EX LIBRIS SOLOMON C AD Y HOLLISTERPRINCIPLES OF THE MECHANICS OF MACHINERY AND ENGINEERING. BY JULIUS WEISBACH, PROFESSOR OF MECHANICS AND APPLIED-M ATHEMATICS IN THE ROYAL MINING ACADEMY OF FREISURG. FIRST AMERICAN EDITION. EDITED BY WALTER R. JOHNSON, A.M., CIY. & MIN. ENG. FORMERLY PROFESSOR OF MECHANICS AND NATURAL PHILOSOPHY IN THE FRANKLIN INSTI- TUTE, AND OF CHEMISTRY AND NATURAL PHILOSOPHY IN THE MEDICAL DEPARTMENT OF PENNSYLVANIA COLLEGE. AUTHOR OF A REPORT TO THE UNITED STATES NAVY DEPART- MENT ON AMERICAN COALS, &C. &C. IN TWO VOLUMES. ILLUSTRATED WITH ONE THOUSAND ENGrRAVTNG-S ON WOOD. YOL. I. THEORETICAL MECHANICS. PHILADELPHIA: LEA AND BLANCHARD. 1848.TA 35° U 42- / S4 S / ( Entered according to the Act of Congress, in the year 1848, by LEA AND BLANCHARD, in the Clerk’s Office of the District Court of the Eastem District of Pennsylvania. PHILADELPHIA .* T. K. AND P. G. C0LLINS, PRINTERS.TABLE OF CONTENTS. PrEFACE TO THE AMERICAN EdITION.......................... AUTH0R’S PREFACE . . ' . Comparative Tables of English, French, and German Measures AND ............................................... SECTION I. PHORONOMY; OR THE PURE MATHEMATICAL SCIENCE OF MOTION. CHAPTER I. § 1—25. Simple motion CHAPTER II. § 26—43. Compound motion SECTION II. MECHANICS IN THE PHYSICAL SCIENCE OF MOTION IN GENERAL. CHAPTER I. § 44—63. Fundamental ideas and fundamental laws of mechanics CHAPTER II. § 64—82. The mechanics of material point . PAGE , vii xi , xv 25 37 50 58IV CONTENTS. SECTION III. STATICS OF RIGID BODIES. CHAPTER I. § 83—97. General laws of the statics of rigid bodies . CHAPTER II. § 98—120. Centre of gravity ....... CHAPTER III. § 121—137. Equilibrium of bodies rigidly connected and supported CHAPTER IV. § 138—153. Equilibrium in funicular machines .... CHAPTER V. § 154—180. On the resistances of friction and rigidity CHAPTER VI. §181—211. Elasticity and rigidity............... SECTION IV. DYNAMICS OF RIGID BODIES. CHAPTER I. § 212—229. Doctrine of the moment of inertia . CHAPTER II. § 230—238. Centrifugal foree ....... CHAPTER III. § 239—252. Of the action of gravity on motions along constrained paths ........................................ PAGE 76 88 109 127 145 182 225 245 259CONTENTS. y CHAPTER IV. § 253—271. The doctrine of impact . SECTION V. STATICS OF FLUID BODIES. CHAPTER I. § 272—283. On the equilibrium and pressure of water in vessels . CHAPTER II. § 284—294. On the equilibrium of water with other bodies CHAPTER III. § 295—302. On the equilibrium and pressure of air . SECTION VI. DYNAMICS OF FLUID BODIES. CHAPTER I. § 303—311. The general laws of the efflux of water from vessels CHAPTER II. § 312—322. On the contraction of the fluid vein by the efflux of water througli orifices in a thin piate CHAPTER III. § 323—336. On the efflux of water through tubes CHAPTER IV. § 337—344. On the resistances of water in passing through con- traction .................................... FAGE 228 309 322 338 350 364 382 401 1*VI CONTENTS. CHAPTER V. PAGE § 345—353 .On the efflux of water under variable pressure . 415 CHAPTER VI. § 354—359. On the efflux of air from vessels and tubes . .418 CHAPTER VII. § 360—371. On the motion of water in canals and rivers . .438 CHAPTER VIII. § 372—382. Hydrometry or the doctrine of the measurement of water ........ 453 CHAPTER IX. § 383—394. On the impulse and resistance of fluids . . 466PREFACE BY THE AMERICAN EDITOR. The want of a good Standard work on those branchesof Mechanics which pertain to Machinery and Engineering, has long been felt as a deficiency in our means of instruction in the higher institutions, as well as in the office of the practical engineer. Any one who has had occasion to direct the study of young men preparing for the duties of the profession, must, in the course of his practice, have encountered the difficulty, on the one hand, of inducing students to extend their researches affer the principies of their profession into the intricacies of the higher calculus, and, on the other hand, of contenting himself with the very limited discussion of principies, found in many works which have been hitherto employed to give the desired basis for pro- fessional knowledge. The difficulty has, of necessity, been met by the expedient of recommending detached parts of several distinet works. The treatises on Natural Philosophy were a very inadequate substi- tute for a practical work on the Mechanics of Engineering. Many of the topies requisite to be handled in a work of this nature, are, in the books now in use, either wholly omitted, or so slightly passed over, as to be of little Service beyond the general statement of certain laws, and perhaps a few illustrations, indifferently applied, and fumishing but feeble helps to the exact understanding of the laws applicable to the subject. In presenting the claims of Prof. Weisbach’s treatise to the con- sideration of Engineers and Machinists of the United States, we can- not do better than adopt the remarks of the English translator. uFrom the well earned reputation of Professor Weisbach as a teacher and original investigator, from the interest which attaches itself, espeeially at the present moment, to ali that pertains to Me-viii PREFACE BY THE AMERICAN EDITOR. chanics or Engineering, from the able manner in which he has treated both the theoretical and practical portion of his subject, and from the variety and abundance of examples which illustrate the principies and formulae, it is presumed that a translation of his excellent work may not be unacceptable to the English reader. ‘The Mechanics of Ma- chinery and Engineering’ has met with deserved favor in Germany, and is not unknown to many of the profession in our own country. The aim and objects of the undertaking are fully explained by the Author in his Preface; and it is hoped that the same end may be ob- tained here, by affording, through the medium of a translation, valua- ble aid to the acquirement of professional knowledge.” In the course of this volume, six great divisions of the Science will be found to be embraced. Of these, the pure mathematical Science of motion, including simple and compound, rectilinear and curvilinear motions, and the laws and expressions relating to the free descent of bodies, have, of course, claimed the first attention. The Physical Science of Motion has next been treated; then the division of forces, and their meas ure,—the density and the state of aggregation of bodies have been duly presented. In the third section, the statics of rigid bodies claim a careful investigation, and are fol- lowed, in the fourth, by the dynamics of the same class of bodies; while, in the fifth, the statics, and in the sixth, the dynamics of fluid bodies are treated with marked ability; and in several parts with an originality and freshness which indicate that the author is here in his peculiar domain—the scene of his chosen labors, and that on which his energies, as an original inquirer, have been very successfully employed. In this branch of his subject, Prof. Weisbach has included the discussion of the mechanics of elastic as well as of non-elastic fluids. We are indebted to him for several new formulae and valuable tables of constants. As students of this work are, from the wide diflusion of French and German science, likely to meet, in other treatises, frequent references to French and German weights and measures, it has been deemed useful not only to present, as in the English edition, the annexed com- parative table of weights and measures, but also to retain from thePREFACE BY THE AMERICAN EDITOR. IX original a few examples eompletely worked out as exercises in com- putation under the different systems. The same has been done in respect to temperatures measured by the Centigrade thermometer. The facility aflorded by the decimal divisions of ali the French weights and measures strongly recommend them both for scientific and practical purposes. The additions of the American Editor are generally thrown into notes suitably designated, or, when embraced in the text, are usually enclosed in brackets. A number of new illusfrations have been added, particularly such as relate to the principies of the it toggle joint,” of rolling and dragging friction, of carriage wheels, and of float- ing docks. A table of co-efficients for the flow of water through wiers, the resuit of some recent experiments, not heretofore published, 'will be found in the sixth section. It has been found necessary to make a very considerable number of corrections of the English copy, especially in the working of examples by the translator, where a departure from the original had been made, on account of a change of weights and measures. Should some few errors stili have escaped notice, it will be no matter of surprise, and of the less importance, perhaps, since the principle of computation is in all cases pointed out, the steps severally indicated, and the general law, therefore, readily applied to each particular illustration. A comparison with the original work in German has been made, and the great care exercised by the learned author in its preparation and publication has been rendered abundantly evident. Justice to the artisans engaged in getting out the present edition requires us to say, that the illustrations here presented will not suffer in comparison with those of either the German or the English edition. In some instances they are decidedly superior to either. Phiudeiphia, March, 1848. Note. In regard to a matter of minor importance, that of the no ation used in this work, a few words raay be proper, in order to apprize the reader, in advance, that the use of the simple point (.) as the sign of multiplication has required the adoption of the commaX PREFACE BY THE AMERICAN EDITOR. (,) to mark the decimal division of numbers. Should any departures from this application be observed, they will be such as are readily understood. In using the Italian (a), and theGreek alpha (a), thelatter ismost commonly applied to ares and angles, and the former to lines orother quantities. This distinction, which is important to avoid confusion, and which is strictly observed in the original, has been sometimes lost sight of in the English edition.AUTHOK’S PREFACE. In giving to the public the First Volume of my Elementary Treatise on Mechanics, for Engineers and Machinists, I feel some degree of diffidence. Although I am conscious that, in composing the book, I have proceeded with the greatest care and circumspection; I am, nevertheless, apprehensive that it cannot satisfy ali, since the views, wishes and requirements of different individuals differ so widely. One reader may probably consider this or that chapter too long and too minutely treated, which another may find too short. Some will miss higher Science in the treatment of certain subjects, which others would have wished to see treated in a stili more popular manner. But many years of study, much experience in teaching, and con- tinued observation, have led me to the method, according to which I have composed the present work, and I consider it the most appro- priate for the intended purpose. My chief aim in writing this work was the attainment of the greatest simplicity in enunciation and proof; and with this to give the demon- stration of ali problems, important in their practical application, by the lower mathematics only. If we consider the great variety of knowledge which the Engineer and Machinist have to acquire, who wish to do credit to their profession, we, their instructors, should make it our duty to render well-grounded study of Science easy by simplicity of explanation, by the use of only the best known and easiest auxiliary Sciences, and by eschewing everything that is un- necessary. I have therefore avoided, in the present work, the use of the dif- ferential and integral calculus; for although there exist now more frequent opportunities, than formerly, for learning these methods, it is stili unquestionable, that without constant practice, the readiness of using them is very soon lost; and that there are, consequently, many excellent practical men who have entirely forgotten how to apply them. Some popular authors give the results of the more difficult problems without proofs. I cannot approve of this, and have pre-xu AUTHORS PREFACE. ferred to give the proofs of practically important problems in an elementary way, although this may sometimes appear rather long and tedious. There are, therefore, but few formulae in the work unae- companied by their derivation. A general acquaintance with some doctrines of natural philosophy, but especially an intimate knowledge of pure elementary mathematies, are of course necessary for the understanding of the present work. I have taken especial pains to preserve the right medium between generalizing and individualizing; for, although I am fully aware of the advantages of generalization, I must stili adhere to the opinion, that in an elementary work, too much generalizing is to be avoided. Simple cases are, in practice, of more frequent occurrence than com- plicated. It is also undeniable, that in treating a general case, the knowledge which might be gained by the treatment of a specific case, is frequently lost, and that it is not unfrequently easier to deduce the compound from the simple, than to eliminate the special from the general. The “Mechanics of Engineering and Machinery” must not be mistaken for a work on the construction of machines, but it is to be considered merely as an introduction to or preparatory Science for this. This Treatise of Mechanics is to stand in the same relation to the construction of machines, as descriptive geometry stands to the drawing of machines. After mechanics and descriptive geometry have been learned, the instructions on the construction and the draw- ing of machines may with advantage be united in one course. The propriety of dividing the Mechanics of Engineering into a theoretical and practical part, may perhaps be doubted. If it be borne in mind, that this work is to furnish instructions on all me- chanical relations, in architecture and the Science of machines, the utility, or rather necessity of this division must be apparent. In order to form a complete opinion of a building or machine, the most various doctrines of mechanics {i. e. the doctrines of friction, strength of materials, inertia, impact, efflux, &c.), must be taken into con- sideration; the material for the study of the mechanics of a building or machine must, therefore, be collected from all parts of mechanics. Now, as it is much more useful practically to be able to study the doctrines relative to every individual machine in connection, than to have to collect them from all departments of mechanical Science, the utility of the adopted division seems to be beyond all doubt. Having practical application always in view, I have endeavored to illustrate each doctrine, as much as possible, with appropriateAUTHOR’S PREFACE. XIII examples, taken from practice ; and I can truly assert, Ibat this book excels most works of a similar nature, in the great number and appropriate choice of worked out examples. The large number of carefully executed figures will, no doubt, likewise assist the attain- ment of the above-mentioned object; and I cannot omit here to express how greatly satisfied I feel with the manner in which the publishers have performed their part in the getting up of the book. Finally, it is necessary to point out to the reader of the work, that he will find much that is new and peculiar to the author. Passing over smaller matters, which occur in almost every chapter, I will mention only the following more comprehensive subjects. A general and easy method of determining the centre of gravity of plane sur- faces and even-sided polyhedra, will be found in the paragraphs, 107, 112, and 113; an approximate formula for the catenary in the paragraphs 147 and 148; supplements to the theory of the friction of axes in the paragraphs 167, 168, 169, 172, and 173. The doctrine of impact especially has received essential additions by- the para- graphs 262 and 263, since the impact of imperfectly elastic bodies has been hitherto insufficiently treated ; and the case, where a per- fectly elastic body comes in contact with a partially elastic body, has been passed over entirely. The largest number of additions, and some entirely new laws will be found in the hydraulics, the author having made this branch of the Sciences his particular study during a number of years. The laws of the incomplete contraction of the fluid vein, discovered by the author, appear here for the first time in a course of mechanics. The chief results of the author’s experiments on the flow of water through slides, cocks, clacks and valves, are likewise given, as well as his principal observations made during his latest experiments on the flow of water through prolonged oblique tubes, and through angular, curved, and long straight tubes, though he has not yet been able to publish the third number of his “Unter- suchungen im Gebiete der Mechanik und Hydraulik,” which is to contain the results of these experiments. The chapters on flowing water and the gauging and impulse of water have likewise received additional matter from the author. But now, after completion of the first volume, I cannot refrain from wishing that several subjects had been treated in a different manner, a though I have not been able to disco ver any essential errors or im- perfections in it. Should the reader here and there find omissions, I must refer him to the second volume, which will contain supplements, 2xit AUTHOR’S PREFACE. most of which have been intentionally reserved for that volume, as has been mentioned in several passages of that now published. It will give me great satisfaction and pleasure, if the pnrpose which I had in view in writingthis work has been attained in some measure. I wished to supply the practical man with useful advice, the in- structor in mechanics with a guide for teaching, and the young engineer and machinist with a welcome auxiliary for the acquirement of the science of mechanics. JULIUS WEISBACH. Fusurae, Marchy 1846.'COMPARATIVE TABLES OF ENGLISH, FRENCH, AND GERMAN MEASURES AND WEIGHTS. I. Measures of length. IV. Measures of capacity. V. Weights. English (and Rus- sian) foot French metre. Paris foot Prussian, Danish, and Rhenish English gallon, =277.27384 cubic inches. French litre, = .001 Prussian quart, = 64 cubic inches. English French. Prussian, Hanoverian, Brunswick, and Hessian foot. cubic metre. pound, avoirdupois. Livre, poids Kilogram. de mare. pound. 1 .3047945 .9382928 .9711361 3.280899 1.065765 1.029722 1 .3248394 .3138535 3.078444 1 0.9661806 3.186199 1.035003 1 1 .2200967 .2520176 4.543458 1 1.145031 3.967977 0.8733386 1 1 2.204597 1.079163 .4535976 1 .4895058 .9266439 2.042879 1 .9698245 2.138072 1.046599 II. Square meafeure. Bushel, = 8 gallon s. Hectolitre, = 1001itres. Scheffel, 1.031114 .4677110 .9554758 1 Square foot Sq. metre. Square foot. Square foot = 3072 cubic inches. Hundred weight. Quintal me- trique. = 100 kilo. Quintal, =100 livres Centner, 1 10.76430 .09289969 1 0.8803934 9.476817 0.9431053 10.15187 1 2.751208 0.3634767 1 .6613296 1.819455 = 112 lbs. avd. (old mea- sure.) = 110 lbs. 1.135856 1.060327 .1055207 .09850405 1 0.9335049 1.071232 1 1.512105 0.5496150 1 1 1.968390 0.5080293 1 1.037841 2.042877 0.9874577 1.943702 IIL Cubic measure. 0.9635386 1.012702 0.4095058 0.5144821 1 1.051023 0.9514536 1 Cubic foot Cubic metre. Cubic foot. Cubic foot. L. G. 1 35.31658 1.210556 1.091842 .02831531 1 .03427727 0.9019342 0.8260668 29.17385 1 .03091584 .9158836 32.34587 1.108728 1 The United States Standard gallon has a capacity of 231 cubic inches, and contains 8.3388822 avoirdupois pounds, or 58372.1754 Troy grains of distilled water at 39°.83 Fah., the barometer being at 30 inches. The U. S. Standard bushel contains 2150.42 cubic inches, and 77.627413 pounds avoirdupois of distilled water at 39°.83 Fah.—Am. Ed. COMPARATIVE TABLES, XVPlease note: These pages were scanned as they appear in the original (no pages were skipped in the scanning process)PRINCIPIES OF THE MECHANICS OF MACHINES,Y AND ENGINEERING. SECTION I. PHORONOMY; OR, THE PURE MATHEMATICAE SCIENCE OF MOTION. CHAPTER I. SIMPLE MOTION. § 1. Rest and Motion.—Every body occupies a certain position in space; a body is at rest when it does not change its position; and is in motion, on the other hand, when it successively passes from one position into others. The rest and motion of a body are either abso- lute or relative, according as we refer its position to a space which itself is at rest or in motion, or considered to be in either state. Upon the earth there is no rest, for ali bodies upon the earth share in its motion about the sun, and about its own axis; but if we suppose the earth to be at rest, then ali those terrestrial bodies are at rest which, with reference to the earth, do not change their position. § 2. Kinds of Motion.—The continuat succession of positions which a body in its motion gradually occupies, forms a space which is called the trajectory, or path of the moving body. The path of a moving point is a line; that of a geometrica! body, is another body; but by this latter is generally understood that line which a certain point of the body, viz. the centre, describes in its motion. When the path is a straight line, the motion is rectilinear; when a curved line, the motion is curvilinear. With reference to time, motion is uniform or variable.26 UNIFORM MOTION. § 3. A motion is uniform, when equal spaces are described by it in equal and arbitrarily small times; and variable, when this equality does not take place. When the spaces, described in equal times, increase continuously with the time, a variable motion is called— accelerated; and when the spaces decrease,—retarded. Periodic differs from uniform motion in this, that equal spaces are described within certain intervals only, which are called periods. The apparent diurnal revolution of the fixed stars, and the progres- sive motion of the hands of a watch are instances of uniform motion. Falling and upwardly projected bodies, the sinking of the surface of water by its flow from vessels are instances of variable motion. The oscillations of a pendulum, the play of the piston of a steam engine, &c., are illustrations of periodic motion. § 4. Uniform Motion.—Velocity is the rapidity or magnitude of a motion. The greater the space is which a body describes in a given time, the more rapid is its motion, and the greater is its velocity. In uniform motion, the velocity is invariable; and in a variable one, it changes at every instant. The measure of the velocity at any de- terminate point of time, is the path which a body actually describes, or would describe in a unit of time or second, if from that moment the motion were to become uniform, and the velocity to remain invariable. In general this measure is called simply—velocity. § 5. When a body describes the path a at each particle of time, and a second consists of very many (n) such particles; the path dur- ing one second is the velocity, or rather the measure of th£ velocity: c = n. « 25 + 5 X 2.5 « 25 + 12.5 tm 37.5 feet, &c.; proeeeding from Ute point uniformly, it will pass over 37.5 feet in every second.—3. A with a 30 feet velocity is so retarded, that in each second it loses 3.5 feet of ve vi_ acceleration is — 3.5; its velocity, therefore, after 6 seconds is ® oa 30-* • 30 — 21 = 9 feet § 10. Uniformly accelerated Motion.—The velocity of every motion28 UNIFORMLY ACCELERATED MOTION. raay be regarded as invariable within a small particle of time t. We may, therefore, put the space described in such time <* «= v . t, and we obtain the whole space in the finite time t, by the measurement of these small spaces. Now for ali these small spaces, the time * is one and the same; this sum, therefore, may be put equal to the pro- duct of these particles of time, and the sum of the velocities corre- sponding to equal intervals. In uniformly accelerated motion, the sum (0 + v) of the velocities, in the first and last moment, is as great as the sum pt + (v — p t) of the velocities in the second and last but one; also, equal to the sum 2 p t + (v — 2 p t) of the velocities in the third and last but two; and this sum is equal to the terminal velocity v. Here, therefore, the sum of ali the velocities is equal to the product of the teripinal velocity r, and half the number of ali the particles of time. The space described is the product ~ of the terminal velocity v, and half the number and magnitude of the particles. Now the magnitude (t) of such a particle, multiplied by the number, gives the time t; the space, therefore, described in the time tfwith a uniformly accelerated motion is s = £ The space, therefore, described in uniformly accelerated motion is as in uniform motion when, in the latter case, its velocity is half as great as the terminal velocity of the former. Example l. If a bodyin 10 seconds has acquired a velocity v by uniformly accelerated motion of 26 feet, the space described in that time is *—. ^ X 10— feet—g. A 2 carriage which, in its accelerated motion, goes over 25 feet in 2J seconds, proceeds at 2x25 50x4 the end with a velocity v = ——— = ——:— = 22.22. J 2.25 9 § 11. The two fundamental formulae of uniformly accelerated mo- tion : v t I. v = pt and II. s * -jp which express that the velocity is a product of the acceleration and the time; and the space, half the velocity and the time ; include two other principal formulae which are obtained, if from both equations v be eliminated once, and t twice. It follows that: p/2 p* III. s = andiv. s = s— 2 2p From this, the space described is a product of half the acceleration, and the square of the time; and it is also the quotient of the square ve*oc^y ky twice the acceleration. Ihese four formulae give, by inversion, after one or other of the magnitudes contained ha ve been separated, eight other formulae. Example 1. A body moving with an acceleration of 15,625 feet, describes in secondUNIFORMLY ACCELERATED MOTION. 29 15.625 X (1.5)« ^ x 17.578 feet—2. A body transported with 2 ’ 8 a space 2 8 . . ___________________________________________________ an acceleration p = 4.5 into a velocity v = 16.5 feet, has described a space t ^-51’= 30.25 feet 2.4,5 § 12. By a comparison of two uniformly accelerated motions, we arrive at the following: , The velocities are v = pt and vx = pjn the spaces on the other hand are $ = and ** 5 from this it follows: «i M M* ®A If we put t, = f, we haye — = — = — ;the spaces described are *i *i A . ■ . to each other as the terminal velocities; or, as the acceleration». If further, we take p, = p, it gives — — — and — = tj = ~i» 80 that, in like accelerations, and also in one and the same uniformly accelerated motion, the terminal velocities are proportional to the times and the spaces described to the squares of the times, as also to the squares of the terminal velocities. Further, if = v gives — == —, and — = in equal velocities Pi * si h . . the accelerations are inversely, and the spaces directly proportional to the times. Lastly, sx = s gives ?— = ^; with equal spaces, the ac* celerations are inversely as the squares of the times, and directly as the squares of the terminal velocities. § 13. For a uniformly accelerated motion commencing with a velor city (c) we have § 9: I. v=c+pt, ... and as the space c t belongs to the invariable velocity (c), and the space ^ to the acceleration p: II. s pf* “ + T- If we eliminate p from both equations, we have: III. s C + V t, and suhstituting the value of t, IV. ss ExampU 1. A body propelled with an initial velocity e « 3 feet,^and with an ration p = 5 feet, describes in 7 seconds, a space # * 3.7+ 5. — sas 21 + 122.5 - - 8* t>2—c230 FREE DESCENT OF BODIES. 143,5 feet.—2. Another body which in 3 minutes = 180 seconds, changes its velocity lrom feet into 7J feet, performs in this time a distance 2.5+ 7.5 j 8o —. 900 feet. § 14. For a uniformly retarded motion with an initial velopity c, these formulae are applicable: I. V SB» C--p t, II. mc + v . . s= —+-. t, IV. 5 = 2 C2— V2 2p they are derived from the former §, when p is made negative. Whilst in uniformly accelerated motion, the velocity increases without limit, in a uniformly retarded one, the velocity at a certain point of time becomes null, and afterwards negative, i. e. it goes on in an inverse direction. If in the first formula we put v = 0,pt = c, the time at which the velocity becomes null is, t== —; if we substitute this value of t in p the second equation, we have the space which the body has described at the point of time =—. 2p c c2 If the time be greater than -, the space is less than — ; if it be 2c • s —, the space becomes null, and the body returns to the point P 2c from which it set out. If the time be greater than —,then s becomes . Pm negative, and the body is on the opposite side of its initial point. Exampk. A body which rolls up an inclined plane with an initial velocity of 40 ft, by 40 40* which it suffers a retardation of 8 feet per second, ascends only = 5 seconds and —- = 100 feet in height, then rolls back and returns affer 10 seconds with a velocity of 40 feet to its initial point; and after 12 seconds, arrives at a distance 40 X 12 —4 X (12)* s 96 feet below this point if the plane extend itself backwards. § 15. Free Descent of Bodies.—The free or vertical descent of bodies in vacuo, offers the most important example of uniformly ac- celerated motion. The acceleration of this motion brought about by gravity is designated by the letter g, and has the mean value of: 9,81 metres 30,20 Paris feet. 32,22 English feet. 31,03 Vienna feet. 31,25 Prussian feet. If either of these values of g be substituted in the formula: and * - >/2^7,FREE DESCENT OF BODIES. 31 ali questions, with reference to the free descent of bodies, may be answered. For the English measure: v wm 32,2 . t = 8,02 v/ - 16,1. t* — .0155 and t «=» 0,031 v = 0,249 \/ s. Exanrple 1. A body acquires in its free descent of 4 seconds a velocity t; as 32.2 X 4 * 128.8 feet, and describes in this time a space * = 15.625 X 4* ss 250 feet.—2. A body falling from a height s sss 9 feet, has a velocity v = 8.02 y/ 9 = 24.06 feet—3. A body projected vertically with a velocity of 10 feet ascends to a height t ;= 0.016 X 10* = 1.6 feet, and requires for it a time t = 0.031 X 10 = 0.3, or about one-third of a second. § 16. The following table will show the relations of the motion to the time in the free descent of bodies. Time in seconds. 0 1 2 3 4 5 6 7 8 9 10 Velocity. 0 Ig 2g 3g *g A? 6g 1g 8g 9-? lpg Space. 0 1 £. aI. 9&- 16^ 2bi- 36il 49-£ 64£. 8lJjl00f Difference. 0 2 lJL 2 3il 2 5— 2 7-H 2 9^ 2 ll£- 2 13L 2 151- 2 nM- 2 19§ 2 2 2 2 2 2 2 2 2 ■2 The last horizontal column of this table gives the spaces which the freely falling body describes in the single seconds. We see that these spaces are to each other as the odd numbers 1, 3, 5, 7, &c., whilst the times and velocities are as the natural numbers 1, 2, 3, 4, &c., and the spaces fallen through as their squares 1,4, 9, 16, &c. For example, the velocity after six seconds, is 6g = 193,2 feet, that is, the body would, if it proceeded from this time uniformly upon an horizontal plane, offering no impediment, pass over in each second a space 6gz=s 193,2 feet. This space it describes in the course of the following and seventh second, but not in reality, for according to the last column it amounts to 13.^« 13 x 16,1 =* 209,3 feet, in the A eighth second it is 15. ^ — 15.16,1 s 241 feet, &c. Remarks.—Many writers designate the space of 16 feet, which a body freely desoend- ing will describe in one second, by g, and term it properly the acceleration of gravity. They have then for the free descent of bodies, the following formula: r = 2^ = 2 y/gt, -"-r '-¥-v/T Iliis custom, which is met with in Germany only, is disappearing by degre«^, ani m consequence of its being frequently misxmderstood, and the many mistakes w thereftom, this is much to be desired.32 FREE DESCENT OF BODIES. § 17. If the free deseent of a body go on 'with a certain initial velo- city (c) the formulae are of the following kind:___ »=c + fi'<=c + 32,21; also» = x/ c» + 2gs = ^£,+6M.s; /3 »J-C2 s = ct+g-=ct-\- 16.1t2, alsos 0,0155 (»2—c2). 2 , - 2 g If, on the other hand, the body be projected vertically to a height with the velocity c, then: _______________________________ » = c—gt = c— 32,21; also» = c* — 2gs= hy tand * hy *>*+* -«(*+now (t+tf - p + * 9or because t is small =* t% + 2 t *9 it therefore follows S + * - a? + lastly, t, „ ± = 2 at. By (t + WC ^ea™ ^r0m Ae last formula v + * = 2 a x ' + 2 a t9 so that * = 2a t and the acceleration p = t ^ a' kave, therefore, in this way found from the formulae for the spaces, formulae for the velocity and acceleration. § 20. The velocity c » — differs from the velocity — of an 'I™' and j? given when the space, which in a certain time itsplf *a m°tion is described, is divided by the time thaf 1S.'? , ^ mean velocity, and may be also regarded as in a oclty which a must ^ave *n or(ler to describe uniformly vana)^17611^1111^ ^ a 6*ven#sPace which, in reality, is described y. fco, for example, in uniformly variable motion9 the velocity is equal to half the sum of the initial and terminal velocities: for, according to § 13, the space is equal to this sum /c -j multi- ^ W ^ **me W* hile a handle turns uniformly in a circle, the load attached to it,34 GRAPHICAL REPRESENTATION. the piston of an air or water-pump, for instance, moves variably up and down; the velocity of this at its lowest and highest point is at a minimum, viz., null; at half the height a maximum, viz., equal to the velocity of the handle. In half a revolution, the mean velocity equals the whole height of ascent, i. e., the diameter of the circle which the handle describes, divided by the time of half a revolution. The diameter = 2r and the time = t, then the mean velocity of the 2 7* load = y The handle in this time describes the semicircle * r; its velocity, therefore, y, and consequently the mean velocity of the load = — = g = 0,6366 times as great as the invariable velo- city of the handle. § 21. Graphical Representation.—The laws of motion found above may be expressed by geometrical figures, or, as it is said, graphically represented. Graphical representations especially facilitate the con- ception, sustain the thoughts, prevent mistakes, and serve not unfre- quently for the discovery of a quantity, and on that account are of great use in mechanics. In uniform motion the space (s) is the product (ct) of the velocity and the time, and in geometry the area of a Fig* !• rectangular figure is the product of the N____________c height and base. We can, therefore, repre- \ sent the space (5) uniformly described by a rectangle AB CD> Fig. 1, whose base A B is the time (t) and whose height (AD = B C) is the velocity (c), provided that the time A Jf be expressed in the same unit of length as the velocity, and that the second of time and the foot be represented by one and the same line. § 22. Whilst, in uniform motion, the velocity (MN) at any other time (A M) of the motion is one and the same, it differs at every in- stant in a variable one; this motion, there- fore, can only be represented by a quadri- lateral figure A B C D, Fig. 2, which has A B the time for base, and for the other limits, three lines A D, B C, C D, of which the first two are equal to the initial and ter- minal velocities, and the last is determined by the extremity (J\f) of the different velo- cities at the intervals (M). The line C D is either straight or curved, according to the different kinds of varia- ble motion from the commencement, ascending or descending, or lastly concave or convex towards the base. But in every case the area of this figure must be put equal to the variably described space (s); for each area of space A B C Dy Fig. 3, may be considered as decom- Fig. 2. V CTHE MEAN VELOCITT. 35 posable into many small rectangular strips, Fig. 3. like M 0 P J\f, of which each is a product of a part (M 0) of the base, and its corre- sponding height (M N or O P), and the spaces described in a certain time com- posed of particles of which each is a pro- duct of that particle and its corresponding velocity. § 23. In uniformly variable motion, the increase or diminution (v—c) of the velocity (=pt, § 13) is propor- tional to the time. If in the Figures 4 and 5, the line DEbe drawn Jd o B Fig. 4. Fig. 5. parallel to the base A B> and B E and Jlf O = to the initial velocity A D be cut off frpm the lines M N and B C> there remain the lines C E and J\T O for the increase or diminution of the velocity, for which from the above we have the proportion NO: C E = D O : D E. Such a proportion requires that N as well as each point of the line C D lie in the straight line connecting C and D, and also that the line C D limiting the different velocities (M N) be a right line. In consequence of this, the uniformly accelerated and uniformly retarded space described may be represented by the area of a trape- zium A B C D9 which has for the height A B> the time (t), and for the parallel bases the initial and terminal velocities A D and B C. The formula of § 13, s — C—V. t is in perfect accordance with this. 2 In uniformly accelerated motion, the fourth side D C ascends from its initial point, and in uniformly retarded motion descends. In a uni- formly accelerated motion beginning with a velocity null, the trape- zium becomes a triangle whose area is J B C x A B = J ct. § 24. The mean velocity of a variable motion is the quotient of the space divided by the time; multiplied, there- fore by the time, it gives as a product the trajectory, and consequently may be also eonsidered as the height A F = B E of the parallelogram A B E F Fig. 6, which has the time (tf) for the base A JB, and an area equal to the four-sided figure A B C N D which measures the trajectory or space Fig. 6. N36 THE MEAN VELOCITY. passed through. The mean velocity is, therefore, likewise obtained by transforming the four-sided figure A B C J\T D into a parallelo- gram A B E F of the same length. Its determination is of import- ance, particularly in periodic motions,which occur in nearly all machines. The law for these motions is represented by a curved line C DE F GHKy Fig. 7. If the line L M running parallel Fig. 7. M-------f------^-------------&-------------2* with A B cuts off the same space as the curved line, and is, as it were, the axis round which C D E F. . . . coils itself, then the distance A L = B M between the two parallel lines A B and L M is the mean velocity of the periodic motion, whilst A C, 0 E> B K> &c., is the maximum, and N D,P F, &c., the minimum velocity of a period A O, 0 Q, Q B9 &c. § 25. The acceleration also, or the increase of velocity during a second of time may be easily shown in the figure. In the case of uniformly variable motion it remains unchangeable; it is hence the difference P Q, Figs. 8 and 9, between two velocities O P and M JV*, Fig. 8. Fig. 9. the one of which appertains to a longer time by one second (M O) than the other. If the motion is not uniformly variable, and the line of velocity C D therefore a curve, then for each second of time (M) the acceleration varies, and is, consequently, not the real difference P Q between the two velocities O P and M N = O Q Figs. 10 and 11; but it is the increase R Q of the velocity M JV*, which would occur if commencing at the moment M the motion became uniformly accele- rated, and the curved line of velocity N C passed into the straight line J\T E. Now the tangent or line of contact JV* E, is that straight line in the direction of which a curve D JV* proceeds, when from a certam point (JV*) it ceases to change its direction; hence the new c?^nf^es w^h the tangent, and the perpendicular line O R which cuts it, is accordingly the velocity which would take placeCOMPOUND MOTION. 37 after the lapse of a second, supposing the motion to have become uni- formly accelerated from the commenceinent of that period, and lastly, Fig. 10. C/ Fig. 11. D. A____I_L M O JL_ ir~o the difference R Q between this velocity and the primary velocity (M N) is the acceleration for that moment which is determined by the point M in the time line A B. CHAPTER II. COMPOUND MOTION. § 26. Compound Motion.—One and the same body may at the same time have two or more motions; every (relative) motion con- sists of the motion within a certain space, and of the motion of this space within, or in relation to, a second space. Each point upon the surface of the earth has thus two motions, for it revolves daily once round the axis of the earth, and simultaneously with the earth once yearly round the sun. A person walking on board a ship has two motions in relation to the shore, his own motion and that of the water; water flowing from a hole in the bottom or side of a vessel, whilst the latter is moving along in a carriage, has two motions, the motion from the vessel and the motion with the vessel, &c. Hence we distinguish simple and compound motion, Those recti- linear motions are called simple, of which other rectilinear or curvili- near motions, consequently called compound, are made up, or may be imagined to be made up. The combination of several simple motions to form one single mo- tion, and the resolution of a compound motion into several simple motions, will be treated of in the sequel. § 27. When simple motions occur in the direction of one and the38 PARALLELOGRAM OF THE VELOCITIES. same straight line, their sum or difference gives the resulting com- pound motion, the former, when the motions take place in the same direction ; the latter, when their directions are opposite. The truth of this axiom becomes directly obvious, when the contemporary spaces of the simple motions are United into one. The contemporary spaces cx t and c2 t correspond with the uniform motions and their velocities cx and c2; if these motions go on in the same direction, then after t seconds the space becomes s = cx t + c2t = (cx + c2) ^ an^ conse- quently the resulting velocity with which the compound motion pro- ceeds is the sum of the velocities of the simple motions. When the directions of both motions are opposite, then s = cx t — c2t == (cx — c2) t, here, therefore, the resulting velocity is equal to the difference of the simple velocities. Example 1. To a person moving with a velocity of four feet upon the deck of a ship, in the same direction with the motion of the ship itself, which has a velocity of six feet, the objects on the shore appear to pass by with a velocity of 4 -|- 6 = 10 feet.—2. The water which flows from the lateral opening of a vessel with a velocity of 25 feet, whilst the vessel containing it is moved in an opposite direction with a velocity of 10 feet, has, in relation to the other objects at rest, only a velocity of 25 — 10 = 15 feet. § 28. The same relations obtain wTith variable motions. If one and the same body have, in addition to the primary velocities, cx and c2 the constant accelerations px and p2, then the corresponding spaces t2 t2 . are cx tv c2 tv px -, p2 -, if the velocities and the accelerations are m 2 2 the same direction, the whole space corresponding to these simple motions, will be: 5 = (C, + Ct) t + (px + P2) Y’ 12 If cx + c2 = c and px + p2 = P> then obtain s = ct + p —, and 2 it follows, consequently, that not only the velocity of the resulting or compound motion is made up of the sum of the simple velocities, but that also the sum of the accelerations of the simple motions gives the resulting acceleration. Example. A magnet falis more quickly to the earth than another body, when a mass of iron is immediately below it. The acceleration which the magnet experiences, in eonsequence of this iron, may be considered invariable when the height from which it falis is small and the mass of iron very considerable, viz., an extensive layer of mag- netic iron ore. If this acceleration were 5 feet, then the magnet would fall with an increased velocity of 31,25 5 = 36,25 feet in the first second, therefore it would fall 18$ feet instead of 15f feet. § 29. Parallelogram of the Velocities.—If a body has at the same time two motions differing from each other in direction, it will assume a medium direction between them; and if these motions are of differ- kinds, viz«, the one, uniform, and the other uniformly increasing, then the direction will vary in every part of the motion, and the mo- tion itself become curvilinear. The place O, Fig. 12, which a body moving simultaneously in the directions A X and A Y -will occupy after a certain time (t), is foundPARALLELOGRAM OF THE VELOCITIES. 39 Fig. 12. when the fourth eorner of the parallelogram A M 0 JF, determined by the contemporaneous trajectories A M = x and A N = y, as well as by the angle X A F, or the distance by which the directions of motion deviate from each other, is known. The correctness of this mode of procedure becomes evident when the trajectories x and y are supposed de- scribed one after the other, and not at the same time. In compliance with the one motion, the body describes the trajectory A M * x; and in compliance with the other, the trajectory proceeding from M in the direction of A F, therefore in a line M Z parallel to A F, or the trajectory A JF = y. If M O = A JF, then 0 is the position of the body corresponding to both motions x and y simultaneously, which, in accordance with the construction, is the fourth eorner of the parallelo- gram A M O N. We may likewise imagine that the space A JW = x is passed over in a line A X, which with ali its points proceeds at the same time in the direction A F, and therefore carries with it M in a parallel direction to A F, and causes this point to perform the trajectory M O = A N = y. § 30. If both the motions in the directions A X and A Y take place uniformly and with the velocities cx and c2, then the spaces will be- come after a certain time (t): x = cxt and y == c2t; their relationship v c . - = — is, therefore, the same at ali times, a peculiarity which is only X Cj proper to the straight line A O, Fig. 13. Hence it follows that the compound motion proceeds in a straight line. If, with the velocities A B = cx and A C = c2, the parallelogram A B C D is constructed, its fourth eorner gives the po- sition D, in which the body will be placed after the lapse of one second. But as the resulting motion is rectilinear, it follows that it must always occur in the direction of the diagonal of that parallelogram which is constructed by the velocities. If the trajectory A O which is actually passed through in the time t be = $, then, on Fig. 13. AD AB J .AD , and it consequently follows that this trajectory s Jl D A D.t. In accordance with the last equation, the trajectory in the diagonal is proportional to the time (0» th® itself consequently uniform, and the diagonal A JD The diagonal, therefore, of a parallelogram formed bytwo velocum,t40 PARALLELOGRAM OF THE VELOCITIES. and the angle which they make with each other, gives the direction and magnitude of the actually resulting motion. This parallelogram is called the 'parallelogram of velodties, the simple velocities are called the components, and the compound velocity the resultant. § 31. By the use of trigonometrical formulae, the direction and mag- nitude of the mean velocity may be ascertained by calculation. The resolution of one of the equal triangles, viz., A B D, of which the parallelogram of velocities A B D C (Fig. 14) is composed, gives the mean velocity A D = c by means of the com- ponents A B = cx and A C = c2, and the angle BA C = a formed by their directions by the formula: c = \/ c* + c22 + 2 c2 c% cos. o, and the angle B A D = $, included by the mean velocity, and the velocity is expressed by the formula sin. $ = - 2 , 0r tang. __ —C2 S^'7}' *—# jf velocities c, and C1 + C2 C0S• a c2 are equal, and their parallelogram conse- quently a rhombus, then we obtain in a more simple form, in con- sequence of the diagonals being at right angles to each other: c ss 2 cx cos. J a and = Ja. Lastly, if the velocities enclose a right angle, then likewise we ob- tain more simply: ________ £ c = v/c2+c2 an(l tang. q> = 12 ci Example 1. The water flowing from a vessel or a machine has a velocity cl = 25 fi, whilst the vessel is moved with a velocity c2 = 19 feet in a direction forming an angle a° = 130° with the direction of the flowing water. What is the direction and magni- tude of the resultant, or as it is also called, the absolute velocity? The required resulting velocity is c = \/252 -f- 192 + 2.25 . 19 cos. 130° = y/625 -f- 361-^50.19 cos. 50° = ^/986—950 cos. 50° = ^/986—610,7 = ^375,3 = 19,37 feet, 19 sin. 130° Moreover, sin. =-1^37--*** 9,9808 sin. 50° = 0,7513, and consequently the angle by which the resultant differs from the velocity Cj <;> = 48°, 42' and the angle which it makes with the direction of motion of the vessel: a— =81°18'. 2. If the former velocities were acting at right angles to each other, then cos. a =cos. 9o° = 0, thence the mean velocity c = ^/935 = 31,40 feet; for its direction we should 19 have tang. 9 = — = 0.76, and consequently, its deviation from the first velocity: = 37°, 14'. § 32. Any given velocity may be supposed to consist of two com- ponents, and can consequently be resolved into them, in accordance with certain conditions. If, for instance, the angles D A B = $ and DA C = 4, Fig. 14, are given, and enclose the velocities required together with the mean velocity A D—cy then draw through the ter- mmal point D other lines which represent the degrees corresponding to c, parallel to the directions A X and A Y: the points of section wul then cut off the required velocities A B=cx and A C= c2. Fig. 14.COMPOSITION OF THE ACCELERATIONS. 41 Trigonometry expresses these velocities by the formula ct — c sin. * „ c sin. * In the usuai practical cases, the two sin. (4 -f* 4') 2 sin. velocities are at right angles toeach other, and then -f 4 = 90°, sin. (* + 4)=1> and it follows: cx = c cos. 4 and c2 = c sin. $. Therefore, with one component (cx) and its angle of direction ($), the direction and magnitude of the other component may be estimated. Lastly, from the velocities c, cx and c2 alone their angles of direction may be determined, as the three angles of a triangle may be com- puted by the three sides. Example. Suppose velocity c = 10 feet is to be resolved into two components which deviate from its direction by the angle ^ ass 65° and 4 == 70°. These velocities will be: 30 sin. 70° sin. 135° 9,397 10 sin. 65° 9,063 ’ si». 45° Fig. 15. ; 13,29 feet and c2 — ^50 — 0,7071 ~~ 12,81 ^ § 33. Compos&on and Resolution of Velocities.—By repeatedap- plication of the parallelogram of velocities, any number of ve ociles may be reduced to one. By constructing the parallelogram AB JJ C3 Fig. 15, the mean velocity AD to cx and c2 is obtamed; by con- structing the parallelogram A D F JE, we get the mean velocity A F to A^D and A E = c3; and in like manner by constructing the parallelogram A FH G, the mean velocity A H = c to A F and A G s c4 is obtained, and thus the mean of cv c2, c3, and c4. The simplest method of obtaining the mean velocity in question, is by the con- struction of a polygon A B D F H, the sides of which A B, BDy DFy and F H, are drawn parallel and equal to the given velocities cv c2, c3, and c4; the last side A H is then always the resultant velocity. In the case, also, in which the velocities are not in the same plane, the mean velocity may be ascertained by repeated application of the parallelogram of velocities. The mean velocity AF = c (Fig. 16) of three velocities A B = cl9 A C = c% and A E = c3, which are not in the same plane, is the diagonal of a parallelepipedon B C ® the sides of which are equal to these velocities. The parallelepipedon of velo- dties is, therefore, also a term in general use. § 34. Composition ofthe Accelerations.— Two uniformly accelerated motions, begin- ning with null velocity, produCe, when combined, a uniformly accelerated motion ln a straight line. If the accelerations of these motions, proceedmg 4*42 COMBINATION OF VELOCITY AND ACCELERATION. in the directions A X and A Y (Fig. 17) are pl and p2, then, at the close of tne time t, the spaces will be A M Fig. 17. X 2 , and AN—y- s ^ , and their 1 relation x_ P* ^ — —1 is in nowaydepend- V P»*2 P* , x, ent upon the time; and, consequently, the trajectory A 0 of the compound motion is rectilinear. If A B is made —pl and B D =A C=p2, we obtain a parallelogram A B D C, which is similar to the parallelogram AMON, and from which A D Ji Jti Pi P; therefore, A 0=- A D . t2. 2 ’ 9 2 From this equation it ap- pears that the trajectory A 0 of the compound motion is proportional to the square of the time; the motion itself, therefore, uniformly accele- rated, and its acceleration is the diagonal A D of the parallelogram constructed by the simple accelerations p1 and p2. In the same manner, therefore, as velocities can be composed or resolved by the parallelogram of velocities, and, according to pre- cisely the same rules, accelerations may be united into one, or broken up into several others by a parallelogram, which is called the paraU lelogram of accelerations. § 35. Combination of Velocity and Acceleration.—By the combina- tion of a uniform with a uniformly accelerated motion, an entirely variable motion is produced when the directions of the motions do not coincide. In a certain time t, with the velocity c in the direction A F, (Fig. 18,) the trajectory A N=y=c t will be described, and in the same time with an unchange- Fis*18* able acceleration, and a direction A X at right angles to the former, P t2 the trajectory^?will be described, and the body arrives at the terminal point O of the pa- rallelogram composed of y=ct and Xzsz ? With the aid of 2 these formulae, the position of the body can be determined for any given time, but it is not always in one and the same straight line, for if from the first equation we take and place this value in the second, we obtain the equation ^ . In accordance with this, the trajectories (#) in the direction of the second motion do not cor-PARABOLIC MOTION. 43 respond with those in the first, but with the squares (y2) of thosein the first; and, consequently, the trajectory of the body is not recti 1- near, but is a certain curved line, known in geometry by the name o parabola. Retnark. Let ABC, Fig. 19, be a cone with a circular base A E B F, let D E F be a section of it paraliel to the side B C and at right angles to the section ABC, and let O P N Q be a second section paraliel to the base, and, consequently, also circular. Then let E F be the line of section between the base and the first section, and O N that between both sec- tions; imagine, then, in the triangular section ABC, the paraliel diaineters A B and P Q, and in the section D E F, the axis D G. Then, for the half chord of the circle, M N = M O, the equation applies M N* = P M X M Q; but M Q = B G and for P M we have the propor- tion P M: M D = A G: D G; hence, it follows M N*=B G X—-. But, in like man- D G ner, G E2 = B GX A G; if orie equation is divided by the other, we obtain, therefore, D M M N2 , r ——sss------; the portions cut off from the ■D G G Et axis (abscissa) bear, therefore, the same pro- portion to each other as tlie squares of the cor- responding perpendiculars (ordinales). This law agrees completely with the law for motion lound above; this motion, therefore, takes place in a curved line D N E*, which can be shown to be a section of the cone (Conic Section). § 36. Parabolic Motion.—In order thoroughly to comprehend mo- tion produced by the combination of velocity and acceleration, we must be abi e also to indicate the direction, velocity and the spacepassed through during any length of time (t). The velocity paraliel to A Y is invariable and = c, that which is paraliel toAXis variable and = pt. If with these velocities, O Q = c and 0 P == p ty the parallelogram O P R Q is con- structed, Fig. 20, we obtain in its diagonal O R the mean, or that velocity with which the body at O follows the parabolic curve A O U. This velocity itself is v = v'd*+pt\ In like manner, O R is the tangent or direction in which the body at O proceeds for a single instant, and we obtain for the angle POR=XTO =

supposing that the body has required the time t tomove from the vertex Cto the real commencing point A. Fig. 21. We have, therefore, c. cos. a=pty consequently t = c2sin.acos.a c1 sin.2 a C COS. a and 1. A B = a = c sin. a. t- 2. B C=b = lL = 2 C2 COS.a2 2P 2 p P , and If by these distances we have found the vertex of the parabola C, then, beginning from thence w~e can find for any required time the position 0 of the body. Moreover we have: making C M= x and M 0 = y the general formula x — Pf 2 016'202 « 6>4 feet> ^«ains a ss 2 6 4 «in. iqoo * «TZ M ) == 5,34 feet, and has a range of projection * M nn* 132 * 2*6»4 48° = 9,51 feet The time which each particle of water requires to perform the whole parabolic curve AC D is = Ifl« 2.20. sin. 66° & fi 19fi »1.17 seconds. The height corresponding to the horizontal distance •A 3 feet, is, as y —— 3 0,73 » 4,61 feet 1,755, NO 1,755* 5’34-4OT^66°)ir 5,34 —46 CURVED MOTIONS IN GENERAL. 2. The jet of water flowing from a horizontal tube has at a height of 1 j feet a range (half a range of projection) of 5^ feet, what is the velocity of the water ? From the formula x — —_2_________it follows h if x in this case = 1,75 and y = 2 c2 4h ~~ 4x 5,25* 5,25, then h as = 3,937 feet, and the velocity corresponding to this height is c = 15,58 feet § 39. Curved Motions in General.—By the combination of several velocities and several invariable accelerations, a parabolic motion is likewise produced, for not only the velocities but the accelerations also may be United into a single one ; the resuit is, therefore, the same as if there were only one velocity and one acceleration, i. e. only one uniform and one uniformly accelerated motion. If the accelerations are variable, they can just as well be United into a mean as if they were constant, for it is admissible to consider them invariable within the limits of an infinitely small space of time (*); and the corresponding motions, therefore, during that space of time, as uniformly accelerated. Of course the resulting acceleration is variable, as are its components themselves. If this resulting ac- celeration be combined with the given velocity, it is possible to deduce a small parabolic curve, according to which the motion is effected during the small portion of time. If again the velocity and mean acceleration are determined in the same manner for the next small portion of time, we are enabled to obtain a new curve belonging to another parabola; and if this be farther repeated, we at last obtain the whole course. § 40. Any minute portion of any curve may be considered as the arc of a circle. The circle to which this arc belongs is called the circle of curvature, the radius pertaining to it is the radius of curva- ture. The course of a moving body may, in the same manner, be composed of the ares of circles, and thus a formula for its radius established. Let AM(Fig. 23) be a very small trajectory described with a uniformly accelerated motion x Fig. 23. Vt2 . s7—in the direction AXy and 2 let AN bea very small uniform- ly described trajectory y = cr, and Othe fourth terminatingpoint of the parallelogram constructed from x and y, i. e. the point which a body proceeding from A would occupy at the end of the short time (*). Let AC be drawn at right angles to A F, and let us observe from what point C in this line, a small arc of a circle through and O can be drawn. On account of the smallness of the arc A O, we may assume that not only C A, but also C O P is at right angles t0 r > that> therefore, in the small triangle N O P the angle NP OCURVED MOTIONS IN GENERAL. 47 is a right angle. The solution of this triangle gives us 0 P= 0 N sin. OJYP^AMsin. XAY = L sin. <*, and the tangent AP = A J\T+ JVP= c tcos.a = (c+~~ cos. a) may be made =cf, because ^ cos. a, on account of the infinitely small factor *, is inap- preciable with respect to c. But now according to the property of the circle AP2 =POx(PO + 2 C O), or, as P O vanishes when com- pared with 2 CO, AP^^PO x 2 CO ; we have, therefore, the de- sired radius of curvature, CA-CO-r-iP.-Zt-------------------------2- 2 P O p t2 sin. a p sin. a By the aid of the same formula, the radii of curvature of ali the ele- ments of curves may be found, when the respective velocities (c) and the acceleration (p) are inserted, and also the angle <* which the acce- leration makes with the velocity, or with the direction of motion indi- cated by the line of contact. Example. For the parabolic path caused by the acceleration of gravity, we have r = c® 0,031 —:- and in the vertex of these curves, where * = 90°, therefore, sin. «=1, sin. a it results that r s= 0,031 c*. With a velocity* of 20 feet, it would therefore be found that r ** 12,4 feet; the fhrther, however, the body is removed from the vertex, so much the sm&ller a becomes, and so much the greater, therefore, the radius of curvature. § 41. Proceeding from a point A (Fig. 24), where the acceleration is effected at right angles to the direction of motion A Yf if, therefore, r3 a sss 90°, we obtain the radius of curvature C A = r = —, and the P velocity at the following point 0 is composed of c and of p *, hence v * y/ c2 -f* p212 = c —, because r is infinitely small compared with c. If we make i> = c 4. i?! *. we may then consider — T 2c J *c as acceleration, and ^ . t as the corresponding increase of velo- y. But as t is infinitely small, the acceleration —— becomes ai^ *n one secon^ of time we have an infinitely form 5» jFeaSe ofvelocity, a»d may therefore consider the motion uni- i°nn, and consequently make v J c. chanJT^ of motion, the direction of acceleration also thpn «rS> 1v“ese romain constantly at right angles to eachother, rem ni 6 . a^ays have v = c; the velocity of motion, therefore, ^ ins mvariably the same as it was at the commencement, namely f?* acceleration such as this, which is always at right angles ..e ®o«on, or causes the body to deviate at right angles from the onal direction, is called normal acceleration, and we hence know48 CURVED MOTIONS IN GENERAL. Fig. 24. that it alone never causes a change of velocity, but only a deviatioa straiSht direction. According to the formula above, r ** ~ we must make the normal aeceleration p = — = the square ofthe velocity divided by the respective radius of curvature. In the circle A O D (Fig. 24) the ra- dius of curvature (r) is the radius of the circle CA= C 0 itself; hence, when motion occurs in it, the aeceleration p = — is invariable. An invariable ac- r celeration, therefore, which constantly causes the body to deviate at right an- gles from its motional direction, obliges it to revolve in a circle. Kxample. A body which rotates in a circle of 5 feet diameter, for each revolution, it requires 5 seeonds of time, has a velocity c in such a manner, that, 2 ir r ____2 ir . 5 2 . ir = 6,283 feet, and a normal aeceleration p —- __ 7,896 feet j viz., in erery 5 second it will deviate from a straight line by ^ p = £ X 7,896 s 3,948 feet § 42. In the simultaneous motions of two bodies9 a con- stant change is taking place in their relative position, dis- tance, &c., but with the aid of the foregoing formulae it may be found for any given moment of time. In Fig. 25, let A be the point of application of the one body, B that of the other; the first advances in the direction A X in a certain time (t) to M, the second in the direction BY in the same time to J\T; we then have in this line the relative position and distance of the bodies A and B at the end of this time. If we draw AO pa- rallel with JlfJV*, and also make AO= MJ\T, then will the line AO likewise give the opposite position of the bodies A and B. If further we draw ONy we obtain a parallelogram in which OJ\T is also = AM. If finally we make BQ parallel and equal to JTO, and draw OQ, we have then another parallelogram BNOQ, in which one side BJ\T is the absolute path (y) of the second body, and the other side BQ the path (x) of the first body, described in the opposite direction. The fourth corner 0 is the relative position of the second body, in so far as it is referred to the position of the first body, which is considered as invariable. The relative position O of a body (B) in motion is also found if we add to the body, besides its own proper motion (BJ\T), that AJWof the body (A) to which we refer its position BQ, but in an inverse Fig. 25. ICURVED MOTIONS IN GENERAL. 49 Fig. 26. direction, and then resolve these motions by the parallelogram BNOQ in the usual manner. § 43. If the motions of the bodies A and B are uniform, we may substitute for AM and J9JV* the velocities c and cv i. e. the spaces de- scribed in one second. We obtain, therefore, the relative velocity of the one body, when we add to the same in an opposite direction, besides its own absolute velocity, that of the body to which the first velocity is referred. The same re- lation takes place with the accelera- tions. Example. A locomotive train sets out upon the line A X, Fig. 26, from A with a velocity of 35 feetj another simultaneously from B upon the line B Y, which makes with the former the angle B D X = 56° with a velocity of 20 feet If now the initial dis- tances A C = 30,000 feet, and C B = 24,- 000 feet, how great is the distance A O of the twotrains at the endof a quarter of an hour? From the absolute velocity B E = cl = 20 feet of the second train, the inverse velocity -B E = c = 35 feet of the first, and the in- cluded angle E B F = * = 180° — B D C ==180° — 56°=124°. The relative velocity of the second train BG=s/c2+ c 2+2cclcos.a = V 35* + 20* — 2.35.20. cos. 56^= v/1225 -p 400 — 1400 cos. 56° = \/T625 — 782,9 = \/842fZ = 29,02 feet For the angle G B F= = 38° 4(y 340} 5(y = 73°, 307, and the distance of the two trains afler 157: ’A 0 = y/jZB*+B& — 2AB7b O cos. A B _________ = \/3841924- 26ll8*—2.38419.26118 cos. 73°,30' = \/1588190000=39852 feet. 550 MECHANICS—FORCE. SECTION II. MECHANICS, OR THE PHYSICAL SCIENCE OF MOTION IN GENERAL. CHAPTER I. FUNDAMENTAL PRINCIPLES OF MECHANICS. § 44. Mechanics.—Mechanics is the Science which treats of the laws of the motion of material bodies. It is an application of phoro- nomics to the bodies of the external world, in so far as the latter is concerned with the motion only of geometrical bodies. Mechanics is a part of natural philosophy, or of the doctrine of laws according to which changes take place in the material world, viz., that part which considers the changes in bodies resulting from measureable motions. § 45. Force.—Force is the cause of motion or change of motion in material bodies. Every change of motion, viz., every change in the velocity of a body must be regarded as the effect of a force. For this reason we measure the force called gravity by a body falling freely, because the same incessantly changes its velocity. On the other hand, rest, or the invariability of the state of motion of a body, must not be attributed to the absence of forces, for opposite forces destroy each other and produce no effect. The gravity with which a body falis to the ground stili acts, though the body rest upon a table; but this action is counteracted by the solidity of the table or of the support. § 46. A body is in equilibrium, or the forces acting upon a body are in equilibrium, when there is no residuary effect, no motion pro- duced or changed, or when each neutralizes the other. In a body suspended by a thread, the strength of the thread is in equilibrium with gravity. In forces, equilibrium is destroyed, and motion arises if one of the forces be removed, or in any way counteracted; for in- s ance a steel spring, bent by a weight, enters into motion when the ta^en away> because the force of the spring, called elasti- city, then comes into action. Staticsiis that part of mechanics which treats of the equilibrium of orces. ynamics, on the other hand, treats of forces in so far as they produce motion.DIV1S10N OF FORCES—PRESSURE—EQUALITY OF FORCES. 51 § 47. Division of Forces.—According to their effects, forces are either moving forces or resistances; that is, as motion is brought about or impeded. Gravity, the elasticity of a steel spring, &e., belong to motive forces. Friction, the solidity of bodies, &c., are resisting forces or resistances, because by them motion is either diminished or destroyed, and can by no means be brought about. Moving forces are divided into accelerating and retarding; the first produces a posi- tive, the second a negative acceleration; by the one an accelerating, by the other a retarding motion is produced. Resistances are re- tarding forces, but a retarding force is not always a resistance. Gravity, for example, acts upon a body projected vertically upwards to retard it; but gravity, on this account, is no resisting force; for, by the consequent falling down of the body, it then again becomes a motive one. There is a distinction between constant and variable forces. While constant forces always act in the same way, and, therefore, produce like effects in like particles of time, i. e. equal increments or decre- ments of velocity, the effects of variable forces are different at different tR11 fS ’ W^e f°riner bring about a uniformly variable motion, to the latter corresponds a variably accelerated or a variably retarded one. § 48. Pressure.—Pressure and traction are the first effects of forces upon material bodies. By means of them, bodies are compressed and ex ended, and especially changed in their form. The pressure in rac ion brought about by gravity, acting vertically downwards, which the support of a heavy body, or the string to which a body is attached has to sustain, is called the weight of the body. ressure and traction, and weight also, are magnitudes of a parti- cular kind, which can only virtually be compared with each other, as 6J!0*1011 °* *orces serves for their measurement. The simplest, and weight aCC0Un* mos* Senera^ means of measuring forces is by § 49. Equality of Forces.—Two weights, or two pressures, or trac- ions, and also the forces which correspond to these last, are equal, w en one may be replaced by the other, without producing different C /T* *or exampfo> a steel spring be bent by a weight G, as by another then are these weights, and therefore the gravities in both odies, equal. If a loaded balance be made to vibrate as much by a weight G as by another Gv substituted for G, these two weights G, 2 are e V 10,4 § 57. Specific Gravity.—Specific gravity or specific weight is the relation of the density of a body to that of the density of some other, generally water, taken for unity. Now the density is equal to the weight of a unit of volume: hence the specific gravity is also the relation of the weight of one body to that of another, viz. water, under the same volume. In order not to confound the specific weight with that which belongs to a body of a certain magnitude, the last is usually called the abso- lute weight. If y be the density of matter (of water) to which we refer the den- sity of other matter, and 7l the density of any one kind of matter, whose specific gravity we will designate by £, then the formula *=-^l and yx = s. y. y holds good, and the density of a substance is equal to its specific gravity into the density of water. The absolute vreight G of a mass of volume V and specific gravity * is: G = V7l = Vey. Example.—1. The density of puresilver is 653,368 lbs. and that of water = 62,38 lbs.. CKO OCO consequently the specific gravity of the former = ’—10,474: i. e. each mass of 62,38 silver is 10£ times as heavy as a mass of water filling the same space.—2. The specific gravity of quicksilver = 13,598; its density, iherefore, is = 13,598 X 62,38 =848,24 lbs.; a mass of 35 cubic inches, therefore, weighs: G=848,24. V= = 17,18 lbs. IV 2 o Remark. In these calculations the use of the French measure and weight has this advantage, that in order to eflect the multiplication of g and y, it is merely requisite to advance the decimal point; because a cubic centimetre of water weighs one gramme, and a cubic metre a million, or one thousand kilogrammes. The density of quicksilver, according to the French measure and weight = 13,598 X 1000 = 13598 kilog.; i e. a cubic metre of quicksilver weighs 13598 kilogrammes. § 58. The following table contains the specific gravities of certain bodies constantly coming into application in mechanics: Mean specific gravity of dry laurei wood . . = 0,659 “ saturated with water . • * Mean specific gravity of dry pine wood . • = u saturated with water . • Quicksilver . . . . . • = 13,598 Lead . . . . == 11>33 1844^ ** °n th<6 AbsorPtion of Water by Wood.”—Polytechnisches MttheUungen, Part iv.56 STATE OF AGGREGATION—DI VISION OF FORCES. Copper, cast and compact “ forged .... Brass .... Iron, cast, white “ “ gray .... “ “ medium “ bar iron Zinc, fused .... “ rolled .... Granite .... Gneiss .... Limestone .... Sandstone .... Brick ..... Masonry, with lime mortar of quarry stone : fresh dry “ “ “ of sandstone: fresh dry “ “ u of brick: fresh dry Earth, loamy, hard stamped, fresh dry Garden earth, fresh dry ... Dry, poor earth .... . = 8,75 . = 8,97 . = 8,55 . - 7,50 . = 7,10 . = 7,06 . = 7,60* . = 7,05 . = 7,54 2,50 to 3,05 2.39 to 2,71 2.40 to 2,86 1,90 to 2,70 1.40 to 2,22 . = 2,46 . = 2,40 . = 2,12 . = 2,05 1,55 to 1,70 1,47 to 1,59 . = 2,06 . = 1,93 . = 2,05 . = 1,63 . = 1,34 § 59. State of Aggregation.—Bodiesappear to us, according to the different cohesion of their parts, under three principal conditions, which we term states of aggregation. They are either solid or fluid, and in the latter case, either liquid or gaseous. Solid bodies are those whose parts adhere so strongly together that a certain force is required to change the form of these bodies, or to effect their division. Fluid bodies, on the other hand, are those whose parts may be displaced about each other by the smallest force. Elastic fluid bodies, whose representant is atmospheric air, are distinguished frora the liquid re- presented by water, in as much as there is inherent in them an en- deavor to dilate themselves more and more, which is not the case with water, &c. While solid bodies have a proper form and determinate volume, liquid or aqueous bodies possess only a determinate volume without any proper form, and the elastic extensible fluid bodies have neither one nor the other. § 60. Division of Forces.—Forces are different according to their nature; we will here mention the principal: 1. Gravity, by means of which all bodies tend to approach towards the centre of the earth. * Rolled boiler piate iron has a sp. gr. from 7.6013 to 7,7922, or a mean of 7.7344, the amount o variation being jJ.^th part of the mean density. By seventeen trials of ham- mered bar iron, its mean sp. gr. was found to be 7,7254. See “ Report on Strength of Ma- terials for Steam Boilersf p. 232. Also Journal Franklin Inst1837.—Am. Ed.ACTION AND RE-ACTION. 57 2. Theforpe of inertia, which manifests itself when changes inthe velocity of inert masses occur. 3. The muscular force of animated beings; the force exerted by the muscles of men and animals. 4. Elastidty or springforce, which bodies exhibit in a change of their form or volume. Th e. force of heat or caloric, in consequence of which bodies expand or contract by a change of temperature. b. The magnetic force, or the attraction and repulsion of magnets. 7. The cohesive force, the force by which the parts of a body are kept together, and resist separation. 8. Adhesion, the force with which bodies brought into dose contact attract each other. The resistances of friction, rigidity, solidity, &c., arise mainly from the force of adhesion. § 61. In reference to forces we have to distinguish: 1. Its point of application, that point of a body on which the force immediately acts. 2. Its direction\, the straight line in which a force moves forward lts point of application, or strives to move it forward, or to impede its motion. The direction of a force, like every direction of motion, has two senses, it can take place from left to right, or from right to ett, from above to below, and from below to above. The one is termed positive, the other negative. As we write from left to right, ann +k0ln a°°^e *° below, it would be most convenient were we to cah these motions positive, and those in the opposite direction, nega- 3. The absolute magnitude or intensity of a force, which, as above stated is measured by weights, as pounds, kilogrammes, &c. § 62. Aetion and re-action.—The first effect which a force produces in a body, is a change of form or volume combined with extension or contraction, which begins at the point of application, and from thence i uses itself further and further. By this inward change of the body, 1 u3 * * 6/?* eIasticity *s ca^ed into action, puts itself into equilibrium wi h the force, and, therefore, is equal and opposed to the force. Ac- ion and re-action are equal and opposed to each other. This law not only prevails in reference to forces produced by contact, but also m the so-called forces of attraction and repulsion amongst which the niagnetic force and gravity itself may be ranked. The more strongly a magnet attracts a bar of iron, the more strongly is the magnet itself attracted by the iron. The force with which the moon is attracted towards the earth (gravitation) is equal to that with which the moon reacts upon the earth. The force with which a weight presses upon 1 v*slJPPor^: &*ven back in an opposite direction; the force with which a workman draws or pushes at a machine, &c., reacts upon ^he workman and strives to move him in the opposite direction. When a body impinges against another, the pressures are reciprocally equal on each of the bodies. § 63. Division of Mechanics.—The whole subject of mechanics58 THE MECHANICS OF A MATERIAL POINT. may be included under two principal divisions, according to the state °f aggregation of bodies. 1. The mechanics of solid bodies, which is also well named geo- mechanics. 2. The mechanics of fluid bodies, hydromechanics or hydraulics ; the last is subdivided into: 1. Into the mechanics of water and liquid bodies especially, hy- dromechanics or hydraulics. 2. Into the mechanics of air, and other aeriform bodies, especially, aeromechanics, the mechanics of elastic fluids. If we now have regard to the division of mechanics into statics and dynamics, we have the following parts: 1. Statics of solid bodies, or geostatics. 2. Dynamics of solid bodies, or geodynamics. 3. Statics of fluids, or hydrostatics. 4. Dynamics of fluids, or hydro dynamics. 5. Statics of aeriform bodies, or aero statics. 6. Dynamics of aeriform, aerodynamics, or pneumatics. CHAPTER II. THE MECHANICS OF A MATERIAL POINT. § 64. A material point is a material body, whose dimensions are indefinitely small in comparison with the space occupied by it. In order to simplify the representation, we will in the following consider only the motion and equilibrium of a material point. A finite body is a continuous union of an infinite numberof material points. If the single points or elements are all perfectly equal, i. e. move equally quick, in parallel straight lines, we may then apply the theory of the motion of a material point to that of the whole body, because, in this case, w’e may assume that equal parts of the mass of the body are impelled by equal parts of the force. § 65. Simple constant Force.—If {p) be the acceleration with which a mass (M) is impelled by a force, we have, from § 53, the force : P =s Mp> and inversely, the acceleration, p = G If, further, we put the mass M = —, where G is the weight of the body, and g the acceleration of gravity, we have the force : P * • ^ ~ g an£l ^ie acceleration: P P=Gg- 2.THE MECHANICS OF A MATERIAL POINT. 59 We find, therefore, the force (P) which impels a body with a cer- tain acceleration (p) when we multiply the weight of the body (G) by the ratio of its acceleration, to that of gravity. Inversely, the acceleration (jp), with which a body is moved forward by a force (P) is given, when the acceleration (#) of gravity is multi- plied by the ratio of the force and weight of the body. Example. Let us suppose a body lying on an horizontal and perfectly smooth table, which presents no impediment to the body in its course, but counteracts the effect of gravity upon it. If this body be pressed upon by a force acting horizontally, the body will give way to this influence, and move forward in the direction of this force. If the weight of this body be G = 50 lbs., and if P= 10 lbs. presses uninterruptedly upon it, it . . P 10 will enter into a uniformly aceelerated motion with the acceleration p =—• 8= — X Cr DU 32,2 = 6,44 feet. On the other hand, if the acceleration with which a 42 lb. heavy body becomes aceelerated by a force (P) = 9 feet, then will this force P = • G = X 42 =0,031 x 378 = 11,7 lbs. § 66. If the force which acts npon a body is constant, there arises a uniformly variable motion, and indeed a uniformly aceelerated one, if the direction of the force corresponds with the initial direction of the motion; and, on the other hand, a uniformly retarded one, if the direc- tion of the force is opposite to that of the initial direction of motion. If we substitute in the formulae (§ 13 and § 14) for p, the value P P — = — £*, we obtain the following: I. For uniformly aceelerated motions: 1. v — c — gt9 otv = c + 32,2 — 2. s = ct or s= ct 4- 16,1 — t2. ^ G 2 9 ^ G II. For uniformly retarded motions: 1. v = c------— gt = c — 32,2 — t, 2. s = ct _ L.&L = ct— 16,1 Z t2. G 2 9 G With the help of these formulae ali those questions may be answered which can be proposed relative to the rectilinear motions of bodies by a constant force. Example.—1. A carriage weighing 2000 lbs. goes with a 4 feet velocity upon a hori- zon tal line, offering no impediments to it, and pushed forward by an invanable force o^ 25 lbs. during 15 seconds, with what velocity will it proceed after the action of this orce. This velocity v = c + 32,2 ~ t, but c=4,P= 25 lbs., G = 2000,and* = 15; hence ir it follows, v = 4+ 32,2. .15 = 10,03feet.—2. Undersimilarcircumstaneesa car- riage, weighing 5500 lbs., which, setting out with a uniform velocity, has_traversed 950 feet in 3 minutes, is so impelled forward by a force acting oontinuously ior secon , that it afterwards passes over 1650 feet in 3 minutes; what is this force . Here me60 MECHANICAL EFFECT. . . . , . 950 1650 lnitial velocity c = — ^ = 5,277 feet per second, and the terminal velocity, v = P 3 889 G = 9,166 feet; therefore — gt = v — c = 3,889, and the force P = —---= 0,031 X tr gt 3,889 X -^y- = 0,12056 x 22,10 lbs.—3. A sledge, weighing 1500 lbs., sliding forward with a 15 ft. velocity, loses, through friction, upon its horizontal support, its whole motion in 25 seconds: how great is this friction? Here the motion is uniformly Pt Gc ietarded, and the terminal velocity v = 0; hence c = 32,2 and P = 0,031 — = 0. 031 X -500X15 = 0,031 x900 = 27,9 lbs. the friction demanded.—4. Another 25 sledge, of 1200 lbs. and 12 feet initial velocity, has to overcome by its motion a friction of 45 lbs.: what velocity has it after 8 seconds, and how great is the distance described? 45 X 8 The terminal velocity is v s 12 —32,2 X i200~ “ 12 — 9,66 = 2,34 feet} antJ the distance described s = ^^ X 8 =57.36 feet. § 67. Mechanical Effect.—The work done, or mechanica! effect, is that effect of a force which it produces in overcoming a resist- ance: as that of inertia, friction, gravity, &c. Work is performed when loads are lifted, a great velocity imparted to masses, bodies changed in their form or divided, &c. The work done, or the me- chanical effect produced depends not only on the force, but also on the distance through which it is made to act or to overcome the resist- ance ; it increases, of course, simultaneously with the force and the distance. If we lift a body slowly enough to allow of our neglecting its inertia, the labor expended is then proportional to its weight; for 1, the effect is the same whether m (3) times the weight (m G) is lifted to a certain height, or whether m (3) bodies of the single weight (G) are lifted to the same height; it is, namely, m times as great as the effort necessary for the lifting of a single weight to that height; and, again, 2, the work is the same, whether one and the same weight be raisedto7i(5)times the height (n A),or /i (5) times through the height, and it is of course n (5) times as great as if the same weight were raised to a single height (A). The work again done by a slowly fall- ing weight is proportional to the magnitude of this weight and the height from which it has descended. This proportionality also holds in every other kind of wnrk done. In order to make a saw-cut of a given depth of double the length, there are twice as many particles to separate as from a cut of a single length; the work, therefore, is twice as great. The double length requires double the distance to be described by the force, consequently the work is proportional to the distance. In like m anner the work of a pair of mill stones increases with the quantity of grains of a certain kind of corn, which they gnnd to a certain degree. This quantity, under otherwise similar circumstances, is proportional to the number of revolutions, or rather to the distance which the upper mill-stone, during the grinding of this quantity of corn, has gone through; consequently the work in- creases in proportion to the distance. § 68. The dependence above shown of the work produced by aMECHANICAL EFFECT. 61 force upon the magnitude of the force and distance described by it, allows us to take that amount of work which is expended in o\er* coming a resistance of the magnitude of the unit of weight (as a kdo- gramme, pound, &c.), along a path of the magnitude of the unit ot length (metre or foot,) as a unit of the mechanical effect, or the dyna- mical unit, and then we may put the measure of this equal to the product of the force or resistance, and the distance described m the direction of the force whilst overcoming this resistance. If we put the amount of the resistance itself = P, and the distance described by the force, or rather by its point of application, in over- coming this = s> the labor expended is: L =P s units of work. In order to define more clearly the unit of work, for which the single name, dynam, may be used, both factors P and s are generally given; and, therefore, instead of units of work, we say kilogram- metres, pounds-feet; and inversely, metrekilo. and feet-pounds ac- cording as the weight and distance are expressed in kilogrammes and metres, or in pounds and feet. These terms are usually expressed for simplicity by the abbreviations m&, or /cm, Ib.ft., or ft* lb. Example.—1. In order to raise a stamper 210 lbs. 15 inches high, the mechanicai effect L = 210 x ~ = 262,5 ft lbs. is necessary.—2. By a mechanical effect of 1500 ft. lbs., a sledge, which in its motion has to overcome a friction of 75 lbs., is driven forward a space * = — = 3-^22 —B 20 feet. P 75 § 69. Not only in an invariable force or constant resistance is the labor a product of the force and distance, but also the labor may be expressed as a product of the distance and force, when the resistance whilst being overcome is variable, if a mean value of the continuous succession of forces be taken as the force. The relation is here the same as that of the time, the velocity, and the space; for the last may be regarded as a product of the time by the mean value of the velocities. The same graphical representations are here also appli- cable. The mechanical effect produced or expended may be con- sidered as the area of a rectangular figure, JlBCDy Fig. 27, whose Fig. 27. Fig. 28. base AB is the space described (s), and whose height 1 ~ ^ invariable force (P) itself, or the mean of the diflferen v forces. In general, the work may be represented by whose figure ABCD, Fig. 28, which has for its base the space ( )»r j- height above each point of the base is equal to the force P S 662 PRINCIPLE OF LIVING FORCES. with each point of the path described. If the figure ABCD be trans- formed into a rectangular one ABEF of like area, we have the height AF = BE for the mean value of the force—the mean effort. § 70. Arithmetic and georoetry give different methods for finding a mean value from a constant succession of magnitudes. Amongst these, Simpson’s rule is that which is the most frequently applied in practice, and it combines a high degree of accuracy with great sim- plicity. In every case it is necessary to divide the space AB=s (Fig. 29) into n (the more the better) equal parts, Fig. 29. as AE=zEG= GJ, &c., and to measure the forces £F=P,, GH=PV IK=P^ &c., at the ends of these parts of the distance. If, then, we put the initial force AD= P0 and the force at the other end PC=PB, we have for the mean force: P=(hPo+Pi + P2+P3+ •*.. +Pn^ +iP^n, and, therefore, its work is: Ps=(*P0+^.+^+ ••• +P-i+hP-)-?~- 71 If the number of parts (n) be even, viz., 2, 4, 6, 8, &c., Simpson’s rule gives stili more accurately the mean force: P=(P0+4P1 + 2P3+4P3+ .... +4 P n-j+P n)-^3 n, and, therefore, the corresponding wrork: Ps=(P0+4P1+2P2+4P34- .... +4 P^+P.jJL. Example. In order to find the meehanical work which a draught horse performs in drawing a carriage over a certain way, we make use of a dynamometer, or measurer of force, which is put into communication on one side with the carriage, and on the other with the traces of the horse, and the force is observed from time to time. If the initial force P0 == 110 lbs., the force, after describing 25 feet = 122 lbs.; after 50 feet = 127 lbs.; after 75 feet = 120 lbs., and at the end of the whole distance of 100 feet = 114 lbs.; then the mean value, according to the lirst formula: P = (£ . 110 —|— 122 —f- 127-f- 120-|-£ . 114) -f- 4 = 120,25 lbs., and the meehanical work: P s = 120,25 X 100 = 12025 ft. lbs. from the second formula: P= (110+4.122 + 2 . 127 + 4.120+114)-f- 3 X 4 ___ 1446 ___ ^0,5 lbs., and the meehanical work 12 P s = 120,5 X 100 = 12050 ft. lbs. §71. Principle of the Vis Viva, or Living Forces.—If, in the formula of (§ 13) s = —_— 2p P or ps = ——-- we substitute for the acceleration p, its value — g, we thus obtain Ps=(—----—\ G, or G \ 2g / if wre designate the heights due to the velocities and by h and hl: . . P s = (h—AJ G. It we interpret this equation, so useful in practical mechanics, wePRINCIPLE OP LIVING FORCES. 63 find that the work* (P s) which a mass either acquires when it passes from a lesser velocity (c) into a greater (t>), or produces, when it is compelled to pass from a greater velocity into a less, is constantly equal to the product of the weight of this mass, and the diflerence of the heights due to the velocities l—-------—V \2g 2g/ Example 1. In order to impart to a carriage of 4000 lbs. weight, upon a perfectly smooth railroad, a velocity of 30 feet, a mechanical work jP t as — G = 0,0155 c® G « 2 g 0,0155 X 900 X4000 = 55S00 ft. lbs. is required; and just so much work will this car- riage perfbrm if a resistance be opposed to it, and it be gradually brought to rest—2. Another carriage of 6000 lbs. goes fbrward with a velocity of 15 feet, which is trans- formed by a fbrce acting upon it into a velocity of 24 feet, how great is the work aequired by this carriage, or done by the fbrce? To the velocities 15 and 24 feet correspond the heights due to velocity ht = L- = 3,49 ft, and h = — s» 8,928 ft.: from this the me* , 2g 2g chanical work P t = (h—hx) G = 5,441 X 6000 = 32646 ft lbs. If, now, the distance be known in which this change of velocity goes on, the fbrce may be fbund; and when this is known, the distance may be determined. In this last case, for example, let the distance of the carriage amount to 100 feet, and, whilst describing this, the velocity passes from 15 into 24 feet: we have the force P s (h—h.)— s » 326,46 lbs. ’ V 1 s 100 Were the force itself 2000 lbs., the space t would be s (h—hj 2- = 326-4-lss 16,323 P 2000 feet—3. If a 500 lbs. sledge has entirely lost, through friction on its path, its velocity of 16 feet, after describing a space of 100 feet, then is the resistance of friction P = — SOO * = 0,0155x 16*X -jqp «0,0155 X 256 X 5» 19,84 lbs. § 72. The formula found for the work in the foregoing paragraph: Ps={h-hx)G ^ is not only good for constant, but also for variable forces, if, instead of P, the mean value of the force (from § 70) be introduced; for if the whole space (s) of motion be considered as consisting of equal and uniformly accelerated parts described then we have the amount of work for these: ■(t)-* % ’g *—V, -G, -G, &c., in so far as «t, vv v3, &c., stand for the velocities acq . 8 end of these parts of space; and by th,e addition of a ve]0city we have the whole work required for the transformatum o c into v: J>»« (P,+P4+P,+ . •.)—G, becausefor an infinite num- 8 «D i e. Working power.64 COMPOSITION OF FORCES. ber (n) of forces (P1+P2+P3+.. .)+n, it transforas itself into a mean force, and because the members on the right hand of the equation G and — G, as also G and — G, &c. are opposed to 2g 2g 2g 2g each otber, so that the members— G and G, determined by the 2g 2g terminal velocity v and the initial velocity c, only remain. The formula P s = G = (A—*,) G is not used merely for the determination of the work, but not unfrequently, also, for the mea- surement of the terminal velocity. In the last case h is put = + If by the constant motion of a body, the Ps I o , 0 Ps terminal velocity v = the initial velocity c, the work done = zero, i. e. as much work is performed by the accelerated, as is expended by the retarded part of the motion. Example.—A carriage of 2500 Ibs. proceeding upon a railroad without friction, has acquired by an augmentation of its velocity, which at the commencement amounted to 10 R., a mechanical work of 8000 lbs., its velocity affer this work will be: v —J 10M-64,4 . 2222 = v' 1004*206= 17,49 feet. Remark. The product of the mass M ~ — and the square of the velocity (v2) : Mv3 is g called, without attaching to it any definite idea, the living force (vis viva) of the moved mass: and hereafter, the mechanical work which a moved mass acquires, may be put equal to half of the vis viva of the same. If a mass enters from a velocity c into another r, the work performed is equal to half the difference of the vis viva at the commencement and end of the change of velocity. This law of the mechanical performance of bodies by means of their inertia, is called the principle of living forces, or the vis viva. § 73. Compositiori of Forces.—Two forces Px and P2 act upon one and the same body, in the same or in an opposite direction, the effect is the same as if only one force acted upon the body, which is the sum or difference of these forces; for these forces impart to the mass M p p the acceleration, px and p2 = consequently from § 28, the acceleration p=Pl+p2=z resulting from both, is p -f- p —-2., and accordingly the force corresponding to this, is : Mp = PX±P2. The equivalent force P derived from these two is called the resultant; its constituents P1 and P2 the components. ab^hu”^”^'1* ^ body lying flat upon the hand presses so long only upon it with its bodv • h aS tJlG band is at rest’ or *s moved up and down uniformly with the if it be sudd ) nd be raisec* quickly, it suffers a greater pressure; on the other hand, the hand ^ “ropped, the pressure is then less than the weight; it becomes null if rawtl oack with the acceleration of gravity. If the pressure on the hand j e body falis with a force G—P, whilst its mass M =rs —; if we put the accelera-PARALLELOGRAM OF FORCES. 65 tion with which the hand falis = p, G—P = —p, and therefore the pressure P G g t G=( 1 — P) G. If the body on the hand be raised with the acceleration p, —p *s * V 8' / p\ then opposed to the acceleration g, therefore the pressure upon the hand P = \ ' g / G. According as a body ascends or descends with a 20 feet acceleration, the pressure upon the hand= ( 1 — = (1-0,62) G==0,38, ofthe weight of the body,or = 1+0,62 = 1,62.—2. If with the flat hand I throw a body of 3 lbs. 14 feet perpendicu- larly upwards, whilst I urge it on with the hand for the first 2 feet, the mechanica wor^ performed is P s = G h = 3 X 14 = 42 ft. lbs., and the pressure upon the hand, r — ~ = 21 lbs. Whilst the resting bodypresses with 3 lbs., it reactsupon the hand dunng the projection with 21 lbs. § 74. Parallelogram of Forces.—When a material point M, Fig. 30, is acted upon by two forces, P 1? P2> whose directions MX and MY make, with each other, the angle XMY= a, these lines generate the accelerations in these directions, P P pY = —1 and p2 = —^, and from their M M union, there arises a mean accelera- tion (§ 34) in the direction MZ, both of which are given by the diagonal of a parallelogram formed from pv p2, and the angle a; this mean or resultant ac- celerationp= \Zp12+p22+2p1p2cos. and for the angle which its direction makes with M X of the one acceleration px: . o0 sin. o sin. $ =0---------. p If we substitute in these formulae the above values of px and p2• ?-J(b),+(b)'+2(b)(s) - (S)' COS. a and sin . $ sm. If we multiply the first equation by Jlf,____ M p = v/P?+P22+2 P, p2 C05. a, or, since J[Ip is the force corresponding to the acceleration. 1. P = s/P12+P22 + 2P1P2COS.a. P 2. sin. = * 2 Sin. a Thus, Me resultant force is determined in ^ yhm /rom the component forces exactly as the resultant acceie the component accelerations. # j these lines be lf we represent the forces by straight hn^? * weights, as drawn, bearing the same proportions to each o 6*66 RESOLUTION OF FORCES. pounds, &c., the mean force may be represented by the diagonal of the parallelogram whose sides are formed by the lateral forces, and one of whose angles is equal to that made by the directions of these lateral forces. The parallelogram which is constructed from the lateral forces, and whose diagonal is the mean force, is called the parallelogram of forces. Example. Wlien a body of 150 lbs. weight, resting upon a perfectly smooth table (Fig. 31) is acted upon by two forces P, = 30 lbs. Fi_ 3| and P2 = 24 lbs., which make with each other an’ angle P, M P2 = * + B = 105°: in wliat direc* tion, and with what acceleration, will the motion take place ? Since cos. (« + 8) = co*. 105° = —_ cos. 75°, the mean force: P — v/302-f-24*—2~X30X24 cos. 75° = ^/900+576-^1440 cos. 75° = v/1476—372,7 = 33,21 lbs., the acceleration corresponding with it is: P_Pg 33,21 X32,2 71001A rri P~M I 150 The direction of motion makes with the direction of the first force an angle «, which is determined by: 24 sin. « =----- sin. 105° = 0,7224 sin. 75° = 33,22 0,6978, or * =44°, 15'. Remark. The mean force P depends, from the formula* found, only on the component forces, and not on the mass of the body upon which the forces act. For this reason, we find in many works on mechanics, the correctness of the parallelogram of forces proved without regard to the mass, but with the assumption of some funda- mental law. § 75. Resolutiori of Forces.—By help of the parallelogram of forces, not only two or more forces may be reduced to a single one, but also given forces under given relations may be resolved into two or more forces. If the angles a and £ are given, which the components M Px = Pv and M P2 = P2, make with the given force MP = P, the com- ponents may be found from the formulae: p P sin. /3 p _ P sin. a 1 sin. (a+$), 2 sin.(a+fl) If the components are at right angles to each other, a+£=90o, and sin. (a+p) = 1, and Pj = P cos. a and P2== P sin. a. If p and a be equal to one other, P2=P1, viz: p P sin. a. P p ^2=~—“i—-------------= sin. 2 a 2 cos. a Example 1. What is the pressure of a body M upon a table A B, Fig. 32, whose weight G = 70 lbs. and upon which a force P = 50 lbs. acts, and whose direction is inclined to the horizon at an angle P M P, = a = 40°? The horizontal component of P is P, = P cos. a = 50 cos. 40° = 38,30 lbs , and the vertical component P2 = P sin. « = 50 sin. 40° =32,14 lbs.; the latter strives to draw the body from the table, tliere remains then for the pressure : G—P2 = 70—32,14 = 37,86 lbs.—2. If a body of 110 lbs. is so moved along an horizontal way, Fig. 32.RESOLUTION OF FORCES. 67 by two forces, that it describes in the first second a space of 6,5 feet, in a direction nc deviates from the two directions of force by an angle a = 5*2° and B = »1 e orcef themselves are given as follows. The acceleration is twice the space in the first secon , so that/) = 2 X 6,5 = 13 ft. Now the mean force is P =^—=0,031 X 13 X P sin. 77° __44,33 sin. 77°^^$ 44,33 lbs., therefore the one component P2 = lbs., and the other P9: 44,33 sin. 52° sin. 51° sin. (52° + 77°) 51° = 45,59 lbs. § 76. Forces in a Plane.—In order to find the mean force P for a System of forces Pv P2, P3, &c., we may adopt exactly the same method (§ 33) as that followed in the compo- sition of velocities, viz : by the repeated application of the parallelogram of forces, we may resolve them two and two and so on, till but a single force remains. The forces P1 and P2, for ex- ample, give from the paral- lelogram M Pj Q P2, the mean force MQ= Q, if this be joined to P3, we have from the parallelogram MQRP3, MR = R ; and this last again forms a parallelogram with P4 and gives the force MP = P the last, and the resultant of the four forces Pv P2, P3» P4* It is not necessary, in this wayof composing forces, to complete the parallelogram, and draw its diagonal. We may form a polygon MPX QRP, whose sides MPV PjQ, QP, PP, are parallel and equal to the given components Pv P2, P3, P4, the last side MP completing the polygon will be the mean force sought, or rather its measure. Fig. 33. this metlinr? ,?r« ^ l,®e*V t(J sn,ve mechanica] problems by construction also; though Inind from «rr n0t lni! °^ suc^ accuracy as that of calculation, it is free on the other the forcpq mA t cri‘or®» and may therefore servre as proof of the calculation. In Fig. 33 20', and p Tpe = Qooe4?/m,erlthe give? an*,es P« MP2 = 72°, 30' ; P2 3/Pa = 33<\ ’ Qr /p* • 4 ’ 7^ ’ an< are 50 drawn that a pound is represented by a line or J*o n* ( m7 lnch' The Pi = 11,5 lb., P2 = 10,8 lbs., P3 = 8,5 lbs., P4 = 0 900 ? the0re/r exPressed by sides of 11,5 lines = 0,958 ... inches, 10,8 lines = carelul Prm^t!08’ §’5 ’!Ie? = °’708 ••• inches, 12,2 lines == 1,016 ... inches in length. A ]4 r* ii ", ac lon °* P°iygon of forces gives the magnifude of the mean force P = S( ljo and ti,e varialion of its direction iliP from the direction J\IP1 of the first force= if each of thp ^vU^tant ^ *S ^etermined niore simply and clearly ino- tn t - -bi en comPoner^s P2» P3, &c., be resolved accord- each other 7ta c^lrec^ons and FF, Fig. 34, at right angles to other, mto component forces as Q[ and Rv Q2 and P2, Q3 and * The Prussian inch (see § 15) is equal 1.031 English inches. A>r. Ed.68 RESOLUTION OF FORCES. /f3, &c., the forces lying in the same direction of axis, added together, and the resultants in mag- nitude and direction of these two rectangular forces be then sought for. If the angles Px MXy P2 MXy P3 MXy &c., which the direc- tions of the forces Piy Pv P3> make with the axis XX =an a2, a3, &c., we have the com- ponents Qj = Px cos. 0l, Rx = Px sin. a,, Q2 = P2 cos. a2’ sin. a2, whence it foliows from Q = Qx + Q + Q3+ ..., 1. Q=P1COS.a1 + P2 COS. a2 + P3 COS. a3 + , and from jR = Rx + R2 4* R3 + ..., 2. R = PX sin. Oj + P2 sin. a2 + P3 sin. a3 + ... From the two components Q and R so found, the magnitude of the resultant sought, is: 3. P = s/Ql+R2 and the angle PMX=$, whose direction with XX is given by 4. tang. In the algebraical addition of the forces, regard must be had to the sign, for if it be different in two forces, i. e. if the directions of these be upon opposite sides of the point of application My this addition then becomes arithmetical subtraction (§ 73). The angle is acute, as long as Q and R are positive; it is between one and two right angles, when Q is negative and R positive; between twx) and three, when Q and R are both negative, and lastly, be- tween three and four, when R only is negative. Example. What is the magni- tude and direction of the resultant of the three components P] = 30 lbs., P2 = 70 lbs., P3 = 50 lbs.y whose directions, lying in a plane, make between them the angles PxMP2 = 5b° and P2 M P3 = 104 ? If we draw the axis XX Fig. 35.FORCES IN SPACE. 69 in the direction of tbe first force, we have a, — 0, a2 — 56, and “3 ^ 39 14 160°; hence, 1. Q = 30 X cos. 0» + 70 X «>*• 86° + 50 X «>..160» = 30+ 39,14 - 46,98 = 22,16 lbs.; and 2. K=30 X sin. 0» + 70 X sin. 56» + 50 «». 160 - °+58,03+ 17,10 = 75,13 lbs. Hence, 3. tang.

sin. 73° 34' = —l13 = 78,33 lbs. 0,9591 § 78. Forces in Space.— If the directions of the forces do not he m one and the same plane, we must draw through the pomt of appnca- tion a plane, and resolve each of the forces into two others, one ying in the plane, and the otherat right anglesto the plane ; we must then find the resultant of the components so obtained in the plane, irom the rule in the foregoing paragraph, and add together the componen s at right angles tothe plane, and from the two rectangular components thus obtained, their resultant may be found according o e rule (§ 74). , r Fig. 36 puts the above mode of proceeding more clearlj e ove u , let MP1 = Pv MP2 = P2, MP3 = P3 be the separate forcesthe plane (of projection) and ZZ the axis at right angles to it.^ From the resolution of the forces Pv P2, &c., the forces Sv S2 are given in plane, and those of JV*2, &c., in the normal to it^ZZ. These are again resolved according to two axes XX and YY into the lateral Fig. 36. forces Q„ Q2, &c., Rv R2, &c., and pve the componen^^ ^ {he of which the resultant S consists, wbich, j required. normal forces, JVy JV2, &c., gives P the resultant requ70 FORCES IN SPACE. If we put j3j, p2, for the angles at which the directions of force are inclined to the plane AB or to the horizon, the forces in the plane are given, Sv—Pl cos. S2=P2 cos.p2,&c., and the normal forces, JV'1 = P1 sin. pv JV*2 = P2 sin. p2, &c.; lastly, if we designate the angles which the projections of the directions of the forces lying in the plane AB, make with the axis XX, by av a2, we obtain the three following forces, forming the sides of a rectangular parallelopiped. Q= cos. 0j + S2 cos. a2 4- S3, cos. a3, or 1. Q= Px COS. 0J COS. Oj +p> COS. P2 COS. a2+ . . . , 2. R=P1 cos. sin. ax +p> cos. p2sin. a2+... 3. J\T=P1 sin. P1 + P2 sin. p2+... From these three follows the final resultant: 4. P= s/ Qf+B^+N2, further the angle of inclination to the plane of projection PMS=^, from 5. tang.4=—=—^= S Vqp+R*, lastly the angle SMX= we then obtain72 FORCES IN SPACE. P . J\IJV= P,. JV/JV, + P2. MNr Likewise also OJV, P2_ P, MO ’ P2 0JV>a„d* OJV , therefore MO ~ P MO P . OJV=Px. 0^ + P2. 0JV2. These equations stili hold good, if P the mean force be made up of three or more forces Pv p» p„ because generally Q= Qi+ Q2+ Q3+ • • • R=^R1 + R2+R3+ . . . and, therefore, generally we may put: 1. P .mjv=p1.mjv1 + p2.mjv2+p3.mjy3+ . . . , 2. p. ojv=p, . ojv,+p2 . ojv;+p3 . ojy3+ ... In both equations the mean force P rausi correspond to the forces Pj, P2, P3, and from these equations, not only the magnitude, but also the direction of this force may be determined. § 80. If the point of application M move in a straight line towards O, or if we imagine this point to have described the space JV/0=s, then the projection of this space MJ\r=s3 in the direction of the force MP is called the space of the force P, and the pro- duct Psx of the force and its space, the work or ejficiency of the force. If we substitute in the equation (1) of the last (§) these designations, we have Ps= PiSl+P2s2+ Pj*3+ . • • > or the work, or mechanical effect, of the resultant is equivalent to the sum of the works, or mechanical effects, of the com- ponents. In the summation of the mechanical effects, as in that of the forces, we must have regard to their signs. If a force (Q3) of the forces Qv Q2, &c., of the last § acts in an opposite direction to the rest, we must Fig. 38. Fig. 39.TRANSMISSION OF MECHANICAL EFFECT. 73 introduce it as negative, but this force Q3, Fig. 39, is the component of a force P3, which, acting in the circumstances set forth in the former §, opposed to their proper motion MN# we are, therefore, obliged to consider that force opposed to the motion MN, Fig. 40, as negative, and that one P, Fig. 41, acting in the direction of motion MJST as positive. If the forces are variable in magnitude or direction, the formula Ps=sP1$1+P2sa4-P3$3+ .. . is only correct for infinitely small spaces &c. The spaces of the forces whilat the mass itself M describes s, we then have from the foregoing: ft»PiS1+PJs1+ ... and, therefore, the following gener&i formula: *A+*A+ • • which expresses that the sum of the mechanical effects of the single forces is equal to half the gain of vis viva of the mass talnng up these forces. If the velocity during the motion be invariable, that is and the motion itself be uniform, we have then v#—c**»!), consequently neither loss nor gain of vis viva, and, therefore: P1S1 + P2*2 + P3S3 +---«0, *• e. the sum of the mechanical effects of the single forces 0. If inversely the sum of the mechanical effects ® 0, then the forces do not change the motion of the body in the given direction, nor im- part to it in the given direction any motion which it had not before. If the forces are variable, the variable velocity v after a certain time again passes into its initial velocity c, which takes place in ali periodic motions as they present themselves in many machines. Now v c gives the effect » 0; therefore within a period of the mo- tion the loss or gain in mechanical effect is null. -Ea:ampU. A carriage, of the weight O « 5000 lbe* Fig. 42, i» moved fcnmd npon a oonzontal turfaoe by means of a force Pt» 660 lbsn ascending under an «agi® * * ** , na has during its motion two resistances to overootne; one, horiaontal / * corresponding to the frietion; and a resistance P% mm ZOO lb^+ettog downwards, and74 CURVILINEAR MOTION. inclined to the horizon at an angle /3 = 35°. What work will the force (P,) perform,in order to convert the two feet initial velocity of the carriage into a ve- locity of 5 feet? If we put the distance of the car- riage MO = s, we then ha ve for the work of the force P, = Pj. MN. — Px s cos. a = 660 X * ros. 24° = 60*2,94 . further, the work of the resisting force = (— P2) . * —- — 350 . s; lastly, the work of P = (—P3 . AfiV3 = — P3 * co«. /3 = — 230 X * cos. 35° = — 188,40. s. There then remains for the Work of the effective force: P«=P, s cos. a.—P2* cos. 0—P3 s cos 0 = (602,94—350—188,40) — 64,54 . # R. lbs. The mass, however, requires for the change of its velocity, the mechanical effect: (ZjrfL) G = ( 5J~2’) x 5000 = 0,0155 X (25—4) X 5000 = 1627 rt. lbs. If now we equate both mechanical effects, we then obtain 64,54.«= 1627, conse- 1627 quently the distance of the carriage: 8 = - -■ =25,26 feet; and lastly, the mechanical 64,54 effect of the force P :Pl s cos. a = 602,94 X 25,26 = 15230,2 ft. lbs. Fig. 42. § 82. Curvilinear Motion.—Provided that the spaces cr, . Fig. 49. Exarnple. There act upon a body the forces P,=20 lbs. and Pa = 34 Ibs. whose directions meet under an angle P, CP2 = a = 70°, and are distant from a certain point 0=0 N]=al = 4 feet, and O N2 = a2 = 1 foot; what is the magnitude, direction, and position of the resultant ? The magnitude of the resultant is: P = v/20a+34*+2X20X34 ros. 70° = y/400-f- i 156-f1360 X0,34202 = y/2021,15 = 44,96 lbs.; further, for its . 34 X sin. 70° direction, sm.

P2> P3» * 53, act upon a body at the points •#2, */?3, two of these forces (Pl9 P2) v the last rule must be joined, and their resultant CQ =* Q found, this agam joined to the third force (P3)> and the parallelogram DR2RR3 constructed trom the forces DR2=CQ and DR^A^«Itant 0f P,, P2> The diagonal - P is the requtted resultant from this easy to see how the ~ resultant might be found ^ if a fourth force P4 were to be intro- duced. In this composition of the forces, the magnitude and direc- tion of the resultant is as accurate- ly found as if the forces acted in one single point (§ 77); the rules of calculation (§ 77) are, there- fore, applicable for finding these twofirst elements ofthe resultant; but in order to find the third, viz., the position of the resultant or its line of action, we must make use of the equation between the statical moments. Here, also, Ar *ue arms of the OJV,- fll, OJV2 = a%, OK\ = «3. with reference three component* P* P„ K and of their resultant ", to an arbitrary point O. So that: Pa as Q. OK+P3a3, and , resultant of Px an^ and &K' YirSSgSZ« e^Uone we * - e,«,+ £,»,+andf“ »«£() «e forces; if, further, v °o&c-’ are ,he an® vy ■ ’ intersected by the &c., imder which an arbitrarily chosen axis " g qNON2, directions of force, and if av a2, <*& designa e section O of both OJV*3, of these forces with regard to the poin o axes XX and YY> we have from §§ 77 and 87:80 COMPOSITION OF FORCES IN A PLANE. Fig. 54. 1. The component parallel to the axis XX: Q= P1 cos. ttj + P2 cos‘ a2 4- P3 cos. o3..., 2. The component parallel to the axis YY: R = Pl sin. oj + P2 a2 + P3 sin. ay .. , 3. The resultant of the whole system : P = 4. The angle which the resultant makes with the axis by tang. * 5. The arm of the resultant or the diameter of the circle to which the direction of the resultant is a tangent : n — PIai^^2a2+- • • p If this resultant be replaced by an equivalent opposite force (—P), then the forces P„ P2, P3.... (—P) are in equilibrium. Example. The forces P, = 40 lbs., P2 = 30 Ibs., P3 = 70 Ibs., Fig. 55, intersect the axis XYnt angles ax = 60°, «2 = — 80°, «s = 142°, and the distances of the points of intersection Dv D2, Dv of the directions of the forces with the axis: />, Z>2 = 4 ft., Fig. 55.PARALLEL FORCES. 81 and D2 D3 = 5 ft Required the elements of the resultant. The sum of the compo- nent forces parallel to XX is: Q = 40 co*. 60°+30 cos. (— 80°)+70 cos. 142° = 40 cos. 60°+ 30 cos. 80°—70 cos. 38° = 20+5,209—55,161 = —29,952 lbs. The sum of the components parallel to YY: R = 40 sin. 60°+30 sin. (—80°)+70 sin. 142° = 40 sin. 60°—30 sin. 80°+70 sin. 38 == 34,641 —29,544+48,096 = 48,193 lbs. The resultant sought is therefore: P = x/Q*+& = v^'29,9522+48,193* = ^/3219,68 = 56,742 lbs. The angle f, which it makes with the axis, is further determined by: tang. — A — 48J93 ----- 1,6090, it is therefore * = 180°—58° 8' = 121° 52/ Q _____29 952 The arm OiV, of the’ force P,is = OD, a, = (4+5) 60° = 9X0.86603 = 7,794 feet; the arm ON2 of P2 = OD2 sin. a2 = 5 sin. 800 = 4,924 feet: lastly, the arm ON3 of P, = O, when the point of application O is transferred to Dr The arm of the resultant is finally given by: a _ 40X7,794—30X4,924 _ 311,76—147,72 _ 164,04 _ 2 g91 feet 56,742 56,742 56,742 § 89. Parallel Forces.—If the forces Pv P2, P3, &c., Fig. rieid svstom nrp tiqpqIIoI a rigid system are parallel, the arms ONiy OJV2, OJY3, are in the same straight line; if now we draw through the point of ap- plication O an arbitrary line XXy the directions of the forces cut off the parts ODv OD2,OD3, &c., which are proportional to the arms r\ */• p t»'* ^ Fig. 56. . . -------------— OJV,, OJV*2, OJV3, &c.,be- 1 ' cause a OD^ooiOD^coaOD^. Ifthe angle be designated by o, &c., the arms OJVj, OJV2, &c., by «,»a2, &c.,the abscisses ODv OZ>2, &c., by bv 62, &c., we then have : a1=b1 cos. a, a2=b2 cos. a, &c. If, lastly, these values be substituted in the formula: Pa= Pax+P2a2+. . . , we then obtain: Pb COS. a= PJ>1 COS. a+P2b2 COS. a +• • • > or if the common factor cos. a be left out: pb=Plb1+p2b2+... ln every system of parallel forces it is allowable to replace the arms by the distances ODiy OZ)2, cut off from any line XX. Because the magnitude and direction of the resultant is the same, the forces may act at one or at different points; hence the resultant of a system of parallel forces has the same direction with the single forces, and is equivalent to their algebraical sum. Therefore 1. iwi+p2+p3+... 2. n — Piai+Pia» + • • * ^1 + ^2"!" • • • and or also:82 COUPLES. 3. b PA±£A± ' P. + P2+- ___32 lbs., -Pj = 25 lbs., and their directions § 90. Couples.—'Two parallel, equal and opposite forces, Pi an(i —P„ Fig. 57, have the resultant p = p + (—P,) = P, — A = 0, with the arm * 00 Fig. 57. Fig. 58. For restoring equilibrium to such a couple, according to this, a single finite force P acting at a finite distance, is not sufficient but twogsuch couples may easily hold each other in equilibrium. If P and _P, and -P3and P2, Fig. 58 ani two.such couples, and OM — a oM = OM. — MtN‘l = a, —61? «f further, and OJV*2 = OM —M2J\'2=?a2—b2 are the arms taken from a certam point 0, we have for equilibrium: . _ . P—* P2 »* . . Two such couples are, therefore, in equilibrium if the product of one force, and its distance from the opposite force, are as great m the one couple as in the other. _ . . , . . A pair of equal opposite forces is called s.mply a andthe product of one of the forces and its normal distance from the other force, the moment of the couple. From the above, two couples acting in opposite directions are in equilibrium, if they have equal momenta. If we substitute in the formula (§ 87) for the arma of the resultant: P.a. + PAH------------------------------- a =- P = 0 without the sum of the statical moments becoming null; we obtain likewise a = oo, a proof that in this case a i re- sultant, but only a couple, possible. Sxampk.Ua couple consists of the dUtenee ^8oo^fcuiUbnun.,tbe normal distance of the 25 X 3 second must amount to =-----—— = feet. 18CENTRE OF PARALLEL FORCES. 83 § 91. Centre of Parallel Forces,—If the parallel forces lie in dif- ferent planes, their union may be effected in the following manner. If the straight line Ax A2, Fig. 59, which unites the points of application of two parallel forces P1 and P2, be prolonged to the plane XY be- tween the rectangular axes MX, MY, and if the point of inter- section K be taken for the initial point, we shall in this manner obtain for the point of application A of the resultant (Px + P2) of these forces, Fig. 59. (P1 + P2).KA=P1. KAx + P2 . KAr As now B, Bx, and B2 are the projections of the points of applica- tion\ Ay Av A2, on the plane XY, we have: AB : AXBX: A B =KA : KA' : KA2, and therefore also a r> W5 “esi^nate by zv z2, z3, &c., the normal distances AXBV A2B2, 3, V c,> ^ points of application from the principal plane XY, two forces1 • at ^ from the same plane, we have for the / p1 i" *** f°r t^lree or more, and generally ^ 1 2+ 3 + • • •) ^=P1^14-P2^2+P3z3 . . . Consequently 1. 10 = "P ^2^2 "b • • • A+p +777”’ likewise the distances and AD of the point of appli- then obtain 6 ^ ^ pIaneS *Z and FZ = » and u’ *e 2. v = ^>1^1 • • • and Pt + P~. 3. » - Pixt + ... P +P + amnle6 w t^e Principal planes, as for ex- the noint q r °?r *7° s^e waHs of a room, fully determine lelomn^rt 9 ?T 2t e^th terminating point of the paral- there h ^onstru.ctec* from u, v, w, consequently, in such a system As ih* one Sln&^e point of application of the resultant, the fnrr.o ^or!nIu^aB f°r u9 v> do not contain the angles which indenpn f 7lt{l Principal planes, the point of application is svsfprn ^ese f°rces> and also of their directions; the whole ceasin + I?ltS’ therefore, of being turned about this point without its ,• „gt° „ ~Point of application, provided only that in this turn- mg the parallehsm of the forces be preserved. j* 7 suc a System of parallel forces the product of a force, and the ance of its point of application from a plane or line, is called the84 FORCES IN SPACE. moment of the force with reference to this plane or line, and gene- rally, the point of application of the resultant is called the centre of parallel forces. The distance of the centre of a system of parallel forces from any plane or line whatever, (the latter when the forces lie in the same plane,) is obtained, when the sum of the moments is divided by the sum of the forces. Example. If the forces are Pn the distances xn Vn “ “ za The moments are Pn xn “ “ P 1! 1 n Vn “ “ Pn Zn 5 —7 10 SUM. 3 lbs. l 2 0 9 ft. 2 4 5 3 “ 8 3 7 10 “ 5 —14 0 36 ft. lbs. 10 —28 50 12 “ 40 —21 70 40 « Now, if the sum of the forces =19 — 7 = 12 lbs., the distances of the Central point of this system from the three principal planes are consequently: 5±3e_H 27 = 9 12 12 4 ’ 10+50+12-28 _44_11 12 12 3 >/?_ 40+70+4^21 _j29_43_ , 12 12 3 § 92. Forces in Space.—If it be required to unite a system con- stituted of differently directed forces, a plane must be carried through the system, the different points of application transferred to this plane, and each force resolved intotwo component forces, the one coinciding with the plane, the other at right angles to it. If 01? p2... are the angles under which the plane is intersected by the directions of the forces, then the normal forces are Px sin. /j, Px sin. /3. .., and those in the plane Px cos. pl9 P2 cos. /32, &c. The latter from § 88, and the former from the last § 91 may be combined to a resultant. In gene- ral, the directions of both resultants will nowhere intersect each other, and accordingly a composition of these is impossible, but if the result- ant of parallel forces passes through a point Ky Fig. 60, in the direc- tion AB of the resultant of the forces in the plane (the plane of the paper) a composition is then possible. If we put the distances OC — DK = uy and OD = CK = v for the point of application of the first resultant, on the other hand the arm OJY of the second = a, and the angle BAOy at which it intersects the axis XX = a, the condition for the possibility of a composition is: usin. a + v cos. a= a. If this equation is not satisfied, if, for example, the resultant of the normal forces passes through Kv the reduction of the whole system of forces to a resultant is then impossible, but it readily admits of Fig. 60.PRINCIPLE OF VIRTUAL VELOCITIES. 85 being reduced to a resultant P, Fig. 61, and a couple P,— P, if the resultant JY of the parallel components is resolved into the forces— P and R, of which the one is equal, and directed parallel and opposite to the resultant P of the forces in the plane. § 93. Principle of Virtual Ve- locities.—If a system of forces Pv P2, P3, acting in a plane, Fig. 62, is progressive, i. e. moves forward so that ali the points of application A,2, A3, . . . pass through equal parallel spaces Ax Bv A2 P2, A3 P3, the effect of the resultant (in the sense of § 80) is equivalent to those of the components, and in a state of equilibrium therefore = 0. If the projections Ax A2 JV2, &c., coinciding with the directions of the forces of the common spaces Ax Bv = A2 B2, &c., = sl9 s2, then the mechanical effect of the re- sultant is: Fig. 61. This law follows from one of the for- mulae of § 88, accord- ing to which the com- ponent of the resultant running parallel with the axis XX is equal to the sum Qx+ Q2+.. . &c., of the similarly running components of the forces Pl9 P2 ; now from the simi- larity of the triangles W and AXPiQlf there follows the pro- portion Ql_A1JV1_ Sl Ps=P1s1+P2s2+... Fiff. 62. ABi AB and from this: Q,« £is', Q, = &c., AB V AB Fig.63. we may, therefore, in place of Q=Qi+Q2+. • • Put Ps= P1s1-\- P2s 2+* • • § 94. If the system of forces Pv P2, &c., Fig. 63, be made to revolve a very little about the point O, the law of the principle of virtual velocities enunciated above in § 80 and § 93 holds equally good, as may be proved 8 in the fol-86 PRINCIPLE OF VIRTUAL VELOCITIES. lowing manner. From § 86 the moment P. ON of the resultant is equivalent to the sum of the moments of the components, so that: Pa= P.o^ P3a3~^~ • • • . , The space AlBl corresponding to a revolution through the small angle AfiB^ or the arc is perpendicular to the dia- meter OA, therefore, the triangle ABC»*hich is formed if a per- pendicular line BXCX be let fallon thedirectmn onhje force issimilar to the triangle OAxA\ determined by the arm OJV\—a„ and accord- ineIy M\ A& OAx AXBX If the virtual velocity AxCx=, we then obtain OAx.ax OAv$ — also a2 = &c. * T T If these values be substituted in the above equation for av a2, we then have P # * or, as t is,a common divisor, P<,=Pl(yi+P *2+.. .&c„ the same as in § 80. So that, for small revolutions the mechanical effect (P«) of the re- sultant is equivalent to the sum of the mechanical effects of the components. § 95. The principle of virtual velocities holds likewise for ar- bitrarily great revolutions, if instead of the virtual velocities of the points of application, the projections NXDX, N2D2> &c.9 Fig. 64, of the spaces commencing at the points JYV J\T29 be intro- duced, and their values BXCX= OBx sin. JYxOBx^ax sin. *, P2C2= OB2 sin. N2OB2=a2 sin. *, &c., be substituted for , . Pa=P1a1+P2a2+..., the known equation for statical moments. This principle is correct also for finite revolutions, if the directions of the forces revolve simultaneously with the system, or if, while the point of application incessantly changes, the arm ONx=z OBx remains invariable, then from Fig. 64. Pa=Pxax+P2a2, and multiplying by <*>, we have Pa $= Pxax

B0, &c., of the points N> Nv &c. 1 1 2 2 § 96. Every small motion or displacement of a body in a plane naay be regarded as a small revolution about a movable centre, and paay be proved in the following manner. Let two points A and B, ig. 65, of this body (this surface or line) be advanced by a small motion to Ax and Bx, let also AXBX= If at these points we draw per- pendiculars to the small spaces de- scribed AAX and BBX, they will intersect at a point C, from which as a centre AAx and BBX may be considered the circular ares described. Now from the equalities AB=AXBX, AC=AXC, and BC=BxC> the triangles ABC and AXBXC are equal, therefore, also the *BxCAx = £BCA and the z ACAX = z BCBx. If we make AXDX=AD, we obtain from the equality of the zs . and E AC, and from that of the sides CAX and CA in CAXDX and CAD, again two congruent trian- gles in which CDX= CD, and zAxCDx=z ACD. Consequently any arbitrary point D in AB, by its small advancement, describes a cir- cular arc DDX. If lastly E be any point without the line AB, and TiguUy connected with it, the small space EEX may be regarded as he^°f ai c'rcl^.from c as a centre, for if we make the z EXAXBX i distance AxEx=AEy we again obtain two congruent rmngles and E AC with equal sides CEX and CE, and equal z.s. ACE, and the same may be shown for every other point nguily connected with AB. We may consequently regard every small motion of a surface rigidly connected with AB, or of a rigid y, as a small revolution about a centre, which is given when the point of intersection C is determined, in which the perpendiculars to other3^8 anc^ *wo P°mts of the body intersect each § 97. From § 94, for a small revolution of a system of forces, the mec amcal effect of the resultant is equivalent to the algebraical sum 0 i s components; from § 95, every small displacement may be re- gar e as a small revolution: hence the law of the principle of virtual ^eocities above enunciated is, therefore, applicable to every small motion of a rigid body or system of forces. li ®9ui^brium obtain in a system of forces, i. e. if the resultant be nuiJ, the sum of the mechanical effects must be also null for a small arbitrary motion. If inversely for a small motion of the body, the um ot the effects be null, equilibrium does not from this necessarily 0 ’. e ®um for all possible small displacements must be = 0, if qui ibrium is to take place. Since the formula expressing the law Virtual velocities only fulfils one condition of equilibrium, it is re- quisite for equilibrium that this law be satisfied, at least for as many Fig. 65.88 CENTRE OF GRAVITY. motions as can be made from these conditions for example, in a Sys- tem of forces in a plane, for the three motions indepenuent oi each other. CHAPTER II. CENTRE OF GRAVITY. § 98. Centre of Gravity.—The weights of the parts of a heavy body form a system of parallel forces, whose resultant is the weight of the whole, and whose centre may be determined from the three formulae of § 91. This middle point of a body or system of bodies is called the centre of gravity, and also the centre of the mass of the body or system of bodies. If a body be turned about its centre of gravity, this point does not cease to be the Central point of gravity, for if the three planes, to which the points of application of the separate weights are referred, revolve at the same time with the body, the position of the directions of force to these planes alone changes by this revolution, the distances of the points of application from these planes remain invariable. The centre of gravity is, there- fore, that point of a body in which its weight acts vertically down- wards, and which must be, therefore, supported, and fixed, in order that in every position the body may remain at rest. § 99. Every vertical straight line in which this point lies is called the line of gravity; and every plane passing through the centre of gravity, a plane of gravity. The centre of gravity is determined by the intersection of two lines of gravity, or that of a line of gravity and a plane of gravity, or by the intersection of the planes of gravity. Since the point of application may be displaced at will in the direction of force, without changing the action of the force, so a body is in any position in equilibrium Fig. 66. if a point in the vertical line pass- ing through the centre of gravity is fixed. If a body M, Fig. 66, be sus- pended by a thread CA, in its prolongation AB we have a line of gravity, and if it be similarly suspended by a second line, we get a second line of gravity DE. The intersection S of both lines is the centre of gravity of the body. If the body be suspended upon an axis, or be brought uponDETERMINATION OF THE CENTRE OF GRAVITY. 89 a sharp edge (knife edge) into a state of equilibrium, we sball obtain in the vertical plane passing through the axis, or through the knife edge, a plane of gravity, &c. Experimental determinations of the eentre of gravity, as just pointed out, are rarely applicable; we have generally to make use of geometrical rules, which will presently be given for the determination of this point with accuracy. In many bodies, for example, in rings, the eentre of gravity falis without the mass of the body. If such a body is to be fixed in its eentre of gravity, it is necessary to connect a second body with the first, in such a manner that the centres of gravity of both may coincide. § 100. Determination of the Centre of Gravity.—If a\, x2, x3, yl9 y2, y& z» Z3, &c., be the distances of the parts of a heavy body from the three planes xz9 yz, xy9 and the weights of these parts be pv p» &c., we then have the distances of the centre of gravity from these three planes, x as ^1X1 P*** ^3^3 + » » » P1+P%+P3+ • • • P&1 + p*y*+ p&+ • • • p1 + p2 + p3+... 9 * _ PlZl + P^> + P3Z3 + . . . Py *4“ Pj ”1* Pj “f* . . . If the volumes of these parts be Vv V%, V3, &c., and their densities v» y*» y3> &c., we may put therefore x fc Vrtixi + • • • Kft + ^y* + ... ’ If the body be homogeneous, t. e. ali parts of the same density y, then: (*>i+ V& + .. •)y . L (^+n+...)y’ or since the common factor y above and below is cancelled: 1. -- FA + fA + • • • ^+FS+... ’ 2. y —c • • • & tt ; i7 T” > 3. z + vt + senarau?^?^S°f ,0^ weights, substitute the volumes of the eravitv n^f ki j^ereby make the determination of the centre of gr5Ejr \pr?.blem of PFe geometiy. thin nhUc e68 a-e a btt^e extended inone or in twodimensione, as and L; S’ f6 WIIles’ ^C-> tbey may be regarded as surfaces or fines, 1I C/Dtre? of.^avity likewise detemined with the help of the SSfe/0^’lf f°r the VOlumes V» V*>the or leaStbs be eenfii^ir 1° re&u*aT figures the centre of gravity coinddes with the «f ngnre, as in dice, cubes, spheres, equilateral triangles, cir- 8*90 CENTRE OF GRAVITY OF LINES. cies, &c. Symmetric figures have their centre of gravity in the plane or axis of symmetry. The plane of symmetry ABCD divides a body ADFE, Fig. 67, into two congruent halves; the portions on both Fig. 67. Fi8* 68‘ sides of this plane are equal; the moments also on the one side are equal to those on the other, and, consequently, the centre of gravity fatis within this plane. Because the axis of symmetry EF cuts the plane surface ABCD, Fig. 68, into two congruent parts, here the portions on the one side are equal to those on the other; the moments also on both sides are equal, and the centre of gravity of the whole lies in this line. Lastly, the axis of symmetry KL of a body ABGH, Fig. 69, is its line of gravity, because it arises from the intersection Fig. 69. Fis*70* of two planes of symmetry, ABCD and EFGH. For this reason, the centre of gravity of a cylinder, of a cone, and of a surface of revolu- tion, or of a rotating body formed on the potter’s wheel, lies in the axis of these bodies. § 102. Centre of Gravity of Lines.—The centre of gravity of a straight line lies in its middle. The centre of gravity of a circular arc AB = ft, Fig. 70, lies in the diameter CM, and passes through the middle M of the arc, for this lameter is the axis of symmetry of this arc. But in order to find the is ance CS=x of the centre of gravity Sfrom the middle point, or cen re oj ke circle, the arc must be divided into many elementary parts, and statical moments of these, with reference to an axis XXCENTRE OF GRAVITY OF LINES. 91 passmg termined. through the centre C and parallel to the chord JB-s, be de- If PQbe a part of the arc, and PJVbe its distance frornXX, then the statical moraent of this portion of the arc — v* ' nn now the radius PC=MC=r be drawn, and QR! parallel to JUS, we obtain the two sirailar a8 PQR and CPN, for which. PQ:QR=CP:PJV\ frora which the statical moment of the elementary arc PQ.. QR. CP = QR .r is determined. . . , Now, for the statical rnoments of all the remaining ares, radius r is a common factor, and the sum of all the projectioris Q of the elementary ares is equal to the chord correspon ing o e Pr° jection of the whole arc ; it follows, therefore, that t e momen o whole arc is also = the chord (s) times the radius r. If this mo- ment be put equal to the arc (b) times the distance *, and therefore bx = $r, we then obtain: and*-". So that the distance of the centre of gravity, from the middle poin is to the radius, in the ratio of the arc to the chord. If the angle at the centre ACB of the arc b be = /3, the arc c responding to the diameter 1 is >3 == j^qo" • We have then b = , . 3 2 sin. i fi. r /3r, and s = 2 r sin. ^; whence it follows that, x =-^ * 3 2 For the semicircle /3 == * and sin. | = 1; therefore, x = - r = 0,6366 .. .r= JL r nearly. § 103. To find the centre of gravity of a polygon or a connection of lines ABCD, Fig. 71, we must seek the distances of the middle points H, K, Jkf, of the lines AB=^LV BC=L2, CD—L3y &c., from two axes OX and OF, viz : HHl = HH2 = xl% KKX = y2> KK2 =s x2, &c. ; the distances of the centre of gravity sought from these axes are then: SS = x — + - * * 2 - L1+L2+... 9 SS, = y= *L#i+La*+ ::• Lj+L2+ ... For example, the distance of the r ^2 centre of gravity S of a wire bent into the form of a a AB , &• > from the base is: * , a + b " Fig. 71. + hbh92 on juttv OF PLANE FIGURES. CENTRE OF GRAVITY O* if the sides opposite to the angles C be designated by a, b, and the height CGby h. ................................... Fig. 72. If the middle points M, of the sides of the triangle be connected with each other, and in the triangle so obtained a cir- cle be described, its centre will coincide with the centre of gra. vity S, for the distance SI) froin one side H K is = DJT — SN= -— 2 ch = e a+6 + < Fig. 73. Fig. 74. 2 2(a+6 + c) . Hidtances SE and SF from the other sides. the § 104. Centre o/ Gravity of Plane Figures. —The centre of gravity of a parallelogram ABCD, Fig. 73, lies in the point of inter- section of its diagonals, for all strips, such as KL, which are formed by drawing lines parallel to one of its diagonals are bi- sected by the other diagonals each of the diagonals, therefore, is a line of gravity. t » /IRCFie. 74, every line CD from one angle to the In a plane a ABC, JMg.^, ^ J the opposite side ^ is a Kne of gravity, for the same bisects all the ele- ments KL of the a which are given when lines parallel to AB are drawn. If from a second angle A a second line of gravity be drawn to the middle E of the opposite side BC, the point of intersection of the two will ffive the centre of gravity of the whole a. Because BD = BC = ^ DE is parallel to AC and = and a DES ■ -- - similar to the a CAS, and lastly, CS= 2 SI), ■f further „ ,dd SD, i, •»« The centre of gravity S lies at J of the line CD from the middle point D of the base, and at § of the same from the angle C. If CH and SN be drawn perpendicular to the base, we have also SLAT = J CH; the cen- tre of gravity S is at £ of the height from the base of the a. The distance SS1 of the centre of gravity of a b JiBC, Fig. 75, from an axis XX is = DDX + J Fig. 75.CENTRE OF GRAV1TY OF PLANE FIGURES. 93 (CCj— JDDj), but DDj = £ (AJj + BJ3J, consequently, ar = ^ = —i.e.* the arithmetical mean of the distances 3 9 of the three angular points. Since the distance of the centre of gravity is determined in the same manner by three equal weights at the angular points of a a, so the centre of gravity of a plane triangle coincides with the centre of gra- vity of these three equal weights. § 105. The determination of the centre of gravity S of a trapezium ABCD} Fig. 76, may be made in the following manner. The straight Fig. 76. line J\IJYy which connects the middle points of the two bases AB and CD with each other, is a line of gravity of the trapezium ; for lines drawn parallel to the bases decompose the trapezium into elementary parts, whose middle points or centres of gravity lie in MN. Now to determine corapletely the centre of gravity Sy we have only, therefore, to find its distance SH from a base AB. Let B represent the one, and b the other of the parallel sides AB and CD of the trapezium, h the height or the normal distance of these sides. Let DE be now drawn parallel to the side BC, we shall then obtain a parallelogram BCDE of the area bh, and whose centre of gravity is Sv and distance from AB = h 2’ and a a ADE of the area ^ ^ and centre of gravity S2, and whose distance from AB 3 The statical moment of the trapezium, about the line AB, is there- fore = bh.fl+(j^h-.h-=(B+2b)*, it the area of the trapezium is = (5+i) it follows, therefore, at the normal distance of the centre of gravity S from the ase is rro _ h(B+2b)h*_B+2b A; - ^ (B-\-b)h B-\-b 394 CENTRE OF GRAVITY OF PLANE FIGURES. To find the centre of gravity by construction, letthe^o bases be prolonged, the prolongations CG madeb a strai’ght lin^ . , , , rrcr B + 2b _ it foliows that of gravity sought; for, from iio_ 3 MS= and NS=z B+2b B+b * 3 MS B+2b $B+b 2 B+b B+b MA+AF MJY 3 MF and ‘CG+JYC ’ JYS = 2B+b B+\b ... . which actually arises from the similarity of the triangles MSF and 5^06 To find the centre of gravity of any other four-sided figure v ' ABCD, Fig. 77, we may decompose it by the diagonal AC into two triangles, and from the foregoing, determine their centresof gravity S, and S2, and thereby a line of gravity Fig. 81, is given, if we put the moment of the sector ACBM equal to the sum of the moments of the segment and the moment of the triangle ACB. If r be the radius CAy s the chord AB, and A the area of the segment ABMy the moment of the sector = the sector x arc chord 2 1 „ 2 r = g sr2,fur- CS, = arc ther the moment of the triangle = triangle x CS2 = ~ ^ r2—CENTRE OF GRAVITY OF CURVED SURFACES. 97 2 I ST2 . g I r------— = — — _, and from this the moment of the segment: •#. CS = = i sr2— ^--------=s JL; consequently the dis- tance sought is x = 1 8r* For the semi-circle s = 2 r and ^2 = — ^» hence a; = 2 12.^- 4 7* = —, as found above. Oyf In like manner we may find the centre of gravity S of a portion of a ring ABDE, Fig. 82, which is the difference of two sectors ACB and DCE. If the radii be CA=r and CD=rv and the chords AB=s and DE =*$!, the statical moments of the sectors fi 5 r2 are: and -1 , therefore the stati- 3 3 cal moments of the portion of ring: = sr2—s r.2 f . s, r. \ . ----—-1 1 , or (since —^- = —) is = 3 s r / r3—Tj —-— . But the area = — 3 r 2 Qy ^ /T2 T 2\ • = /3 (—\, provided that /3 represents the arc corresponding to the angle at the centre ACB; the centre of gravity, therefore, of the portion follows from the distance CS =x== mome^ ==^—. area r*— 2 s __ 2 / r3—/-j3 \ chord __ 4 ww. J /3 r3—rt3 3‘ r^“3 Vr2—/y* / arc = 3 r*—r,2"' Example. The radii of the surfaces of a dome are: r=5 ft, r, = 3£ ft, and the angle at the centre, 0° = 130°, then is the distance of the centre of gravity of these surfaces from their Central point: ____4 sin. 65° 5»—3,5» 4.0,9063 125—42,875 3,6252X82,125 ~~ 3 arc. 130« * 59—3,59 = 3 • 2,2689 * 25—12,25 “ 6,8067X12,75 = 3,430 feet. § 110. Centre of Gravity of Curved Sur- faces.—The centre of gravity of a curved sur- face (envelope) of a cylinder ABCD, Fig. 83, lies in the middle S of the axis MJV of this body, for ali the annular elements of the cylin- drical envelope which are obtained by sections drawn through the body parallel to the base, are equal, and their centres of gravity lie in the axis; these centres of gravity form a uni- form line of gravity. For the same reason the 9 Fig. 83.98 CENTRE OF GRAVITY OF BODIES. Fig. 84. centre of gravity of the surfaces of a prism lies in the middle pomt of the straight lines connecting the centres of gra\ i y o both the The centre of gravity of the envelope of a right cone Fig. S 84, lies in the axis of the cone, and is one-third of this line from the base, or two-thirds frora the vertex- for this curved surface may be decom- posed into an infinite number of small triangles by straight lines, which are called the sides of the cone whose centres of gravity form a circle HK, which is distant two-thirds of the axis froni the Vertex, and whose centre of gravity or centre S lies in the axis CM. The centre of gravity of a spherical zone ABDE, Fig. 85, and likewise that of a spherical cup lies in the centre S of its height ; for from the rules of geometry the zone has the same surface as a cylindrical envelope FGHK, whose height is equal to that of MJY, and whose radius is equal to that of the radius CO of the spherical zone; and this equality also exists in the annular elements, which are ob- tained by carrying an infinite number of planes parallel to the circular bases through the same; according to this the centre of gravity of the zone coin- cides with that of the cylindrical envelope. Remark The centre of gravity of the surface of an oblique cone or oblique pyramid liefrTtout oneThird of the height from the Use, but not m the atra.ght hne passing ftom the vertex to the centre of gravity of the t^useufao. Pa™»e. ,o the base decompose the surface into rings, which vary m breadth at different parts of their surface. _______ Fig. 85. S 111 Centre of Gravity of Bodie The centre of gravity of a § 111. Centre oj g j Fig. 86, is the centre S of the straiffht line which connects the centres of gravUy M and of both bases AD and for the prism may be decomposed by sections narallel to the base into exactly congruent slices, whose centres of gravity 1,e in MJf, and by their superposition make the line MJY a uniform line ot gravity. - For the same reason the centre of gravity of a cylinder lies in the middle of its axis The centre of gravity of a pyramid ADF, Fig. 87, lies in the straight line MFfrom the ■vertex F to the centre of gravity M of the base, for all slices as , have from their similarity with the base, their centres of gravity m this ine. If the pyramid be triangular as ABCD, Fig. 88, each of the four Fig. 86.CENTRE OF GRAVITY OF BODIES. 99 angular points may be considered as vertices, and the opposite sur* faces as bases; the centre of gravity S is determined by the intersec* Fig. 87. tion of two straight lines drawn from D and A to the centres of gra- vity M and JV*of the opposite surfaces ABC and BCD. If the straight lines EA and ED be given, we then have from § 104 EM =% EA and EJ\T = £ ED ; therefore MJV is parallel to AD and = J AD, and the a JWJY^similar to a DAS. Again from this similarity we have MS =■ J DS, or DS = 3 MS, also MD = SD + MS = 4 MS, and inversely MS = \ MD. Hence the centre of gra- vity is found to be one-fourth of the line joining the centre of gravity M of the base with the vertex D. Further, if the heights DH and SG be given, and HM be drawn, we then obtain the two similar a8 DHM and SGM, in which from the foregoing SG = £ DH. We may, therefore, say that the distance of the centre of gravity S of a triangular pyramid from the base is equal to one-fourth, and that from the vertex three-fourths of the height of the pyramid. As every pyramid, and also every cone, is made up of an infinite number of three sided pyramids of the same height, the centre of gravity of every pyramid and cone is a fourth of the height from the base and three- fourths from the vertex. We may, therefore, find the centre of gravity of a pyra- mid or cone, if a plane be drawn parallel to the base at a distance one-fourth from the base, and the centre of gravity of the section or its intersection with the line joining the vertex and the centre of gravity of the base be determined. § 112. If the distances AAv BBiy of the four an-100 CENTRE OF GRAVITY OF BODIES. gles of a triangular pyramid ABCD, Fig. 89, from a plane HK be known, the distance of the centre of gravity S from this plane is found from the mean value SS — -Mi+BBi+cct+DDu The distance of the centre plane is (§ 104): MMX = of gravity M of the base ABC from this AAX + BBX + CCX ---------3--------’ and that of the pyramid S is: SSX—MMX + \{DDX—MMX), where DDX is the distance of the vertex: hence it follows by com- bining the two last equations, that: SS, = = ^ The distance of the centre of gravity of four equal weights applied to the angles of a triangular pyramid, is equivalent to the arithmetical AA. -f BB. -t* CC. -f- DD mean------i-J---l—-----1~~----- 4 y consequently the centre of gra- vity of the pyramid corresponds with that of the system of weights. Remark. The determination of the volume of a triangular pyramid, from the co-ordinates of its angles, is simple. If we draw planes XY} XZ, FZ, through the vextex O of such a pyramid ABCO Fig. 90, and represent the distances of the angles ABC from these planes by z z3, Vi, V* Vy and x„ *3, the volume of the pyramid will be v= i Gw3 + s. + _ which will be given, if the pyramid be considered as an aggregate of four oblique prisms. The distances of the centre of gravity of these pyramids from the three planes are: y __*i + *2+*3 «_vi + Vi+y* t — +J*+h. ~ 4 4 4 Fig. 90. § 113. Since every polyhedron as ABCDO, Fig. 91, may be decom- posed into triangular pyramids ABCO, BCDOy we may also find its centre of gravity S if we calculate the volumes, and the statical moments of the single pyramids. If the distances of the angles Ay B, C, &c., from the co-ordinate planes passing through the common vertex O of all the pyramids, are xv x2> xv &c., yv y y &c., z z , z &c., the volumes of the single pyramids are: £“±* (W*++ xj/iZ , , (x^z4+^4Z3—*3y2c4— and the distances of their centres of gravity:CENTRE OF GRAVITY OF BODIES. 101 Fig. 91. .. __Xl + X2 + X3 _Jfl ■ J2 ■ .73 t/1 W1 — - > I ~ - 71 y, + y2 + y3 zi+z» + z3 —--------,- 4 „ x2 + *3 + ar4 y2 + y3 + y4 _ z2 + z3 + z4 &c. «2 = -----4----, V3------4 ’ w* - 4 ’ From these values the distances of the centre of gravity of the whole body may be finally calculated by the formula: „ _ *>,+ *>2+ • * • V2V2+ • * • - F1+F1+... * ..’ „ ... *>i+*>,+••• K1+F,+ ... * Example. A body bounded by six triangles ADO, Fig. 91, is determined by the fol- lowing values for the co-ordinates of angles; whence the co-ordinates of its centre of gravity may be found. Given co-ordi- nates 20 45 12 38 Six times the area of the triangular pyramids. Four times the co or- dinates of centre of gravity. Twenty-four times the statical moments. e S ■'T a TP § •24 F. U« 24 Va Va 24 Vn IPa ( 20.29.28 y C20.40.30y 6Fl= ? 41.45.40 C —*? 23.28.45 C=31072 (23.30.12y £41.12.293 C45.35.28y ( 45.40.20 ) 6 F2== 29.20.12 C — ? 29.28.38 } =17204 77 92 99 2392544 2858624 3076128 95 104 78 1634380 1789216 1341912 C30.38.40y ( 30.12.35 ) - Sum: 48276 4026924 4647840 4418040 three planes FZ, XZ, and XFfollow. 4026924 1 u = — . 4 48276 s 20,853, 9*102 CENTRE OF GRAVITY OF BODIE9. 1 4647840 _ 24tm, 4 48276 Waml 82,879. 4 48276 § 114. The centre of gravity of a truncated pyramid JlDQJY(Ylg 88), lies in the line MG, which connects the centres of gravity of the two parallel bases; in order to deterjnine the distance of this point from one of the bases, we must determine the volumes and moments of the entire pyramid ADF} and the supplementary pyramid JYQF. If the areas of the bases AD and JVQ -» G and g, and the normal dis- tance of both — A, the height of the supplementary pyramid will be given from the fbrmul©: ____ _ or i + 1 - Jj, and a?- .~p/# , as also h + a; g xr x g v G—Vg hy/G The moment of the whole pyramid with reference to the base G is now G{h + x) 3 h+x ~T~ 1 h3 G3 — J2 ' * ^at suPPlementary pyramid - f(*+j)-| * follows that the moment of the truncated pyramid: - nG--4^g£±3|j), *. 12 (G—2^Gg+g). 12 8) Now the solid contents of the truncated pyramid are: (G + SG~g+g)-, hence it follows finally that the distance of its centre of gravity iSfrom the base is ___ A G+2F^Fv and the whole height or distance of the outermost parallel plane is =A; the volume of the body, therefore, according to Simpson’s rule (approximately) is: V = (F0, +4 Ft+2Ft+A F3+F4)A. If we multiply in this formula each of these vojumes by their dis- tance, we obtain the moment: 7 A A Fy=(0 .F0+ 1.4 F,+2.2 F2+3. 4 . - ^; lastly, by dividing one expression by the other, we get the distance required: (0 . F0+l . 4 F,+2.2 . 4F,+4£)A -V-----------P0T4FI + 2F2+4F3 + F4 4 If the number of elementary slices = 6, wTe have: 1AF1 + 2.2F2+3AF,1-{-4.2Fa + 5AFs~\-6.F6 A F0+4F1 + 2F2+4F3+2F4-{-4:F5-\-F6 6 It is easy to understand how this formula may be altered w en number of slices is different from the above. This rule requires on y Ficr. 96.106 CENTRE OF GRAVITY OF BODIES. that the number of the slices should be even, and, therefore, that of the surfaces uneven. In most cases of application, the determination of one distance is enough, because, besides this, a line of gravity is known. The bodies commonly raet with in practice are solids of rotation, generated in a lathe whose axis of rotation is the line of gravity. This formula, lastly, is applicable to the determination of the centre of gravity of a surface, in which case the sections jF0, Fv F2, become lines. Example.—1. For the parabolic conoid ABC. Fig. 97, which is generated by the revo- lution of a parabola ABMeboat its axis AM,xve obtain by making the section DNE, the folio wing : The height AM= A, the radius BM = r, AN= NM= , and bence the radius DN = r ^J-L. The area of the section through A is F0 = 0, of that through N= Fx = ____ m ra DN* = and of that through M = Fi= v r*. Hence the volume of this body is: V = A CO + 4 r, + FJ = A (2 * r" + w r”) = J r’ A = J F, h; o o A’ on the other hand, the moment is = — (1 .2 a-r1 + 2. *• r3) = £ w t-3 A* = £ jFaA3; lastly, the distance of the centre of gravity 5 from the vertex, is: AS $F9h' tF,h :*A. Fig. 97. Fig. 98. Example 2. A vessel ABCD, Fig. 9S, has its mean half breadths, r0 = t inch, r, = 1,1 incb, ra = 0,9 inch, r3 = 0,7 inch, r4 = 0,4 inch, with a height MN= 2,5 inch. The sec- tions are F0= 1. Fx = 1,21 . w, Jr3 = 0,81 . *, F3 = 0,49 . *-, FA = 0,16 . tt ; hence, the distance of the centre of gravity from the horizontal plane AB, is: MS — °- J *+ 1.4.1,21 .*- + 2.2.0,81 *■+ 3.4.0,49*-+ 4 0,16 .ir2,5 1 ir + 4.1,21 * + 2.0,81 *• + 4.0,49 *-+ 0,16 *• 4 14,60 2,5 9,58 * T -TITT -- 0,9502 inches. 38,32 The capacity, therefore, is = 9,58 w . — 6,270 cubic inches. 1 ^ ^n^efesting and sometimes very useful application of the aws o the centre of gravity is the properties of Guldinus, or the ba- rocen ric method. According to these, the volume of a body of revo- u ion (or of a surface of revolution) is equal to the product of the genera ing surface (or generating line), and the space described by itsGULDINUS’ PROPERTIES. 107 centre of gravity during the generation of the body or surface of revo- lution. The correctness of this propositiori may be made evident in the following manner. Guldinus’ Properties.—If the plane area ABC, Fig. 99, revolve about an axis XX, each element FXF%, &c., of the same will describe an an- nulus; if the distances FXGX, F2G2, &c\, of these elements from the axis of revolution XXx be = rx, r2, &c., and the angle of revolution AMAX = a°, therefore the arc corresponding to the radius 1 = a, the circular paths of the elements will be == rxa, r2a, &c. The spaces described by the elements Fp &c., may be considered as curved prisms having the bases Fx, F2, &c., and the heights rxa, r2a, &c., and the volumes Fxrxa, F2r2a, &c., and therefore the volume of the whole body ABCBXAXCX: V= Fxrx a + F2r2a... = (Fxrx+ F2r2 + ...)• a. If MS= x be the distance of the centre of gravity S of the generating surface from the axis of revolution, we have also (Fx + F2+ ...)x= Fxrx + F2r2 +..., consequently the volume of the whole body V = (Fx + F2 + ...) x a. But Fx + F2 + ... are the contents of the whole surface F, and xa the circular arc w SSX, described by the centre of gravity S; consequently, V = Fw, as above enunciated. This formula holds good also for the revolution of a line, because it may be considered as a surface made up of infinitely small breadths; .Fis namely = Lw: i.e. the surface of revolution is a product of the generating line (L) and the path (w) of its centre of gravity. Example.— 1. In a half ring of an elliptical section ABED, Fig. 100, let the semi axis of the section be CA = a and CB = 6, and let the distance CM of the centre C from the axis XX = r; then the elliptical generating sur- face F = traby and the path of the centre of gravity (C) w = tc r; hence the volume of this half-ring tr1 abr, and that of the whole ring = 2 w3 abr. If the dimensions be, a = 5 inches, ^ = 3 inches, r = 6 inches, the volume of one- fourth of the ring = J . *■*. 5.3.6 = 9,8690 .5.9 = 4 44,132 cubic inches. Example.—2. For a ring with a semicircu- lar section ABD, Fig. 101, if CA= CB = o, represent the radius of this section, and MC = r that of the bollow space or neck, the volume is Fig. 100. Fig. 99. V = ^■2n(r+T^=7ra\nr + Ta)- Example.—3. To find the surface and volume of a cupola ADB of the dome of a con- vem, Fig. 102, half the width MA = MB = a, and the height MD = h aregiven. From ^lmensions il foIIows that the radius CA = CD of the generating circle = r — —2^~' and the anSle ACD subtended at the centre by AD=cL°,if we put the sin. a =108 GULDINUS’ PROPERTIES. —The centre of gravity of an arc DADX == 2 AD is determined by the distance CS = r. chord.^Z) = r .**n'_ • further, CM=rco$.a, consequently the distance MSof the arc AD a centre of gravity S from the axis MD = r — r cos. a = r —• «>*• « Y and A \ A ' the path described by the centre of gravity in the generation of the surface ADB = 2 irr. — cos. a) . The generating line DADt = 2 ra; and since it only is required to determine the half ADB, this line = rA, and consequently we must put the whole surface O = r a . 2 ie r^****— — cos. a^ = 2 n r3 («V». a— a cos. a). Fig. 101. Fig. 102. Very commonly a° = 60<>; therefore, a = JH, sin. a = £ ^/3, and the cos. a = £; hence O it follows that O = ar r2 3 — = 2,1515. r3. For the segment i)ADt =z A = r1 (a — J sin. 2 a) the distance of the centre of gravity from the centre C is = (2 • —= — . ** **?'** hence the distance from the axis 12 A 3 A MS = CS — CM = iL . — r cos. a ; finally, the path of this centre of gravity 3 A described in one revolution is: «o = 2-?r~- (Jr*sin.A8— Acos, a) = ^fJsin.A3— [a—£sin.2 a]cos.a). «^1 A. The volume of the whole body generated by the segment DAD„ is given if this path be multiplied by A, and the volume of the dome found by taking the half of this: there- fore, V = tt r3 ($ sin. a3 — [a — J sin. 2 a] cos. a). For example, a° = 60° = a O sin. a = £ y/ 3, and cos. a = i; hence: V=vr> (L^/T —11 = 0,3956.r». \ 8 6 Remarlc. Guldinus’ properties find their application in those bodies which arise when the generating surface so moves that in every position it remains perpendicular to the path of its centre of gravity, because we may assume every small part of a curvilinear motion to be circular. From this we may find the solid contents of the threads of screws, and sometimes also calculate the masses of earth, heaped up or removed, as in the case of canals, roads, railroads, &c. § 120« An°ther application of the doctrine of the centre of gravity, nearly allied to the last rule, is the following. vV e may assume that every oblique prismatic body ABCHKL, Fig. 103, consists of an infinite number of thin prisms, similar to Fx GvKINDS OF SUPPORT. 109 If Gj, G2 are the bases, and hv h2 the heights of these elementary prisms, we have for their solid contents Gx hv G2 A2, &c., and the volume of the whole oblique prism V = G1h1+ G2h2 + .. . Now an element Fx of the oblique section HKL is to the element Gx of the base ABC as the whole oblique sur- face F to the base G ; therefore, K. G, - yT. *«• «~1 r-y(KK+rA+-)- And because F1hl 4- F2h2 + • • • is the statical moment Fh of the whole oblique section, it follows that: V = — . Fh = i. e., F Fig. 103. the volume of an oblique prism is equal to the volume of a per ec prism, which stands upon the same base, and whose height is equa to the distance SO of the centre of gravity S of the oblique sur ace from the base. - In a right or oblique triangular prism, if hv A2, A3, be the edges o the sides, the distance of the centre of gravity of the oblique surface from the base h = 4* ^2 ^ A 3 V= G Zhence the volume (Ai 4- h2 -f h3) CHAPTER III. fquilibrium of bodies rigidly connected and supported. f ^n(^s °f Support.—The rules developed in the first chapter ° i*1S ®ec*10n> on *he equilibrium of a rigid system of forces, are ^pp icable to that of rigid bodies acted upon by forces, if we consider e weight of the body as a force applied to its centre of gravity, and achngverticallydownwards. odies balanced by forces, are either freely movable, t. e. yield to byeoth^10Ib0d^O^CeS, °r are fixe(^ one or more points,or supported If a point of a rigid body is fixed, any other point may take up a ouon whose path lies in the surface of a sphere, described from the X 10°mt aS 3 Centre ^ distance of the other point as radius. If,110 KINDS OF EQUILIBRIUM. on the other hand, two points of a body are fixed in every possible motion, the paths described by the remaining points are circles, which are the intersections of two spherical surfaces described from the fixed points. These circles are parallel to each other, and perpen- dicular to the straight line joining the two fixed points. The points of this line remain immovable ; and the body revolves about this line, which is called the axis of revolution. The radius of the circle in which each point moves, is found by letting fall from the point a perpendicular upon the axis of revolution. The greater this is, the greater also is the circle in which the point revolves. If three points of a body, not falling within the same line, be fixed, the body can in no sense take up motion, because the three spherical surfaces, in which a fourth point must move, intersect each other in a point only. § 122. Kinds of Equilibrium.—If a body, fixed at one point, be balanced by one force or by the resultant of several forces, the direc- tion of this force must pass through the fixed point; for a point is fixed when every force passing through it is counteracted. If this force consist merely of the weight of the body, it is then necessary that its centre of gravity should lie in the vertical line passing through the fixed point. If the centre of gravity coincide with the fixed, or the so called pointof suspension, we then have indiflerent equilibrium, because the body is balanced, in whatever direction it may revolve about the fixed point. If a body AB, Fig. 104, be fixed or sustained at a point C lying above its centre of gravity S, it then finds itself in a Fig. 104. Fig. 105. condition of stable equilibrium, because, if this body be brought into any other position, the component J\T of the weight G tends to bring it back into its first position, whilst the fixed point C counteracts the other component P. On the other hand, if a body ABy Fig. 105, be iixed at a point C below its centre of gravity Sy the body is then in a state of unstable equilibrium ; for if the centre of gravity be drawn °f k v^**ca* Passing through C, the component JY of the weight ? . ? 0 ^ ^ not onty does not bring it back into its former position, u raws it more and more out of that position, until the centre of gravity at last comes below the fixed point.PRESSURE ON THE AXIS. 111 The same reasoning will also apply to the case of a body fixed by two points, or by an axis ; it will be either in indifferent, stable, or unstable equilibrium, according as the centre Fig. 106. of gravity lies vertically above or vertically below the axis. § 123. Pressure on the Axis,—If a body, acted upon by forces in space, be fixed by two points or by a line, relations then take place, which we will now investigate. We may reduce, according to § 92, every system of forces to two, viz., one running parallel to the fixed axis, and the other acting in the plane normal to this line. Let AN=N, Fig. 106, be the first, parallel to the axis XX, passing through the fixed points C and D; and OP=P, the second force acting in the plane YZYat right angles to the axis XX. If we introduce other forces, as BJ\T = — •AT, CN1 = JVj, and DN2 =— JV2, we change nothing in the condi- tion of equilibrium or of motion, because these forces are entirely taken up by the axis. Now the forces N and — JV form together a first couple, and the forces JY1 and — J\Tiy acting in the plane XY and perpendicular to XX, a second couple ; we may, therefore, so manage, that these shall perfectly replace each other. If EO is the normal distance between the force N and the axis XX = y, and CD that of the fixed point = x; from § 90, we have the momentsof both couples = Ny and Nxx, and these are equivalent to each other, if JVy=N1x. We may also assume inversely that the force N is entirely taken up by the axis XX, whilst the axis has to sustain in its proper direction the pressure J\T, and the forces N1 == ^-JV, and — Nx = — ^ Napplied perpendicularly to it at the points C and D. That the body may be in a state of equilibrium, it is necessary ha the direction alsoof the resultant acting in the normal plane i pass through the axis. This force P may be replaced by t^o par forces P and P2 applied at the points C and D, ina\ e mined,Jf we put P,. CD = P . DO and 1\ . CI) = P .CO,tbt axis XX will have, therefore, besides the forces BN— * 1 *^2 P •Ar1 and DJY2 = — Nl9 also to react against the forces Px — x and P2 — . P,which may be calculated from the distances CD=x, X OC=xiy and OD=x2.112 EQUILIBRIUM OF FORCES ABOUT AN AXIS. § 124. From the results of the investigations of the foregoing para- graph we may easily calculate the forces sustained by the axis and the fixed points C and D. First, the axis has a pressure to sustain equivalent to the force JV* in itsown direction, which may be entirely resisted by one or other of the two fixed points. Secondly, from the forces JV. = 1 A*, P. = 3. P and—JV. =—V- JVand P2= %-P, act- XX X 2 ing in planes normal to XX9 and applied at the points C and D9 there arise the resultants Rx and R%9 which must be also sustained by the fixed points C and D. If we put the angle POY9 which the direction of the force P makes with the plane XY containing the axis XX and the direction of the force J\T=* o, the angle J\TX CPX is also = a; on the other hand, N2DP2 —- 180°—a, and the resultant pressures are therefore given by: Rx = x/J\ri2+P12+2J\T1P1 cos. a9 and R2 = vA/V^+P/—2*/V\P2co5.a. Kxample. A set of forces of a body fixed by its axis XX, is resolved into a normal force P = 36 lbs.,and aparallel force JV= 20 lbs.; the distance of the last from the axis is y = 1J feet, and the distance CD = x = 4 feet. To find the forces sustained by the axis, or by the fixed points in it, with the condition that the direction of P deviate by an angle « = 65° from the plane XY, and its point of application O be distant by CO = x, = 1 foot from the fixed point C? The force iV= 20 lbs. imparts to the axis along its direction a thrust X= 20 lbs. ; besides, it generates also the forces N, = — Nss= . 20 = 7,5 lb. and — N, = — 7,5 lb., against which the fixed points C and D react. From the force P arise the lorces Pt =— P = -—1.36*=27 lbs. and Ptss — P =— . 36 = 9 lbs. x 4 x 4 and by substitution of these values we have the resultant forces:____ Rt = 7,59-f 27a-f 2.7,5.27 . cos. 65° = ^/56,25+ 729+171,160 = s/$5>b,410 = 30,926 lb., and i?a = 7,53+9a—2.7,5.9 . cos. 65° = ^/56,25+81—57,054 = v/80,196 =8,955 lbs. § 125. Equilibrium of Forces about an Axis.—The force P is the resultant of all those component forces whose directions lie in one or more planes normal to the axis. But now in these cases, from § 86, the statical moment Pa of the resultant is equivalent to the sum Pxax + P2a?+ ... of the statical moments of the components, and for the condition of equilibrium of the fixed body the arm a of the resultant = 0, because this passes through the axis; hence the sum is also: 2>1a1+P2a2+.... = 0; 4. e. a body fixed by its axis is in a state of equilibrium, and remains also without revolving, if the sum of the moments about this axis = 0, or if the sum of the moments of the forces acting in one direc- tion of revolution, is equivalent to the sum of the moments of those acting in the opposite direction. By the help of this last formula we may find either a force or an arm for an element of a system of forces in equilibrium. he f°^Ces of rotation Pt = 50 lbs., and P3 = — 35 lbs., act upon a body P f p u”? a^°ut an ax*s at arms at = 1$ foot, and a, = 2£ feet; required, * w ich shall act at the arm a% = 4 feet, in order to restore the balance, i. e. to prevent rotation about the axis? Itis:THE LEVER. 113 50 . 1,25—35.2,5+ 4 P3 = 0, hence § 126. The Lever.—A body capable of turning about a fixed axis, and acted upon by forces, is called a lever. If we imagine it to be devoid of weight, it is then called a mathematicaly but otherwise, a material or physical lever. It is generally assumed that the forces of a lever act in a plane at right angles to the axis, and that the axis is replaced by a fixed point, called the fulcrum. The perpendiculars let fall from this point on the direction of the forces, are called arms. If the directions of the forces of a lever are parallel, the arms form a single straight line, and the lever is called a straight lever. If the arms make an angle with each other, it is then called a bent lever. The straight lever acted upon by two forces, is either one-armed or two-armed, according as the points of application lie on the same or on opposite sides of the ful- crum. There is a distinction made of levers of the first, second and third order ; the two-armed lever is termed a lever of the first order; the one-armed, of the second or third order, according as the weight acting vertically downwards, or the power acting vertically upwards, lies nearest to the fulcrum. § 127. The theory of the equilibrium of the lever has been already fully laid down; we have now, therefore, only to treat of each specially. In the two-armed lever, ACB, Fig. 107, if the arm CA of the power P be designated by «, and the arm CB of the weight Q by h, from the general theory : Pa = Qby i. e. the moment of the force is equal to the moment of the weight; or also, P : Q=b : a, i. e. the power is to the weight inversely as the arms. The pressure on the fulcrum is R=P+Q. In the one-armed levers ABC, Fig 108, and BACi Fig. 109, the same relation takes place between the power P and the w^eight Q, but here the direction Fig. 107. Fig. 108. Fig. 109. 10*114 THE LEVER. of the power is opposite to that of the weight, and therefore the pres- sure on the fulcrum is their difference, and in the first case R = Q — P, and in the second, R = P — Q. Also in the bent lever ACB, with the arms CW==aand Fig. 110, P : Q=*b : here the pres- sure on the fulcrum is equivalent to the diagonal R of the parallelogram CPflQ^ which may be constructed from the power P, the weight Q and the angle P^Q^PDQ—*, which their directions make with each other. Let G be the weight of the lever, and CE=e, Fig. 111? the distance of the fulcrum C from the vertical line SG, passing through its centre of gra- vity; we shall then have to put Pa-f Ge= Qby and the plus or minus sign before G, according as the centre of gravity lies on the side of the power P, or on that of the weight Q. Remark. The theory of the lever finds its ap- plicatiori in many tools and machines, viz. in the different kinds of balances, crow-bars, the brakes of pumps, wheelbarrows, &c. The second part will treat fully of these. Example.—1. If we press down the end A of a crow-bar ACB, Fig. 112 with a force P=60 lbs., and with the* arm CA of the power equal to 12 times that of thearm CBof the weight then will this, or rather the force ex- erted at P, be = Q = 12 times that of P = 12 . 60 = 720 lbs.—2. If a load Q, hanging from a pole, Fig. 113, be car- ried by two men, which pole the one lays hold of at A and the other at B, we may readily find out what weight each has to sustain. Let the load Q = 120 lbs., the weight of the pole G = 12 lbs., the distance AB of both points of application = 6 ft., the dis- tance BC of the load from one of these points = 2i feet, the distance of the centre of gravity 5 of the pole from this same point BS = feet. If we take B for the fulcrum, the power Pt has to balance at A the weights Q and G, therefore Pt. BA = Q . BC -f- G . BS, i. e. 6 P, = 2,5 . 120 + 3,5 . 12 = 300 -j- 42 = 342; hence, P, == ___= 57 lbs. On the other hand, if 6 Pa . AB = Q . AC + G . AS, and in 10 = 450; hence, the power Pa of the of the forces Pt-\-P^ acting upwards, ol -f- 75 132 lbs., is exactly equal to the sum of the iorces acting downwards, Fig. 111. Fig. 112. A be considered as the fulcrum, we must ] numbers, 6 Pa = 3,5 . 120 -f 2,5 . 12 = 420 second man is P, = = 75 lbs. als0) the Fig. 110.PRESSURE OF BODIES ON ONE ANOTHER. 115 Q+ G=120+12=132 lbs.—3. In a bent lever, ACB, Fig. 114, of 150 lbs. wt. the vertically pulling force_Q = 650 lbs., and the arm CB = 4 ft., but the arm CA of the power P = Fig. 113. Fig. 114. p<— JL s /V !f’ * GJ 6 ft, and that of the weight CP= 1 foot, what is the amount of the power P, and the pressure on the pivot R required to restore the balance ? CA . P — C ° ^ * G, i. e., 6 P = 4.650 + 1 . 150 = 2750; consequently, the power P = = 4581 lbs.; the pressure on the pivot consists of the vertical force Q -f- G = 650 4“ 150 = 800 lbs., and the horizontal power P = 458$ lbs., and is therelore: = j” = y/(«nd _ 5171. (»J -f /3J 1 57-71. (04 + /3J lastly, froin the equality JY = P sin. /3 Pj 577?. j31 5771. (a + /3) 5777. (c^ + /*i) Example.—Wlmt resolution of the forces takes place if a body Mv Fig. 116, sustained by a sup- port DE, be pressed npon by another, capable of revolving about an axis C with a force P = 250 Ibs., the angles of direction being the following: PAN s=z a = 30°, PAS = B = 480, PlAlNl=zetl=z650, PtAtSt = /J, = 50°. From the first formula the normal pressure be- tween the two bodies is determined by — y _ ^«w- # — 250 480 ___ ** 1 “ JM. («+*) sin. 830 ~~ 187,18 lbs.; from the second tlie pressure on the axis, or on the point <7, is P sin. a 250 sin. 35® S = sin. («+/?) «w. 830 = 144,47 lbs.; and bycombining the third and fourth equation, there follows finally for the component opposed to DE: Nttin.a, _ 187,18 sin. 65° _ 22I46 ,bs sin.Bi sin. 50° § 129. Stabiliti/.—When a body pressing against a horizonta] plane is acted upon by noother force than gravity, it has no tendency to move forward, because the weight acting vertically downwards is exactly sustained by this plane; nevertheless, a revolution of the body is possible. If the body ADBFy Fig. 117, rests at a point D upon the horizontal plane HR, it will re- main at rest, if its centre of gravity S be supported, i. e., if it Jie in the vertical line passing through D. If a body is supported at two points on the horizontal surface of another, it is requisite for its equilibrium that the vertical line of gravity should inter- sect the line connecting the two points. Lastly, if a body rests at three or more points on a horizontal plane, equilibrium subsists if the vertical line containing the centre of gravity passes through the triangle or polygon which is formed by the straight lines connecting the points of support. In bodies which are supported, we raust distinguish between stable and unstable equilibrium. The weight G of a body AB, Fig. 118, raw s its centre of gravity downwards; if no resistance be opposed Fig. 117.STABILITY. 117 to this force it will cause the body to tum until its centre of gravity has attained its lowest po- sition, and equilibrium will then be restored. We may mention that the equilibrium is stable when the centre of gravity is in its lowest pos- sible position, Fig. 119, and unstable when in its highest, Fig. 120, and indifferent, when the centre of gravity in every position of the body remains at the same height, Fig. 121. Fig. 118. Fig. 119. Fig. 120. Fig. 121. 122. Example.—1. The homogeneous body ADBF, consisting of a hemisphere and a cylin- der, Fig. 117, rests upon a horizontal plane HR. What height SF = h must its cylin- drical part ha ve, that the body may be in equilibrium ? The radius of a sphere is per- pendicular to the oorresponding plane of contact: now the horizontal plane is such a one; consequently the radius SD must be perpendicular to the horizontal plane, and the centre of gravity of the body lie in it. The axis FSL of the body passing through the centre of the sphere is its second line of gravity; the point S, the intersection of thetwo lines, is therefore the centre of gravity of the body. Let us now put the radius of the sphere and cylin- der SA = SB = r, and the height of the cylinder SF = BE = A, we then ha ve for the volume of the hemisphere : Vt = $ 7r H, for the volume of the cy- linder V1 = 7r HA; for the distance of the centre of gravity of the sphere S,: SSl = § r, and for that of the cylinder Sa : SSa = J A. That the centre of gra- vity of the whole body may /ali in S, the moment of the sphere wH.fr must be put equal to the jnoment of the cylinder, *r H A . £ A; from which we ha ve: A* = J r», i. e, A = r s/i= 0,7071. r. <6. 1 ne pressure which each of the three legs, A, ■?’ C, Fig. 122, of any loaded table has to sustain, is t etermined in the foliowing manner. Let S be the centre of gravity of the table with its j°ad, and SE, CD, perpendiculars upon J1B. If G be the weight of the whole table, and R the pressure on C, we may, considering jIB as the axis, put the moment of R = to the moment of G, i. e., R. Ci)=G. SE, and we then obtain R = SE " *JBS • G; likewise also the pressure on B s Q — A A CS &BCS AjSBC *ABC CD G A ABC G, and that on A = P = G. § 130. Let us now take the case of a body having a plane base resting on a horizontal plane. Such a body possesses stability, or is118 FORMULA OF STABILITY. in stable equilibrium when its centre of gravity is supported, i. e. when the vertical line containing the centre of gravity of the body passes through its base, because, in this case, the tendency of the weight of the body to cause it to turn is prevented by its own ngidity. When the line of gravity passes through the edge of the base, the body is then in unstable equilibrium, and when the line passes out- side the base, no equilibrium subsists. The body falis to one side and overturns. The triangular prism ABCDE, Fig. 123, is, according to the above, stable, because the vertical SG passes through a point Jf of the base. The parallelopiped ABCG, Fig. 124, is in unstable Fig. 123. Fi8* 124# equilibrium, because SG intersects a side CD of the base. The cylinder ABCD, Fig. 125, is without sta- bility because SG no where intersects the base CD. Stability is the power of a body to pre- serve its position by its weight alone, and to oppose resistance to any cause tending to overturn it. If we have to choose a measure of the stability of a body, we must distinguish whether this has reference to a displaceraent or to an actual overturning of the body. Let us now take into considera- tion the firstonly of these circumstances. § 131. Formulce of Stability.—A force, P, not directed vertically, tends not only to overturn a body ABCD, Fig. 126, but also to push it forward ; let us assume in the mean time that a resistance is opposed to the pushing or pulling forw^ards, as it may happen, and have regard only to its revolving about one of its edges C. If we let fall from this edge C a perpendicular CE = a upon the direction of the force, and CJV* =x upon the vertical line SG passing through the centre of gravity, we have Fig. 126. Fig. 125.FORMULA OF STABILITY. 119 only to consider a bent lever ECN, for which Pa= Gx, so that P = ~ G; if the external force P be greater than x ^ the body revolves about the point C, and, therefore, loses its stability. Hence the sta- biiity depends upon the product (Gx) of the weight of the body, and the shortest distance between a side of the perimeter of the base and the vertical line passing through the centre of gravity; Gx may there- fore be regarded as a measure of the stability, and for this reason is properly called the stability itself. Hence we see that the stability increases simultaneously with the weight G and the distance x, and may conclude that under otherwise similar circumstances a body twice or thrice as heavy does not pos- sess more stability than one of the single weight with twice or thrice the distance or arm xy &c. § 132.—1. In a parallelopiped ABCF, Fig. 127, of the length AE=l, breadth AB= CD=b, and height AD=BC=h, the weight G= Vy— bhly, and the stability S= G . KJV = G, \ CD == — = ^ b*hly, provided y represent the density of the mass of the parallelo- piped. 2. In a body ACFH consisting of two parallelopipeds, Fig. 128, the stabilities about the two edges of the base C and E are different Fig. 127. Fig. 128. arrrn°!!r "et VS*ake ,he heiShts -SCand EF=h and an I r* _r6A; /^S ~ ^ and tbe weights of the parts G b LlVlrW'yiand Wr ’ then the arms about C wil1 be KJYi = i Th“i bl + 1 b" and those about = + * and £ br he stabilities accordmgjy are: first for the revolution about C, t5,~g &+!*) +iGA = (iV^+M^+jPA) ly. formpr rter stab,bty ls ab<>ut *$t—S =* (A—A ) greater than the theTe m 11'6 Vsh !° increase the stability of a wall AC by offsets, ofrpvpi!- bePlacfd on that side of the wall towards which the force revolution (wind, water, pressure of earth, &c.) acts.120 dynamical stability. 3. The following is the stability of a wall Fig. 129, battering on one side. The upper breadth AB=b, the height BC = h and the length CH=l, and the batter = i. e. upon = a height of 1 foot; feet or inches of batter, therefore upon h feet ED = nh. The weight of the parallelopiped ACF is G, = bhly, that of the three sided prism A DE = G2 = \nh. hly, the arms for a revo- lution about E are = DE + f- ^6 and $ DE=%nh, consequently for the sta- bility ve have S= Gx (nA+Jft) + f G2nh = {\b2+nhb + \n2h2) hly. A parallelopipedical wall of equal vol- ume has the breadth b + % nh, hence the stability is: S, = \ (b+lnh)*hly= (i b2+\nhb + %n2h2) hly; its stability is, therefore, about S—Sx = (b+j5.2nh). \ nh2ly, less than that of the battered wall. For a wall sloped upon the opposite side, the stability is S2 = (b2+nhb+\n2h2) . \ hly, less also than S, and indeed about S~S2 = (6+\nh) . \nh2ly, as well as about S2—Sx = j1? n2hHy less than the stability of the parallelopipedical wall. Exanrple. What is the stability for each foot in length of a stone wall of 10 feet in height, and 1$ feet of upper breadth with batter of 1 in 5 on the back? The specific gravity of this wall (§ 58) is taken at 2,4, its density y is, therefore, = 62,5.2,4, = 130 Ibs.; now /= 1, A = 10, 6 = 1,25, and n = £ = 0,2; hence it follows, that the sta- bility sought is: S= ($ . ri,25]a-f 0,2.1.25.10 + i. [0,2]\ IO5) 10.1.130 = (0,78125+2,5+ 1,3333) 130 = 4,6146.130 = 603,4 fi. lbs. With the same quantity of material, and under otherwise similar circumstances, the stability of a parallelopipedical wall would be: Sx = (i. [l,25]a + i . 0,2.1,25.10 + j . 0,2*. 10a) . 130 = (0,78125+ 1,25 + 0,5) . 130 =2,531.130 = 329 fi. lbs. The same wall, with a sloping front, would have the stability : 5, = (+ [1,25]* + i . 0,2.1,25.10 + i. [0,2]*. 103) . 130 =(0,78125 + 1,25 + 0,666...) . 130 = 2,6979.130 = 350,7 ft. lbs. Remark.—It is evident from the foregoing that it allows of a saving of material to batter walls, to construet them with counteriorts, to give them offsets, or to place them upon plinths, &c. The second part will give a further extension of this subject, when we come to treat of the pressure of earth, and of vaults, chain bridges, &c. § 133. Dynamical stability.—We may distinguish from the mea- sure of stability treated of in the last paragraph, stili another to a certain degree dynamical measure of stability, when we consider the effect which is to be expended in order to overturn a body. Now the mechanical effect of a force is equal to the product of the force and the space, but the force of a heavy body is its weight G, and the space equal to the vertical projection of that described by its centre We ma^ consequently take for the dynamical measure of the stability of a body the product Gs, if s be the height to which the cen re of gravity of the body must ascend in order to bring the body trom its stable condition into an unstable one. Fig. 129.DYNAMICAL STABILTTY. 121 Let C be the axis of revolution and S the centre of gravity of a body ABCD, Fig. 130, whose dynamical stability we wish to find. If we cause the body to revolve so that its centre of gravity comes to Sv i. e. vertically over C, the body will be in unstable equilibrium, for if it only re- volve a little further it will fall over. If we draw the horizontal line SJV> this will cut off the height NS^s to which the centre of gravity has ascended, from which the stability Gs is given. If now CS = CS1 = z, CM = SN = x, and the height CN = MS = y, it fol- lows that the space StN = s = z — y = ___y, and the stability in the last form of expression is S= G(^x* + y2 — y). If the body is a prism with a symmetrical trapezoidal trans\erse section, as Fig. 130 represents, and if the dimensions are the follow- ing: length = l, height MO = h, lower breadth CD = biy upper breadth AB = &2, we then have MS = y = • ^ (§ I®®) bx + b2 o and CM = x — %biy hence CS = and the dynamical turn it: stability, or the mechanical effect, required to over- Fig. 130. r\,Mf a w / s r r / , a + 2h\2 K+zh = GN Uv +\~ktk's;- Fig. 131. Example.—What is tbe dynamical stability or the mechanical effect necessary for the overturning of an obelisk ABCD, Fig. 131, of granite, if its height h = 30 it, its upper length and breadth /, = 1J, and bt = 1 ft, and lower length and breadth Z, = 4 ft, 6a = 3£ fl. ? The vo- lume of this body is (§ 115): v= (2 M,+2ft.i+ak4+U)i = (2.f.l + 2.4.J+1.4+|.J) V = 40,25.5 = 201,25 cubic feet. Now a cubic foot of granite weighs = 3.62,5 = 187,5 lbs.; the whole weight of this body is: G = 201,25 . 187,5 = 37734,3 lbs. The height of the centre of gravity above the base is: y &A+3*A+V.+ ftA h _ 4-} + 3. J.l+1.4+|. I 30 27,75.15 ~ 40,25 * Y= 40,25 = 10,342 ft* Provided it be a revolution about the longer edge of the base, the horizontal distance of the centre of gravity from this edge will be: x==:i^a = i.J = Jft.; hence, the distance of the centre of gravity from the axis will be: CS = 2 = v/x#+ya = ^/(1,75)’+ (10,342)3 = v/110,002 = 10,489: and the height to which the centre of gravity must be raised to bring about an overthrow will be.122 THEORY OF THE INCLINED PLANE. i = z — y = 10,489 — 10,342 = 0,147 ft ; lastly, the corresponding mechanical effect or stability will be: Gs = 37734.3.0,147 = 5547 ft Jbs. Remark. The factor s = y/ x* + y* — y gives for y = 0, t =s x. for y as ^ * = x 1) = 0,414 x, for y = nx, * = (y/n* + 1 w) x» aPPro*imately =- (n -|- ^---n) x = also for y = 10 x, t * ^ ant* for y = oo, * = —- = 0; the dynamical stability is iherefore so much the greater, the lower the centre of gravity lieg) and it approximates more and more to null, the higher the centre o gravity lies above the base. Sledges, carriages, ships, floating docks, &c., must on t ns account be so con- structed and loaded, that the centre of gravity may lie as ow as possi e, and besides, be situated over the middle of the base. § 134. Theory of the Inclined Plane.—A body Fig. 132, rest- ing on an inclinod plane, that is, on one inclined to the horizon, may take up two motions; it may slide down the inclined plane, and it may also revolve about one of the edges of its base and overturn. If the body is left to itself, its weight G is resolved into a force JV normal, and to a force P parallel to the base, the first is resisted by the reaction of the plane, and the last urges the body down the plane. Let the angle of inclination FHR of the inclined plane to the horizon = », we have therefore the angle GSJY = a, and hence the normal pressure: JV = G COS. a, and the force parallel to the plane: P = G sin. a. If the vertical line SG passes through the base CD as in Fig. 132, a sliding motion only can take place, but if this line passes outside the base, as in Fig. 133, an overturn ensues, and the body, therefore, is Fig. 133. without stability. Besides, a body AC resting on the inclined plane FH> Fig. 134, has a stability different from that of one on a horizontal plane. If DM=x and MS=y are the rectangular co-ordinates of \fA^entre of grav*ty we have the arm of the stability DE= DO— ==x cos. a—y sin. a, while, if the body is on a horizontal plane, l 11 X' ,ce x >x cos. a—y sin. o, the stability with reference o e ower edge D comes out less for the inclined than for the hori- Fig. 134. Fig. 132.THEORY OF THE INCLINED PLANE. 123 zontal plane; it is null for x cos. a = y sin. a, i. e. for tang. a=-. When a body that is stable G x on a horizontal plane is transferred to an inclined one, whose angle of inclination corresponds to the ex- pression tang. <*=-, it will lose its stability. On the other hand, a body may acquire on an inclined plane the stability which is wanting to it on a horizontal one. For a turning about the upper edge C, the arm CE^ C01 + MJV=xl cos. a+y sin. a, whilst in its position on the horizontal plane it is = xv If now xY is negative, the body has no stability so long as it remains on a horizontal plane, but if it rests on an inclined one, for whose angle of inclination tang. a is the body is stable. , , JD/in If another force besides gravity acts upon the body JiBGD.hxg. 135, its stability continues if the direction of the resultant JVot the weight G and the force P intersects the base CD of the body. Example. The obelisk in the example of the preceding paragraphs has x j fl. and y = 10,342 ft, and will lose its stability, consequently, if transferred to an inclined plane, for whose angle of inclination: tang. a = 7 s 7000 = 0,16922, and inclination a = 9° 36'. 4.10,342 41368 § 135. As the inclined plane only counteracts that pressure which is directed perpendicularly against it, the force P which is necessary to prevent a body supported upon an inclined plane from overturning, is determined by the condition that the resultant J\T of P and G, Fig. 135, must be at right angles to the inclined plane. From the theory of the parallelogram of r , P sin. OJYP ., forces we have — = ————now the G sin. POJY z PJVO = z GOJ\r = FHR = a, and z POJV= POK+ KOJY= $ + 90°, in so far as w?e represent by p the z PEF= POKy by which the direction of the force deviates from the inclined plane; hence we have Fig. 135. P sin. a . P ___________ sin. a G sin. (90 -f fJ)’ G cos. p therefore the force which maintains the body on the plane is: G sin. a P = -----------• cos. /3 For the normal pressure N ~ = ~0GJV, but the / 90° — (a+^) and G sin. ONG OJYG= PON= 90 -f /3, hence it follows JV _ sin. [90°-----(a + /3)] _ COS, (a + p) G sin. (90° — /3) cos. fi124 PRINCIPLE OF VIRTUAL VELOCITIES. and for the normal pressure againsl the plane jy_ G cos. (q+j?) COS. /3 If the force P is parallel to the plane, /3 = 0 and cos. /3=1, since P = G sin. a and JV = G cos. a. If P acts vertically o+/3 = 90°, hence cos. /3 = sin. a, cos. (a+/3) = 0 and P = G and JV* = 0, the inclined plane has then no control over the body. Lastly, if the force acts horizontally, /3 = — o, and cos. /3 = cos. «, hence „ G sin. a „ , j ■»/• G cos. 0 G P —----------= G tang. a; and J\ =---------------- =»-----• COS. a COS. a COS. a Example. To maintain a body of 500 lbs. upon an inclined plane of 50° inclinatiori to the horizon, a force is applied whose direction makes an angle of 75° with the horizon, what is the magnitude of this force, and the pressure of the body against the plane i The force is: P = — ° nn• 50° = —J.nn’ 50° = 422,6 lbs.; and the pressure on the plane: cos. (75—50) cos. 25° N=z—.’ co*:.7-.° = 142,8 lbs. cos. 25° § 136. Principle of Virtual Velocities.—If we corabine the prin- ciple of the equality of action and reaction set forth in § 128, with that of Virtual velocities (§ 80 and § 93), the following law transpires. If two bodies and M2, Fig. 136, hold each other in equili- brium, then for a finite rectilinear or infinitely small curvilinear mo- tio n of the point of contact or pres- sure A, the sum of the mechanical ejfects of the forces of the one hody is equivalent to the sum of the me- chanical ejfects of those of the other. If Pj and Sx be the forces of the one body, and P2 and S2 those of the other, then, for a displacement of the point of contact from A to B, the respective distances de- scribed are ADVAEVAD2 and AE2, and according to the above law: Px. AD, + S.AE, = P2. AD2 + S2.AE2. The correctness of this proposition raay be proved in the following manner. As the normal pressures N, and JY2 are equal, there is also equilibrium between their mechanical effects, J\r,.AC and JV2 . AC, with this difference, that the mechanical effect of the one force is po- sitive, and that of the other negative. Now from what has preceded we have the mechanical effect N,. AC of the resultant JV, equivalent to the sum P, AD, -f- S,. AE, of the mechanical effects of iis compo- nentsP, and S„ and likewise JV2.AC=P2. AD + S2.AE2; hence also P,. AD, + Sr AE, = P2 )aD2 -f S2! AE2. Fig. 136.THEORY OF THE WEDGE. 125 The applicatiori of the principleof Virtual velocitiesthus made more general possesses great advantage in statical investigations, as by it the evolution of algebraical expres- sions becomes much simplified. If, for example, we move a body A up an inclined plane FH, Fig. 137, a distance AB, the corresponding path of the weight G, = AC = AB sin. ABC = AB . sin. FHR = AB sin. a. On the other hand, the path of the force P is AD = AB. cos. BAD = AB . cos. /3, and lastly, that of the norraal force JV = 0; now the me- chanical effect of JV is equivalent to that of G + that of P, hence we have to put JV. 0 = — G . JiC 4- AC ^ Gsin.a P . JW, and so we find P = _. quite in accordance with the former paragraph. In order to find the normal pressure N, we must move lorwar e inclined plane//F, Fig. 138, through a space AB at right angles to the direction of the force AP, to deter- mine the corresponding paths of the forces, and again put the meehani- cal effect of JV equivalent to that of G + the mechanical effect of P. The path of JV* is AD = AB cos. BAD = AB cos. /3, that of G is AC = AB cos. BAC = AB cos. (a + /3) and that of P = 0, hence the mechani- cal effect JV. AD= G. AC + P. 0, and JV G.AC = G cos. (o 4- £) ^ just as was foun(j jn the former pa- AD cos. & ragraph. § 137. Theory of the Wedge.—Affer this the theory of the wedge comes out very simply. The wedge is a movable inclined plane, Fig. 139. *11126 THEORY OP THE WEDGE. formed by a triangular prism FHR, Fig. 139, generally the iovceKP is = Py and at right angles to the back FR of the wedge, and holds in equilibrium another force or load AQ *= Q> which presses against its lateral surface FH. If FHR * a be the angle measuring the sharpness of its edge, and further, the angle by which the direction of the force KP or AD deviates from the surface FHy therefore FHK » HADy = a, and lastly the angle LAHy the deviation of the direc- tion of Q from this same surface, = 0f then the paths will be giyen which are described by the advance of the wedge from the position FHR into that of F^R^ in the following manner. The path of the wedge is AB * FF1 =* HHl9 and that of the force is = AD = AB cos. BAD * AB cos. (BAH— DAH) = AB cos. (a—a); further, the . t u aT , i . an AB sin. ABC AB sin. a path of the bar AL or load is AC = sin. ACB sin. HAC ABjin. a ^ simuitaneous path of the normal pressure J\T be- sin. 0 tween the wedge and the foot of the bar = AE «= AB sin. o. By the advance of the wedge a distance ABy the normal pressure N produces the mechanical effect JV.AE = J\T. AB sin. o, the force, however, develops the mechanical effect P. ADss P. AB cos. (o— StTl* o and the resistance the mechanical effect, Q.AC «=* Q.AB— *-, hence sin. 0 JVAB sin. a =* P • AB cos. (a—8) i. e. N sin. a = P cos. (o—a), as also JST. AB sin. a = Q. AB , i. e. N sin. a = Q —*W*a, and from sin. $ sin. 0 these equations the equation between the power and resistance sought is given: / _ \ Qsin.a P cos. (»—8)=—.-------, or v sin. /3 Q sin. a sin. 0 cos. (a—a)’ which may likewise be obtained by the decomposition of the forces. If the direction of the force is parallel to the base or lateral surface HRy a=a, hence P= and if, further, the direction of the load sin. 0 is perpendicular to the side FHy /3=90°, and P foIlows = Q sin. a. Example. The edge FHR of a wedge = « = 25°, the force is directed parallel to the base HR, therefore, s= a, and the weight Q acts at right angles to the side FH, therefore 0 = 90°, in what proportions are the power and weight to each other ? P is = Q rin. therefore ~ s sin. 25° s 0,4226. For a weight Q of 130 Jbs. the power P comes out = 130.0,4226 = 54,938 lbs. In order to drive forward the weight or bar 1 foot, the wedge must pass over the space jiB = — 1 2,3662 feet. p sin* 0,4226 * e t^ieor^es of the inclined plane and the wedge will be more fully deve- P m tne nflh chapter, where the effect of friction is taken into accountFUNICULAR MACHINES—KNOTS OR NODES. 127 CHAPTER IV. EQUILIBRIUM IN FUNICULAR MACHINES. § 138. Funicular Machines.—We have hitherto assumed that bodies, on which forces act, do not change their form in consequence of this action ; we will now take up the equilibrium of such bodies as suffer a change in their form by the smallest forces. The former are called solid or rigid, the latter flexible bodies. In truth there is no body perfectly flexible; many of them, however, such as strings, ropes, cords, &c., and in some respects chains also, require so small a force to bend them that they may in many cases be regarded as perfectly flexible. Such bodies, which are moreover extensible, will be the subject of the following investigations. We understand by a funicular inachine, a cord or a connection of cords (the word cord taken in its general sense) which becomes stretched by forces, and in this chapter we will consider the theory of the equilibrium of these machines. That point of a funicular machine to which the force is applied, and where the cord forms an angle with the direction of the force, is called a knot or node. This may be either fixed or movable. Tension is the force which a stretched cord transmits in the direction of its axis. The tensions at the ends of a straight cord or portion of a cord are equal and opposite, § 83; also a straight cord cannot transmit other forces than the tension acting in the direction of its axis, because it must otherwise bend, and, therefore, cannot remain straight. § 139. Knots or Nodes.—Equilibrium obtains in a funicular ma- chine, when there is equilibrium at each of its nodes. Hence we must next find what are the relations of equilibrium at any one node. Equilibrium takes place at a node K, which a portion of a cord AKBy Fig. 140, forms, when the resultant KS of the tensions of the cord KS1 = Sx and KS% = S2 are equal and opposite to the force P applied at the node Kyior the tensions &i and S2 produce the same effects as equal and opposite forces, and three forces hold each other in equi- librium, if one of them is equal to and acts opposite to the resultant of the other two (§ 75). The resultant P of the force P and the first tension &i is equal and opposite to the second tension S2, &c. In every case this equation may be used to find out two of the quantities to be determined, viz. the tension of the cord and Fig. 140.128 KNOTS OR NODES. its direction. Let, for example, the force be P, the tension < and the /. between the two JKP= 180°— JKS= 180°—a, we have for the other tension. ___________ S2 = v/P2+^,2 - 2 PSt cos. », and for its direction or deviation from anc* Example. If the cord AKB, Fig. 140, is fixed at the extremity P, and at the extremity A stretched by a weight G = 135 lbs., and the middle K by a force P = 109 Ibs., whioh pulls upwards under an angle of 25° j required the direction and tension of the portion of oord KB. The magnitude of the tension is : S9 = ^/109*4- 135* — 2.109.135 cos. (90° — 25°) = ^/11881+ 18225 —29430.cos. 05° = ^ 17668,3 == 132,92 lbs. For the angle 6, «'». fi = 65° Xog. «n. |S = 0,904017 — 1,hence fi S, 132,99 5 = 67° 0/, and the inclination of the portion of the cord to the horizon = a + £ —90° = 65° + 67° — 90° = 42°. Fig. 141. § 140. If the node if is a running or movable one, or the force P acts by means of a ring running along the cord AKB, Fig. 141, the resultant S of the tensions Sx and S2 is equal and oppo- site to the force P at the ring; besides this, the tensions are equal, for if the cord be drawn a certain space s through the ring, each of the tensions Sx and S2 wiil pass over the space s> and the force P over a space = 0; consequently, provided there is perfect flexibility, the mechanical effect P . 0 = Sx. s — Sx . sy i. e. Sx s = S2 s and St = S2. From this equality of the tensions there follows the equality of the angles AKS and BKS, by which the resultant S deviatesfrom the directions of the cords. If we put these angles = a, the resolution of the rhomb KS1 SS2y gives S = P = 2 Sx cos. a and inversely s1 = s2------------ 2 COS. a A and B are the fixed points of a cord AKB of given length (2 a) with a mova- ble node K, the place of this node may be found by constructing an ellipse, whose foci are A and 2?, and whose major axis is equal to the length of the cord 2 a, and if a tangent is drawn to this curve at P # right angles to the given direction of the ’ e r^su10ng point of contact is the place of the node, because tnrp °1 e^Pse KS makes equal angles with the radii vec- cord s ind S ’ “ d0eS the resultant S «th the tensions of the Fig. 142.KNOTS OR NODES. 129 If AB be drawn parallel to the given direction of the force, and BB be made equal to the given length of the cord, ^Dbiseeted at M and the perpendicular MKbe raised, the place of the node K may likewise be obtained without the construction of an ellipse, for since the z AKM = z BKM and AK = BK, it follows that z ARS also = z BKS and AK + KB = DK + KB = BB. Kxample. Between the points Zlandil, Fig. 143, a rope of 9 feetin length is stretcie by a weight G of 170 lbs. suspended to it by aring; the horizontal distance AC of the two points is ft., and the vertical distance BC = 2 ft.; to find the po- sition of the node, the tensions and directions of the rope. From the length AD = 9 ft. as hypothenuse and the horizontal line AC = 6^ ft.; it follows that the vertical CD = y/92 — 6,52 = ^/81 — 42,25 = v/38,75 = 6,225 feet; and from this the base BD of the equilateral triangle BDK, = CD — CB — 2 = 4,225 ft. The similarity of the triangles DKM and DAC gives DK = BK = DM DA ; 4,225.9 DC 2.6,225 = 3,054 ft.; hence it follows, that AK =9 — 3,054= 5,946 feet; and for the angles a, by which the sides of the rope are inclined to the vertical: COI. . = BM = 2,1125 _ 0,6917; BK 3,054 ’ hence, a = 46° 14'j and lastly, the tension of the rope 5, = S7 = — = 122,9 lbs.• 170 2.0,6917 * If the demonstrations applied in the text to the simple funicular machine, where a single weight is represented as sustained by means of two parts of a flexible cord, attached to two fixed supports, be applied to the case of two rigid planes hinged toge- ther at a middle point, and also joined by hinges to two other planes capable of sliding to and from each other, but in opposite directions, then will the principies of the formulaj above given, be found to afford the relation between the force applied and the resistance which it is capable of overcoming, in the well-known machine called the tricardo, vulgarly the “ toggle joint,” which has been much applied of late years in the construction of printing, coining, and other presses. Wlien two ropes hang parallel to each other, the whole gravitating power of the iveight is divided between them, and equally so between the points of support which sus- tain their upper extremities. The limit of the weight is the absolute strength of the ropes, and, in case of the tricardo, the force which could be applied to the planes would, in that posilion, be limited by the crushing force of the materials of the planes. In the funicular machine, the question generally relates to the tension on the cords, not to the force tending to bring together the points of support, while, in the tricardo, the * effort to separate the opposite extremities of the movable planes is the thing to be calcu- lated. The following figure (143#) may render this more evident.130 FUNICULAR POLYGON. § 141. Funicular Polygon.—The relations of equilibrium in the funicular polygon, i. e. in a Fi&-144- stretrhed cord which is acted upon by forces applied to dif- ferent points, are in accordance with thoseof the equilibrium of forces, which are applied toone point. Let JiKB, Fig. 144, be a cord stretched by the forces pv pv P3, Pv P2 act at A, let Px and 3 at K, and P the P, and P5 at B. Let us put tension of the portion AK=z Sx and that of BK = S2, we shall then obtain Sx for the resultant of Px and P2 applied to Ay and if we carry the point of appli- catiori A of this tension from A to K, we shall again get S2 for the resultant of Sx and P3, or of P P»P5 ; lastly, if we transport the point of application of S2 from K to B, we shall then obtain in S2, P4 and Ps, or since S2 is the resultant of Pv P2, JP3, also in Pv P2, P3, Pv P5 a set of forces balancing each other. We may accordingly assert that, when certain forces Pv p^ P3, §*c., hold a funicular polygon in equilibrium, they will hold each other in equilibrium also9 if applied at a single point C, their direction and magnitude remaining invariable. If the cord AKX K2. .. B, Fig. 145, be stretched at the points or nodes, Kv K2 by weights G„ G2 . .. and the extremities A and B by the vertical forces Vx and F», and the horizontal forces Hx and Hny the sum of the vertical forces will be : Vx + Fn— (G1 + G2 + G3 -f- ...) and of the horizontal forces: Hx — Hn. The condition of equilibrium requires that both sums = 0; therefore 1. vx+ K= Gx+ G2+ G3 + ... and 2. Hx= Ha; i. e. In a funicular polygon stretched by weights, the sum of the vertical forces or vertical tensions at the extremities or points of suspension is c and d, supposed to be capable of moving freely to and from each other along the plane MN. The hinge of a and b at P being supposed to be acted on by the small constant force P, the practical question is the relation of the resistances P„ P2 to this constant force P, in the different positions of the two planes a and b. If the angle PAC or PBC = tt represent the angle of divergence of the planes a and b from the straight line QQV it is evident that the force P will be represented by CE = 2 CP = 2 sin. a, and the forces Pt Tf1* y* eac^1 = 2 cos. a.. Hence P : Pt = sin. a: cos. a; or as tang. a : rad. - shown that the force applied at the Central hinge of the “togglc joint” has t^e or^ which resists the thrust of the planes, the relation of the sine to the cosine *nc'hnation, or, what is the same, that of tangent to radius; or, in the This movable planes forming one and the same plane, that of 0 to 1, or 1 to oo . thp trip»rSJU ibund more fully treated of, and illustrated with figures of tute vnl ii;0,JnQrPaPerby l*ie wr^er of this note in the Journal of the Franklin Insti- tute, voi. in. p. 3o4, for May, 1S29.—Am. Ed.FUNICULAR POLYGON. 131 equivalent to the sum of the suspended weights, and the on~ sion at the one extremity is equal and oppositely directe o zontal tension at the other extremity. Fig. 145. If the directions of the tensions Sx and Sn at the cords A and B be prolonged to their intersection C, and the points of application of these tensions be transferred to this point, we shall then have the single force P = Vx -f Vn, because the horizontal forces Hx and Ha counter- act each other. Since this force holds in equilibrium the sum Gt + G2 + G3 + • • • °f the suspended weights, the point of application, or centre of gravity of these weights must, therefore, lie in the direction of the same, t. e., in the verlical line passing through the point C. § 142. From the tension Sx of the first portion AKX whose angle of inclination S1AH1 = av the vertical tension follows; V1 — sin. av and the horizontal Hx = S1 cos. ar If, now, wre transfer the point of application of these forces from A to the first node Kv the weight Gx acting vertically downwards meets these tensions, and now for the following portion KY K2, the vertical tension V2 = V1 — G, = Sx sin. ai ~ Gv for which the horizontal tension H2 = Hx = H remains un- changed. Both forces united give the tension of the axis of the second portion S2 = x/ V22 + H2 and its inclination o2 by the formula . F S.sin.a.— G, tang. a9 = —- =r -i—-----l----* 6 2 " S.cos. ax H i, i. e. tang. a2 = tang. cq — Gj H If the point of application of the forces V2 and//2 r new yer- Kx to K2, we obtain in the weight G mectmg portion of the tical force, and therefore the vertical force of th cord132 FUNICULAR POLYGON. Vi=v3-Gi=vi — (G, + G%) = $ sin. a, — (G, + G#), whilst the horizontal force H3 remains = H. The whole tension of the third portion is S3 = s/V* + H2, and for its angle of inclination a3, we have V, S,s£.*t-(Gl+ G,) , # tang. a~ = —— = -1-----^1------------9l' €- 5 H S1cos.al 4 4 @1 G 2 tanga3 = tang. ax------ H For the angle of inclination of the fourth portion of the cord, tang• «4 = tang. Oj----L-7_ ?-----3, &c. H Besides, the tensions 3^, *%, &c., as well as the angles of incli- nation ax, «2, a3, &c., of the separate portions of the cord may easily be represented georaetrically. If we make the horizontal line CA*=* CBy Fig. 146, = the horizontal ten- sion H and the vertical CKX = verti- cal tension V1 at the point of suspen- sion A, the hypothenuse.AK'1 gives the whole tension Sx and the z CAKV also its inclination to the horizon; if now further we apply the weights Gv G2, Gv &c., as parts KXK2, K2K3, &c., of CKy and draw the transversal lines AK2y AK3y &c., we shall have in them the tensions of the successive portions of the cord, and in the angles K2ACy K5ACy &c., the angles of inclination o2, a3y &c. of these portions. § 143. From the investigations of the preceding paragraph, the law for the equilibrium of cords stretched by weights, comes out thus: 1. The horizontal tension is at all points of the cord one and the samey viz.: H= Sx COS. a, = Sn COS. an 2. The vertical tension at any one point is equal to the vertical ten- sion at the other extremity ahove it, less the sum of the intermediate suspended weights, therefore Vm= Vx— (Gx + If the angle ax be known and the horizontal tension H, the vertical tension at the extremity A is*known; Vx = H. tang. al9 and accord- ingly that at the extremity B:Vn=(G1+G2 + ...+Gn)—V1. If, on the other hand, the angles of inclination at and a* at both points of suspension A and B are known, the horizontal and vertical tensions are given at the same time, viz.: » tang. au ^ anj therefore, vi tang. ax Vn = Vt tang' °n tang. axFUNICULAR POLYGON. 133 ,y 4 t« Since Vx + Vn = G, + G2 + ;.., t\*. /tang. ox + tang. on\ p Gt + G2... , it follows that: \ oq / 1 . 1 p __ (Gt + G2 + ..* ) tang. 1 ox + tang. an Vn = (^i + G2 + »■») tang. an^ anj £rom thjg. tang.a« H= Vl cotg. ax = F» co/gj. an. __ rr __ If both sides have the sarae mclmation «„ = ai> tiien v i n Gi + G2 + •.. + Gn anj the one extremity A supports as much as. 2 . the other B. For the rest, these laws hold good also for the funicularpolygon, especially when stretched by forces, if the directions o t e orces are substituted for the verticals. ExampU.The funicular polygon AKt K, K3 Fig. 147, is stretched by three weights Gt = 20, Ga = 30, and G3 = 16 lbs., as well as by the horizontal force Ht pig. 147. = 25 lbs.; required to find the ten- sions of the axis and the angles of in- clination of the sides, in the hypothe- sis that the endsof the string have the same inclination. Here the vertical tensions are equal, viz., Vt = F4 = G, -}- Ga G3 20 -j- 30 16 2 2 33 lbs. The vertical tension of the second portion of the string is Fa = Ft —G,= 33 — 20= 13 lbs., that of the third Fs = F4— Gaor (Gj-|~ Ga _ Ft) = 33 — 16 = 17 lbs.; the an- gles of inclination at and a4 of the ends are determined by tang. a, = tang. a4 — Yjl — ^ — 1 32; that of the se- H 25 cond and third portions by the tang. a^ = tang. a, — tf = b32 — ^7 = ant* tang. a3 = tang. a4_^ = 1,32 — 15 —. 0,68 ; hence at = a4 = 52° 51'; «a = 27 28 H 25 ______________________ a3 = 34° 13'; lastly, the tensions of the axis are St = S4 = y/ Ft* -|- i/2 = y/ 33- + 2,)2 = y/Im = 41,40 lbs., S, = y/VJ+IP = y/132 + 25* = y/794 =28,18 lbs., and S3 = y/r3*+H* = y/173 + 25a = 30,23 lbs. § 144. The Parabola as Ca- tenary. — Let us suppose that the string ACB> Fig. 148, is stretched by equal weights Gv G2, &c., suspended at equal horizontal distances from each other. Let us represent by bx the horizontal distance AM be- tween the point of suspension A and the lowest G, but the ver- tical distance CM by a. Let 12m FUNICULAR POLYGON. us put further for another point 0 of the polygon, the corresponding co-ordinates OJY = y and CJY = x. If, now, the vertical tension of A be = Vt that of 0 will be = - . V, and hence for the angle of in- 6 clination to the horizon, NOT= ROQ = $ of the portion of the string OQ, we shall have tang. $> = — . 21, where H is the constant of the o H horizontal tension. Hence Qfl = OR . tang. , OR . t . 21 is the vertical distance b H of two adjacent angles of the funicular polygon. If we substitute for y OR, 2 OR, 3 OR, &c., the last equation will give the corresponding vertical distancesof the first, second, and third angles, &c., reckoned from below upwards; then, if we add together ali these values, whose amount may be = m, we shall obtain the height CN of the point O vertically above the lowest point C, viz.: X = CJf = 21 • - (OR + 2 OR + 3 OR +... + «• OR) V_ H OR1 (1 “f* 2 -b 3 “b ... bi) 325 V m(m + 1) OR* ~T’ b v- ■ ~ -............. - H i . 2 in accordance with the theory of arithmetical series. Lastly, if OR be put 223 we shall have: V m(wi + l) y* x ~W 2 m* ’ T' If the number of weights be very great, m + 1 may be taken =b m, whence we shall have: , 21 £ H ' 2b‘ For x = a, y = b, hence also: a sss — . and more simply: H 2’ - — which is the equation to a parabola. a b% If, therefore, a string devoid of weight be stretched by infinitely many weights applied at equal horizontal distances, the funicular polygon will pass into a parabola. For the angle of inclination $ we have besides: tang. , «. o b 2y. ® -2y * - H?, as also b* f y , 2a tang. * 3= __ b Therefore the tangent OT cuts the axis of the abscissae, so that CT 9 C JY w* x. If the chains and ioda of a chain bridge, Fig. 149, were withoutCATENARY. 135 weight, or light enough in respect to the weight of the loaded bridge DEFy which only is to be taken into consideration, then the chain ACB would form a parabola. Example. The whole load of a chain-bridge in Fig. 149, =320000 lbs.; the span AB — 25 = 150 feet, and the height of the arch CM = a = 15 feet; to find the tensions anri o*Vio- «-oiatinn. ryf nhniti*. The inclinations of the ends of the chain to the hori- and other relations of the chains. The inclinations — ^ = — = - = 0,4, therefore, a = 21 b 75 5 zon is determined by the formula, tang. a s 48'. The vertical tension at each point of suspension is Vl = h weight 160000 lbs.; the horizontal, H= F, colg. a = 160000.= 400000 lbs.; lastly, the whole ten- sion at one end: S = v/r2-f- H2 —: v^/l + cotg. a5 = 160000 = 100000 = 80000^/29 = 430813 lbs. § 145. Catenary.—When a perfectly flexible and extensible string suspended from two points, or a chain consisting of short links, is stretched by its own weight, its axis forras a curved line, to which the name of catenary has been given. The imperfectly elastic and ex- tensible cords, ropes, bands, chains, &c., met with in practice, give curved lines which approximate to the catenary only, but may usually be treated as such. From the foregoing, the horizontal tension of the catenary is equally great at all points, on the other hand, the vertical tension is equivalent to the vertical tension of the points of suspension lying above it, less the weight of the portions of the chain above. Since the tension at the vertex, where the catenary is horizontal, is null, the vertical tension, there- fore, at the point of suspension, is equivalent to the weight of the chain from that point to the vertex; and the vertical tension at each place also equivalent to the weight of the portion of the rope or chain lying below it. If equal lengths of the chain be equally heavy,we have then the common catenary, which only we will now consider. If Fig. 150.136 CATENARY. a portion of the rope, or chain one foot in length, weighs y, and if the arc corre&ponding to the co-ordinates CM*=a and MA=by Fig. 150, AOC*szly we then have the weight of the portion of the chain AOC = Zy; if, on the other hand, the length of the arc (Z) corresponding to the co-ordinates (CjW*x, and JV*0=y) =s, we have the weight of this arc =s y. If we put the length of a similar portion, whose weight = Hy = c, (the horizontal tension,) we have further i/=c y, and, therefore, for the angles of inclination o and at the points A and O: tang, a — tang. SAH= — = = -, and H C y C tang.

= sin. $ + - sin. $>3 + — sin. o 40 V5 + . ..; if we have regard only to the two first members, it therefore follows that: 2x 1 /2x\3 2x 4 /x\3 y 6 \y / y 3’ \y/ ‘ Now the arc AC = s = r $ = —.t; hence : 2x ^ 2 2* S — V and 2 x* r 2/xyn -J + j-y-sLl + aQ} But inversely, y = s 1 + y = s -10] , which may be put: and on the other hand : X= !| y{s-y). r the rest is not known, Example. The width of a very compressed arc, °forei is: is 2 6 = 3,5 feet, and the height a = 0,25 feet; lts length, __« 543 r 2 /025\n . 2 n 14^—3 54-3,5.0,0136 = 5,048 it. 2 l = 3,5 [1 + | . (g) ] =3,5 (1 + j • 0,143*)-3,5 + 12*138 CATENARY. § 148. We will now apply the formula s = y e* -1 (di for Fig. 153. the length of a compressed arc to a strongly stretched catenary JlCBy Fig. 153, while we put the vertical tension at a point O, = V = Sy — y 4- ? J . r, and therefore for the angle made by the tan- gent TOJV = *, tang. * = L = ± £l + | g) J. If we divide the ordinate y into m equal parts, we find the portion RQ = NU of the absciss x corresponding to one such part ORy when we put RQ = OR . tang. $ = OR . ^£l + m Since x is small in comparison with y,RQ is approximately = OR . V-. If now we put OR = X and successively for ^, &c c m * we obtain by degrees the several parts of x, whose sum, therefore, is cm* cmr (§ 144) and which corresponds with the equation to the parabola. 2c But if we wish to attain greater accuracy, we must put QR=OR . ?[l + ? substitute for x its value last found J^-, and we shall then obtain: Let us again successively put y = —, —, —, &c., and for OR m m m likewise we shall then find the several values of x, and the sum m itself: X-s[w<1+2+3+-• )’(1’+2,+3>+.. Now for a very great number of members, the sum of the natural numbers from 1 to m = and the sum of their cubes = ac- cordingly: c (,2+6 c1 ' 4 / 'CATENARY. 139 12 c* 1 ,_y* , V* _y* f, , J_ /y\S"j,theequationofastrongly -X-2~c + 2T?-2~cL+ 12 WJ stretched catenary. 4 c* x* By inversion it follows that y2= 2 c a: — 5=5 ® c x a?2 = 2 cx----—, therefore: 2. ^2cx—y> or approximately — ^/2cx^l — —^ The measure of the horizontal tension is further given * y3 , y* , y* 4x* ------- 2x^ 2x. 12 c* 3. «_£+£ 2a: 6 The angle of the tangent * is determined by: yTi + ?f-Vl . yr, L 3\y/J "c'c 4. Lastly, we must here place the formula of rectification found in the former paragraph: 5- »-»[i + l(*)’]-!'[i + §(?)’]• Example,—1. For a span 2 b =s 16 feet and height of arc a = 2 J feet, the length 2 l is s 16 £ 1 + ^(~y J ■* 16 + 16.0,065 « 17,04 feet, the length of the portion of chain which measures the horizontal tension: c=== — +?.:=: — + —=== 12,8+0,417 Qn~« * ' J2 «« O D J2 - 13,217 feet; the tangent of the angle of suspension: tang. «=^[1 + I /i)*] 1 1/5 \ #“l 5.103255 ^ ^ 8 L 3" \i(5/ J®*------^----- = 0,6453 ..., the angle of suspension, therefore, 32 A chain of 10 feet length and 9J span, has the height of its arc a= f-n—ja5 __ /3 (10—9f) 9$ /3 19 /57 ^ 2 N2 2 2 *J 2 9 16*432 1j7812 1,335^ feet, and the measure of the horizontal tension: 2« *6 4,75* + 1,335 B 8,673 feet 3 Jf , '6 2,1,335 1 g------------j— *—” 20 lbs I"6 ^ ^Get an<* we*ghing 8 Ibs., be stretched horizontali/ b/ a force of ** v^^VertiCal—i<>n r==i G==4 lb8‘i ^e horizontal force H * v**-*"* v' ** \/384 as 19,596 lbs. The tangent of the angle of suspension: tcifur a r 4 0 g ** ~1~9[59Q ™= 0>20412, the angle ? itself «« 11° 32'; tbe measure of140 THE PULLET. tr o qn the horizontal tension c ss — =: H -f- — as — H 30 73,485 feet; the span 2 6 -2‘ [‘-T- -aW9s», and the height of the arc a = b (l—6) = 3 29,792.0,208 2.2 = ^/29,792.0,078 = 1,524 feet § 149. The higher calculus gives the following general formulae for the catenary, and which hold good for all tensions. 1. s m* x/ 2 c x + x3, and inversely, x =» %/ c2 + s2 — c and s2 — ar* c = —---------. 2. s =~ yi^—e *y inversely y*=cLn where e is the base 2,71828 of the natural system of logarithms, and L n the logarithm = 2,30258 times the common logarithm. 3. ,=cLn(c+x+ in^elyx- |(e?+;T)-,, 4. y = 2x The use of these formulae is very troublesome, especially in com- plicated problems, where a direct solution is generally not possible. Example. The two coordinates of a catenary are x =s 2 feet, and y — 3 feet; required the horizon tal tension c of this curve? Approximately from No. 3 of the former paragraphs = — -f- — = 2,58. From No. 3 of the present para* graphs y isexactly s c Ln * 4* cx-\- ^ 3 = c Zn 2.~tl If c be here put = 2,58, we then have the error / = 3 — 2,58 Ln ^458 + 2\/3,58\ \ 2,58 / ^ / 8,3642 \ = 3- 2,58 Ln ^ ---~J = 3 — 3,035 == — 0,035; but if c be put = 2,53, we then have the enor /, = 3 — 2,53 Ln /?■£!+ 2 vA53 \ _ 3_ 2,53 Ln V 2,53 / V 2,53 / * 3 — 3,002 ss — 0,002. In order noW to find the triie value of c, if, according to a known rule, we put sstJL^s.2P—^ = 17,5; in this mariner it will follow that: o—2,53 /j 0,002 ’ 16,5 . c ss 17,5.2,53 —2,58 =s* 41,69; therefore : c * —-ff ■ — 2,527 feet 16,5 Remark. Practical applications of the catenary will be given when, in the Second Part, we come to treat of the eonstruction of vaults, chain-bridges, &c. § Tfe Pulley.—Ropes, cords, &c., are the usual means by which forces are transmitted over the wheel and axle. We will here develop what is most general in the theories of these two arrange- however, taking into account friction and rigidity of A pulley is a circular disc, ABC, Fig. 154 and Fig. 155, turning about an axis on whose circumference lies a cord or stnng, and whoseTHE PULLEY. 141 extremities are stretched by the forces P and Q. In a fixed pulley, the block in which the axis or pivot reposes is immovable; in a free pulley, on the other hand, it is movable. Fig. 154. Fig. 155. In the condition of equilibrium of a pulley, the forces P and Q at the extremities of the string are equal; for every pulley is a bent lever, the arms of which are equal in length, which we may obtain if we let fall perpendiculars CA and CB from the axis C on the directions of the forces, or of the strings DP and DQ. It is ciear that the forces P and Q in any revolution about C describe the sanie space, viz. r $>, if r be the radius CA=CB and $>° the angle of revolution; and that fromthiswe may infer the equality between P and Q. From the forces P and Q there arises the resultant CR=R, which is taken up by the block and is dependent on the angle ADB=a, which the directions of the string include; and moreover it gives as the diagonal of the rhomb CP1RQ1 constructed from P and a : R = 2 P cos. |. §151. In the fixed pulley, Fig. 154, the force Q consists of the weight to be overcome or raised at one extremity of the string; here, therefore, the force is equal to the weight, and the application of this pulley effects nothing but a change of direction. In the movable pulley, Fig. 155, on the other hand, the weight on the hook R acts at the extremity of the block, whilst the one extremity of the string is fastened to a fixed object; here, thereforce, the force P is to be put = ------— If we represent the chord AMB, which corresponds to the 2 cos. * 2 arc over which the string passes, by a, the radius CA^= CB, as before = r, then a == 2AM = 2 . CAcos. CAM = 2 CA cos. ADM = 2r cos. -,hence - may be put = --------—, and likewise — From z a J r a R a 2 cos. - 2142 THE WHEEL AND AXLE. Fig. 156. this, therefore, the powerin the movable pnlley is as the radius of the jmlley to the chord of the arc over which the string passes. ... If a = 2 r,the string passes over a semicircle, Fio- 156 the force then is at a minimum; viz. pt fft.ifa = r, that is 60» of the part of ,he pulleyover which the string passes, we have = . the smaller, therefore, a becomes, the greater is P, and for « infinitely small, the force P becomes in- finitely great. An inverse proportion takes place in the spaces; if s is the space of P, which cor- responds to a space we have then Ps = Rh, therefore, 1 = -. h r The movable pulley is thus a means of modifying force; for example, a given weight may by this means be raised by a smaller force, but in propor- tion as there is gain in force, there is loss in space. Remarlc. We shall treat of the composition of putleys and systema of pulleys, as well as of”the resistances arising from friction and rigidi.y, more fully m a subsequen, Part. & 152 The Wheel and Axle.—The wheel and axle is a rigid con- nection of two fixed pulleys or wheels, capab e «f j^^^bout a »■»»»" axis ABFC, Fig. 157. The ihsse .b«l. ,s Fig. 157. greater one the wheel. The round extremities E and F> on which this arrangement rests, are called gudgeons. The axis of revolution of the wheel and axle is either horizontal, or vertical, or inclined. Here we shall only speak of the wheel and axle which revolves about a hori- zontal axis. We shall also here suppose, that the forces P and Q, or the power P and the weight Q act at the ex- tremities of a perfectly flexible string, which passes round the cir- cumference of the wheel and axle. The questions to be answered are, in what relations the powers and weights are to each other, and what pressures the gudgeons E and .Fhave to sustain? Let us imagine a horizontal plane passed through the axis CD and the points of application A and B of the power P, and the weight Q transferred to this plane, and therefore P and Q applied at Ax and Bx. If the angles AAXC and BBxDy which both forces make with the143 THE WHEEL AND AXLE. horizon s=5 a and p, these forces may be replaced by the horizontal forces R = P cos. », Q cos. p, and by the vertical forces PX=*P sin. o, Qj = Q sin. p. The horizontal forces are directed towards the axis, and being applied at C and D, become perfectly counteracted by the axis. The vertical forces Px and Qlf on the other hand, tend to turn the wheel and axle about its axis. If K be the intersection with the axis of the line connecting the points Ax and Bv KAX and KBX are the arms of Px and Qx, and equilibrium subsists about K, and also about CTJ, if: Pi • KAl = Q, . KBV or, since EB - if pi 9i Q P . CMl = DB DB* Q,. Mu or, as sinCeSB/ £«g|,and P CAX CA CAX QDB . ])B, %. e. CAX 1 J)BX P . CA = Q . DB, or Pa = Qb, an(* ^ rePfesent the arms of the power and weight, or the radii ot the wheel and axle. In the wheel and axle, therefore, as in every ^v.er*^ke Hioment of the power is equivalent to the moment of the ^orces Qi give at K a vertical pressure Px+ Q., with wjuch must also be associated the weight G of the whole wheel and axle applied at the centre of gravity S. The supports of the §>* »^r?n,S J?* ^ ^ave to sustain the vertical pressure }i 7T i3=5 j Mn‘ ° ^ & + G. If we put the whole length oi the wheel and axle measured from E to F= L, the part XC=- llt 7i = l2) therefore X = and the distances ES and ot the centre of gravity S from the supports d. and «L therefore also X = dx + d3t we shall obtain since r BE_______P*1 DC1\+Q for the vertical pressure Xx at the gudgeon E: Xx.EF=* G .FS+(Pt + QJJFK, x>= Gdj + (P, + Qi) (l* + pi+ • *) i.e. Gda + (P, 4- Q,) h + Pt* L On the other hand, for the vertical pressure Xt at F: X3 . EF= G . ES+ (P,+ Q,) EK, i, e. Gdx + (P, + Q,) (lx 4- p^Q •144 THE WHEEL AND AXLE. The horizontal forces R and S have the moments about F, R t FC = R (/ + /2), and S . FD = S . l2, and about E : S . ED = $ {l + J,), and R . EC = Rlx ; if, therefore, we put the horizontal pres- sures upon E and F effected by them = Yx and F2, we shall obtain: Yx . FE=* R . FC—S. FD, as Y.. FE— 5. ED—R .EC, as v _ (^+^i)—fix x*~ L From X, and F, the total pressure at E is: Zx = v^X,* 4- F,2, and likewise from X, and F2, the same at P. v/Xa2+ F,2. Lastly, if ? and 4. be the angles which the directions of these pres- sures make with the horizon, we shall then have ^ X X tang. t = -^ and tang + = Example. The weight Q of a wheel and axle pulls perpendicularly downwards, and amounts to 365 lbs.; the radius of the wheel a = lf ft. *, that of the axle b £ ft.; t^e weight of the machine itself is 200 lbs.; its centre of gravity S lies distant from E and F, dx s 14, and d2 = 2$ ft.; the middle of the wheel is about l% = J ft. from the gud- geon J5, and the vertical plane in which the weight acts is about /2 =2 2 ft. from the gudgeon F. Now if the force P necessary for restoring the equilibrium at the wheel inclined to the horizon at an angle 50° = a, pulls downwards, what will tliis be, and what will be the pressures on the gudgeons ? Q = 365, 0 = 90°, consequently Q, = Q sin. 0 = Q and S = Q cos. 0 = 0; further, P being unknown, and a = 50°, conse. quently P, =s P sin. a = 0,7660 .P and R = P cos. * = 0,6428 . P; but now a = 1$ sss l and b = f, it follows, therefore, P = — Q = 4 • 365 = 156,4 lbs., P, = 119,8 * a and P = 100,5. Further, because G = 200, dt as |, d3 = £, — $, / — 2, Z = 3 4, f “ ant* l *= L— = 4— i1 = f» so that the vertical pressure at E is •* v- __200 . | -f- (365 -f- 119,8). 2 + 119,8 . | 1619,35 Ai -----------------------4----------------------- —j— = 404,8 lbs. and that at F: 200 . 1+ (365 + 119,8) . j + 365 . j 1119,85 4 4 Both of these forces together give : Xx -f* X2 = Q -f" G "h P\ = 684,8 lbs. The horizontal force at E is: = 280,0 lbs. Y, 100, 5 . (| + 2) — 0.2 81,7 lbs., and that at F: v_0.(f+ f)- x2-------- 100,5 . J 18,8 lbs. the sum of these is exactly = R + S =2 100,5 lbs. , The pressure at E is inclined at an angle to the horizon, for whic we tang. , Log. tang. $ = 0,69502, 4 = 78° 35 . Yt 81,7 The pressure itself: Zx = ^L. =413,0 lbs. »». 9 Vm On the other hand, for the inclination 4 of the pressure at i* : tang. 4 = A* = ,?8M, Zog. ftrng. 4 = 1,17300, 4 = 86°, 9', 5 ; Jfa 18,8 v and the pressure: Z2 --------2— = 280,6 lbs. cos. 4RESISTANCES OF FRICTION AND RIGIDITY. 145 CHAPTER V. ON THE RESISTANCES OF FRICTION AND RIGIDITY. § 154. We have hitherto assumed that two bodies can only act upon each other by forces at right angles to the plane of contact. If the surfaces at the point of contact were perfectly mathematical, *. e. not interrupted by the smallest irregular elevations or depressions, this law would also be fully confirmed by experience; but becausei everybody possesses a certain degree of elasticity or softness, and because the surface of every body, even if it is smoothed or polished in a high degree, has stili some small elevations or indentures, and in consequence of the porosity of matter, no continuity; therefore, by the reciprocal action of two bodies in contact, reciprocal impres- sions and partial penetration of the parts take place at the point of contact, by which an adhesion of the two bodies is caused, which can only be overcome by a distinet force, whose direction coincides with the plane of contact. This adhesion, produced by the impression and partial penetration of the bodies in contact, and the resistance on the plane of contact arising from it, has obtained the name of friction. Friction presents itself in the motion of bodies as a passive power or resistance, because it only impedes and checks motion, but never produces nor promotes it. It is introduced into investigations in mechanics as a force which is opposed to every motion, whose direction lies in the plane of con- tact of two bodies. In whatever direction we move forward a body resting on a horizontal or inclined plane, friction will always act opposite to the direction of motion; for example, it will impede the ascent as much as the descent of a body on an inclined plane. The smallest addition of force produces motion in a system of forces in equilibrium, so long as friction is not called into action; but when the same exerts its effect, a greater addition of force, dependent on the friction, is required to disturb the equilibrium. § 155. On overcoming friction, the parts in contact are compressed, and those which protrude, bent down, tom, or broken off, &c. Fric- tion is not only dependent on the roughness or smoothness of the sur- faces in contact, but also on the physical properties of the bodies them- selves. Hard metals, for instance, cause less friction than soft. We can, however, lay down no general rules a priori of the dependence of friction on the physical properties of bodies; it is, on the contrary, necessary to make experiments on friction with bodies of different substances in order to find out the friction which takes place under various circumstances between bodies of the same substance. The unguents which are applied to the rubbing surfaces exert a particular influence upon the friction and on the abrasions arising 13146 KWDS OF FRICTION. from the contact of bodies. The pores are filled up and other aspe- rities diminished, and in general, the further penetration of the bodies prevented by the fluid or semi-fluid unguents, such as oil, tallow, fat, soap, &c., for which reason these occasion a considerable diminution of friction.* Friction must not, however, be confounded with adhesion, i. e% with that holding together of two bodies, which takes place when they come into contact at many points without reciprocal pressure. Adhesion increases with the size of the surface in contact, and is in- dependent of the pressure, whilst the contrary is the case with fric- tion. If the pressure be slight, the adhesion will be considerable in proportion to the friction; but if the pressure be considerable, then it will constitute but a small part of the friction, and, therefore, gene- rally may be neglected. Unguents, like ali fluid bodies, increase the adhesion, because they increase the number of the points of contact. § 156. Kinds of Friction.—Two kinds of friction are distinguish- able, viz., the rolling and the sliding. Sliding friction is that kind of resistance which is given out when a body so moves that ali its points describe parallel lines. Rolling friction, on the other hand, is that resistance which arises from rolling, i. e. that motion of a body which moves progressively and rolls at the same time, and whose point of contact describes as great a space upon the body in motion as upon the body at rest. A body M supporting itself upon the plane HR, Fig. 158, for instance, moves sliding over the plane, and conse- quently has to overcome sliding friction when its points, A, B, C de- scribe parallel spaces AAX, BBX, CCX, &c., and therefore all these Fig. 158. Fig. 159. points of the inoving body come into contact with others of the sup- port. The body M, Fig. 159, on the other hand, rolls upon the plane HR, and has to overcome rolling friction, when the points A, B, & c., of the surface so move that the space ABX = AB = AXBX, likewise AD = AE, and BXE = BXDX, &c. The friction of axles is a particular kind of sliding friction, which arises when a cylindrical axle revolves in its bearing. We distin- guish two kinds of axles, the gudgeon and the pivot. The gudgeon rubs against its support or envelop, whilst its other points always successively come into contact with the same points of the support. f * ^ he “anti attrition metal,” composed of copper 1 lb., antimony 2 lbs., and tin 3 lbs., (anerwards tempered or softened byre-melting with more tin,) now very generally used by machimsts, in the United States, performs the 9atne office, and prevents the heatine of gudgeons, boxes, &c.—Am. Ed.LAWS OF FRICTION—CO-EFFICIENT FRICTION. 147 The pivot, on the other hand, presses with its circular base against its support, where its points revolve in concentric circles. Further, particular frictions arise when a body oscillates upon a sharp edge, as in the balance, or when a vibrating body reposes upon a point, as in the magnetic needle. Lastly, we distinguish the friction of quiescence which is to be overcome, when a body at rest is put into motion, from the friction of motion which opposes itself to the transmission of motion. § 157. Laws of Friction.—The general laws to which friction is subject, are the following: 1. Friction is proportional to the normal pressure between the rub- bing bodies. If a body be pressed against another by a double force, the friction is as great again; three times the pressure gives three times the friction. If in small pressures this law varies from ob- servation, it must be attributed to the proportionately greater effect of adhesion. 2. Friction is independent of the extent of the surfaces of contact. The greater the surfaces are, the greater is the number of parts which rub against each other; the smaller the pressure, the less the friction of each part; the sum of the frictions of all the parts is the same for a greater as for a less surface, in so far as the pressure and the other circumstances remain the same. If the side surfaces of a parallelo- pipedical brick are of the same quality, the force necessary to push it along a horizontal plane is the same, whether it rest upon the least, the mean, or the greatest surface. With very large side surfaces and with small pressures, this law has exceptions, in consequence of the effect of adhesion. 3. The friction of quiescence is indeed generally greater than that of motion ; the last, however, is independent of the velocity; it is the same in small as in great velocities. 4. The friction of greased surfaces is generally less than that of un- greased, and depends less on the rubbing bodies than on the unguents. 5. The friction of gudgeons revolving on their bearings is less than the common sliding friction; the friction of rolling is in most cases so small, that it need hardly be taken into account in comparison with the sliding friction. § 158. Co-efficient of Friction.— From the first law laid down in the former paragraph, the following may be deduced. A body AC, Fig. 160, presses against its support, first with the force JV*, and requires to draw it along, i. e. to overcome its friction, the exertion of a certain force F, and se- condly with the force Nv and requires the force F, to cause it to pass from a state of rest into one of motion. From the foregoing we ha\e . ~ = and therefore F — ^ . JV*. t j Jv1 Fig. 160.148 THE ANGLE AND CONE OF FRICTION. If, bV experiment, we have found the f»ctl°" f, corresponding t0 a certain pressure JV„ we hence find, if the rubbing bod.es, and the other circumstances are the same, the to another pressure JV when we multiply th.s pressure by the ratio of the values Ft and JV, corresponding to the first obser- VaThne' ratio of the friction to the pressure or the friction for a pres- sure lunity, a pound, for instance, » called theleo^hmt Zn, and will in the sequel be expressed by/, wheretore we may ge. nCThe TefficiTnfof friction is different for different substances and different conditions of friction, and must therefore be found out by experiment for each particular case. When a body AC is drawn a distance over a surface, there is a mechanical effect Fs to perform; the mechamcal effector work re- quired to overcome friction is, therefore,/JVs, equal to the productof the co-efficient of friction, the normal pressure, and the distance along the plane of contact. When the plane is also moving, we must then understand by s the relative distance. Exampk— 1 If by a pressure of 260 lbs., the friction amounts to 91 lbs., the corre- sponding concient of friction is / = ^ = 0.35.-2. To draw a 8,edge of 500 lbs. weight along a horizonml.and -^oTs^Tf TeffiS ^ friction is/—0,04, the d roaj j9’ 0,45 and the load amounts to 600 lbs., the s.»- u -/* -».«• •» ■»»- A. lbs. & 159 The Angle of Friction and the Cone of A body §io». g j ^c,Fig. 161, lies on an mclined plane FH, whose angle of inclination FHR = a, its weight G resolves itself into the normal pressare N= G cos. a and into the parallel force P = G sin. a. From the first force there arises the friction F f G cos• Oj which is op- posed to every raotion upon the plane, , wherefore the force to push it upwards on the plane = F + P = fGcos. a + a - G, on the other hand, the force to push itdownwards is-F—• P = {f cos. a — sin.a) G; the last force is null, e. the body is sustained by its friction on the plane, when sin. a = f cos. a, i. e. when the tang. a =/. As long as the inclined plane has an angle of mclmation, whose tangent is less than f, the body remains at rest on the plane, but when the tangent of this angle is a little greater than/, the body immediately begins to slide down. The angle, whose tangent is equal to the co-efficient of friction, is called the angle ot friction orthe angle of repose. The co-efficient of friction is given by observing the angle of friction p (for the friction of repose), when f is put = tang. p. Fig. 161.EXPERIMENTS ON FRICTION. 149 In consequence of friction, the surface FH, Fig. 162, reacts not °nly against the normal pressure JVof another body AB; but also against its oblique pres- sure P, when the deviation NBP = $> of the direction of this pressure from the normal BJY does not exeeed the angle of friction, for since the force P gives the normal pressure BN = P . cos. $>, and the lateral ortangential pressure BS = S = P sin. , and there arises from the normal pressure P cos. $ the friction f P cos. <*> opposed to every motion in the plane FH, S will therefore be unable to give rise to motion, and will remain in equilibrium so long asfP cos. d>> P sin. , or f cos. $>> sin. $,i.e. tang. * is < p. If the angle of repose CBD = p be made to revolve about the normal CB, it will describe a cone, which we may call the cone of friction or resistance. The cone of resistance includes ali those directions of force by which a perfect counteraction o eo lique pressure takes place. Example. To draw a filled cask weigbing 200 lbs. up an inclined wooden plane of 50°, the force required with a co-efficient of friction /=0,48 is = P = (/«»*• *"r*1”-a) = (0,48 cos. 50° + sin. 50°) . 200 = (0,308 + 0,766) . 200 = 215 lbs.; to let it down, or to prevent its sliding down, the force required, on the other hand, is: (P =/co*. a nn. a) G = — (sin. 50° — 0,48 . cos. 50°) 200 = — (0,766 — 0,308) . 200 = — 91,5 lbs. § 160. Experiments on Friction.—Experiments on friction have been made by many philosophers, the most extensive of which, and on the greatest scale, are those of Coulomb and Morin. To find out the co-efficients of friction for sliding motion, these two made use of a sledge sliding on a horizontal surface, which was pulled forwardby a cord, passing over a fixed pulley, from which weights were sus- pended, as in Fig. 163, where AB represents the way, CD the sledge, E the pulley, and G the weight. To obtain the co-efficient of fric- tion for different substances, the surfaces in contact, not only of the sledge, but also of the way forming the support, were cov- ered with the smoothest possible pieces of the substances under experiment, such as wood, iron, &c. &c. The co-efficients of the friction of repose were given by the weight which was necessary to cause the sledge to pass from a state of rest into one of motion, an the co-efficient of the friction of motion by the time t, which t e s

^ f^ere fore follows from § 65, that the acceleration of the umformly accele- 13* Fig. 163. Fig. 162.150 EXPERIMENTS ON FRICTION. rated motion arising, is : p = an<^ inversely, the co-effi- i T w cient of friction f = L — . £. But * = K (§ 11), there- P+ 2 _ G * g, r. . To measure the co-efficient of friction, for axle friction, a fixed fore, p =_,and/ = — Fig. 164. pulley JCB, Fig. 164, w made use of over which a cord passes, which is stretched by the weiehts P and Q. From the sum of the weights, the pressure P + Qis given, and from their dif- ference P— Q the force at the circumference of the Dullev, which is in equilibrium with the fric- tion of the axle, + Q),if n°w = a the radius of the pulley, and CD=r that of the axle, we have from the equality of moments C P_ O) a = Fr =f(P + Q) r> antl therefore ^ - — Q , the friction of repose; J q * “* on «ha band, for that of motion, if the weightP faUa » «paee a in the time (£), and Q rises as much, f = q giy r Remark. Before Coulomb, Arnontons, Camus, Bulffinger, Muschenbroek, Ferguson, Vince, and others turned their atteption to and made experiments on friction. The re- sults of ali these investigations are of little value in practice, because tliey were con- ducted upon too small a scale. The experiments of Ximenes, which were made about the same time as those of Coulomb, also fail in this respect. The results are to be found in a work, “Teoria e Pratica delle resistenze de’ solidi ne’ loro attriti,” Pisa, 1782. The experiments of Coulomb are fully described in his work, “Theorie des Machines sim- ples,” 1821. The latest experiments upon friction are those of Rennie and Morin. Rennie used for his experiments partly, a sledge upon a horizontal surface, and partly upon an inclined plane, from which the bodies were allowed to slide down, and by which the amount of the friction was deduced from the angle of friction. Rennie’s ex- periments extend to substances of various kinds met with in practice, as cloth, leather, wood, stones, and metals; they give important results upon the abrasion of bodies, but from the apparatus and the mode of conducting these experiments, we cannot rely upon them for that accuracy which those of Morin appear to have attained. The experiments of Rennie are to be found in the “ Philosophical Transactions” of 1818. The most ex- tensive experiments, and promising a high degree of accuracy, have been completed by Morin, although it cannot be denied that they leave some doubts and uncertainties, and somewhat to be desired. This is not the place to describe the methods and apparatus of these experiments; we can only refer to the author’s writings, “Nouvelles Exp^riences sur le Frottement,” par Morin. An excellent article on Friction, and a full description of all the experiments upon it, especially those of Morin, is given by Brix in the “Trans- actions of the Society for the Promotion of Manufacturing Industry in Prussia,” 16 and 17 Jahrgang, Berlin, 1837-8 * * A series of experiments on the resistance of friction, P^rt^.u^r^332 chap. 6 • Vid. way cars, will be found in “Wood’s Treatise on Railways, 2d .} > ) Smith’s Am. ed., pp. 171-228.—Am. Ed.EXPERIMENTS ON FRICTION 151 § 161. The following tables contain a condensed summary of the co-efficients of friction the most useful in practice. TABLE I. CO-EFFICIENTS OF THE FRICTION OF REPOSE. Nature of the surfaces and unguents. JtAMES OF BODIES. b Q Damped with water. ‘o © o S £ ■s J * o H d 1 b p •i & 'd c 03 -d © 1 £ Greasy and wetted. Wood upon wood - • C least, J mean, j greatest, values, 0,30 0,50 0,70 0,65 0,68 0,71 — 0,21 0,14 0,19 0,25 0,22 0,36 0,44 0,30 0,35 0,40 — Metal upon metal - ■ fleast, J mean, ] greatest, L values, 0,15 0,18 0,24 — 0,11 0,12 0,16 0,10 0,11 — 0,15 — Wood upon metal - - 0,60 0,65 0,10 0,12 0,12 — 0,10 — Hempen ropes, twisted or matted, upon wood' (least, 1 mean, | greatest, ^ values, 0,50 0,63 0,80 0,87 Thick sole leather,upon wood or iron ( high at the edges, ( flat or smooth, 0,43 0,62 0,62 0,80 0,12 0,13 — — — — 0,27 Black strap leather, upon pulleys ( of wood, ( of iron, 0,47 0,54 — — — — — 0,28 0,38 Stones or bricks upon stones or bricks, smooth worked C least, < greatest, ( value, 0,67 0,75 Stones and wrought iron C least, < greatest, ( value, 0,42 0,49 ■ Oak upon muschekalk - 0,64 7 152 INCLINED PLANE. TABLE II. CO-EFFICIENTS OF THE FRICTION OF MOTION Nature of the surfaces and unguents. HAMES OF BODIES. Wood upon wood - j r least, mean, greatest, t value, 1 f least, Metal upon metal - < 1 mean, | greatest, L value, Wood upon metal - j f least, [ mean, | greatest, [ value, Hemp, cords, twists, J kc. ( j on wood, j on iron, Sole leather, smooth, upon wood or metal ' ( raw, ? compressed, l greasy, The same, high at the ] edges, &c. i fdry, i greasy, Q Water. Olive oiL Lard. Tallow. Lard and black lead. I § t jj £ Dry soap. Greasy and wetted. 0,20 0,06 0,06 — — 0,14 0,08 0,36 0,25 — 0,07 0,07 — — 0,15 0,12 0,48 — — 0,07 0,08 — 0,16 0,15 0,15 0,06 0,07 0,07 0,06 0,12 - 0,11 0,18 0,31 0,07 0,0tt 0,09 0,08 0,15 0,20 0,13 0,24 — 0,08 0,11 0,11 0,09 0,17 — 0,17 0,20 0,05 0,07 0,06 __ . 0,10 0,42 0,24 0,06 0,07 0,08 0,08 0,10 0,20 0,14 0,62 — 0,08 0,08 0,10 — — — 0,16 0,45 0,33 — — 0,15 — 0,19 0,54 0,36 0,16 — 0,20 0,30 — — 0,25 0,34 0,31 0,14 — 0,14 0,24 Remark. The co-efficients of friction for porous masses will be given in the Second Part, in the theory of the pressure of earth. § 162. Inclined Plane.—The theory of sliding friction has its chief application in the investigation of the equilibrium of a body ACy on an in- clined plane FH, Fig. 165. If in accordance with § 135, FHR = a, the angle of inclination of the in- clined plane, and POS^p, the angle which the force P makes with the in- clined plane, we have the normal force arising from the weight G of the body JV = G cos. a, on the other hand, the force for sliding down = S= G sin. a, further the force with which P strives to draw the body down the plane is = P sin. p,INCLINED PLANE. 153 and the force Sl with which it pushes the body up the plane = P cos. fi. The remaining normal pressure is: N—Nt = G cos. a—P sin. fi, consequently the friction F=f(G cos. a—P sin. fi). If it be required to find the force P drawing the body up the plane, then there will be friction to overcome, and it must therefore be Sx = S + F, i. e. P COS. /3 as 'Gr sin. a + f{G COS. a — P sin. /3). But if the force, which is to prevent the body from sliding down is to be determined, then friction comes to its assistance, and the force is: + F = S, i. e. P cos. fi + f (G cos. a — P sin. fi) = G sin. a. From this the force may be determined: For the first case: P = ^n-a+ fc°s-* . G, cos.i3 + fsin.il For the second: P = S*n' -—• G. cos. fi—Jsin.fi If the angle of friction pbe introduced, whilst we puty* = tang. p = sin. p cos. p' , we shall obtain P = sin. a . COS. p + COS. a. sin. p sin. /3. cos. p + cos. /3. sin. p sin. (a + p) G, or from the known rules of trigonometry: P = —. G, and the cos. (fi + p) upper signs are to be taken, when motion is to be brought about; the lower, on the other hand, when motion is to be impeded. The last formula is found by a simple application of the paral- lelogram of forces. Since a body counteracts that force of another body, which deviates by the angle of friction p from the normal to its surface (§ 159), equilibrium in the foregoing case can subsist if the resultant OQ = Q of the components P and G makes with the normal ON the angle NOQ = p. If now we put in the general formula^- = GOK = GOJV + JYOQ = » + p, and POQ = POS1 + «SjOQ = fi + 90° — p, we then have = sin. (o + p) sin. (a + p) , r . , * „. -------------^t-------------4, and for a negative value of p. . / , *\ cos. (fi—p) 6 Sira.h3—p+^j ^ p; ■P sin. (o>■ p) t < G = cos. (fi + p)’ quite in accor^ance with the above. If the body reposes on a horizontal plane a = 0, therefore, the force to push P forwardis: P = f G _ f G sin, p Tf t. c cos. fi+fsin.fi cos. (fi—f)’ e torce acts parallel to the inclined plane, then fi =» 0, and therefore, P = (sin. a -j^f Cos. «) G, = ° jl p ) _ q, (compare § 159)- If the force acts horizontally fi = — «; cos. fi = cos154 WEDGE. tang. (<» +p) (*• (sin. a +^f COS. a 0 and sin. p = — fin «, therefore, -= 7«. . ~+ fsin. “ G> ato 1 ApdMbe f°rce to pnah a body V.eforflhTd1re«L of force <]'”■>’<'! b, the angle of friction from the inclined plane, the force itself is the least and = (* + P) • sin Fig. 166. — \ > / ~ What nressure on the axis has the prop AE, Fig. 166, to sustain, in order Example. What pressure o ^ a block of stone (a wall) ABCD, of 500011*. weight from slipping down the inclined plane CD supposing the angle of the prop to the horizon to be 35°, that of the inclined plane CD, 50°, and the co- efficient of friction / = 0,75 ? Here G = 5000, «, =* 50°, B 35° — 50° = — 15°, and/= 0,75; therefore the formula gives: sin. a—f cos. a „ sin. 50°—0,75 cos. 50° Pz=cos.B-fsin.B ’ “cos. 15°-t-0,7-5000 0!766-0^ ^ = ^ 224 0,966+0,194 1,160 If the prop were horizontal, we should have B =_ 50°, and tang. p = 0,75; hence p = 36° 52'; lastly, P==G tang. («—P) = 5000 tang. (50°-36° 52') i 5000 tang. 13° 8' = 5000.0,2333 = 1166 lbs. To ,________________________ push up the same wall upon the supporting one by a horizontal force, under otherwise similar ^c»m*ta"c^-a f°r“^sT“ld ta nece88ary = G tang. (»+(.) = 5000 tang. 86° 52' = 5000.18,2670 = 91338 lbs. S 163. Wedge,— In the wedge, friction exerts a considerable in- 3 ° fluence upon the statical relations. The section of a wedge forms an isosceles triangle Fig. 167, with the edge FHR = a, the force Pacts at right angles to the back and the weight Q at right angles to the side FH. If we drive the wedge upon the base HR a space s = = = RRlt the weight Q is raised through a space CC\ = DDl = HL = . sin. HHt L = s sin. a, and force passes over HK = HHt. cos. /f, HK == s cos. according to the principle of virtual veloci- ties, and without regard to friction, P . HK = Q • DDlf i. e. P $ cos. fL fl 2 a r\ • ., _ t-\ Q sin. a - = Qs sin. a, therefore P = - - = Fig. 167. „ . a a Q sin. 2 -C0S-2 a 2 ‘"•i = 2 cos.WEDGE. 155 Q ^n• which also follows from the formula in § 137, if \ve put in it sin. 0 = 1, and cos. (o—5) = cos. /i There are now, however, three frictions which come into play, viz., the friction against the sides HF and HR, and the friction of the body ABCD in its constrained motion. As the directions of the force on both sides of the wedge deviate equally, the pressure against both is equal, namely = Q, and the friction arising = / Q. The spaces of these frictions, however, are different. For the fric- tion upon HR : s = HHV for that upon HF= HXL = s cos. «; accord- ingly the mechanical effects of both frictions are: =/ Q s +/ Q s cos. a =/ Q s (1 + cos. a) = 2f Q s ^cos. . Lastly, the friction between CD and FH presses upon the body ABCD at right angles to its direction, and there produces the friction/, ./Q, if/, represent the co-efficient of friction for its constrained motion. This friction, however, has the same space as the weight Q, viz., DD1 = s sin. a; and to it corresponds the mechanical effect/, / Q s sin. a. In order now to find the extreme limits of the condition of equilibrium, we must put the mechanical effect of the force P equal to that of the weight Q, plus the mechanical effects of the friction, therefore, Ps cos. ^ = Qs sin. a+2 Qfs ^cos. ^ + ffx Qs . sin. a, and we obtain the force: P = 2 Q (sin. | +/cos. + In a wedge ABCy Fig. 168, as it is used Fig.168. for the splitting asunder and compression of bodies, the force at the back corresponding to the normal pressure Q against the sides AC and BC, is P = 2 Q ^sin. ^ +/cos. which is given if we put the sum of the vertical com- ponents of Q and F = / Q, i. e. 2 F, = 2 Q 2 an(l ^ V2 = 2 f Q cos. ^ equivalent to the force P. Example. The load of the wedge Q in Fig. 167 = 650 lbs., the edge * = 25°, the co-effiei^1 of friction /, =/ = 0,36. Required, the mechanical effect necessary to move loa(i Q forward about * a foot. ,01 The force is P == 2 . 650 [sin. 12*° + 0,36 cos. 12}° + (0,36)* sin. 12* J = 1300 . (0,2164 + 0,36.0,9763 + 0,1296.0,2164) — 1300 . (0,2164+0,3515 + 0,0281) == 1300.0,5960 = 774,8 lbs. For,tothe space of the load CC, = * foot, corresponds the space of the force HK=s = a ' cos‘ 2 * ~— = ---------= 1,155 feetj therefore the mechanical effect and weight ia 2 sin. — 4.0,2164 2156 AXLE FRICTION. P* — 774,8 . 1,155 = 895 ffc. lbs. Without regard to friction, it would only be 650 . £ = 325 ft. lbs. In consequence of friction, the mechanical effect expended would be nearly tripled. § 164. Axle Friction.—In axles, the friction of raotion only is of importance, on which account experiments on this only exi st. From the following table very important results for practice may be drawn, with axles of wrought or cast iron, moving in bearings of cast iron or brass, coated with oil, tallow or hog’s lard, the co-efficient of friction is By continuous greasing «* 0,054, In the usual manner =» 0,070 to 0,080. The values found by Coulomb vary partially from the annexed. TABLE III. CO-EFFICIENTS OF AXLE FRICTION, FROM MORIN. KAMES OF THE BODIES. Nature of the surfaces and unguents. Dry or a little greasy. Greasy and wetted with water. Greased and wetted with water. Oil, tallow, or lard. Very soft and purified carriage grease. Hog’s lard with plumbago.. l ! — In the usual way. Continuously. Bell metal on the same _ — 0,097 — — Cast iron upon bell metal — — — 0,049 Wrought iron upon bell metal 0,251 0,189 — 0,075 0,054 0,090 0,111 Wrought iron upon cast iron — — — 0,075 0,054 — — Cast iron upon cast iron — 0,137 0,079 0,075 0,054 — — 0,137 Cast iron upon bell metal 0,194 0,161 — 0,075 0,054 0,065 — 0,166 Wrought iron upon lignum vitae.... 0,188 ... — 0,125 Cast iron upon lig- num vitae 0,185 _ 0,100 0,092 _ 0,109 0,140 Lignum vitae upon cast iron 0,116 , 0,153 Lignum vitae upon lignum vitae.... — — — 0,070 § 165. If we knowthe pressure Ubetween an axle and its bearing, and if further the radius r of the axle, Fig. 169, be given, the mecha- nical effect which the friction of the axle counteracts in every revolu- tion may be calculated. The friction F*/jR, the space correspond- ing to it, the circumference 2 h r of the axle; it therefore followsAXLE FRICTION. 157 that the mechanical effect lost by friction in each revolution is = f R • 2 * r = 2 ,if Rr. If the axle raakes one revo- lution per minute u, the mechanical effect ex- pended in each second = 2 rt/Rr. — = nufB. r = 0 J60 30 The mechanical effect consumed by friction increases, therefore, with the pressure on the axle, in proportion to the radius of the axle and the number of revolutions. It is, therefore, a rule in practice, not to augment unnecessarily the pressure on the axis in rotating machines by heavy weights, to make the axles no stronger than the solidity required for durability, and likewise not to make a great many revolutions in a minute, at least, not unless other circum- stances require it. By the application of friction wheels, which are substituted for the bearings, the mechanical effect of friction is much diminished. In Fig. 170, AB is a wheel which reposes by its axle CEEX on the circumferences Ei/, Ex Hx lying close to each other of the friction wheels revolv- ing about D and Dx. From the given pres- sure R of the wheel, there follow the pres- D sures JV* = JVj = --------, if a be the angle 2 cos. — 2 DCDl which the Central lines, or lines of pressure, CD and CD, make between them. From the rolling friction between the axle C and the circumferences of the wheels, these latter revolve with the axle, and there arise at the bearings D and D1 the frictions f JV and f Nv which together amount to If the radius of the wheel a COS. ~~ 2 Fig. 170. Fig. 169. DE=zDx Ex be represented by av and that of the axle DK*=DX Kx by rv we shall have the force at the circumference of the wheels, or at the circumference of the axle C resting upon these, which is requisite to overcome £.R ; Zl .■ /R whilst it will be if a a, a i COS. - 1 cos. -> 2 2 the axle C rest immediately in a socket. If we disregard the weights of the friction wheels, the mechanical effect of the friction by the ap- plication of these wheels is ——— times as great as without them.158 AXLE FRICTION. If we oppose to the pressure of the axle a single friction wheel GH,Fig. 171, and prevent any accidental lateral forces, bythe fixed cheeks Kand L, a = 0, cos. -= 1, and the above relations = Ii. 2 «. Example. A wheel weighs 30000 lbs., its radius a ss; 15 ^ and that of its axle r = 5 inches, what is the amount of force at the circumference of the wheel necessary to overeome the friction of the axle, and to maintain it in uniform motion and what is the corresponding expenditure of mechanical effect if it makes 5 revolutions a minute ? We may assume the co-efficient of friction / here = 0,075, wherefore the fric. tion/R ^ 0,075.30000 = 2250 lbs. Since the diameter nf 10 . 12 192 . . 1 Fig. 171. the wheel is _______ _____ = 38,4 times as great as the — 5 5 diameter of the axle or the arm of the friction, the axio frio tion reduced to the circumference of the wheel =Z£a= 2250 __ 5g}59 lbs. The circumference of the axle is ? ‘ ^ • * -------- = 2,618 feet; consequently the path of the friction in one second__ 2,618 . 5 __ Q 2182 feet, and its mechanical effect during one second 0 2182 / R ==^02182.2250 = 491 ft. lbs. If the axles of this wheel rest upon friction wheels whose radii are 5 times a, great as those of the axle, and therefore £ * ^ the power expended, referred to the circumference of the wheel, wUl only bel. 38,4 = 491 7,68 ft. lbs., and the mechanical effect of friction expended only — = 98,2 ft. lbs. S 166. The friction of an axle AFig. 172, which presses on its bearing in one point A only, is less than Fig. 172. that of a new axle resting on ali points of the bearing. If no revolution takes place, the axle then presses on the point , through which passes the direction of the raean pressure R; but if revolution begins in the direction AB, the axle by its friction will rise just so high in its bearing until the sliding force cornos into equi, librium with the friction. The mean pressure -----------------r ;s decomposed into a normal pressure JV and a tangential S; JVpassesinto the bearing and gives rise to F = /J\r, acting tangentially; S puts itself in equihbnum with «is there- fore = fN.According to the Pvthagorean doctnne, R is=JV*±Jp, therefore B* is here = inversely the normal pressure JV= —- , and the friction F = -£> or> an^e friction s/l + f* s/1 +f* p be introduced,jf = tang. p : F ®—--------. R = tang. p cos. p R = R sin. p. \/l 4- tang.p2 If the axle begins to move, the point of pressure B moves forward in the bearing by the angle ACB = p in the opposite direction.AXLE FRICTION. 159 If no forward motion took place, F would be — fR — R tang* f p; consequently the friction is the cos. p times as great after moving forward as before the motion. Generally,/ = tang. p no quite JL and cos. p > 0,995, therefore the difference is not quite = —1 ; therefore, in ordinary cases of application, we need have 200 no regard to the effect of this motion. If the wheel AB revolves in a nave or eye, Fig. 173, about a fixed . axis AC, the friction is the same as if the axis moves in a roomy nave, only the arm of the fric- tion is the arm of the nave, not that of the fixed axle. § 167. If the axle lies in a prismatic beanng, there is greater pressure, and consequently more friction, than in a round bearing. If the bearing ADB, Fig. 174, is triangular, the axle lies ontwo points A and B, and at each there is the same friction to overcome, the mean pressure R is de- composed into two lateral forces Q and Qv and each of these gives a normal pressure JY and JYV and each a tangential force F = f JY and Fx — f JYX equal to the friction. According to the former §, these frictions may be put = Q sin. p and = Qx sin. p, we have then the whole friction = (Q + QJ sin. p. The forces Q and Qj are given by the resolution of a parallelogram constructed from Q and Qj with the aid of the mean pressure JR, the angle of friction p, and the angle ACB = 2a, which corresponds to the arc ABy lying in the bearing. If QOR = ACD — CAO = a— p, QxOi? = BCD + CBO = a + p; lastly, QO^ = a —p + a + P=2a. The application of the formula § 75, gives: «n.(»-p) _ R and q _ Mn-(»+p) R. sin. 2 a sin. 2 a hence the friction sought is: „ _ -.Rsm. p F + F, = (Q + QJ sin. F = {sin. [a — p] + sin. [a + f]) Fig. 173. # dn* (a + f) = 2 sin. a cos. p and sin. 2 a * sin. a cos. a, therefore F + F — ^ dn* a R sin. p cos. P & R sin. 2 0 1 2 dn- 0 cos'* ~2~cosTaT9 ^rom smallness of P may be put =* —p. cos' ® friction of a triangular bearing is from this —— times as great as thiif- r «. cos. o> ot a cylwdrical one. If, for example, ADB = 60°, ACB160 AXLE FRICTION. 180° ___ 60® = 120°, and ACD = « = 60®, we then have: 1 = twice the friction of that of a cylindrical bearing. C°$i 168 With the aid of the last formula, the friction may be found foranew round bearing which the axle touches at ali points. Let tor a new rounu Fi 175f be such a bearing. Let us pig. 175. divide’ the arc JDB, in which lie the axle and the bearing, into many parts, such as JYO, &c., which have equal projections on the chord AB, and let us suppose that each 0f these parts supports an equal amount of the whole pressure R, viz., = —, (n being the number of parts) of the axle on the bearing, According to the former §, the friction of two opposite parts NO and NlOl = — . NCD' ®Ut C°S’ = cos. ONP = where NP represents the projection of the part NO, and NP = chord^ABIt therefore foliows that the fric- tion corresponding to the parts NO and N101 — — — n . NO _ R sin.2 P ^ j^q jn order now to find the friction for the thole arc ADB instead of NO, we must put.in the arc AD^ \ADB, because the sum of all the fnctions is times the sum of all the parts of the arc, it foliows that the friction in a new , • • p R «n 2 p arcAlBif we put the angle ACB bearing is: F=R stn. & p . chord JB> subtended .< the centre b, Whicb corresponds to the arc of the bearing, = 2 a°, therefore the chord AB = 2 AC . sm. a. _ B siw. 2 p' __a_> or 2 p = 2 * 2 * sin.a taken approxiraately F = R sin. p. a sin. a - If a bearing be of an acute wedge-shape it is — such wedge-shaped cavity would create an additional friction; but as iriction depends on tveight, not on extent of surface, the demonstrations of this section relative to bearingg in cylindrical cavities in which there is no wedging or tightening, but only a distribution of the uxight over the several parts of the surface of contact, the effect of the moving force in a “ new bearing” will be to relieve the pressure on one part and transfer it to another at the same distance from the centre of motion, or centre of the axle. To a certain extent, that is, to the depth of the semi-diameter of the axle, this surface increases, but if there be a distribution of pressure over a greater number of elements or units of surface, so is there a less amount of pressure on each point rubbed. This law of friction, according tc pressure, and not according to surface rubbed, agrees with all experiments.—Am. Ed. conceivable that an axle pressed intoAXLE FRICTION. 161 From this the friction is the greater the deeper the axle lies in its bearing. If, for instance, the bearing is half the circumference of the axle, a is then = J a and sin. a 1, we then have F = - • R and because ^ = 1,57, therefore 1,57 times as great as that of the free bearing. In an axle which does not rest deep in its bearing, a is small, therefore the sin. a may be put = a—= a ^1 — whence it follows that: F = ^1 + R sin. p, or = R sin. p, if a be very small. § 169. The axle pressure R is given generally as the resultant of two forces P and Q, directed at right angles to each other, and is therefore = y/ P2 + Q2. Provided we require it only for the de- termination of the friction /R =/ y/ P% + Q3, we may be satisfied with an approximate value of it, partly because the co-efficient f can never be so accurately determined and depends upon so many accidental circumstances, and partly because the whole product of the friction f R is mostly only a small part of the remaining forces of the machine resting upon the axle bearing, as the lever, pulley, wheel and axle, &c. The doctrine which teaches us to find an approximate expression of y/ P* -f- Q2 is known under the name of Poncelefs theorem, and may be developed in the following manner: y/ P2 + Q2 = P^ 1 + ^ = P y/1 + x2, whereby x «= (“q)’ which supposes that P is the smaller force, therefore, a; is a mere fraction. We may now put: x/1 + x2 = p + vxy and determine the co-efficients p and *, answer- ing certain conditions. The relative error is: V = y/l + a* — p- pX 1 — + pX ■y/l + a? ^1 +** For the smallest value of x, viz., x -» 0, y = 1 —p, and for the great- est, viz. x = 1, we have y = 1 ** If we make these errqrs, y/2 corresponding to the limits of x, equal, we then obtain an equationof 0,414 . a*. If we take condition i fl + V or v \ x/2 x = -, the resuit is, that y*=*l — + 1,3 =* ^ 9 ^ as a negative error, is greater than any other which arises by ing x «= -+ a, that is, a little greater, or a little less than ^ , P— the latter case we have 14*162 AXLE FRICTION. -( -a -Q ________p2 + v2 ______ x/p2 v2 Hh 2 /i * A 4" /a2 A2 (/A2 4- *3 * a)2 2 4" *3 db ^ f*v & 4“ p2 a2 f*2 + ** — JA4 A2 -)■ /a2 4- *2 If now we make this greatest negative error equal to the greatest positive error, we shall then obtain the following second equation of condition: s/p2 4- ** — 1 = 1 —p; or p + x/p2 4- v2 = 2. But the first equation gives v = 0,414 p; it, therefore, follows that /a (1 4. x/1 4- 0,4142) = 2, i. e. „ =---------2 = 0,96 and - => 0,414.0,96 = 0,40. 1 + v/1,1714 _____ We may, therefore, put approximately x/1 + x2 = 0,96 4- 0,40 . x, and in like manner the resultant R = 0,96 P 4- 0,40 Q, knowing that we thereby commit, at most, the error +y = 1 —/a = 1 — 0,96 = 0,04 s four per cent. of the true value. This determination supposes that we know which is the greater of the forces; but if this be unknown, we may assume x/1 4- x2 = u d . x\ p (1 4- x), and so obtain y =1------^ - • Here not only the limit x/1 4* x1 x = 0 gives the error = 1 — p, but also the limit x = oo , the same error =1 — == 1 —p; but if we put x = - = 1, we then ob- x p tain the greatest negative error and by making these errors equal: 1 —p = p \/2— 1, therefore, „ = ^_________?________L_ _ 0,825, for which 0,83 may be put. * 1 + ^/2 “2,414“ 1,212 In the case where we do not know which is the greater of the forces, R may be put = 0,83 (P 4- Q), and we know that the greatest error committed will be + y = 1 — 0,83 = 0,17 per cent. s= i of the true value. If, lastly, we know that x does not exceed 0,2, we may disregard it altogether, and write x/P2 4- Q2 = P, but if x exceeds 0,2, thenLEVER. 163 \/P2 -f Q2 is more accurately = 0,888 . P + 0,490 . Q; in both cases the greatest error is about two per cent.* , . § 170. Lever.—The theory of friction above developed hnds lts applicatiori in the material lever, the wheel and axle, and in other machines. Let us, in the first place, treat of the lever, and take the general case, viz., that of the bent lever ACB, lig. 176. Let us re- present as before (§ 127) thearm CA of the power P by a, the arm CB of the weight Qby b, and the radius of the axle CH by r, let us put the weight of the lever = G, its arm CE = s, and the angles APK and BQK, by which the directions of the forces deviate from the horizon = a and 0. The power P gives the vertical pres- sure P sin. <*, and the weight = Q sin. fi; the whole \ertical pres- sure is, therefore, V= G+ P sin, fi. The power gives further the horizontal pressure P cos. a, and the weight Q an opposi e pressure Q cos. fi: since there remains for the horizontal pressure, H= P cos. a— Q cos. fi, we may put the whole pressure on the axle . R = V+v H= ^ (G+P sin. a+ Q sin. fi) + r(P cos. a—Q cos. fi), of which the second part v (P cos. a—Q cos. fi) must never be taken negative, and, therefore, in the case where Q cos. fi is >■ P cos. a, the sign must be changed, or rather P cos. a must be subtracted from Q cos. fi. In order to find that value of the power which corresponds to unstable equilibrium, so that the smallest addition of force produces motion, we must put the moment of power equal to the moment of weight, plus or minus the moment of weight of the machine (§ 127) plus the moment of friction, therefore, Pa= Qb+ Gs+fRr = Qb+ Gs+/(fi V+v H) r, from which follows Qb+ Gs+/[fi(G+ Q sin. fi) + Q cos. fi] r P =----=------------;---------------------- a—ix fr sin. a+v f r cos. o. If P and Q act vertically, R is simply = P+Q+G, therefore, Pa= Q6+ Gs+/(P+ Q+ G) r. If the lever is one-armed, P and Q act opposite to each other, then R = P—Q+ G, and consequently the friction is less. Besides R must be put constantly positive in the calculation, because the friction f R only impedes, but does not pro- duce motion. From this we see that a one-armed, is mechanica y more perfect than a two-armed lever. Fig. 176. c h __ 4 A , -= i ft. and Example. If the arms of a bent lever, Fig. 176, ar®\a""" ' the weight Q=5600 r = l£ inches, the angle of inclination a = 70°, 0 = 50 , and fur estore the unstable lbs., and that of the lever G = 900 lbs., the power required to r ___ ^ |heref 0,2 V, therefore, more correctly: R = 0,888. H+0,490 F = 0,888.8627 + 0,490.2349 = 8811, and it follows that the moment of friction =/ri? = 0,009375.8811 = 82,6 ft ik. . . , . D 22400 — 450 + 82,6 ‘ M and lastly, the power P =--------------------!----- = 6 3672 lbs., which value varies little from the above. § 171. Pivot Friction.—When in the wheel and axle a pressure takes place in the direction of the axis, as in the case, for example, of upright axles, in consequence of their weight, there is a friction on the base of the one axle. Because pressure is there exerted on points between the pivot and its step, this friction approximates to the simple sliding friction, and to the axle friction which we have hitherto con- sidered, and we must put for it the co-efficients of friction given in Table II. To find the mechanical effect absorbed by this friction, we must know the mean space which the base AB, Fig. 177, of such an upright axle describes in a revolution. Let us assume that the pressure R is equally distributed over the whole surface, let us also suppose that on equal parts of the bases the frictions are equal. Let us further divide the base by radii CD, CE, &c., into equal sectors or triangles DCE; to these wiH correspond not only equal amounts of friction, but also equal moments, therefore, it will be neces- sary only to find the moment of friction of one of these triangles. The frictions of such a triangle may be regarded as parallel forces, for they ali act tangentially, t. e. at right angles to the radius CD, and since the centre of gravity of a body or a surface is nothing more than the point of apphcation of the resultant of the parallel forces equally distributed over this body or surface, accordingly the centre of gravity »S of the sector or triangle DCE is here the point of application of the result- ant arising from its different frictions. If now the pressure on this sector = i* f and the radius CD — CEy the base — r, it follows n that (from § 104) the moment of friction of this sector — CS . n Fig. 177.POINTED AXLES. 165 = - r.IUl, and lastly, the moment of the entire friction of the axle 3 n 2 /R2 -n = 3 n =3^Rr' Sometimes the rubbing surface is a ring .1B E D, Fig. 178. If its radii are CA=r1 and CJJ=r2, we have then to de- termine the centre of gravity S of a portion of a ° 2r3____r3 ring, and from § 109, obtain the arm CS =- 12 0 7*1 T2 2 //• 3_T 3\ therefore, the moment of friction = ^"X2—fh)' If we introduce the mean radius = r> an<^ the breadth of the ring rx—r2 = b, we^obtain this moment of friction also — f R (r + The mechanical effect of friction for a revolution of the axle is m the second case = 2 * . \fRr=z\1tfRr> anc^ *n ^rst 3 3 — 7t f R ( Tl r% \. Here we easily see that to diminish this loss of 3 \r*—r22/ r . mechanical effect, the upright axle or shaft must be made as lignt as possible, and that a greater loss of mechanical effect would arise if, under otherwise similar circumstances, the friction were to take place in a ring instead of a complete circle. Kxample. In a turbine making 100 revolutions a minute, and 1800 lbs. weight, the size of the pivot at the base, is £ inch; how much mechanical effect does the friction of this pivot consume in one second? The co-efticient of friction being taken = 0,1 we have the friction / i? = 0,1 . 1800 = 180 lbs., the space per revolution r = i. 3 14._L = 0,1745 ft. lbs., hence the mechanical effect per revo- 3 3 24 , lution = 180 . 0,1745 = 31,41 ft. lbs. But now this machine makes m a second 152. = 2 of a revolution, hence it follows that the loss of mechanical effect sought 60 3 = ilill = 52,3 ft lbs. § 172. Pointed Jlxles.—U the axle J1BD, Fig. 179, has conical ends, the friction comes out greater than if it has plane ends, because the pressure of the axle R is resolved into the normal forces N, JVj, which produce the friction, and which together are greater than R alone. If the half of the convergent angle ABC — BBC = a, we have 2 N = ■ , and sm. a consequently the friction of the pointed axle = f r~“— • Let the radius of the axle CA mi. a = CB at the entrance into the step be re- Fig. 178.166 POINTED AXLE8. presented by r,, we shall then have as before the moment of friction = ?l. Let this axle dip a little only into the step, the me- chanical effect of this axle will be less than that of an axle with a plane base, and on this account the apphcation of the pointed axle is of Service. When, for exarople, -A- - therefore, r, = £ r sin. the pointed axle of the radius r, causes only half the loss of mecha- nicJf effect through friction which the truncated ax.s of the radius r does. If the pivot forms a truncated cone, Fig. 180, friction takes plac as well at the envelop as the truncated surface, and the momen! \ f jJL - * V O rf* » 111 01 friction comes out sin -r 2 3 \ 2 f R g -jT> if r be the radius of the place of entrance into the step, and = DE that of the base, and a° the half of the convergent angle. Fig. 180. Fig. 181. Fig. 182. Lastly, the pivot or upright axles (Figs. 181, 182) are very often rounded. Although by this rounding, the friction itself is by no means diminished, there arises nevertheless a diminution of the mo- ment of friction, from the extremity not dipping far into the step, If we suppose a spherical rounding, we obtain by the aid of the higher calculus for a semi-spherical step, the moment of friction . Rr: but for that of a step having a less segment 2 B | £ 1 _|_ 0,3 ^A) r being the radius of the sphere = MB, r, the radius of the step CA = CB. Example.If the weight of the armed axle of a horse capstan = 6000 lbs., the radius of the conical pivot = r = 1 inch, and the angle of convergence of the cone 2 • = 90°, then the moment of friction of this pivot = — ./. —-= — • 0,1 . —-—PP__ 3 sm. a 3 sin. 450 . -L = = 47,1 ft. lbs. This axle makes during the lifbng up of a ton from a 12 shaft or mine = u = 24 revolutions, then the mechanical effect which is expended at the pivot in this time by friction =2 ir u . ^ —, = 2 it . 24.47,1 = 7103 fl. lbs. sin. a § 173. Points and Knife Edges,—To avoid as much as possible the friction of the axle, rotatory bodies are supported on pointed pivots, knife edges, &c. If we had only to do with perfectly rigidPOINTS AND KNIFE EDGES. 167 and inelastic bodies, no loss of labor would arise through friction by this method of support or suspension, because no measurable space here is described by the friction; but since every body possesses a certain degree of elasticity, by the resting of such a body on a point or knife edge, a slight penetration takes place, and a rubbing surface is thereby caused, upon which a space is described by the friction which gives rise to a certain loss of labor, although very small. In rotations and vibrations long sustained, bodies supported in this raanner, present similar surfaces of friction arising from the abrasion of their points or knife edges, and the friction must then be estimated by what has been already mentioned. For these reasons this mode of support is applicable only to such instruments as the compass, the balance, &c., where it is of importance to diminish the amount of friction, and where motion is only allowed from time to time. Experiments on the friction of a body resting upon a hard steel point, and revolving about it, have been made by Coulomb. From these, it results that the friction increases somewhat more than the pressure, and varies with the thickness of the supporting pivot. It is least for a granite surface, greater for one of agate and of rock crystal, greater stili for a glass surface, and greatest of all for a steel one. For a very small pressure, as in the magnetic needle, the pivot may be pointed to 10° or 12° of convergence. But if the pressure is great, we must then apply a far greater angle of convergence, viz. 30° to 45°. The friction is less when the body having a plane sur- face reposes upon a point than when it lies in a conical or spherical step. Similar relations take place in the knife edge as applied to the balance, and the beams of balances, that are intended for heavy loads, require sharp axes of 90° convergence, while an edge of 30° is sufficient for the lighter ones. Remark. If we assume that the needle AB, Fig. 183, rests on the point DCE of the pivot FCG, of the height CM = h, and radius DM = r, and suppose that the vol- ume £ ir r* h is proportionate to the pressure -R, the amount of friction may be found in the following manner. If we put J r3 h = [a R, where fx is a number resulting from experience, and introduce the angle of convergence DCE = 2 a, and, therefore, Fig. 183. put h = r cotg. a, we obtain the radius of the base r RJ (f i. friction on the pivot increases equally with the cube root of the fourth power of the pressure, and the cube root of the tangent of half the angle of convergence. The amount of friction of a beam AB, Fig. 184, which oscillates on a sharp edge CCV may be found in like manner. If a be half the angle of convergence DCM, l the length CCX of the edge, and R the pressure, this amount is given = • 3|3 fxRU S * R tang. < [, and frR = We must, therefore, assume that the 184.168 ROLLING FRICTION. Fig. 185. K___,F 1*1 ir § 174 Rolling ^£&% means firmly grounded, \*e ^ for a larger diameter of the and that it is greater for a S,™ebraical dependence this friction stands rolling body; but in what g rolling body, cannot as yet be to the pressure and femeterrfflj few experimenta considered as determined. thick, of lignum vitae and elm with rollers, from two to tw which were rolled along a surface of oak, by means of a thin thread passing over the roller whose extremities were stretched by un- equal weights P and Fig. 185. From the results of these experi- ments, rolling friction appears to increase directly with the pressure, and inversely with the diameter of the roller, so that the force neces- sary to overcome this friction may be expressed by F =f. —, if Rhe the pressure, r the radius of the roller, and / the co-efficient of fric- tion derived from experiment. If r be given in inches, then from these experiments QQ For rolling upon compre,S«l «ood/ - 0,0189, These formulas suppose that the forceFacts at the circumference of the roller, but if the force be apphed to the axis of the rolling bodies, by which, as in every descnption of carnage, axle friction en- sues, the required force is 2 F,because here the arm CH is only half that of KH with respect to the pomt of applica ion H. A body MBS is moved forwards, Fig. 186, lying on the rollers C J and D, the required force P here Fig. 186. comes out very small, because two rolling frictions only, viz., that between AB and the rollers, and that between the rollers and the way HR are to be overcome. The progressive space of the rollers is only half that of the load Q, and on this account for farther progression, the rollers must be replaced under it from before, because the points of contact A and By by virtue of the roll- ing, recede as much as the axis of the rollers advances. If the roller AH has revolved about an arc AO, the roller has then moved over a space AAX equal to this arc, and 0 comes into contact with Op the new point of contact Ol has, therefore, receded by AOx = AO behind the former (A). If the co-efficients of friction are JROLLING FRICTION. 169 and /„ the power necessary to draw the load forward is P — + Q2— 2 PQ cos.aT If further we put P == Q + F and P2 approximately = Q2 + 2 QF, we then obtain: F = f v/ Q2+ 2 QF + Q2 — 2 Q2cos. a — 2 FQ cos. « of which the cylinder and the bed, EHC, are composed; this allowance would again cause a new pressure and friction, and thus a decreasing geometrical series of weights must be added at the point C, having for the first term such a part of W as is expressed by the relation of pressure to friction, in the case of the given materials, and for a com- mon ratio of the progression, the fraction expressing the same relation. The sum of ali the terms, continued to zero, will be the actual amount of P at the moment when motion commences. The sum of ali the terms following the first, will be found by multiplying together the first and second terms of the series, and dividing the product by their difference. The quotient added to the first term gives the sum of the series required. The applicability of a similar method of computation to the friction on the gudgeons of water wheels, moved by the gravity of water, is too obvious to require demonstration. If instead of applying a weight at P only, we should apply, as above supposed, two equal forces, one in the direction of CP, and the other in that of EM, the amount of friction caused by the former would be relieved by the latter, and consequently, there would remain only the friction of the cylinder. The same would be true if the forces were to take either the directions iV/and JF/T, or KO and DS respectively. Supposing the cylinder to be placed on an axis smaller in any given proportion than its own dia- meter, as GF, then the whole effect of gravity would be transferred to this axis, and if this were to be caused to revolve by a forceapplied tangentially to the axis itself, it must be of the same magnitude as that which had before been applied to the cylinder when placed in the groove. But if applied to the exterior of the cylinder, it must be as much less than before as the diameter of the axis is less than that of the cylinder. In other words, the difficulty of overcoming friction at the axle, is to that of overcoming the same at the outer periphery, when, confined in a bearing, as the diameter of the axle is to that of the cylinder. If D be the diameter of the cylinder, d of the axle, and F the relation of weight to friction, we shall ha ve the proportion D: d = F • = the force required D ’ to overcome the friction on the axle. This subject may be stili further illustrated by Fig. 1863, where the horizontal plane Hh, is represented as fumished at equal distances, with small balls or prominences so attached to its surface as to present equal obstructions to the dragging of heavy bodies along that surface. Fig. 1863. he exterior of the wheel ED is likewise represented as furnished with equa promi nences at equal distances. When, therefore, the wheel is compelled to ma e one revo- ution without advancing, as many prominences would be broken from its penp lery as would be dislodged from the plane surface while it advanced, without revolving through a space equal to its circumference. This applies to a locomotive slipping its172 ROLLING FRICTION. = /^/2(1— cos.»)((?+ QF)=2fsin.| s/Q* which again may be put - 2/sin. “ (Q+ i n if we have regard t0 the tw° 2 wheels on the miis in the one case, and sliding w.th whee s clogged or engine revergi in the other. But when the wheel is allowed both to revolve and to advunce in such a manner as to apply its periphery to a lengtli of plane just equal to the space tra- versed by the centre, the prominences will be geared toK«ther 1 «the teeth of a rack and pinion But in the latter case the prominences may all remain unbroken. But when the wheel rests with its whole weight on an axle, as the number of promi. nences which can be disposed at die same distances as before, oo the circumference of the axle, will be diminished, in proportion as the radius of the axle is smaller than ED% that of the wheel. When a wheel or cylinder rolls on a surface as nearly plane as it is possible for art to produce the arnount of friction, being no more than is due to the moment of inertia, is extremeiy small compared with that of dragging, but the observations already made and the examples cited, will be sufficient to show that the actual advancement tends, by a loree equal to that which produces the rotation, to break down the prominences of the surface, for if we consider the cylinder rolled forward by a fine thread unrolled from its upper side, we may consider also the plane beneath to oppose a force tending todraw it in the opposite direction, and this force is friction. In experimenting with wheel carriages, or cars descending by their own gravity along inclined planes, to ascertain the ratio of friction to weight, we have to determine separately the rolling friction of wheels and axles with various weights and diameters, and then their influence combined with that of the insistent weight of cars and loads, which latter can alone produce sliding friction at the axle. The weight of the wheels resisted only by the slight arnount of rolling friction at the periphery, tends to accelerate the velocity of the car and load. If we suppose the wheels and axle only to be placed on apiane ngo, Fig 186«, so little inclined as just to continue their rolling motion, and afterwards on another npo so much inclined as to allow a car to descend with all the friction at its axles, we shall readily conceive that over the latter plane the wheels would, by them- selves, have desccnded with a constantly accelerated mouon, and consequently, that they Fig. 186 .ROLLING FRICTION. 173 first members of the square root only. Now if F—/ F sin. t is given = 2fQsin. then the friction sought is would, to the extent of their accelerating force, overcome a portion of the resistance which friction opposes to the motion of the car. Thus, in every case where we would compute the effect of friction by comparing the actual distance passed over by a carriage, with the theoretical descent as caused by the inclination of the plane, we must oonsider the weight of the car and load as the cause of friction on the axle, and the gravitating power of the wheels (and that of the axles when they revolve with tbe wheels), as aiding to overcome the friction occasioned by the load. To compute the effect ofanyobstacle of given height which a rolling body is compelled to surmount, as dependent on the diameter of the wheel, we may tajte two wheels EAD and ead, Fig. 1864, of different heigbts, intended to surmount the equal obstacles TA and ta. Let the weights JFand w be the same for both wheels, and the powers P and p be such as to produce an equilibrium in the wheels DAE and dae respectively. Then since (§75 and 139) three forces are in equilibrium, where each is represented by the tine of the angle comprehended bctween the directum* of the other two, the direction of gravity and that of the horizontal line of traction being at right angles to each other, the sine of the angle com- prehended between their directions is equal to rodius, and is, therefore, represented by CA or ca. Again, the horizontal loree is represented by the sine of the angle BCA or beo, which is the line BA or bo, while the vertical force or gravitating power of W is represented in the two cases by the sines of BAC and hac respecti vely, which are the lines BC and bc. These two forces multiplied respectively by the distance at which they act, in a perpendicular direction from the point A or o, about which the wheel must re- volve in order to surmount the obstaele, ought to give equal moments to those of Wmul- BJjCW BC tiplied by BA and ba. Hence BA X BC X P» or P s Jhff* xy TDfF bc X/>i«P «s-------- - . In order, therefore, to know the absolute values of P and p, we must determine the actual lengtha of BA and BC, of ba and bc. BC is easily found by subtracting the perpendicular height of the obstaele from the radius, and BA is found by subtracting the square of BC from that of AC, and extracting the square root of the remainder. Thus CA2 — jBC* = BjP. But BC=CA — BD, therefore BC* ** CA*— 2 CA X BD + BD?. Substitutio g this value of BC* in the equation ~CAP— &&*=* BA*, we obtain 2 CA X BD—BDP^BA?; hence BAn* y/2 CA X BD—BJP. The value of P, therefore, must be BD? ^ now ©xpressed in terms of BC, or (CA—BD) the radius and versed sine. Substituting B for tbe radius of the laiger wheel, and r for that of the smaller, as also A for the perpendicular height of the obstaele in both enses, the above expresskm of the value of P beoomes sonmg precisely similar, we obtain p P P ** y X x/2M—P WXs/Irh—J? ’ and by a eourse of W' Hence _____________________, .. y/3 Bk—V . telMRxf divided bf eosines of the nma angtas. And «Heproportion i>“ ^ ^ ment . r ----------- We may be assured that this state- P°se the forcePorJ?fL ?®n*lder. ^at ^e centre of gravity of the wheel, where we sup- commence its ♦? jf. I?U8t ^esCfibe the curve CNYorcny, and that it must wia be required wh«n or co, Fig. 186^ and that the greatest effort e*presses the lemrth «#■ 6 .ce”tre at ^ a**d c respectively. But the tangent CO, which 8an*e magnitudfl n. toe mc“n®d plane which the weight begins to surmount, is of the ** “lore nearlycoiiw*ue angle DCA, and the successive plane* wiii / uciaeot wuh the horizontal line. Having, on the fcreyoiqg principies, 15*174 ROLLING FRICTION. 2 fQsin.*- / F =----------which generally is = 2 f Q sin. _ ( 1 + 1—/sin.- and indeed very often =* Q sin. In ordtr to draw the cord determined the effect of obstacles which oppose the rolhng of a wheel, we may proceed to consider tlie influence of that resistance on the quantity of friction at the axle. As carriage wheels are ordinarily constructed, arui as roads, and «•pecially railroads «. ---m0,lp th« resistance of obstacles at the circumference is inucli more easii J are commonly made, the resistance ofobstacles overcome than that of friction at the nave. Ihus, in t ig. more easily 1S06, where the wheel Fig. 1S6,. I vnDTE turns on its axle, «0, the line or spokesD becomes the proper representative of a suspended lever, impelled at the upper end in tlie direction of it or t.r' by the resistance of friction at #, and at the lower by the resistance of opposing obstacles, (the amount of which resistance has just been stated,) in the direction DH. It generally happens, how- ever, that, except on very rough roads, the resistance from the latter cause is less than that from the former. Hence, the wheel cornmences its forward motion sooner than the axle begins to slide in the box; so that its sliding motion is not (except in cases of great resistance at the periphery), in the direction of tlie tangent d, but in that of some otherROLUNG FRICTION. 175 over the edge, a force P — Q + F = 1 + 2/sin. ^ Q is necessary; \l—fsin.y and inversely, to prevent the descent of the weight Q by the cord, a line, as IT. The axle, then, ought to be found bearing not on the bottoni, but on a part back of the lowest Une of the cyhnder. Having been leti to tbis eonclusion from the theory now developed, the writer was induced to inquire of several wheelwrigbts, coach- makers, carriage-smiths, and keepers of livery stables, whether they had e ver noticed the fact, or whether they supposed it to be true, that the axle did rest in its box elsewhere than on its lowest part; ali, a fier a momen t'8 reflecti on, answered, that as a force was applied to draw it forward, it must press and be most worn against its front and lower side; but upon exarnining the old axles in their possession, they ha ve uni fotmly found the above views to be confirmed by evidence which they could not doubt. In moving a carriage, then, the animal exerts his strength to bring the axte into such a position that it willdescend by the gravityof the load along an inclined plane asif to follow somedirec- tion, as IT. If the axle be smaller than the box, so as to leave qonsiderable space be- tween them, the centre may retreat from the vertical C,D, ascending at the same time from Cx to where it exercises a gravitating force due to the weight, and in the direc- tion Ca Diy acting of course on the arm of the lever equai in length to Dl and D, and which would, if the force P were relaxed, cause the wheel to retreat and again depress the point m towards z, describing the portion mz of a cycloidal curve. This effect is often observed to take place. This position of the axle likewise accounts for the retrograde rota- tion of a wheel which is sometimes observed to take place througb a portion of a revo- lution, when a heavily loaded car first passes from rough ground to smooth ice. The gravitating force, when the centre takes the position C„ may be resolved into Ct I, per» pendicular to the side of the box, and the inclined tangent Ig\ then the force which presses the suriace causes the friction and opposes motion, is less than when it lies on the horizontal plane #1, and the friction is diminished in proportion as C,I is less than C9 g. Again, as the force P now acts in the direction C« P,, it tends to relieve even the remaining portion C* /, in the ratio of that line to C%K. When the force necessary to surmount the obstacle becomes infinite, the centre of the axle will take the position C3, but this can happen only when the height of the obstacle is equai to the radius of the wheel. The tangent of the angle formed at the centre will then be infinite also, and the expression before given, viz. P = — tan%' will be as applicable to this extreme R ease as to any other where the height of obstacle is less. moving^ve^a^ori^nr11/101? l^ie f°regoing remarks, are, that the friction of a roller, peripherv and tha^hf sufface’dePends on the reiation between the velocity of the tial velocitvand that ft Centre 8ravuy; also, that this reiation between the tangen- Again tra"sPortation, wiil depend on themoment of inertia of the cyhnder. ployed to slide nm °^»a wbef* over a riedge, where the same materials are em- oj L XittZofZ^ " 11,088 which compose 1116 >« “d M ^ by the reteuon portional^1!^ °f ^ the axIe to *** overcome by the moving force will be pro- the Drom-A»« tl,e we,?ht tbe ,oad> but will depend also on the obstacles which oppose equai tha wiW attain its when the height of the obstacle is mount th» ,„1?8 tbe. wheel, at which moment the advantage of the wheel to sur- Th A obstacIe 18 a Animum or zero. over whi I*1?8®® a wbeel to overcome any obstacle of a given height when the plane taneenf Jf moves, and the line of draught, are both horizontal, will be at the anoth*r tLIt!LangU/ormed ° draum from the centre of the wheel doumwards, and wheel aWn ^rom ^ 9ame point to the top of the obstacle, dtvidtd by the radius of the car!^L« m?re fuU exP°«bon of the views of the writer of this note on the friction of *ag wheels, see Journal of the Franklm Institute, voL v. p. 57.—Am. Ed.176 R0LL1NG FRICTION. force Pj= provid- P into Q, and Q into P, we obtain P, — (l + 2/«». g) ed only a motion in the direction of Q is to be prevented. The friction F= P—Q is in the first case = + 2/ sin. “^ —1J Q, and in the second = Q — P, = Jj^l + 2/sin. 1J Pl = J^1 — (i + 2/sin.I)—DJ Q-R0LLING FRICTION. 177 The same formulae are applicable to Flg' a body winding round a cylinder, and consisting of members^as, for instance, a chain ABE, Fig. 18$, where n is the number of links in contact, the length AB of a link = /, and the distance CA of the axis A of a link from the centre of the arc covered = r, \ve then have . a / sin. - = —. 2 2 r Example. What is the amount of friction at the circumference of a wheel 4 feet in diameter, if twenty links of a chain, five inches long and one - inch thick, pass over it, one end of which is fixed, and the other stretched by a force of 50 lbs.? Here P, = 50 Ibs.n = 20, «in. 1 = _i_ = let us now put for / the i 4o-j-1 , * mean value 0,35, we then obtain the friction with which the chain acts agamst the wheel in its revolution: x ^ P=[(l+ 2.035.^)“-l].50=[(l+~) 1 J . 50 = £ (i!)20-! J . 50= 2,974.50 = 149 Ibs. § 176. A stretched cord ABy Fig. 190, lies about a fixed and cylindrically rounded body ACB, the friction may be likewise found from the rule of the former paragraph. Here the angle of deviation EDB = a° = the angle ACB at the centre subtended by the arc of the cord AB; if we divide this into equal parts, and consider the arc AB as consisting of n straight lines, we have then n corners, each with a de- a° viation of —, and consequently the equation be- n tween the power and weight, as in the former §: p = ( 1 + 2 /sin. From the smallness of — we may put the sin. 2n Fig. 190. whence 2 7i 2n P = +/^nQ. If further we make use of the binomial series, obtain: / /• \3 \ . _/• . -I-HIM1 ■ »(»-l)(’—a)(Zl)-4-•••) 1200 _ _ » = 7,0901 __2,5918 = 4,4983 . Log. Q = 1,9536. Q = 89)9 lbs. § 177. of Chains.—U ropes, or other similar bodies, &c., are placed over a pulley, or on the circumference of other cylinders revolving about an axis, the cord or chain friction considered in the foregoing paragraph ceases, because the circum- ference of the wheel has the same velocity as the rope; but now the force of bending by the winding of the rope about the pulley, and also that of unbending by the unwinding, becomes perceptible. it it is a chain which winds round a drum, there arises the resistance of the winding and unwinding manifested in a friction of the chain pins, while these last are revolving through a certain angle. If ABy h ig. 192, is one link, and BG the one lying next, if, further, C is the axis of revo- lution of the wheel on which the chain stretched by the weight Q winds itself, if, lastly, CM and CJV are let fall perpendicularly to the longer axes of the links AB and BG> MCN= a° is the angle through whichRIGIDITY OF CORDS. 179 the wheel revolves whilst a fresh link is laid on, FBG = 180°—iBE is the angle by which the link BG with its bolt BD revolves abeut the link AB. If now BD = BE= rx is the radius of the bolt, the point of friction or pressure D describes an arc DE=rxa, and the mechanical effect of friction fx Q hereby produced at the point B is f Q . r}a. The force Px ex- pended in overcoming this friction, acting in the direction of the longer axis BG, describes the simultaneous space CN times the arc of the angle MCNmz CN. a, and the mechanical effect = Px . CN. a; by equating both labors we havePx . CN a =/ Qr.a, and the required force, if • j---- Fig. 192. . a=7\ . yr,o. ana me rcquucu 1 a represent the radius of the drum CJY increased by half the thickness of the chain: P1=f1Q . h.. CL Without regard to friction, the force for a revolution of the wheel would be P = Q, having regard to the friction in the winding up of the chain P = Q+P1== (l + fi Q. If the chain unwinds itself from the drum, an equal resistance takes place; if, therefore, a wind- ing on one side, and an unwinding on the other take place, the force P = ^1 Q, or approximately: -(* + Lastly, if the pressure on the axle => R, and its radius = it follows that the force, taking into account all resistances, is: Example. What is the magnitude of a force P at the ex- tremity of a chain passing over a pulley ACB, Fig. 193, if the Weight Q drawing vertically downwards = 110 lbs., the weight of the pulley with the chain 50 lbs., the radius of the pulley measured to the middle of the chain = 7 in., that of the axle C J inch, and that the chain bolts = j in. ? The co-efficients of friction / = 0,075 and fx = 0,15, therefore from the last formula we obtain the force: p— (l+2.0,15 . g3_) . 110+0,075 .JL(110+50+P), or> if we assume P on the right hand nearly = 110 lbs. P = 1,016.110 + 0,0067 . 270 = 111,76 + 1,81 + 113,6 lbs. Fig. 193. or The § 178. Rigidity of Cords.—In bending a co j^otion. I wheel, rigidity comes in as a resistance °PP , g unrolling from same takes place, but in a far less degree, mpasuring the amount cylinders. Amontons and Coulomb set a ou180 RIGIDITY OF CORDS. Fig. 194. of this resistance by experiment. The resiilts obtained by them are by no means satisfactory; partly because the,n°t in^ sufficient accordance with each other, and partly e^au J extension so desirable for practical appima ' • ^ 1 s Coulomb, which are those only of wh.ci we hall made with hempen cords of foTwVcan kno^whTfs the resistance of rigidity of ahempen rope of from 2 to 3 inches thick, when wrapped round a drum of from 1 to 6 feeHn height; and also what is the amountof this resistance tn the case of the wire-ropes, now come generally «to «se.» caseotxne v Coulomb made his expenments m tw0 ways; at one time with the apparatus of Amontons, Fig. 194, where J1R is a roller, with two cords winding round it, the tension is effected by a weight Q, and the rolling down of the cylinder by a second one P, which pulls, by means of a thin string at this roller; at another time, with a cylinder, which was allowed to roll upon a horizontal line, and round which a cord was wound, and from the difference of the weights sus. pended at both extremities, which effected a slow rolling forward, and afler abstraction of the rolling friction, the resistance of the rigidity __________________was deduced. ..... It results from the expenments of Coulomb, that the rigidity in. creas^s equally^with the tension of the winding cord ; that U consists, moreover, of l constant part K, which is no more than m.gh be ex- pected, because a certain force is necessary to bend an unstretched cord. It also appears that this resistance increases m verse ly as the of the nullev- that it is, therefore, with twice the diameter o^tTe nuUevhonfy half as great; with three times the diameter, one- thi3 r ^e relation between the thickness and the rigidity of the cord is only approximately given from these expenments s.nce he rigidity dependi upon the quality of the matenals, the twisting of the strintrs &c For new ropes, the rigidity was found proportional old being the dimeter ofth, rope. I, ia, therefore, only an approximation, when some assume hat th.s resist- ance increases proportionali)' with the thickness, others with the square of the thickness of the rope. , , § 179. The rigidity of cords raay be therefore expressed by the formulae: g (K v Q), where d is the thickness of the cord, , .h?for .hef»,Ce P rena & J1. E = x: l9 it therefore follows 1. That P = y F . E,and inversely, 2.* = . The same formulae are also applicable to a body AC, Fig. 15)7 0f the 11,e sam ^h jD = />and the transverse sectioni » p Fig. 197. jf jj become shortened a length x by the compression of a force P• . , . „ Bv the aid of these formulae we may calculate from the change of volume (x) the corresponding force i>, or from the force the quantity of the extension or compression — - j wd?° incl‘e», X ___ JL inch consequently - = further /'= ___ 0,7854 1 2 /J_V = 0,0218 square inches, the required force accordingly is = _L_ . 0,02i8 14626000 — 442 lb« — 2. The modulus of elasticity of iron wire is 263250000 lbs . ‘fan^clTaineOreet long and 0,2 inchcs thick, be -etch-xlbya force of same will be increasedby a length X = 0,7854 . (fy2)* * 26325000U 31416726325 0,7854 . (0,2)* * 263250000 = 0,013 inches = 0,156 lines. § 184. Modulus of Working Load and Strength ' 1 r i-onsvprse section unity accui 5 184 Modulus of Working hoaa anu —The force T, which a body of the transverse section umty accumulate» when its extension attains the limit of elasticity, is eas.ly determ.ned from the modulus of elasticity E and the elongation * thl. limit, for T: E = x : 1, therefore, E. Ihis is the strain beyond which materials used in construction not be loaded if they are to maintain sufficient safety together with dura, bility. If the transverse section of a body, which has to sustain a tensile strain P be = F, we have then ^ 1. P = FT, and 2. F - y The force Tby which we judge of the working load of bodies, may be introduced into calculations under the name of modulus of working load. ______________________ * See Appendix.STRONGEST FORM OF BODY. 185 The modulus of strength K, which expresses the force by which a body of the transverse section unity becomes ruptured, is entire y i - ferent from this modulus. If the transverse section of a prisma ic body, or its least section ■* F, it follows that the force, for the rup ture of this body, is: 1. Px = FK9 and inversely, 2. Generally the strength of materials of construction and parts of machines are calculated by the co-efficient K, which is divided tor security’s sake, by one of the numbers 3, 4 to 10. This makes little difference in the resuit, as we may see froin a companson of the values found in the succeeding table, but the supposition is incorrect, or to be justified only in so far as the modulus of strength is from 3, 4 to 10 times that of the modulus of tenacity, or generally bears a constant relation to it. If the section of the body be a circle of the diameter d, we have therefore, ____ .—- as Fy so that d = 1,128 %/F= 1,128 ^ -yr and hence, from the load or strain P on a body, and the m^.uV*s tenacity T of its material, the strength may be found, for wmcn body will not be strained beyond the limit of elasticity. Example. What load will a column of fir sustain, if it be 5 in oh es in breadth and A inches in thickness ? The modulus of tenacity being taken at 3000 lbs. and the sec on F being as 5.4 as 20 square inches, we obtain P = 20.3000 =s 6000 lbs. for te power of tenacity of this coiumn. But if we take the modulus of strength K — 12UOU lbs., and assume a triple security, we obtain P = 20 . ss 80000 lbs.; but to maintain security for a long period, we must only take one-tenth of JT, and we shall then have P = 20 .1200 s 24000 lbs.—2. A round and wrought iron pump*rod isto sus* tain a weight of 4500 lbs.; what diameter ought it to have ? Here T * 20000 lbs., WU 1» WClj^lU Mi iww li/o. J TT IIWV .. _____„ _ . ______ 'herefore, d = 1,128 = 1,128 . /^- = 0,535 feet The modulus of strength ’ ’ +J 20000 ’ *J40 for wrought iron of the medium kind a= 58000 lbs., and if we take one-sixth for the : 10000 lbs., and d s 450 - s 0.756 ineh, the 10000 security, we then obtain K s re qui si te thickness of the rod, tic^hodv' ^ronSes^ ofBody.—If a vertically suspended prisma- be or example, a pole or cord, is very long, its weight G must Z pt '? the f?r,ce of ™d, therefore, P+ G must be put cuhio v ?°w ^ °f the body, and y the weight of a b c lnch of its mass, we have then G^Fly, and. thereforefp- F (T~~ as ‘nversely F = If a body ABC.. G, Fig. 198, cons,sts of equal portions, each of the length l, its successive trans- sections are as follows. The section of the first portion is as P - - before F, r=Ty* ivuv n ----- # For the second portion, whose section is , 16*186 STRONGEST FORM OF BODY. and weight ly, P + F,Zy + Ftly - FJ, hence F% - = F>+ F,Zy „ / h \ For the third portion it follows that ji j-----+ y_____________ly/ t y / ly \ /i , , for the fourth F = p F, = r,(i + J^) » F, (i + T=V / (y \ f /j + y, and generally for the nth por- V + T— ly)' , __ P /. *1 tion ~ T-—ly' /, /y v~ ; F„ = F,. (l + y^) or F“ “ r-/y ( + r-Zy) ’ the corresponding section. . , If Z is very small, the portions therefore very short, we may then put • - P/- . Zy\“ F. _ £(l + I)' 1 \ J- / If the number of portions is very great, or if the thickness of th body JiGy Fig. 199, increases uniforraly from below upwards 6 may then (from the reasons in § 175,) put the cross section ’ We F„ = -.c T = r ' e ” 7 Fig. 198. Fig. 199. A where e represents the base 2,71828 . . . of the Naperien logarithms, and L the entire length of theAbodY of uniform thickness to have the same tenacity throughout, must have a transverse sec- „ P If iy is small as compared tion i =■ fZ-Ly ■with F, — is a small fraction, so that we may put: and 1 + ^ + m . ©■} further, the weight of the first body is and that of the second = F. Ly - - mu- -[ i+^+ be called L' * T ’ \I7Ji hence the prismatic body is heavier, and on that account more costly than one having at each point in its length a cross section corre- sponding to the load it has to bear, and which may therefore 1 a body of uniform resistance, or a body of the strongest form Examplet.—1. What cross section ought a wrought iron shaft 100 feet long to have, when besides its own weight it has to sustain a load .P=75000 lbs. ? The modulus of tena* city or strain is taken at T=IK= 10311 lbs., and the weight of a cubic inch of wrought 7 60 62 4 ” P iron y = -j- = 0,27444 lbs. The section sought is F =s —-— = 12.12.12 1 •" -L/yELASTICITY AND STRENGTH. 187 __15^027444- square inches, and the weight of the shaft G = jF . Ly ^>J1 . 1200.0,27444 = 2473 Ibs.—2. If we were to give to this shaft the form of a of uniform resistance, we should then obtain for the least section F = — = 75000 T 10311 7,28 square inches; for the greatest section 2',n = 7,28 .e0’**444 °* 18= 7,Q8 . e ' = 7.513 square inches, and the weight a f.7,28+7'6L3) . 329,3 =2435,5 lbs. (approxi- mately). 2 § 186. Numerical Values.—In the following table are given the mean values of the different moduli, of elasticity, tenacity, and strength of the materials most commonly occurring in construction. TABLE I. THE MODULI OF ELASTICITY AND STRENGTH. 2tAMXS OF THE 8UBBTAXCE8. Extension at the ' limits of the elasticity. X T' Modulus of elasticity. E. Modulus of working load. T. Modulus of strength. K. Modulus of safety. Box, oak, ftr, firm Scotch fir - 1 600 1856005 3094 12373 1237. Iron in wires 1 1250 26808964 21650 87645 14436 Iron in bars 1 1520 29902306 20622 59805 10311 Iron in plates Cast iron 1 1200 26808969 17528938 14436 56712 19592 9280 3094 Steel 1 835 30933420 37120 123700 20622 Hard cast Steel Copper ^°pper wire ...... Brass ........ 1 4500 1 1320 45369016 97955830 98987 7218 150543 38151 75271 18560 24740 6187 12370 3093 Brass wire . k . . . r 1 742 149511530 20622 75271 12370 Bell metal 1 1590 48462358 3093 35058 5774 Lead ..... • • f • • • 1 477 721779 1547 928 329; h^aden wire Marble ^Pes under l inch’ I 1 I 1 — 3 inches ... above 3 «... . ktrap8 ... 1 1500 1031114 2680896 722 2062 2062 9280 7218 5156 351 206 3093 2371 1753 299188 FLEXURE OF BODIES. The values containcd in Ihe uecond vertical column of thi, tahle, of the relative extensio» (|) at the limita «t elaaticity, gi.e likewiae the relatio» 1 of the vah.ee of the fourth »»d third columna. The aixth column is derived from ' bl ^Theltrength of wires is always stmngerthanthetr nnc^oa.^ bo(lj, Fig. 200, § 187. Flexure J .t fixed at one extremity, for fo- Fig. 200. stance, imbedded in a wall, and at the other extremity acted upon by a force P; strains then take place in this body, in consequence of which, one part is extended, and the other compressed, and the whole becomes deflected. If we imagine the whole body to be decomposed into thin laminai by planes parallel to the axis, and at right angles to the direction of force, we may then „ *w tViprp is a certain mean lamina which is called the assume thaJ. the neutral axis of the laminae, which is not strained r‘z2 m 'rrh* ”bi!l,he 1>» by tnis nexure, extension, and those on the concave side the convex a,de undem) ts radius of curvature, we have then DD : KL *= LD : ML, and also or\ . 17r r hjtt • _ r\r\ .1 T n . ~ Tf wn nnW aSSUme 00, : KL= LO : ML;i. e. 00. : 1= f. If we now assume 1 - - - p, and hence * LO we obtain % : l = 1 1 and 001 = v ^ —E = If, finally, we substitute this value of m p l r formula M = WW, we have the moment M = 2^, and inversely, WE = Mf. p The product WE is called the moment of flexure, and hence the190 ELASTIC CURVE. proiluct of tk. nu>me«< M «nd the '°I^TZSti^rtJis KL,W 204. i"‘» * ' P»«». »» Fiir. 204. _ . T iro — Land determine the radii of curvature Lijj L2) ^2^3’ * u’ -vvf Ttrr _ „ &c., corresponding to these parts, the angle MLi - Pi’ J’ o 2TJ0X = t2°> &c-> which evefy two r»dii curvature include, ure know», vir. U, “ „ ~♦» - - &e„ und therefore ^ ^ ^ If’ fura,"> ___ - .1 1 X • of of , . . I VE_ WE ^ we then obtain *, =- vre substitute p, = ♦'j — ^ T1 t &c>; and by the suramation of ali these angles leiJ th.”a^l LOK - by »bich a greater porlion, or the whole neutrat ““uj*!'jf „e supp<»e a atuall ftexure, »e tnay § 189. Elasti ■ parallel to the initial direction of the take the projectiori of the beam itself, anj undeflected beam and ^ual to in^ f al t0 the parts likewise the projections LDV clc., 4 r 1 1 2* q /».1 . 1 • ^ and we obtain the moments &c., of the neutral axis, 1. e. =*—> anu wc uu n JV/, = —, M - 2 f/, JVf, = 3 Pl> &c. If we substitute these values i^ the forraute for *„ &” ., then the measures of the angles of curvature are given: ^ p% 2 pp 3 . — n2 WE ’ *2 ~n2 WE' *3 n2 WE'ELASTIC CURVE. 191 and by addition, the measure of the whole angle of curvature KOL = a of the neutral axis : Pl2 a ~ n2 WE ^ * + 2 + 3+ ...+ n) = With the assistance of the last for- mula, we may now find the equation to the curve formed by the neutral axis, KL, Fig. 205. Let us divide the absciss LJY = x, commencing at the point Z, into m equal parts, and find the parts of the ordinate NQ = y cor- responding tothem. Since the radius of curvature QR is perpendicular to the part of the arc QQp the angle QQ1[/= QRK = a2, and thereforethe part QU of the ordinate yl = QYU. tang. a2, or Qjl/being put = — and tang. a2 = a2, QU Xa m m Now a2 = LOK — LMQ = a —a, Pl2 2 WE X Px1 therefore, that QU =— . m 2 WE 2 WE 2WE P (P — x2); it follows, (P—x2). If for i3 we substitute suc- , we then obtain by the last formula ali the parts of y, and by the addition of these, the whole ordinate : (w)'+ • • • ] - ~ • SWE O - + 2’ + 3'+ .. + m2 1 . Px /„ A J, 1- e. y — 2 “ t> height of the arc CK = a. Pl This last is: PP Therefo " "" 2 WE' '~~ t) " WWE' the length. * ^ height of the arc increases as the force and the cuheof i? (—) »i? ftl\\ n n n x/ l ' 21 ^ &c., because the centres of gravity of the loads q . -, q . q . ^ &c., lie in the middle of i ^ £-*, the arms are, therefore, * . . J , ii, Hence we obtain ” W oP i 2*. qP _ _ i 3a. qP s ♦i " i • ~tfWE’ ** “ * ' »3 WE’ *3 * ’ n3 IFE ’ C* And therefore, 3 - i (1,+2‘+3'+ •''+"!) - dwt • t - Se, and likewise, - _ ?** , and a, * q— (P — **)• From this last measure of the angles an element of the ordinate x x ? (73 — x3), and now for x3 putting successi vely «2 = m m 6 . Q mP — (^) (1’ + * H-+ ”’)] “ m ‘ 6 WE ■[--©•f]"('"T)- 6 the equation of the curve sought. If again we take x = /, we obtain the height of the arc ji ,p__??___________«L__» 6ra ,f _ sire _ sra ■*' 3K®’’-"' a = fths as great as if the load Q were suspended at the extremity of the beifthe beam is loaded by a weight Q, uniformly distributed, and by a force P at the extremity, the height of the arc is then pp , QP _(L 4. a = 3 WE + 8 WE \ 3 + 8 / WE If a beam JIMB, Fig. 206, is supported at both extremities, and loaded in its middle by a weight P, both the extremities are deflected upwards by the reactions J P and J P, as was in the former case (§ 189), the one extremity downwards, the formula then found here holds good, if instead of P, we put —, and instead of the whole length LL =* /, half the length KL » -. Hence the height of the arc is: 2RECTANGULAR BEAMS. 193 «- LLM, A . 3 WE — = a sixteenth of o WE the height of the arc of the beam, which is loaded at its extremity. If, lastly, the load Q = ql is uniformly distributedover the body./?#, Fig. 207, sup Fig. 206. Fig. 207. ported at both extremities, we must put m the formula a = + I) Win pIaceof/’i’ in place of P, P_+_Qand 2 for Q,—because with respect to K, the weight — at Z the arm - is opposed to the 4 P + Q re ac t ion 'Z i. Consequently at the arm (P±S-2\-L- = (p + *q) 16/ 8WE V^8*J Q/3 For P = 0, a = £ . 48 WE' ; the load is, therefore, uniformly dis- 48 WE tributed over the whole arc, and the height of the arc is £ times as great as if the weight acted at the middle of the beam. § 191. Rectangular Beams.—Inorderto givethe relations of flexure °f a beam or other prismatic body, and the elastic curve formedby its neutral axis, the transverse section of the body must be known, an the moment of flexure WEy calculated from it. If the section of the beam be a rectangle ABCDy 208, of the width AB = CD = 6, the height AD r" BC = hy the moment of flexure WE = (Fx z,2 + Fiz22 + ...) E will be known if we decompose this cr°ss section by lines parallel to the neutral axis JVO mto 2 n equal laminae, each having the area b . — = bh 1 2^ > and determine the moments of these laminae, anu Fig. 208. 17194 REDUCTION OF THE MOMENT OF FLEXURE. 1 2 3 add them together. Ifwe put successively - • 2’» * 2 °f * m “ 2. E « shall then obtain tb. momenta of the l.miniB On «ne nide rfthe neutra, axis; but ifwe rl.uble theiraum.we have «he moment of flexure ... Ax* -i _ bh WE = 2 2w \UflV = M . /i-Vtl2+2*4-3*4-.. 4-»*)£=-^-£ = „ 71 \27i/ 4.3 12 * The moment of flexure, therefore, of a rectangular beam incre as the width and the cube of the depth of the beam. QSes - wr» • A /• 1 D/3 *ormi! IU n . r/A2\ , /^\# + (jT-) +•••"] [(r„) + fc)+W , J 5 the vnain anu, frfKr --v D/3 If we put this value of WE into the formula = we shall obtain u 4 . 3 _______, but if into the formula = _L ------------ 6A3£ 48lTg- of § 190, then a = Inversely, the modulus of elasticity _ API* follows from the height of the arc a E = Ior the one, and E — ?Jl-,for the other case. 4 abh Example.—1. A wooden beam, 10 feet = 120 inches in length, 8 inches in width, and 10 inches in height, is to be supported at both its ends, and bear a umform load Q = 10000 lbs, what flexure wUl it undergol The height of the arc i» « = } 120* 50000 ■ 12* _ 1350000 jtfow E being put = 1800000 ih„ 8 . 10* . £32.8 £ 4 .£ it follows that a = _1E_- = 0,1875 inches.-2.If a rectangular castironbar, 2 inche, wide and A inch thick has been deflected J inch by a weight P =18 lbs. lying in the middle of t, whilst the distance of the supports amounts to SJeeMhe modulus of eia9. ticity of cast iron will be £ ' “ I *^^ 16000=- 15552000 lbs. __ 5 — 3* 10000 Fig. 209. 552000 lbs. § 192. Redudion of the Moment of Flexure.—If we know the moment of flexure of a body, J1BCD, Fig. 209 about an axis JV,0„ lying without the centre of gravity, the moment about another axis passing through the centre of gravity S, and running parallel with the former, may be found. If the distance HH1 == KKl of both axes = d, and the distances of the elementary surfaces Fl Ft, &c., from the neutral axis JVO = 2„ z &c., we shall have the distances from the axis JVjO,, = d + + z2> &-c., and the moment of flexure will be = [£, + z,)2 -f p (d + z.)* 4-...]£ = [£, (d3 4- 2dz, 4- z* f + F* (d2+2dzi + z\) + ...] £= [ and b the breadth AB of the triangular section ABC, we have the height of these laminae = their lengths = &c., to — and n n n n nJ their distancesfromJVjOj = &c., to From these the n n n n areas of the laminae are Fx = F2 = F3 = ^ and their Fig. 210. moments Fxzx = b h? , Fx* = 2*. F,z? = 3’. b n'n' and the moment of flexure about the axis A^ 0,: bh3 /i .t i cvt i m ■ i . bh3 , &c., W. ~(F + 23 + 33 + .. + «») = . Z. = ™ ' «4 A. 4 0 The distance of the centre of gravity S from the point is = 1 h, and the area of the whole triangle = therefore = h-^Ah* 2hh* A>v 2 2 9 — g—> and the moment of flexure about the neutral axis 'nro sought is: WE = ( — Fd2) E — (™L _ 26A3\ 1 = a third of 9/ 3 12 the same dpnil!n0mient-i>( ^exure °f the rectangular beam, which has has doublt* *L * i W1 . as B>e triangular one. But since this beam circumstann e ^°iume, it then follows, that under otherwise similar the rectangu]S> ^ tnan^u^ar beam has § of the moment of flexure ofK>| >- 196 hollow beams. Fig. 211. We may find in the same manner the mo- ments of flexure of many other bodies used ™ construction. For the transverse section of a T-shaped body J'B'C/ib ^ r1’ bc, - 1 1___4 1 2 1^ ^ 1 ^ ^ as follows, if we consider each of these rectangles as the half 0f rectangles having double the height with the neutral axis Now the area AXCXD — bh — bfaand lts moment Fd = _ bA . A = I (bh2 — 6jV)i hence Jt follows that the arra MS 2 2 = d= — ~AAj1, the moment Fd2 = J — ijA,*)*(M—i 2 (6A — ^i^i) . . and the moment of flexure about the neutral axis passing through the centre of gravity S: W= W1 — Fd1= — ~ h'-1- — i (bh* — — b^) 4 (M3 _ 6,v) (&*-W-3(M« — wy — 12 (6A — A,A,) (W—bA*)1— 4 iA Mi (A — hif ~ 12 (6A — bA) S 194. tfotfow Beams.—The moment of flexure of a hollow rectan- gular beam A, Fig. 212, is determined, if we deduct from the moment of the complete beam that of the hollow part. is the external breadth, and = the height, and A1B1 = 6, the internal breadth, and 5,C, = a the height, we then have the moments of flexure 0f both = h~ and and by subtraction we Fig. 212. 12 12 get the moment of flexure of the hollow beam W = W * A A 3 ■W rr “ i2 We may find in an exactly similar manner the moment of flexure of a body ABCD, Fig. 213 hollowed out at the sides. AB = b is the outer breadth and BC = A the height; and if __________ A1B1 = bv and BXCX = hv the sum of the breadth and the heights of both hollows, by subtraction we have again: W — 12CYLINDERS. 197 The moment of flexure of a body ABCD, Fig. 214, o a cr shaped section, may be obtained in the same manner. Here AB = b the width, and BC = h the height of the middle piece, and if AlBl — AB = bx and AXDX = hx are the sum of the breadths and the height of the side ribs; by addition we have the moment of flexure: W = + bxhx ~ 12 It is besides easy to see, that deep, hollow, and ribbed or flanged sections of the same area have a greater moment of flexure than square sections. Because this moment increases with the transverse section F and the square (z2) of the distance from the neutral axis, one and the same fibre affords, therefore, a greater resistance to flexure, the further it is distant from the neutral axis. If, for examp e, e eig h of a massive rectangular beam be equal to double lts brea , i s b . (2 b)3 Qi4 __ 2 b . b3 moment of flexure will be either W = -------—— = ^ o or — ^ = J b4, according as we put up the beam with the lesser breadth b, or the greater 2 6; in the first case, therefore, the moment of flexure is four times greater than in the second. If we replace the massive beam of the cross section bh by a hollow one, whose hollow bh is equal to the massive part of the section bxhx — bh, if, therefore, bxhx — bh — bh, i. e. bxhx = 2bh, or bx = %/2 and hx = h \/2, we shall obtain the moment of flexure of the last ==^ ^2 (A v/2)—------ 12 12 = t3£ bh3, i. e. three times as great as for the first. § 195. Cylinders.—The moment of flexure of a cylinder is deter- mined in the following manner. Let AOBN, Fig. 215, be the cir- cular transverse section, and NO the neu- tral axis of the cylinder. The diameter AB, divides this section into two equal parts, having equal moments of flexure, and the moment of flexure of the whole may be found by doubling the moment of the half ANB. The half may be divided by sections DE, FG, &c., parallel to AB, and at right angles to NO into thin lamina, which may be considered as rectangular. The moment of flexure of such a portion DEFG, = KL • — Now CA= the radius of the circular 12 section, a quadrant ^JVhas, therefore, the area ,andif we . i r< __ The this into n equal parts, any such part DG = - • -g- 2 n Fig. 215. 17*198 CYLINDERS. projectiori parallel to CJV*, GH= corresponds to this part, and may be determined by putting, GH: GD = GK : CG, and, there- fore GH = — ' — = JL. . GK.Hence we have for the mo- ’ CG2« ment of flexure of the part nK)3 DEFG-&-GK. 3 n (GK)4. /S71 * ~ If we put the variable angle corresponding to the section GF, jqCG=t>° we shall obtain the ordinate GK—-r cos. $>, and for the last . « 4 3 + 4 cos. 2*4- cos 4* moment of flexure = ^ r4 cos. t = r------------------------- The moment of flexure of the half cylinder will be now found, if f0r . 1 h 2 n 3 h o . n n * we successively put the values &c.,t0 - —, and add the results. But ~ = ■ —- is a common factor; we o 71 o z^kn have, therefore, only to consider the sum of such values, as 3 + 4 cos. 2 t + cos. 4 *. The number 3 added n times gives 3 n- the sum of ali values of the cos. 2 $ which present themselves, when i> is made to increase from 0 successively to and, there- fore, 2 * from 0 to n, equal to 0, because the cosines in the second quadrant are equal and opposite to the cosines in the first; lastly, the sum of all the cosines of all angles from 0 to 2 * = 0, hence the sum of all values of 3+ 4 cos.2 *+cos. 4 f taken between the hmits * =- 0 and t = - is = 3 n,and the measure of the moment of flexure ^ 4 4 of the half cylinder = .3 n = and, lastly, that of the whole cylinder: W= j r4 =0,7854 r4. For a tube or hollow cylinder with the outer radius r, and the inner r. TV= j (n4—V). 4 Fig. 216. 4 To find the moment of flexure of a body having a semi-circular transverse section ADB, Fig. 216, we may make use of the rule found in § 192, from which the moment about the axis JYO passing through the cen- tre of gravity S is equivalent to the moment about the diameter ABy con- sidered as a second axis, less the trans- verse section F (= J rtr2) times the square of the distance CS of both axes.RELATIVE STRENGTH. 199 From this we obtain the moment sought 1 * < 1 _____ 2 * 4 r ~2* ^ “ f - s "* • (0 «108) - “ ^ (s - A) “ °’110' ^ § 196.—Relative Strength,—When we know the moment of flexure of a prismatic body, we may determine from it by simple multi pii- cation the working load and the absolute strength of the body. If a single fibre, or layer of fibres, is extended or compressed to the limits of elasticity, the body has then attained the limits of its tenacity. If we again represent by T the modulus of tenacity and the distance of the furthermost fibre from the neutral axis by e, we shall have T = - E> and y, or the relative elongation, = i, hence ll p Et T E — ~ If we substitute Lfor — in the formula for the moment of flexure, it will then give the statical moment of the tenacity • We have Px » SW = therefore, also, Px = It is evident that p e this moment is a maximum when x = /, or when the arm = l; from this we may conclude, that at the extremity where the beam is fixed, the greatest flexure ensues, and the limit of elasticity is first attained. Accordingly, the working load of a beam is determined by the formula p-™: e l In like manner, the strength, or the resistance to rupture of the beam, may be determined. If a fibre is strained to the point of rup- ture, the breaking of the whole beam takes place, because the beam has now a section smaller by the section of these fibres, and there- fore a greater deflexion ensues, and thus a rupture of the succeeding fibres or layer of fibres follows. If we put the modulus of strength E K = K, wehave — = —, and, therefore, the force for the rupture of the beam: P P = KW el' In a uniform rectangular beam, the distance of the outermost lamina of fibres from the neutral axis » —, hence the formula Pl — T, 2 E bh3 ~ • 12 (§ 191) gives the resistance to rupture _ 2 K bh3 bh* K P~T"l2lTr** ... If the beam is hollow, as in Fig. 212, we have P » -* g ’'A’ so that the formula also holds good for a body, as in Fig. 213, hol* lowed out at the sides.200 EXPERIMENTS. I» a prismatic body of « trianml.r cross sectio», as i» Fig. 510, K bh3 _ bh_ K' According to this, rectan- e = $K hence P = ^ M l gular beams for a simLr sectio» haee twice the tenaci, of tria„galar beFor a cylinder of radius r, e - r, therefore, T ft/r4___r 4\ If the cylinder is hollow, we have Pl = j If we substitute the modulus of the working load T for that of the strength, or for K, an aliquot part, i. e. Tyh, the working load is given by the formula already found. § 197. Experiments*—To find the deflexion and tenacity 0f beams, we may make use of the experimental values for E and T in § 186; but as concerns the strength of beams, it is safer to replace the modulus of strength there given and derived from experiments on tensile strain, by those values of K which have been found from experiments on compression. A perfect accordance cannot exist between the moduli found by these two methods, because in rupture, not only an extension, but also a compression takes place, and both of these not only in the direction of the axis, but also in the trans- verse section, though here not to the same amount. Besides, many other circumstances affect the elasticity, tenacity and strength of bodies, on which account, considerable variations in the results always present themselves. Timber, for example, is stronger at the core and at the root than at the sap and the top. Timber will also bear a greater strain when the force acts perpendicular to the annual rings, • See Appendix.MODULUS OF RELATIVE STRENGTH. 201 than when parallel to them. Lastly, the soil and the situation where it has grown, temperature, dryness, age, &c., affect the resistance of woods. Besides, the deflexion of a body after it has been loaded for a long time, is always somewhat greater than on the immediate appli- cation of the load. Experiments upon elasticity and strain were made by Eytelwein and Gerstner, with the apparatus represented in Fig. 217. AB AXBX are two tressels, C and Cx two iron supports. DDX the rectangular beam for experiment resting upon them. The load P for the flexure of the body lies upon a scale-pan EEX suspended to a stirrup whose upper and rounded extremity lies in the middle M of the beam. In order to find the deflexion corresponding to a load P, Eytelwein applied two fine horizontal threads FFl and GGX and likewise a scale M resting upon the middle of the beam; vonGerstner 9 on the other hand, availed himseif of a long one-armed delicate lever OK, whose fulcrum was at Jkf, and whose extremity, like the hand of a watch, indicated upon a vertical scale KKX the deflexion of M to fifteen times its amount. Remark. Experiments on elasticity, &c., ha ve been made by Banks, Barlow, BufFon, Burg, Ebbels, Eytelwein, Finchan, von Gerstner, Gauthey, Muschenbroek, Rennie, Ron- delet, Tredgold, &c. An ample summary of these, and besides a theory somewhat dif- ferent from the above, is given by Burg in the 19th and 20th yols. of the “ Jahrbucher des polytechnischen Instituts in Wien.” The experiments of Eytelwein and von Gerst- ner are described in Eytelwein’s “ Handbuch der Statik fester Korper,” vols. ii., and in von Gerstner’8 “ Handbuch der Mechanik,” vol. i. The Treatise printed from the trans* actions of the Association of Prussian Industry, “ Elementare Berechnung des Wider- standes prismatischer Korper gegen Biegung,” by Brix, has been used for the preparation of the foregoing article. § 198. Modulus of Relative Strength,—The following table con- tains the mean values of the modulus of rupture for several bodies met with in the arts. To find, with the assistance of these, the pressures which bodies can sustain with safety for a long duration, we must put for wood the tenth, for metals and stones, from the third to the fourth of K* TABLE II. THE MODULUS OF FRACTURE OR MODULUS OF STRENGTH FOR THE FLEXURE OF BODIES. Names of Substances. Modulus of Fracture K. Names of Substances. Modulus of Fracture K. Box Oak Pine Scotch Fir - - . Dea i 10000 to 24000 8000 “ 24000 8000 “ 13000 7000 “ 17000 7000 “ 14000 Elm Cast Iron - - - - Limestone • - • Sandstone - - • Brick - - - - - 6000 to 12000 24000 M 56000 700 " 1700 600 “ 800 ISO « 340 According to this, we may assume for wood as a mean K ■» 12000 Rnd for cast-iron K m* 40000 pounds, and we sball then obtain for a See Appendix.202 MODULUS OF RELATIVE STRENGTH. rectangular beam imbedded in a wall at one extremity and loaded at the other: 1. Pl = 200 . bh2y if it consist of wood, and tenfold security be allowed. 2. Pl =s 1000 . AA2, if the beara be of cast-iron, and fourfold security be given. If the body be cylindrical, we then have for wood 3. Pl = 950 r3, and for cast-iron 4. Pl = 4700 r3. P, /, b, A, r, have the denominations hitherto used. For wrought iron K is taken 20 per cent. less, because this bends more than cast iron ; here therefore we must put Pl = 800 bh2 = 3600 r3. If the load Q be uniformly distributed over the beam, the beam will bear as much again, wherefore the above co-efficients must be doubled. If the beam rest at its extremities on points of sup- port, whose distance is /, and if the load P act in the middle be- P l tween these points, then for P we must put — and for /, —, where- 2 2 Pl fore Pl becomes —, and the the tenacity quadrupled. But if the 4 load between the points be uniformly distributed over the beam, we then shall have for the pressure which acts from below upwards at a point of support, the moment Q and for the opposite pressure_as the half of the load pulling downwards at the centre of gravity, the moment----------= Ql 8 hence there will remain as the pressure for rupture at the middle, the mo- ment —______— z=z —, and therefore Ql = 8 . K, also = 8. ^ br3 Ky 4 8 8 6 4 therefore the strength or tenacity is twice as great as if the load acted at the middle, and eight times as great as if it pulled downwards at one extremity whilst the other remained fixed. If a beam, Fig. 218,* is irabed- ded in a wall at both extremities, or if its extremities are fixed, then the beam sustains as much again as if it rested freely at its extremities; for in this case the greatest flexure is not only in the middle, but like- wise at the extremities; the beam, therefore, breaks at the same time in the middle and at the extremi- * See Appendix.STRONGEST BEAMS. 203 ties ; whilst at the intermediate points C and D, where the convexity passes into concavity, no flexure at all ensues. Consequently, for a p i P portion AC, the pressure = —, the arm = -, and the moment = — 2 4 * • ^ = ~. If, finally, in this last case the load Q is uniformly distri- buted over the beam, the moment presents itself = because we may suppose, that the one half of Q is immediately sustained by the points of support, and that the other half acts in the middle of Q. The weight G of a beam acts exactly as if the load Q were distri- buted uniformly over the beam; for a beam fixed at one extremity, therefore, the moment = Pl + A Gl; but for a beam resting on both P l G l G extremities and loaded in the middle, it is = — . - + — . - — —. i = (P+4G) l, &c. Example.- rest on both 20 feet; what load, suspended from the middle, will it sustain? 6 = 7, A = y, * — feet = 240 inches; hence 240 . P =s 4.200. 7. 93; consequently this load P = 70.27 = 1890 lbs.—2. A round wooden water-wheel, and its axle, 10 feet long, is to sustain at the wheel, together with its own weight, a uniformly distributed load Q= 10000 lbs.; what diameter must the wheel ha ve ? Ql = 10000. 120 = 1200000, = 8.960. r*, or r« 1200000 3____ = ■ g Q5Q = 167,9; hence the radius sought r = ^157,9 = 5,4 inches, and the dia- meter of the axle 2 r = 10,8 inches, for which we may assume one foot.—3. To what height may the corn in agranary be heaped up if the bottom rest upon beamsof 26 feet m length, 10 inches in breadth, and 12 in depth, the distance between the axes of any two beams = 3 feet, and one cubic foot of corn weighs 48,6 lbs. ? If we apply the formula Q/= 16.200 .bh7, we mustput b = 10, h = 12, /==25 12 = 300 • consequently ^ 16.200.10.144 k ' , , Q=------------------= 15360 lbs. A parallelopioed. 25 ffi«t lnnrr 3 feet broad, x feet s 15360 lbs. A parallelopiped, 25 feet long x . 48,5 lbs.; hence, if we put this value = Q, it follows that 300 deep, weighs = 25 . 3 15360 " *---5 = 4>22 feet» the requisite height to which the grain may be heaped up. § 199. Strongest Beums.—Bodies of equal section very often pos- sess different relative strengths; the formula PZ —* ~ • AA* showsthat the strength increases, as the breadth, as the square of the depth, and tnversely as the length of the beam. The depth has consequently a greater influence upon the tenacity than the breadth; a beam of double the breadth bears twice as much, i. e, as much as two single beams; on the other hand, a beam of double the depth, four times thatof a beam of the same depth. For this reason beams are made, namely, when they are of cast iron, much deeper than broad; they are hollowed out near the middle, and what is taken away replaced by parts at a greater distance from the neutral axis; but this rule must be particularly attended to, viz., always to lay the beam on the least side, or rather so to lay it, that the pressure may act in the direction of the greater side.204 STRONGEST BEAMS. The strength of a round trunk, or of any other cylindrical body, is P=zj,— K, that of a square with equal breadths and depths 2 r, * v _p_ 2 r • (2 rf K -,if we compare both pressures / 6 3/ with each other, — = - . - = 0,588 ; the cylindrical body has, P 4 4 therefore, only about 59 per cent. the strength of a beam having a square transverse section. Wooden beams are hewn or cut frona round trunks of trees, and thereby are much weakened. But the question now is, which is the strongest form of beam that can be cut from a cylindrical trunk ? Let ABDE, Fig. 219, be the section of the trunk, AD = rf its diameter, further AB = DE = b the breadth Fig* 219* and AE = BD = h the depth of the beam! Then b2 + h2 = rf2, or h2 = (P — b2, and the moment of rupture. pi = £ . - ( *). The problem amounts to making b (rf2 — b2) = bd? — b3 as great as possible. If instead of b, we put b + x, where x is very small, we then obtain for the last expression (b + x) d?—{b +_ x)3 = bd2—b3 +_ (rf2—3 b2) x—3 fo2, provided we neglect x3, and the difference of the two = (+ rf2—362) x + 3 bx2. That the first value^rf2—b3 may in every case be greater than the last, the difference + (rf2—3 b2) x + 3 bx* must be put positive, whether we take b greater or less than x. But this is only possible if rf2—3 b2 = 0, for the difference then = 3 bx2, therefore positive, whereas, if rf2— 3 b2, is a real positive or negative value, 3 ix2 may be neglected, and the difference may be put == + (rf2—3 b2) x, which if x has the same sign, is at one time positive, at another negative. But if we put rf2 — 3 b2 = 0, we obtain the breadth sought b = d J, and the corresponding depth h = v' rf2—b2 = rf h v/ 2 therefore, the ratio of the depth to the breadth: - = —^ = 1,414 or b \/l about J. The trunk must be so fashioned that it shall produce abeam whose depth to its breadth is as 7 to 5. To find the section corre- sponding to greatest strength, let us divide the diameter AD into three equal parts, raise at the points of division M and N perpendiculars MB and JVE, and finally connect the points of intersection B and E rf ^rc*e Ite extremities A and D of the straight line AD. ABDE is the section of greatest resistance; for since AM: AB = AB : AD and JJJY AE = AE : AD, therefore ^ which is actually requisite. Remark. The trunk has the moment of rupture Pl = * — • r3? but the beam ofgreatest 4 8 K L resistance formed from it Pl = ^ . d $ d? = == *** ^ trunk, therefore, loses by squaring about 1-— . — = 1 — 0,65 = 0,35, i. e. 35 per ^/243 » cent. of its strength. To spare this loss, the trunk is oflen hewed not quite square, but the corners rounded off. A beam with a square section formed from the same trunk, has the moment Pl ____ — . d s/ £. —, because here the breadth = the depth = d = “6 2 0,707 d, hence the loss here = 1 8 .1=1 8 6.2 \/2 3 «^ an(j^ therefore, the modulus of Fig. 224. E B 1 i 11 elasticity K must be taken at P COS. a less; therefore, we must substitute for K, K — P COS. a whence it follows that: P sin . / __ P COS. i. e. if the pressure P acts in the middle of the beam, P becomes a minimum, because, if we put lx = - + x and l2 = - — x, the pro- duct forming the denominator lx l2 — - — x2 is always less than 4 4 whether - be made somewhat (x) greater or less. 2 A beam, therefore, supported at its extremities, sustains least when the load is applied at its middle, and one so much the greater the nearer the load approaches one of the points of support. If a load Q be uniformly distributed over the length c, the centre of which is lx and /2 distant from the points of sup- port A and P, Fig. 226, we shall thenPLANE OF RUPTURE. 209 have to take the difference \ for the moment of rupture, LZ , Q because the pressure Qx = at the arm Zlf and half the ^eig t ^ l acting at the arm - is opposed to it. Therefore K bh2. Example. What load does a hollow cast iron beam sustain, if its outer depth and re addi amount to 8 inches and 4 inches, and inner breadth and depth 6 inches and j Ulc“es j and if further, the middle of the load, uniformly distributed over 3 feet in ength, is distant from one point of support 4, and from the other 2 feet? It is = I.:512-2'216 = 202; further, « _ £ = fili _ 12 = K 8 ' I 8 \ 6 8/ 2 inches; hence, |q= 1000.202; and consequently, Q = 17565 lbs. § 203. Plane of Rupture.—If the beams are not prismatic, if they have different transverse sections at different places, the plane ot rup- ture, i. e. the plane in which rupture will ensue, will no longer e the same as for prismatic bodies, because this place is not only e- pendent on the arm x, but also on the transverse section. If we sup- pose a rectangular section of variable breadth Wy and height z, and assume the beam to be fixed at one extremity, and at the other acted upon by a pressure P, and the distance of the transverse section wz from the extremity where the pressure acts = x, we must then put jr 2 2 P = — . and find the minimum value of in order to de- 6 a: x termine the weakest part, or plane of rupture of the beam. Here many cases present themselves; let us consider only the fol- lowing. Let the body ABEG> Fig. 227, be a truncated wedge, or have the form of a prism with a trapezoidal base, let the breadth DE = FG at the ex- tremity = 6, the depth EF = DG = k, and the distance UK of the edge cut off from the terminating surface EG, = c. Let us now assume that the plane of rupture NL is distant UV = x from the terminat- ing surface, we shall then ob- tain for it the depth ML = z =zh+-h=h(l + wrhilst the uniform breadth is The value (l + £\’ =, b(i + - + f) increases and x x \ c / ^ diminishes simultaneously with- + and is> therefore, also a mini x c Fig. 227. •18210 BEAMS OF THE STRONGEST FORM. mum, when this latter term is of the last value. But if in place of x, we put c + u, where u is a small nurnber, we shall then obtain for it: c + u c + u » i (2 — - + — — . . .) + — — - + ~a' As now in this last c \ c c* / c* c ' cr expression u appears only as a square, it follows that every other value, which is obtained when the distance x is assumed greater or less than c, gives a greater value than for x ** c, that con- sequently for x ** c, i —-> and, therefore, also--------------- 5^2 ^ ' x c* x /I 4. ? 4- I\ « — is a minimum. From hence it follows, that \c c c/ c the magnitude of the surface of rupture « b . 2 h 2 bh, and is dis- tant from the terminating surface EG » bh as much again as the edge HK of the portion cut off. In a similar manner, the distance of the plane of rupture from the terminating surface of a truncated pyramid or truncated cone is equal to half the height of the supplementary pyramid or supple, mentary cone. § 204. Beams of the Strongest Form.—A beam, which opposes an equal resistance to rupture throughout ali its sections, of which, there- fore, each may be considered as a plane of rupture, is called a beam of the strongest form. Of all beams of equal strength, the body of equal resistance at each point of its length has the least quantity of material, and is, therefore, the most suitable, and that which should be selected for architectural construction, and for machines, not only out of regard to economy, but also, that the weight may not be increased unnecessarily. If we put the distance of a plane of rupture from the further ex- tremity = x, and the roeasure of the moment of flexure for that section ~ W9 we then have the pressure requisite for rupture P W WK e x As K is a constant factor, a beam of the strongest form — must be ex constant also, i. e. it must be of the same value for every possible section. If for a beam of a rectangular section the variable breadth •* «, and the depth * v; but the breadth at the origin, or end supposed fixed « uv* , 12,9ade b, and the depth there W t, hence P = — . 2’ x K h, we have generally , and for the origin, for 6 which x has become /, P *= If we make these two values of P equal, we obtain the equation bh* * ----mm — for die beam of the strongest form. In a beam of equalBEAMS OF THE STRONGEST FORM. 211 breadth u = b is — = therefore, — = ?, which is the equation x l h2 l to the parabola (§ 35, Remark), and points out that the longitudinal section ABE, Fig. 228, must have the form of a parabola whose vertex is the extremity or point of suspen- sion E of the load. If the beani AB, Fig. 229, rests upon its extre- mities, and sustains a load in its middle C., or if a beara AB, Fig. 230, is supported in its middle C, and two pressures, balancing each other, are applied at the extremities A and By then the longitudinal pro- file has the form of two parabolas meeting in the middle. The last Fig. 229. Fig. 230. Fig. 228. case occurs in balances, which, as they are weakened by the holes at the points Ay C9 By are provided with ribs, or have a middle piece AB given to them. If the depth v = h is constant - = * or - = for x l b l the breadth u is proportional to its distance from the extremity, the horizontal projection, therefore, of the beam ABDy Fig. 231, forms a triangle BCD, and the whole beam a wedge with a vertical edge coinciding with the direction of force. If the body ABD, Fig. 232, is to have similar transverse sections, we shall then have -= ^,hence h b —l* e. ^ = j, therefore, the breadth and the depth in- Fig. 231. Fig. 232.212 BEAMS OF THE STRONGEST FORM. crease as the cube roots of the corresponding arms. For example, a section eight times further from the outer end than a given section, would only have double the height and breadth of that of the given section. If a beam be uniformly loaded, we have the variable load Q = and its arm = -, hence, instead of Pxy we must put xq . - = jf!? 2 2 2 whence = uv2. — and also 1^- must be taken = bh2 and 2 6 2 b consequently uv~ = Were the breadth invariable, that is u = b bh2 l2 9 we should have — = —, therefore, also - = and, therefore, a tri- angle ABE for the longitudinal section, and a wedge Fig. 233, for the body of the strongest form. If we take a uniform depth Fig. 233. Fig. 234. V = h, we then obtain £ = p and, therefore, for the plane a surface BDC, bounded by a parabolic arc, as in Fig. 234. If we agam make similar transverse sections, then £= ~ so that we have both in the vertical as in the horizontal profilc, a cubic parabola, in which the cubes of the ordinates increase, as the squares of the abscisses. If a body AB supported at both extremities, Fig. 235, is uniformly loaded over its whole length, we have for the moment of rupture at a distance from a point of support AO = x: ■§■ . x—qx . | = 1(^—0, on the other hand for the middle point: R L—R L= 91 = ^ ~2 ' 2 2 * 4 8 8 * If we suppose the body to be of Fig. 235.THE THICKNESS OF AXLES. 213 K andT bh2 uniform breadth, we have to put | (Ix—£*) = bv2 . ^ . and by division ^ or v2 = Were h = J ly v2 would be = Ix—x*, and therefore the longitudinal section would be a circle ADXB described with 2 as a radius, but because Ix—x2 must stili be multiplied by in order to obtain the square v2 of every ordinate OMy this circle passes into an ellipse ADBy whose semi-axes are CA = a = £ / and CD = /3 = h. The same relations exist for bodies with circular sections as for those with similar rectangular sections. In the case of a beam im- bedded in a wall at one extremity, and loaded at the other i. e. the radii increase as the cubes of the distances from the point of application. § 205. The Thickness of Axles.—In the parts of machines, as the shafts, axles, &c., flexures may prejudicially affect the working of machines, by giving rise to vibrations and shocks; and it is here, therefore, often more desirable to determine the sections, not ac- cording to their strength, but according to their degree of flexure. Gerstner and Tredgold maintain that a beam of wood, supported at both extremities and loaded in the middle, may suffer a deflexion o = ggg • ^ without disadvantage, and that such a beam of cast or wrought iron can only undergo a deflexion or height of arc a = ^ But now from § 190: a = an(* ^rom § 191: B'' = —» hence follows the height of arc: a = and ~ = . If now we ° 4 bh? E l 4 bh? E put — ss * , and E = 1800000, we obtain for wooden beams the v l 288’ tenacity or strength at the middle: _ a 4 bh? E 1 4 bh3 . 1800000 bh? l P 288 P P For cast iron =* —and E = 17000000, hence: P* J_ . . 17000000 _ 142000 . ~ 480 P P If further we take for cast irOn , and £»29000000 lbs., I 4oU we obtain for a rectangnilar beam of this material: 5 bh3 P « 242000 . -ijL. The co-efficients 25000,142000, 242000 must be multiplied by214 RUPTURE BY COMPRESSION. 3 h = 9,42, and h and b be replaced by r, for cylindrical beams as round axles, &c. The following table gives the dimensions of the transverse sections, l being expressed in feet, 6, A, r in mches, and P in pounds. Substances. Rectangular section. Circular section. Wood pn pp r‘“* 1600 Cast iron - - - - u>-±± 980 . PP 9250 Wrought iron - - 1680 15800 If the load Q be uniformly distributed over the beam, P must be replaced by £ Q, § 190, and if the weight of the beam be taken into account, by P + £ G. If it be the case of a beam which is fixed at one extremity and loaded at the other, P and l must then be doubled, therefore, PP to be multiplied by eight; if, lastly, the beam fixed at one extremity sustains a load Q uniformly distributed, for PP> we must substitute f . 8 QP *= 3 QP for PP* Examples.—1. What load will a wooden beam, 20 feet long, 7 inches thick and 9 deep, reposing on both its extremities, sustain for a length of time ? This load is p = j7o . = 170.<* 1190 . — = 2170 Ibs. p 20* 400 In § 198 P was found s=r 1890 lbs.—2. What thickness must an iron axle, 12 feet long, be cast, if the same has to sustain a uniformly distributed load Q == 40000 Jbs., without SQP . - , . 5 40000.12» any detrimental flexure? r4 = ■ t^ere^ore kere ^ = g* * —i5§qq— =228, and r ss ^/ 228 = 3,89 inches; consequently, the thickness of the axle 2 r = ot about 7j inches. By the formula for strength, if the modulus of tenacity of wroughtiron be taken at ^ times that of cast iron: = .----—-------ss ob «e * 8 . y . 4700 8.14.4700 y8’5S s_____ hence, r y/ 98,5 = 4,62 inches, and 2 r * 9,24 inches. & 206 Rupture by Compressum.—If prismatic bodies are so strongly compressed in the direction of their axes, as to aroount to rupture, their resistance to compression hastobeovercome Thumpture may take place in two ways. If the body be short, if it approximales to a cube, it will fall to pieces without undergoing flexure, but if the body is longer than it is broad and thick, flexure similar to that which takes place will precede the rupture. The one kind of rupture con- sists in a crushing, bruising, transverse strain, or splittmg asunder of the bodyor its parts; the other, in a fracture or destruction of a section of the body. Hence a distinction is made between the crushing strength and strength of rupture under compression, The resistance to crushing is, for similar sections, proportional to their areas ; for regular sections, however, somewhat greater than for irregular, and greatest of ali for circular sections. It is besides inde-RUPTURE BY COMPRESSION. 215 pendent for the most part of the length of the body. Short wooden prisms split asunder in the direction of their length, or form a bulge; stones break into several pieces or separate along an inclined plane. Ten times the absolute strength is given to wood and stones; to iron, only one of live times; and to walls of rough stones, twenty times. If K be the modulus of resistance to crushing, and F the transverse sec- tion of the bodies, the working load will be P = FKX and F = where for Kv i K to ^ K must be substi- ti tuted. TABLE OF THE MODULUS OF RESISTANCE TO CRUSHING* Names of substances. Modulus JT. Names of substances. Modulus K. Basalt - 27000 Brick - 580 to 2200 Gneiss . 5100 Oak 2800 w 6800 Granite • 6000 to 11000 Pine 6800 “ 8000 Limestone 1500 “ 6000 Fir 2000 Marble • 3200 “ 12000 Cast iron 146000 Mortar - 450 “ 900 Wrought iron 72000 Sancis tone 1400 “ 13000 Copper - 60000 The values of K contained in the preceding table are not unfre- quently, especially for wooden columns, applicable even when the bodies are very long, only it has been found necessary to diminish these values by one, two, or three-sixths, when the columns are twelve, twenty-four or forty-eight times as long as they are thick. Accordingly, for a column of oak, one foot thick and twenty-four long, K must be taken at from 2800 (1 — §) = 1900 Ibs. to 6800 . f *■» 4500 lbs. The formulae developed in § 185 for the transverse section of bodies of considerable weight, and of bodies of the strongest form, here find their application.* Examples.—1. What load can a round column of piney 12 feet long and 11 inches dia- _ «p meter, sustain? F as —I--= 95 aquare inches; if we now take for K a meanvalue 4 _ 6800 -f 8000 2 ~ 888 7400’ and diminish tbe TOlue one-sixth, becaose the length is 13 2“ 'hal °flhe *hiCkne8Sl “d therefore Pu* K « 7400 . f s 6200«*,- .ndgive.leo- ^ es secunty, we shalt then have P T. =- 620.95 s 58900 lbs,—2. How orner JtenwhTand 40^!°^ buUding of 20000000 lbs. weight, 60 feet gneiss ? Let r. Ka th* * hreadth, if for this purpose we use well finished blocka of breadth • th *u re TTT 7if — P DJ^s &C., hence we have the measures Mx = P • •'“a — • 2 p 777" 7-7- of the angles: „ - If ™ r?'i1;,1- r-» sm.ll*curva,n«, we may then write : ii, _ D L T T E2L2 = i i. = ; if further we di- vide the en*, height of the a,c CK _ » into » equal parta, we may then put: J),i, -= CC, = E,L, = C,C„ &c. = but i, = J!, j) l „ ^ Sic and by the substitution of these values it follows 3 5 n> *> that: p a a n na Pa* P.£f. ~ JT£:n2ffl » t2= w a I __ WE 2 Pa2 IKE n2a1 ’RUPTURE UNDER COMPRESSION. 217 4-3 = 3 P WE &c., or 4, a = WE ^*«x = 2 P WE n2 3 P 2_ W^,&c' The sum ^a + fj ^ a, + . .. = Pa2 WE n2 (1 + 2 + 3+.. + «) = —--------!L = ——, and may be also found, ifa be divided WE.n22 2 WE mto m equal parts, and any such part —, be put = $ =$„ = $,, &c. m WeshaJlthen obtain^a + f2 ai + *3 a2 +• • • = — • a+ a------- m m \ m/ + — (a — —\ + ... -i. — . — by taking out the common factor m \ m) T m m , and writing it in an inverse order = (1 + 2+... +m) = and by making these two sums equal to each other , an equation between the angle of curvature LOK = o°, Pa2 ~WE and the height of the arc CK = a. For the equation of the elastic line LK, Fig. 238, let us take LJV = x and JYQ = y as co-or- dinates, and put the correspond- ing angle of curvature LMQ =ar In the last equation, if we put a for y, and y for a, we then have to replace the sum 0 + t2ai+ t3«2 + ...by-----K——-Phenee, if we represent the supplementary angle QSK = a — ax by a2, we aflerwards obtain a2 — a„2 = Fig. 238. iy , or a. , and so a2 P __________ v. - ----- auu w u--- I___ WE * WE S WE since o,° = / QQ,T, and tang. QQtT TQ TQ1 have A — element 6 of the ordinate element « of the absciss WE • v' a2—y2. If in a rectangular triangle ABC, Fig. 239, with the hypothenuse AB = a, the angi increases by a small amount BABX = 4°, the perpendicular BC= y increases by the amount 19218 COLUMNS. DBX = $, and the base AC decreases by the amount D2?== CXC, then DB. j3C . S x/a2 — , , « /Zi---5 = ——, i. e. — = —-*L; and hence — = \/« —y2. Bv -nnn^a 4 J BB, AB' i . ____ * I P -------------------- 6 comparing the two expressions — = . >/a%—y1 and = v/a2 — y2, we obtain the equation 4 = • ^or = ^ Therefore the ratio of the element f of the absciss to the element of the 1 jyE arc 4 is invariable, and = I — , and hence, also, the ratio of the absciss x to the whole arc A = > *"• = and 1 1 “ A = x JL-. If, finally, we substitute this value of in the equa* tion BC= AB sin. A\ i. e. y — a sin. , we obtain the equation sought: y = a sin. (xJww)- With the assistance of this last formula we may find the ordinate NQ = y corresponding to any absciss LN = xy Fig. 240. Jf in this we put x = l and y = a, we then obtain a = a sin. (' Jwe)1 ’• “•1 “ J w)' "h Fig. 242, fixed at one ex- See Appendix.220 TORSION. tremity, is acted upon by a force whose direction lies in the plane normal to the axis, and, therefore, endeavors to turn the body about the axis, or when two forces of revolution P and Q act in different normal planes upon a body ABy fixed by its axis, Fig. 243, the fibres running parallel to the axis undergo a wrenching or torsion, the amount of which we wish to determine. Let AB, tig. 242, be a Fig. 242. Fig. 243. fibre before, and AD the same fibre during the torsion, and theref let the extremity of the fibre B be advanced by the force of torsioni D. If now l be the initial length AB> and x its extension, theref ° / -f xthe length AD during the torsion, and if s be the correspond*0^ torsion BD, we have after the Pythagorean law to put lnS AD2 = AB2 + BD2 [l + x)2 = l2 + s2, or /2 + 2 Ix + x2 = l2 + $2, may be put approxi- mately = x = If further F be the section of such a fibre, we then have for the force required to produce this extension in the direc- tion of the fibre S = ^. F . E.But this force or tension {S) 0f a fibre is only a component of the force of torsion R, which produces besides a further pressure JY, normal to the fibres. From the simi- larity of the triangles RDS and RD./1, it follows that S: R ~ s : lf hence S= —, and by equating both values of S: l * = k‘F-E' Therefore the force of torsion of a fibre increases as the torsion (5), and the transverse section F, and inversely as the length (/) 0f the fibre. To find the force of torsion of a cylindrical axle CBA, Fig. 244, let us divide its radius r into n equal parts, and suppose concentric circles passing through the points of divi, sion, so that the transverse section becomes decomposed into annular elements of theTORSION. 221 thickness I, and radii The solid contents of these n n n n n elements are F, = 2 x . - . - = 2 */-) , = 2 *. -L . = 1 n n \n/ n 71 4 * (Vj\ F, = 2 x . — 6 x Q2, &c. If all the fibres are twisted by the angle BCD = a°, they have the corresponding tor- sions s, = L a, St = — O, s3 = — », and hence the forces of torsion "■ * “ia•*■©**-?©’R’"^©'*'■- 9a^ /r\3 -j— f -) E, &c. If further we multiply these forces by the arms —, —, and add together the values so obtained, we have for the moments of torsion Pa = ~ (l3 + 23 + 33 + ... + n3) E, t. e. = -y- ~ . jE, and inversely, the measure of the angle of torsion: 41.Pa a ~ If the axle be hollow and have radii rx and r2, we have then K-XiE(r' - 'Atherrfore • “ . The application of hollow axles gives also with respect to torsion a saving in material, for if we put r2 = r, and rx = r \/2, we then ob- tain for the hollow axle, which has the same section as a solid one, the moment of torsion: 01 fi E f A 4 4^ Q a ^ ^ ^ f p - ~TT ('}"3 * "T thrice as great as for the solid axle. r «• § 210. For a shaft or axle of a rectangular section J1BVA, ng. 245, the moment of torsion is found in the fol- lowing manner. If we divide half the breadth J9G = b into n equal parts, and carry through the points of division the parallel planes HLy MJYy &c., we obtain elements of equal sections, each ss - . hy where h represents half the height n J&F = GC of the section. If now we divide one of these elementary strips into m equal parts, we have for its area — . — = — Let the normal distance Cote strip HL from the”centre C, and the distance KH of the element 19*222 T0R810N. K from the normal CH, = e, then the distance of the element from the axis is CK — accordingly the arc of torsion » ° and the moment of torsion sf7~+~? . E - m (c* + e*) E. a >/c* + e* bh mn 21 mn ^ ~ 1 -------2mnl' 12 3 If now we successively put e *s —A, — A, — A, &c., and sum the m m m m ' m ' m results, we have the moment of the strip: M A* 4A3 . . 9A* HL-^(c* + ^ + * + — + Pa~ 31 E' For a cylindrical axle 5 if we Put 6 “ r we then obtain i*,®,* the moment of the square is, 3 h 4 * d* therefore, = — ™ 1,756 times as great as that of the round axle. 7 3* But if we make 4A* = nr*, and, therefore, both sections equal, we then obtain Pa = a • *V E = — . - . Pfix — 5 A0!» therefore the 4.3/ 4.3 * 3,. . . , , square shaft is only a little stronger than the cylindrical axle. If the axle be hollow, and the outer and inner radii be 2bx and 2bv we then have: Pa 4a E (V-V). 3/BREAKING TWI8T. 223 § 211. Breaking Turist.—When the torsion exceeds a certainlimit, the fibres are torn asunder, and the cylindrical axle is twiste asun ^ For the moment of rupture of the fibre furthest from the axis ^ , hence it follows that ^ but is also as —— —- --y .* -— - * l 2P 2 P l The statical moment of twisting for the round axle is: 'e-*1* 1ke- 4 2 K E * Pa I 2 K # E _ r \ E * 4 2 \ 2 but for the square shaft, where the greatest distance of a fibre is half the diagonal b y/ 2, it follows that . ab IX. ——— since — m [_ i 2/* l V E K E and Pa — y/KE. 3 Since the fibres are not only extended by torsion, but also com- pressed, and as we have only had regard to extension in our develop- ment, so it may be expected that the formulae found do not in their quantitative relations quite correspond with experiment and, therefore, it is necessary to take the constants E and s/ KE from experiments made especially to determine them. If a be given in degrees, such observations admit of our putting for the torsion: Substances. Circular section. Square section. Wood .... Cast iron .... Steel and wrought iron Pa = 3500 . *0r< l Pa — 160000 “0r\ / Pa ss 280000 *Qr4. 1 Pa = 5800 HOlt. «°M J>as= 280000 If-' m°b* Pa b 470000 In what relates to the strength of torsion, numerous experiments made upon cast iron have given JXE- = 30000 to 66000 lbs., if therefore, a five-fold security be taken,then is ~ J KE a 12600 lbs. 22?*’ r?und cast iron axle Pa = 12600 r», and for the same formulae hold good for axles of wrought iron, but for mT.!.n °n,eS.We may Put Pa - 1260 r3 and = 1600 bst ». e. the moaulus of strength = ^ that of iron axles. The modulus of strength for Steel J KE must be taken at twice that of iron, and gun . , \ 2 metal at onehalf.* See Appendix.224 BREAK.ING TWIST. Exemplet.—1. The iron upright axle of a turbine exerts at the circumference of a toothed wheel of 15 inches radius reposing upon it, a force of 2500 lbs.; what thick» ness must be given to it? Pa = 2500.15 a=s 37500, and if we put r8*— 37500 375 we shaii obtain r ss /_____— = 1.44 inches: hence, the thickness of 1 126 If the distance of the " ’ ' o .__ 12600 ~~ 126 the axle 2r = 2,88 inches, for which 3 inches may be assumed, toothed wheel from the water wheel is 60 inches, the torsion of the axle Pal 37500.60 375.6 14,06 QO o/ „ 160000 r* 160000.1,44« "* 160.4,88 “ 4,28 ’ *** 00nsiderable. —2. On a square axle of fir, a force P s 500 lbs., acts at an arm of 20 feet, whilat the load is applied at an arm of 2 feet, the distance measured in the direction Qf th * axis hm 10 feet; how thick must this axle be made, and how great is the torsion ? It 6 JPo s Qb > 800.2.18 ■ 120000 inch lbs.; but the load Q «■ f P=*5000 lbe- half pa the side b of die axle is determined byfrmm j 120000 1500 80; hence 6 as ^5 4 31 inches, and the whole side - 8,62 inches. The torsion amounts to = pal 120000.12.10 ______________ 144000 ___7o. therefore, here very considerable. In 5800 . 6« 5800. 4,31« 58.345 genera], less torsion is allowed, and therefore the axles are made much stronger. Ge rally, this angle does not amount to $ a degree. If, bowever, we put «° » 4° for 1 A A AAA 4 ______ 3 ’ * «IIS case we shall obtain 6« 144000 58 . i : 4965, hence b * ^/4965 as 8,4 inches, and 26 sss 16,8 inches. According to Gerstner, the angle of torsion of an axle ought not to amount to more than 0,1°. Remark. If an axle has to sustain relative elasticity and that of torsion, we must make the calculation for both, and apply the greater ratio of the dimensions found.KINDS OF MOTION—RECTILINEAR MOTION. 225 SECTION IV. DYNAMICS OF RIGID BODIES. CHAPTER I. DOCTRINE OF THE MOMENT OF INERTIA. § 212. Kinds of Motion.—The motion of a rigid body is either one of translation, or one of rotation, or both. The spaces described by the particles of a body in motion of transla- tion or progression are parallel and equal to each other; on the other hand, in motion of revolution or rotation, the particles of a body describe concentric circles about a certain straight line, which is called the axis of revo- lution. Compound motion may be regarded as a motion of revolution about a moving axis. The latter is either uniform or varia- ble. The piston DE, or the piston-rod BF of a pump or steam-engine, Fig. 246, has a motion of translation, whilst the arm or the crank JIC has one of rotation, and the con- necting rod AB a compound motion; for one °f *ts extremities B has a progressive, the other A a rotatory motion. The axis of revolution, in a rotating cylinder, is invaria- ble ; that of the connecting rod AB, on the other hand, is variable; for it is the inter- sectum M of the perpendicular to the direc- tiori of the axis of the rod BM, and the prolongation of the crank CA. § 213. Rectilinear Motion.—The laws of motion of a material point found in § 81, &c., have their direct application in rectilinear Progressive motion. The particles of the mass Mx, M2, M3, &c., of a °dy, moving with the acceleration p by their inertia, offer resistance to motion with the forces Mj), Mj), M&c. (§ 53); and sincethe motions of ali these take place in lines parallel to each other, the226 MOTION OF ROTATION. directions, therefore, of these forces are parallel; hence, the resultant of ali these forces arising from the inertia is equivalent to the sum Mj, + + ... = + M2 + M +. ..)/> = where M represents the mass of the whole body, and its point of application coincides with the centre of gravity of the body. In order, therefore, to change the motion of an otherwise freely inoving body of the mass M or of the weight G = Mg into one rectilinear and progressive, a force P = Mp = whose direction passes through the centre of g gravity S of the body, is requisite. If, in virtue of the action of the force P, the velocity c acquired while the body describes the space <* increases to the velocity v, then the mechanical effect accumulated the mass (§ 71) is : * - (tt) ' - (tt) g G- J in Example, The piston of a pump, with its rod or that of a steam engine or blowing machine, has a variable motion ; it has no velocity at its highest and lowest positions and at its mean position its velocity is a maximum. If the weight of the piston and rod* be = G, and its maximum velocity at the middle of its ascent and descent = Vy ^ eifect which it will accumulate by virtue of its inertia in the first half of its path, and will give out again in the second half = —— G. For G 800 lbs. and v s=s 5 ft. this effect = 0 0155.5*. 800 = 310 ft lbs.; were half the path of the piston t = 4 feet^ We should then have the mean force which would be requisite to ^celemte the motion of the piston in the first half of this path, and which it would exert in the second half by its retardation: G = = 77 lbs. 2 gs 4 .. ,. nf notatiori — If the inoving force P of a body AB, § 214 Mohon tfRotahon. ^ centre gravity the bod; Fig. 247, does not P j ^ke up a rotatjon about this point, and this motion will go on as if the force were directly applied to it, as may be proved m the following manner. From the centre of gravity let a per- pendicular SA be drawn to the direction of the force ; let this be prolonged backwards and the prolongation SB be made equal to the perpen- dicular, and let two forces, balancing each other and acting parallel to , the one + £ P, and the other — i P, be applied to the point The force 4- i P will give, when combined with half the force P applied at A, the force applied at the t g p — 1 P + i P = P, for which the force — l p wiflform l couple with the second half of P applied at A; there will resuit, therefore^ from the excentrically acting force P, a force P acting through the centre of gravity, which wil move forward this point, to- crether with the whole body, and a couple (i P,—i P), which will cause fhe body to rotate about the centre of gravity without producing a messure on it. The statical moment of this couple wiU be = P + SB . i P = SA . P equivalent to the statical moments of theMOMENT OF INERTIA. 227 force applied at A; consequently the resultant rotation will be the same as if the centre of gravity S were fixed, and P acted alone. According to this, therefore, every arbitrarily directed force com- municates two motions to a body; one of translation and one of rota- tion ; and hence it is necessary to study the laws of this latter. If, finally, a body be constrained by a path or directrix to assume a motion of translation, an excentric force will then produce the same effect as if it were applied to the centre of gravity, because the forces of rotation will be taken up by the directrix. § 215. Moment of Inertia.—In the rotation of a body AB> Fig. 248, about a fixed axis C, all its points describe equal angles in equal times. If the body rotate in a certain time 0 through the angle <|>0 or arc t . *, the elements of the body Mv M2 . . . will describe at the distances CMX = yv CM2 = y2, &c., from the axis, the spaces Wv If the angular velocity, i. e. the velocity of those points of the body which are distant a unit of length (a foot) from the axis of revolution = u, then the simultaneous velocities of the elements of the mass at the distances yv y2, &c., will be = **yv «y2, &c.; hence their vis viva is («yi)2 Mv (wy2)2 M2> &c., and the sum, or the vis viva of the whole body : («y,)2 Mx + {»y2)2 M2 + ... = «* (Mj* + MlV2 +. ..). The sum of the product of the particles of the mass and the squares of their distances from the axis of revolution, is called the moment of inertia of the body, the moment of rotation, and the moment of the mass. If we represent this by Ty and therefore put Nj* + A/2y22 + .. •, we then obtain u,2T for the vis viva of a body revolving with the angular velocity w. Hence, to communicate to a body, already in a state of rest, an angular velocity w, a mechanical effect Ps = % w2 T must be expended ; and inversely, a body produces this effect when it passes from this angular velocity into a state of rest. If, generally, the initial angular velocity of a rotating body = *, and the terminal angular velocity = w, we then have for the mechanical (^2 2 v — ■ —) T; and inversely, the terminal velocity corresponding to an expended or accumulated mechanical Fig. 248. 2 Ps Example. If a body, AB, Fig. 248, originallyat rest, and capable of turning a u fixed axis C, possesses a moment of inertia of 50 ft. lbs., and by means of a cor yi g over a pulley is made to rotate in a path s = 5 feet, the acquired angular velocity o b body will be » = / 2 P* __ /2 • 20 • 5 = ^/4 = 2 feet, »*.«., each point at the dis- T J 50 . tance of one foot from the axis of revolution will describe, after the accumulation o t is22S REDUCTION OF INERT MASSES. mechanical effect, two feet in a second. 60 _ 60 seconds, and the number of revolutions per minute u = —--------31TT6 • 2 feet, t 2tt The time of a rerolution = — = 3,1416 19,1. If the SCUUIIUS, tUJU. U1C UU1UWVI V» --* [ «I4’4'' „ • tn tu velocity * = I feet’then thls mass Pro- angular velocity found « =2 feet, passes in 50^ / 25 = 55 25 duces the mechanical effect Pt*»— I 2*— V,T/ -» 2 ' v 16 . » t P of 10 lbs., raised 8,593 feet high. = 85,93 ft. lbs-; for instance,! Zrt Ma 'sses.—If the angular velocities of two §216. Reduchono/I instance, these raasses belong to masses JW, and M3 beequa , , viva- will be to each other one and the same m at.ng body^the ^ ^ = ^ ,f ^ as their moments o , ’ 1 w}n p0ssess equal vires viva. Two bC eqUaU°veeathe; °lreX"ual influende on thJ state of motion of a revolv ng body^ and one may be replaced by the other without causing anv change in the condition of motion, if they possess equal moments any cnang 3 , M 2 and are inversely as the squares of the r B, help of the fomnln M, - M v \ a mass may be reduced from one distance to another; u, Tmass Jtf may be found, which, at the: distance y„ has the same m- fluence onthe condi,ion of motion of the revoBonj; body^s a g,ve„ mass Jlf, at the distance y,; namely, M, « —pV ~i' «• «•, «e „ rsdacsd to tko dista** , istu fotioot of tU moment o/inerti, ofths mass **d tU s^mrs wrights, Q and Q„ attached axle, Fig. 249, with the arms CB == 6 and CB = fl> have an equal influence, invirtue of their masses, on the motion of the wheel and axle, if Q, a1 = 5 therefore, Q, «. !53| Q&* Hence, if a force act at the arm a CA — C = and cause rotation in a mass of the weight Q at the distance CB = we shall have then to reduce the latter to the arm a of the force P; therefore, for Ob2 Q substitute Q, = and the mass set into motion by J> will be = (J> + + S i »»* Fig. 249. motion of the weight: force V P = Pa2 Pa2+ Qb2 ‘ gy and the angular acceleration ^ • g- Example. The weight of a rotating ma9S Q = 360 lbs., its distance from the axis of rotation b =2,5 feet, the weight constituting the moving force P = 24 lbs., and its arm a = 1,5 feet, the inert mass put into accelerated motion by PREDUCTION OF THE MOMENT OF INERTIA. 229 g =0,031 ^24 + — 36o) = 0,031.1024 = 31,75 Ibs.; and henoe the accelerated motion of the weight :p = _ 0,756 feet; on the other hand the acceleration of the motion of the mass Q, = -.p=^> = = 1'26 ft - and the anSular acoele* ration s= — = 0.504 feet. After 4 seconds the acquired angular velocity will be « = 0.504.4 = 2,016 feet, and the corresponding distance = —— --------- = 4,032 feet; con- sequently the angle of revolution *» = • 180° = 1,28. ISO» = 230» 24'; conse- * p(* 0,756 .4* f quently the space described by the weight r - g — ? e • § 217. Reduction of the Moment of Inertia.—When the moment of inertia of a body or system of bodies about an axis, passing through the centre of gravity S of the body, is known, the moment of inertia about any other axis running parallel to it, may be easily found. Let Fig. 250 be the first axis of revolution, passing through the centre of gravity S, D the second axis, for which the moment of inertia of the body is to be deter- mined; further, let SD=e be the distance of the two axes, and let SJY1=x1 and JV1M1=y1 the rectangular co-ordinates of a particle of the mass MxX>f the body. Now the moment of inertia_of this particle about B=Ml. DM2 = Mi= [(e+*il and about S = Ml . SM* = M1. (SJY* + MJIt») = Mi (*.2 + 5 hence the difference of both moments: = MX (e2+2 exl+x*+y12)—M1 (x2+y*) = 2 Mxex. For another particle of the mass M2, it is = M2e2 + 2 M2 ex2, for a third = M3e*+2 M3 ex3y &c., and for all the particles together* = (Mx+M2+M3+ . . .) e?+2e (M1x1+M2x2+M3x3+ . . .)• But Mx + M2+ ... is the sum M of all the masses, and Mxxx + •M2x2-{- . . . the sum Mx of their statical moments; hence the differ- ence between the moment of inertia Tx of the whole body about the axis B and the moment of inertia T about S: Tx— T=Me1+2 eMx. But since, lastly, for every plane passing through the centre of gravity, the sum of the statical moments of the particles on the one side is as great as that of those on the other, the algebraical sum of all the par- ticles therefore = 0; we have, also, Mx = 0, and, therefore, Tx— T = Me2; i. e. Txmm T + Me2. The moment of inertia of a body about an axis not passing through the centre is equivalent to its moment of inertia about an axis running parallel to it through the centre of gravity, increased by the product of the mass of the body and the square of the distance of the twot ^es. It is also seen from this that of all the moments of inertia about parallel axes, that one is the least whose axis is a line of gravity of the body. Fig. 250.230 THE ROD. § 218. It is necessary to know the moments of inertia of the Prin- cipal geometric bodies, because they very often come into application in mechanical investigations. If these bodies are homogeneous, as in the following we will always suppose to be the case, the particles of the mass M., M., &c., are proportional to the corresponding parti- cles of the volume V„ V„ &c., and hence the measure of the moment of inertia may be replaced by the sum of the particles of the volume, and the squares of their distances from the axis of revolution. In this sense the moments of inertia of lines and surfaces may also be f<> If dthe whole mass of a body be supposed to be collected into one point, its distance from the axis may be determined on the supposi- tion, that the mass so concentrated possesses the same moment of inertia as if distributed over its space. This distance is called the radius of gyratiori, or of inertia. If be the moment of inertia, .\[ the mass, and r the radius of gyration, we then have and hence r = I—. We must bear in mind that this radius by no means \jM gives a determinate point, but a circle only, within whose circum- ference the mass may be considered as arbitrarily distributed. If into the formula Tx = T + Me3, we introduce T = Mr2 and Tx = Mr2, we obtain rt2 = r2 + e2, i. e. the square of the radius of gyration referred to a given axis = the square of the radius of gyra- tion referred to a parallel line of gravity, plus the square of the dis- tance between the two axes. . § 219. The Rod,—The moment of inertia of a rod AB, Fig. 251, which turns about an axis XX through its middle point C, may be determined in the following manner. The cross section of the rod = F, its half length CA = /, and the angle which its axis includes with the axis of rotation ACX= o. If we divide half the length into n parts, we then obtain n portions each of the contents —: the distances of these portions from the middle are l, n n n n’ &c., hence their distances from the axis XX are MJ\T, = — sin. a , • 2 ” n 31 . „ , ,, / A/lsin.a\2 sin. a — sin. a, &c., and these squares = ^—-— J > 4 ^—_—\ # 9 &c- By the multiplication of these with the contents of FI an element —and by the addition of ali the products, we have the n moment of half the rod: r - ? [ (~y + ^ (^)'+o +• • i Fig. 251.RECTANGLE AND PARALLEL0P1PED. 231 FP sin. a2 (l2 + 22 + 32 +. .. + n2), or, l2 + 22 + 32 + . .. + n2 = 11 t — 3’ 1 "" FP sin. or But as the volume of half the rod FI is to be Fig. 252. considered as the mass M, it follows, finally, that T = J MI2 sin. a.2, The distance of the end of the rod from the axis XX is AD = a = l sin. o, hence it follows more simply that T = £ Ma2, which formula is also to be applied to the whole rod, if M be the mass of the whole. A mass M at the extremity A of the rod, has the moment of inertia M^2, hence if we make Mx = \ M, it has then the same moment of inertia as the rod. Whether, therefore, the mass be uniformly dis- tributed over the rod, or its third part be collected at the extremity A, it comes to the same thing. If we put T = Mr2, we obtain r2 = | a2, and hence the radius of gyration of the rod: r = a v/ 3 = 0,5773 .a. If the rod stands perpendicularly to the axis of rotation a = l, therefore, T = J MI2. If, lastly, the rod AB, Fig. 252, be not in the same plane with the axis of ro- tation, and if the shortest distance between the two axes be CE = e, wre shall then have from § 217, the moment of inertia T. = T + Me2 = M (e2 + £«2). § 220. Rectangle and Parallelopiped.—The mo- ment of inertia of a rectangular piate, ABDEy Fig. 253, which turns about an axis XSfpassing through its middle C, and parallel to a side, is as for a rod = J MI2, but if the axis YY stand perpendicular to the plane, the moment of inertia is then determined from the former paragraph in the following manner; the half AEFG is divided by lines parallel to the side AE into strips of equal breadth, such as KL, the moments of these strips determine, and then added together. If the half length FE = GA = l, half the breadth CF = CG = b, and the number of Fig. 253. i 2 bl parts = n, the area of a part =-.26 = n n The distances from C of these strips are in the series &c., therefore their squares ^-^2> 4 ® ’^c' hence moments of inertia are: 2 blT (l\2j. b2l2 Wrwh* , 63~| 2 6/r0//\2 , *2_1 ~LU + IJ’— L4(;) +i]’ —[ W +sJ and the moment of inertia of half the piate ; =mo' (1 + 4 + 9 + ... + «*) + n ' 3j232 PRISMS AND CYLINDERS. _ 2 blr /Z\a n* n^*j = 2bl(P + &*) = £ Jtf (Za+ fta), because 2 6/musfbe considered as the mass of half the piate As the semi-diagonal CA = d “ mass, the fori Fig. 254. mula holds good for the moment of in- ertia of the whole piate. Since, further, a parallelopiped Fig. 254, may be I~-jgdecomposed by parallel planes mto equal I rectangular plates, the above formula is I H also applicable to this, if the axis of ro- JBg tation pass through the middle points of 1 any tw0 opposite surfaces. It follows, besides, from this formula, that the mo- ment of inertia of the parallelopiped is equivalent to the moment of inertia of the third part of its mass applied at a comer A. . „ , . From the formula for the moment of inertia of a parallelopiped, that of a triangular prism may be also calculated. Ihe diagonal plane ADF divides the parallelopiped into two equal triangular * prisms with rectangular and triangu- lar bases Fig. 255; hence, for the rotation about an axis passing through the middle C and K of the hypothenuse, the moment of inertia = ^ If now we make use of the proposition § 217, we ob- tain the moment of inertia about an axis passing through the centres of gravity S and St of the bases : T Fig. 255. = £ Md? — M. CS2 = [j-(i C5)2] “ (i) ] = $ Md\ and it follows also that the moment of inertia about a side T — T+ M. SB* =■ 3 MtP + M ($ d)3 = 9 Mtp — § Mip, where i always represents half the hypothenuse of the tnangular base S 221. Prisms and Cylinders.—lor a prism Flg- 256, with isosceles triangular bases^the mo- ment of inertia about an axis XXy which connects the middle points of the bases, T = § Md2, if d represent half the side AD of the surface of the base, because this surface may be decomposed by the line of the height AB into two equal rectangular triangles. If now the height AB of the isosceles triangular base = h, we have then for the moment of inertia Fig. 256.CONE AND SPHERE. 233 Fig. 257. of this prism about the axis YY passing through the centres of gravity of the base: T = | Md2 — M = M ($ d2— b^2) = \ (2 d2 — £ h2), and finally, the moment of inertia about the edge passing through the points A and F of the bases: Tl=T+M($h)2=M(^- — ~ + -9-) = Hence the moment of inertia of a right and regular prism revolving about its geometric axis may be found. Let hbe the height , Fig. 257, of one of the supplementary tnangles, = d-r the radius of the base or of a supplementary tnangle, and the mass of the entire prism. We have then by the last formula: The regular prism becomes a cylinder when h = r, hence the moment of inertia of a right cylinder about its geometric axis is: T= + = * Jtfr». The moment of inertia of a cylinder is, therefore, equivalent to the moment of inertia of half the mass of the cylinder collected at its circumference, or equivalent to the moment of inertia of the entire mass, at the distance r s/% = 0,7071 . r. If the cylinder ABDE be hollow, Fig. 258, the moment of inertia of the hollow space must be subtracted from that of the solid cylinder. If the outer radius CA = rv and the inner CG = r2, we then have from what has preceded the moment of inertia of the hollow cylinder: T=4 (Mjrf-Mjrfl-l * (r* . r?— r. Fig. 258. —hH (ri—r*) iri+r2) )=£ * (ri4 r24) \ M(r~2+r2), —s " v#i —7'2 ) v'i 1 ' * / 2 v' 1 ■ -1 /' 2. Tfrbe because the volume considered as the mass = rt r2 r the mean radius and b the breadth of the annular surfa > 2 we then have T = M ^r2 + § 222. Cone and Sphere.-The moment of inerti as well as that of a sphere, may be calculated t 259, for the moment of inertia of a cylinder. Ee pg __ r be a cone revolving about its geometric a*is\ T:th the axis, the radius of its base, and CD = h its height co.ncidmg «ith 20*234 CONE AND SPHERE. If \ve raake n slices Fig. 259. and as l4 + 24 + Fig. 260. parallel to the base at equal distances, we then • %» T T T obtain thin discs of the radii 2 - , 3 — n n n n and of the comraon height The halt vol- n 71 .. /r\2 h /2 r\2 umes of these discs are n , n \ ~\ . A, * . (—)* . A, &c., and hence their mo- 2n \n/2n ments of inertia: the sum of these values gives, finally, the m0- ment of inertia of the entire cone: T = _ l4 + 24 + 34 + 2n5 4- . . + n4 = we have 5 + n T = nr*h ttr^h . 3 ,, T’ “Io'** 3 To” ” io For the right pyramid ACE, Fig. 260, with rectangular base, under the same circum- stances T = \ Md2, if d represent the semi- diagonal DA of the base. Also by subtraction of the two moments of inertia, the moment of inertia of a right truncated cone with the radii r and r2and the heights A, and h2 may be ob- tained: T= A(r^-r/A2) = jA(r/_ 10 or, since the mass Af= ^ (riA r*ha) = r23), T-Tc M In a similar manner, we find the moment of inertia of a sphere, revolving about one of its diameters DE = 2 r. Let us divide the hemisphere ADB, Fig. 261, by sections paral- lel to the base ACB,^into n equally thick cir- cular slices, as GKH, &c., and determine their moments. The square GK2o{ the radius of any such slice is: = CG2 — CK2 = r2 — CK\ hence its moment of inertiaCONE AND SPHERE. 235 = ?L(r4 — 2 r2 2 + 2n K r 2r 3r Let us put in succession for CK to —, and addthe r n n n n results, we shall then have the moment of inertia of the sphere: T = — Vn . r4 — 2r* (-) (l2 + 22 + 32 + . . + n2) + 2 n L W . . , /r\* „ "1 2 0 (l4 + 24+ 3*+ •* +”)]=2^L ^'3 + Now the sohd contents of a hemisphere M = | «r2, hence we may ^ T = i • * ap3 . r2 = f Jtfr2» and if we take M for the whole sphere, the formula will hold good for The formula T = f JWr2 is true also for a spheroid whose equa- torial radius = r (§ 117). # . .. . ^ ;fo If the sphere revolves about another axis at the dis ance e centre, the moment of inertia is then T= M (e2_ + f r2). The radius of gyration = r \/§ = 0,6324 . r; two-fifths of the mass of the sphere at a distance from the axis of rotation equal to the radius of the sphere, have the same moment of inertia as the whole sphere. § 223. The moment of inertia of a circular disc ABDE, Fig. 262, revolvine: about its diameter BE, is, as for the mo- 7tr4 Mr2 Fis*262# ment of flexure of a cylinder, (§ 195) = —- = , consequently the radius of gyration = r y/ \ ^ r, t. e. half the radius of the circle. From this we may now find the moment of inertia of a cylinder ABDE, Fig. 263, which revolves about a diameter FG> passing through the centre of gra- vity S. If l be half the height and r the radius of the cylinder, we then have the volume of half the cylinder = *r2/, and if we make Fi&- 263- equi-distant sections parallel to the base, we decompose this body into n equal parts, each of which = the first is distant n - from the centre of gravity S., the second 21 31 —, the third —, &c. In virtue of the for- n n # niula, § 217, the moments of inertia of these circular s ices are.236 SEGMENTS. «H r. . /3i\H -r^+(-)!- n L ' n 'A a,- nf half the cylinder: &c., their sum gives the moment of in * “l Ttr*l r rvA , /*_\ (l2+22+32+ • • +n )J T----7L4 V»/ 3v /r2 „hich h.Id, goodlikewise for .h. «hole cylinder, if JM repant, i„ “wi find i» like manner for ,he righ, through its centre of gra- vity, and is perpendicular to the geometrical axis CD r-*Ar(»*+£). For a piate ABC, Fig. 265, in the form of a rectan- gular triangle, the moment of inertia about an axis passing through the centre of gravity S> and parallel to the side ACy is according to § 193: Fig. 264 Fig. 265. T= 6A3 A2 = 1*5 Fig. 266. Fig. 267. bh 1 = — 2*18 , ,, , 6„ , t. tllP ax;s 0f revolution, and the height if the breadth bparal This formula holds good even for an perpendicular to it be g • runs paraHel to the base, and h oblique angled triangle, trianffie. From this the moment of represents the height o /WFF 266, may be found, if inertia of a tin«r/of gravi,,'! the axis of revolution XX p an(j jg parapej to ^ side DE of the surface of the base, it follows from the same method as that adopted for the cylinder, that ■where l represents half the length of the prism. § 224. Segments. — The moment of inertia i-of a paraboloid of revo- lution, BAD, Fig. 267, which turns about its axis of rotation AC, isSEGMENTS. 237 n h 1 n A £ a fl O 2 C = - . * . - a2, - * . - a , - . * - a2, &c., n ~ 1 71 determined in a similar manner to that of a sphere. Let the radius of the base CB = CD = a, the height CA = A, and the body consist ot n slices, each of the height we have then the contents of these £ - . x - n n n n n n because the squares of the radii are as the heights. From this the moments of inertia are given _A «‘ 1 and hence, finally.lt follows that the moment of inertia of the whole paraboloid is ,, 3 ai ai na4h n3 *a4h «ah m2+224-32+ . . +n*) =- 2 7l3 4 a4 h rt 9 a4 « ^9nm2ml?~9nm2m n»9 ’’ 2/i3 _ *^A ^ = 1 Jtfa2 ~ 2 3 5 (Ji^h because the volume of this body is JVf = ——• The same formula is applicable to a small segment of a sphere, ut if the height h is not very small compared with a, we have to put the moment of inertia of slices = 5* . a4 = 5* . A2(2 r— A)2 = (4 r2 A2—4 rh3+h*), 2n 2 n ' 2n where r represents the radius of the sphere. If now we take suc- cessively for h the values & c., we then obtain for the mo- ri n n ment of inertia of a segment of a sphere = (20 r2—15 rh+3 The contents of the segment are M= nh2 (r—J A), hence T = Mh»(r— * A) • y (»—15 A+5 ® • = (r—y5j A+jV . Generally T= § Mh (r— T#s A) is sufficiently correct. This formula finds its application in pendulum-bobs. The moment of inertia of the surface of a parabola ABD, Fig. 268, which revolves about an axis XX, passing through the mid- dle C of the chord BD, is found if the sur- face be decomposed into equally broad stripes, such as EF, and their moments added together. Let AC=l be the length, and CB = b half the breadth of the sur-238 WHEEL AND AXLE. face, CF = xthe absciss, and EF = the ordinate oMength of an element. Its moment of inertia is then = - y ^ ’ ^utas 3:8 hzy,therefore y-/(l— »* follows that lhis moment b2 l ' 3 -(i—-J + i3(i—J*) J- If now x be successively put b 2b 3b &c<> and the results added, we obtain the moment of inertia of half the surface of the pa™bola : r-w[£(w-n+*n~?)+*-*J _ - | w (A P i - i •**(* ** + &s), " \3.35 3.5/ „ horause the surface of the parabola \s M= f M. This formula, which holds good for the entire surface of the para- bola, is also applicable to a prism havmg a parabohc surface for the bT’22Lm^:rSbS-The theory Of the moment of inertia finds its most frequent application in machines and Instruments, be- cause in these rotatory motions about a fixed axis are those which generally present themselves. Manyapphcationsofthis doctrine 3 be met with in the sequel, hence it will suffice to treat of only a fe "oPl«aS eand r.“”on . «heel and axln JCDB, Fig. 269, ° with the arms and DB=b through w - ««i the medium of a perfectly tlexible string, and if the radius of the gudgeons be s6 small that their friction may be neglected, it will remain in equilibrium if the stati- cal moments P • CJl and Q . DB are equal, and therefore Pa= Qb. But if the moment of the weight is greater than that of Q, therefore, Pa > will de- scend and Q ascend; if Pa < Qb, P will ascend and Q descend. Let us now ex- amine the conditions of motion in one of the latter cases Let us suppose that Pa > Qb. The force corre- Ipondingtothe'weight Q and acting at the arm b generates at the arm a a force which acts opposite to the force corresponding to a the weight P,and hence there is a residuary moving force — Fig. 269. Qb acting at A. The mass — is reduced by its transference from the g Qh distance b to that of a to — ,hence the mass moved by P---------- is ga2 «WHEEL AND AXLE. 239 M={p + g, or, if the moment of inertia of the wheel and ' a / n.z Qyl axle b —^ , and, therefore, its inert mass reduced to A = j> we g ga have more exactly : M = (p + +* = (Pa*+ QJ* + Gy) + ga’. From thence itfollows that the accelerated motion ofthe weight P, together with that of the circuraference of the wheel, namely p—— „ _ moving force _____ « ____ mass Pa—Qb Par+QP+Gy* 'ga ga\ Pa* + Qb% + Gy* on the other hand, the accelerated motion of the ascending weight Q, or of the circumference of the wheel is: b Pa—Qb h 9 ~ aP ** Pa*+ Qb*+ Gtf' S ' The tension of the string by P is S =■ P — — =s P ^1 —^ (§ 73), that of the string by Q: Q j. —t s Q fl + -\ hence the pres- ,, g \ g/ sure on the gudgeon is: S + 3P=P+ Q — PP+9l= P + ,. L ,* ‘* gj * P&3 + Qb* + Gy* the pressure, therefore, on the gudgeons for a revolving wheel and axle is here less than in a state of equilibrium. Lastly, from the accelerating forces p and j, the rest of the relations of motion may be found. After t seconds the velocity of P is v =s pt, of Q : vx =» qt> and the space described by P : s =* J pt2> by Q : ^ 3 | qt*. Examplt. Let the weight P at the wheel be as 60 Ibs., that at the axle Q aas 160 lbs., the arm of the first CA as a as 20 inches, that of the second DB s b os 6 inches; fur- ther, let the axle consist of a solid cylinder of 10 lbs. weight, and the wheel of two iron rings and four arms, the rings of 40 and 12 lbs., the arms, together, of 15 lbs. weight j lastly, let the radii of the greater ring AE sa 20 and 19 inches, that of the less FG * 8 &nd 6 inches. Required, the conditions of motion of this machine. The moving force at the circumference of the wheel is: P— - Q B. 60 _ i.. 160 60 — 48 b. 12 lbs., , a 2° e moment of inertia of the machine, neglecting the masses of the gudgeons and the strings, is equivalent to the moment of inertia of the axle ss as P*08 2 2 '/) 12 . (8*+ 6*)_— 600% plus the mo- 2 " 2 s 40 • (20a+ 19a) ^ 15220, pius the moment of the arms, **) _ ■* (f,‘+ rft . 1« • Cty + 19—L±il» 2885; 3 the moment of the smaller ring ment of the larger ring approximatelyss ^_________ hence, collectively, Gy* a« 180 -f- 600 -f* 15220 + 2885 3 18885, or for foot measure240 WHEEL AND AXLE. 18885 --144_ = The collective mass, reduced to the circuraference of the whee] -('+aEiat)+-[»+'«(S),+!si>. = ( 60 + 160.0,09 4- . 0,031: ' ^ 400 / 1121,61 X 0,031 =377 lbs. Accordingly, the accelerated motion of the weight P, together with that of the circum* ference of the wheel, is: P—1Q a p^p+WTW' 12 1 3,77 3,183 feet; on the other hand, that of Q : q ^ b a* . A . 3,183 = 20 t 0,954 feet; further, the tension of the string by P is 0—I) p =: ^1 — ® 54,07 lbs.; that by Q, on the other hand, Q == A ^ q \ = (1 + 0,925.0,031) . 160 = 1,030.160 = 164,8 lbs.; and consequentlv th» I Q on the gudgeons S -f- T = 54,06 -f" 164,80 = 218,86 lbs., or inclusive Of, I,n l)rcssufe the machine = 218,86 + 77 = 295,86 lbs. Alter 10 seconds, P has acq'Veight of lMirgQ tliA velocity pt = 3,084 . 10 s 30,84 feet, and described the space t_vt_0r. 2 — 30>84.5 = 154,2 feet, and Q has ascended a height f.* = 0,3 . 154,2 = 46,26 feet a & 226. The weight P which communicates to the weiobf ^ , 4 , t. Pab-Qb3 . 'gM Qthe accelerated motion q = -p~1+Ql)1+ &? • S ®ay also be replaced by another weight Pv without changing the acceleration of the m tion Q, if it act at the arm a, for which: m°' P1al—Qb _ Pa—Qb />,’+ Qb*+ Gf ~ Pa3+ Qb3+ Gy3 ’ The magnitude ^presented by k, and we obtain a3 — kax =_______ Qb(b+fc) + Gy ^ an£j arm jn qUestjon . ** 1__________________________ Qbjb + kj + Gy3 We «ay a.» ,heKfore- -¥+4(t)+s^ If we represent the statical moments of bo ^ y r, we must then substitute for the moving force P - Q> *be value P __whence the acceleration of Q comes out aATTWOOD’8 MACHINE. 241 (Pa—Fr) b—Qb1 , 9 “ Pa3+Qb*-rGy* 'g™d Qb+Fr J( Qb+Fr^ , Qb*+ Gy1 > / + P Examplet.—1. The weight» f = 30 lbs. Q = 80 lbs. act at the arma a •« 2 feet, and ® 2 footof a wheel and axle, and their moments of inertia Gy“ amount to 60 lbs. j then the accelerated motion of the ascending weight Q is: 30,2 ■ $-80 ■ (j)> 30-90 32 2 - J?L - 1 61 *eL But 30. 2a+80 . ($)a+60 *S 120+20+60 * * «00 ’ if a weight Pt ss 45 lbs. gene rates the same aoceleration in themotion ofQ, the inK of Pi is then: J 25 _ 32 5 + i 11,358*5+ 3,786 : 60—40 8,786 feet, or 1,2 U feet—42. Thevsoale- rated motion of Q comes out greatest if the arm of the foroe or radias of the wheel amount to: 4+ y/ 40 i . 80 30 and g is bs 20+60 30 30.1,7207—20 JO' + i+ / 2! + Hl 3TJ 9T 9 \ 31,621 )g = 3,4413 fcet, _ . g = 8,339 feet—3. The «Mieal 30 . (3,4415)’ + 80 / ° 435,32 moment of the friction, together with the rigidity of the string, i» Fr = 8; then, inttead of Qb, we muat put Qb -f- fr = 40 -f- 8 = 48; whence it fbllows thet: • = — -f- 30 loree q 5 S‘+i 1,6 + y/ 5,22 7 «3,886, and the correspondent maximum «ocelerating 30.1,943—8 . i—20 g = 2^??.32.2 = 2,071 feet 533 30 . (3,886)^+80 § 227. Attwootfs Machine.—The formulae found in § 225 for the wheel and axle hold good also for the simple fixed pulley, for if the wheel and axle becomes either a pulley or an axle. Retaining the other denominations of the paragraph mentioned, we have then for the accelerating motion withwhich P descends and Q ascends: p = q " {P+tya*+Gtf ' 8* °r haV‘ng rCgard t0 f"Ction! (P—Q) a*—Far P~ (P+Q) (P+QW+Gif+G^«T. because the inert masses of these wheels reduced to the circmnference of the friction wheels, or to the gudgeon of the wheel, ■■ ‘ inversion we obtain the accelerating force of gravity: 21242 attwood^ machine. y 27*j (P+ Q) a2+ G + Gt Fig. 270 g (P—Q) a1—Far For a small difference of the two weights t ‘he accelerating force p comes out small» hence the motion goes on slowly, and if the resistance opposed to the weight by the air be inconsiderable, with the assistance of experiments upon the descent of weights in such an arrangement, the accelerating force of gravity may be measured with tolerable accuracy, which by a body falling freely it is impossible to do. Experiments of this kind were first in- stituted by Attwood,* whence this arrange! ment is known by the name of Attwood’s machine. To determine the spaces fallen through, there is a scale HK along which the weight descends. From the space fallen through s, and the corresponding time t, it followsof course that p =. ~t\ if, how- ever, during the descent, the moving force be removed, while an equal weight LL forming a hollow ring is taken up by a fixed narrower ring the rernaininf — ---------- Gxetl narrower — .—......g •11 be described with a umtorm motion, and part of the space s, wm d clock the velocity will be given» having the time observed by a g ^ ^ Jf = _ !l, and the accelerating force p - f ^ P — 4 ’ . i•______»1.. and nv mittinor tl,- Fig. 271. *•* O - i frl>| riment gives directly p - and by putting the "lue found in the above formula, it will give the accelerating force of gravity. aC r 228 The acceleration of the motion of the ? i tc " p and Q, which are suspended to a rlfi l-inA . l-d .-*■**«- a ^voV»1p nulley EG, 1S following manner PLet the weights of the pulleys AB and rp 'a and G„ their moments of inertia Gy* d Gv2and the radii CA = « and DE = «f the re fore the masses reduced to the circumference of the wheel v» . G, y? ju = — • — tan(l *^iig by" fflT&j* SSl* (Q + CJ i; if b, this descent acquires the velocity o, then 2 _______________________________ • Attwood'. Trea.iae on Rectilinear and Rotary Motion.ATTWOOD’8 MACHINE. 243 Q + Gt takes the velocity and the pulley J1B at its circumference the velocity v, and the pulley EG, since in rolling motion the pro- gressive and rotary motion are equal to each other, at its circumfer- ence the velocity The sum of the vis viva corresponding to these masses and their velocities is — . v* + -. (^) + —• • v% + g g \2/ a* -~L . , and if their halves be equated to the mechanical effect expended, we shall obtain the equation: Hence, the velocity corresponding to the space $ descnbed by Pi (r - ~T7"Q+cTT^T^yi5 p + —— + ~*+~w For the acceleration ps ; hence, here p— Q+Gt />+ 5±°l+ 4 a,* The acceleration of Q+ G, is « and the rotary acceleration of G, is equal to it. The tension of the string BE> connecting both pulleys, is P — ^P -f because the force ^P + - is expended upon the acceleration of the motions of P and G. The tension of the fixed string Gif, on the other hand: S — p Si a* 2g because the pulley EG is put into rotation by the difference S i of the tensions of the strings. ExampU. In a gyatem of pulleys, Fig. 271, the weights P ■« 40 lbs. and Q are suspended, and each of the solid pulleys weighs 6 lbs.; requhred, tne ^ ^ g of the motions of these weights. The moving force is P — —" * 40 — 2 Of = ga,a 2g 4 lbs., the mass of a pulley reduced to its circumference 1». 3 * — (S 221), and the aggregate of the inert masses: S q\ 247 = (p + «±^ + ^ + ^)+,-(40 + 18 + 3+4) + ^-47’244 ATTWOOD’S MACHINE. hence the accelerated motion of the descending weight * - . 4 = -1? ' £___ 247 247 16.32,2 _ 515,2 ( 2 086 fee|. on the Qther han(j> the accelerated motion of the 247 247 Fig. 272. ascending weight: ^ = 1,043 feet. The tension of the string BE is. S=zp_ P = 40 — 43.?!^= 40 — 2,782 = 37,218 lbs.; that of \ ' 2 / g 32,2 _ TT e G p q7 018 — 3 1,04j- = 36,247 lbs. the string GH, = £-— . E- = 37,218 — § 229. The motion is more complicated, if the pulley EGy Fig. 272, be suspended only by a string passing round it. Let us assume that Pdescends with the accelerating force p, and Q ascends with the accelerating force q, we then obtain the acceleration of the rotary mo- tion at the circumference of the loose pulley qx -= ~ — 9 (§ 42). Let us now put the tension of the strings at AE = S> we then obtain P — S = (P + ^_) L;further, — (Q + G,) = (Q 4. g,) 1, since from § 214, it may be assumed that S acts g at the centre of gravity D of EG; and lastly, S = # 9i^ since it may also be assumed that the centre of gravity D is fixedfand th pulley put into rotation by S. The last three formulae give the 6 celerating force P — S /S—(Q+GJx . Sa2 p' JIR g'q “ ( Q + g, ) *ani1 - -557. * *«d a2 all three being put into the equation qx = p — q, we obtain &»,» p-S z S-(Q±G±) lO n - .3 © G^ „ Gy2 P + —r Q + G, vhence the tension of the string follows 2 2 + Gy2 5 = Uy, 2 + Q + + Gy2) + \wjyi t «r , . r ct 1 1 The accelerating forces are given from the value of S by the appli. cation of the above formulse. . . If we neglect the mass G of the fixed pulley, and also put Q = 0, we obtain simply: c 2Pa2.G,y-1 3 PG,y,2 p (a,2 + y,2) «2 + Ga,2y,2 G y,2 + P K* + y,2)' If the extremity of the string AE, instead ot passing o\er the pulley is fixed, we have then the accelerating force p = 0, hence ^ = — <7, and consequently the tensionCENTRIFUGAL FORCE. 245 s (Q + Gt) G,y,» (Q + Gi) < + s - G'y' for Q 0: Gy* If the rolling body G, be a solid cylinder, we have then _ i G„ and for the first the tension S 2 PG 3 P+Gt and for the Q. second S = _l. If in the first case the weight P descends, p is 3 then negative, therefore: S > P; t. e. 2 PGlVl* > PGlVl* + P8 (ax8 + y,8), ^ s simply -jl > 1 4. ; further, that G, may descend, it is necessary that S < G., therefore ^ 1 — Pi Vi ExampU. If when in a system of pulleys,Fig. 271, the string GH suddenly brealcs, the string BK at the oommencement beeomes stretched by the force Si 2 P + Gy» 2.40 + 3 (*+■??■) +1 (t+^)(40+3) + 1 83.72 »976 !{.210Ibs 25.43+ 72 1147 The accelerated motion of the deecending weight P will be: p = / P~S A g = /40—5,21°\ 32 8 M 34,79 322 — 86 023 feet. \P | ) v 404-3 y 43 Further, that of the descending pulley: » - CW) • - which we have become acquainted If the radius ?mics Hncler the name of the normal accelerating force. “ r, and it» ° ,!yature «t «ny place of the path of the moving point velocity ~ v, we have then for the normal accelerating force r r IS )• l the first cause of the point chang g normal act upon the position. If no other tangenti ore hence the normal point, its velocity vwill be invanable and = c, a f eP = * will be dependent only on the curvature at each mo- the curvature or .« rad.ue to, , as for , si„8le tallins instance, the non»»1| -J, M is constrained by a horizontal of curvature. If a ™ ., a^UI.ve it will have, disregard- path, Fig. 273, to des ing the friction at all places, the same velocity c, and will exert at each place a pressure equivalent to the normal force against the concave surface. During the description of the arc AB, the pressure = -== > during that of v/vl BD — , for the arc DF it = EB Mc2 and for the arc FH = -==, if CA, Kr EB, GD, and KFare the radii of curvature of the portions of the path AB, BD, ^tnTand Centrifugal When a material point § 231- Cenin^taland^nj^^^ ^ or body moves in tripetai forc whilst the force which the body whence.tiscalled the ' J which acts radially outwards, opposes, by vutue^nts Ztrifugal forceThe centripetal force * has received*directly upon the body; the centnfugal is the reacting fowe^fthe body. Each is equal in amount and opposite in direction t0 Tn^he^evoluUon of the planets about the sun, the attractive force nf thesun is the centripetal; but were the body constrained by a °{-“ trix Fie 273, to describe its circular orbit, this directnx would act, by its rigidity, as a centripetal force, and opposed to the centri- fuga! force of the body. , , If lastly, the revolving body be connected by a thread or by a rod with the centre of revolution, the elasticity of the rod will then be in equilibrium with the centrifugal force of the body, and thereby act as a centripetal force. , , f Let G be the weight of the revolving body, therefore its mass jyj __ lL,the radius of the circle in which it revolves = r, and the velocity of revolution = v; from the last § we have the centrifugal force:CENTRIPETAL AND CENTRIFUGAL FORCE. 247 P = . 9* = 2 . t. . (2, therefore also P : G - 2 . £ : r, r gr 2g r *g i. e., the centrifugal force is to the weight of the body as double the height due to the velocity is to the radius of revolution. If the motion be uniform, which always takes place if no other force (tangential force) than the centripetal act upon the body, the velocity v = c may be expressed by the time of revolution T, if we put c = -P?**1 B — and hence we shall obtain for the centrifugal time T ’ force: P = ~ . Mr _ . Gr. r 0,031, we \T ) r T% gT2 Since 4=» 39,4784, and for the measure in feet — g then have more conveniently for calculation p_Jiiiy.jifr-M24.gr. The number n of revolutions per minute is often given, and there^ fore Tis replaced by whence it follows: n P “ “* Mr “ °»010966 n% Mr — 0,000331 n* Gr. Hencc, for equal times of revolution, or for an equal number of Srt ofX1»'" ^^n «me, the centrifugal force incretes as the pro- tional nth« -aSS ^ radias.of rev<>lution, and is inversely propor- revol ,Honh n C1™“s.tances be!ng alike> to the squares of the times of revolution, or directly proportional to the squares of the number of revolutions. As is the angular velocity «, we may finally put: P - »2. Mr. pefS^Z^, arb°°f 50 l,'?S,Wei«ht describe 8 «hole of 3 feet radius 400 times per minute, its centrifugal force will then be: If ,k- ? ■ °-000331 • 400*. 50.3 = 52,96.50.3 = 7944 lbs Mreneth^h^^fr^R^^,^8- b/ “ hemPen <*>"»■ «be modulus of g Of the cord (§ 186) be 700 lbs., it will foilow that 7944 =s 7000 . F; bence the section of this cord will be : F Z)-*- __ /4,5468 N » V 3,1416 radfos 1,2 “ 1,2 • 2.’732 “ 2»077 inches.—2. From the eerth’i ss 24 «n in mi i1?!?8 °S feet> anc*the time of rotation or length of a day r»24 h. * * ®6400^» lhe centrifugal force of a body at the earth’s equator is -P 1,224 G • 20,250000 2478 1 . „ 66400®---" 86F G== 300'-' «>“* wer* *• of ay T,th par, only, then: ~ as 1 h. 24 m. 42 sec. this force wouid be 17*»289 times fore^uTe*’cem^f?.^ ^S1™1 *° about *•>« weight of the body. A» *be equator, there- neither rise nnr rfn *?rce Would he equivalent to that of giavity, and the body wouid is counteracted K V revolution of the moon about the eartit, its centrifugal force distance fmm vk tbe ?ttraction of the earth. Let O be the weight of the moon, r its castanee from the earth, and T its time of re vote*»! the «35^ «we of this = as 1,1342 square inches, and its radius 7000 v/1,44 * 1,2 inch. But for a threefold security D248 CENTRIFUGAL FORCES. Fig. 274. W, _ ■ *» Lel ■> be ihe - »• •"> “ “““ "* /< :*a centre incrcflses iuversely n.s a that the force of gravity at different distances the m00n or the attractive force power of these distances; we have then the g y rf the earth = G£)", and if we equate both forces to each other, we then obtatn Now «=±, r = 1250 nti.Uon feet,a„d T = 27 days, 7 hours, 42 «mute. - 39342 • 39342.60 seconds; hence it follows: minutes - / 1 \n 1,229.1260 __ I nearlv = ( —) (—) — "3934* 36 3600 . o V , the gravitating force of the earth is m an inverse ratio to the and hence, n = 2; t. & .or thed^ Forcesof Extende* Mas For any system § 232. Geninj ^ Qf finite extension, the formula above found of masses, or fo directly applicable, because we know for the centnfugal force is no^ hefo/ewhat radius of gyration we have to introduce into the calculation. To find this, we proceed in the following manner. In l1 ig* 274, let be the axis of revolution, and its two rectan- gular co-ordinate axes, further let be a particle M,, and MK = x, = and MJY = z, its distances from the co- ordinate planes and F. As the centrifugal force acts radially, its point of application may be transferred to its point of intersection O with the axis of revolution. If now we resolve this force ---- in the direction of the axes and CY, « fSrt Lheo“”Sr.nfro*OQ0°oVit which oy : ur — ^ ^ Q x P and R = - Pywhere r represents the distance OM of the • \ c nJis of revolution. Let us proceed in a similar particle from the a manner with ali the particles, and we Fi 275. shall obtain two systems of parallel forces, one in the plane XZ, and the other in the plane YZ, but each acting perpendicularly to the axis CZ. For distinction, let us avail our- selves of the index numbers I, 2, 3, &c., and, therefore, put for the par- ticles of the mass, J\f,, and for their distances x,, x2, x3, &.c., we shall obtain the resultant of the one system, Fig. 275^ ^OF EXTENDED MASSES. 249 «2 . (M1x1 + M2x2 + . . .) and that of the other R = R1 + R2 + • • • = w2 . (Mly1 + Mjy2 + ...). Let us finally put the distances of the Ndei from the plane XY, COv C02, &c., = zlf z2, &c., we shall obtain for the points of application of these resultants the dis- tances CU = uy and CV = v by the equations (Q, + Qa + • • •) u = iT1 4" • • • an^ (-®i 4- R2 4* • • •) v = P\z\ 4* R2~2 4* • • •» Phenee it follows : u s=z 4* *M2x2z2 4* . . » 1 Qi+ Q24- • • MlXl+M^+... ’ ana y z=_ 4-*Mgya~2 4" * • * + . . -MM + -M&+ .. . ffence, therefore, in general, the centrifugal forces of a system of bodies, or an extended body, may be reduced to two forces, which, so long as u and v are unequal, cannot be resolved into a single one. Example. The masses of a system are Mx = 10 Ibs., M2 = 15 lbs., M3 = 18 lbs., M4 = 12 Ibs., and their distances x. = 0 inches, xQ = 4 inches, = 2 inches, x. = 6 inches. V. = 3 “ y3= 1 “ y3= 5 “ y4 = 3 ‘‘ 2, = 2 • xa= 3 « q we have then the following mean centrifugal forces Q = . (10.0-f 15.4+ 18.2+ 12 . 6) = 168 . (10.3 4- 15 . 1 + 18 . 5 + 12.3) = 171 }>oints of application from the origin C: 10.0.2-f 15.4.3+ 18.2.3+ 12.6.0 = 3 , and «> 0 u = 10 10.0 4- 15.4 4.18.2 4- 12.6 . 3.24- 15 . 1 . 34- 18 . 5.34- 12 . and the distances of their == 1,714 inches and 168 7 ____________________________3'0 _ 375 __ 125 __ 2,193 inches. 10 . 34-15 . 1 4-18 . 54-12 . 3 171 57 The difference of these two values of u and v shows that the centrifugal forces cannot be replaced by a single force. § 233. If the particles of the mass lie in a plane at right angles to the axis of revolution, Fig. 276, their centrifugal forces may be re- duced to a single force, because their directions intersect at a point in the axis. Retaining the denominations of the former §, we shall obtain the resultant centrifugal force in this case: P = v~W+R2 = *2 v/[ + .. ,)s 4- (-W,y, + Mjy% +..)*]• Now CK — x, and CL = y, are the Fig. 276. --- — j----------- & T - co-ordinates of the centre of gravity of the system M = Jtf, + «M, + we then have: M1xl + JWjXj 4- ... = hence the centrifugal force: /> = j> ^ + 3iy Mx and Mly1 + ^^2 + * • • ® «2 Mry x = w’ V JU’X- f Juy = J1 M x/ x2 4- y2 — provided further that r = \/ x2 + y2 representjlie distance 0 e centre of gravity from the axis of revolution CZ. . For the angle PCX = a, which this force includes with the ax250 CENTRIFUGAL FORCES. centre of gravity. Fig. 277. Fig. 278. R My_ y .hm CX, tang. a — ~ ji|- * force is exactlythe same as if the collective For a disc AB at right angles to the axis of i *;„n 7Z Fie- 277, the centrifugal force reVf°'“ this J «Ar, if Jtf represents its mass, its centre of „„ity o r tR» avis To find the centrifugal force ff Ser body Fig. 278, we must decompose it into elementary discs by planes „t riffht ancles to the axis ZZ, find their centres of gravity 3,, S„ &and determ ne the cen- trifucal forces by help of these last decompose each of them in the direction of the axes CX «nd CYinto component forces, and reduce the forces in the plane ZCX to a resultant Q, and Aose in the plane ZCFto a resultant R. If the centres of grayty ,aggregate discs lie in a line parallel to the axis of revo- lution, * - - ^-&c” and y = ^= V» &

reduced to its centre of gravity, fugal force of J.h Hcation and centre of gravity coincule. From this and its point« aPP f u rotary bodies, whose geometnc axes run thC u f» th?he axis «f rotation may be found. If the geometric axis of "any such body coincide with the axis of rotation, the centrifugal force is equal to nothing. nmnle —The dimensions, the density, and strength of a millstone, ABDE Fig. 279, Exampu.—lue are g,ven; it is required to find the angu- Fig. 279. ---0----, < --Q- lar velocity w, in consequence of which rupture will take place in virtue of centri- fugal force. If we put the radius of the millstone = rt, the radius CK of its eye = rv the height AE = GH = /, the density = y, and the modulus of strength = K, we obtain the force required for rupture = 2 (rt—ra) /if, the weight of the stone G = v (fj9—r*%) /y, and the radius of gyration of each half of the stone, i. e. the distance of its centre of gravity from the axis of rota- A 4OF EXTENDED MASSES. 251 moment of rupture, the centrifugal force of half the stone is equivalent to tlie strengtli ; we hence obtain the equation of condition ** . J — = 2 (r,—ra) IK, i. e. ot1. $ (ri* rj) ly S = 2 (r,—r2) IK} or leaving out 2 / on both sides, it follows that /3* (r,-r,)A:_ / 3 gK (r,3—ra3) y (r* + r, rf r*) 7 If r, = 2 feet = 24 inches, r3 = 4 inches, K = 750 lbs., and the specific gravity of ^ . 62,5 X 2,5 tne millstone = 2,5, therefore the weight of a cubic inch of lts mass = —— = 0,0903 lbs., it follows that the angular velocity at the moment of rupture is: /3.12.32,2.780 _ /869400 = incheg v/ 688.0,0903 W 62,1264 If the number of rotations per minute = n, we have then m = ■ ; hence7 inversely, n = ' but here = 30'= 1070. Tlie average number of rotations of such a » it millstone is only 120, therefore 9 times less. § 233. If the collective particles Mv M2, of a system of masses, Fig. 280, or the centres of gravity of the elements of a body lie in a plane passing through their axis of revolution, the cen- trifugal forces will then form a system of parallel forces, and these may be reduced according to the rule to a single force. The distances of the particles or the ele- ments from the axis of revolution ZZ, are OlMl == rt, 02M2 = r2, &c., we obtain for their centrifugal forces: Pj = w2 Mlrv P2 = w2 M2rv &c., and hence the resultant centrifugal force: P = w2 (Mlrl + M2r2 + . .) = w2Jlfr, r representing the distance of the centre of gravity of the mass M from the axis of revolution. Therefore, here also the distance of the centre of gravity from the axis of revolution must be regarded as the radius of gyration. But to find the point of application O of the resultant centrifugal force, let us put the distances of the particles of the mass from the normal plane: CO. = CO, = &c., into the formula: rn- :_ Mirizi + M*r2Z2 +••• Mxrj + M2r% + . . . By help of the formula P = w2 Mry the centrifugal forces of rotary bodies and other geometric bodies may be found, 'when their axes and the axis of revolu- tion lie in one plane. The centrifugal force of a right cone JJDBy Fig. 281 > may be found, if the distance SJY of its centre of gravity S from the axis of re- volution ZZ be put as r into the formula. If the height of the cone CD =* A, the distance DF of the point D from the axisFig. 282. 252 CENTRIFUGAL FORCES. of revolution = a, and the angle CGE which the geometrical axis deviatesfromthe ari^of^ta^Z^-^e £ + lh stn. ». tora rod [Jc\\n*xon ABZto the axis of revolution = we have r - <&V= «*• •» therefore, the centrifugal force: , P = „* IJIf/riti.»; but to hnd the point of application 0 of this force, in the expression «* . x sin. a x cos. = ws . — * sin. o cos. a 71 M for the moment of the element — of the rod, let 71 l 21 q / us put for x successively the values _, lf r &c., and add the results, in this manner we * *t — A--o «n/l • VVV/‘) —----- shall obtain the moment of the entire rod: Pz = j— sin. o cos. * — (l*+2*+3*+ n n* . + n») _ l JMP sin. o cos. o, . o , „„„ Z\JMF sin. a cos. a: i-2 JW stnVo = iLC0S-a’ and Fig. 283. Fig. 284. 5tn. a cos. a: sin' a — ? * ana the distance of the point of application 0 from the extremity of the rod B lying in the axis, Bif7h! rod AB, Fig. 283, does not reach the axis, we then have: . . p __ i sin. o—^ - F/, ^n. a = ^ p "//2 / 2\ an(l the moment: stn. a (Ij —*2 b 3 »,N p~ = ^ «2 F sxn' a C05* a (v—^3)> because the mass of CA - the cross section into the length, = and the mass of CP = P/2, ■ I hence it follows that the distance of the pomt of ■J application O from its intersection with the axis — 3 I 2___/ 2 12/ 1 where / expresses the distance CS of the centre of gravity, but /, the length of the rod AB. ,, This formula is also applicable to a rectan- eular piate ABDE, Fig. 284, which is di- vided into two congruent nght angles by the plane of the axis COZ, because the centri- fWal force acts at the middle of each of the eiements, which are obtained by sections normal to CZ. Therefore the distances CF and CG of the two bases AB and DE from the point Cof the axis^, are ^ and l2, we have here also CO = — • -=^OF EXTENDED MASSES. 253 Fig. 285. § 234. In the case where the particles of the body neither lie m a plane normal to, nor in a plane passing through the axis of revolution, the resultant centrifugal forces, Q = «2 + Mjc2 + • .) and R = + Mj/2 + . •) canno be reduced to a single force, nevertheless it is possible to replace these forces by a force acting at the centre of gravity: p = ^ Q*+ R2 = Mr, and by a couple composed of Q and R. If, namely, we apply to the centre of gravity S> four forces + Q and — Q> + R and — R balane- ing each other, the positive parts will give a resultant P= <*2 y Q*+ R2, and the negative parts on the other hand, Q and R, will form the couples (Q, —Q) and (R, — R) with the centrifugal forces applied at U and F, which may be reduced to a single couple. In order to make ourselves acquainted with this reduction of the centrifugal forces of a rotary body, let us take the following simple case. Let the bar ABy Fig. 285, which turns about the axis ZZy lie parallel to the plane FZ, and rest with its extremity B on the axis CX. Let the length of this bar = /, its weight = G, the angle BAD at which it is inclined to the axis of rotation = a, and its distance CB from the plane FZ, which is also its shortest distance from the axis ZZ = a. Let M now E be an element — of the bar, n and BE = x its distance from the ex- tremity By we shall then have the pro- jection BN = x sin. o, and hence the components of the centrifugal force Px of this element: Q, = . CB = «* . — a and R, = «.*. . BJV = «*. — 1 71 71 71 n x sin. a, and their moments about the principal plane XCY: Q.z. b . — . CB . EN = «2 . — ax cos. a and R1zl = «2---------** • n n 71 . sin. a cos. o. The several components parallel to the plane XZ give the resultant Q = Qi + Q2 + • • • “ n • 2X5 an(* *tS moment Qu = Qgzl + Q2z2 + ... or, as Xj is to be taken ^ 71 Qu t M * . — a COS. a 71 .-(l + 2 + 3+ .. + n) . — acos.«(*,+ xi +••')’ n &C., n . M 2 x3 71 a COS. a 22254 FREE AXES. / 7>2 1 - . — = - w2 . Mal cos. a; the distance, therefore, of the point of n 2 2 applicatiori of this component from the plane XYis: LS=U= klcos. ., Ma i. e. it coincides with the centre of gravity of the bar. The cona- ponents, which act parallel to YZ, give the resultant R = RA + R + . . . = . — sin. a (xt + a?s + . . .) - «* • — ^ T n » n 2 M =■ J , andjet the co-ordinates of the particles referred to the axes XX and ZZ: be xv x% . ., z19 z2. M therefore for the particle Mt : MXK = xx and MXL = zv From this we easily obtain: xx—MXA=SR+RL= SO cos. <$> -f- OJWx sin. q>=u1 cos. $ -f- vx sin. $ z1=M1L = — OR+ OF = —SO sin. 0M1 cos. = —uxsin. $ + v\ cos. ; and hence the product: xizi=(ui cos. t + sin. $) (—ux sin. $ + v1 cos. t) = — (ux —v*) sin. cos. $ + u1v1 (cos. *2 — sin. 4>2) or, since sin. cos. <1»=^ sin. 2 and cos. $>2— sin. $2=zcos. 2 t, xxzx = — h (ui — v.2) sin. 2a + u.v. cos. 2 t, and hence the moment ot the particle Mx : ^ixizi = — (ui2 — vi) s^n* 2 t + Mlulvl cos. 2 The mo- <6 ment of the particle : -VjXjZj = — ^2 — t'j2) sin. 2 j + M2u2i\ cos. ^4 2 t, &c., and256 FREE AXES. the sum of the moments of ali the particles, or the moment of the entire mass: r/w . , 0 M.x.z. + M2x2z2 + . . . — — i «'»•2 * [ (Miui + + • .) — (JW^,2 + M2v22 +••)]+ cos. 2 + M2u2v2 + . .). That XX may become a free axis, its moment from the former paragraph must be = 0; hence we must put i lf 2 V i %n. 2 , [ (M,u- + JW,V +•••)- «'■ + + • •) ] _ cos. 2 * (JM.u,», + M2u2v2 + • •) = 0» and from this we obtain the equation of conditioni sin. 2 t__________2 (M,u,v, + M2u2v2 tang- 2 * = " (Mtu2+M2u2*+.. twice the moment of the centrifuga! force = difference of the moments of inertia By this formula, two values for 2 * are given, which vary 180° from each other, and therefore also two values of *, which vary 90° fr0m each other; on this account, not only is theaxis XX determined by this angle *, a free axis, but also the axis perpendicular to it. § 237. The free axes of many surfaces and bodies are known without any calculation. In symmetrical figures, for instance, the axis of symmetry is a free axis, the perpendicular to the centre of gravity is a second, and the axis perpendicular to^ the plane of the figure a third free axis. The axis of rotation ZZ of aj-otary body AB, Fio-. 287, is a free axis, so is every normal XX> YY. . to this, passing through the centre^of gravity^. Lvery diameter of a sphere is a free axis, the axes XX, FF, ZZ, of a right parallelopiped ABDy Fi*. 288. Fig. 287. Fig. 289. Fie 288 bounded by six^ctangles, passing through the centre 0f r lg. .400, uuuiiuc jvjt„ s, and normal to the sides BD, and AD, or running par- allel with the edges, are free axes. Let us now determine the free axes of an acute angled parallelo- graro I ig. 289. Let us draw through its centre of gravity S, the co-ordinate axes UU and VV at right angles to each other, so that one of the sides AB of theFREE AXES. 257 parallelogram may run parallel to it, and let us decompose the parallelo- gram by parallel lines into 2n equal strips, such as FG. If, now, one side AB == 2 a, the other AD = 2 b, and the angle ADC between the two sides = 0, we then obtain for the strip FG, distant from UU, SE*=x, the length of one part: EG =* KG + EK=* a + x cotg. o, and that of the other EF — a — x cotang. », and since - sin. o is the breadth of both, the area of these strips ** b **n,-(a+x cotg. ») 7 71 and 6 ”n- tt (a—x cotg. «); the measure of the centrifugal forces w about the axis VV is therefore: _ b sin' ° (a + X cotg. a) . i («+* cotg. o) = («+* C0^- a)* n __ and b CTW‘ ° (a—x cotg. o)2, and their moments about the axis UU: 2n (a+x cotg. o)2 x, and 6 («-* cos. a)2 x. As both the forces about VV act opposite to each other, the uniting of their mo- ments gives the difference: 6jr^stn. a cotg. a)g—(a—x cotg.)2] = ? a6x* cos. a. If we substitute in this formula for x the values: b sin. a 2 b sin. a 3 b sin. a « -------,--------->-------------, &c., n n n successively, and add the results, we shall obtain the measure of the moment of the centrifugal force of half the parallelogram: 2 ab — _ __ . cos. a. (l2+22+32+ . . +n2) -2 ab2 st». a2 cos.a . JjL n n2 J 3n3 2 = ~ ab3 sin. a2 cos. a, and, therefore, for the whole parallelogram, or o 4 'Mlulv1+M3u3vi+ ... = - ab3 sin. o2 cos. ». The moment of inertia U of a strip FG about the axis VV is: _ b sin. a /(a+x cotg. a)3 ( (a—x cotg. a)2\ “ “» V 3 + 3 / =* Sln* a. (a3+3 ax* cotg. as) =* - — sin. a (a*+3 x2 cotg. a2); if Jn 3 n now we substitute in succession for x: b sin. a 2 6 sin. a 3 6 sin. a « —-—, -------------,--------y cfcc., and sum the resulting values, we shall have the moment of inertia of a half - * ab sin. a (a2 + b3 cos. a2), and hence that of the whole «•258 FREE AXES. = - a6 sin. a (a2+i2 cos. a2). On the other hand, the raoment of inertia of the parallelogram about the axis of revolution UU is = 4 sin. a . ^ 5171,01 = ^ a3 (§ 220); hence the diflerence of the 3 3 moments of inertia sought, i. e. (M1u*+M3u2*+ . . .}—Mlv2+M2v2+ . . .), — 1 ab sin. a (a2+b* cos. a2) — ^ ab3 sin. a3 3 3 s ^ ad sin. a [a2+&2 (cos. a2—sin. a2)] 3 s — ab sin. a (a2+A* cos. 2 o). 3 _ LastJy for the angle USX-*t, which the^ree axis XX roakes with the co-ordinate axis UU or the side AB from § 236: tang. 2 *__________2 (M1u1vl+Miu1v2+ , .) * (Mlu*+M2u*+ . ^-(M^+M.vZ+T:) 2 - ab3 sin. a2 cos. a 3 b2 sin. 2 < 4 * • / a , L2 o \ a2+b2 cos. 2 a - ab sin. a (a2 + b2 cos. 2 a) 3 In the rhombus a = 6, hence , 0 sin. 2 a tang. 2 $> = 2 sin. a cos. a 1 + cos. 2 o 1 + cos. a2—sin. a2 2 sin. a cos. o , 2co-s 3------^ therefore 2 $ == 0, and ♦ = As this angle gives the direction of the diagonal, it follows that the diagonals are free axes of the rhombus Exampk. The sides of the acute angled parallelogram, ABCD, Fig.289, AB =s 2 a_ 16 inches, and BC =s 2b = 10 inches, and the angle of the perimeter ABC = a ^ gJS’ what directions have its free axis ? ’ ______ A 5a. sin. 120° _ 25 . sin. 60° 25.0,86603 _____ ~ 8a + 5*. cos. 120° 64—25 cos. 60° “ 64—25.0,5 0,42040 = tang. 22° 48', or tang. 202° 48'. From this it follows, that f = 11° 24' and 101° 2d' are the angles of inclination of the two free axes to the side AB. The third free a • stands at right angles to the plane of the parallelogram. These angles determine af*3 the free axes of a right parallelopiped with rhomboidal bases. 30INCLINED PLANE. 259 CHAPTER III. OF THE ACTION OF GRAVITY ON MOTIONS ALONG CONSTRA1NED PATHS. § 238. Inclined Plane.—A heavy body-may be impeded in various ways from falling freely, and in the following we shalJ consider only two cases, the one where a body is supported on an inclined plane, and the other where it revolves about a horizontal axis. In both cases the paths of the body are contained in a vertical plane. If the body rests on an inclined plane, its weight raay be resolved into two components, of which the one is directed normal to the plane and resisted by it, and the other parallel to the plane, and acts upon the body as a moving force. If G be the weight of the body J1BCD, Fig. 290, and a the inclination of the inclined plane FHR to the horizon, Fi8*290, we shall then have from § 134, for the normal pressure ; = G cos. «, and the moving force P = G sin. a. The motion of the body may be either sliding or rolling, let us next consider the first only. In this case all the parts of the body equally participate in its motion, and hence have a com- mon motion of accelerationpy which is given by the known formula: force P Gsin. a ^ mass’ M G ' 8 — g*• Therefore p : g = sin. <* : 1, i. e. the accelerated motion of a body on an inclined plane is to the accelerated motion of free descent as the sine of the angle of descent to unity. In consequence of the friction wThich takes place, this formula is rarely sufficiently accurate, hence it is necessary in many cases of application to take this into account. If the body moves on a curved surface, the accelerating force is variable, and at each place equal to the accelerating force, which cor- responds with the plane of contact to the curved surface. § 239. A body slides with the initial velocity 0 down an inclined plane, without friction, from § 10 the final velocity after t seconds is : v ™ 8 s*n• « . t =s 32,2 sin. a . t ft., and the space described1: s = £ 8 f17*- a • P =* 16,1 sin. » . t2 ft. In free descent vx =* gty anc[ si J5 2 gt, hence we may put: v : vx = s : sx = sin. a : 1, *• e* velocity and the space of descent down the inclined plane are to the final velocity and space of free descent as the sine of the angle of in- clination of the inclined plane to unity. Fig. 290.260 INCL1NED PLANE. Fig. 291. The perpendicular of a right-angled triangle, FGH Fig. 291, with vertical bypothenuse = FG sin FGH - F sin. FHR = FGsin if 0 is the angle of inclination of this perpend.cu ar to the horizon, hence FH: FG - »«• a : 1, and a body scribi the vertical hypothenuse FG and the in- clined side FH in one and the same time. The siace of free descent corresponding to the space of descent down the inclined plane may be found flm this and the latter from the former by construe- . mptpr fG, Fig. 292, are nght angles, the semicircle on &c., on the dia cuts off from ali the inclined planes, commencing at F, the spaces described with this diameter, and therefore, in equal times, F F/f2,&c., hence it is asserted, the chords of a circle and its diameter vnll descend isochronously. This isochronism is besides true, not only for the chords FH2, & c., which have their origin at the highest point F of the circle, but also for the chords KtG, K2G, &c., which commence at the lowest point G, for chords FK1, FK2, &c., may be 1^^— ----- drawn through F, which have like posi- ,io„s ..d equal lengths «i it follows that * ■' 1 h. «mintinn S ^ c » o Fig. 292. § 240. From the equation s = ^— = _— * *P 2 g = —, and, inversely, v = s/2 gs sin. a. But now s sin. s sin. ql Fig. 293. 6* 7 is the height of the inclined plane or the vertical projection s, of the space FH = s upon it, hence the final velocities of bodies wliich descend with an mitial velocity 0 down planes of equal heights , F,Ht, &c and of different inclination, l ig. 293, are equal, and also equal to the velocity which a body would acquire if it feli freely I" -------------- - from the height FR of these planes From the equation s = h g *•• * * f°1,0WS the f°rmUla f°r the time' .-ft- i i jtrm 1 = J7777 = ^ \ ^ 8 But for a free descent through the height FR the time is. tx = it follows accordingly 1 : sm. a = : FR, the time of descent down the inclined plane is to the time of fiee descentINCLINED PLANE. 261 from the height of this plane as the length of the inclined plane is to its height. Examples.—1. The initial point F of an inclined plane Fig. 294. FH, Fig. 294, is given, and the final point H in a given line AB- required to determine the descent down the plane so that it may take place in the shortest time. If the horizontal line FG be drawn through F to its intersection with AB, and GH be made = GF, we shall obtain in H the point sought, and, therefore, in FH the plane of quickest descent; for if through F and H a circle tangent to FG and FH be carried, its isochronously described chords FKt, FK2, &c., will be shorter than the lengths FH„ FHV &c., of the corresponding inclined planes; consequently, there- fore, the time of descent for these chords will be less than for these lengths, and the time of descent for the inclined plane FH, which coincides with a chord, will be the shortest. 2. Required the inclination of that inclined plane FH„ Fig. 295, down which a body would fall in the same time as if it originally feli freely from the height FR, Fig. 295. and then proceeded with the acquired velocity horizon- tally to //,. The time of falling down from the vertical height FR = sx is tx = /^, and the acquired velo- ____ N S city at R: v = ^/2 gsx. If now no loss of velocity ensue in transition from the vertical to the horizontal motion, which would follow if the corner R were rounded, the space RHX = $x cotg. a will be uniformly described, and in the time t, = f' = A ro,g, . /2., \/'2 g*i the inclined plane is t = _L_ IllL- sin. a %J g ’ 1 . . , The time of descent down hence, if we put * = ta, we shall obtain the equation of condition ■. x— j cotg. a, whose solution will give tang. a = i. sin. a In the corresponding inclined plane, accordingly, the height is to the base and to the length as 3 is to 4 is to 5, and the angle of inclination is a = 36° 52' 11".—3. The time for sliding down an inclined plane of a given base a is 4 a 1 2* _ I 1 2« / / g sin. a nJ g sin. a cos. a +J = 90°, or «° = 45°. inclination. f g sin. 2 a * quickest when sin. 2 a is a maximum, t. e. = 1; therefore 2 »° Hence, water falis down in the shortest time from roofs of 45° § 241. If the motion on an inclined plane proceeds with a certa_ initial velocity c, we shall then have to apply the formulae found § 13 and § 14. According to these the terminal velocity of a boy ascending an inclined plane is v = c g sin. a . t, an . ,]• described s = ct — \ g sin. a . t2; on the other hand, for a y down the inclined plane: . /2 v = c + g sin. a . t, and s = ct + \ g sin. o • In both cases of motion the formula is true: v2—c2 . v2 — c2 v2 s = ---------—, or s sin. a == —------ = op* 2 g sin. a ** g ® /0\ jpcrribed The vertical projectiony therefore, (5 sin. a) of the space ( ) . ^ along the inclined plane is always equal to the difference oj 0 due to the velocity.262 1NCLINED PLANE. Fig. 296. If two inclined planes FGQ and Fig. p^n °ft^r in a rounded edge, no impulse will take place in the passage from one plane to the other, and for this reason, no loss of velo- city ensue; the rule for the descent of a body down this combination of two planes is also true, height of descent (FR) is equal to the diference of the heights due to the velocity. It is easy to ascertain that this rule is correct also for the ascent or _________ descent on any system of any number of planes, and for the ascent or descent on curved lines or surfaces. (Compare § 82.) _1 A bo.lv ascends with a 21 feet initial velocity an inclined plane of 22« inctodon what is the amount of its velocity and its space descnbed in li seconds? The 1*}°°'%£2„„ 22° . 1.5 = 21 — 32,2.0,3746.1,5 = 2,906 feet; and the space: r = 21 _ 32,2 ««+22 . + 2>906 3 _ 23,906.3 = 17>928 ^ 3 = ~2~-‘ — § 2 4 2 How high does a body, with an initial velocity of 36 feet, ascend an inclined plane of 48» acclivity ? The vertical heigh, is = £= 0,01550 * = 0,0155.36’ = 21,638 ® g 21,638 feet; hence the whole space up the inclined plane. s ^ a *in. 28,494 feet. The time required is: v 2 . 28,499 28,494 = 1583 seconds 36 1« S 242 Sliding friction exerts a considerable influence upon the ascent and descent of a body along an indmed plane. From the weight G of the body, and from the angle of mclination a of the in- clined plane, C'L U “e »bt,«.ThU from the force".? = G sin. o, with which gravity urges the body down the plane, there then remains for the mov.neforce,-7 ° f'1 — f G cos. o, and the accelerating force of the body sliding down the plane is known : the force G sin. a —J G cos^a ^ __ 0 —y co$ ^ ^ P = —---------=-------- Gr ° 7 ° The movfngforce of the body ascending the inclined plane is nega- tive and Josin.a +f.G cos. a, hence also the accelerating force p is negative and = [sin. a + jr cos. a) g. If two bodies are supported on dif- ferent planes FG and FH, Fig. 297, by perfectly flexible stnngs connected ■with each other, passing over a roller C, it is then possible for one of the two bodies to descend and pull Up the other. If we represent the weights of these bodies by G and G,, and the angles of inclination of the inclined planes along which they move by Fig. 297.R0LL1NG MOTION. 263 a and 0l, and if we assume that G descends and draws G, upwards, we shall then obtain as the moving force: G sin. a—G. sin. <*.— f G cos. a—f G. cos. a.= G (sin. a—f cos. a) G 4- G — Gj (sin. a x+ f cos. aj, and the mass moved = -----S hence the & accelerated motion with which G descends and G1 ascends: n G (sin. a — f cos. a) — G, (sin. ax + / cos. aj ^ G+Gi . . Since friction as a resisting force can generate no motion, it is requi- site for the fall of G and the rise of Gx> that G (sin. a —f cos. a) be > Gx (sin. a, + f cos. aj, therefore 9^ ^ s*n' ai + fcos. ai ion the other hand, Gx descend, and G Gj sin. a — f cos. o be drawn up, then must: Gx ^ ^ Mn. a + f COS. » Qr ^ < °i —fC0S• ai G sin. Oj—fcos. a’ y Gx sin. a + f cos. o G • So long, however, as — lies within the limits: Gi sin ttj + f COS. cq j sin ax — J cos. ax sin. a — f cos. a sin. a + f COS. a so long will the friction resist motion. Examples.—1. A sledge moves down an inclined snow plane, 150 feetlong and 20° in- clination, and when arrived at the bottom, proceeds along a horizontal one until friction brings it to rest. If the co-efficient of friction between the snow and the sledge be taken = 0,03 feet, what space will the sledge describe along the horizontal plane,neglect- ing the resistance of the air ? The accelerating force p = (sin. a — / cos. a) g = (sin. 20°) . 32,2 = (0,3420 — 0,03.0,9397) . 32,2 = 0,3138 . 32,2 = 10,104 feet; hence, the final velocity of descent is, v = v/ 2 ps = v/2 . 10,104.150 = 3031,2 == 55,54 feet. On the horizontal plane the accelerating force isp, =— fg = — 0,03.32,2 = 0,966 feet; hence, the space st = —— as as 1630 feet. The time of descent is t = — — ^0_ = 5,22 2 fg 1,932 v 55,54 Fig. 298. 2 s 3260 seconds, and that for sliding onward tx = —1 = -------= 58,6 seconds: hence the whole v 55,54 time of the course t -f- U = 63,82 seconds = V 3,82".—2. A filled tub K, Fig. 298, of 250 lbs. ciear weight, is drawn up an inclined plane FH, 70 feet long and of 50° inclination, by a de- scending weight G of 260 lbs.; what will be the time required for tliis, if the co-efficient of friction of the tub along its path amount to 0,36. The moving force is as G — (sin. * + / «&*• “) K =260 — (sin. 50°+0,36 . cos. 50°) . 250 = 260 — 0,9974 . 250=10,6 lbs.; hence, the accelerating force p=-----——----- 250 + 260 I 10 6 = 'sio = 0>0208 lbs.; further, the time 1 = /— = /-= ^/6731 = 82,04 seconds = 1' 22") V P «J 0,0208 V and the final velocity v = — = ]4Q — 1,70 feet. * 82 § 243. Rolling Motion.—When a carriage rolls down an inclined plane, the friction of the axle chiefly acts in opposition to the acce e- rating force; if r be the radius of the axle, and a that of the wheel,264 rolling motion. Fig. 299. the friction will amount to ^ N == — a, and hence the ac- a d fr celerating force p *= (sin. a---cos. a) G. If a round body AB, a cylinder or sphere,forexampleroll down an inclined plane FH9 rig» we have to consider a progressive and a rotatory motion at the same time. Generally the accelera- tion of the progression is equal to that of the rotation (§ 156); since if we put the moment of inertia of the rolling body = and the radius of the cylinder = we shall then ob- ___ tain for the force AK = K, with which the cylinder is set into revolution by virtue of the penetration of its parts into those of the inclined plane :• K=p. f G cos. a, i. e. tang• a>J\ .... , . efficient to communicate to the body a velo- the friction is no longer -t of progression; hence the accele- city of rotation equal to dsvefocty P^B u. ration of progression, as for siiu & __________ —to 178, note.CIRCULAR PENDULUM. 265 ^ G sin. o —f G cos. <* , • s p =--------------^-------------g = (sin. a—/ and that of rotation: cos. o) g, Pl = /* G COS. a —: •£____________o* = Gf+a2 ■g=f^gcos. For a carriage of the weight G with wheels of the radius a, and with the moment of inertia Gwe have: if G y2 . A = p —— and p s G sitt. a — f — G COS. a a K g (sin. a P G -f- COS. a) a g> i- e. 1 + GlV2 Ga2 Examples.—1. A loaded wagon of 3600 lbs. weight, with wheels 4 feet high, and moment of inertia 2000 fl. lbs., rolls down aninclined plane of 12° inclination, what will be its accelerated motion, if the co-efficient of axle friction = 0,15, and the thickness of the axles of the wheels amounts to 3 inches? Gy _ 2000 5^ = 0 139 and f r = „ 15 _ J_ = 0i0094 36 a 4.4 It Goa 3600.29 hence the accelerating force sought is p = 32~2 18° ^ = 32,2 (0,2079 - 0 0094.0,978) = 32,2 . 0,1987 = wilI be the 1,139 1,139 accelerating forces of a solid cylinder rolling down an inclined plane of a 40° angle of descent ? The co«efficient of the sliding friction of the cylinder on the plane = 0,24, we have then / ^ 1 + = 0,24 (1 +2) = 0,72 ; but now the tang. 40° = 0,839, hence the tang. a is greater than / ^ 1 4* and the acceleration of the rolling motion less than that of the progressive. The last is p = (*inf « — f cos. a) g s (0,6428—0,24 • 0,7660) . 32,2 =0,459.32,2 = 14,78 feet, but the first only jpt= 0,24.2.32,2 cos. 40° == 11,85 feet. § 244. Circular Pendulum.—Equilibrium subsists in a body sus- pended to a horizontal axis so long as its centre of gravity lies ver- tically below the axis; but if its centre of gravity be drawn out of the vertical plane containing the axis, and the body be left to itself, it will take an oscillatory motion; that is, it will move up and down in a circle. In general, however, a body oscillating about a horizontal axis is called a pendulum. If the oscillating body is a material point, and its connection with the axis of revolution be made by a line devoid of weight, we then have the mathematical or simple pendulum; but if the pendulum consists of a body having dimen- sions, or of several bodies, we have then a compoundy physical, or material pendulum. Such a pendulum may be regarded as a c*)n- nection of simple pendulums oscillating about a common axis. Ihe simple pendulum is an imaginaiy one only, but its assumption pos- sesses great advantage, because it is easy to reduce the theory of the motion of the compound to that of the simple pendulum. 23266 CIRCULAR PENDULUM. If the pendulum suspended at Fig. Fig. 300. 300, be drawn out of its vertical position ---* CM into the position and then left to itself, it will go back by virtue of its gravity with an accelerated motion towards and its mass will arrive at its lowest point with a velocity whose height — is equal to the height of descent In virtue of this velocity, it will now describe on the other ^_____________ side the arc — and will thereby ascend to the height DM. From it will affain fall back to M and A, and so it will go on successively describ- ine the circular arc AB. If the resistance of the air and fnction were entireiy set aside, this oscillation of the pendulum would go on in. definitely; but because these resistances can never be done away with, the amplitude of the oscillation will become smaller and smaller, and the pendulum come at last to a state of rest. The motion of the pendulum from A to B is called an oscillation, the arc AB the amplitude, the angle measuring half the amplitude by which the pendulum is distant from either side of the vertical CM, the angleof elongation or angle of deviation. Lastly, the time in which the pendulum makes an oscillation, is called the time of oscillation. & 245 On account of the frequent application of the pendulum t0 s ’ the purposes of life, to clocks namely Fig. 301. it is of consequence to know the times 1=1 of oscillation, hence the determination of these is one of the principal pro. blems in mechanics. With the view of solving this problem, let us put the length of the pendulum AC = r, Fig. 301, and the height of ascent or descent corresponding to a complete oscillation MD => h. Let us assume that the pendulum has fallen from t0 G, and let the height of fall DH = x correspond to this motion, we may then put the acquired velocity v = and the particle of time in which the particle of space GK is dei scribed r ^ = -2L- If» now» from the centre 0 of MD V y/ ‘Zgx = h and the radius OM = OD- 4 *, we^scfri^,e ^e. «circle MJYD, we then have a portion of this arc JVP of the height PQ ^ KL = RH equal to GK, which is in a simple ratio to this particle of space GK. From the similarity of the tnangles GKL and CGH, and from the similarity of the triangles JVPQ and OJYH, KL GHCIRCULAR PENDULUM. 267 ‘y— = hence, if we divide these two equations by each other, PQ NH and bear in mind that KL = PQ, we then obtain the ratio of the said portion of arc: . 37/ From the properties of the circle, JVP GH. OJY and from the theorem of the mean proportional, GHa = MH (2 CM— MH) and NH* = MH. DH; hence it follows: _ GK _ CG . -yDH ry/x NP ~ OJV. /2 CM—MH “ £ kx/2 r — (k—x) and the time for describing an element of space is: r y/x” _ NP_______________________2r_______ 1 ss $ Ay/2 r—(h—x) y/2 gx 17 A*P lg Ay/2^[2r—(A—x)] JVP 2r In most cases of application, a sraall angle of deviation is given to the pendulum, and for this reason -A, as also —> and, therefore, 2r 2r h___x m is so small a quantity that we may neglect it as well as also 2r its powers, and now put t Jj The duration of a semi- JVP . . . . 1* h oscillation, or the time in which the pendulum describes the arc AM, is equal to the sum of ali the particles of time corresponding to the elements GK or NP, or as I. I- is a constant factor, equal _ * Sg toJJH times the sum of ali the elements forming the semicircle DNM; i. e. = ^ I- times the semicircle \ itself, therefore hyg_ \ 2/ 1 Jr fth n \r ~ h\g * T ~ 2 *Jg' The pendulum, however, requires the same time for ascending, for here the velocities are the same, and only opposite in direcfion, and for this reason the duration of a complete oscillation is twice as great, i. e. t ■Jr accuracv duration of an oscillation with greater let us transform thenexpressmWhere ^ an^es of osciUation are ,ar£e>26S CIRCULAR PENDULUM. i___/j_____h i. j h—x ~ \ 2 r ) 2r 1 + i • 2r / into the series i. y 2r ' M 2r ) + JV7» A put and we shall obtain the time for an element of space^ If we put the angle NOM, subtended at the centre by NJW% —. . , shall then also obtain ~ *’We MH = h—x = NO (1 — cos. £ A (1^— cos. $); hence • If we divide the semicircle DNM into n equal parts, and if we each = NP = we shall obtain 2n , = (i+i.in^*)+..\ it.jl. \ 4 r / y g 2n by substituting successively for , — . . . to and i n n n n 7 u «uci- ing the results, we shall then obtain half the time of an oscillation • t = (n + — (n—the sum of ali the eosines) + . .\ |!L * \ 8r / \ i? * 2n ' But the sum of the eosines of all the angles from * 0 to * • r ^ « IS = 0; hence, we have more correctly . * IU. If we have regard to more members of the series, we shall obtain • the last formula but one is, however, generally sufficient. If tf pendulum oscillates in a semicircle, we then have h = r; hence th6 duration of an oscillation: e t = (j + i + ^ 225 256 + 18432 + From the angle of elongation «, it follows that r-A , * therefore, - - 1 — cos- “i and h*nce, * COS. a ss ---- = 1 —■ ^ f Qr * J 1 — cos- a sin from this’ consecluently> the Cor. h ' ’En '’ — = £ / m. 1 = 0,00426; on the other hand, for o = 50 . A ' ‘ 8?CIRCULAR PENDULUM. 269 = 0,00047; for the latter angle of elongation, therefore, the time of oscillation is t = 1,00047 . * We may, therefore, for a deviation under 5°, put tolerably accurately the time of oscillation: t = rt I— = —=y/ r = 0,562 \/ r. S g s/g § 247. As the angle of deviation does not appear m the formula t » h it follows that the small times of oscillation of pen- dulums are independent of this angle, and therefore that pendulums of equal length, but of different angles of deviation, vibrate isochro- nously, or perform their oscillations in equal times. A pendulum deviating 4° has the same time of oscillation as a pendulum de- viating 1°. If we compare the time of oscillation t with the time of free descent, we shall then arrive at the following. The time of free descent from the height r will be tx = ^— = y/H . , hence t : tx = * : y/2; the time of an oscillation is, therefore, to the time in which a body of the length of the pendulum freely descends, as * to the square root of 2, or since tx is also = 2 ^ , the time of oscillation is to the time of descent of half the length of the pendulum as ft is to 2. If we put the times of oscillation t and tv corresponding to the lengths of the pendulum r and rv we then obtain t : tx = s/r : s/rx\ therefore, for one and the same acceleration of gravity, the times of oscillation are as the square roots of the lengths of the pendulum. On the other hand, if n be the number of oscillations which a pendulum makes in a certain time, one minute, and nx the number which ano- ther pendulum makes in the same time, we then have t ; tx == - : —> hence, inversely, n : nx = \/rx : ^/r, i. e. the number of oscillations is in an inverse ratio to the square roots of the lengths of pendulums. A pendulum four times the length gives, therefore, half the number of oscillations. A pendulum is called a seconds pendulum, when its time of oscilla^ tion is one second. If we put / =z 1 into the formula t = * 9 we obtain the length of the seconds pendulum r ** ** 39,13929 inches « 0,9938 meters. 23*270 CYCLOID. From the formula follows by inversion that Fig. 302. #=(-Yr; the acceleration of gravity may be found, therefore, from the length of a pendulum, and from ds time of osedlation t. This method is both simpler and safer than that of Attwood s ma- C^ne- u . , , pendulurn {orce which is £enerated by the diurnal rotation of the earth aWt “u «l and .o the incmase of the eartlfs radius from the poles to the equator. The centrifugal forne at the equator diminishes gravity by — of its value (§ 231), wbilst at „ it i, nuii. If/? be the latitude of the place of observation, the accelerating forcp KviVf^m Polulum observa,ions will o2S fore at the equator where 8 = 0; therefore co*. 2 8 1, 8 32,2 (1 0,00259) 32 11 feet, aml at the poles, where 0 = 90«; therefore co.. 2 fi = res 180« = J. , P = 32,2 ! 1,00259 = 32,283 ft. For the rest g is less on mountains and m mines than at the level of the sea. & 248 Cycloid.—We may in an infinite number of ways set a body into vibration, or into an oscillating motion, and we call every bodv in this condition of motion a pendulum, and distinpish accord- inely several kinds of pendulums, for example, the circular pendulum, 8 * which we have al’ ready considered, and the cycloidal where the body, by virtue of gravity, oscdlates to and fro m a cycloidal arc and the torsion pen- dulum, where the body vibrates by virtue of the torsion of a thread, or wire &c We shall here speak only of the cycloidal pendulum The cycloid AD, Fig. 302, is a curved l.ne descr^ed by a point^ of a circle APB which rolls along a straight line If this gene. rating circle has rolled forward BB, = and, therefore, come into thp nosition AB it has then also revolved through the arc Ap = jl pP= BB = PP , consequently the ordinate corresponding to any absciss MP,'= ordinate MP of the circle plus the arc of revolution AP. In this rolling the generating circle revolves about the point of contact at each instant with the base line, if, therefore, it be in it will then revolve about Bv and describe thereb) the elementary arc PlQl of the cycloid; consequently the chord B will be the direc- tion of the normal, and the chord JixPx that of the tangent to the cycloid at the point Px. The prolongation PQ of the chord AP reach- ing to the ordinate OQY is, therefore, equaltothe element of the cycloid PjQj, as further the space of revolution is equal to the space RQ 0fCYCL01DAL PENDULUM. 271 progression, PQ is then the base line of an isosceles triangle PRQ, and equal to double the line PJV*, which the perpendicular RN cuts off, but PN is the difference of the two contiguous chords AP> AR, and consequently the eleraent of the cycloid PXQX = twice the dif- ference of the chords (AR—AP). As the continuous elementary ares make up together the whole arc APX> and likewise the aggregate of the differences of the chords, the whole chord AP, the length of the cycloidal arc APiy is, from this, equal to double the chord of the circle AP, appertaining to it. To the semi-cycloid APXD> corresponds the diameter as a chord of a circle, hence the length of the half of the cycloid is equal to double the diameter of the generating circle. § 249. Cycloidal Pendulum.—From the above known properties of the cycloid, the theory of the cycloidal pendulum, or the formula for the time of oscillation of a body vibrat- ing in a cycloidal arc, may be easily developed. Let AKM9 Fig. 303, be the half of the cycloidal arc in which a body ascends or descends, or oscillates, and ME the generating circle, therefore, CE = CM = r its radius. If the body has described the arc AG, it has, therefore, fallen from the height DII = x (§ 246), it has then acquired the velocity v = \/ 2 gx> with which it describes the elementary arc CK in the time * GK GK J Fig. 303. v y/ 2 gx GK But from the similarity of the triangles GLK and —— = KL ™ or as FJiF= MH. ME, ^ = \/_MH . ME_ x^ME ^ 'KL ^ MH’ the similarity of the triangles JYPQ and , — orsince PQ JVH* = MH. DH, — = ______. PQ v/ MH. DH it follows by division: GK ,/~ME s/MH.DH x/ ME . DH NP OM ’ Now KL == PQ, hence or, since ON is half the height of descent = ME = 2r, and DH = x: GK x/ 2 rx2 s/ 2 rx MP h272 CYCLOIDAL PENDULUM. By puttintr GK = 2 ^ 2 — . JYP into the formula * = — . b h ^2^ we obtain: ______ __ _ 2_S_2rxjyp = £ lr ^jyp. v/ 2 gx. h The time of falling from A to M is the sum of all the values of - which are obtained; if for JVP all the particles of the semicircle 2 I r DJYMbe successively substituted, therefore, = times tbe semicircle DJVM($,h)’ In this raanner we obtain the time for falling through the arc ___ -i» ing the ai time of dei = 2 H I - = * I —• \ g \ '• ^ e ^ .. and as the time for ascending the arc is likewise as great, the time of oscillation, or the time o£describing_the whole arc I /t. I As this quantity is quite independent of the length of the arc, it follows that, mathematically speaking, the times of oscillation for all ares of one and the same cycloid are equal, the oscillations of the cycloidal pendulum are, therefore, perfectly isochronous. If we com- pare this formula with the formula for the time of oscillation of a cir- cular pendulum, it follows that the times of oscillation for both kinds of pendulums are equal, if the length of the circular pendulum }s equal to four times the radius of the generating circle of the cycloidal pendulum. Remark 1. It may be proved by the higher calculus that the cycloid has, besides the property of isochromsm or tautochronism, also that of brachistochronism, which is that line between two given points in which a body falis in the shortest time from one point to the other. Remark 2. In order to make a body, suspended to a perfectly flexible thread, vibrate in a cy- cloidal arc, and thereby represent the cycloidal pendulum, we suspend the body between two cy- cloidal ares CO and COu Fig. 304, so that the thread for every deviation unwinds from the one arc and winds round the other. By this windine and unwinding of the thread COP, its extremity P describes a curve similar to the given cycloid and it may be similarly represented that the evo-’ lute of the cycloid is a similar cycloid in an inverse position. As the length of half the cycloid COjJ CZ) = 2 AB, we have likewise the arc = to the straight line evolved OP; but the arc OA = o chord AF= 2 GO, hence also PG = GO^jjp and HN=zAE. If now we describe upon DH a semicircle DKH and draw the ordinate iVP, we then have KH = PG - and hence also PK = GH = AH—4G = AH—FO =- arc AFB — arc AF = arc BF = arc DK; and lastly, the ordinate NP = the ordinate NK of the circle, plus the corresponding arc DK; therefore NP is the ordinate of a cycloid, and DPA the cycloid corresponding to the generating circle DKH.COMPOUND PENDULUM. 273 For the application of the cycloidal pendulum to clocks, see Jahrbilcher Institutes in Wien., vol. xx. art. 2. § 250. Compound Pendulum.—To find the time of oscillation of the compound pendulum, or that of any other body Fig. 305, oscillating about a horizontal axis C, let us first seek the centre of oscillation, i. e. that point K of the body, which if it oscillates of itself about C, or forms a mathematical pendulum, has the same time of oscillation as the whole body. It is easily seen from this explanation that there are several centres of oscillation in a body, but in general, that point only is meant which lies with the centre of gravity, in one and the same perpendicular to the axis of revolution. From the variable angle of deviation KCF = t, the accelerating force of the isolated point K, = g sin. $>, because we may suppose that it slides down an inclined plane of the inclination KHR = KCF. But if My2 be the moment of inertia of the entire body or set of bodies AB, Ms will be its statical moment, i. e. the product of the mass, and the distance CS= s of its centre of gravity S from the axis of revolution C, and r the distance CK of the centre of oscillation K from the axis of revolution, or the length of the simple pendulum which vibrates isochronously with the material pendulum AB, we have then the mass reduced to K = and the force of revolution reduced to r2 Fig. 305. this i M g sin. ;consequently the accelerating force = = r mass - Mg sin.f-i- . g sin. $>. That this pendulum may have the same time of oscillation as a mathematical one, it is requisite that both should have their motion in every position equally accelerated, that therefore, . g sin. $ = g sin. . Now this equation gives: r __ My2 __ moment of inertia Ms statical moment * We, therefore, find the distance of the centre of oscillation from the centre of gyration, or the length of the simple pendulum, which has a time of oscillation equal to that of the compound one, if we divide the moment of inertia of the compound pendulum by its statical moment. If we substitute this value in the formula / = n we for the time of oscillation of the compound pendulum the formula 1 = = * Jfs’ 0r m0re accurate1^ = i1 + &) Inversely, the moment of inertia may be found from the time of oscil- lation of a suspended body, if we put: My 2 = \ • Mgs, or = gs-COMPOUND PENDULUM. ExampUt,— 1. For a uniform prismatic rod Fjg. 306, whose centra of oseillation is distant CA = /, and CB = I, from the «Mmitiea A an,l B, we have (§ 219) for the moment of inertia: My -it (t* + ^s) and the statical moment A/s = J F ^ Vrnnm^v with T °f lhe mathemaucal pendulum which vibrates isochronously wnh this rod is , /,»+V _ r+3 ±t£L r ~ i G/ 4- iC (/ + f) , . . n n« ii,s the bob 1,5 Ibs., further, the length of the rod If, now, the wire is °>° ’ . j l5 inches, we then have the distance of 1 foot, and the radius of t 'iulum from the axis of rotation : ,he centre of o£.Ua ion of [P°n 2,4 + 260,177 _ 262,577 m +0.05.1»+l.S^^l±ir----------------- 0,3+ 19,725 “ lo^T Wk r=“ ^053+ !>5'13)15 ’ 262,577 ’ mm = 13 112 inches. Disregarding the rod, r would - — = 13,312 , U heine reduced to its centre, r would =13,15 inches; and the inert mass £ £**£ inches. Ti.e_t.me o ** _____ ^ ^ ^77^6“. = 0,5874 seconds. ^ g f cicnension and centre of oseillation of a ma- § 251. The centre of s P . g the one may be interchanged terial pendulum are r.e.c‘P tupr ’an’a the pendulum may be suspended •with the > oseillation, without the time of oscil- - “ f,!<“ proof of this propositio,, ‘r bo ^n bySd of § 2n> followiog m,„. y 1/T he the moment of inertia of the cora- ner'nrl nendulum J1B, Fig. 308, oscillating about pound P ity s, we have then for a revolu- + «*« 1 «is b, distant CS - s from the cen- tion abou f = T + JWs2, hence the distance ofthe centre of oseillation from the axis of revo- lution C: rp T + T — ,, Jlfs U A- ionce KS = r — S of the centre of oscil. If now we represent the distanc we then obtain the sim j lation from the centre of gravity oy p i equation _ L, in »hich s and s, app*»' i” » ”>“"«• >”d hence may be snTstituted one for the other. This foranda is no, only Fig. 308.COMPOUND PENDULUM. 275 Fig. 309. true for the descent, if s represents the distance of the centre o osci - lation from the centre of gravity, but also inversely, if s expresses e distance of the centre of oscillation, and $, that of the cen- tre of gyration from the centre of gravity, and C will there- fore serve for the centre of oscillation if K serve for the centre of suspension. We avail ourselves of this property in the so-called convertible pendulum AB, Fig. 309, first proposed by Bohnenberger, and afterwards applied by Kater, which is furnished witli two knife edges C and K, which are so situated with regard to each other, that the times of oscillation remain the same whether the pendulum oscillates about one or the other axis. In order that the axes may not be displaced with regard to each other, two sliding weights P and Q are applied, the smallest of which is attached by a fine screw. If by the shifting or adjust- ment of these weights, the time of oscillation comes to be the same, the pendulum may be suspended at C or at K, we shall then obtain in the distance CK of the two edges, the length r of the simple pendulum which vibrates synchronously with the convertible pendulum, and we shall now obtain the time of oscillation by the formula t = * jJL. \ g § 252. The swinging or rocking of a body with cylindrical base may be compared with the oscillations of a pendulum. This rocking, like every other rolling motion, is composed of a progressive and a rotary motion, but it may be assumed that it consists of a simple rotary motion with a varia- ble axis of rotation. This axis of rota- tion is the point of support P, by which the vibrating body ABC, Fig. 310, rests on the horizontal base HR. If CD = CP is the radius of the rolling base ADB = r, and the distance CS of the centre of gravity of the entire body from the centre C of this base = s, we have then for the distance corresponding to the angle of rotation SCP = $, SP = y °£ the centre of gravity from the centre of gyration: y2 = r2 -f s2 — 2 r s cos. = (r — s)% + 4 r s ^ sin. ^ ; hence, if further we represent the moment of inertia of the entire body about the centre of gravity S by Mk2, we shall obtain the moment of inertia about the point of support P: T= M(Ar* + tf) = M[/fcJ + (r— +4 (sin. which for small angles of vibration may be put = M [k2 + (r — s)2 + r s 2],COMPOUND PENDULUM. 27o , _ fV»p moinent of torcc = G , or only M [F+ (r — ^l- Since" lfolloWS that the angular ac SN = Mg . CS sin. ♦ = ■«* * F**’ celeration for the rotation ^ t _g£«7M_ . , = = sPF31 moment ot ine «n. * represent is length* «^ -_££? .. * - F+C^)3 ,r‘the balance is from this: The time of the vibra u —s)2 t = t*I—^ == * \ Ss , of olso be applied to a pendulum Fig. This theory may • lrotation if for the radius 311, with a rounded axis substituted. If, instead of of curvature of ™ ««D be applied, the time of the .rounded axis, a & vibration will then e ■ft^2 = rt i**+ {s tx = H 1---~J)7f~ \ 8 (S x) „.;„f f e edge from the centre of the round the distance tu m Both pendulums have equal axis being represented Dy x times of vibration if ^ _ fc»+(s—x)* , or — — x = — 2 r. s—X If we write shall then obtain x Vis. 312. 5 F x approximately, and neglect r2, we 2rs3 F-F S—mK , , In the Second Part, under the article “Regulator,” the Jtemark 1. 1m, mentioned. conical Pf"dul"l" ic Pendulum.— Bodies orelikewise very often set Reniark 2. IM* elasticity. A string, or fine wire, AB, Fig. into vibrawry motwn byela ^ y = ^ If this ight u ^ 312, is stretched J C to A the strwg is thereby stretched from the P0UJ\f h weight be afterwards left to itself, it will, by CD r5 uand “ t^itv of the string, be raised again to C; it will virtue of the e asu j yelocity> and a9Cend by its ru viva to £ arrive there with -n fau to D and C. In this manner the from whence it wi time in the space DE = 2 CD = 2 r weight wiU osciiiai n now i9j a9 to its duration of oscillation. to and fro, and m '4 ; tran9verse section F and modulus of From thew^fthestring, it follows, § 183, that the force to extend elasticity 0 p___* hence, the mechanical effect it a length CM = * 19 r - / i ;t the length L is = — = -r • — F E. Let required to extend it me n n l n . -lvr___ 1, ?H, &c., and add the cor- usnowputsuccessively*- n ’ *’ 1 oniral effects, we shall then obtain the whole 3S for the extension of the string.COMPOUND PENDULUM. 277 CD -r:X = ^.F£(i + 3!l+...)-5iF£(1+2 + --4'") n/ \n n ' t* _,r If now, f2 wi =* -— BB. — =s — . BB: and for the extension Cia = x . — n*/ 9 2/ * 2/ inversely, the string be contracted by DJtf, therefore the weight D ascend from D to M, /-a___jrS\ i. e. r — x} it will give the mechanical eifect L — X» = y———j FE, and communicate to the weight G a velocity v corresponding to the vis viva v9 M = — . G \ whence we *hall have to put — G = f9* ^ FE, and the variable velocity of oscillation will be 2 g \ 21 / V = / ^/r*___But now v/rrf—r* may be put equal to the ordinate MK = y of \l MI ----- I FE a semicircle described upon DE; hence it follows, more simply, that v = J— . y, and tLe instant for describing the particle of space MN: r = —— . Jjrg- From the simi: * ru EM • ilfiV y MN KL • ‘arity of tbe triangles KLH and KCM, — = »• <-^- = 7’or — “— hence. i» follows that r = — /— : and lastly, the whole time of oscillation, or the r «J-F-E ___ «me of describing the space D£ is : I = i l-^i times the sum of ali the elements r V BB the semicircle = _1 Ml times the semicircle FE irr / Af/ /—■ 7'JfE~~ JfE s/g +JFE Tr r VBB ^ « *» lor example, an iron wire, 20 feet long and 0,1 tnch thick, be stretched by a weight G = 100 Ibs. and set into Iongitudinal vibration, the duration of tne osciilations will then be, since from § 186 E = 26000000, Fig. 313. J 100.20 (0,1)*.«J2600 26000000 s 0,553 1-1-=0; 65 . w 0,05464 seconds. n/ oo . «r * i -r « Remark 3. We have a torsion pendulum it string or wire CD, Fig. 313, turns about an arm AB and is brought out of its natural position into the position AB, and then left to itself. 1 «e rod or arm AB is set into vibration by virtue ol the torsion of the string, which extends to an enu** distance on both sides of MN, so that AM = AtM. If we put the force of torsion for the distance (1) and for the arc of vibration (1) = K, it will then be, for the angle of vibration MCP — ♦ > “ A *3 jyf * 9 of velocity of tke mass Mx in a given finite time *x+ *a+*3 + ... = ^p ^ ..) -^-y as also the consequent change of velocity of the mass 5of the magnitude M%: *** +• .) In the following or impinging body Jly the pressure acts opposite to the velocity ev consequently here a decrement of velocity takes place; and after a certain time, the residuary velocity of the body iS; * ex — (Px + P2 + . 0 i in the Preceding or impinged body jg on the other hand, the pressure acts in the direction of motion- hence there is an increment of velocity c3, and^it is converted into ,0, + ^ +*,+ ••) ' 'M If w e eliminate from both equations (P, + Pa+ ..)*, there will then remain the general formula: M M T jtf fc_« ■)= JML (».—Ct), or 3f,r, + =■ JlljC, + The productof the mass and velocity of a body is called the mo. mentum of the body, and it may therefore be enunciated, thatforeack instant of the time of impact, the aggregate of the momenta of the two bodies is as great as before impact. _ , , At the instant of maximum compression, both bodies have an equal velocity; hence, instead of and v, we may put this value into the equation found; then Mxv+M%v will remain = Mxcx+Maca, and the velocity1 of the two bodies at the instant of maximum compression willbe: Mtcx + Mact v—mjmt: If the two bodies A and B are inelastic, they exert therefore no power after compression to re-expand themselves and the communi- cationof a change of motion will then cease, if both bodies are com- pressed to a maximum; and hence the two will go on after impact with a common velocity: Af.c.+Af/, Exempla,—1. An inelastic body B of 30 lbs. weight, moves with a 3 feet velocity, and is struck by another inelastic body A having a 7 feet velocity, the two will then proceed, after the blow, with the velocity v as 30.7+30,3 3509011«5J feet—2. To cause a body of 120 lhs 60+30 80 b *ELASTIC IMPACT. 281 weight to pass from a velocity r2 = lj feet into a 2 feet velocity t?, it is struck by a body of 50 ibs. weight, what velocity will the body acquire? Here (t>—c2) 2 i (2—1,5) . 120 = 2 4- — = 3,2 feet. ^50 * Mx § 256. Elastic Impact.—If the impinging bodies areperfectly elas- tic, they will then expand themselves after compression in the first period, gradually again in the second period of the time of impact, and when they have resumed the former shape, they will proceed in their motions with different velocities. But, since the mechanical effect wrhich is expended on the compression of an elastic body is equal to the effect which the same gives out again by its expansion, no loss in vis viva will take place from the collision of elastic bodies, and hence the second following equation will be also trne for this case: IL Mtv*-f M2v22*= M^*+M2c22, or MAc2-v2)^M2{v2-c2). From the equations I. and II., the velocities v1 and v2 of the bodies after impact may be found. First, it follows by division that C 2_« 2 2_c 2 A-----L — —ZSLf i, e. c. + vl = v1+ ct,or»,—v1 — c2—c,; if now Cl~Vl v—c? . . _ we put the resulting value of v2 = ct + vi—c,, into the equation I: M1vl+Mivl+Mt(c1—c2) =.JMlc1+JWlc2, or, c,—2 —c2), from which we have the value: 2 M. g*■(<,-___________2.W,(C-C,) 1 2 1 M a M ““ M1+Mi M1+Mi Whilst for inelastic bodies the loss in velocity of the one body is C—V = c, — Mic*+M*c* = Jtf.+Jlf, M2+Mt * for elastic bodies it comes out twice as great, namely: 1 1 Mt+Mt ’ and while the gain in velocity of the other body for inelastic bodies is: ______________r __ JU1c1-\~M2c2__ M1 (ct c2) * Mi+Mi Ci Mt+M^ for elastic bodies it is v.—c9 = likewise twice as great. 8 Ml+M2 Example. Two perfectly elastic spheres, the one of 10 Ibs. the other of 16 lbs. weight, impinge with the velocities 12 and 6 feet against each other, what will be their velocities after impact ? Here M, = 10 and c, = 12 feet, but M2 = 16 and r9 = —6 feet, hence the loss of velocity of the first body will be C, - „ ll16 (12+ 6) _ 2 -16 -,18,22,154 feet, 10 + 16 26 and the gain in velocity of the other- v. — c. B 2'10'18 s 13,846 feet. From this 8 26 the first body after impact will recoil with the velocity r, -* 12 — 22,154 = — 10,154 leet; and the other with thfkt of — 6+ 13,846 mm 7,846 feet. Moreover,the measure of 24*282 PARTICULAR CASES. vis viva of the two bodies afler impact * 4" ® ^ • i?*1?4, 7’®4^ = 1031 + 985 = 2016, as likewise of that before impact, namely: Af,c, -+- -«W =« io. i2* 4- 16.6*= 1440 4- 576 = 2016. Were these bodies inelastic, the first would only ]08e in velocity Ct~"ri = 11,077 feet, and the other gain — c% 6,923 feet; the first be velocity 12—n,u w =0,923 feet, and the second ac- 0 923, and the loss of mechanical eflfect would be (2016—. 2 ' 2 would stili retain, alter impact, the velocity 12—11,077 quire the velocity—64“ 6,923=0,923, and the lossof mt (10 4- 16) 0,923*)-*- 2g *= (2016 — 22,2) . 0,0155 *= 29,35 a lbs, § 257. Particular Cases.—The formula developed in the foregoin paragraphs, for the final velocities of impact, hold good also in case where the one body is at rest, or where both bodies move opposH to each other, or where the mass of the one is indefinitely great co ^ £ared with the other. If the mass M% be at rest, we then have c ^ ence for the inelastic body MlCl 0, 0 + Jtf, + M» 2 «AfjCj ' M, + Mt 2 Mlcl and for the elastic: Mt — Ms M. + M, 2 M, ‘2 c„ and ci* M. + M, + M, If the bodies meet, c, is therefore negative, and for an inelastic , . Jtf, c, an(j for an eiastjc one body it will follow that » = ss Cj •— i/ * jyf , • momenta be equal, M3c3 - M3c3, for the inelastic If in this case the m n 4 each other tQ rest> b(lt for body then y = 0, *• e- U1C u elastic bodies: g ^ + Jtf.c,) _ c 2 e, - — c„ and v- = Cl--------^7+ M3 ^ 2 (M,ct — — — c, + 2 c2 — + c ; »s = — «» + -ptfT+ M, , , j- An„na after impact with opposite velocities. If on tZttr KdSthe^iasses are equal, we have then for inelastic bodies c, - c and for elastic v3 = c„ t. e. the masses V = z rebound with their velocities interchanged. If the masses again meet in the same direction, and if the prec H’ mass M% be indefinitely great, we shall then have for inelastic bc^^ M%c% c%y and for elastic vx 2 (c2 c2) = 2 ____ _ - * " ci> M* O r iUa velocity therefore of the indefinitely great mass v* " c» + cVuhe.Je^ollision of the finite mass. If, now, the in- not be nltered b, th<“fore, c _ 0, we ehell then hnve defimtely greatmass be^nt te^‘“5^ _ _ e„ e, _ 0, the for inelastic bodies v » remain at rest, but m the first case, indefinitely great mass jose its velocity, and in the second the lmpingmg body will entireiy io* j case this will be converted into an oppLOSS OF MECHANICAL EFFECT. 283 2 M% 1 n Examples.—1. With what velocity must abody of 8 lbs. impinge against another atrest of 25 lbs., in order that the last may have a velocity of 2 feet ? Were the bodies ine‘asUc> we should then have to put: v == —^-1—, i. e. 2 = hence c, = —- = ^eet> M, + M2 8 +25 2 Af c, ^ the required velocity; butwere they elastic, we should have r2 = 1 5 hence,cx Az, iw2 33 Y = 4 i feet.—2. If a sphere Mt, Fig. pig# 318. 318, strike against a mass at rest M2 = n^i with the velocity c,, the second, a ^hird mass Mz = n M2 = n3 Ml with the velocity communicated by the im- pact, thisagain another mass MA =nMa = w3 &c., we shall have from the perfect elasticity of these masses, the velocity t, _ 2 Mt _ 2 _ ________ Af, +nAf, 1 1 + n 3 Af2 + n M2 vA = ^ 2 y ^ ^ jf> for example, the weight of each mass be half as great as that of the succeeding one, and we have therefore the exponents of the geometrical series formed by the masses: n = J, it will follow that •* = J *»’* - (j)' C» ^ •’ (i) C' = 13,32 • § 258. Loss of Mechanica1 Effect.—In the collision of inelastic lasses, a loss of vis viva constantly ensues, whence the masses afler impact have not the power of producing so much mechanical effect, as before impact. Before impact the masses Mx and M2 proceeding 'with the velocities cx and c2, contain the vis viva, Myc2 + M2c22, but afler impact the masses proceeding with the velocity v = have the vis viva M.v* + M2v7 ; hence the subtraction Mi + M2 °f these forces will give the loss in vis viva by the collision: K = Mx (ci2 — v2) + M2 (c22 — v2) = Mx (cx + v) (Cj — v) — M2 (c2 + v) - ca),but Mt) Cl—v) = Mt (v — c2) = » hence K ~r (C1 + v—c2—v) .^LMAc-c2) (Cr-C^T M1 + M2 “ Jtf, + Jfcf2 J_ , JL' If the weight of the masses are Gx and G2, Mx is, therefore, = G P ^pand we shall from this have the loss in mechanical effect: L = (-C' Ca)2 . . Wp pq]1 G2 Gl+ G2 G1+ G3 T-l— the harmonic mean Of G, ,„dV amTW maj from this °^*Xu2X™^ effect which is produced by the impact of wo i uivalent to the 'which is expended upon the disfigurement o e > height offoll product of the harmonic mean ofboth masses, andoj which is due to the difference of the velocities oj shall have the If one of the masses, for example M2, be a r ,284 PILE DRIVING. loss in mechanical effect L = -w~> ant^ mass m°ved 2g* G1 + ^2 M, be very great in coroparison with the one at rest, G, will vanish as compared with G2, and there wili remain !(J= Jf.Tc,' — (ei‘ — »’) ” M'$.r 2ci” + »’ + 2 c,„ = JW1 >c‘ _ „)* + ‘JM, (c, — »)*, because JW, (c — v) = .W2 (» __ Cj). From this therefbre, the vis viva lost hy melastu impacts is equiva- lent to the silm of the products of the masses and tfu: squares oftheir loss ot gain in velocity. Exampla.—1. Ifin a machine, 16 blows per minute talte place between two inelastic bodies M, = lbs. and M, = — lbs., with the velocities c, = 5 feet, and r1==2 feet, then the lofs in mechanical effect from these blows will be: 16 (5—2)» 1000 ■ 1200 _ 4 .9 .0.0155 £222 = 0,558 . i22- = 20 oo 60* 2/r * 2200 15 11 11 ft. lbs. per second.—2. If two trains upon a railroad of 120000 lbs. and 160000 lbs. Weight come into collision with the velocities c, = 20, and — 15 feet, there wiU ensue a loss’ ot mechanical effect expended upon the destruct.on of the locomotives and carriage8 which, in the case of perfect inelasticitjr of the impmgmgparts, w.ll amount to (20 + 15)» 120000 ■ 160000 _ 35, 0,0155 1220000 _ 13020(X) ft ,bg =-------^ 280000 28 S 259 Pile Driving.—The effects of impact are very often appliej to ram or drive one body B, Fig. 319, into another E, a soft mass, for instance. If the resistance which the latter mass opposes to the pene- tration of the former be constant and = P, and the depth of penetra* tion bv one blow = s, a mechanical effect will be then expended. If, on the other hand, this resistance at the commeneement be = 0, and if it increase si- multaneously with the depth of penetration, so that at the end, after the body has penetrated the second mass a depth s, it be = P, the me- chanical effect expended will be then only (0+P) s _ ^ pIf; lastly, the initial resist Z ’ ’ • y Fig. 319. ance be = P„ and and increase si multaneously with the space, ‘so that, after describing a space . . r, ch„u then have to put the mechanical effect = sy it becomes P2, we shall tnen n i (p± ±p>) If Ihe body B, whose mass may be begins with the velocity „ , ” 2 .. . • veiocity of penetration increase, it to penetrate a mass, and it tms veiu j ,r , will, in virtue of its vis viva, have produced the mechanical effect ^ = |L G,if G = Mgrepresent its weight. z zg , j KPILE DRIVING. 285 v2 When the resistance is constant, we must put: = —- G; on the 2 other hand, when the resistance beginning from nought gradually increases: Ps = — . 2 G . :and when it increases gradually from % PX to P> -(P, + PJ « - £ . 2 G. • The initial veloeity v is generated if a third mass Ay whose magni- tude may be = Mx and weight = Gv be allowed to impinge upon the second mass P, with a certain veloeity c. If, now, these masses nre inelastic, we then have the veloeity with which the two proceed after impact, and begin to penetrate the mass E: Mxc __ Gxc v M + Mx G + G, In the drivingof a pile or post, Fig. 320, B consists of a pile shod with iron, and A of a e*v.Y body falling from a certain height, called a ram, or block of iron. If — « 11 i Fig. 320. k-( ( G' \2 ( G. V \G + Gr ) ‘2g \G + G/ H, uie met the veloeity v - (A)' GH, GH; °r,2.Ps = -|£|-, \ vr T and that of the pile and ram together Hnt if the resistance of the bed of earth be constant, the mechanical effect expended in tbe penetration of the pile will be = Ps, hence 'we shall have to put: l.Ps—( Gi V A „ \o + oJ - ■ - , . , .. the first if the ram does not, during penetration, remain upon the pile, and the second if both go down together. The weight G + Gx produces, in penetrating, the mechanical effect (G + GJ s, we may then more correctly put: (P—G—Gx) s -S1 ; but G + Gt is small compared with P, and may gene- rally be neglected. Hence, were the impact perfectly elastic, we should have to put: Ps = ( JlG1 V . GII. Were G small, compared with G„ as for * \G Gj/ . n TJ instance, in the driving of a nail, we should have either rs or Ps = 4 GH.286 PILE DR1VING. . . . , . „ uv |nst round of 20 blows of a Example. A pile of 400 lbs. weight is n(jeepelr) what resistance wil| 700 lbs. ram, falling from a height of S leeb witliout penetrating deeper? *• •*»* °ff«< 0r What ,0Bd W‘U t P" . k- 0,025 feet, wbereby it U Here G = 400, G, = 700 lbs., H = 5, and - 2Q far for each blow. From the flrst formula supposed thaMhe goooo = 32400 lbs.; and from the second: ^== \700 -j- 400/ 0,025 700*. 5 4900 200 = 89100 lbs. ii-r. — •” 1,1 ” A °r 'u“ sttength' § 260. The formulae found above are applicabfe to the breaking 0f bodies by descending weights or balls. Let Fig. 321, be a prismatic body of the mass M, or weight G = Mg, supported at its extremities, which is bent a depth CD = by a weight G, falling from a height jJD — upon its middle, and in this manner broken; the conditions under which this is possible are to be determined. From § 190, the deflexion, or the P/3 Fig. 321 — height of the arc, is given _____ 48IPE» from the pressure P in the middle of the beam, and from its length BB = l and if further its moment of flexure HL is known; there- fore, inversely, the pressure corresponding to a certam deflexion s iS: p _ 48 WEspresSure however is not constant, but increases simultaneously with s, hence the mechanical effect expended in the de- flexion by a depth s, not - Ps, but only J P>, i. e. i . = This mechanical effect may now be equated to that which the falling body communicates to the beam. S.nce the beam rests on its ex- tremities, we must (from § 219) consider only the third part of its mass as inert, and hence put for this mechanmal effect: u g + gJ y + Gi The flrst, if the weight G, flies back after its descent, and the second if it remains on the beam during the fract,’re- If we suppose a rectangular beam of the depth h, and breadth b, we shall then have to put: W = 72 ’ an(^ —j- hence pi = y . bhK2 • accordingly, the mechanical effect for the rupture w t*HARDNESS. 287 of the beam will be G, 51# P1 P _ ~ We 1. GH gj- v 4G + gJ WE bhlK2 6 E , and we may now put: , or 2. G*H _ bhlK2 i G + Gt “ 18 £ ' Example.—From what height must an iron weight G, of 100 lbs. be allowed to feli to reak a cast-iron piate, 36 inches long, 12 inches broad, and 3 inches thick, in itsmiddle? i q^vvJ10^11^8 eIasticity °f cast-iron E = 17000000, and the modulus of strength K = iy000, hence it folio ws that: bhlK1 __ 12 . 3.36 . 19000* _ 216 . 19* _ 216.361 _ ^0? T 6 £ = 6.17000000 17 ““ 17 “ ° 1 now a cubic inch of cast iron weighs 0.275* lbs., the weight of a piate G will then be 558 12 • 3.36.0,275 = 1296.0,275 = 356,4 lbs.; hence: G ( \ = 356,4 . \ = 74,44; on the other hand, VJG+O,/ \218,8/ 12^^.= 45,70. Hence the height of fall required is: 4587 JG+G, 218,8 H = 4587 = 61,6 inches, or H 74,44 3.45,7 : 33,5 inches. Fig. 322. § 261. Hardness.—When the modulus of eIasticity of the impmg- ing bodies is known, we may then find the force of compression and its amount. Let the transverse sections of the bodies A and B, Fig. 322, be Fj and P2, the lengths lx and /2, and the moduli of eIasticity Ex and E3. If both impinge against each other with a force P, the compressions effected will be from § 183^ Pl. ■ = -JT-F-» and x, = FiEi and their ratio: Pt* f3e; • /3 T/I f, t Jpl JLt 11 tor simplicity we represent by H> we obtain xx = —, and l H. x P r/ 2 ~~ tt» as well as = _2. If, afler the example of Whewell,f h; ~ ...........*2" h; we call the quantity ^ the hardness of a body, it follows that the depth of compression is inversely proportional to the hardnes If a mass M = — impinges with the velocity c upon an ble or indefinitely great mass, it then expends its 'who J 8 _ JVfc2 c* q But now the upon the compression, hence $ Ps = ^ ^ space s is equal to the aggregate of the compressions xx a 2 * A nearer statement of its weight is, 0,2604 lbs. avoirdup t The Mechanics of Engineering, § 207.288 ELASTIC AND INELASTIC IMPACT. p P . and x, a —; hence it follows that ", Ht as, inversely, P ■ HA Hy+H% • 2g «A Hl+Ha G, therefore, s, and the equation of condition £ JHi + H, G g from which P, x, and x, may be calculated. ExatmpU. If a wrought-iron hammer, of 4 aquare inches boae and 6 inch strikes with a velocity of 50 feet upon a piate of lead, of 2 aquare inches hase and*i • thick, the following relations preaent themselves. The modulus of elasticitv 1 *nch iron is Ex *= 29000000, and that of lead E% » 700000: hence, the hardnL. ^^ght bodies is: H, = OjEl a t-il^P00000 B 19333333, and H, ^ 2 •700000*® ‘i 8 /, ~ j-----ss 1400000. If we put these values into the formula t » c + H* G sl^fJC'7’ 8ub*ti, tute for the weight of the bammer =«4.6.0,29 wm 7 lbs.; therefore, SL ^ 7 0,217, we shall then obtain the apace of the hammer in the compression: ^ / 20733333.0,224 / 0,46443 . t ss 50 /---------------- s 50 /----------= 0,0207 inches ss 0 249 19333333.1400000 V 2706666 lines. pr< thia the force of impact or pressure follows: jy, jy9 19333333.1400000 orn 0,0207 as 27037 lbs.j further, the 00,11 pres* P ** wTirfr • * 20733333 . .. P ^ 27031_ * 0,0014 inches « 0,016 lines, and that sion of the hammer18 * 19333333 at P 27037 ^ 0,0193 inches *« 0,233 lines, of the leaden piate: iaawwi 1400000 § 262. Elastic and Inelastic Impact.—If two.masses M and M move with the velocities c, and c,, the common velocity ofShe t * at the moment of maximum compression will then be fronj § 256 M^+MjCj an(j the mechanica] effect expended on the „ M^Ma tne COfa- pression from § 259: t _(c-c>Y A A (£rt)* Gi G> 2 + 2g 'G.+ g; This mechanical effect may be also put: - i Ps - \ /»(*, + S) - i • consequently the sum of the compressions of both masses will be- 7, Hl + H2 G2) Ht Ht from which the compressing force P, compressions of the separate masses x. and xa, may be found. If the masses are inelastic, these compressions will remain after impact, but if only one of the two bodies be inelastic, the oth^r willELASTIC AND INELASTIC IMPACT. 289 again recover its form in the second period, and produce a mechani- cal effect which will generate a new change of velocity. If, for ex- ample, Mx = be elastic, the mechanical effect in this second period „f impa« i A, - i ■ % - jL (j^)V - ^ ("* C* TT 1 . . • —*_ .________?__ will be given out; hence we shall have in this <*i + G2 Ht + H3 B case for the velocities v1 and v2 a fler impact the formula: M1v1 + M2v2 = + M2ca, and + 3^5 • Tf-TW' M M. , « Jlf, Jtf, /£ +(c* Cj) • jif, • 1 M M H i- e. • 3^+%; If the loss of velocity —tq be put = x, we shall then have the gain of velocity v2—c2 = and the last equation will assume the JVlq, form: x*—2 (c.-c,) x+ty.-c,)* . ” " = 0, or, JVf. Jtf, + JM, * Ht + Ht and * — be put If this be multiplied by +~j^ “uu h, + H2 = 1 _ Ht we shall then obtain the quadratic equation: (X—(C—C*)M +* jtf) “ (Cl^Cl)a (jtf, V-w) ‘ Bt + H* whose solution will give x, or the loss in velocity of the first body: C- (c-CS) i1 + ^H^Th) and the gain in velocity of the second i („,_C, - C'-*J H7+TT, (‘+ -Jif[TTr) Example. If we assume the iron hammer in the example of Pr^tajn the loss in to be perfeetly elastic and the piate of lead inelastic, we s a since c, = 0 and velocity of the 7 lbs. hammer, descending with a 60 feet veloci y, "*“*• , , „ v / , rSuooooN c.-r.-r, (l + " 50 ( + >/20733333/ 25290 TMPERFECTLY ELASTIC IMPACT. * c _ 63 * 50 — 63: s —13 = 50(1 +0,26) a= 63 feet; hence, the velocity of the hammer afler the blow is : v, * feet. The velocity of the supported piate of lead * 0. § 263. Imperfectly Elastic Impact.—If the bodies impinging against each other are imperfectly elastic, they only partially recover their figure in the second period of the time of impact, them mva expended in compression during the first period, will not, therefore, again be completely given out. If, agam, x, and x, «re the depths of pene- tration, and P the pressure, we shall then have the loss of mechanical effect by the compression — i f\* an , */V “ dunng the expansion the Mth part of this, or generally during the expansion of the one body the *,tb, and during that of the second the ^,th be given back, there will remain the aggregate loss of mechanical effect alter impact: P p L~IP [(1 — Ml) + 0—#•») Sl or - fj> «nd S = ~, *“l «j P* f1 *** + * *** ~1 But from the former paragraph- L Ht H J ------------_ Hfts M.M. p - ~Hf+k'and s “ (c,~Ca) fL tix-\- n2 fiMx + M2 * HXH% hence the loss of the mechanical effect in question is known: ' *• TI TT /1 -I r (ci —CJ L----------r- i* H,H3 .(tu* V hx + l=£s\ y * ■ M, + Mt * H, + h3 \ H; r //; , Now, in order to find the velocities vx and v2 after impact, we h to combine them with each other and to solve the equations: ave Mxvx + M2v2 =s Mxcx + M2c2> and Mxv2 + M9v* — Mxcx% + M c * MXM^_ (l-^)ga + (l-.^fl " - r? ■—. — («i—c*)s Mt + M3' if, + i/2 In the same manner as in the former §, the loss of velocity 0f the first body is given: ct— v. !/!/«»• -____________________________________________________________ (Cx C3) + „ (1 + V if, + if3 /’ ..■i, + Jffj . and the *ain of velocity ofthe body precedtng: -------------- , Mt /, j. kif, + f,if2\ + N ff, + ff, A These two general formula: also embrace the laws of perfectly elastic and perfectly inelastic impact. If in them we put i», - m* » 1, we then obtain the formula already found above for the impact of per. fectly elastic bodies, but if we assume t*x — £2 we then obtain the formula for inelastic impact, &c. II u°th bodies have the same degree of elasticity, therefore, px =* we aave simply: c. — V, — (C,—C2) C1 + ^ )* and c. '■ (ci—c>) M. + M3 •M, Mt + M, (1 + ✓ *•).OBLIQUE IMPACT. 291 If, further, the raass M2 is at rest, and infinitely great, it then fol- lows that: cx — vx = cx (1 + x/p), i. e. vx = — Cx x/fxy as inversely, /x = • If now ^1 be allowed to fall from a height h upon a similar mass M2, and if it reascend to a height hxy we may then find from the two the co-efficient of imperfect elasticity, by the formula p = Newton has already found in this manner for ivory '8\2 h M = Q* = |4 = 0>79; for glass = = 0,9375s - 0,879; It must be Fig. 323. for cork, steel, and wool, fi = ^ = 0,5552 — 0,309. here supposed that the impinging, or striking body, is spherical, and the body impinged upon, or the support, flat. Example. What velocities will two steel plates acquire after impact, if they possessed Wore impact the velocities c, = 10 and c4 = — 6 feet, the one weighs 30 the other 40 ibs.? Here ^ -Vl * (io + 6) 12 (l -f 1) = 16. i . - = — = 14,22 feet; hence, the velo- '70 \ 9/ 7 9 9 i ,nRA cities sought are t>. = c, — 14,22 = 10 — 14,22 = — 4,22 feet, and v2 = c,+ 10,66 = — 6-f-10,66 = 4,66 fee*. § 264. Oblique Impact.—If the directions of motion SlC1 and °f two bodies A and B, Fig. 323, de- viate from the normal JVJY to the plane of contact, the impact is then oblique. We may reduce the theory °f this to that of direct impact if we Resolve the velocities S1C1 = cv and o2C2 s c2y in a normal and tangential direction; the lateral velocities in the direction of the normal NN commu- fiicate a certain impact, and hence are altered to the same amount as for centric impact; the velocities, on the other hand, parallel to the plane of contact, communicate no im- pact, and hence remain unaltered. If we join the normal velocity of a body changed in accordance with the laws of centric impact to the remaining unchanged tangential velocity, we shall obtain the resultant velocities of these bodies after impact. If we represent the angles which the directions of motion make with the normal by ax and a2, then CXSXJY = a2, and C2S2JV = a2, we shall obtain for the normal velocities and S2E2 the values cx cos. ax and c2 cos. a3, for the tangential velocities on the other hand SXFX and S2F2: cx sin. «,an p2 Sln» <*2. The first velocities suffer alteration from the effect o e impact, and the one passes into vi = Cl C0s- ». — (c, cos. Oj — c2 COS. a2) (1 + 'S?), and the second into:292 OBLIQUE IMPACT. M — V* = C* C0S- »2 + (C1 COS. a, — c2 COS. o2) ^ ‘jyO + ^V)» •Wj and Jlf2 representing the masses of the bodies. The resultant velocity Sx Gx of the first body is given by vx and c sin. olf wl = x/v * + c* sin. a*, and the velocity S2 G2 of the second body by v2 and c2 sin. »2; w2 = x/v22 + c* sin. o2*; the deviations from the normal are also given by the formula: c, sin. o, . . c sin. a2 tang.*, = ----% and tang. *2 =* —----------> V\ V2 *x representing the angle Gx Sx N and the angle G2 S2JV. Example. Two spheres of 30 and 50 Ibs. impinge against each other with the ties c, = 20 and c, = 25 feet, which deviate from the normal by the angle a VC °C** 35' and «a = 65o20', in what directions and with what velocities will the two proceed alter impact? The uniform component velocities are : ct «»... = 20 35' = 7,357 feet, and <•, nn. «,= 25.»m. f>5°2°'J(y ‘f‘(‘t^er feet. If the bodies are inelastic, then /u = 0; lience, the altered normal velocities r, = 18,598 — (18,598 — 10,433) — = 18,598 — 5,103 = 13,495feet, and tl=- ,0^ 3 + 8,165. = 10,433 + 3,062 = 13,495 feet. The resultant velocities are now • wt = ^13,495»+ 7,357’ = y/236,24 = 15,37 feet, and tra=v/13,495a4- 22,719»= ^/698,27 = 26,42 feet j and 7,357 , we have for their directions the tang. *, a log. tang. = 0,73653 ___. j 22 719 28° 36' and tang. f, = log. tang.= 0,22622, = 59° 17'. § 266. If a mass Fig, 324, strike» F‘S- 324. against anolher mass, indefinitely . or against an immovable resistance ftft’ we have c, = 0 and = oo , it then fol ’ lows that Vi = cx COS. ax — Cx COS. ax(l + ^ = — cx cos. ax and v2 = 0 + cx cos. Oj. jf» ^ ^ ^m) F X -N : -E x^jl X 00 = 0 + 0 = 0; if now, further, p = 0, vx will also = 0; but if m - V»«wil] - — c» ».*’• e. in inelastic impact, the normal velocity . , n/w nther hand, it is changed in the oppo- is entirely lost; in el°fhc’™ b ;vhich the direction of motion after site direction. hor tne angic uy impact deviates from the normal tang. *2 - c. sin. a. ci5in* ai___ = — tang. a. - 5 * ""0,005. ^ 1 1 tang. ax ^ _ for inelastic bodies the tang. * i is therefore = q — cc9 i. e. ti = 90°, and for elastic tang. t, - — • *• = — AfterOBLIQUE IMPACT. 293 the impact of an inelastic body against an inelastic resistance, the first proceeds with a tangential velocity cx sin. ax in the direction SF of the plane of contact; after the impact of an elastic body against an elastic resistance, the body proceeds with a uniform velocity in the direction SG, which lies in a plane with the normal NN and the initial direc- tion XS, and makes with the normal the same angle GSN as does the direction of motion with it before impact, but on the opposite side. The angle XSN which the direction of motion before impact makes with the normal or the vertical, is called the angle of incidence, and the angle GSN, which the direction of motion after impact makes with the same, the angle ofreflexion; and it may therefore be enun- ciated, that in perfectly elastic impact, the angles of reflexion and in- ddence lie in the same plane with the vertical, and are equal to each other. In imperfectly elastic impact, the ratio \/ of the tangents of these angles, is equal to the ratio of the velocity given back by the expansion to that of the velocity lost by the compression. By the aid of this law, the direction may be easily found in which the body./?, Fig. 325, must impinge against the immovable resistance that after impact it may pursue a cer- tain direction SY. If the impact be elastic we must let fall from a point Y of the given direction the perpendicular YO on the in- cident perpendicular NN, prolong the same until the prolongation O Yx is equal to the perpendicular : SY\ is then the direction of the impact in question, for from this construction, the angle NSYt = NSY. If the impact ^e inelastic, we may make OY1 = s/ p • OY, then Yfi will be like- _ 9L. and also Fig. 325. wise the initial direction sought, since tang. , or’ p. f a perpendicular YR be let fall upon the line SR parallel to the plane of contact, and its prolongation be made = RY, we Incidence so^ugh^ta*n> ^ ev*^en* reasons’ 1° SX the direction of the (See Theorie oblique impact hasespecml applicatiori to the game of billiards. Corioli. in qU® d®?..effet3 jeu de billard,” par Coriolis.) According to velocity to thar nf ,hln^ a bilhard-ball against the cushion, the ratio of the reflected With the assiln'nr\nK en! “ ?’5 to °’6} therefore ju = 0,5» = 0,25 to 0,6» = 0,36. must strike aimin » value, ^le direction may now be known in which a ball A point K W« i f f"8blon that il may be reflected from this towards a given • may let fall from the given point Fthe perpendicular YR to the line of vity 0f the ball parallel to the cushion, prolong it by RX= ' ' 10 10 */± asi^to ~ of its __ _ _ • ^^ ^ o n it the place where the value, and draw the straight line YtX: the point of intersectio 25*294 OBLIQUE IMPACT. tali A must be struck that it may rebound to By «he motion of rotation of the tali, this ratio will be somewhat altered. § 267. In oblique impact a friction takes Pla?* b?‘We,e" ‘he >m' pinging bodies, which changes the latera ve ^‘ determinp of the plane of contact. The friction of impact >s deermined hke ,h« of the t • ’ .1 4kof lil^p impact, it acts only durin liiat m uiv iiiuuw*» i-- co-efficient of friction, it is F = f 11 ls ulo“^ ‘ "Y *he friction of pressure in this; that hke impact, it acts only dunng a very short time. The changes of velocity produced by it are not immeasurably small, for the pressure , and consequently the partof it f P, is generally very great. If we represent the .mpmging roass by J\f.and the normal acceleration generated by the pressure P, by « we shall then have P = Mp, and hence f =/Mp as well as the retardation or the negative acceleration due to friction dunng the im. pact ” ‘ * -------♦ o« the normal pressure. Butthe Fig. 326. ion or the negauvc au^iv.----- pact — =fp;i- e./times as great as the normal pressure. But the * J\f two pressures have equal durations; hence, therefore, ) cos. a. The lateral velocity, therefore, after impact c sin. a, passes into _ _ C sin. a—fc( 1 + : tang. = —s 1,2548 j therefore, f = 51° 27', and the velocity after impact will 5,833 remain =— ___— 9,360 feet. Besides, the ball will have further a velocity oi cos. 51° 27' rotation | f. 1,65 . 10,607 = 8,220 feet about its vertical line of gravity. Since the ball moves with a rolling and not a sliding motion, we must assume that besides its progres- sive velocity c = 15 feet, it possesses an equal amount of velocity of rotation, and this may likewise be resolved into the components c cos. a = 10,607 and c sin. a = 10,606. The first component answers to a rotation about an axis parallel to the cushion, and Passes into c cos. a — f / (1 + x/r) c cos. a = 10,607 — 8,220 = 2,387 feet, the other component c sin. a = 10,607 feet answers to a rotation about an axis normal to the cushion, and remains uniform. § 267. Rotary Bodies.—If two bodies, Jl and B, capable of rotat- ing about two fixed axes G and Ky lig. 327, strike against eae other, changes of velocity ensue, which may be determined from the Flg* 327# moments of inertia Mxyx and Mjy2 °f the masses of these bodies about the fixed axes, with the assistance of the formulae already found. If the perpendiculars GH and KL, let fall from the axis of rotation upon the line of impact, are ax and a2, we then have the inert masses reduced to the points H and L where the perpendiculars meet the line of im- pact = and and if these values be substituted for M1 ai ai and M2 in the formula for free centric impact, we obtain the changes of the velocity of the points H and (§ 264) = ici‘—c2) (1 + '/r) MlVl2 ai+M^2 ■+• a2 (C1 Ci) (C1 ^2) 2 (1 + y/ /and = (c—c2) MlVl2 -i- ax MlVia2 (i + (i + ✓?). r A ■ + M^2ai . . . . ci and c2 representing the velocity of these points before impact. But it we introduce the angular velocities, and represent these be-296 ROTARY BOD1ES. fore impact by f. and »s, and after impact by and we shall have to put c, = a\ ,,, c? = «, fj, and shall obtain for the loss in angular velocity of the impinging body: * _____n + s//*)> - «1 («1 *l «2 '*) My2a* + and for the body irapinged upon, the gain of the same: , „ \____________________(1 + s/Z). = «2 (fli ‘i a* '*) Mj/fa' + consequently the angular velocities themselves^after impact, will be: = '1—«i (a> *i—(1 + ^+ Vy/a,* and “2 = + a* (ai “2 (1 + ^ If both bodies are perfectly elastic, we shall have m = 1, therefore 1 + = 2, and if inelastic, /. = 0, therefore 1 + vV — 1. ln the latter case, the loss of w viva produced by impact , „ a, Jtf.y,2. - K «1 a2 *2l • MjW + M^X2' Example. The armed axle jJG, J?\fT 328, has the moment of inertia about 'f axis of rotation G, = M\ y? = 40000 g, and the tilt hammer BK one about it* axis K, = 150000 g; the arm GC 0f the axle is 2 feet, and the arm KC of the hammer 6 feet, and the angular velocity ^ of the axle at the inornent of impact on -----V-X the hammer == 1,05 feet; what is this velocity after impact, and what effect is lost at each blow, if tl.ere is an entire absence of elasticity ? The angular velocity of Ure axle sought is: / An \ _r< 4 • b05 • 1500<^-= 1,105 l _ = 1,05.0,706 _ 1,05 — tt^7i -irt 4- 150000 7* _V 204/ Fig. 328. 40000.36 + 10UUUU ‘2.6. 1,05.4 = 0 247 feet> it only Qne- P41 feet, and that of the hamnter _ _ _h H„w ___0^1 feet, and that ol me 204 " \ •, The loss ofmechanical effect by each blow ,s: third that of ‘he aile. 1 05 j 50000 = 0.0155. (2,1)’. -■ , (2.1,05)* Z—2__—TTSmTi 144 + 60 L = b—------ 40000.36 + 150000.4 150000 _201 ft- lb*. = 0,0155.4,41. C OCR 4 bodv JI, in a state of free and pro- Fig’ **' gressive' motionf Fig. 329 impinges against a f u RPK capable of rotatmg about a fixed axis ry b? fcrL- if>in 2„J«,and «, i" „ and instead of the mert mas a£JB ,r ,1,- fi«t body, the other denominatione re- — miiig the aatn.J Hence, the velocity of the first mass after impact is:ROTARY BOD1ES. 297 »X — Cj — (c, — a2 H) (1 + vV) • and the angular velocity of the second : "2 — f2 + fl2 (C1 °2 (14-■%/(«)• If the mass Mt be at rest, therefore vi= ci — ci (! + 'Sr) • Mxa* 4* ^%y* Mx Mxa22 4- 0, we have: •^$2---- and "2 — a2Cl( 1 + ^ /*) + M2y* M. ' Mxa2 4- . , If, on the other hand, Mx is at rest, that is, the oscillating body the impinging one, we shall have Cj =» 0, and hence a. t. (1 + v'/<*)• ■Mj/i Mxaf + Mjj, JU_ and ?)• The velocity communicated by impact to another at rest, fPe? S not only on the velocity of impact and of the masses of the bodies, but^also on the distance KL = a% at which the direction of impact •^•Wis distant from the axis K of the rotary body. If the free mass be the impinging one, the rotary mass will assume the angular velocity _v Mxa2 ar»d if the oscillating mass strike against the free, this will acquire the velocity , _ n , /-i M#* • «2 1 - tj(1 + ^ Mxa2 4- hut both velocities will be so much the greater, the greater TfT---a*_____or________L-______is, and therefore the less *«' + **.• + ’ «2 4- Mj/I is. a2 If for a we put a + x, when x is very small, we shall obtam the value of the last impression: Mx («±*) + « M.a±Mlx + %2l (I + - + 5±* a Hhx « a a or in consequence of the smallness of the powers of x, - -w.« 4- ^ 4- (Mx - + ■ * * If now a correspond to the least of all the values of \a% ~~T~> the member + (M, — x will disappear, because the addition of the quantity (x) will give to it a different sign to that of a diminuti ve (— x.) Therefore :298 BALLISTIC PENDULUM. (M1__X must be = 0, i. Mjh2 M,t consequently: 5 y2 J M* M, \ M1 “X -“i If at this distance one body impinge against the other, then will the latter take the greatest velocity, and be in fact _v 1 I = Cj (1 + v/ju) g 1 y2 \ in the case where the rotary body is impinged upon; and V = i y%>% (! + 3., M, WlTL^uffiheh?ronmpi°rt of the distance corresponding to .V .CvdocitT or of the arm a, is sometimes improperly called t SetnmpS but more properly thepoin/ JP The6d'smn"e'gof «his point from «he fixed axis of the ia.t body: a = ‘ffOOO = fMfi - 7,906 feeu If «he impac, be inelastic, and «he block .«rike agai„8l J 16 f latter will receive the velocity v = the sphere with the velocity . = 3 feet, tu L . 7,906 = 11,86 feet. Kallistic Pendulum.-An application ofthe laws lajd down, Ballisiicre .g found in the theory of the balhstic pen- dulum, or the pendulum of It consists of a mass MII, turning about a horizontal axis C, Fig. 330, which is set into oscillating motion by a ball projected against it, which serves for the measure- ment of its velocity. That as inelastic a blow as possible may ensue, there is an opening made on the further side, which from time to time is filled by fresh wood or clay, &c. The ball remains after each projection sticking in this mass, and os- cillating in common with the whole body. For the measurement of the velocity ‘of the ball, it is requisite to know the angle i __________0f elongation of this pendulum, on which . , .U „ra,h.-ited arc BD applied, and an index £ account there is further a g pendulum, which slides along fixed to the centre of gravity ot Ine * with the former. tve angular velocity of the ballistic From the foregomg paragrapn, «n b pendulum after the impact of the ball is. the mass of the ball, Mj,,'the moment of inertia of the pendulum,CENTRE OF PERCUSSION. • 299 cx the velocity of the ball, and a2 the arm CG of the impact, or the distance of the line of impact JVJV* from the axis of revolution of the pendulum. If the distance CM of the centre of oscillation M of the entire mass, together with the ball from the centre of suspension C, i» e. the length of the simple pendulum, which oscillates in equal times with the ballistic, = /, and the angle of elongation BCD = <*, we have the height of the isochronously oscillating simple pendulum: h = CM—CH =1—1 cos.« = /(1— cos. a) = 21 = ( sin. ^ ; and hence the velocity at the lowest point of its path: v = 2gh = 2 and „ o /^1 + \ s . a T V 2 Jf the pendulum makes n oscillations per minute, the time of oscil- lation \g' n nn and the required velocity of the ball c_^ + °1---17 M2 120 gs M, n n a9 sin. -. ExampU. If a ballistic pendulum, of 3000 lbs. weight, whose angle « amounts to 15°, is set into oscillation by the projection of a 6 . » ^ projec- distance . of the centre of gravity from the axis = 5 feet, «ion from this axis = 54 feet, and lastly, the number of oscHat.ons per minute from the above formula the velocity of the ball at the moment of the impact w , 3006 120.31,25.5 . »,0 _ 501.3750 .«m. 7 30 = 1774 feet ~T • 40 ■ 3,1416.5^5 ^ H-----------4473,1416 § 270. Centre of Percussion.—If a body turning about ne_ C is impinged upon by another, a reaction from .e ,„n„n(jent prin- rally take place upon the axis of the body, which is^ . P ^ an(lthat cipally upon the distance between the direction o pressure in of the axis. Let us determine this reaction or. 18 n(i:Cular tothe the simple case, when the direction of the blow is PerPFig. 331 gQQ * CENTRE OF PERCUSSION. pu* p,ssi„g through the axis of re«ol».ion and ,h« centre „f gra.ity ofthebody. throuirh the axis of revolution Let BDbe a plane o |ravyJ)“ perpendicular axis in this XX of the body m t ig. * anffles to this plane of gravity. plane, and ZZ a third «*“» * detf„nined by the co-ordinates = An element Mx of the bo j ^ __ y ancl LM = in this Sys- tem of axes intersecting in the point and another element by the co-ordinates x , J/»» *»» &c* If * be the angular ac- celeration, we shall have the force of inertia of the element : Q, = . x . KM, and if these be resolved into the component forces parallel, and S at right angles to the imaginary plane of gravity, the similarity of the triangles KML and QMR or MQS, gives . *2t = * Af,z,, and S = From this the aggregate of the component forces parallel to the plane = * (AT,*,* z + ..) and that of the component forces at right angles to this plane: x ,yt Mj/2 + . .). 1-==— ---------- S;nce the plane BD passes through the . „„m nf the moments Mtzl 4- 4- .. = 0, centre of gravity, the , .. sum 0f the forces »(Af,y, 4- M2y2 -f ..). hence there remains oniy anj W the resistance, or the reac- If, now, P be a bloW inPthe first place, have to put: tionoftheaxis^we^aH.mAe + ^ } The statical moment of the loge. KM1, = JVfj x . KM — Mi*. b ' _ x the moment of the or the distance KM«P* » x &c., hence the stati- force of another Fu rire inerti * Wn’ .). If cal moment of the * i\TO of t^e direction of the impact from the now we put the distance ^ ^ave the moment of percussion direction of the axis — > fW Q. We may hence also put: about XX = Pb, whilstthatof IV # + . Pb =f * {M/i + * * equations: and obtain byfteb^l”at+ 4- • 0 f. e. the reaction sought: TF = P (1---------------- • • / „ , , . *v j:'lance JiN of the direction of the impact If, lastly, we represent the distance ^ ‘ , • ,1 distance J2Uof the pomt of applica- further the moment Pa = mom. Wu 4- mom. x .)»CENTRE OF PERCUSSION. 301 and the distance of the point of application sought will be: Pa — x + M2x^2 + ..), ,• „ W ’ ' * u a + M2r22 + . •) — (Mix1y1 + + • •) Mj* + M2r* + .. — b (M1yl + Mj/t + ..) -11 , —2'2 ' i 2*^2 • * y „ The reaction W= 0, if A (MlVl + + ..) = -*W + + . + M2r2 + .. inornent of inertia 6. 1,0 =3= —— - -... . - -------- < i ■ 7 . ^iVl + -^2 + • • and also lts moment = 0, if statical moment Pu * (Mlx1yl -f* •M2x3y2 -f- . .), i. e. 2. a = + + • • Tk . ’ . ^ t + : • J ne point O determined by these co-ordinates a and bf in the plane of gravity containing the fixed axis, is called the centre of percussion. ~ery blow passing through this point, and at right angles to the plane of gravity, is completely taken up by the mass, without leav- jng any residuary effect upon the axis, or producing any pressure. formula (1) shows that the centre of percussion is at the sanie distance from the axis of revolution (compare § 251) as the centre of That a hammer may not jar by its blow the hand ^hich j|olds i, or react upon the wrist about which it turns, it is requisi e blow pass through the centre of percussion. Examples — 1. In a prismatic bar CA, Fig. 332, which turns about one of its extreme points, the centre of percussion lies about CO = b = = -f l — $ CA from tlie If» therefore, the bar be fixed at one extremity, and be struck at ® Pjint O at the diston CO = | CA, then no jar wili be felt-2. In a parallelopiped BDE, Fig. 333, which Fig. 332. Fig. 333. turns about an axis XX running parallel to its ftmr sides, and^^Ifssion O from the from the centre of gravity, the distance AO of the centre of percussion i/ axis 6 = *±**, where d is the semi-diagonal of the lateral surfaces through which the axis XX passes (§ 220). If the fbrce of the blow P were to pass throu0 of gravity, the reaction would be: pjz Pdl 26302 excentric impact. Fig. 334. § 271. Excentric Impact.— Lastly, lej ™ Jsperfectjyfree* simple case of excentnc impact, w en . upon each other When two bodies A and BEimpact JVJV passes tCgh hetntre Of gr.M of the one Wv and beyond the centre of gravrty S, 0f the other body, the impact with respect to the first body is centric, and with respect to the other excentnc. The effects of this excentric impact may be found from the propositmn, c 214 if we assume first, that the second body is free’, and that the direction of impact passes through the centre of gravity S„ and secondly, that this body is fixed at lts centre of gravity, ______________________ an(l that the percutient force acts as a force of revolution. If now c, be the initial velocity ' . nf the Centre of gravity of and if the two velocities «.ftom the eLt of the blow. .here «,tl as i» § 4. jvf v* = Mxcx 4" M%c%. , v. the riritiaUngular velocity of the body BE, and in its If further . be the m ^ Jssingthrough the centre of gravity M1vl + Mjv2 = Mlcl + M.cv If further . be the initial angular velocity of the bouy *>*,, anu in its revolution about the axis passing through the centre of gravity, and perpendicular to the plane JVJVSL and if this velocity pass from the effect of the blow into and M#,' represen the inornent of inertia of this body about S2, or s the d.stance S Kof the centre of gravity from the^irection of impact we shall then also have Mj/i M,v, d—' ' * • then the points of contact of the two If both bodies are inei > ve,ocities, therefore, further v, will will have at the ena ^ the above equations v, and „ by v + su.If we dete j d int0 the last equation, we shall then * Vl) M£iSiZZ VA+ si, ----r» 5 = + and °btam: p » + c, + \ jetermined the loss in velocity of the first and from this may be determ body; ci—vi = (mT+MJ Vt + ■ 1velocity of the second: the gain in progressive^ ^ s.) v*—c* = and the gain in angular velocity • w_, = CZSa^fr (,'\+5eYlstic impact, these values are double ; jwrs ■<£*• ** ” <*+'V— as great.EXCENTRIC IMPACT. 303 Example. An iron ball A, of 65 lbs. weight, strikes a parallelopiped BE of fir, ori- ginally at rest, with a 36 feet velocity; the length of this body is 5 feet, its breadth 3 feet, and thickness 2 feet, and the direction of the impact NN deviates by S.2 Ksss iss feet from the centre of gravity 5^,, then the following velocities after impact are given. The specific gravity of fir may be taken = 0,45, the weight of the parallelopiped is therefore = 5 X 3 x 2 x 62,5 x 0,45 = 843,75 lbs. The square of the semi-diagonal of the lateral surface parallel to the direction of impact is : bence the velocity of the ball after impact is: further, the velocity of the centre of gravity of the parallelopiped: 65.1,75. 36 2,379 feet: Mx sc (Afl + A/9)y9+Af1*3 7130,06 — ss 0,574 feet.304 FLUIDITY. SECTION V. STATICS OF FLUID BODIES. CHAPTER I. ON THE EQUILIBRIUM AND PRESSURE OF WATER IN VESSEES. & 272. Fluidity.—We regard fluid bodies as systems of material points, whose cohesion is so feeble, that the smallest forces are suf- fic ent to effect a separation, and to move them amongst each other (t bl) Many bodies in nature, such as air, water, &c., possess this property of fluidity in a bigh degree; other bodies, on the contrary, such a/oil, fat, soft earth, &c., are fluid m a [ow^degree.. The former are called nerfedly fluid, the latter tmperfectly fluid bodies Certam JSies, as,Vinstance, paste, are intermed.ate between solid and fluid b°Perfectly fluid bodies, of which only we shall subsequently speak, are at the same time perfectly elastic, i.e., they may be compressed bv external forces, and will perfectly resume their former volume after the withdrawal of these forces. The amount of the change of volume corresponding to a certain pressure, is different for differ«it fluids ; ln ZiddVbodJthis is scarcely percept.ble, while m aenform bodies, which, on this account, are also called elasicfluids, i is very great. This sliffht degree of compressibihty of hquid bodies is the reason why in most mvestigations in hydrostatics (§ 63) they are considered and treated as incompressible or inelastic. As water, of ali hquids, ,s the one most generally diflused, and the most useful for the purposes of life, it is taken as the representant of all these fluids, and m the ,n- vestigations of the mechanics of fluids, water only is spoken of, whilst it is tacitly understood that the mechanical properties ot other liquids are the same as those of water. . From a similar reason in the mechanics of the elastic tiuid bodies ordinary atmospheric air is only spoken of. Remark.—A column of water of one aquare inch transverse section is compressed by a weight of 15 Ibs., which corresponds to the atmospheric pressure, by alx>ut 0,00005 or 50 miilionths of its volume, while the same column of air under this pressure would be compressed to one half of its original volume.PRINCIPLE OF EQUALITY OF PRESSURES. 305 Fi^> 33o. § 273. Principle of Equality of Pressures,—The charactenstic property of fluids, which essentially distinguishes them irom so i bodies, and which serves as a basis of the laws of the equili rium fluid bodies, is the capability of transmitting the pressure which is exerted upon a part of the surface of the fluid in all directions un- changed. The pressure on solids is transmitted only in lts proper direction (§ 83); while, on the other hand, when water is pressed on one side, a tension takes place in the entire mass, which exerts ltse on all sides, and may be observed at all parts of the surface. o satisfy ourselves of the correctness of this law, we may make use ol an apparatus filled with water, as is shown in the honzontal section in Fig. 335. The tubes AE and BF> &c., equally distant, and at an equal height above the horizontal base, are closed by perfectly movable and accu- rately fitting pistons; the water presses, therefore, by its weight, as strongly against the one piston as against the other. Let us do away wTith this pres- sure, and regard the water as devoid of weight. Let us press the one piston with a certain pressure P against the water, this pressure will then be trans- mitted by the water to the other pistons B, C, D. and for the restoration of equi- , ... librium, orto prevent the pushing back of these pistons, i is reqm that an equal and opposite pressure P act against each of these pis- tons. We are, therefore, justified in assuming, that the pressure acting upon a point A of the surface of the mass of water, produces in it a tension, and not only transmits this in the straig me , but also in every other direction BF, DH, &c., to every equal area of the surface C, B, D. „ , . If the axes of the tubes BF,CG, &c., Fig. 336, are parallel to cach other, the pressures which act npon their pistons may be united by addition into a single pressure; if n be the number of the pistons, then the aggregate pressure upon these amounts to Px = nP, and in the case represented in the figure Px = 3P. But now the areas Fx of the pressed surfaces B, C, D} are equal to n times the pressed surface F of the one piston, hence n may not only P F P F. be put = —l, but also = therefore = ^r- P r 007 and if wTe If the tubes B, C, D, form a single one, as in> an(j p close it by a single piston, Fx then becomes a smg > 1 Fig. 336.306 THE FLUID SURFACE. is the pressure aeting upon it, hence there follows this general law, the pressure which a Jiuid body exerts upon different parts of the sides of a vessel, is proportional to the area of these parts. Fig. 337. **• 338- This law corresponds also to the principle of Virtual velocities. If the piston AD= F,Fig. 338, moves inwards through a space ./M =S) it then presses the column of water Fs from its tube, and if the piston BE=Fv it passes outwards through the space £5,=«,, it then leaves a space F1s1 behind. But since we have supposed that mass of water neither allows of expansion nor compression, its volume then by this motion of the piston must remain unaltered, that is, the increase Fs must be equal to the decrease F, s,. But the equation F, st = Fs gives Zi = —, and by combining this proportion with the proportion ■F si „ F? = Ex. it follows that Li = —, hence, therefore, the mechanical P F F s, effect P1s1 = mechanical effect Ps (§ 80). Example. If the piston JD has a diameter of 1J inches, and the piston one of 10 inches, and each is pressed by a force P of 36 lbs. upon the water, this piston exerts a pres- sure P — £.p _ . 36 s= 1600 lbs. If the first piston is pushed forwards 6 Fi 1i** F 9.6 27 inches, the second. will only go back by », = — * = "4Q0" = 200 = l’^33 *n’ Remark. Numerous applications of this law will come before us in the hydraulic press, or water column machines, in pumps, &c. & 274 The Fluid Surface.— The gravity inherent in water causes ali its particles to tend downwards, and they would actually so move unless this motion were prevented. In order to obtain a coherent mass of water, it is ne* cessary to enclose it in vessels. The water in the vessel ABC, Fig. 339, is then only in equilibrium if its free surface HR is perpen- dicular to the direction of gravity, and there* fore horizontal, for so long as this surface is curved or inclined to the horizon: then there are elementary portions£, F, &c., lyinghigher, which, from their extreme mobility in virtue of their gravity, slide down on those below them, as if it were on an in* clined plane GK.THE FLUID SURFACE. 307 Since the directions of gravity for great distances can no onger e regarded as parallel, we must, therefore, consider the free sur ace,. °r the level of water in a large vessel, as for example, in a grea a e, no longer as a plane, but as part of a spherical surface. lf any other force than that of gravity act upon the par it es o water, the fluid surface in the state of equilibrium, will be perpen- dicular to the direction of the resultant arising from gravity and tne Fig. 340. concurrent force. If a vessel ABC, Fig. 340, is moved forward honzontally by a uniformly accelerating force p, the free surface of the water in it will form an inclined plane I)Ffor in this case every element E of this surface will be impelled downwards by its weight G, and horizontally by its inertia P = - G, there will then be a resultant R, g . . which will make with the direction of , ootYI<3 gravity a uniform angle REG = <*. This angle is a e «. the angle DFHwhich the surface of the water makes with the horizon. T . 6 P p it is determined by tang. o = — = -• If, on the other hand, a vessel , Fig. 341, rotates uniformly about its vertical axis XXy the surface of the water then fornis a hollow surfaceAOC, whose sections through the axis are parabolic. If « be the angular velocity of the vessel and the water in it, G the weight of an element of water JS, and y its distance ME from the ver- tical axis, we shall then have for the centri- fugal force of this element F= 231), and hence for the angle REG = TEM = t, which the resultant R makes with the vertical or the tangent to the water profile with the horizon: F Jy tang. t -q - ~ From this, therefore, the tangent of the angle which the line of contact makes with this ordinate, is proportional to the ordinate. As this pro- perty belongs to the common parabola (§ 144), the vertica ;sec. AOC of the surface of water is also a parabola whose axis coinci Fig. 341. with the axis of revolution XX. . „ uniformly If a vessel ABH be moved in a vertical circle, 1 ig- ’water about a horizontal parallel axis C, the surface o ..jj- we form in it a cylindrical surface with circular sec ion the prolong the direction of the resultant R of the 8^7 Q with the centrifugal force F of an element E to the in n vertical CK passing through the centre of reA o u >308 PRESSURE ON THE BOTTOM. Fig. 342. obtain the similar triangles ECO and EFR, for which OC _ FR __ G EC “ EF F' but now, if we put the radius of gyration EC = y, and retain the last notation, F —- —it follows that the line co = ^ = 32,2 /30\2 2936 (?)-■ Fig. 343. (3,1416) if u represents the number of revolutions per minute. As this value of CO is one and the same for ali the particles of water, it follows that the coraponents of ali the - particles forming the section DEH are directed towards O, and hence the section perpendicular to the direc- tions of these forces, is a circle described from O as a centre. Ac cordine to this, the surfaces of water in the buckets of an overshot wheel form perfect cylindrical surfaces, corresponding to one and the same horizontal axis. c R 275. Pressure onthe Bottom.-The pressure of water m a ves- sel ABCD, iig. 343, lmmediately un- der the water level is the least, but be- coroes greater and greater in proportion to the depth, and is greatest immediately above the bottom. To prove this gene- rally, let us assume that the level of the water H0R0, whose area may be F0, is uniformly pressed by a force P0, for ex- ample, by the superincumbent atmo- sphere, or by a piston, and let us sup- pose the whole mass of water divided by horizontal planes, as HXRV H2R2, into equally thick strata ot water. It now x be the thickness or the height of «d> » of water, we where P0 represents the external pressure on the unit of surface. 4he pressure o he suc- ceeding horizontal section H R2 may be determined as exactly as the pressure of the stralurn HXRv if we take into consideration that the initial pressure upon the unit is now = p0 + whilst it was then only p0. The pressure in the horizontal stratum H2R2 then follows: p2 = Pi + — Po + + *r = Po + 2xr; likewise the pressure inPRESSURE ON THE BOTTOM. 309 the third stratum H3R3 = p + 3xy, in the fourth = p0 + 4xy, and in the nth = pQ + nxy. But now m. is the depth G0Gn = h of the wth stratum below the level of the water, hence the pressure upon each unit of surface in the nth horizontal stratum may be put: p = p0+hy. The depth h of an element of surface below the water level, is called the head of water, and the pressure of water upon any unit of surface may from this be found, if the externally acting pressure be increased by the weight of a column of water whose base is this unit, and whose height is the head of water. The head of water h on a horizontal surface, for instance, on the bottom CD, is at all places one and the same; hence the area of this surface = F, and the pressure of water against it is: P = (p0 + hy)X F = FpQ phy -- Po 4- Fhy, or if we abstract the outer pressure: P == Fhy. The pressure of water against a horizontal surface is there- fore equivalent to the weight of the superincumbent column oj water Fh, This pressure of water against a horizontal surface a horizon- tal bottom, for instance—or against a horizontal part of a lateral ^all, is independent of the form of the vessel; whether, therefore, the vessel AC, Fig. 344, be prismatic as a, or wider above than below as or wider below than above, as c, °r inclined as d, or bulging out as e, &c., the pressure on the bottom will be always equal to the weight of a co- lumn of wTater w hose base is the bottom and whose height is the depth of the bottom below the level of the water. As the pressure of water transmits itself °n all sides, this lawr is therefore ap- plicable when the surface, as BC, Fig. ^45, is pressed upon from below up- wards. Every unit of surface in the stratum lying in BC is pressed by a column of water of the height HB = % h; consequently, the pressure against CB = Fhy, F being the area of the surface. Fig. 345. Fig. 346. It further followTs from this, that the water in with each other ABC and DEF, Fig. 346, when stands equally high, or that the two levels AB and E ar310 LATERAL PRESSURE. Fig. 347. the same horizontal plane. For theJ“b*,sten^rpiblv^nresMd^own! requisite that the stratum of water be as o y , ' wards by the superincumbent Weut as in both cases the wards by the mass ofwaerlymgbe ^ head of water in both surface pressed is one and tne sa , must stand h cases be one and the same, theretore tne ie>c above HR as the level EF. above for th § 276. Lateral „0t directly applicable to a of water against a hcnz^ ^ horiz<;n; for in this case the heads of plane surface are different. The pressure = on each water at ^«fferen P ^ stratum of water, which lies a JSrtTSSSfS*-» i» »11 ai-ec.ions(§273),»nd eous,- depm « uci quently also perpendicular to the fixed lateral walls of the vessel, which (from § 128) perfectly coun- teract it. If now F,be the area of anelement of a lateral surface Fig. 347, and A, its head of water FH, we shall then have the normal pressure of the water against it: Pt s Fj. A,y; if F3 be a second ele- ment of the surface, and A2 its head of water, we shall then have the normal pressure on it: = F2A2y; and for a third element P3 = y, &c. These normal pressures form a system of parallel forces, whose resultant P is the sum of these pressares; therefore P = (F h +Fh + V•)/• But now, further, FA.+ F& + tlfiateTS i FA>representing the area of the whole surface, and Ae depth 80 k its centre of f^lanfsurdet*P- X’ the aggregate “^“‘P^o^watTr of a surface, the depth SO of S^ntreof!gravfty belowthe level of the water- the generalrule, its centre oi g J messure of water against a plane surjace is u ae and whose height is the head °fwa^f °f the water is not I, mus. furthe, b« sta.ed, .ha. .h,s 'q“ water which is before or below the pressed surface, that therefore, for example, a flood-gate, AC, Fig. 348, under otherwise similar cir- cumstances, has to sustain the same pressure, whether the water to be dammed up be that of a small sluice Fig. 348.CENTRE OF PRESSURE. 311 ACEFy or that of a larger clam ACGH, or that of a great reservoir. From the breadth AB = CD = b and height AD = BC = a of a rec- tangular flood-gate, F = aby and the head of water SO = hence Mf the pressure of water P= ab.-7 = ba*by. 27 2 Therefore the pressure increases as the breadth, or as the square of the height of the pressed surface. Example. If the water stand 3$ feet high before a board of oak 4 feet broad, 5 feet high, and 2$ inches thick, what will be the force required to draw it up ? The volume of the board is 4 5 A. = — cubic feet. If now we take the density of oak saturated ’ 24 6 with water from § 58 at 62,5 X 1,11 = 67,3 lbs., the weight of this board will be: G =- ?? . 67,3 = 280,5 lbs. The pressure of the water against the board, and also the 6 pressure of this last against the guides will be: P s- I . (Zy . 4 , 62,5 = 49.30,25 = 1531,25 lbs.,; if now we take the co-efficient °f friction for wet wood from § 161, / = 0,68, the friction of this board against its guides will be F =/ P = 0,68 . 1531,25 = 1041,25 lbs. If to this be added the height of the board, we shall obtain the force required to pull it up = 1041,25 + 67,3 = 1108,55 lbs. Fig. 349. § 277. Centre of Pressure.—The resultant P = Fhy of the coi- lective elementary pressures Flhlyy KKy> &c., has, like every other System of parallel forces, a definite point of application, which is called the centre of pressure, Equili- brium will subsist for the whole pressure of the surface, if this point be supported. The statical rao- nients of the elementary pressures F2h2y, &c., with respect to the plane of the level OHR, Fig. 349, are: F^y . hx = FJifyy F2A22y, &c.; therefore, the statical rnoment of the whole pressure with respect to this plane is: (FJi * + t • FA2 + • • •) y. If we put the distance KM of the centre of this pressure from the level ofthe water = 2, we shall then have the roo- ment of pressure = Pz = (FA + F • • •) anc^ by equating o moments, the depth in question of the centre below the sur ace. n i b . T, 1 « . JP J* 2 _L_ JP h * -4» • • 1. z = F.h* + F2h2* + or FA2 + W + “ ~Fh if, as before, F represent the area of the whole jet’ermine this depth of its centre of gravity below the surface. from another pressure completely we must know further its t of the e]e. plane or line. If we put the distances F.yl9 P2 v determines ments of the surface f\F2, &c., from the line AC which determines312 CENTRE OF PRESSURE. the angle of inclination of the plane = yxy2, &c., we shall then have the moments of the elementary pressures with respect to this line = FAy#, F9hj/y, &c., therefore, the moment of the whole sur- face = (F1h1yl + F2hjj2 + . .) y; and if we represent the distance MJY of the centre JY from this line by v, we shall then have the mo- ment also = {Fxhx + F2h2 + . .) vy; if, lastly, we make both moments equal, we shall obtain the second ordinate : 2 r ± -, or- +_^ Flhl + F2h2 + .. Fh If a be the angle of inclination of the plane ABC to the horizon, and xlf x2, &c., the distances FtRItF2R2, &c.,of the elements FlfFt &c., as likewise u the distance of the centre of pressure from the line of intersection AB of the plane with the level of the water, we shall then have: A, = xx sin. a, h2 = x2 sin. o, &c., as well as z = u sin. a- and if these values be put into the expressions for z and vy we shall then obtain: F,x* + J>22 + .. _ moment of inertia anH U — Fxxx + F2x2 + .. statical moment ’ F,xAyx + F^r2y2 + . . __ centrifugal moment V — Fx + F2x2+ .. statical moment We may, therefore, find the distances u and v of the centre of pressure from the horizontal axis JtYy and from the axis JiX formed by the line of fall, if we divide the statical moment of the surface with respect to the first axis, once by its moment of inertia with respect to the same axis, and a second time by its^ centrifugal moment with respect to both axes. The first distance is at once the distance of the centre of suspension from the line of intersection with the line of the water (§ 251). It is easy to see that the centre of pressure coin- cides perfectly with the centre of percussion, determined in § 270, if the line of intersection J1Y of the surface with the level, be regarded as the axis of revolution. If the pressed surface is a rectangle ACy Fig. 350, with horizontal base CD, the centre of pressure M will be found in the line LK let fall upon CD bisecting the basis, and will be distant § of this line from the side AB in the surface of water. If this rectangle does not Fig. 350. Fig. 351. Fig. 352. reach the surface as in Fig. 351, if further the distance KL of the lower base CD from the surface be lv and that of the upper baseCENTRE OF PRESSURE. 313 = l3, we then have (he distance of the centre of pressure from the fluid surfaee: 1 -- 2 For the case of a right-angled triangle ABCy Fig. 352, whose base AB lies in the fluid surfaee, the distance KM of the centre of the triangle^ which pauses from the point O to the middle point of the base, NM = v = £ ft, where b represents the base AB. If the point C lies in the surfaee, as in Fig. 353, therefore, the If the whole triangle ABC, Fig. 354, be under water, if the base yB is at a distance AH = /2, and the point a distance CH = lx trom the surfaee HR. we then have the distance MK from the sur- lace HR: pressure may be determined for other figures. Example. What force K must be expended to FF, Fig. 355 f Let Tis lengtii = 1± feet, its breadth EF s= 1 $ feet, its weight = 35 lbs.; further, the distance CK of the axis of revolu- tion C from the surfaee HR, measured in the plane of the door, == 1 foot, and the angle of inclination of this plane to the horizon s= 68°. height BC of the triangle, and the distance of the same point from the other lee, as this point in every case lies in the line CO bisecting 4.1 ^ . . . * « .i * • A ^ Mll * ^ PA. base AB below this point, we have — ~—- - — i--* KM = «= f/and JYM = v = | * = f % FI * Fig» 3o3. Fig. 354. In a similar manner the centres of Fig. 355. draw up a trap-door J1C lurning about an axis 27314 CENTRE OF PRESSURE. The pressed surface is F = j ^ ~ square feet, and the head of water or the depth of its centre of gravity below the surface, h = HS sin. a = (HC-\-CS) sin. a = (//C-f-J -^C) sin. . = (l + i . sin.68° = — «». 68» = 13 -°f— => 1-5067 *«, hence, ,he \ ~2 4/ 8 8 pressure of water on the surface is: P = = — • 1*5067 • 06 186,45 Ibs. The arm of this force about the axis of revolution is the distance CM of the centre of pres- sure M from this axis; therefore = HM — HC^ V from this axis; luerent —/ 4 \3 2 j -1.1*}---1 =4* * "Ir” * 3 (|)'-(t) 6 81~ 64 16 ' ■ 1 Fig. 356. , i moment of the pressure of water = 186,45 . 0,705 = ^S.^ ITetheecentre of gravity S of the trap-door lies about half the length cs ± 1 — 1 feet from the axes of revolution, the arm CD of the weight of the revolving ddor will be = CS«.. . = | • «>*• 68° = 1 . 0,3746 = 0,2341 ft., and i .nant r\f this weiaht s= 35.0,2341 = 8,19 ft. Ibs. By the addition hr>!! hhe 9'at‘^ we obtai/the wbole moment for drawing up.the trajwtoor = 131,46 % $5 ««• * «* «*•effect ac‘atarm ^=».25 feet, its amount will be = = 112 lbs' & 279 If water presses against both sides of a plane surface V- 7W tbprp nrises from the resultant forces corresponding to the Fig. 356, there anses trom^m ^ # new resultant> which is ob, tained by the subtraction of the former, because these two act oppositely to each other. „ . , If F is the area of the pressed portion on the one side of the surface AB, and the depth AS of its centre of gravity below the level of the water; further, F, the area of the portion At B, on the other side of the surface, and h, the depth A^ of^ its cen- tre of gravity below the corresponding level of the water, we then have for the resultant sought, = =IHhemoment of inertia ofthe fint portion of the fluid surface with respect to the line in which the plane of the surface intersecas that of thewater, = FV, the statical moment ofthe pressure of water of the one side is therefore = Fz3 . y; if, further, the moment of inertia ofthe second portion with respect to the line of lntersection with the second surface of water = F1x,*, the statical moment of the pressure of water of the other side about the axis lymg on the second surface is then = F.x\. Further, if the distance AAX of the axes = a, we then obtain the augmentation of the last moment mits transit from the axis^j to the axis.#, = Fxhx a y, and hence the statical moment ofthe pressure of water with respect to the axis in the first surface = + FA . « . y = (W + W r- From this, then, it follows that the statical moment ot the dinerencePRESSURE IN A DEFINITE DIRECTION. 315 of both mean pressures = (Fx2 — Fixi — a FjAJ y, and the arm of this latter force, or the distance of the centre of pressure from the axis in the first surface of water is: Fat — Frf — aF^ ~'■ Fh— If the portions of surface pressed are equal to one another, which takes place when, as Fig. 357 represents, the entire surface AB is below the water, we have then more simply P = F (A — A,) y and u = A; the last, be- cause A — hl = a, and x* = x2 — 2 a h + a2 * * * (§ 217). In the last case, therefore, the pressure is equivalent to the weight of a co- lumn of water, whose base is the surface pressed, and whose height is the diflerence of altitude RH1 of both surfaces of water, and the centre of pressure coincides with the centre of gravity S of the surface. This law is also further correct if both surfaces of water are besides further pressed by equal forces, for ex- ample, by a piston or by the atmosphere. For this pressure upon each unit of surface = p, and therefore the corresponding height of a column of water x = (§ 275), we have then to substitute for A, y h + x, and for A,, A, + x; and by subtraction, we have the residuary force P = (A -f x — [hx + x] ) Fy = (A — A,) Fy. For this reason, the pressure of the atmosphere in hydrostatic investigations is gene- rally left out of consideration. Example. The height AB of the upper surface of water in a canal, Fig. 356, amounts to 7 feet, the water in the lock stands 4 feet high at the sluice-gate, and the breadth of the canal and of the lock measure 7,5 feet, what mean pressure has the sluice-gate to sustain ? It is F = 7 . 7,5 = 52,5, and Ft = 4.7,5 = 30 square feet. Further, h = • 7 * = — and h. — i. = 2 feet, a = 7 — 4 = 3 feet,x2 = -i . 79 = ~ and X* = 2 2 1 -2 3 3 t 16 • 49 — —; hence it follows, that the mean pressure sought is: P = (Fh — FJiJ y ^ 3 = (52,5.1 _ 30.2) . 62,5 = 123,75 . 62,5 = 7734,375 Ibs.; and the depth of its Fip. 357. point of application below the surface of the water is: 52,5.1? _ 30 . — 3 . 60 3_________3 52,5 . 1 _ 60 2 517,5 , 1CO r . =------— = 4,182 feet. 123,75 § 280. Pressure in a Definite DIn many cases it u> of importance to know only one part of the pressure actmg 10a i e direction upon a surface. In order to find this compone n, . the normal pressure jMP = P of the surface * r the given direction MX, and in the direction a be into two component pressures MPX = Pl9 ant^ ,.2 4 2\fx mabs the angle PMX, which the normal in the given direc 1 n __ p with the component, we shall then obtain for the compo , 1316 PRESSURE IN A DEFINITE DIRECTION. Fig. 358. cos. a and P2= P sin. —---------2 Let a pro- jection Ax Bl CD of the surface AB be made on a plane at right angles to the given direction MX, we shall then have for its area Fv the formula F1 =» F. cos. ADAV or since the angle of inclina- tion ADAX of the surface from its pro- jection is equal to the angle PMX = a between the normal pressure P and its component Pv we then have F = F F cos. a, or inversely : cos. a zJ and F F hence Px = P. -J. But as the normal pressure P = Fhv, it follows finally that P, = pressure with whichwater pressesagainst a surface in a given , equal to the weight of a column of water which has for hase the of the surface perpendicular to the given ,and for , the depth of the centre of gravity of the surface below that of the water. It is important, in most cases of application, to know oniy the verti- cal or the horizontal component of the pressure of water against a surface. Since the projection at right angles to the vertical direction is the horizontal, and the projection at right angles to the horizontal direction, a vertical projection, the vertical pressure of water against a surface may be found, if the horizontal projection or its trace be con- sidered as the surface pressed, and on the other hand the horizontal pressure of the water in any direction may be also found, if the verti- cal projection or the elevation of the surface at right angles to the given direction be considered as the surface pressed, but in both cases the depth of the centre of gravity of the surface below that of the water taFonraaSpri^maticFig. 359, the longitudinal profile r for the horizontal pressure of the water and the horizontal projec- tion EL of the surface of water for the vertical pressure must be regarded as the surfaces press- ed. Hence, if the length AG of the dara = /, the height AE = hy and the front slope EF = a, We have then the horizontal pressure of the water = /A . = I A2 Zy> Fig. 359. and its vertical pressure = al If now, further, the upper breadth of the top of the dam = the slope at the back CD = av and the density of the mass of the dam == we then have thePRESSURE OS CURVED SURFACES. 317 idpiaccmeiii, wc *»»»< I^/y=gay+(6 weight of the dam - (b + and the whole vertical pres- sure of this against the horizontal bottom = |«ttr + (& + «r.- [§ay + (b + ~Hr~!) Vl] If we put the co-efficient of friction =/, then the fnction or force to push the dam forward is: r.]^- In the case where the horizontal pressure of the water is to effect this displacement, we have: g ~2 a )y‘ or more s*mP^ : *==/{"«+ ^26 + a + Therefore, in order that the dam may not be pushed away by the water, we must have: h ^2 b + a + 0^, For safety we assume that the base of the dam is quite permea e, on which account there is further a counter pressure from below up- wards = (6 + a + a,) lh t0 abstract, and we may put h / (2 *+ «).y Aceording W experience, a dam will resist a ^ slope and breadth at the top are equal to one another j if m the last form P hsss b = a, then / =k J-, whence we must in other cases put:— 3 i . h = -L £(2 b 4- a + a,) (?! — l) — a, J, and for clay dams especially, h m* — (a ® + o), and inversely, 6 = If the height of the dam be 20 feet, and the of slope * = 36°, the slope a will be = h cotg. a as 20. cotg. 36° ss 20.1,3764 = 27,53 feet, 00 — 27,53 and hence the upper breadth of the dam b — • 2 * § 281. Pressure on Curved Surfaces.—-The law ftmnd iR paragraph on the pressure of water in a definite direction is J for plane surfaces, or for the separate elements of curved su a » not for curved surfaces in general. The normal pressures i t i separate elements of a curved surface may be resolved into coraponents parallel to a given direction, and into others acting plane normal to it; these components form amystem of parallel , whose resultant gives the pressure in the given direction, ana ese »2731S PRESSURE ON CURVED SURFACES. components may be reduced to a resultant, but the two resultants admit of no further composition when their directions donot intersect. It is not possible in general to reduce the aggregate pressures against the elements of a curved surface to a single force, but particular cases present themselves where this composition is possible. Let Gv G2, G3, &c., be the projections, and hv h2> A3, &c., the heads of water of the elements Flf F2> F3, &c., of a curved surface, then have the pressure of water in the direction perpendicular to we the plane of projection : Pl==(GA + G2h2 + GA + .. .) r> and its moment with respect to the plane of the surface of water. Pxu = (GA2 + GJi* + GA2 + . . .) y. 1 f the curved surface pressed upon can be decomposed elements which have a uniform ratio to their projections, We then put F F F _± = _£ = _J., &c., = n, we then have: into may G, G, G, r, or, since the ratio of the entire curved surface F to its projection F . G, i. e. — is = w, G p_ Fh y = Ghy; in this case we have, as for every plane sur- 1 n % face, the pressure in any direction equivalent to the weight of a prism of water, whose basic surface is at right angles to the projection of he curved surface in the given direction, and whose height is equal to the depth of the centre of gravity of the curved surface below the surface of water. So, for example, the vertical pressure of water against the envelope of a conical vessel ACB, filled with water, Fig. 360, is equal to the weight of a column of water which has the bottom for its base, and two-thirds of the length of the axis CM for height, be- cause the horizonta] projection of the envelope of a right cone upon its base, as likewise the en- velope, may be resolved into exactly similar tri- angular elements, and because the centre of gravity S of the surface of the cone is distant two-thirds of the height of the cone from the vertex (§ 110). If r be the radius of the base, and h the height of the cone, we shall then have the pressure against the bottom = and the Fig. 360. vertical pressure against the envelope = ^ *r2Ay> but as bottom is o rigidly connected with the sides, and both pressures act opposed to each other, the force with which the vessel is pressed downwards by the water is:H0R1Z0NTAL AND VERTICAL PRESSURE. 319 = (l — 0 «r2h7 = i «r2hy = Fig. 361. Fig. 362. to the weight of the whole mass of water. If the bottom be separated by a fine cut from the envelope, this will then press with lts full loree nr^hy downwards, or on its support, and on the other hand it would be necessary to hold down the envelope with a force - *r2%toprevent its being raised ofT. Remark.From this the force which the steam of a steam- engine or the water of a water-column maehine exerts on he piston, i» independent of the form of the piston. Whether the surface of pressure be augmented by being hollowed out or rounded, the pressure with which the steam or water pushes forward the piston is equivalent to the product of the cross section or horizontal projection of the piston and the pressure on a unit of surface. The pressure on the larger surface of a funnel-shaped piston AB, Fig. 361, whose greater radms = CB=r and lesser radius GD = GE = r„ is = and the reaction upon the envelope = «• (r1 ' jV) ence’ 1 ’e residuary effective pressure is = *“ * V ri)P = the cross section of the cylinder multiplied by the pressure on a unit of surface. § 282. Horizontal and Vertical Prensure.— Whatever may be the form of a curved surface, AI5, Fig. 362, the horizontal pressure of the water against it is always equivalent to the weight of a column of water, whose base is the vertical projection A1Bl of the surface perpendicular to the given direction of pressure, and whose height of pressure is the depth CS ot the centre of gravity S of the projection below the surface of water. The correctness of this follows directly from the formula P,. = (pA + GA + • • 0 Tt when we consider that the height of pressure hiy h2% &c., of the elements of the surface are also the heights of pressure of their projections, that, there- fore, GA + GA + ... is the^ statical moment of the whole projection, i. e. the product Gh of the vertical projection G and the depth h of its centre of gravity below the surface of water. We have here, therefore, again to put Px = Ghy, and to consider h as the height of pressure of the vertical projection. The vertical section which divides a vessel containing water into two equal or unequal parts, is at once the vertical projection of the two parts, but the horizontal pressure on one part of the wall ot e vessel is proportional to the product of its vertical projection and to the depth of its centre of gravity below the surface of the water, con- sequently the horizontal pressure on a part of the wall of the vessel is exactly equal in amount to the oppositely acting horizontal pressure on the part opposite, and consequently the two forces balance each320 THICKNESS OF PIPES. Fig. 363. Fig. 364. other in the vessel; the whole vessel is therefore equally pressed by the enclosed water in all horizontal directions. If an opening 0 be made in the side of a vessel tig. 363, P 8 the part of the pressure corresponding to the section of this opemng disappears, and the pressure on the oppositely situ- ated portion of the surface F now comes into action. Whilst, therefore, the water flows out at the lateral aperture, an equal distribution of the horizontal pressure no longer takes place over the whole extent, _______ and there ensues a reaction opposite to the motion of the flowing water: P = Fhy, F being the projection ?f the aperture, and hthe height of pressure of its projection. By this reaction the vessel may be set mto motion. The vertical pressure of water is P, — G^y against an element of «nrface FFie. 364, of the side of the vessel, since the horizontal 1 projection G, may be regarded as the transverse section, and the height of pressure A, as the height, and therefore Glhl as the volume of a prism,«equivalent Io the weight of a column of water HFX incumbent on the element, and reach- in therefore, to avoid any transverse rent in the pipe or in the boiler, the thickness of the sides must be made e > longitudinal rents, AE, LH, &c., those running diametrically, such as AE, take place the most easily, because they have the smallest area, whence we must only take these into account. Let us consider a portion of a pipe of the length l, and let us have regard to the occurrence of a rent of the length l, we then obtain a transverse section of the surface of separation = /e, and hence the force for rupture in this surface leK. For two oppositely situated rents this force is = 2 whilst the pressure of water for each half of the pipe is proportional to the trans- verse section 2 rl, and hence is = 2 rlp. By equating the two ex- pressions, it follows that 2 leK = 2 i. e. = rp, therefore the thickness e =* r^~. To provide against longitudinal rents, the sides niust be made as thick again, as to provide against transverse rents. From the formula e = VL = it follows that the strength of K K similar pipes is as the widths and as the heads of water or pressares upon a unit of surface. A pipe three times the width of another, which has five times the pressure to sustain on each unit of surface that the other has, must have its sides fifteen times as thick. Hollow spheres, which have to sustain a pressure p from withm on each unit of surface, require a thickness e = because here the projection of the surface of pressure is the greatest circle rf/*2, and the surface of separation the ring 2 «re (l 4 ^, or approximately for a smaller thickness = 2 ure. The formula found give for p = 0, also e = 0, for this reason,322 BUOYANCY. therefore, pipes which have no internal pressure to sustain, may be made indefinitely thin; but as each pipe mu st sustain a certain pres- sure from its own weight, we must stili give to it a certam thick- ness e., to obtain the strength of a tube which will resist under all circumstances. Hence, for cylindrical pipes or boilers we must put e ^ e + rh? , or more simply, if d represents the interior diameter of the pipe,»the pressure in atmospheres, each corresponding to a colunm of water 33 ft. high, and /» a number from expenment e = c, + hnd. From experimenta made we must .takefejpipes of Iron piate Cast iron Copper . Lead Zinc Wood . Natural stones Artificial.stones e = 0,00086 nd -|- 0,12 inches e wm 0,00238 nd + 0,33 “ e = 0,00148 nd + 0,16 “ 0,00242 nd + 0,20 “ e = 0,00507 nd + 0,16 “ e ss 0,0323 nd+ 1,04 “ e sss 0,0369 nd +1,15 ii e = 0,0538 nd + 1,53 “ t Example. If.a perpendicular water colunrin machine has cast-iron pipes of 10 inches inner width, how thick must these be at 100, 200, and 300 feet deptlis? From the for- mula, for 100 feet pressure, this thickness is: = 0,00238 . -1— . 10 + 0,33 == 0,07 + 0,33 0,40 inches ; ’ 33 for 200 feet, = 0,14 + 0,33 = 0,47 inches; and for 300 feet pressure, ==: 0,22 + 0,33 — 0 55 inches. Cast iron conducting pipes are commonly proved at 10 atmospheres, for which reason, e = 0,0238 . d+ 0,33 inches; therefore, for pipes of 10 inches width, the . thickness e = 0,24 + 0,33 = 0,57 inches must be given. Remarks. The thickness of the sides of steam-boilers will be consjdered m the Second Part Concerning the theory of the strength of pipes, a treatise by Brix, in the “ Ver- handlungen des Vereins zur Beforderung des Gewerbfleiszes in Preuszen,” Jahrgang 1834, may be consulted. The technical relations and the provmg of pipes are fully treated’ *of in Hagen’s “Handbuch der Wasserbaukunst,” vol. i., and in Gemeys’ “Essai sur les ‘moyens de conduire, &c., les eaux. ' [For a view of the general principies governing the construction and strength of cylindrical steam-boilers, the editor may refer to his paper on that subject read before the Franklin Institute, July 26,1832, and published in its Journal, in which the relation stated in the text, be- tween the strength retjuired in the direction of the curvature and that in the direction of the length of the tube or boiler, was pointed out, accompanied by a table of diameters and thicknesses of boilers, with the tenacities per inch of iron required in each direction for a given pressure. See, likewise, American Journal of Science and Arts, vol. ;xxiii. No, 1.] CH APTER II. ON THE EQUILIBRIUM OF WATER WITH OTHER BODIES. § 284. Buoyancy.—A body immersed under water is pressed upon fby the water on all sides, and now the question arises as to theBUOYANCY. 323 Fig. 366. Fig. 367. amount, directiori and point of application of the resultant of all these pressures. Let us imagine this resultant to consist of a vertical and a horizontal component, and determine these forces according to the rules of § 282. The horizontal pressure of the water against a sur- face is equivalent to the horizontal pressure against its vertical pro- jection, but now every projection of a body, Fig. 366, is at the same time the projec- tion of the fore part ADC and the back part ABC of its surface; hence, also, the horizontal pressure of water against the back portion of the surface of a body is equal in amount tothat °f the front portion, and as both pressures are exactly opposite, their resultant = 0. As this relation takes place for every arbitrary hori- zontal direction, and the vertical projection corresponding to this, it follows that the result- ant of all the horizontal pressures is nothing; that, therefore, the body AC below the water is equally pressed in all horizontal directions, and for this reason exerts no effort to move forward in a horizontal direction. | To find the vertical pressure of the water against the body BCSy Fig. 367, let us suppose it made up of ver- tical elementary prisms, AB, CDy &c., and determine the vertical pressures on their ter- jninating surfaces A and B, C and D. If the lenghts of these prisms are /2, &c., the depths of their upper extremities B, D, &c., below the surface of water HR : A,, h2, &c., and the horizontal transverse sections F,F2, &c-> vve then have for the vertical pressures acting from above downwards against the extremities, B, Dy &c., = F,Air, F2A2y, &c.; °n the other hand, the pressures acting from below upwards and against the extremities Ay C, &c., = F, (hx + lY) y, F2 (A2 + /2) y> &c*; and it now follows, from a composition of these parallel forces, that the resultant P = Fi {K + /,) y +F2 (A2 + /,) r + • • • — FAr — F2A2y —. . < . ** (F,^ + F2/2 + •..) y = Fy, y represents the volume of the immersed body or the water dis- placed. Therefore the buoyancy or the force with which the water strives to push a body immersed from below upwards, is equivalent to the weig i °f water displacedy or to a quantity of water which has the same volume as the submerged body. Further, to find the point of application of this resultant, let ns Pu the distances AAV CCV &c., of the elementary columns ABy C JJy c., from a vertical plane HJY: a.y a2, &c., and determine the moments ot the forces with respect to this plane. If S is the point of applica ion °f the upward pressure, and SSX =* x its distance from that pnncipai plane, we shall then have: •*324 BUOYANCY. Vy . X = F, «i + + *Vi + Fih ai + F2 hy • ffl2 + >U + and hence, if Flf ^2» &c., re- Fig. 368. - 1 -1 • *, + V2 + ' ’ . present the contenls of the elementary columns. i>ince (trom § 100) the centre of gravity is accurately determined by the sarae formula, it follows that the point of applicatiori of the upward pressure coincides with the centre of gravity of the mater K 285 The weight G of the body.actmg in an opposite direction associates itself with the buoyancy of the body immersed or under water, and from the two there anses a resultant — Fy or = (c__i) VY, if * be the specific gravity of the body. If the mass of the body be homogeneous, the centre of gravity of the displaced water will coincide with that of the body, and hence this point will be the point of application of the resultant R; but if there be not homogeneity, then these centres of gravity do not coincide, and the point of application of the resultant R deviates from both centres of gravity. Let us put the horizontal distance SH Fig. 368, of both centres of gravity from each other* = b, and the horizontal distance of the point of application A sought from the centre of gravity S of the displaced water = a, we shall have the equa- tion Gb = Ra, from which is given: Gb Gb ii w — ^ n q___p Tf the immersed body be left to its own gravity, the three following cases may present themselves. Eitherthe specific gravity of the body k eoual to that of the water, or it is greater, or it is less than the sneTfic eravity of the water. In the first case the buoyancy ,s eqnal, specinc gra y the thir(j ,t 1S greater than the weight in the second 11 Ae first case, equilibrium subsists between of the water. » he bod must in the second case sink the weight and the ^»7^ ^ = (, _ i) Vy, and, in the third case, rise with the force G = (1_,) Vy. The rising goes on only as long as the mass of water V , cut ott' from the plane of the surface and displaced by the body, has the same weight as the entire body. The weight G = V ,y of the body BB Fig. 369, and the buoyancy 1 — VlV now constitute a couple, by which the body is made to revolve until the directions of both coincide, or until the centre of gravity of the body lies in one and the same vertical line with the centre of gravity of the displaced water. The line passing through the centre of gravity of the fioating body and through that of the displaced water, is called the axis of floatation ; and on the other hand, the section of the body formed by the plane of the surface of the water, the plane of floatation. Every plane which divides a body, so that one part is to the whole as the specific gravity of the body to that of the fluid, and that the centres of gravity of the Fig. 369.DEPTH OF FLOATATION. 325 two parts lie in a line norraal to this plane, is a plane of floatation of the body. § 286. Depth of Floatation. If the figure and weight of a floating body be known, the depth of immersion may be calculated beforehand, with the help of the F‘g- 37°* previous rule. If G be the weight of the body, we may then put the volume of the displaced water V = — \ if we combine with it the ste- reometrical formula for the volume V, we shall obtain the equation of condition. Hence, for the prism ABC, Fig. 370, with vertical axis, for example, V = Fy, if F represent the sec- tion and y the depth BD of immersion, Fy = G G — and y = -------- For a pyramid ABC, Fig. 371, whose vertex y Fy floats under the water, V = if f represents the section at the distance of unity from the vertex; hence it follows, that: and hence the depth CE = y = s|^~* y \J y Fig. 371. F»g- 372. For a pyramid ABC, Fig. 372, floating with its base below the water, the distance is given CE = yx of the vertex from the surface, from the height h of the entire pyramid, if we put:_________ For a sphere AB, Fig. 373, with the radius CA = r, Fig. 373. Fig. 374. 28326 DEPTH OF FLOATATION. V =* * y2^ r— hence we shall have to solve the cubic equation tf — 3rf + — = 0, to find the depth of immersion DE of the *y sphere. For a floating cylinder JlG, with horizontal axis, Fig. 374, of a radius BC = DC = r, if a° be the angle BCD subtended at the centre by the are immersed, the depth of immersion rfh = y —- r (1 ___ cos. 1 «), but to find the arc immersed, we must put the volume of the water displaced — tothe segment ___less the triangle r* sin. » mu]tip]ied by the length GK = / of the cylinder; therefore, 2 2 C . EV ‘r (a — a) _ -s- and solve the equation 2 y sin. approximation, with respect to a. Examples.—1. A wooden sphere, of 10 inches diameter, floats 4$ inches deen tl volume of water displaced by it is then: *5 16 ».81 .7 567.i : 222,66 cubic inches, r di whilst the solid contents of the sphere are —------ o r. 10® 6 523,6 cubic inches. From this, 523,6 cubic inches of the mass of the sphere weigh as much as 222,66 cubic inches of water, and it follows that the specific gravity of the former is: 222^^0,425.. 523,6 . , 2. How deep will a wooden cylinder of 10 inches diameter and specific gravity c = 0425 sink» = », = 0,425.»=1,3352 ; now a table of seE. ’ 2 Ify ments gi ves for the area —---== 1,32766 of a circular segment, the angle subtended at the centre by the are «°= 166°, and for -—= 1,34487, the same angle = 167°; hence, simply, the angle subtended at the centre corresponding to the sljce 1,3352 is: „«,a “° = 166°+ 10=1,660 + ! “ethe depth of immersion: y = r (1 — eo». 4 «) = 5 (1 — cos. 83° 13') = 5.0,8819 = 4,41 inches. § 287. The determination of the depth of immersion occurs chiefly in the case of ships, boats, &c. If these have a regular form, the depth may be calculated from geometncal formulae; but if this regu- larity fails, or the law of configuration is not known, or if the form is very complex, the depth of immersion must then be determmed by experiment. , An example of the first case is in the body ACLEG(a pointed scow), represented in Fig. 375, bounded by plane surfaces. It consists of a pa- rallelopiped^CE, and of two four-sided pyramids J?r Gand CEL, form- ing the head and the stegn, and its plane of floatation is composed of a parallelogram and two trapeziums MO and STJy and cuts off a bulk of water, consisting of a parallelopiped MCS, and two triangularDEPTH OF FLOATATEON. 327 prisms PJYR, and two quadrilateral pyramids BQP. If we put the length AD of the middle portion = /, the breadth AF = b, and the Fig. 375. depth AB = h; further, the length GW of each of the two en ds = c, and the depth of immersion, i. e , BM = CT = y, the immersed part Jl/CW of the middle portion will be: = MJV x MI" x = /6y. The base of the quadrilateral pyramid BQP is BM. BR, and the height PJ> hence the solid contents of this pyramid = £ BM. BR • P J. But now: and likewise: PJ-n0W-l‘- x- hence the contents of both pyramids are: -2 i i/ The transverse section of the triangular prism A^Ois £ . P J == £ y . ^ and the side hence the solid contents of both prisms are: By addition of the three volumes found, the volume of the water dis- placed is known: V~ bly + i W + (l+ £Jf_ 1 . 5 A* ^ A A2 \ A 5 A3/ ^°w the gross weight of the boat = G, we then have to put: A QI2 ^ /A2 y3 —3 hy2-------y -f 3 h2G bcy = 0. C OCy "Ihe depth of immersion y is determined from the loading by the solu- tion of the last cubic equation.328 STABILITY. Examples.—1. If the length of the middle portion / = 50 feet, the length of each end c = feet, the breadth 6 = 12 feet, and the depth h = 4 feet, with a depth of immer- gi011 y = 2 feet, the whole weight amounts to G= [50+15 .f-|. 15. (f)’]. 12.2.62,5 = (50+7,5—1,25) . 24.62,5 = 87235 Ibs. 2. If the ciear weight of the former boat amount to 50000 Ibs., we shall have for the depth of immersion: y3— ]2ya—160y + 202,02 = 0. By trial, it is easily found that this equation may be answered pretty accurately by y = 1,17, whence the depth of im- mersion sought may be taken as great. Remark. To know the weight of the load of a ship, a scale is attached to both sides, which is called a water-gauge. The divisions are made from experiment, while it is ob- served what loads correspond to delinite imrnersions. » § 288. Stability.—The floating of bodies takes place either in an upright or an oblique position ; and further, with or without stability. A body, a ship, for example, floats uprightly, if one plane throughthe axis of symmetry is a plane of symmetry of the body; and a bo^ly floats obliquely if it is not divided by any of the planes, which may be car- ried through the axis of floatation into two congruent halves. A body floats with stability, if it strives to maintain its state of equilibrium (compare § 130); if, therefore, mechanical eflect isto be expended to bring it out of this position, or if it returns of itself into a position of equilibrium after having been drawn out of one. On the other hand, a body floats without stability if it passes into a new position of equi- librium after having been brought out of one by a shock or blow. If a body ABC, Fig. 376, floating at first uprightly, is brought into an inclined position, the centre of gravity S of the water displaced passes from the plane of symmetry EF, and assumes a position Sx on the larger half immersed. The buoyancy applied at S: P = Vy9 and the weight applied at the centre of gravity C of the body, viz., G = — P form a couple by which (§ 90) a revolution is produced. About what e ver point this revolution may take place, the point C, yielding to the weight G, will always go down, and S19 or another point M of the vertical SXP, obedient to the force P, will rise, there- fore the plane of symmetry, or of the axis EF of the ship, will be drawn downwards at C, and upwards at Af, and hence it will remain upright if A/, as in the figure, lie above C, or incline itself stili more Fig. 376. Fig. 377. as in Fig. 377, if Af lie below C. From this, then, the st abdit y of a floating body, or ship, is dependent on the point Af, in \\ 1C e v er- tical through the centre of gravity Sx of the displace vva er in er- sects the plane of symmetry. This point is called e me acen re.STABILITY. 329 Fig. 378. It follows, therefore, from this that a ship or other body floats with stability if its metacentre lies above the centre of gravity of the ship, and without stability if it lies below, lastly if the two points coincide, it is in a state of indifferent equilibrium. The horizontal distance CD of the metacentre M from the centre of gravity C of the ship, is the arm of the force of the couple consti- tuted of P, and G = ____P, and hence the moment of the last is the measure of its stability = P .CD. If we represent the distance CM by c, and the angle of revolution SMS1 of the ship, or of the plane of its axis, by *°, we obtain for the measure of stability S = Pc sin

and the couple (Q,—Q), or what comes to the same thing, an opposite force applied at Sx keeps in equilibrium a force—P applied at S together with a couple (Q,— Q), or more simply, the couple (P,—P)is in equilibrium with the couple (Q,— Q). If now the transverse section HER—H1ER1 of the part of the ship iramersed = P, and the section HOHl=RORl of the space by which the ship is drawn up on the one side, and down on the other = Fx; if, further, the horizontal distance KL of the centre of gravity of these spaces = fl, and that of MT of the centres of gravity S and tf^or the horizontal projection of the space SSl which S describes during the rolling = s, we have then from the conditions of equilibrium of the two couples: ^ Fs = hence s = L a and SM = IfL = = tth; F sin. 4> sin. f t sin. * The line CM = c, appearing as factor in the measure of t e sta- bility is = CS + SM; hence, if further, we represent the distance CA 28*330 STAB1LITY. of the centre of gravity C of the ship frora the centre of gravity of the displaced water by e, we obtain the measure of the stability S = Pc sin. $ = P (^yr + e If the angle of revolution be small, the transverse sections HOHx and RORx may be regarded as equally small triangles; if we repre- sent the breadth HR = HXRX of the ship at the place of immersion by 6, we may then put Fx = \ .\b. \b$ = J i2 f > and KL = a = 2 . § — = § 6, as also sin. = $>, from which the stability is: s-f<*Y+e,>~(£? If the centre of gravity C of the ship coincides with the centre of gravity S of the displaced water, we then have e = 0, hence: ° b'* 1 S = _ . Pand if the centre of gravity of the ship lies ^elow that of the displaced water, we then have e negative ; hence S = — e^j P t- It follows also that the stability of a ship is nothing, + e b3 Fig. 379. if e be negative and at the same time e = p* It is seen from the results obtained, that the stability comes out greater, the broader the ship is, and the lower its centre of gravity lies. Example. A rectangular figure Fig. 379, of the breadth AB = 6, height AE = h, and depth of immersion EH= y, F = &y, and e = — ^ ^ ; hence, the amount of stability is:S=P&>( —------^-4- —Y or if the specific gravity Y\12 by 2 ~ 2 / 7 of the mass of the body be put = i, s=Pf(——— (i—«) Y \12 Ai 2 ’ Hence, the stability ceases if 6’ = 6 h1 i (1 — ■), i. e., if A = v/6.(1 For , = 4 1 = = v/T = n h 1,225; if, therefore, the breadth is not 1,225 of the height, the body will float without any stability. [The principies explained in this section apply not only to the con- struction and use of vessels of every description, and to the ballasting and lading of ships themselves, but likewise to the loading of floating docks with vessels, including their cargoes or armaments. The form of the floating mass, and the position of its centre of gravity, together with its absolute weight, must be taken into account, as well as the density of the liquid in which it floats. Thus, a floating dock Dy Fig. 379^ in the form of a rectangular prism, capable of being closed at the ends after having received the ship S9 and of being freed from water, will be subject to exact calculation, if its weight and centreSTABILITY. 331 of gravity be known, together with the weight of the vessel and its centre of gravity. Example.—Adraitting that Dis 90 feet wide, 36 feet Fig. 379t. high frorn b to t> and 250 feet in iength—that its weight, including ballast, is 3860 tons, and that its centre of gravity c is fourfeet above the bottom d; also that the ship S, weighing 5200 tons, has been re- ceived and securely shoared in place, having its centre of gravity cx 25 feet vertically above the bottom of the keel, supposed to be 1 foot above c, then the coramon centre of gravity C is ^ x 5200 ^ 14 92 6 J 9060 ft. above the point c, and 18,92 feet above the bottom of the dock. The total displacement in sea water(641bs. per cubic foot) will be 337500 cubic feet; and, consequently, in a state of repose, it will sink to the depth of bH= ScSr = 15feet- wiH be the “load line,” and 250,90 the common centre of gravity of ship and dock C will be 3,92 feet above it. Should any force acting in a horizontal direction careen the dock so as to make the angle AwH = = 22° 25', depressing the side AHy 18 feet, what force must be applied 25 above the water line to keep the dpck in this position, i. e. what is now the restoring power/ The centre of gravity of the original quadrangular prisrn of dis- placement HRfb is at O, that of the new triangular prism EAb is at Or The weight of ship and dock C acting at the point C is 9060 tons, which is also the force P of the prism of water acting upward at332 OBLIQUE FLOATATION. Oj—tending to restore the position of the dock. The original centre of displacement 0 is 18,92 — 7,5 = 11,42 feet below the centre of gravity of ship and dock. The area of the immersed section which is transferred by the careening from one side to the other is 401,5 sq. feet, and the distance transferred ii1 = 58,5 feet, hence = = 3=5 *he horizontal distance, the centre of the whote displacement has been removed by the inclination supposed ; and sin. 22^ 25 . 17,4 = rad. : 45 63 ft. = height of the metacentre above the original centre of buoy- ancy O. Again, putting r, = 11,42, we have r, : *> 11,42 : 4,35 feet; hence, eet, the “ equilibrating lever,” or distance apart of CG and PO, is 17,4 —4,35 = 13,05 feet, and the statical moment is, therefore, 9060.13,05 = 116058,6 ft.-tons; which, for a distance of 25 feet from the centre of oscillation, C gives a stability or re- storing powTer of 4642 tons.] § 290. Oblique Floatation.—The formula S— P e sin. pj for the stability of a floating body may be also applied to find the dif- ferent positions of floating bodies, for if we put = 0 we obtain the equation for a second position of equilibrium, whose solution leads to the determination of the corresponding Fig. 380. angle of inclination. The equation, therefore, « sin.

. The transverse section of a parallelo- piped AD, Fig. 380, is F= HRDE = BBiaMp HfliDE = by, if b be the breadth AB SjiBfBBBlJ = HR, and y the perpendicular depth EH = DR ; further, the transverse sec- tion Fx = IIOHl = RORl as a rectan- gular triangle with the leg OH = OR = \ b, and the leg: HH, = RR, = h btanS- ♦. = i tmS- If, further, the centre of gravity F is distant from the base FU = J b tang. f, and if from O about OU = % OH= £ b, it fol- lowrs that the horizontal distance of the centre of gravity F from the middle O, = OK = OJY + JVK = OU cos. * + f = £ cos. J b tang. sin. , and the arm: a = KL = 2 OK = % b cos. * + sin. cos. $ According to this the equation for the oblique position of equili- brium is: i tang. $ b cos. 2 -f £ b sin. $2) --------2----------------—-A-------L2— € sin. t = 0, by cos. $> Sln’ * = tang. (p being substituted, or, cos. T i»*) 6J—ey] = 0; which equation will be satisfied by: sin. t = 0 and by tang, $ 1. The first equation, when 9 * 0, corresponds to upright, and the second to oblique floatation. The possibility of the latter requires that If now A be the height of the parallelopiped, and « its specific gravity, we then have: y ssx t h and e =— ^ ^ (1—«)-, hence it follows that 4 yo * (1—*) h% tang. f — s/21------*-----------1, and the equation of condition of oblique floatation is: A I 1 b \ 6 i (1-----i) Examplet.—1. If the floating parallelopiped is as high as it is broad, and has a spe- cific gravity c sas J, then the tang. f is «as y/ 2 y/ 3 . $ — 1 = >/3 — 2 = 1,* hence, = 45°. 2. If the height A « 0,9 of the breadth 6, and the specific gravity J, we have then tang. f » ^/3.0,81 — 2 «= ^/0,43 as 0,6557; hence, f = 33° 15'. § 291. Specific Gravity,—The law of buoyancy of water raay be applied to the determination of the density, or the specific gravity of bodies. From § 284, the upward pressure of water is equal to the weight of liquid displaced; hence if V is the volume of a body and yx the density of the liquid, we then have the buoyancy P = Vyx. If now yt be the density of the mass of the bodies, we then have the G weight of the body G=*Vy%; hence the ratio of the densities ^ , 7i # P i. e. the density of the body immersed is to the density of the fiuid as the absolute weight of the body to the buoyancy or loss of weight by immersion. Therefore, y2 G t — n. and 7l _P G y**> or if y be the density of water, *x the specific gravity of the liquid, and that of the body, Q p then will 7l =* fly, and ya *= «2y, *2 = — §x, and tx=s — «2. If, there- fore, the weight of a body or its loss of weight by immersion is known, then the density or the specific gravity of the mass of a body may be found from the density or specific gravity of the liquid, and inversely, the density or specific gravity of the first, from the density or specific gravity of the last, If the fluid in which the solid body is weighed is water, we then have #1 =■ 1, and yx *■ y » 1000 kilogrammes, or 62,5 lbs., accord- ing as we take the cubic metre or cubic foot for unit of volume, hence for this case the density of the body is:334 SPECIFIC GRAVITY. r, - -S- r = $SOlute weight times the density of water, tr loss ot weight and the specific gravity: G __ absolute weight *2P loss of weight To estimate the buoyancyor loss of weight, as well as to determine the weight C, we make use of an ordinary balance, only that below one of the scale pans of this balance there is appended a hook, to which the body may be suspended by a fine thread or fine wire, whilst it dips into the water contained in a vessel underneath. A balance arranged for the weighing of bodies in water is commonly called a hydrostatic balance. If the body whose specific gravity we wish to determine is lighter than water, we may connect it mechanically with another heavy body, so as to make it sink. If this heavy body loses the weight P2, and the system the weight Pv the loss of weight of the lighter body: P-- Px—P2y now if G represents the loss of weight of the lighter body, we have then its specific gravity: G G P- p-P; If the specific gravity of a mechanical combination, or a composi- tion of two bodies, and the specific gravities of their constituents f and i2 are known, from the weight of the whole, the weights Gx andX G2 may be estimated. In every case Gx 4- G2 = G, and also the volume—1 + volume —1 =* volume —, therefore: GL G «2r *r By combining these equations we have : «■-«(HKHH «.-«(HMK)- Examples.—1. If a piece of limestone, weighing 310 grains, becomes 121,5 grains lighter when under water, its specific gravity is i = = 2,55.— 2. To find the 121,D specific gravity of a piece of oak, round which a piece of lead has been wrapped, and which has lost by being weighed in water 10,5 grains; if now the wood itself weighed 426,5 grains, and the system under water was 484,5 grains lighter than in the air, the specific gravity of the mass of wood would be: ___ 426,5 _ 426,5 __ n Q * “484,5 — 10,5 474 ’ ’ 3. An iron vessel, completely filled with quicksilver and perfectly closed, has a net weight of 500 lbs., and has lost 40 lbs. in the water; if now the specific gravity of cast- lron =s 7,2, and that of quicksilver is 13,6, the weight of the empty vessel is: Gt = 500 ____L-W-________ \500 13,6/ * \7,2 13,6/ *■ 500 (0,08 — 0,07353) -j- (0,1388 — 0,0735) _ 500.0,0<7647 3235 **— — - ss 49,5 lbs., 0,0653 65,3AREOMETER. 335 and the weight of the enclosed quicksilver: G,=500 (0,08 — 0,1388) : (0,07353 — 0.13S8) = 500.0.0588 0,0653 _2940_ 4 )bs 6,53 Remark 1. For the determination of the specific gravities of liquids, meal, corn, &c., the mere weighing in open air is sufficient, because we may give to the bodies any volume at will, by filling vessels with them. If an empty bottle weighs = G, and the same filled with water G., and the weight G, if it contain any other substance, we shall 1 Q _ Q then have the specific gravities of masses of these : * = —?——. For example, to find G j — G the specific gravity of rye (not rye grains), a bottle is filled with the grains, and after tnuch shaking, then weighed. After deduction of the weight of the empty bottle, the weight of the rye was = 120,75 grms., and the weight of an equal quantity of water J 120 75 = 155,65; the weight of the rye is aocordingly = — - s=r 0,776; and, therefore, 1 cubie fbot of this grain weighs a 0,776.62,5 = 48,5 lbs. Remark 2. The problem solved by Archimedes of finding the ratio of the constituents from the specific gravity of a mixture, and from the specific gravity of its constituents, admitsonlyof a limited application to Chemical combinations, metallic alloys,&c., because a contraction or expansion of the mass generally takes place, so that the volume of the mixture is no longer equal to the sum of the volumes of the constituents. Remark 3. The further extension of this subject, namely, its application to the measure- ment of volume, &c., belongs to physics and chemistry. § 292. Areometer.—Areometers are principally used to determine the density of liquids. These instruments are hollow bodies, formed about a symmetrical axis, whose centres of gravity lie very low, and by floating perpendicularly in liquids, give their density. They are made of glass, brass, &c., and are called, ac- cording to the various purposes for which they are intended, hydrostatic balances, salzometers, hydrometers, alcoholo- meters, &c. There are two kinds of hydrometers, viz., the weight and the scale hydrometer. The first are often used for the determination of the weights, as was the specific gravity of solid bodies. 1. If Fbe the volume of the portion of a hydrometer ABC, Fig. 381, floating freely, and immersed up to a certain mark O in the water, G the weight of the whole balance, P the weight placed upon the piate while float- ing in the water, whose density may be = y, and P1 the weight required to be put on to make it float in any other liquid of the density yv we shall then have Vy = P + G and VyJ = + G; hence, Vi = Pi + G 9 Tf D ^ P G ^he weight which must be put upon the piate to make C* bf ro*1?eter MC% Fig. 382, sink up to a mark O, and Px the weig which must be put upon A, together with the body to be ^eig e , to obtain the same immersion, we shall then have simply the weight of this body Gx = P— Pv But if P1 must be augmented y 2, when the body to be weighed is put into the cup D under the Fig. 381.336 AREOMETER. Fig. 382. Fig. 383. surface, to preserve the depth of immersion unchanged, P2 will then be the buoyancy, and hence the specific gravity of the body: i 2 x 2 Those hydrometers which have a cup suspended below for the determination of the specific gravities of solid bodies, minerals for instance, are called Nicholson’s hydrometers. 3. Let the weight of a hydrometer ABC, Fig. 383, == Gy and the volume immersed, if this balance floats in water, = Vy then G = Vy. If the balance rise by OX = Xy when immersed in a heavier liquid, for the transverse section F of the stem, the volume immersed is = V — FXy and hence G = (V — Fx) 7l; the two formulae, divided by one another, give the density of the liquid: y* = V—Fx ‘ y =7^(1 “ TX) = **). If the liquid in which the hydrometer is immersed, be lighter than the water, it will sink in it to a depth x, for which reason, G — (V + Fx) y, and hence we must put y, = y-s-(l + /**)• To find the co-efficient p = —, the balance is loaded with a weight P of quicksilver, which is poured in and takes the lowest position, so that while floating in water, a considerable length l of the stem to which the scale is applied, sinks lower down. If now we put = , we shall then obtain : F _ P _ P 11 ~ T~ TTV~ Gf Examples.—1. If a Nicholson s hydrometer weighs 65 grains, 13,5 grains must be taken olf the piate, that it may sink to the same depth in alcohol as it does in water; the specific gravity of alcohol is = ^---—----= 1—0,208 = 0,792. 2. The normal weight of a Nicholson?s balance is 1500 grains, i. e. 1500 grains require to be put on to make the instrument sink to 0; from this 1030 grains must be taken by the weighing of a piece of brass placed upon the upper piate, and 121,5 to be added if this body is placed on the lower piate. The absolute weight of this piece of brass is therefore = 1030 grains, and* its specific gravity = 1030 = 8,47.—3. A scale areometer, of 1162 grains weight, after • 121,5 having been lightened by 465 grains, rises 6 inches, and has therefore the co-efficient u, 465 ' = -r~1AO - = —— = 0,00686. After complete filling and restoration of the weight of i lo^.b 6772 1162 grains, it ascends, when floating in a saline solution, 2 yg inches; hence, the specific gravity of this is : = 1 H- ( 1 — 0,00686 x??) = l-r 0,983 = 1,02. Remark. The further extension of this subject belongs to physics, chemistry, and tech-LIQUIDS OF DIFFERENT DENSITIES. 337 § 293. Liquids of Different Densities.—If several liquids, of dif- ferent densities, are in the same vessel, without their exerting any Chemical action upon each other, from the ready displacement of their particles, they arrange themselves above each other, according to their specific gravities, viz. the densest below, then the less dense, and then the lightest. The limiting surfaces are also in a state of equili- brium, as likewise the free surface horizontal; for as long as the surface of limitation EF be- tween the masses M and JV, Fig. 384, is in- clined, columns of fluid, of different densities, like GK, G^, rest on the horizontal stratum HR, and hence the pressure on this stratum will not be everywhere the same ; and lastly, no equilibrium will subsist. In communicating tubes AB and CD, Fig. 385, the liquids arrange themselves one above the other, accordingto their densities, only their surfaces A and D do not lie in one and the same level. If F be the area HR of the transverse section of a piston, Fig. 386, in the one branch AB of two communicating tubes, and the height of pressure or Fig. 3S6. the height EH of the surface of the water in the second tube CD above IIR, = A, we then have the pressure against the surface of the piston P = F hy . On the other hand, if we replace the pressure of the piston by a column of liquid AH, Fig. 386, of the height AH = A, and the density y„ we then have P = F\ 7l; and equating both expressions, we obtain the equation hx 7l = A y or the proportion There/ore, the heights of pressure in communicating tubes, for the subsistence of equilibrium between two different liquids, or the heights of the columns of liquid measured from the common plane of contact, are inversely as the densities or specific gravities of these liquids. As mercury isof about 13,6 times the density of water, a column of mercury, in communicating tubes, will hold in equilibrium a column of water of 13,6 times the heidit. 29 5 Fig. 384.338 TENSION OF GASES. CHAPTER III. ON THE EQUILIBRIUM AND PRESSURE OF AIR. § 294 Tension of Gases.—The atmospheric air which surrounds us as well as all kinds of air or gases, possesses, in virtue of the repulsi ve force of its parts or molecules, a tendency to occupy a greater and greater space; hence, we can obtain a hmited mass of air only by confining it in perfectly closed vessels. The force with which gases endeavor to dilate themselves is called their elasticity, tension, or expansive force. It exhibits itself by pressure against the sides of the vessels which enclose it, and so far differs from the elas- ticity of solids and liquids, that it manifests its action in every condi- tion of density, while the elasticity of the last-mentioned bodies in a certain state of expansion, is nothing. The pressure or tension of air and other gases is measured by the the manometer, and the valve. The barometer is chiefly used for determining the pressure of the atmosphere. The common, or as it is called, the cistem barometer, Fig. 387, consists of a glass tube, closed at one end and open at the other which, when filled with mercury, is inverted, and its open end immersed in a cistem likewise containing mercury. By the in- version of this instrument, there remains in the tube a column of mercury BS, which (§ 393) is sustained in equilibrium by the pressure of the air on the sur- face of mercury HR. The space JIS above the mer- curial column is deprived of air, or a vacuum; hence, there is no pressure on this column from above, for which reason, the height of the mercurial column above the surface of mercury HR in the cistem, serves for a measure of the air’s pressure. To measure this height with precision and convenience, an accu- rately divided scale is appended, which runs lengthvwse along the tube. A more particular description of the different barometers, and an explanation of their uses, &c., belong to the department of physics. § 295. It has been found by the barometer, that for a certain mean state of the atmosphere, and at places very little above the level of the sea, the air’s pressure is held in equilibrium by a column of mer- cury, 76 centimetres, or about 28 Paris inches = 29 Prussian inches = 30 English inches nearly, (29,994 exactly.) As the specific gravity of mercury is nearly 13,6 (13,598), it follows that the pres- sure of the air is equivalent to the weight of a column of water,MANOMETER. 339 0,76 . 13,6 as 10,336 metres ars 31,73 Paris feet = 32,84 Prussian feet = l?’598 • 29)998 12 33,988 English feet. The tension of the air is very often measured by the pressure it exerts upon a unit of surface. Since a cubic centimetre of mercury weighs 0,0136 kilogrammes, the pressure of the atmosphere, or the weight of a column of mercury 76 centimetres high on a base of 1 centimetre square, =» 0,0136.76 =» 1,0336 kilogrammes, and since 66 13 6 a cubic inch of mercury weighs ——’ == 0,5194 Prussian Ibs., X / 62,5 . 13,6 0,491 lbs. English, the mean pressure of the atmo- or 1728 sphere is then *= 29.0,5194 = 15,05 Prussian lbs. on the square inch, = 2167 lbs. on the square foot, and ir\ English measure = 30.0,491 sa 14,73 on the square inch, = 2131,12 lbs. avd. on the square foot. 14,76 lbs. per square inch is the Standard usually adopted. In mechanics, the mean pressure of the atmosphere is commonly taken as unity, and other expansive forces referred to this and assigned in atmospheric pressures, or atmospheres. Hence, to a pressure of n atmospheres corresponds a mercurial column of 30 . n inches, or a weight of 14,73 lbs. on each square inch; and inversely, to a mercurial column of h inches corresponds a tension of A -- 0,03571 h or = A * 0,0333 h atmospheres, and to a pressure of p lbs. on the square inch, a tension of jjAg * 0,0644 p or -A^ == 0,0678 atmospheres. Th' e'>',ati°n,te of reduction h = 1,8604 p inches and p =* 0,5375 h lbs., or h = 2,036 p inches, and p = 0,491 h lbs. English. For a tension h inches = p lbs., the pressure against a plane surface of F square inches: P = Fp = 0,491 Fh lbs. English, or — 0,5375 Fh lbs. Prussian. Examples. 1. If the water in a water-pressure engine stands 250 feet above the sur- face of the piston, the pressure against the surface will then be _ 7 35 at. ^essur^on^verv °1& cy,indrical be,,ows has a tension of 1,2 atmospheres, its I ssure on every square inch = 1,2 . 14,73 = 17,676 ibs., and on Ihe surface of the piston of 50inches diameter =1^21.17,676*34707 lbs. As the atmosphere exertsa P sure - . 14,73 ss 28922,3 lbs., the pressure on the piston is «= 34695— 28922,3 s= 5784,7 lbs. in ^e fens,°n of gases or vapors enclosed whir*h or« * I?iS *uments similar to the barometer are made use of, rnrv nr w *a e<^ n^nome{ers» These instruments are filled with mer- j ater, and are either open or closed; but in the latter case,340 MANOMETER. the upper part is either a vacuum or full of air. The vacuum mano- meter, Fig. 388, differs little from the ordinary barometer. To measure by this instrument the tension of air in a reservoir, a tube GK is fitted in, one end of which G passes into the reservoir, and the other K projects above the surface of mercury CE in the cistern of the instrument. The space EFHR above the mercury is hereby put into communication with the air-holder, and the air in it assumes the tension of the air in the holder, and forces into the tube a column of mercury OS, which sustains in equilibrium the pressure of the air which is to be measured. The siphon manometer, ABC, Fig. 389, open above, gives the Fig. 388. Fig. 389. Fig. 390. excess of tension above the pressure of the atmosphere in the vessel MN) because the pressure of the atmosphere on Sy joined to that of the mercurial column RS, is in equilibrium with the tension. If J be the height of the barometer, and h that of the manometer, or the difference of heights RS of the surfaces of mercury in both branches of the manometer, we shall then have the tension of the air com- municating with the shorter branch measured by the height of a column of mercury: bx = b + A, or the pressure measured on a square inch p = 0,491 (b + h) lbs.; or if b be the mean height of the barometer, p = 14,73 + 0,491 h lbs. Cistern manometers, Fig. 390, ABCE, are more common than siphon manometers. As the air here acts through a greater quantity of mercury or water, as it may be, upon the column of fluid, its oscil- lations do not so quickly affect the column of fluid, and its measure- ment, when thus at rest, is rendered both easier and more accurate. For the sake of convenience of measuring by, or reading off from the scale, a float is not unfrequently attached to it, wThich rests on the mercury, and is connected with an index hand, accompanying the scale by means of a thread passing over a small roller.LAW OF MARIOTTE. 341 The expansive force of a gas or va- por enclosed in MN may be likewise determined, but with less accuracy, by the help of a valve DE, Fig. 391, if the sliding weight is so placed that it is in equilibrium with the pressure of the air or vapor. If CS = s be the distance of the centre of gravity of the lever from the fulcrum C, CA *= a the arm of the weight, and Q the weight of the lever with its valve, we then have the statical moment wTith which the valve is pressed down by the weight *= Ga + Qs ;* if now the pressure of the gas or vapor from below = P, the pressure of the atmosphere from above = Pv and lastly, the arm CB of the valve = d, we then have the statical moment with which the valve strives to lift itself up = (P — Px) d> and by equating the moments of both : Pd— Pxd = Ga + Qsy and P = Px + » If r represent the radius J DE of the valve, p the internal and px the external tension, measured by the pressure on a square inch, we then have: P = h r*p and Px = * i^px; hence, p = px + Fig. 391. Examples.—1. If the height of mercuryof a manometer, open above, is 3,5 inches, but that of the barometer 27 inches, the corresponding expansive force is then A = 6 —J— A, = 27 + 3,5 = 30,5 inches, or p = 0,491 . h = 0,491.30,5=14.97 lbs.—2. If thej height of a water-manometer is 21 inches, the expansive force corresponding to this, with the height of the barometer at 27 inches, is : h = 87 + -|L — 28,54 inches =14,01 lbs. English. 13,6 3. If the statical moment of an unloaded safety-valve is 10 inch lbs., the statical mo- ment of a 10 lbs. sliding weight 15.10 = 150 inch lbs., the arm of the valve measured, from the valve to the fulcrum, 4 inches, and the radius of the valve 1,5 inches, then the difference of the pressures on both surfaces of the valve is: „ „ 150+ 10 160 -AA1. P ~Px = iTo,5)^74 = 97 = ’ lbS- Were the pressure of the atmosphere p, = 14 lbs., the tension of the air below the valve would from this amount top= 19,66 lbs. § 297. Law of Mariotte.—The tension of gases increases with their density; the more a certain quantity of air is compressed or condensed, the greater is its tension; and the greater its tension, the more it is allowed to expand or become rarefied, the less does its ex- pansive force exhibit itself. The ratio in which the tension and the density, or the volume of the gases, stand to each other, is expressed by the law discovered by Mariotte, and named after him. This law assumes that the density of one and the same quantity of air or gas is proportional to its tension ; or, as the spaces which are occupied by one and the same mass are inversely proportional to the densities, If the weight of the lever and valve be counterpoised by a weight attached to a cord, passingupwards and over a pulley above S. the statical moment is reduced to Ga. Am. Ed. 29*342 LAW OF MARIOTTE. that the volume of one and the same mass of gas inversely as its pansive foret. Accordingly, if a certain quantity of air becomes com- pressed to one-half its original volume, its density is therefore doubled, its tension is also as great again as^ at first \ and on the other hand, if a certain quantity of air be ex- panded to one-third of its original bulk, therefore, its density reduced one-third, its elasticity "will be equal to one-third only of its original tension. If atmo- spheric air, for example, under the piston of a cylinder AC, Fig. 392, be supposed to press with 15 Ibs. on every square inch, it will press on the piston with a force of 30 lbs., if this piston be pushed to E,FV and the enclosed air compressed to one half its original volume, and this force will amount to 3. 15 = 45 lbs., if the piston come to and describes two-thirds of the whole height. If the area of the piston be 1 square foot, the pressure of the atmosphere against it will amount to = 144.15 = 2160 lbs.; hence, to press down the piston one-half the height of the cylinder, it will require 2160 lbs., and to push it down F ig. 393. two-thirds of this height 2.2160 = 4320 lbs. to be exerted. The law of Mariotte may be likewise proved by pouring mercury into the tube GH communicating with the air of a cylinder AC, Fig. 393. If a co- lumn of air AC be originally enclosed by the quan- tity of mercury DEFH, which has the same tension as the external air, and afterwards be compressed to one-half or one-fourth its volume by the addition of fresh mercury, we shall then find that the distances of the surfaces of GxHlt G2H2, &c., of mercury are equivalent to the single and treble height of the ba- rometer b, &c., that, therefore, if we add to this the single height, corresponding to the external pressure of the air, the tension will be twice or four times as great as that due to its original volume. The tensions are h and h, or p and y and y1 the corresponding densities, and V and F, the volumes appertaining to one and the satne quantity of air, we then have, according to the law laid down: y r," V K P_ Pi hence Vi From this the density and also the volume may be reduced from one tension to another. Example*.—1. If in a blowing machine, the manometer stand at 3 inches, . . 28 -p o __ **1 ___ i barometer is at 28 inches, the density of the wind is = —^ 2^ > whilst the d07 timesLAW OF MARIOTTE. 343 that of the external air.*—2. If a cubic footof atmospheric air, with the barometer at 28 inches, weighs _, lbs.; 770 * 62.5 770 with the barometer at 34 inches it will weigh: 34 _ 21250 28 21560 0,985 lbs. § 298. The mechanical effect vvhich must be expended to condense a certain quantity of air to a certain degree, and the effects which the air by its expansion will produce, cannot be directly assigned, because the expansive force varies at every moment of condensation or ex- tension; we must therefore endeavor to find a special formula for the calculations of this value. Let us imagine a certain quantity of air AF, enclosed in a cylinder ACy Fig. 394, by a piston EF, and let us inquire what effect must be expended to push for- ward the piston through a certain space EEX = FFV If the original tension = p, and the original height of the capacity of the cylinder = s0, and the tension after describing the space EEl = pv and the residuary volume of air = sv the proportion Px '• p = $0 1 si tben holds true, and gives Pi = ^-P- fi While describing a very small space EXE% = x, the tension px may be regarded as invariable, and hence the mechanical effect to be expended is = Apxx = when A represents the surface of the . si piston. It follows from the properties of logarithms, that a very small mag- nitude y = hyp. log. (1 + y) = 2,3026 Log. (1 + y\ if hyp. log. represents the hyperbolic, and Log. the common logarithms; we may consequently put Jlps0 — = Jps0 hyp. A + —\ \ s, / = 2,3026 Aps0 log. -f But now: hVP- l°g-(\ + J^j=hyp. log. = hyp- l°g- (s!+x}—hyp. log. i,; hence the elementary mechanical effect is: =*Aps [hyp. log. (sx+x)—hyp. log. ij. L»et us imagme the whole space EEX to be made up of very small parts, such as x, and therefore put EEX = nxy we shall find the me- chanical effects corresponding to all these parts, if in the last formula we substitute for “b + 2 xy sx + 3 xy . . .10^ + (n—I) x> an(^ Sx + Xy Sx + 2 Xy Sx + 3 Xy 8iiC.y tO Sj + 7lXy Of SQy Fig. 394. • For comparisons with the manometer, the division of a barometer ought not to be into either inches or metres, but into lOOOth parts of 1 atmosphere.—Am. Ed.344 STRATA OF AIR. and by summation the whole expenditure of mechanical efTect in de- scribing the space s0—s1: (hyp. log. {sx+x)—hyp. log. s, \hyp. log, (Sj+2x)—hyp. log. (sx+x) jhyp. log. (s,+3x)—hyp. log. (s,+2 x) L = J3ps0 / [hyp. log. (s.+nx)—hyp. log. [>,+(«—1)*] = jips0 [hyp. log. (*,+»*)—%>• log. s,] *=Jlps0 {hyp. log. s0—hyp. log. sj =* Aps0 hyp. log. (^~jt since one member in the one line always cancels one member in the following one. Since -f®. = bl. = £*, this mechanical effect may be put: h p L = Aps0 hyp. log. (A) = Jps0 hyp. log. If we put the space described by the piston s0—s, =» s, we shall hence find that the mean force of the piston p condensing the air is in the proportion A b. ii, P = — = Ap hyp. log. (£). h p s « W. . Let ./3 = 1 (square foot) and «0= 1 (foot), weobtam the mechanical eflect produced L = p hyp. log. = 2,3026 p log. This formula gives the mechanical effect which must be expended to convert a unit or cubic foot of air of a lower pressure or tension p into a higher tension p„ and to reduce it thereby to the volume cubic feet. On the other hand: L - Pl hyp. log. = 2,3026 Pl log. expresses the effect which a unit of volume of gas gives out or pro- duces when it passes from a higher pressure p1 to a lower p. To reduce by condensation a mass of air of the volume V, and ten- sion p to the volume Vv and the tension px = ^ py the mechanical ' 1 effect requisite to be expended is Vp hyp. log. j, and when, in- versely, the volume Vx at a tension^ is converted by expansion into the volume V9 and into the tension p = ^Pv the effect: v p hyp. log. ^ = VlPl hyp. log. will be given out, 1STRATA OF AIR. 345 Kxamples.—1. If a blast oonverls 10 cubic feet of air per second, of 28 inches tension, into air of 30 inches tension, the effect to be expended upon this for every second will be= 17280.0,491.28 hyp. log.(55) = 237565 15— 14) = 237565 (2,708050 — 2,639057) = 237565 . 0,068993 r= 16390 inch lbs. = 1365.8 ft. lbs — 2. If a mass of vapor in a steam engine below the surface of a piston A = wS* = 201 square inches, stands 15 inches high, and with a tension of three atmospheres, pushes UP the piston 25 inches, the mechanical effect evolved, and which is expended on the piston, is: £=201.3.14,73.15 hyp. log = 133232 hyp. log. } = 133232.0,98083 = 130567 inch lbs. = 10881 feet lbs., and the mean force of the piston, without regard to its friction and the oounter pressure, is : 130567 25 : 5222 lbs. § 299. Strata of Air.—Air enclosed in a vessel is at different depths of different density and tension, for the upper strata press together the lower on which they rest, so that there are only one and the same density and tension in one and the same horizontal stratum, and both increase with the depth. But in order to discover the law of this increase of density downwards, or the decrease upwards, we *aust adopt a method very similar to that of the former paragraph. Let us imagine a vertical column of air AE, Fig. 395? 0f the transverse section AB = 1, and of the height^F= s. Let the density of the lower stratum = y, and the tension = p, and the density of the uPper stratum EF = yl9 and the tension = pv we shall then have If x is the height EEX of the r P stratum ExFy we have its weight, and hence also the diminution of its tension corresponding to: y = 1 . x . yx a Xy?-L, and inversely, X = p . 1, or as in the former paragraph : y P x = p hyp. log. (l + ?) = P- Vyp- + ^-hyp- Pl]- y ' Vv y Let us put for p eaeceesivdy to + (,_1))r> and add «.f btigbU of the strata or valnes of *, and we shall then obtain the height of the entire column of air, former §. s = ^ (hyp. log. p—hyp. log. px) = hyp. log- als0 * = ^ hyp. log. = 2,302 £ log. if b and bx are the heights of the baroraeter corresponding to th sions p and px. # A rlpnsitv If, inversely, the height s is given, the expansive force of the air corresponding to it may be calculated. It 1S*346 GAY-LUSSAC’8 LAW. Py —iv — a» e ; , therefore y. = ye P , where e =« 2,71828 P1 i is the base of the hyperbolic system of logarithms. Remark. This formula is applicable to the measurement of heights. Leaving the tem- perature out of consideratiori, we may put* = 58604 . Log. J for the English measure s 60000 Log. Examples.__1. If the height of the barometer at the foot of a mountain be 28 inches, and at the top 25 inches, the height of this mountain will be: s = 58604 . Log. Qp) s 58604 . 0791813 = 4640 Prussian feet.—2. The density of the air on a mountain 10,000 feet high is: Log. — as - - ■ = 0,1706 ; yt 0o0O4 hence, Z = 1,481, and Z = _J— = 0,675; it is therefore only 67$ per cent. of y, y M81 the density of that at the foot. § 300. Gay-Lussatfs Law.—Heat or temperature has an essential influence on the density and expansive force of gases. The more air enclosed in a vessel becomes heated, the greater does its expansive force exhibit itself, and the higher that the temperature of the air enclosed by a piston in a vessel is raised, the more it expands, and pushes against the piston. From the experiments of Gay-Lussac, which in later times have been repeated by Rudberg, Magnus and Regnault, it results that for equal densities the expansive force, and for equal expansive forces the volume of one and the same quantity of air increases as the temperature. We may place this law by the side of that of Mariotte, and name it, for distinctiones sake, Gay-Lus- sac’s law. According to the latest experiments, the expansive force of a definite volume of air increases by being heated from the freezing to the boil- ing point, by 0,367 of its original value, or for this increase of tem- perature the volume of a definite quantity of air increases, the tension remaining the same, by 36,7 per cent. If the temperature is given in centigrade degrees, of which there are 100 between the freezing and boiling point, it follows that the expansion for each degree is = 0,00367, and for t degrees temperature = 0,00367 . t; if we make use of Fahrenheit’s thermometer, which contains between the freez- ing and boiling point 180°, for each degree the expansion is = .002039, and for t degrees temperature = .002039 . t. This co-efficient is true only for atmospheric air; slightly greater values correspond to other gases, and even for atmospheric air, this co-efficient increases slightly with the temperature. If a mass of air of the original volume F0, and of the temperature (centigrade) 0°, be heated t degrees without assuming a different ten- sion, the new volume is then F= (1 + 0,003671) V0, and if it acquire the temperature t19 it will then assume the volume: Vx = (1 0,00367 tj) V0, and by dividing the ratio of the volumes: V _ 1 + 0,00367 t Vx 1 + 0,00367//DENSITY OF THE AIR. 347 on the other hand, the corresponding ratio of density: y _ Vx _ 1 + 0,00367 tx yx V “ 1 + 0,00367 t * If, moreover, a change take place in the tensions, if p0 is the tension at zero, p that at the temperature ty and px that at tv we then have: F= (1 + 0,003671) ^ V0, further Vx — (1 + 0,00367 tx) h. V0, P Pi hence : Z = 1 + 0,00367t and y_ = 1 + 0,00367 . tt p Qr Vx i + 0,00367 tx ’ p ’ r, 1 + 0,00367 . t ' p* ’ 7 _ 1 + 0,00367 . tx b ^ ~ 1 + 0,00367 . t ' 6,' Example. If a mass of air, of 800 cubic feet, and of 10 lbs. tension, and 10° (centi- grade) temperature, is raised by the blast, and by the warming apparatus of a blast-fur- nace to a tension of 19 lbs. and to a temperature of 200°*, it will at length assume the greater volume: V. = 1+0,00367.200 15 #800 = 1,734 12000 = 1Q56 cubic feet (Prussian). 1 + 0,00367.10 19 1,0367 19 § 301. Density of the Air.—By aid of the formula at the end of the former paragraph, y may now be calculated by the density cor- responding to a given temperature and tension of the air. By accurate weighings and measurements we have the weight of a cubic metre of atmospheric air at a temperature of 0°, and 0,76 metre height of baro- meter = 1,2995 kilogrammes. Since a cubic foot (Prussian) = 0,030916 cubic metre and 1 kilogramme = 2,13809 lbs. The density of the air for the relations given is: = 0,030916.2,13809 . 1,2995 = 0,08590 lbs. If now the temperature is = P cent., the density for the French measure: y =_____ 1 + 0, and for the Prussian measure y 1,2995 kilogrammes; ,00367 t 0,08590 lbs > and for the English : y 1 + 0,00367 . t 0,081241 lk Tr ^ —, ^ . lbs. 11 now the expansive force vanes 1 + 0,00204 t from the mean, if, for example, the height of the barometer is not 0,76 metres, but 6, we shall obtain: 1,2995 6 _ 1,71 . h 1 + 0,00367 . t ’ 0,76 “ 1 + 0,00367 t g# or if 6, as is commonly the case, be given in Paris inches: „ _ 0,003058 . b lk_ 7 “ 1+0,003671IbS* Very often the expansive force is expressed by the pressure p, on a square centimetre or square inch, for this reason the factor 1 0336 9 °r i4^73> °r 15*05 mus* ^ntr0(luce^> an(^ ** ^en ^ows that: 9 9 * 10° C. s=s 50 F., and 200° C. « 392° F.,—the co-efficient will then be .002039.— Ax. Ed.348 DENSITY OF THE AIR. 7 = 7 = 1,2995 1+0,00367. 0,08565 P = _________1>257jg— kilog. or 1,0336 P 1+0,00367. t 0,005691» ,, D 1+0,00367 . t ' 15,05 ~ 1 +0,00367 . t S' russian' For the same temperature and tension, the density of steam is f c the density of atmospheric air, for which reason we have for steam: r\ » _. n 78!S77 7 = 0,8122 P 1 + 0,00367 t 0,05353 = 1+0,00367 . t 0,050775 1+0,00204 t ' 14,73 _ 0,78577--- or -- 4 . r\ AAOO"7 * 1 + 0,00367 t 1,0336 P _ 0,003557p jbs prussian 15,05 1 +0,00367. t JP__ - °>00344>_ Jbs. English. 1+0,00204 t5 Examples._1. What weight has the wind contained in a cylindrical regulator of 40 feet length and 6 feet width, at a temperature of 10° and 18 Ibs. pressure ? The density of this wind is: _ 0,005691.18 __ 0,10244 = 0,0988 Ibs. (Prussian); ““ 1,0367 1,0367 the capacity of the regulating vessel is = w. 3a.40 = 1131 cubic feet; hence, the quan- tity of wind =0,0988.1131 = 112 Ibs—2. A steam engine uses per minute 500 cubic feet of vapor, of 107° C. temperature and 36 inches pressure, how many pounds of water are required for the generation of this steam? The density of this steam is: 0,05353___ 36 __ 0,05353.36 _ Qj0494 Ibs. ; ~~ i-f0,00367.107° * 28 1,393 • 28 hence, the weight of 500 cubic feet, or the weight of the corresponding quantity 0f water, = 500 .0,0494 = 24,7 Ibs. i 302. By aid of the results obtained in the last paragraph, the theory of the air manometer may be explained. This instrument consists of a barometer tube of uniform bore AB, Fig. 396, filled above with air and below with mercury, and of a vessel CE like- wise containing mercury, which is put in communi- cation with the gas or vapor whose tension we wish to find. From the height of the columns of mercury and of air, the expansive force may be estimated as follows. The instrument is commonly so arranged, that the mercury in the tube stands at the same level as the mercury in the cistern, when the temperature of the enclosed air t = 50° (10° C.), and the tension in the space EH equal tothe mean atmospheric pres- sure b = 0,76 metres = 30 inches. But if for a height of the barometer b, from EH a column of mercury h, has ascended into the tube, and the length of the column of the residuary air is hz, we have then its tension Fig. 396. = A-»-+ b, and hence b, = + • If a change of temperature takes place, the temperature from obser- vation of hx and h2 is not as at first = t, but tv we then have e ten- sion of the column of air:LAW OF MARIOTTE. 349 qS - hi+ K 1+0,00367 . tx ' 1 + 0,00367 . t A and hence the height of the barometer in question: _ , hx + h% 1+0,00367 . tx = A, + 1+0,00367 . t . b. For b = 28 inches (Paris), and t = 10° C., it follows that bx m. hx + 27 (1 + 0,00367 tx) * whereby h = hx + h2, h% represents the whole length of the tube measured to the surface of mercury HR. From the height of the barometer bv it follows that the pressure on the square inch (Prussian) is p = hx + 15,6 . g (1 + 0,00367 < * - 0,538 A, 28 28 A, + 14,51 (1+0,00367 t) * lbs. Example.—If an air manometer, of 25 inches length, at 21° temperature C., indicates a column of air of 12 inches in length, then the corresponding height of the barometer is: > o* 25 fclS= 25 —12+ 27(1 + 0,00367.21) 12 = 13+ 9. 1,07707.-- = 13 + 60,58 as 73,58 inches, and the pressure on a square inch = 0,538.73,58 = 39,59 lbs. 30350 EFFLUX. SECTION VI. DYNAMICS OF FLUID BODIES. CHAPTER I. THE GENERAL LAWS OF THE EFFLUX OF WATER FROM VESSELS. § 303. Efflux,—The doctrine of the efflux of fluids from vessels constitutes the first principal division of hydro-dynamics. We dis- tinguish first between the efflux of air and the efflux of water, and then again between the efflux under uniform and that under variable pressure. We next treat of the efflux of water under constant pressure. The pressure of water may be assumed as constant when the same quantity of water is admitted on one side as flows out from the other, or when the quantity of water flowing out in a certain time is small compared with the size of the vessel. The chief pro- blem, whose solution will be here treated, is that of determining the discharge through a given orifice, under a given pressure, in a defi- nite time. If the discharge in each second = Q, we then have the expendi- ture, after the lapse of t seconds, under variable pressure: Qx = qt. But to obtain the efflux per second, it is necessary to know the dimensions of the orifice, and the velocity of the particles of fluid issuing from it. For the sake of simplicity of investigation, we shall for the present assume that the particles of water flow out in straight and parallel lines, and on this account form a prismatic vein or stream of fluid. If, now, F be the transverse section of the vein and v the velocity of the water, or of each particle of water, the discharge per second will form a prism of the base F and height v, and, therefore, Q will be = Fv units of volume and G = Fvy units of weight, y being the density of the water or the effluent liquid. Examples.—1. If water flows through a sluice, of 1,7 square feet aperture, with a 14 feet velocity, the discharge will be Q = 14 . 1,7 = 23,8 cubic feet, and hence the dis- charge per hour will be = 23,8 . 3600// = 85680 cubic feet.—2. If 264 cubic feet of water were to be discharged through an orifice of 5 square inches in 3 minutes and 10 seconds, the mean velocity of efflux would amount to ^ 204 264 . 144 Ft 5 ~144 = 40 feet. 190" 5 . 190VELOCITY OF INFLUX AND EFFLUX. 351 § 304. Velocity of Efflux.—Let us suppose a vessel AC} Fig. 397, filled with water, having a horizontal exit orifice, Fy rounded from the inside, which forms but a small part of the transverse section or bottom sur- face CD. Let the head of water FG, supposed invariable during the efflux, = A, the velocity of efflux = v, and the discharge in each second = Q, and therefore its weight = Qy. The me- chanical effect which this mass of water produces by sinking from the height A, is = Q A y, and the mechanical effect which the effluent mass Qy accumulates in its transit from a state of rest into v2 that of the velocity r, is — Qy (§ 71). If noloss of mechanical effect take place in its passage through the orifice, both mechanical effects will be equal, and therefore A Qy = Qy; i. e., A = and, in- 4 ______ 2g 2g _ versely, v = \/2 g A, or in feet, h = 0,0155 v2, and v = 8,02 h. The velocity, therefore, of water issiung through an orifice is equi- valent to the final velocity of a body falling freely from the height oj the water. The correctness of this law may be proved by the following expe- riments. If we apply an orifice directed upwards to the vessel AC, Fig. 398, the jet FK will ascend vertically, and nearly attain the level HR of the water in the vessel, and we may assume that it would ex- actly attain this height, were ali resistances, such as those of the air, friction at the sides of the vessel, disturbances from the descending water, &c., entirely removed. But since a body ascending to a perpendicular height A, has the initial velocity v = 2 g h (§ 17), it ac- cordingly follows that the velocity of efflux is v == y/2 g h. For a different head of water hv the velocity is == \/ 2 g hv hence we have v : vx = v/ h : y/ hx \ therefore, the velocities of efflux are to each other as the square roots of the heads of water. Examples.—1. The discharge which takes place in each second through an orifice 10 inches square, under a pressure of 5 feet, is: Q=zFv=\0.12 y/2 g h = 120.8,02 ^/5 = 962,4.2,236 = 2151,9cubic inches. . That 252 cubic inches may be discharged through an orifice of 6 square inches m eae i second, the head of water required is: A = (-|y= 0,0155 (2¥Ly= 0,0155 (42)a = 27,34 inches. § 305. Velocity of Infiux and Efflux.—If water flows in with a certain velocity c, the mechanical effect —, corresponding to the ve- Fig. 398. Fig. 397.352 VELOCITY OF INFLUX AND EFFLUX. locity due to the height hl = —9 must be added to the mechanical 2g effect h . Qy; hence we have to put: (h + A ) Qy = 1. Qy,or A + A, = iL, , ., . 2g 2g- and therefore the velocity of efflux: v = ^2 g(A -f A,) = v^2 Since the quantity of water flowing into a vessel kept constantly full is as great as that Q which flows out, we may put Gc = Fv, where G represents the area of the transverse section HR (Fig. 397) of the water pouring in. Accordingly if we put = v, we shall Or then obtain: v2 / F\* v3r / F\2“| h~Yg \g) Yg~ L1 — (g') J *g' and hence :v = y/2gh J1-( -V g) According to this formula, the velocity increases, the greater the ratio of the sections— becomes; the velocity is least, viz., = \/2gh, G if the transverse section F of theorificeof efflux is small compared with the transverse section G of the orifice of influx, and it approximates more and more to an infinite- ly great velocity, the smaller the difference is between F these orifices. If F = G, therefore — = 1, then v = Or Fig. 399. v/^_ = oo , and therefore also c = oo. This infinite value must be understood to express that the water must flow to and from a bottomless vessel ACy Fig. 399, with an infinite velocity, in order that the stream of fluid CF may entirely fili up the orifice of discharge F. If we put v = Gc —, we shall then obtain : -[( G_\* f) l"l —, hence F— J2'g G J' + 2gh which expression indicates that the transverse section Foi the stream flowing out, for an infinite velocity of influx, is constantly less than the transverse section G of the stream flowing in, and hence, that the discharging orifice is not quite filled when it it is greater than GVELOCITY OF EFFLUX, PRESSURE AND DENSITY. 353 Remark. The accuracy of this formula, given by Daniel Bernoulli, --------V'2**-------, has of late been brought into doubt by many philosophers. j-t&y have endeavored to prove how unfounded are the objections made, in an article “ Efflux, in the “ Allgemeinen Maschinenencyclopadie.” , Example. If water runs from a circular orifice, 5 inches in width, in the bottom o a prismatic vessel of 60 square inches transverse section, under a pressure of 6 feet, the velocity is then: 8,02 x/h _ 8,02.2,449 _ ____19,641 __ 19,641 r== " ^/6,8931 y/l—(0,327)a 0,945 = 20,78 feet. Fig. 400. § 306. Velocity of Efflux, Pressure, and Density.—The above for- mulae are only true if the pressure of the air on the fluid surface is as great as its pressure against the orifice; but if these pressures are different from one another, we have then to extend these formulae. If the upper surface HR, Fig. 400, is pressed by a piston K with a force Pv which case, for example, presents itself in that of the fire-engine, we may then imagine it to be replaced by the pressure of a column of water. If hx be the height of this co- lumn, and y the density of the liquid, we may there- fore put Px = Gh$. If we substitute for h the head of water augmented by h. = P P 7T> h + hl — h -l we then obtain for the velocity of efflux : Gy v = ^ 2 g(h + 7}y)’ when, moreover, we suppose ^ to be very small. If, further, we represent the pressure on each unit of surface p of G by pv we have more simply _» = pv and hence Cr v = ^ 2 g Again, if we represent the pressure of water at the level of the orifice by p> we may then put P = (h + —^ y; therefore, A + ^ = -, whence v = I 2 g -• \ y / y r S r The velocity of efflux, therefore, increases as the square root of the pressure on each unit of surface, and inversely as the square root of the density of the fiuid. Under equal pressures, therefore, a fluid of a density represented by 4, runs out half as fast as one of density 1. Since the air is 770* times lighter than water, it would, ^According to the experiments of Prout, 100 cubic inches of air at 60° temperature and 30 inches barometer, weigh 31,0117 grains troy, and consequently at 32° a cubic foot will weigh 31,0117 X ~ X 17,28 = 560,1 grains; and since a cubic foot of water weighs 62,5 lbs. avoirdupois = 62,5 X 7000 = 437500 grains, the relative weights of 30*354 VELOCITY OF EFFLUX, PRESSURE AND DENSITY. if it were an inelastic body, flow out 770 = 27f faster than water. If the water does not flow freely, but under water, in conse- quence of a counter-pressure, a dirainution of the velocity of efflux then takes place. If the mouth of the vessel j$C, Fig. 401, is the depth FG = h below the surface of the upper water HR, and FG1 = h1 below the surface 0f the lower water, we then have the pressure downwards p = hy} and the counter pressure upwards px =■ hx y, hence the force of efflux is: p—p1 = (A— hx) y, and the velocity of efflux » = J 2 ^ ” A,). For efflux under water the difference of level A—hx between the surfaces raust be regarded as the head of water. If the water on the side of the outer orifice be pressed by the force p, and on the side of the inner orifice or of the surface of water by the force pv we have then generally :____ v~J2s( J + Fig. 401. Examples.—1. If the piston of a 12 inch cylinder, or that of a fire-engine, were pressed down with a force of 3000 lbs., and there were no obstacles in the tubes or pipes, the water would then issue through the mouth-piece of the tube and be directed vertically upwards with a velocity: ---- -------- 1 ---------- f 600.4 tpwaiuo wiiu c* f V/ • ___ -- it 12,5 = 62,697 feet, and ascend to the height h = 0,0155 . v“ — 60,93 feet 2. If water rushes into a rarefied space; for example, into the condenser of a steam-engme whilst it is pressed from above or on its exposed surface by the atmosphere, the last formula v_ / 2 g ( hI for the velocity of efflux is then to be applied. If the head of water h =3 feet, and the external barometer stand at 27 inches, and the internal at 4 n . . , , „ i,flVP P* —P 27 — 4 = 23 Paris inches = — 1,035 Paris inches, we shall now nave-------------^ 1 ° = 1,9837 Prussian feet = 2,042 English feet, or a column of water = 13,5 . 2,042 = 27,57 feet, and the velocity of the water rushing into the vacuum v = 8,02 /3 i_27 57 = 44 34 feet.—3. If the water in the feed-pipe of a steam-engine boiler stmids 12 feet above the surface of the water in the boiler, and the pressure of steam be 20 lbs., and the pressure of the atmosphere only 15 lbs. on the square inch, the velocity with which the water enters into the boiler will be: 5. V = 8,02 J 12 + (15—20) . 144 62^ — 8,02 Jl2. 144 62,5 _ = 8,02 v/0,48 = 5,55 feet. air and water at that temperature is 566,1 = 1 . At 60° the relation will be 535,88 ___ 437500 772 * : ^ == 1:616* as y/ 816 = 28,5, the velocity of efflux will, under this condition, 1)6 97 7f, Part more rapid than in that supposed in the text.—Am. Ed.HYDRAULIC PRESSURE. 355 § 307. Hydraulic Pressure.—When water enclosed in a vessel is in motion, it then presses more feebly against the sides than when at rest. We must, there- fore, distinguish the hydrodynamic or hydraulic j pressure from the hydrostatic pressure of water. If pl be the pressure on each unit of surface HrRj = Gv Fig. 402, p the pressure without the orifice F\ and h the head of water FGV we then have for the velocity of efflux_ » = J*g (* + m 4+5=£_[1-(|)1]giif>fur,her> . in another transverse section H2R2 = G2, which lies at a height FG2 = /tj above the orifice, the pressure = p2, we then have like- wise: -(0]| If we subtract one expression from the other, it then follows that: or, if the head of water GlG2 of the stratum H2R2 = G2 be repre- sented by h2, the measure of the hydraulic pressure of water at H2R2is: r .- „ JFv F But now — is the velocity r, of the water at the surface Gv and ^ the velocity v2 of the water at the section G2; hence more simply we may put = ?± + h — ---- y r . %/ Therej'ore, from this it follows that the hydraulic head of water ^ at any place in the vessel is equivalent to the hydrostatic head of 7 water + h2 diminished hy the difference of the height due to the _ . y velocity at this pointy and at the place of entrance. If the upper sur- face of the water Gx is great, we may neglect the velocity of influx, and hence may put + h2______and the hydraulic head of y y 2jg* water is less hy the height due to the velocity than the hydrostatic head of water. The faster, therefore, water flows in conduit pipes, the less it presses against the sides of the pipes. From this cause, pipes very often burst, or begin to leak, when its motion in them is checked, or when the pipes are stopped up, &c.356 hydraulic pressure. Fig. 403. By means of the apparatus of efflux ABCD, Fig. 403, we may have ocular demonstration of the diflerence between hydraulic and hydrostatic pressure. If we carry upwards a small tube ER from the transverse section G?, it will become filled with water, which will ascend in it above the leve! of the fluid surface if G2is G,, therefore v2 2g other hand, the transverse section G3 be < Gv and the water there- fore flow through G3 quicker than through Gv we shall then have the height of the column of water in the small tube Ex whose inner orifice less than A3, and hence it will not is at G3, y = h3 — 3 3 W V 2g 2g, reach to the level HR of GvAgain, if G< be very small,^and there- fore the corresponding velocity v4 very great, then — may be > h .and hence the corresponding hydraulic head of water z may be negative, i. e. the air may press more from without than the water from within. A column of water will therefore ascend in the tube E^K, which is inserted below, and whose outer onnce is under water, which in conjunction with the pressure of the water, will balance that of the external atmosphere. If this small tube be short, the water, which may be colored for this purpose, will ascend fronn the vessel K underneath, through the tube, enter the reservoir of efflux, and will arrive at F and be discharged. Fig. 404. Remark. If the discharging vessel ACE, Fig. 404, consists of a wide reservoir AC and of a narrow vertical tube CE, the hydraaijc pressure at all places in this tube is then negative. If we disregard the pressure of the atmosphere pu the pressure of the water in the vicinity of the mouth F may be put = 0, because the whole head of water here GF = h will be expended in generating the velocity v = y/2g h; on the other hand, at a place DtEt at the height G,G = h, below the surface of water, the hydraulic pressure = A, — h = — (A—A,) negative ; if, therefore, a hole be bored in this tube, no water will run out, but air will rather be drawn in, which will arrive at F and flow out. This negative pressure will be greatest directly below the water, in G because h2 is there least. § 308. By means of the formula Q=Fv=F^/2g h, the discharge issuing in one second can only then be calculated directly when the orifice is horizontal because here only the velocity throughout the wholeHYDRAULIC PRESSURE. 357 transverse section F is the same ; but if the transverse section of the orifice has an inclination to the horizon, for example, if it is at the side of the vessel, the particles of water at different depths will then flow out 'with different velocities, and the dis- charge Q can no longer be con- sidered as a prism, and hence, there- fore, the formula Q=Fv=F \/2g h cannot be applied directly. The most simple case of this kind is pre- sented in the efflux through a cut in the side of a vessel, or in what is called a weir, Fig. 405. This cut forms a rectangular aperture of ef- flux EFGH, whose breadth EF = GH is represented by b, and height EH = FG by h. If we divide this surface bh by horizontal lines into a great number n of equally broad laminae, we may suppose the velocity in each of these to be the same. Since the heads of water of these laminae from above down- wards are &c., wre then have the corresponding velocities n n n Fig. 405. I 2 g ~9 I 2 g. I 2g . — ; and since, further, the area of a \ n \ n \ n lamina = b. ^ = —9 we then have the discharges: bh I 9 „h bh I 2h bh I 0 3h e V yj 2 g«’ T yj 2 S* V T yj2 8 * IT’ &c*5 the discharge through the entire section: = ^Ji^2gh (^l ' + ^2 + ^3 + . . . + v/n). Tly/n But now: x/1 + %/2 + v/3 + . . + \/n, or 1* + + 3^ + .. + = ------ = £ n2 = § n y/n; hence, the discharge required is: q __ _h\/2gh ^ ^ n __ ^j)hx/2g h = $ b y/2gh*. n^/n If by the term mean velocity (v) be understood that velocity which must sub sis t at all places, that as much water, in consequence, does issue as with the variable velocities of efflux within the whole profile ; we may then put: Q= bh . v, and, consequently, v = § \/2ghy i. e.358 HYDRAULIC PRESSURE. the mean velocity of water issuing through a rectangular cut in the side of a vessel is § of the velocity at the sili or lower edge of the cut. It’the rectangular aperture of efflux KG, Fig. 406, with horizontal sili does not reach the surface of the water, we may find the discharge by regard- ing the aperture as the difference of the two cuts EFGH and EFLK. Hence, if hx is the depth HE of the lower, and KE == h2 that of the upper edge,wethen have the discharge from these apertures § b ^/2 g h* and § b \/%gh23> and hence the quantity of wa- ter for the rectangular orifice GHKL : Q = § b-v/2 gh3 —§ b y/'2g h23 = § — h2* ), and the mean velocity of efflux : v = Q b(K—K) 3 3 ______ h 2____h 2 hx + h2 If h is the mean head of water l- ^—?, or the depth of the centre of the orifice below the surface of water, and a the height of the orifice HK = hx— hv we may then put: v = § Wg (h +1) _ ( 2) or approximately: -[*■-•'"(!)’] Example. If a rectangular orifice is 3 feet wide and Ij feet high, and the lower edge lies 2} feet below the surface of water, the discharge is then: Q = $. 8,02.3 (2,751 _ 1,5?)= 16,04 (4,560 — 1,837) = 16,04.2,723 = 43,67 cubic feet. From the formulae of approximation the mean velocity of efflux is: » = [i_>V . 8,02 /ij® .= (1 — 0,0036) .11,685= 11,685—0,042 = 11,643 feet, and hence the discharge Q = 3. |. 11,643 = 43,65 cubic feet. Remark.—If the cut in the side is inclined to the horizon at an angle 31, we shall then have to substitute the height of the aperture ^I_—— for its vertical projection, whence ___________________________ sin. $ we must put Q = f —4 ( y/h.3 — */A23). ^ the transverse section of the reservoir sin. $ parallel to the aperture be not considerably greater than the section of the aperture, we shall then have to take into account the velocity vx = v with which the water flows to it, and for tliis reason put:TRIANGULAR LATERAL ORIFICE. 359 § 309. Triangular Lateral Orijice. — Besides rectangular lateral orifices, we have in practice tri- angular and circular. Let us first consider the efflux through a triangular orifice EFG, Fig. 407, with horizontal base, whose vertex E lies in the surface of the water. Let the base FG = b, and the height EF = A, let us divide the last into n equal parts, and carry through the points of division lines parallelto the base, we then resolve the entire surface into small elements of the areas: b h 2b h 3b h « — . —, — . —,— . — 9 au;., n n n n n and the heads of water: h 2h 3A o —, —, n n n The discharges for these are: Fig. 407. y I*,* *** uhg.M&c. „■ * n ^ * n »* 5 n and we obtain the discharge for the whole orifice: Q = —- I 2 g - + 2>/2 + 3\/34- •• + n >/ w) n2 \ n = (! + 22 + 3* + . . + «b» n2 s/n or since the series in the parenthesis 3*1 = - f Q - f bhs/Tgh = § v/2 2 • 1 If the base of the orifice EGK lies in the surface and the vertex lower by A, we then have the discharge f 6A v/ 2 g/i flowing through the rectangle EFGK, Qj = | bh y/ 2 gh—§ bh 2 gh — T45 bh 2 gh. Through the trapezium ABCD, Fig. 408, whose upper base AB = bv lies in the surface of the water, and whose lower base is CD = A2, and height DE = A, we may find the discharge by regarding the orifice as composed of a rectangle and two triangles, viz., Q= W sTZgh + TV (b—bjh (2 + 3i2) ✓ 2 *A. Fig. 408. Fig. 409. Further, the discharge for a triangle CZLE, Fig. 409, of the base DE = Aj, and of the height hl9 and whose vertex C is distant A from360 CIRCULAR LATERAL ORIFICES. the surface of water: Q = discharge through less the discharge through AE 68 = t45 bh s/Jgh-,\ (26 + 3 s/^gh, — '/~%g[2 6 (h$— h%)'—3 6,A,^]. As the breadth AB = 6 may be determined by the proportion 6 : 6t = A : (A—AJ, it follows that O = - ‘ b (A*—A12) q ^ f \ y 15 l A—Aj 3 "l / _ 2 ^/~2gbl (2 K*—5 A A,* + 3 A *\ 15 \ A—Aj )' Lastly, for a triangle ACD, Fig. 410, whose vertex lies above the base, the quantity discharged is Q-f ^.6, (A*—A i) 2 v/2g . 6, 15 _ 2 >/2g . 6, /2 A*—5 A A,* + 3 15 V A—At /3 A*—5 A, A* + 2 A,*\ \ h-h, r Fig. 410. Fig. 411. Example. What quantity of water flows through the square ABCD, Fig. 411, whose vertical diagonal AC = 1 foot, if the angular point A reaches the surface of the water ? The upper half of this square gives the expenditure: Q = § b y/~sTgh?== | 1 .8,02 ^£ = 0,4.8,02.0,353 = 3,21.0,353 = 1,1331 cubic feet, but the lower water expenditure: n2 b /2 A& - 5 h A,* + 3 h} _2.8,02 /2—5 ($)* + 3 (i)i\ ' 15 V A— h,) 15 \ 1—J / = JL2’08 (2—1,7678 + 0,5303) = 32,08 ‘ °’762- = 1,6307 cubic feet; the discharge 15 15 through the entire orifice is Q = 1,1331 -f- 1,6307 = 2,7638 cubic feet. § 310. Circular Lateral Orijices.—The dis- charge through a circular orifice AB, Fig. 412, may be determined by an approximate formula in the following manner. Let us decompose the orifice by concentric circles into equally small annuli, and each annulus into very small elements, which may be regarded as parallelo- grams. If now r is the radius of such an an- nulus, b its breadth and n the number of its ele- ments, we have the magnitude of an element Fig. 412.CIRCULAR LATERAL ORIFICES. 361 K, = If h is the depth CG of the centre C below the sur- n face of water HR, and $ the angle ACK, by which an element K is distant from the highest point A of the annulus, we have then the head of water of this element: KF = CG — CL= h — r cos. *, and hence the discharge of this element = — r ^ 2g (h—r cos. $). Now \/ h— reos. $ n = v'h £l — £ £ cos. * — i ^0* cos. *2 + . = y/h £l — i ^ cos. 4» — ylg ^0 (1 + cos. 2 *) + . .. J ; hence the discharge of an element: = v/ 2gh £l—£ . J cos. 4. — T>5 (0 (1 4- cos. 2*)—.. .J. The discharge of an entire annulus is now known, if we put in the parenthesis for 1, n . 1 = w, for cos. $ the sum of all the cosines of 9 from $ = 0 to $ = 2 *, and for the cosine of 2 the sum of all the cosines of 2 <*> from 2* = 0to24>=»4rf. But as the sum of all the cosines of a complete circle is = 0, these cosines vanish, and the discharge for the annulus : = 2 nrb 1 — T»s ^0* — . .. If now for b we substitute and for r, —, to —, we then m m m m m obtain the discharge of all the annuli which make up the circular sur- face, and lastly, the quantity of efflux of the whole circle Q — 2 tt r \/2 g h p—_ (14"2 + 3 + «*4" w) Lm2 ' 13 (l3 + 23 + 33 + . . + m3)"l rrrk2 J o —T / r m2 - r3 m4\ 2 ftr %/ 2 gh . I —- . —-y s —• ~r) \m2 2 m4 A2 4 / = H r2 y/ 2gh £l ---5*2 or more accurately: Q=.-••]• If thp circle reaches the surface of the water, then „ Q * * r* s/TgJ - 0,964 F y/ 2 gh, it i represents the area of the circle. It is besides easy to conceive that in all cases where the head of362 DISCHARGING VESSELS IN MOTION. water at the centre is equal to or greater than the diameter, we may put the whole series = 1, and take Q = F s/ 2g*A. This rule may also be applied to other orifices, and, therefore, in all cases where the centre of gravity of an orifice lies at least as deep below the fluid sur- face as the figure is high, the depth h of this point may be regarded as the head of water, and Q put = F s/ 2 gh. If we consider that the mean of all the cosines of the first quadrant = and that all the cosines of the second = — and likewise that 4’ 4 the mean of the first and of the second vanishes, we may then, by the method adopted above, find the discharge of the upper semicircle: :*-nG and that of the lower: Q2 = ^ y/^gh [ _1 + ^G Q = p 12 . Example. What quantity of water flows hourly through a circular orifice 1 inch in dia- meter, above which the fluid surface stands j1^ inch high ? = hence (y)1 = i! = °>735 i further, 1 — ^ = 1 — 0,023 = 0,977, and consequently the discharge per second : 8,02 / 7— .0,977 = . 8,02 . 0,977 ^/T = 16,26 cubic inches, 144 4 which, per minute, = 973, and per hour = 33,78 cubic feet. § 311. Discharging Vessels in Motion.—The velocity of efflux varies if a vessel previously at rest or in uniform motion changes its condition of motion, because in this case every particle acts by its own weight, as well asby its inertia against the surrounding medium. If we move the vessel AC, Fig. 413, upwards with a vertical ac- celerating foree, whilst the water flows Fig. 413. through the bottom by the hole jF, an increase takes place, and if it be moved downwards vertically by an accelerat- ing force, a diminution of the velocity of efflux ensues. If p is the accelerat- ing force, each element of water M presses not only by its own weight Mg, but also by its inertia Mp; consequently the force of each element in the one case, must be put (g + p) M, and in the other (g—p) M, therefore instead of g> From this it follows then that — = {gj~p) h, and hence for the 2 velocity: v = \/ 2 (g+p) h. Ii' the vessel ascends with the accelerating force then isDISCHARGING VESSELS IN M0T10N. 363 v = y/2.2 gh = 2 y/ gh, therefore the velocity of efflux 1,414 times that of a vessel at rest. If the vessel falis by its own weight, therefore, with the accelerated motion g, there is v = y/ 0 = 0, no water therefore flows out. If the vessel moves uniformly up or down, there remains v = %/ 2 gh, but if it ascends with a retarded motion, then will v = y/ 2 {g—p) h, and if it descends with the same retardation, then v = y/ 2 (g -r p) h. . If the vessel moves horizontally, or at an acute angle to the hori- zon (§ 274), the fluid surface will be inclined to the horizon, and a change in the velocity of flow will take place. By the rotation of a vessel AC, Fig. 414, about its vertical axis XX, the concave surface forms a parabolic funnel AOB, hence there will be over the middle F of the bottom a lesser head of water than at the edges, and hence the water will flow through the orifice F, in the axis, more slowly than through any other orifice K at the bottom. If h represent the head of water in the middle, then the velocity of efflux at the middle will be = y/ 2 gh, if y be the distance FK = ME of any other orifice K from the axis, and w the angular velocity, we shall then have the corresponding elevation of the water above the middle: _ , co2 y w2 OM = \ TM = i ME cotang. T = \ y . —J^ £* ^ $ & if w be the velocity of rotation of the ori- fice K. Hence then the velocity of efflux for this is ______ ’,-J2Hh+^)-J2sh+w’- This formula is true for every arbitra- rily shaped vessel, and also for one closed above, as AC,~Fig. 415, so that the fun- nel cannot be formed. Its application to wheels of reaction and to turbines will be found in the sequel. Exampleg.—1. If a vessel full of water AC, Fig. 413, weighs 350 lbs., and by means of a rope passing over a roller K is drawn by a weight G of 450 lbs., it will ascend wit an accelerating force p == '****? . g = J_g = £ g, and hence the velocity of efflux F 450+350 6 800 will be t? = ^/ 2 (g+7)A = \/2 . f -gh = y/fgh. Were the head of water h = 4 feet, the velocity of efflux would be r = l y/V7g = 3 y/32,2 = 16,01 feetr-r2. If a vessel AC, Fig. 415, full of water revolves so that it makes 100 re,voluuons per » if the depth of the orifice F below the surface of water in the middle ainounts ° L ’ and the distance from the axis XY, 3 feet, then the velocity of efflux is » = = Jc4,4.2+ (3' = v/128.8 = 11.34 feet. Fig. 41o.364 CO-EFFICIENT OF VELOCITY. CHAPTER II. ON THE CONTRACTION OF THE FLUID VEIN BY THE EFFLUX OF WATER THROUGH ORIFICES IN A THIN PLATE. § 312. Co-efficient of Velocity.—The laws of efflux developed in the preceding chapter accord almost entirely with experiment, so long as the head of water is not small compared with the width of the orifice, and as long as the orifice gradually widens inwards without forming corners or edges, and is close at the bottom or sides of the vessel. The experiments made by Michelotti, by Eytelwein, and by the author on this subject, with smoothly polished metallic mouth- pieces, have shown that the effective discharge, or that which actually flows out, amounts to from 96 to 98 per cent. of the theoretical quantity. The mouth-piece ADy Fig. 416, represented in half its natural size, gave for a head of water of 10 feet, 97,5 per cent., for 5 ft. 96,9 per cent., and for 1 ft. 95,8 per cent. Since for this efflux the fluid vein has the same transverse section as the orifice, we must then assume that this diminution of discharge is accompanied with a loss of velocity, w hich is caused by the friction or adhesion of the water to the inner circum- ference of the orifice, and by the viscidity of the wrater. In what follows, wTe shall call the ratio of the effective velocity of efflux to that of the theoretical = v x/2 gh, the co-efficient of velocity, and represent it by $>. From this, therefore, the effective velocity of efflux in the most simple case is vY = $ v = y/2 gh, and the discharge: Q = Fvx = $ Fv = $ F x/ 2 gh. If we substitute for 4» the mean value 0,97, we then obtain for the quantity in feet Q= 0,97. F s/ 2 gh = 0,97.8,02 F x/h = 7,779 F s/H. A vis viva ^ . v*, is inherent in a discharge Q issuing with the velo- Clty vi* vb*tue of which it is capable of producing the mechanical effect Qy . _i_. But since by its descent from the height h = the weight Qy produces the mechanical effect Qy . h = Qy ^-9it follow s that by the efflux of the wrater, this suffers a loss Fig. 416.CO-EFFICIENT OF CONTRACTION. 365 = 0,059 Qy . or 5,9 per cent. Therefore, the effluent water produces by its vis viva 5,9 per cent. less mechanical effect, than does its weight by falling from the height h. § 313. Co-efficient of Contractiori.—If water flows through an orifice in a thin piate, a considerable diminution of the discharge under otherwise similar circumstances takes place, whilst the particles of fluid rushing through the orifice move in convergent directions, and in this way give rise to a contraction of the fluid vein. The measure- ments of the vein made by many, and especially of late by the author, have shown that the vein at a distance which is about equal to one- half of the width of the orifice, has the greatest contraction, and a thickness equal to 0,8 that of the diameter of the orifice. If F1 is the transverse section of the contracted vein, as also F the transverse section of the orifice, we then have from this F1 = (0,8)2 F = 0,64 F. The ratio _i of these transverse sections is called the co-efficient of F contraction, and is represented by a, and accordingly, the mean value for the efflux of water through orifices in a thin piate may be put : a = 0,64. As long as we possess no more accurate knowledge on the con- traction of the fluid vein, we may as- sume that the stream flowing through Fig. 417. a circular orifice ABy Fig. 417, forms ; Orifices made after this figure of the contracted vein give pretty nearly veloehy of discharge vl = 0,97 . v. The contraction of the fluid vein is caused by the water which lies irectiy above the orifice flowing out together with that which comes to it from the sides. There takes place, therefore, in the interior of the vessel a convergence of the filaments of water, similar to that represented in the figure, and the contraction of the fluid vein con- MA == MF = / of the generating arc AF by means of the equation: a body of rotation ABEFy whose en- velope is generated by the revolution of a circular arc AF about the axis CD of the stream. Let the diameter AB of the orifice = dy and the dis- tance CD of the contracted section EFy = \ d9 we then obtain the radius : EFy = \ dy we then obtain the AJY2 = FJY (2 MF~ FN)y or 31*366 CONTRACTION OF THE FLUID VEIN. sists in a mere propagation of this convergence. We may convince ourselves of this motion of the water in the vicinity of the orifice by means of a glass apparatas of efflux; if we drop into the fluid minute substances which are either heavier or lighter than water, for exam- ple, such as oak saw-dust, bits of sealing wax, &c., and allow them to pass out with it from the orifice. § 314. Contractiori of the Fluid Vein.—If water flows through triangular or quadrilateral orifices, and in a thin piate, the stream then assumes particular figures. The inversion of the jet, or the altered position of its transverse section with respect to that of the orifice, is very striking to the eye, in consequence of which a comer of this section comes to coincide with the middle of one side of the orifice. Hence, from a triangular orifice ABC, Fig. 418, the section of the stream at a certain distance from the orifice forms a treble star-like vein DEFy from a quadrilateral orifice ABCD, Fig. 419, a star of Fig. 418. Fig. 419. Fig. 420. four veins EFGH, from a five-sided orifice ABCDE, Fig. 420, a star FGHKL, consisting of five veins. These sections vary at different distances from the orifice: at a certain distance they diminish, and at a successive one again increase ; hence the vein consists of plates or ribs of variable breadth, and thereby forms, when the efflux is observed under great pressure, bulges and nodes, similar to what is seen in the cactus. If the orifice ABCDy Fig. 421, is rectangular; at a lesser distance from the orifice, the section will then form a cross or star; and at a greater one, it will again assume the form of a rectangle EF. Observations on various kinds of orifices have been made by Bidone, and accurate measurements of the vein from square apertures also by Poncelet and Lesbros.* The last measurements haveled to a small co-efficient of contraction 0,563. The measurements of water issuing through lesser orifices, give us, however, greater co-efficients of contraction; they show, moreover, that these are greater for elon- Fig. 421. £ See Allgem. Maschinen-encyclopiidie, article Ausfluss.CO-EFFICIENT OF EFFLUX. 367 gated rectangles than for rectangles which approximate more to the square. § 315. Co-efficient of Efflux.—If in the flow of water through orifices in thin plates, the effective velocity were equal to the theo- retical v = y/2 gh, we should have the effective discharge: Q = a F . v = a F y/2 gh, because a F represents the transverse section of the vein at the place of greatest contraction, where the particles of water move in parallel directions. But this is by no means the case: it is shown rather by experience that Q is smaller than a F y/2 gh, that we must therefore multiply the theoretical discharge F y/2 gh by a co-efficient which is less than the co-efficient of contraction, in order to obtain the effective discharge. We must hence assume that for efflux from an orifice in a thin piate, a certain loss of velocity takes place, and therefore in- troduce a co-efficient of velocity *, and hence put the effective velocity of efflux vx = v = $ y/2~gh. From thisthen we have the effective discharge: Qx = Fx . vx = a F. $ v = a$ Fv = a$> F \/2 gh. Again, if we call the ratio of the effective discharge to the theoretical or hypo- thetical quantity, the co-efficient of efflux, and represent it in what follows by /i, we then have: Q^rQ—rFv^pF y/2 gh, hence p = a $, i. e. the co-efficient of efflux is the product of the co- effidents of contraction and of velocity. Multiplied observations, but chiefly the measurements of the author, have led to this, that the co-efficient of efflux for orifices in thin plates is not constant; that for small orifices and for small velocities, it is greater than for large orifices and for great velocities: and that it is considerably greater for elongated and small orifices than for orifices which have a regular form, or which approximate to the circle. For square orifices of from 1 to 9 square inches area, with from 7 to 21 feet head of water, according to the experiments of Bossut and Michelotti, the mean co-efficient of efflux is p = 0,610; for circular ones of from J to 6 inches diameter, with from 4 to 21 feet head of water, = 0,615, or about T^. The single values observed by Bossut and Michelotti vary considerably from one another, but we cannot discover in them any dependence between the dimensions of the orifice and the magnitude of the head of water. From the author^ experiments at a pressure of 24 inches, the co-efficient for an orifice of 0,3937 inches or 1 centimetre diameter is p « 0,628 0,7874 “ 2 centimetres “ * 0,621 1,1811 “ 3 « « s 0,614 1,5748 << 4 « u as 0,607. On the other hand, at a pressure of 10 inches for the round orifice of 1 centimetre diameter *— 0,637 2 centimetres “ 0,629 3 “ « q* 0,622 4 “ “ *368 RECTANGULAR LATERAL ORIFICES. From these it is manifest that the co-efficient of efflux increases when the dimensions of the orifice and the head of water decrease. If for we take the mean value = 0,615, and for o == 0,64, we obtain the co-efficient of velocity for the efflux through orifices in a thin piate, = - = 0,96, therefore, nearly as great as for efflux a through rounded or conoidal orifices. Remark 1. BufFs experiments (see Poggendorfs Ann. Band 46), show that the co-effi- cient of efflux for small orifices and for smaii headsof water or velocities is considerably greater tlian for large or mean orifices and velocities. An orifice of 2,084 lines diameter, gave for inch pressure, /a s 0,692, for 35 inches /a = 0,644; on the other hand, an orifice of 4,848 lines for 4£ inches pressure /u = 0,682, and for 29 inches /u =s 0,653. Remark 2. According to the authors experiments, the co-efficients for efflux under water are about lf per cent. less than for efflux in air. § 316. Rectangular Lateral Orifices.—The most accurate experi- ments on efflux through large rectangular lateral apertures are those made at Metz by Poncelet and Lesbros. The widths of these orifices were two decimetres, (nearly 8 inches); the depths, however, varied from one centimetre to two decimetres. In order to produce perfect contraction, a brass piate of four millimetres, = 0,1575 inches, thick- ness was used for these orifices. From the results of their experi- ments, these experimenters have calculated by interpolation the tables at the end of this paragraph for the co-efficients which may be used for the measurement or calculation of the discharge. If b be the breadth of the orifice, and if Ax and A2 are the heads of water above the lowest, and above the uppermost horizontal edge of the orifice, we then have, from § 308, the discharge: Q = § b y/ 2 g (h^____A/). But if we substitute the height of the aperture a, and the mean head of water A == hl \ ~ > we then have approxi- ».*!, Q _ ( I _ -|L) discharge Qx = ^ Q = ( 1 ■ ab \/ 2gh, and hence the effective r ab \/2gh- If,further,weput (> 96 h2) /ij, we have then simply Qx = ^ab \S2ghy and in order to allow of our calculating by this simple or general formula of efflux, not only the values of m, but also those of u, are riven in the following tables. Since the water in the vicinity of the orifice is in motion, it stands lower directly before the aperture than at a greater distance from the piate in which the aperture is made; on this account two tables have been compiled, the one for heads of water measured at a greater dis- tance from the orifice, and the other for those measured immediately at the side in which the orifice lies. It may be seen, moreover, from both tables, although with certain variations, that the co-efficients of efflux increase the lower the orifice is, and the less the headof water.RECTANGULAR LATERAL ORIFICES. 369 If the orifices have different breadths, we are compelled, so long as we have no further experiments, stili to use the co-efficients of these tables in like manner for the calculation of the discharge. If, further, the orifices are not rectangular, we must determine their mean breadth and mean depth, and introduce into the calculation the co-efficients corresponding to these dimensions. Lastly, it is always preferable to measure the head of water at a certain distance from the side in which the orifice lies, and to use the first table, than directly at the orifice where the surface of water is curved and less tranquil, than a little above it. Examples. 1.—What quantity of water flows through a rectangular aperture. 2 deci- metres broad and 1 decimetre deep, if the surface of water is l£ metre above the upper edge ? Here 6=0,2; a=0,l,h = ht = 1,55; hence the theoreticai 2 2 discharge Q = 0,1.0,2 y/Jg . y/ljrt = 0,02.4,429.1,245 = 0,1103 cubic metre. But now Table I. gives for a = 0,1 and A, = 1,5, = 0,611, hence the effective discharge Qx = 0,611.0,1103 = 0,0674 cubic metre—2. What discharge corresponds to a rectan- gular orifice in a thin piate of 8 inches breadth, 2 inches depth, with a 15 inches head of water above the upper edge?* The theoreticai discharge is Q = ■£. ^ . 7,906 y/ £ = 0,8784.1,1547 = 1,014 cubic feet. But now 2 inches is about 0,05 metre, and 15 inches alxmt 0,4 metre; hence, according to the table a = 0,05 and A, = 0,4, the corresponding co-efficient /t*, = 0,628 is to be taken, and the quantity of water sought is Q, = 0,628 . 1,014 = 0,637 cubic feet.—3. If the breadth = 0,25, the depth =0,15, and the head of water A, = 0,045 metre, then is Q = 0,25.0,15.4,429 . ^/0,12 =0,166.0,3464 = 0,0575 cubic metre. To the height 0,15 corresponds for A, = 0,04, the mean value: 0,5824-0,603 0,5854-0,605 n ... /ucj=—----4_J----- — 0,5925, and Aa = 0,05, /xl = —----3LJ----— 0,595; but since h2 0 5925 1 0 595 is given = 0,045, we must then substitute the new mean —_______L_1_____=0,594 for the 2 co-efficient of efflux, and we therefore obtain the discharge sought: Qt = 0,594.0,0575 = 0,03415 cubic metre. * In using the following tables, the English measures will be furnished with the pro- per co-efficients by employing the first, or left hand column, in which to find the height A, and the column under the number of inches answering to the height of orifice a.— Aar. Ed.370 CO-EFF1CIENTS FOR TUE EFFLUX TABLE I. The co-efficients for the efflux through rectangular orifices vertical piate, from Poncelet and Lesbros. The heads o wa measured at a certain distance back from the orifice, or a p where the water may be considered as stili. Head of water, or distance of the sur- face of water from the upper side of the orifice. HXIGHT OF OBJFIC1. Eng. in. Metres. 0,20« 0,10® 0,05« 0,03« 0,02 or 8 in. or 4 in. or 2 in. or 1,18 in. or ,8 in. 0,00 0,000 tt u tt tt tt 0,19 0,005 It tt tt tt tt 0,39 0.010 tt u 0,607 0,630 0,660 0,57 0,015 u 0,593 0,612 0,632 0,660 0,78 0,020 0,572 0,596 0,615 0,634 0,659 1,18 0,030 0,578 0,600 0,620 0,638 0,659 1,57 0,040 0,582 0,603 0,623 0,640 0,658 1,97 0,050 0,585 0,605 0,625 0,640 0,658 2,36 0,060 0,587 0,607 0,627 0,640 0,657 2,75 0,070 0,588 0,609 0,628 0,639 0,656 3,14 0,080 0,589 0,610 0,629 0,638 0,656 3,54 0,090 0,591 0,610 0,629 0,637 0,655 3,93 0,100 0,592 0,611 0,630 0,637 0,654 4,72 0,120 0,593 0,612 0,630 0,636 0,653 5,51 0,140 0,595 0,613 0,630 0,635 0,651 6,29 0,160 0,596 0,614 0,631 0,634 0,650 7,08 0,180 0,597 0,615 0,630 0,634 0,649 7,87 0,200 0,598 0,615 0,630 0,633 0,648 9,84 0,250 0,599 0,616 0,630 0,632 0,646 11,81 0,300 0,600 0,616 0,629 0,632 0,644 15,75 0,400 0,602 0,617 0,628 0,631 0,642 19,68 0,500 0,603 0,617 0,628 0,630 0,640 23,62 0,600 0,604 0,617 0,627 0,630 0,638 27,56 0,700 0,604 0,616 0,627 0,629 0,637 31,49 0,800 0,605 0,616 0,627 0,629 0,636 35,43 0,900 0,605 0,615 0,626 0,628 0,634 39,37 1,000 0,605 0,615 0,626 0,628 0,633 43,30 1,100 0,604 0,614 0,625 0,627 0,631 47,24 1,200 0,604 0,614 0,624 0,626 0,628 51,18 1,300 0,603 0,613 0,622 0,624 0,625 55,11 1,400 0,603 0,612 0,621 0,622 0,622 59,05 1,500 0,602 0,611 0,620 0,620 0,619 62,99 1,600 0,602 0,611 0,618 0,618 0,617 66,93 1,700 0,602 0,610 0,617 0,616 0,615 70,86 1,800 0,601 0,609 0,615 0,615 0,614 74,80 1,900 0,601 0,608 0,614 0,613 0,612 78,74 2,000 0,601 0,607 0,613 0.612 0,612 118,11 3,000 0,601 0,603 0,606 0,608 0,610 0,01» or ,4 in. 0,705 0,701 0,697 0,694 0,688 0,683 0,679 0,676 0,673 0,670 0,608 0,666 0,663 0,660 0,658 0,657 0,655 0,653 0,650 0,647 0,644 0,642 0,640 0,637 0,635 0,632 0,629 0,626 0,622 0,618 0,615 0,613 0,612 0,612 0,611 0,611 0,609THROUGH RECTANGULAR ORIFICES 371 TABLE II. Co-efficients of efflux through rectangular orifices in a vertical piate, from Poncelet and Lesbros. The heads of water are measured directly at the orifice. Head of water, or distance of the sur- face of water from the upper side of the orifice. HEIGHT OF ORIFICE. Eng. in. Metres. 0,20» o © 9 0,05" 0,03- 0,02" 0,01* or 8 in. or 4 in. or 2 in. or 1,18 in. or ,8 in. or ,4 in. 0,00 0,000 0,619 0,667 0,713 0,766 0,783 0,795 0,19 0,005 0,597 0,630 0,668 0,725 0,750 0,778 0,39 0,010 0,595 0,618 0,642 0,687 0,720 0,762 0,57 0,015 0,594 0,615 0,639 0,674 0,707 0,745 0,78 0,020 0,594 0,614 0,638 0,678 0,697 0,729 1,18 0,030 0,593 0,613 0,637 0,659 0,685 0,708 1,57 0,040 0,593 0,612 0,636 0,654 0,678 0,695 1,97 0,050 0,593 0,612 0,636 0,651 0,672 0,686 2,36 0,060 0,594 0,613 0,635 0,647 0,668 0,681 2,75 0,070 0,594 0,613 0,635 0,645 0,665 0,677 3,14 0,080 0,594 0,613 0,635 0,643 0,662 0,675 3,54 0,090 0,595 0,614 0,634 0,641 0,659 0,672 3,93 0,100 0,595 0,614 0,634 0,640 0,657 0,669 4,72 0,120 0.596 0,614 0,633 0,637 0,655 0,665 5,51 0,140 0,597 0,614 0,632 0,636 0,653 0,661 6,29 0,160 0,597 0,615 0,631 0,635 0,651 0,659 7,08 0,180 0,598 0,615 0,631 0,634 0,650 0,657 7,87 0,200 0,599 0,615 0,630 0,633 0,649 0,656 9,84 0,250 0,600 0,616 0,630 0,632 0,646 0,653 11,81 0,300 0,601 0,616 0,629 0,632 0,644 0,651 15,75 0,400 0,602 0,617 0,629 0,631 0,642 0,647 19,68 0,500 0,603 0,617 0,628 0,630 0,640 0,645 23,62 0,600 0,604 0,617 0,627 0,630 0,638 0,643 27,56 0,700 0,604 0,616 0,627 0,629 0,637 0,640 31,49 0,800 0,605 0,616 0,627 0,629 0,636 0,637 35,43 0,900 0,605 0,615 0,626 0,628 0,634 0,635 39,37 1,000 0,605 0,615 0,626 0,628 0,633 0,632 43,30 1,100 0,604 Q,614 0,625 0,627 0,631 0,629 47,24 1,200 0,604 0,614 0,624 0,626 0,628 0,626 51,18 1,300 0,603 0,613 0,622 0,624 0,625 0,622 55,11 1,400 0,603 0,612 0,621 0,622 0,622 0,618 59,05 1,500 0,602 0,611 0,620 0,620 0,619 0,615 62,99 1,600 0,602 0,611 0,618 0,618 0,617 0,613 66,93 1,700 0,602 0,610 0,617 0,616 0,615 0,612 70,86 1,800 0,601 0,609 0,615 0,615 0,614 0,612 74,80 1,900 0,601 0,608 0,614 0,613 0,613 0,611 78,74 2,000 0,601 0,607 0,614 0,612 0,612 0,611 118,11 3,000 0,601 0,603 0,606 0,608 0,610 0,609372 WIERS. § 317. Wiers.—If water flows through wiers, or through notches in a thin piate, as, for example, FB, Fig. 422, the fluid vein then suffers a contraction on three sides, by which a diminution of the dis- charge is effected, since the quantity discharged from these orifices is Qj = § ju b h y/ 2 gh. But here the head of water EH = A, or the head of water above the sili, of the wier must not be measured imme- diately at the sili, but at least two feet before the piate in which the orifice lies, because the fluid surface before the opening suffers a depres- sion, which becomes greater and greater the nearer it is to the ori- fice, and in the plane of the orifice amounts to a quantity GR of from 0,1 to 0,25 the head of water FR, so that the thickness FG of the stream in this plane is only 0,9 to 0,75 of the head of water. Expe- riments instituted by many philosophers on the flow of water through notches in thin plates, have afforded a multiplicity of results, but not always of the desired accordance. The following short table contains the results of the experiments of Poncelet and Lesbros on wiers of two decimetres, or about 8 inches breadth. TABLE OF THE CO-EFFICIENTS OF EFFLUX FOR WIERS OF 2 DECIMETRES, = 7.87 INCHES BREADTH, ACCORDING TO PONCELET AND LESBROS. Head of water h. metrs. 0,01 or ,4 in metrs. 0,02 ,8 in. metrs. 0,03 1,2 in. metrs. 0,04 1,6 in. metrs. 0,06 2,4 in. metrs. 0,08 3,2 in. metrs. 0,10 4 in. metrs. 0,15 6 in. metrs. 0,20 8 in. metrs. 0,22 9 in. Co-efficient of efflux f*i=f 0,424 0,417 0,412 0,407 0,401 0,397 0,395 0,393 0,390 0,385 From the average of determinations, we may here put til = 0,4. Experiments on wiers of greater breadth gave Eytelwein the mean ^ = f m = 0,42, and Bidone ^ = $ . 0,62 = 0,41, &c. The most extensive experiments are those of d’Aubuisson and Castel. From these, d’Aubuisson asserts that for wiers whose breadth is no more than the third part of the breadth of the canal or side in which the wier lies, the mean of y. is = 0,60, therefore, we may put § u =0,40: but, on the other hand, for wiers which extend over the whole side, or have the same breadth as the water-course: p = 0,665, therefore, T3 ^>444; lastly, for other relations between the breadth of the wier and that of the canal, the co-efficients of efflux are very different, and lie between 0,58 and 0,66. Experiments made by the author, reduce the variability of these co-efficients to certain laws (§ 322). [Dunng his investigations in the summer of 1845, to determine the relative value of the several sources for supplying water to the city of Boston, the editor had an opportunity of making extensive series ofWIERS. 373 experiments on the passage of water through wiers of 1,2, and 3,01 feet in breadth, and from 0,066 foot to 2,087 feet in depth above the bottom of the notch. The water was measured in a cubical box, 6 feet on a side, to which was attached, on the exterior, a glass gauge tube with a scale extending to the top of the receptacle. In like man- ner, a gauge tube was inserted in the dam which contained the notch, and several feet distant from it, with the 0 of its scale accurately ad- justed to the level of the bottom of the notch. A scale sliding ver- tically was placed immediately over the centre of the wier by which the depth over the edge of the notch-board could be ascertained. The reservoir from which the water was drawn was at least 6 times as wide as the opening of the notch. The following are some of the co-effi- cients = f p) for the several breadths of wier: 1.—Wier 3,01 feet in breadth, cut in 2 inch planks— Co-efficients of discharge ” = ■§/*• 0,3667 0'l89 “ . . 0,3794 0,280 “ . . 0,3973 0,316 “ . . 0,4211 0,360 “ . . 0,4307 0,480 “ . . 0,4349 0,545 “ . . 0,4376 0,689 “ . . 0,4301 0,755 “ . . 0,4294 0,801 “ . . 0,4208 1,023 “ . . 0,4129 Full depth A, over bottom of notch. 0,075 feet Depression of surface at the notch. 0,021 feet 0,040 “ 0,070 “ 0,079 “ 0,086 “ 0,097 “ 0,120 “ 0,149 “ 0,155 “ 0,158 “ 0,167 “ 2. Wier 2 feet wide, in 0,199 feet 1,020 “ 1,062 “ 1,232 “ 1,280 “ 3. Wier 1 foot wide, in 0,329 feet 0,333 “ 0,339 ‘‘ 0,352 “ 0,360 “ 2,060 “ 2,087 “ a 1 inch board— 0,4195 0,4344 0,4408 0,4477 0,4460 inch board— 0,4144 0,4166 0,4191 0,4265 0,4265 0,4149 0,4130 uuu. 0,196 “ 0,206 « 0,228 “ 0,230 « unc. ‘ unc. ‘ unc. ‘ 0,068 ‘ 0,070 ‘ 0,118 ‘ 0,125 ‘ The ‘‘ full depth over the notch” here signifies, of ^^n^achcase general level of the reservoir, above the edge of the wi • jn. it will be observed that the above co-efficients \nc CTa(iually as crease of depth up to a certain Point>/ndAthe*JdlfleXpenments were far as the observations were extended. As tnes r ^ econo- made with a view to determine an important pra believed mical question, they were conducted with great care, 32374 MAXIMUM AND MINIMUM OF CONTRACTION. to be worthy of reliance as the basis of computation for works on an extended scale.] Examples.—1. A wier, 0,25 metre broad and 0,15 head of water, gives per second the discharge Q = 0,393 . bh y/2gh =» 0,393.4,429.0,25 . (0,15)* = 0,435.0,0581= 0,02527 cubic metres.—2. What breadth must be given to a wier which, with a head of water of 8 inches, will allow 6 cubic feet per second to pass through ? It is J _ Qi =_______________6 __________1______= 3,436 feet. 0,4.8,02 v/ 3,208.0,5443 If, according to Eytelwein, we take /u, = 0,42, it follows that : b 6 = 3,368.0,5443 3,27 feet. § 318. Maximum and Minimum of Contraction.—In the flow of water through orifices in a plane side, the axis of the streara is per- pendicular to the surface of the side, and therefore the amount of the contraction is a mean, but if the axis of the orifice or of the fluid streara forms an acute angle with the portion of the side containing the orifice, the contraction will be less; and if the angle between this axis, and the inner surfaces of the edges of the aperture, be obtuse, the contraction will be stili greater. The one case is represented in Fig. 423, and the other in Fig. 424. Without doubt this difference of contraction is caused by the particles of water, which flow towards the orifice frora the sides, deviating less from their direction in the one than in the other case, when they pass through the orifice and unite to form a fluid stream. Fig. 423. Fig. 424. The contraction is a minimum, i. e. nothing, if bythe gradual con- vergence of the side which embraces the orifice, the lateral flow is entirely prevented, and a maximum if the side has a direction opposite to that of the fluid st ream, so that certain particles of water must re- vohe 180 before arriving at the orifice. Both cases are representedPARTIAL CONTRACTION. 375 in Figs. 425 and 426. In the first case, the co-efficient of efflux is about 1, viz. 0,96 to 0,97; and in the second from the measurements of Borda, Bidone and the author, a mean of 0,53. Changes in the co-efficients of efflux through convergent sides very often present themselves in practice ; they occur in dams, which are inclined to the horizon, as in Fig. 427. Poncelet found for a similar opening the co-efficient of efflux p = 0,80, when the board was inclined 45°, and on the other hand, p = 0,74 only for an inclination of 63J°, that is, for a slope of For similar wiers, Fig. 428, where, as in the Fig. 427. Fig* 428. Poncelet sluice-board, contraction takes place at one side only, the author found p = 0,70, therefore, ^ ^ = 0,467 for an incli- nation of 45°, and n = 0,67, therefore, p. = 0,447 for an inclination of 63£°. Example. If a sluice-board, inclined at an angle of 50°, which goes across a channel 2£ feet broad, is drawn up £ foot high, and the surface of water stands 4 feet above the bottom of the channel, the lieight of the aperture may be put a = ^ sin. 50° = 0,3830 feet, the head of water h = 4 — £ . 0,3830 = 3,8085 feet, and the co-efficient of efflux /u = 0,78 ; hence, the discharge Q, = 0,78 . 2,25.0,3830 . 8,02 3,8085 = 10,49 cubic feet (English). § 319. Partial Contraction.—We have only hitherto considered those cases where the water flows from ali sides towTards the aperture, and forms a contracted vein around, and we must now investigate others, where the water flows from one or more sides to the aperture, and therefore produces a stream only partially contracted. To dis- tinguish the circumstances of contraction, we will call the case, where the vein is contracted on ali sides, generat; and the case, where it is only contracted in one part of its circumference, partial, or imperfect contraction. Partial contraction is induced when an orifice in a plane thin piate is confined by other plates in the direction of the fluid stream at one or more sides. In Fig. 429, are represented four orifices of equal size a, b, c, d, in the bottom of a vessel. The con- traction by efflux through the orifice a in the middle of the bottom is general, because the water can flow to it from all sides; the con- traction from the efflux through i, c, d, is partial, because the water can only flow to them from one, two, or three sides. Like- wise, if a rectangular lateral aperture goes to the bottom of the vessel, the contraction is then partial, because it falis away at the bottom side, if, further, the aperture of the Fig. 429.376 PARTIAL CONTRACTION. dam reaches the bottora in the lateral walls of the channel, there is then only a contraction on one side. Partial contraction is remarkable in two respects; first, by giving an oblique direction to the stream; and, secondly, by increasing the quantity of discharge. If the lateral aperture JF, Fig. 430, reaches a second side (1D, so that no contraction takes place there, the axis FK of the fluid stream becomes deflected by an angle KFG of about 9° from the normal FG to the plane of the orifice. The obliquity of the stream is much greater if two adjacent sides of the orifice have projecting borders. If the orifice has borders in two oppositely situated sides, and contraction at these prevented, such a deviation of course will not take place, but at the other side, the vein at some distance from the orifice, will spread out more than if the border were not there. Although a greater discharge is obtained by a partial contraction, we must, as a rule, endeavor to avoid this, because the fluid stream, in consequence, suffers a deviation in its direction and a greater exten- sion in its breadth. Experiments on the efflux of water with partial contraction have been made by Bidone and by the author. They allow us to assume that the co-efficients of efflux increase simultaneously with the ratio of the contracted part to the whole perimeter, though it is easy to per- ceive that this relation is different, if the perimeter is almost or en- tirely restricted, and the contraction almost or entirely suppressed. Let us put the ratio of this restriction to the entire perimeter = n, and let us represent by *, any number deduced from experiment, we may then, although only approximately, put the ratio of the corresponding co-efficient of efflux /unof partial contraction to the co-efficient of efflux of perfect contraction: ^ = 1 + x w, and consequently = (1 + * n) p0. i“° Bidone’s experiments give for circular orifices * = 0,128, and for rectangular % = 0,152; the author’s, however, give for the last, * = 0,134. Rectangular orifices with borders, are those which are most frequently met with in practice; we will assume for them the mean value * = 0,143, and hence put *tn = (1 + 0,143 . n) /*0. or a rectangular lateral orifice of the depth a and breadth £, b 71 JT(a + b)9 ^ contraction on one side b is suppressed; if, for instance, this side lies in the plane of the bottom; again, n = J, if a side a and a side b are bordered, and n = ^ a if on one side . J L 1 • 2(fl-f-^) 0, and both sides a, the contraction is prevented; if, for example, the Fig. 430.IMPERFECT CONTRACTION. 377 orifice takes up the whole breadth of the reservoir, and reaches the plane of the bottom. Example. What quantity of water flows through a 3 feet broad and 10 inch deep ver- tical aperture of a dam at a pressure of 1^ feet above the upper side of the aperture, if the lower one coincides with the bottom of the channel, and hence there is no contraction at the bottom'? The theoretical discharge is: Q = jj . 3 . 8,02 y/1,5 + /g- = f . 8,02 1,9166 . . = 28,11 cubic feet. According to Poncelet’s table for generai contraction fx = 0,604, we have, therefore, 3 9 n sss--------- = ________= A: hence, for the above cases of partial contraction 2(3+ jj) 18+5 fxn (= 1 + 0,143 . . 0,604 = 1,056.0,604 == 0,638, and the effective discharge is Qx == 0,638 Q = 0,638.28,11 = 18,14 cubic feet § 320. Imperfect Contraction.—The contraction of the fluid vein depends, further, upon whether the water before the orifice is tolera- bly stili, or whether it arrives before it with a certain velocity. The quicker the water flows to the orifice, the less contracted does the vein become, and the greater is the discharge. The relations of con- traction and efflux above given and investigated, have reference only to the case where the orifice lies in a large side, and it can only be assumed that the water flows to it with a small velocity; hence we must know the relations of contraction and efflux, when the trans- verse section of the orifice is not much less than that of the affluent water, and when, consequently, the water arrives at the orifice with a considerable velocity. In order to distinguish these two cases from one another, we shall call the contraction, where the superincumbent water is stili, perfect; and that where it is in motion, imperfect con- traction. The contraction, for example, is imperfect in the efflux from a vessel ACy Fig. 431, because the transverse section F of the orifice is not much smaller than that of G of the arriving water, or the area of the side CD, in which this orifice lies. If, on the other hand, the vessel had the form ABClDv and, there- fore, the area of the bottom surface C1D1 much greater than the transverse section F of the orifice, the efflux would then go on with per- fect contraction. The imperfectly contracted vein is besides distinguishable, not merely by its greater thickness from the perfectly con- tracted fluid vein, but also by its not having so transparent and crystalline an appearance. If the ratio of the area of the orifice F> and the side containing the orifice G, therefore, =z n, the co-efficient of efflux for perfect contraction = and that for imperfect = /*n, we may with greater accuracy, according to the experiments and calculations made by the author, put: 1. For circular orifices: = M0 [1 + 0,04564 (14,821 *—1)], and 32* Fig. 431.378 C0RRECTI0N8 OF THE CO-EFFICIENTS OF EFFLUX. 2. For rectangular orifices: #**- * [1 + 0,0760 (9"-l)].* To render the calculation easier in cases of application, tne cor- rections ~ of the co-efficient of efflux on account of imperfect contraction are compiled in the following short tables. TABLE I. CORRECTIONS OF THE CO-EFFICIENTS OF EFFLUX FOR CIRCULAR ORIFICES. n 0,05 0,10 0,15 0,20 0,25 0,30 0,35 0,40 0,45 0,50 Mn—f*0 0,007 0,014 0,023 0,034 0,045 0,059 0,075 0,092 0,112 0,134 n 0,55 0,60 0,65 0,70 0,75 0,80 0,85 0,90 0,95 1,00 f* 0 0,161 0,189 0,223 0,260 0,303 0,351 0,408 0,471 0,546 0,613 TABLE II. CORRECTIONS OF THE CO-EFFICIENTS OF EFFLUX FOR RECTANGULAR ORIFICES. n 0,05 0,10 0,15 0,20 0,25 0,30 0,35 0,40 0,45 0,50 0,009 0,019 0,030 0,042 0,056 0,071 0,088 0,107 0,128 0,152 0,55 0,60 0,65 0,70 0,75 0,80 0,85 0,90 0,95 1,00 A*o 0,178 0,208 0,241 0,278 0,319 0,365 0,416 0,473 0,537 0,608 ' The different values of the ratio of the transverse sections n » _ Gr stands above in these tables, and immediately below additions to the • “ Experimenu on the Imperfeot Contraotion of Water,” ite., Leip.ic, 1843.EFFLUX OF WATER IN MOTION. 379 co-efficients of efflux, on account of imperfect contractiori; for example, for the ratio of the transverse sections n = 0,35, i. e., for the case where the area of an orifice is 35 hundredths of the area of the whole side of the orifice, we have for circular orifices —---------— = 0,075, and po for rectangular orifices = 0,088; therefore, the co-efficient of efflux for perfect contraction in the first case is to be made about 75 thou- sandths, and in the second about 88 thousandths greater to obtam the corresponding co-efficients of Were the co-efficient of efflux p0 first case /*0 35 = 1,075.0,615 = 1,088.0,615 = 0,669. Example.—What discharge does a rec- tangular lateral aperture F, li feet broad and £ foot deep, give if it be cut in a rec- tangular wall CD, Fig. 432, 2 feet broad and 1 foot deep, and the head of water EH = h in stili water amounts to 2 feet? The theoretical discharge is Q = 1,25.0,5.8,02 %/2~= 5,012 . 1,414 = 7,086 cubic feet, and the co efficient of efflux for perfect con- traction is, according to Poncelet, /u0 = 0,610 ; but now the ratio of the transverse sections n = Z_ l'25 • °’5 _0,312, and for » = G 2.1 0,312, from Table II., —= 0,071 + (0,088 — 0,071) = A* o that the co-efficient of efflux for the present = 0,6557, and the discharge Qt = 0,6557 . l efflux for imperfect contraction. = 0,615, we should have in the 0,661, and in the second, p0 35 = Fig. 432. 0,071 -|- 0,004 =0,075; hence it folio ws, ise is fxQ,3,2 = 1,075 . fxQ ^ 1,075.0,610 = 0,6557 . 6,987 = 4,581 cubic feet. § 321. Efflux of Water in Motion.—We have hitherto assumed that the head of water has been measured in stili water; we must now, therefore, consider the case when only the head of water in motion, and flowing with a certain velocity towards the orifice, can be measured. Let us suppose the case of a rectangular lateral orifice, and represent its breadth by 6, and the heads of water with respect to both horizontal edges hx and A2, the height due to the velocity c of the affluent water by k, we shall then have the theoretical dis- charge : Q = § 6 v/2 gIA+ k) i — (K+ A:) 2]. This formula is not directly applicable to the deterromation o discharge, because the height due to the veloci y. * _ £ _ i («\’ U ». dependent o» Q, and fnrthe, transforma- 2 g2 g\G) ... hence it is far tion leads to a complicated equation of a higher » and under* simpler to put the effective discharge ® n’e especially stand by ^ not the mere co-efficient of efflux, { frequently, dependent on the ratios of the transverse sections- vater flowing this case presents itself when the object is to me380 EFFLUX OF WATER IN MOTION. in canals and courses, because it is sel- dom possible in this case to dam up the water so high by a transverse sec- tion BCy Fig. 433, containing the orifice of discharge, that the orifice EF becoraes only a small part, compared with the transverse section of the stream of water flowing to it; and, hence, the velocity of the last very small com- pared with the mean velocity. From experiments made by the author on this subject with Poncelet orifices, where the head of water is measured one metre above the plane of the orifice, the expression : ~ 0,641 *= 0,641 . n*, /*0 VG/ F may be taken as tolerably accurate, when n = f- is the ratio of the G transverse section, which, however, should not muchexceed\\ further,p0 represents the co-efficient for general contraction, taken from PonceletJs table corresponding to the present case. If b be the breadth, a the depth of the orifice, B the breadth and A the depth of the fluid stream, and h the depth of the upper side of the orifice below the surface of water, we have, accordingly, the effective discharge: Q, _ [l +0,641 (A)”] * . «4 J * 641 • 0,296') f40 = 1,056.0,602 =0,6357, and the effective quantity discharged = 1^,93.0,6357 = li,3j cubic feet § 322. Imperfect contraction veryoften occursinthe efflux through wiers, as in Fig. 422. Wiers may take up a part onlyof the breadth of the reservoir or canal, or the whole breadth. In the latter case, contraction at the sides of the aperture does not take place, and for this reason more water flows through them than through wiers of the first kind. The author has made experiments also on these circum-CORRECTIONS FOR WIERS. 381 stances of efflux, and deduced from the results formulas by which the corresponding co-efficients may be estimated with tolerable certainty , . . Cr h b rr with the assistance of the ratio of the sections n = — = jj-jf 11 we retain the denominations of the former paragraph, we then have for the Poncelet wiers: = 1,718 (*$ — 1,718 . n*, V G/ and for wiers occupying the entire breadth of the canal: **n **» M 0,041 + 0,3693 »*, 1*0 , hence, in the first case, the discharge is: l[ 1 + 1,718 (ii)*] * • b And in the second: ___ Q, * |^1,041 + 0,3693 * . b f°r most sim- ple values of n are put down. TABLE I. CORRECTIONS FOR THE PONCELET WIERS. n 0,05 0,10 0,15 0,20 0,25 0,30 0,35 0,40 0,45 0,50 * e 0,000 0,000 0,001 0,003 0,007 0,014 0,026 0,044 0,070 0,107 TABLE IL CORRECTIONS FOR WIERS OVER THE ENTIRE SIDE, OR WITHOUT ANY LATERAL CONTRACTION. n 0,00 0,05 0,10 0,15 0,20 0,25 0,30 0,35 0,40 0,45 0,50 0,041 0,042 0,045 0,049 0,056 0,064 0,074 0,086 0,100 0,116 0,133 1 1 1 ' 1 ' 1 wacanalS feettooad,a Example. To determino the quantity of water «irnea o J the water is allowed waste board is applied, with an outward sloping edge, ove 0f the canal is 3 j to flow after it has ceased to rise; the headof water Q ss } . 5 . 8,02 . feet, and above the edge lj feet, hence the tkeoretical aiac argo382 SHORT TUBES, OR MOUTH-PIECES. 48,12 cubicfoot. The co-efficient of efflux is, since —- ll5 3,5 $ and fxQ *= 0,577, /u J == [1,041 +0,3693.(|)9]. 0,577 = 1,110.0,577 =0,64,hence the effectivedisclmrge Q, = 0,64 . Q = 0,64.48,12 as 30,79 cubic feet. CHAPTER III. ON THE EFFLUX OF WATER THROUGH TUBES. § 323. Short Tiibes, or Mouth-pieces.—If water is allowed to flow through short tubes, or mouth-pieces, other relations take place than when it flows through orifices in a thin piate, or through outwardly sloping orifices in a thick piate. When the tube is prismatic, and its length 2\ to 3 times that of its width, it then gives an uncon- tracted and opaque stream, which has a small distance of projection, and hence, also, a smaller velocity than that of a jet flowing, under otherwise sirailar circumstances, through an orifice in a thin piate. If, therefore, the tube KL has the same transverse section as the orifice F, Fig. 434; and if also the head of water of both is one and the same, we then obtain in LR a troubled and uncontracted, and, there- fore, a thicker jet, and in FH a ciear and contracted, and, therefore, thinner one; and, it may be observed, that the distance of the pro- jection ER is less than that of DH. This ratio of efflux only takes Fig. 434. Fig. 435. place when the tube is of a given length; if it is shorter, or scarcely as long as it is broad, then the jet KL, Fig. 435, will not touch the sides of the tube, the tube will have no influence on the efflux, and the jet will be the same as through orifices in a thin piate. Sometimes in tubes of greater length, the fluid stream does not entirely fili the tube, naraely: when the water is not allowed to come into contact with the sides of the tube; but if in this case we close the outer orifice by the hand or by a board for a few moments, a stream will then be formed which will entirely fili the tube, and the so-called full flow will then take place. Contraction of the fluid vein takes place also in the flow through tubes, but the place of contrac-CYLINDRICAL TUBES. 383 Fig. 436. tion is here in the interior of the tube. We may be convinced of this, if we avail ourselves of glass tubes, such as KLy Fig. 436, and color the water, for in this case we shall remark, that there is progressive motion only in the middle of the transverse section G close behind the place of entrance K, but not at the outside of it, and that it is a sort of eddying motion which takes place. But it is the capillarity, or the adhesion of the water to the sides of the tube, which causes the fluid entirely to fili the end FL of the tube. The water flowing from the tube has only a pressure equal to that of the atmosphere, but the contracted section G is only a times the size of the section F of the tube, and for this reason the velocity in it - times as great as the velocity of efflux v \ hence the pressure of the water in the vicinity of G is -S-K) the atmospheric pressure. 1 "1 iL (§ 307) less than at its exit, or than J # If we bore a narrow hole in the tube at G, no discharge will pass through it, but there will be an absorption of air rather; the full discharge and the action of the tube will at last entirely cease if the hole be made wider, or more holes bored. § 323. Cylindrical Tubes.—Numerous experiments have been made on the flow of water through cylindrical additional tubes; but the results vary considerably from one another.* The co-efficients of Bossut are those, which from their smallness (0,785) have been found to vary most from others. From the experiments of Michelotti, with tubes from £ to 3 inches width, and with a head of water of from 3 to 20 feet, the mean of this co-efficient is: p = 0,813. The experi- nients of Bidone, Eytelwein and d’Aubuisson vary very little from this. The mean, however, which may be adopted, and which cor- responds particularly with the author’s experiments on the discharge through short mouth-pieces = 0,815. As we have found this for orifices in a thin piate 0,615, it follows that, under otherwise similar r oi e circumstances and relations,---- = 1,325 times as much water 615 flows through cylindrical additional tubes, as through round orifices in a thin piate. These co-efficients, moreover, increase as the width of tubes becomes less, and but slightly with the increase of the head of water or velocity of efflux. According to the author’s experiments, under a pressure of from 9 to 24 inches for tubes three times as long as broad : through adjutages was, * A considerable series of experiments on the flow of wa er ^ which yet await some years since, performet! by a committee of the Franklm n ’ a proper reduction to render the results available.—Am. Ed.384 CO-EFFICIENT OF RESISTANCE. at i 2 3 4 centimetres width, or ,4 in. or ,8 in. or 1,2 or 1,6 inches width. M = 0,843 0,832 0,821 0,810 According to this table, therefore, the co-efficients increase con- siderably as the width of the tubes decreases. Buff found for tubes 2,79 lines wide, and 4,3 lines long, the co-efficients of efflux gradually to increase from 0,825 to 0,855, when the head of water sank from 33 to inches successively. The author found a co-efficient of efflux of 0,819 for the flow of water through rectangular additional tubes. If the additional tubes KL, Fig. 437, are on the inside partially confined; if, for instance, one side is contiguous to the bottom, and if a partial contraction is produced thereby, then the co-efficient of efflux, from the author’s experiments, does not perceptibly increase, but the Fig. 437. Fig. 438. water flows away at different parts of the section, with different velo- cities, and, of course, from the side BC faster than from the side opposite to it. If the inner anterior surface of a tube does not coincide with the side surface, but projects, as «, b, c, Fig. 438, then this tube is called an intemal additional tube. If the anterior surface of this tube is at least Jth as broad as the tube is wide, as for example a, then the co-efficient of efflux will remain the same as if this surface were in the plane of the side, but if the anterior surface be less, as b9 c, the co-efficient will then be less. For a very small and almost vanishing anterior surface, according to the experiments of Bidone and the author, this amounts to 0,71 if the vein filis the tube, and 0,53 (compare § 318) if it does not quite fili the inner sides of the tube. In the first case (6) the fluid stream is broken and divergent, like a brush ; and in the second (c) strongly contracted and quite crystalline. § 324. Co-efficient of Resistance.—As water flows without contrac- tion from prismatic additional tubes, it follows that, in its efflux through these mouth-pieces, the co-efficient of contraction = unity, and the co-efficient of velocity = the co-efficient of efflux /a. A dis- charge Q with the velocity v, possesses a vis viva v2, and is ca-CO-EFFICIENT OF RESISTANCE. 385 pable of producing the mechanical effect — Q y (§ 71). But now 2g the theoretical velocity of efflux = -, hence the mechanical effect * ^ 2 — . —. Q y corresponds to the mass of water flowing out, and the discharee Q accordingly loses by efflux the mechanical effect \t* 2g 2g) \** ) 2g For efflux through orifices in a thin piate, the mean of 9 = 0,97, hence the loss of effect here amounts to for efflux through short cylindrical tubes 9 = 0,815, and the corre- sponding loss of effect „ [7-J_V - ll - « r - 0.605 ? Q r, L\0,815/ r % i. e. eight times as great as for efflux through orifices in a thin piate. In rendering available the vis viva of flowing water, it is consequently better to let the fluid flow through orifices in a thin piate, than through prismatic tubes. But if the inner edges in which the tube meets the sides of the cistem are rounded, and by this a gradual passage from the cisterns into the tube effected, the co-efficient of efflux will then rise to 0,96, and the loss of mechanical effect will be brought down to per cent. In shorter adjutages, accurately rounded, having the form of the contracted fluid vein = 9 = 0,97, and hence the loss of mechanical effect as for orifices in a thin piate = 6 per cent. A head of water (— — 1 Q y is due to the loss of mecha- \9a /2g nical effect (—___1 \ hence we may also suppose that from ' W / 2g the obstacles to the efflux, the head of water suffers the loss (-y — 1 1 —, and assume after deduction of this loss, that the resi- * /2g duary part of the head of water is expended in generating the velocity. We may call this loss (—----1 \ —, which increases with the square \9* / 2g of the velocity of efflux, the height due to the resistance, and the co- efficient -i. _ 1, with which the height due to the velocity is to be multiplied to obtain the height due to the resistance, the co-efficient of resistance. We shall represent in what foliows, the co-efficient, ex«- pressing the ratio of the height of resistance to the head of water, by the letter ?; therefbre, the height due to the resistance itself may be expressed by f . g. By the formula f « -1 — 1 and ♦ - -==, -/1+f 33386 OBLIQUE ADDITIONAL TUBES. the co-efficient of resistance may be calculated from the co-efficient of velocity, and vice versa. y/ 1,4 ir s 0,02182 square feet; consequently, the quantityof water sought is Q = 0,263 cubic feet.—2. If a tube of 2 inches width, under a pressure of 2 feet, deliver in a minute 10 cubic feet of water, its co-efficient of efflux, or of velocity, is then f r Fy/ 2 g h 10 1 ---- ss 0,674, the co-efficient of resistance 60.0,02182.8,02 . 2 1,050 ^ 2 ( 1 \ — 1 = 1,201; and lastly, the loss in head of water produced by the resistances \ 0,674/ of the tube: = 1,201 — = 1,201.0,0155 (S.X =0,0186__________l----*= 1,085 feet. 2g \F/ 0,1309* § 325. Oblique Additional Tubes.—Obliquely attached or obliquely cut tubes give a smaller quantity of water than rectangularly attached, or rectangularly cut additional tubes, because the direction of the water in them becomes changed. Experiments conducted upon an extensive scale, have led the author to the following. If S be the angle which the axis of the tube KL, Fig. 439, makes with the normal KJY to the plane AB of the inner orifice of the tube; and if f be the co-efficient of resistance for rectangularly cut tubes, we shall then have the co-efficient of resistance for the inclined tube: f 1 = ? + 0,303 sin. S + 0,226 sin. 62. Let us take for f the mean value 0,505, and we shall obtain: Fig. 439. for *> = 0 10 20 30 40 50 60° The co-efficient of resistance = The co-efficient of efflux fxx = 0,505 0,815 0,565 0,799 0,635 0,782 0,713 0,764 0,794 0,747 0,870 0,731 0,937 0,719 From this, for example, the co-efficient of resistance of an addi- tional tube deviating by 20° from the axis is f = 0,635, and the co-efficient of efflux = — A = 0,782, and for 35° deviation, the v^l.635 first = 0,753, and the last = 0,755. In genera], these inclined and additional tubes are larger than we a\e itherto assumed, and they should be longer too, because the water would not otherwise perfectly fili the tube. The preceding tormula represents only that part of the resistance which is due toIMPERFECT CONTRACTION. 387 that portion of tube at the inner orifice, which is three times as long as the tube is wide. The resistance which the remaining portion of tube opposes to the motion of the water, will be given subsequently. Example. If the plane of the inner orifice AB of a horizontally lying pond sluice 1TZ, Fig. 440, as likewise the interior surface of the pond dam, is inclined 40° to the horizon, then the axis of the pipe makes, with the normal to this plane, an angle of 50°, and hence the co-efficient of resistance for efflux through the portion of the interior orifice of this tube is = 0,870; and if now the co-efficient of resist- ance 0,650 were due to the remaining and longer portion, the co-efficient of resistance of the entire tube would thenbe = 0,870 + 0,650 = 1,520, and hence the co-efficient of efflux =--------- - = ——.............. ■ = ^/\ + 1,520 v/ 2,520 0,630. For a 10 feet head of water and 1 foot width of tube, the following discharge would be given: Q = 0,63 . — . 8,02 y/To = 12,55 cubic feet. 4 Fig. 440. § 326. Imperfect Contractiori.—When a cylindrical additional tube KLy Fig. 441, is inserted into a plane wall AB, whose area G does not much exceed the transverse section F of the tube, the water then comes to the place of insertion with a velocity which must not be dis- regarded, and it then issues into the tube with imperfect contraction only, on which account the velocity of efflux is again greater than when the water before entrance into the tube is to be Fig. 441. assumed as stili. Again if __= 7i is the ratio of the section of the G tube to that of the area of the side, and further, p0 be the co-efficient F of efflux for perfect contraction, where — may be equated to 0, we (r shall have, according to the experiments of the author, to put the co- efficient of efflux for imperfect contraction, or for the ratio of the sections n: = 0,102 n -f 0,067 n2 + 0,046 n3 or ^0 h* = (1 + 0,102 n + 0,067 7i2 + 0,046 n3). If the transverse section of the tube occupies the sixth part of the whole surface of the side, there is: nh = (1 + 0,102 . i + 0,067 . * + 0,046 . 5 jg) = ^o(l + 0,017 + 0,0019 + 0,0002)= 1,019 ^0,or ™ \\TQg.put = °’815» = °>815 • b019 = 0,830. ihe following useful and convenient table gives somewhat more accurately the values for correction —____— n388 CORRECTION FOR IMPERFECT CONTRACTION. TABLES OF CORRECTION FOR IMPERFECT CONTRACTION, BY EFFLUX THROUGH SHORT CYLINDRICAE TUBES. n 0,05 0,10 0,15 0,20 0,25 0,30 0,35 0,40 0,45 0,50 f*0 0,006 0,013 0,020 0,027 0,035 0,043 0,052 0,060 0,070 0,080 n 0,55 0,60 0,65 0,70 0,75 0,80 0,85 0,90 0,95 1,00 0,090 0,102 0,114 0,127 0,138 0,152 0,166 0,181 0,198 0,227 f*o By efflux through short parallelopipedical tubes, these corrections are nearly the sarae. These co-efficients are especially applicable to the efflux of water through compound tubes; for ex- ample, in the case represented in Fig. 442, where the orifice of the short tube KL enters the wider short tube CKy whose orifice again lies in the cistern AC. Imperfect contrac- tion takes place at the entrance of the water from the wider into the nar- rower tube, and hence the co-efficient of efflux must be determined by the last rule. If we put the co-efficient of resistance corresponding to the co-efficient of efflux = the co-efficient of resistance for the entrance into the wider tube from the cistern = C, the head of water = A, the velocity of efflux = vy and the ratio — of the section of the tubes = n, therefore, the velocity of the water in the wider tube the formula gives: nvy then 4 - £ + f ■• ^ + f. a + vr ■+ y £ And hence: 2g 2g v' 1 + n* f + i’ Exampk.What discharge will the apparatus delineated in Fig. 442 deliver, if the liead of water h = 4 Icet, the width of the narrower tube 2 inches, and that of the wider one 3 inches? n = (})» = <, hence = 1,069 . 0,818 = 0,871, and the corresponding co-efficient of resistance £, = ( 1 V — 1 = 0,318: but now we V1 \ 0,871 ) have ( = 0,508 and n’ . £ = J} . 0,505 = 0,099; hence it follows, that 1 + n* { +CONICAL TUBES. 389 = 1 + 0,099 + 0,318 as 1,417, and the velocity of efflux v = ^/1,417 1,417 = 14,34 feet Again, since the transverse section of the tube = 0,02182 square feet, the discharge will be Q = 14,34.0,02182 = 0,313 cubic feet § 327. Conical Tubes.—Additional conical tubes give a discharge different from that of prismatic or cylindrical tubes; they are either conically convergent, or conically divergent; in the first case the outer orifice is smaller, and in the second case larger than the inner orifice. The co-efficients of efflux for the first tubes are greater, and for the last, less than for cylindrical tubes. One and the same coni- cal tube gives more water when the wider orifice is made the exit orifice, as K in Fig. 443, than when it is turned inwards, as L in the same figure, except that it does not give a greater quantity in propor- tion as the wider orifice exceeds the narrower. If many, as Venturi and Eytelwein, give for conically divergent tubes, a greater co-effi- Fig. 444. cient of efflux, than for conically convergent tubes, it must always be borne in mind, that they take the narrower transverse section for the orifice. The following experiments instituted at pressures of from 0,25 to 3,3 metres, with a tube AD 9 centimetres long, Fig. 444, bring before us the effect of conicalness in tubes. The width of these tubes at one extremity amounted to DE =* 2,468, at the other AB = 3,228 centimetres, and the angle of convergence, i. e. the angle AOB, which the oppositely situated sides AE and BD of a section in the direction of the longer axis include = 4°, 50'. By efflux through the narrower orifice, the co-efficient was = 0,920, but by efflux through the wider it was = 0,553, and if in the calcula- tion, we take the narrower entrance orifice for the transverse section, it will give = 0,946. In the first case, when the tube was applied as a conically convergent adjutage, the fluid vein was little con- tracted, thick and smooth; but, in the second case, when the tube served as a conically divergent adjutage, it was strongly divergent, broken, and spouting. Venturi and Eytelwein have experimented further on efflux through conically divergent tubes. Both philoso- phers have applied these conical tubes to cylindri- cal and conoidal adjutages, made after the form of the contracted fluid vein. By such a connection as is represented in Fig. 445, where the divergent por- tion KL of the outer orifice is between 12 and 2l£ lines wide, and 8{ § of an inch long, and the angle 33* Fig. 445.390 RESISTANCB OF FRICTION. of convergence estimated at 5°, 9'. Eytelwein found /4 » 1,5526 when he took the narrow end for the orifice, and, on the other hand, p a 0,483 for the wider end, in which he was right. Through this combined adjutage there certainly flows ** 2,5 times as 9 1 5526 n much as through a simple orifice in a thin piate, and =» 1*9 times as much as through a short cylindrical tube. With small velo- cities and greater divergence, it is scarcely possible, even by pre- .viously closing the tubes, to bring about a full flow. § 328. The most ample experiments ha ve been made by d5Aubuis- son and Castel on effiux through conically convergent additional tubes. The tubes for this purpose were of great variety, of different lengths, widths, and angles of convergence. The most extensive experiments were those made with tubes of 1,55 centimetres width at the dis- charging orifice, and of from 2,6 times greater, i. e. of 4 centimetres in length, for which reason we will here communicate the results in the following table. The head of water was, throughout, 3 metres. The discharges were measured by a special gauge-cistern; but in order to obtain besides the co-efficients of effiux, those of the velocity and contraction, the amplitude of the jet, due to given heights, were measured, and from these the velocity of effiux (see § 38, Ex. 2) cal- v culated. The ratio — , of the effective velocity v to the theo- ______ */2gh. * retical y/ 2 g h gave the co-efficient of velocity 9, as also the ratio Q m ______ ——===== of the effective discharge Q to the theoretical F %/2 g h F y/ 2 g h gave the co-efficient of effiux /*, and, lastly, the ratio of both co-effi- cients, t. e. —, determined the co-efficient of contraction a. $ TABLE OF THE CO-EFFICIENTS OF EFFLUX AND VELOCITY FOR EFFLUX THROUGH CONICALLY CONVERGENT TUBES. Angle of convergence Co-efficient of effiux. Co-efficient of velocity. Angle of convergence Co-efficient of effiux. Co-efficient of velocity. 0° Fig. 446, is horizontal, inclines, or ascends, if only by the head of water is understood the depth RL of the middle point L of the orifice of the tube below the surface of water HO of the efflux reservoir. If h is the head of water, hx the height due to the resistance for the orifice of entrance, and h2 that for the remaining portion of the tube, we then have: v2 _ l i t l . v2 %g' If f represents the co- k — (A1 +h2) = 2g> or h = K + efficient of resistance for the portion of tube next the cistern, and the co-efficient of the resistance of friction of the rest of the tube, we then have , v2 l v2 v2 - 1 ■ 5 + 1‘ ■ d • Tg + l \ V2 or, !•)*-( 1 + f + d) 2g' and 2.)v ■■ x/2gh J From the last formula the discharge Q = Fv is given. For very long tubes 1 + ? is small compared with fj -, whence, d simply, h = f, L. and inversely, d 2g________________ s § 330. The co-efficient of friction, like the co-efficient of efflux, is not quite constant; it is greater for small, and less for great veloci- ties; i. e. the resistance of the friction of water in tubes does not in- crease exactly with the square of the velocity, but with some other power of it. Prony and Eytelwein have assumed, that the head of water lost by the resistance due to friction ought to increase as theRESISTANCE OF FRICTION. 393 simple velocity and as its square, and have given for it the expres- sion h- (a v + j3 v2) where o and /3 are co-efficients deduced from experiment. To determine these co-efficients, 51 experiments, which, at various times, were made by Couplet, Bossut, and Du Buat, on the naotion of water through long tubes, were made use of by these hy=- draulicians. ^ l Prony found from this, that hx = (0,0000693 v + 0,0013932 v2) Eytelwein hx = (0,0000894 v + 0,0011213 v2) d’Aubuisson as- sumes hx s= (0,0000753 v + 0,001370 v2) ^ metres. A formula, discovered by the author, agrees more accurately with observation. It has the form *■-(•+ and is based on the hypothesis, that the resistance of friction increases simultaneously as the square, and as the square rootof the cube of the velocity. From this we have the co-efficient of resistance £’,«« + ——, and the height due to the resistance of friction h. = C, .-- >/r 5 1 1 d2g For the measurement of the co-efficient of resistance or of the auxiliary constants o and 0, not only the determinations of Prony and Eytelwein from the 51 experiments of Couplet, Bossut, and Du Buat were used by the author, but also 11 experiments made by him, and 1 experiment by Gueymard in Grenoble. The older experiments extend only to velocities of from 0,043 to 1,930 metres; in the expe- riments of the author, however, the extreme limit of velocity reached to 4,648 metres. The widths of the tubes, in the older experiments, were 27 mm. = 1.06 in.; 36 mm. =* 1.95 in.; 54 mm. = 2.12 in.; 135 mm. = 5.31 in.; and 490 mm. = 19.29 in.; later experiments were conducted with tubes of 33 mm. =» 1.29 in.; 71 mm. = 2.79 in. ; and 275 mm. = 5.31 in. By means of the method of least squares, it has been found from the 63 experiments laid down; f, = 0,01439 + ; therefore, A, — (0,01439 4 0,0094711 \ l -----1 - . — metre: f r» • v' w / 2g lor Prussian measure: A, = (0,01439 + \ 1 . Ei feet, , /® or lor English measure : A, = (0,01439 4. 0,Q17^63\ i . feet. s/ v * d 2g § 331. For facilitating the calcuiation, the following table of the co-efficients of resistance has been compiled. We see from this, that394 LONG TUBES. the variability of these coefficients is not inconsiderable, as for 0,1 metre velocity it is =* 0,0443, for 1 metre =» 0,0239, and for 5 metres « 0,0186. TABLE OF THE CO-EFFICIENTS OF FRICTION.* lOths of a metre. vft 0 1 2 3 4 5 6 7 8 9 v or4 in. or 8 in. 12 in. 16 in. 20 in. 24 in. 28 in. 32 in. 36 in. it in. 0 5° GO 0,0443 0,0356 0,0317 0,0294 0,0278 0,0266 0,0257 0,0250 0,0244 3.4 Si 0239 0234 0230 0227 0224 0221 0219 0217 0215 0213 6.7 62 0211 0209 0208 0206 0205 0204 0203 0202 0201 0200 9.10 |3 0199 0198 0197 0196 0195 0195 1 0194 0193 0193 0192 13.0 I4 0191 0191 0190 0190 0189 0189 0188 0188 0187 0187 We find in this table the co-efficients of resistance due to a certain velocity, when we look for the whole metre in the vertical, and the tenths in the first horizontal column, then proceed from the first number horizontally, and from the last vertically to the place where both motions meet; for example, for v =» 1,3 metre -» 0,0227, for v - 2,8, Zx = 0,0201. For the Prussian measure we may put: v 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 o,9 a 0,0679 0,0522 0,0453 0,0411 0,0383 0,0362 0,0346 0,0333 0,0322 r 1 1* i 'h 2 3 4 6 8 12 20 ft. t. 0,0313 0,0296 0,0282 0,0263 0,0242 0,0229 0,0213 0,0204 0,0192 0,0182 § 332. Long Tubes.—With respect to the motion of water in long tubes or conducting pipes, the three following fundamental problems present themselves for solution. 1. The length / and the width d of the tube and the quantity of water Q to be conducted are given, and the head of water is required. We have first to calculate the velocity v = ~ k 1,2732 . then to look for the co-efficient of friction corresponding to this value, in one of the last tables, and, lastly, to substitute the values of * To apply this table to English measures, the velocity, if below 40 inches, will be talcen out from the numbers under w lOths of a metre,” and if above 36, the number of feet and inches in the left hand column will be used with the inches at the head. Thus, for r = 7 feet 11 inches we have on a line with 6 feet 7 inches at the left, and under 16 inches at top, the co-efficient, 0205.—-A*. Ed.LONG TUBES. 395 dy 1, Vy f and where ? represents the co-efficient of friction for the portion of the interior orifice, in the last formula 2g 2. The length and width of the tube, as well as the head of water or fall, are given to determine the discharge. We must here find the velocity by the formula s/2gh j1 + c + (t.r but since the co-efficient of resistance is not quite constant, but varies somewhat with v, we must know v approximately beforehand, in order to be able to find out From v it tllen follows that Q = v =» 0,7854 cP v. 4 3. The discharge, the head of water, and the length of the tube are given, to determine the requisite width of the tube. As v = therefore v1 = \ . -L \ n ) d* we then have 2 gh = ^1 4- f + ^ ‘ jp or Ssi' (tq)[ “ (1 + {>* + F °' 2 gh . ds = (1 + ?) d + l, hence the width of the tube d_ ,GT+g?+M But now ^ = 1,6212, 1 + f as a mean = 0,0155, hence we may put: _________ d = 0,4787 (1,505 .d + ^l)^- feet. This formula can only be used as a formula of approximation, be- cause the unknown quantity d, and also the co-efficient dependent t 5f(l + !)d + Stl m /4 \ 2gh \ h ) 1,505 and — on v: 4 Q « d? y appear in it. . Examplei.—1. What head of water does a oonducting pipe, of 150 feet length and o inches width, require, if it is to carry off 25 cubic feet of water per minute? Here * =* 1,2732 . : 12 as 3,056 feet, hence we may put = 0,0242, and the head of 60.59 Water or the entire fall of the pipe will be: h = (1,505 + 0,0242 . IffL.l?) . 0,0155.3,056* = (1,505 + 8,712) . 0,0155.9,339 am 1,479 feet, (EngU»h) 2. What discharge will a oonducting pipe, 48 feet kmg and 2 inches wide, with a D feet head of water, deliver ? It wiU be:396 BENT TUBES. v _______8,02 v/ 5 J 1,505 -f £, 48/ }1 _______17,88 y/ 1,505 + 288 . £~ If we previously take = 0,020, we shall obtain v = 17,88 2,7 6,6, but v 6,6 gives more correctly = 0,0211, hence we shall have more correctly: 17 88 17 88 v = —......- — = — — = 6,50 feet, and the quantity of water y/ 1,51)5 + 288.0,0211 y/ 7,582 Q = 0,7854 . I—\ . 6,50 = 0,137 cubic feet = 236,7 cubic inches. 3. What width must be given to a conducting pipe, 100 feet in length, which at a head of water of 5 feet, delivers half a cubic foot of water per second? Here d = 0,4787 ^/(1,505 d -f 100 £,) . j ($)* = 0,4787 ^/0,075 d + 5 If we set out with £, = 0,02, we obtain 130 = 0,1337 cubic feet = 231 cubic inches. 2. If the curved buckets of a turbine form channels 12 inches Fig. 450. l°ng> 2 inches broad, and 2 inches deep, as ABC, Fig. 450, and if the water flows through them with a velocity of 50 feet, and the mean radius of curvature R of this axis of the channels amounts to 8 inches, then is _!L = the co-efficient of the resistance of cur- R vature = 0,127; further, — = Ll = — = 0,4774 ; and lastly, tr 8 7T the height due to the resistance corresponding to the curvature of the scoopJETS D’EAU. 399 = 0,127.0,4774 — = 0,0606 = 0,0606.0,0155.50’=2,34S feet. 2 g2 g lherefore, by the resistance of curvature, 2,348 feet in fall are lost. Remark. The earlier formula given by Du Buat, Gerstner, and Navier for the resist- ance of curvature, are quite useless. An extended account of the experiments of the author on this subject will be published in the third number of his “ Investigations in Mechanics and Hydraulics.” § 335. Jets cPEau.—A conducting tube either discharges into the air or under water. The discharge under water is applied when the tube at its outer orifice is so wide that the entrance of air raay be feared. Here of course the head of wa- ter RCy Fig. 451, must be taken from the surface H of the upper water to that of C of the lower water. If the tube, for exam- ple, KLMy Fig. 452, discharges into the open air, it will give a stream of water OR, which, when allowed to ascend, is call- ed a jet d?eau. We shall here consider what is most required for these jets. That a jet may ascend to the utmost possible height, it is necessary that the w^ater should flow from the adju- tage with great velocity; hence such adjutages must be applied which offer the fewest obstacles to the water in its passage, to which, therefore, the greatest co-efficients of velocity are due. Orifices in a thin piate, short tubes fashioned like the contracted fluid vein, and long and conically convergent ones, are those which give the greatest velocities of efflux. Orifices in a thin piate are little suitable, because a jet formed by them presents nodes and bulgings, and, therefore, is sooner scattered by the external air than the prismatic jet. The same takes place in a certain degree with short mouth-pieces, shaped like the contracted vein. Hence, for fountains and fire-engines, mostly long and slightly conically convergent, similar to those which d’Au- buisson used for his experiments, are very properly made use of. Sometimes entirely cylindrical jets are used. Where these mouth-pieces, as, for ex- ample, KL, Fig. 453, are screwed on to the con- ducting tube AB, they should gradually widen, that no contraction may occur in passing into them. these mouth-pieces or discharging tubes are ^ry long, like those of fire-engines, the friction o the water in them will then cause a considera- ble loss of pressure, because the wrater has here Fig. 453.400 JETS D’EAU. a great velocity. For great velocities we may properly put the co-efficient of resistance f — 0,016, and, therefore, the loss of head of water «= 0,016 _ . H_. If now the length of a hose is twenty times as great as the mean width, we shall then obtain the height of the resistance due to friction -0,016.20.^-0,32.^; thus, from this cause, above 32 per cent. of the height of ascent is lost. These tubes are generally much Jonger, hence this loss is greater. The velocity with which water passes out of a mouth-piece or hose, and on which the jet or the height of ascent principallydepends, may be estimated by means of the above principies. If we put this velo- city of efflux — v, the width of the mouth-piece at the exit orifice — rf, anu the mean width of the conducting tube — dv we shall then obtain If f represent the co-effi- jj the vek>city of the water in it vt — — v. 1 d 2 cient of resistance at the inner orifice of the tube, that of the resist- ance of friction in the pipe, and ?2 the co-efficient for the knees or cur- vature of the pipe, the height due to the resistance for the motion of water in the conduit pipes will be: *-(t + s,‘j + U g- (C + t,j + U% ■ §• It is seen from this, that the resistance to the water is less, the wlder the conduit pipe is. It is hence an important rule, to employ as wide pipes and hoses as possible, for leading water to jets Fig. 452, of the outer orifice O below the surface of water H in the reservoir — A, the formula *- [i +f, + (r+c, J + «y nolas true, and hence the velocity of efflux is: \/2g A ^ + (f + ^ ~ + ^ ir the jet weie to rise perpendicularly and in vacuo, the height of ascent would be:ABRUPT WIDENING. 401 ' 2g l+(f + ^+f,)£+?3 but because the air and the descending water offer impediments to the ascent, and to the direction of the jet, as is the case in fire- engines, the effective height of ascent is somewhat less. Accord- ing to d’Aubuisson’s conclusions from the experiments undertaken upon this subject by Mariotte and Bossut, the effective height of ascent is sx = s— 0,01 . s2 = s (1 — 0,01 . s) metres, or for English measure = s (1 — 0,00305 s) feet. We see from this that in great ascents proportionately more height is lost than in small velocities. Thick jets ascend somewhat higher than thin ones. In order to diminish the resistance of the descending water, the jet raust be directed a little inclined. As to the height and amplitude of oblique jets, see § 38. Example. If the conduit pipe for a fountain be 250 feet long, and 2 inches diameter, if the co-efficient of resistance corresponding to the mouth-piece = 0,32, if the entrance orifice at the reservoir be sufficiently rounded, and the bends that occur have sufficient radii of curvature to allow of our neglecting the corresponding co-efficients of resistance, to what height will a jet ^ inch thick, under a head of water of 30 feet, rise? If we take the co efficient £x of friction = 0,025, we shall then obtain the entire height due to the resistance: A = o + °-02S • • (|)<+ °’32) g =1>47 • fg' ” 2 l OA hence, the height due to the velocity * =--=-------= 20,41 feet, and the effective J 1,47 1,47 beight of ascent = 20,41 (1—0.00305.20,41) = 20,41 — 1,27 = 19,14 feet. CHAPTER IV. ON THE RESISTANCES OF WATER IN PASS1NG THROUGH CONTRACTION. § 337. Abrupt Widening.—Changes in the transverse section of a tube, or of any other reservoir of efflux, produce changes in the velo- city of the water. The velocity is inversely proportional to the trans- verse section of the stream. The wider the vessei is, the less is the velocity, and the narrower the vessei, the greater the velocity of the water flowing through it. If the transverse section of a vessei be suddenly altered, as, for example, in the tube ACE, Fig. 454, there then ensues a sudden alteration of the velocity, and this is accompanied by a loss of vis vivay or connected with a corresponding diminution of pressure. This loss may be as accurately mea- sured as the mechanical efleet in the impact of inelastic bodies (§ 258). Every particle of water which passes 34* Fig. 454.402 ABRUPT WIDENING. f from the narrower tube BD into the wider tube DG, strikes against the slowly moving mass of water in this tube, and, after impulse, joins itself to and proceeds onwards with it. It is exactly the same with the collision of solid and inelastic bodies; these bodies go on likewise after impact with a common velocity. Since we have already found that the loss of mechanical effect by the impact of these bodies is T _ (Vl — V*Y Gl G2 L- 2g so we may here, as the impinging particle of water Gx is inde- finitely small compared with the impinged mass of water G%, put: L ~Gi9 and, consequently, the corresponding loss of head* There aris es, therefore, from a sudden change of velocity a loss of head, which is measured by the height due to the velocity corresponding to this change. If now the transverse section of the one tube AC, = Fv and that of the other CE, = F, the velocity of the wrater in the first tube Fv and that in the other = v, we then have vx hence the = *V loss in head of water in the passage from one tube to the other is A, II II — 1 V . —, / % Fig. 455. The experimentsundertaken by the author on this subject accord well with theory. That the tube DG may be filled with water, it is requisite that it be not very short, nor much wider than the tube AC. This loss vanishes, when, as represented in Fig. 455, a gradualpassingfrom one tube into the other is accomplished by the rounding of the edges. Remark. The head of water found ht s (— — 1 ^ cannot, of course, be utterly , \Fl J 2g lost, we must rather assume that the mechanical effect produced by it is expended on the separation of the previous continuity of the particles of water. Example. If the diameter of a tube, of the construction in Fig. 454, is as great agam as that of another tube, then is JL. = (-p) = 4> t1©006 the co-eflicient of resistance { = (4 1)* =*s 9, and the corresponding height due to the resistance on passing from the narrow into the wide tube = 9 . If the velocity of the water in the latter tube s= 10 feet, the height due to the resistance is then = 9.0,0155.10» = 13,95 feet § 338. Abrupt Contraction.—A sudden change of velocity alsoABRUPT CONTRACTION. 403 occurs when water passes from a cistem Fis*456- Fig. 456, into a narrow tube DG, especially when there is a diaphragm at the place of entrance CD whose orifice is less than the transverse section of the tube DG. If the area of the contraction is = Fv and a the co-efficient of con- traction, we have then the transverse sec- tion F2of the contracted fluid vein =aFx; and if, on the other hand, F is the transverse section of the tube and v the velocity of efflux, we then find that the velocity at the contracted F section E is, v2 = ------v, and hence the loss of head in passing from a ^ and F2 into Fy or from t>f into v: A = -l) " 2g \»F1 / 2g ✓ f \2 the corresponding co-efficient due to the resistance: f = (—p —1) . Without the diaphragm, we have a mere short tube, Fig. 457; hence, Fig. 457. If we assume <* = 0,64, we then ob- tain: ? = (L=^i)S =(*)*= 0,316. But the co-efficient due to the resist- ance for the transit through an orifice in a thin piate is about 0,07; hence, here, where the water flows out - times as fast as from the contracted a transverse section, the corresponding height due to the resistance ”0,07'(r) v3 0,07 i_ = 0,07 . -L . _ 2g O.2 2g 0,41 . — = 0,171 . —. _ - . _ . ?g By combining these two resistances, we obtain the entire height due to the resistance for efflux through a short tube: = 0,316 £ + 0,171 £ = 0,49 . H 2 2s 2g whilst we before found it = 0,50 —. 2g Experiments on the efflux of water through an additional tube, with a narrow inner orifice, as in Fig. 456, have led the author to the following results. The co-efficient of resistance for transit through a diaphragm, and for a contraction at the wider tube, may be expressed F by the formula f -CV1)' , but there must be put:404 EFFECT OF IMPERFECT CONTRACTION. I F for —! F 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0 a 0,616 0,614 0,612 0,610 0,607 0,605 0,603 0,601 0,598 0,596 Lnd it folio W: S that: f 231,7 50,99 19,78 9,612 J 5,256 J 3,077 J 1,876 1,169 0,734 0,480 From this, for example, the co-efficient of resistance in the case where the narrow transverse section is half as great as that of the tube, is C = 5,256, i. e., for transit through this contraction, a head of water is required which is 5| times as great as the height due to the velocity. Example. What discharge will the apparatus delineated in Fig. 456 give, if the head of water is l£ feet, the width of the circular contraction lj, and that of tube 2 inches? We ha ve here^J —: : Ci)9 = = 0,56, hence a = 0,606, and (= ( 16 jV _ ( 16 — 5,454 y /10.546V V 9.0,606 / \ 5,454 / \ 5,454/ 4- t) ____» we shall l^en obtain the velocity of efflux: 2 g v = ^ ^ = 4}56 feet, and consequently the quantity discharged: v/l+C for a, we must G substitute variable numbers which are greater the greater the ratio — between the transverse section of the contraction and that of the conduct- ing tube AB. If the diaphragm CD lies in a uniform tube AG, Fig. 459, then the same condition holds, only here the co-efficient a depends Fx on_L. Fig. 459. Fig. 458.SLIDES, COCKS, VALVES. 405 From experiments undertakenby the author,we must put into thefor- (F \2 -------1) for the co-efficient of resistance, aFx / for T? = F 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0 a 0,624 0,632 0,643 1 0,659 ^nd it 0,681 follow 0,712 s: 0,755 0,813 0,892 1,000 t 225,9 47,77 17,50 7,801 3,753 1,796 0,797 0,290 0,060 0,000 These losses are diminished when, by rounding off the edges, the con- tractiori is diminished or counter- acted, and they may be entirely neg- lected, if, as is represented in Fig. 460, a gradually widening tube MN is put on. Fig. 460. Example. What head of water is requisite tbat the apparatus in Fig. 461 may deliver 8 cubic feet of water per minute “? If the width of the diaphragm Fx = 1^, the width of the efflux tube DG = 2 inches, and the lower width of the afflux tube^C= 3 inches, we shall then have £l = (l£)' =i,hence» = 0,637; further, £. = (—V = ( })’ G \ 3 / Ft \1^/ :== and the co-efficient of resistance: / 16 iV /10,267\9 = 3 207. The velocity of efflux is now V 9.0,637 / V 5,733 / v=_ * Q_________^ ^ — — *^2 ss 6,112 feet: hence, the head of water in question is w d* 60.» (j)* ir h = (1 +{) 4,207.0,0155.6,112’= 2,43 feet. 2 6 § 340. Slides, Cocks, Valves,—For regu- lating the flow of water from pipes and cis- terns, slides, cocks, valves, &c., are used, by 'which contractions are produced which offer obstacles to the passage of the water, and these may be determined in a manner similar to the loss estimated in the last paragraph. But since the water here undergoes further changes of direction, divisions, &c., the co- efficients a and f cannot be determined direct- ly, but special experiments are necessary for this purpose. Such experiments have been also made,* and their principal results are communicated in the following tables. Fig. 461. , . , i-undertaken and cal* * Experiments on the efflux of water through valves, sii , 2^.^ der Mechamk culated by Julius Weisbach, under the title “ Untersuchunge und Hydraulik,” &c. Leipzig, 1842.406 SLIDES, COCKS, VALVES, TABLE I. THE CO-EFFICIENTS OF RESISTANCE TO THE PASSAGE OF WATER THROUGH SLIDING VALVES IN RECTANGULAR TUBES. Ratios of transverse n F F section 1 Coefficientof | resistance i 1,0 0,9 0,8 0,7 0,6 0,5 0,4 0,3 0,2 O o o 0,09 0,39 0,95 2,08 4,02 8,12 17,8 44,5 0,1 193 TABLE II. THE CO-EFFICIENTS OF RESISTANCE TO THE PASSAGE OF WATER THROUGH SLIDES IN CYLINDRICAL TUBES. Height of place # - 0 1 i 1 1 5 S 1 i Ratio of transverse section. 1,000 0,948 0,856 0,740 0,609 0,466 0,315 0,159 Co-efficient of re- sistance f. 0,00 0,07 0,26 0,81 2,06 5,52 17,0 97,8 THE CO-EFFICIENTS OF RESISTANCE FOR THE PASSAGE OF WATER THROUGH A COCK IN A RECTANGULAR TUBE. Angle of position. 5° 10° 15° 20° 28° 30° 35® 40° 45° 50° 55° 66} Ratio of transverse section. 0,926 0,849 0,769 0,687 0,604 0,S20 0,436 0,352 0,269 0,188 0,110 0 Co-efficient of resistance. 0,05 0,31 0,88 1,84 3,45 6,15 11,2 20,7 41,0 95,3 275 00 TABLE IV. THE CO-EFFICIENTS OF RESISTANCE FOR THE PASSAGE OF WATER THROUGH A COCK IN A CYLINDRICAL TUBE. Angle of position. 5° 10° 15° 20° 25° 30° 35° Ratio of transvexee section. 0,926 0,850 0,772 0,692 0,613 0,535 0,458 Co-efficient of resistance. 0,05 0,29 0,75 1,56 3,10 5,47 9,68SLIDES, COCKS, VALVES. 407 Angle of position. 40° 45° 50° 55° 60° 65° 82i° Batio of transverse section. 0,385 0,315 0,250 0,190 0,137 0,091 0 Co-efficient of resistance. 17,3 31,2 52,6 106 206 486 00 TABLE V. THE CO-EFFICIENTS OF RESISTANCE FOR THE PASSAGE OF WATER THROUGH THROTTLE VALVES IN RECTANGULAR TUBES. Angle of position. 5° 10° 15° 20° 25° 30° 35° Batio of transverse section. 0,913 0,826 0,741 0,658 0,577 0,500 0,426 Co-efficient of resistance. 0,28 0,45 0,77 1,34 2,16 3,54 5,72 Angle of position. 40° 45° 50° 55° 60° 65° 70° 90° Batio of transverse section. 0,357 0,293 0,234 0,181 0,134 0,094 0,060 0 Co-efficien< of resist- ance. 9,27 15,07 24,9 42,7 77,4 158 368 00 TABLE VI. THE CO-EFFICIENTS OF RESISTANCE FOR THE PASSAGE OF WATER THROUGH THROTTLE VALVES IN CYLINDRICAL TUBES. Angle of position. 5° 10° 15° 20° 25° 30° 35° Batio of transverse section. 0,913 0,826 0,741 0,658 0,577 0,500 0,426 Co-efficient of resistance. 0,24 0,52 0,90 1,54 2,51 3,91 6,22408 SLIDES, COCKS, VALVES. Angle of position. 40° 45° 50° 55° 60° 65° 70° 90° Ratio of transverse section. 0,357 0,293 0,234 0,181 0,134 0,094 0,060 0 Co-efficient of resist- ance. 10,8 18,7 32,6 58,8 118 256 751 00 § 341. By means of the co-efficients derived from the above tables, we may not only assign the loss of pressure corresponding to a cer- tain slide, cock, or position of a valve, but also deduce what position is to be given to this apparatus that the velocity of efflux or the resistance may be of a certain amount. Such a determination is, of course, the more to be relied on, the more the regulating arrange- ments are like to those used in the experiments. The numerical values given in the tables are only true for the case where the water, after its transit through the contractions produced by means of this apparatus, again filis this tube. That this full flow may take place for small contractions, the tube should have a considerable length. The transverse sections of the rectangular tubes were 5 centimetres (1,97 inch) broad and 2£ (,98 inch) deep. The transverse sections of the cylindrical tubes had, however, a width of4 centimetres (1,57 inch). By the slide, Fig. 462, there is a simple contractiori, whose transverse section forms in the one tube a mere rectangular Fl9 Fig. 463; in the second, however, a lune, Fv Fig. 464. In the case of cocks, two Fig. 462. Fig. 463. Fig. 464. contractions present themselves, and also two changes of direction; on this account the resistances are also very considerable. The transverse sections of the greatest contractions have very peculiar hgures. The stream in throttle valves, Fig. 466, divides itself into Fig. 465. Fig. 466.VALVES. 409 two portions, each of which passes through a contraction. The transverse sections of these contractions are, in the throttle valve of rectangular tubes, rectangular, and in cylindrical ones lunar-shaped. The following examples will suffice to show the application of the above tables. Examples.—1. If a sliding valve is applied to a cylindrical conducting pipe 500 feet long and 3 inches wide, and this be drawn up to $ of its entire height, and therefore close $ of the pipe, what discharge will it deliver under a pressure of 4 feet? Theoo- efficient of resistance £ for entrance into the pipe may be put from the above at 0,505, and that of the resistance of the slide from Table ii. = 5,52, hence the velocity of the 8,02 . y/T 8,02.2 ______16,04 flow V J j 1,505 + 5,52 + :,1 x/7,025+500.4 t2 ^/7,025 + 2000 f, d If ^ 16,04 = 2,12 put the co-efficient of friction £a = 0,025, we shall then obtain r = ^/57,025 feet But the velocity v = 2,12 feet, gives more correctly £a = 0.026; hence, more accurately, v = 16,04 — 2,08 feet, and the discharge per second = — . 9 . 12 . ^/59,025 4 2,08 = 56,16 . % = 177 cubic inches.—2. A conducting pipe, 4 inches wide, delivers, under a head of water of 5 feet, 10 cubic feet of water per minute; what position must be given to the throttle*valve applied, that it may afterwards deliver only 8 cubic feet? The velocity at the beginning is = ss 1 * 1,91 feet, and after putting on 60.* (*)* the valve » . 1,91 = 1,528 feet The oo-efficient of efflux is . 1,91 1 y/2 g h 8,02 5 0,106, hence the co-efficient of resistance =-------------1 1 =88; the co- 0,106* efficient of efflux for the second case is = T*d . 0,106 = 0,0848; hence the co-efficient of resistance =_______i_______1 = 138,0, and consequently to produce the co-efficient of 0,0848» resistance of the throttle-valve: ( = 138 — 88 = 50. But now, from Table vi., the angle of position a = 50°, ( = 32,6, and the angle of position a = 55°, £ = 58,8; hence . 5° = 53°, the desired 26,2 * we may be allowed to assume, for a position of 50° -(- quantity of discharge may be obtained. If we consider, further, for a change of velocity of 1,91 feet to 1,528 feet, the co-efficient of resistance passes from 0,0266 into 0,0281; 281 then, more correctly, {= 138,0 — 88 --------= 138,06—92,96 =45,04,and,accordingly, /, 266 the angle of position = 50° 4- • 6° = 52°. 26,2 § 342. Valves.—The knowledge of the resistance produced by valves is of great importance. Experiments have been made by the author on this subject. The conical and clack, or Jlap-valves, Figs. Fig. 467. Fig. 468. 467 and 468, are those which most frequently are met with in prae- tice. In both, the water passes through the aperture formed by a410 VALVES. ring RG; the conical valve KL has a guide rod, by which it is fixed in guides, and admits of an outward push only in the direction of the axis; the clack valve KL opens by tuming like a door. It is easily seen in both apparatuses that a resistance is opposed to the water, not only by the valve ring, but also by the valve piate. For the conical valve with which the experiments were undertaken, the ratio of the aperture in the valve ring, to the transverse section ot the whole tube was 0,356, and, on the other hand, the ratio of the surface of the ring for the opened valve to the transverse section ot the tube 0,406; hence, for the mean, we may put —1 — 0,381. Whilst the efflux in different positions of the valve was observed, it was found that the co-efficient of resistance dimimshed when the valve slide was greater, and that this diminution was almost lnsignm- cant when it exceeded half the width of the aperture. Its amount was in this case ** 11, therefore, the height due to the resistance or the loss of head of water =* 11 . v being the velocity of the 2 g water in the full tube. This nuraber may be also used for deter- mining the co-efficients of resistance correspondinjj to the other ratios of the transverse sections. Let generally f — ("Tj?---* ) 9 we ^en obtain for the observed case «■ 0,381, f ■■ 11, and 11 «* (o3§n~-1)*hence 0,608, * " 0,381 (l + vof) 4,317.0,381 and, lastly, in ganeral: If, for example, the transverse section of the aperture is one half that of the tube, the co-efficient of resistance will accordingly ■» (1,645 . 2 — l)s — 2,29* - 5,24. For the clack, or trap valre, the ratio of the transverse section of MfJ the aperture to the tube was » 0,535; but the following table F shows in what degree the co-efficient of resistance diminishes with the size of the aperture. TABLE OF THE CO-EFFICIENTS OF RESISTANCE FOR TRAP VALVES. Angle of aperture. 15° 80° 88° 30° 35° 40» i 46» 60® 55« 60® 65« 70® Co-efficient of resist- ance. 90 62 42 30 20 14 9,5 6,6 4,6 3,3 2,3 1,7COMPOUND VESSELS. 411 The co-efficients of resistance of these valves may be calculated approximately with the help of this table, even when the ratio of the transverse section is any other. The same method must be adopted as that followed for conical valves. Example. A forcing-pump delivers by each descent of the piston, 5 cubic feet of water in 4 seconds, the width of the tube of ascent, in which lies the valve opening upwards, is 6 inches, the aperture of the valve-ring 3£ inches, and the greatest dia- meter of the valve inches; what resistance has the water in its passage through the valve to overcome? The ratio of the transverse section for the aperture is (~~6^ ) = (ti)* = 0,34, and that of the annular contraction to the transverse section of the tube =1 — ^ ^ = 1 — (£)2 = 0,44, hence the mean ratio of section ___ 0,34 -f- 0,44 __ q gg an(j coj-regponding co-efficient of resistance 2 ___ / 1,645^ — 1^ == 3,223 = 10,4. The velocity of the water is : V 0,39 / 5 20 v =__________-— = = 6,37 feet, the height due to the velocity = 0,629 feet; and, 4.Z..(i)2 * consequently, the resistance due to the height = 10,4.0,629 = 6,54 feet. The quan- tity forced up in one second weighs | . 62,5 = 77,6 lbs.; hence the mechanical effect 'vvhich by the transit of the water through the valve is consumed in this time = 77,6 • 6.54 = 507,5 a lbs. § 343. Compound Vessels.—The principies already laid down on the resistance of water in its passage through contractions, find their application in the efflux of water through compound vessels. The apparatus AD represented in Fig. 469, is divided by two partition walls containing the orifices F' and jF2, and on this account forms three vessels of communication. Were there no partition walls, and the edges, where one vessel passes into the other, rounded off, we should then have as for a single vessel the velocity of flow through F : v = h repre- x/1 +f senting the depth FH below the surface of water, and f the co-efficient of resistance for the passage through the orifice F. But since obstacles are to be overcome on passing through JPX and F2, we then have to put v = I------------— r . \l+f + f! + f2 # s2, the co-efficients of resistance corresponding to the contractions Fx ^nd F2. If we represent the transverse sections of the vessels CD, DC and ABy by G, Gx and G2, we may further put (§ 338): hence also: Fig. 469. and to substitute for fx and v = y/2 gh 412 COMPOUND VESSELS. Exactly the saine relations take place in the compound apparatus of discharge represented in Fig. Fi&* 470- 470, except only that the friction of the water in the tube of com- from AC to GL: munication CE has perhaps to be taken into account. If l is the length, and d the width of this tube, but f, the co-efficient of fric- tion, and vx the velocity of the water in the tube of communica- tion, we then have the height which the water loses in passing or, since the velocity is to be put: If this height be deducted from the whole head of water A, there will remain the head of water in the second vessel h2 = h the velocity of efflux: hv and hence Fig. 471. y/2gh Fx is hx This determination is very simple with the apparatus represented in Fig. 471, because the transverse sections G, G„ Ga, of the cisterns may be made indefinitely great with respect to the transverse sections of the orifices F, Fiy F%. Hence the first difference of level OHy or height due to the resistance in passing through: %F>2 1 /VA2 _ / aFy v2^ tgw ypJ;v and likewise the second diflerence of level Ox Hiy or the height due to the resistance in passing through wiiere a, a„ a2, represent the co-efficients of contraction for the orifices F, F1 and F2. It accordingly follows that: \/2gh _____ v = J1 dh . "+&) +■&) and the quantity of discharge:PIEZOMETERS. 413 Q = aF s/ 2 gh ___________\/ 2gh__________ ■J (zf) + (^fJ + (^f)‘ It is easy to perceive that compound reservoirs of efflux deliver less water, under otherwise similar circumstances, than simple ones. Example. If in theapparatus, Fig. 470, the whole head of water or depth of the centre of the orifice F below the surface of water of the first cistern is 6 feet, the orifice S inches broad and 4 inches deep, the tube connecting both reservoirs 10 feet long, 12 inches broad, and 6 inches deep, what discharge will this reservoir give ? The raean width of the tube s ^ ^ = $ ft., hence L = ?.*— = 15 ; let us now put the co- 2 . 1,5 d 2 efficient of friction £, = 0,025, and it follows that f, . L — 0,025.15 = 0,375; if the d co-efficient of resistance for entrance into prismatic tubes be here put 0,505, we obtain l + (i- iy+t, 1=1 + 0,505 + 0,375 = 1,88. As ^ = ^l-4 =0,2845, the co-efficient of resistance for the entire connecting tube = 1,88.0,28459 = 0,152, and the co-efficient of resistance for the transit through = 0,07, we then obtain the velocity of efflux: t> =_____^6 - = ^ = 17,77 feet. The contracted section is ^07+ 0,152 >/ 1,222 0,64.1 . £ = 0,32 square feet, hence the discharge = 0,32 . 17,77 = 5,68 cubic feet. § 344. Piezometers.—The loss of pressure which water suffers in conduit pipes from contractions, friction, &c., may be measured by columns of water, which are sustained in vertically placed tubes, which, when used for this purpose, are called piezometers. If v is the velocity of the water at a place B, Fig. 472, where a piezometer is applied, l the length, d the width of the portion of tube ABy h the head of water or the depth of the point B below the surface of water; if, further, f is the co-efficient of resist- ance for entrance from the reservoir into the tube, and the co-efficient of fric- tion, we then have for the height of the piezometer measuring the pressure at By If the length of a portion of the tube BC = /x> and its fall — hv we then have the height of the piezometer at C: 35* Fig. 472.414 PIEZOMETERS. h + h. -( »+f+f.j+&i)£ and hence the difference of these two heights: hence, inversely, the height of the portion of tube J3C, due to the resistance: y l. v2 Ax + z — Zj the fall of the portion of tube plus the difference of the heights of the piezometers. From this it is seen that piezometers are applicable to the mea- surement of the resistance which the water has to overcome in con- duit pipes. If a particular impediment is found in the tubes; if, for instance, some srnall body is found fixed there, this will immediately be shown by the falling of the piezometer, and the amount of the resistance produced, expressed. The resistances which are caused by regulating apparatus, such as cocks, slides, &c., may be likewise expressed by the height of the piezometer. The piezometer, for ex- ample, stands lower at D than at C, not only in consequence of the friction of the water in the portion of water CD, but also in conse- quence of the contraction which the slide S produces in this tube. If for a perfectly opened slide the difference NO of the height of the piezometer =a A,, and for the slide partlv closed — Aa, the new differ- ence or depression A, i* hx gives the height due to the resistance which corresponds to the passage of the water through the slide. Lastly, the velocity of efflux may be also estimated by the height of the pie- zometer. If the height of the piezometer PQ — z, the length of the last portion of tube 1)E «■ /, and its width « d> we then have: rl •V-*—■ Wramplt. If tbe height of the pi&ometer PQ ■*= z, Fig. 471, j foot, and the length of the tabe DE, meaaured from the pi&ometer to the discharging orifice, = 150 feet, the width of the tube 3$ inchea, the yelociqr of efflux then follows: »»■8,03 . /_____3'.? _ . °'21 B 8,03.0,2415 = 1,937 feet, and the discharge V 150.13 0,035PRISMATIC VESSELS. 415 CHAPTER V. ON THE EFFLUX OF WATER UNDER VARIABLE PRESSURE. § 345. Prismatic Vessels.—If a cistern from which water flows through an orifice at the side or bottom, has no influx to it from any other side, a gradual sinking of the surface of water will take place, and the cistern at last empty itself. If, further, the quantity of influx Q, be greater or less than the quantity of efflux ^ F 2 g h, the surface of water will then rise or fall until the head of water h = — /-9 \ , and after this the head of water and the velocity of 2g\fiF/ efflux will remain unaltered. Our problem, then, is to find how the time, the rise and fall of the water, and the emptying of vessels of given form and dimensions, depend on each other. The efflux from a prismatic vessel presents the most simple case when it takes place through an opening in the bottom, and when there is no efflux from above or below. If x is the variable head of water FGX, F the area of the orifice, and G the transverse section of the vessel ACy Fig. 473, we have then the theo- retical velocity of efflux v = \/ 2 g x, the theo- retical velocity of the falling surface of the water p p ________ = — v = — v' 2 gxy and the effective velocity G G Fig. 473. = r.— 2 gx. At the commencement: G ^ = FG = h, and at the end of the efflux x = 0, therefore, the initial velocity is: n _______ c = 2 gh, and the final velocity cx = 0. G It is seen from the formula vl = 2 g x, that the motion of the surface is uniformly retarded, and the measure of the retarda- tion p = g, hence we also know (§ 14), that this velocity =0, and the discharge ceases, when f _ 2Gs/h y.Fx/2g We may also put: t = 2 Gh 1 &gh : T~’ n F\ 2Gh Q ' a A • t. /* Fv/ 2gh ana, accoraing to this, assume that double the time is required for the416 VESSELS OF COMMuNICATlUfl. efflux of the discharge Gh through the orifice at the bottom Fy under a head of water decreasing from A to 0, than under a uniform pres- sure* • As the co-efficient of efflux p is not quite constant, but is greater for a diminution of pressure, we must, therefore, in calcu a lons o this kind, substitute a mean value of this co-efficient. Example. In what time will a rectangular cistern, of 14 square feet section, empty itself through a round orifice at the bottom, of 2 inches width, if the original head of water amount to 4 feet ? The time of efflux would be theoretically: * = -2-:.-14^4 = 2-141442 = = 320" b5 min. 20»ec. 8,02 . — Q)a 8,02 * n 8,02 • w At the end of half the time of efflux, the head of water will be = (J)’. A = 1.4 = 1 ft. Now the co-efficient of efflux, which corresponds to the head of water = 1 foot is 0,613, hence the effective time of discharge will be = _521" = 8 minutes, 41 seconds. 0,013 § 346. Vessels of Communication.—Since for an initial head of water hv the time of efflux tx = 2 G y/ \ ^ an(j for an initial head of P F \/2g water A2this time t2 = it then follows by subtraction, that 1 . . . . p F • the time within which the head of water passes from hY to K and the surface of water sinks A. — h2 is: 2 q __ _________ t = ----------- — (y/ h. — v/ h2), or for the English foot measure: 1* F. 1/ 2g t = 0,249 — (-/ K — ht). f*F . Inversely, the depression of the surface corresponding to a given time of efflux is s = hx— A2, and is given by the formula: or, s The same formulae are further applicable, when a vessel CD, Fig. 474, is filled by another AB in which the water maintains a uniform height. If the transverse section of the tube of com- munication, or of the orifice = Fy the transverse section of the vessel to be filled = G, and the original level G Gx of the two surfaces of water = A, we have then, since here the surface of water Gx in the second vessel is uniformly retard- ed, the time of filling likewise, or the time within which the second surface of water comes to the level HR of the first: Fig. 474.VESSELS OF COMMUNICATIONS 417 t = 2 G s/h fi F. \/ 2g and likewise the time in which the height of level hx passes into h2, and, therefore, the surface of water ascends to: GGX = s = hx — h2. t------— sT2). fi F. x/2g Examples. 1.—How much will the surface of water in the vessel of the last example sink in two minutes ? /i, = 4, t = 2 . 60 = 120, ~ = —OLi— and if we assume, fur- G 14.144 ther, fx = 0,605, it follows then A,= (y/A, —f* • 2 S • ^ * V 2 G / ?)-( 2—0,605.8,02. 0,605.8,02 . w. 120' 2.14.144 pression sought is «= 4 — 2,393 = 1,607 feet. 2. What time does the water in the 18 inch wide tube CD, Fig. 475, require to run over if it communi* cates with a vessel AB by a short lj inch wide tube, and the rising surface of water G stands, at the begin- ning, 6 feet below the uniform surface of water A, and 4 i feet below the head C of the tube. It is: t =------—-------- (v/X - v/X), 5 . ■ 168 ■\a J = 2,393 feet, and the de- Fig. 475. f* • F t = 6, A2 = 6 —4,5 =1,5,^-= (ii) =144 and E \1,5 / — 0.81, whence it follows that: h. t = 2.144 -r 288.1,2248 0,81.8,02 = 54,3 sec. (x/ 6 -v/M) = 0,81.8,02 If the first vessel AB, Fig. 476, from which the water runs into the other, has no influx, and its section Gx also not to be considered as indefinitely great compared with the section G of the subsequent vessel CD> we have then to naodify the condition. If the variable dis- tance Gx Ox of the first surface of water from the level HR at which both surfaces stand at the end of the efflux = x, and the distance GO of the second surface of water from this same plane = y, we have then the variable head of water == x + y> and the corresponding velocity of efflux: v = x/2g (x + y\ and the quantity of water: G,x_cs,,„_^(l+ C),, f he velocity with which the surface of water in the second vessel ascends is now: Fig. 476. V- F Io „ / 1 G418 NOTCHES IN A SIDE. consequently the retardation: and the time of efflux: _______2 G s/y__ >FJ2*(1 + §;) Let us substitute for x and yy the initial height of level hy and therefore put : (Q v 1 + q) y = h> and we then obtain : y = + « G. , and the time in which the tvvo surfaces of water come to a level: 2 G x/h t = 2 GG1 y/h ,r{i+°yrt -n®+«.>*'»» The time within which the level falis from h to hiy is, on the other hand: twm 2 GG, (s/h— y/AJ f>S2~gF(G+ Gx) Example. If the section of a cistern from which water flows is 10 aquare feet, and the section G of the recipient cistern 4 square feet; if, further, the initial level h of the two surfaces amounts to 3 feet, and the cylindrical tube of communication is 1 inch wide, then the time in which the water comes in both vessels to the sanie level is: # 2 . 10.4 . y/T _ 320.72 . y/T ~ 0,82.8.02.'.Jl~ t».»» • 8,02 .IV 4 144 : 275,9 sec. § 348. JYotches in a Side.—If water flows through the notch or cut DE of a prismatic cistern ABCy Fig. 477, to which there is no influx, the time of efflux may then be estimated in the following manner. Let us represent the transverse section of the cistern by G, the breadth EF of the notch by by and the depth DE by hy and divide the whole aperture of efflux by horizontal lines into small slices, each of the breadth b and depth At a constant n pressure the discharge per second willNOTCHES IN A SIDE. 419 be, Q = £ ^ 6 y/2gh3, if we divide this into the area of a stra- Gh n Gh $l*nb \/2gh3’ tum of water, we shall then obtain the time of efflux t = 'which we may write: 2 m n b \/ 2g Now, to obtain the time of efflux t for a quantity of water G or to determine the time in which the head of water above the line DE =* h sinks to DEX = hv let us make hx » — A, and n — s therefore A, to consist of m parts, and let us now substitute for A successively: and finally add the results obtained. In this manner we shall obtain •(?) ■ the time required: * - ct;C( j)_!+*f,+~+Gn = „ 3°* .*_^ + (m+l)“ *+...+»“**') 2 pnb 2 g n—§' -_12£±_r(r 2 j» n ^6 ' i—! + 2 ^ +3“ 2+...+ » __ 3 __ 3 __ 3 --3-1-9 3-1- 3-- 3 — ^1 2 + 2 2 + 3 or> from the “Ingenieur,” Arithmetic, § 28: t _ 3 GA 2 /w-^1 m~2+1\ »—h /a; '—i + 1 —i + 1/ + ...»» )} 2 fin * b y/2g 3 GA* (’ .2 (m -i-n-i 2 t* b -✓*=•)]. kYi / } remained uniform. Exampk. In what time will a pond, whose surface has an area.0^ deepest place, empty i^elf, if there be a conduit 15 feet below the surface, and at the deepest^ pi^ * which forms a channel 15 inches wide and 50 feet long? ie ’ 36422 SPHERICAL AND OBELISK-SHAPED VESSELS. efflux is *==£.. F^/Zgh = |§. 765000.15 V©' . 8,02 ^/15 19584000 — 200848 sec. w. 8,02 ^15 But nowthe co-efficient of resistance forentrance into the channel, inclined about 45°, is, £ = 0,505 + 0,327 (see § 323) = 0,832, and the resistance of the conduit due to friction = 0,025 L . 11 = 0,025.52 . HL = hence, the complete co-efficient of efflux d 2g f 2g 2g for the channel is: p —_____________* - — * — 0,594, and the time of efflux demanded : v/ i 0,832 -f 1 \/ 2,832 t = 200848 -i- 0,594 = 338128 = 93 hours, 55 minutes, 28 seconds. § 350. Spherical and Obelisk-shaped Vessels.—By means of the formula of the last paragraph, we may now find the times of efflux for many other vessels, such as spherical, pontoon-shaped, pyramidal, &c. For the emptying of a spherical segment JlBy Fig. 481, we obtain: t = 4 nTh% _ 2 ~\Fs/2gh (10 r— = 15 H --------=—9 pF s/ '2g therefore, for the emptying of a full sphere, where = 2 r, Fig. 481. _ 16 x r1 ■»/ and for that of half a sphere where : h _mf 14 h — rt9 =--------=l . \bpFS2g Here the horizontal stratum HXRX corresponding to the depth FGt ~ \ h 2 rt rhx 7thz* r = x = G. = 7t x (2 r—x). - ---------------------, therefore: n n n t = 2 »< r ^____ x A # x! . n ^ Fs/ 2 g np F*/ 2 g as the first part of this expression agrees with the formula for the emptying of a prismatic, and the second part for the emptying of a pyramidal vessel, if we put first 2 rtr h in place of W, and secondly n h2 in place of G, we shall obtain by means of the difference of the times of emptying of a prismatic and pyramidal vessel, found in the former paragraph : t = $ ■ ~lhL__, and t = f . -9ih=, pFy/2gh fiFx/2gh the time also of the emptyingof a spherical segment. The above formula may be likewise applied to the case of an obelisk or pontoon-shaped vessel ACDV Fig. 482, since this is com- posed of a parallelopiped, two prisms, and a pyramid. Let b be theSPHERICAL AND OBELISK-SHAPED VESSELS. 423 Fig. 482. breadth at top AD, and bx the breadth^ Dx at bottom, l the length at top AB, and lx the length at bottom AXBV and lastly, h the height of the vessel, we have then for the area of the surface AC: bl= bxl. + + h(b — bl)+ (/-/,) (4 —bx), of which 61/1 belongs to the parallel- opiped A^EG, bt (/ — /,) + lx (b—bj to the two prisms CFBXCV and AFBlAl and (Z — ZJ (b — to the pyramid BF Bx. But now the time of efflux for the parallelopiped, whose base is bxll9 is tx = 2Mi y/h. further, that for the two triangular prisms #* Fn/ 2 g _ t | [4, (/-/,)+ /,(4-4,)]^ A 2 ?Fy/2g and, lastly, for the pyramid: f _ ,(*-*,)(4-4.)>/A. iiF\/2g hence the time of discharge for the whole vessel is: t = t1 + t2 + t3 =[30 4,/,+lQ 4, {l—h)+ 10 h (4—4,)+6 (4—4,)] s/h 15pFs/2g = [3 4/+8 4,^+2 (4fj+4,/)] 2 y/h 15 pF */2g If A. = A, we have then a truncated pyramid to consider. Let , h * the one base bl = G, and the other bxlx = Gv we then obtain: t = (3 G+8 G.+ 4y/~GG.) _______2'/h______ 15pFS2g It would be easy to show that this formula holds true also for every trilateral or multilateral pyramid. Kxample. An obelisk-shaped water-cask is 5 feet long, and 3 feet broad at top, and at the depth of 4 feet, that is, at the levei of a short horizontal discharge-tube, 1 inch in width, and 3 inches in length, it is 4 feet long and 2 feet broad, what time will be re- quired for the water in the full cask to sink 2^ feet ? The time for emptying is, f* being taken = 0,815: '=[8.4.2+3.5.3+2 (3.4+5 . 2)]----------^------------- 15 . 0,815 .*.(±Y 4 \12/ 8,02 _ 153 . 4 . 4 . 144 = 153 2304 = 153.7,481 = 1145 sec. As the 15.0,815.8,02 * ' 12,225 . 8,02 ^ , , ^2» feet, hence levei 4 — 2$ = l£ feet above the tube l=lx + §— 4 J and b »+* * the time for emptying if the vessel be filled only up to this levei,_is_. 11S2v/1,5 <, = [8.4.2 + 3. Y • V+2(2-V+4’ V)] • 15 . 0,818 ."s)» * = 131,672424 1RREGULAR VESSELS. 4,5779=602,76 sec. The difference of the times found gives the time in which the surface of water originally at the top of the vessel sinks 2$ feet. § 351. Irregular Vessels.—When we have to find the time of efflux for an irregularly formed vessel HFR, Fig.483, we must apply Sirnp- Fig. 4S3. sonJs rule as a method of approximation. If we divide the whole raass of water into four equally thick strata, and the heads of water G0y Gv Gv G3, Gv corresponding to the horizontal slices, represented by ho9 hv h2y h3, h4, the timeof efflux will be given by Simpson’s rule t = K-K_ (go | ■, 2Gl i 4Gl \ G±_\ 12 P F y/ 2g '\/ h0 y/ hx y/ h2 y/ h3 y/ hj In assuming six strata: t = K-K_( Gq_ 4G, 2Gi+ 4^ ( 2G,+ 4^ | G^v l8pFy/2g\y/ h0 y/ hx y/ h2 y/ h, \/ h4 y/ h& y/kj The discharge in the first case is: Q = -°.~T—i (G,+ 4G1 + 2G,+ 4GJ+ GJ, in the second : 12 Q = hzik(G0 + 4 G, + 2 G, +4 G3 + 2 Gt + 4 G, + G#). IO When the form and dimensionsof the vessel of efflux are not known, we may then calculate very nearly the discharge by the heads of water noted in equal intervals of time. Let t be one such interval, wre have then for apertures at the bottom and sides: Q = + 4 + 2 x/Tt + 4 v/ + v' h4), O and for divisions or notches in a side: __ Q = | *bty/~2^ (y/V0 + 4 v/Fx + 2 */A*2 + 4 v/A1s + Example. In what time will the surface of water in a pond sink 6 feet, if the sluice fornis a half cylinder, 18 inches wide, 9 inches deep, and 60 feet long, and the surfaces of water have the following areas ? Gp at 20 feet head of water, = 600000 square feet. Gv « 18,5 “ w =495000 “ G„ “ 17,0 “ “ =410000 “ G& 11 15,5 « “ = 325000 “ Gv « 14,0 « “ = 265000 “ C|)’ = ~ = 0,8836 square feet Let the co-efficient of resistance for the emrance = 0,832, and that for the friction: 0,025 . 0,025.60.1,091 = 1,6356, then is the co-efficient of efflux 1 i t* = —.. .... =------------- = 0,537, y/1 -f 0,832 + 1,6365 */ 3,4685INFLUX AND EFFLUX. 425 0,537.0,8836.8,02 a* 3,8054. Now G, 495000 115090, and /t F ^/2g s _ 600000 _ i34170) _ \/ho v/ 20 v/A, 18,5 -4 = IS = 99440, = = 82550, %/A» v/17 ga 265000 fs_. s/ 15,5 \/K \/ 14 = 70S30; hence, then, the time of efflux fq^iows: t =------_----(134170 + 4.115090 + 2.99440 4- 4.82550 + 70830) 12.3,8054 = 156940 sec., = 43 hours, 35 min. 40 sec. 7,6108 The discharge is: Q = (600000 + 4.495000 + 2.410000 + 4.325000 + 265000) as i965000 —- 2482500 cubic feet 2 § 352. Influx and Efflux.—If the vessel during the efflux from below has an influx to it from above, the determination of the time in which the surface of water rises or falis a certain height becomes more complicated, so that we must be satisfied generally with but an approximate determination. If the discharge per second Q2 is ^ fi F 2 ghy then there is a rise, and if Qx < /* F y/ 2 gh9 a fall of the surface. Moreover, a state of permanency occurs whenever the head of water is iticreased or decreased by k = — (. The 2g\p F/ time *, in which the variable head of water x increases by the small amount {, is given by the equation Gi*= Ql* — l*F*/2gx . ^ ? and, on the other hand, the time in which it sinks the height by Gx i = n F v' 2 gx . t — Qx r. Hence we have in the lirst case t =-----, and in the Qx— p F 2 gx By the application of Simpson’s rule second * =_____ fi F \/2gx— Q1 we then obtain the time of efflux, during which the lowering surface from G0 to Gv G9 and the head of water from h0 to hv t. _ K—K r G„ 4 G, 2Gt 12 LpFx/2ghe—Qt liFs/2ghl—Q1 pFx/2gh—Ql + 4 G. - -1 2gK—Q%^ i*F s/ 2ghs— Q, pFy/2ghi or, more siraply, if we represent ^ by vf k, pF v' 2g 4 *-A~Vr _g!_= , 4G»_ , f JG»_ \ UpF^gly/^—^k y/hl—Sk V*,—*'* + J&_T y/ht—y/k y/hA’—%/ife J 36*r* wuL UaU,w. - -Jljpt l, ^ v> . I fc+Yk&JYT-Y*))j Yx-Q. ~/t?fiJ Yv+Yu' J v.~Yl = I rzC-4 j 4 A7h f %~Yk - 7 A rK/Jk .' osr - 2$*) rX; .tt?/ 4vt) 426 LOCKS. If the vessel is prismatic, and has a uniform transverse section G, we then have: 2 G / .-r . .v , , /x/h— x/k \ t = _ — (s/h—s/hx + x/k . hyp. log. (^—----—) )> p F ^/2g\ \ y/hx — x/k' ' the time in which the head of water passes from A to hx. Since for: l 7 x/ A — x/ k \/ h — x/ k hl =k, ——------ =---------= oo, v/A,—v/#t 0 it follows that the condition of permanency takes place indefinitely late. The following formula is the resuit of investigation for a wier or notch in a side. , = °±[hyP.log. + 3 Q, L (yh1—^kf{h + v/AA' +_£)_ -rr (v/A—^/A.) v/12 + v/12 . arc (tang. = --—--J-1---— 3 k-f~ (2 A -f- ) (2 \/Aj -f- k^ •where k = (-hyp. log. represents the hyperbolic \lub S 2 gt >)]' p b %/ 2 g' logarithm, and arc {tang. = y) the arc whose tangent = y, According as k is ^ A, and the inflowing quantity of water: Ql > § ^ 6 v^2 g A3, there is a rise or fall of the fluid surface. The condition of permanency occurs, when hx = k9 and the time corre- sponding becomes oo. Example. In what time will the water in a rectangular tank 12 feet long and 0 feet broad rise from 0 to 2 feet above the edge of a notch $ foot broad, if 5 cubic feet of water flow in per second? We have here h = 0; hence, more simply : t = — T hyp. log. h\+\/h'k±]t + ^/L2 arc (tang. =* A>---)"]. 3(2*L (x/tl-x/lc)' ^ *s/k+s/K'J Now G = 12.6 s 72, Q, = 5, ht = 2, 6 = J, and /u = 0,6, £ — M =* 2,1338, and the time sought is: Vf.0,6.i. 8,02/ _____ ! [hyp. log. 4^±vA^ - ^ ____✓ •_____ L ^ (1,4142—1,4607)* v * 1,4142+2,9214/J 72.2,1338 I 3 . 5 10,242 f hyp. log. 6,L^-------x/ 12 . arc (fang. — 6 )"] L ^ 6 0,002162 4,3356/J 0,002162 4,3356 > = 10,242 (7,961 — 1,781) = 10,242.6,18 = 63,*9 sec. § 363. Locks.—A very useful application of the doctrines hitherto treated of may be made to Fig. 484. the filling and emptying of canal locks. We distinguish two kinds of navigation locks, single and double. The single lock, Fig. 484, consists of a chamber B9 which is separated by the upper gate HF from the upper reach A} and by theLOCKS. 427 The double lock, Fig. 485, Fig. 485. lower gate RS from the lower reach C. on the other hand, con- sists of two chambers, with the upper gate KLy the middle one HFy and the lower one RS. Let the mean horizon- tal transverse section of a simple lock chamber = G, the distance of the middle of the sluice in the upper gate from the upper surface HR of the upper reach = hv and from that of the lower reach = A2, and, lastly, the area of the aperture or sluice opening = F, we then obtain the time of filling up to the middle of the Gh aperture t = -------==, and the time for filling the remaining 1 F x/2ghx A space, where a gradual diminution of the head of water takes place, ^ ^ 1——; consequently, the time for filling the single ti F x/2ghx sluice is; t = L + *2 = (A2 + 2 hx)G n F s/2ghl If the aperture in the lower gate is entirely under water, then while emptying, the head of water gradually decreases from hx + h2 to zero, hence the time for emptying or running off is: __ 2 G \/ hx + h2 ^ F x/ 2g If, on the other hand, a part of the aperture stands above the lower water, we then have two discharges to take into account; the one flowing above and the other below the water. Let the height of the part of the aperture above the water = a,, and that under the water = av the breadth of the aperture = 6, we then obtain the time of efflux from the expression : t =_________________2 G{hl+ h2)_________________ "b (ai + K -| + «, chamK^°U^Le* J°c.ks’ heacl °f water gradually decreases in the intn „er WaiC ,ls cI°sed from the upper reach, during the discharge ofth f second chamber. If G is the horizontal transverse section sinlr f c“a.m^er> an(l Ifie original head of water hx in this chamber of ih ° X’ *^e water in the second chamber rises to the middle 6 a^r^re s^u^ce> we have then the corresponding time t ___ 2 G _________ _ ~F ^/2~g — \/x). Now the quantity of water423 EFFLUX OF STILL AIR. G (hx — x) = G,A2, hence x =* hx-----—1 A2, and G 2 G pFs/2g' \ G / pFs/2g The time in which the water rises as high in the second as in the first chamber, and in which, therefore, it comes to the same level in both, may be found frorn §_347: ______________ t _ 2 cg. _ gg.v/Gy/GA,— * " ?F(G+ G.)*/2g " F(G + GJ y/Yg and the whole time for nlling: / = tf + t% - ^ f%/2g GAl “ G+G, '/GA> ~~ GlA*)' Example. What time is required for the filling and running off of the following single lock chamber? The mean length of the lock = 200 feet, mean breadth ms 24 feet, therefore G ss 200.24 aes 4800 square feet, distance of the centre of the aperture of the sluice in the upper gate from the two surfaces of water 5 feet, breadth of both apertures 2£ feet, height of the aperture in the upper gate 4 feet, and of that in the lower gate (entirely under water) 5 feet. Let t = A, = 5, h, = 5, G = 4800, p > 0,615, F = 4 . 2j «= 10, v77j t* F x/2 ght = 8,02, we tlien obtain the time of filling: 3.5.4800 t =-----------; 14400 652,85 seoonds. If we substitute in the 6,15.8,02 v/5____1,23.8,02 ^/5 formula t s= 2 ^ fo, G bb 4800, ht -f* h% b 10, F 5.2^ b 12,5, we then nFyflg obtain the time for emptying of the sluice: t = 2 ‘ 4800 ^10 a- 491,78 sec. = 8 min. 21,78 sec. 0,615 . 12,5.8,02 CHAPTER VI. ON THE EFFLUX OF AIR FROM VESSELS AND TUBES. § 354. Effiux of Stili Air.—Condensed air does not flow from vessels quite in accordance with the law which regulates the flow of water, because an expansion takes place during its discharge, wrhich is not the case with water. But in order to discover a similar law for air and other gases, let us make the mechanical eflect Qy which a 2g quantity of air Q of the density y requires to pass from a state of rest into that of the velocity r, equal to the mechanical eflect Q p hyp. l°g- (j) found in § 298, which the same quantity of air producesEFFLUX OF STILL AIR. 429 when it passes from a greater pressure px to a less p. If, therefore, Pi be the elastic force of air enclosed in a vessel, v its velocity of efflux for the tension of the external air, and y its density, then Qy . ~ == Qp hyp. log. therefore, the height due to the velo- city: | %. (&) - 2,3026 i %.(&), and the velocity itself v J2#y hVP-l°Z‘ (y )' When the tensiols p and p, differ little from each other, when Pi—pis < T>0p, then we may puts hyp. log. = hyp. log. (l + £=?) = and hence v= I 2g (P~)‘ But the height of In external column of air which isjn equilibrium ty its weight with the pressure p1—p (§ 294), is A = ; hence w e may put the velocity of efflux t> = %/2gA, and a perfect analogy with the efflux of water will hereby subsist. For h.gh pressure this for- Riula is not of course sufficient, for in this case: k„.Ug. (a.) _£rf- 4 (*=*)'.«1-. Hence, then, more accurately : - -i -v) *• 0r if we represent the height of the barometer by b. v = by. and • * T1 ___1 Fig. 486. If the discharging orifice F of the vessel Fig. 486, is accurately and smoothly rounded, the particles of air then flow in parallel lines, and hence the quantity of air flowing through the orifice in each second, and measured by the height of the external barometer, is: Q = Fv = F(l — or more accurately: _ = F J2 gb hyp- log- (~T~) • § 355. The above formulae do not admit of d r re ct applicat ion, be^ cause we cannot measure the internal or the extern p Y430 EFFLUX OF STILL AIA. length b 4- A, and b of the columns of air. These pressures are gene- rally measured by columns of water or mercury. As regards the quotient^L = _jl^, it is immaterial whether b and h be expressed in p b columns of air, water, or mercury, because each reduction of b and h leaves the fraction - constant, except tliat the quotient P = 6, is stili dependent on the temperature of the effluent air, and varies for different kinds of gas. For atmospheric air (§ 301), if p represent the pressure of air on one square centimetre, and y the weight of a cubic metre of air, and t the temperature in degrees centigrade, we have P ** 1 + 0,00367 .t Qn the other hand, for steam, y 1,2572 9 9 9 p 1 + 0,003671 y 0,7857 If we substitute these values in the general formula for r, we shall obtain for atmospheric air: 395 J(i+° ,00367 . t) hyp. log. (-y-) metres, or being small: 395 ,J(1 + 0,00367 o=500,6^(1 + 0,00367. t) - metres, and for steam b t) hyp. log. metres. The theoretical discharge as estimated under the external pressure is Q = Fv, but if this is to be estimated at the internal pressure, we must then make Q.p,®* Qp>hence Qx=*— Q* t-tK* Reduced to the 1 1 px o + /i temperature of zero, the quantity discharged is: e,- Q 1+0,00367. t , therefore, for atmospheric air 395 F f hyp. log. {b+h)—hyp. log. b cubic metres V 1+0,00367 . t If equa] masses of air of different temperaturos issue from different orifices Pand jP, at the same tension, we then have: _ /T+0,00367 f "" \ 1+0,00367 t ’ If, for example, t = 0 and tx = 150° C., we then have: Fi - s/T,5605 . F=* 1,245 F. If, therefore, a blast furnace is to be supplied with heated air of 150°, we must apply nozzle pipes, which have a one-fourth greater transverse section at the discharging orifice than if cold air were to be used.EFFLUX OF AIR IN MOTION. 431 For Prussian measure, and centigrade scale of temperature: v = 1258 . ^ (1 + 0,00367 t) hyp. log. and for steam v = 1595 . ^ (1 + 0,00367 t) hyp. log. ^ For English measure, and Fahrenheifs scale of temperature: v = 1295 . ^ (1 + 0,00204 t) hyp. log. and for steam = 1642 . J(1 + 0,00204 t) Example. In a large reservoir, air at 120° C., temperature is enclosed, which corresponds to the height of a mercuria 1 manometer of 5 inches, whilst the external barometer stands at 27,2 inches; what quantity of air will flow from tliis through a round aperture H inch wide? Itis: hyp. bg. (t + = hyp. log. (|HrL) = hyp. log. 32,2 — log. 27,2 = 5,77455 — 5,60580 = 0,16875, hence the velocity of efflux is: v = 1258 . (i -f 0,00367.120) 0,16875 == 1258 . 1,4404.0,16875 = 620,2 Prussiari feet. Now the area of the orifice = — ($)*=—— =0,01227 square feet; hence it follows 4 256 that the discharge Q = 0,01227.620,2 = 7,61 cubic feet. Estimated at the interior 272 pressure, it is = . 7,61 = 6,43 cubic feet, and reduced to the mean height of the barometer, 28 inches and 0° temperature (30 English inches and 32° temperature F.), the quantity discharged is: = 7,61 . J?72. . 1 — 5,13 cubic feet. 280 1,4404 § 356. Efflux of Air in Motion.—The formula of efflux given: suppose the pressure px or the height of the manometer h to be mea- sured at a place where the air is at rest, or has a very slight motion, but if px or hx is measured at a place where the air is in motion, if, for instance, the manometer communicates with the air Flg* 487* in a conducting tube CFy Fig. 487, we shall then have to take into account the vis viva °f the arriving air. If now c be the velocity of the air pass- ing the orifice of the manome- ter we shall accordingly have to make: 4f w lit, , B 1 Qr • ~ = Qr ■ ^ + Qp hyp. log. or if F be the transverse section of the orifice, and that of the tube, or of the air passing the orifice of the manometer, according to the law of Mariotte, ~c = J?L, or Gcp. = Fvp, therefore, Fv px C Qr [l - (SV i-Qr hyp. log. (b\432 EFFLUX UNDER DECREASING PRE6SURE. and the velocity of efflux in question: **•(&) The velocity of efflux is, therefore, here exactly like that of water F from vessels, the velocity is greater, the greater the ratio of the G transverse section of the orifice to that of the tube or the arriving cur- rent of air. From this it is evident, that under otherwise similar cir- cumstances, the height of the manometer px is so much the Jess the narrower the conducting tube is, or the greater the velocity of the air issuing from it. Examplet.—1. A mercurial manometer, placetl upon an air tube 3$ inches wide, stands at 2£ inches, while the air flows from its conical extremity through a round orifice 2 inohes in diameter; witli what velocity wili the current move? If the external barometcr pt 27$+2J 30 stand at 27$ inches, we shall then have — = ——-j— ■■ ■and p 27£ 27,0 ^ P /2\ i] 16. 11 44 . hence the theoretical velocity of efflux at 147 1258 y/1,0367.0,087 396 Prussian feet.___ Lli-f-LX.u-. G px \ 3,5 / 49 . 12 a temperature of the air 10° C.: __ 1258 . ^/1,0367 . hyp. bg. (| f j ’ ^ x/0,9104 2. The tension pg in the air regulator, where the air is without motion, is given by the fbrniula, . vl >>yp■ kg- (y) hyp. bg. (y)*= ■ -j, oi hyp. bg. p, «* hyp. bg.p+-7~#>y therefore, in the present case, =* hyp. log. 27,5 + 0,087 0,9104 ; «3,3142 + 0,0965 *3,4107. Hence it follows that p% as 30,3 inches. § 357. Efflux under Decreasing Pressure.—If an air reservoir has no influx, whilst an uninterrupted efflux goes on, the density and ten- sion gradually diminish, and hence the velocity of efflux becomes less and less. We may determine in the following manner in what ratio this diminution is to the time and to its discharge. Let Fbe the volume of the reservoir, h0 the initial height of the manometer, and hn the height of the manometer at the end of a certain time t9 b the height of the external barometer. Then the quantity of air in the reservoir at the commencement reduced to the external pressure * and at the end of the time t, b =* —and, consequently, the quantity discharged in the time t, and at the external pressure is: A * Vn - V(b+kQ) V(b + hn) _ V(^—hn). b b---- ~ ----b------ ’EFFLUX UNDER DECREASING PRESSURE. 433 and, inversely, the height of the manometer corresponding to the dis- charge Vn is: hn-h0--!£.b. If we take four intervals, and the initial height of the manometer ho9 and at the end of the time t = hv and K = K — k° 4 k*> K=K—x (K ~K)> and h3 = h0 — J (A0 — h4), we shall then obtain by Simpson’s rule the time , _ V(K-K)__ 12 Fbj2gE\ %>• loS- ) Jhxp.lV. (*+**) J %. log. ( J hyp. log. ( Jfiyp.log. (*-±^ )j For moderate pressures or heights of the manometer: isp.bg. (t+*)-l( consequently ^ hyp. log. y—j = ^1 — and (‘+ A) Ji' Jisp.bg.(^) If we now take n intervals, and therefore the discharge for one inter- val: ^ , we shall then obtain the corresponding element of time: - J r- *»-• (*-^) V(h,—h.) (‘ + 4i) Ji nb •---- FJ2*; /i-i.hh _ V(K-hn)V +TbJ *-Fj2gl£ 37434 CO-EFFICIENTS OF EFFLUX. Now if we substitute for A; A0, A,, A,, . . . . h„, we shall then ob- tain the sum of all the (~^!L) h~^ = 2(A0*-An*) - 2(v/ A0—s/ and the sum of all the (= |(A^—A* - f (>/V-----), whence the sum of all the small intervals of time, or the whole time in which hn passes into and the quantity of air Vn which flows out, is: t--------JLL= t (^T0 x/hn)+ or J 2 ghV- 2 V . t- (v' A0-^ An)^l + “^gjj ^ Fhgbl ~ r approximately: _______2 F F^2gbl Exampte A 50 feet long and 5 feet wide cylindrica1 air regulator of a blowing machine is filled with air; the height of its manometer h = 10 inches, and the ther- mometer stands at 6° C. If now a flow of air takes place m a space where the height of the barometer is 27 inches, through a 1-inch wide ciicular onfice, then the question arises, in what time will the height of the manometer fall to 7 inches, and what will be the corresponding discharge ? The yolume of the cbambens for Prussian measures,— = - 5* 50 s 1°50 - = 981,75 cubic feet, hence the discharge, measured at the 4 '4 external pressure, is F, = (-^-?) (^) ’ 98 *>75 “ 109’08 cubic ««• Now J QgP.^ 1258 v/1 + 0,00367.1 s» 1258 */ 1,02202 a 1272, and F—*( l.y______ * __0 005454 aquare feet, hence the time of efflux in question i» 4 576 , = _ 8•981'75 ( fio 0,005454.1272 V J 27 N 27/ \ 8.27 / = - 1963’5 . 0,0994.1,079 s 30,3 seconds. 5,454.1,272 § 358. Co-efficients of Efflux.—The phenomena of contraction, which we have considered in the efflux of water from vessels, occur also in the efflux of air. If the orifice of efflux be cut in a thin piate, the air passing through it has a smaller transverse section than the orifice, and on this account the discharge is less than the product Fv of the transverse section jFof the orifice and the theoretical velocity v. Let -^} be the ratio of the transverse section Fx of the blast to that of the orifice T, = /i,we then have the effective discharge as for water:CO-EFFICIENTS OF EFFLUX. 435 Q, - „ Q - - pFv = p hyp. log.(&). From the author’s reduction of Koch’s experiments at pressures of the manometer of from 3 to £ of an atmosphere, we may take the mean of = 0,58. The effective discharge in the issuingof air through short cylindrical adjutages, is likewise less than that determined theoretically; we have, therefore, to multiply this latter by a number deduced from experi- ment, the co-efficient of efflux, in order to obtain the former; only F v here /t is not the ratio of the transverse section —i, but the ratio -i of F v the effective velocity of efflux vx to the theoretical v. Koch’s experi- ments give for the above pressures, in the flow of air through cylin- drical adjutages, which were nearly ali six times as long as wide, as a mean *■ 0,74. Conically convergent adjutages, similar to the nozzles of bellows, give a stili greater co-efficient df efflux; a tube of 6° lateral converg- ence in the experiments of Koch, gave when five times as long as wide, the mean co-efficient ^ = 0,85. From this, therefore, the effective discharge for the flow of air through orifices in a thin piate, measured at the external pressure, is Qj = 751,1 — +0,00367<)|:cubic feet (Eng.), for efflux through short cylindrical adjutages: Q, = 958,3 F(l _ A.) J (1 + 0,00367 t) £ cubic feet, and through conical adjutages of 6° convergence. Q1 = 1090,7 .F^i___^ (1 + °>00367 l) ~ cubic feet.* * Experiments on the efflux of air have been undertaken by Young, Schmidt, Lagerhjelm, Koch, d’Aubuisson, Buff, and in later time, by Pecqueur, Saint-Venant, and Wantzel. For an account of the experiments of Young and Schmidt, we may refer to Gilbert’s “ Annalen,” vol. 22, 1801, and vol. 6, 1820, and to PoggendorfTs u Annalen,” vol. 2, 1824; for those of Koch and Buff, to the ‘‘Studien des gottingschen Vereines bergmannischer Freunde,” vol. 1, 1824; vol. 3, 1833; vol. 4, 1837, and vol. 5, 1838; also in Poggendorff’s “Annalen,” vol. 27, 1836, and vol. 40, 1837. The experiments of lagerhjelm are described in the Swedish work, “ Hydrauliska Forsok of Lagerhjelm, Forsellesoch Kallstenius,” 1 vol. Stockholm, 1818. D’Aubuisson’s experiments are to be fbund in tlie “Annales des Mines,” vol. 11, 1825; vol. 13, 1826; vol. 14, 1827, and likewise in his “Trait6 d’ Hydraulique.” The latest experiments instituted in France are reported in the “ Polytechnischen Centralblatt,” vol. 6, 1845. Most of these experi- ments were tnade with very narrow orifices, and, therefore, scarcely answer the purpose in practice. The experiments of d’Aubuisson and Koch deserve mosi consideration; and next to them, perhaps, those of Pecqueur; but the most extensive are those of Koch. The wished*for accordance is hardly to be met with in the results of all these experi- ments ; the co-efficients of efflux found by d’Aubuisson vaiy considerably from those calculated by Koch. The grounds for my placing the most confidence in the co-efficients of Koch, are given in the “ Allgemeinen Maschinenencyclopadie,” under the article “ Ausfluz,” and in a Memoir of mine in PoggendorfTs “Annalen,” vol. 51, 1840. [For calculations of the above, and all similar cases, the co-efficient of t for the Fahrenheit’s thermometer is 0,002039 instead of 0,00367; (see above, p. 346;) but the degrees com- puted are actually t — 32 on that scale.]—Ax. Ea.436 FLOW THROUGH TUBES. Example. If the two orifices of a bellows together possess an area of 3 square inches, if, further, the pressure of the manometer is 3 inches, the external barometer 27$ inches, and the temperature of the air 15°, then is the discharge: Q, = 1069 . TJ A--------1—\ /( i_j_ 0,00367.15) _L- V 4.27,5/ >/ T ' 27,5 207 ________ _______________ = =21,66 0,1151 = 7,34 cubic feet. ^ § 359. Flow through Tubes.—If the air issues through a long tube GFy Fig. 488, it has then the resistance of friction to overcome in Fig. 488. the same manner as water; this resistance may also be raeasured by the height of a column of air, which has for expression l 2 hH = f . - . —, where, as in the conducting of water, v represents d 2g the velocity, / the length, d the width of the tube, and f a co-efficient of resistance to be determined by experiment. Numerous experiments of Girard, d’Aubuisson, BufFand Pecqueur, lead to the mean value f = 0,024. From this, therefore, the resist- ance generated by the friction of air in tubes may be measured by " 7 2 the height hn = 0,024 _ . — of a column of air, or by the height b 1 2 hn = 0,0000023 _ . ?- of a column of quicksilver, and the manome- d 2g ter will stand at this much less height at the end of the conducting tube than at the beginning. If at the end of a conducting tube of the width d, the manometer stands at h2, whilst the air flows through an orifice of the width d, then from what precedes, the velocity of discharge will be: J2 g £. hyp. hg. (t+!b) . : but if hx be the height of the manometer at the beginning of the con- duit, we shall then have: 7 (t4) - [1 +(>£>,)’(1)‘+ °’024 5 (7)'] because the velocity in the tube = i- v; hence in this case d2 v =FLOW THROUGH TUBES. 437 V = 12g — %»• l°g- (4^) If, lastly, the height of the manometer h is measured in the reser- voir at the beginning of the conduit, where the air may be regarded as at rest, we then have: 4 I fi 4 + 0,024 lA d5 If, further, we put the co-efficient of resistance f for entrance into the tube, which when p, = 0,74 amounts to 0,826, and, further, join to it the co-efficient of efflux /t for the outer adjutage, we then obtain for the velocity: J2 g Z. Kyp. log. 4 + S + 0,024 d5 or 1294 p ^ (1 + 0,00367 t) hyp. log. + 4 1 + f + 0,024 Id,4 d5 feet (Pruss.) According as the point of the interior orifice lies s lower or higher than the point of the exterior orifice, we have to add + s to the quan- tity under the radical in the denominator. Moreover, other hindrances may present themselves in the tube, sueh as curvatures, contractions, and widenings, &c. Satisfactory experiments on these obstacles do not exist, but we may assume with great probability that these resist- ances are not much different from what takes place in the case of water, because the co-efficients of efflux, and the eo-efficient of firic- tion are nearly the same for air as for water. As long, therefore, as no further experiments are made on this sub- ject, we may avail ourselves with tolerable safety of the co-efficient of resistance found for water in investigations on the motion and flow of air. Example. In the regulator at the head of a 320 feet long and 4 ineh wide air-con- ductor, the mercurial manometer stands at 3,1 inch, whilst the extemal barometer is at 27,2 inch; further, the width of the orifice of the conically contracted extremity of the conductor is 2 inches, and the temperature of the air 20° C., what quantity of air will this conductor deliver ? It will be: 1 + f + 0,024 ^ = 1,826 + 0,024. -2H2.. (|)‘ ■* 1,826 + 0,024 . - ^ 3 =. 1,826 + 1,44 k 3,266; further, (1 + 0,00367 /) hyp. bg. 37*438 RUNNING WATER—DIFFERENT VELOCITJES. = (1 + 0,00367 . 20) hyp.log. ^ = 1,0734 . (5,7137—5,6058) = 1,0734 .0,1079 = 0,1158 j if now, furtlier, we introduce tlie co-efficient of elllux, and hence also: Fxc t + F2c2 + Fl+F%+.. Besides the mean velocity, the mean depth of water has to be mtro- duced, that is, the depth a which a section must haye at all points that it may have the same area as it actually has with the variable Hence, therefore, __ F area of section b breadth of section depths av a2, a3, &c.440 MEAN VELOCITY. If the separate parts of the breadth biy b2, b3, have the correspond- ing mean depths av av &c., Fig. 491, we then have: f = "f“ ^2^2 "F • • • f and hence also: albl + a2b3 + • - bt + bt + • . Lastly: "I" aJ>tc* "h • * afiy + aJ)% + • • * and if the portions bv b2, &c., be of equal size, _ qi ci + %c% + . • ai + «2 + • • A river or brook is in a state of permanency when an equal quan- tity of water flows through each of its transverse sections in an equal time; when, therefore, Q or the product Fc of the area of the section and the mean velocity throughout the whole extent of the stream is a constant number. Hence this simple law comes out: in the perma- nent motion of water, the mean velocities in two transverse sections are to each other inversely as the areas of these sections. Fig. 491. Examples.—1. At the section of a canal, ABCD, Fig. 491, it was found that the Portions of the breadth - - - bx =. 3,1 feet, b% s 5,4 feet, bs * 4,3 feet Mean depth - • - - - a, = 2,5 “ a2 = 4,5 M a3 = 3,0 M Corresponding mean velocities - - r, = 2,9 w == 3,7 u c3 = 3,2 “ Hence the area of these profiles F = 3,1.2,5 + 4 . 4,5 -f- 4,3 . 3,0 s= 44,95 square feet, and the discharge: Q = 3,1 . 2,5.2,9 + 5,4.4,5.3,7 -f- 4,3.3,0.3,2 == 153,665 cubic feet, and the mean velocity c = S_ 153,665 44,95" = 3,419 feet. * 44,95 2. When a cut is to conduct 4,5 cubic feet of water with a mean velocity c of 2 feet, we must then give to it a transverse section of = 2,25 square foot area.—3. If one and the same stream has a mean velocity of 2$ feet at a place 560 feet broad and 9 feet mean depth, it will then have, at a place 320 feet broad and 7,5 feet mean depth, the mean velocity c 560 • 9_ . 2,25 =3 .£51 = 4,725 feet. 320.7,5 120 § 363. Mean Velocity.—If we divide the depths of water at any point of a flowing stream into equal parts, and raise ordinates upon them corresponding to the velocities, we shall then obtain a scale of the velocity of the current J1B, Fig. 492. Although it may be granted that the law of this scale, or of the dinerence of velocity is expressed by some curve, as according to Gerstner by an ellipse, yet it is allowable, without fear of any great error, to substitute for this a straight line, or assume that the velocity diminishes uniformly with the depth, because the diminution of velocity downwards is always very small. From the experiments of Ximenes, Briinnings, and Funk, the mean velocity in a perpendicular cm = 0,915 c0, where c0 repre- Fig. 492.BEST FORM OF TRANSVERSE SECTION. 441 sents the velocity at the surface, or the maximum velocity. The velocity, therefore, diminishes from the surface to the middle M by cQ — cm « (1 — 0,915) c0 = 0,085 c0, and, consequently, the velocity below or at the foot of the perpendi- cular may be put cu = c0 — 2 .0,085 c0 = (1 — 0,170) c0 — 0,83 c0. If, now, the whole depth = a, we then have, by assuming a straight line for the scale of the velocities, the corresponding velocity for a depth AN = x, below the water a \ a/ Further, let c0, cv c2, . . . be the superficial velocities of a whole transverse profile of not very variable depth, we have then the corre- sponding velocities at a mean depth: 0,915 ^,0,915^,0,915 c2, and hence the mean velocity in the whole profile: c = 0,915 (c0 + C1 + c2 + • • •. ,c”). Lastly, if we assume that the velocity diminishes from the line of current towards the banks, as it does according to the depth, we may then again put the mean superficial velocity ■(c0-+ C1 + • • + c*) = 0,915 c0, n and so obtain the mean velocity in the whole profile: c asss 0,915.0,915 . c0 = 0,837 . c0, i. e. from 83 to 84 per cent. of the maximum velocity, or of that of the line of current. Prony deduced from Du Buat’s experiments conducted with very small channels, and for these cases perhaps more correctly: /2,372 + ca \3,153 + cj c0 metre , . v_______ = c feet English. ,153 + c0/ u \10,25 + cj 6 Tor medium velocities of 3 feet it hence follows that cm » 0,81 c0. Example. In the line of current of a brook the velocity of the water is 4 feet, and the depth 6 feet, we have then the mean velocity at a corresponding perpendicuiar cm = 0,915.4 = 3,66 feet, and that at the bottom = 0,83.4 = 3,32 feet; tb® velocity 2 feet below the surface is v = (1 —0,17 . f) 4 =(t 0,057) 4 —- , ee > lastly, the mean velocity throughout the profile is, c = 0,837.4 = 3,348 feet, and ac- cording to Prony, c s 4 = “ 3,29 feet’* § 364. The Best' Form of Transverse Sections.—The resistance which the bed opposes to the motion of the water in virtue o i s ac hesion, viscosity, or friction, increases with the surface °f ,con tween the bed and the water, and therefore with the perime p the water profile, or of the portion of the transverse sec io comprises the bed. But as more filaments of water pass g * This and the following subjects have been fully treated of^ ? experiments gnng des Wassers,” in tlie “ Allgemeinen Maschinenencyclop • ^ prfohriinirsresi»l- and new views may be found in the fofiowing writings: La meye n 8W:cj. 1^45 tate uber die Bewegung des Wassers in Fluszbetten und Kan than the first 1+ “ (2 d £ it is necessary that the member with the factor x should vanish, and therefore that may become a minimum, we must have £ 2 v/7»a+1—n 1 , F -----F--------ST"0» or smce: n = cotang. 0 and y/ n2 + 1 2 y/n%+l— n „ F sin. 0 «71.0 2 COS. 0 Hence, therefore, the most appropriate form of profile correspond- ing to a given angle of slope 0, and a given area is determined by I Fsin.e A k F * a = and & = —----a cotang. 0. • cos.0 Example. What dimensions must be given to the transverse profile of a canal, whose banks are to have 40° slope, and which is to conduct a quantity of water Qof 75 cubic Q 75 feet, with a mean velocity of 3 feet? JF 8 == — == 25 square feet, hence the depth 25 sin. 40° 2 — cos. 40° /0,0 4i “ 5J|,23£ 04279 ,23396 3,609 feet, the lower breadth 6 25 3,609 3,609 cotang. 40° = 6,927 — 4,301 = 2,626 feet the slope or cut of the banks aa 3,609 . cotang. 40° = 4,301, the upper breadth s= 6,927 -f- 4,301 = 11,228 feet, the perimeter p 2a 7,218 *s 6 4- ----s 2,626 4- -—rr= = 13,855 feet, and the ratio determining the friction nn. e ’ * sin. 40° p 13,855 F ™ 25 0,5642. § 366. The dimensions of the most suitable profiles which corre- spond to different anglesof slope and to a given profile are to be found in the following table.UNIFORM MOTION. 445 Angle of slope. 9 Relative slope. Dimensions of transverse profile. Quotient Depth a. Lower breadth b. Absolute slope na. Upper breadth b *|* 2 ti a. P F 90° 0 0,707 y/F 1,414 y/F 0 1,414 v/f 2,828 y/F 60° 0,577 0,760 y/F 0,877 y/F 0,439 y/F 1,755 y/F 2,632 y/F 45° 1,000 0,740 y/F 0,613 y/F 0,740 y/F 2,092 y/F 2,704 y/F 40° 1,192 0,722 y/F 0,525 y/F 0,860 y/F 2,246 y/F 2,771 y/F 36° 52' 1,333 0,707 y/F 0,471 y/F 0,943 y/F 2,357 y/F 2,828 y/F 35° 1,402 0,697 y/F 0,439 y/F 0,995 y/F 2,430 y/F 2,870 y/F 30° 1,732 0,664 y/F 0,356 y/F 1,150 y/F 2,656 y/F 3,012 y/F 26° 34' 2,000 0,636 y/F 0,300 y/F 1,272 y/F 2,844 y/F 3,144 y/F Semicircle 0,798 y/F 1,596 y/F 2,507 y/F We see from this table that the quotient ~ is least for the semi- r circle, namely, = ^9?; greater for the semi-hexagon, and greater y/ f1 stili for the half square, and the trapezium of 36° 52', &c. Example. What dimensions must be given to a profile, which has for an area of 40 square feet, a slope of its banks of 35° ? From the preceding table, the depth a = 0,697 y/40 =4,408, the Iower breadth =0,439 y/40 = 2,777 feet, the absolute slope = 0,995 ^/40 = 6,293 feet, the upper breadth = 15,363, and the quotient 2,870 y/40 = 0,4538. § 367. Uniform Motion.—The motion of water in beds is for a certain tract either uniform or variable; it is uniform when the mean velocity at ali transverse sections of this length remains the same, and therefore, also, the areas of the sections equal; and variable, on the other hand, when the mean velocities, and therefore, also, the areas of the sections vary. We shall treat first of uniform motion. In the uniform motion of water along the distance J1D =* l, Fig. 489, the whole fall HD = A is expended in overcoming the friction of the water in the bed, because the water flows on with the same velocity with which it arrives, therefore, a height due to a velocity is neither taken up nor set free. If we measure this friction by the height of this column of water, we may then make the fall equal to this height. But the height due to the resistance of friction increases446 UNIFORM MOTION. with the quotient with l and with the square of the mean velocity F c (§ 329); hence then the formula holds good : 1. h = f. lt. F 2g in which f expresses a number deduced from experiment which may be called the co-effident of the resistance of friction. By inversion it followsj 2. c lp .2gh. In determining the fall, therefore, when the length, the cross section and the velocity are given, and inversely, in deducing the velocity from the fall, the length and the cross section, we must know the co- efficient of friction f. According to Eytelwein’s reduction of the ninety-one observations of Du Buat, Briinings, Funk and Waltmann, f as 0,007565, and hence h = 0,007565 ,ll F 2g If we put g = 9,809 metres or 31,25 feet (32,2 feet English), we have for the metrical measure ^ . c2 and c ■ h = 0,0003856 T and for the foot measure: A=0,00011726 !l.c* and c=92,35 F i Id 41 IP For conduit pipes -L., F Fh pl English measure. , hence this formula gives for 0,03026 - . —, whilst we have found more correctly for d 2g J l I «a pipes h these (§ 331) for mean velocities h _ 0,025 1 . 2L. d 2g The friction, therefore, as might be expected, is greater in the beds of rivers than in metallic conducting pipes. ExampU». 1. What fall must be given to a canal of the length l = 2600 feet, lower breadth 6 = 3 feet, upper breadth bt = 7 feet, and depth a = 3 feet, if it is to conduct a quantity of water of 40 cubic feet per second ? It is: (7+3)3_ 2 = 34.2^2*4.3*« 10,211, F 2 = 15 and c=— : 15 }, hence the fall - 1,476 feet.—2. sought, A=0,0001173 2600.10,211 __ 0,305 . 10,211.64 15 15.9 What quantity of water does a canal 5800 feet long, having a 3 feet fall, 5 feet deep, 4 feet lower and 12 feet upper breadth ? Here: P_ F hence the velocity c = 92,35 /— 3 : 0,42015.5800 and the quantity of water Q = p 4 + 2 5* -f 4’ 5.8 92,35 11^1^ —0,42015; 40 92,35 02.35 ' 28,5 3,24 feet, v/0,14005. 5800 812,29 Fc 8 40.3,24 ss* 129,6 cubic feet, English measure.CO-EFFICIENTS OF. FRICTION. 447 § 368. Co-efficients of Friction.—The co-efficient of friction for rivers, brooks, &c., the mean value of which, in the foregoing para- graphs, we have taken at 0,007565, is not constant, but, as in pipes, increases somewhat for small and diminishes for great velocities. We have, therefore, to put: f_f1(i + £)orf1(i + _^). The author of the work alluded to in § 363, finds from 255 experi- ments, the greater part of them undertaken by himself, for the Prussian measure f = 0,007409 ^1 -+- , and hence it follows for the metre f = 0,007409 (l + °>00A39 ^ and for English measure 007409 ^1 + 0,0308 \ It is raanifest that these formulae, for a velocity c = feet, give again the above assigned mean co-efficient of resistance { = 0,007565. The following useful table of the co-efficients of resistance in the me- trical measure serves for facilitating calculation. Velocity c. 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 Meter. Co-efficient of resistance {=0,00 811 776 764 758 755 753 751 750 749 Velocity c. 1 1 1,2 i 1,5 2 3 Meter. Co-efficient of resistance II o 8 748 747 746 744 743 The following table serves for the Prussian or English measure: Velocity c. 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 1* 2 3 5 10 ft Co-efficient of resistance { = 0,00 815 797 785 778 773 769 766 763 759 752 749 745 743 These tables find their direct application in all cases where the velocity c is given and the fall to be found, and where the formula No. 1 of the former paragraph is applicable. But if the velocity c is unknown, and its amount to be determined, these tables will then only admit of a direct application, when we have already an approximate value of c. We may set to work in the simplest manner by deter- mining c approximately by the formula c 50,9 Fh pl and from448 VARIABLE MOTION. this a value of f, taken from the table, and the value so obtained put into the formula c* h F IF 0 . pr- = - . — or c = __ . 2 gh. %g f lp \C//> From the velocity c, the quantity of water is then given by the formula Q = Fc. If, lastly, the quantity and the fall are given, and, as is often requisite in the construction of canals, it be required to determine the transverse section, we may put E. = —— (see Table, § 366) and * F c — ~ into the formula h = 0,007565 -a . and write, therefore, A =* 0,007565 , and accordingly determine: 2gF* F = ^0,007565 i. e., for the met re E = 0,0431 or the English foot measure E» 0,0268 Hence it follows, approximately, that c = -?; if we take a correspondent value of f from F one of the tables, more accurately F = . ra~ ; and hence, more exact values for c = Qy p = m Fy as also for ay by &c. F Examples.— 1. What fall does a canal 1500 feet long, 2 feet lower and 8 feet upper breadth, and 4 feet depth require to give a discharge of 70 cubic feet per second? It is ________ 70 p = 2 -f- 2^/ 4a -f- 39 = 12, F = 5.4 = 20, c s __= 3,5, hence f = 0,00748, and h =0,00748 . 1500 . 12 20 — 6,732.0,1902 = 1,2 8 ft. (Eng.)—2. What discharge does a brook 40 feet broad, 44 feet mean depth, and 46 feet water profile, if it has a feli of 10 JTo a~R iTj 92,35 ---!—’—!---= —-.......= 6,089 46 . 750.12 x/ 230 feet, and hence { = 0,00745. Hence we obtain, more correctly: — = —=___________4’5 • 40 • 12--«* —?— = 0,5844, and c = 6,119 feet. Lastly, ■ig (lp 0,00745.46.750.12 1,7112 the corresponding discharge is Q = 4,5.40.6,119 = 1101 cubic feet, (Eng.)—3. A trench 3650 feet long is to be cut, which for a total fall of 1 foot is to carry off a discharge of 12 cubic feet per second, what dimensions are to be given to the transverse profile, if it is to preserve a regular semi-hexagonal figure? Here m = 2,632 (see Table, § 366), hence, approximately, F = 0,0268 (2,632 . 3650.144)^ = 7,665 square feet,and c = -JiL = 1,539 feet. Hence £ is to be taken = 0,00758, and = ^0,00758.2,632 . ^ * s 7,67 square feet. There/ore the depth must be made: a a 0,760 \/ F ="^,104 feet, the lower breadth = 0,877 ^//^ = 2,428, and the upper breadth = 2.2,428 = 4,846 English feet. § 369. Variable Motion.—The theory of the variable motion of water in beds of rivers may be reduced to the theory of uniform mo-VARIABLE MOTION. 449 Fig. 499. tion, provided the resistance of friction for a short length of the river may be considered as constant, and the corresponding height, in like manner, as = f .ij? . 2L. But, besides this, regard must be had to the vis viva of the water, which corresponds to a change of velocity. Let ABCD, Fig. 499, be a short ex- tent of river, of the length AD = /, the fall DH = hy and let v0 be the velocity of the arriving, and vx that of the depart- ing water. If we apply the rules of efflux to an element D of the surface, we shall obtain for its velocity *l = h+V-l' 2g + 2g’ as regards an element E belowT the surface, it is true that on the one side it has a greater height of pressure AG = EH\ but as the down- stream water reacts with a pressure DEy there remains for it only the fall DH = EH—ED, as pressure inducing motion, and so, for this or any other element, the formula: A- •v, ® answers: and if, further, the resistance due to friction be added, we then obtain: i 2 /n + 2g , l_P ? ‘ F'2g* where py F and v are the mean values of the wetted perimeter, trans- verse section, and velocity. If F0 is the area of the upper, and Fx that of the lower section, we may then put: and Q= F0v0= F1 vv whence it follows that: H'_ 1 (Q 2g V* F 1. v+p,! 2gL\Fj l = (L + ±) i W^ F?) Fo + Fl we obtain: “ [^2“F*+ f fJ + JF?)] Q2 2g as also 2. Q = */2gh I i___L + f V F* F*+ F0 + F, ^ F,+ Fl The corresponding fall h may be calculated by means of the for- mula 1, from the quantity of water, the length and transverse section of a river or canal; and, inversely, the quantity of water from the fall, the length and the transverse section, by formula 2. To obtain greater accuracy, we may make the calculation for several short por- tions of the river, and take the arithmetical mean. If the total fall 38*450 VARIABLE MOTION. only is known, we must substitute this at once for h in the last for- mula, and put 1111 r; F 2 A n * o where Fn denotes the area of the last section, and in place of f. Ip \ (f.* + t;») the sum of ali similar values of the separate lengths of the river. Excmyk. A brook has for a distanoe of S00 feet a fall of 9,6 inches, the mean peri- meter of ita water profile is 40 feet, the area of the upper transverse profile 70, that of the lower 60 aquare feet; what quantity of water does this brook discharge ? It is 8,02 v/08 Q» ------------------------- !—■—- sj 60* 70* 7,173 + 0,00756* . 3 y3b— (-glj- + ^r) 7,173 130 : 354,43 cubic feet. The mean velo- ^/0,0000731 -f 0,0003365 v*),0004096 2 Q 708 8 lodty is ---———------- s 5,452 feet: hence, more accurately, £ must be taken ■ F9+ F 130 ’ ’ 7 s = 0,00745 in place of 0,007565, and therefore more nearly: ^ 7,173 Q = — ------= 357,5 cubic feet. If the same brook, with the same V^O,0000731 + 0,0003314 head of water, had for a length of 450 feet, a fall of 11 inches, and if its upper trans- verse profile had an area of 50 and its lower of 60 square feet, and the mean perimeter of the profile measured 36 feet, we should then ha ve: ________8,02 0,9167_________________ " /-i_______X^^45. 450 17 • 60* 50* ' 110 V 603 ^ 50*/ 8,02 r Wo,o 0,9167 110 \ 603 308 cubic feet. >,0001222 + 0,0007436 The mean of these two values is Q =s 3£>7.5 -|- 308 2 : 332,75 cubic feet. § 370. In order to obtain a formula for the depth of water, let the upper depth = a0 and the lower = av the slope of the bed = a, con- sequently the fall of the bed *= l sin. a. We then obtain the fall of the water h = an — ax + l sin. «, and there results the equation: F-)2g LF.+ F,\F‘^ " W0 a°~ar(l\2 hence l ■■ / 1 1 \ Q» fl° fl| \F2~F2) Tg ^6 F2) 2g sm. a.J /, J_ , _L\ Q2_ Sin. a F.+ F^F* " F*/2g The length l which corresponds to a difference a0—ax of the depth of water, rnay be determined by this formula. But if the reverse pro- blem m to be solved, we must do it by the method of approximation, and first determine the distances lx and l2 corresponding to the assumed depressions a0—al9 and ax—av and from these calculate by a propor- tion, the depression corresponding to a given distance /.* * See “ Ingenieur,” Arithmetik, § 16, v.FLOODS. 451 The formula is further capable of simplification when the breadth i of the running water is constant, or may be considered as such. In this case we put: F*) 2g F * F 2 -1 # •* i 2g and likewise: 2g F,) (F.+ F,) V F? 2g _ 9 (ao—fli) • f »0 2g 2 ——r • ir~> hence 2g’ , ^('-7 4) s. p a0b . ----sin. a 2g , and hence l ? . JL, . — sin. «. «06 %g_______ 1_A.5l ‘ % 2g The difference (aB—aj of the depth corresponding to a given extent l may be calculated directly by this formula. Example. In a horizontai trench, 5 feet broad and 800 feet long, it is desired to carry off a 20 cubic feet discharge, and to let it flow in at a depth of 2 feet, what depth will the water at the end of the canal ha ve ? Let us divide the whole length into two equal portions, and determine from the last formula the fall for each of them. 800 Here the sin. « = 0, / =-----= 400, and b = 5; for the first portion v = 20 —- = 2, hence { = 0,00752, also aQ = 2; sincep as 8}, it follows that «„ —at = 2.5 0,00752 . M . JL' 1—1. 2 —-—-1.400 = = 0,1692 feet. Now, for the second half, a, = 2? 2—0,1692= 1,8308, and j>,=8,2, t», = second portion: /0,00752 . . 2,184?.\ 1 9,154 2g 1 20 9,154 2,1848, and the depression of the 2,1848* 0 1997 . 400 = -J ^^9 = 0,2173 feet, hence the 1,8308 2g , whole depression = 0,1692 -f- 0,2173 = 0,3865, and the depth of water at the lower end = 2 — 0,3865 » 1,6135. § 371. Floods.—When the depth of water in rivers and canals varies, variations in the velocity and discharge take place likewise. A greater depth of water not only involves a greater section, but also a greater velocity, and hence, for two reasons, a greater quantity of water, and likewise a diminution of the depth of water, gives a dimi- nution of the section and the velocity, and hence also a decrease of the discharge. If the original depth a, and any increased depth452 FLOODS. =s av the upper breadth of the canal = 6, then the augmentation of the section may be put =s b (ax—a), and hence the section afterwards ^Fl = F+ b (ax—a), it also follows from this that Jy' approximately - 1 + If further p be the original, px the increased perimeter of the water profile, and 0 the angle of slope of the banks, then hance £l _ 1 + »(«.-"), ,„j y sin. 0 p _____ p sin. 0 i + v -, as also £ $tn. 0 \ Pi p $in. 0 Now the velocity with the first depth of water is £=s 92*35 I—, and with the second c. = 92,35 f?JL . A, NP* SPi l hence we may put: il_ jE fP - A + &K— «OW. c ~ S F ’ S Pi V /\ therefore the relative change of velocity: 1. c-~c On the other hand, the ratio of the discharge is: Q, FlCl / ^ 6 (a, — a)\ n . /. -n / * Fc 1 + (a, — a) -------------------1—V \2F p sin. 0/ and the relative increase: 2. _ k _«) /y_____________!_v Q ' \2F p sin. 0/ Less accurately, but in many cases, especially in broad canals with little slope, we may put F =» ab, and neglect -—1—_, whence it fol- , and p sin. 0 lows more simply that: ci — c c Qi— Q _ » a, — a . Q a From this, therefore, the relative change of velocity is and the relative change in the quantity of water that of the relative change in the depth of water. Examples.—1. When the head of water increases T*j of its original amount, the velo-GAUGES. 453 city is then and the quantity greater than its original value.—2.—Wlien the depth diminishes 8 per cent., the velocity then diminishes 4, and the quantity 12 per cent—3. From the more cor- Fig. 500. rect formula: Q>-Q Q , \ /3 *> = («, — a) (---- 3 \2F 1 .2 F p sin. ©/ a scale of the depth of water KM, Fig. 500, may be con- structed, on which the discharge of a canal corresponding to any depth KL, may be read off, when the quantity of water for a certain mean depth is once known. If b = 9 feet, bt = 3, a = 3, and © = 45°, we then have F = (9 + 3) 3 _ 2 hence: Q,— Q____/ 3.9 1 = 18 square ft., p = 3 -f- 2 . 3 y/2 = 11,485 and sin. © = = 0,707, ■(— \2 . 1 (a — a) = (0,750 — 0,123) (a, — a) = 0,627 18 11,485.0,707 / ' ' V ' (a, — a). If the quantity corresponding to a mean head of water Q = 40 cubic feet, we then have Q, = 40 4- 40.0,627 (a, — a) = 40 4- fll a. If a, — a = 0,04 T ’ v ' ^ 0,04 feet = 5,76 lines, it follows that Q, = 41; a] — a = 0,08 feet == 11,52 lines, we then have Q, = 42 cubic feet; if, further, o, — a = — 0,04, then is Q, = 39 cubic feet, &c. A scale, therefore, whose intervals are LM = LN = 5,76 lines, gives the discharge accurately to a cubic foot. Of course the accuracy is the less, the more the head of water differs from a mean value. Remark. The conducting and carrying off of water in canals, as well as the subject of weirs and dams, will be fully treated of in the Second Part. CHAPTER VIII. HYDROMETRY, OR THE DOCTRINE OF THE MEASUREMENT OF WATER. § 372. Gauges.—The quantity, which a stream discharges in a certain time, is determined either by a gauge, by an apparatus of efflux, or by an hydrometer. The most simple way of measuring water is by the guage, i. e. by the use of a graduated vessel, but this method is only applicable to small discharges, carried off by pipes or small brooks, or drains. The gauge vessel is generally made of wood, and of a rectangular form, and to increase its strength is bound round with iron-hooping. The water is conducted into it by a trough EF, Fig. 501, at whose ex- tremity there is a double valve GH, by which the water may be made to flow at will into the vessel ACy or by the side of it. To obtain the exact depth of the body of water in the vessel, a scale KL is fur-454 EFFLUX REGULATORS. ther applied. If before measureraent, the index Z be moved down to the surface of the water, already in the vessel, and merely covering the bottom, and the head of water read off frora the scale, we shall obtain the height ZZX of the gauged water by subtraction of this from the head of water which the scale indicates when the index hand Zx is brought into contact with the surface of water at the end of the observation. Before measurement, the valve must be so placed that the water may flow off outside the vessel. When we are convinced that the efflux in the trough is in a state of permanency, and, wratch in hand, have noted a certain moment, the valve must then be turned, so that the water may run into the gauge vessel, and after it is either partly or entirely filled, a second interval is noted by the watch, and the valve again brought into its first position. From the mean sec- tion F of the vessel, and the depth ZZX = a of the body of wrater, the whole quantity = Fa may be estimated, and again from the time of filling t, given by the difference of the times observed, the quantity of water per second Q = Rcmark. To determine a variable quantity of efflux at each period of the day, we may make use of the apparatus represented in Fig. 502, as applicable especially in salt works. There are here two gauge vessels, A and B, which alternately fili and empty themselves, and the water which is conducted by thepipe F passes through a short pipe CGy which is rigidly connected with a lever DE, re- volving about C. When one vessel A becomes filled, the water then flows through a short tube H into the little vessel Af, this draws the lever down again on one side, and the pipe CG comes into such a position that the water is conducted into B. The draw- ing up of the valves O and P takes place by means of strings passing over pulleys, whose extremities are con- nected with the lever, and sustained by iron balls, which impart a final impulse to the descent of the lever. The vessels Mand N have small efflux orifices,by which tliey empty themselves after each reversion of the lever. An apparatus is besides applied, by which the number of strokes may be read ofF at any time. § 373. Efflux Regulators.—Small and medium discharges are very frequently determined by means of their flow through a definite oiifice, and under a known head. From the area F of the orifice, the head of water h, and the efflux co-efficient /*, the discharge per second Q = p F v/ 2 gh is given. The Poncelet orifices are those best adapted for this purpose, because the co-efficients of efflux of these under different heads of water are known with great accuracy (§ 316), stili they are applicable only to certain medium discharges. The author availed himself of four such orifices for his measurements, one of five, one of ten, one of fifteen, one of twenty centimetres depth, but all of twenty centimetres width. These orifices are cut out of brass piate, Fig. 502.EFFLUX REGULATORS. 455 and fixed to a wooden frame AC, Fig. 503, which is fastened by four strong iron screws to each wall. In many cases, indeed, greater orifices, the co-efficients of efflux for which are not so accurately determined, and soraetimes wiers must be used, which admit gene- rally of a stili less accuracy. In ali cases, how- ever, the rule holds good, that w’e must endeavor to get as complete and perfect a contraction as possible, and for this reason must give to the orifice, if it is in a thick piate, a slope outward. The corrections which must be applied for incomplete and partial contraction, have been sufficiently distinguished in paragraphs 319, 320, &c. To mea- sure the water of a brook we must set the frame with its orifice, and wait for the moment when the head of water is permanent. For the measurement of the head of water we must avail ourselves of the index scale, Fig. 500, or of the movable scale EF, Fig. 505. If wTe would note the efflux directlyfrom the apertures of sluices, it isbetter to fix before hand a pair of brass scales BC and BE, Fig. 504, with Fig. 503. Fig. 504. Fig. 505. their indices F and G to the slide, and to the sluice-board A, in order to read off more safely the height of the aperture. It is generally better for the purpose of measuring wTater, to put on a new sluice- board with its guide, and with the requisite slope outwards. The simplest means of measuring water in a channel, consists in putting in a board CD, with its upper edge sloped off, Fig. 505, and measur- ing the fall produced by it. If the channel is long, and there is little rise, it is generally some time before the condition of permanency takes place, and it is for this reason good, before measurement, to put on a second board, so as to impede the efflux of w?ater for a long time, in order to accelerate the rise to a height corresponding to a state of per- manency. To measure the quantity of water of a brook? we may dam it up with posts and boards as in Fig. 506, and let the water C run off through an aperture, or we may avail our- selves of a simple overfall or wier, but of this wre shall treat in the second part. § 374. But as it is often long before a state of permanency occurs in water dammed up by this construction, we may adopt with Fig. 506.456 THE WATER-INCH. advantage the following method, first proposed by Prony. We may first close entirely the aperture by a sluice-board, and let the water rise to some height, or as high as circumstances will admit, then draw it so far up that more water may flow in than out, and measure the heads of water at equal and very short intervals; lastly, the aperture of the sluice must be again perfectly closed, and the time t in which the water rises to the first height, further noted. In each case, then, during the whole time of observation t+tlf as much water flows in as out, and hence the quantity flowing in, in the time t+tv may be ex- pressed by the quantity flowing out in the time tv If the heads of water during the depressions are A0, hv Aa, A3, and A4, we have then the mean velocity of efflux : v = + ^4) (&e § 351), and if the area of the aperture = Fy we have then the quantity of efflux in the time t: V= ——^ + A2 + 4 >/ h5+ y/ *J, and the quantity flowing in per second : + %^ho + hi + 2v'K + K+ Examplt. To measure the water of a brook used for the driving of a water-wheel, which has been dammed up by a sluice, Fig. 506, after opening the rectangular aper- ture, the following is observed: the original head of water is 2 feet, after 30" 1,8 feet, after 60" 1,55 feet, after 90" 1,3 feet, after 120" 1,15 feet, after 150" 1,05 feet, and after 180" 0,9 feet, breadth of the aperture 2 feet, depth J foot, time of rising to the first height with closed aperture = 110". The mean velocity of efflux is: 8,02 _ — ---------------- _ _____________ _________ _________ 9 8=8 (y/2 4- 4 v/l,8 +2^1,55+4 ^1.3+2 ^1,15+4^/1,05 + = 0,440(1,414 + 5,364+2,490 + 4,561+2,145 + 4,099 + 0,949) = 0,440.21,022 = 9,248 feet; but now ^ = 2 + = 1 square foot, hence it follows that the theoretical dis- charge is = 9,248 cubic feet. If the co efficient of efflux is taken = 0,61, we finally ob- tain the quantity of water sought: _ 0,61 . 180 9 248 = 3,5015 cubic feet, (Englisli.) ^ 180+ 110 § 375. The ici + «Ac2 + aAc3 + • • •» and, lastly, the mean velocity: C = — = a2^2C2 • • F a1b1 + a2b2 Example. In a tolerably straight and uniform extent of river, we have at the middle points of portions of the breadth : 5 feet, 12 feet 20 feet, 15 feet, 7 feet, The depths • - - . 3 « 6 “ 11 “ 8 “ 4 u The mean velocities - - 1,9 “ 2,3 “ 2,8 u 2,4 “ 2,1THE TACHOMETER. 461 Hence we may put: The area of the profile .F = 5 .3 + 12.6 + 20.11 + 15.84-7.4 = 455 square feet. The quantity of water Q = 15 . 1,9 + 72.2,3 + 220 . 2,8 + 120 . 2,4 + 28.2,1 = 1156,9 cubic feet The mean velocity c = f * = 2,54 feet. 455 § 378. The Tachometer.—The most eligible hydrometer is the tachometer of Woltmann, Fig. 516. It consists of a horizontal axle ABy with from two to five vanes F, placed at an inclination to the direction of the axis, and gives, when immersed in the water and held at right angles to the direction of motion, by the number of its revolutions in a certain time, the velocity of the running water. To read off the number of these revolutions, the axle has a few turns of a screw C, and these work into the teeth of a wheel D, upon whose lateral surfaces numbers are engraved, which give, by means of an index, the number of revolutions of the wheel. But to be able to register a great number of revolutions upon the axle of this toothed Fig. 516, wheel, there is a pinion which works into the teeth of the wheel Ey by which, like the hands of a watch, several multiple revolutions may be read off. If, for example, each of the two toothed wheels has fifty teeth, and the trundle ten, then the second wheel revolves one tooth whilst the first advances five teeth, or the vanes make five revo- lutions, if the index of the first wheel points to 27 = 25 + 2, and that of the second to 32, the corresponding number of revolutions of the vanes is accordingly: = 32.5 + 2 = 162. The entire instru- ment is screwed to a staff having a tin vane attached, to admit of easy 39*462 THE TACHOMETER. immersion in the water, and of being kept opposed to the current. But that the wheelwork may only revolve during the time of observa- tion, the axis is connected with a lever GO, which is pressed down by a spring, so that the teeth of the first wheel are thrown into gear with the screvv only when the lever is drawn up by a string. The number of revolutions of a wheel in a certain time, for exam- ple, in a second, is not exactly proportional to the velocity of the water, hence we cannot put v = au, where u is the number of revo- lutions, v the velocity, and a a number deduced from experiments; but rather: v = v0 -f o u, or more correctly t) = D0 + au + ^u2,.,, or stili more correctly: v = a u + >/v2 + P u2, where v0 is the velocity, at wdiich the wrater is no longer able to tum the wheel, and a and p are co-efficients from experiment. The constants v0, a and /3, are to be determined for each instrument in particular. With their assistance the velocity is known from a single observation, nevertheless it is always safer to make at least two, and to substitute the mean value as the correct one. Example. If for a sail-wheel v0 = 0,110 feet, « = 0,480, and £ = 0, therefore v •= 0,11 + 0»48 and we have by an observation with this instrument found the number of revolutions 210 in 80", then the corresponding velocity is; oin v = 0,11 + 0,48 . = 0,11 -f 1,26 = 1,37 feet. 80 Remark 1. The constants r0, a and 0 depend principally on the magnitude of the angle of impact, i.on the angle which the plane of the vane makes with the direction of motion of the water, and therefore, also, with the direction of the axis of the wheel. To observe with tolerable accuracy small velocities, it is well to have a large angle of impulse, i. e., one of 70°. For the rest, it is desira- ble to have vanes of different sizes and with different angles of impulse, and to use the vane with small angles of impulse for great velocities, and a smaller one for shallow water. Remark 2. To find the velocity of the surface of water, a small tin wheel may be used, as represented in Fig. 517, and its under part allowed to dip into the water. The number of its revolutions may be determined by a system of wheels, as in the tacliometer. § 379. To find the constant or co-efficient of the tachometer, it is necessary to set this instrument in a stream, wrhose velocity is known, and to note the corresponding number of revolutions. Although as many observations only are required, as there are constants, it is stili safer to have as many observations as possible, and especially for very different velocities, because w7e may then apply the method of least squares, and thereby eliminate the effect of accidental errors of ob- servation. The velocity of the water may be found by the floating ball, or by receiving the water in a gauge vessel, and dividing the measured discharge by the transverse section. In using floating balls, the air should be stili, and the tract of water straight and uniform. The tachometer is to be held at several places of the space described by the floating ball, and it is also requisite for accuracy, that the diameter of the ball should be equal to that of the tachometer. The second method of determination has several advantagcs when Fig. 517.THE TACHOMETER. 463 the water in which the instrument is immersed is received into a gauge vessel. For this purpose, and especially for adjusting the hydrometer, it is well if the engineer can erect a proper hydraulic observatory, consisting of a vessel of efflux, a gauge reservoir, and a channel of communication between the two. With such an arrange- ment, we may impart to the water any arbitrary velocity, because we can not only regulate the entrance into the channel, but also the motion by means of boards placed in at pleasure. During ob- servations we must keep the tachometer at different parts of the transverse section of the channel, measure the depth of this section by a scale, and, lastly, gauge the water running through in a definite time, in the lower reservoir (§ 372). We obtain the area F of the transverse profile by multiplication of the mean depth with the mean breadth, and the quantity of water Q is found from the mean trans- verse section G of the gauge measure, and the height (s) of the quan- Gs tity which has flowed in during the time by the formula Qs— ; but t O Gs the mean velocity of the water: v = — = _ follows from Q and F. F FI The corresponding number of revolutions u of the wheel is the mean of all the revolutions which are obtained when the instrument is immersed at different breadths and depths of the measured profile. If from a series of experiments we have found the mean velocities vv v29 v3> an^ *he corresponding number of revolutions, we then obtain by substitution in the formula v = v0 + <& w, or in the more correct one: v = au + s/v2 + (3u2 as many equations of condition for the constants v0, a, /3, as there have been observations made, and we may from these find the constants, if these equations are divided into as many groups as there are unknown constants, and these added together for as many equations of condition as are requisite for deter- mining v0, «, and also /3 when required. Remark. If we adopt the more simple formula with 2 constants,1 we may then, after the method of least squares, put: v __ 2 (y)9 3 (*) — * Qy) 2 (y) unA m __ I c*9) 2 (y) — i (*y) z (*) ° 2(^)2 (y3) - [2 (ay)T 2 M 2 (y9) — [2 (sy)p ’ wliere r _* and y = —, and the sign 2 represents the sum of all successive similar V V 11 values, for example, Z (^) = —I-h — H-----» % vs vt va v2 V% v% Example. For a small tachometer the velocities are: 0,163; 0,205; 0,298; O,3$0^ 0,610 metres, the number of revolutions per second: 0,600; 0,835; 1,805; 3,142 required to determine the constants corresponding to this wheel. given in the remark it follows, that: 0,600 1 (*) = _i— -f ----------f- ..= 18,740, 2 (y) = -^-5- + • 0,163 ' 0,205 ^ °>163 2 (^) = (-L .V 4. /_JL_V + . . = 82,846, I (y5) = 105,223, and ' \ 0,163/ ~ \ 0,205 / 2 = . . °’600___|- °’835____f. . . = 80,961, From the formula : 22,759 (0,163)a ‘ (0,205)*464 PITOT’S TUBE. = 105,223.18,740 80,961 . 22,759 _ 129,5 ^ and 0 ^ 82,846. 105,223 — (80,961)* 2162 a = — = 0,1703, hence for tliis instrument the formula v = 0,060 + 0,1703 u. 2162 If in this we put u = 0,6, we then obtain • v = 0,060 + 0,102 = 0,162; further, u = 0,835, v =s 0,060 -j- 0,142 = 0,202; further, u = 1,467, v == 0,060 -f 0,249 = 0,309, u = 1,805, v = 0,060 -j- 0,307 = 0,367; lastly u = 3,142, = 0,060 -f- 0,535 = 0,595; therefore, the calculated values agree very well with the observed. § 380. FitoVs Tube.—Other hydrometers are not so satisfactory as the tachometer, for they either admit of less accuracy, or they are more complicated in their use. The most simple instrument of this kind is PitoVs tube. In its simplest form it consists of a bent glass tube ABCy Fig. 518, which is held in the water in such a manner that its lower part stands hori- zontally, and is opposed to the water. By the percussion of the water, a column of water is sus- tained in this tube, which stands above the level of the exterior fluid surface, and the elevation DE of this column is greater, the greater the percus- sion or the velocity of the water generating it; this elevation or difference of level may hence serve inversely for a measure of the velocity of the water. Let this elevation DE above the external surface ot water = h, and the velocity = v> then h =»---------, where u is a J W __________________ number derived from experiment, and we have inversely, v = n y/ 2gh> or more simply: v = 4 y/ A. To find the constant 4, the instrument is immersed at a place in the water where the velocity v1 is known ; if the elevation is here = h.y we then have the constant 4 = . -1-,. which is ✓ A, to be applied in other cases, where the velocity is to be determined with this instrument. To facilitate the reading off of the height A, the in- strument consists of two tubes, as shown in Fig. 519, and from the one a small tube F is directed against the stream, from the other two small tubes G and Gx at right angles to the direction of the stream, both tubes are connected with a single cock if, by which the water can be retained in them. When the instrument is drawn out of the water, we may conveniently read off on a scale attached to both the tubes, the difference CD = k of the two columns of water. That the water in the tube may not oscillate much, it is necessary to make the exterior orifices of the tubes narrow, and that the closing of them may take place quickly and safely; the cock is Fig. 519. Fig. 518.HYDROMETRIC PENDULUM—RHEOMETER. 465 providet! with an arm and an even rod HK, which terminates above, near the handle of the instrument. § 381. Hydrometric Pendulum.—The hydrometric pendulum has been used in preference by Ximenes, Michelotti, Gerstner and Eytelwein for the measurement of the velocity of running water. This instrument consists of a quadrant ABy Fig. 520, divided into degrees and smaller parts, and a metallic or ivory ball K of from two to three inches diameter, sus- pended by a thread from the centre (7, the velocity of the water is given by the angle ACEy at which the thread when stretched by the ball deviates from the vertical, when the plane of the instru- ment is brought into the direction of the stream, and the ball submerged in the water. As the angle rarely amounts to forty or more degrees, this instrument has often the form of a right angled triangle given to it, and the divisions made on its horizontal arm. For the placing of the index or zero line in the vertical, it is best to use a spirit level on the horizontal arm of the instrument, or the ball itself may serve for this purpose, by letting it be suspended out of the water, and the instrument revolve until the thread coin- cides with the zero line of the division. For velocities under four feet we may use the ivory ball, but for greater velocities the hollow metal ball. On account of the constant undulations of the ball in the direction of the motion of the water, as also at right angles to the direction of the current, the reading off is somewhat difficult, and leaves a good deal of uncertainty, for which reason this instrument cannot be relied upon for the more exact numbers. The dependence between the angle of deviation and the velocity of the water may be determined in the following manner when the ball is not very deeply immersed. From the weight G of the ball and from the impulse of the water P = Fv2, increasing simulta- neously with the square of the velocity v and the section of the ball F, the resultant jR, whose direction the thread assumes, follows, and is determined by the angle of deviation /3, for which the tang. /3 = ~ = ** } hence also inversely : G G v2 = ~ ta7)g- P ? an(j v I . y/ tang. /3, i. e. v = 4 %/ tang. /3, n F \ pt if 4 represents a co-efficient derived from experiment, which must be obtained before use, according to the above-mentioned instructions. § 382. Rheometer.—The remaininghydrometers, such as Lorgna’s water lever, Ximenes’s water vane, Michelotti’s hydraulic balance, Briinning’s tachometer, PolettFs rheometer, are more complicated in their use, and not altogether to be relied on. The principle of ali these instruments is the same, they are composed of a surface of im- pulse and a balance, and the last serves for the purpose of giving the466 IMPULSE AND RESISTANCE OF VVATER. percussion P of the water against the former, but since this = ^ Fv*, we then have inversely: = J-^r = 4 %/ P, where 4 denotes a constant deduced from ex- \ t*F periment dependent on the magnitude of the surface of impact F. The rheometer, which was lately proposed by Poletti, and does not materially differ from the hydrometric balance of Michelotti, consists of *a lever AB, Fig. 521, turning about a fixed axis C, and an arm CD to which the surface of impulse, or according to Poletti, a mere impulse-staff is screwed. To maintain equilibrium with the percussion of the water against the surface, the boxes suspended at the extremity A of the lever are loaded with weight or shot, and to put the empty balance in equilibrium in stili water, a counterpoise is placed at B, which makes up the outermost end of the arm CB. From the weight put on G, the impulse P is found by means of the arm CA = a and CF = b from the formula Pb = Ga, whence, therefore, Fig. 521. P = ® G,and v = I — _ [ — = + x/ b yj where 4 is a constant derived from experiment. Remark. With respect to the last hydrometer, ample details will be found in Eytel- \vein's “ Handbuchder Mechanikfester Korper und der Hydraulik;” further, inGerstner’s Handbuch der Mechanik,” vol. 2; in Briinning’s uTreatise on the velocity of rutining water;” in Venturolfs “Elementi di Meccanica e d’Idraulica,” vol. 2. Concerning Poletti s hydrometer, we must refer to Dingler’s “Polytechn. Journal,” vol. 20, 1826. The hydrometer described in Stevenson’s treatise on Marine Surveying and Hydrometry is the tachometer of Woltmann, see Dingler s “Journal,” vol. 65, 1842. CHAPTER IX. ON THE IMPULSE AND RESISTANCE OF FLUIDS. § 383. Impulse and Resistance of Water.—Water or any otherfluid imparts a shock to a rigid body, when it meets it in such a manner that its condition of motion is thereby altered. The resistance which water opposes to the motion of a body, does not essentially differ from impulse. The investigation of these two forms the third principal division of hydraulics. We distinguish from each other: 1. The impulse of an isolated stream. 2. The impulse of a limited stream. 3. The impulse of an unlimited stream.IMPACT OF ISOLATED STREAMS. 467 Fig. 5QZ. An impulse of the first kind takes place when a body, for instance, the float board of an over-shot water-wheel, is opposed to a stream of water issuing from a reservoir ; an impulse of the second kind occurs where water, in a canal or in a water-course, impinges against a body whioh entirely filis up its transverse section, as for instance, against the float board of an under-shot wheel; the third kind, lastly, presents itself, when running water strikes against a body immersed in it, whose transverse section is only a very small part of that of the cur- rent of water, as, for instance, against the float boards of a floating mill-wheel. We must distinguish the impulse of water against a body at rest and against a body in motion, and further, the impulse against a curved and against a plane surface, and in this last again, between the perpendicular and the oblique impulse. Let us consider at once the general case, namely, the impulse of an isolated stream against a surface of rotation which moves in its proper axis, and in the direction of motion of the stream. § 384. Impact of Isolated Streams.—Let BAC> Fig. 522, be a sur- face of rotation, AX its axis, and FA a fluid stream meeting it in this direction. Let the velocity of the water = c, that of the surface = v, and the angle BTXy which the tangent DT at the extremity B of the generating curve or of each of the filaments of water BT leaving the surface, includes with the direction of the axis BE = a; lastly, let us further assume that the water in running off from the surface loses nothing in vis viva by friction. The water strikes against the surface with the relative velocity c—v% and hence leaves the sur- face with this, and therefore quits it in the tangential directions TB, TC, &c. From the tangential velocity BD v c —1>, and the velocity of the axis BE =* v, the absolute velocity BG =* cx of the water after impinging against the surface is found by the known formula: C% «= y/ (C—Vf + 2(c—v) V COS. a -f V2. But now a quantity of water Q is able to produce by virtue of its vis viva the mechanical effect — . Q y, if its velocity c is fully impart- ed; accordingly the residuary effect of the water: =s -i- . Qy; consequeatly the mechanical effect distributed over the ‘Ig surface is: n.. ^ ~ O.,468 IMPULSE AGAINST PLANE SURFACES. 2 c v—2 v2—2 (c—v) v a Qy, Qy, i. e. and the force or the impulse of the water in the direction of its axis is: P = (1—COS. <*) Qy. £ If the surface raeets the water with the velocity v> we then have: P =* (1—COS. a) . (Ci'V) Qy, < s and if this is without motion, therefore, v = 0, the impulse or hydrau- lic pressure of the axis comes out: P = (1—COS. o) -. Qy. £ It follows from this, that the impulse of one and the same mass of water under otherwise similar circumstances is proportional to the rela- tive velocity c+vofthe water. From the area Fof the transverse section of the fluid stream, it follows that the quantity discharged is Q = Fc; hence P = (1—COS. a) (c+v^c. S and for v = 0: P = (1—COS. o) — Fy. S . For an equal transverse section of the stream, the impulse against a surface at rest increases therefore as the square of the velocity of the water. § 385. Impulse against Plane Surfaces.—The impulse of one and the same fluid stream depends principally on the angle a, under which the water, after the impulse, leaves the axis; it is nothing if this angle == 0; and, on the other hand, a maximum, namely, = 2 Q y, if this angle is 180°, therefore its cosine = — 1, g where the water, as represented in Fig. 523, leaves the surface in a Fig. 523. Fig. 524. direction opposite (o that in which it impinges. This is generally greater for concave surfaces than for convex, because the angle is thereMAXIMUM EFFECT OF IMPULSE. 469 oblique, therefore the cosine negative and 1 — cos. a becomes 1 + cos. a. Most frequently the surface, as represented in Fig. 524, is plane, and hence a = 90°, therefore cos. a = 0, and the impulse P = . Q y; for a surface at rest: & p = -iQr = ^y = 2. fFv = 2 Fhy. g 9g . 2g The normal impulse of water against a plane surface is therefore equivalent to the weight of a column of water which has for base the transverse section F of the stream, and for altitude, twice the height • c^ due to the velocity 2 h = 2 . — H The experiments made on this subject by Michelotti, Vince, Langs- dorf, Bossut, Morosi, and Bidone, have nearly led to the same results when the transverse section of the impinged surface was at least six times as great as that of the stream, and when this surface was twice as far from the plane of the orifice as the thickness of the stream. The apparatus which was used for this purpose consisted of a lever, similar to that of Poletti’s rheometer, which received upon one side the impulse of the water, and whilst its other side was kept in equilibrium by weights. The instrument which Bidone made use of is represented in Fig. 525. BC is the surface impinged on by the stream FA> G is the scale-pan for the reception of the weights, D the axis of rota- tion, KL counter-weights.* § 386. Maximum Effect of Impulse.—The mechanical effect of impulse: Pv = ( 1 — COS. a) {c—vyoQ y depends principally on the velocity v of the impinged surface; it is, * The latest and most extensive experiments on the percussion of water are those of Bidone. See “ Memorie de la Reale Accademia delle Scienze di Torino,” vol. 40, 1838. They were performed with a velocity of at least 27 feet, and on brass plates of from 2 to 9 inches diameter. In general, Bidone found that the normal impulse against a plane surface was somewhat greater than 2 F hy, yet this variation is perhaps to be attributed to an augmentation of the leverage which is produced by the falling back of the water. See Uuchemin’s “ Recherches experim. sur les lois de la resistance des fluides.” When the impinged surface was quite near the orifice, Bidone found that P was only 1,5 F h yj when, further, the surface had a transverse section equal to that of the stream, in which case the water only deviated by an acute angle a, then, after Du Buat and Langsdorf, P was only = F h y. Lastly, it has been deduced by Bidone and others that the impulse is in the first moment nearly as great again as the permanent impulse. 40470 IMPULSE OF A LIMITED STREAM. for example, nothing, not only for v = c, but also for v = 0; hence there is a velocity for which the effect of the impulse is a maximum. It is manifest that it only depends on (c—v) v becoming a maximum. If we consider c as half the perimeter of a rectangle, and v as its base, we have then its height = c—v and its area = (c—v) v. But of ali rectangles the square is that which has for a given perimeter 2 c the greatest area, hence also (c—v) v is a maximum, wrhen c—v = v, i. e. v = _, and we therefore obtain the maximum value of the me- 2 ehanical effect of the impulse tohen the surface moves from it with half the velocity of the water, and indeed Pv = (1—COS. a) . J . --- . Qy = (1—COS. a) . £ Qhy. If nowa =s 180°, and if, therefore, the motion of the water be reversed by the impulse, we then have the effect equal to 2 . ^ Qhy = Qhy. But if a = 90°, i. e. if it impinges against a plane surface, this effect is then only \ Qhy, therefore, in the last case, the half only of the whole disposable effect, or that which corresponds to the vis viva of the water, is gained or brought to bear upon the surface. Examples.—1. If a stream of water, of 4Q square inches transverse section, delivers a quantity of 5 cubic feet per second, and strikes normally against a plane surface, and escapes with a ] 2 feet velocity, the effect of impulse is then: P= (c — v'> Qy— (l• 144 _ 12^ . 0,031 . 5.62.5 = 6.0,031.312,5 = 58,12 lbs., g \ 40 / and the mechanical effect brought to bear upon the surface Pv = 58,12 X 12 == 697,44 ft. lbs. The greatest effect is for v = — = h • ** * = 9 feet, and indeed : 2 40 == i . £. . Q y = £ . 18* .0,0155.5.62,5 = 81 . 0,0155.62,5 = 784,68 fl. lbs.; the corresponding impulse, or hydraulic pressure = -- 87,18 lbs.—2. If a stream Fd, Fig. 526, of 64 square inches section, strikes with a 40 feet velocity against an im- movable cone, having an angle of convergence BAC = 100°, then is the hydraulic pres- sure in the direction of the stream: P= (1 — co*.») -L Q y = (1 — cot.50°) 40.0,031 . _ . 40.62,5 g 144 = (1—0,64279) . 1 . 24 . = 0,35721 . 1377,7 = 492 . 13 lbs. Fig. 526. Fig. 527. § 387. Impulse of a Limited Stream.—If we add borders BD, CE, to the perimeter of a plane surface BE, Fig. 527, which project from the side impinged upon by the w^ater, then will the wrater deviate from its direction at an obtuse angle, in a similar manner as fromOBLIQUE IMPULSE. 471 concave surfaces, and hence the impulse will be greater than for plane surfaces. The effect of this impulse depends principally on the height of the border and the ratio of the transverse section between the stream and the part confined. In an experiment, where the stream was 1 inch thick, the cylindrical enclosure 3 inches wide and 3^ lines deep, the water ran off almost in a reversed direction, and the impulse amounted to 3,93 — Fy; in every other case this force was less. In consequence of the friction of the water at the surface and the sides, c2 the theoretical maximum value never reaches 4 — F y. In the impulse of a limited stream FAB, Fig. 528, a rising at the edges takes place; this rising occu- pies only a portion of the perimeter, and extends itself, on the other hand, simultaneously to the impinged surface and the fluid stream. The impinging water takes the direction of the un- bordered portion of the perimeter, and here, therefore, becomes deflected 90 degrees, whence the formula above found for the isolated stream P = Q y holds good; yet this a may also be deduced in the following manner. If we assume that the velocity c of the arriving water by the impulse against its surface is changed into the velocity v of the surface, we may then also assume (c—r)* * Fig. 528. that a loss of mechanical effect 2g Q y (similar to that in § 337), expended in the division of the water, is connected with it. But now c2 the effect due to the vis viva of the arriving water = — Q y and to that 2g v2 of the water going on = — Q y, hence it follows that the mechanical 2g effect imparted to the surface is: Pv = [«*__(«_„)•_**] 1 Qr = Q 2g. § 388. Oblique Impulse.—In oblique impulse against a plane sur- face, we must distinguish whether the water flows away in one, two, or in all directions in the plane. If, as in the impact of limited water, the surface AB, Fig. 529, is confined at three sides, so that the water can run off only in one direction, we have then the hydraulic pressure against the surface in the direction of the stream P = (Teos. a) g * This formula will be found applicable hereafter, when we come to the theory of water-wheels.472 OBLIQUE IMPULSE. Fig. 529. Fig. 530. But if the impinged plane BC> Fig. 530, is only bordered on two oppositely situated sides, the stream then divides itself into two un- equal portions; the greater portion QA takes the small deflexion a, and the lesser Q2, the greater deflexion 180 — a; hence, the whole impulse in the direction of the stream is: P = (1 — COS. a) . C.—~. Qj y + (1 + cos. o) . -------------- Q2 y = g g (~Y~)7 [(1 _ cos‘o) (1 + cos'o) Q’]' Now the equilibrium of the two portions of the stream requires that the pressures (C~.P) y (1 — COS. a) Q, and (c ~ ”)y (1 + COS. a) Q2 g g Q, between them should be equal; hence, also: (1 — cos. a) Qj = (1 + cos. a) Q2, or since Qt + Q2 = (1 — COS. a) Qj = (1 + COS. a) (Q — QJ, i. C.y Q, = Q, and Q,- Q, so that the whole impulse in the direction of the stream is finally: p = (±=?L) y . 2 (1 - COS. a) (L± = (*-*)* (1 - COS. a2) Q, Q y- i. e. c — v . 2 P = ---------- sin. a g Besides the parallel impulse P, acting in the direction of the stream, we distinguish, further, the lateral impulse Sy acting at right angi es to the direction of the stream, and the normal impulse JYy composed of these two, and at right angles to the surface. In every case P = JV sin. a, and S = N cos. a; hence, inversely, c —v sin. a Qy and S sin. Qy- sin. a %g " ~ 2g The normal impulse, therefore, increases as the sine, the parallel impulse as the square of the sine of the angle of incidence, and the late- ral impulse as double the same angle. Lastly, if the inclined surface impinged on is not bordered, then the water can spread over it in all directions; the impulse is then greater, because of all the angles by which the filaments of water are deflected, a is the least; and hence, each filament which does not move in the normal plane, exerts aACTION OF AN UNLIMITED STREAM. 473 greater pressure than the filament in this plane. Let us assume tha a portion Q1 corresponding to the sectors AOB and DO E, Fig. 531, is deflected by the angles a and 180° — a, and another Q2, correspond- ing to the sectors A OD and B OE, by 90°, and that both portions exert a parallel impulse, we may then put: P = C— Q1 v sin. a2 + £=? Q2y, , sin. a2 g S = Q2, and Q, -f- Q2= Q; hence it follows, that Q,(i + sin. a2) = Q, and the whole parallel impulse P = c—2Qysin. a2___ 2 sin. a2 c—v n g / 1 + sin. a2 1 + sin. a2 g 7 Although this hypothesis is only approxi- mately correct, it tolerably well agrees, never- theless, with the latest experiments of Bidone. § 389. Aetion of an Unlimited Stream.—If a body moves progres- sively in an unlimited fluid, or if a body is put into a fluid which is in motion, it then suffers a pressure which is dependent on the form and dimensions of this body, as well as on the density and on the velocity of the one or the other mass, and in the one case is called the resistance, and in the other the impulse of the fluid. This hydraulic pressure arises principally frorn the inertia of the water, whose con- dition of motion is altered bystriking against the solid body, and also, further, from the force of cohesion of the particles of water, which are hereby partially separated from one another, or pushed aside. If a body AC moves against running water, Fig. 532, it pushes away Fig. 532. Fig. 533. before it a certain quantity with an augmented pressure. Whilst this mass of water, by the further advance of the body, always increases on the one side, on the other a constant flowing away takes place, while the particles lying near the anterior surface assume a motion in the direction of this surface. If the moving mass of water strikes against a body at rest, Fig. 533, then is there likewise an increased pressure produced in front of it, which causes the particles before the body to deviate from their original direction, and to run off at the surface AB. When these particles have reached the limits of the surface, they then turn and flow away by the lateral surfaces until they come to the back, when they then again immediately unite, but 40* Fig. 531.474 THEORY OF IMPULSE AND RESISTANCE. assume an eddying motion. It is manifest that the general circum- stances of motion of the particles surrounding the body are the same in the impact of moving water as in the resistance of a body moving in water, except that in the eddies a difference so far takes place, that with short bodies the eddy in the latter case occupies a less space than in the former. In both cases the velocity of the particles in- creases more and more from the middle of the anterior surface to its limits, attains its maximum at the commencement of the lateral sur- faces, where, for the most part, a contraction takes place, gradually diminishes in the tfater which passes away at the sides, and lastly, attains its minimum when the water reaches the back and passes into a whirling motion. § 390. Theory of Impulse and Resistance.—The normal pressure raiies at different points of the body; it is greatest at the middle of the anterior, and least at the middle of the posterior surface, and, next to that, at the parts of the sides nearest this; because, in respect to the body, there is at the one place rather a flow to, and at the other a flow from these surfaces. If the body be symmetrical, as we shall suppose it to be, with respect to the direction of motion, then the ag- gregate pressures in this direction counteract each other, and hence only the pressures in the direction of motion are to be taken into ac- count. But now the pressures on the posterior surface are opposed to those on the anterior; hence the resultant impulse or resistance of the water may be equated to the difference of pressure of the anterior and posterior surfaces. If we cannot assign the amount of these pressures a priori, we may, nevertheless, from the great similarity of the circumstances to the im- pulse of isolated streams, assume that at least the general law for the impulse of unlimited water does not differ from that of the impulse of isolated streams. If, therefore, F is the area of a surface, which is impinged on by an unlimited current whose density is y, with a velo- city t>, then the corresponding impulse or hydraulic pressure may be m2 put ?■ f — Py, where C represents a number deduced from expe- 2!gr riment, dependent on the form of the surface. But this expression is nal only applicable to action against the anterior, but also to that agmmt the posterior surface, only that in this last, when the water has a tendency to flow away, it consists of a draught or negative pres- sure. If now Fh y is the hydrostatic pressure (§ 276) against the front and back surface of a body, the whole pressure against the front is: Pj ax Fhy + . JL Py, and that against the back : P,** Fhy . 2s ?2 • 2^. ^and resultant impulse or resistance of the water is then found: p=pi—p2 “ (WJ . |! Fy - f. £ Fy, if f, + f, - f. This general formula for the impulse of unlimited water is applicable to the percussion of the wind or to the resistance of the air. Besides theIMPULSE AND RESISTANCE AGAINST SURFACES. 475 difference of aerodynamic pressure at the front and back, there is fur- ther a difference of aerostatic pressure, because the air in front, in consequence of its greater elasticity, has a greater density (y) than that at the back. For this reason, in high velocities, as those of cannon- balls, the co-efficient of the resistance of air is greater than that of water. Remark.—The adhesion of a certain quantity of air or water to the body, is a peculiar phenomenon of the impulse or resistance of an unlimited medium (water or air),whose influence is particularly remarkable in the variable motion of bodies, as, for example, in the oscillations of the pendulum. For a ball, the air or water adhering to the moving body is equal to 0,6 of the volume of the ball. For a prismatic body moved in the direc- tion of its axis, the ratio of this volume = 0,13 -f- 0,705 where l is the length, and F the transverse section of the body. These relations, discovered by Du Buat, have been fully confirmed by the later observations of Bessel, Sabine, and Baily. § 391. Impulse and Resistance against Surfaces.—The co-efficient of resistance f, or the number with which the height due to the velo- city is to be multiplied to obtain the height of a column of water mea- suring this hydraulic pressure, varies for bodies of different figures, and onJy for plates which are at right angles to the direction of mo- tion is it nearly a definite quantity. According to the experiments of Du Buat and those of Thibault, we may put f = 1,85 for the im- pulse of air or water against a plane surface at rest*and, on the other hand, assume, but with less accuracy, for the resistance of air or water against a surface in motion f = 1,40. In both cases, about two-thirds of the whole effect are expended on the front, and one-third on the back. The resistance which the air opposes to a surface revolving in a circle, has been found by Borda, Hutton, and Thibault to vary a good deal, but may be expressed by a mean of ? bb 1,5. If the surface does not stand at right angles to the direc- tion of the motion, but makes with it an acute angle », we may . 2 f sin . then, with Duchemin, substitute for f, ——■ with tolerable cor- 1 + sin. a2 rectness. The impulse and resistance of unlimited media are also augmented when the surfaces are hollowed out or have projecting edges at their perimeters, but we have arrived at no general results on this subject. Example. If the wind impinges with a 20 feet velocity against a firmly fixed wind«milJ wheel, which consists of four wings, of which each has an area of 200 square feet and 75° inclination to the direction of the wind, then is the impinging force of the wind in its direction, or in that of the axis of the wheel: P — 1,85 . -2 (**"• 75)* . H2!. 4.200.0,081 s 1,85.0,965 . 6,81 . 800.0,081 1 + (rin. 75)’ 2g = 718,4 ft, lbs., when the density of the wind is (from § 301) taken at 0,081 lbs. Remark. Views, with respect to the impulse and resistance of unlimited fluids, entirely at variance with these, are put jorward in the above-mentioned worlc of Duchemin. It is there maintained, for instance, that the impulse and resistance against the front sur- face of a thin piate amounts to 2 . — F A, and is not negative at the back, that the 2g impulse = 0,136 — Fy, and the resistance = 0,746 — Fy. It would be too circum- is 2g ^ . stantial here to give a detail of the reasons why the author cannot agree with the views of Duchemin, but more with reference to this will be found in Poncelet’s u Introduction A la m^canique industrielle,” 2d edition, 1841.476 IMPULSE AND RESISTANCE TO BODIES. § 392. Impulse and Resistance to Bodies.—The impulse and resist- ance of water to prismatic bodies, whose axis coincides with the direction of motion, diminishes when the length of the body is con- siderable. From the experiments of Du Buat and Duchemin, the impulse of the front surface is invariable, and only the effect against the back surface variable. To this corresponds the co-efncient = 1,186, for the total effect, however, with the relative lengths l f =* 1,86; 1,47; 1,35; 1,33. - For stili greater ratios between the length / and the mean breadth s/ Foi the body f diminishes, owing to the friction of the water at the lateral surfaces of the body. From the resistance of the water, reverse relations take place. Here, from Du Buat, for the effect on the front surface, fx = 1 invariably; for the total effect, however, with f®l,25; 1,28; 1,31; 1,33, so that, for a prism which is 3 times as long as broad, the impulse is the same as the resistance. * The experiments undertaken by Borda, Hutton, Vince, Desaguil- liers, Newton, and others, with angular and with round bodies, leave stili much uncertainty. In what relates to spheres, it appears that for moderate velocities the mean co-efficient for motion in air or water = 0,6. For a greater velocity and for motion in air, accord- ing to Robins and Hutton, for the velocities n— 1, 5, 25, 100, 200, 300, 400, 500, 600 metr. ? — 0,59; 0,63; 0,67; 0,71; 0,77; 0,88; 0,99; 1,04; 1,10. Duchemin and Piobert have given particular formulae for the rate of increase of these co-efficients. For the impulse of water against a sphere, Eytelwein found S — 0,7886.* jExample. If, according to Borda, we put the resistance and impact at right angi es to the axis of a cylinder at half as great as that against a parallelopiped which has the same dimensions, we then obtain for the resistance { = £ . 1,28 = 0,64 and the impact ** i • 1,47 == 0,735. If we apply these values to the human body, whose section has an area of some 7 square feet, we then find for the resistance and impulse of air against it, the values: P = 0,64.0,0155.7.0,081 v* s 0,00562 ©9, and P 7= 0,735.0,0155.7.0,081 v* = 0,00646 v2. Hence the resistance of air for a velocity of 5 feet is only 0,00562.25 = 0,1405 lbs.; and the corresponding mechanical effect per second = 5.0,1405 = 0,70 ft. lbs.; for a velocity of 10 feet this resistance is four times, and the effect expended eight times as great, and for a velocity of 15 feet, the resistance is 9 times and the effect 27 times as great as for a 5 feet velocity. If a man, with a 5 feet velocity, moves against wind having a 50 feet velocity, he has then a. res2*tairicJf .55* 8 19,54 lbs. to overcome, corresponding to the relative velo- city 50 -f- 5 s 55 foet, and thereby to produce the m<*»-h«.ni«w.l effect of 19,54.5 = 97,7 ft. lbs. (English.) * P°nc©let, in his work above cited, and Duchemin and Thibaultin their “ Recherches experimentales,” have treated very fully of these circumstances. In the Second Part we shall treat of the resistance to floating bodies, especially to ships, &c., as also the im- pact of the wind on wheels. &c.MOTION IN RESISTING MEDIA. 477 § 393. Motion in Resisting Media.—The laws of the motion of a body in a resisting medium are rather complex, because we have here to deal with a variable force, i. eone increasing with the square of the velocity. From the force Px which urges the body forward, and 2 from the resistance P2 = ? . — Fy, which the medium opposes to the motion, the motive force is: P=P1-Pi=P1-?.gfV, Gr but since the mass of the body =* M = —, the accelerating force is: __r v* ]?y p s£F*yM-(1 ? ),y> -i-0 or if we represent JjL- by—. 2gP, «r2 _J g. But the velocity v is accelerated in the instant of time r by * = p hence: x = J g t, and inversely: G, [WQ] g Now to find the time corresponding to a given change of velocity, let us divide the difference vn—vo9 of the final and initial velocity into n parts, let any such part ---— = *, and let us calculate the n velocities: J>1 * *\> + *, v2 wm v0 + 2 *, r3 * v0 + 3 *, &c., and substitute these values in the formula of Simpson. In this man- ner, by taking four parts we shall obtain the time sought 1.1 = 9.. »«—V 1 , 4 , pi i2s 'i— + 1-t©, + 1-f(S)'+w#’ Further, the small space described in any instant *(§ 19), is o = v*, therefore, P _*L. By the application of Simpson’s rule, we or since *=-, e P VX a 478 % MOTION IN RESI8TING MEDIA. shall now find the space which is described while the velocity v passes into that of vn. 2u G T v„—v, 12 + 2u„ e ^-f©’ + wQ)'+ w -{Gr) >-'©).■ 1—r Of course the accuracy is greater, when we take six, eight, or moreparts. This formula takes into account the variability of the co-emcients of resistance, which in considerable velocities is neces- sary. For the free descent of bodies in air or water Pl =* G, and for motion on a horizonta! plane Pl = 0, is more correctly equal to the friction/ G. Since this is a resistance, we have then to introduce it as negative into the calcuiation, whence P - — (P, + P,), and p — [l + f (^)2] Pu As it cannot be a question here of an increase, but on tion of velocity, we have then to substitute in the above formula Vq— Vn for Vn— VQ. In the case, where the body is urged by a force, by its weight for instance, the motion apptoximates more and more to a uniform one, so that after the lapse of a certain time, it may be considered as such, although not so in reality. The accelerating force p = 0, when (2fP1 dg• y of a diminu- jFy = P.* when, therefore, v J- SjCFr The velocity of a falling body approximates, therefore, to this limit more and more, without ever actually attaining it. Example. Piobert, Morin, and Didion found, for a parachute' whose depth was 0,31 that of the diameter of its opening ( = 1,94 . 1,37 as 2,66. Hence, from what heiglit in Prussian feet will a man, of 150 lbs. weight, be able to descend with a similar para- chute, of 10 lbs. weight and 60 square feet transverse section, without acquiring a greater velocity than that which he would have acquired by jumping from a 10 feet height, without a parachute'? The last velocity is v aa 7,906 y/ 10 = 25 feet, the force is Px s G = 150-f- 10 ss 160 lbs., the surface F = 60 square feet, the density y a 0,0859, and the co-efficient of resistance f = 2,66, hence : — — 60 • °>085£ _ 0,000515, and { . — = 2,66 . 0,000515 . 25’ = 0,85625. If, 1 1 u? 69*5. 160 w therefore,‘ we take 6 parts, we then obtain for these: 1 — C • ■ 0,97621; 0,90486 ; 0,78593; 0,61944; 0,40537; 0,14375, and for 0; 4,268j 9,210; 15,905; 26,910; 51,393, and 173,913: from Simpson^ rule the mean value is: *= (1 • °4*4.4,268-f2.9,210 4-4.15,905+2.26,910-f4.51,393-fl. 173,913)-:- 3. 6 : 29,58 ; and from this the space of descent sought: 532,42 18 vn — ro times the mean value of . 25— 0 1—f. — 31,25 . 29,58 =a 23,6 feet.PROJECTILES. 479 The corresponding time of descent is. since the mean value of_* . = (1.0+ 4.1,024 + 2. 1,105+4 . 1,272 + 2 . 1,614 + 4,2,467 + 1 . 6,957) - = 1,747, t 25 18 31.25 . 1,747 == 1,4 sec. Remark. For a constant co-efficient of resistance, the higlier calculus gives us: v = (—-T 1 \ I 2g -J-— and s = — L n u- /_ V e^+ 1/\ ‘ \ 4efxt -)• I P P v where /t* = / 2 g £ - *, *->e being the base of the hyperbolic system of powers, and Ln the hyperbolic logarithm. § 394. Projectiles.—We have already investigated the motion of projectiles in vacuo (§ 38), and found this motion to be parabolic; we may now obtain a more exact knowledge of motion in a resisting medium, and consider that, for instance, of a shot. In F,s*534- no case is the path AGNy Fig. 534, of a body passing through the air a symmetric curve; the portion GN in which the body descends is rather shorter, and, therefore, less inclined than the portion AG in which the body as- cends, because the resist- ance of the air operating in the direction of motion tends always to shorten the por- tions of its path ACy CEy EGy &c., more and more; if, therefore, the first portion of the path ACy for motion in the air is only a little shorter than it would be in vacuo, the last portion LN is considerably shorter in the first motion than it is in the last. The construction of the path in a resisting medium by means of circles of curvature may be accom- plised in the following manner. From the initial velocity vv and the angle of elevation BAN = 0l it follows that the z ABC =90 — al9 and sin. ABC = cos. aiy from § 40 the radius of curvature O.A = Ofi = r.= —L__, g COS. Oj hence with this we may approximately describe the portion of arc AC. If now we assume the angle subtended at the centre AOx C = therefore AC = s1 = rt we then obtain for the succeeding particle of space CE the angle of inclination a°2 = a°j — $°i« Let further, the height of fall BC = hiy and the measure of the retardation due to the air’s resistance Z . — F* beingr 2 g5 1 V G % \ °v\ i* Ao2 480 PROJECTILES. ?' ^ S “ * V*' therefore f • 2% “ * from the principle of mm inW, we then obtain for the velocity v% at the initial point of the second portion of arc: v% V1 l /vi*+v*\ M . \V2 (i—*?,) vf—igK v% i =(1 — j»Sj) -i---A„ and hence t>, ■* I ^ i Since now the height of fall ht «« \g*% ” > it !+#*, follows that: (1—rt)®/- (f)‘ I/- coj. a,* If we substitute these values of a, and t>2 in the equation: V 2 -l—, we then obtain the radius of curvature OaC — 0,E of __ ------, „V —VM VV11MM «H« V‘ gcos.*i the sacceedmg portion of arc CEf and if we assume an angle of re- Yolution CO% E » *, it again follows from this that the angle of inclination in the vicmity oi point : s a2 *2, and the velocity at this V 1—P*t fi- cos. 1+#»»■* t. It u therefore easy to see how the entire path of the projectile may be successively composed of circular ares. Ertimplt, A casfciion ball, of 4 iodiei diameter, is shot off at an angle of elevation of 50° with a vekMtty of 1000 foet, requiied its path, if only approximately, according to Prussian weights and measnrea. The radias of curvature of the iirst portion of arc = —— =s —tOOOOOO — 49783 ft. As the density of the air = 0,0859, and g cos. a 31,25 cor. 50° that of cast-iron = 470 lbs^ we have then (a . £; now for v = 1000, { n 0,90, hence /u = •then obtain the velocity at the end of it: Fy 3.3.0,0859 =0,00041122 2 O ’ 4.470 0,0003701. If we take an arc of 1° only, we 1000 £ —0,0003701.49783.0,017453—(0,017453 wt. 50°)a 1+0,0003701.49783.0,017453 sssss 7697 feet. and the radius of curvature for a second portion of arc : (769,7)9 r-3f,k^49° = 28897 ^ For v9 sss 769,7 feet, £ = 0,81, therefore fx = 0,0003331. If, therefore, we describe with the last radius, an arc f2s2°, the velocity at its ending point will be v% « 769,7 / 1 — 0.33598 — 0,002831 __ 54^47 feet >f 1,33598 For a third arc Qs, the radius of curvature rs» 13757 feet, and if, therefore, we assume { a* 0,75, we shall then obtain at the end of a length of arc of 4°, the velocity v4 s= 398,85 feet. The radius of curvature fbr a fourth arc may be likewise found t4 =3 6960,5 by assuming £ = 0,72, and we shall then obtain the velocity vs = 288,85 feet, at the end of an arc of 8°, from which a fiith radius of curvature r8 a 3259 feet may be calculated. Proceeding in this manner, we shall obtain, by degrees, the collective ele- ments for the construction of the line of projection in question.INDEX A. Abrupt widening, 402 Absolute magnitude, 57 -----resistance, 183 Acceleration and combination of velocity, 41 -----, normal, 47 Aetion and re-action, 57 ■----of an unlimited stream, 480 Adhesion, 57 Aeriform bodies, 58 Aerodynamics, 58 Aerostatics, 58 Aggregation, state of, 56 Air, density of the, 346 -----, efflux of stili, 428 -----in motion, efflux of, 431 -----, strata of, 345 Angle of friction and the cone of friction, the, 148 Angular tubes, 396 Areometer, 335 Attwood’s machine, 241 Axes, free, 254 Axis, equilibrium of forces about an, 112 -----, pressure on the, 111 Axle and wheel, 238 -----friction, 156 -----, pointed, 165 -----, the thickness of, 213 -----, wheel of an, 142 B. Balance, hydrostatic, 334 Ballistic pendulum, 298 Beams, hollow, 196 -----—i hollow and elliptical, 205 -----, rectangular, 193 -----, strongest, 203 -----, the strongest form of, 210 Bodies, aeri form, 58 -----, centre of gravity of, 98 -----, dynamies of solid, 58 -----, flexure of, 188 -----, floating, 458 -----, free descent of, 30 | 41 Bodies, impact of free, 279 ------, mechanics of solid, 58 ----------------------- fluid, 58 ------, perfect elastic, 279 ------, rotary, 295 ------, staties of aeriform, 58 ------, staties of solid, 58 ------, strongest form of, 185 Breaking twist, 223 Buoyancy, 322 — ■■ ' ■ centre of, 332 C. Catenary, 135 Centre of buoyancy, 332 ------of gravity, 88 ------------------- of bodies, 98 ------------------ of curved surfaces, 97 ------------------- of plane figures, 92 -------------------of lines, 90 ------of parallel of forces, 83 ------of percussion, 299 ------of pressure, 311 Centrifiigal force, 245, 246 ------------------of extended masses, 248 Centripetal force, 246 Chains, rigidity of, 178 Circular lateral orifices, 360 ------pendulum, 265 Co-efficient of efflux, 367, 434 ---------------------- through rectangular ori* fices, Tables I. and II. from Poncelet-and Lesbros, 370, 371 ■ ■■ ■ —■- for circular orifices, table of corrections of the, 378 ------of contraction, 365 Co-efficient of velocities, 364 ----------of discharge by \viers, 373 ----------of resistance, 384 . ■■ ■ table of efflux and velocity, 390 ----------of friction, 147, 447 __________of friction, Table of the, 394 ----------. of the friction of repose and mo- tion, Table of, 151, 152 ------------- of resistance of curvature in tubes, 398 Cobesive force, 57482 INDEX. Columns, 218 Combination of velocity, 42 Communication, vessels of, 410 Components, 64 Compositiori and resolution of velocities, 41 ------------of forces, 64 ----------------------in a plane, 79 Compound motion, 37 ■ --------pendulum, 273 ----------vessels, 411 Compression, rupture by, 214 ____________________under, 216 Cone of friction, tbe, 148 Conical tubes, 389 Constrained paths, 259 Cords, friction of, 170 -----rigidity of, 179 Contraction, effect of imperfect, 408 ——, co-efficient of, 365 -------, imperfect, 377, 387 -------, maximum and minimum of, 374 --------of the fluid vein, 366 -------, partial, 375 Corrections for the Poncelet wiers, Table of, 381 ■ ■ ■ for wiers over the entire side, or without any lateral contraction, 381 --------of the co-efficients of efflux for circular and rectangular orifices, 378 Coulomb’s experiments, 180 Couples, 82 Crushing, Table of the modulus of resist- ance to, 215 Curve, elastic, 190 Curved motions in general, 46 -------tubes, 397 Curvilinear motion, 74 Cycloid, 270 Cycloidal pendulum, 271 Cylinders, 197 ■ and prisms, 232 Cylindrical tubes, 383 D. Deau, jets, 399 Declivity of water, 438 Densi ty, 54 -------and pressure, velocity of efflux, 353 -------of the air, 347 Depth of floatation, 325 Descent, free, of bodies, 30 Determination of the centre of gravity, 89 Different velocities in the transverse section, 438 Discharging vessels in motion, 362 Division of forces, 56 ......of mechanics, 57 Docks, floating, 330 Dynamical stability, 120 Dynamics of aeriform bodies, 58 ----------of fluids, 58, 350 ----------of rigid bodies, 225 ----------of solid bodies, 58 E. Edges, points and knife, 166 Effect of imperfect contraction, 404 Efflux, 350, 425 ------, co*efficient of, 367, 434 ------of air in motion, 431 ------of stili air, 428 ------of water in motion, 379 ----------------through tubes, 382 ------, regulators of, 454 ------under decreasing pressure, 432 ------, velocity of, 351 ------, velocity of pressure and density, 353 Elastic body, perfect and imperfect, 279 ------curve, 190 ------impact, 281, 288 ------pendulum, 276 Elasticity and rigidity, 182 ------and strength, 183 ------, modulus of, 183 ------, or spring force, 57 Elliptica! beams, hollow and, 205 English, Frencli, and German measures and weights, comparative tables of, xv. Equality of forces, 51 Equilibrium, kinds of, 110 ------of bodies rigidly connected, 109 ------of forces about an axis, 112 ------in funicular machines, 127 ------of water in vessels, 309 ----------------with other bodies, 32 2 ------and pressure of air, 338 Excentric impact, 302 Experiments on beams, 200 ------on friction, 149 ------of Rennie, 150 Eytelwein, 201 F. Fall of water, 438 Flexure of bodies, 188 ------, reduction of the moment of, 194 Floatation, depth of, 325 ------oblique, 332 Floating bodies, 458 ------docks, 330 Floods, 451 Flow through tubes, 436 Fluidity, 304 Fluid surface, the, 306 ------vein, contraction of, 364 Force, 50 ------about an axis, equilibrium of, 112 ------, centre of parallel of, 83 ------, centripetal andcentrifugal of extended masses, 246 ------, cohesive, 57 ------, compositiori of, 64 ------, direction of a, 57 ------, di vision of a, 51, 57 ------, equality of, 51 ------, in a plane, 67INDEX. 483 Fore e, in a plane, composition of, 79 -----, in space, 69, 83 -----, living, 62 -----, magnetic, 57 -----, measure of, 53 -----, muscular, 57 -----, normal, 245 -----, of inertia, 57 -----, of heat, 57 -----, parallel of, 81 ■ ---, parallelogram of, 65 -----, resolution of, 66 -----, simple constant, 58 Formulae of stability, 118 Free axes, 254 -----descent of bodies, Table of, 30, 31 French, English, and German measures and weights, comparative tables of, xv. Friction, axle, 156 ------------, Table of co efficients of, (from Morin,) 156 -------, co-efficient of, 147,447 -------, experiments on, 149 -------, kinds of, 146 ■ -----, laws of, 147 -------, of cords, 170 -------, of motion, 152 ■ -----, of repose, Table of the co-efficients of the, 151 •------, pivot, 164 -------, resistance of, 391 -------, rolling, 168 -------, the angle and cone of, 148 Funicular machines, 127 -------, polygon, 130 G. Gases, tension of, 338 Gay-Lussac's law, 346 Geodynamics, 58 Geomechanics, 58 Geostatics, 58 German, English, and French measures and weights, comparative tables of, xv. Gerstner, 201 Graphical representation, 34 Gravity, specific, 55, 334 Gravity, 57 -------, action of, along constrained paths, 259 -------, centre of, 88 -------, determination of the centre of, 89 ---■---of lines, centre of, 90 -------of plane figures, centre of, 92 -------of curved surfaces, centre of, 97 Gauges, 453 Guldinus’s properties, 107 H. Horizontal and vertical pressure, 319 Hydraulic pressure, 355 Hydrostatics, 58 Hydrodynamics, 58 -------balance, 334 Hydrometers, 335 Hydrometry, 453 Hydrometric sail wheel, 461 Hydrometrical pendulum, 464 I. Impact, doctrine of, 228 -------of isolated streams, 467 -------, elastic, 281 -------, excentric, 302 -------, ine Iastie, 279, 288 -------, in general, 278 -------, imperfectly elastic, 290 -------, oblique, 291 Imperfectly elastic impact, 290 Imperfect contraction, 371, 387 ----------------------, effect of, 404 Impulse, maximum effect of, 469 -------and resistance, theory of, 466, 474 -------oblique, 471 -------of a limited stream, 470 ---------and resistance against surfaces, 475 ----------------------to bodies, 576 ----------------------of water, 466 Inclined plane, 152, 259 -------, theory of the, 122 Inelastic impact, 279, 288 Inert masses, reduction of, 228 Inertia, 52 -------, force of, 57 -------, momentof, 227 -------, radius of, 230 -------, reduction of the moment of, 229 Influx, velocity of, 351 -------and etflux, 425 Intensity of a force, 57 Irregular vessels, 424 J. Jets d’Eau, 399 K. Kinds of support, 109 -------of equilibrium, 110 -------of friction, 146 -------of motion, 225 Knife, edges of, 166 Knots, 127 L. Lateral pressure, 310 Laws of friction, 147 ---of Mariotte, 341 ---of Gay-Lussac, 346 ---of staties of rigid bodies, 76 Hardness, 287 Hollow beams, 196 ■--------and elliptical beams, 205484 INDEX* Lever, mathematica], 113 -------, equilibrating, 332 -------, material or physical, 113 -------, theory of the equilibrium of the, 113, 163 Living forces, 62 Liquids of different densities, 337 Loading beyond the middle, 208 Locks, 420 Long tubes, 394 M. Machines, Attwood’s, 241 -------, funicular, 127 Magnetic force, 57 Manometer, 339 Mariotte’s law, 341 Mass, 53 ---, reduction of inert, 228 Masses, centrifugal forces of extended, 248 Mathematical pendulum, 266 -------lever, 113 Material pendulum, 266 Matter, 51 Maximum and minimum of contraction, 374 ----------effect of impulse, 469 Mean velocity, 440 Mea sure of forces, 53 Mechanics, 50 -------fundamental laws of, 50 -------division of, 57 -------of sol id bodies, 58 -------of fluid bodies, 58 -------of air, 58 -------of a material point, 58 Mechanical effect, 60 -------, transmission of, 73 Media, motion in resisting, 477 Metacentre, 328, 332 Modulus of elasticity, 183 _______of working load and strength, 184 _______of elasticity and strength, 186 -------of relative strength, 201 Moment of inertia, rotation and mass, 225, 227 Morin’s experiments, 151 Motion in resisting media, 477 -----, accelerated, 27 -----and rest, 25 -----, compound, 37 -----, efflux of air in, 431 -----, in general, curved, 46 -----, kind of, 25, 26, 225 -----of water, permanent, 439 -----, of rotation, 226 *----, mean velocity of a variable, 35 -----, parabolic, 43 -----, rectilinear, 225 -----, rolling, 263 -----, simple, 25 -----, uniform, 26, 445 -----, uniformly variable, 27 -----, variable, in particular, 33 Motion, variable, in water, 448 -----, curvilinear, 74 Mouth pieces, 382 Muscular force, 57 N. Nodes, 127 Normal acceleration, 47 --------force, 245 Notches in a side, 418 Numerical values, 187 O. Obelisk shaped vessels, spherical and, 422 Oblique impact, 291 --------pressure, 207 --------lioatation, 332 --------additional tubes, 386 --------impulse, 471 Orifice, triangular lateral, 359 --------rectangular lateral, 368 --------circular lateral, 360 Oscillation, timeof, 266 P. Parallelogram of the velocities, 38 -■ ■■ of accelerations, 42 — -■ of forces, 65 Parallelopiped of velocities, 41, 119 --------, rectangle of, 231 Parallel forces, 81 ----------, centre of, 83 Parabola, 43 Parabolic motion, 43 Particular cases of impact, 2S2 Partial contraction, 375 Pendulum, hydrometrical, 465 Pendulum, circular, 266 -------, ballistic, 298 -------, elastic, 276 Percussion, centre of, 299 Permanent motion of water, 439 Perimeter, 438 Permanency, 439 Phoronomy, 25 Physical lever, 113 Piezometers, 413 Pile driving, 285 Pipes, thickness of, 320 Pitofs tube, 464 Pivot friction, 164 Plane of rupture, 209 -----, composition of forces in a, 79 -----, forces in a, 67 -----, inclined, 153, 259 -----, theory of the inclined, 122 Pneumatics, 58 Points and knife edges, 1G6 Pointed axles, 165 Polygon, funicular, 130 Poncelet and Lesbros, two tables of co-effi*INDEX, 485 cients of efflux, through rectangular ori* fices, 370, 371 Pontoon, a, 102 Pouce cTEau, 456 Pressure, 51 --------, horizontal and vertical, 319 *--------- in a definite direction, 315 --------on the axis, 111 —-------of bodies on one another, 115 --------of oblique, 207 --------of efflux under decreasing, 432 --------on curved surfaces, 317 Pressures, principies of equality of, 305 --------and density, velocities of efflux, 353 --------, centre of, 311 --------of water in vessels, 309 --------, hydraulic, 355 --------, lateral, 310 ■■ on the bottom, 308 Principle of equality of pressures, 305 —— , ■ of the vis viva, 62 --------of Virtual velocities, 85, 124 Prismatic vessels, 415 Prisms and cylinders, 232 Profile in river, 438 Projectiles, 479 Properties of Guldinus, 107 Pulley, the, 140 Pyramidal-shaped vessels, wedge, &c., 420 R. Radius of gyration or inertia, 230 Rectangle and parallelopiped, 231 Rectangular beams, 193 --------lateral orifices, 368 Rectilinear mofion, 225 Reduction of the moment of flexure, 194 --------of inert masses, ■' ' of the moment of inertia, 229 Regulators of efflux, 454 Relative strength, 183, 199 Rennie’s experiments, 151 Rest and motion, 25 Resistance and impulse against surfaces, 475 -— to compression, 183 --------to torsion, 183 —, impulse and, to bodies, 476 ---------, co-efficient of. 384 — of friction, 391 ■-------—---------and rigidity, 145 - of water, 466 ---------, theory of impulse, &c., 474 Resolution of velocities, 41 --------- of forces, 66 Restoring power of floating docks, 331 Resultant, 64 Rheometer, 465 Rigid bodies, statics of, 76 Rigidity of chains, 178 --------and friction, 145 --------of cords, 179 Rod, the moment of inertia of, 230 • Rolling motion, 263 ----------friction, 168 ■ - ■ ■ and dragging friction, 169 Rotary bodies, 295 Running water, 438 Rupture, plane of, 209 --------by compression, 214 ■ under compression, 216 S. Sail wheel, hydrometric, 461 Section, transverse, different velocities in the, 438 Segments, 236 Short tubes, 382 Simple constant force, 58 Slope of water, 438 Space, forces in, 69, 84 Specific gravity, 55, 333 Sphere and cone, 333 Spherical and obelisk-shaped vessels, 422 Spring force, 57 State of aggregation, 56 Stability, 116, 328 -------, formulae of, 118 -------, dynamical, 120 -------of floating docks, 331 Statical moment, 78 Statics of solid bodies, 58 -----of rigid bodies, 75 -----of fluid, 58, 309 -----of aeriform bodies, 58 Strata of air, 345 Stream, action of an unlimited, 473 Streams, impact and isolated, 467 -------, impulse of a limited, 470 Strength and elasticity, 183 ■ ■ - - ■ modulus of working load and, 184 -------the moduli of elasticity and table, 187. -------relative, 199 -------modulus of relative, 201 Strongest form of body, 185 --------beams, 203 --------form, beams of the, 210 Support of kinds, 109 . T. Table of co-efficients of the friction of re- pose, 151 -----comparative, of English, French and German measures and weights, 15 _________________________of motion, 152 _____of co-efficients of axle friction from Morin, 156 _____I., the moduli of elasticity and strength, 187 ----- II., the modulus of strength for the flexure of bodies, 201 -----of the modulus of resistance to crush- ing, 215486 INDEX. Table of co-efficients of efflux through rect- angular orifices, 370 -----of co-efficients of efflux for wiers, 372, 373 ----- of the co-efficients of resistance for trap valves, 410 -----of the co-efficients of friction, 394 -----of the oo-efficients of the resistance of curvature in tubes, 398 -----of the co-efficients of resistance for the passage of water through a cock in a rectangular tube, 406 -------------in a cylindrical tube, 406 --------------------- through throttle-valves in rectangular and cylindrical tubes, 407 -----showing the relations of the motion to the time in the free descent of bodies, 31 ----- of corrections of the co-efficients of efflux for circular and rectangular orifices, 378 --------for the Poncelet wiers, Twist, breaking, 223 U. Uniform motion, 26, 445 Uniformly accelerated motion, 28, 42S ---------variable motion, 27 Unit of weight, 52 y. Valves, 408 --------, table of the co-efficients of resistance of traps, 410 Variable motions in particular, 33 --------motion, mean velocityof a, 35, 448 Velocities, co-efficient of, 364 ■, combination of, 42 438 381 ■-------------------for wiers over the entire side, or without any laterai contraction, 381 -----of correction for imperfect contraction by efflux through short cylindrical tubes, 388 -----of the co-efficients of efflux, 390 Tachometer, 461 Tension of gases, 338 Theory of the inclined plane, 122 --------of impulse and resistance, 474 --------------------wedge, 125 Thickness of axles, the, 213 --------of pipes, 320 Time of oscillation, 266 Toggle joint, note on, 129 Torsion, 219 Traction, 51 Transference of the point of application, 76 Transmission of mechanieal effect, 73 Transverse section, different velocities in the, 438 —-------------------the best form of, 441 Tredgold, 213 Triangular laterai orifice, 359 Tricardo, 129 Trigonometric expression, 41 Tubes, angular, 396 -----, conical, 389 -----, curved, 397 -----, cylindrical, 383 -----, flow through, 436 -----, long, 394 -----, oblique additional, 386 -----, of Pitot, 464 -----, short, 382 -----, table of co-efficients of the resistance of curvature in, 398 -, composition and resolution of, 41 in the transverse section different, -, mean, 440 —, parallelogram of the, 38 —, parallelopipedon of, 41 -, principle of Virtual, 85, 124 - of etllux and influx, 351, 352 -, of efflux, pressure and density, 353 -, Virtual, 85 Vertical and horizontal pressure, 319 Vessels, compound, 411 -------in motion, discharging, 362 -------, irregular, 424 of communication, 416 -, prismatic, 415 -, spherical and obelisk-shaped, 422 -, wedge and pyramidal-shaped, 428 Virtual velocities, 85 Vis viva, 62 # W. Water-inch, 456 -----, flow of, through wiers, 372, 373 -----in motion, efflux of, 379 -----, permanent motion of, 439 -----running, 438 -----, slope of, 438 Wedge and pyramidal-shaped vessels, 420 ---------, the, 154 -------, theory of the, 125 Wheel, and axle, the, 238 -----, hydrometric sail, 461 -----and axle, the, 142 -----carriages, note on, 174 Widening, abrupt, 401 Wiers, 372 —, table of the co-efficients of efflux, for 372, 373 —, table of corrections for the Poncelet, 381 —, over the entire side, table of correc- tions for, 381 END OF VOL. I.ERRATA. Page 30, line 5 from bottom, for 32,22 read 32,2. “ 31 “ 6 for 15,625 read 16,1, and for 250 read 257,6. “ “ “ 8 for 0,016 read 0,0155. “ “ “ 10 from bottom, for 241 read 241£. “ 32 “ 9 from bottom, for 0,480 read 0,465, for 0,320 read 0,310, and for 0,160 read 0,155. t: “ “11 from bottom, for 0,032 read 0,031, and for 0,480 read 0,465. “ 38 “ 9 for I5f feet read 15J feet Prussian measure. “ “ “23 from top, for ct tt, ca tt read c, t, c9 t. “ 55 “ 5 for 205 read —, for 0,4083 read 0,4219, also for 0,4083 read J 502 485,8 0,4219, and for 705,54 read 729,04. “ 73 “ 4 from top, for Fig. 40 read Fig. 41, and in line 5, for Fig. 41 read Fig. 40. “ 75 “ 3 for (h — hf) M read (h — hx) G. “ “ “ 6 for mass read weight. “ 100 “ 17 from bottom, for x3y2 zt read x2 y3zv “ 153 “ 9 from bottom, for GOK read GOQ. “ 163 “ 14 from bottom, for -f Q cos. 0 read -f- * Q cos. 0. “ 184 “ 10 from bottom, for 1 read l, and for \ E read A E. u 189 “ 17 from top, for zt . FtS z, read zt . F Szv “ 257 “ 13 from top, for cos. a read cotg. a. “ 264 “ 2 from top, for G read g. “ 331 “ 6 from top, for shoared read shored. “ 394 “ 15 from bottom, for 11^| 1 | 2 | read 11| 11£| 2 |. “ 414 “ from bottom, for 01,274 read 0,1274. “ 422 “ 15 from bottom, for h = rt, = read h = r,t == . “ 426 “ 12 from bottom, for 63,29 read 63,89.