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UNITED STATES OF AMEEICIA.
THE
GRAPHICAL STATICS
MECHANISM.
A GUIDE FOR THE USE OF MACHINISTS, ARCHITECTS,
AND ENGINEERS;
AND ALSO
A TEXT-BOOK FOR TECHNICAL SCHOOLS.
BY
GUSTAV HERRMANN,
PROFESSOR IN THE ROYAL POLYTECHNIC SCHOOL AT AIX-LA-CHAPELLE.
TRANSLATED AND ANNOTATED BY
I
A. P. SMITH, M.E
NEW YORK:
D. VAN NOSTRAND, PUBLISHER,
23 Mueuay and 27 Wirren Streets.
1887.
SU
Copyright, 1887,
By W. H. FARRIXGTON.
RAND AVERY COMPANY,
ELECTKOTYPERS AND PRINTERS,
BOSTON.
PREFACE.
Since the appearance of Culmann's work, which
marked an epoch in the history of graphical statics,
the graphical method has attained pretty general
dissemination in engineering circles ; its advantages
over the analytical have been recognized more and
more, and its further development kept constantly
in view. It is universally applied in the design-
ing of stationary structures, — such as bridges, — for
determining the requirements of the individual parts.
In machine design, also, the graphical method gives
valuable aid in finding the moments to which the
machine parts are subjected, and in determining di-
mensions. Accordingly courses in graphical statics
have been introduced in all technical schools. The
graphical method has also been adapted with advan-
tage to certain departments of dynamics, as in
Radinger's " Dampfmaschinen mit hoher Kolbenge-
schwindigkeit," and Pröll's " Versucht einer graphischen
Dynamik." *
* Radinger' s Steam-engines with High Piston Speed, and Pröll's
An Attempt at Graphical Dynamics.
iii
IV PREFACE.
In all these determinations, however, friction and
the special hurtful resistances to motion have not
been taken into account. Heretofore all attempts to
ascertain these hurtful resistances in machines, and
to determine the efficiency which is dependent upon
them, and is of such great importance in practice,
have been confined to the analytical method, which
is often awkward and at times utterly inapplicable.
No method is as yet known to me by which the
frictional resistances and efficiency of any desired
mechanism can be graphically determined. In my
lectures on machinery in the polytechnic schools of
this place I have endeavored to show the relations
existing between the forces in mechanism in a simpler
form than that offered by the analytical method. Out.
of that endeavor has grown the following treatise,
which in reality amounts to nothing more than a
wider application of the long known but little used
angle of friction.
My object in the present treatise was principally
to facilitate study for the students of the technical
schools, upon whose time and industry increasing
demands are made from day to day ; perhaps the
work may also be of interest and value to those more
advanced.
GUSTAV HERRMANN.
Aix-la-Chapelle, May, 1879.
TRANSLATOR'S PREFACE.
The following translation was undertaken from a
belief that a knowledge of Professor Herrmann's
work on — w^e might almost say discoveries in — this
subject should not be bounded among English and
American engineers by an ability to read German or
French; the treatise having already been translated
into the latter language. The original has been fol-
lowed faithfully. There is one word that perhaps
needs explanation. The German expression im Sinne
seems to have no technical equivalent in English, and
has been literally translated "in the sense of," and
coupled with " direction," the latter being used loosely.
"Acting in the sense of a force Q" means producing
a similar result to Q. Thus, suppose a force P applied
to a crank tangentially to the crank-circle at every
point of revolution. The load Q is suspended from
a rope wound on the windlass drum turned by the
crank. Then if the rope is being wound on to the
drum, the force P acts at every instant in an opposite
" sense " to Q, while there is only one point in each
revolution at which it acts in an opposite direction.
VI TRANSLATOR'S PREFACE.
To complete the definition, suppose the windlass to
turn stiffly in its bearings, and the weight Q to be
so small that when the crank is released the weight
will not run down. To lower the weight, then, a
force (P) would have to be applied to the crank to
turn it in the opposite direction. This force (P)
would then act in the same " sense " as Q at every
instant, though it would have the same direction as
Q but once in each revolution. " Sense," then, has
reference simply to the effect produced through no
matter what amount of intervening mechanism, and
is entirely distinct from " direction" in its stricter
meaning.
A few of the terms of Professor Kennedy's transla-
tion of Reuleaux's " Kinematics " have been employed,
because there are no other well known equivalents,
but they will be easily understood even by those who
are not as yet familiar with that work.
The great advantage which the method presents is
its simplicity. By the use over and over again of a
few easily mastered principles, the most complicated
problem may be solved. No knowledge of higher
mathematics is required in its mastery, and no hand-
ling of lengthy and involved algebraic formula is
necessary in its use. A few lines are drawn in accord-
ance with easily understood rules, and the result
stands out so clearly on the paper that every bright
mechanic ought to understand it at a glance. This,
TRANSLATOR'S PREFACE. Vll
with the additional merit of rapidity and accuracy,
should soon render its use common in every class-
room and drafting-office.
The respectful thanks of the translator are due
to Professor J. F. Klein of Lehigh University and to
Mr. T. M. Eynon, recently of the same institution, for
their critical revision of the manuscript.
A. P. SMITH, M.E.
United States Patent Office,
Washington, D.C.
CONTENTS.
PAGE
§ 1. — The Efficiency of Mechanisms 1
§ 2. — The Equilibrium of Mechanisms 14
§ 3. —Sliding Friction 23
§ 4. — Journal Friction . 39
§ 5. — Rolling Friction 76
§ 6. — Chain Friction 84
§ 7. —Stiffness of Ropes 102
§ 8. —Tooth Friction 110
§ 9. — Belt Gearing 122
§ 10.— Examples ...'*'* 140
Concluding Remarks . 150
ix
THE
GRAPHICAL STATICS OF MECHA
§ 1. — THE EFFICIENCY OF MECHANISE
The object of every mechanism is to transfer 3
in a certain way, from one particular machine
another. In this transfer the magnitude and di
of the motion may remain unchanged, or the re:
motion may vary in either or both of these res
and the rate of variation in either magnitude 01
tion may be constant or not. In the steam-e
for instance, the straight-line motion of the pis
communicated continuously to the cross-head
piston-rod unchanged in magnitude and
while this right-line motion generates a motion i
crank-pin, through the agency of the connectin
whose direction varies constantly, and whose magi]
differs in its relation to the magnitude of the
head's sliding motion at every instant. In the ca
two cylindrical toothed wheels of equal diameter, ^
ing together, the rotation of one is transformed ii
rotation of the other, equal, but in an opposite direc
The ratio between the lengths of the paths thn
which the two pieces move depends upon the :
which exists between the force P working upon
driving member of the machine, and the resistanc
1
HE GRAPHICAL STATICS OF MECHANISM.
3 driven member encounters. If no hurtful
?s or hinderances, such as friction, appeared in
)ii under consideration, the mechanical work of
Qg-force, for any period of time, would exactly
t done during the same period in overcoming
ance presented to the driven member, under the
on of uniform motion. This follows from the
wn principle of virtual velocities. If p and q
he distances over which the points of applica-
P and Q travel along the directions of these
md if we let Pp and Qq — i.e., the product of
ito distance — represent the mechanical work
r the respective forces, we have the equation
Pp = Qq, or Pp - Qq = 0.
jr the supposition, on the other hand, that one
. of the machine parts does not have a uniform
according to the law of kinetic energy, again
ill hurtful resistances, the equation becomes
Pp - Qq = L,
L represents the amount of work which the
I mass M has either absorbed or given out in the
*ed time. According as the above value of Pp —
positive or negative, the moving masses have been
crated or retarded ; and from Pp — Qq z= there
- r s evidently a uniform motion during the assumed
i. From the preceding it follows, that, always
Aug hurtful resistances, in no machine construction
hatever form is there either a loss or gain in
lanical work ; for if, in the case of an inequality
THE EFFICIENCY OF MECHANISMS. 6
between Pp and Qq, the masses M are accelerated, this
acceleration represents a storing-up of mechanical work,
which, later in the period, is available to its full amount.
But there can be, in reality, no movement between
two bodies in contact without certain hinderances aris-
ing ; which hinderances, since they always have the char-
acter of an impediment to the desired motion, are gener-
ally known as hurtful resistances, in opposition to the
useful resistances in the overcoming of which the useful
work of the machine consists. In every case, such hurt-
ful resistances arise only where one body has a motion
relative to another: as, for instance, the friction of a
journal turning in a bearing which is stationary, or
rotates at a different speed ; the resistance of the air to
arms or wings in rapid motion, etc. These resisting
forces disappear with the cessation of relative motion
between the two bodies assumed. Therefore, we must
admit that no loss of work occurs when tensile or
compressive forces act on a solid body, so long as the
elastic limit is not exceeded, since their effect is only to
increase or diminish the distance between the molecules
of the body, — there being no contact between these
molecules, — and the contraction of the body gives out
again the work used in expanding it; at any rate,
experience has not yet proved the contrary. That no
one may bring forward the example of the resistance
due to the stiffness of ropes as an exception to the
above generalization, it should be remarked, that, when
a rope is bent around a pulley there occurs a relative
motion between any two threads or strands which are
at different distances from the axis of the pulley.
A certain amount of mechanical work, Ww, is necessary
to overcome the hurtful resistances TT during a period
4 THE GRAPHICAL STATICS OF MECHANISM.
of motion in which the distance w is traversed. This is
not only lost as far as the work proper of the machine
is concerned, but exercises a further and evil influence,
in that it is transformed into molecular work, whose
effect is evident in the wear of the moving parts.
Taking into account the hurtful resistances the equa-
tion becomes
Pp — Wto = Qq + -L, or Pp — Wiv = Qq,
if, in the last equation, we suppose the useful work Qq
to include the positive or negative quantity L repre-
senting the work done in the acceleration or retardation
of the masses M. The last assumption, which will be
taken as the basis of all that follows, is always admissi-
ble, because, as before remarked, ail work stored up in
the acceleration of the masses will be given out again
in the performance of useful work when the velocity of
the masses sinks to its previous value. But these regular
or periodic accelerations and retardations characterize
the state of permanency (or normal running condition)
of the machine, which is the only state or condition
taken into consideration in the present case.
From what precedes it follows without further ex-
planation, that in all machines, without any exception
whatever, the useful work done is less than the work
expended by the driving-force in producing motion, by
an amount exactly equal to the work required to over-
come the hurtful resistances. The smaller these latter
are, the more completely will the machine accomplish
its object ; that is, the work of the driving-force will be
transferred to its point of useful application with the
greater economy. We may therefore regard the ratio
THE EFFICIENCY OF MECHANISMS. 5
of the accomplished useful work Qq to the work Pp
expended by the driving-force as the efficiency of the
machine ; and the ratio
Qq_
Pp- 71
is known by this name, or by that of " useful effect."
The efficiency of a machine — which will always be
represented by rj hereafter — is, according to what pre-
cedes, less than unity in every case except the ideal
one, in which all hurtful resistances, TF, disappear. The
value of r) will evidently grow smaller as the number
and magnitude of the hurtful resistances increase; and,
since the latter occur wherever there is relative motion,
it follows that, in general, the simpler a machine is in
its construction the more economical will it be in work-
ing, since every additional moving piece reduces the
efficiency by the addition of a new hurtful resistance.
Tt is hardly necessary to remark that the value of the
useful effect depends not only upon the number, but
upon the magnitude, of the hurtful resistances, which
magnitude varies greatly according to the kind of
resistance offered. Thus there are not a few machines
whose efficiency, on account of the resistances incurred,
sinks to a small percentage ; in fact, the value of 77 may
even become equal to zero or negative. The numerator
of the fraction __l = y becomes zero in all cases where
Pp
a weight Q is moved in a horizontal plane, — for instance,
in the rolling of an object along a horizontal track, in
the turning of a crane, of a turn-table, etc., — because the
distance q passed over in the " sense, 1 ' or direction, of
the load Q (i.e., the line of direction along which it
(j THE GRAPHICAL STATICS OF MECHANISM.
would -tend to move if left without any support what-
ever) is equal to zero. In such cases the entire work
of the driving-force P is employed in overcoming the
hurtful resistances W; and we have for these cases, for
such mechanisms, the equation
Pp = Ww.
If, further, we suppose that in a certain case Ww>Pp,
there follows, from the general equation,
Pp — Ww = Qq,
a negative value for Qq ; i.e., in order to render motion
possible in such a case the force Q must work in a
direction, — y, opposite to its usual " sense " or direc-
tion.* The force Q is no longer a resistance to be
overcome through the agency of the machine: it is, on
the contrary, to be regarded as one of the forces
necessary to produce motion, and working in the same
" sense," or direction, as P. There can no longer be
any question of useful work, and machines of this
nature are never used to produce motion : they find
their application in those cases where it is required to
hinder an undesired motion. Every screw the pitch-
angle of whose thread is not sufficiently great to cause
it to turn backwards under the influence of the load
upon its nut, is a machine with negative efficienc}' for
the case of backward motion ; and the same is true of
the differential jail ley as generally constructed. The
* Thus, if it was a weight to be lifted, the attraction of gravitation
would have to be reversed, or overcome by an opposite and greater
force before motion would occur. — Trans.
THE EFFICIENCY OF MECHANISMS. 7
clamps, so common in practice, which never release
their hold under any strain whatever, and of which we
shall treat more particularly hereafter, may always be
regarded as machines with negative efficiency.
We may define efficiency in another way which is
often used, and in many cases is more convenient. If
in any mechanism where, as before, a driving-force P
expends the work Pp in order to move a load Q along
the path q, we neglect the hurtful resistances, there
would evidently be necessary to accomplish the useful
work Qq, under this supposition, only a driving-force
P , of a less value than P, but working along the same
distance /?, at the same point of application, and in the
same direction. In the case of this force P we have of
course, through the neglect of all hurtful resistances, the
equation
from which the expression for the efficiency becomes
v — pp - Pp — p-
If, for brevity, we call P the theoretic, and P the
actual, driving-force, the efficiency becomes the ratio of
the theoretic to the actual driving-force ; and this value
accordingly represents that percentage of the driving-
force which serves to overcome the useful resistance.
If, on the other hand, under the assumption of fric-
tionless motion we regard the force P as working along
the path p, it would overcome a resistance Q along the
path q, which theoretic resistance Q would have a larger
value than the useful resistance Q actually overcome ;
8 THE GRAPHICAL STATICS OF MECHANISM.
and, from the equation Pp = Q q, we should have the
efficiency,
__Qq_ __Q± _ Q_
*- Pp~ Q q~ Q >
that is to say, equal to the ratio which the actual useful
resistance Q bears to the theoretic one Q .
If, in the case of any particular mechanism with the
useful load Q, the driving-force becomes less than P,
there can be no motion in the " sense " of P; the
mechanism remains at rest. The same thing continues
until the driving-force approaches the theoretic P in
value. If, then, there were no hinderances to motion to
be taken into account, the slightest further diminution
of P below the value P would result in a backward or
reverse motion of the mechanism, the load Q becoming
the driving-force. But since certain hurtful resistances,
which may be represented by ( IF), oppose the backward
motion, that backward motion can only occur from that
moment when the force _P, through further diminution
below P , has sunk to a value (P), for which the
relation
Qq — ( W)w = (P)p
holds. The force (P) which is just able to prevent the
backward motion, or running down, of the mechanism,
is smaller than the theoretic driving-force P by an
amount which increases with the magnitude of the
resistance (IT) offered to the backward motion of
the mechanism. Thus, in speaking of the efficiency
for backward motion, we will hereafter understand the
ratio
« =
From (,) = i|) =
it follows that (P) =2 0, and therefore
since (P)jp + QW)w — Qq.
On the other hand, for forward motion,
Pp — Qq + Ww.
If the hurtful resistances (TT) for the backward
motion are equal to those W for the forward, the
substitution of the former for the latter in the last of
the above equations would give
Pp = Qq + (Tf> = Qq + Qq=z 2Qq ;
from which
Pp 2Qq
THE EFFICIENCY OF MECHANISMS. 11
The assumption that W ■= (TF) is a sufficiently close
approximation, since the hurtful resistances are the
result of both P and Q* Moreover, since TT> ("FT),
it follows that 2Qq < Pp, and consequently that the
efficiency rj must always be less than |. Therefore we
may lay down the principle that every self-locking
machine has an efficiency of less than fifty per cent, a
conclusion which experiments with screw hoisting-gear
and differential pulley-blocks confirm.
Machines met with in practice consist generally of a
collection of elementary mechanisms, so that the driving-
force P r necessary to overcome the useful resistance Q
of the first mechanism must be regarded as the useful
resistance Q' of the second mechanism, for the overcom-
ing of which in the second mechanism another driving-
force P" is required, and so on. Take, for example, an
ordinary warehouse hoist : here the rope with its drum
is one mechanism, in which the load hanging from the
hook on the end of the rope is the useful resistance $,
to overcome which a force P f must be applied at the
circumference of the toothed wheel connected with
the drum. As regards the toothed gearing which forms
the second mechanism, the force P f becomes the useful
resistance Q\ which must be overcome by a force P"
applied at the circumference of the second large gear
* The force of this reasoning lies in the fact that
W\ (W) :: friction of (P and Q) : friction of (P) and Q,
in which Q is generally much the larger factor; and hence the differ-
ence between IF and (W) is much less than it would be if they were
dependent solely upon P and (P), and the relation W : ( W) :: friction
P : friction (P) existed. — Trans.
12 THE GRAPHICAL STATICS OF MECHANISM.
upon the axis of the pinion. In the same way, this
force P" is the resistance Q" for the third mechanism,
represented by the second pair of gears and the crank.
To overcome this resistance Q", a force P is necessary
at the crank-handle. When, as in the case of a crane,
the load Q does not hang directly from the drum, but
the rope, or chain, is first led over one or more pulleys,
each of these pulleys is to be regarded as a separate
mechanism. In the same manner, every machine,
however complicated, can be resolved into its simple
elementary mechanisms. Such a resolution greatly
simplifies the determination of efficiency of machines,
inasmuch as the number of common mechanisms is
small, while the diversity between complete machines
is almost endless.
One general law may be established for the efficiency
of a machine consisting of any desired number of ele-
ments. If we let Q again represent the load, and q the
path described by it in a slight movement, and let P v
P 2 . . . P n denote the driving-forces for the separate
mechanisms, and p v p 2 . . . p n the corresponding paths
of these forces, we may understand by rj v rj 2 . . . rj n the
efficiencies of the individual mechanisms. For the first
machine, we have
-&- = Vl .
For the second, in which P x is the useful resistance,
and P 2 the driving-force, we have
THE EFFICIENCY OF MECHANISMS. 13
and so on for each separate mechanism. From the
multiplication of all these equations, there results
QV P iPi PjiPn -Pi.-iPi.-i - v v v v
or Qq
jr = v = v t . v 2 • % • • • %.
That is to say, the efficiency of a machine composed of
any number of mechanisms is equal to the product
of the efficiencies of all the separate mechanisms.
From the fact that the efficiency of a simple mech-
anism is always less than unity it follows, as before
remarked, that in general the useful effect of a machine
decreases as the number of its constituent elements
increases.
Since, furthermore, the above product cannot be
negative as long as all the factors are positive, it
follows that a machine can only be self-locking when
this property belongs to at least one of its elementary
mechanisms.
14 THE GRAPHICAL STATICS OF MECHANISM.
§ 2. — THE EQUILIBRIUM OP MECHANISMS.
Although mechanisms, by their very nature, can only
effect their object while in motion, or by virtue of the
same, yet, for the ascertaining of the relation existing
between the various forces, we may always assume as a
basis that condition of equilibrium which corresponds
to the limit where the slightest increase of the driving-
force would produce a motion in the "sense*-" or direc-
tion, of that force. In what follows P will again
represent the driving-force and Q the useful resistance.
Neglecting for the present any acceleration of the
masses, we will suppose a uniform motion in which,
at each instant, the work of the force during a small
portion of time is just sufficient to overcome the useful
resistance Q after the hurtful resistances W have been
disposed of. It will then easily appear in what way the
accelerating force working upon the mass M in the case
of variable motion can be ascertained.
The exterior forces P and Q working upon any
mechanism, call forth certain internal forces, or re-ac-
tions J2, between the members of the machine wherever
two parts come in contact. These re-actions are to be
regarded as two equal and opposing forces occurring at
every surface of contact. Every pair of forces thus
arising at the same point is, therefore, in equilibrium.
We must imagine such re-actions wherever two bodies
come in contact, whether the bodies move relatively to
each other or not. We can, therefore, in every case
neglect the bodies in-contact and think only of the
THE EQUILIBRIUM OF MECHANISMS. 15
re-actions offered by those bodies. Under this supposi-
tion, any member of a machine which is acted upon by
certain exterior forces P and Q, and which is supported
at certain points by neighboring bodies, must be under
the influence of the exterior forces P and Q, and of the
re-actions Ä, which are sufficient to replace the imagined
supports, in order to be in the supposed limiting con-
dition of equilibrium. The conditions of equilibrium
furnish us, in general, with a means by which from the
known elements, — direction and magnitude of indi-
vidual forces, — we may ascertain the unknown. In
the majority of cases the intensity of the re-actions of
the supports is unknown ; of the exterior forces, there
is, as a rule, one element — the direction, or intensity,
of one force — unknown at first. As regards the direc-
tion of the re-action replacing a support, it is deter-
mined empirically by the condition that it shall be
inclined to the supporting surface at a certain determi-
nate angle whose magnitude depends upon the nature
of the two bodies in contact, as to smoothness, hardness*
etc. The hurtful resistances to motion, W, which, as
previously remarked, arise only at the point of contact
between two bodies (i.e., at the supporting surfaces),
depend on the nature of the material, and of the sur-
faces constituting the supports. The size of the angle
at which the surfaces of contact will be cut by the
direction of the re-action existing between them de-
pends closely, as will be shown in what follows, upon
the amount of hurtful resistance generated between the
surfaces.
If we suppose, in the next place, that no hurtful
resistance W exists, — a condition of affairs which, of
course, never occurs in practice, — the angle formed by
16 THE GRAPHICAL STATICS OF MECHANISM.
the direction of re-action and the supporting surface
would be a right angle ; in other words, when there are
no hurtful resistances a supporting surface can only
re-act, at each of its points, in the direction of a normal.
With the assumption which thus entirely ignores all
hurtful resistances, it is a simple matter to determine
the proportion of power and load at every point in a
machine ; and, for this purpose, a graphical method can
be used to good advantage. A few examples will serve
to illustrate this procedure.
Let J-, B, and (Fig. 1, plate I.) be the centres of
the three pins on a bell-crank, the middle one of which,
6 7 , turns in the fixed bearing C v while the end pivot A is
enclosed by the eye or end bearing A x of the rod A X A V
and the other pivot B is attached in the same way to
the rod B X B V If, now, a force Q acts upon the lower
end A 2 of the rod A l A 2 in a certain direction, the rod
A 1 A 2 must, according to the preceding principles, be in
equilibrium under the influence of the force Q, and the
re-action M 1 replacing the pivot or journal A ; and this
latter re-action, being perpendicular to the surface of
the journal, must pass through its centre. Two forces,
however, can only be in equilibrium when they are
equal, and work in opposite directions along the same
straight line ; from which it follows immediately that
the line of direction of the force Q must pass through
the centre of A, or, if it does not, there wiH be a turning
of the rod A 1 A 2 about the journal A until this condition
is fulfilled. In the same way, it follows that the direc-
tion of the force P acting on the rod B X B 2 must pass
through the centre of JS, and that the journal B must
exert upon the rod B X B 2 a re-action R 2 which shall be
equal and opposite to P. The rods A X A 2 and B X B 2 will
THE EQUILIBRIUM OF MECHANISMS. 17
be subjected to tension by the forces Q, M 1 and P, P 2 ,
respectively ; i.e., there will be called forth internal
elastic stresses in the material of each section of the
rods, which will be in equilibrium one with another, and
with the exterior forces at the ends of the rods. These
interior strains are of great importance in determining
the dimensions of cross-sections of the separate machine
parts, but have no direct influence upon the conditions
of equilibrium of the machine. We shall not, however,
go into the determination of dimensions here, or in
the remaining portion of the treatise. For all that the
reader is referred to the well-known works on machine
construction and the resistance of materials. Regard-
ing now the lever ABC alone, we have the external
forces Q and P acting upon it at the points A and B ;
and we can also suppose the bearing C x to be replaced
by a re-action P 3 , whose line of direction passes through
the centre of C. For the condition under consideration
these three forces must be in equilibrium. This can
only occur when the three forces intersect at the same
point ; and therefore the re-action P 3 , exerted by the
bearing C\ upon the journal C, must also pass through
the intersection of the lines of the forces P and Q.
Moreover, the relative intensities of the three forces, P,
$, and P 3 , are easily determined if we let OD == Q, the
load, and complete the parallelogram ODFE, whose
other side falls upon OP, and whose diagonal coincides
with 00. We have then in OE the force P, and in
OF the pressure of the journal O upon its bearing, and
the equal but opposite re-action P 3 of the bearing C x
against the journal C.
The relative intensities of P, Q, and P 3 correspond
to the oft-mentioned limiting condition in which the
18 THE GRAPHICAL STATICS OF MECHANISM.
slightest increase of Q or P would cause motion in the
"sense/ 1 or direction, of the force so increased, as can
be at once seen from the figure ; for, if we increase Q
until it is equal to OD\ the diagonal 0F\ which repre-
sents the pressure of the journal upon its bearing goes,
to the left of the centre Ö. This points to a motion of
the lever in the "sense," or direction, of Q. On the
other hand, an increase of P to the value 0E f gives a
journal pressure OF", in consequence of which there
would be a right-handed revolution of the lever.
As a further example, the slider-crank mechanism
(Fig. 2, plate I.) may be adduced. The force acting
from the piston-rod AA' upon the pin A of the cross-
head is transferred through the connecting-rod A 1 B 1 to
the crank-pin B; and, under the supposition of entirely
frictionless motion, the force T must pass through the
centres of A and B since it is perpendicular to the
surfaces of contact of the journals A and B with their
bearings. Since the forces P and T have different
directions, the pin A cannot be in equilibrium under
their influence alone : a re-action It 1 exerted by the
guide jPjDg upon the cross-head J) is necessary. For
the condition of equilibrium this force acting jierpen-
dicularly upon the supporting surface D^D^ must pass
through the intersection A of the piston-thrust P and
the connecting-rod resistance T. From this we can
easily find the forces R x and T by drawing AF equal
to P, the piston-thrust, and completing the triangle
AFGr, in which FGr is parallel to R v i.e., perpendicular
to the guide D X D T If, now, the axis C of the crank
meets a resistance Q at the distance CV7,— as though a
gear-wheel, with the radius CJ was on the axis C and
meshed with another gear-wheel JK, whose resistance
THE EQUILIBRIUM OF MECHANISMS. 19
would be represented by Q, — it must be in equilibrium
under the influence of the connecting-rod thrust acting
at jB, the resistance Q acting at J, and a re-action R. 2 of
the bearing O v The direction of the latter again coin-
cides with the line joining and (7, if represents the
intersection of Q and T. By constructing upon the line
AG, already determined as the value and direction of
the force T, the triangle AGH, whose sides AH and
GH are parallel to the direction of the resistance Q,
and of the re-action R v respectively, we have in GH
the value of the re-action iü 2 , and in AH that of the
resistance Q overcome at J, at the instant under consid-
eration. It will be seen that the ratio of P to Q varies
for every instant, and that, with a constant piston-
thrust the amount of resistance Q overcome at J will
vary between zero at the dead points and a maximum
at some intermediate position. When, therefore, as in
the ordinary case, the resistance which the gear JK
presents to the motion of the crank is constant, this
resistance must have a mean value between zero and
the maximum value of Q ; and there will result an
acceleration or retardation of the masses (of fly-wheel)
according as the value of Q, determined as above, ex-
ceeds, or falls short of, this average value of the resist-
ance between the gear-wheels. This peculiarity of
slider-crank mechanism is well enough known to render
a further discussion of it here superfluous.
In the same way as in the two examples shown we
can obtain in every case the ratio of the force P to
the resistance Q. In nearly all cases the graphical
method offers great advantages over the analytical on
account of its simplicity and plainness. The analytical
method frequently leads to involved expressions, espe-
20 THE GRAPHICAL STATICS OF MECHANISM.
cially when it is attempted to bring the hurtful resist-
ances into the calculation ; when, in other words, it is
no longer a question of determining P or $ , but of
P or Q. Since the economic value, or efficiency, of any
machine depends directly upon the magnitude of the
hurtful resistances connected with it, it is evident that
the determination of the ratio actually existing between
the forces when these resistances are taken into account
is of vastly more importance in practice than any deter-
mination of the merely theoretical forces. We have
been accustomed in the past to the use of only the
analytical methods. The graphical methods for the
determination of the friction in, and hence the effi-
ciency of, mechanisms have hitherto been rarely em-
ployed ; at any rate, in all the text-books only the
formulse for these determinations have been given.
How complicated such investigations often became
even in the simplest machine as soon as any exact
calculation was attempted, is well known. Thus, for
example, we could only obtain an expression for the
journal friction of a bell-crank, as in Fig. 1, plate I., or
of a pulley where the ropes were not parallel, through
a long radical, — a circumstance which compelled the
assumption of parallelism in all cases of pulley friction,
even when there was a marked inaccuracy in the same.
For the same reason it is the custom to assume an
infinite length of connecting-rod in all cases of crank-
gear, in order to render less unwieldy the expressions
in which the angle of crank to piston-rod occurs. From
the well-known advantages which the use of the graph-
ical method offers in the designing of machine elements,
as in the determination of the moment to which axles,
cranks, etc., are subjected, arose the idea of finding an
THE EQUILIBRIUM OF MECHANISMS. 21
expression for friction by the same method. From that
idea has sprung the following treatise. In the course
of the same, it will be shown that we can obtain a
graphical determination of the actual proportion exist-
ing between the forces in the same simple and sure way
as was done in the preceding examples where all friction
was neglected; and it is evident, that, by comparing
the values obtained for P or Q with those of P or # >
we have immediately an expression for the efficiency, — ■
' , = £.= £
The methods used do not differ from those indicated
in the determination of the theoretical forces ; and in
nearly every case the drawing of force polygons suffices
to attain the desired result. The main question will
therefore be to express the separate hurtful resistances
graphically. The solution of this problem will be at-
tempted in what follows.
The hurtful resistances in mechanisms which are to
be taken into consideration are few in kind and consist,
if we neglect the resistance of the medium in which
they work, only of friction, which may occur as sliding
and rolling friction, journal friction, chain friction, and
the friction of toothed wheels. The stiffness of ropes
may be considered as equivalent to chain friction. In
the following pages these simple resistances will be
taken up one by one. The resistance of air or water is
here neglected because in ordinary machines it may
be left out of account as insignificant, and in most
cases is not regarded. In individual cases and in
particular machines where such neglect is not allow-
22 THE GRAPHICAL STATICS OF MECHANISM.
able, where (as in ventilators) the moving of this
medium is the principal object, its resistance should
be determined by the rules of hydraulics. It has
ceased to be a hurtful, and become a useful resistance.
SLIDING FRICTION. 23
§3. — SLIDING FRICTION.
A load which presses upon a horizontal surface (7(7
(Fig. 3, plate I.) with its weight Q = AB calls forth,
while at rest, an infinite number of re-actions from points
in that surface, whose resultant is equal to the weight
Q, and opposite in direction. This re-action passes,
therefore, through the same point I) in the supporting
surface as the weight Q (the term " weight " being here
used in its meaning of a resultant of the forces of
gravitation acting upon each particle of the body ^4).
Now, it is known that a force P = pQ, acting parallel
to the plane (7(7, is necessary to produce a horizontal
sliding of the body A along that plane, where fx repre-
sents the co-efficient of friction. If, now, a force P,
represented in the figure by A C, acts upon the body J.,
the body is under the influence of the two forces P and
Q, which must be in equilibrium with the re-action R
of the supporting surface ; it being always remembered
that we are assuming that limiting condition of equi-
librium where the slightest increase of P would cause
motion. This condition of equilibrium requires, there-
fore, a re-action R = EA of the supporting surface
equal to the resultant of P and Q, and opposite in
direction. From the equation P = ^Q is determined
the angle $ = EAB which this resultant makes with
the normal AB to the supporting surface, since
/* = — = tan ;
24 THE GRAPHICAL STATICS OF MECHANISM.
and this is called the angle of friction for the materials
composing the bodies A and GG. If we suppose the
force P to increase gradually from zero to the value
AC, the re-action R, given forth by the supporting
surface, would gradually deflect from its original direc-
tion BA to HA; and for all positions between these
two the conditions of equilibrium would be satisfied.
In this way the point of intersection D of the re-action
Avith the surface would move from D to F; but in
every position it is to be regarded as the point of
application of the resultant of all the elementary sur-
face re-actions. These relations evidently hold what-
ever the direction of the horizontal force P. For
example, it is true that when P has the direction
AC V the re-action of the supporting surface coincides
with the line E X A. By a complete revolution of the
force P in the horizontal plane the re-action would
describe a conic surface concentric to AB. This is
called the cone of friction, and limits the space within
which the re-action of the surface GG may exist with-
out motion resulting. We must therefore regard the
supporting surface as re-acting against the supported
body in certain directions whose angles of intersection
with the normal are less than the angle of friction ;
motion commencing from that moment at which this
angle of intersection exceeds, by the slightest amount,
the angle of friction. We may employ this property of
surfaces in the graphic representation of sliding friction
under the following rule : —
If two bodies having plane surfaces in contact under-
go relative motion along that plane of contact, we may
completely replace each of these supporting planes by
a re-action which is inclined to the normal at an angle
SLIDING FRICTION. 25
equal to the angle of friction, and so situated that its
component parallel to the plane of contact will work in
the direction of the motion which the supporting sur-
face has relatively to the body supported.
The necessity and correctness of the last part of this
can easily be seen from the figure. If, for instance, the
supported body A is moved by the force P in the direc-
tion from A to (7, the relative sliding of the support, or
bearing, GG to the body A is evidently in the opposite
direction CA; and the re-action R, by which GG is
replaced, has, according to previous demonstration, the
direction EA, whose component CA works in the direc-
tion of the sliding of GG upon the body A.
By virtue of this general law we can easily obtain
in every case the value of the sliding friction called
forth by the relative motion of one body upon another.
If, for instance, in the case of the slider-crank gear
(Fig. 2, plate I.), we wish to determine the influence of
the friction existing between the slipper-block D of the
cross-head and the cross-head guide, we have only to
draw the re-action R v passing through A, in the direc-
tion E X A) making the angle E X AE = <£ with the normal
to the cross-head guide. If, then, we draw through F
the line FG X parallel to E X A, we have in AG X the actual
thrust T of the connecting-rod ; while the value pre-
viously obtained by neglecting friction was AG = T .
The thrust of the connecting-rod has been reduced,
through the sliding friction of the cross-head i>, by the
amount GG X ; and we have the value
AG, T „
AG T V
for the efficiency of the right-line motion in the slider-
26 THE GRAPHICAL STATICS OF MECHANISM.
crank gear. It is unnecessar}^ to explain that the value
of this sliding friction and its dependent efficiency
varies for every position of the gear.
By the use of the angle of friction in the manner
discussed above, the efficiency of a great number of
machines may be easily determined, as will be shown in
a few examples. Let ABC (Fig. 4, plate I.) be the rack
of an ordinary jack, upon whose lug, or claw, A the load
Q = Io l acts vertically downward ; while at the pitch-
line DD of the rack, the force P acts vertically upward.
In consequence of the fact that the load Q acts in one
direction only, the rack is continually pressed against
the casing at B and C; and we may regard it as being
supported by the resultants b and c of the re-actions R x
and 7t 2 . The four forces Q, P, R v and R 2 must be in
equilibrium, which can only be the case when the
resultant of any two is equal to that of the other two,
and opposite in direction. If, therefore, o x is the inter-
section of Q and R v and o 2 is that of P and R v the line
o x o 2 is the direction of the resultant of Q and R 1 as well
as of P and R 2 . From the given load Q, we can now
determine the forces P, R v and R 2 by drawing the
force polygon. Make Io x = Q; draw III parallel to
R v and intersecting o x o 2 ; and then construct the tri-
angle II III o x by drawing II III parallel to P, and
III o x parallel to R 2 . We have, then, in II III the
force -P, which must be applied at the pitch-line of the
rack to lift the load Q. Without friction, Po = Q, and
therefore the efficiency of this mechanism
- ^2
V P
is known.
SLIDING FRICTION. 27
The figure has been constructed with a co-efficient
of friction ^ = 0.2, and a corresponding angle of fric-
tion > = 11° 20', and gives P = 122.3 for Q = 100;
therefore
100
122.3
= 0.817.
It should be here remarked, that in this and all
following cases the numerical results are not taken
from the lithographed plates, but from the original
drawings of the author, which were on a much larger
scale. In drawing the direction of the re-action it is
not necessary to know the angle of friction in degrees
and minutes ; a knowledge of the co-efficient of friction
is entirely sufficient, and both accuracy and simplicity
recommend the construction of the angle of friction,
through the geometric method, from its tangent == (jl
rather than taking it direct from the table. The effi-
ciency rj as determined above is, of course, only the
efficiency of the rack in its casing. In order to get at
the efficiency of the entire jack we must take into
account the friction of the gears and of their journals,
in the manner hereafter to be explained.
For the case of backward motion in the jack, that
is, when the load Q is the cause of motion, the same
construction holds, with the assumption that the re-
actions (.ßj) and (i£ 2 ) are inclined to the opposite side
cf the normal by an amount equal tQ the angle <£,
and we have the force polygon / 2 3 o v shown in
broken and dotted lines in the figure, where the
diagonal 2o 1 is parallel to (oj) ( 2 ), and the line 2 3
gives the amount of force (P) which must be applied
as a brake to prevent the accelerated downward motion
28 THE GRAPHICAL STATICS OF MECHANISM.
of the load Q. For the same value of /x = 0.2, we
have
(P) = 79.1 .-. 0?) = ^p = 0.791.
o
In this connection it should be remarked that the
points b and , in which the bearing surfaces of the rack
are pierced by the re-actions R x and i£ 2 , are not com-
pletely determined by the geometrical character of the
mechanism ; and, consequently, it is necessary to start
with the supposition that the points of application of
the forces are at the middle of the supporting sur-
faces. Any variation upon this point would affect the
efficiency of the mechanism.* In the preceding case of
the movement of the cross-head (Fig. 2), such an inde-
termination was excluded by the requirement that the
re-action of the guides must pass through the inter-
section of the forces T of the connecting-rod, and P of
the piston-rod thrust. In the present case (Fig. 4),
equilibrium is also possible if R 1 and R 2 do not pass
exactly through the central points b and c of the sup-
porting surfacco. But with any other position of b and
c the values of R x and R. 2 would be changed ; and it is
easily seen that R x and R 2 will have the smallest value,
* Tims, if, in the figure b and c were brought nearer together by
raising the former, and lowering the latter, the points of intersection
0| and o-i would be correspondingly raised and lowered, and the diag-
onal O] o 2 would be inclined more from the perpendicular direction
of (<), so that the line I II would have to be produced farther before
intersecting it; the result being that the sides of the parallelogram
which represent the forces B u P, and R 2 would be increased, and the
efficiency
Po
diminished. — Trans,
v = p
SLIDING FRICTION. 29
and friction be least appreciable, when the vertical dis-
tance between the points b and c is the greatest possible.
If the guiding surfaces were perfect planes, and the
material would not wear away, it would be correct to
assume as points of application for R x and i£ 2 , the outer
edges b x and c v With this assumption the frictional
resistance would be the least possible ; and this choice
of points would correspond with the law, so well estab-
lished by experience, that nature always works along
the line of least resistance. But since the outer edges
would soon become rounded away by wear, on account
of the great pressure concentrated upon them, the pre-
vious assumption, by which the re-actions act at the
centres of the surfaces, corresponds best to the actual
condition and assures a sufficient degree of accuracy of
determination.
In the same way, we may determine the force P
(Fig. 5, plate I.) which must be applied to the rope
attached at the point D in order to lift the platform
of an ordinary elevator as it occurs in grist-mills. In
this case also, the platform at the bearings B and C is
pressed against the guide EE by the force acting at A,
which is the centre of gravity of the load and platform
combined ; and, as before, these supporting surfaces
may be replaced by the re-actions R t and R 2 making
the angle > with the horizontal. Taking the points of
application of these re-actions at b and c, the centres
of the sliding surfaces, we have again, in the line con-
necting the point of intersection o l of Q and Ii 1 with
the intersection o 2 of P and R v the position of the re-
sultants of these two pairs of forces. If, then, we make
Io x equal to the total load, and draw I II parallel to
R v and o x III parallel to R v and through the point of
30 THE GRAPHICAL STATICS OF MECHANISM.
intersection II the vertical II III, we have in the last
the necessary driving-force P, and in II /and o 1 III
the re-actions R x and H T The force Z =: o x II is the
one which acts directly upon the platform. For back-
ward motion the necessary brake force (P) is given by
the line 3 2 of the diagram 1 2 3 o v From the propor-
tions given in the drawing, with a co-efficient of friction
ix — 0.16, we have, for
Q
— P = 100, P = 105.9 ;
therefore
V = 105.9 = °- 944 '
and
(P) = 93.5 ;
therefore
0» = ~ = 0-935.
It can easily be seen that the friction grows less, and
the efficiency increases, as the horizontal distance be-
tween the forces P and Q becomes less, and that between
the sliding surfaces B and C greater. In practice, there-
fore, the height BO should not be taken too small, and
the line of force P should be brought as near as possible
to centre of gravity A by bending outwards the iron D
to which the hoisting-rope is attached. For the same
reason, in lifting a weight it should be placed as near
the guide EE as possible, while in descending the strain
can be partially taken off the brake by placing the
weight far away from the guide.
The influence which the ratio of the distance between
P and Q to the distance between the guiding surfaces
SLIDING FRICTION. 81
B and C has upon the relative magnitudes of P and Q
can be seen in the simple clamping device shown in
Fig. G, plate L, a mechanism commonly used in saw-
mills to hold the block upon the carriage. The log is
held firmly upon the carriage, or table, simply by the
wrought-iron clamp AB which rests upon it and runs
loosely up and down the fixed standard DE. If from
any cause, as the upward stroke of the saw, a force Q
acts vertically and tends to slide the clamp AB upward
along the cylindrical standard DE, there is a resultant
tendency to rotate on the part of AB which presses the
outer edges b and c of the eye hard against the standard.
The latter re-acts at these points with the forces R x and
i? 2> which, under the supposition of an actual slip of
the bushing, act downward at an angle to the hori-
zontal, and oppose such slipping motion. The force R x
acts from b toward 0, and B 2 from toward . Their
resultant passes through 0, the point of intersection.
In the case of motion just beginning, when B v B 2 , and
Q are in equilibrium, the point mast fall on the line of
force Q. If, therefore, from any point A in the line
of Q we draw the two lines Ab' and Ac parallel to the
re-actions B 1 and R v we have in V and c' the points
of application at which R x and R 2 must act when the
desired condition of motion is fulfilled ; that is, the eye
of the clamp must have an axial dimension equal to
Vc as shown in dotted lines in the figure. In that case
even the smallest force Q would cause a loosening of
the clamp's hold upon the log.
But when, as shown in the figure, the eye of the
clamp has a less height, so that the point of intersection
of the re-actions falls between the standard and the
line of force Q, there can only be motion when a fourth
32 THE GRAPHICAL STATICS OF MECHANISM.
force P is added to Q, R v and R 2 in such a way that
the resultant of P and Q passes through o. Such a
force may be applied in numberless ways. If we take
the force P parallel to Q, and suppose it to act along
the axis of the sleeve or eye, it can be easily deter-
mined. Since the resultant of P and Q must pass
through o, we have by the well-known law for parallel
forces P . Bo = Q . Ao ; and by construction we get P in
JB if BF = AGr = Q, and a line is drawn through F
and o intersecting Q in IT; after which JB is made
equal to ÄA, since by similar triangles
AH or JB : FB or AG:: Ao: Bo;
from which
P . Bo = Q . Ao.
This force P evidently must work in the direction of
Q as long as the point of intersection of the resultant
forces lies between P and Q ; and in such case, there-
fore, no single force Q, however great, can loosen the
clamp ; but to accomplish this object a special force P
acting in the " sense," or direction, of Q is necessary.
It is clear, also, that the slightest force acting at B
alone will lift the clamp if it can overcome its weight.*
* This statement is slightly inaccurate, since there would be friction
to overcome in lifting the clamp when the lifting force is applied at
B\ for the weight of the clamp would act at its centre of gravity, a
point to the right of B, and would form a couple with the lifting
force. This couple would be balanced by that formed by the resist-
ances Pt l and J?2, which would act on the opposite sides of the stud
ED, and be inclined to the opposite side of the normal from that
shown in Fig. 0. The determination would, in fact, be exactly the
SLIDING FRICTION. 33
For fixing the log in j^lace, a slight pressure or blow
upon the clamp is sufficient.
That kind of mechanism which we have previously
designated as self-locking finds extended use in practice
where it is the object to prevent an undesired motion
by means of clamp-gearing. The well-known hold-fast
of the planing-bench works upon this principle, as do
also certain kinds of ratchet-gear.
Sliding friction plays an important part in wedges,
whose efficiency is greatly reduced thereby, becoming
less as the wedge grows thinner ; i.e., as the angle of the
wedge becomes more acute. For an example we will
choose the wedged device for adjusting the pivot-bearing
of the upright shaft W shown in Fig. 7, plate I. The
bearing L supports the shaft W and rests upon the
wrought-iron key K, while any side motion is prevented
by the sides of the casing N. The re-actions R x of the
wedge, and R 2 of the casing, will be called forth by
the load Q = IA ; all three forces acting upon the cap
i, and necessarily in equilibrium. Of the two re-actions
R x and i2 2 , we only know the directions, which must be
inclined to the normals of the surfaces in contact at
the angle of friction <£, but we do not know their points
of application. If we assume that the re-action R x of
the wedge against the bearing acts at the point A,
where the weight Q is supposed to take effect, the
re-action R 2 of the casing must also pass through this
point. In the present case it does not matter, as far as
the result of the construction is concerned, where the
same as that shown in Fig. 5; and as there proved the lifting-force
would have to be more than sufficient to "overcome the weight" of
the clamp before motion could result. — Trans.
34 THE GRAPHICAL STATICS OF MECHANISM.
points of application A and C of R i and R 2 are taken.
If we assume the point of application of the re-action
R A in a the re-action R 2 would lie along the line cb
which passes through 6, the point of intersection of R l
and Q. The value of these re-actions is given in any
case by the sides of the force polygon A II and II I,
supposing that I A is made equal to Q, and the sides
mentioned are drawn parallel to R x and R 2 respectively.
If a horizontal force P acts upon the wedge K by
means of the screw S, the wedge must be in equilibrium,
for the case where motion is about to commence, under
the influence of the re-action — R x of the cap X, the
re-action R s of the support H and the force P. The
position of the force R s is determined by the require-
ment that it must pass through the intersection o of P
and R r To get the value of P we have only to com •
plete the force polygon by constructing upon II A, or
— R v the triangle II III A, in wdiich the side A III
parallel to P represents that force, and III II the
direction and value of the re-action given forth by
the support H against the wedge K. In order to
determine the theoretical force P it is only necessary
to draw R x in the direction A II perpendicular to the
surface of the wedge, and R 2 in the direction I II
perpendicular to the axis of the shaft. If we then
construct the triangle II III A, whose sides are paral-
lel to P and perpendicular to HII respectively, we
have in A III the force P which would suffice to
raise the bearing if there were no friction. In the
same way we find the forces acting in the backward
motion of the wedge from the force polygon I 22 A, in
which A 2 is drawn on the opposite side of the normal
A II to the wedge surface at the angle 2 A II Q = .
SLIDING FRICTION. 35
Also the re-action (i2 2 ) of the casing which holds in
equilibrium the forces Q and (Ü^) must come from the
opposite side of the casing in the direction from ((7) to
A. Further, the position of the re-action (i2 8 ) of the
support H is fixed by the condition that it must pass
through (0), the point of intersection of the re-action
(iüj) with the force (P), which acts along the line of
P, but in an opposite direction. To complete the force
polygon for backward motion we draw A 3 parallel to
(P), and 2 3 parallel to (i£ 3 ) ; and in the former we
have the value of the force (P) necessary to withdraw
the wedge from under the bearing. Since this force
(P) acts from A toward 3 in the same "sense," or direc-
tion, in which the load Q tends to move the wedge it
is evident that a backward motion of the wedge cannot
result from the action of Q alone, and we must regard
the efficiency (jf) = ^— - as being negative. With a
value /a = 0.16, and a taper of 1 in 9 for the wedge,
we get from the drawing, for
Q = 100, P » 11.1, P = 45,
and therefore
V = ^ = 0.247, also (P) == -19.8,
Hence
W - TiT = - 1 ' 78 '
It may be remarked here, to avoid repetition, that, as
in Fig. 7, the lines of force and the force polygon will
36 THE GRAPHICAL STATICS OF MECHANISM.
be drawn in full lines for the forward motion, those for
backward motion in broken and dotted lines, and those
for the theoretical force P in broken lines. All con-
struction lines will be simply dotted.
The resistance which the foot-journal of an upright
shaft encounters from the plane surface FF (Fig. 8,
plate I.) upon which it rests may be deduced by the
methods of sliding friction. We may suppose the ele-
ments of friction which arise at every point in the sur-
face of contact to be concentrated in the circumference
of a circle A x A 2 , whose radius A A 2 = §r, r being the
radius of the journal A F. If we imagine the load Q to
be replaced by two equal forces CA, each equal to | Q,
which act at diametrically opposite points A x and A 2 of
the circumference, we can replace the re-action of the
bearing by two forces R l and R 2 at these points A x and
A 2 . These forces will be inclined to the axis AC by
an amount equal to <£, the angle of friction, and will lie
in planes perpendicular to the plane of the forces ~
passing through A x and A 2 . The horizontal projections
A X E X and A 2 B 1 represent the resistances of this species
of pivot friction concentrated at A x and A 2 . To over-
come these an equal and opposite couple with forces
E x A x = DiA 2 = P must be applied at the extremities
of the lever-arm A 1 A 2 . When, as is always the case
in practice, the driving-force P is applied only at one
point the journal will press against one side of the
bushing in which it runs on account of its one-sided
working, and there will result a certain amount of neck-
journal friction beside the pivot friction already taken
into account between the end of the shaft and its sup-
porting surface. The determination of this neck fric-
SLIDING FRICTION. 37
tion corresponds to journal friction, which will be dis-
cussed in another chapter.
In the same way we can ascertain the amount of
friction in the thread of a screw. Suppose SS (Fig. 9,
plate I.) to be a screw provided with both right and
left handed threads, as occurs in certain forms of coup-
ling for railway-cars ; and suppose that each of the nuts
M t and M 2 is drawn outward with a force Q = AB;
Ave may then obtain the force necessary at the lever N
to turn the spindle of the screw in the following way :
Each of the two nuts is regarded as acting upon the
screw in two diametrically opposite points of a central
helix (or pitch line) whose diameter d is equal to the
arithmetic mean 1 "t" — 2 of the inner and outer helices
2
of the screw. And letting A represent that point of
the first pair, and C that point of the second pair,
nearest the observer, we have in the two lines AO
and CO drawn at the angle $ from the normals A0
and C0 to the direction of the screw-thread Aa and
Cc, the direction of the re-actions at A and C. Now
draw from the point of intersection the line OJ
parallel to the axis and equal to ^Q. Then in the
line KL drawn perpendicular to JO we have the value
of the force P which is in equilibrium with the two
re-actions of the nuts M x and M 2 against the screw.
Since the same construction holds for the two other
points diametrically opposite to A and 6 y , it follows
that for the turning of the screw a couple of forces,
each equal to P = KL, is necessary ; the arm of the
couple being twice the radius r of the helix in which
the action of the nuts upon the screw is supposed to
be concentrated. Without friction we have the force
88 THE GRAPHICAL STATICS OF MECHANISM.
P =£ 2f L i the re-actions being assumed in the direc-
tion of the normals O A and C. For a co-efficient
of friction /x = 0.1, and a pitch of screw n = ^ we
have, with %Q = 100,
and
P ~ 37.8, P = 16.67,
^^ = 0.441.
The construction remains the same when the pitch of
the two screws is different, as in differential screw-gear-
ing ; or when the pitch of one screw equals zero, as in
the much-used tension mechanism (Fig. 10, plate I.)
where the thread of one screw is merged into a ring
and swivel in which end-journal friction only occurs.
Here, as before, we make OJ = J$, and have in KL
one force of the couple necessary to turn the nut M ; it
being understood that the line of re-action OK is drawn
at an angle = the angle of friction to the axis of the
rod S v Without friction the force P would be given
in JL Q . For a pitch n = T V, and a co-efficient of fric-
tion ix = 0.1, the construction gives, for \Q = 100,
P = 28.9, P s 8.33,
and
v = ^ = 0.288.
The application of the turning-force at one side only
of the mechanisms (sketched in Figs. 9 and 10, plate I.)
causes a certain amount of neck friction which may be
determined by methods to be explained in the following
chapter.
JOURNAL FRICTION. 39
§ 4. — JOURNAL FRICTION.
If a cylindrical journal (Fig. 11, plate I.) is pressed
bj' a force BE = Q perpendicular to its geometric axis
into a bearing A v that bearing re-acts at B with a force
equal to Q and opposite in direction, the same as any
other supporting surface would do. If the journal
revolves in the direction of the arrow a certain force
is necessary which does not go through the axis. Let
BG be such a force acting at i?, and of such value P
that it is just sufficient to hold the journal in equilib-
rium, and by the slightest increase to cause a turning.
Under this supposition the journal is in equilibrium
under the two forces P and Q, and of the re-action R
exerted upon it by the bearing. This last re-action
must be equal and opposite to OJ, the resultant of P
and Q. The bearing A i acts upon the journal at the
point If with the force BK == JO. By that we do not
mean that the re-action is actually exerted at the point
K, but that the resultant of all the re-actions of the
elements of the bearing against the journal passes
through the point K of its surface. . We have to
assume that the resistance to turning offered by the
bearing is friction at the point K. This friction is
exerted in the direction of motion of the supporting
surface, as previously laid down in a general law, and
has a value //JV; N being the normal pressure at if, and
/x the co-efficient of friction. If we resolve the re-action
BK into components at right angles, one normal and
40 THE GRAPHICAL STATICS OF MECHANISM.
the other tangent to the surface of contact at K, we
have in the component &fiT the normal pressure, and in
the tangential component LS the friction al resistance
T V
at the circumference of the journal. Since, now, — — =
J SK
fj, — tan (f) we find that the angle LKS or OKA which
the re-action makes with the radius drawn to its point
of application equals the angle of friction ,
where r is the radius of the journal. The same value
for this arm TA will be obtained in every case wherever
the turning-force P is applied ; and if Ave suppose the
force F to occupy, one after another, all possible posi-
tions about the axis A the re-action of the bearing will
in each case be tangent to a circle described about the
centre A with a radius AT = p = r sin >.
We can therefore regard the journal bearing, in its
action upon the journal, as entirely replaced by a re-
action which is tangent to a circle drawn about A with
the radius r sin > ; and the direction and situation of
this re-action will be known as soon as any other condi-
tion is settled, as, for instance, in the present case, that
it must pass through the point 0. This circle of a
radius p == r sin >, which for brevity will be called the
Friction circle through analogy to the nomenclature, fric-
tion angle, friction cone, etc., oilers a convenient means
JOURNAL FRICTION. 41
for the graphic expression of journal friction. Since,
under the supposition of an entirely frictionless motion,
the re-action of the bearing passes through the centre
of the journal, we may regard it as being tangent to
the friction circle, which has become equal to zero in
this case.
It is evident from the figure that the second tangent
drawn from to the friction circle, shown in the dotted
line OIF, corresponds to a revolution of the journal
opposite to the arrow, and that in this case, when the
turning-force acts in the direction OB^ the point of
application of the bearing against the journal is to be
assumed at W. Either If or IF may represent the point
of support of the journal, according to circumstances.
If we further assume that the journal A is fixed,
and that the bearing is acted upon by the forces P and
Q like the hub of a wheel running loosely upon an axle,
we must then regard the point K A or W x as the point
at which the hub is supported by the fixed axle ; the
point K 1 corresponding to a left-handed revolution like
that indicated by the arrow, and the point W 1 to an
opposite revolution. The amount and direction of the
re-action R are not affected by these changes, there
being merely a transfer of the point of application from
if to if x , and from W to W x .
On the contrary, if we suppose the force P and Q,
and consequently their resultant, to act in the opposite
directions OF\ 0H\ and OJ\ the point of support will
fall upon K 1 , while the journal will have a revolution in
i\ right-handed * direction opposite to the arrow. We
* In general right-handed revolution is to be understood hereafter
as meaning revolution in the direction of the hands of a watch.
42 THE GRAPHICAL STATICS OF MECHANISM.
must determine, therefore, in every case, which of the
two tangents is the line of direction for the re-action in
that case. This determination is rendered much less
difficult here, as in the case of sliding friction, if we
keep resolutely before us the principle that each of the
two pieces exerts upon the other a re-action which coin-
cides in direction with the motion of that piece relative
to the other. Even if both parts, journal and bearing,
have absolute motion, as is the case in all link connec-
tions, it is still not difficult to determine the relative
motion of one part with respect to the other. Later
remarks will serve for the further elucidation of this
law.
To determine graphically the friction circle of a jour-
nal, we have only to draw any radius AB (Fig. 12,
plate I.) of the journal, and lay off the angle of friction
> := BAC. When this angle is not given directly, but
only the co-efficient of friction /x, draw BO perpendicu-
lar to AB and equal to /x . AB. Then draw BE through
-Z), the point of intersection of the hypothenuse AG
with the circumference of the journal, parallel to AB.
We have then, in the circle drawn about A tangent to
the line DE, the desired friction circle of a radius
AE = p = r sin >.
Rods, or links, provided with two pins, or eyes, con-
necting them with other machine parts, often occur in
mechanisms. Such a rod, or link, as AB (Fig. 13, plate
I.) would, in the absence of friction, simply transmit
force from journal to journal along the line AB con-
necting their centres. A force applied to either journal,
AC iov instance, would call forth in the other, BD, an
opposite force with which it would be in equilibrium.
Since, when friction is neglected, the pressure of the
JOURNAL FRICTION. 43
journal upon its bearing in the rod can only be perpen-
dicular to the surface of contact, both these equal and
opposing forces must necessarily pass through the cen-
tres A and B. This would also occur in reality when
there is no motion of the journal relative to its bearing,
as, for instance, in the joints of a linked chain which,
loaded with a weight, is drawn vertically upwards.
When, on the contrary, a turning of the journal rela-
tive to its bearing occurs friction enters into the consid-
eration of the motion ; and according to preceding prin-
ciples force can only be transferred from journal to
bearing along lines which are tangent to the friction
circle. Therefore, in the present case, a transfer of force
between the journals AC and BD can only take place
along one of the four possible tangents to both friction
circles AE and BF. We can easily determine which
of the four tangents is to be regarded as the line of re-
action in any particular case by the application of the
rule previously given. With it we have only to decide
in what direction, whether to the right hand or to the
left hand, the turning of the journal occurs in its bear-
ing, and in what direction the journal acts upon the rod ;
i.e., whether the latter is in tension or compression.
The action of the link on a pin or journal must then
have such a direction that, in consequence of this action,
the eye of the link will assume relatively to the pin the
rotation which actually does take place. This rule
holds also for the action of the pin on the link ; for
not only the direction of the force, but the direction of
rotation, will be reversed in this case.
In Fig. 13 the four tangents are denoted by the
figures 1, 2, 3, and 4, for brevity. For still greater
clearness there are shown in Fig. 14, plate I., four sepa-
44 THE GRAPHICAL STATICS OF MECHANISM.
rate mechanisms in which a rod AB of the type under
discussion unites two swinging levers MA and NB. In
these four cases, I., II., III., and IV., the heavy arrow
drawn at one lever denotes not only the motion of that
one lever, but of the whole system ; the other lever is
therefore always a resisting body. By the little arrows
drawn at each joint is shown the motion of the rod, or
link, relative to that lever with which it is there united.
We may regard the journal as forming a part either of
the link or the lever, since, as before explained, such
assumption has no effect on the direction and amount
of the re-action, merely shifting the point of application
to another position on the same line. If we suppose
the link AB to be in tension in the cases I. and III.,
and in compression in II. and IV., it will be readily
seen that tangents 1, 2, 3, and 4 in Fig. 13 correspond
respectively to cases L, IL, III., and IV. in Fig. 14.
This correspondence of Fig. 13 to Fig. 14 holds also for
the opposite motion on condition that the driving-force
is applied at the same lever, for then both the nature of
the force acting in the link and the direction of the
turning will be reversed.
In a similar way we can determine the direction of
re-action of a journal upon its bearing in every particu-
lar case. This is in reality all that needs explanation
or demonstration in the method of calculating journal
friction, since the further graphical determinations con-
sist simply of the application of well-known principles
in regard to the resolution and composition of forces.
The method above referred to may be shown in a few
examples for the sake of clearness.
Let ABC (Fig. 15, plate II.) again represent a bell-
crank for which we are to determine the force P that
JOURNAL FRICTION. 45
must be applied to the arm B to lift the load acting
upon the pin A of the arm CA. Drawing the friction
circles for the journals A, B, and C we have, in the
tangents ao and bo parallel to the lines of force Q and
P, the lines of direction for these forces. Because of
the turning of the bell-crank to the right as shown
by the arrow, the rod grasping the journal A has a left-
handed turning about that journal, and the tangent oa
must be drawn, according to the principles previously
established, touching that side of the friction circle
farthest away from C; so that one might say that the
arm of the resistance to be overcome is increased by
journal friction. It follows, in the same way, that the
line of direction for the re-action of journal B against
its rod, which also has a left-handed turning, must be
tangent to the inner side of the friction circle, so that
the arm of the force P is shortened by journal friction.
The re-action of the supporting bearing against the
journal C must pass through o, the intersection of oa and
ob. There being a turning of the bell-crank to the right
hand the re-action R of the bearing to the journal C
will lie along the line oe, passing through o and tangent
to the friction circle of 0. Since P, P, and Q are in
equilibrium we have, by making ol ~ Q and completing
the parallelogram ol II III the force P in oIII and
in oil, the re-action of the bearing equal and opposite to
the journal pressure. Without friction we should obtain,
as in Fig. 1, plate I., the force P = III ; i.e., if the
direction of the forces passed through the centres of
the pins and OI — Q. In the case under consideration,
with a co-efficient of friction /* = 0.1, the drawing gives,
for Q = 100,
P = 91.3, P = 87.8,
46 THE GRAPHICAL STATICS OF MECHANISM.
and r, = ^0 = 0.962.
For another example we will choose the ordinary
slicler-crank train (Fig. 16, plate II.) as used in steam-
engines. The downward pull P of the piston is exerted
by the piston-rod JTupon the pin A of the connecting-
rod. We shall endeavor to determine how great a
resistance Q at the distance CE from the shaft can be
overcome by this piston force in the position of the
mechanism shown by the drawing, friction being taken
into account. By the force P we do not represent the
entire pressure of steam upon the piston, but simply
that really acting upon the cross-head after piston and
stuffing-box friction have been deducted. The pressure
variations caused by the acceleration and retardation of
the piston will also be taken into account in the value
of P. The line of this force P must pass through the
centre A of the cross-head pin, since this pin is rigidly
fixed to the cross-head and piston-rod, and no relative
turning can occur between them. There is such turn-
ing, however, between the pin A and the end A 2 of the
connecting-rod A 2 B 2 . The line of direction of the force
S acting in the same will therefore lie along the tangent
ab to the friction circles of the two journals A and B.
(Which of the four possible tangents is to be here taken
is shown by the little arrows which indicate the direc-
tion of turning of the connecting-rod ends about the
journals A, B, and enable us to apply the principles
explained in Figs. 13 and 14, plate I.) The cross-head
pin cannot be in equilibrium under the influence of the
two forces S and P alone, which have different direc-
tions. Equilibrium requires a third force which must
JOURNAL FRICTION. 47
be exerted by the guide D in the re-action li 1 . The
direction of this re-action is fixed by the condition that
it must be inclined at the angle > to the normal to the
guide, and its situation by the requirement that it must
pass through the intersection o x of the forces P and S.
Accordingly the guide D exerts upon the cross-head a
re-action along the line do 1 . Further, the force S will
be transmitted without loss by the connecting-rod from
the pin A to crank-pin B along the line ha ; and we find,
as in the case of the bell-crank (Fig. 15), that the force
iS'and the resistance Q acting at E are held in equilib-
rium by a re-action R 2 exerted by the bearing upon the
shaft C. We have the direction of the latter in the line
o 2 c which is drawn from the intersection o 2 of the forces
S and Q tangent to the friction circle of the shaft 0.
To determine each force we have only to draw the force
polygon in which Io 1 — P,I II is parallel to o { d, II III
is parallel to o 2 c, and o 1 177 is parallel to Q or FE, The
line o^II ■=. Q gives the resistance acting at E, and
II o x gives the tension S in the connecting-rod, while
II I represents the re-action of the guides, and II III
the pressure upon the bearing of C; the determination
of the latter forces being of especial importance in pro-
portioning the parts under consideration. For the
determination of the theoretical resistance Q = o x III^
it is only necessary to draw the re-action R 1 normal to
the guide 7>, and the directions of aS^ and R 2 through the
centres of A, 7?, and (7, as is shown in the force polygon
drawn in broken lines. With the assumption of a co-
efficient of journal friction /x r= 0.1, and sliding friction
at the guides /x z= 0.16, the drawing gives, for P = 100,
Q = 48.4, Q = 52.5 ;
48 THE GRAPHICAL STATICS OF MECHANISM.
therefore
v = Q- = 0.922.
It is, of course, self-evident that for any other position
of the mechanism there would be a different efficiency.
If we turn now to the diagram for the oscillating
engine (Fig. 17, plate II.) it will be observed that the
force S acting in the piston-rod B X D is tangential to
both journals B and C\ as in the preceding case, and
falls along the line be. The demonstration of this fact
is as follows : If P is the force exerted b}^ the steam
pressure upon the piston E, and which acts along the
line of centres BC, this force must be in equilibrium
with the other forces which act upon the piston-rod.
There are besides P only three forces to be considered :
the re-action >S' of the crank-pin B, the only known con-
dition of which is that it must be tangent to the friction
circle of i?, and the re-action ll x exerted by the stuffing-
box D 1 against the piston-rod at the point rf, together
with that, R 2 , of the cylinder against the piston E at the
point e. These re-actions are inclined at an angle to
the normal to the geometric axis of the cylinder, and
act from the cylinder toward the piston and piston-rod
respectively.
If we now regard the relative motion of cylinder to
piston-rod only we may imagine the piston and piston-
rod to be held fast, while the cylinder with its bearing
moves a short distance along the piston-rod under the
pressure of the steam upon the cylinder-head. Here
also the various forces acting upon the moving cylinder
must be in equilibrium. These forces are the following:
First, the pressure of the steam upon the cylinder-head,
JOURNAL FRICTION. 49
which is of course equal to the steam pressure upon
the piston, and acts along the line of centres CB, but in
the opposite direction, and is therefore to be denoted by
— P ; next, the re-actions which formerly acted upon
the piston and piston-rod at e and d now act at the same
points, but in opposite directions, against the cylinder
and stuffing-box, and are to be denoted by — R 1 and
— i? 2 ; finally, there is the re-action Z exerted by the
bearing C x against the trunnion-journal (7, the only
known condition of this re-action being that its line of
direction must be tangent to the friction circle of 0.
Since, then, the condition of equilibrium exists between
the four forces P, R v R v and S\ and also between — P,
— R v — P 2 , and Z, it is evident that P, R v and R 2 will
balance — P, — R v and — R 2 respectively, thus leaving
S = — Z; i.e., the forces /S r and Z are equal and act
in opposite directions. The force &\ therefore, must
coincide with the tangent cb to both friction circles.
For the condition of equilibrium between the forces
P, R v R 2 , and S, it is necessary that the resultant of
any two, as P and P x , shall be equal and opposite to
the resultant of the other two, S and R 2 . By joining
o v the intersection of P and P 1 , with o 2 , the intersection
of # and P 2 , we get in o l o 2 the direction of these result-
ants. If we make C I = P, and draw C II parallel to
JK 1 , ///parallel to o 2 o v /////parallel to Jc, and I III
parallel to P 2 , the line II III gives us the value of the
force S which tends to pull the crank-pin around. Of
equal value, as has been demonstrated above, is the re-
action which the bearing C\ exerts upon the journal C
of the trunnion. The further determination of the
acting forces is carried out in the usual manner. If a
resistance Q acts at the radius AF from the shaft we
50 THE GRAPHICAL STATICS OF MECHANISM.
have in the line o s a drawn from the intersection o s of Q
and S tangent to the friction circle of A the direction
of the re-action i? 3 exerted by the bearing upon the
shaft journal A ; and to determine the value of this re-
action and of the resistance Q we have only to resolve
II III into U IV and III IV parallel to the directions
o s a and o s F of the re-action and resistance Q respec-
tively. Without friction we should have P = S, and
the triangle CI IV would give immediately the value
of Q = I IVq, CI being resolved in the directions of
Q and OA.
For a co-efficient of journal friction fx = 0.1, and of
friction in the stuffing-box and cylinder a = 0.16, the
drawing gives, for P = 100,
Q = 61.0, Q = 64.2,
and
= -£
Q,
0.950.
It should be remarked that the friction here consid-
ered as existing between the piston and cylinder, and
between piston-rod and stuffing-box, is only that arising
from one-sided or lateral pressure. The friction winch
is caused by the pressure of piston and stuffing-box
packing must be estimated in other ways, and deducted
directly from the piston pressure.* We have, therefore,
* The " one-sided or lateral forces " to which the author here refers
are those resulting from the non-coincidence of the opposing forces S
an 1 P. They arc. in other words, the re-actions which keep the
piston-rod and cylinder in the line BC, their tendency being to arrange
themselves along the line of tension be. If Ave imagine the piston,
piston-rod, and cylinder to be made of some elastic material which
would bend at the slightest application of a deflecting force, they would
so arrange themselves when the steam pressure P was applied that
JOURNAL FRICTION. 51
in the value of y\ given by the figure, not the efficiency
of the steam-engine, but only the efficiency ot those
parts composing the mechanism.
The same slider-crank motion which lies at the basis
of the oscillating engine finds frequent application in
planing-machines for the production of a quick return
motion. Fig. 18, plate IL, represents an arrangement
of this kind, where the slide HGc bearing the tool M is
moved back and forth in a prismatic guide by the link
F 1 F 1 which receives its reciprocating motion from the
lever D l E l oscillating about the fixed journal D. The
oscillation of the latter is produced by the crank AB,
whose crank-pin B works in the block C sliding along
the slot C v of the oscillating lever F X I) V The resistance
Q offered by the material to the cutting-tool M produces
the re-actions M 1 and li 2 at the points g and h ; and these
three forces, Q, M v and Ii v must be in equilibrium with
the force 8 acting in the reciprocating link E X F V The
the line passing through the centres of piston and cylinder-head would
coincide with the line be of tension between trunnion and crank-pin.
There is another class of "one-sided or lateral forces" which the
author has omitted to mention; namely, those arising from the oscilla-
tions of the cylinder. Taking the case when the crank is upon either
dead-point, with the engine running at a high rate of speed, Ave have
the entire mass of the cylinder, by no means small in practice, in rapid
motion in one direction; passing to the other dead-point we find it in
equally rapid motion in the opposite direction. Between the two the
inertia of this mass moving at this rate of speed has twice to be over-
come byre-actions of the nature of Tt 1 and R>, but vastly greater than
II C and I III. The resultant of these re-actions applied at the crank-
pin is of the nature of those accelerations and retardations to which
reference is made in the first chapter, and have no effect on the work
done, since the work stored up in the quadrant on one side of the dead-
point is given out in the quadrant on the other side; but the friction
at the points e and a, or their opposites, represents a loss of energy
never given back again. — Trans.
52 THE GRAPHICAL STATICS OF MECHANISM.
direction of the force S is given by the tangent ef to
the friction circles of E and F. If we therefore connect
o x , the intersection of Q and _B X , with o 2 , the intersec-
tion of S and i2 2 , we have in the side II III of the
force polygon o x I II III, the value of the force S acting
in the link B 1 F 1 , under the supposition that o x I ' = Q,
1 II is drawn parallel to B x , II III parallel to ef, and
o x III parallel to B 2 . The crank-pin B acts upon the
sliding bearing C, and through it upon the guide C 1 of
the oscillating lever, in a direction o 3 c which must be
tangent to the friction circle of B and inclined at the
angle > to the normal to the slot C v The line of this
re-action T is therefore o s c, and similarly the line do s
drawn from the intersection o s of T and S tangent to
the friction circle of D is the direction along which the
journal D re-acts against the lever B l B 1 . If, further,
the driving-force P is applied at the end if of the radius
AH from the shaft A we can get the direction of the
re-action exerted upon the crank-shaft A by its bearing,
in the line o^a drawn tangent to the friction circle of A
from 6> 4 , the intersection of T and P. Completing the
force polygon by resolving the force # = II III into
II IV Mid IV III, parallel to o%c and o s d, and drawing
IT ["parallel to o 4 a, and IV ^parallel to o 4 ff, we have
in V IV =. P the force which must be applied at H to
overcome the resistance Q = o x I at 31. Assuming a
co-efficient of journal friction /x = 0.1, and of sliding
friction ^ ^= 0.16, the construction gives, for Q = 100,
P = 84.1, P = 66.6,
and
v = 5) z= 0.792.
JOURNAL FRICTION. 53
In the case of a steam-engine with single beam
(Fig. 19, plate II.) the investigation is pursued as
follows: The steam pressure P acting upwards upon
the piston-rod AB coincides with the geometric axis of
the cylinder, and passes through the centre of the jour-
nal A. On the other hand is the thrust S of the rod,
i.e., the re-action of the beam upon A, which must lie
along a line tangential to the friction circle at A. These
two forces, S and P, cannot be in equilibrium, since
they do not act along the same line. For this a third
force is necessary, which is found in the re-action Il 1 of
the stuffing-box against the piston-rod. By drawing
through a middle point b in the stuffing-box the direc-
tion of M 1 at an angle with the normal we have, in
the intersection o x of R i with P, the point through
which the thrust S must pass; this force then acts
along the line o x a. If we make o Y I ' = P, and draw
through I a parallel to P x , we have in o x II the force
S exerted upon the journal A of the beam. It may not
be uninteresting to remark that the stuffing-box P a is
exposed, according to the above demonstration, to a
certain side thrust II I This side thrust is not the
result of an inaccuracy in the parallel motion, as is
the case in approximate motions like that of Watt,
for the Evans motion here represented is well known for
its absolute accuracy. This side thrust is directly the
result of the journal friction occurring at A. For a
clearer understanding of this fact we may imagine the
piston-rod to be bent to the left by a force equal to
the journal friction of the beam applied at its upper end,
and tending to revolve toward the right. Such a force
would evidently call forth in the stuffing-box the re-
action which we have been considering. It is also
54 THE GRAPHICAL STATICS OF MECHANISM.
evident that the side thrust R x will act in an opposite
direction and from the other side of the stuffing-box
when the piston makes a down-stroke, since at each
change of motion the direction of the journal friction at
A is reversed. On the other hand, the piston-rod thrust
or pull S in upward and downward stroke remains
always in the line of the tangent o x a, since at each
reversal of motion, i.e., at each dead-point, the direction
of turning of the journal A, as well as the direction of
the force S, is reversed. We may therefore regard the
effect of journal friction at A as being in every case to
diminish the lever-arm of the force S. Beside the force
S three other forces act upon the beam ; they are the
re-actions of the radius-rod DE, of the connecting-rod
FG, and of the guides K X K V We find the directions of
the force L in the radius-rod, and T in the connecting-
rod, according to previous rules, in tangents to the
friction circles at D and E, and at F and G. Recol-
lecting that the radius-rod is under compression while
the connecting-rod is in tension, and noticing the direc-
tion of turning at each journal as indicated by the
arrows, we decide upon cle and fg as being the desired
tangents. The sliding-block C x exerts upon the journal
C a re-action which must be tangent to the friction
circle, and upon the guide K X K 2 a re-action which must
be inclined at an angle $ to the normal ; these two con-
ditions fix the position of R 2 as coinciding with the line
he. In Fig. 19a the sliding-block C x is shown in detail.
We first observe that during up-stroke the lower guide
I^K.2 exerts the re-action, and during down-stroke the
upper guide K^K^ is under pressure. In further ana-
lyzing the re-actions exerted upon the sliding-block O x
we must remember that it has a double reciprocating
JOURNAL FRICTION. 55
motion ; that is, it makes a complete stroke out and
back for every half-stroke of the piston. Beginning
then at the lower dead-point with the beam in the posi-
tion CA 1 we have, during the first half of the up-stroke,
a motion of C x toward the right, and right-handed turn-
ing of the journal (7, which gives k 1 c 1 as the direction
of the re-action R 2 . When the beam reaches the posi-
tion CA 2 the radius-rod DU is parallel with it, and any
further motion will, by the action of the radius-rod DE,
cause G x to move to the left. During the second half
of the up-stroke, therefore, k 2 c 2 is the direction of the
re-action R 2 . At the beginning of the down-stroke,
when the beam is in the position CA^ the sliding-block
C x again reverses its motion and moves to the right;
but it now presses against the upper guide K x f K 2 \ and
the direction of turning at the journal has also been
reversed, and the re-action lies along the line from Jc s
to c 2 or Jc 2 . Similarly, during the last half of the down-
stroke, there is sliding toward the left, and left-handed
turning of the journal ; so that k 4 c v or k v is the line
of re-action. In every case, therefore, the re-action is
so applied as to lengthen its lever-arm and render the
force #less effective, which is entirely in keeping with
the obstructive action of friction.
In order that the four forces S, L, T, and R 2 acting
upon the beam shall be in equilibrium the resultant of
the two forces L and R 2 intersecting at o 2 , and the
resultant of the two forces T and S intersecting in 3 ,
must both lie along the line o 2 o s uniting these points.
In the figure, as a result of the proportions assumed,
the point of intersection falls beyond the limits of the
plate. In determining the direction of the line o 2 o s
the following construction may be employed to advan-
56 THE GRAPHICAL STATICS OF MECHANISM.
tage. Draw any line, o 2 a, intersecting 8 and T in the
points a and ß; draw any other line, o-w, parallel to o 2 a,
and intersecting S and T in the points o- and r. If,
now, we assume the third point, w, on the line o-w, so
that the proportion
a/3 : a*T : : /3o 9 : tw
is true, we know from geometry that the three lines o 2 o>,
yör, and ao- will intersect in one and the same point : but
the intersection of ßr (the line of the force T~) and ao-
(the line of the force >S r ) is the "desired point o 3 . There-
fore the line o 2 w gives the desired direction o 2 o s . The
point w can easily be located by drawing any line, aS,
making a8 equal to 1 . Taking these re-actions at e and d
inclined at the angle (/> to the normal we have again, in
the line joining the point of intersection o x of B Y and Q
with that o 2 of R 2 and S, the direction of the resultants
of these pairs of forces. If we therefore make o x I — Q,
and draw / II parallel with o x o^ and II III parallel to
be, and I III parallel with R 2 , we have in III II the
thrust aS' exerted in the eccentric-rod. If, further, the
motion of the shaft A is caused by a force P applied at
the end F of the lever-arm AF we have again in ao« the
direction of the re-action B s exerted by the bearing
against the shaft A ; and by resolving III II into
III IV and II IV parallel to ao z and P we get in
IV lithe value of the force P which must be applied
58 THE GRAPHICAL STATICS OF MECHANISM.
to overcome the resistance Q. For /* == 0.1 and /x = 0.16
the drawing gives, for Q ~ 100,
P = 36.2, P = 22.8,
and
v = ^ = 0.630,
a small value for 77, due to the large radius of the friction
circle of the eccentric B.
An interesting application of the crank-train is found
in the ordinary Blake crusher shown in Fig. 21, plate
III. The crank-rod BC is here connected with two
links DE and FG, forming a knee, by the pressure of
which the materials fed in at L are crushed, through the
intervening plate JH swinging from the centre II. It
is evident that as turning occurs at both ends of each
link the pressure can only be transmitted through them
along the tangents de and fg to the friction circles.
Furthermore, the force S of the crank-rod or pitman
which is tangent to the friction circle at B must also
pass through the intersection o 2 of the forces jPand T x ;
i.e., act along the line o 2 b. And also the re-action R x
against the journal H of the swing-plate must pass
through o v the intersection of the thrusting-force T in
the link DE with the working-resistance Q. If the
motion of the crank-shaft A is caused by a force P
applied at iT wo have, according to well-known methods,
the re-action of the bearing against the shaft in the line
o 3 a tangent to the friction circle of A. Then draw
the force-polygon as follows: Make o 1 I equal to Q, the
crushing-resistance of the material; draw / II parallel
to öjA, and IIo l is the thrust T sustained by the link
With a
co-efficient /x = 0.1
Q = 100,
P = WA, P
and
JOURNAL FRICTION. 59
BE. In III II we have the thrust T x in Fa, if III II
is drawn parallel to fy, and öjZZZ parallel to 6o 2 . Finally,
by resolving IIIo 1 = #, the strain in the pitman, into
Oj/F parallel to o 8 a, and III IV parallel to P, we have
in III IV the value of the force P which must be
applied at K to crush the material at L. The broken
lines again indicate the construction by which the theo-
retic force P = HI Q IV is obtained.
the drawing gives, for
= 12.3,
rj = ^ = 0.80.
Another example of toggle-joint mechanism is the
hand-punch shown in Fig. 22, plate III. Two bent
levers, A 1 CD l and B 1 CE 1 , are here connected by a
hinge or bolt at (7, so that when their ends D t and E x
approach each other under the action of the screw FGr
the head HJis forced down, and the punch L forces the
metal under it through the die K. To produce this
result the screw FGr is turned by a long wrench applied
at its square head F, and its right and left threads draw
the nuts D 2 and F 2 together with a certain force P, as
was the case in the coupling shown in Fig. 9, plate I.
On account of the turning of the levers the nuts are
connected to them by the journals D and E, from which
it follows that the force P by which they are drawn
toward each other acts along the tangent de to the two
friction circles of D and E. If we now suppose one of
the levers (the under one, for instance) to be at rest
it must be in equilibrium under the forces acting
60 THE GRAPHICAL STATICS OF MECHANISM.
upon it. The forces consist of the driving-pressure
P acting along de, and the two re-actions R 1 at and
R 2 at B. For the first re-action R 1 exerted by the
upper lever A 1 CD X we have the direction ac tangent to
the two friction circles of A and G 7 , the lever A 1 CD l
turning toward the right hand ; while the direction of
R 2 must pass through o v the intersection of P and R x ,
and be tangent to the friction circle of B. It lies, there-
fore, along the line o x b. If, then, w T e make o x I = P,
and resolve it parallel to the two re-actions by drawing
I II parallel to o x b, we have in IIo x the re-action JR 1
exerted by the upper lever upon B x OE x , and in I II
the re-action R 2 offered by the journal B.
The head HI must, in its turn, be in equilibrium
under the influence of the force R 2 acting along bo x ,
the resistance Q which the metal offers to punching
acting along the axis of the punch Z, and the two re-
actions R s and i2 4 of the guide-bushing H^^ If we
assume h and i at a sufficient distance (say 5 mm., or
\ inch) from the edges as the points of application for
the re-actions Ii s and R 4 which are drawn at the angle
of friction <£ to the normal, the line connecting o 2 , the
intersection of JR S and R 2 , with o 3 , the intersection of
72 4 and $, furnishes us with the means of completing
the force polygon in the usual way. Resolving II I
— R 2 into III I parallel to R s and II III parallel to
o 2 3 , and then II III — o 2 o s into III IV parallel to R ±
and II IV parallel to Q, we have in IV II = Q the
resistance which can be overcome by the application of
the force J? upon the nuts D 1 and JE V It is also evident
that the side IV Q II of the force polygon drawn in
broken lines normal to the surfaces, and passing through
the journal centres, is the value of the theoretic' resist-
JOURNAL FRICTION. 61
anee # which should be overcome by the same force P.
With the usual values /* = 0.1 and /x x = 0.16, the draw-
ing gives, for P = 100,
Q = 194, Q = 350,
and
V = -^ = 0.554.
Vo
It will be seen that the ratio of P to # would remain
the same whether we use the arrangement shown, or
whether we suppose the force P to act only upon one
arm B l U 1 ^ while the nut D 2 is replaced by a cylindrical
eye and spindle so arranged that no sliding can occur
in the direction of the axis of the spindle, as in the
swivel, Fig. 10, plate I. The ratio between P and Q
would not be changed, because if the driving-force P
was applied only at JE X there would arise at i) 1 an
opposite equal re-action which would be transmitted by
the swivel-ring to D. The only difference between the
two arrangements is that by the movement of both
levers the space traversed by the head HJ for any
given portion of a revolution of the screw FGr is double
that which would result if only one lever moved while
the other was held fast. The work done by the turning-
force for this portion of a revolution is, of course, twice
as great in one case as in the other. The above remarks
apply exactly only in the case of frictionless motion ;
for with the arrangement giving a re-action — P of a
iixed point, as in the swivel, a new friction enters which
must be determined in the same way as in Fig. 10, plate
I. The investigation of the present mechanism has not
included the determination of resistances arising in the
62 THE GRAPHICAL STATICS OF MECHANISM.
screw ; such a determination would be made in the
manner shown in Fig. 9, plate I.
The amount of resistance Q which can be overcome
by a certain force P, as well as the efficiency of the
mechanism where knee-joints are employed, depends
upon the angle which the centre lines of the links form-
ing the joint make one with another. It will be readily
seen that this resistance Q becomes greater, and the
efficiency ?] smaller, as this angle approaches 180 de-
grees. If we assume this last value, or one which differs
from it by an infinitely small amount, as in Fig. 23,
plate III., a force P acting upon the journal C would
be able, in the absence of friction, to overcome an
infinite resistance
M=^w =cc )-
On account of journal friction, however, these lines
of force are to be found in the tangents ca and cb to
the friction circles ; and we therefore find the actual
forces in the parallelogram acblii we let cl = P. The
greatest resistance Q which can be overcome is therefore
given by the equation
Q = P tan ^#,
where 2w is the obtuse angle acb of the two directions
of pressure. The efficiency rj = -¥- in this case where
Q = oo is equal to zero.
If we further suppose tlie knee to be in the condition
of backward motion, i.e., if we assume that the tendency
of the re-actions at ^1 and B is to force the joint C out
JOURNAL FRICTION. 63
to one side or the other, it is evident that such backward
motion can only begin at that instant in which these
re-actions coincide in one straight line. If, therefore, we
draw such a case (Fig. 24, plate III.) in which the tan-
gents ac and be to the friction circles fall upon the same
straight line we have the limiting position at which the
knee is self-locking. The angle 2(w) = ACB, which
differs from the angle 2w in Fig. 23 only by an inappre-
ciable amount, determines on both sides the limiting
position within which a backward motion, i.e., an open-
ing of the press by the re-action offered by the material
within its jaws, is impossible. As Qtv) depends on the
proportions of links assumed, the limits within which
the mechanism is locked become greater as the distance
from journal to journal becomes smaller, and as the
radius of the journals increases. The knowledge of
these proportions is of special importance in the design-
ing of mechanisms in which the knee-joint is employed
to grip an object and hold it fast, as in certain forms of
vise.
The methods heretofore employed in determining the
efficiency of machines can also serve the purpose of
determining friction as applied to useful ends in many
machines and processes. Thus friction serves to produce
the necessary tension in all spinning-machinery, and is
employed also in sewing-machines and water-frames or
throstles.
Let Fig. 25, plate III., represent the ordinary spindle
with the Arkwright flyer 00 which rotates with the
rapidly moving spindle. The thread F passing through
the stationary glass eye at D with a certain velocity u,
and leading to the loose spool L after several turns
about the arm of the flyer, serves as a driver to the
64 THE GRAPHICAL STATICS OF MECHANISM.
spool, which is caused by the thread to revolve in the
same direction as the spindle and the flyer. The spool
holds back on account of the frictional resistance offered
to it, and at each instant the portion of thread running
out is unwound by this difference between spool revolu-
tion and flyer revolution. The friction of the spool by
which the tension in the thread is determined occurs
principally in two places, — at the circumference of the
spindle as journal friction, and as pivot friction where
the under surface of spool at Gr rests upon the bobbin-
frame EE which slowly rises and falls. This friction
must attain a certain value in order that the tension
of the thread shall be sufficient for a certain amount of
twist, and in order that the thread shall not belly out
between B and I) under the influence of centrifugal
force, and become entangled with the thread of the
neighboring spindle. The tension S of the thread may
be determined as follows: If 6r = HA is the weight of
the spool with the quantity of yarn already upon it, it
produces friction upon a ring-shaped portion of the
surface of the bobbin-frame EE. We may therefore
suppose the bobbin-frame EE to be replaced by re-
actions which are uniformly distributed over an average
circle of contact whose diameter is EE* all these re-
actions making the angle <£ with the normal in the
direction prescribed by the motion. As was shown in
* The radius p of the circumference of contact is =
where r, and r 2 represent respectively the radii of the outer and inner
circumference of the supporting surface.
JOURNAL FRICTION. 65
the case of pivot friction (Fig. 8, plate I.) we can here
suppose all re-actions to be concentrated in two diamet-
rically opposite points, a\ and a 2 . From the parallelo-
gram AJHK we get in AJ and AK the re-actions
exerted by the bobbin-frame, and in their horizontal
components a x i and a 2 k the frictional resistance offered
by these re-actions to the revolution of the spool. If
the thread draws the spool in the direction fa at a cer-
tain moment with a tension S the spindle re-acts with
equal force along a line ab drawn parallel to the direc-
tion fc, and tangent to the friction circle of the spindle.
The question then is of the equilibrium of the spool
under the influence of the two frictions a x i and a 2 k,
the tension S, and the re-action of the spindle along the
line ab. Joining the intersection o^ of the thread-
tension S with the friction a 2 k, with that o 2 of the
friction a x i with the spindle re-action, we have the ten-
sion $ given in value by the line IIo 2 if Io 2 = a x i and
I II is drawn parallel to o x o 2 .
So far in these investigations it has been tacitly
assumed that the journal friction was only exerted
upon one bearing. This is never the case in practice.
Every shaft has at least two supporting points or bear-
ings, and at these the forces P and Q will call forth
certain pressures and re-actions of a value proportional
to the distance of the point of application of the forces
from the bearings. If we suppose the shaft supported
by the bearings A t and A 2 (Fig. 26, plate III.) to
encounter a resistance Q acting through a wheel or
pulley placed at C, with a radius q — AG, we can
resolve this fpvce into two components parallel to Q,
and having the same lever-arm q, lying in planes passing
through the points A x and A v and normal to the shaft.
66 THE GRAPHICAL STATICS OF MECHANISM.
The forces are determined by the well-known rela-
tions
Qi = q aa~ and Q * = q ja:
12 12
and can easily be found by construction.
If the same construction is carried through for the
determination of the driving-force acting at each bear-
ing we have in P x and P 2 the forces which, lying in the
planes through A x and A 2 , act upon the parallel lever-
arms AH = p to overcome the resistances Q x and Q 2 .
It follows that under the supposition of equal journal
radii and equal co-efficients of friction, i.e., with equal
friction circles, at A x and A 2 these forces P 1 and P 2
must be in the same ratio one to another as Q Y to Q 2 ;
so that if P x and P 2 were compounded in one result-
ant P it would have to lie in the same plane, passing
through 6 y , in which Q acts. Therefore we should obtain
by this method the same value of P which has hereto-
fore been determined directly from Q by the employment
of friction circles.
It follows from the above that the preceding construc-
tions for the determination of journal friction can be
entirely accurate only when the driving-force P lies in
a plane perpendicular to the same axis as that to which
the plane of the resistance Q is perpendicular, and when
the diameters of the journals are equal. In reality the
first condition is seldom fulfilled, and the journals also
are seldom of the same size in a shaft. It therefore
remains to bring this influence within the scope of cal-
culation by force polygons.
With this object in view let us suppose that the
driving-force P (Fig. 26, plate III.) which is to over-
JOURNAL FRICTION. 67
come the resistance Q is applied at P, a point outside
the bearings A 1 and J. 2 , as is frequently the case in
practice. It is further assumed that the resistance Q at
C acts with a lever-arm A Cr = q, and the force P with
a lever-arm AH = p at the point P. If no friction
came into account we could assume P Q and Q as work-
ing in the same plane, and get
p = Ol p
in the well-known way by uniting the intersection o t
with centre A, and then resolving o x Q into this direction
and that of P. Then imagine Q to be resolved into the
forces Q x = Q~r~ 2 ~i~ ac ting at A v and Q 2 = Q *
A 2 A 1 A^A 2
acting at ^4 2 . Then lay off these forces equal to o 1 Q 1
and o 1 Q 2 in the direction of Q. In the same way P
can be resolved into two parallel forces, P x and P 2 ,
acting at the points A x and A 2 with the lever-arm
AH — p. Their values are given by the equations
Pi = p °f 2 f and p « = p »f a f-
^l 1 yi 2 ^1^2
Since P acts outside of the supporting points ^L x and
J. 2 , P 1 and P 2 act in opposite directions o 1 P 1 and 0^2
along the line of P. We find the resultant of P 1 and
Q x in the diagonal o 1 Z>, and this force acting in the
plane through A x will call forth an equal and opposite
re-action of the bearing A v This re-action R x does
not act along the same line o 1 D as the resultant, how-
ever, since among the forces acting at A x alone equilib-
rium does not exist. The position of R x is found by
68 THE GRAPHICAL STATICS OF MECHANISM.
drawing the line a x a x tangent to the friction circle at
A x and parallel to o x D. Similarly we get the journal
pressure at A 2 in the diagonal o x E, and in the line a. 2 a 2
parallel to o x E and tangent to thp friction circle at A 2 ,
the position of the re-action R 2 which the bearing A 2
exerts. We now see that the shaft is in equilibrium
under the couple o x D and R v and the couple o x E and
R 2 . Since we may suppose the couples to be slid along
the axis until they lie in the same plane we can immedi-
ately find P by uniting o x and the intersection o 2 of
R x and R 2 , and resolving the resistance o^ Q parallel to
the diagonal o x o 2 and the direction of P. We have
therefore in o l P x the necessary turning-force P. In
Fig. 27, plate III., the friction circles of A x and A 2 are
drawn to a larger scale, and we see that the direction
of R coincides nearly to the direction of a line drawn
through o x tangent to a mean friction circle shown in
dotted lines. We can therefore employ this simple con-
struction with sufficient accuracy in the generality of
cases, and especially in those where both P and Q fall
within two bearings not far apart. But for exact deter-
mination, and in cases where a force is applied outside
of the bearing, the full construction is necessary. It
will be noticed that the latter has still a slight inaccu-
racy, since the component forces P x and P 2 are obtained
from P instead of P. The error is quite inappreciable,
however, and a correction unnecessary; though such cor-
rection could be obtained by determining P x and P 2
anew from the value of P as deduced, and repeating the
construction.
We can now determine all the friction al resistances
which occur in a screw. In Fig. 28 a , plate IV., S 2 S S is
the direction of a helix at a mean distance from the
JOURNAL FRICTION. 69
axis. S r s x and S v s 2 are the directions of re-action at
two diametrically opposite points of the helix, so drawn
as to make the angle BS l s 1 = BS x s 2 = $ -f- a with
the axis of the screw, a being the pitch-angle of the
screw. We then obtain the resistance q x = CB acting
perpendicularly to the axis at each of these points by
making BA X = Q, the load upon the screw, and draw-
ing through A x and B the lines A X B and DB parallel
respectively to S l s l and S x s 2 , and intersecting at B.
The load Q also presses the nut M down upon the
standard, and produces friction against the ring-shaped
surface of contact A 2 A S . As pivot journal friction we
can suppose this concentrated at a mean circumference
A 2 A S , and acting at two diametrically opposite points.
If, therefore, we draw the corresponding lines of re-
action A l a 1 and A x a 2 inclined at the angle of friction
> to the axis A 1 B we have in CE = q 2 the amount of
friction at each of the supporting points in the ring-
shaped surface A 2 A S if the line BE is drawn parallel
to A 1 a 2 . We can imagine these couples q x q x and q 2 q 2
as acting at the points bj) 2 (Fig. 28 6 , plate IV.) and
c x c 2 respectively, and by compounding them get the
resultant couple
d l e 1 — q s and d 2 e 2 = q s .
If the nut M is revolved by worm-gearing, we must
represent the worm TFas acting upon it along the line
w x w 2 with a force w applied at a mean helix of the
worm. This one-sided working of the force w presses
the nut M up against the sides of the standard K
where the latter is bored out above A 2 A S , and calls
forth a re-action parallel to w 2 w x and tangent to the
70 THE GRAPHICAL STATICS OF MECHANISM.
friction circle of M in mm. To find w we join o x arid
2 , the points of intersection of the couple q s q s with m
and w respectively, and making o 2 e = , a l being the pitch of the
screw on the worm.
The worm W in its turn is forced by the load to
against its bearing i, and friction results along the
mean circumference of the ring L 1 L r This friction
can be supposed to act at two opposite points of this
circumference, as at A 2 and A s in the case of the screw
S. We have, therefore, a resisting couple TJ = p 2
(Fig. 28 c ) where TJ is obtained by drawing FJ and
OJ through and F at the angle to OF.
Finally, the load w upon the worm will also call forth
frictional resistances at the bearings of the two neck-
journals L and iV, since the one-sided action of w thrusts
the journals against their bearings with a certain press-
ure, the value of which, equal and opposite in the two
cases, we have from the equation of moments
w . W l w 1 = p s . iiV,
JOURNAL FRICTION. 71
from which
p s : w : : W x ii\ : LN : : LK X : K X K V
From the last proportion we see that we can get p s
by drawing through F (Fig. 28 c ) a parallel to LK 2
(Fig. 28 6 ), when OiT will represent the value of p s , the
neck re-action exerted at L and N.*
In order to determine the force P to be applied at a
crank W s to the worm W, we must first unite the one-
sided resistance p x and the two couples p 2 p% an d p^Pz-
This can be done in the following way : First draw the
two tangents l x and l 2 to the friction circle of the jour-
* Throughout this discussion the author has assume! that the
thrust w of the worm is applied along the line w x w, parallel to its
axis. This is neither true in practice, nor does it correspond to that
case shown in the figure. The thread of a rack is not square, but has
the sides of its profile inclined at an angle of 75 degrees to the axis.
The normal to the surfaces of contact would then be inclined at an
angle of 15 degrees to the axis. The line of re-action would fall short
of this inclination by an amount equal to > — 5° 43' with a co-efficient
ß = 0.10, so that its actual direction would be inclined 0° 17' to the axis.
The re-action Oiin of the nut would be parallel to it, and the shape of
the parallelogram o 2 ejf would be changed, and a different value for
of = iv obtained. But since that component of this real value of to
parallel to the axis WW would only differ from o 2 /by an inappreciable
amount, which increases and decreases with the radius of the friction
circle at A, and since the component perpendicular to the axis merely
adds to the throat-friction at L an amount which it takes from that at
N (its tendency being, of course, to thrust the worm bodily over to
the right), the accuracy of the final result is practically unaffected.
Moreover, as the gear may work in either direction, and as in the case
of backward motion the condition of affairs would be exactly the
reverse of that pointed out in the first part of this note, it is probable
that the author assumed the average position, i.e., the one parallel to
the axis, in order to have a general discussion applicable to all cases.
— Trans.
72 THE GRAPHICAL STATICS OF MECHANISM.
nal L representing the direction of re-action of p%, and
then at a distance equal to the radius of the mean helix
of the worm draw the vertical tangent tv s . Now lay-
off from 6> 3 , the intersection of w s and Z x , the distance
oJ: x = p x (= FH'm Fig. 28 c ) and o^K l = p 3 (= OBT
in Fig. 28 c ) ; then the diagonal o 3 x gives the resultant of
the forces p x and jp 3 , which latter acts along the tangent
? x . This resultant, when compounded with the other
force 2h acting along l 2 in the opposite direction, gives a
force passing through o 4 parallel and equal to o^i x = _p r
We see from this that the influence of the two frictions
p z produced by the neck-journal re-actions only causes
the resistance to turning of the worm to act in the same
direction and with equal force, but at a longer lever-arm,
since it passes through o> 4 instead of o 3 . This corre-
sponds to the well-known principle that the composition
of a force and a couple merely effects a parallel shifting
of the force unchanged in value. In the same way, by
uniting the force p x going through # 4 with the couple
P2P2 w hich corresponds to the friction offered by the
ling-shapecl supporting surface L X L 2 (Fig. 28 6 ), we ob-
tain merely a shifting of the force p x always parallel
with itself from o 4 to o Q . It is done as follows : Draw
the two forces p 2 p 2 (TJ in Fig. 29 c ) as two parallel
tangents ? 3 / 4 to a circumference of the diameter L 1 L 2 .
Then lay off from the point of intersection o~ of one of
these tangents with the force p x now passing through
o 4 the distance o b i = p 2 and o b h 2 = p v and the diag-
onal o 5 y cuts the second tangent Z 4 in the point o 6
through with the force p x must pass. The further
construction is familiar. Through the intersection o 1 of
the resistance p x with the driving« force P the re-actions
of the journal-bearings L and N must pass. This re-
JOURNAL FRICTION. 73
action R is therefore acting along the line o 7 l 5 drawn
tangent to the friction circle at L. If we then make
o 7 ä 3 = p t = 0.1 for journal
and thread friction, the construction gives, with a pitch
n — tan a = Jg of the screw & for (> = 100,
iv = 67, w = 11.8,
and
ri t —^ = 0.176,
w
where to is the force to be applied at the pitch-line of
the worm. With a pitch n = tan a t = ^ f or the worm,
for w = 67,
P = 12.56, P = 4.3,
and
, 2 = 5q = 0.342.
For the efficiency of the entire jack we have then
t; :zz rj 1 rj 2 = 0.060.
In the same way we can determine the efficiency for
backward motion by finding the force (P), which must
be exerted in the same " sense," or direction, as Q to
produce a sinking cf the load.*
* To fix the method to be followed in every case firmly in mind, it
miy not be amiss to briefly sketch here the general aspect of the
74 THE GRAPHICAL STATICS OF MECHANISM.
problem encountered in all these examples, giving the known quanti-
ties in every case, and the way in which they are to be combined to
determine the unknown, so that the student, in attempting to solve
an outside problem, will know just what he has to work with, and
just how to set about that work. First, by means of the friction angle
and friction circle we can always draw the direction of the forces
transmitted longitudinally through links joining two turning pairs,
acting at sliding surfaces, or in links joining a sliding and a turning
pair. Sometimes in the last case the exact position of the force is
fixed, after its general direction has been determined by the angle of
friction, by the requirement that it shall pass through the intersection
of two others, instead of being tangent to a friction circle as in the
simplest case. An example of this is the ordinary cross-head, Fig. 2,
plate I. The line of the re-acting force at a lever or crank or bell-
crank bearing is found by drawing a line from the intersection of the
two other forces acting upon the lever, crank, or bell-crank tangent to
the friction circle at its journal. The directions of P and Q are always
given as the force of gravity, a piston-thrust, etc. We then have
given, or can determine by these elementary methods, the direction
of all the forces in any problem. We also have given the intensity of
either P or 0.
With these data the problem is solved as follows: Draw the force-
polygon of all the forces acting upon the same piece as the known
force, and dependent upon it. These can only be three in number if
there is circular motion of the piece to which it is applied, and four
if there is right-line motion. In the first case it is a question of
drawing a triangle of forces, knowing the directions of all and the
amount of one. In the second case combine the forces two and two,
join the points of intersection thus obtained, getting in the line thus
drawn the direction of the common resultant of the two pairs. Then
resolve the known force in the direction of that force with which it
is paired, and of the common resultant which here represents the com-
bined effect of the other two forces. Having thus obtained the value
of the resultant resolve it in the direction of the two forces making
up the second pair, and all the forces are known in direction and
amount. One of these becomes the known force acting on the next
link in the mechanism, and by repeating the process all the forces
acting throughout the machine may be determined.
Whenever there are more than four forces acting on one piece they
will be of such nature that they can be reduced to four; and generally
where the limit is exceeded, as in the condensing beam-engine, and
still further in the compound condensing beam-engine, it will be the
JOURNAL FRICTION. 75
result of a compounding of several chains of mechanism, and in each
simple chain either P' or Q' will be given, so that its resultant action
on the common link can be determined, and combined with the P or
Q of the main chain, their resultant action being regarded as one force.
In the case of the condensing beam-engine the resistance of the air-
pump (/would be known, and, as shown in plate VII., combined with
the thrust of the piston to get the resultant force acting upon the
beam. In the compound engine, P', the steam-pressure in the second
cylinder, would be known. — Trans.
7G THE GRAPHICAL STATICS OF MECHANISM.
§5.— ROLLING FRICTION,
The resistance which is opposed to the rolling of a
cylinder along a smooth path is of such small value that
it may be left out of account in most cases in comparison
with sliding and journal friction. We are accustomed,
when it is taken into consideration, to assume it propor-
tional to the pressure Q with which the roller is forced
down upon the bearing surface, and inversely propor-
tional to the radius r of the roller. For rollers and
surfaces of iron and hard wood the formula
P = 0.02^
will generally give the resistance, r being expressed in
inches. If r is expressed in millimetres the formula
becomes
P = 0.5^.
In order to get a graphic representation of this resist-
ance let A (Fig. 29, plate IV.) be the centre of a cross-
section of a cylinder, with radius AB ~ r, which is
supported at B by a horizontal track. Let the load
resulting from its own weight, and acting at the axis A,
be represented by the vertical line AC — Q. To cause
d rolling of the cylinder a horizontal force P = c . ~
must be applied at the axis A, AI) representing the
ROLLING FRICTION. 77
intensity of this force. If we assume, as heretofore,
the ' limiting condition of equilibrium for which the
slightest increase of P will cause motion, the cylinder
must be in equilibrium under the influence of the
exterior forces P and Q, and of the re-action R offered
by the surface GG. This is only possible if the re-
action R is equal to the resultant of P and Q, and acts
along the same line in the opposite direction. The
surface GG then re-acts upon the cylinder with a force
whose direction and value are given by the line PA.
The cylinder, therefore, will remain at rest as long as the
surface GG re-acts upon it along any line inclined to
the normal at a less angle than that of PA as was the
case in sliding friction. The plane surface then opposes
the same character of resistance to the motion of the
cylinder as in sliding friction, with this difference, how-
ever, that while in the case of sliding friction the great-
est possible deflection angle of the re-action depends
only upon the nature of the material, and is constant
for a given material, being, of course, the angle of
friction for the same, in the case of rolling friction it
depends both on the material and on the form, i.e., upon
the size of the cylinder. From the expression given
for the resistance to rolling
r
it will be readily seen that the co-efficient € is capable of
geometric representation, since it follows that
Q : r : : P : «,
giving us directly in the figure BF = .e; which is, in
78 THE GRAPHICAL STATICS OF MECHANISM.
other words, the greatest possible distance between the
point of application F of the re-action and the theoretic
point of contact B of roller and track. While in the
case of sliding friction we have a constant angle for
the friction angle <£, in the case of rolling friction we
must deal with a linear value € which, for the same
material, remains the same for rollers of all sizes. If it
was of sufficient interest we could follow out the parallel
still farther, showing that the friction cone in the case
of sliding friction corresponds to the wedge-shaped
space whose cross-section is FAF V whose edge is the
axis A, and whose sides, shown in projection at FA and
F X A, cut the supporting surface at the distance c to
each side of the perpendicular AB* It follows that
with P acting in the opposite direction the track would
re-act from the other side of AB in the direction F 1 A.
We can make this connection clear if we assume that
in reality the roller is not supported on a line passing
through B parallel to the axis, but upon a surface of a
width e from each side of the normal AB, produced by
a flattening of the roller and a corresponding indenta-
tion of the supporting surface under the pressure of the
load Q. This view corresponds also to the assumption
of a fixed fulcrum for the roller at the constant distance €
from the normal plane (that is, at the point F), and this
should be kept in mind during the following discussion.
The value of e, according to the above, is from 0.02 to
0.03 in inches, or from 0.5 to 0.75 in millimetres, for
metals and hard wood. In the case of yielding materi-
* For the connection between sliding and rolling friction see O.
Reynolds, Philosophical Transactions, vol. 1C0, and Zeitschrift des
Vereins deutscher Ingenieure, Jahrg., 1877, S., 417.
ROLLING FRICTION. 79
als, as, for instance, carriage roads, the value of € is
much greater, and in all such cases the correct value
must be estimated.
If a body KK (Fig. 30, plate IV.) rests upon a roller
A which rolls along the horizontal track GG rolling
friction occurs at both KK and GG. If Ave therefore
draw through the centre A of the cross-section of the
roller the normal DB to the two surfaces, we have,
according to what precedes, the point of support of
the fixed track in F at the distance BF = e from B,
and also in E at the distance DE =z e from D the
point at which the downward pressure of the moving
body KK acts. If, therefore, EC := Q denotes the
load upon the roller A, and P denotes the force acting
in the horizontal plane KK necessary to move the body,
these forces P and Q must be in equilibrium with the
re-action R exerted by the track GG through the roller
A upon the body KK, which re-action of courses takes
the direction FE. We then have in the side EJoi the
parallelogram ECHJ the necessary force P to produce
motion.
Equally well can be determined the force P (Fig. 31,
plate IV.) necessary to move the load Q upon the wagon-
wheel AB, by means of the value BF = e for rolling
friction upon GG, and the friction circle of the journal
AE upon which the load Q rests. The direction of the
re-action B of the track GG against the axle-bearing
of the wagon is along the line Fa drawn from F tan-
gent to the friction circle of A, Therefore, by making
AC = Q, and drawing through C a parallel CD to
Fa, we get the driving-force
P = AD.
80 THE GRAPHICAL STATICS OF MECHANISM.
The investigation is practically the same when the
track GG for the wheel has any desired inclination to
the horizon, as in Fig. 32, plate IV. If we here draw the
line AB through the centre A of the axis, and perpen-
dicular to the track GG, make BF = e, and draw
through .J 7 the tangent Fa to the friction circle of A, we
have in this tangent the direction of re-action of the
track GG against the axle-bearing. If, then, we make
CA = Q, draw through C a parallel to the line of re-
action R, and through A a parallel to the line of P, we
get in AE the value of driving-force necessary,
P = AE = DC.
Without hurtful resistances the re-action of the track
would lie along the normal BD , and we have in D C
the theoretical driving-force P . With the grade or
inclination of 1 in 3 for the track the drawing gives,
for Q ad 100,
P = 34.8, P = 33.3,
and
v = £j = 0.957.
In the manner shown the resistance to running-gear
of every description, upon both horizontal and inclined
tracks, can be easily determined. As a further example
we may take the roller-bearing for a swinging crane
(Fig. 33, plate IV.). In this case the cylindrical sur-
face of the stationary post or mast A serves as a track
for the rollers B and C united to the brace L. If we
connect the centres B and C with A, and make FI) z=
GE = e, the lines Db and Sc drawn through B and E
ROLLING FRICTION. bl
tangent to the friction circles of B and G 7 , and intersect-
ing one another in o v will give the directions of the
re-actions R t and R 2 of the mast. If, now, a turning
of the brace is produced by a force P acting in the
direction HJ, and intersecting the line of pressure Q of
the boom in o 2 , we have only to connect o x and o 2 , make
Io 2 = Q, and draw through / a parallel to o x o 2 in the
well-known way. We have thus determined in o 2 H
the necessary turning-force P. In order to get the re-
actions R 1 and R 2 of the mast against the rollers, resolve
the resultant I //parallel to o x b and o x c^ and we have
R x = III I and R 2 = II III
A later example will show in what way the re-action in
the bearing at the upper part of the mast is to be con-
sidered.
As already remarked rolling friction in most cases
of mechanism is quite inappreciable as compared with
other hinderances.
At first thought it may not be clear how the geometric diagrams
(Figs. 29 and 30, plate IV.) will always give the value FB constant
and equal to e for any one material, whatever the load or size of
roller, as stated on p. 78. The following analysis will render it
clear that such a result does follow from the assumed relation
r
and the supposition that the roller is always a solid homogeneous
cylinder. FAB being the angle made by the re-action R to the
normal AB we have
(1) tan FAB = - = — .
From the second value we get
(2) FB ^ r tan FAB.
82 THE GRAPHICAL STATICS OF MECHANISM.
There are three ways in which the conditions may change : First,
the size and consequent weight of the roller may remain constant,
but a varying superimposed load acting upon a surface as KK
(Fig. 30) may be applied. If in this case Q' represents the sum of
the varying load and the constant weight of the roller we have,
from the fundamental relation,
r
and, from equation (2),
F'B = r tun F' AB,
F' AB being the angle the new re-action R' makes with the normal.
By supposition both e and r are constant, and therefore P' varies
directly as Q' ; or, to put it in another form,
F _. e P
Q' r Q
Evidently
tan F 1 AB = — = - = tan FAB,
Q' Q
and therefore
F'B = rtan FAB = rtan FAB = FB.
That is, F'B, the distance of the intersection of R' with line GG
measured from the foot of the normal, is equal to the value first
obtained, FB or c.
In the second case there is no superimposed load, but the roller
varies in size and consequently in weight. The weight will vary
as the cross-section of the cylinder, and r will vary as the square
root of the cross-section or weight. If Q" is the varying weight
we have
O" P"
P"= e.M-, tun F" AB = — ,
Q" 7
and
F'B = r" tan F" AB.
If m is the ratio of Q" to Q we have
r" = \/mr y and Q" = mQ.
ROLLING FRICTION. 83
Substituting these values in the three equations above,
P"
*=..J*SL=*-fa* to>F»AB = e
sJTn r r — r^m
7)1 Q
F"B - rSjm . ~= = e.
r\m
In this case also there is no variation from the original value e or
FB.
Thirdly, where both superimposed load and size of roller vary
we have
Q! = mQ,
the former variable quantity, and
Q" = m'Q,
the latter ; while
Q» = Q' + Q",
their sum. As before
r'" - r\[m\ and Q'" = mQ -f m'Q.
Substituting in the three equations we have
D ,„ Q'" wzQ + w?'Q Q (m + m')
r'" r
£
tan F'" AB = -
Q (m + m')
r ' \jm' £
Q(m + m') r\Jm!
and F ,„ B _ rV /- £
the same result as in previous cases.
This may seem like begging the question, since if £ is assumed
to be a constant, and a certain line FB is found once to be its
graphic equivalent, this intercept FB must always remain the
same ; but the analysis may be of use in showing how the diagram
adapts itself to this requirement under all conditions. — Trans.
84 THE GRAPHICAL STATICS OF MECHANISM.
§ 6. — CHAIN FRICTION.
When a chain is wound on to or off of a drum or
pulley there occur certain hurtful resistances on account
of the change of direction in the links, which resistances
may be determined in exactly the same manner as jour-
nal friction. Let A (Fig. 34, plate IV.) be the journal
of a chain-pulley whose radius AB = AD is represented
by a. At the left side a weight Q is attached to the
chain BC; and we are to ascertain the force P which
must be applied to the other portion DE in order to
cause a revolution of the pulley in the direction of the
arrow, and a lifting of the weight Q. If we neglect
friction of the journal A, it is evident that on account
of the equality between the lever-arms AB and AD the
forces P and Q must also be equal for the condition of
equilibrium if there were no hurtful resistances at the
points B and D where the chain winds on to and off of
the drum, as would be the case if we suppose the chain
replaced by an infinitely fine thread of perfect pliability.
In this case there would be no slipping of the strands
of the thread over one another, since the thread has no
appreciable thickness ; and therefore the causes of fric-
tion would be wanting, because the latter can only
occur, as remarked in the introduction, where there is
motion of two elements relative to one another. But
as the links of the chain have a certain thickness, rela-
tive motion will occur at the point of connection of two
links at the instant of winding on or off, at the point B
CHAIN FRICTION. 85
or D respectively, and this motion must be regarded as
a turning. If we imagine the link B X B to be in motion
from C to jB, it has evidently no relative motion toward
the preceding link B 2 B S in which it hangs as long as
the latter rises in the same straight line CB. At that
instant when the preceding link B 2 B S , adjusting itself to
the circumference of the pulley, begins to share in the
latter's motion, there occurs relative motion between
the links B 1 B and B. 2 B^ which is, as shown by the
arrow in the figure, a left-handed turning of the link
BB X about the link B 2 B S . In this turning the end of
the link B X B serves as a journal, and the eye or loop
of B 2 B S as a bearing. It is now clear, according to the
laws of journal friction, that these links can only act
upon one another along the tangents to the friction
circles of these journals. Since the portion of chain
i?G 7 is subjected to tensile strain we have the direction
of this re-action in the tangent he which touches the
friction circle at B on the opposite side from the centre
J.. In other words, the lever-arm of the weight Q is
increased through chain friction by an amount equal to
the radius x of the friction circles at the chain joints.
In the same way the link D 2 D Z about leaving the pulley
at the point D on the other side undergoes left-handed
revolution about the link DD X still upon the pulley, as
shown by the arrow. Therefore the force P in the
portion ED 2 of the chain will act along the tangent de
to the friction circle at D. In other words, the lever-
arm of the driving-force P is shortened through chain
friction by an amount equal to x-> the radius of the
friction circles. The two forces P and Q acting verti-
cally downwards must be in equilibrium witli the re-
action R offered by the bearing A i to the journal A.
86 THE GRAPHICAL STATICS OF MECHANISM.
This re-action on account of journal friction can only
act tangent to the friction circle of A, and on the same
side as P. The investigation is now resolved, therefore,
into the determination of two parallel forces P and Q,
which have such relative values that their distances
from the resultant lying between them shall be
a — p — x anc l a + P +X
respectively ; in which expressions p is the radius of
the friction circle for the journal A, and x the radius
of friction circles of the chain. We have, according to
this,
p = Q a + P + *
This value can be readily constructed by drawing at
any point the horizontal line (?5, making GrK — Q,
drawing the horizontal line KJ, and then a line through
J and L to M. We then have in MG = NH the
necessary driving-force P, and in KM the re-action of
the bearing R — P -j- Q.
Since the links rub one another while in a dry condi-
tion we assume a co-efficient of friction /x = 0.2. With
this assumption, and that of /* = 0.1 at the journal, the
figure gives, for Q = 100,
P = 105;
and since P = Q = 100 we have, for the fixed pulley,
v = £) = 0.952.
CHAIN FRICTION. 87
The investigation is the same if the directions of the
chains are not parallel one to another. For the guide-
pulley ABO (Fig. 35, plate IV.) we draw first the
medial lines 00 and OB of the chain, and then the lines
ob and oc along which the tension of the chain acts. We
then have in the line ao drawn through the intersection
c>, and tangent to the friction circle at A, the direction
of the re-action R of the bearing. So that by making
ol equal to Q, and drawing I II parallel to oc, we get
III=P and oIIzzzB.
For /x =: 0.1 and /x 2 = 0.2, the figure gives, for Q = P
= 100,
P = 104.4 and ^^^^ 0.958.
In the same way may be determined the friction of
an idler, or guide-pulley, which serves merely to support
the chain and keep it from sagging (Fig. 36, plate IV.),
and is often so employed in cranes. Draw the lines
BE and BF (shown in dotted lines) along which the
chain rolls off and on to the pulley, and draw parallel
to them, at a distance equal to x-> the radius of friction
for the chain, the lines of tension oc and od. Then the
re-action of the journal A is given by the tangent oa to
its friction circle passing through the point of intersec-
tion 'o. By making ol equal to the resistance Z 1 acting
in the portion BD of the chain, and drawing through I
the parallel I II to oc, we have in I II the force Z 2
which is transmitted to the portion BO of the chain.
In the case of the loose pulley (Fig. 37, plate IV.)
upon whose journal the load Q hangs, and where one
88 THE GRAPHICAL STATICS OF MECHANISM.
end of the chain is fastened at -Z), we have to determine
the vertical force P acting at the other end of the chain
CE. To do this we draw^ the directions of tension bd
and ce parallel to the chain, and at the distance x of the
friction radius of the chain, and also the vertical tangent
ag to the friction circle at A Then draw through the
centre of the journal, or at any other convenient place,
the horizontal line BAC, make AG = Q, draw through
Gi the line b^ parallel to BC, join B with c x , and draw
through H the line JK parallel to BC We now have,
as will be readily seen, in e x K = b x J the tension Z of
the fast end of the chain, and in KG the force acting
in the portion CE of the chain.
By means of friction circles for journal and chain,
whose radii will be denoted as heretofore by p and x
respectively, the proportions of the various forces in all
kinds of block and tackle and pulley gearing can be
easily determined, as a few examples will show.
Fig. 38, plate V., represents an ordinary block and
tackle with two blocks, within each of which are three
pulleys of equal size, and ranged side by side on the
bolts A and B. When, by raising the load, the pulleys
are turned in the direction shown by the arrows, and
the chains wind on at E and D, and off at F and (7, it
is evident that the pull of the load hanging on the hook
üTacts upon the journal B of the lower block along the
vertical tangent o 2 b to the friction circle at J5, while
the re-action transmitted by the support Gr to the jour-
nal of A acts along the tangent o x a to its friction circle.
The forces of tension in the chains also will act at a
distance x, the radius of friction of the chain, nearer to
the centre of the pulleys at F and C, and at a distance
increased by the same amount at E and D. Since the
CHAIN FRICTION. 89
lower block B swings free, and opposes no resistance to
side motion, it will shift its position a distance 2\ to the
left when motion begins, so that the line of tension in
the chain shall be vertical ; otherwise equilibrium could
not exist. In the figure such a side movement of the
block B is supposed to have taken place, so that the cen-
tres A and B do not lie in the same vertical line, as
would be the case when they are at rest. It is equally
evident that with an opposite motion (that is, with a
sinking of the load) a corresponding shifting of the
block B to the opposite side must occur. We now
denote the tension in the separate portions of chain by
Z v Z 2 . . . Z v in such way that Z x is the tension of the
first portion which hangs from the stationary block A
and winds on to the first pulley of the block B at E,
while Z 1 is the force which is to be applied to the free
end of the rope to raise the load Q. It is then evident
from foregoing principles that the relation existing
between the tension of each portion of the chain and
that of the next following is
Z n {r + p -f x) = Z„ +1 (r - p - x)
if r is the distance from the centre of the chain to the
centre of the pulley.
If, now, we draw at any convenient point a horizontal
line which cuts the directions ce and fd of chain tension
at J and K, and the directions of journal re-action at o l
and o 2 , we know from the figure that
JK — 2r, Jo { == o 2 K = r — p — x,
and
Jo 2 = o x K = r + p + x-
90 THE GRAPHICAL STATICS OF MECHANISM.
From this follows immediately the construction given
below for the determination of Z*. or P v Make J I
equal the tension of the first portion Z x , draw through
Zand o 1 the line cutting KD in II; K His then the ten-
sion of the second portion Z 2 . From II draw through
o 2 the line cutting off the distance J III, which is the
tension in Z s . In the same way the lines IIIo^IV,
IVo 2 V, Vo x VI, and VIo 2 VII give the tensions
Z, = KIV, Z, = J V Z, = K VI,
and
z 1 = j vn.
The load to be lifted is given by the equation
Q = Z, + Z 2 + Z 3 + Z 4 + Z 5 + Z 6 ,
while Z 1 is the force P to be applied to the free end of
the rope in order to lift it. In the figure the sum of the
tensions from Z x to Z Q is shown by NL, and MO = Z 7
is the force P. By the construction here chosen we
start with a value of Z v while in reality Z x is yet
unknown, since only Q is given; but the method lends
itself with equal ease to the solution under the latter
condition. "We can assume Z x = J I oi any convenient
length, and get, in the manner shown,
NL = Z, + Z 2 + Z 3 + Z 4 + Z 5 + Z 6
and
MO = Z r
CHAIN FRICTION. 91
We then find from the given value of Q, and the pro-
portion between NL and MO, the force
v ivr
which in reality only amounts to the assumption of a
particular scale of force. On the contrary, a construc-
tion direct from the value of Q would be unnecessarily
tedious.
This construction also holds for the backward motion
of the tackle if only we regard the force P applied at
the end of the rope to prevent any accelerated motion
of the load Q as the tension in the first portion of
the chain, and every following tension as increasing
in the ratio — "*" p ~*~ . Then regarding J I as (P) or
r — P — X
Z v we have by the same construction the tension Z Q ,
Z h . . . Z v each one of which is greater than its prede-
cessor, so that the tension Z x of the chain attached to
the stationary block A is now the largest. Here also
Q = Z, + Z 2 + Z s + Z 4 + Z 5 + Z fl ,
and
Z, = (P).
In the figure the sum of the series Z^ to Z 6 is shown by
SB, and (i>) = Z 1 by TU.
For forward motion, and Z x = 100, the figure gives
§ = 682.7, P = 134.6,-
92 THE GRAPHICAL STATICS OF MECHANISM.
and since P = — = 113.8
v = 5d = 0.845.
For backward motion, for Z 7 = (P) = 100,
£ = 717.4;
and since P = — = 119.6
(,) = iEi. = 0.837.
^0
If the pullej^s in the blocks are not of equal size, and
therefore the ropes are not parallel, as when the pulleys
are arranged on different centres in the block, the above
determination for parallel ropes will in general have
sufficient exactness on account of the slight divergence
from parallelism in the case under consideration. An
absolutely correct determination can be made, however,
by taking into account the points of intersection of the
ropes produced. An example in which this is done is
shown in Fig. 40, plate V.
We will next take up the differential pulley (Fig. 39,
plate V.). Here also the load Q, hanging upon the hook
ÜT, acts in the direction of the vertical tangent bit to the
friction circle of B, while the re-action of the support
G- to the journal A lies along the tangent ag. The
direction of tension in the portion Z x of chain unwind-
ing from the smaller pulley at 6\ and winding on to the
loose pulley at L\ h again given in direction by ce, as is
CHAIN FRICTION. 93
also the direction of tension in the portion Z 2 winding
on and off at D and F respectively, by the linefd. The
two tensions Z x and Z v which in the ordinary differ-
ential pulley can be assumed as parallel, stand in the
following relation one to another :
ZJr - p - x) = Z^r + p + x)
where r is the radius of the pulley JEF, and
Q = Z x + Z 2 .
If we then draw through any convenient point o of
fd the line oM perpendicular to fd, and make MI = #,
we have in the intersection b 2 of the line ol with the
direction b x b of the load Q a point which will give us
the proportion of Z l to Z 2 ; for, drawing through b 2 the
horizontal b%N, we have the proportion
bfa : Nil : ob x : b x M : : (r - p - X ) : (r + R + x) 5
from which we get
b x b 2 = MN= Z x and N I = Z v
These two forces 2\ and Z 2 must be in equilibrium
with the tension Z z or force P applied to the free end
of the chain JK along the line iK, and the re-action R
given forth against the journal A along the line aj.
We can most easily find the condition of equilibrium
for these parallel forces through the drawing of an equi-
94 THE GRAPHICAL STATICS OF MECHANISM.
librium polygon.* For this purpose let us regard o as
the pole of the force polygon MNI, then draw through
£ the line a£ parallel to oM, and ßt, parallel to oN\ then
draw through ß the line ßti parallel to ol, and we have
in the closing line a8 of the polygon the direction of the
polar ray o II, which gives us in I II the driving-force
P applied at i, and in II If the re-action of the journal
A against the pulley. f
The theoretic driving-force P can be determined by
a similar construction, or more easily from the relation %
p = Q^nA = q AD - al
v 2E 1 V DJ
In order to determine the proportion between the.
forces for backward motion we have only to remember
* For a more complete explanation of the principles here employed
see Fart I. of The Elements of Graphic Statics by Karl Yon Ott. —
Trans.
t The figure &fö« is the funicular or equilibrium polygon, and it
will be readily seen that the forces P, Z,, Z 2 , and the re-action at A
acting upon the vertices r, C, /3, and a respectively of the polygon would
keep it in equilibrium. — Trans.
t This is, of course, derived as follows: Without friction
*, = *, = $
so that the equation of moments about A becomes
PoB, + { hi 2 = S ßl
transposing
P oRl = Q(^i ~ J* 2 ) or Po = qäi^zÄ2 t
— Trans. .
CHAIN FRICTION. 95
that the chain tensions and journal re-actions coincide
with the broken lines in the figure. For backward
motion also
Q = (Z,) + (Z a ),
but now
(Z 2 )0 + P + x) = (Z x )(r - p - x) ;
therefore the tension (Z{) of the chain CE is exactly
equal to the tension Z 2 of the chain FD for forward
motion, and also
(Z 2 ) = z v
If, then, we make MQN) = NI in the force polygon,
and draw the polar ray o(N), we can construct the
funicular polygon (a) (£)(/?) (8) for backward motion
by drawing (")(£) parallel to' oM, (ß)(0 parallel to
o(N), and 08)(8) parallel to oL Then (a)(8) is the
closing line of the funicular polygon, and the polar ray
o(II) drawn parallel to this gives us the force
which must be applied at J during backward motion.
Since this force is here acting upwards it is self-evident
that the tackle cannot of itself commence backward
motion, but that it is self-locking.
The figure gives, for fi - 0.1 for journal friction, and
fx x = 0.2 for chain friction, with a ratio - 1 =
and for Q = 100,
R l== AD _ 10
R'AL' 9 '
P = 12.8 and (P) = -2.7.
96 THE GRAPHICAL STATICS OF MECHANISM.
Since P z= 5 we have
— ^ = 0.391 and (y) = ^ = -0.54.
F P
As another example we wall investigate the arrange-
ment of pulleys often employed in hydraulic lifting
machinery.* In this case the chain upon which the
load Q hangs is first led over guide-pulleys supported
by the roof timbers at A and B (Fig. 40, plate V.), from
which it hangs down in a loop containing the loose
pulley (7, and then, after passing around the fixed pulley
i>, comes back and is attached to the journal of C.
Under the supposition of parallelism between the chains
this arrangement would cause, for any distance trav-
elled by (7, an elevation of the load Q through exactly
three times that distance. The downward motion of
is produced by aid of a second chain, one end of which is
fastened to the journal of (7, and the other, after pass-
ing around the loose pulley E and the fixed pulley F,
returns again to the journal of E. The movement of
the journal E is produced by the aid of the vertical
piston-rod of a hydraulic cylinder, not shown in the
drawing as not entering into the calculation. With
the supposition again of parallelism between the ropes
a downward motion of the piston w r ould cause the
pulley C to traverse three times as great a distance,
and the load Q to be raised by an amount equal to nine
times the piston travel. Without hurtful resistances,
therefore, the piston force F would equal 9$; and it is
* See Weisbach, Ingenieur- und Maschinenmechanik, III., Theil;
also, Rühlman, Allgemeine Maschinenlehre, IV., Bd.
CHAIN FRICTION. 97
evident that this arrangement would find its only appli-
cation where, as in hydraulic apparatus, the travel of
the driving-force P is necessarily small, while the force
itself is of great power. The proportion given above
for travel of power and load under the assumption of
parallelism would in reality vary but little from the
actual result. On account of the non-existence of
complete parallelism we will, however, follow out the
investigation, taking this inclination of the chains one
to another into account for the sake of showing the
general method.
The direction of the forces Z v Z 2 . . . Z 7 acting in
the separate portions of the chain can easily be deter-
mined by increasing the lever-arm by the radius x of
chain friction at every point where the chain winds on
to a pulley, as at A x , B x , C\, D v E^ and F v and by
diminishing it by the same amount wherever the chain
unwinds from a pulley, as at A 2 , B 2 . . . F 2 .
The direction of re-action R 1 of the bearing of the
fixed pulley A is given by the line o x a drawn through o x
tangent to the friction circle at A. Its value is ascer-
tained by making 01 = Q, and drawing through a
parallel to o x a, and through I a parallel to o x o 2 . We
have, then,
III=Z V
the tension in the chain A 2 B V In the same way the
bearing re-acts against the journal B of the second
guide-pulley along the line bo 2 ; therefore a resolution
of the tension Z x = I II in the direction of bo 2 and o 2 o s
will give in I III the tension Z 2 in the portion B 2 C\ of
the chain, and in III II the re-action R 2 of the bearing
at B. There are now acting upon the loose pulley C
98 THE GRAPHICAL STATICS OF MECHANISM.
four forces ; namely, the tension Z 2 of the chain in the
direction o 3 o 2 , that Z z of the portion G 2 D l in the direc-
tion o 5 d v that Z± of the end of the chain in the direction
o 5 d 2 , and finally the tension Z 5 of the second chain in
the direction ce x . The lines of the two last, Z 4 and Z h ,
in ust evidently be tangent to the friction circle at C.
Z and Zo intersect in o 3 , Z 4 and Z. in o 4 . The line
OoO
3 V 4
joining the two is therefore the line of direction for the
resultant of Z 2 and Z 3 as well as of Z± and Z 5 . If we
then draw through I the line /.ZF" parallel to o s o±, and
through ZZ7 the line III IV parallel to 2 d^ we have
III IV = Z^
and in IV I the resultant of Z± and Z 5 . By resolving
this resultant IV I into IV V parallel to d 2 o 5 , and / V
parallel to ce v we get in VI V the tension Z 4 between C
and D 2 , and in / Fthe force Z 5 with which the second
chain pulls down upon the journal of the loose pulley
C. We also have the re-action of the bearing against
the journal D of the fixed pulley in V III, the resultant
of Z s and Z 4 .
We also know that the four forces Z 5 , Z 6 , Z 7 , and P
acting upon the loose pulley E must be in equilibrium.
The force Z b acts along the line ce v that of Z 6 along
f 1 e 2 - Their intersection is at o Q . The force of tension
in Z 7 and the piston-force P are tangential to the fric-
tion circle of E. These last two forces intersect in o v
The line o Q o 7 is then the direction of the resultant of Z 6
and Z & and that of Z 7 and P. We therefore draw
through /the line I VI parallel to o Q o 7 , and V VI par-
allel to f x e 2 , to get in VI V "the tension Z 6 of the portion
E 2 F X of the chain; while VI I represents the resultant
CHAIN FRICTION. 99
of Z x and P 8 . Revolving this force VI /parallel to f 2 o s
or Z v and parallel to P, we have in VI VII the tension
Z 7 , and in 2" VII the force P acting upon the piston-rod.
The re-action of the bearing upon F is given by VII V,
the resultant of Z 6 and Z v
In apparatus of this kind the directions of the ten-
sions in the chains vary but slightly one from another,
so that their points of intersection o often fall without
the limits of the drawing. This difficulty can be sur-
mounted by the employment of the method used in
Fig. 19, plate IL, in which the direction of the lines is
obtained without having their points of intersection
located. We must proceed, according to this method,
when the points <9 3 , o 5 , o-, and ö 8 , fall beyond the limits
of the drawing. When the position of the chains
approaches parallelism there is the difficulty also that
the point of intersection of two such lines cannot be
determined with any degree of exactness. A sufficient
degree of accuracy can, however, be obtained with the
aid of the following construction : If the force Z 2 = I III
is to be resolved into the directions III IV parallel to
J? 3 , and I IV parallel to # 3 6> 4 , we can imagine this system
of forces to be acted upon by two opposite and equal
forces aß and a l ß l along the line C X C V which would
not disturb the equilibrium. Let la represent a/3, and
Ilia be the resultant of this force, and
z 2 = mi
Compound also the opposite force a 1 ß 1 with the yet
unknown force Z 3 , and there will be another resultant
whose direction my be determined. For when the
resultant of Z 2 and aß is compounded with that of Z z
100 THE GRAPHICAL STATICS OF MECHANISM.
and a 1 ^8 1 they will give a resultant which must coincide
with o s o±. If, therefore, we draw through a a parallel
to Ilia, to the point of intersection w with o s o^ the
resultant of Z 3 and a 1 ß 1 must also pass through w, and
is given by <*>a 1 . So that by resolving the force Ilia
parallel to o s o^ and wa 1 , by drawing the triangle IIIm 1 ^
we get the point a 1 from which the point IVcslu be
accurately determined by drawing through a x a line
parallel to a 1 ß 1 intersecting a line from / parallel to
o 3 6> 4 . The auxiliary forces aß and a^ may be chosen
of any convenient value, but should be assumed so that
the lines which are to determine the desired point by
their intersection should be nearly at right angles one
to another.
In the same way the point of intersection VI can be
determined by the application of two opposite and
equal forces along the line E X E^ Instead of resolving
the force Z 5 = V I directly parallel to o Q o 7 and Z 6 , the
force F"S is substituted for Z b and resolved in the direc-
tions o Q o 7 and wßy
In this and similar ways we can employ upon every
diagram constructions in which the lines will diverge
sufficiently to determine accurately the points of inter-
section.
For Q ~ 100 the figure gives
P = 1295;
and since, with the assumption of parallelism, the theo-
retic force I J = 900 ~ 9$, we have
7) = :5j} = 0.G95.
CHAIN FRICTION. 101
In this value of the efficiency no account is taken of
the resistances within the hydraulic cylinder. These
latter must be determined in each case by a special
investigation. It is evident, moreover, that the value
of the piston and stuffing-box friction must be added to
the force P already found to get the necessary pressure
to be exerted upon the piston by the water from the
accumulator.
Any calculation for backward motion would have to
take into consideration the weight of the chains, loose
pulleys, and piston ; and we would get in the force
which would have to be applied at the free end of the
chain A x the counter-weight, which in this species of
hoisting-gear is attached to the hook in order to render
the backward motion automatic. In some cases the
weight of the platform or cage is sufficient of itself to
do this.
102 THE GRAPHICAL STATICS OE MECHANISM.
§ 7. — STIFFNESS OF ROPES.
When pliable ropes and cords are wound on and off
a pulley or drum there are certain hurtful resistances
called forth, in part by friction between the strands of
the rope produced by bending it, and in part by the
resistance of these fibres to the expansion and contrac-
tion which they are compelled to undergo. We can
bring this resistance into the calculation in the same
manner as chain friction by assuming that, in conse-
quence of it, the lever-arm of the load is increased at
the point where the rope winds on, and that lever-arm
of the power where the rope runs off the pulley is
decreased by a similar amount. This value has to be
determined by a special investigation, and is generally
expressed by an empirical formula. By such investi-
gation it has been proven that this resistance is propor-
tional to the tension in the rope, that it is inversely
proportional to the radius of the pulley, and that it
increases with the thickness d of the rope, not in the
simple ratio, but as some higher power. For hemp
ropes we assume that the resistance increases as the
square of the thickness, and is given by the equation
d 2
S = k-Z
r
where Z is the tension in the rope, and k is a constant
co-efficient.
STIFFNESS OF ROPES. 103
That the resistance due to stiffness of ropes really
causes a lengthening and shortening of the lever-arms
of the load Q and the force P respectively can be proven
as follows : Let A (Fig. 41, plate V.) be the centre of a
pulley, and AB = AC = r be the radius of the same
extended to the centre of the rope. A load Q hanging
on the rope at D produces a certain tension in the ele-
ments of any section of the rope which w r e majr assume
as equally distributed over the cross-section, so that the
resultant of all these elementary tensions acts in a result-
ant passing through the centre of the cross-section, and
coinciding with the geometrical axis BD of any portion
of the rope. In the condition of rest, therefore, as long
as there is no turning of the pulley, the rope BD will
arrange itself in such a position that the line of force Q
will pass through the centre M of the section at B. If,
however, we suppose an exterior force to be applied to
the pulley which tends to cause a revolution of the
same in the direction of the arrow, and a winding-up of
the rope at B, there will be a bending of the rope at
that point, and only the strands in the neutral plane
M 1 31 2 will retain their original length, while those lying
in the outer semicircle M X 3I 2 will be stretched, and
those in the other semicircle 3I 1 31 2 J will be shortened
by a force of compression. Each strand in the half 031
will now receive, beside the strain due to Q, a certain
elastic tension o- which is proportional to the distance of
that strand from the neutral plane 31 X 3I 2 . We may
suppose all the elementary tensions in the half-section
M 1 3I 2 to be combined in one resultant s which shall
be applied at some point a. In the same way each
strand of the inner half-section M 1 M 2 J will experience
a certain compression which will also be proportional to
104 THE GRAPHICAL STATICS OF MECHANISM,
its distance from the neutral plane M X M^ and if these
are combined in one resultant we have a certain press-
ure p acting at some point ß. In the bending of rigid
bodies, as beams, it is known that the assumption s = p
is made. It is not necessary here, however. We know
from the preceding consideration that the rope BD upon
which the exterior force Q acts at D is under the influ-
ence of three forces at the section B: namely, the ten-
sion Z acting upwards at the centre B, the force 6* also
acting upwards at a, and that. p. acting downward at /?.
These three forces must have a resultant which is equal
and opposite to the force Q. It will be readily ^een that
the resultant S of Z, *, and p which is to be equal to
Q must be at a greater distance from A than that at
which the centre 31 of the rope section is ; and this fact
is proven when we combine Z and g, and then com-
pound their resultant with p. From the relative posi-
tions of B and a the resultant of Z and s must lie at
a greater distance from the axis A than the radius r
of the pulley, and the composition of this resultant with
the opposite pressure p gives the point of application b
of the final resultant S still farther off. For all pur-
poses of the following demonstrations it is sufficient to
represent the distance IB simply by o- which evidently
corresponds to the value x used in chain friction.
A demonstration similar to the preceding would show
that friction shortens the lever-arm of the force acting
in the portion CE of the rope. In that case also there
would be acting in the section at C : First, a tension Z x
opposite to P, and beside that a force s 1 acting upward
at some point 8 of the inner semicircle X l N. 2 K, and the
force p l acting downward at some point 8 of the outer
semicircle. It follows, then, that by the unwinding of
STIFFNESS OF ROPES. 105
the rope from the pulley at C the strands of the inner
half-section N^^K are stretched, and those of the
outer half N l N 2 L are pressed together. Therefore
the resultant aS^ of Z v s v and p v equal to P, must
have its point of application c between C and A, and
the lever-arm of the acting-force is reduced by an
amount
Go = o-.
If we suppose the ends of the rope BD and CE
to swing free with weights suspended from them the
weights will be shifted to one side or the other, accord-
ing to the direction of revolution, by an amount a- as
was shown for a similar case in the chain pulley. In
every case of rope pulleys where p represents the radius
of the friction circle at the journal the equation
Q(r + P + cr) = P (r - P - cr)
is true, which becomes the same as that for chain friction
when x is substituted for o-.
The investigation of all rope-gearing proceeds, there-
fore, in the same way as for chain friction, and it only
remains to get a graphic equivalent for the quantity o-.
The way in which this value o- is to be applied can easily
be determined in every case where the direction is known
in which the forces act by remembering that the lever-
arm of the unwinding rope is always shortened, and
that of the winding-on rope always lengthened by this
amount cr. If, for instance, a load Q (Fig. 42, plate V. )
hanging from the drum CITis to be raised by a revolu-
tion of the pulley A we have the directions cle,fy, and
hh for rope tensions at once.
106 THE GRAPHICAL STATICS OF MECHANISM.
We have now to obtain a graphic representation for
a- from some of the empiric formulae for the stiffness of
ropes. Of the different formulae * the one most gener-
ally used in practice is that of Eytelwein, which is
simple in form and sufficiently accurate, at least in the
case of hempen ropes and with large forces. This for-
mula will then be assumed as the basis of our calcula-
tions. It should be remarked, however, that in special
cases, as with wire ropes, other formulae should be used,
from which the value of o- can be determined in the
same way as from Eytelwein's formula.
According to this formula the entire resistance S due
to stiffness of the rope at both winding-on and unwind-
ing points is given by the expression
d 2
S = 0.0186-0
where d is the thickness of the rope, and r the radius
of the pulley, both in millimetres, and Q the tenison of
the rope. S merely represents the force which is suffi-
cient to overcome the stiffness of the rope at both sides,
omitting journal friction. In order to produce motion,
therefore, a force
P = Q + S = Q(l + 0.0186^)
must act upon the other end of the rope.
Now, according to foregoing principles, in the absence
* See Weisbach, Lehrbuch der Ingenieur- und Maschinenmechanik,
Theil 1.
STIFFNESS OF ROPES. 107
of journal friction there is the following relation between
Panel Q:
Q(r + a) = P (r - which the
forces make one with another. If we imagine this
pressure R of one wheel upon the other to be laid off
on the lines CD and CE we have in the diagonal CF of
the parallelogram the necessary driving-force P. Under
the supposition again of equal pressures on the two
pairs of teeth this resultant CF is parallel to the com-
mon normal GH, and lies at a distance
£ = \a x a 2 . tan 4>
from the latter. If we now make the permissible
assumption a x a 2 = £sina, a being the angle of the
normal GH to the line of centres X 2 , we get, by
substituting ^ = tan$, the desired distance
fit .
£ = -g sin a.
The value of £ is therefore independent of the posi-
tion of the two points of contact <( 1 b 1 and a 2 b 2 with
reference to (9, it being supposed that they lie on oppo-
site sides of 0, so that the lines of pressure may intersect
at an angle 2$. According to these considerations we
114 THE GRAPHICAL STATICS OP MECHANISM.
may conceive of tooth friction acting in the following
way : While in the case of Motionless motion the force
is transmitted from one wheel to the other in the direc-
tion of a normal to the surfaces in contact, from which
it results that the force P must be exactly equal to the
resistance Q acting along the same line, in the case
where friction is taken into account there is a parallel
shifting to one side of the driving-force through a dis-
tance £: so that the lever-arm of the driving-force with
respect to the axis of the driven wheel B is shortened
by an amount £, while with respect to the axis of the
driving-wheel A it is increased by an equal amount.
Since the line FC represents the re-action of the load Q
for the wheel ^4, and OF = P is the driving-force with
reference to the wheel B, we may say, as in chain and
rope friction, that the arm of the power is diminished,
and that of the resistance increased, by the amount £.
In order to get a graphic representation of tooth friction
Ave have only to determine the value £, and to then shift
the line of force to one side of the theoretic line, so
that it shall be parallel to the latter, and at the distance
£ from it.
It will not be difficult to show that the result of this
shifting corresponds to that which we have been accus-
tomed to find in practice by calculation, and which is
given by the formula
z = qM~ + -).
Here Z is the force which" must be applied at the
point of contact of the pitch circles to overcome tooth
friction alone, and n 1 and n 2 are the number of teeth on
TOOTH FRICTION. 115
the wheels respectively. To show this coincidence let
r t be the radius of the pitch circle A, and r % that of B,
and a = GrOO^ be again the inclination of the normal
pressure to the line of centres. A resistance Q acting
at along the line 0(7 to the wheel B has the lever-arm
r 2 shia, to overcome which a force CF must be applied
at G 7 , which, from the relation r 2 sin a and r 3 sin a — £
between the lever-arms, would be given by the equation
_ n r 2 sina
X =Q
r Sill a
This force X must be exerted by the driving-wheel
A at the point C in the direction CF, with a lever-arm
rjsina -f- £; therefore, there must be applied at the
point of the wheel A a force in the direction Off
given by the ratio of lever-arms in the equation
V — Y r ! sin <* + £ _ q r 2 s ^ n a r i sin a -f- £
r x sin a r 2 sin a — £ r x sin a
Substituting here
— l, r 9 = -h% and £ = ^-<
2tt 2 2tt 2
we get, after reduction,
y = £_^l_ . w i + p* = q( 1 + ff . ff Y
>* 2 — /X7T w x \ w x ' w 2 /
Since without friction § = P it follows that the
116 THE GRAPHICAL STATICS OF MECHANISM.
value of the tooth friction, the force which must be
applied at in the direction of Q, is
z = r - p = qJ± + IX
\n l n 2 /
It may be remarked here that the latter formula,
which is commonly used in calculating tooth friction,
is deduced * under the supposition, employed in the pre-
ceding investigation also, that the pressure is trans-
mitted equally by two pairs of teeth. In the analytic
determination of tooth friction it is customary to employ
as Q the resistance which acts at the contact point of
the pitch circles, normal to the line of centres, in order
to avoid the value sin a. Even if such an approximation
is permissible (sin a = sin 75° = 0.9659 approaching
unity) it will be seen that it is unnecessary in the
graphic method, as that loses nothing of its simplicity
by drawing Q in its proper direction. It follows also
that the construction remains the same when teeth of
any other than the involute form are employed. If we
only know the profiles of the teeth in contact the re-
action is always inclined at the angle (f> to the normal
to the surfaces where contact is taking place. In the
case of profiles laid out by auxiliary circles the chord of
the auxiliary circle passing the point of contact of the
teeth, and that of the pitch circles at 0, is the desired
normal. Of course in cycloidal and most other than
involute profiles the normal will vary in its direction ;
* Weisbach-Hermanu, Ingenieur- und Maschinenmechanik, Theil
III., 1.
TOOTH FRICTION. 117
for such cases we should assume that position of the
teeth for the determination of £ at which the angle a of
the normal with the line of centres has an average
value. The variations of £ will only be small, how-
ever, and can well be neglected as compared with
the uncertainty which clings to all co-efficients of fric-
tion.
So far the investigation has been carried out on a
basis of the equal transmission of power by two pairs of
teeth. If there is contact between only one pair, which
occurs when the arc within which the teeth mesh is less
than f, the friction is somewhat less. In explanation let
us suppose the teeth in Fig. 44, plate V., to be limited
by the circles a 2 a s and b^^; then contact would exist
between only one pair at a time, beginning in a 1 J 1 at
the moment it ceased at a 2 b 2 . ^ that instant the tooth
of the wheel A would act at the point a x with a force
R in the direction a x C upon the wheel B. As long as
motion continued the force R would remain parallel to
a x C until the point of contact between the teeth came
into the line of centres at 0, at which instant there
would be no friction. In this case, as in all preceding
ones, the arm of the power is diminished by an amount
£, and that of the load increased by the same amount.
Here we understand by £ the perpendicular distance of
the point from the line a x G \ that is,
£ z=z a ! 0. tan <£ ~ ^e x
if the distance a x = e . Since this value of £ grows
less and less, becoming equal to zero when the point of
contact is at 0, we see that the friction has its maximum
value when contact occurs at a x b v and its minimum
118 THE GRAPHICAL STATICS OF MECHANISM.
when at 0. If we wish a mean value we have such an
one at a point midway between and a 1 in
£i = ¥i^
In the same way for motion from to a 2 we find the
friction starting at zero, and reaching its maximum
value at a 2 , where
£ = Oa 2 tan ^ = /xe 2
if e 2 = 0a 2 . For this also we have the mean value
Q 2 — 2/^2*
There is this difference, however, in that after the point
is passed the pressure acts along the line Ca 2 . If we
suppose a x = 0a 2 , and consequently ^n^- |£sin a,
we get for an average value of £ the equation
This value of £ is only half as large as that
£ = ±[d sin a
obtained for the preceding case, where two pairs were in
contact ; and, further, the arc of contact was twice as
groat, =z 2f, in the first case discussed as here where it
TOOTH FRICTION. 119
equals t. If, therefore, we denote the entire arc of con-
tact on both sides of the line of centres by o> we have
the general equation for both cases
£ = l/xwsilla.
This equation holds for all intermediate values of the
arc of contact between t and 2t, when the transmission
of force occurs part of the time through one pair and
part through two pairs of teeth.
We can then find the value of £ in every case by the
following simple construction : Lay off on any straight
line the distance OA — -J« (Fig. 45, plate V.) where w
is the length of the arc of contact between the wheels ;
draw the line OB at an angle a with the normal 00 x to
OA, a being 75 degrees for involute teeth, and 80
degrees for cycioidal profiles ; then draw AB perpen-
dicular to OB, and lay off the angle OBJD equal to the
angle of friction $. We then have in the perpendicular
OB to BO the value of £. We may assume ^ between 0.1
and 0.12. By making OA = ^w — ^t we get the same
value for the friction that the formula /xr.( 1 )
\n t n 2 /
gives.
Having gotten the value of £ the tooth friction is
readily determined. Suppose the load Q (Fig. 46, plate
V.) to act upon the drum AB by means of a rope at
B; it is then required to find the force P which must
be applied in the direction EF to the crank EB in
order to turn AB by means of the gear-wheel AC and
the pinion CB, and lift the load. First find the direction
in which Q acts, which is, of course, along the line o x b
drawn at a distance o- from the mean circumference of
120 THE GRAPHICAL STATICS OF MECHANISM.
the drum AB. The line of pressure Z between the
teeth is given by the line o x c drawn at an angle of 75
degrees or 80 degrees to the line of centres AD, and
intersecting the latter at a distance £ from the point of
contact of the pitch circles. Then draw through o v
the intersection of Z and Q, the line o 1 a tangent to the
friction circle at A, and giving the direction of re-action
at that journal. In a similar way we get the re-action do 2
of the journal D. The force polygon can now be drawn
by making o x I = Q, and drawing I II parallel to Z,
then resolving I II = Z in the direction of P and o 2 d.
I III gives us the value of the force P which must be
applied at the crank. To determine the theoretic force
P we have only to draw the broken lines, as shown,
through the centres of the journals, and perpendicular
to AD through C. The drawing gives, for /x = 0.1, and
Q = 100,
P = 30.4, P = 28.1 ;
and therefore
^^ 0.924.
Tt will be readily seen that if in Fig. 44 the driving
was done by the wheel B, but in such a manner that the
points of contact remain at a 1 5 1 and a 2 b 2 (that is, if
the motion was in an opposite direction to that shown
by the arrows), the direction of pressure between the
teeth would remain parallel to GH\ but the force Z
would lie on the opposite side of this line, and would
pass through C x between GrH and the axis of A, at a
distance £ from the former. This case corresponds to
the backward motion of the windlass ( Fig. 40) under the
TOOTH FRICTION. 121
influence of Q. When, however, the wheel B (Fig. 44)
is driven in the opposite direction to the arrows by the
wheel A the investigation is similar in all respects,
except that the direction of pressure is now along the
line H'G'.
122 THE GRAPHICAL STATICS OF MECHANISM.
§ 9. — BELT GEARING.
The principal source of loss in the transmission of
rotary motion from one axis to another by means
of belting is the friction which the axes suffer from
being pressed against their bearings by the tension in
the belt, since the resistance clue to stiffness where the
belts wind on and off the pulleys is so small that it
cannot be taken into consideration. But, on the con-
trary, the journal friction is much greater for the trans-
mission of a given force than by toothed gearing, since
in belt gearing only the difference of tensions in the
two portions of the belt represents force transmitted,
while journal friction is caused by the sum of these two.
The investigation of this resistance, with the determina-
tion of the tension in the belt, is pursued in the follow-
ing manner : —
Let the shaft A (Fig. 47, plate V.) be driven from
the shaft B by means of the belt and pullej^s CD and
EF. We are to find the force P which must act on B
with the lever-arm BK to overcome the resistance Q
acting upon A with the lever-arm AL. If the belt
surrounding the two pulleys is stretched to a certain
tension this tension S is the same in both portions of
the belt DE and CF while at rest. If, then, the pulley
EF is acted upon by a force tending to revolve it in
the direction shown by the arrow, the tension in the
BELT GEARING. 123
belt OF increases to a value S v and at the same time
that in DE diminishes to a value $ 2 , until
#1 — ' ^2
is sufficient to overcome the resistance Q at the axis A,
It being supposed that no slipping of the belt upon the
pulleys occurs, the condition for such slipping is given
by the formula
S 1 = S 2 e» a ,
where S r is the tension in CF, S 2 that in DE, e is the
base 2.71828 of the natural system of logarithms, jm is
the co-efficient of friction between belt and pulley, and
a the arc of contact between the same, the radius being
unity. The greatest resistance W, therefore, which can
be overcome with a tension 8% of the belt ED, when it
(the resistance W) acts with a lever-arm AC from the
axis A is
W = S t -S 2 = 8 2 (e^- 1).
The tension 8 in the belt when at rest must be deter-
mined according to this relation, and is usually assumed
8 = l(S, + SJ,
If we then suppose the two tensions S x and 8 2 to have
a resultant Z we can draw the force polygon, as in pre-
vious cases, by substituting this resultant for the tension
in the two portions of the belt. The direction of the
resultant is obtained by determining the proportion
124 THE GRAPHICAL STATICS OF MECHANISM.
of the two tensions for the limiting condition of slip-
ping
—1 z= e^'
S 2
so that by laying off from the intersection of the belts
the distances OGr and OH of any convenient lengths,
but in the ratio
OH S x
OG ~ S 2 ~ etX%
we have in the diagonal OJof the completed parallelo-
gram the resultant Z which may be substituted for the
belt tensions themselves. To determine the value of
this resultant, and of the belt tensions S x and S 2 , draw
through o x and o. v the intersections of the resultant
with P and Q, the tangents o x a and o 2 b to the friction
circles of the journals A and B. Then make ol = (),
and by drawing I II parallel to o x a we get the resultant
Z = ojl
of the belt tensions, and those tensions S 1 and # 2 , by
resolving o x II in the directions o x IV parallel to CF,
and II IV parallel to DE. From o x II = Z we get also
the value of P by drawing o i III parallel to o 2 K, and
II III parallel to o 2 b. In order to find the theoretic
force P which would be sufficient to overcome Q in the
absence of friction we can either employ the direction
OJ of the resultant Z, and draw the re-actions o x A and
<> 2 B through the centres of the journal, or take the
intersections X and 2 of the forces Q and P with OF,
BELT GEARING. 125
regarding the latter as a force acting at and F with-
out friction.
In Fig. 48, plate VI., the logarithmic spirals are drawn
for the commonly occurring co-efficients of belt friction
,1 = 0.12, 0.18, 0.28, 0.38, and 0.47,
in order to give a graphic representation of the ratio of
belt tensions
If we take the radius of the circle OA as unity, the
radius vector OB drawn to the spiral, constructed with
the particular co-efficient of friction /x, at the angle
AOC = a, the angle of contact, gives the value e^ a .
Therefore, if the tension S 2 of the slack side is repre-
sented by OA, we have in OB the tension S 1 of the
driving side, and in
CB = aSj — >&2
the force transmitted.
From the preceding it is easy to draw a comparison
between the relative efficiencies of belt and toothed
gearing. For this purpose we shall investigate the
motion of a millstone, since these are as frequently
driven by belts as by gearing. Let A (Fig. 49, plate
VI.) be the mill-spindle, and BD the pulley on the same,
which has the arc BMD in contact with the driving-belt.
With the aid of the spiral for ^ = 0.28, as given in
Fig. 48, plate VI., we find the ratio
S 2 :S 1
126 THE GRAPHICAL STATICS OF MECHANISM.
from that of OA : OB, having laid off the angle of con-
tact
BAD = a
from the radius OA. Laying off these distances in KL
and OK (Fig. 49) we have in OL the direction of the
resultant Z of the two belt tensions which act upon
the mill-spindle A as driving-force P. Since the resist-
ance of the millstone upon its grinding-surface is exactly
equal to this we must represent this resistance by a
couple of forces Q and Q acting in the directions _ff/and
FGr. This couple will not cause any side pressure of
the mill-spindle against its bearing ; the re-action of the
bearing will be that called forth by the resultant Z, and
will therefore be equal to that resultant, and opposite in
direction. We then draw R in the tangent ao 2 of the
friction circle at A parallel to OL. For the existence
of equilibrium between the four forces Q, Q, R, and Z,
the resultant of any two must be coincident with and
opposite to that of the remaining two. Uniting o x and
o v making o x I = Q, and drawing I II parallel to o x o v
we have in ^ 77 the resultant of the two tensions S x and
aS y 2 ; and by resolving o x II into IIIo x , and II III parallel
to the directions of the belt, we get
IIIo x = S x and II III = S 2 .
Without friction we should assume the direction of re-
action at the bearing parallel to OL, and passing through
the centre of A along the line A0 2 ; then drawing I II
through /parallel to o x 2 , we should get
P ^ 11,0,.
BELT GEARING. 127
From the figure, with the assumed value /* = 0.28, and
a journal friction of 0.1, we get, for Q = 100,
P = 401.6, P = 379.2,
and
77 = *~* = 0.944.
If, on the other hand, we suppose the stone to be
driven by gearing from the upright shaft B (Fig. 50,
plate VI.), A again being the mill-spindle, Ave have the
pressure transmitted by the gearing along the line o x c
inclined at the angle of 75 degrees to the line of centres,
at the distance £ from (7, and in the parallel line o 2 a the
direction of re-action at the bearing of the spindle.
Making Io 1 z=z Q, and drawing ///parallel to x ö 2 , we
find P in o x II. To determine P we assume the thrust
of the gearing along the perpendicular to AC passing
through (7, and the re-action of the bearing parallel to
this and passing through A. Then, if I O 1 = Q, we
have in O 1 II the theoretic force P acting perpendicu-
lar to AC through C In order to determine the effi-
ciency we must compare, not the force P acting along
o x c, but that component P f parallel to P , with the
latter force. Let w be the point of intersection of
the two directions, then we must resolve
o x II = P
in the direction of 1 C and wJ5. Draw o x III parallel
with O x C, and II III parallel to uB, and we have in
128 THE GRAPHICAL STATICS OF MECHANISM.
o x IH the force I" which must act at C perpendicular
to the line of centres AB, thus giving the efficiency
17 = ^A
1 P'
From the drawing we get, for Q = 100,
P' = o x III = 226.4, P = 217.8,
and
v = 5; = 0.962.
From these two examples we draw the conclusion
that belt gearing is less economical than toothed gear-
ing, as would have been expected from previous consid-
erations. It is also evident that the result for belt
transmission would be more unfavorable if the co-
efficient of friction of belt upon pulley is less than
0.28, or if the arc of contact between the belt and the
pulley of the mill-spindle subtends a less angle a, since
the belt tension and resulting journal friction would
then be much greater. In the foregoing comparison the
journal friction of the main shaft is omitted in both
cases, because in the general arrangement of several
stones about one central shaft the opposing tensions or
pressures would counteract one another, and the shaft
would run freely in its bearings. If this were not the
case, but only one stone was driven from the shaft, then.
on account of the greater belt tension, the friction of the
shaft in its bearing would be greater than in the case of
toothed gearing.
BELT GEARING. 129
The tensions in brake bands are to be estimated in
the same way as with belt gearing. Here also there are
two different tensions S ± and S 2 in the two ends of the
band which have the relation
one to the other, the greater tension S x being that which
opposes the sliding of the brake pulley within the band.
We can therefore introduce the resultant Z of the two
tensions S 1 and S%, and regard it as a force preventing
the motion which tends to occur. Let A (Fig. 51,
plate VI.) be the axis of a drum on which is the brake
pulley BC, whose brake band is fastened at one end to
the stationary bolt E, and at the other to the bolt D of
the brake-lever EDG which turns about E. We get the
ratio of S x to S 2 from the spiral in Fig. 48, which corre-
sponds to the co-efficient of friction for brake bands
fi = 0.18,
by making the angle a == CMB the arc of contact, and
lay off the distances so determined along the direction
of the ends of the band from their intersection in
OUT and OJ. The diagonal OK of the completed par-
allelogram gives the direction of the resultant Z. As
regards the direction of tension of the brake band we
see that the end CE fastened at E pulls in a line passing
through the centre of E, since there is no relative turn-
ing of the band about this point. But the line of
tension in the end DB attached to the moving journal
D would be tangential to the friction circle at D.
If, now, a force Q acting with a lever-arm AF tends
130 THE GRAPHICAL STATICS OF MECHANISM.
to turn the shaft A in the direction of the arrow the
direction of journal pressure at A will be given by
the line o x a drawn through the intersection o 1 of Q and
Z tangent to the friction circle at A. Therefore, by
making o x I = Q, and drawing through I a parallel to
OK, we get in I II the necessary resultant Z of S x and
>S f 2 . Resolving this parallel to HO and JO, we have
III II = S 1 and IUI = S 2 .
To determine the force P applied to the brake-lever G
to produce the tension
S 2 = IUI
in the band BD we first draw from o 2 , the intersection
of S 2 and P, the line o 2 e tangent to the friction circle at
E to get the re-action of that bearing ; then a resolu-
tion of I III = S 2 into I /{^parallel to o 2 e, and IV III
parallel to o 2 G or P, gives us in IV III the force P
which must be applied at Gr.
The determination of the driving and brake forces in
a whim or windlass, such as is shown in Fig. 52, plate
VI., is of especial interest. Here two ropes FD and
EC lead from the windlass-drum G- in such manner,
that, if the drum is revolved in either direction (say,
that indicated by the arrow), the rope EC is wound
up, and that BD unwound ; and, in consequence, the
weight Q consisting of useful load, and the weight Gr of
cage, hooks, etc., hanging upon the rope ECB, is lifted,
while the weight G of the empty cage hanging on
the rope FDB sinks, and thus aids the revolution
of the drum. If there were no hinderances to this
motion we should at once assume that the working of
BELT GEARING. 131
the entire apparatus amounted simply to lifting the
useful load Q, since the cages balance one another and
could be left out of consideration. Such an assumption
is, however, not permissible on account of friction ; and
Ave must regard the machine as a combination of two
hoisting-gears, one of which burdened with the load
Q -f- G- on the rope ECB is in forward motion, while
the other is running backwards under the influence
of the load G attached to the rope FDB. Accordingly
the investigation would be pursued as follows : The line
of tension in the ascending rope BCE must be assumed
on account of stiffness in the rope along the lines b 1 o 1
and ce in such manner that the lever-arms at B and E
shall be greater than the radii of the pulley and drum
respectively by an amount o-, and at C less by an equal
amount. On the other hand, the directions of tension
for the rope FDB are given in fd and o 2 b 2 ; the lever-
arms at D being increased, and those at F and B dimin-
ished, by an amount o-. The direction of re-action at
the bearing of the pulley A CB which revolves toward
the left would be given in the line a x o x , while that of
ABB revolving in the opposite direction would be along
o 2 a 2 . Making o x I = Q -f- Gr, and o 2 III = Gr, and draw-
ing through I a parallel i" II to ce, and through III a
parallel III IV to df, we have in
III = S t
the tension in the rope EC, and in
IV III = S 2
the tension in FD, The resultant Z of these tensions
132 THE GRAPHICAL STATICS OF MECHANISM.
is given by V I if we draw II V equal and parallel to
III IV, and complete the triangle. This resultant
passes through the intersection o z of ec and fd, and
Ave must therefore draw o 3 <9 4 parallel to / V
If the drum is driven by a pinion HJ meshing with
the gear-wheel GrJ we draw the line ii inclined at an
angle of 75 degrees to GrH, and at a distance £ from J,
for the direction of pressure between the teeth, and then
draw from o 4 , the intersection of this line with that of
Z, the tangent o±g to the friction circle at Gr. This
gives us the direction of re-action at the bearings of Gr.
Then resolve the force / V into I VI parallel to o±g,
and VI V parallel to o 4 i, and we have in VI V the
pressure on the teeth of the pinion HJ, or the necessary
driving-force P.
To determine the theoretic force P we have, as pre-
viously remarked, only to lay off a distance equal to the
useful load Q along the medial line of the rope EC from
the intersection of that line with the common tangent
JO to the pitch circles, so that OI = Q. Then draw
through the radius 0(7, and through I the line I U
parallel to the latter, and we have
II = P .
To determine the efficiency we must again compare P
with that component of P obtained by resolving
nv=p
in the direction of P and W Ä This force P f is given
by VI VII if VI VII is drawn parallel to OJ and V VII
BELT GEARING. 133
parallel to wff. From the figure we get, for Q ~ 36 and
G = 24,
P f = VI VII = 47.3, P Q = 40.1,
and therefore
r, = £} = 0.848.
For convenience, in order to use the same lines in
the figure, we will suppose in determining the brake
force that the loaded cage is now upon the rope FDB.
Accordingly we lay off the broken and dotted lines
o 2 (III) equal to (Q + Gr), and o { (I) equal to Gr; draw
(ill) (IV) parallel to fd, and (I) (II) parallel to ce;
we then have
(JF)(1ZZ) = (£,) and {1-){I1) = (^).
And by drawing tf 3 (JO equal to (III) (IV), and mak-
ing (F)( VI) parallel and equal to (I)(II), we get in
(VI)o s the resultant (Z) of (aS^) and (S 2 ). If the
brake is applied by means of the brake-lever NLKM
turning on the fixed point K, and attached to the ends
of the brake band at iVand L, we first draw the lines of
force W x n and W 2 l tangential to the brake jmlley T^IF 2
and to the friction circles of the bolts N and L. From
the intersection o 5 lay off the distances o h TJ and o 5 T
proportional to the tensions s 2 and s t in the brake band,
which have been determined from Fig. 48. The diag-
onal of the completed parallelogram then gives the
resultant z of these tensions. If we now draw from o 6 ,
the intersection of z and (Z), the tangent o Q to the
friction circle of Gr, and resolve the force (Z) = ( JT)tf 3
134 THE GRAPHICAL STATICS OF MECHANISM.
parallel to o 5 o$ and o^g, we get in (TTI)( VI) the value
of the resultant z which may be substituted for the two
tensions s t and s. 2 in the brake band. To determine the
force p to be applied to the brake-lever at M draw from
o 1 , the intersection of this force with o 6 o 5 , the tangent
o 7 k to the friction circle of K, and resolve the force
(IT/) (FT) = z in the direction of o 7 k and Mo v thus
getting
(F//j)(r/)= P ,
the force to be applied at the brake-lever.
It has been heretofore tacitly assumed that the driving-
force P has been applied in just sufficient amount to
overcome the resistance of Q. This condition would be
fulfilled, for instance, if a water-wheel acted upon by
equal impulses at equal moments drove a millstone
whose working resistance is constant. It is also ful-
filled in most hoisting apparatus. But in many cases
in practice either the moment of the power, or the
moment of the load, or both, vary periodically ; but in
such a way, if continuous motion is supposed, that for
any such period the work done by the driving-force
exactly equals that overcome in the shape of useful and
hurtful resistances. This is always the case in the
slider-crank motion. Thus in the ordinary steam-engine
the steam pressure transmitted through the piston-rod
and connecting-rod to the crank-pin has an extremely
variable moment, which is reduced to zero at the dead
points, and reaches a maximum value somewhere be-
tween these two. On the contrary, the resistance Q
which probably acts at the circumference of some pulley
or wheel keyed upon the shaft has a constant value.
It is therefore clear that the moment of the resistance
BELT GEARING. 135
Q must have a mean value between the greatest and
least moments of the power if the motion is to be con-
tinuous. Consequently there will be, on account of the
periodical variation of the moments, an alternate accel-
eration and retardation of the moving masses of the
machine in such way that, during the time when
the moment of the power exceeds that of the resist-
ance, the excess of work done by the former is stored
up in the moving parts in the form of living force, from
which it is again given out when the moment of the
power sinks below that of the resistance. A similar
state of things exists when, on the contrary, a resistance
acts upon a crank with varying moment, while that of
the power is constant, as when pumps are driven by
water-wheels.
The manner in which the motion of such a mechanism
would be investigated may be learned from the example
of a jig-saw (Fig. 53, plate V.l.). Here the shaft A
which gives the saw-frame EF a reciprocating motion
through the crank AC and the connecting-rod CD is
driven by a belt on the pulley B^B^ ; it also has the ffy-
wheel U keyed on to it, which tends to render the
motion uniform, as shown in the preceding paragraph.
Let the vertical resistance which the teeth Z of the saw
encounter from the log K on the downward stroke be
represented by Q, and laid off in o x L The connecting-
rod acts upon the grate in the direction dc of the com-
mon tangent to the friction circles of D and (7; and,
furthermore, the bearings Gr x and H x of the grate act
upon the guide GH at the angle of friction from the
normal with the re-actions H x and J? 2 . Joining o v
the intersection of Q and R v with ö 2 , the intersection
of li 2 and T, we get the various forces by drawing o^III
136 THE GRAPHICAL STATICS OF MECHANISM.
parallel to dc, and ////parallel to R v ///then equals
R v II III R 2 , and o x III the necessary force T in the
connecting-rod. Next, suppose the belt which half
encircles the pulley B X B 2 , and whose ends are therefore
parallel, to have in the slack side the tension
$2 — ^2^2'
and in the other the tension
S x = B 1 L V
these tensions aS^ and S 2 having been determined by the
aid of Fig. 48. The resultant Z then equals
BL = S x -f- S 2 = B l L 1 + B. 2 L 2 ;
and its position is determined, according to the laws of
parallel forces, by making
B 2 l 2 = B l L 1 and B 1 l 1 = B 2 L 2 ,
and drawing l x l 2 , when the intersection I of this line
with B 1 B 2 gives the point of application of the result-
ant Z. Now lay off from o 1 of the force polygon the
line o x IV parallel and equal to 5/, and we have in
the line IV III the resultant of the belt tension Z,
and the pull T of the connecting-rod, which together
act upon the shaft A. For equilibrium there must be a
re-action R of the bearing upon J., whose direction and
BELT GEARING. 137
magnitude is given in III IV. The position of this re-
action It is then given by the tangent aa to the friction
circle of A parallel to III IV. This line aa cuts the
force T in 0, a different point than that B, the intersec-
tion of Z and T, from which we see that the three forces
Z, T, and R cannot be in equilibrium ; for in that case
the force Z must act along OJ instead of BL. The
moment of the belt tension Z is here less than that of
the resistance, and therefore at this moment living force
must be given out by the fly-wheel to render motion
possible. On account of the equable distribution of the
masses of the fly-wheel about its axis we are not to
regard its action in the light of that of a single force,
but as a couple whose forces M, M are applied at any
two diametrically opposite points m 1 and m% of the cir-
cumference, whose radius Am x = Am 2 is equal to the
radius of gyration p of the fly-wheel. The value of
the forces M, M is determined by the condition that the
resultant of the couple and the belt tension Z or BL
must be a force passing through 0, parallel to BL, and
equal in value to
Z =0J = BL.
To construct this value of Mwe join o s , the intersection
of BL and m^M, with ö 4 , that of OJ and m 2 M, and
then resolve the force Z = BL in the direction of 3 o 4
and of M. We therefore draw through IV in the force
polygon a parallel to o s o 4 , and through o x a parallel to
M (M being here assumed as acting vertically, parallel
to $), and have in o x V the value of the two forces of
inertia 31, which, applied at m x and m 2 in the fly-wheel,
aid the motion of the shaft in the moment under con-
138 THE GRAPHICAL STATICS OF MECHANISM.
sicleration. It will be readily seen that the same values
of M would be obtained in whatever direction they
were supposed to act, if always their distance apart
m x m t := 2;j. But if we assume this distance to be of
another value 2 Pl we get a different value itfj for the
forces of inertia, so that the moment of inertia is always
constant,
•23f P = 2M lPv
From the preceding it follows that the action of moving
masses may be represented by couples when they are
not arranged eccentrically, and that the influence of
such a couple is to produce a parallel shifting of the
driving-force Z or P without any consequent increase
of journal pressure and friction. It is also evident that
the couple acts in the opposite direction, opposed to the
driving-force, if the moment of the latter becomes greater
than that of the resistance. This would be the case if
in the present example the arm Al of the belt driving-
force was greater than AN. In that case the action of
the couple would result from an acceleration of the
motion ; while under the condition investigated above,
where the fly-wheel acts in the " sense, " or direction, of
the desired motion, there is a retardation of the masses.
It is well known that acceleration and retardation occur
in regular periods of continuous but non-uniform motion.
The amount of these momentary accelerations and re-
tardations in the masses can be readily ascertained for
every case by this method of couples, but as such an
investigation lies without the object of this work we
will not pursue it farther.
Two examples corresponding tu commonly occurring
BELT GEARING. 139
machines will serve, in closing, for the further discussion
of those conditions more fully investigated in the pre-
ceding chapters, which are principally influential in
determining the efficiency of machines. These exam-
ples are the crane in plate VII., and the beam-engine
in plate VIIL
140 THE GRAPHICAL STATICS OF MECHANISM.
§ 10. — EXAMPLES.
In the crane (Figs. 54 to 58, plate VII.) the revolving
jib or outrigger turning about the post or mast L has
at its outer end the guide pulley B (Fig. 57), over
which the hoisting-chain is led ; one end of the latter
is fastened at (7, while in its loop hangs the pulley A
supporting the load Q. From B the chain is led over
the supporting pulleys D and D x to the drum EE X ,
which receives motion from the crank UK (Fig. 54) by
means of the gearing at F and J. Drawing the direc-
tion of the forces according to well-known laws we
have the load Q acting along the tangent aa (Fig. 55)
to the friction circle of the journal A of the loose
pulley, while the tension S x of the chain fastened at C
is along the line a 1 a 1 , and the tension S 2 of the portion
A 2 B X of the chain acts along the vertical a 2 a 2 , so that
the lever-arm at A x is lengthened, and that at A 2 short-
ened, by an amount x* If«» then, I II =; Q, and Ia x is
drawn parallel to IIa v the line adjoining these points
gives, by its intersection III with Q, the tension
11111= S x
in A X C, and
IIII = # 2
in A 2 B V Drawing in the well-known way the direction
of forces b x o x and b 2 o x in the chain over the pulley B
EXAMPLES. 141
we get the re-action of its journal in the tangent o x b to
the friction circle at B. A resolution of the chain
tension
IUI = &,
in the direction of o x b and b 2 o 1 gives us in
TV III = S s
the tension in the portion BD of the chain. The direc-
tion of the portion of the chain passing over the pulley
D is shown (Fig. 56), which is drawn to a larger scale.
From this we see, without further explanation, that the
journal re-action of this pulley * is in the direction do r
We therefore draw through the point III in the force
polygon the parallel III V to do 2 , and from IV the
parallel IV V to e x o^ and get
IV V = S„
the tension in the portion of the chain between D
and the drum E v which has such a direction e 1 d l
(Fig. 54) as to increase the lever-arm at E x , and
decrease it at Z), by an amount \. Draw now at the
distance £ from F, the point of contact of the pitch
circles, the direction of pressure fo s between the first
pair of gear-wheels at 75 degrees to the line of centres
EGr, and also the tangent z x e to the friction circle at E,
* The slight resistance of the pulley D, is neglected. If it was
desired to bring it into the calculation it could be done as shown in
Fig. 36, plate IV.
142 THE GRAPHICAL STATICS OF MECHANISM.
so that it shall pass through the point of intersection of
d 1 e 1 and fo s ; and we can then resolve
IV V = s±
parallel to fo s , and e x e into
IV VI = z v
the pressure between the teeth at F, and V VI equals the
re-action of the journal bearings of the drum. It should
be remarked here that the direction e^ 1 of the journal
re-action is drawn by the aid of the auxiliary construc-
tion previously given, since the intersection of fo s and
d 1 e 1 lies beyond the limits of the drawing. We draw
through E a line /3e8, and any parallel line j8 1 8 1 ; then
locating C]L , so that
/?€ : c8 : : ß 1 z x : ^Sj,
we have in e x a point in the desired line of re-action,
c is here the point of intersection of the line /38 with
the friction circle, this being a sufficiently close approxi-
mation if the line ße is drawn perpendicular to what is
judged to be about the direction of the re-action e x e.
In the same way we draw the line of pressure o s i
between the teeth of the second pair of gear-wheels
HJ and GrJ^ and from o s the tangent to the friction
circle of G-. Then a resolution of
11 IV = Z x
EXAMPLES. 143
in the direction of o s i and o z g gives us in
VI VII = z 2
the pressure with which the pinion HJ acts upon the
wheel GrJ along the line o s i. Finally, draw from o 4 ,
the intersection of P and Z v the tangent oji to the
friction circle of H, and through VII a parallel to P,
and through F7a parallel to o 4 h, and we have
VIII VII = P,
the force which must be applied at the crank K in the
direction Ko±. Since this value becomes very small in
the figure, the triangle VI VII VIIUs drawn on a scale
five times greater in VI VII VIII\ in which
vi vir = 5 . vi vii,
according to which
p = i - . vnrvir.
To determine the theoretic force P we can suppose
the tension in the chain as it winds on to the drum at
E x equal to -—, and therefore make VIV = \IIL Then
draw, as before, the force polygon IV V VI VII VIII ,
taking the pressure between the gear-wheels perpen-
dicular to the line of centres at F and J", and the journal
re-actions as passing through the centres of the journals
144 THE GRAPHICAL STATICS OF MECHANISM.
at E. G\ and H. Thus is constructed the polygon shown
hi broken lines, which gives
p = riii vn = i . viii ( ;vii ( ;.
The drawing gives, for Q = 100,
P = 1.155, P = 0.833,
and therefore
r, = £> = 0.722.
To determine the brake force p which is to be applied
at W in the lever WU we must take the backward
motion of the crane into consideration. In this case it
is evident that the load Q = I II hanging upon the
loose pulley causes, in sinking, a tension S { of the chain
fastened at C equal to III I, and a tension
S 2 = III II
of the portion between A 2 and B x ; in other words, the
strains in the two chains are reversed by backward
motion. It will also be readily understood that the
lines of tension in the chain at the pulleys B and D
and at the drum will be those shown in broken and
dotted lines ; that is, (o x ) (b^) and (o^) (6 2 ) at B,
(fl 2 ) (^2) an d (°2 ) ( e \) at A an d ( e \) (^1) a * the drum,
while the journal pressures at B and B take the direc-
tion (oj) (7>) and (0 2 )OO- TJ ie direction in which the
pressure (Z x ) of the teeth of the wheel ET now acts
upon the pinion GF is given by the line (/) (/), which
EXAMPLES. 145
cuts the line of centres EG at 75 degrees, and at a dis-
tance £ from the point of contact F of the pitch circles
toward the side of the driven axis G-, as was shown
formerly in the discussion of tooth friction. The direc-
tion of journal re-action at E corresponding to (/) (/)
would vary but little from that, e£ X , found for forward
motion ; it may therefore be assumed as the same.
Taking these things into account the force polygon
would be drawn as follows : Make
(I)(III) = III II = (£,) ;
draw (J) {IV) parallel to (oj) (ft), and (III) (IV)
parallel to (o{) (6 2 ), also (IV)( V) parallel to (o 2 ) (, are supported by the pivot L in the head
of the mast or post. This pivot has a side thrust upon
it from the bearing L v while the two friction rollers
W x and W 2 press upon the base T of the post. The
pressures at L and T are determined from Fig. 57 as
follows: Suppose S to be the centre of gravity of all
the swinging portion of the crane, windlass, pulleys,
chain, etc., and suppose G the weight of all these parts
EXAMPLES. 147
acting downwards at S. This force may be combined
with the weight Q hanging at A into one resultant
force,
acting at M. Now lay off on this line the distance
in = Q + a,
and draw the horizontal re-action R of the cylinder T
through the middle of the rollers W. The re-action R x
of the pivot L must then pass through the intersection
of R and Q + (?, and must take the direction OL drawn
through the centre of the bearing L v To determine R
and R x we therefore resolve I II in the direction of
OT and OL, and have
IUI - R,
the re-action against the rollers TT, and
II III = R v
the combined re-actions at L. This latter may be
resolved back into horizontal and vertical components,
/ZZ7 and II L The vertical re-action equal to G -f" Q
produces end journal friction upon i, which may be
regarded as the action of a couple, each force of which
equals
MQ + #)*
and whose arm equals |c/, if d is the diameter of the
journal. The horizontal component of the re-action R x
148 THE GRAPHICAL STATICS OF MECHANISM.
is equal and opposite to the re-action R of the post
against the rollers in the direction TO, so that these
two forces form a couple in equilibrium with that
formed by the vertical re-action Z, and the weight
Q -\- Gi at M. To determine the turning-force p w r e
draw, in Fig. 58, the horizontal pressure 1 2 equal to
IUI in Fig. 57, and in such direction that it shall be
tangent to the friction circle at L. Also lay off the
two end journal frictions
F = F X = l-^Q + G)
at the distance \d on each side the centre of X, and in
opposite directions, so as to bring this end journal fric-
tion into the calculation. The compounding of this
couple F, F l with the horizontal re-action 1 2 will
simply result in a parallel shifting of that re-action
undiminished in value to the position 4 5. To fix the
amount of this shifting make 2 3 = jP, draw 3 1, and
through o v the intersection of 1 2 and F, draw a parallel
to 3 1 ; this will cut F x at a point o 2 , through which the
resultant 4 5 of 1 2 and FF X must pass. The correct-
ness of this construction is shown by the equilibrium of
the four forces 1 2, F, F v and 5 4. Now draw the re-
actions u l w 1 and u 2 w 2 of the post against the friction
rollers in such way that they shall be tangent to the
friction circles of W 1 and TF 2 , and shall pass to one side
of the contact points L\ and U 2 of the rollers at the
distance e determined previously for rolling friction. If
the turning of the crane is produced by a pinion on the
vertical shaft V, and working with a circular rack or
internal gear Y on the base plate, we draw finally the
direction vv of pressure between the teeth at 75 degrees
EXAMPLES. 149
to the line of centres VL, and at the distance £ on the
outer side of the contact point of the pitch circles.
This line intersects the resultant of journal pressure 4 5
in o 3 , the two re-actions of the post against the rollers
intersect in o 4 ; therefore o s o 4 must be the common
resultant of these pairs of forces. Draw 5 6 parallel to
vv, and through 4 a parallel to o 4 o s , and we have
5 6 = Z,
the pressure which the teeth of the pinion on the shaft
V must exert along the line vv. The method by which
the force necessary to turn V by means of a crank
would be determined is sufficiently well known from
previous examples.
In conclusion we will explain the diagram for the
condensing beam-engine shown in plate VIII. Let
I II = P, the force acting upon the piston (Fig. 59).
This force acts along KC\ the geometric axis of the
piston-rod, since the latter is rigidly fixed to both
the piston and the cross-bar (7, and there can be no
relative motion between those elements. If, then, we
consider the piston, piston-rod, and cross-bar C to form
one piece, we have at the point C three forces, P the
piston force, and S x and S. 2 exerted by the link BC and
the parallel rod EC respectively. In the present posi-
tion of the engine these forces are in compression, and
act along the tangents hc x and e^ shown on a larger
scale in Fig. 60 a . The intersection of these forces is at
o v and since P does not pass through this point the
three forces P, S v and S 2 cannot be in equilibrium.
There must therefore be a fourth force, which with P
will give a resultant passing through o v and which must
150 THE GRAPHICAL STATICS OF MECHANISM.
tend to produce right-handed revolution about o v since
P alone would produce left-handed turning. This
fourth force is the re-action R x exerted by the stuffing-
box against the piston-rod ; it passes through the centre
of the stuffing-box, is inclined at an angle cf> below the
horizontal, and acts from the right toward the left.
This re-action cuts the piston force in the point o 2 ; and
therefore, as frequently shown before, the line o x o v not
drawn in the figure, must be the common resultant of
S x with aS' 2 , and P with R. So that, drawing through
/ a parallel to R v and through II a parallel to o 1 o 2 ,
we get in
IUI = B 1
the re-action of the stuffing-box against the piston-rod.
This indicates that in spite of the right-line motion
there is a certain side thrust against the stuffing-box, as
shown in Fig. 19, plate II., which is due to journal fric-
tion on the cross-arm C\ and not to any inaccuracy in
the parallel motion. This side thrust, which, as pre-
viously shown, reverses its angle at every change in the
direction of piston motion, is, however, so small that it
is generally left out of account. Draw, next, through
II and III the parallels II IV and III IV to e x e 2 and
c x b, and we get the forces
S 2 = II IV and S 1 = III IV.
The last force
S x = III IV
is transmitted directly to the beam through the link
CB. A second force, partly due to the resistance of
EXAMPLES. 151
the air-pump L attached at F, and partly to the influ-
ence of the radius-rod ED, acts upon the beam at G.
This force is the next to be determined. Let II V
represent the resistance L of the air-pump, which acts
upon the link FG in the tangent to the friction circle
of F passing through the point of intersection o s of the
piston force L of the air-pump and the stuffing-box re-
action R 2 . This small re-action R 2 , and its inappreci-
able influence on the resistance L, will not be further
taken into account.
Looking at the link FGr we have acting upon it three
forces : L the air-pump resistance in the direction fo s
(see also Fig. 60^) ; the pressure aS y 2 of the parallel rod
OF in the known direction c 2 e v and of the known value
II IV; and finally a force aS y 3 exerted by the radius-rod
ED upon the pin E, and which acts in the known direc-
tion e 2 d v tangent to the friction circles at E and D.
These three forces must be held in equilibrium by a
force S 4 exerted by the journal G upon the link GE,
and which remains to be determined. For this purpose
we combine the two known forces
of the air-pump, and
II V
S 2 = IV II
of the parallel rod GE, in a resultant IV V. This
resultant must pass through the intersection o 4 of its
components (Fig. 60 6 ) ; and by drawing through o 4 a
parallel to IVVwe get in o 5 , its intersection with the
force # 3 acting along e 2 d l in the radius-rod DE, a point
152 THE GRAPHICAL STATICS OF MECHANISM.
through which the desired re-action of the journal Gr
must pass. If we draw the tangent o b g through o 5 to
the friction circle of (2, the direction of # 4 , the resultant
of S 2 , aS' 3 , and L is then found. In the force polygon
we now resolve IV V, the resultant of L and # 2 , in the
direction IV VI parallel to o 5 #, and VI V parallel to
e % d x , thus getting
V VI = s s ,
the tension in the radius-rod, and
IV VI = s v
the pressure of the link EG upon the journal Gr of the
beam. There are now acting upon the beam from
the two links the forces
S\ = III IV
in the direction c x b, and
# 4 = IV VI
in the direction go rQ . The resultant S of these two is
given in value and direction by the line III VI of the
force polygon. To determine the position of this re-
action we must draw through the intersection of c x h
and o h g a parallel to III VI Since this intersection
lies beyond the limits of the drawing we can draw a
funicular polygon derived from the force polygon in the
well-known way to determine the position of S. Choose
EXAMPLES. 153
any convenient point for the pole of the force poly-
gon, and draw the polar rays IV, VI, and III,
and construct the corresponding funicular polygon *
a/38. Then the vertex 8 of the polygon is a point of
the resultant $ which must be drawn through 8 parallel
to VI III To check the construction a second funicular
polygon ei 1 /5 1 8 1 has been drawn, the line 88 x giving
the direction of the resultant S of the pressure S\
in the link OB, and the tension &\ in that GrE. If the
construction is accurate 88 x will of course be parallel
to Villi
The other end of the beam acts upon the crank MJ
with a force T in the connecting-rod HJ, the line of
force T coinciding with hi, the common tangent to the
friction circles at H and J, To determine this force T
we must know the direction in which the bearing of the
beam journal A re-acts against it. This re-action passes
through the intersection of T and S, and is tangent to
the friction circle at A. Since this point lies beyond the
limits of the drawing we must employ the construction
shown in Fig. 19, plate II. For this purpose draw a
line through the centre of the journal A perpendicular
to the supposed position of the desired re-action, cutting
the directions of T and S in S and c, and the friction
circle of A in v. Then draw a parallel to this line,
cutting T and S in 8 X and e v and so locate the point v x
that
Oji/| * €^v^ • • o:
V
We then have in v x a point in the direction of the re-
* The principles of the funicular polygon referred to here and in
previous chapters are clearly set forth in Karl von Ott's little book on
Graphic Statics. — Trans,
154 THE GRAPHICAL STATICS OF MECHANISM.
action R which may be drawn through v 1 tangent to
the friction circle at A. To locate the point v 1 so
that the above proportion shall be true we have only
to draw through 8 X and ^ any two parallel lines S 1 8 2 and
e 1 € 2 , make
8-^2 = 81/ and
E l€2 — C] /,
and the intersection of 8 1 € 1 and S 2 *r 2 is the desired point
v v Having thus obtained the direction of the re-action
It we resolve the force
S = IIIVI
in the direction VI VII parallel to /«', and VII III
parallel to v x a, thus getting
T = VIIVI and B= VII III.
If X is the point of contact of the two pitch circles
of the wheels MX and NX by which the driving-force
is transmitted from the engine shaft M to a line of
shafting N we draw first the direction of pressure
between the teeth in the line xx at 75 degrees to the
line of centres MN, and at a distance £ from X. Taking
the weight of the fly-wheel, which acts downward and
increases the journal friction at M, into consideration
we lay off in the force polygon the weight Gr equal to
VI VIII, and have in VII VIII the resultant of con-
necting-rod thrust T and weight Gr in direction and
value. By drawing through o 6 , the intersection of the
forces T and G, a line parallel to VII VIII, we get
the intersection o 7 of their resultant with xx, the line of
EXAMPLES. 155
pressure Z between the teeth of the gear-wheels. The
tangent o 7 m drawn through o 7 to the friction circle of
M gives the re-action of the fly-wheel bearing. We
have finally to resolve the force VII VIII into VIII IX
parallel to xx, and VII IX parallel to # 7 m, thus getting
Z -IX VIII
the force exerted by the teeth of the gear-wheel MX
upon those of NX along the line xx. If r is the lever-
arm NN 1 of this force Z with reference to the shaft iV,
and if this shaft N encounters at the moment under
consideration a resistance whose moment is Zr, the
power of the steam-engine is sufficient to overcome
that resistance. If, on the contrary, the moment of
the resistance to be overcome has any other value Zr x
there will be a moment
Z(r — r x )
either expended in the acceleration of the moving masses
when r — r t is positive, or given out by those masses
if r — r x is negative. The discussion and investigation
ill the case of the saw-grate (Fig. 53, plate VI.) apply
also here.
156 THE GRAPHICAL STATICS OF MECHANISM.
CONCLUDING REMARKS.
The graphical determination of forces and of the
resulting efficiency, as explained and applied in the
foregoing chapters, will not present great difficulties in
any form of mechanism, since in each case it is merely
a question of determining the direction of the forces
involved, and from them drawing the corresponding
force polygon. This method possesses all the general
advantages of graphical determinations over the ana-
lytical methods until now alone employed. It rests
upon the elementary laws of mechanics, does not re-
quire a knowledge of analytics, leads to the desired end
by a quicker and surer route, and has in nearly every
case a sufficient degree of accuracy. To meet the last
requirement it is only necessary to draw the diagrams
to a sufficiently large scale, and in practice it would
be well to employ a scale from six to ten times larger
than that of the diagrams attached to this work. For
the sake of clearness, and in consequence of the reduced
scale of these plates, certain dimensions were chosen
greater than they should be ; for instance, most of the
friction circles will be found to be larger than the cor-
responding co-efficient of friction would give. And
therefore, as before remarked, the numerical results
given for forces and efficiencies in the various examples
were not taken from the figures in the plates, but from
much larger drawings in which all dimensions were cor-
rectly proportioned. That with large but manageable
CONCLUDING REMARKS. 157
drawings sufficient accuracy for practical purposes is
obtained is evident when we consider the uncertainty
there is about the co-efficient of friction itself which
is determined empirically. A co-efficient of friction is
never given with certainty beyond two decimal places,
as a glance over the tables of these co-efficients shows,
and it is safe to assume that in the average case there
is an uncertainty of several per cent. In the light of
these facts how worthless is the determination of forces
carried out to many decimal places, to hundred-thou-
sandths even, as is the case in many analytical deduc-
tions ! At any rate it follows from what has been said,
that with moderately large drawings and with fine lines
the attainable accuracy is all-sufficient. Supposing, for
instance, that an untrained e} r e would permit an error
of one millimetre in laying off distances, the error would
be at the most only one per cent if the shortest line of
force was one decimetre.
In reference to the accuracy of results obtained by
graphic methods we may also remark that in such
determinations there is no temptation to employ certain
approximations which are generally introduced in ana-
lytic calculations to render the formulae less unwieldy,
and, in fact, often have to be employed to render further
progress possible. Many of the preceding examples
correspond to just such cases. Thus in pulleys where
the ropes are not parallel such parallelism has to be
assumed in determining the journal pressure ; in slider-
crank gearing the inclination of the connecting-rod is
neglected, etc. All such assumptions are unnecessary
in graphic methods. In the determination of the force
acting upon the lever of a brake-band the friction of
the journal by which the band is connected to the lever
158 THE GRAPHICAL STATICS OF MECHANISM.
is neglected in analysis, while in the graphic method it
simply amounts to drawing the friction circle of that
journal, and the determination is in no way complicated
by thus taking it into consideration. Even though, as
it may be urged, such small resistances may be safely
neglected as inappreciable, it goes to strengthen the
assertion that the graphic method is more accurate in
such cases than the analytical.
A special advantage of the graphical method is its
great clearness. This is of the highest importance to
the designer or engineer. Take, for instance, the
graphical analysis of a hoisting-gear. It presents to
the eye the position, direction, and intensity of all the
forces at one glance ; and having these it is not difficult
to determine by the laws of graphical statics the mo-
ments at any section of any piece, and from that the
necessary dimensions. All such determinations are
without the scope of the present work, however ; they
will be found in the standard works on machine design
and graphical statics.
Plate I.
Plate
Plate Ml
W~^f
Plate IV.
_D_ Fig. 30 TTi K . 31
Plate V.
Plate VII.
Plate VIII.
*
42
43
45
44
46
47
49
48
50
51