of height K L = h of the columns
of water upon the scale placed between them. In
order to prevent the water from oscillating in the
tubes, it is necessary to make their mouths narrow;
and in order that the cock may be shut quickly and
certainly, it is provided with a crank and a rod
HS, which is represented in the figure principally
by a dotted line and terminates near the handle of
the instrument.
REMARK-1) Although Pitot's tube is not so accurate
as the tachometer, yet, on account of its simplicity, it
can be highly recommended. The author has discussed
this instrument at length in the "Polytechnisches Cen-
tralblatt, 1847," and gives there a series of numbers, de-
termined by experiment, and the values of the coefficient

راح
deduced from them. With fine instruments, when the
velocities were between 0,32 to 1,24 meters, we found
v = 3,545 vī meters.
2) Duchemin recommends Pitot's tube with a float.
Since the latter must be pretty wide, it dams the water
back to a certain extent, so that it cannot be employed for narrow canals
(see Duchemin: "Recherches expérim. sur les lois de la résistance des
fluides"). Boileau describes in his work, cited in § 412, a new kind of
Pitot's tube, which is provided with a small gauged vessel; the velocity
is measured by the quantity of water pressed above the surface of the
water.
§ 493. Hydrometric Pendulum.-The hydrometric pendu-
lum (Fr. pendule hydrométrique; Ger. Stromquadrant or hydro-
1000
[§ 493.
GENERAL PRINCIPLES OF MECHANICS.
metrisches Pendel) was principally employed by Ximenes, Michelotti,
Gerstner, and Eytelwein to measure the velocity of running water.
FIG. 839.
L
E
A
This instrument consists of a quadrant
A B, Fig. 839, divided into degrees and
parts of a degree, and of a string attached
to its centre C, at the other end of which is
fastened a metal or ivory ball K, 2 or 3
inches in diameter. The velocity of the
water is given by the angle A CE formed
by the stretched string with the vertical,
when the plane of the instrument is placed
in the direction of the stream, and the
ball is immersed in the water. Since the angle cannot easily exceed
40°, this instrument often has the form of a right-angled triangle,
and the graduation is then marked upon the base. In order to
place the zero line vertical, we can either place a level upon the
instrument or we can employ the ball itself by allowing it to hang
out of the water and then turning the instrument until the string
corresponds with the zero line. For velocities less than 4 feet we
can employ an ivory ball; for greater velocities, however, we must
use heavy balls of metal. On account of the vibrations of the ball,
not only in the direction of the motion of the water but also in
that at right angles to it, it is always difficult to read off the angle,
and the result is never free 'from uncertainty; this instrument
cannot therefore be considered to be a perfect one.
The dependence of the angle of deviation, for a ball that is not
deeply immersed, upon the velocity of the water can be determined
in the following manner. The weight G of the ball and the im-
pulse of the water P = µ F v², which increases with the cross-
section F of the ball and the square of the velocity v, give rise to
a resultant R, which is counteracted by the string and is deter-
mined by the angle of deviation d, for which we have

P μ
tang. S = G
F v²
G
or inversely
v²
G tang. S
u F
G
and v = 1
Vtang. 8,
μπ
F
I.E.,
v = 4√tang. 8,
in which is an empirical coefficient, which must be determined
$ 494
1001
HYDROMETRY, ETC.
in the manner stated above (§ 491) before the instrument can be
used.
§ 494. Rheometer.-The remaining hydrometers, such as
Lorgna's water-lever, Ximenes' water-vane, Michelotti's hydraulic
balance, Brunning's tachometer and Poletti's rheometer, etc., are
difficult to use and partially uncertain. The principle of all of
them is the same; they consist of a balance and of a surface,
which is subjected to the impact of the water; the former serves
to measure the impulse P of the water against the former, but
since the impulse is μ Fv3, we have inversely
in which
v =
VP
μ F
= 4√P,
is an empirical constant, dependent upon the magni-
tude of the surface subjected to the impulse of the water.
FIG. 840.
L
B
A
G
The Rheometer, which has been lately proposed by Poletti, does
not differ essentially from Michelotti's balance and consists of a
lever A B, Fig. 840, movable about a fixed axis C, and of a second.
arm CD, upon which a surface, or, according to
Poletti, a simple rod, which is to be subjected
to the impact, is screwed. In order to balance
the force of impact of the water, shot or weights
are put into the sheet iron box, which is sus-
pended at 4 upon the lever, and to balance
the empty apparatus in still water, weights
are hung at B, the extreme end of the arm
CB. From the weights added at G and the
arms of the lever CA = a and C F = b, we
obtain by means of the formula P b = G a the
impulse

D
α
P
b
& G and v = 1
P
M. F
a G
mb F
= &NG,
in which
denotes an empirical constant.
A hydrometer constructed upon the same principle, in which
the impulse of the water is balanced by the force of a spring (hy-
dromètre dynamométrique) is described by Boileau in his treatise
upon the measurement of water.
REMARK 1.—The last-mentioned hydrometers are discussed at length in
Eytelwein's "Handbuch der Mechanik," Vol. II, in Brunning's “ Abhand-
lung über die Geschwindigkeit des fliessenden Wassers," in Venturoli's
1002
[§ 495
GENERAL PRINCIPLES OF MECHANICS.
"Elementi di Meccanica e d'Idraulica," Vol. II. Concerning Poletti's
Rheometer, see Dingler's Polytechn. Journal, Vol. XX, 1826. Stevenson's
hydrometer is Woltmann's tachometer, see Dingler's Journal, Vol. LXV,
1842. The water-meters and gas-meters constructed like reaction wheels
will be treated in the following chapter.
REMARK 2.-A work to be particularly recommended for practical
purposes is the "Hydrometrie oder practische Anleitung zum Wasser-
messen von Bornemann, Freiberg, 1849." Boileau's work has already been
mentioned several times (see § 412, etc.).
CHAPTER IX.
OF THE IMPULSE AND RESISTANCE OF FLUIDS.
§ 495. Reaction of Water.-The total pressure of the still wa-
ter in a vessel is, according to § 362, reduced to a vertical force equal
to the weight of the mass of water; but if the vessel AF, Fig. 841,
FIG. 841.
has an opening F, through which
the water issues, this force under-
goes a change not only because a
portion of the wall of the vessel is
absent, but also because the water,
which issues from the orifice, like
every other body, which changes
its conditions of motion, reacts by
virtue of its inertia. The change
in the motion of a body may consist
either of a change of velocity, or
of a change of direction, and, there-
fore, the reaction (Fr. réaction; Ger. Reaction) of the issuing water
may be due not only to an acceleration but also to a constant
change in the direction of the water, which is approaching the
orifice.

E
W
V Z D
We can make ourselves acquainted with the complete reaction
of the water in a discharging vessel in the following manner.
Let c be the velocity of the water, which is issuing from the
orifice F, c, the relative velocity of the water at the surface A,
§ 495.]
1003
THE IMPULSE AND RESISTANCE OF FLUIDS.
G the area of this surface and h the head of water A D at the ori-
fice.
Then we have
C²
h +
2 g
2 g
and the discharge
Q
= Fc = G c₂.
If we imagine the vase A F, Fig. 841, to move forward in a
horizontal direction with a velocity v, we must put for the absolute
velocity c₂ of the water entering the vessel
2
C₁₂² = c₁² + v³,
and if the angle of inclination of the axis of the stream to the
horizon is E F c = a, we have for the absolute velocity w of the
effluent stream
w3
= c² + v² - 2 c v cos. a.
Now the actual energy of the water before efflux is
2
L₁ = ( 09/29
g
+ π π1) Q Y =
(
c₁² + v²
2 g
+ h) 2:
Q Y
(
十
​'C² + 2)²
2 c v cos. a
a)2Y;
Q
2 g
and that after efflux it is
W²
L₂ = Q r
2 g
hence the energy withdrawn from the water and transmitted to
the vessels is
L
L₁ -
L₂
Liz
(
2 c² + 2 c v cos. a
+1) Q2
n)
+ h) Q Y,
ولا
2 g
c²
or, since
2 g
2
2 g
=
h,
L =
cv cos. a
9
Q r.
The horizontal component of the reaction of the water is
L
c cos. a
H=
Q Y.
V
Since Q
Fc, we have also
c²
H =
F y cos, a = 2.
9
Fy cos. a = 2 h F y cos. a,
29
and therefore, when the direction of the stream is horizontal, as in
Fig. 842,
H = 2 h Fy.
Therefore, the reaction of a horizontal stream is equal to the
weight of a column of water, whose cross-section is that of the stream
and whose height is double that (2 h) due to the velocity.
1004
[$ 496.
GENERAL PRINCIPLES OF MECHANICS.
FIG. 842.
REMARK.—Mr. Peter Ewart, an Englishman, has recently made experi-
ments to prove the correctness of this law (see "Memoirs of the Manchester
Philosophical Society," Vol. II, or the "Ingenieur, Zeitschrift für das ge-
sammte Ingenieurwesen," Vol. 1). He hung
the vessel HR Fupon a horizontal axis C,
Fig. 842, and measured the reaction by a bent
lever A D B, upon which the vessel acted by
means of a horizontal rod A G, which pressed
against the vessel exactly opposite to the ori-
fice F. For efflux through an orifice in a thin
plate, he found

R
B
D
A P
=
P 1,14.
FY.
2g
If we put the cross-section
F₁ = 0,64 F
1
and the effective velocity of discharge
(see § 405), we obtain by the theoretical formula
V
1
01 0,96 v
v2
2
P
2
•
1
F₁ Y 2. 0,962. 0,64 .
FY
=
1,18
F
2 g
2g
29
v2
29
or about the same that was given by experiment.
With an orifice shaped
like the contracted stream, he found P 1,73 Fy, and the coefficient
of efflux or velocity 0,94.
=
Since in this case F.
v2
we have theoretically
22
P=2.0,942
2 g
F y = 1,77
•
FY,
2 J
F and v
0,94 v,
1
which agrees very well with the result of the experiment.
§ 496. If we imagine the discharging vessel 4 F, Fig. 843, to
be moved vertically upwards with a velocity ", we have for the
absolute velocity of the water which
enters it
W
FIG. 843.

Co
Z D
C₂ = V
C19
and, on the contrary, for that of the
water issuing from it (the same no-
tations being employed as in the
foregoing paragraph)
w² = c² + v² + 2 c v cos. (90° + a)
c² + v² 2 c v sin. a.
Hence the total energy of the
volume of water Q per second is
2
L₁ = ((" + c )² + h) Q 7,
= = 29
and, on the contrary, that of the water discharged is
L₂ = (c² + v² — 2 c v sin. a) Q: 2 g
L2
§ 496.] THE IMPULSE AND RESISTANCE OF FLUIDS.
1005
vessel is
L L₁ - L2
consequently the mechanical effect imparted by the water to the
2
2 v c₁ + c₁² c² + 2 c v sin. a
h)
+ h) Q Y,
2 g
C²
or, since h =
2 g
2 g
(c sin. a — c₁)
L
QY,
g
L
c sin. a
C1
√ =
G¹) QY =
(sin.
α
ย
F) I
C
QY
F
(sin.
•
in. a
and the corresponding vertical force is
g
41) 2
F\ c²
8-5 F y = (sin. a - 77)
G g
G
g
2 h Fy.
If the orifice of efflux is small, compared to the surface G, we
have 0, and, therefore, the vertical component of the reaction
G
F
V = 2 h Fy sin. a.
According to the foregoing paragraph the horizontal compo-
nent of this force was
H = 2 h Fy cos. a;
hence the total reaction of the water is
R
√ V² + H² = 2 h Fy,
and its direction is exactly opposite to that of the motion of the
effluent water.
If F = G, I.E., if the water flows through a pipe of uniform
width, we have
F
1, and therefore
G
V = (sin. a 1). 2 h Fy = - (1 sin. a). 2 h Fy;
in this case V does not act upwards but
reaction is
A
D
R = √ √² + H² = √ cos. a² + (1
FIG. 844.
E
•
B
downwards, and the total
sin. a). 2 h FY
= √2 (1 sin. a). 2 h F y
= 4 h F y sin. (45°

For a =
90°, I.E., when the pipe forms a
semicircle, R 4h Fy
=
If a = + 90°, we have the case represented in
Fig. 844, where H= 0 and
V
=
(c — c₁)
g
Q r = (1 - 77)
2 h FY,
1006
[§ 498.
GENERAL PRINCIPLES OF MECHANICS.
F
consequently, for
= 0, we have
G
V=R = 2 h Fy.
The total weight of the water in the vessel will be diminished
that much, when the water is allowed to flow out.
§ 497. Impulse and Resistance of Water.-Water or any
other fluid, when it impinges upon a solid body, imparts a force or
impulse to it, and thus produces a change in its state of motion.
The resistance (Fr. résistance; Ger. Widerstand), which water
makes to the motion of a body, is not essentially different from im-
pulse. The examination of these two forces constitutes the third
chief division of hydraulics. We distinguish from each other first,
the impact of an isolated stream (Fr. choc d'une veine de fluide;
Ger. Stoss isolirter Wasserstrahlen); secondly, the impact of a
bounded stream (Fr. choc d'un fluide défini; Ger. Stoss im be-
grenzten Wasser oder Gerinne); and thirdly, the impact of an unlim-
ited stream (F. choc d'un fluide indéfini; Ger. Stoss im unbegrenz-
ten Wasser) Impact of the first sort takes place when a stream
discharged from a vessel encounters a body, as, E.G., the bucket of
an overshot water-wheel; impact of the second sort occurs, when
the water in a canal or trough strikes against a body which en-
tirely fills the cross-section of the latter, as, E.G., the float of an
under-shot water-wheel. Finally, impact of the third kind occurs,
when running water strikes upon a body immersed in it and the
cross-section of the latter is but a small part of that of the stream,
as, E.G., the float of a wheel in an open current.
We distinguish also impact against bodies at rest and bodies in
FIG. 845.
D
B
B
-D
P
motion, against curved and plane
surfaces; the latter may be either
direct or oblique.
We will now consider a more
general case, viz. the impact of an
isolated stream against a surface
of revolution, moving in the direc-
tion of the motion of the stream,
which coincides with the direction
of the axis of the surface.

§ 498. Impact of an Isolated
Stream.-Let B A B, Fig. 845, be
§ 498.] THE IMPULSE AND RESISTANCE OF FLUIDS.
1007
=
a surface of revolution, A P its axis, and FA a stream of water
moving in the direction of the axis of the latter and impinging
against it; let us put the velocity of the water c, that of the
surface v, and the angle B TP, which the tangent D T to the
end B of the generatrix or each fibre B D of the stream of water,
which leaves the surface, makes with the direction B E of the axis,
= a, and let us assume that the water does not lose any vis viva in
consequence of the friction while passing over the curved surface.
The water impinges upon the surface with the velocity c
v and
then passes over the surface with that velocity and leaves it in a
tangential direction T B, T B, etc., with the same velocity. From
the tangential velocity B D = cv and from the velocity BE
v in the direction of the axis, we obtain the absolute velocity
B Cc, of the water, after it has impinged upon the surface, by
the well-known formula
v) v cos. a + y².
c₁ = √(c
Tc 2)² + 2 (c
Now a discharge Q can produce by its vis viva a mechanical
effect
Qy, when it loses its entire velocity c; hence the energy
2 g
remaining in the water is =
2
Qy, that transmitted to the sur-
2 g
face is
P v =
Q x
Q x =
2 g
{c² — (c —
2 J
c² - c₁²
2 g
Q r
v)²
2 (c — v) v cos. a
v]
Qr
2 g
2 c v― 2 v²
—
2 (c - v) v cos. a
QY, L.E.
P v = (1
-
cos, a)
(c
2 g
v) v
QY,
g
and the force or impulse in the direction of the axis is
C
P =
(1 — cos. a)
Qx.
g
If the surface moves with a velocity 2, which is in the opposite
direction to that of the water, we will have
P = (1 cos. a)
(c + v)
g
QY,
and if the surface does not move or if v = 0, the impulse or hydrau-
lic pressure in the direction of the axis is
C
P = (1 − cos. a
Q Y.
1008
[§ 499.
GENERAL PRINCIPLES OF MECHANICS.
From this it follows that the impulse of one and the same mass
of water, when the other circumstances are the same, is proportional
to the relative velocity cv of the water.
If the area of the cross-section of the stream is F, the volume
of the impinging water is F (cv); hence
(cv)²
P = (1 -
cos. a)
Fy;
g
or for v = 0,
c²
P = (1
cos. a)
Fy.
g
If the cross-section of the stream remains the same, the impulse
against a surface at rest increases with the square of the velocity of
the water.
§ 499. Impact against Plane Surfaces.-The impulse of
the same stream of water depends principally upon the angle a, at
which the water moves off from the axis after the impact; it is
null when this angle 0, and, on the contrary, a maximum and
= 2
(cv)
g
Q r
when this angle is 180° or when its cosine =
Fr 846.
B
FIG. 847.
W
P
►P
and 1 cos, a becomes 1 + cos. a.
1, in which case, as is
represented in Fig. 846, the
water quits the surface in a di-
rection opposite to that in which
it struck it. In general the im-
pact is greater against concave
than against convex surfaces;
for in the former case the angle
is obtuse and its cosine negative


Usually the surface is, as is represented in Fig. 847, plane and
therefore a = 90° or cos. a = 0 and the impulse
P
(cv)
g
Qx.
C²
Fy 2 Fhy.
2g
When the surface is at rest, we have
C
P Q y = Fy = 2.
g
C²
g
The normal impulse of water against a plane surface is equal to
the weight of a column of water, the cross-section of whose base is
equal to the cross-section of the stream, and whose height is twice
that due to the velocity (2 h = 2.
C
2g
§ 499.]
THE IMPULSE AND RESISTANCE OF FLUIDS.
1009.
FIG. 848.
OL
Κ
E
The results of the experiments made upon this subject by
Michelotti, Vince, Langsdorf, Bossut, Morosi and Bidone were
about the same, when the cross-section of the impinged surface was
at least 6 times that of the stream
and when this surface was at a
distance not less than twice the
thickness of the stream from the
orifice. The apparatus employed
consisted of a lever like Poletti's
Rheometer (§ 494), upon one end
of which the stream impinged,
the impulse was balanced at the
other end by weights. The ap-.
paratus employed by Bidone is
represented in Fig. 848. B C is the surface subjected to the action
of the stream, G the scale-pan for receiving the weights, D the axis
of rotation, and K and L are counter weights.

B
A
REMARK.— The most extensive experiments upon the impulse of water
were made by Bidone (see "Memoire de la Reale Accademia delle Scienze
di Torino," T. XL, 1838). They were made with a velocity of at least 27.
feet and with brass plates of from 2 to 9 inches in diameter. Bidone gen-
erally found the normal impulse against a plane surface somewhat greater
than 2 Fhy; but this increase is to be ascribed to the increase of the arm
of the lever, in consequence of the falling back of the water. See Duchemin :
Recherches expérimentales sur les lois de la résistance des fluides (translated:
into German by Schnuse). When the impinged surface was very near the
orifice, Bidone found P to be only 1,5 Fh y. When the impinged surface
was of the same size as the stream, in which case the angle of deviation a
is acute, according to du Buat and Langsdorf, P is only
Fhy.
Bidone
and others have found that the impulse during the first instant was nearly
twice the permanent impulse. Comparative experiments upon the impulse
and reaction of water have been made by the author with a reaction wheel.
See his "Experimentalhydraulik" and the "Civilingenieur," Vol. I, 1854.
=
By more recent experiments upon the impact of isolated streams of air
and water (see Civilingenieur, Vol. VII, No. 5, and Vol. VIII, No. 1), the
author found the effective impulse of an isolated stream of air or water
against a normal plane to be 92 to 96 per cent. of the theoretical force P
QY
that, on the contrary, the impulse of such a stream against a hollow.
surface of rotation by which the direction of the stream is made to deviate
an angle d 134°, is but 83 to 88 per cent. of the theoretical force P =
c Q Y
c (1 — 008. 8)
Q Y
g
64
1010
[$ 500.
GENERAL PRINCIPLES OF MECHANICS.
J
§ 500. Maximum Work done by the Impulse.—The me-
chanical effect
P v = (1
cos. a)
(c — v) v
g
Q r
depends principally upon the velocity v of the impinged surface;
E.G. it is null not only for v = c, but also for v = 0; hence it fol-
lows that there must be a velocity, for which the work done by the
impulse is a maximum. It is evident that this is the case when
v) v is a maximum. If we consider c to be half the periphery
of a rectangle and v to be its base, we have its height = C v and
its area = (c v) v; now the square is that rectangle, which has
the greatest area for a given periphery; hence (c — v) v is a maxi-
mum, when (c — v) = v, I.E., v
=v, = = and we obtain the maximum
C
2'
mechanical effect of the impulse, when the surface moves in the
direction of the stream with half the velocity of the latter; the
work done is then
C³
P v =
(1
cos. a). !
2g
•
1
Q y = (1 - cos. a). Q hy.
Now if a = 180°, I.E., if the motion of the water is reversed by
the impact, we have the work done
= 2.1 Qhy = Qhy;
but if a = 90°, I.E., if the stream strikes against a plane surface,
the work done is butQhy, in this case the water transmits to
the surface but one-half of its actual energy, or but one-half of the
mechanical effect corresponding to its vis viva.
EXAMPLE-1) If a stream of water, the area of whose cross-section is 40
square inches, delivers 5 cubic feet per second and strikes normally against
a plane surface, which moves away with a velccity of 12 feet, the impulse is
(c
P =
Q x =
g
(
5. 144
40
= 58,125 pounds,
12). 0,031. 5. 62,5 = 6 . 0,031 . 312,5
and the mechanical effect transmitted to the surface is
P v = 58,125. 12
697,5 foot-pounds.
The maximum effect is obtained, when
с
5. 144
v =
=
9 feet,
2
40
and it is
L =
Q y = . 182. 0,0155. 5. 62} = 81. 0,155 . 62,5 784,6875.
2 J
foot-pounds;
the corresponding impulse or hydraulic pressure is
P =
784,6875
9
87,19 pounds.
§ 501.]
1011
THE IMPULSE AND RESISTANCE OF FLUIDS.
2) If a stream FA, Fig. 849, the area of whose cross-section is 64 square
inches, impinges with a velocity of 40 feet upon an immovable cone,
whose angle of convergence B A B = 100°, the
hydraulic pressure in the direction of the stream is
FIG. 849.

с
F
P = (1.
cos. a) - Qr
9
P
64
= (1
cos. 50°). 40. 0,031. 40.62,5
+
144
10000
= (1 — 0,64279) . 1,24 . 9
=
0, 35721. 1377,8 492,16 pounds.
=
§ 501. Impact of a Bounded and of an Unlimited
Stream.-If we surround the periphery of a plane surface B B,
Fig. 850, with borders B D, B D (Fr. rebords; Ger. Leisten), which
FIG. 850.
D
project beyond the surface struck by the water, the
water will be deviated from its course at an obtuse
angle as in the case of concave surfaces, and the
impulse is greater than when the surface is plane.
The action of this impact depends principally upon
the height of the border and upon the ratio of the
cross-section of the stream to that of the enclosed
surface. In an experiment, where the stream was
one inch thick and the cylindrical border 3 inches in diameter and
3 lines high, the water flowed from the surface in nearly the oppo-
site direction and the impulse was

B
c²
3,93 FY;
2 g
in all other cases this force was smaller. It is impossible ever to
c²
attain the theoretical maximum value 4 Fy in consequence
2g
of the friction of the water upon the surface and upon the border.
In the case of the impact of the bounded stream F A B, Fig.
851, there is also a border; it is, however, only partial and includes
FIG. 851.
B
but a portion of the periphery; it
limits, moreover, both the stream and
the impinged surface. The imping-
ing stream is turned in the direction
of the portion of the periphery, which
has no border, and is therefore de-
viated 90° from its original direction;
hence the formula, which we found for the isolated stream,

1012
[§ 502.
GENERAL PRINCIPLES OF MECHANICS.
P =
(c — v)
Q Y =
(゜​_゜​)
g
g
༧)
c FY,
holds good here. If the surface B B, Fig. 847, against which the
stream strikes, moves away with a velocity v in a direction, which
forms an angle & with the original direction of the stream, the ve-
locity of this surface in the direction of the impact is
v₁ = v cos. §;
hence the impulse is
(c v cos. d)
P =
Qr
g
(c
v cos. d) v cos. s
б
L P v₁ =
Qx.
g
and the work done by it per second is
The principal application of this formula is to the impact of an
unlimited stream, in which case
Q
P
=
F(c
v cos. d), and therefore
2
(c
v cos. d)²
g
Fy.
§ 502. Oblique Impact.-There are several cases of oblique
impact, viz.: where the water after impact flows away in one, in
two or in more directions. If, as in the case of the impact of a
bounded stream, the surface A B, Fig. 852, has a border upon
three sides so that the water can flow away in one direction only,
we have the hydraulic pressure of the water against the surface in
the direction of the stream
(c
- v)
P
(1
cos. a)
Q Y.
g
FIG. 852.
FIG. 853.


B
B
F
A
P
v P
N
But if the impinged plane B C, Fig. 853, has a border upon
two opposite sides only, the stream divides itself into two unequal
parts, the angle of deviation a of the larger part Q, is less than
that 180° a of the smaller part Q, and the total impulse in the
direction of the stream is
1
C
บ
C
V
P = (1
cos. a).
Q₁ y + (1 + cos. a)
Q2 Y
g
g
§ 502.]
1013
THE IMPULSE AND RESISTANCE OF FLUIDS.
= (°² 7 ³) [(1 cos. a) Q₁ + (1 + cos. a) Q₂] y.
g
But the conditions of equilibrium of the two portions of the
stream require that the pressures
(c — v)
g
(1
-
cos. a) Q₁y and
γ
(c — v)
g
(1 + cos. a) Q₂ Y
shall be equal to each other; hence
(1
cos. a) Q₁ = (1 + cos. a
a) Q2,
or, since Q = Q1 + Q2, we can put
(1 − cos. a) Q₁ = (1 + cos. a) (Q – Q1), I.E.
a
Q₁ = (1 + cos. α) 2 and 2₁ = ( − &
cos. a
a) 2,
1
2
Q2
2
so that the total impulse in the direction of the stream is
(1 + cos. a) Q
P =
(c — v)
g
2 (1
—
cos. a)
2
(c — v)
(1
2
cos. a) Qy, I.E.
QY',
g
γ
P =
с
g
V
2
sin.² a QY.
Dividing the work done by the impulse in a second
L = P v =
(c — v)
9
v sin,² a . Qy
by the velocity A v₁ = v₁ = v sin. a, with which the surface recedes
in a normal direction, we obtain the normal impulse
which consists of the parallel impulse
N:
(c — v) v sin.² a
g v sin. a
Q Y
=
(c — v')
g
sin. a. QY,
(c — v)
P= N sin, a =
sin. a. QY,
g
SN cos. a =
(c — v)
g
C
V
sin. a cos. a . Q y =
sin. 2 a Qy.
2g
and of a lateral impulse
The normal impulse is proportional to the sine, the parallel im-
pulse to the square of the sine of angle of incidence, and the lateral
impulse to the sine of double this angle.
If, finally, the oblique surface, which is struck, has no border,
the water can flow away in all directions and the impulse is still
greater; for a is the smallest angle which the fibres of water can
make with the axis; hence every fibre which does not move in the
normal plane exerts a greater pressure than those which do. If we
assume that the angles of deviation of one portion Q₁, which corre-
1014
[§ 503.
GENERAL PRINCIPLES OF MECHANICS.
sponds to the sectors A O B and D O E, Fig. 854, are CO Fa
a, that those of the other portion Q, which cor-
and CO G
=
180°
FIG. 854.
F
K
B
H
responds to the sectors A O E
and B OD, are C 0 K=COH
90°, and that the two por-
tions produce equal parallel im-
pulses, we can put

P
+
C
g
V
c - v
Q₁y sin.² a
Q2 Y,
g
and that the total parallel impulse is
and, since Q, sin.' a = Q, and
Q
=
Q1 + Q2, it follows that
Q, (1 + sin.³ a) = Q,
P
()
v) 2 Q y sin.² a
1 + sin.² a
2 sin.¹ a
C ༡
er..
1 + sin.² a
9
Although this assumption is only approximatively correct, yet
the results of the latest experiments by Bidone agree very well
with it.
REMARK.-Prof. Brock, in his Mechanics, page 614, finds for oblique
impact against a circular surface
P=
(
0 )
-a tang. a
a
g
I . (5)
N = tang. a l, cotg.
2
Q Ÿ r
and
27.
§ 503. Impact of Water in Water.-If a certain quantity
Q of water discharges with a velocity A c = c into a vessel D E,
Fig. 855, which is moving with a velocity Avv, a part only
FIG. 855.
D
B
E
L₁ =
In
Q c²
2 g
Qc,
2
2g
y of its actual energy L. =

y will be expended in producing
and maintaining the eddy A B,
which is due to the loss of velocity
c₁. If we denote by a the angle v A c,
made by the direction of the stream
with that of the motion of the ves-
sel, we have
c₁² = c² + v² — 2 c v cos. a,
§ 504.] THE IMPULSE AND RESISTANCE OF FLUIDS.
1015
and, therefore, the mechanical effect lost in consequence of the
eddy
Q (c² + v² — 2 c v cos. a)
L₁
2 g
As the volume Q of water participates in the motion of the vessel,
its velocity v is the same as that of the latter, and the energy,
which it still possesses, is L,
transmitted to the vessel and
L= L₂- L₁ - L₂
Q v²
2 g
expended in moving it forward, is
y; hence the energy which is
C²
(c² + v²
2 c v cos. a)
2 c v cos. a−2 vª
QY
Q r =
Q Y
2g
2 g
(c cos.a-v) v
QY,
and the force with which the vessel is urged forward in the direc-
tion of its motion by the water which flows into it is
Now the
L
P =
v
༠ ་
c cos. a
g
2)2 r.
discharge per second, which impinges against the
vessel, is Q Fc, F denoting the cross-section of the stream at
its entrance; hence we have
P =
(c cos. a
g
v) c
FY,
and for the case when the vessel is at rest, or when v = 0,
P =
c² cos. a
9
c²
Fy = 2
Fy cos. a= 2 Fhy cos. a,
2 g
c²
in which h denotes the height
due to the velocity.
2 g
The mechanical effect is a maximum for v =
¦c cos. a and it is
c² cos.² a
L₂ = =
Im 12
Q Y
=
iQhy cos.² a.
2 g
If the direction of the stream is the same as that of the motion
of the vessel, a = 0, and we have
(c — v) v
L =
g
L₂ = Qhy.
Qy and
γ
In this case but half the total energy Q hy of the water is utilized
(compare § 500).
§ 504. Experiments with Reaction Wheels.-The best
method of proving the above theory of the impact and reaction of
water is to make use of a reaction wheel A A B, Fig. 856, with a
1016
[§ 504.
GENERAL PRINCIPLES OF MECHANICS.
vertical axis of rotation CD (see the author's "Experimental-Hy-
draulik," § 48, etc.). The water which turns the machine enters
into the receiver A 4 of the wheel nearly tangentially through two
FIG. 8

I
A
A
S
B
F
D
R
K
H
lateral canals E, E, and is discharged through two lateral orifices
F, F'in the ends of the revolving tubes R, R. In order to maintain
the efflux of water constant and the rotating force invariable, the
pipe which conveys the water to the reservoir G is provided with a
cock H; from the reservoir the water is conveyed by the pipe K L
to the chamber E E, into which the canals E, E open. While the
machine is in operation, the cock H must be turned in such a
manner that the surface of the water in the reservoir G shall
always touch the end of the pointer Z
When we wish to determine the reaction of the effluent water,
§ 505.] THE IMPULSE AND RESISTANCE OF FLUIDS . 1017
a thin string S, to one end of which a weight is attached, is passed
over a pulley and then wrapped round the central tube R. The
quantity of water discharged is measured in the reservoir, from
which the water flows into the pipe with the cock H, by observing
the area A of the surface of the water and the distance a which it
sinks during the experiment. If the duration of the observation is
=t, we have the discharge per second
Q
=
A a
t
and if the fall, I.E. the vertical distance between the surface of the
water in the reservoir G and the orifice of discharge of the wheel
=h, the total energy of the water discharged per second is
L = Q h y
A a hy
= t
Now if the machine has raised the weight G a distance s in the
time t, the work really done by the wheel in a second is
L₁
Gs
t
and we can now compare these two values, the second of which is
always the smaller.
§ 505 Theory of the Reaction Wheel.-The total fall h
in such a wheel consists of the fall h, from the surface of the water
to the point E, where the water enters the wheel, and of the fall
hy from the latter point to the orifice, by which the water leaves the
wheel. From h, we calculate, by means of the formula c, 2 gh,
the velocity with which the water enters the wheel, and from he,
according to § 304, by means of the formula.
1
c = 1 2 g h₂+212
vi
2
the velocity with which it quits it, when the velocities of rotation
v₁ and v of the wheel at the points of entrance and exit are known.
Since the direction of this reaction of the water, which acts as the
rotating force, is opposite to that of the velocity of discharge, the
absolute velocity of the water upon leaving the wheel is
and its square
W
2',
-
w² = c² - 2 c v + v² = 2 g h − 2 c v + 2 v² – vi²;
3
1018
[§ 505.
GENERAL PRINCIPLES OF MECHANICS.
hence the energy of the effluent water is
w²
3
L₂ = QY. QY (1₂ -
•
2 g
2
(c — v) v
V₁
9
2 g
The water, which enters the wheel with the relative velocity
w₁ = c₁v₁, loses (according to § 436) by the impact the energy
2
(c₁ V₁)²
L₁ = QY
29
and consequently of the total energy
Q x
(72.
1
C1 VI
9
+
2
'),
Qh y =
Q (h₁ + h₂) Y,
only the portion
L = Qy (h — h− h) + Qy ((© — v ) v + c 2 ) = Q v
v) VI
c—v)v, c₁v₁
+.
+
g
g
g
g
is transmitted to the wheel.
In order to obtain the greatest amount of work from the wheel
0 or v = c and w₁ = 0 or v₁ = c₁, and therefore
we must have w = 0 or v = c and w₁
2
v₁
2 g
vi
2 g
=
h₂ or v₁
√2 g h₂, as well as
h₁ or v₁ = √2 g h₁.
In this case, therefore, h,
maximum effect of the machine is
Im
Q Y
g
h,h and the corresponding
QY. =
2 Q h₁ y = Q h y,
g
I.E., equal to the total energy of the water.
If r, denotes the distance of the point of entrance and r that of
the orifice of exit of the wheel from the axis, we have
V₁
V
r
whence v₁ =
v,
ľ
,
jo
and, in general, the rate of work of the wheel
L
Q
Υ
x (c -
v +
C₁
go
:)
V
9
so that the rotating force, measured at the distance r, is
L
P =
V
Q x
g
+ 1a).
v +
If the arm of the suspended weight or load is a, which in the ap-
paratus represented is very nearly the radius of the central tube B,
we have G a = P r, and, therefore, the weight to be attached and
to be raised during the rotation of the wheel is
r
G = P
a
Q Y
[(cv) r + c₁ r₁],
да
V19
or for cv and c₁ = v₁9
§ 505.]
1019
THE IMPULSE AND RESISTANCE OF FLUIDS.
G
Q Y
да
Q Y
C₁ r₁ =
Vi ri.
ga
If F denote the area of the orifices of efflux and F, that of those
of influx, we have
Fc = F₁ c₁, and therefore
Q
=
Q
F₁
=
C₁
Q
√2gh
and
"
2 g h₁
FN
2
Vi
2 g h ₂ + v²
vi
1
ih, we have
F=
Q
Q
с √2 g h₂ + v²
For vc and v₁ = c₁, in which case h₁ = h₂
Q = F v, and therefore
Q h y
P =
= Fhy;
บ
on the contrary, for v = 0, Q
=
FV2 gh, and therefore
P =
FCY
g
(0+10).
If we allow the water to enter the wheel slowly, we can put
c₁ = 0 and h₁ = 0 and the force of the reaction in the last case
becomes
Р
F c² Y
g
2 F c² Y
2 g
= 2 Fh₂ y = 2 Fhy,
as we found above.
Since in these calculations we neglected the passive resistances,
the experiments with the machine represented do not give the
values for the force found above, but values which are a few per
cent. less. However, the results of experiments carefully made
with such a wheel agree very well with the theory just demonstrated.
When we wish to make use of this machine to test the theory
of the impact of water, we begin by removing the chamber E E so
as to allow the water to enter near the centre without any velocity
of rotation, and we then fasten opposite to the orifices in the re-
volving tubes the plates 0, 0, small vessels, etc., which are sub-
jected to the impact of the water discharged. The rotating force
is then equal to the difference between the reaction within the
wheel and the impulse without it. We find, in accordance with
the theory, that the wheel stands still, when the stream issuing
from it impinges upon a plane plate at right angles to the direction
of the water, or when it flows into a vessel filled with water. If the
stream strikes obliquely against plane-plates or against convex sur-
faces, the wheel moves in the direction of the reaction, and if it is
received by a concave surface, the wheel turns in the direction in
which the water issues from the orifice.
1020
[§ 506.
GENERAL PRINCIPLES OF MECHANICS.
§ 506. Water-meters. More recently water-meters (Fr.
compteurs hydrauliques; Ger. Wassermesser) have been much
used for measuring running water. They are put in motion by
the reaction of the water discharged, and consist essentially of a
reaction wheel or turbine. An ideal representation of the cross-
section of such a wheel is given in Fig. 857. The water to be
measured flows through a tube A into the centre of the wheel B B,
FIG. 857.
I
D
B
E
W
19 1 3 18
П
D
B
B
E
and passes through 4 ca-
nals C B, C B ... to the
exterior circumference,
where it is discharged
into the case D E, from
which it is conveyed
away by a tube EF. The
shaft W of this wheel
carries a pointer Z, or
rather a train of wheel-
work, which indicates the
number of revolutions
of the wheel, and by it
the volume of the water,

which flows through it in any given time; for this volume is pro-
portional to the number of revolutions. If h denotes the height of
a column of water which measures the loss of pressure of the water
in passing through the wheel, Q the discharge per second, c the ve-
locity of efflux, and v the velocity of the wheel in the opposite
direction, we have c² = 2 g h, and the rate of work of the
v³
wheel
(c — v)
L:
vQy (see § 505).
9
If R is the resistance of the wheel, in consequence of the fric-
tion on the bearings, etc., we can put L = R v, and from it we
obtain the formula
R =
Q Y,
g
if F denotes the sum of the areas of all the orifices of efflux, so
Q
F
c= we can put
or,
that Q
Fc or c
Q
R =
(2 - v ) 2 2
v
ข
g
Q Y
from which we obtain
g R
Q Y
Q
F

§ 506.] THE IMPULSE AND RESISTANCE OF FLUIDS.
Q
1021
F'
or
If R were null, or at least very small, we could put v =
assume the velocity v of rotation to be proportional to the discharge
Q, which indeed it should be. If, on the contrary, R= v, or if
the resistance of the wheel increase with v, we will have
v=
Q
F (1 + 1/2)
QY
v+
4gv
QY
=
Q
F
or
approximatively = (1-2).
F
If, then, the resistance R of the wheel is not very small, the
velocity of rotation of the wheel is less than when R is null or
negligible, and the instrument indicates too small a discharge.
If we put v = 0, we obtain for a discharge Q, the correspond-
ing velocity of efflux
Co=
g R
Q. Y
and we can then put, approximatively at least,
v = cc, and
π Fru
Q = F(v + c):
=
30
+ Q₂ = µu + Qor
r denoting the radius of the wheel, u the number of its rotations
and a coefficient to be determined by experiment.
Within the last few years Siemens's water-meter has come into
very general use; its principal parts are represented in cross-
section in Fig. 858. The water which enters from A passes
FIG. 858
ww
B
B
k
C
E
E
1022
[$ 506.
GENERAL PRINCIPLES OF MECHANICS.
through the pipe B B into the wheel CC and is carried by the
revolving tube D D into the case E E, from which it is carried off
by the pipe F. The shaft W of the wheel passes upwards through
a stuffing-box and sets a train of wheel-work in motion by means
of an endless screw fastened to its end. The wings k, k upon the
wheel assist in regulating its motion of rotation by the resistance
which they experience in moving in the water.
The reaction wheel can be constructed in such a manner that
every time it makes a revolution it will allow a certain quantity of
water to pass through. To accomplish this object, the wheel B A B,
FIG. 859.
h
A
Fig. 859, is partially immersed in water,
so that, when turning, the spiral tubes
are alternately filled with air and water.
Here also the water is conducted by a
pipe into the centre of the wheel, and
from thence by spiral pipes into the
free space of the case E F, from which
it flows away through the pipe F. The
surface of the water in the interior of
the wheel is at a distance h above that
of the water in the case; hence, if the
wheel turns in the direction indicated
by the arrow, as soon as the orifice D
arrives at the level of the water in the
interior, the water begins to discharge, and in so doing reacts with
a certain force P, by which the rotation of the wheel is main-
tained. If V is the volume of the water contained in one of the
spiral pipes, and n the number of these canals, the discharge per
second, when the number of rotations per minute of the volume of
n u V

B
F
the water is u, is ↓
60
REMARK.-An account of Siemens' water-meter is given in the "Zeit-
schrift des Vereines deutscher Ingenieure," Vol. I, 1857, in which Jopling's
water-meter (in which the water is gauged) is also described. See also the
paper: "Siemens and Adamson's Patent Water Meter." A very peculiarly
constructed water-meter of the nature of a reaction wheel is described in
the "Génie industrielle," Tome XXI, No. 126, 1861, under the name:
Compteur hydraulique pour la mesure d'écoulement des liquides par
(
· Guyet." Two water-meters are described in the English work "Hydrau-
lia," by W. Matthews. A compteur hydraulique used at the railroad station
at Chartres is described in the "Bulletin de la Société d'encouragement,"
51 year (1852) Uhler's apparatus for measuring fluids is treated of in
§ 507.] THE IMPULSE AND RESISTANCE OF FLUIDS.
1023
Dingler's Journal, Vol. 161. A description of an apparatus for measuring
the quantity of spirit made in distilleries is contained in the "Mittheilun
gen des Gewerbevereines for Hannover," new series, 1861.
For a description of several kinds of water-meters, see "The Transac-
tions of the Institution of Mechanical Engineers," 1856 (Tr.).
§ 507. Gas-meters.-The so-called wet gas-meters (Fr. comp-
teurs à gaz; Ger. Gasmesser or Gasuhren) are, like certain water-
meters, small wheels with spiral canals, which are more than one-
half immersed in water and are put in motion by the reaction of
the gas passing through them; each spiral canal transfers a certair.
volume of gas from the inside to the outside. The essential parts
of such a gas-meter are shown in the two sections of Fig. 860.
B
F
G
A
FIG. 860.
G
H
G
B
G
The gas, which arrives,
enters by a bent pipe A
into the interior of the
measuring wheel B B,
in which it depresses
the surface of the water
a certain distance h,
which depends upon
the tension of the gas
passing through the in-
strument. From this

central chamber it enters successively the spiral canals, fills them
almost entirely and, finally, passes out through the orifices at the
circumference into the case G G, from which it is conducted by a
pipe H to the point, where it is to be used. As we wish every
spiral canal of the measuring wheel to carry over a certain definite
quantity of gas at each revolution, we must so arrange the appa-
ratus that at least one of the orifices of a canal shall always be
under water; for in that case, when the gas is filling the canal,
there is no efflux, and during the efflux no gas can enter it. The
volume of gas V, passed by one spiral canal, is consequently a defi-
nite one, and we can, therefore, put the discharge per minute
Q
=
n u V
60
when the wheel makes n revolutions per minute. If we denote the
height of the harometer in the gas leaving the machine by b, that
in the gas entering it is b + h, and, therefore, according to Ma-
riotte's law, the quantity of air in one spiral canal, measured at the
pressure of the gas after it has left the measuring wheel, is
1024
[§ 507.
GENERAL PRINCIPLES OF MECHANICS.
V₁
b
(b + h) v
consequently the quantity of gas, which passes from the wheel into
the exterior case when the outlet of one of the spiral canals rises
from the water, is
–
V₁ - V = 1/1/
h
V.
b
When this quantity streams into the case the mechanical effect
set free is
h
'b
(0 + +
+ h)
A
=
Vpl (
b
(see § 388), and since is small, we can put
b
h
1 ( ² + 1 ) = 1 (1 + 1 ) = 3/3
(+)
:
hence, if the heaviness of the substance, with which the manometer
is filled, is y, we have p = (b + h) y=by, and therefore AVhy.
One portion of this mechanical effect is expended in turning
the wheel, and the rest in producing an eddy. The first portion
is determined by the expression
A₁
(c — v) v h
g
b
√ Y 19
in which h denotes the mean height of the manometer, c the mean
velocity of efflux, v the velocity of the wheel at its circumference
and y, the heaviness of the gas discharged. If R is the resistance
of the wheel, reduced to its circumference, and r its radius, we
have the required mechanical effect
R.
A₁
(c — v) v h
g
b
2 π r
and therefore we can put
N
60 v
2 π r
VY!
R, or since 2 π r =
N
VY 1
60 R
;
N U
с
g
21
h
b
hence it follows that the velocity of rotation, corresponding to the
distance h between the two surfaces of water, is
g b 60 R
ข
V = c
h V Y ₁
N U
and that the number of revolutions of the meter per minute is
30
u =
π r
60 g b R
n u V h Y i
Approximatively we have c =
N
hy
√2 g
when y
denotes the
"
Y₁
$ 508.] THE IMPULSE AND RESISTANCE OF FLUIDS.
1025
heaviness of the substance with which the manometer is filled.
The volume of gas passing per minute is
Q
n u
60
V,
and it is proportional to the number of revolutions u.
§ 508. Newer Gas-meters. Instead of placing the spiral
canals of a gas-meter in a plane perpendicular to the axis, we can
wind them round it like the thread of a screw. The action of
such a gas-meter is shown by the two sections I and II, Fig. 861, in
which D D represents the surface of the water at the front and E E
E
B
D
C
A
FIG. 861.

、་་
E B
ΤΟ
B
B
B=3
4

P
B
II
2
B
4
B
E
D
that at the back of the measuring wheel, which is a horizontal
drum. The orifice A of the spiral canal A O B opens into the
chamber, which is in front of the drum, and receives the gas, which
is arriving; the orifice B, on the contrary, delivers the gas into
the chamber at the back of the drum, from which it is carried off
by a pipe. In Fig. 861, I, the different positions of a spiral canal,
viewed from in front of the wheel, are represented. Fig. 861, II,
on the contrary, represents the various positions of the canal as
seen from the rear of the wheel. In consequence of the rotation
of the wheel, in the direction indicated by the arrow, around the
horizontal axis C, the inlet orifice A in (I, 1) is just emerging from
the water in front, while the outlet B is just entering the water in
the rear, in (1, 2) and (I, 3) the arcs A O, A O of gas have entered
through the orifice A, and in (I, 4) the orifice has re-entered the
water, so that after a certain quantity V has been received into the
canal, the entry of the gas is cut off. Shortly afterwards the orifice

1026
GENERAL PRINCIPLES OF MECHANICS.
[S 508.
B rises, as is represented in (II, 1), from the water in the rear of
the drum and the discharge of the gas, which had previously been
taken in, begins, and it is in full operation in the positions (II, 2)
and (II, 3). When a new revolution begins, B re-enters the
water in the rear of the drum, as is represented in (II, 4), and the
gas again begins to fill the canal. During half a revolution of the
spiral canal A O B, an arc of gas A O (1, 4), which is at the greater
tension b+h, enters the former and during the second half of
the same it is transferred to the space beyond the wheel, where the
pressure is less. In passing from the greater pressure to the less,
the mechanical effect A = V hy is set free; a portion of this is
expended in moving the wheel, as was shown in the foregoing
paragraph. The general arrangement and action of such a gas-
meter can be better understood from the ideal representation in
Fig. 862. The gas is first introduced by means of a bent tube A
into a chamber B B, which communicates in the middle around
the axis of rotation C with the water in the case E F G, but upon
the exterior circumference, where the spiral tubes enter it, it is air-
tight. The drawing shows the spiral canal H K to be receiving
gas from B B and the canal L M, which a short time before had
received a certain volume of gas, to be discharging it at M into the
upper space in the case E F G, from which it is carried away by
the pipe F. By this arrangement of the meter the gas in the first
chamber is cut off entirely by the water from that in the rear
chamber, and, therefore, the packing, which causes great loss of
force, is rendered unnecessary. The other end D of the axis CD
of the wheel has a couple of turns of a screw cut upon it, by means
of which the train of wheels of the counting apparatus is set in
motion.
FIG. 862.
G
F
B
M
H
Ba
FIG. 863.
Cg
A
B
E
K
D
C2
C
A₂

§ 508.]
THE IMPULSE AND RESISTANCE OF FLUIDS.
1027
Crossley's gas-meters, which have come into very general usc,
are constructed according to the principles explained above; but
their spiral canals are not tube-shaped, but real chambers or cells
with spiral partitions and with triangular inlet and outlet orifices,
which are made by bending out the end surfaces. Fig. 863 is a
perspective view of such a wheel with the cover removed; it con-
sists of 4 pieces of sheet iron like that represented in Fig. 864.
A1, 4, A3, A, are the inlet orifices, B₁, B.... the outlet orifices
and C1, C2, C3 ... the partitions of the measuring wheel which
turns around the axis D D. Fig. 865 is an elevation of the gas-
meter with the exterior drum or case; we observe at K the bent
tube, which conducts the gas into the chamber, and at Z the pipe,
W
F
f
G
B
FIG. 864.
FIG. 865.
O
K
C
W
A
which carries off the gas
from the upper space A A
of the case of the meter.
The gas does not flow di-
rectly into K, but the pipe
E carries it first into a cham-
ber F, from which it passes
through the conical valve i
into the chamber G, where
it enters the upper part of
the vertical pipe H, through
which it is conducted into
the bent tube K. The sur-
face of the water in the
chamber G reaches exactly
to the top of the pipe H,
through which the super-
fluous water overflows into
a reservoir L. In order, on
the other hand, to prevent
the water from sinking too
low, a float is placed in the
chamber, which, when it
sinks, carries the valve i with
it and closes the opening,
when the float has sunk a
certain distance. The dis-
charge of gas then ceases en-
tirely, and we are thus noti-

1028
GENERAL PRINCIPLES OF MECHANICS.
[$ 508.
fied that it is necessary to fill the meter with water through an
orifice M, that opens into a chamber N, which communicates, at
the bottom only, with the water space.
Fig. 866 is transverse elevation of the front of such a meter, in
which are to be seen not only the chamber N with the orifice M,
but also the clockwork of the counting apparatus, which is set in
motion by an endless screw upon the axle of the drum and a ver-
tical shaft with a cog-wheel upon it.
An important resistance to the motion of Crosley's gas-meter is
that occasioned by the entry and exit of the water through the
narrow triangular orifices. We can calculate from the area F of
an inlet or outlet orifice and from the discharge per second, which
can be put equal to the volume Q of the gas, the velocity of exit
FIG. 866.
F
G
H
W
S
M
N
P
Q
F
and entrance v₁ = and consequently the corresponding loss of
mechanical effect per second
L₁ = 2
" QY = (2) QY
29
g
REMARK.-Particulars upon the subject of gas-meters can be found in
Schilling's "Handbuch der Steinkohlengasbeleuchtung," and Heeren's
article "die Einrichtung der Gasuhren " in the "Mittheilungen des Ge-
werbevereins für das K. Hannover," year 1859. A new gas-meter by Han-
sen is described in the "Journal der Gasbeleuchtung," 1861.
§ 509.]
1029
THE IMPULSE AND RESISTANCE OF FLUIDS.
§ 509. Action of Unlimited Fluids. If a body has a mo-
tion of translation in an unlimited fluid, or if a body is placed in a
moving fluid, it is subjected to a pressure, which is dependent upon
the form and size of the body as well as upon the density of the
fluid and the velocity of one or other of the masses; in the former
case it is called the resistance and in the latter the impulse of the
fluid. This hydraulic pressure is principally due to the inertia of
the water, whose condition of motion is changed when it comes
into contact with a rigid body, and also to the force of cohesion of
the molecules of water, which are partially separated from and
moved upon each other.
If a body AC, Fig. 867, is moved in still water, it pushes a
certain quantity of water, the pressure of which is increased, before
it. As the body progresses the quantity of water on one side is
increased, while upon the other it is constantly flowing away, and
the particles lying immediately contiguous to the surface A B
1
FIG. 867.
D
FIG. 868.


BC
D
B
C
assume a motion in the direction of this surface. If a stream of
water encounters an obstacle 4 C, Fig. 868, which is at rest, the
pressure of the water in front of it is increased, the molecules of
water are diverted from their original direction and move along
the front surface A B. When the particles of water have reached
the edges of the front surface, they turn and follow the sides of
the body, until they arrive at the back surface, where they do not
immediately reunite, but assume first an eddying motion. We see
that the general relations of the motion of the molecules, which
surround the body, are the same for the impulse of water as for the
resistance to a body moving in the water; but there is a difference.
in the eddies, when the body is short; for in the latter case the
eddies occupy less space than in the former. The velocity of the
molecules of water increases gradually from the centre of the front.
surface to the edges, where a contraction generally takes place and
where the velocity is a maximum, it decreases as the water passes
along the sides and becomes a minimum when the water arrives
at the back surface and begins its eddying motion.
1030
[$ 510.
GENERAL PRINCIPLES OF MECHANICS.
§ 510. Theory of Impulse and Resistance. The normal
pressure of still or moving water upon a body moved or immersed
in it is very different at different points of the body. This pres-
sure is a maximum at the centre of the front surface and a mini-
mum in the centre of the rear surface and at the beginning of the
sides; for at the first point the water flows towards the body, and
at the latter points it flows away from it. If the body is, as we
will suppose in what follows, symmetrical in reference to the
direction of motion, the pressures at right angles to this direction
balance each other, and we must, therefore, consider only the
pressures in the direction of the motion. But since the pressure
upon the rear surface acts in an opposite direction to those upon
the front surface, it follows that the resulting impulse or resistance
of the water is equal to the difference between the pressures upon the
front and back surfaces.
Although we cannot determine a priori the intensity of this
pressure, yet, as the circumstances are very similar to those of the
impact of an isolated stream, we can at least assume that the gen-
eral law of the impact of an unlimited stream does not differ very
much from that of an isolated stream. If Fis the area of surface
which an unbounded stream, whose heaviness is y and whose velo-
city is v, encounters, we can put the corresponding impulse or hy-
draulic pressure
v2
P = 5
FY,
2 g
in which denotes an empirical number dependent upon the shape
of the surface. This formula can be applied not only to the front,
but also to the rear surface. But in the latter case, where the
water tends to separate itself from the body, the expression becomes
negative. Now if Fhy is the hydrostatic pressure (§ 690) against
the front and against the back surfaces of a body, the total pressure
against the front surface is
and that against the back surface is
v³
P₁ = Fhy +5₁.
FY,
·
2 g
v²
P = P₁ - P₁ = (51 + 52) •
hence the resulting impulse or resistance of the water
v3
+52) Fy = 5.
P₁ = Fhy - 52 •
Fy;
2 g
is
FY,
2
J
29
when we put ₁ + 52 = 5.
This general formula for the impulse and resistance of an un-
limited stream is also applicable to the impulse of wind and to the
§ 511.] THE IMPULSE AND RESISTANCE OF FLUIDS
1031
of the
resistance of the air. Here, however, besides the differe
aerodynamic pressures upon the front and rear surfaces, a difference
in the aërostatic pressure also exists, which is due to the fact that
the air at the front surface has a greater heaviness (y), in conse-
quence of its greater tension, than that at the rear surface. For
this reason, at least when the velocities are great, as in the case of
musket and cannon balls, the coefficient of resistance of the air is
greater than that of water.
REMARK.-A peculiar phenomenon attends the impulse and resistance
of an unlimited medium (water or air), viz., a certain quantity of water or
air attaches itself to the body, the influence of which is shown by the vari-
able motion of the body, which, E.G., is very evident in the oscillations of
a pendulum, The quantity of air or water which attaches itself to a sphere
is 0,6 the volume of the sphere. For a prismatic body, moving in the di-
rection of its axis, the ratio of these volumes is
= 0,13 + 0,705
F
}
in which denotes the length and F the cross-section of the body. This
ratio, which was first determined by du Buat, has been fully confirmed by
the later experiments of Bessel, Sabine and Bailly.
§ 511. Impulse and Resistance against Surfaces.-The
coefficient of resistance, or the number by which the height
22
2g
due to the velocity must be multiplied, in order to obtain the height
of the column of water which measures the hydraulic pressure, is
very different for bodies of different form; it is determined approx-
imatively only for plates, which are placed at right angles to the
stream. According to du Buat's experiments and those of Thi-
bault, we can put for the impulse of water and air against a plane
surface at rest = 1,86, while, on the contrary, we can assume with
less certainty for the resistance of the air and water to a plane sur-
face in motion 1,25. In both cases about two-thirds of the
action is upon the front and about one-third upon the rear surface.
The values, found for the resistance offered by the air to a body re-
volving in a circle by Borda, Hutton, and Thibault, vary much
from each other. The latter found with a rotating plane surface,
the area of which was 0,1 square meter, the resistance
P = 0,108 Fv, whence
29
<= 0,108.
=
0,108.
Y
19,63
1,25
1,70.
This coefficient is, according to these experiments, almost con-
1032
[§ 511.
GENERAL PRINCIPLES OF MECHANICS.
stant, when the angle a formed by the surface with the direction.
of the motion is not less than 45°. When the angle is less than
45°, the coefficient diminishes with this angle of impact, and for
a = 10°, is only 0,53. According to the researches of Didion,
etc., we have for the resistance of rotating plane surfaces, whose
areas are 0,2 . 0,2 = 0,04 square meters,
5
=
5 = (0,1002 + 0,0434 v−²)
ら
​in which v must be given in meters.
2 g
1,573 + 0,681 v²,
γ
For a plane surface, whose area was one square meter, Didion
found, when the motion was vertical, the coefficient of resistance
2 g
Y
5=(0,084 +0,036 v²). = 1,318 + 0,565 v2,
while Thibault, on the contrary, found for such surfaces, when
their area was 0,1 to 0,2 square meters,
<= (0,1188 +0,036 v²).
2 g
=
1,865 + 0,565 v−².
γ
The foregoing formulas hold good only when the motion of the
surface is uniform; if the motion is variable, they require an ad-
dition. If the velocity of a body which is moving in a resisting
medium changes, the quantity of the fluid moved by the body or
carried along with it varies; the resistance is, therefore, dependent
upon the acceleration p. According to the experiments of Didion,
etc., with a surface whose area was 1 square meter, and with one
whose area was square meter, which were moved in a vertical line,
the resistance was
P = (0,084 v² + 0,036 + 0,164 p) F; hence
<=
· [0,084 + (0,036 + 0,164 p)v−²].
1,318 + (0,565 + 2,574)v−².
2 g
γ
We must also remember that for variable motion the mean
square of the velocity is different from the square of the mean
velocity.
The impulse and resistance of an unlimited medium is increased
when the surfaces are hollowed out or provided with borders; but
we have as yet no general data concerning the subject.
For a parachute, whose cross-section was 1,2 square meters and
whose mean diameter was 1,27 meters and whose depth was 0,430
meter, Didion, etc., found for an accelerated motion, during which
the hollow surface was in front,
P = (0,163 v² + 0,070 +0,142 p) F, whence
5 = 2,559 + (1,099 + 2,229 p) v˜³.
§ 512.1
1033
THE IMPULSE AND RESISTANCE OF FLUIDS.
§ 512. Impulse and Resistance against Bodies.-The im-
pulse and resistance of water against prismatical bodies, whose axis.
coincides with the direction of motion, decrease when the lengths
of the bodies increase. According to the experiments of du Buat
and Duchemin, the impulse upon the front surface is constant, and
the action upon the rear surface alone is variable. The coefficient
₁ = 1,186 corresponds to the former; but when the relative
lengths are
= 0, 1, 2, 3,
SI
the total action is
VF
51,86; 1,47; 1,35; 1,33.
If the ratio between the length and the mean width √F be-
comes greater, the coefficient again increases in consequence of
the friction of the water upon the sides of the body. The reverse
is true of the resistance of the water. In this case, according to
du Buat, the constant action against the front surface is ₁ = 1, and
the total action for
NF
5
0, 1, 2, 3, is
= 1,25; 1,28; 1,31; 1,33;
so that for a prism three times as long as wide the impulse of the
water is the same as the resistance.
=
The experiments of Newton, Borda, Hutton, Vince, Désaguil-
liers and others with round and angular bodies leave much uncer-
tain and undetermined. It appears that for moderate velocities the
coefficient of resistance of spheres can be put 0,5 to 0,6. But
when the velocities are greater and the motion takes place in the
air, we can put, according to Robins and Hutton, for the velocities
v = 1, 5, 25, 100, 200, 300, 400, 500, 600 meters,
5 = 0,59; 0,63; 0,67; 0,71; 0,77; 0,88; 0,99; 1,04; 1,01.
Duchemin and Piobert have given particular formulas for the
increase of this coefficient of resistance. According to Piobert the
resistance to a musket ball in the air is
P = 0,029 (1 + 0,0023 v) F v² kilograms, whence
5 = 0,451 (1 + 0,0023 v).
For the impulse of water against a ball, Eytelwein found
<= 0,7886,
while, on the contrary, according to the experiments of Piobert,
etc., made with cannon balls 0,10 to 0,22 meters in diameter, the
resistance to the balls in water is
P = 23,8 Fv kilograms; hence we can put
5 = 0,467.
5=
1034
[§ 512.
GENERAL PRINCIPLES OF MECHANICS.
The coefficients of resistance for bodies partially immersed are
different from those for bodies entirely surrounded by water. For
a floating prismatic body five to six times as long as wide and mov-
ing in the direction of the axis, should be put equal to 1,10. If
the body is sharpened in front by two vertical planes like A B C,
Fig. 869, increases with the angle A CA ẞ, and we have
5

for B
5:
= 180° 156° 132° 108° 84° 60° 36° 12°
1,10 1,06 0,93 0,84 0,59 0,48 0,45 0,44
| | |
If, on the contrary, the rear portion A C B, Fig. 870, is sharp-
ened, and if the angle B C B
FIG. 869.
ẞ, we have
FIG. 870.

P
B
for ẞ =
180°
138°
96°
48°
24°
Š
1,10
1,03
0,98
0,95
0,92
When both front and rear portions of the floating body are
sharpened, becomes still smaller. For river steamboats,
to 0,20 and for large ocean steamers, Š
=
0,05 to 0,10.
= 0,12
REMARK. This subject is treated at length by Poncelet in his "Intro-
duction" cited above, and by Duchemin and Thibault in their "Recherches
expérimentales, etc." The subject of the resistance to floating bodies, par-
ticularly ships, and also that of the impulse of the wind against wheels,
will be treated in the second and third volumes.
EXAMPLE.-If, according to Borda, we put the resistance and impulse
at right angles to the axis of a cylinder that against a parallelopipedon,
which has the same dimensions as it, we have the coefficient of resistance
5 = }} . 1,28
and the impulse against the same
=
0,64
==
= .1,47
=
0,735.
If we apply these values to the human body, the area of the cross-
section of which is 7 square feet, we find the resistance and impulse of the
air against it
.
P = 0,64 . 0,0155 . 7 . 0,086 v² = 0,00597 v²
and
P = 0,735 . 0,0155 . 7 . 0,086 v²
0,00686 v².
§ 513.]
1035
THE IMPULSE AND RESISTANCE OF FLUIDS.
For a velocity of 5 feet, the resistance of the air is, therefore, only
0,00597 . 25 = 0,1492 pounds, and the corresponding work done per
second is = 5 . 0,1492
= 5 . 0,1492 = 0,746 foot-pounds; for a velocity of 10 feet, the
resistance is 4 times and the expenditure of mechanical effect 8 times as
great, and for a velocity of 15 feet the resistance is 9 times and the work done
27 times greater. If a man moves with the velocity 5 feet against a wind,
whose velocity is 50 feet, he has to overcome a resistance 0,00686. 552
20,75 pounds, which corresponds to a velocity of 50 + 5 55 feet, and
to perform an amount of work equal to 20,75 . 5 = 103,75 foot-pounds.
§ 513. Motion in Resisting Media. The laws of the mo-
tion of a body in a resisting medium are not very simple; for the
force in this case is variable, increasing with the square of the
velocity. From the force P, which is drawing the body onwards,
22
and from the resistance P₁ = 5. Fy, offered by the medium,
we obtain the motive force
2 g
P₁ = P − P₁ = P − 5.
v³
P.
-
•
FY,
2 g
G
but since the mass of the body is M =
its acceleration is
9
v²
P-5
FY
P
M
p = j = (P-51 Fy): M =
༧°
2 g
9,
G
g
or if we denote
SFY
2 g P
1
2 g P
by
or put
=w, we have
was
SFY
p =
V
W
The maximum velocity which the body can assume is
P
g.
V = W =
2 g P₁
1
SFY
If the motive force P is constant, the motion approaches grad-
ually a uniform one; for the acceleration becomes smaller and
smaller as v increases.
Now the velocity v increases, when the acceleration is p, in an
element of time 7 a quantity = p т, hence we can put
K =
T=
P
[1-(9)] 7
G
P
K
9 7, or inversely
9 [1 − (~~)]
พ
1036
[§ 513.
GENERAL PRINCIPLES OF MECHANICS.
In order to find the time, corresponding to a given variation of
velocity, let us divide the difference vn v, between the initial and
the final velocities in n parts and put such a part
Vn
N
༠
K
and then calculate from it the velocities
21 = V。 + K, V₂ = v。 + 2 K, V3
0
v。 + 3 K, etc.,
substituting these values in Simpson's formula, we obtain the
required time, when we assume four divisions,
4
G
Vn
Vo
1
1) t =
+
P
12 g
1
(1)
2
1
พ
2
4
1
+
+
V 2
1
1
го
V4
1 - (1)
2
The space described in an element 7 of time (§ 19) is
σ = v T, or since we can put т =
К
+
2
23
W
V K
σ =
σ
Ρ
or
V K
G
1
(
Pg
го
By employing Simpson's rule we find the space described, while
the velocity changes from ", to v, to be
G
In
2) s
P
12 J
+
2 V2
2
2
1
го
1
W
473
4 21
20
+
2
1
20
บ.
4
+
V3
4
1
1
201
20
2
The calculation is of course more accurate when we make 6, 8,
or more divisions. This formula allows us to take into account
the variability of the coefficient of resistance, which is necessary for
high velocities. For the free fall of a body in air or water P = G,
the apparent weight of the body, and for motion in a horizontal
plane P = 0, or more correctly, equal to the friction f G. Since
this is a resistance, it must be introduced as a negative quantity in
the calculation; hence we must put
$513.] THE IMPULSE AND RESISTANCE OF FLUIDS.
1037
2 P
P。 =
(P + P₁) and
p
- [ 1 + (-)² ] /
(4)] G
W
g.
Since in this case there can be no question of an increase, but
only of a decrease of velocity, we must substitute ",", in the
above formulas instead of vn
0
When a body is impelled by a force, such as its own weight, the
motion approaches more and more to a uniform one, and after a
certain time it may be considered as such, although it never will be
really so. The acceleration p becomes = 0, when 5 Fy=
29
P.
or when
v = √
2 g Po
5 F Y
W.
A body falling freely in air approaches more and more to this
result without ever attaining it.
=
=
EXAMPLE.-Piobert, Morin and Didion found for a parachute whose
depth was 0,31 times the diameter of the opening, the coefficient of resist-
ance 1,94. 1,37 2,66. From what height can a man weighing 150
pounds descend with such a parachute weighing 10 pounds and with a
cross-section of 60 feet, without assuming a greater velocity than that he
attains when he jumps down 10 feet? The latter velocity is v =
25,377 feet, the force P G = 150 + 10 160 lbs., the surface F = 60
feet, the heaviness = 0,0807 pounds and the coefficient of resistance
<= 2,66, hence
1
202
2,66.60. 0,0807 1,33.3. 0,0807
0,00125
8,025 √10
and
64,4. 160
64,4 . 4
202
0,00125. 25,3773 = 0,805.
If we assume six divisions, we obtain
2
1
w²
V
and
v2
1
202
0,977639; 0,91055; 0,79875; 0,64222; 0,44097; 0,195,
0; 4,326; 9,290; 15,886; 26,343; 47,958, and 130,138;
hence, according to Simpson's rule, we have the mean value
=
= (1.0 + 4. 4,326 +2.9,290 + 4. 15,886 + 2 . 26,343
474,084
+ 4 . 47,958 + 1. 130,138) : (3.6)
18
and the required space, through which he can fall, is
26,338;
Vn
ว
v
8
times the mean of
g
v2
25,377 — 0
32,2
26,338 = 20,76 ft
1
202
1038
[§ 514.
GENERAL PRINCIPLES OF MECHANICS.
The corresponding duration of the fall, since the mean value of
1
22
1
202
is
(1.0 + 4. 1,023 + 2. 1,098 + 4: 1,252
+ 2. 1,557 + 4. 2,268 + 1 . 5,128) : 18 = 1,589, is
t
25,377
32,2
•
1,589
= 1,25 seconds.
REMARK.-If the coefficient of resistance is constant, we obtain by the
aid of the Calculus for the case of a body falling freely
ept +1
eut
ен
lept 1
v =
20
ем
and
8=1
((eμt + 1)²)
4 ept
20²
2 g
= 1
1 (202
202
22
in which
μ =
√29.5
FY
G'
•
203
· 29'
g
-
t
+
1) √ 29
G
g
•
ζ Εγ
G
((ept + 1)²)
ен
(lent
4 eμ t
ем
and e denotes the base of the Naperian system of logarithms and the Na-
perian logarithm.
§ 514. Projectiles.-We have already studied the motion of
projectiles in vacuo and found in § 39 the path or trajectory to be
a parabola. We can now investigate this
motion in a resisting medium, E.G., the
motion of a body projected in the air.
FIG. 871.
R
T

The path of a body projected through
air is certainly not a parabola, as is the case
when it is projected in vacuo, but an un-
symmetrical curve; the portion of the tra-
jectory, where the body is rising, is not so
steep as that where it is falling, as can be
During the instant 7 the body, which is
rising with a velocity v in the direction 4 T, Fig. 871, describes,
in consequence of its inertia, the space
'AZ
M
-X
seen from what follows.
A O = S = VT,
and, in consequence of gravity, the vertical space
O P = h
от
2
I T³
and the first space is diminished by the resistance
C²
Fy of
2 J
the air an amount, which can be determined by the expression
$ 514.]
1039
THE IMPULSE AND RESISTANCE OF FLUIDS.
5
FY
2 g
g
Fg
2,3 73
O Q
5
G
2
2 G
2
FY
If we put
و سلامت
μ, we have more simply
2 G
O Q
212 +2
2
T
The fourth corner R of the parallelogram O P Q R, constructed
with O P and O Q, gives the position which the body occupies at
the end of the time 7, while P is the place which the body would
have occupied at that moment, if the air offered no resistance.
The path A R of the projectile passes, therefore, below the para-
bola, which the body would have described in vacuo.
In like manner we have for a body descending with the initial
velocity v in the direction A T, Fig. 872, the spaces described si-
multaneously in the time
A OUT,
0 P = g and
2
0 Q = μ v² +
2'
and from the above we obtain again the position R occupied by the
body at the end of this time, and the position P which it would
have occupied, if its motion had taken place in vacuo. The path A R
described in this case passes also below the parabolic path 4 P,
which the body would have followed, if the air opposed no resistance.
If the angle of inclination, at which a body rises with the initial
FIG. 872.
FIG. 873.
M
A
0
Y


-8
T
W
W
1
0
N
T
T₁
U1
-X
M
u
X1
velocity v from A, is TAX= a, Fig. 873, the initial co-ordinates.
or velocities in the direction of the axes are
1040
[§ 514.
GENERAL PRINCIPLES OF MECHANICS.
u = v cos. a and
w = v sin. a,
and we have for the position R of the moving body, after an instant
T, the abscissa
A M = x = A Q cos. a
= (1 - 197)
мот
2
= ?! T
T
U v² T
2
cos. a
and the ordinate
y =
VT cos. α,
Q R = (1 — " ₂ T)
V
-
2
-) UT
V T sin, a
g
2
M R = A Q sin. a - Q R =
The velocity in the direction
Ru₁ = u₁ = v cos. a
of the abscissa is
v² ↑ cos. α = (1
μυτ) υ cos. α,
and that in the direction of the ordinate is
Ꭱ
Rw₁ = w₁ = v sin, a μ v² 7 sin. a
Т
g T = (1 — µ v 7) v sin. a—g T.
From the two velocities we obtain the angle of inclination
T₁ R X₁ = a₁ of the path at R by means of the formula
tang. a₁ =
W1
U1
tang. a
I T
(1 — µ v т) v cos. a’
and the velocity in the direction of the curve is
2
2
R v₁ = v₁ = √u₁²+w,² = √(1 −µ v 7 )² v² — 2 ( 1 − µ v 7) v g 7 sin. a + g² 7³.
By repeated application of this formula, we can find the course
of the whole trajectory of the projectile. If, E.G., we substitute in
the above formulas for x and y, instead of a and v the values for a,
and v₁ obtained from the last equation, we obtain the co-ordinates
x, and y₁ of a new point referred to R, etc.
2
4
EXAMPLE.-A massive cast-iron cannon-ball, whose diameter is 2 r =
inches, is projected at an angle of elevation ɑ = 25° with a velocity v =
1000 feet; required the position of the same after 'o, fo, fo, of a second, etc.
Since the weight of a cubic foot of air is 0,080728 pounds and that of a
cubic foot of cast iron is 444 pounds, we have
Fy
2 G
π p² Y
5
π 2" 7'1
3.6
•
0,080728
444
ry 1
0,000409094 5,
and, therefore, for v = 1000 feet, for which ? = 0,9 (sce § 512), we have
µ = 0,0003682.
If we take T = 0,1 seconds, we obtain
X = (1 — 0,0003682. 1000. 0,05) 100 cos. 25°
=
0,98159 . 90,63 = 88,96 feet,
0,01
2
0,16-41,82 feet,
y = 0,98159. 100 sin. 25° - 32,2. =0,98159.42,26
and
§ 514.]
1041
THE IMPULSE AND RESISTANCE OF FLUIDS
tang. a1
=
32,2.0,1
tang. 25°
(1 0,03682). 906,3
0,46631 – 0,00369 = 0,46262;
hence the angle of elevation is
3,22
= 0,46631-
0,96318. 906,3
a = 24° 50′,
and the velocity in the curve is
21
v (0,96318.1000)"
-
2. 0,96318.1000. 32,2 . 0,04226 + (3,22)²
√927716 2621+ 10 = √925105 =
―
961,82 feet.
If we again take 7 = 1 second, we have, since for v = 962 feet, 5 =
0,88, and consequently μ = 0,88. 0,000409094 0,00036,
(10,00036.961,8. 0,05). 96,18 cos. 24° 50'
0,9827.96,18. 0,9075 85,77 feet,
0,9827. 96,18 sin. 24° 50′ — 0,161
39,53 feet,
3,22
21
Y₁
and
tang. a 2
tang. 24° 50′
0,96537. 961,8 cos. 24° 50'
0,46277
0,00382
=
0,45895,
whence
v = √(0,96537 . 961,8)*
c = 24° 39′ and
2.0,96537.961,8. 32,2. 0,04200 + (3,22)*
= √862099 - 2511 + 10 √859598 = 927,14 feet.
-
Assuming once more 7 = 0,1 and v = 927 feet, we have 5
and therefore
Xe
=
μ = 0,87. 0,000409094
0,0003559,
= 0,87
(1-0,0003559.927, 14. 0,05). 92,71 cos. 24° 39′ =0,9835.92,71.0,9089
82,87 feet and
Y₂ = 0,9835.92,71 sin. 24° 39′ — 0,156
ปูน
37,87 feet.
The position of the projectile in reference to the point of beginning is
determined after 0,3 seconds by the co-ordinates
x + x1 + x 2 = 88,96 + 85,77 + 82,87
257,60 feet and
Y + Y ₁ + Y2 41,32 + 39,53 + 37,87 118,72 feet.
x
1
If the air offered no resistance and gravity did not act, we would have
=
= ct cos, a = 1000. 0,3. cos. 25° 300.0,9063 271,89 feet and
X + X₁ + X2 = ct cos. a=
x
1
c t sin. a = 300, sin. 25° 300.0,4226 = 126,78 feet.
Y + Y ₁ + Y ₂ = ct sin. a =
1
If we neglect the resistance of the air only, we have
271,89 feet and
X + X₁ + X 2 =
g t
0,09
Y + Y ₁ + Y ₂ =
1
= 125,33 feet.
66
126,78
126,78 - 32,2.
126,781,449
2
p
APPENDIX.
THE THEORY OF OSCILLATION.
(§ 1.) Theory of Oscillation.-A body has an oscillatory or
vibratory motion (Fr. mouvement oscillatoire; Ger. schwingende
Bewegung) or is in oscillation or vibration (Fr. oscillation; Ger.
Schwingung), when it describes repeatedly the same path backwards
and forwards in equal times. We meet with many examples of
oscillatory motion in nature besides that of the pendulum. The
most general cause of such a motion is a force which attracts or
impels the oscillating body towards a certain point. Thus, E.G.,
gravity sets the pendulum in oscillation. If a body, previously at
rest, can yield without impediment to the action of the force, which
impels it towards a certain point, the oscillation takes place in a
straight line; otherwise it will oscillate in a curve, as a pendulum
does, where the action of gravity is continually interfered with,
the body being united to a fixed point. In like manner, if the
direction of the initial velocity of the body is different from that
of the motive force, the oscillations will also take place in curved
lines.
The simplest and most common case is that where the force is
proportional to the distance of the body from a certain point C. Let
C, Fig. 874, be the seat of the force, I.E.
the position of the body when the force
FIG. 874.

D
R
A
MNP C
=
is 0; let A be the point where the
motion begins, and let M be the variable
position of the body. If we denote the
distance C M by x, and by μ a constant,
B determined by experiment, we have the
acceleration of the body at M
P = μα,
p
§ 1043
00
2.]
THE THEORY OF OSCILLATION.
and since x decreases an amount d x, when the space A Mis in-
creased by the same quantity, we have for the velocity v of the
body (see § 20, III)
11/12 v ²
-Spd x =
μχ
- µ fx d x = − " 2
+ Con.
But at .4, v = 0 and x is a definite quantity C A = a; we have,
therefore,
0 =
μα
2
+ Con., and
or the velocity itself
v² = µ (a³ — x³),
رح
√ µ (a²
x²).
When the body arrives at C, x = 0 and v is a maximum, and
its value is then
v = c = √ µ a² = a √µ.
Upon the other side of C, v gradually decreases, and at the dis-
tance x = C B a from it becomes again = 0; the body
then returns with an increasing velocity to C. This return takes
place in accordance with exactly the same law as the first motion;
at C, v = c, and at A, v = 0. Thus the motion repeats itself in
the space A B = 2 a, which for this reason is called the amplitude
of the oscillations (Fr. amplitude des oscillations; Ger. die doppelte
Schwingungsweite).
(§ 2.) The time in which the oscillating body describes a certain
space A M = x₁, Fig. 875, can be determined in the following
manner. If in the element d t of the time the element of the path
MN d x =
=
d x is described, we have (§ 20, I)
—
d x₁ = v d t, L.E. d x = √μ (a² x²) d t,
and, therefore, inversely
d x
d t = √ µ (a² — x²)
Now if we describe upon A B, with a radius C A C B = a,
a circle A D B, Vaª — x³ will be represented by the ordinate M O
=y, and, therefore, we will have
d t =
d x
Vμ.y
If we put the arc D 0, corresponding to the abscissa C M = x,
equal to s, and its differential O Q - ds, we have, in conse-
quence of the similarity of the triangles O Q R and O C M, in
1044
[§ 2.
GENERAL PRINCIPLES OF MECHANICS.
which O Rdx, OQ=ds, MO = y, and O Ca, the pro-
portion
d x
Y
FIG. 875.
D
and, therefore,
d s
a

d x
d s
; hence it follows that
y
α
d s
Ar
XB
MNP C
t =
-S
d s
νμ. α
d t =
√μ. a
•
and
,
S
+ Con.
μ. α
But at the point A, where the motion begins, t = 0 and s is
equal to the quadrant D A =
πα; consequently
πα
0 =
0:
+ Con.,
νμ.α
and the time required by the body to come from A to M is
πα
S
1
t =
νμ. α
νμ. α
2
(-9).
μ
The period of half an oscillation, I.E. the time required by the
body to pass from the point A to the position of rest C, for which
s = 0, is
t
π
2 V μ
and the period of a complete oscillation, or the time required to
describe the whole distance A B = 2 a, is
After the time
π
t =
2 п
t =
the body has made a double oscillation and returned to the point A.
The time required by the body to describe the space 2 A B =
4 a is the same, no matter from what point M we begin to count;
for the time in which the body goes from M to B and back is
= 2.
arc O B
Vu. a
and that in which it goes from M to A and back is
arc O A
=2.
;
√μ.a
consequently the time required to describe the space 2 MB +
2 M A is
со
1045
3.]
THE THEORY OF OSCILLATION.
= 2.
arc (0 B + 0 A)
αν μ
2.πα
2 π
αν μ
νμ
We see that the period of an oscillation does not depend upon
the amplitude. If we start from the point C, we can put the time,
which corresponds to the distance C M = x,
or, since sa sin.
S
t =
νμ.ά
X
ď
1
X
t
sin.-1
and inversely
νμ
a
x = a sin. (t õ), and
v = √µ Va² — a² [sin. (t √µ)]²= √μ.a V1 — [sin. (t √µ)]²
a cos. (t √μ).
Vua
-
REMARK.—The foregoing theory of oscillation is applicable to the cir-
cular pendulum C M, Fig. 876, if the arcs in which it oscillates are small.
At 4 the acceleration of the point, which is oscil-
lating in the arc A M B,
FIG. 876.
C
p = g sin. A C D =
DA
CA

g.
D
B
M
or, since for small displacements we can put D A
= MA,
Ρ
DA
MA
·
9,
If we denote C A by r and MA by x, we obtain
gx
p =
and by comparing it with the formula p = µ x, we find
g
μ
Hence the period of an oscillation is
П
t
(compare § 321).
νμ
g
(§ 3.) Longitudinal Vibrations.-The most common cause
of oscillatory motion, which is then called vibration, is the elasti-
city of bodies. The most simple case is that presented by a rod,
string or wire O C, Fig. 877, stretched by a weight G. If we move
this weight from its position of rest C a certain distance CA = a
in the direction of the axis of the string and abandon it to itself,
1046
[$ 3.
GENERAL PRINCIPLES OF MECHANICS.
then, in consequence of the elasticity of the string, etc., it will be
raised to C, where it arrives with a velocity c and above which it
ascends, by virtue of its vis viva, to a point B, from
which it falls again, etc.
FIG. 877.
When at rest, the weight G

λ
•
is balanced by the elasticity
ī
FE (see § 204) of the
rod, and consequently the motive force is
λ
B
P
D
C.
1
A₁
λ
ī
FE – G – 0, or FE = G.
But if the weight G is at a lower point N, whose
distance from Cis CN= x, the motive force becomes
B
λ + x
λ
X
P=
FE - G
FE+ FE - G
2
FE
D
X,
N
and if it is at a higher point Q, this force is
λ
X
P = G
FE G
λ
7
X
FE
FE+ FE
7
T.
Z
If we neglect the mass of the rod, the acceleration, with which
the weight G returns towards C, is
p
P
g
G
FE
gx, and consequently we have
G l
μ
FEg
Gl
when we put pμx and denote the length of the rod by 7, its
cross-section by F and its modulus of elasticity by E. As this
formula corresponds to the case treated in the foregoing paragraph,
the period of a simple vibration is
П
t
νμ
Gl
FEg Ng
π
Gl
FE
If instead of F we substitute the weight of the rod G, Fly
E
and instead of E the modulus of elasticity L =
, expressed in
Y
units of length, we obtain
πι
t
G
G, L
g
§ 43
1047
THE THEORY OF OSCILLATION.
If, on the contrary, we observe the period t of the simple vibra-
tions, we can calculate the modulus of elasticity by putting
E=
π² Gl
g t³· F
or L
π² 1 G
gt Gi
These formulas also hold good, when the vibrations of the rod
are produced by simply attaching the weight (at B); in this case
the semi-amplitude on each side of Cis
G
α = λ
1,
FE
while in the other case we assumed a < 2.
A complete vibration is a double oscillation.-[TR.]
EXAMPLE. --If an iron wire 20 feet long and 0,1 inch thick is put in
longitudinal vibration by a weight G = 100 pounds and if the period of
a complete vibration is of second, we have t 1, and consequently the
modulus of elasticity
E = 0,031 . π². 18².
100. 20.4
(0,1)². π
=
0,031. 800000 . 182. π
= 24800. 324 . T = 25000000 pounds.
Π
(§ 4) The foregoing formula is also applicable to the case,
where the weight acts by compression upon a stiff prismatical rod.
It also holds good, when the weight applied at the end of the rod.
has an initial velocity v. According to the principle of mechanical
effect, when the height of fall of G is h, we have
འ”
h
h
FE
Gh + G
FE.
.h', and, therefore,
2 g
2
27
G l
G l
2
2 Ꮐ Ꮣ
22
h =
+
+
FE
FE
FE 2g
After the weight & has described this space, it has lost all its
velocity, and in consequence of the elasticity it rises again to A,
where it arrives with the velocity v. In consequence of its vis
v³
viva G it compresses the rod and rises to a height h, before
2g
returning and beginning a new vibration. For this second dis-
tance we have
FE
2
Gh₂+ h', and, therefore,
v³
G
2 g
27
Gl
h₁
FE
Ꮐ l
2
2 G l v2
+
FE
FE
2 g
By adding hand h, we obtain the
vibration
total amplitude of the
2 a = h + h₁ = 2 √ √ ( 2 ) ².
Gl
2 G l
гов
+
FE' 2 g '
1048
[$ 5.
GENERAL PRINCIPLES OF MECHANICS.
hence the simple displacement is
22
2 G l
+
FE FE 2g
a =
√(G2)² +
Gl
Since in this case also
FE
p =
g x = μx, we have as above
Gl
for the period of an oscillation or simple vibration
π
t
Ng
Gl
FE
If the initial velocity v of the weight G, is caused by a falling
weight G (Fig. 878), we have the case treated in § 348. If the
weight G strikes with the velocity c, and if we suppose
the impact to be inelastic, we have the initial velocity of
G + G₁
G c
FIG. 878.

v = G + Gi
hence the maximum displacement is
G
h
a = √
√ ( ( G + G₁) ) )²
2
2 G* l
c²
+
FE
and the period of a simple vibration is
(G + G₁) F E ' 2 g
G₁
B
π
t
√ (G + G₁) ?
Ng
FE
The elements of the rod also participate in the vibra-
tions of G or G + G₁, but their amplitude decreases as
the position of the element approaches the point of suspension.
For an element C₁, Fig. 877, situated at a distance O C = x from
the point of suspension, the amplitude is
X
Y
α
a;
while the period of its vibration is the same as that of G; for it
does not depend upon y or a. Hence the vibrations of all the ele-
ments of the rod are isochronous, but their amplitudes decrease
gradually from C towards 0.
§ 5. Transverse Vibrations.-The elasticity of flexure and
of torsion cause vibrations of the same nature as those just treated..
If a rod or spring O C (Fig. 879) is fixed at one end and deflected
at the other C by a weight G, we have, according to § 217, the
deflection
HC = α =
P 13
3 WE
§ 5.]
1049
THE THEORY OF OSCILLATION.
inversely the force, with which the rod is bent, is
FIG. 879.
H
Б
C
A
G
P =
3 WE a
73
Now if this force is re-
placed by a weight G, at-
tached at C, and if a is in-
creased or diminished a dis-
tance C A = C B
CB:
= x, we
have the force, with which

the rod will be driven back to its position of rest by its elasticity
3 WE (a + x) 3 WE 3 Π Ε
a =
73
73
hence the acceleration is, when we consider the mass of G alone,
P=
3 WE (a + x)
で
​G =
p =
P
G
9 =
3 WE
G 13
g x, and, since p = µ x,
3 WE
μl =
G 13
g.
x;
The relation between p and x allows us to employ the formulas
of (§ 2), consequently the period of an oscillation or simple vibra-
tion is
π
t =
π
G. 13
3 WE'
If the rod H 0, Fig. 880, is supported at both ends and loaded
in the middle C with a weight G, we have, according to § 217,
H
FIG. 880.
B
A
P 1³
α =
48 WE'
and, therefore, the duration
of a simple vibration

π
G T
t =
Ng 48 WE'
and
If we take the weight G, of the rod into consideration, we must
substitute in the first case, Fig. 879, instead of G, G + G₁,
in the second case, Fig. 880, instead of G, G + & G₁.
From the observed duration of an oscillation or simple vibra-
tion we can calculate the modulus of elasticity, in the first case by
the formula
1
or, if n =
G G
B = ( 7 ) ( ~ 3 + 1 );
g W
73
denotes the number of simple vibrations per second,
(π
E = (7 n)² (G + 4 G¹) r.
1
g W
1050
[$ 6.
GENERAL PRINCIPLES OF MECHANICS.
EXAMPLE.-A pine rod 1 centimeter square is supported at two points
100 centimeters apart, and its centre is deflected a distance a = 3,2 centi-
meters by a weight G = 1,37 kilograms. According to this experiment
the modulus of elasticity of pine is
E
Pl3
48 Wa
1,37. 1000000
48.12. 3,2
107031 kilograms,
while in the table on page 370 we find E = 110000.
The rod was then firmly fixed at one end, was loaded at the other with
a weight G = 0,31, and put in vibration. It was found that the number
of simple vibrations in 35 seconds was 100. The weight of the rod was
G ₁
1
=
= 0,044 kilograms; hence G+ 1 G,
G G
1
1
3,141 321000
•
1
= 0,321 kilograms and
Π
E
( F ). ( F + + F ₁ ) P
2
0,35
80,57.
3 g W
1281000
981
981.12
3
105260 kilograms,
or about the same value of E as was found by the experiment upon flexure.
T
νμ
can
§ 6. Vibrations Due to Torsion.-The formula t
also be applied to the torsion balance or torsion rod (Fr. balance de
torsion; Ger. Torsionspendel), I.E. to a thread or rod D O, Fig.
881, oscillating about its axis, in consequence of its torsion. Gen-
M
B
FIG. 881.
B
A
7
C₁
Α
A₁
M
1
erally the rod is provided with a loaded
arm C C₁, by means of which the origi-
nal torsion of the thread is produced,
by bringing this arm from its position
of rest CC into the position A A.
The torsion drives the arm back to
C C₁, and the latter, by virtue of its in-
ertia, moves further on until it comes
into the position B B,, from which it
returns to C C and A A₁, etc. We
found previously (§ 262) the moment.
of torsion of a prismatic body to be
a W c
Pa =
;
we know, therefore, from this formula, that it is inversely propor-
tional to the length O D = 1 of the rod and directly proportional

G
to the angle of torsion M D
C = a; now if
is the moment of
g
k² G
inertia of the arm CD C₁,
2
is the mass M reduced to the ends
a² g
C and C, of the arm, and the acceleration of this point is
$ 7.]
1051
THE THEORY OF OSCILLATION.
P
a Wc k² G
a a WC g
p =
:
M
Τα a² g
Gkl
If we denote the arc CM = a a, corresponding to the length
of the arm D A = D C = a and to the variable angle of displace-
ment CD Ma, by x, we obtain the expression
p
WC g
G k² l
x, and we can again put p = μx, or
W C g
G k² l'
μ =
The period of an oscillation or simple vibration is, therefore,
π
π
t
νμ
Ng
Gl
WC'
no matter whether the amplitude A CB = A, C, B₁ is large or small.
Inversely, we have
WC = G k² 1,
π²
g t²
and, therefore, the moment of torsion
π²
Pa =
gt
•
a G k².
REMARK.-The above formulas for the vibrations produced by the
elasticity of rigid bodies are not correct unless the displacement during
the vibration is within the limit of elasticity. Great care should be
taken to avoid as much as possible vibrations in the various parts of
machines; for the energy expended upon them is lost to the machine.
For this reason the parts should be united to each other with precision,
and what is known as lost motion is to be avoided, as it gives rise to con-
cussions and vibrations.
M
P
A
1
D
FIG. 882.
A1
M
R
§ 7. Density of the Earth-The
theory of the torsion-rod can be directly
applied to the determination of the mean
heaviness or specific gravity & of the earth.
If we cause a heavy sphere K to approach
the weight G, which is fastened upon the
end of the arm A D A,, Fig. 882, the
latter will be attracted towards the former
a certain distance A M = a; the attrac-
tion R of A balances the force of torsion
P, when G occupies the position M; one
of the above forces can, therefore, be de-
termined from the other. Now if we re-
move the heavy sphere K and allow the

1052
[$ 7.
GENERAL PRINCIPLES OF MECHANICS.
torsion-rod to vibrate, we can observe the period of the vibrations,
and from it we can calculate the force of torsion. According to
the foregoing paragraph, the period of a simple vibration is
П
t
μ t = 2 and p =
Αμ
X
force of torsion
mass of torsion-rod
Pa²
G hoz Is
Gだ
​when. G² denotes the moment of inertia and a the length of the
arm of the torsion-rod; inversely, the twisting or attractive force is
P =
G k² p
g a²
Gだ ​2
µ G k² x
ga²
π²
G k² x
2
π Gh² a
g t
a²
g to
a
and the moment of torsion corresponding to the angle of torsion a is
π
Pa =
It
a G k².
Now if the forces, with which the bodies attract each other, vary
directly as their masses and inversely as the squares of their distances
(see § 302, Example 3), we can compare the attraction P, exerted
upon the body by K, with the weight of the small body which is
placed upon the torsion rod; for the weight is the measure of
attractive force of the earth; thus we obtain
P
Q
K: 82
E: 229
in which s denotes the distance MK of the centres of the two
masses G and K from each other, the radius of the earth and E
its weight. If we solve the above equation, we obtain the latter
weight
and if we substitute E
of the earth
E =
K Q p²
P s²
π r³, ε у, we have the mean heaviness
or if we introduce the length of the second pendulum /=
Y1
Υι = εγ
3 E
4 п роз
3 K Q j·²
4 π P 2.
2
S
3 K Q
4 π Pr 82
3 K Q
4 π r s²
gta²
π³ G k²x²
9
π²
(see
§ 323),
Y1
Υ1 = ε γ =
εγ
3 K l t²
4 π r x s² G k
Qa²
ε
3 K l ť²
4 π r x 8²
Q a²
Gk y
Y
k³
hence the mean specific gravity of the earth is
If we put approximatively G = Q a², we obtain more simply
Klt
3
E
π V X s² Y
Cavendish found in the first place with the torsion rod, or
Coulomb's torsion balance, as it is called, &= 5,48; or, according to
Hutton's revision, ɛ = 5,42.
$ 8.]
1053
THE THEORY OF OSCILLATION.
Reich found afterwards, with the aid of the mirror apparatus of
Gauss and Poggendorff, & 5,43. Baily, on the contrary, found
by experiments upon a larger scale, e = 5,675.
=
When Reich repeated his experiments he found ɛ=5,583. (See
"Neue Versuche mit Drehwage, Leipzig, 1852.") The mean
density of the earth is, therefore, according to these experiments,
about equal to that of specular iron.
REMARK.—The following works may be consulted in reference to the
manner in which the density of the earth was determined: "Gehler's
physikal. Wörterbuch," Bd. III; the treatise of Reich "Versuche über die
mittlere Dichtigkeit der Erde, Freiberg, 1838;" and that by Baily, "Ex-
periments with the Torsion Rod for Determining of the Mean Density of
the Earth, London, 1843."
§ 8. Magnetic Needle. The torsion-balance may also be
employed to find the directing force or the moment of rotation of a
magnet or of a magnetic needle (Fr. aiguile aimantée; Ger. Magnet-
nadel). If we replace the transverse arm of the balance by a
magnetic needle or by a bar magnet M D M, Fig. 883, it will as-
M
FIG. 883.
A
DO
] A 1
M
N
б
sume a position in which the directing force is
balanced by the twisting force. If the non-mag-
netic arm, when at rest in A A₁, forms an angle
ADN a with the magnetic meridian N S,
and if the bar magnet M M, assumes such a posi-
tion that its axis forms an angle MD N
with the meridian N S, we have R₁ R sin. d, in
which formula R, denotes the component of the
directing force R, which is parallel to N S. This
component tends to turn the needle, and is bal-
anced by the force of torsion. The latter force
P, on the contrary, is proportional to the angle of torsion MD A =

α
d, and we can, therefore, put
P = P₁ (a
8);
hence we have R sin. 8 P (ad), and consequently
α
R
(
α
P₁
P₁ = ( 5 ) P₁₂
sin. S
Ꮄ
*when the variation or angle of deviation d is small.
Now according to the foregoing paragraph the force of torsion
is expressed by the formula
Π
π²
P =
gt
G k² x T
g t
Gh³ a (a
a²
Ꮄ
π² G k² (a — 8)
g t
a
and we can calculate from the period t of an oscillation, etc., of the
1054
[$ 9.
GENERAL PRINCIPLES OF MECHANICS.
non-magnetic torsion-rod the directive force of the magnetic needle
by the formula
α
R = (ª
= (a = 0 ) P₁ =
a distance DM
P
α
б
π²
G k³
Ꮄ
Ꮄ g t²
a
The moment of this force, when we assume that it is applied at
a from the axis of rotation and when the varia-
8, is R₁ a = R a sin. d, approximatively, for
tion is MD N =
small variations,
= Ραδ= (a — 8).
π²
I t²
•
G k².
This moment (R a sin. §) is a maximum and = Ra for
sin. 81, I.E., when the magnetic needle is at right angles to the
magnetic meridian, and, on the contrary, a minimum and = 0,
when = 0, I.E., when the axis of the magnet needle coincides with
the magnetic meridian.
§ 9. Magnetism-Since the directive force of the magnetic
needle causes no pressure upon the axis, I.E., the needle has no
tendency to move forward, but only a tendency to turn, when its
axis does not coincide with the magnetic meridian, it follows that
the entire action of the earth upon the magnet must consist of a
R
the maximum moment of which is Ra. Now
couple
R
since every couple
2
2
R
R
2'
2
can be replaced by an infinite number
R.
R₁
2
꽃​) (꼴​,
B), etc., whose moments
of other couples (
2
2
R α, R₁ α1, R₂ α, etc., are equal to each other, it follows that nei-
ther R nor a, I.E., neither the directive force nor the point of appli-
cation, but only the moment R a is determined. This twisting
moment depends, in addition, upon two factors, u, and S, 4, corre-
sponding to the magnetism of the earth and S to that of the bar or
needle; hence we can put
R =₁S and Ra = µ₁ Sa.
The measure of the magnetism of the earth for a needle
vibrating horizontally (the case under consideration) is only the
horizontal component of the intensity of the entire magnetism
of the earth; for the vertical component p. is counteracted by the
support of the needle. If is the angle of dip or inclination or the
angle formed by the magnetic axis of the earth with the horizon,
we have the horizontal component
し
​μl₁ = μl cos. ';
on the contrary, the vertical one
μ₂ = µ sin. i,
§ 10.]
1055
THE THEORY OF OSCILLATION.
and, finally, the twisting moment of a magnetic needle is
Ra sin. S μ cos. L Sa sin. 8,
the maximum value of which is
FIG. 884.
•
Ra = µ. Sa cos.i.
§ 10. Oscillations of a Magnetic Needle.-We can calcu-
late the moment of rotation of a magnetic needle from the period
of its oscillations. If we move the suspended needle M D M, Fig.
884, from its position of rest, where the force of torsion and the
directive force of the magnet are in equilibri-
um, so that its new position shall make a small
Nangle MD C = ☀ with its former one, either
the magnetic directing force R is increased by
Ro and the force of torsion P, is diminished
by P₁ 4, or the reverse takes place; in either
case their resultant
S ·
i
A
Mi C₁ A
1.
D
CM

(R + P₁) $
or its moment
(R+ P₁) & a = (R+ P₁) x
drives the needle back to its position of rest.
If G is the moment of inertia of the needle, the acceleration,
corresponding to this force, is
p =
(R+ P₁) α x
G k²
g;
G
(R+P) ag
if we put it µ x, we obtain
ре
and the period of an oscillation is
π
t =
νμ
G k
π
(R+ P₁) a g
π
Ng
P₁
1
Gh²
(R+ P₁) a
Ꮄ
or, if v denotes the ratio
magnetic force,
T
t =
Ng
R
α
б
of the force of torsion to the
G k
(1 + v) Ra
If we have found t by observation, we can find by inversion the
moment of rotation of the needle, which is
1056
[§ 11.
GENERAL PRINCIPLES OF MECHANICS.
π
Ra =
G k²
g t² ' 1 + v
If the force of torsion is small, I.E., if the position of repose
nearly coincides with that of the magnetic meridian, we can
neglect v and put
t =
Ng
π
G k²
and
Ra
•
G k³.
Ra =
π
g t
We can also substitute for R a its value, which has been given
above, and express the moment of rotation by the formula
µ S a cos. i -
T
2
9 to
•
G k².
For a dipping needle, which oscillates in the plane of the mag-
netic meridian, we have, on the contrary,
μ Sa =
π²
I t²
G k²,
and for a needle, whose axis lies in the magnetic meridian and
which, therefore, tends to place itself in a vertical position we have
Sa sin. 1 =
In the formula u Sa cos.
ㅠ
​π2
9 to
Gh².
Π
g t
·
G k², µ S a cos. is a product
ле
of four factors; however, since the inclination can be determined
by observing a magnetic needle, and since Sa cannot be decom-
posed into two definite factors, we have to only resolve the product.
μ Sa into the factors μ and S a. How this can be done by ob-
serving the declination of the needle will be shown in the sequel.
§ 11. Law of Magnetic Attraction.-The forces, with which
the opposite poles of two magnets attract and the similar poles
repel each other, are inversely proportional to the squares of their
distances from each other. We can convince ourselves very easily
of this fact by observing a small magnetic needle, which has been
set in oscillation near a large bar magnet. The bar magnet is
placed in a horizontal position and in the plane of the magnetic
meridian, its north pole being directed to the north and the south
pole towards the south; we then place a small variation compass
in the prolongation of the axis of the bar magnet. If the distance
s of the pivot of the needle from one pole of the bar magnet ist
& 12.1
1057
THE THEORY OF OSCILLATION.
عالم
much less than its distance from the other pole, we can disregard
the action of the latter upon the needle and we can assume that,
in consequence of the action of the nearer pole, the coefficient of
the magnetic force of the earth is increased a certain amount ¸ or
If the period of the oscillations of the needle is = t, when the
bar magnet is removed, and, on the contrary, if it is t₁, when
the nearer pole of the bar magnet is at a distance s, from the pivot
of the needle, and t, when the latter distance is =S2, we have
π
ה
μ₁ Sa =
g
I to
•
G k³, (µ₁ + k₁) Sa=
π²
g t
Gh² and (µ₁ +₂) Sa=
.Gk',
g ta
μg + K₁
ťo
and
My + K₂
t³
μι
t₁2
22
рех
whence we obtain by division
t²
resolving the last two equations, we obtain
μ₁ and к₂
t
=
( )
2
µ₁, and, finally,
K1
(
't³
ti
2
t₁²
t³
2
t2
2
C
K1: K₂ =
t₁²
2
to²
or, if we substitute instead of t, t, and t₂ the number of oscillations
n =
60"
t
60"
60"
nr =
and n₂ =
t₁
t z
1
K₁ : K₂ =
N₂²
n²: n₂
222.
If the action of the bar magnet upon the magnetic needle is
inversely proportional to the square of the distance, we must have
also
K₁: K.
1
2
N
N
2
n²
ՂՆՋ
2
s²: s,', and therefore
$2
$1
2
29
which is confirmed by the observations.
§ 12. The actions of a bar magnet N S upon a Lagnetic needle
n s are simplest, when the bar magnet is placed at right-angles to
the magnetic meridian in such a manner that the pivot of the
compass n s, Fig. 885, lies either in the prolongation of N S or in the
line which is perpendicular to NS, Fig. 886, and passes through
its middle C. If for the present we put the force, with which a
pole of N S acts upon a pole of n s, when their distance apart is
unity, K, we have in the first case, Fig. 885, when a denotes the
length N S and e the distance Cd between the centres C and d
of the two bodies N S and ns, the force, with which the north
pole n is attracted by S,
67
1058
[§ 12
GENERAL PRINCIPLES OF MECHANICS.
K
K
P
S n³
Sn²
approximatively
and the force, with which n is repelled by N, is
FIG. 885.
FIG. 886.

P
$
d n
P
S

P
SA
p.
P₁
1
Pi
Q₁
K
K
P
N n²
(e + ½ a)²
N
hence the resultant of P and P₁ is
Q = P = P₁ = K
1
1
e a)³
(e + ½ α
R
(e + 1 a)² — (e — a)²
-
(e + ½ α)² (e - a)'
2 a e K
½
(e + ½ a)² (e − a)"
or, if a is small compared to e,
Q
=
2 α e K
e+
2 a K
e³
N
In like manner we find the resultant of the attraction and
repulsion of the south pole s
Q
=
2 a
e³
2 a K
;
2 al K
Q1=
C³
e
3
hence the moment of the couple, formed by these forces, is
when I denotes the distance between the two poles of the needle.
For the second case (Fig. 886), on the contrary, the attraction
and repulsion at s are
K
K
P₁
and those at n are
N s²
S sa
K
K
P =
Sn"
N n²
bence the resultants are
§ 13.]
1059
THE THEORY OF OSCILLATION.
CN
= 2.
•
P₁ =
=
Ns
a P
Ns
a K
and Q
Ns
a K
N n³
3
Now if a andare considerably smaller than e, we can sub-
stitute for N s = S′s and N n = Sn the mean value N d = S d
and for the latter the approximate value Cde; thus we obtain
Q = Q₁
ак
e³
and, therefore, the moment of the couple, formed by Q and Qu
ρι
Q l =
al K
es
I.E., it is one-half as great as in the foregoing case, a result which
is perfectly corroborated by observation.
But the force K is itself a product of the intensity of the
magnetism of n s and the intensity S of N S, I.E., KK S; hence
we have in the first case
Q R S α
Q
and in the second case Q
=
k Sa
€³
e³
§ 13. Determination of the Magnetism of the Earth.-
If in both the above-mentioned cases the magnetic needle n s is
abandoned to the action of the larger magnet, the former will
hence
FIG. 887.
A
R₁
d
N
assume a new position n s, Fig.
887, in which the force Q, with
which the bar magnet acts upon
the needle, is balanced by the
force R, due to the magnetism of
the earth. If d is the variation
Ndn Sds of the needle from
the magnetic meridian, we have
for the components of Q and R,
which balance each other,
Q₁ = Q cos. & and R₁ = = R sin. &;

R
Q cos. & = R sin. ð and
Q
tang. &=
R
or, if we put, according to the last paragraph, either
Q
2 K S a
es
K S a
or Q
=
e³
and, according to § 9 of the Appendix, R= μ, K, we obtain either
1060
[§ 13.
GENERAL PRINCIPLES OF MECHANICS.
tang. &=
2 K S a
Мі кез
2 Sa
M₁ e39
or tang. 8 =
Sa
Мезо
By inversion we obtain the ratio of the magnetic moment of the
bar to the intensity of the magnetism of the earth; for in the first
case we have
Sa
Sa
1 e³ tang. d, and in the second case,
e³ tang. §.
M₁
By observing the period of the oscillations of the bar magnet,
we obtain (according to § 10) the product
π-2
T
μ₁ Sa =
g t
G k²;
by combining the two equations, we deduce the magnetic moment
of the bar, which is
either
or
П
Sa =
√ ½ G k² e³ tang. §
3
t v g
π
Sa =
t v g
√ G k² e³ tang. §,
and the measure of the horizontal component of the magnetism of
the earth, which is either
π
t √ g
√ 2 G k² cotang. S
e³
π
or =
t v g
G k² cotang.S
e³
If we
the first formula being applicable to the case represented in Fig.
885, and the second to the case represented in Fig. 886.
divide by the cosine of the angle of dip or inclination (4), we obtain
the total intensity of the magnetism of the earth
ре
рез
COS. L
In order to obtain a clear idea of the coefficient or measure μ of
the magnetism of the earth, we must put in the formulas.
k S l a
Rau Sa and Q l
a = l = e = 1,
,
e³
and also к = S = 1; thus we obtain Ra
1; thus we obtain Ra = μ and Q 7 =
µ and Q l = 1; hence
1) the measure u of the intensity of the magnetism of the earth
is that moment, with which a magnetic needle, whose magnetic mo-
ment is unity, will be turned by the magnetism of the earth; and
2) the magnetic moment of a magnetic needle is unity when
that needle communicates to another similar and equally powerful
magnetic needle, placed in the position represented in Fig. 886 at
the unit of distance from it, a moment = unity (1 millimeter-milli-
§ 14.]
1061
THE THEORY OF OSCILLATION.
gram). According to Weber, if the acceleration of gravity were
1 millimeter, we would have
in Gottingen μ = 1,774 millimeter-milligrams,
in Munich
in Milan
µ = 1,905
μl = 2,018
66
66
66
but, since the acceleration of gravity in Central Europe is 9810
millimeters, the true values are 9810 = 99 times less.
REMARK.-We would recommend to those who wish to make a more
extended study of magnetism, besides Müller-Pouillet's "Lehrbuch der
Physik;" Lamont's "Handbuch des Erdmagnetismus" (Berlin, 1849), and
Gauss and Weber's "Resultate aus den Beobachtungen des magnetischen
Vereins," Gottingen and Leipzig, 1837 to 1843; also the "Experimental-
physik" of Quintus Julius, and Mousson's "Physik auf Grundlage der
Erfahrung," etc.
§ 14. Waves. In discussing the longitudinal and transverse
vibrations of prismatical bodies, we have heretofore (§ 3, 4 and 5)
neglected the mass of these bodies and considered only that of the
weight, which produced the strain in the bodies. Hereafter, on the
contrary, we will not consider any such weight, but suppose that
the body is put in vibration by a sudden blow or by a force, which
acts for an instant only; we must, therefore, take into account the
inertia of the vibrating body alone. As the most simple case is
that offered by longitudinal vibrations, we will, therefore, treat that
first.
From what precedes, we know that all the parts of a prismatical
FIG. 888.
M
M
M2
Mn
Mg
ᎷᏗ

P
BCD
B C D₁
Bg Cg Dg
BgCgDg
BC.D
K
Bg
I.
F
B
D
J
3.
M1
II.
B
B
D
Bg
Da
T
3
U
NS
Dg
H₁
-B₂
2
3
S
rod B M, Fig. 888, are put in vibration, when this body is extended
or compressed by a force P, acting in the direction of its axis. Not
1062
[$ 15.
GENERAL PRINCIPLES OF MECHANICS.
only the element M at the end, but also every other element
M₁, M2, M3 .... of the rod vibrates back and forth in a certain
space B D, B, D, B. D.... which is called the amplitude of the
vibration; we can also assume, when the rod is very long, that this
space is the same for all the elements. Although the time in
which an element makes a vibration is the same for all parts of the
rod, we cannot, therefore, assume that all these elements M, M₁, M.,
etc., are simultaneously in the same phase of motion, E.G., that they
are all at the same time in the middle of a vibration, but we should
rather suppose that time will be required to communicate the mo-
tion proceeding from M to the succeeding elements, and that the
farther an element is situated from the origin P of the motion, the
later it will enter upon the same phase of motion. It is, therefore,
possible that at the instant, when the element M has made a com-
plete vibration BD forward and back, the element M, has made
but one-half of its forward movement and has arrived at C, and
that the element M, is just beginning a vibration. The latter will
therefore vibrate isochronously with M. The velocity with which
the same phase of motion advances in the body is called the velocity
of propagation (Fr. vitesse de propagation; Ger. Fortpflanzungs-
geschwindigkeit) of the vibrations of the body. The aggregate of
all those elements between M and M, which are in the different
phases of a complete vibration or which are included between two
elements M and M, which are in the same phase, are called a wave
(Fr. ondulation; Ger. Welle) of the vibrating body, and the dis-
tance M M is called the length of the wave. A wave consists of a
back part B D, which contains the returning elements, such as
M₁, M..... and of the wave front D, B4, which comprehends the
advancing elements M, M....; B D is also called the rarefied
and D. B, the condensed portion, since all the elements in B D, are
extended and those in D, B, are compressed.
2
4
2
3
§ 15. The phases of the motion and of the velocity in a wave can
be very well represented by serpentine lines, such as F C, G, C₂ H₁
and B M₁ D₂ N3 B4, Fig. 889, I and II. At the moment when M
begins a new vibration at B, its displacement is a maximum and
its velocity is = 0; at the same time M, is in the position of rest,
and consequently its displacement is 0 and its velocity is a max-
imum; both of these facts are shown by the above curves; for the
first curve (that of the displacement) (I) passes at B at a distance
equal to the amplitude B F = B C above the axis B D, and cuts
§ 15.]
1063
THE THEORY OF OSCILLATION.
2
2
this axis at C₁, while, on the contrary, the second curve (that of the
velocity) (II) cuts the axis at B and at Ci passes at a distance
C₁ M₁, equal to the maximum velocity, above the axis. At the
same moment the element M₂ is upon the other side of the position
of rest C, and at the maximum distance from it, and its velocity,
like that of M, is = 0; this is also shown by the two curves; for
one passes at D, at a distance equal to the amplitude D, G, below
the axis, and the other cuts it at that point, so that the ordinate
which corresponds to the velocity is 0. In like manner the
phases of the motions and of the velocities of the elements M3, M4,
etc., are represented by these curves. Since, E.G., the first curve
cuts the axis at C, and the second passes below that point at a dis-
tance equal to the maximum value C, N, we know that the ele-
ment M, at this moment passes through the position of rest with
the maximum velocity in the positive direction. If we wish to
know the phase of the motion of any other element M, situated
between M, M₁, M₁, etc., at the moment when the element M₁ be-
gins a new vibration, we have only to let fall from it a perpendicu-
lar upon the corresponding curve. The portion R S of this per-
pendicular lying between the curve and the axis corresponds to the
displacement of this element, and the portion TU, between the
second curve and its axis, gives its velocity. Since both ordinates.
are directed downwards, we know that both the displacement and
the velocity are positive, I.E. their direction is that of the velocity
of propagation.
3
3
If the element M were at D, I.E. about to begin its return mo-
tion, the displacements of the other elements of the wave would be
represented by the dotted line J C₁ K, C, LA, and their velocities
M
M1
P
BCD
B,C,D₁
I.
F
B
M1
II.
C
Q₂
FIG. 889.
MA
Ma
Mn
M3

B2 C2 Dg
BgCgDg
B₁CADA
K2
DB
2
-B₂ 2
H
·R-
-B-
DA
ra
8
S
4
B
D4
T
+C3
Bg
U
1064
[$ 16.
GENERAL PRINCIPLES OF MECHANICS.
by the ordinates of the dotted curve D O, B₂ Q3 D4. The period
of a double oscillation or that of a complete vibration, I.E. the time
t, in which the space B D + D B is described, is equal to the time.
in which a vibration is propagated through the length M M₁ = 1
of a wave; if, therefore, c is the velocity of propagation, we have
the total length of the wave
B B₁
= 1 = c. 2 t = 2 c t.
The length of the back part of the wave is
=
B D₂ = 1, B B₂ + B₂ D₂ = c t + 2,
2
=
and that of the wave front is
D₂ B₁ = l₂ = D₂ D₁ — B₁ D₁ = ct - λ,
2
in which 2 denotes the amplitude of a vibration.
4
REMARK.-The phenomena accompanying the interference of waves can
be shown by the aid of the curves of vibration. Let us consider two sys-
tems of equal waves, which are advancing in opposite directions, and let
A B C D E and F G H I K, Fig. 890, be the curves, whose ordinates rep-
R
B
F
M
A
G
FIG. 890.

D
·S...
K
D
resent the displacements. The displacements of an element, which be-
longs to two waves, produce a mean displacement, which is determined in
exactly the same manner as the resultant of two motions (see § 28), that is,
by adding algebraically the two component displacements. Hence at the
two points M and N, where the two curves meet each other, the ordinates
are doubled, and, on the contrary, at the points O and Q, where the curves
pass at equal distances from, but on opposite sides of the axis A E, the or-
dinates cancel each other, and the resultant of the two wave curves is a
third curve FR BO H S D Q K, whose ordinates give the displacements
of all the elements in the axis AE. While the two systems of waves A B C
and F G H are moving towards each other, the position of the wave-curve
FRB 0, etc., of course changes; but it is easy to understand that the
points of no motion 0 and Q do not change; for the ordinates of these
points of the two component curves are always equal and opposite. These
points are called the nodes.
(§ 16.) Velocity of Propagation. The velocity of propaga-
tion of waves can be determined in the following manner.
Let us
imagine the vibrating body B 0, Fig. 891, to be composed of an
infinite number of elements, the cross-section of each being A and
§ 16]
1065
THE THEORY OF OSCILLATION.
its length B C
CDdx, and let us assume that the phase of
the motion of an element BC Adx is propagated completely
=
MN,NN2 O
=
to the following CD Adx
in the elementary time d t, or
that the phases of the motion
are propagated in the direc-
tion of the axis of the body
FIG 891.

B C D
I
шт
with the velocity c =
d x
dt'
Let us assume that the elements B C
=
1
=
and CD oscillate from C to N in the time t, and thus come into
the position M N dx, and NO d x, and let us denote the
corresponding displacement C N by y. If the surface of separation
of the two elements, which before d t seconds was at N₁, comes
after d t seconds to N, the corresponding spaces described by these
elements are
=
N N₁dy, and N N₂ = d y»
and their velocities are
N,
V1
V₁ =
d y₁ and v₂ =
dt
d y z
d t
2
hence the retardation is
Ρ
V₁
d t
V
d yr - d y z
d tº
Since d t seconds before the moment, when the elements B C
and CD occupied the positions MN and N O, N, was in the same
phase as now is, we have C N₁ = DO; and since d t seconds
later N, is in the same phase as 1, it follows also that C N₂ B M.
From these two equations we obtain
=
CN,
1
=
=
N₁O DO DN, DO- (CN, CD) CD and
MN,
CMCN - (BMB C) BC; hence
NN₁ = dy₁ = N, O-NO CD-NO dxdx, and
0 =
NN₁ = dy₂ = MN - MN=BC-MN dxdx.
d x
-
1
=
=
The element dy of the space is equal to the compression
dx, of the element N O, and the element d y, of the space
is equal to the compression dada, of the element M N. If
we denote by E the modulus of elasticity of the vibrating rod, the
strains of the elements M N and NO produced by this compression
d Y A E and
d x
are
d
S₁ = (ã x
d x
dx) A E
(dx = d x )
d x
A E =
d Y A E.
d x
₁
1066
[§ 16.
GENERAL PRINCIPLES OF MECHANICS.
If we subtract the former from the latter, we obtain the retard-
ing force
d x
A E.
If y is the heaviness of the elements B C, CD, etc., of the rod, or
A d x . y the weight, and
acceleration at N, is
P
P = S₂ -
-
Y2
S₁ = (d
(a y — dy₂)
Adx. Y
the mass of such an element, its
g
A E.
I
Adx. Y
g E dy, - d y 2
;
Y
d x²
1
d t²
Y
d x²
- dye whence
d x²
g E
g E
d t²
"
or c²
;
γ
Y
p = {₁ = (dy₁ = d y₂ )
=
M
d x
equating the two values of p, we obtain
dy-dyz g E
2
dy - dy
hence the velocity of propagation of the waves (velocity of sound) is
c =
9 E
Y
= √g L,
in which formula L denotes the modulus of elasticity expressed in
units of length.
EXAMPLE.--If we assume the modulus of elasticity of spruce wood to be
E = 1870000 pounds and the weight of a cubic foot of it to be = 30
pounds, we obtain the velocity of propagation in it
C =
144. 1870000
30
g= √ 48. 187000. g
=
17000 feet,
I.E. about 15 times as great as in air.
REMARK.—This formula for the velocity of propagation is applicable
not only to a stretched string, but also to water and to the air. If p de-
note the pressure of the air upon the unit of surface, we have, according
to Mariotte's law, the tensions corresponding to the ratios of compression
2
d y i
d y z
1
and
d x
d x
S&
p d x
dxe
2
p d x
d x
and Si
p d x
p d x
"
d x 1
d x
d Y z
d Y 1
and, therefore, the motive force upon an element, whose cross-section is A, is
P = A (S₂ — S₁)
d y
now since
d x
dx² and
(dy₁ — dy₂) Apdx
(dx — dy₁) (d x — dy₂)
is a small fraction, we can put (d x d y₁) (d x — dy₂)
1
P = (dy, -dy,) Ap
2
d x
§ 17.]
1067
THE THEORY OF OSCILLATION.
P
This expression agrees exactly with the former one when we substitute
instead of E; hence the velocity of sound in air is
c =
P
Υ
When the theory of heat is discussed in the second volume, it will be
shown that a coefficient must be added to this formula in consequence of
the change of temperature, which necessarily accompanies the change of
density of the air. Since the heaviness of the air is proportional to the
pressure p, they both disappear from the formula and the temperature
alone remains. We generally assume for air
c = 333 √1 + 0,00367. 7 = 1092,5 √1 + 0,00367 . 7 feet.
EXAMPLE.-If (according to the Remark of § 351), when a column of
water is compressed by a force of 14,7 pounds, its volume is diminished
0,000050 of its original volume, its modulus of elasticity is
E
14,7
0,000050
and the velocity of sound in water is
C =
32,2.
294000. 144
62,425
294000 pounds,
32,2.
1693440
2,497
= 4673 feet,
or about 4,3 times that in air.
(§ 17.) Period of a Vibration.-We can now find the period
of a vibration by obtaining the equation, which expresses the de-
pendence of the amplitude of the vibration upon the time and
upon the abscissa a, which determines the position of the vibrating
element when it is at rest. Now y is certainly a function of t as
well as of x; we can, therefore, put y(t) and y = (x).
By differentiating the first equation, we obtain the variable ve-
locity of vibration
V =
dy
d t
=
Pi (t),
d v
Ρ
= 4. (t),
d t
2
and in like manner, by a second differentiation, the corresponding
acceleration
in which ₁ (t) and p. (t) denote other functions of t (compare § 19).
The second function gives the ratio
dy
= 4, (x),
d x
which determines the strain;
from it we obtain the latter
dy
S = AE.
— A E. 4, (x);
dx
x% is
hence the motive force of the element of the mass d M = Ad x is
g
1068
[§ 17.
GENERAL PRINCIPLES OF MECHANICS.
d SAE.
d [4₁ (x)]
d x
A E ₂ (x)
d x
and the corresponding acceleration is
d S
9 E
p
P = d M
Y₂ (x),
Y
in which ₁ (x) and 4, (x) denote other functions of x.
If we equate the two values of p, we obtain
g E
ΦΩ (t)
=
Y½ (x), or, since
g E
c²,
γ
γ
Φ2
P₂ (t) = c². 4, (x).
Φ
Φ
f
The integral of this differential equation is
y = ¢ (t) = ¥ (x) = F (c t + x) + ƒ (c t − x),
in which F and ƒ are undetermined functions of the quantities con-
tained in the parentheses; for
Φι (t)
Φ2
p₂ (t) =
d [o (t)]
dt
d [p₁ (t)]
d t
= c² [F₂ (c t
Y₁ (x):
Y₂ (x) =
d [4 (x)]
d t
d [4, (x)]
d t
= c F₁ (ct + x) + cf. (ct — x),
= c² F₂ (ct + x) + c² f₂ (c t — x)
+ x) + ƒ₂ (c t − x)], and
== F₁ (c t + x) — fi (ctx) and
= F₂ (ct + x) + ƒ₂ (c t − x),
and, therefore, we have really
P₂ (t) = c². 4₂ (x).
Φ2
Although the function.
Y F (ct + x) + ƒ (c t − x)
is an indeterminate one, yet, when we have more definite data in
regard to the vibrating body, it can be employed to determine the
period of the vibrations. A few examples of how this may be done
will now be given.
REMARK.-If we eliminate d t from the formulas dy
=vdt and d x =
cd t, we obtain the expression
d y v
d x
d y
or since
expresses the conden-
d x
V
sation o of the vibrating element of the body, we have σ = the simul-
с
taneous condensation at every point of the vibrating rod is proportional to
the velocity of vibration of that point.
$ 18.]
1069
THE THEORY OF OSCILLATION.
(§ 18.) Determination of the Modulus of Elasticity.-Let
us assume that the vibrating body, whose length is 1, is fixed at
both ends. In this case we have not only for x = 0, but also for
x = 1, y = 0; hence
F(ct) + f (ct) = 0 and F (ct + 1) + f (c t − 1) = 0.
From the first equation we obtain ƒ F, which, when sub-
stituted in the second equation, gives
+
-
f(ct 1) − f(c t − 1) = 0, I.E. ƒ (ct + 1) = f (ct — l),
or, if we put c t − 1 = c t₁,
f
ƒ (ct₁ + 2 1) = ƒ (c t₁).
The function, therefore, assumes the same value when c t, is in-
27
creased by 2 l or when the time is increased by t₁
;
hence the
с
period of a complete vibration or double oscillation is
21
t₁
ειν
C
γ
9 E
If, in the second place, we assume the body to be free at both
ends, we have for x = 0 and x = 1, S′ = 0 and 4, (x) = 0; hence
F(ct)f(ct) = 0 and F₁ (ct + 1) − f, (et − 1) = 0.
We have, therefore,
ƒ₁ = F, and f₁ (c t + 1) = f (c t − 1), or f₁ (c t₁ + 21) = f₁ (c t₁),
and consequently the period of a complete vibration is
t₁
21
C
If the body is free at one end and fixed at the other, we have for
X = 0, y = 0, and for x = 1, S = 0; hence
F (ct) + f (ct) =0 and F (et + 1)f (ct 1) = 0,
from which it follows that f= F and f, F, and therefore
fi (ct + 1) + f (ct 1) = 0, or f (c t₁ + 2 1) = — fi (c t₂).
We see from the latter formula that the body, after the time t,
27
C
will assume the opposite state of motion, and that it will con-
sequently make a complete vibration in double that time, 2 tr
47
с
The period of the complete vibration is, therefore,
to
47
C
Y
47 V
g E'
or double that in the first two cases.
1070
[§ 19.
GENERAL PRINCIPLES OF MECHANICS.
By means of these formulas we can calculate from the period t
of a complete vibration or from the number n of vibrations, which
a prismatical body makes in a given time, the modulus of elasticity
27 Y
and the velocity of propagation or the velocity of
E
(2)²
•
g'
sound in it, c =
21
t
EXAMPLE.-An iron wire, which was 60 feet long and was fixed at both
ends, was put in longitudinal vibration by means of friction in the direc-
tion of its axis. The number of complete vibrations was 1637 in a second;
what was the nodulus of elasticity of the wire and what was the velocity
of propagation in it? According to one of the above formulas, we have
for the modulus of elasticity, expressed in units of length,
L
1
( 2 n 1)²
t
g
(1637. 120)²
32,2.12
= 99870000 inches,
and if a cubic inch of this iron weighs 0,28 pound, the modulus of elasti-
city, expressed in pounds, is
=
E = 99870000.0,28 27960000 pounds (compare the table, § 212).
The velocity of propagation, or the velocity of sound in it, is
.C
√ g L
√32,2.99870000 . 1/2 ✓ 16,1. 16645000
16370 feet,
or, assuming the velocity of sound in the air to be c = 1092 feet, we have
16370
1092
= 15.
(
If the vibrating wire is very long, the period of a vibration depends
upon the length of the wave or upon the distance l between two nodes,
and it is always t
1
27
С
•
This time determines the pitch of the note pro-
duced by the vibrating wire; the greater or smaller t, is, the lower or
higher the note is. The intensity of the sound, on the contrary, increases
with the amplitude of the vibration. For spherical waves, in which sound
propagates itself in air and water, c and t remain unchanged, and it is only
the amplitude of the vibration, or the intensity of the sound, which
diminishes.
(§ 19.) Transverse Vibrations of a String.—The transverse
vibrations of a string or elastic rod can be treated in the same
manner as the longitudinal ones. As the simplest case is that of a
stretched string (Fr. corde; Ger. Saite), we will discuss that first.
Let A D B, Fig. 892, be any position of the vibrating string, A
and B the two fixed points, 7 = A B the length of the string, Gits
weight and the tension, which is to be regarded as constant.
Now if A N = x and NO = be the co-ordinates of any point O of
19.]
1071
THE THEORY OF OSCILLATION
the string, and if we resolve the tension S at it into two components.
K and P, one parallel to A B, and the other perpendicular to it,
FIG. 892.
Pi
S

D
U
K
T
Ki
N
A
SE
P
B
we can regard the latter as the motive force a one end of the
element O Q. If the arc A O = s is increased by the element
O Q = d s, and if the corresponding increase of the ordinate y is
QTdy, P, S, dy and d s are the homologous sides of two sim-
ilar triangles O P S and Q T 0, and we can put
S
Р Q T
O Q
d y
dy
or P =
S.
d s'
d s
RU
d Yv
But another force P₁:
S =
S, which is one of the
Q R
d s
components of the opposite tension, acts in the opposite direction
upon the same element OQ; hence the motive force, which moves
the element OQ back to the axis A B, is
P – P₁
= (dy = dy₁)
S.
d s
The mass M of this element is proportional to its length
O Q = ds; now if we suppose the amplitude y of the oscillation
to be small, we can assume that the mass is proportional to the
element 0 T = Q U = d x of the abscissa, or that M
d x
G
g
If we make this assumption, we have the acceleration with which
the element approaches its position of rest A B
P – P₁ d y
p =
M
d yr g s l
ds.dx
G
or, if we put ds =
d x,
p
dyd y
d x²
g S l
G
dy
Now y is some function of x, E.G.
(x); hence is another
d x
function 4, (2) and dy - dy,
d x
function (x) of this quantity, and
d a y
d x*
d [4, (x)]
is a third
d x
1072
[$ 20.
GENERAL PRINCIPLES OF MECHANICS.
9 8 1
G
p = 4₂ (x).
Since y is also a function of the time t, I.E. y = ¢ (t), the ve-
locity with which the element O Q returns to its position of rest is
(t), and the corresponding acceleration is
V
d y
d t
=
p =
α φ, (t)
d t
= ₂ (t).
Þ₂
If we equate these two values of p, we obtain, as in § 17 of the
Appendix, the differential equation
P₂ (t) = √₂ (x).
42
g S l
G
= c² 4₂ (x),
and we can put here, as we did there,
y = 4 (t) = 4 (x) = F(ct + x) + f (ct — x) and
v = c [F₁ (ct + x) + fi (c t — x)].
Since here also for x = 0 and x = l, y and v = 0, we have
again f F and f (ct + 1) = f (et − 1), or f (c t, + 27) =
f(ct); hence the period of a complete vibration or double oscil-
lation is
21
t₁
C
2 7 4
I SP
G
or,
if we put G
G =
= Aly,
WAY
ΑΥ
t₁
21
I S
The period of vibration of a string is therefore directly propor-
tional to the length l and to the square root of the weight of the unit
of length, and it is inversely proportional to the square root of the
tension S of the string.
EXAMPLE. Since half the period of the vibration corresponds to that
of the next octave, a string will give, according to this formula, the octave
of the fundamental tone, when it is shortened one-half or supported in the
middle, or when it is stretched four times as much, or when it is replaced
by another whose unit of length weighs one-fourth as much as that of the
first one.
(§ 20.) Transverse Vibrations of a Rod.-The period of vi-
FIG. 893.
R
N
B
D
QUIR
bration of an elastic rod or spring
A B (Fr. lame; Ger. Stab), Fig.
893, which is fixed at one end,
can be determined in the follow-
ing somewhat circuitous man-
ner.
According to § 226, if r
denotes the radius of curvature
of the rod at a certain point 0,

§ 20.]
1073
THE THEORY OF OSCILLATION.
determined by the co-ordinates C N = x, and N 0=y₁, the moment
of flexure of the arc A O = s₁ is
WE
M =
j
x
If we put the force, with which an element Q, which corre-
sponds to the co-ordinates CR = 2 and R Q = y, approaches the
axis or position of rest C B, P d x, or its moment
= NR. Pdx = (x,
WE
r
21
for (22
But (X1 - x) P d x =
1
x) P d x, we obtain
x) P d x.
วา
S" P x d x - f." P x d x .
S
21
= Pdx-Padz,
X1
or, if we put "Pdx = P₁, and therefore
1
ƒ"
P
.
d x . x = P₁ x, − ƒˆ P₁ d x,
S" P x d x =
ƒˆ (x, − x) P d x
WE
2
Now we know that r =
P, da; hence we have also
= SP,
troduction to the Calculus),
tion is small, ds = d x,
1
S" P₁d x.
d s³
dx² d (tang. a)
(see Art. 33 of the In-
or, since we can put, when the deflec-
d x
7 =
; hence
d (tang. a)
WE
d
a (tang. a) = f P₁d x,
d x
by differentiating which, we obtain
a))
WE. d (d (tang. a)
d x
4 (x), tang. a =
dy
d x
1
= P₁dx.
d (tang. a)
d x
4₁ (x),
(x), we obtain the equation
If we put y
4½ (x) and d
(d (tang. a)
d x²
P₁
WE. 4, (x),
d Pi
d 4s (x)
P =
WE
by differentiating which again, we find
WEd 3 (x), I.E. P d x =
WEd (x), or
WE, (x).
d x
$
68
1074
[$ 21.
GENERAL PRINCIPLES OF MECHANICS.
In order that the spring shall vibrate symmetrically, we can as-
sume that P is proportional to y, or that P
have
W E 4, (x) = K y, or ¼ (x)
K
when we denote by kt.
WE
4,
Ky; hence we
K
=
WE• Y = k¹y,
k
This differential equation . (x) = k* y corresponds to the equa-
tion y = (x) = A cos. (k x) + B sin. (k x) + C² e² ² + De¯* ²;
for by successive differentiations we obtain
[-
4₁ (x) = k [— A sin. (k x) + B cos. (k x) +
½ (x) = b² [— A cos. (k x) - B sin. (k x)
43 (x) = h³ [A sin. (k x)
√4(x) = k* [A cos. (k x)
so that we have really
-
I
+
C et ²
Cek ² — De* ²],
+ De**],
B cos. (k x) +
+ B sin. (k x) +
C eti
I
44 (x)
= k* y.
I
-k
-k I
De¯**], and
C et = + D ex],
(§ 21.) The period of vibration t of the elastic rod is found, as
above, by substituting p = ₂ (t)
upon an element is
Φ.
force
But the force acting
mass
= Pdx-Kydx WEk' y d x,
-
-
and, when the cross-section is F and the heaviness is y, the mass is
γ
=Fdx ; hence
g
g WEK
Φε (1)
=
FY
Y
µ³ y,
g WEk
by µ³.
Fy
when we denote the expression
This differential equation corresponds to the simple formula
y = 4 (t) = sin. (µ t + T),
in which expresses any arbitrary time of beginning; for by dif
ferentiation we obtain
d y
บ
d t
Φι (t) = μ. cos. (µ t + ↑) and
d v
p =
d t
P₂ (t) = — μ² y.
If in the equation y
y
sin. (µ t); hence for µ t =
quently
= 4½ (t) = — µ³. sin. (µ t + 7), I.E.,
Φ.
sin. (µ t + T) we take 7 = 0, we obtain
2 π, etc., y = 0, and conse-
0, π,
1
§ 21.1
1075
THE THEORY OF OSCILLATION.
t₁
11
π
is the period of a simple vibration and
μ
2 π
2 п
t
ре
FY
g WE
is the period of a complete vibration.
In order to calculate the period of a vibration, we must know
F
not only the quantity k, but also the ratio W
If the rod is cylindrical and its radius = r, we have
F
W
Прог
į π pot
4
4
рог
(see § 231),
F
W
b h
12
1
12
T = b h ³
3
h²
(see § 226).
and if it is a parallelopipedon, whose width is 6 and whose height
is h,
We have, therefore, for the first rod
t =
4 π
rh."
Y
and for the second
9 E'
4 π
h h
3 y
9 E
Ye-hate
The quantity k is found in the following manner from the
equation
y = A cos. (k x) + B sin. (k x) + С ek z + Dekx.
If we substitute in this formula the corresponding values x 1
0, we obtain
and y
1) 0 A cos. (k 1) + B sin. (k l) + C p²² + De¹¹.
If we perform the same operation in the equation
tang. a =
d y
dx
(x), we obtain
l
2) 0 A sin. (kl) + B cos. (kl) + Cek + Dek?.
=
Since the moment of flexure at the end A of the rod = 0 and
consequently the radius of curvature r∞, or y (x) = 0 and
P3 (x) = 0, it follows that
0 A cos. 0 — B sin. 0 + C'eº + De˜º, I.E., A+ C + D = 0
and
0 = A sin. 0 B cos. 0 + C e°
whence
De, I.E.,
· B + C − D = 0,
3) A = C
+ D and
4) B = C
D.
If we eliminate A and B from these four equations, we have
(C + D) cos. (k l) + ( C − D) sin. (k 1) + C' ek ¹ + De-kl
l)
(C + D) sin. (kl) + (CD) cos. (kl) + Ce
0, and
De² = 0;
jehom
7000
Steel voo
YL
}}
1067
80000
1076
[§ 21.
GENERAL PRINCIPLES OF MECHANICS.
from which we obtain by addition
C cos. (kl)
and by subtraction
D sin. (kl) + Ce² = 0,
l
D cos. (kl) + C sin. (k l) + De˜k ² = 0, or
C[cos. (kl) +ek 1] = D sin. (kl) and
D [cos. (kl) + e-k] = C sin. (kl);
hence we have by division
cos. (kl) + ekl
sin. (kl)
sin. (kl)
cos. (k l) + e‍k li
whence
2 + cos. (k l) (ell + e−k ¹) = 0, or
2
cos. (kl)
=
ek l + e-k l°
The smallest of the different values, which correspond to the
different tones that the rod can give out and which depend upon
the number of nodes, is k = 1,8751; the greater are, on the
contrary, nearly
3 п 5 п куп
2 2
k l
2
l
etc.
If we are required to find from the observed period t of the
complete vibration the modulus of elasticity E, we have generally
to consider but the smallest value; we must, therefore, put
1,8751
and k² =
3,516
;
hence for a cylindrical rod
4ج
k
E
Y
2 (1177) =
ΑΠ 2
Y
4 π ľ²
2
Y
= 12,774
g 13,516 r t
I p² t²⁹
γ
3 g
(35
4 п
3,51772)²= 4,2579.
•
g r k
2
and for a parallelopipedical one
E
γ 4
3 g h k²
(177)
ht
REMARK 1.—If we compare with each other the formulas
4 π
t
r k²
γ
g E
1
and t₁ = 2 l₁
27, 1
γ
g E
I h² to
for the transverse and longitudinal vibrations of one and the same rod, we
obtain the proportion
t: t₁
12 3,516
r 2 π
72
l₁, I.E., t: t₁
: 0,5596 7.
1
Wertheim found by experiment that this proportion was correct for
cast steel and brass.
·
REMARK 2.-The transverse vibrations of an elastic rod are discussed
by Seebeck in a "Abhandlung der Leipziger Gesellschaft der Wissenschaften,"
Leipzig, 1849, and also in the "Programme der technischen Bildungsan-
stalt in Dresden," for the year 1846. Wertheim's experiments upon the
elasticity of the metals and of wood by means of transverse and longitu-
§ 22.]
1077
THE THEORY OF OSCILLATION.
dinal vibrations are discussed at length in " Poggendorff's Annalen,"
Ergänzungsband II, 1845.
REMARK 3.—The period of vibration or rather the number of vibrations
of a rod in a given time cannot generally be determined directly on
account of their rapidity; we must, therefore, employ various artifices to
do it. We can determine it either, as Chladni, Savart, etc., did, by the
pitch of the note produced by the vibration, or we can employ the method
first proposed by Duhamel, which consists in causing the rod to describe
by means of a small point a wave-line upon a revolving glass plate, which
is covered with lamp-black. A chronometric apparatus, to which a flying
pinion, such as used in the striking works of town clocks, is attached, is
employed to produce a regular motion of rotation. An account of this
apparatus is to be found in Morin's "Description des appareils dynamo-
metriques, etc., Paris, 1838," as well as in his "Notions fondamentales de
mécanique." Wertheim determined the number of vibrations in a given
time by allowing another body, such as a tuning-fork, whose number of
vibrations was known, to vibrate at the same time with the rod to be ex-
amined. If we cause both bodies to trace wave-lines upon the lamp-black
and then count the number of waves corresponding to the same central
angle, the ratio of these numbers will give the ratio of the numbers of
vibrations. The longitudinal vibrations are generally accompanied by
small transverse ones; the rod describes, therefore, a corrugated wave-
line. By counting the small waves contained in one large wave of the
main wave-line, we can easily compare the number of longitudinal vibra-
tions with the number of transverse ones.
§ 22. Resistance to Vibration. The forces, which cause the
vibrations of a body, are very often accompanied by passive resist-
ances, whose influence must be examined more particularly. If
such a resistance is constant, as, E.G., the friction of a pendulum
upon its axis or that of a magnetic needle upon its pivot, it has no
influence upon the period of the oscillations, but their amplitude
is diminished at every stroke. For the case in § 1 (Appendix), in
which the motive force is proportional to the distance a from the
position of rest or centre C of the motion A B, Fig. 894, we can put
p = µ x = µ (α — x,),
R
FIG. 894.
D
B
MNP C
(a
in which x, denotes the space A M de-
scribed. If we take into consideration
the diminution h of this space, in con-
sequence of the friction, we have, when
the body is describing the first half
A C of its path,

—
p = μ (ak x₁),
1078
[$ 22.
GENERAL PRINCIPLES OF MECHANICS.
and when it is describing the second half C B
p = µ [x, ‒ (a + h)];
the influence of the friction consists, therefore, in this alone,
that for one-half of the path a must be replaced by a k and for
the other by a +k, and that the whole space described in one
oscillation must be changed from 2 a to 2 a 2 h, I.E. the ampli-
tude of the oscillation will be diminished a certain quantity 2 h at
each oscillation. Finally, since the amplitude does not enter into
the formula
t =
γ
νμ
W
k can have no influence upon the period of the oscillations.
The case is different with the resistance of the air. The latter,
when the velocities, as in the case of the pendulum, are small, is
more nearly proportional to the simple velocity than to its square,
as was shown by Bessel's researches upon the length of the simple
pendulum (Abhandl. der Akademie der Wissensch. zu Berlin, 1826).
This is explained by the fact that this resistance is increased prin-
cipally by the condensation and rarefaction of the air in front and
behind the vibrating body, which increase with the velocity v of
the body (see § 510 and Appendix, § 17, Remark). In accordance
with this assumption, we can put the acceleration of the vibrating
body
Ρ
(μ x + v v) or p + v v + µ x = 0,
when we assume the body to be moving from the point of repose
and measure the space from that point.
If we put
d x
d v
x = ƒ (t), v
=
=
= f₁ (t) and p =
d t
d t
= f₂ (t),
we can write also f₂ (t) + vfi (t) + µ ƒ (t) = 0, which corresponds
to the integral equation
ντ
x = [b cos. (4 t √µ) + b, sin. († † √µ)] ez,
in which b and b, denote constants to be determined and
22
1
Now for t = 0, x = 0, whence b =
0; hence we have
Αμ
more simply
ντ
x = b₁ sin. (4 i √µ) e-.
1
Since this value becomes = 0, when t
an oscillation or simple vibration is
=π, the period of
§ 23.]
1079
THE THEORY OF OSCILLATION.
π
π
ψ νμ
Ni
1
1
I.E.
22
1
4 μ
ν
4
times as great as if the resistance of the air were not present.
REMARK.—It is easy to explain why bodies which are set in vibration
make smaller and smaller oscillations and finally come to rest. This effect
is due to two causes, the resistance of the air and the imperfect elasticity
of the vibrating body; in consequence of the latter fact, the contraction
and expansion of the body, particularly within a short space of time, is not
proportional to the forces acting upon it.
§ 23. Oscillation of Water.-The simplest case of the wave
motion of water is that presented by its oscillations in two communi-
cating tubes A B C D, Fig. 895. Let us assume that both have
H
FIG. 895.
R
U
B
L
D
the same cross-section, and let us imagine
the surface of the water in one leg to be
raised a certain distance H A = x above the
position it occupies when at rest, and that in
the other leg to be depressed an equal dis-
tance R D = x. We have here the motive
force

P = A.2xY,
and if I denotes the entire length A B C D
= HB CR of the water, the mass moved is M =
Aly
g
; hence the
acceleration with which the surface of the water rises or falls is
P
2 A x Y 2 g x
p
g=
M
Aly
7
Since this formula corresponds exactly to the law of oscillation
pμx, discussed in § 1 and § 2 of the Appendix, we have for the
period of an oscillation
P
π
t =
= π |
νμ
2 g
Since the period of the oscillations of the simple pendulum, whose
length is
201
7
is
t = π 1
2 g
the oscillations of the water in the communicating tubes are iso-
chronous with those of this pendulum.
If both legs of the tube A B C D, Fig. 896, are inclined, I.E. if
1080
[§ 23.
GENERAL PRINCIPLES OF MECHANICS.
the axis of one of the tubes forms an angle a and that of the other
an angle ẞ with the horizon, the space A H = D R = x, which
the surface of the water describes upwards in one and downwards
in the other leg, corresponds to the difference of level
H
X
z = x sin. a + x sin. ß = x (sin. a + sin. ß) •
FIG. 896.
K
R
J
D
L
hence the force is
P = A y x (sin. a + sin. ß),
the acceleration is

p
g (sin. a + sin. B).x
7
and the period of the oscillations is
B
t = π V
g (sin. asin. B)
1
If, finally, the tubes are of different widths, the determination of
the period of the oscillations becomes much more complicated. Let
A be the cross-section and 7 the length of the middle tube, a₁, Ai
and 1, the angle of inclination, the cross-section and the length of
one lateral tube, and a,, A, and 7, the angle of inclination, the cross-
section and the length of the other; finally, let us suppose that the
surface of the water in the axis of one tube has risen a distance x
and that the surface of the water in the axis of the other has sunk
a distance x. We have then
2
2
A₁ x₁ = A₂ X2, whence x₂ =
1 201
x2
A₁
X,
A₂
P = A₁ (x₁ sin. a₁ + x2 sin. a,) Y
Αγ
A₂
(A½ sin. a, + A₁ sin. aŋ) x1.
and the motive force, reduced to A₁,
to
A
A
The mass of the water in the middle tube is constant and equal
Aly
g
and, since the ratio of its velocity to that of the force is
4, the mass reduced to the point of application is
2
(4)². 47y.
Αιγ.
The mass of the water in the first leg is
A₁ (l + x₁) Y, and that in the second
g
Ag (lą - X₂) Y
A2
(12
$24.1
1081
THE THEORY OF OSCILLATION.
or reduced to the point of application of the force
2
A. (la
= (-4)* 4₂ (1₂ — x₂) Y
Finally the mass moved by P is
M=
Ꭺ Ꭵ
2
2
g
γ
+
• Z G
+
G
Αγ
g A
7
+
= A₁²
g
g
1₁ + X ₁ 12 X2
+
A₁
A
+
l₂ X1
+
1
A2
A₁
A₂ 2
1
and the acceleration is
7.
A₁
G
4 + 4/2 + (-2)²²],
A, A
Z₁
A
(sin. a + sin. a) gx,
A,
1
1
P
A,
p
M 7
Va
1
+
+
A
A₁
A2
+
G-4,
1
X₁
1
A
2
If the cross-sections of the two tubes were the same, we would
have A₁ = A„, and therefore
P
1
(sin. a, + sin. a.) gx,
l₁
G+
A₁
+ 2) 4₁
(sin. a, + sin. a.) g xɩ
A₁
l
+ 12 + 72
A
and the period of the oscillations
t = π
πι
g
1
A₁ l + A (l₁ + 1)
A (sin. a, + sin. a.)
REMARK.-In consequence of the friction and of the resistance due to
the bend in the tube, these formulas must, of course, be modified (com-
pare Appendix, § 25).
§ 24. Elliptical Oscillations. If a body, which is driven with
an acceleration p = µ z = µ . C' M towards a fixed point C, Fig. 897,
W
C M
A
K
FIG. 897.
B
B₁
A1
possesses an initial velocity c, whose
direction differs from that of the
force, the oscillations no longer take
place in a straight line, but in an
ellipse, as we will now proceed to
prove. Let the direction of the mo-
tion at the point of beginning A be
at right angles to the distance СA
a and let the corresponding ve-
locity be = c. If we pass the co-ordi-
nate axes through C, one upon and the other at right angles to

3
1082
[$ 24.
GENERAL PRINCIPLES OF MECHANICS.
CA, and denote the co-ordinates C K and K M by x and y, we
have for the components q and r of p = µz, which are parallel to
the axes, since
p
X
and
j Y
P
q = µ x and r = µ y.
If u and v are the components of the velocity w of the body M,
which are parallel to the axis, we have, according to § 1 of the
Appendix,
and at the same time
W = √μ (a² — x²);
c² — v = z fra y = 2 µ Syd y = y', whence v = √c² — µ y².
µ
Since for y = b, v = 0, it follows that
0 =
=
C²
μ b²; hence c = b √μ and v =
√µ (b² — y²).
But now u =
d x
d t
d y
and v =
and therefore
d t'
a²
x²
d x
d y
, I.E.,
12
NaⓇ
x²
Nb2
y³
a (
2 d x
d y
=
b² — y²
a ()
d
1
or
d
V 1
hence (according to Art. 26, V, of the Introduction to the Calculus)
-1
Y
= sin. + Con.
a
or, since for x = a, y
sin.-1
X
α
0,
α
0
sin.
1
-1
a
b
= sin. + Con., or
sin.-¹ 1 = sin.-¹ 0 + Con., I.E.,
π
= Con. and
2
X
sin.-
-1
18
= sin-
-1
Y Π
+
2'
or
sin.-1
X
sin.
-1
y
П
a
b
2.
When the difference of two arcs is
to the cosine of the other, I.E.,
2
Π
the sine of one is equal
X
√ 1
a
· (3)',
or
( )² + (~)²=
1.
§ 24.]
1083
THE THEORY OF OSCILLATION.
Since this is the equation of an ellipse, it follows that a point,
which is impelled or attracted towards C with an acceleration µ z,
will describe an ellipse, whose semi-axes are C A = a and C B = b.
We have also
d y
d t =
V
d y
√ μ (b² — y²)
; hence the time is
t =
√²sin. 12/
or inversely,
μ
y = b sin. (t Vμ) and x = a cos. (t Vμ).
The time, in which the body will describe a quadrant of the ellipse,
is found by putting y b, and it is
t₁ = √¹
1
Ъ
1
T
sin.-¹
sin.-¹ 1 =
μ
b
μ
The time, in which the body describes half the ellipse, is
π
2 t₁ =
Αμ
and the period of a complete revolution or of a complete vibration is
2 п
4 t₁
or exactly the same as it would be, if the motion were a rectilinear
reciprocating one. It follows also that
- = 1 −
И
a =
√µ (a³ — x²)
a² [cos. (t √µ)]') = ¦µ a sin. (t √μ)
and
µ (a²
v = √µ (b² — y³) =μ b cos. (t vμ);
hence the velocity of revolution is
w = √ u² + v²
Finally, we can put
a + b
2
= μl
V
√ (a sin. t √µ)² + (b cos. t √µ)².
2
a
b
X =
cos. (t vµ) +
cos. (t vμ) and
a + b
а
Ъ
Y
sin. (Và)
2
2
sin. (t õ);
now since the first members
a + b
2
a + b
2
cos. (t Vμ) and sin. († √µ)
correspond to a uniform motion in a circle, whose radius is
a + b
2
and since the two other members correspond to an opposite uni-

1084
GENERAL PRINCIPLES OF MECHANICS.
form motion in a circle, whose radius is a
2
b
[$ 25.
we can also assume
that the elliptical motion of the point is composed of two circular
ones, I.E., that the point describes uniformly a circle, whose radius
a
—
is
2
b
while the centre of the latter moves uniformly in a circle,
whose radius is
a+b
2
If b = 0, the oscillation takes place in a straight line, but we
can imagine it to be composed of two equal opposite circular
motions.
§ 25. Waves of Water.-According to the accurate obser-
vations of the Weber brothers, an example of elliptical oscillation
is presented by the motion of waves of water (Fr. ondes; Ger.
Wasserwellen). Not only every particle on the surface, but also
every particle below it describes in the wave motion an ellipse.
On account of the resistance on the bottom the ellipses below the sur-
face of the water are smaller than those at it, and in general they de-
crease with the distance from that surface. The different elements
in the surface of the water, as well as those in any other plane
parallel to it, are at the same moment in different phases of mo-
tion; while an element A, Fig. 898, is beginning its path at (0),
M
D
FIG. 898.
P
an element B is already at (1), a second C is at (2), a third D
at (3), a fourth E at (4); at this moment the vertical section
of the surface of the water is a cycloidal or trochoidal curve
ABCDEFGH J. Before the wave motion began, the elc-
ments were at the centres K, L... N of their trajectories and
formed the horizontal surface K N of the water; during the wave
motion, on the contrary, part of the elements are above and part
are below this line, and all have, of course, a tendency to return to

$25.]
THE THEORY OF OSCILLATION.
1085
their positions of rest K, L... N. The oscillations are, however,
elliptical so long only as the waves remain unchanged; if they de-
crease gradually in magnitude, the path of each element becomes
narrower and narrower and no longer forms an ellipse, but a spiral
line. On the other hand, when the waves are forming or increas-
ing in size, the elliptical trajectory is formed gradually from a
spiral line.
=
After one instant A has moved in its trajectory to (1), B to (2),
C' to (3), etc., and the wave-form has been moved forward in conse-
quence through the horizontal distance KL between two elements;
after a second instant A is at (2), B is at (3), C' is at (4), and the
wave-form has again advanced the distance KL LM; thus, as
the elements of the water revolve, the wave-form advances more
and more, and when an element has made a complete revolution,
the wave has advanced its own length K N. When an element has
made half a revolution, as is shown in Fig. 899, the place of the
FIG. 899.
E
K
M.
B
H
wave-crest is occupied by a trough or sinus, and that of the latter
by a crest. This advance of the wave-form does not, of course,
consist in any particular motion of the water, but in the forward
motion of the same phase, E.G., in the forward motion of the crest
J (Fig. 898) of the wave to O, P, etc. If the period of a revolu-
tion t of an element of the water and the length A J=s of a wave
are known, we can calculate the velocity of propagation by means of
the formula c=
S
t
The height of a wave, or the sum of the height of the crest and
the depth of the trough is equal to the vertical axis 2 b of the
ellipse, in which the elements of the water revolve; the length CG
of the trough exceeds the half length of the wave by the length 2 a
of the horizontal axis of the ellipse, and the length of the crest is, of
1086
[§ 26.
GENERAL PRINCIPLES OF MECHANICS.
¡
;
course, that much shorter than half the wave length. Hence the
cross-section of the trough of a wave is larger than that of the wave-
crest; now since this is impossible in consequence of the invariabil-
ity of the volume of the water, the centre of the elliptical trajectory
must be somewhat above the surface of the water when it is at rest.
§ 26. Webers' Experiments.-According to Webers' experi-
ments, the path described by a particle of the water at the surface
of a wave is a slightly compressed ellipse; according to Emy, on
the contrary, the particles of water in sea-waves describe upright
ellipses. Both axes of the elliptical path decrease as the depth
below the surface increases, and according to Weber the horizontal
axis decreases more rapidly than the vertical one.
The wave ap-
pears not to be propagated in a vertical direction; elements verti-
cally below each other are, according to the observations of the
Weber brothers, in the same phase at the same time; on the con-
trary, those situated in a horizontal line form a complete series of
the different phases of the motion. From the experiments cited
above, it appears that the period of revolution of an element, or the
time in which a wave is propagated its own length, depends prin-
cipally upon the ratio of the two axes of the path. The greater
the ratio of the horizontal axes 2 a to the vertical one 2 b, the
greater is the period of revolution. The particles, which lie deeper,
describe their paths more quickly than those at the surface; from
this we must conclude that the wave length diminishes towards
the bottom.
The velocity of propagation c =
time of revolution í increases with
S
t
of a wave depends, since the
the ratio, not only upon the
length s, but also upon the height b. If a wave is propagated be-
tween two parallel walls, E.G. in a canal, its width remains con-
stant, its height b diminishes and its length increases in such a
manner that the only change in the velocity of propagation is that
resulting from the friction of the water upon the walls. If, on the
contrary, a wave can propagate itself freely in all directions, and if
it forms a wall which recedes into itself, its length and width are
both increased at the expense of its height, and the wave becomes
gradually flatter and flatter until in a short time the eye is no longer
able to distinguish it. If such a wave is not originally circular it
will gradually approach the circular form as it advances. Accord-
ing to Webers' experiments, the height diminishes in arithmetical
$27.1
1087
THE THEORY OF OSCILLATION.
progression when the wave advances in geometrical progression.
The velocity of propagation of such a wave diminishes gradually,
the farther the wave is propagated. If, on the contrary, a wave is
propagated from without inwards and is contracted more and more
in consequence, its height, length and velocity gradually increase.
There is, therefore, a great difference between the waves of water
and those of sound. In the latter the velocity of propagation de-
pends upon the elasticity and density of the medium alone; in the
former, on the contrary, it is a function of the length and height.
If the undulations of the water are produced by a force which acts
almost instantaneously, E.G., by the immersion and quick with-
drawal of a solid body, the particles of the water describe elliptical
paths which gradually decrease, or rather spiral lines, which draw
themselves together more and more, and the periods of revolution
become smaller and smaller. The origin of a whole series of waves,
which become smaller and smaller, is to be attributed to these rela-
tions of motion. As the waves are propagated farther and farther,
those which follow are increased in size by those which have pre-
ceded them, and those most in advance in a short time become so
flat as to be invisible. This running together of the waves gives
rise to systems of small waves, which present themselves like teeth
upon the front surface of the main wave. These small waves or
teeth advance, according to Poisson and Cauchy, with uniformly
accelerated motion.
§ 27. Hagen's Experiments.-According to the latest in-
vestigations of Geh. Oberbaurath Hagen (see the "Seeufer-und
Hafenbau von G. Hagen, Berlin, 1863," 1 Vol., which forms the
third part of that author's "Wasserbaukunst;" also his treatise
upon waves in water of uniform depth; Berlin, 1862), the particles
of water of waves in deep water describe with constant angular
velocity circles, whose diameters decrease as the depth increases, and
at the bottom they are infinitely small. A filament of water, which
when at rest is vertical, will oscillate, in consequence of the wave
motion, backwards and forwards about this vertical line, its base
remaining fixed very much as a stalk of wheat is moved by the
wind. The line of the wave or the curve which unites the points,
which are in the same phase of revolution and which, when the
water is at rest, is a straight line, is therefore a prolate cycloid,
that becomes more and more prolate as the depth increases; at
the bottom it is nearly a straight line and at the surface it ap-
1088
[§ 28.
GENERAL PRINCIPLES OF MECHANICS.
proaches the common cycloid. From the radius r of the common
cycloid, whose value for high sea-waves rises to 50 feet, we obtain
the length of the wave 2 r, its velocity of propagation
7 = π
•
c = √ 2 g r =
g l
the period of a wave
t
πι
V
2 r
1
с
g
π
g
and the angular velocity with which the molecules of water describe
their elliptical paths, w =
C
The centre of the circle, in which a particle which is situated
lower down revolves, is determined from the radius z of this circle
and from its distance y from the centre of the first circle, whose
radius is r, by means of the formula
y =
= 1'
ጥ
Z
By inversion we obtain z = re, in which e = 2,71828 de-
notes the base of the Naperian system of logarithms. We can
easily understand from this that the circles of oscillation decrease
very rapidly with the depth; for r = 10 feet, at the depth y = 50
feet, z= 10 .e 3,50 feet, and at the depth y = 200 feet,
z = 10.
10. e-0,05 = 0,15 feet.
-0,2
When the waves are of small constant depth, as Mr. Scott
Russel had already remarked, the horizontal motions of the parti-
cles of water, which lie above one another, are equally great; the
filament of water, which was originally vertical, remains so during
the wave motion, but its length and thickness vary. The different
particles describe closed curves of equal horizontal diameters and
of variable vertical ones, which decreases gradually with the depth;
they are, however, ellipses only when we suppose that the height
of the wave is infinitely small compared to the depth of the water.
When the depth of the water is finite and the height of the
waves is great, the laws of the motion of the waves are very com-
plicated.
§ 28. Interference of Waves of Water.-If two water-
waves cross each other, the same general phenomena occur as in
the case of waves of air and other fluids; after they cross each other,
each wave continues its motion as if they had not met; but accord-

§ 28.]
THE THEORY OF OSCILLATION.
1089
ing to Weber's observations, it is accompanied by a small loss of time,
so that a wave requires a little more time to pass from one point to
another when it passes through another wave than when it is prop-
agated freely. If two crests come together, a crest twice as high as
the first is produced, and in like manner when two troughs meet, a
third, twice as deep, is formed. According to Weber's experiments,
the ratio of the height of the simple wave to that of the compound
one is 1: 1,79. When two waves interfere, or when a waye-crest
coincides with a trough of a wave, the two counterbalance each
other, and the point where this occurs remains at the same level as
the surface of the still water. The paths of the single particles,
when two waves meet, become straight lines, which are vertical at
the crest, but at a distance from it their positions are such that
they are inclined towards the crest.
If a wave of water impinges against a solid wall, it will be re-
flected by it as if it came from a point as far behind the wall as
that from which the wave started is in front of it, and the reflected
wave will pass through the one which is arriving exactly in the
same manner as any two waves, which cross each other, do.
In Fig. 900, I, II to V, the phenomena, which are presented
FIG. 900.
I
II
HI
IV
D
DI
EE
Cr
G
D
Di
C
B1
Dr
CC
E
B
Bi
Di
E
BBi
Di
E
C
when a wave ABCDE is reflected by a rigid wall M N, are re-
presented. In I the crest C D E of a wave is arriving at the wall
69

1090
[§ 28.
GENERAL PRINCIPLES OF MECHANICS.
M N and the reflection begins in the form of a wave C, D, E,; in
II the top of the crest D of the wave has arrived at the wall and
has combined with the half D, E, of the reflected crest of the wave;
half a crest C G of almost double the height is thus produced. In
III the trough A B C of the wave has just reached the wall, while
the reflected crest C, D, E, is passing over it; an interference is
thus produced which causes the wave to disappear entirely. In IV
the bottom B of the trough of the approaching wave coincides with
the bottom B, of the trough of the reflected wave; a trough A S
of double the depth is thus formed. Finally, in V the approaching
Wave A B C D E is reflected completely by the wall M N and thus
changed into the wave A, B, C, D, E,, which moves in the oppo-
site direction.
FIG. 901.
IN
E
M
D
Di
E E
C
Ci
G
DI
D
C
Bi
Dr
CC1
B1
D1
E
BB
When the waves are reflected by a wall, the paths of the mole-
cules undergo the same changes as when two waves cross each
other; here also, in the neighborhood of the wall, the horizontal
component of this motion is more and more balanced, and, on the
contrary, the vertical one is increased more and more, so that near
the wall the path becomes a vertical line, and farther from it an
inclined one. If the wave strikes obliquely against the wall, it will
be reflected, like every elastic body, at the same angle at which it
struck. If a wave strikes but partially against an obstacle, the

·
§ 28.]
THE THEORY OF OSCILLATION.
1091
phenomena of inflexion are produced, new waves being formed at
the extreme ends of the obstacle.
Finally, stationary waves of water, like those of a string or any
other solid body, are formed when two waves of the same length,
which originate at two points situated at a distance apart equal to
1, 3, 5, 7... times the fourth part of the length of a wave, cross
each other. Let ABCDEFG H, Fig. 902, I and II, be one, and
A, B, C, D, E, F, G, H, the other wave. At the points K, L, M, N,
where the two systems of waves are at the same distance from, but
on opposite sides of the middle line, the motions counteract each
other and fixed points of interference are produced; on the con-
trary, above and below the points O, P, Q, R, where the two wave-
lines cut each other and the paths are therefore doubled, the tops
of the crests and the bottoms of the troughs are alternately formed.
FIG. 902.
B
F
E
Bi
H
E
B
F
F
Bi
H
H
61
REMARK.-The most complete treatise upon the motion of waves is the
following: "Wellenlehre auf Experimente gegründet, etc.," by the brothers
G. H. Weber and W. Weber, Leipzig, 1825. A good abstract of it is con-
tained in the "Lehrbuch der Mechanischen Naturlehre," by August. Mül-
ler's "Lehrbuch der Physik und Meteorologie," Vol. I, can also be con-
sulted. The treatises of Laplace, Lagrange, Flaugergues, Gerstner and
Poisson are reviewed and criticised in Weber's work. Cauchy's "Wellen-
Theorie" and Bidone's "Versuche" are discussed at length in "Gehler's
Physikalisches Wörterbuch," Art. "Wellen." Emy's wave theory has been
translated by Wiesenfeld and published under the title "Ueber die Be-
wegung der Wellen und über den Bau am Meere und im Meere," Vienna,
1839. Hagen's work has already been cited, § 27. The theory of water-
waves has been treated by Airy in an article upon "Tides and Waves," in
the Encyclopädia Metropolitana.
TRANSLATOR'S APPENDIX.
SINCE the last German edition of the present volume was issued
the author has published in the "Civilingenieur" several articles
upon subjects, which have been treated in the foregoing pages.
As they contain much valuable information and give the results
of a very great number of very careful experiments, a brief abstract
of the matter contained in some of them will be given here. Those
which will first be noticed are three articles upon the efflux of
water, viz.:
(1) the different methods of experimenting upon the efflux of
water under a constant head (Die verschiedenen Methoden der
Versuche über den Ausfluss des Wassers unter constantem Drucke.
X Band, 1 Heft);
(2) experiments upon the efflux of water under a very small
head (Versuche über den Ausfluss des Wassers unter sehr kleinem
Drucke. X Band, 3 und 4 Heft);
(3) the relations of compound efflux, considered theoretically
and illustrated by experiment (Die zusammengesetzten Ausfluss-
verhältnisse theoretisch entwickelt und durch Versuche erläutert.
XI Band, 2 und 3 Heft).
Article No. 1 begins with a description of the various methods
adopted by different experimenters to maintain a constant head in
the main or discharging reservoir. Smeaton returned the water,
which was discharged, to the reservoir by a hand-pump and thus
maintained the water level constant in the former. Christian em-
ployed a large weighted cask, which was suspended by a rope; as
the water was discharged from the reservoir, the cask was allowed
to sink so as to displace exactly the same quantity of water as had
flowed out of the reservoir. In Prony's experiments the escaping
water was caught in a vessel, which was connected with two paral-
TRANSLATOR'S APPENDIX.
1093
lelopipedical cases (made of sheet-metal). The latter floated upon
the water in the main reservoir, and the apparatus was so arranged
that the increase in weight of the vessel caused the floats to dis-
place exactly the same quantity of water as had been discharged.
The impulse of the escaping water will interfere with the working
of this apparatus, unless proper precautions are taken. Hachette
(see his "Traité élémentaire des Machines") passed a hollow tube
through the bottom of the reservoir; by sliding the tube up or
down the level of the water in the reservoir could be changed. If
the volume of the water, which entered the reservoir, exceeded the
discharge, the excess escaped over the top of the tube. A slight
variation of level, of course, took place. The author tried several
different methods of obtaining the same result. The first, which
to a certain extent resembles Smeaton's, was to feed the discharg-
ing reservoir from the main reservoir by means of a pipe, in which
an ordinary cock was placed. An assistant is stationed at the cock,
by turning which he maintains the surface of the water in the
discharging reservoir at a constant level, which is marked by a
fixed pointer in the reservoir. The second method he employed
was Christian's. He used a hollow float made of sheet-metal; its
weight could be regulated by filling it partially with sand. By
allowing the float to sink as the water was discharged, the surface
of the water was maintained at a constant level, which was indi-
cated by a pointer. The volume of the float gives the discharge.
This method is not so accurate as that last described (by means of a
cock), and it is not so simple as it appears at first sight; for the
size of the float must vary with that of the orifice. The floating
syphon gives more accurate results than Prony's apparatus, de-
scribed above. It consists essentially of a T-shaped syphon with
two lateral pipes, by which the water enters, and of a larger central
pipe, by which it leaves the apparatus. Each of the lateral pipes
passes through a water-tight cylinder of sheet-metal, which is open
on top and floats upon the water. These two floating cylinders
support the syphon; by filling them partially with water we can
immerse the inlet orifices of the syphon as deep as we please, and
the outlet orifice can be brought to any desired distance below the
level of the surface of the water in the reservoir. As the surface
of the water in the reservoir sinks, the whole apparatus descends
with it, and the head or distance of the outlet orifice below the
level of the water remains constant.
The author has also applied the principle of Mariotte's flask to
1094
GENERAL PRINCIPLES OF MECHANICS.
maintaining a constant head, or constant velocity of efflux. The
discharging reservoir is a cylindrical vessel, which is provided with
two orifices or openings, but which is in all other respects air-tight.
One of these openings is in the top and the other is upon the side
near the bottom. A tube or pipe, which is open at both ends, fits
in the orifice in the top by means of an air-tight ground joint, in
which it can slide up and down. The orifice in the side was so
arranged that mouth-pieces of various kinds and sizes could be
inserted in it. The vessel is first filled with water through the
upper orifice and the pipe is then inserted and pushed down a cer-
tain distance, depending upon the head we wish to have; the ori-
fice of efflux is then opened and the water in the tube sinks until
air begins to pass under the bottom of the tube and rise to the top
of the vessel. The head is now constant and is measured by the
difference of level between the orifice of efflux and the bottom of
the tube. In order to prevent the air, which enters through the
tube, from causing too much disturbance, the bottom of the tube
is surrounded by a cylinder of wire-gauze. A glass tube, which is
open on top, enters the vessel at the bottom and is turned ver-
tical upwards, serves to measure the pressure. The same principle
can be applied in another form. An air-tight vessel, which is
filled with water, has a pipe inserted in the side near the bottom;
this pipe passes below the level of the water in the discharging
vessel. Another pipe, which is smaller and is made principally of
India-rubber, enters the air-tight vessel near the top, and the other
end of it is placed so as just to touch the surface of the water in the
discharging reservoir. If the level of the water in the latter sinks,
air enters the tube and water is discharged from the air-tight ves-
sel, in consequence of which the surface of the water in the dis-
charging reservoir rises and seals the mouth of India-rubber tube
and the flow of water into the main reservoir ceases. The objec-
tion to this method is the unsteadiness of the surface of the water,
which renders it difficult to measure the head with accuracy. In
order to render it more steady Geh. Oberbaurath Hagen had two
small holes made in the side of the large tube just above the outlet
and in addition employed an intermediate vessel.
A series of experiments, made with the aid of the different ap-
paratus just described, gave the following results. The water was
discharged through an orifice in a thin plate 1 centimeter in
diameter.
TRANSLATOR'S APPENDIX.
1095
Number of the
experiment.
TABLE.


Nature of the head.
Description of the apparatus.
Head in meters.
Value of µ.j
1 Gradually decreasing. Author's ordinary ap-
Q
Constant.
(C
paratus for experi- | h, =0,1700
2
ments upon efflux. .h, = 0,0500 0,6647
Level maintained by
a cock.
a floating body.
Level maintained by
Mariotte's flask.
Level maintained by
h=0,100
0,6776
3
แ
Level maintained by
((
0,6576
66
0,6518
66
0,6654
0,6634
10
5
(6
apparatus last de-
scribed.
•
Average of the above five experiments. .
By the aid of one of the above-described apparatus, experiments
upon efflux with constant influx can be made. The formula to be
employed (see page 923) is
t =
2 G
(VT
µ FV2g
√ Tu + Nk i ( Nh
NTC
Whi
The discharging reservoir which was used in these experiments
was the apparatus represented upon page 927; by means of Mari-
otte's flask, the discharge per second into the former was main-
tained constant during each experiment. In these experiments
the surface of the water in the discharging reservoir either rose or
fell. By preliminary experiments, the coefficients of efflux for the
orifices in both vessels were determined.
In the first experiment the surface of the water in the discharg
ing reservoir rose. The observed duration of efflux was t = 170,25
seconds; that calculated by the above formula from the data given
by the experiment was t 170,5 seconds.
=
In the second experiment the surface of the water sank; the
observed time was t = 213,2 seconds, the calculated was 213,9
seconds:
Another case, which often occurs in practice, is that represented
in Fig. 776, page 908, when the reservoir AC is very large com-
pared to G L. The water passes from the large reservoir A C
1096
GENERAL PRINCIPLES OF MECHANICS.
through a pipe, into the reservoir & L, from which it is discharged
through the orifice F into the air. By prolonging the discharge
pipe of Mariotte's flask so that it will reach below the surface of
the water in the discharging reservoir (Fig. 792), the level of which
surface is variable during the experiment, we obtain an example
of this case.
The formula for the duration of efflux, which must
be employed, is
t
G
[(µs F')² + (µ, F₁)'] 12 g
Nπ - Nπc
N k r ( N h
√
√
√²
+
+
NTc
√ ] ]
V h + Vk V x k)
X
µ F 2 (√h − √x) +
F[2
+ M, R. [2 ( √π − √y) +
√ Te₁ 2 √ π - N k₂ Ny + VT,
V 1 ( V m
Ny Nk₁
У
14, F h −
)])
in which G denotes the cross-section of the main discharging res-
ervoir, F the area of the orifice in the main reservoir, μ its coeffi-
cient of efflux, F₁ the cross-section of the outlet orifice of Mariotte's
flask, μ, its coefficient of efflux, h, the height of the surface of the
water in the main reservoir above the orifice in it, h the height of
the constant water level in Mariotte's flask above the variable one
in the main reservoir, x what h becomes in the time t, y what h
becomes in the time t₁, and h。 h + h₁ =
= x + y; k is the value
of x, when the flow becomes permanent, I.E.
2
(µ, F₁)² ho
2
and
k
(µ F')² + (µ₁ F₁)”
k₁ = h₂-k.
In the first experiment the surface of the water in the main
reservoir sank; the observed value of t was 116,33 seconds and the
calculated value was 116,67 seconds. In the second experiment
the level of the water rose; the observed time was t = 157,5 seconds,
and the calculated value of t was 158,18 seconds.
No. (2.) Experiments upon the Efflux of Water under
a very small head.-From previous experiments by the author
and others, we know that for an orifice in a thin plate one centi-
meter in diameter,
1, when the head is 103,578 meters, u
2,
3,
66
66
0,600
13,574 66 µ = 0,632
66
66
0,909
66
µ = 0,641
4,
CC
66
0,101
CC
μ =
0,665,
TRANSLATOR'S APPENDIX.
1097
and that for a brass tube 1 centimeter in diameter and 2 meters
long, the coefficient of resistance
20,99, is
1, when the velocity is v = 20,99, is 5
66
v =
= 0,01690
12,32, is $ = 0,01784
8,64, is
v = 8,64, is
=
0,01869
2,
66
3,
CC
4,
(C
v =
66
CC
บ
0,57, is
=
5,
2,02, is $ = 0,02725
0,03646;
but we have no experiments which show how the coefficient of
efflux increases, when the head is very small (E.G. 1 to 2 centime
ters). It is also important to know how increases, when the
velocity of the water is very small (E.G. 0,1 meter). In the above-
mentioned article the author gives a detailed description of a very
extended series of experiments, undertaken for the purpose of dis-
covering the above relations. The discharging reservoir was a
wooden trough 2,25 meters long, 0,45 meters wide, and 0,190 me-
ters deep. It was necessary to make the reservoir as long and wide
as possible; for the surface of the water could, of course, sink but
a very short distance during the experiment. The author then
gives a description of the various methods and apparatus employed
to determine with accuracy the cross-section of the orifices and the
head of water. This portion of the article, although of the greatest
interest, would be out of place here.
The table on page 1098 gives the results of the experiments
with orifices in a thin plate and with other mouth-pieces.
The temperature of the water was between 15° and 18° Centigrade.
From the 8 experiments with orifices in a thin plate (No 1 to
No. 5), whose diameters varied from 0,405 to 2,529 centimeters, we
see that the contraction diminishes, when the head is small, as it
does when the head is large, not only with the head, but also with
the diameter of the orifice.
From the data given in the table on page 1098 and at the be-
ginning of the article, the following table has been arranged.


Head h.
0,020 0,101 0,909 13,574 103,578
Coefficient of efflux ¿ 0,711 0,665 0,641 0,632 0,600
The experiments under Nos. 6 and 7 show that in this case also
the coefficient of contraction for an orifice in a thin conically con-
vergent wall is greater than that for an orifice of the same size in a
1098
GENERAL PRINCIPLES OF MECHANICS.
No. of
exper-
iment.
Orifices or mouth-pieces.
Diame- Cross-sec-
ter of tion of the
orifice d.
head h. | of efflux
orifice F.
27.
Mean Velocity Coeffic't
Coeffic't
of resist-
arice S.
of efflux
11.
2,2394
1
0,6628
0,7197
0,405
0,12882-
2,0822
0,6392
0,7287
1,9236
0 6143
0,7410
2
Circular orifices in a thin plate.
1,010
0,80120
2,1907
0,6556
0,7038
3
1,8504 0,6025 0,7177
1,408
1,55700
2,14066 0,6481
0,6861
4
1,985
3,09465
1,92932 0,6153
0,6657
2,529
5,02310
6
Circular orifice in a conically convergent wall; angle of convergence = 100°
1,020
0,81710
2,14915 0,6494 0,6572
2,17415 0,6531 0,8172
1,86349 0,6047 0,8129
77
Circular orifice in a conically divergent wall; angle of divergence
= 100°
1,020
2,18507
0,81710
00
8
Short conical mouth-piece...
1,002 0,78850
9
0,6548 0,6418
1,86786 0,6054
2,17739 0.6289
1,86568 0,5789 0,9589 0,0875
2,2523 0,4582 0,6894 1,1041
0,6519
0,9638 0,0765
0.343
0,09240
2,0837
0,4408 0,6895
1,1037
1.9177
0.4215 0,€874 1,1164
10
2,1737
0,4896 0,7499
0,7781
0,485 0,18475
1,8530
0,4506 0,7478
0.7885
1,5355
0,4067 0,7416
0,8185
11
2,1571
0,5277 0,8118
0,5173
0,728 0,41625
1,8290 0,4753 0,7945
0,5844
Cylindrical ajutages 3 times as long as wide.
1,5102 0,4168
0,7671
0,6995
12
13
1,011
1,402 1,54380
0,80755
2,16678 | 0,5196
0,7983
0.5693
1,81791 0,4761
0,7991
0,5659
2,0907
0,5148 0,80€6
0,5371
1,6155
0,4367 0,7802
0.6427
14
1,927 2,91640
2,8126
0,5859 0,7953
0,5812
1,9240 0,4701 0,7711
0.6818
15
2,487 4,85780
16
Cylindrical ajutages, rounded off at the inlet orifice.
1,014 C,80755
17
Oblique ajutage, angle of deviation d = 30°
1,005 0,79325
18
Interior ajutage..
3,2251
2,0046 0,4645 0,7496
2,17680 0,5424 0,8314
1,86104 0,4965 0,8235
2,14241 0,4460 0,6890 1.1063
1,71583 0,8940 0,€808 1,1576
0,6238 0,7878
0,6112
0,7797
0,4466
0,4747
1,010 0,80120
2,17470 0,4697
0,7203 0,9274
19
Short conically convergent ajutage, angle of convergence = 7° 9′
1,004 0,79170
2,20789 | 0,CC21
0,9162
0,1913
1,90412 | 0,5055
0,9108
0,2056
20
The same, rounded off at the inlet, cylindrical at the outlet orifice.
1,012
0,80440
2,19454 | 0,6043
0.9225
0.1751
1.89447 0,5500
0,5041
0.2233
21
Long double conical mouth-piece.
0,966 0,73290
2,20433 0,5941
0,9047
0.2218
22
The same, shorter.
1,98460 | 0,5498
0,8942
0,2506
1,104 1,54820
2,08975 | 0,5687
0,8913 0,2589
23
Short conically divergent tube, angle of divergence
= 3° 24'.
1,034 0,83970
2,19869 | 0,2605
0.5199
2,3072
**
24
The same, wider, angle of divergence
= 8° 4′
25
The same, rounded off at the inlet orifice.
1,87635
1,541 1,86510 2,11440
0,8358
0,5547
2,2502
0.2607 0.4065
5,0521
1,518 1,80980
2,09467
0,626 | 0,5993
1,7844
TRANSLATOR'S APPENDIX.
1099
thin plate, and that it is less for an orifice in a conically divergent
wall than for the latter. In experiment No. 18 a free contracted
stream could not be obtained. The efflux took place with a filled
tube and the stream pulsated quite violently.
It was also observed that the discharge was not increased as
much by rounding off the inlet orifices of the ajutages, when the
head was small as when it was great.
The table on page 1100 contains the results of experiments with
long tubes made of glass, brass and zinc. Preliminary experiments
were made to determine the coefficients of resistance of the inlet
and outlet mouth-pieces combined. By subtracting the coefficients
thus found from those obtained for the long tube and inlet and
outlet mouth-pieces together, the author deduced the coefficient of
resistance for the tube alone.
These experiments showed the coefficient of resistance of the
water to be very great, when the velocity is small. This coeffi-
cient is nearly the same for glass and brass tubes.
To the table
for v = 20,99, 5
=
0,01690
66
(6
CC
we can now add
12,32,5 0,01784
8,64, 5 = 0,01869
= 2,02, = 0,02725
5
0,485, 0,03453,
5
for v = 0,2028, = 0,0587
(C
=
S
0,0890, = 0,1420.
The third portion of the article is devoted to an account of a
series of experiments upon the flow of water through bends and
elbows under a very small head. The coefficient of resistance for
the inlet and outlet portion was first determined as in the experi-
ments, the results of which are given in the last table. The table
on page 1101 contains the coefficients of resistance for the flow of
water through elbows and bends under small heads.
We see from the last table that the coefficients of resistance of
elbows are much greater than those for bends of the same diameter,
when both cause the direction of the motion of the water to change
90° and when the radius of curvature of the axis of the bend is
equal to the diameter of the tube.
The third article (No. 3), which is cited above, is very long,
covering 68 columns of the Civilingenieur. As it would be impos-
ible to condense the matter contained in it in the limited space
1100
GENERAL PRINCIPLES OF MECHANICS.
Description of the tubes.
a
Diameter of
orifice.
h
Mean head.
V1
Mean velocity
of the water in
the tube.
Coefficient of
friction.
Centimeters.
Centimeters.
Meters.
Length 1₁ = 158,7 centimeters.
5,53096
0,0755
0,3540
Glass tube No. 1
Mean width di
0,29714 centimeters,
0,337
5,41190
0,0730
0,3703
Fall = 0,25 centimeters, Temperature of the water = 14° c.
5,29138
0,0729
0,3630
5,18291
0,0708
0,3769
Glass tube No. 2
Rise
= 157,5 ct. m., dı
0,455 ct. m.,
S
0,47074 ct. m.,
18°.
0,483
8,6985
0,1258
0,1317
3,3650
0,1128
0,1496
I
Glass tube No. 3
Fall
162,0 ct. m., d₂ = 0,71710 ct. m.,
0,055 ct. m., T = 17°.
8,2099
0,2094
0,05604
0,723
2,8847
0,1873
0,06386
2,5614
0,1661
0,07303
Glass tube No. 4
Fall
= 203,5 ct. m., d₁ = 1,06528 ct. m.,
7 =
0,78 ct. m., 16°,5.
1,012
2,8744
0,2351
0,04165
2,3815
0,2075
0,04500
Glass tube No. 5
1. Fall
= 1,70,6 ct. m., dı = 1,4302 ct. m.,
0,73 ct. m., 7 = 18°,5,
2,8065
0,2851
0,04090
1,402
2,2867
0,2212
0,06073
2. Fall =
0,16 ct. m., T = 15°,3.
T=
1,8692
0,1984
0,06579
1,5038
0,1713
0,06747
2,8703
0,1984
0,06758
Brass tube No. 1
II
₁ = 200 ct. m., d₁ = 1,0378 ct. m.
1. Fall = 0,72 ct. m., T = 13°,8,
2. Rise =
= 0,92 ct. m., T =
2,4754
0,1757
0,07099
1,012
2,6672
0,2122
0,04977
= 16°,5.
2,2346
0,1896
0,05269
1,3064
0,0960
0,13243
1,0831
0,0819
0,15171
Brass tube No. 2
= dı
298,1 ct. m., d₁ = 1,4336 ct. m.,
T =
Fall 0,54 ct. m., 18°,5.
1,402
2,5440
0,2236
0,03872
1,9178
0,1883
0,04434
Zinc tube No. 1
Fall
Zinc tube No. 2
=
III
Zinc tube No. 3
Zinc tube No. 4
= 342 ct. m., d= 3,22683,
T
0,02 ct. m., 7 = 15°,5.
311 ct. m., d, 2,4799,
T
Rise 0,34 ct. m., 7 = 18°,5.
=
673,8 ct. m., di
ct. m.,;
Fall 0,86 ct. m., r = 18°,5.
17, 1016 ct. m., di = 2,49498 ct. m.,
α,
Rise 0,82 ct. m., 7 = 19º.
3,315
3,3549
0,3600
0,03434
2,445
2,2259
0,2418
0,04251
2,49908,
2,445
8,0394
0,2076
0,04316
ར་
2,445
2,5240
0,1451
0,05187
TRANSLATOR'S APPENDIX.
1101
Num'r of the
experiment.
Nature of the tubes.
α
Diameter of
the orifice in
centimeters.
h
Mean head in
centimeters.
ช
Mean velocity Coefficient of
of the water
in meters.
friction for
the elbow or
bend.
1
Short cylindrical ajutage, rounded off inside with an elbow of 90° and an-
other cylindrical mouth-piece beyond the elbow.
2,2105
1,012
0,3160
2,5137
1,9058
0,2893
2,6315
d
The same with a bend of 90°, instead of an elbow.
1,012 {
2,1762
0,4227 0,5657
1,7774
0,3681
0,7444
3 Two bends, one 90° and the other 180°, with a cylindrical ajutage, inter-
mediate pipe and outlet pipe.
1,012 {
2,1828
0,3285
2,0923
1,7456
0,2834
2,3786
4
A wider elbow
1,402
2,1994
0,3290
2,2349
1,9017
0,3025
2,3173
with ajutage and mouth-piece like No. 1.
5
A wider bend
1,402 {
2,1384
0,4204
0,6324
1,7288
0,3658
0,7832
6
A simple zinc tube, with bend at the end (90°). Fall = 0,23 ct. m.
2,445
2,3259
0,2338
0,2608
7
"6
double bend at the end (180°). Fall=0,725 ct. m.
2,445
2,7958
0,2508
0,6734
8
แ
rectangular elbow (90°). Fall = 0,35 ct. m.
2,445
2,4910
0,2263
1,4566
1102
GENERAL PRINCIPLES OF MECHANICS.
'
•
which is at our disposal, we will content ourselves with an enumer-
ation of the subjects treated. They are-
(1.) The simultaneous discharge of water through two orifices,
when the head diminishes.
(2.) The variable discharge of water from one vessel into a sec-
ond, in which the orifice is submerged, while a constant quantity
of water is continually discharged into the first vessel.
(3.) The variable efflux of water through a notch, either with or
without influx.
(4.) Efflux of water from a prismatical vessel, with free influx
into the latter from another prismatical vessel.
(5.) Efflux of water from a prismatical vessel, with influx under
water from another prismatical reservoir.
These cases are treated at length; the formulas are first deduced
and then tested by very careful experiments. Any one interested
in the subject of hydraulics will find this article worthy of his
most attentive perusal.
We would also call attention to the following articles by the
author upon subjects connected with hydraulics.
"Hydrometric experiments upon the application of the formulas
of Daniel Bernouilli (page 804) and Charles Borda (page 884), as
well as upon the use of a new water-meter; also upon the friction
of water in conical pipes and upon the play of jets d'eau" ("Hydro-
metrische Versuche über die Anwendung der Formeln von Daniel
Bernouilli und Charles Borda, so wie über den Gebrauch eines
neuen Wassermessers (einer Wasseruhr); ferner über die Reibung
des Wassers in conischen Röhren und über das Spiel von springen-
den Wasserstrahlen," Civilingenieur, Band XIII, 1 Heft). "Com-
parative hydrometric measurements by means of a tachometer, a
large rectangular orifice of efflux and a large overfall extending
across the whole wall,” (“Vergleichende hydrometrische Messungen
mittels eines hydrometrischen Flügelrades, einer grösseren rec-
tangulären Ausflussmündung und eines grösseren über die ganze
Wand weggehenden Uberfalls," Civilingenieur, Band XIII, 5 and 6
Heft).
The latter article contains an account of Schwamkrug's water-
divider, mentioned upon page 986.
I. "The quicksilver differential piezometer and its application
to the determination of the difference of the pressure of the water
in a set of conduit pipes."
TRANSLATOR'S APPENDIX.
1103
II. "The water piezometer with a micrometer, as well as its
application to the determination of the pressure of gas in pipes,
etc."
III. "A supplement to the article cited above upon the differ-
ent methods of experimenting upon efflux under a constant head."
("I. Das Quecksilber-Differentialpiezometer, etc. II. Das Wasser-
piezometer mit Mikrometer, etc. III. Eine Ergänzung der Ab-
handlung über die verschiedenen Methoden der Ausflussversuche
unter constantem Drucke." Civilingenieur, Band XV, 2 Heft.)
The translator would also call attention to two articles by the
author upon "experimental mechanics," which form a part of a yet
unpublished work upon that subject. The titles of the articles are:
(1.) "Experiments to accompany lectures upon the elasticity
and strength of solid bodies" ("Versuche bei Vorträgen über Elas-
ticität und Festigkeit fester Körper," Civilingenieur, Band IX,
5 Heft), and
(2) "Experiments to accompany lectures upon Mechanics"
("Versuche bei Vorträgen über Mechanik," Civilingenieur, Band
XIV, 6 Heft).
The first article contains a description of the apparatus used
by the author in experimenting before the students at Freiberg
upon flexure and torsion. By means of this apparatus, which is
very simple and easily constructed, the professor can show to the class
almost all the phenomena of flexure and torsion. He can also de-
termine the moduli of rupture and of elasticity not only by observ-
ing the deflection and angle of torsion, but also by allowing the
body to be experimented upon to vibrate and counting the number
of vibrations. The modulus of resilience and that of fragility can
also be determined. No. (2) contains an account of some modifi-
cations of the above apparatus, by means of which experiments
upon the theory of couples (including their composition and de-
composition) can be made. This is followed by the description of
a simple reversable pendulum, by means of which the value of g
can be determined in the lecture-room with little difficulty. The
author then takes up the subject of the elasticity of rigid bodies.
He discusses four cases of double flexure: first, that of a prismati-
cal rod of a rectangular cross-section, bent by a force, whose direc-
tion forms an angle d (which is not 90°) with one of the sides of
the cross-section; secondly, that when the cross-section of the rod
is a right-angled triangle and the direction of the force is perpen-
dicular to the base of the triangle; thirdly, that when the rod is
1104
GENERAL PRINCIPLES OF MECHANICS.
acted upon by two forces, whose lines of action do not lie in the
same plane; and fourthly, that when the beam is bent in the shape
of an elbow and loaded at the extreme end with a weight (the
crank is an example of this case). The article closes with an ac-
count of some experiments with compound girders.
Those engaged in teaching will find the last two articles full of
valuable information; but a translation of them would occupy too
much space here.
In conclusion, we would mention an article upon "the flexure
of a homogeneous prismatical measuring rod, supported in two
points, as well as the shortening of its length, produced by it, dis-
cussed in as elementary a manner as possible" ("die Biegung eines
in zwei Punkten unterstützten homogenen prismatischen Mess-
stabes, sowie die durch dieselbe hervorgebrachte Verkürzung seines
Längenmaasses, auf möglichst einfache Weise ermittelt von Julius
Weisbach," Civilingenieur, Band XII, 4 Heft).
INDEX.
The figures give the number of the page.
A.
Aberration of the stars, 152.
Abscissas, 34.
Acceleration, 108, 113, 124.
66
along the abscissas, 146.
ordinates, 146.
normal, 143, 607.
of gravity, 113, 159.
Adhesion, force of, 163, 762.
plates, 762.
<<
Aerodynamics, aerostatics, 165.
Aggregation, state of, 162.
Air balloon, 798.
efflux of, 932, 934, 939.
"heaviness of, 795.
66
layers of, 787.
manometer, 796.
pressure of the, 777.
pump, 790.
Amplitude of an oscillation, 649, 1043.
Angular acceleration, 576.
velocity, 576.
Antifriction pivots, 349.
Aperture of efflux, 800.
Apparatus for hydraulic experiments,
926.
Application, point of, 163, 192.
Arc, length of an, 85.
Archimedes, principle of, 757.
Areometers, hydrometers, 758.
Arithmetical mean, 97.
Arm of the lever, 195.
Ascension, vertical, 116.
Asymptote, 49, 51, 52.
Atmosphere, pressure of the atmo-
sphere, 777, 787.
Attraction, the law of magnetic, 1056.
Atwood's machine, 599.
Axes, free, 624.
<
principal, 624.
Axis, neutral, 410.
of a couple, 205.
Axis of revolution or rotation, 205,
248, 573, 629.
Axis, pressure upon the, 250.
Axles, friction on, 311, 316.
B.
Balance, hydrostatic, 756.
torsion, 1050.
Ballistic pendulum, 693.
Barometer, 776.
measurement of heights
with the, 788.
Beam, 418, 422, 427, 430.
66
subjected to a tensile force, 559.
Bed of a river, 955.
Bending, flexure, 409.
rupture by, 452.
Bends, curved pipes, 896.
Bent lever, 256.
Binomial function, 57.
series, 57.
Bodies, material, 154.
46
of uniform strength, 387, 498,
504, 539.
rigid, flexible, elastic, 280.
Boilers, thickness of, 738.
Bottom of the channel, 955.
pressure on the, 721.
Brachystochronism, 659.
Brittle, 372.
Buoyant effort, upward thrust, 742,
797.
C.
Capillarity, 762.
Capillary tubes, 772.
Cataract, 876.
Catenary, 293; common catenary, 299.
Central impact, 667, 669.
Centre of gravity, 213.
1106
INDEX.
Centre of mass, 213, 574.
CC
CC
oscillation, 661.
parallel forces, 205.
percussion, 637, 692.
pressure of water, 725.
Centres, 349.
Centrifugal force, 608.
of water, 719, 720.
work done by, 610.
Centripetal force, 608.
Chain bridge, 292.
friction, 358, 361.
Cinematics, 154.
Circle, 34.
แ
Contraction, coefficient of contraction,
822, 944.
Contraction, complete and incomplete
or partial, 837.
Contraction, perfect and imperfect, 840,
858, 887.
Contraction, scale of, 836.
Convexity, 39, 55.
Coordinates, 34.
66
Cosine and cotangent, functions of, 71.
Couple, 200, 412.
(C
oblique, 79.
axis of a, 205.
Crank, 121.
centre of gravity of an arc of Cross-section, 376, 676, 801, 955.
osculatory, 87, 142, 415.
a, 216.
Circular functions, 70.
Cistern barometer, 776.
manometer, 779.
Clack valves, 900, 905.
Cloistered arch, 243.
Cocks, 900, 903.
Cohesion, 371, 762.
force of, 163.
Collar bearings, 347.
Columns, proof load of, 532.
Combined elasticity and strength,
373, 547.
Communicating pipes, 723, 761.
Components, 129, 174, 177, 1071.
Component velocities, 129.
Composed forces, 174.
motions, 126.
Composition and decomposition of ve-
locities and accelerations, 131, 132.
Composition and decomposition of
forces, 174, 177, 179, 195, 207.
Composition and decomposition of
couples, 202.
(C
Compound discharging vessels, 907.
pendulum, 661.
Compressed air, work done by, 783,
936.
Compression and extension, 374.
CC
CC
elastic and permanent,
376.
strength of, 372, 373.
Concavity, 39, 55.
Conduit pipes, 874.
Conical pivots, 347.
،،
tubes or pipes, 872.
valves, 905.
Connecting rod, 537, 573.
Constant factors, 41, 61.
(C
force, 166.
members, 41, 61.
quantities, 33, 41.
Contracted vein or stream of water,
821, 823.
weak, dangerous, 495.
sudden variation of, 883.
Curvature, radius of, 87, 142, 413.
Curve, elastic, 414, 417.
Curved surfaces, 40.
Curves, convex, concave, 39, 44, 54
CC
quadrature of, 78.
rectification of, 85.
Curvilinear motion, 141, 145, 189.
Cycloid, cycloidal pendulum, 655, 656,
Cylinder, hollow, 443.
Dam, 732.
D.
Daniel Bernouilli, 804.
Decomposition and composition of
couples, 202.
Decomposition and composition of
forces, 174, 177, 179, 195, 207.
Decomposition and composition of
velocities and accelerations, 131, 132.
Density of bodies (specific gravity), 161.
Dependent variable, 33.
Deviation, angle of, 895.
Differential, 38.
ratio or quotient, 39.
Directive force of the magnetic needle,
1053.
Discharge, 800, 933.
Discharge-pipe of a dam, 858, 922.
Displacement, angle of displacement,
530, 649.
Diving-bell, 783.
Ductility, 372.
Dynamics, 155, 165.
E.
Earth, magnetism of the, 1054, 1059.
Efflux, coefficient of, for water, 824.
air, 944.
66
(C
"
from moving vessels, 817.
INDEX.
1107
Efflux of air from vessels, 932, 934, | Forces, 154, 155, 163, 205.
แ
3
939, 941.
of different fluids, 805, 930.
of moving water, 842.
of water under water, 806.
of water from vessels, 800.
under variable pressure, 910,
952.
velocity of, 800.
with filled tube, 853.
Elastic curve, 414, 417, 522.
66
extension, 375, 404.
fluids, 712.
Elasticity, 163. 371, 1045.
CC limit of, 371, 376.
tr
modulus of, 378, 407, 1049.
Elbows, 894.
Elevation, angle of, 136.
Ellipse, 50, 284.
Ellipsoid, 594.
Elliptical oscillation, 1081.
Emptying of a vessel, 910.
Energy, 168.
of discharging water, 801.
Envelope, 139.
Equality of forces, 156.
Equilibrium, 155.
Evolute, 88.
kinds of, 249, 250, 264.
indifferent, 250, 266.
Expansive force of steam, 35.
Expansion by heat, 793.
coefficient of, 793
of the air, 781
Exponential function, 63.
Extension, elastic and permanent, 375,
394.
experiments upon, 393.
F.
Fall of a stream, 955.
(C
of bodies, 35, 113, 639, 659.
Filling and emptying locks, 924.
Final velocity, 108.
Flexure, 409.
CC
strength of, 373, 450.
moment of, 412, 414, 432, 436.
Flotation, axis of, plane of, 746.
Floating, depth of floatation, 745, 749,
756.
bodies, floating spheres, 989.
staff, 990.
Fluids, 162, 712.
Force, direction of a, 163.
equality of, 156.
Fragility, modulus of, 383, 453.
Free axes, 624.
Freshet or flood, 973.
Friction, resistance of friction, 309.
<<
angle of, 314
CC
''
E
(C
35
CC
ic
balance, 317.
coefficient of, 313.
coefficient of, of air in pipes,
949.
coefficient of, of water in pipes,
864.
coefficient of, of water in riv-
ers, 965.
cone of, 314.
height of resistance of, 864.
kinds of, 310.
laws of 311.
of axles, 311, 316.
rolling, 353.
upon inclined plane, 323.
wheels, 336.
work done by, 313, 335.
Fulcrum, 256.
Function (), 44.
Functions, 33.
Funicular machine, 280.
polygon, 286.
G.
Gases, aeriform bodies, 776.
Gas-meters, 1023.
Gauging, 976.
Gay-Lussac's law, 793.
Geostatics, geodynamics, geomechan-
ics, 165.
Girder, 418, 422, 427, 430, 464.
"C
hollow and webbed, 437, 477.
Goblet, hydrometric, 986.
Gram, kilogram, 157.
Graphic representation, 34, 122.
Gravity, 113, 154, 163.
C6
centre of, 213.
determination of the centre of,
214.
plane of, line of gravity, 213.
specific, 161.
Gudgeons, 311.
Guldinus, properties of, 241.
Gyration, radius of, 581, 608.
Force, living, 171, 173.
H.
Forces, measure of, 158.
moment of, 195, 414.
CC
"
normal, 143, 607.
**
tensile, 374.
Hard, 372.
Hardness, 676.
Head of water, height of water, 722
801, 809.
1108
INDEX.
Heat, force of, 163.
Heat, work done by, 936.
Heaviness, 160.
mean, of the earth, 1051.
of air, 795.
66
steam, 795.
water, 160.
Height due to the velocity, 115, 809.
of rise, height of fall, 116, 878.
Horizontal and vertical pressure, 732,
736, 742.
Hydraulic observatory, 995.
Hydraulics, 165.
Hydrometers, Hydrometry, 976, 989.
Hydrometric goblet, 986.
pendulum, 999.
Hydrostatic balance, 757.
Hydrostatics, hydrodynamics, 165.
Hyperbola, 51, 80.
I.
Impact, different kinds of, 667, 668.
direct, 667.
K.
Kater's pendulum, 665.
Kilogram, 157.
Knee lever, 257.
Knife edges and points, 352.
Knots, 281.
L.
Law of Gay-Lussac, 793.
Mariotte, 37, 780.
Laws of nature, 35.
Length of a wave, 1064, 1085.
Lesbros' experiments, 846.
Lever, arm of, 195.
(C
bent, 257.
kinds of, 255, 256, 343.
Limit of elasticity, 371, 376.
Line of current, mid-channel, 956.
gravity, 213.
impact, 667.
rest, 743.
CC
<<
(6
support, 743.
Load, proof, 379.
duration of, 668.
"(
elastic, 668.
4:
friction of, 685.
imperfectly elastic, 680.
line of impact, 667
oblique, 668, 682.
strength of, 702, 705.
Impulse, 1002, 1006.
64
of air or wind, 1030.
water, 1006, 1011, 1029.
Incidence, angle of, 624.
Inclination, angle of, 314, 639.
Inch, water, 983.
Inclined plane, 272, 274, 639.
Inertia, 157.
"C
force of, 157, 163, 574.
moment of, 576.
Inflexion, 1091.
CA
point of, 55, 424.
Integral, integral calculus, 60.
66
formulas, 73.
Integration by parts, 76.
Intensity of a force, 164.
the earth's magnetism,
1060.
Interference of waves, 1064, 1089.
Interpolation, 98.
Isochronism, 640, 658, 659.
eccentric, 480.
Locks, 924.
Logarithm, 64.
Longitudinal vibration, 1045.
Loss of mechanical effect in impact,
674, 883.
M.
MacLaurin's series, 57.
Magnetic force, 163, 1056.
needle, 1053.
Magnetism, 1054, 1059.
of the earth, 1054.
Malleability, 372.
Manometer, 776, 778.
Mariotte's law, 37, 780.
Mass, 158.
moment of, 577.
Material pendulum, 661.
point, 165.
Matter, 156.
Maximum and minimum, 53.
16
contraction,
"
834.
tension, 515.
Mean, arithmetical, 97.
J.
Jets of water, 876.
Journals, trunnions, gudgeons, axles,
305, 311, 345.
"
harmonic, 675.
Mechanical effect, 168, 187, 209.
(6
CC
loss of, during im-
pact, 674, 883.
of compressed air,
783, 936.
of friction, 313, 335.
INDEX.
1109
Mechanical effect of heat, 936.
of inertia, 171, 577.
of the centrifugal
force, 612.
Mercury, efflux of, 930.
Metacentre, 751.
Metal springs, 506.
Method of least squares, 95.
CC
CC
interpolation, 98.
Mid-channel, line of current, 956.
Modulus of elasticity, 378, 407, 1049.
logarithms, 65.
CC
CC
CC
proof strength, 380, 457,
529.
"resilience
383, 453.
and fragility,
rupture, or of ultimate
strength, 380, 452.
Molecular action, 762.
Molecules, molecular forces, 163, 762.
Moment, magnetic, 1054, 1060.
Obelisk, centre of gravity of, 234.
Oblique coordinates, 79.
Observatory, hydraulic, 995.
Oil, efflux of, 930.
Ordinates, 34.
acceleration along the, 146.
velocity along the, 145.
Orifices in a thin plate, 821, 930, 944.
inlet and outlet, 875, 880.
of efflux, 800.
CC
rectangular, 812, 828, 842, 846.
Oscillation, 649, 1042.
66
amplitude of an, 649, 1043.
centre of, 661.
period of an, 649, 1043, 1067.
of a pendulum, 649.
of the magnetic needle,
1055.
of water, 1079.
Overfalls, notches, weirs, 811, 833, 844,
849, 914.
of a couple, 200, 201.
“
inertia, 577.
parallel forces, 207.
statical, 195.
Momentum of a body, 670.
P.
Parabola, 3, 87, 133, 291, 302.
Parabolic motion, 134, 141.
Motion, absolute and relative, 1C5, 149. | Paraboloid, 591, 720.
པ
པ
<<
་
accelerated, retarded, 106.
curvilinear, 141, 145, 189.
in resisting media, 1035.
kinds of, 573.
of air in pipes, 950.
(C
water in channels, 955, 969.
water in pipes, 869.
translation, 573.
phases of, 1062.
rectilinear and curvilinear,
105.
simple and composed, 126.
uniform and variable, 106.
N.
Naperian logarithms, 64, 80.
Natural philosophy, 154.
Nature, laws of, 35.
Neil's parabola, 86.
Neutral axis, surface, 410.
Nicholson's hydrometer, 759.
Normal, 87.
acceleration, 143, 607.
force, 189, 607.
Notches, overfalls, weirs, 811, 914.
Numbers, natural series of, 59.
0.
Obelisk, efflux from an, 919.
Parallel forces, 199.
C
Parallelogram of accelerations, 132.
*C
<<
plates, 770.
CC
forces, 17.
"motions, 127.
velocities, 128.
Parallelopipedon of velocities, 132.
Pendulum, ballistic, 693.
bob of a, 591.
compound, 649, 661.
hydrometric, 999.
<<
Kater's, 665.
*
ご
​oscillation of a, 649.
reversable, 665.
rocking, 665.
simple, mathematical, 648,
661.
Perfect fluids, 712.
Percussion, centre of. 637, 692.
point of, 692.
Period, periodic motion, 106, 121.
Permanency, state of, of running wa
ter, 957.
Permanent extension or set, 375, 394.
Phoronomics, 105, 154.
formulas of, 119.
Piëzometer, 779, 881.
Pile driving, 698.
Pipes, long, 863.
thickness of, 738.
Piston rod, 538, 573.
Pitot's tube, 998.
Pivots, friction of, 345.
1110
INDEX.
Plane, inclined, 272, 323.
of revolution, 248.
Pneumatics, 165.
Point of application, 163, 192.
แ
inflexion, 54.
<<
CC
suspension, 249, 664.
Polyhedron, centre of gravity of, 231.
Poncelet's orifice of efflux, 828.
theorem, 341.
Position, 105, 150.
CC
Reaction, 164.
of effluent water, 1002.
wheel, 1015.
Rectification of curves, 85.
Reduction of a force, 255.
masses, 578.
"the moment of flexure,
432.
the moment of inertia,
580.
relative, relative motion, 150. Reflection, angle of, 684.
Pound, 157.
Powers, natural series of, 64.
Pressure, hydraulic, hydrodynamic,
808.
hydrostatic, 713, 723, 724.
in water, 724.
of the atmosphere, 777, 787.
on the bottom, 721.
vertical, horizontal, 732.
Principal axes, 624.
Principle of equal pressure, 713.
Profile, longitudinal and transverse,
955.
transverse, of running water,
955.
Projectile, path of a, 1038.
Projectiles, height attained by, range
of, 136.
"
motion of, in the air, 136.
motion of, in vacuo, 1038.
Prony's method of measuring water,
982.
Proof load, proof strength, 379, 451.
moment of, 451, 472.
เ
CC
Proof strength, modulus of, 380, 457,
529.
Propagation, velocity of, 1062, 1085.
Properties of Guldinus, 241.
Prosaphy and synaphy, 763.
Pull, traction, 156, 374.
Pulley, fixed and movable, 303, 304,
368, 601.
Puppet valve, 905.
Regulating apparatus, 900.
Representation, graphic, 24, 122.
Resilience, modulus of, 33, 453.
Resistance, coefficient of, 856, 884.
height of, 856.
of water, 1028.
to buckling or breaking
across, 535.
to compression, 376, 392.
Resistances, 155, 809.
passive, 1077.
Rest, absolute, relative, 105.
Resultant, 174, 177, 194.
Revolution, axis of, 205, 248, 573, 629.
plane of, 248.
solids and surfaces of, 238,
241, 242, 593, 626.
Rheometer, 1001.
Rigidity of cordage and chains, 261, 363.
of hemp and wire ropes, 364,
366.
River, bed of a, 955.
Rocking, rocking pendulum, 665.
Rod, vibration of a, 1072.
Rolling down an inclined plane, 646.
friction, 353.
of bodies, 605.
Rotary motion, 210, 211.
Rotation, axis of, 205, 248, 573, 629.
plane of. 248.
(C
64
time of, 609.
Running water, 955.
Rupture by breaking across, 535.
modulus of, 381, 452.
Co
plane of, cross-section of, 495
Q.
Quadrature of curves, 78.
Quantities, constant and variable, 33.
Quicksilver, efflux of, 930.
Quotient, 93.
differential of a, 43.
R.
Radius of curvature, 87, 142, 413.
gyration, 581, 609.
(4
Ram, 698.
S.
Scale of velocities of a stream, 957.
Set, permanent extension, 375, 394
Sheering force, 412, 510.
66
strength of, 373, 406.
Shoots, efflux through, 848, 850.
Short pipes, conical, 861, 891.
tr
(C
CC
แ
་
conical convergent, 861.
conical divergent, 862.
cylindrical, 853, 888.
efflux through, 852, 854
INDEX.
1111
Short pipes, inclined, 857.
interior, 855.
CC
Simpson's rule, 81.
Sine, curve of, 71.
(C
function of the, 70.
Sliding, 310, 639.
down an inclined plane when
friction is considered, 643.
Slope of a stream, 955.
Soft, 372.
Sound, velocity of, 1066.
Sounding rod, sounding chain, 991.
Specific gravity, 161, 755.
Sphere, 227, 236, 588, 605, 646, 747, 918.
Spheroid, 237, 588.
Springs, spring dynamometer, 506.
force of, 163.
Statics, 155, 165.
Stability, 250, 264, 269.
. of floating bodies, 750.
Steam, expansive force of, 35.
heaviness of, 795.
Torsion balance, 1050.
elasticity of, 373, 523.
moment of, 524.
pendulum, vibrations due to
torsion, 1050.
strength of, 373, 528.
Traction, pull, 156, 374.
Tractrix, 350.
Translation, motion of, 573.
Transverse vibrations, 1048, 1070.
profile of running water,
955, 959.
Trigonetrical functions, 70.
lines, 72.
Twisting couple, 564.
Tubes, conical, convergent, 861.
(C
(C
divergent. 862.
short, efflux through, 852, 854.
CC
CC
CC
"C
conical, 861, 891.
cylindrical, 853, 888.
inclined, 557.
CC
interior, 855.
Steel springs, 506.
"C
long or pipes, 863.
tempered and annealed, 402.
Stereometer, 788.
U.
Straight line, 49.
Strength, 372.
<<
(*
of buckling or breaking Ultimate strength, modulus of, 380,
across, 535.
ultimate, 379, 380.
452.
Unguents, 310.
String, vibrations of a stretched, 1070. Uniform motion, 106.
Subnormal, 87.
Subtangent, 40, 66, 292.
Surface, neutral, 410.
แ of water, 719.
Surfaces, curved, 40.
Symmetrical bodies, 215.
Syphon manometer, 778.
Symmetry, axis of, plane of, 215.
Uniformly accelerated, uniformly re-
tarded motion, 107, 108,
112.
Unit of weight, 157.
CC
varied motion, 107.
work, 169.
Upward thrust, buoyant effort, 742,
797.
V.
T.
Tachometer, Woltmann's, 992.
Tangent, tangential angle, 39, 47, 146.
function of, curve of, 71.
plane, 40.
Tangential acceleration, 144.
velocity, 146.
Valve-gate, 900, 903.
Valves, 776, 779, 904.
clack, 900, 905.
puppet. 905.
throttle, 901, 903.
Variable, variable quantity, 33.
3
dependent, 33.
independent, 33.
motion, 106, 117.
"
force, 189.
Tantochronism, 659.
Temperature, 793.
Tension, 281, 775, 776, 793.
horizontal and vertical, 287.
Velocity, 107.
Theorem, Poncelet's, 341.
Thickness of boilers and pipes, 738.
Throttle-valve, 901, 903
Top. 610.
Torsion, 372, 523.
angle of, 524.
of running water,
969.
along the abscissas, 146.
along the ordinates, 146.
coefficient of, 824, 944.
final, 108.
height due to the, 115, 809.
initial, 108.
1112
INDEX.
Velocity, mean, 121, 124, 956.
<<
of propagation, 1062, 1085.
of running water, 956.
of sound, 1066.
sudden variation of, 885.
virtual, 187, 209, 212, 275.
Vibration of a stretched string, 1070.
of an elastic rod, 1072.
"C
Virtual velocity, 185, 209, 212, 275.
Vis viva, principle of, 171, 174.
Volume, 156.
Volumeter, 789.
Water, waves of, 1084.
Waves, 1062.
crest and trough of, 1085.
height of, length of, 1085.
of water, 1084.
Web, 478, 479.
Wedge, 277, 329, 496.
Weight, absolute, 156, 159, 161.
(6
unit of, 157.
Weir, overfall, notch, 811, 833, 844,
849, 914.
Wheel and axle, 305, 567, 595.
Work done by a force, mechanical ef
W.
""
(C
fect, 168, 187, 209.
"C
friction, 313, 335.
heat, 936.
Water, apparatus for measuring, 976.
"
こ
​68
efflux of, 800.
heaviness of, 160.
height of in communicating
tubes, 723, 761.
hydraulic pressure of, 808.
hydrostatic pressure of, 722.
inch, 983.
jets of, 138.
meters. 1020.
running, 955.
stream of, 801, 821.
surface of, 718, 765, 767.
inertia, 171, 577.
unit of, 169.
Working load, 380.
X
Ximenes' experiments on friction, 318
"
water vane, 1001.
Z
Zore, 393.

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