THE LIBRARY 
 
 OF 
 
 THE UNIVERSITY 
 OF CALIFORNIA 
 
 LOS ANGELES 
 
 GIFT OF 
 
 U. of California 
 Berkeley 
 
/#i4 /, /?/> ~ J ***- a z -"- ?' / / / '^ r 
 
 /*! 
 
A TREATISE ON 
 
 HYDROSTATICS 
 
 VOL. I 
 
 CONTAINING THE MORE ELEMENTARY 
 PART OF THE SUBJECT 
 
 BY 
 
 GEORGE M. MINCHIN, M.A., F.R.S, 
 
 LATE PROFESSOR OF APPLIED MATHEMATICS 
 IN THE KOYAL INDIAN ENGINEERING COLLEGE, COOPER'S HILL 
 
 SECOND EDITION, REVISED 
 
 OXFORD 
 
 AT THE CLARENDON PRESS 
 1912 
 
HENRY FROWDE, M.A. 
 
 PUBLISHER TO THE UNIVERSITY OF OXFORD 
 
 LONDON, EDINBURGH, NEW YORK, TORONTO 
 MELBOURNE AND BOMBAY 
 
Library 
 
 Q/\ 
 
 VI 
 PREFACE TO VOL. I 
 
 THE present edition of this work is divided into two 
 volumes, the first of which covers the course of hydro- 
 statics required of students who compete for scholarships 
 at the Universities. The book has been, in great part, 
 re- written, and the examples have been very largely 
 increased in number. 
 
 Very much of this subject of hydrostatics is easily 
 and profitably treated without the use of the differential 
 and integral calculus not that the calculus is evaded by 
 artifices more difficult than the principles of the calculus 
 itself. For example, nearly all the practically useful 
 work relating to centres of pressure, and much of that 
 relating to floating bodies, is more easily treated by 
 simple geometry and algebra than by the calculus. 
 
 Hence the first volume contains very little of the 
 
 1 differential and integral calculus. The fundamental 
 
 principles of certain forms of turbine have been intro- 
 
 duced, as they involve no mathematical difficulties and 
 
 are of great practical importance. 
 
 In the revision of proof-sheets I have had the benefit 
 of the advice of so able and competent a mathematical 
 physicist as Mr. Pidduck of Queen's College. 
 
 GEORGE M. MINCHIN. 
 
 o 
 
 OXFORD, 
 
 September, 1912. 
 
 u- 
 
 6- 
 
CHAPTER I 
 
 PAGE 
 
 NATUBE or FLUID PRESSURE ..... i 
 
 CHAPTER II 
 THEOREM OF PLANE-MOMENTS . . . . 19 
 
 CHAPTER III 
 LIQUID PRESSURE ON PLANE SURFACES . . .26 
 
 CHAPTER IV 
 
 PRESSURE ON CURVED SURFACES : PRINCIPLE OF 
 
 BUOYANCY . . . . . . . .67 
 
 CHAPTER V 
 GASES . . ...... 102 
 
 CHAPTER VI* 
 HYDRAULIC AND PNEUMATIC MACHINES . . .129 
 
 CHAPTER 
 
 STEADY MOTION UNDER THE ACTION OF GRAVITY . 151 
 INDEX . 197 
 
CHAPTER I 
 NATURE OF FLUID PRESSURE 
 
 1. Experimental Illustration of Pressure. Let a vessel 
 of any shape be fitted with a number of weightless pistons 
 of different areas moving in cylindrical tubes without any 
 friction, and let this vessel be filled with a liquid suppose 
 water or mercury. We shall suppose also that the piston 
 fittings are perfectly liquid-tight, so that no liquid can 
 escape through the piston tubes. 
 
 Fig. i. 
 
 Then especially if the vessel has considerable height 
 and the liquid is mercury we shall observe that, for the 
 equilibrium of the liquid, each piston requires to be pressed 
 in with a particular force the magnitude of which depends 
 on two things: (i) the area of the piston, and (2) the position 
 of the piston in the vessel. 
 
 The forces which urge the pistons out are due, of course, 
 
 1424 B 
 
2 Hydrostatics VOL. i 
 
 to the weight of the liquid. In the figure A is the highest 
 point of the liquid, and the pistons are fitted at B, C, ... . 
 Let P, Q, R, . . . be the forces which have to be applied to 
 keep the pistons B, C, D, . . . at rest, and let the areas of 
 these pistons be a lt 3 , 3 , ... ; then forces per unit area 
 exerted by the liquid on the pistons are 
 
 These are called the intensities of pressure at 7?, C, D, ... ; 
 and in the ideal case which we have supposed (absence of 
 friction, &c.) we should find that the intensity of pressure 
 at a point is greater the greater the vertical depth of the 
 point below the highest point, A; and also that the 
 intensity of pressure is the same at any two points whose 
 vertical depths below A are the same. For example, if 
 
 p 
 B and C are in the same horizontal line, : - will be 
 
 OR fll 
 
 found to be equal to , and will be greater than 
 
 P 2 ?3 
 
 . We shall now vary the experiment. Having supplied 
 
 a 
 
 to each piston the proper force from without so that there 
 is equilibrium everywhere, let us fit a piston at A, its weight 
 being also negligible. This piston requires, of course, no 
 force to keep it in position. Let s be the area of this 
 piston. Now to any one of the pistons suppose A let an 
 additional force, F, be applied to move it in ; then we shall 
 find that each of the other pistons will require a special 
 additional force to keep it in position, and we shall find the 
 following simple result : if the intensity of the pressure, 
 
 F 
 
 -, applied at A is denoted by/, the additional force required 
 
 to keep B in position is/x a l ; the additional force required 
 at D is/x c 3 , and so on ; that is, the intensity of pressure 
 
CH. i Nature of Fluid Pressure 3 
 
 applied at A is transmitted without loss to every other 
 point, ,C,D,G,.... 
 
 Precisely the same result is obtained if we produce the 
 additional intensity of pressure, f, by additional force 
 applied at the piston 1), or any other piston : this intensity 
 is transmitted, undiminished, by the liquid to every point 
 of the vessel. This result is known as Pascal's Principle, 
 which may be thus enunciated : if intensity of pressure of 
 any amount is applied at any point of the surface of a liquid, the 
 liquid acts as a perfect machine for transmitting this intensity 
 of pressure, unaltered in amount, to all points of the liquid. 
 
 The reason why we began by applying special forces, 
 P, Q, H, ... to the several pistons in order to keep them in 
 position, before applying the intensity of pressure f } is that 
 we desired to eliminate the effect of gravity and to deal 
 with a liquid throughout which no bodily force (such as 
 gravity) acts ; and this method enables us to do so as 
 regards points on the surface of the containing vessel ; but 
 since the size and shape of the vessel may be altered at 
 pleasure and the result still holds, we can conclude that 
 Pascal's Principle holds for points inside the liquid as well 
 as for points at its surface ; but a complete proof of this 
 will be given presently. 
 
 If the vessel were filled with a very light gas, such as 
 hydrogen, and it were possible to keep it in without leakage, 
 it would not be necessary to begin by equilibrating the 
 pistons. 
 
 Intensity of pressure is & force per unit area. Thus if 
 the area of the piston A is -25 square inches, and the force 
 ^applied to the piston is 10 pounds' weight, the intensity 
 
 of pressure, f, is , or 40, pounds' weight per square inch. 
 25 
 
 2. Perfect Fluid. A body such that, whatever forces 
 are applied to it, there is no friction between any two 
 
 B a 
 
4 Hydrostatics VOL. i 
 
 particles in contact is called a perfect fluid. The action 
 between particles consists wholly of force acting perpen- 
 dicularly to their surface of contact. Such bodies are, of 
 course, only abstractions, which for many practical purposes, 
 however, may be assumed to exist : water, mercury, gases, 
 and many liquids may be taken as such. 
 
 It is usual to divide fluids into liquids and gases, and to 
 define liquids as fluids that cannot be compressed, or that 
 can be compressed only by applying very great pressure to 
 them ; gases are fluids that can be easily compressed. 
 
 3. Stress in bodies. If a solid body of any kind is 
 acted upon by any forces, the interior of the body will be 
 strained. Suppose Fig. 2 to represent such a body ; and at 
 
 Fig. 2. 
 
 any point inside it consider the particles at both sides of 
 a very small plane run. Owing to the forces acting upon 
 the body of which its weight will be one the particles 
 at the side A of the plane mn exert forces on those at 
 the side B which are in contact with them. These forces 
 may be of the nature either of pressure or of temion 
 according to the way in which the forces P, Q, ... are applied 
 to the body ; and the total force exerted on the particles at 
 the B side will be a force which we have represented by f 
 (acting towards the B side). As represented, this force is, 
 on the whole, a pressure, but it has a component along the 
 
CH. I Nature of Fluid Pressure 5 
 
 plane mn, which is called a shearing component. The parti- 
 cles at the A side of mn experience from those at the B side 
 exactly the same force /re versed. This force, or stress, has, 
 then, two aspects : it acts in one direction or in the opposite 
 according- as we consider the action of the A side on the B 
 side of mn, or vice versa. 
 
 If the body which we are considering is a perfect fluid, 
 this stress f will always be normal to the plane mn, whatever 
 position this plane may have in the fluid. This we shall 
 assume as the definition of & perfect fluid. 
 
 But in the case of any other kind of body, a great deal 
 depends on the position of the plane mn. We may imagine 
 it turned round its centre so as to occupy various positions ; 
 and then, as a rule, with each position both the magnitude 
 and the direction of the stress f will vary : in some positions 
 of mn the stress/ 1 may lie in the plane, having- no normal 
 component. The stress in this case would be pure shear. 
 Again, for some positions of mn the normal component of 
 f may be pressure, and for others tension. In the case of a 
 perfect fluid/ is always normal pressure. 
 
 From the fact that fluid stress on a small plane surface 
 acts perpendicularly to the surface we can prove a most 
 important property of this stress namely, that its intensity 
 is the same for all small planes at the same point, P, in the fluid. 
 
 Let P, Fig. 3, be any point in the fluid (whether liquid 
 or gas) ; let cdfb be a small plane 
 at P, and cdea be the same plane 
 turned through any angle, 2 0, about 
 the line cd ; complete the prism by 
 the triangular faces, abc, efcl, per- 
 pendicular to the edge cd. This Fig- 3- 
 prism contains a small volume of 
 
 the fluid which we can separate in imagination from the 
 rest of the fluid, and we can, regard it as kept in equilibrium 
 
6 Hydrostatics VOL. i 
 
 by the stresses produced by the surrounding fluid on its faces 
 and by its weight, if gravity is the external force acting on the 
 fluid. 
 
 Now, whatever be this external .force, its magnitude is 
 proportional to the volume of the prism. Let the total 
 stress on the face cdbf have a magnitude q : it will act 
 perpendicularly to the face at its middle point. Let the 
 stress on cdea be a force r, and that on abfe a force s. 
 There will also be stresses acting on the faces abc and efd, 
 and each of these acts parallel to cd. 
 
 Let Pmn be the cross-section of the prism through 
 P, which point we can take as the middle point of cd. Let 
 cd = /, ab = zc, cb = ca = a ; then the areas cdfb and cdea are 
 each I. a, and the volume of the prism is area abcxl, or 
 c . a cos Q x /. The external force (weight, or other) acting 
 on the prism can be represented by 
 kxc .a .Icosd ; 
 
 and if p and // are the intensities of the stresses on the faces 
 cdfb and cdea, we have 
 
 q=pxl.a, r=p'xl.a. 
 
 Now express the fact that the forces acting on the prism 
 have no component parallel to mn, and suppose the external 
 force to make any angle, $, with mn \ then we have 
 
 p' x /. a . cosQp x I . a . cos + k x c . a . /cos cos </> = o. 
 or // p + kxc cos = o. 
 
 If now we diminish the size of the prism indefinitely, 
 c vanishes while all the other quantities remain finite, so 
 that the third term, depending on the external force, 
 vanishes, and we have simply 
 
 / = P, 
 
 whatever 6 and < may be ; that is, the intensity of pressure 
 on the plane cdfb is constant whatever be the position of 
 
CH. i Nature of Fluid Pressure 7 
 
 the plane at P; and observe that this equality of the 
 pressure intensities depends on the fact that the stress 
 always acts perpendicularly to the plane : if q and ; did not 
 act normally to the planes on which they act for all 
 positions of these planes, the result would not follow. 
 
 4. Equality of Fluid Pressure round a point. The 
 student must avoid the fallacy of supposing that pressure 
 acts on a point. Strictly speaking-, nothing can act ' on a 
 point ' : pressure acts on a surface ; and when we speak of 
 intensity of pressure at a point, we mean that pressure acts 
 on a small surface placed at the point. We see now that if 
 we imagine any small plane area, say I square millimetre, 
 placed at a given point P of a fluid, the stress or pressure 
 exerted on this area by the particles at one side of it 
 on those at the other acts normally to the area and has the 
 same magnitude however the position of the little area is 
 varied at that point P. This is the meaning of the 
 expression ' equality of fluid pressure at a point '. No such 
 result holds for any body other than a perfect fluid, as has 
 been already stated. 
 
 To each point, then, in a fluid acted on by any forces 
 belongs a special intensity of pressure the amount of which 
 can be calculated if we know the magnitude of the stress,/ 7 , 
 exerted on any small plane area, s, at the point, no matter 
 what the position of this area at the point may be : 
 the pressure intensity is 
 
 This pressure intensity will, of course, be different for 
 different points in the fluid. 
 
 If a fluid is acted upon by no external forces, such as gravity, 
 but only pressure applied somewhere on its surface, the intensity 
 of pressure is the same at all points within it. 
 
 Let a perfect fluid be contained within the surface 
 
8 Hydrostatics VOL. I 
 
 ABCD (Fig. 4), and suppose pressure to be applied over 
 its surface so that the intensity of this pressure at A is p 
 pounds' weight per square inch. At A take a very small 
 area, .? square inches, represented by 
 A A', and on this little area erect a 
 
 A[A 
 
 right cylinder, A P, of any length. 
 Now consider the separate equilibrium 
 of the fluid contained within this cy- 
 linder. This fluid is held in equili- 
 brium by the force p . s pounds' weight 
 acting on AA' } a pressure on the base 
 at P, and a series of pressures all over 
 its curved surface. Resolving forces in the direction A P, we 
 have p . # = the pressure on the base at P, since the pressures 
 on the curved surface are all at right angles to A P. But 
 the area of the base is also * square inches, therefore if 
 p' intensity of pressure at P, 
 
 j).S=ff.S . (l) 
 
 .-. P = p'- 
 
 Again, the pressure intensity at every point on the 
 surface is also p. For, let the base at P be turned. round 
 through any angle, and on its new position construct a 
 right cylinder cutting the surface obliquely at B. Let 6 
 be the angle between the normal to the surface at B and 
 the axis, P , of the cylinder ; let p' be the intensity of 
 pressure exerted by the envelope at B on the fluid, and 
 consider the separate equilibrium of the fluid in the 
 cylinder P B. The area of the normal cross-section of the 
 cylinder being s, the area cut off from the surface at B is 
 * sec 0, and the total pressure on this is // . s sec Q. Now 
 resolving along the axis P B for the equilibrium of the 
 enclosed fluid, we have 
 
 j) . s =. p'.s sec . cos 0, 
 .-. p'-p. 
 
CH. I Nature of Fluid Pressure 9 
 
 This may also be seen by constructing on the area s at A 
 a tube, AA'B B f , of any form whatever and of uniform 
 normal cross-section. The fluid inside this tube is kept in 
 equilibrium by the terminal pressures on A A' and BB' 
 together with the pressures of the surrounding fluid which 
 are all normal to the sides of the tube. Hence (except 
 that the terminal forces at A and B are pressures and not 
 tensions) this fluid is in the same condition as a flexible 
 chain stretched over a smooth surface and acted upon by 
 two terminal forces only, in addition to the continuously 
 distributed normal pressure of the smooth surface ; and it 
 is obvious, by considering the equilibrium of each elementary 
 length of the tube, that the forces per unit area at A and 
 B are equal. 
 
 Even when the fluid is acted upon by bodily force, such 
 as gravity, the Pascal Principle of the transmission of 
 pressure still holds ; that is, any intensity of pressure 
 applied at the surface of the fluid is transmitted undiminished 
 in amount to all parts of the fluid, but in addition to this 
 transmitted pressure there is pressure of different intensities 
 at different points produced by the action of the bodily force. 
 
 5. Principle of Separate Equilibrium. The following 
 principle is very largely employed, and has been already 
 used in the previous pages, in the consideration of the 
 equilibrium or motion of a fluid, or, indeed, of any 
 material system : 
 
 We may always consider the equilibrium or motion of any 
 limited portion of a system, apart from the remainder , provided 
 we imagine as applied to it all the forces which are actually 
 exerted on it ly the parts imagined to be removed. 
 
 Thus, suppose Fig. 5 to represent a fluid, or other mass, 
 at rest under the action of any forces, and let us trace out 
 in imagination any closed surface enclosing a portion, M, 
 of the mass. Then all the portion of the mass outside this 
 
io Hydrostatics VOL. I 
 
 surface may be considered as non-existent, so far as M 
 is concerned, if we supply to each element of the surface of 
 M the stress which is actually exerted on it by the mass 
 outside it. The stresses exerted on the elements of surface 
 of M, when the body is a perfect fluid, are represented by 
 the arrows in the figure. 
 
 The portion M is, then, in equilibrium under the action 
 
 of these pressures and what- 
 ever external forces (gravity, 
 <fec.) also act upon it. 
 
 Again, it is evident that, 
 having traced out in imagi- 
 \ f ^Tl x / nation any surface enclosing 
 
 Fig. 5. a mass, M, of the fluid, we 
 
 might, without altering any- 
 thing in the state of this mass M, replace the imagined 
 enclosing surface by an actual material surface, and then 
 remove all the fluid outside this surface ; for the enclosing 
 material surface will, by its rigidity, supply to M at each 
 point the pressure which is exerted at that point by the 
 surrounding fluid. 
 
 6. The Hydraulic Press. A machine the action of 
 which illustrates the Principle of Pascal is the Hydraulic 
 Press, represented in Fig. 6. 
 
 It consists of a stout cylinder, A, in which a cast iron 
 piston, or ram, P, works up and down. This piston has 
 a strong iron platform fixed on the top ; on this platform 
 is placed a substance which is to be subjected to great 
 pressure between the platform and a strong plate, 1), fixed 
 to four strong vertical pillars. The pressure is applied 
 at the bottom of the piston P by a column of water which 
 is forced into the cylinder A through a tube, f, which 
 communicates with a reservoir of water, B. The water is 
 driven out of B by a force-pump whose piston, p, has a 
 
1 I 
 
 CH. I Nature of Fluid Pressure 
 
 diameter much smaller than that of the piston P ; the 
 piston p is worked up and down by means of a lever L, and 
 the cylinder in which p works terminates inside the vessel 
 B in a rose, r, the perforations in which admit water while 
 preventing the entrance of foreign matter. 
 
 It is easy to see what an enormous multiplication of 
 force can be produced by this machine. If F is the force 
 
 w 
 
 applied by the hand to the lever L, n the multiplying ratio 
 of the lever, and * the area of the cross-section of the piston 
 p, the intensity of pressure produced on the water in the 
 
 nF 
 
 vessel is ; so that if S is the area of the cross-section 
 if 
 
 of the piston P, the total force exerted on the end of 
 this piston by the water in A is 
 
 nF. 
 
 8 
 
12 Hydrostatics VOL. i 
 
 Thus, if S i oo s and the ratio, n, of the long to the 
 short arm of the lever L is 5, the upward force exerted on 
 the piston P is 500 F, so that if a man exerts a force of 
 100 pounds' weight on the lever a resistance of nearly 
 50,000 pounds' weight can. be overcome by the piston. 
 
 In order to prevent the intensity of pressure in the 
 vessel B from becoming too great, a safety-valve closed by 
 a lever loaded with a given weight, W^ is employed. 
 
 The Hydraulic Press remained for a long time com- 
 paratively useless, because the great pressure to which 
 the water was subject drove the liquid out of the cylinder 
 A between the surface of the piston P and the inner 
 surface of the cylinder. This defect was remedied in a 
 very simple and ingenious manner by Bramah, an English 
 engineer, in the year 1 796. In the neck of the cylinder A 
 is cut a circular groove all round, and into this groove is 
 fitted a leather collar the cross-section of which is repre- 
 sented at c in the form ^\. This collar is saturated with 
 oil, in order that it may be water-tight, and it will be seen 
 that it presses with its left-hand and upper portion against 
 the cylinder A, while its right-hand portion is against the 
 piston P. When, by pressure, the water is forced up 
 between the surface of the piston and the surface of the 
 cylinder, this water enters the lower or hollow portion of 
 the inverted U-shaped collar and firmly presses the leather 
 against both the piston P and the surface of the groove, 
 thus preventing any escape of water from the cylinder. 
 
 In consequence of this great improvement in the machine, 
 it is very commonly called Bramah' s Press. 
 
 In order to prevent the return of the water from the 
 cylinder A on the upward stroke of the piston p, there is a 
 valve, represented at i in Fig. 6, and shown more clearly at 
 i in Fig. 7, which is a simple sketch of the essentials of the 
 force-pump. When the piston jp rises, a valve, e, in the 
 
CH. I 
 
 Nature of Fluid Pressure 
 
 pipe dipping- into the reservoir B, opens upwards and 
 allows the water to fill the cylinder / and to flow through 
 o to the valve i. When p descends, the water closes the 
 valve e and is forced to open the valve i which is pressed 
 down by a spiral spring. When the piston p moves 
 upwards, the water which has passed the valve i into the 
 cylinder A cannot return into the cylinder / because it 
 obviously assists the spring in closing the valve i. The 
 safety-valve is represented at v in Fig. 7. 
 
 Fig. 7. 
 
 The piston p works in a stuffing-box in the upper part of 
 the cylinder /, this stuffing-box playing the same part as 
 the leather collar round the ram i. e., preventing leakage. 
 The piston must not fit the lower part of the cylinder J 
 tightly, because when/? in its downward motion passes o, the 
 cylinder would be burst if the water above the closed valve 
 e could not escape round the piston and out through the 
 valve i. 
 
 Another machine depending essentially on the same 
 principles and illustrating the Principle of Pascal is the 
 
14 Hydrostatics VOL. I 
 
 Hydrostatic Bellows, which is formed by two circular boards 
 connected, in bellows fashion, by water-tight leather, the 
 boards being the ends of a cylinder the curved surface of 
 which is formed by the leather. One of these boards being 
 placed on the ground, the other lies loosely on top of it. 
 A narrow tube communicates with the interior of this 
 cylinder. If this tube is a long one and held vertically, 
 when water is poured into it at its upper end, the upper 
 board of the bellows, and any load that may be placed on 
 it, will be raised by the pressure of the water, the intensity 
 of which pressure depends (as will be subsequently explained) 
 on the height to which the narrow tube is filled. 
 
 7. Specific Weight. The specific weight of any sub- 
 stance means its weight per unit volume. If w is the weight 
 of a homogeneous body per unit volume, the volume of the 
 body being V and the weight W, 
 
 W V .w. 
 It will be useful to remember that 
 
 i cubic foot of water has a mass of about 62^ Ibs. 
 i ' 1000 ounces, 
 
 j ,, inch of mercury -491 Ibs. 
 
 These numbers are, of course, only approximate, because 
 the mass of a cubic foot of any substance depends on the 
 temperature of the substance. 
 
 A gramme is defined to be the mass, or quantity of 
 matter, in i cubic centimetre of water when the water is at 
 its temperature of maximum density ; this temperature is 
 very nearly 4 C. ' 
 
 A term in frequent use is the specific gravity of a sub- 
 stance, which ought, apparently, to signify the same thing 
 as its specific weight ; but it does not. The specific gravity 
 of any homogeneous solid or liquid means, in its ordinary 
 employment, the ratio of the weight of any volume of the 
 
CH. I Nature of Fluid Pressure 15 
 
 substance to the weight of an equal volume of distilled 
 water at the temperature o C. Thus, for example, in the 
 following table of specific gravities : 
 
 gold 19-3 
 
 silver 10-5 
 
 copper .... 8-6 
 
 platinum .... 22-o 
 
 sea- water. . . . 1-026 
 
 alcohol .... -791 
 
 mercury .... 13-596 
 
 the number opposite the name of any substance does not 
 tell us the weight of a cubic foot, or of any other volume, 
 of the substance ; it merely tells, with regard to platinum, 
 for example, that a cubic foot of it, or a volume V of it, is 
 22 times as heavy as a cubic foot, or a volume F, of distilled 
 water. A table of specific gravities is a table of relative 
 weights of equal volumes. In the C. G. S. system, since 
 the unit of weight is that of i cubic centimetre of water, 
 and since water is the substance with which in a table 
 of specific gravities all solids and liquids are compared, 
 the number (specific gravity) opposite any substance ex- 
 presses the actual mass, in grammes, of i cubic cm. of the 
 substance. 
 
 If s is the specific gravity of any substance and w the 
 actual weight of a unit volume of the standard substance 
 (water), the weight of a volume V of the substance is given 
 by the equation 
 
 W= Vsw. 
 
 The term density is used, as has been said, to denote the 
 mass per unit volume of a substance. Thus if mass is measured 
 in grammes and volume in cubic centimetres, the density of 
 silver is 10-5 grammes per cubic centimetre ; the density 
 of mercury is 13-596 grammes per cubic cm. If mass is 
 
1 6 Hydrostatics VOL. I 
 
 measured in pounds and volume in cubic inches, the density 
 of silver is -3797 Ibs. per cubic inch and that of mercury 
 491 Ibs. per cubic inch. These latter numbers are, of course, 
 proportional to the former. 
 
 The term density has no reference to gravitation. If 
 silver and mercury are taken from the Earth to a position 
 in interstellar space in which there is felt no appreciable 
 attraction from any Sun or Planet, it is still true that silver 
 has a mass of 10-5 and mercury a mass of 13-596 grammes 
 per cubic cm. Neither would, in this position, have any 
 specific weight^ since there is no external force of attraction 
 acting on them ; but the moment th'ey are taken to the 
 surface of any planet each acquires weight, and the ratio 
 of the weights of equal volumes of them is the ratio, 
 10-5 : 1 3'596, f their densities. If, for example, they were 
 carried to the surface of the planet Jupiter, the weight of 
 a cubic cm. of each would be nearly i\ times as great as it 
 is on the surface of the Earth ; but a table of relative 
 weights of substances on Jupiter would be exactly the same 
 as a table of relative weights on the Earth. 
 
 If any given volumes of a number of homogeneous sub- 
 stances are mixed together in such a way as to make a 
 homogeneous mixture whose volume is the sum of the 
 volumes of the separate substances, the specific weight of 
 the mixture is easily found. For, let v l and w l be the 
 volume and specific weight of the first substance ; v z and 
 w 2 those of the second ; and so on. Then if w is the 
 required specific weight of the mixture, since the weight of 
 the mixture is equal to the sum of the separate weights, 
 
 w 
 
 Such a mixture is called a mechanical mixture as, for 
 
CH. I Nature of Fluid Pressure 17 
 
 instance, a mixture of sand and clay. But when a chemical 
 combination takes place between any of the substances, the 
 volume of the mixture is not equal to the sum of the 
 volumes mixed as when sulphuric acid is mixed with 
 water. If for any chemical mixture V (which must be 
 specially measured) is the volume of the mixture, it is 
 evident that we have, as above, 
 
 2 vw 
 
 '"= 
 
 EXAMPLES. 
 
 1. A cask A is filled to the volume v with a liquid of specific 
 weight w ; another cask, B, is filled, also to the volume v, with 
 
 v v 
 
 another liquid of specific weight s ; - is taken out of A and - 
 
 also out of B, the first being put into B and the second into A , 
 and the contents of each cask are shaken up so that the liquid 
 in each becomes homogeneous. The same pi'ocess is repeated 
 again and again : find 
 
 (a) the specific weight of the liquid in each cask after m 
 such operations; 
 
 (b) the volume of the original liquid in each cask. 
 
 Result. If w s is denoted by d, and if w m , s m are the specific 
 weights of the liquids in A and B, respectively, after m opera- 
 tions, 
 
 d 
 
 d ( / 2N 7 ") 
 
 = *+ -i 1 - (' ) [ > 
 
 2 ( n' ) 
 
 and the volume of the original liquid in either cask is 
 
 [N.B. The liquids are assumed not to enter into chemical 
 combination.] 
 
 2. In a hydraulic press an effort of 20 pounds' weight is applied 
 
1 8 Hydrostatics VOL. I 
 
 at the end of a lever, at a distance of 9 inches from the fulcrum, 
 moving the plunger of a force -pump, the point of attachment of 
 the plunger being 4 inches from the fulcrum. The diameter 
 of the plunger is i inch and that of the ram 8 inches ; find the 
 thrust exerted by Ihe ram. 
 
 if tons' weight. 
 
 3. In a hydraulic press the diameter of the plunger is i inch 
 and that of the rani i foot ; the effort is applied at a distance of 
 10 inches from the fulcrum, and the point of attachment of the 
 plunger is 5 inches from the fulcrum ; calculate the magnitude 
 of the effort required to produce a thrust of 2 tons' weight by 
 means of the ram. 
 
 15! pounds' weight. 
 
CHAPTEK II 
 
 THEOREM OF PLANE-MOMENTS 
 
 8. IF any number of particles whose masses are M lt 
 m 2t m z , ... are at distances z l5 r 2 , z 3 , ... from any plane, it 
 is known that the distance, I, of their centre of mass, or 
 centre of gravity, from the plane is given by the equation 
 
 _ _ 1 9t 9a ... , 
 
 and the same expression gives the distance of the ' centre ' 
 of a system of parallel forces P 1} P 2 , P 3 , ... from a plane, 
 the distances of their separate points of application from 
 the plane being z lt z 2 , z z) ... (see Statics, vol. i, chap. xi). 
 
 In either case if distances measured from one side of the 
 plane are taken as positive, distances measured from the 
 other side are to be taken as negative ; and, in addition, if 
 parallel forces acting in one sense are taken as positive, 
 those forces which act in the opposite sense are negative. 
 
 The result (a) holds, of course, for particles occupying 
 positions in space of three dimensions, as well as for 
 particles lying in one plane ; and similarly for parallel 
 forces. 
 
 Owing to the importance of this equation in all calcu- 
 lations, we give a diagrammatic representation of the way 
 in which the student will find it convenient for actual 
 
 c 2 
 
2o Hydrostatics VOL. i 
 
 calculation. It is to be observed that no such result as (a) 
 holds for distances measured from an axis, or a line ; but in 
 many cases the result is applicable, by accident, to distances 
 measured from a line. Thus, if all the particles, or points 
 of application of parallel forces, happen to lie in one plane 
 (P), the above will give the distance of their centre from 
 any line (L) in the plane, but only because the distances 
 z l} s z . ... measured from (L) are really measured from a 
 plane through (L) perpendicular to (P). 
 
 We shall refer to (a) as the theorem of mass-moments, or 
 theorem of plane-moments. 
 
 It is obvious that the mass-moment of any system with 
 respect to any plane passing through its centre of gravity 
 is zero. 
 
 It is also evident that the distances ,? 1 ,.z 2 ,...need not be 
 perpendiculars ; they may be oblique distances all, of 
 course, measured in the same direction. 
 
 The work of practical calculation is often facilitated by 
 forming tables of masses, distances, and products, in columns, 
 as in the following example. 
 
 EXAMPLE. 
 
 At the vertices, A, B y C (Fig. 8) 
 of a triangle and at the middle 
 points, a, 6, c, of the opposite sides 
 act parallel forces whose magnitudes 
 and senses are represented in the 
 _____ figure ; find the position of the centre 
 
 // of the system. 
 T6 V The position of the centre will be 
 
 ~ known if its distances from any two 
 
 sides, AB and AC, are known. 
 
 To find its distance from AB, let the length of the perpen- 
 dicular from C on AB be p ; and form a table of forces and 
 distances of their points ot application from AB as in the fol- 
 
CH. II 
 
 Theorem of Plane-Moments 
 
 21 
 
 lowing scheme, taking as positive those forces which act in the 
 sense of that at A : 
 
 Forces. 
 
 Distances 
 from AB. 
 
 Products. 
 
 Distances 
 from A C. 
 
 Products. 
 
 10 
 
 
 
 o 
 
 O 
 
 O 
 
 -16 
 
 %P 
 
 -Bp 
 
 O 
 
 O 
 
 8 
 
 P 
 
 Sp 
 
 O 
 
 
 
 - 6 
 
 P 
 
 -3P 
 
 <? 
 
 - 3? 
 
 12 
 
 o 
 
 o 
 
 9 
 
 12 q 
 
 14 
 
 o 
 
 
 
 \<l 
 
 1<1 
 
 2 
 
 
 -3P 
 
 
 -Sq 
 
 The sum of the first column answers to 2, m, the denominator 
 of (a), p. 19, while the sum of the third column answers to 2 mz, 
 the numerator, so that the perpendicular distance of the centre 
 from AB is 
 
 or 
 
 Drawing, then, a line parallel to AS at a distance f p (above 
 (7), we know that G lies somewhere on this line. 
 
 Denoting the perpendicular from B on AC by q, forming a 
 table (column 4) of distances from AC, aud a column (number 
 5) of corresponding products, and dividing the sum of these 
 products by the sum of the forces, we have the distance of G 
 from AC equal to 
 
 - 2 <7 
 
 , or 49-. 
 
 Hence G lies on a line to the right of B distant 47 from AC. 
 The point of intersection of this with the previous line is G. 
 
 Thus also to find the position of the ' centre of gravity ' of a 
 trapezium ABCD whose parallel sides AB and CD are 2 a, 26, 
 and height h, divide it into two triangles by the diagonal A C ; 
 then the areas ABC, ADC are ah, bh, and they may be taken as 
 represented by a, b, since the m's in equation (a) may all be 
 multiplied by any common factor. Take mass-moments about 
 
22 
 
 Hydrostatics 
 
 VOL. I 
 
 AB; the distances of the centres of gravity from AB are 
 ; therefore the diagram is 
 
 mosses 
 
 distances 
 from A B 
 
 products 
 
 CL 
 
 to 
 
 Wh 
 
 b 
 
 *& 
 
 ^bh 
 
 a+b 
 
 
 ^(a*zb)h. 
 
 Fig. 9. 
 
 and dividing the sum of the third column by the sum of the 
 first, we have 
 
 _ a + zb h 
 
 As the centre of gravity lies on the line joining the mid points 
 of AB and CD, it is therefore found. 
 
 9. Simple cases of Continuously Distributed Forces. 
 We shall now find the centre of a system of parallel forces 
 distributed continuously over a plane area, in a few simple 
 cases which do not require the application of the Integral 
 Calculus. 
 
 (i) If normal pressure acts all over any plane area 
 in such a way that its intensity is the same at all points, 
 the resultant pressure acts at the centre of area (' centre of 
 gravity,' so called) of the figure. For, if at any two 
 points, A, B, in the given figure we take any two elements 
 of area, the pressures on them are directly proportional 
 to the areas themselves, and the resultant of these forces 
 acts at a point in the line AB dividing it into segments 
 inversely as the forces, i. e., inversely as the areas ; hence 
 it acts at the centre of area of the two elements of area. 
 
CH. ii Theorem of Plane-Moments 23 
 
 The process, therefore, of finding the point of application 
 of the resultant of the whole system of pressures acting on 
 the indefinitely great number of elements of area into 
 which the given figure can be broken up is precisely 
 the same as that of finding the centre of area of the 
 figure. 
 
 (2) If parallel forces act at all points of a right line, 
 AS, Fig. 10, in such a way that the force at any point P 
 is directly proportional to the distance, PA, of P from one 
 extremity, A, of the line, the 
 resultant force acts at the / 
 
 point on AB which is of / 
 
 the length AB from A. f 
 
 For, imagine AB to be 
 
 \F 
 
 A. 
 
 broken up into an indefi- f~ 
 
 nitely great number of small / 
 
 equal parts, PQ; describe / 
 
 any isosceles triangle, MAN, M " "~"B 
 
 having AB for height, and 
 
 from all the points, P, Q, . . . of division of AB draw parallels 
 
 to the base MN, thus dividing the area MAN into an 
 
 indefinitely great number of narrow strips. The area of 
 
 any strip, QF, is simply proportional to the distance, PA, 
 
 of the strip from A. Hence the areas of the strips are 
 
 exactly proportional to the given system of parallel forces ; 
 
 but the centre of area of the strips, or centre of area of the 
 
 whole triangle, is f AB from A. This point is, then, 
 
 the centre of the given system of forces. 
 
 (3) If parallel forces act at all points of a right line, 
 AB, Fig. 10, in such a way that the force at any point 
 P is proportional to the square of the distance, PA, of 
 P from one extremity, A, of the line, the resultant force 
 acts at a point on AB which is f of the length AB 
 from A. 
 
24 Hydrostatics VOL. I 
 
 For, imagine AB to be broken up, as before, into equal 
 elements, such as PQ ; describe any solid cone having- AB 
 for its axis, and let this cone be represented by MAN. 
 From all the points of division of AB draw planes parallel 
 to the base MN of the cone, thus dividing the cone into 
 an indefinitely great number of thin circular plates. The 
 volume of the plate at P is irPF 2 X PQ, and since the 
 thicknesses of the plates are all equal to PQ, the volume 
 of the plate is proportional to PF 2 , i. e., to PA 2 . Hence 
 the volumes of the plates vary exactly as the forces of the 
 given system, and therefore the centre of volume of the 
 plates is identical with the centre of the force system ; but 
 the former (centre of volume of the cone) is f AB from A ; 
 therefore, &c. 
 
 (4) If parallel forces act at all points of a right line 
 AB, in such a way that the force at any point P is propor- 
 tional to the product of the distances PA, PB of P from 
 the extremities of the line, the resultant force acts at the 
 middle point of AB. 
 
 For, taking a point, P', whose distance from B is equal 
 to that of P from A, the forces at P and P' are evidently 
 equal ; their resultant therefore acts at the middle of AB. 
 Hence the system of forces from A to this middle point 
 is the same as the system from B to this point ; the re- 
 sultant, therefore, of the whole system acts at the middle 
 o -V point of AB. 
 
 (5) If each infini- 
 tesimal element of any 
 plane area is acted upon 
 by normal pressure pro- 
 portional conjointly to the 
 magnitude of the area and 
 to the distance of the element from a given plane, the 
 magnitude of the resultant pressure will be proportional to 
 
CH. ii Theorem of Plane-Moments 25 
 
 the product of the whole plane area and the distance of 
 its centre of area from the given plane. 
 
 For, let yOx, Fig-, n, represent the given plane, and 
 ABCD the given plane area. Take any point, P, in this 
 area, and round P describe a very small closed curve whose 
 area is s. Let PN, the perpendicular from P on the plane 
 yOx, be denoted by z ; then, by hypothesis, the amount of 
 force, /, on the element at P is given by the equation 
 
 f=k.sz, 
 
 where k is a given constant. If *', /',... are any other 
 elements of area whose distances from the given plane 
 are /, z'\ . . . the resultant pressure, being equal to the sum 
 *&fif r *f"i of the individual pressures on the elements, 
 is equal to 
 
 But if A = the area of the whole plane figure, and z is 
 the distance, GQ, of the centre of area, G, from yOx t we 
 
 have , , ,, 
 
 Az = sz + / / + *" /'+.... 
 
 Hence if P is the resultant pressure, 
 
 P = k.A.z ....... (a) 
 
 The student must be careful to observe that the resultant 
 pressure does not act at G, but evidently at some such point 
 as 1, whose distance from the plane yOx is greater than the 
 distance of G from the plane. 
 
 In this case, then, the mean intensity of pressure on the 
 area is that which exists at G. 
 
CHAPTER III 
 
 LIQUID PRESSURE ON PLANE SURFACES 
 
 10. Intensity of Pressure produced by Gravity. Let 
 ACS, Fig. i a, be a vessel of any shape containing water 
 or other homogeneous liquid. Then at each point, P, of 
 
 the liquid the action of 
 
 tltllNlll'I'l I fa gravity produces a certain 
 
 intensity of pressure, the 
 magnitude of which we 
 
 proceed to find. At P 
 
 x x^ ^" draw an indefinitely small 
 
 c -pj I2 horizontal element of area 
 
 s square inches, suppose 
 
 and on the contour of this area describe a vertical cylinder, 
 PN. Consider now the separate equilibrium (Art. 5) of the 
 liquid in this cylinder. 
 
 If PN is z inches in length, the volume of the cylinder 
 = z . s cubic inches, and if the specific weight of the 
 liquid is w pounds' weight per cubic inch, the weight 
 of the cylinder = wzs. This cylinder is acted upon by 
 a vertically upward pressure on the base s at P and a 
 system of horizontal pressures round its curved surface, 
 in addition to its weight omitting, for the present, the 
 surface pressure at N produced by the atmosphere or any 
 other cause. 
 
 If ;; pounds' weight per square inch is the intensity 
 
CH. in Liquid Pressure on Plane Surfaces 27 
 
 of pressure at P, the upward pressure on the base s is p s - 
 Resolving forces vertically, we have, then, 
 
 p . s = 10 z . s ; 
 .-. p ws, (a) 
 
 which gives the required intensity of pressure. 
 
 If the surface intensity of pressure is j) pounds' weight 
 per square inch, this will be added to the value (a), by 
 Pascal's principle ; hence the complete value of p is given 
 by the equation 
 
 p - wz +p (/3) 
 
 Observe that we have not assumed the bounding surface 
 AB to be horizontal. 
 
 Without any reference to the shape of the surface AB, 
 we can see that the intensity of pressure is the same 
 at all points P, Q, ... which lie in the same horizontal 
 plane. 
 
 For, draw PQ ; at P and Q place two indefinitely small 
 equal elements of area, s. perpendicularly to PQ ; form a 
 cylinder having PQ for axis and these little areas for 
 bases, and consider the separate equilibrium of the liquid 
 enclosed in this cylinder. The forces keeping it in equili- 
 brium are its weight, a system of pressures all round 
 its curved surface, and the pressures on its bases at P and Q. 
 Resolving forces along PQ for equilibrium, neither the 
 weight nor the system of pressures on the curved surface 
 will enter into the equation ; therefore the pressure on the 
 base * at P = the pressure on the (equal) base # at Q ; that 
 is, the intensity at P = the intensity at Q. 
 
 From this it follows that the bounding surface AB on 
 which at all points there is either no pressure, or pressure 
 of constant intensity, must be a horizontal plane. 
 
 For, take any two points, P, Q, in a horizontal plane, 
 and let their vertical distances below AB be z and /. 
 
28 Hydrostatics 
 
 Then by (/3), we have 
 
 VOL. I 
 
 z = z. 
 
 that is, all points in the same horizontal plane are at the 
 same depth below the surface AB which proves AB to be 
 a horizontal plane. 
 
 It is usual to speak of the surface, AB, of contact of 
 the liquid with the atmosphere as the free surface of the 
 liquid. It is simply a surface at each point of which 
 the intensity of pressure is constant, the constant being the 
 atmospheric intensity. 
 
 The result at which we have arrived may be also stated 
 thus all points in a heavy homogenemis liquid at which the 
 
 intensity of pres- 
 sure is the same 
 lie in a horizontal 
 plane ; and from 
 this it follows 
 that if a mass 
 of water partly 
 enclosed by sub- 
 terranean rocks, &c., has access to the atmosphere by any 
 number of channels, the level of the water will be the same 
 in all these channels. It is to be observed that z in (a) and 
 (ft) is the depth of the point P (Fig. 13) lelow the free surface 
 not the distance, PD, of the point P from the roof of the 
 cavity in which the water is partly confined. 
 
 We may here, if we please, consider the separate equili- 
 brium of a small vertical cylinder of the liquid terminating 
 at D, and we shall have simply the result (/3) in which, 
 however, p Q would now mean the vertical downward com- 
 ponent of the pressure intensity of the roof of the cavity at 
 D on the water. But the result (ft) holds for the intensity 
 
CH. in Liquid Pressure on Plane Surfaces 29 
 
 of pressure at P if PH is put for z, where // is the foot of 
 the perpendicular from P on the plane of the free surfaces 
 ab, cd, ef of the water ; for, nowhere in the liquid will the 
 state of affairs be altered if we imagine the roof of the 
 cavity to be removed, and the space 6Dc to be filled with 
 water up to the level lie. In this way we shall have 
 a vertical cylinder, PH, unobstructed by the roof, and 
 terminating on the free surface. 
 
 It is usual to illustrate the fact that all parts of the free 
 surface of a liquid lie in a horizontal plane by taking 
 a vessel, ABC, of any shape and fitting into it tubes 
 or funnels of various forms, and then pouring water in 
 through any one of these tubes, the visible result being 
 that the water stands at the same level in all the tubes. 
 This is, indeed, nothing more than the principle of separate 
 equilibrium (see end of Art. 5) ; for, these variously shaped 
 funnels may be supposed to have been surfaces traced out 
 in imagination in a large vessel of water whose free surface 
 was af, these imagined surfaces being then replaced by 
 material tubes, and the outside liquid removed. The level 
 of the liquid in each tube would still be af. 
 
 11. Superposed 
 Liquids. If in a vessel, 
 AOB, Fig. 14, several 
 liquids be placed as 
 layers, one on top of 
 another, there being no 
 chemical combination 
 between them, the 
 common surface of each 
 
 Fig. 14. 
 
 pair of liquids is a horizontal plane. Let the specific weights 
 of the liquids be w l , w 2 , w 3 , ... . The free surface, AB, has 
 been already proved to be a horizontal plane (Art. 10); and 
 the same process will prove CD, the surface of separation 
 
3 Hydrostatics VOL. I 
 
 of w^ and w 2 , to be a horizontal plane. For, in the liquid 
 w 2 take any two points, Q, Q' (as in Fig-, iz, p. 26), in the 
 same horizontal plane. Then by taking a slender horizontal 
 cylinder having QQ' for axis, we prove that the intensity 
 of pressure at Q = that at Q'. Now taking a vertical 
 cylinder Qmn, at Q, considering its separate equilibrium, 
 we find that if p is the intensity of pressure at Q, and 
 Qm = x, mn = y, 
 
 p = W. 2 ll' + 10 'j I/. 
 
 Similarly if Q,' mf ri is the vertical line at Q', and 
 Q'm' = a', m'n' = /, 
 
 p = 
 Hence ^ (a , _ ^ = 
 
 But Qn = Q'n', i.e., x+y x' +y', .'. x x' = y' y ; 
 so that unless x x' = o and y f y = o, equation (i) will 
 give w l = w z , which is not the case, by hypothesis. 
 
 Hence we must have 
 
 Qm = Q'm', and mn = mn', 
 
 and since this holds for all points Q, Q' in the same 
 horizontal plane, all points, m, m', . , . in the surface CD are 
 at the same height above the same horizontal plane ; 
 therefore CD is a horizontal plane. Similarly, by taking 
 two points in the same horizontal plane in the liquid w 3 , 
 we prove that EF is a horizontal plane. 
 
 If h^ and /i 2 are the thicknesses of the layers w l and w 2 , 
 and if R is a point in w 3 ^,t a depth z below the surface, 
 El\ of #> 3 , the intensity of pressure, jo, at R is given by the 
 equation 
 
 p = w l h 1 + 2V 2 /i 2 + w,^z, ..... (2) 
 
 to which, if atmospheric (or other) pressure acts on the 
 uppermost surface AB, must be added /? , the intensity of 
 this surface pressure, so that 
 
 W^Z ..... (3) 
 
CH. in Liquid Pressure on Plane Surfaces 31 
 
 Similarly for any number whatever of superposed layers. 
 
 Each layer of liquid, in fact, acts as an atmosphere, 
 producing an intensity of pressure on the next layer below 
 it equal to 
 
 wh, (4) 
 
 where w is the specific weight of the layer and h its 
 thickness. 
 
 If the /^'s are measured in centimetres and the w's, in 
 grammes' weight per cubic centimetre, the above equations 
 express p in grammes' weight per square cm. 
 
 The method of regarding any layer of liquid, even when 
 there in only one liquid in question, as an atmosphere pro- 
 ducing an intensity of pressure given by (4) on the layer 
 on which it rests, this intensity being then transmitted un- 
 altered to all points below (by Pascal's principle), is one 
 which we shall frequently employ in the sequel. 
 
 From the general principle (Statics, vol. i, Art. 121) 
 that, for stable equilibrium, any system of material particles 
 acted upon by gravity only must arrange themselves into 
 such a configuration that their centre of gravity occupies 
 the lowest position that it can possibly occupy, it follows 
 that in a system of superposed liquids of different densities 
 they must arrange themselves so that the density of each 
 liquid is greater than that of any one above it. 
 
 Again, if ABC, 
 Fig. 15, represents 
 a vertical section 
 of a vessel of any 
 shape into which 
 are poured two dif- 
 ferent liquids, AB 
 
 and C, which do not mix, the system will settle down 
 into a position in which the centre of gravity of the whole 
 mass occupies the lowest position that it can occupy, and 
 
32 Hydrostatics VOL. i 
 
 the vertical heights, h, h', of the free surfaces, A and C, 
 above the common surface, B, of the liquids will be inversely 
 proportional to the specific weights of the liquids. 
 
 To see the latter, we may take any point (most con- 
 veniently a point in the common surface B) and equate 
 the intensity of pressure produced there by everything 
 at one side of the point to the intensity of pressure pro- 
 duced by everything at the opposite. Thus, let w and w' be 
 the specific weights of the liquids AB and BC, respectively; 
 select a point, P, in the common surface B. Then if h is 
 the difference of level between P and A, the intensity of 
 pressure produced at P by the liquid AB and the overlying 
 atmosphere at A is 
 
 wh +p . 
 
 Also, h' being the difference of level of P and C, the 
 intensity of pressure at P produced by the right-hand 
 liquid and the atmosphere above C is 
 
 There is only one intensity of pressure at P ; hence these 
 must be equal ; 
 
 . . w . k = i<f . h', ...... (5) 
 
 which shows that the heights of the free surfaces above the 
 common surface, B, of the liquids are inversely as their 
 specific weights. 
 
 Thus, if AB is mercury and BC water, the surface C will 
 be 13-596 times as high above B as the surface A is. 
 
 As an example, let two liquids, AB, BC, Fig. 16, be 
 poured into a narrow circular tube held fixed in a vertical 
 plane, the lengths of the arcs occupied by the liquids being 
 assigned ; it is required to find their positions of equi- 
 librium. 
 
 The figure of equilibrium will be defined by the angle, 6, 
 
CH. in Liquid Pressure on Plane Surfaces 33 
 
 which the radius, OB, to the common surface of the liquids 
 makes with the vertical, OD. 
 
 Let the angles, AOB, BOG, sub- 
 tended by the liquid threads at the 
 centre of the circle be a, a' ; let 
 their specific weights be w, w\ re- 
 spectively ; and let r be the radius 
 of the circle. 
 
 Equate the intensity of pressure 
 produced at B by the one liquid to 
 
 that produced by the other. The difference of level between 
 B and A is 
 
 r {cos 9 cos(0 + a)}, 
 
 and this multiplied by w is the pressure intensity at B due 
 to the first liquid. The difference of level of B and C is 
 
 r (cos 6- cos (a -6}\. 
 Hence 
 
 w (cos0 cos(0 + a)} = w' {costf cos (a'-0)}, . (6) 
 
 , . a . a 
 
 w sin" 5 w sin j - 
 
 .-', tanfl = 2 7-,-V- _*.... (7) 
 w sin a + iv sin a 
 
 The equation (6) can be shown to express the fact that 
 the centre of gravity of the system of two liquid threads 
 has, in the position of equilibrium, the greatest vertical 
 depth below that any geometrical displacement of the 
 two liquid threads could give it ; and this is a particular 
 case of the general property that the equilibrium position 
 of any system of particles subject to any frictionless con- 
 straints assumes the lowest position that any arrangement 
 of the particles can give it. 
 
 12. Pressure on a Plane Area. Let AECD, Fig. 1 1, p. 24, 
 represent a plane area occupying any assigned position in 
 a heavy homogeneous liquid whose free surface is xOy. 
 
34 Hydrostatics VOL. i 
 
 Then if w (pounds' weight per cubic inch, suppose) is 
 the specific weight of the liquid, and z (inches) is the 
 depth, PN, of any point below xOy, the intensity of pres- 
 sure at P, due solely to the weight of the liquid, is ws. 
 Hence (case 5, p. 24) the resultant pressure on one side 
 
 of the area is 
 
 A . z . iv, (a) 
 
 where A (square inches) is the magnitude of the area, and z 
 is the depth, GQ (inches), of its centre of area below the 
 free surface. 
 
 If on the free surface, xOy, there is intensity of pressure 
 (atmospheric or other) of jo (pounds' weight per square 
 inch), this pressure will produce its resultant, Ap , at G, 
 and the total pressure on one side of the area is 
 
 4(zw+p ) (P) 
 
 As before pointed out (p. 25) the pressure (a) due to the 
 liquid does not act at G t but at some point lower down. 
 
 If a plane area, S, Fig. 14, p. 29, occupies an assigned 
 position in a liquid on the surface of which are superposed 
 given columns of other liquids, the resultant pressure on 
 the area is easily found. For, if z is the depth of the 
 centre, G t of area lelow the surface EF, of the liquid w 3 , 
 the pressure of this liquid is Azw 3t where A is the magni- 
 tude of the area. Also the column AD produces a resultant 
 pressure equal to Ah^ w l , where h^ is the thickness of the 
 column ; the second column produces Ah z w 2 ; so that the 
 total pressure on S is 
 
 A (h l w l + h^ w t +zw 3 ); (y) 
 
 and similarly for any number of liquids, the resultant 
 pressure will be 
 
 A(h l w l +^w 2 +h a w 3 + ...+zw n ), ... (5) 
 
 where z is the depth of G below the surface of the liquid, ic n , 
 in which the area lies. 
 
CH. in Liquid Pressure on Plane Surfaces 35 
 
 EXAMPLES. 
 
 1. If a plane area, occupying any position in a liquid, is 
 lowered into the liquid by a motion of translation unaccom- 
 panied by rotation, show that the point of application of the 
 resultant pressure on one side of the area rises towards the 
 centre of area, G, the more the area is lowered. (See Fig. 19.) 
 
 Draw the horizontal plane CD, touching the boundary of 
 the area at its highest point, and consider the pressures due 
 separately to the layer between CD and the free surface, AB, 
 and to the mass of liquid below CD. Since there is no change 
 in the position of the area relative to the liquid below CD, 
 this latter pressure will always act with constant magnitude 
 and point of application, 7 ; but the pressure of the super- 
 incumbent layer, always acting at G, increases in magnitude 
 with x, the distance between AB and CD. Hence of the two 
 parallel forces at / and G the first remains constant, while 
 the second continually increases ; their resultant, therefore, gets 
 nearer and nearer to G as CD is lowered. 
 
 2. Calculate in pounds' weight per square inch the intensity 
 of pressure at a depth of 100 feet in water, neglecting atmospheric 
 pressure. 
 
 43-4- 
 
 3. A vertical cylinder i foot in diameter communicates by 
 a tube with a vertical cylinder i inch in diameter ; and both 
 contain water. A load of i ton is placed on the surface of the 
 water in the large cylinder; what force must be applied to 
 the surface of the water in the small one so that the water may 
 stand 20 feet higher in the latter than in the former cylinder 1 
 
 Ans. About 8;f pounds. 
 
 4. If the load on the water in the small cylinder is removed, 
 how much higher will the water stand in this cylinder than in 
 the larger one ? 
 
 Ans. 45-6 feet. 
 
 5. Calculate in pounds' weight per square inch the intensity 
 of pressure at a depth of 8 inches in mercury, neglecting 
 atmospheric pressure. 
 
 3-93- 
 
 D 2 
 
36 Hydrostatics VOL. i 
 
 6. The ram of a hydraulic accumulator is i foot in diameter, 
 and it descends 7 feet in a minute while the water is working 
 an engine and developing 12 horse-power; what is the mean 
 intensity of pressure in the water? 
 
 Ans. 500 pounds' weight per square inch. (Take TT = 2 T 2 
 and i horse-power =550 ft.-lb. weight per second.) 
 
 7. A triangular area of 100 square feet has its vertices at 
 depths of 5, 10, and 18 feet below the surface of water; find 
 the resultant pressure on one face of the area, the atmospheric 
 intensity being 1 5 pounds' weight per square inch. 
 
 127-12 tons' weight. 
 
 8. Find the depth of a point in water at which the intensity 
 of the water pressure is equal to that due to the atmosphere. 
 
 About 34^ feet. 
 
 9. A rectangular vessel i foot high, one of whose faces is 
 6 inches broad, is filled to a height of 4 inches with mercury, 
 the remainder being filled with water ; find the total out- 
 ward pressure against this face, the atmospheric intensity being 
 15 pounds' weight per square inch. 
 
 About 1117^ pounds' weight. 
 
 10. Into a vessel containing mercury is poured water to a 
 height of 8 inches above the mercury. If a rectangular area 
 6 inches in height is immersed vertically so that part lies in the 
 mercury and part in the water, find the length of the area im- 
 mersed in the mercury when the liquid pressure on this portion 
 is equal to that on the portion in the water. 
 
 Nearly 1-46 inches. 
 
 11. A beaker containing liquid is placed in one pan of a 
 balance, and is 'counterpoised by a mass placed in the other pan. 
 If a solid body suspended by a string held in the hand is then 
 immersed in the liquid, what will be the effect on the balance 1 
 
 If the string sustaining the solid is attached to the arm from 
 which the pan containing the beaker is suspended, and the 
 system counterpoised by a mass in the other pan, will the state 
 of the balance be the same whether the body is immersed in the 
 beaker or not ? 
 
 Ans. In the first case the pan containing the beaker will 
 descend ; in the second case there will be no change. 
 
CH. in Liquid Pressure on Plane Surfaces 
 
 OT 
 J/ 
 
 . 12. A circular tube whose plane is vertical (Fig. 16) contains 
 a column of liquid of specific weight w; on the ends of the 
 column rest two piston-heads of weights P and Q, fitting the 
 tube exactly and freely movable along it ; find the position of 
 equilibrium. 
 
 Let a be the angle subtended at the centre by the thread of 
 liquid and 6 the angle made with the vertical by the radius 
 drawn to the plug P ; then 
 
 _ W(i cos a 
 
 ~ 
 
 sn a 
 
 IF sin a + a(P+Q cos a)' 
 where W is the weight of the liquid. 
 
 13. A straight glass tube of small bore is bent so that the 
 two portions, AB, BC, are at right angles ; it is held in a vertical 
 plane with the point B downward, and the branch BC inclined 
 at the angle a to the horizon ; into AB is poured a liquid of 
 specific weight w, the length of the column being I ; into EG is 
 poured a liquid (w f , I') ; find the position of equilibrium. 
 
 The length of the branch EG occupied by the liquid of 
 
 . wl w'l' tan a 
 
 specific weight w is -. r- 
 tv(l+tan a) 
 
 14. If the bent tube is rotated in a vertical plane about B t 
 find the locus described by the point of contact of the two 
 liquids. 
 
 Result : if P is this point and PB = r, = angle made by 
 PB with the horizon, P being assumed to lie in the branch AB, 
 the locus is the curve 
 
 it} 
 
 r(sin + cos 0) = I' cos 6 -- -.1 sin 0. 
 
 w 
 
 15. A rectangular area, 
 LMRS, Fig. 17, whose 
 plane is vertical, has one 
 side, LM, in the free sur- 
 face of water; show how 
 to divide the area, by hori- 
 zontal lines into n strips 
 on each of which the water 
 pressure shall be the same. 
 
 Let LM =a, LR = h, 
 
 Fig. 17. 
 
38 Hydrostatics VOL. i 
 
 w = specific weight of liquid, and measure the depths. x v x z ,x s ,... 
 of the successive lines of division, / x m 1 , Z 2 w 2 , l z m s , ... each from 
 the surface. 
 
 Then the pressure on the rectangle Lm 1 = - (pressure on LS); 
 
 tv 
 
 ii Lm t = ( ) ; 
 
 7Z- 
 
 ?! si ii -^3 = - ( ,i ) ; 
 
 and so on. Thus, instead of calculating the pressures on the 
 
 separate strips, Lm l) L^vti^ L t m 3 , ... and equating them to - 
 
 n 
 
 of the pressure on LS, we take successive rectangles each having 
 one side L Af in the free surface. This is simpler. 
 
 Now the pressure on LS is o/t .-./; the pressure on Lm^ is 
 ax 1 . .w; pressure on Lm 2 is ax t . . w ; hence 
 
 ft 
 
 16. A triangular area, ABC, has its vertex A in the surface 
 of water, its plane vertical, and its base BC horizontal ; divide 
 the area by horizontal lines into n strips on which the water 
 pressures shall be equal. 
 
 Ans. The depths of the successive lines of division are 
 
CH. in Liquid Pressure on Plane Surfaces 39 
 
 17. A parallelogram is immersed vertically in water with 
 one side horizontal and at a depth of 6 feet below the 
 surface, the opposite side being at a depth of 14 feet; find 
 the horizontal line which divides the area into two parts 
 equally pressed. 
 
 The line is about 4-8 feet below the upper side. 
 
 18. ABCD is a parallelogram placed vertically in water with 
 the side AB, which is 9 feet long, at a depth of 6 feet, and the 
 side CD at a depth of 18. P is a point in AB distant I foot 
 from A. Show how to draw through P a line across the area 
 dividing the area into two parts equally pressed. 
 
 The required line cuts CD at a distance of 2 feet 
 from C. 
 
 19. ABCD is a trapezium whose parallel sides are AB and 
 CD ; the first is placed in the surface of water, and the second 
 is below at a depth of 15 feet, the plane of the figure being 
 vertical. AB = 14 feet, CD = 16. Find a point P in the 
 area such that the pressure on PAB should be equal to that on 
 PCD. 
 
 P lies anywhere on a horizontal line at a depth of 
 12 feet. 
 
 20. ABCD is a trapezium immersed as in the last case; find 
 a point P in the area such that the pressures on PAD and PBC 
 shall be equal. 
 
 In this case the areas PA D and PBC must be equal ; hence 
 the locus of P is the right line joining the middle points of 
 AB and CD. If the pressure on PAD is n times that on PBC, 
 the locus of P is still a right line. 
 
 21. ABCD is a parallelogram with the side AB in the surface 
 of the liquid and the side CD below. Show how to draw from 
 the corner D a right line across the area dividing the area into 
 two parts equally pressed. 
 
 If the line cuts BC in P, we have 
 
 PC: BC = 3-^3: 2. 
 
4o Hydrostatics VOL. i 
 
 22. In the last show how to draw the line through D so 
 that the pressure on the triangle DPC shall be to that on the 
 trapezium ABPD as 5 to 7. 
 
 P bisects BC. 
 
 23. A trapezium whose plane is vertical has one of the 
 parallel sides in the free surface of a liquid ; divide the area by 
 a horizontal line into two parts on which the liquid pressures 
 are equal 
 
 Let a, b be the parallel sides, the former lying in the surface ; 
 let h = height of trapezium ; let c = b a, and x = depth of 
 the required line ; then x is given by the equation 
 
 2c(2x 3 h s ) + 3ah(2x*~- h 2 ) = o (i) 
 
 The root of this equation which is relevant lies between 
 h . h 
 -y and j, which, respectively, correspond to the cases of 
 
 22 23 
 
 examples 15 and 16. 
 
 When b = o, or the area is a triangle with its base in the 
 
 surface and vertex down, the values of x in ( i ) are - , - ( i 
 
 the first of which alone is relevant to the problem, since 
 the latter two give, respectively, a value of x which is > h, 
 and a negative value of x both of which are physically 
 impossible. 
 
 24. A cube is filled with a liquid, and held with a diagonal 
 vertical ; find the pressures on one of the lower and one of the 
 upper faces. 
 
 2 i 
 
 --7=. W and 7= W, where W = the weight of the liquid 
 V3 Vs 
 
 in the cube. 
 
 25. A circular area is immersed in a homogeneous liquid, a 
 tangent to the circle lying in the free surface, A being the 
 highest point of the circle; draw a chord, BC, of the circle 
 perpendicular to the diameter through A so that the pressure 
 on the triangle ABC shall be a maximum. 
 
 The distance of BC from A is f of the diameter. 
 
CH. in Liquid Pressure on Plane Surfaces 41 
 
 26. A triangle has its base, BC, in the free surface of a liquid, 
 and its vertex A, down ; find a point, 0, in its area such that 
 the pressures on BOC, CO A, AOB shall be proportional to three 
 given numbers. 
 
 If the pressures on these areas are to the pressure on ABC 
 in the ratios ft : p 8 : p s , and h is the depth of A, the point 
 is the intersection of a horizontal line at a depth h */ '/?, with 
 a line drawn from A to a point, P, in BC such that 
 
 PC = f a ' 
 
 27. A semicii'cular area is placed in water with its diameter 
 in the surface ; show how to divide it into n sectors about the 
 centre on each of which the pressure is the same. 
 
 Divide the diameter into n equal parts ; the extremi- 
 ties of ordinates at the points of division determine the 
 sectors. 
 
 28. A triangular area, ABC, occupies any position in a liquid; 
 find a point, 0, in its area such that the liquid pressures on the 
 parts BOC, CO A , and A OB shall be proportional to three given 
 numbers. 
 
 Let a, /3, y be the depths of A, B, C below the free surface ; 
 let the ratios of the pressures on the above areas, respectively, 
 to the pressure on the whole triangle ABC be p l} p a , p a ; let z 
 be the depth of 0, and put x for z + a + fl + y; then x is deter- 
 mined from the cubic 
 
 Pi Pz PS r / \ 
 
 . . . 
 
 x a x ft xy a + fi + y 
 
 Assuming a>/3>y, the value of x in this equation which is 
 >a is the only one relevant, because the values which are 
 between a and /3 and between /3 and y give negative values of z. 
 
 The position of is completely defined by its areal co- 
 ordinates, i. e., by the ratios of the areas BOG, COA, AOB to 
 the area ABC. If these ratios are I, m, n, respectively, the 
 equations are 
 
 *(*+/3+y) = ft(a+/3+y), .... (2) 
 
 and two similar, where z is 2a + mfi + ny. When z is known 
 from (i), I is found from (2); &c. 
 
42 Hydrostatics VOL. i 
 
 13. Centre of Pressure. Hitherto we have been occu- 
 pied with the calculation of the magnitude of the resultant 
 pressure on one side of a plane area. We have now to 
 consider the point of the area at which this resultant 
 pressure acts. Except in the case in which the plane of 
 the area is horizontal, this point which is called the centre 
 of pressure is always lower in the area than G, the cen- 
 troid, or ' centre of gravity ', of the area. 
 
 The position of the centre of pressure on a given area 
 varies with the position (depth, orientation, &c.) of the 
 area in the fluid ; and before determining its position in 
 a few simple and frequently occurring cases, we shall lay 
 down a general principle, founded on the remark near the 
 middle of p. 31, which is often of great assistance in calcula- 
 tion. When a plane area or, indeed, any surface whatever 
 occupies any position in a liquid, we may draw any hori- 
 zontal plane whatever in the liquid and consider the column 
 of liquid above this plane as playing the part of an atmo- 
 sphere i.e., as producing at all points below the plane 
 a constant intensity of pressure, which is transmitted in 
 virtue of Pascal's Principle. The most convenient hori- 
 zontal plane for this purpose is one through the highest 
 point of the given area. 
 
 Thus, for example, if nrm y Fig. 18, is any plane area 
 
 whose plane is vertical in a 
 liquid, and we wish to find 
 the magnitude and point of 
 action of the resultant pres- 
 sure on one side of this area, 
 we may draw a horizontal 
 plane, CD, touching the con- 
 tour of the area at its highest point, n, and then consider 
 separately the pressures due to the layer of liquid between 
 AE and CD and to the body of liquid below CD, 
 
CH. in Liquid Pressure on Plane Surfaces 43 
 
 With regard to the layer ACDB, if x is its thickness, we 
 know that it produces at all points on CD and at all points 
 below (Art. n) an intensity of pressure equal to 
 
 w . x ; 
 
 and since this pressure is uniformly distributed over the 
 area n r m, its resultant is (case I , Art. 9) 
 
 A w x acting- at (?, (5) 
 
 where G is the centre of area of nrm and A the magnitude 
 of the area. 
 
 Hence, if we knew the magnitude and point, 7 , of 
 application of the pressure of the liquid below CD, we 
 should have the magnitude and point, /, of application of 
 the pressure of the whole liquid below AB on the area by 
 a simple composition of two parallel forces acting at G and 
 7 . This we shall presently illustrate by a few simple 
 examples. 
 
 Thus we obtain the following construction for the centre 
 
 of pressure, 7, on a plane area A B 
 
 (Fig. 19) occupying any posi- 
 tion in a liquid : through the 
 highest point, n, on the con- 
 tour of the figure, draw a 
 horizontal plane, CD, the free 
 surface of the liquid being 
 AS ; from the centroid, G, (or 
 
 ' centre of gravity ') of the figure draw a vertical line meet- 
 ing these planes in P and Q ; suppose 7 to be the (known) 
 position of the centre of pressure if the surface of the liquid 
 were CD ; draw QT , and from P draw PI parallel to QI , 
 meeting GT in 7. This point 7 is the required centre of 
 pressure on the area. We shall presently proceed to illus- 
 trate this method by some simple examples. 
 
44 Hydrostatics VOL. I 
 
 14. Special Cases of Centre of Pressure. 
 
 (i) To Jind the position of the centre of pressure on a 
 plane parallelogram^ whose plane is vertical, with one side in 
 the free surface. 
 
 Let ABDC, Fig-. 20, be the parallelogram. Let the area 
 be divided into an indefinitely great number of indefinitely 
 narrow strips, of which mnsr is the type, and let E and F 
 be the middle points of the sides AB and CD. Then the 
 middle point of every strip lies on the line EF. Also if x 
 is the depth of the strip ms below AB, and w the specific 
 weight of the liquid, the intensity of pressure is the same 
 at all points in the strip and (Art. 10) equal to wx, and the 
 
 resultant pressure on the 
 strip acts at its middle point, 
 i.e., at the intersection, f, 
 of ?ns with EF. Hence the 
 resultant pressure on the 
 whole parallelogram acts at 
 some point on EF. Also, 
 since the areas of the strips 
 are all equal, the series of pressures on them are simply 
 proportional to their distances from AB ; therefore (case 2, 
 p. 23) the point of application of the resultant pressure is f 
 of FE from E. Denote this point by T (see Fig. 21). Then 
 
 ET=%FE. (a) 
 
 If h is the height of the parallelogram, and p the 
 perpendicular distance of the centre of pressure, T, from 
 the surface p-\h (a'} 
 
 If the plane of the parallelogram is not vertical, the 
 same point, T, will still be the centre of pressure. For, if 
 the area is inclined to the vertical at any angle, 0, and x is 
 the perpendicular distance off from AB. we have 
 x =/E . cos0 ; 
 
CH. in Liquid Pressure on Plane Surfaces 45 
 
 and as 6 is the same for all the strips, the pressures on them 
 will still be proportional to their distances fE, &c. 
 
 (2) To find the position of the centre of pressure on a 
 plane triangle having one side in the free surface, and vertex 
 down. 
 
 Let ABD be the triangle. Divide the area, as before, 
 into an indefinitely great number of strips, of which ts is 
 the type. Let x be the perpendicular distance of this strip 
 from the base AB. Now compare this with another strip, 
 #/, whose perpendicular distance from D is also x. Let h 
 be the height of the triangle, a = AB, k = the indefinitely 
 
 ll OC 
 
 small breadth of each strip. Then tn = . a ; so that 
 (Art. 1 2) the pressure on this strip is 
 
 ~x(h-x)w ...... (i) 
 
 But this is also the pressure on the second strip, t'*'. 
 
 CG 
 
 For, t'n' ja, and the depth of t'ri \&hx\ therefore (i) 
 
 is the pressure on this strip. Since each strip is pressed at 
 its middle point, and since all the middle points lie on ED, 
 the resultant acts at some point on ED. Also we have just 
 seen that the pressures along ED are equal at two points 
 such that the distance of one from .Z7=the distance of the 
 other from D. Hence (case 4, p. 24) the resultant pressure 
 acts at the middle point, M, of ED (see Fig. 21) ; that is, 
 
 EM = | ED. . ' ..... (/3) 
 If Jo is the perpendicular distance of M from the surface 
 
 Also, whatever be the angle of inclination of the plane of 
 the triangle to the vertical, the same point, M, is the centre 
 of pressure. 
 
46 Hydrostatics VOL. i 
 
 (3) To find the position of the centre of pressure on a 
 plane triangle having a vertex in the free surface and its 
 base horizontal. 
 
 Let ACD (Fig. 21) be the triangle. Then a combination 
 of the two results just proved will enable us to find Q, the 
 centre of pressure. For, complete the parallelogram A BDC. 
 Then the pressure on the parallelogram is the resultant of 
 the pressures on the two triangles ACD and ADB. Let 
 AF bisect CD and let DE bisect AB. Let T be the centre 
 of pressure on the parallelogram, and M that of the triangle 
 ADB. Then the force at T is the resultant of one at J/ 
 and one at Q. Join the point, Af, of application of one of 
 the two parallel forces to the point, T, of application of the 
 
 B 
 
 resultant, and produce MT to meet AF in Q. Then Q is 
 the point of application of the pressure on ACD. 
 
 OF FT 
 
 Now ==* *=*'. 
 
 (y) 
 
 If h is the height of the triangle, and p the perpendicular 
 distance of Q from the free surface, 
 
 * = !*; ....... </) 
 
CH. in Liquid Pressure on Plane Surfaces 47 
 
 and, as before, the point, Q, of application of the resultant 
 pressure is the same whatever be the inclination of the 
 plane of the triangle to the vertical. 
 
 The result might have been deduced directly from case 3, 
 p. 23. For, if the area be divided into strips, we have 
 
 /Yl 
 
 mt = -ra, where a = CD, and x is the perpendicular from 
 
 A on mt. Hence the pressure on the strip mt is j wx 2 , so 
 
 that the pressures along AF are proportional to the squares 
 of the distances of their points of application from A. The 
 resultant, therefore, acts at a point f of the way down 
 along AF. 
 
 These three simple cases, combined with the principle 
 (see p. 31) of regarding any column of liquid as an atmo- 
 sphere, producing its resultant pressure at the centre of 
 area, will suffice for calculations concerning the centres of 
 pressure of many plane polygonal and other figures occupy- 
 ing any positions in a liquid. 
 
 Thus, let the area be nrm, Fig. 18, p. 42 ; and suppose 
 that, if all the liquid above the horizontal plane CD is 
 removed, we know the depth, J? , of the centre of pressure, 
 J , of the remaining liquid below CD. Then, if Z Q is the 
 depth of G below CD, A = magnitude of the area, w = 
 specific weight of the liquid, the pressure, P , at / is 
 
 A Z Q w. 
 
 Let x = the thickness of the column AD. Then the 
 pressure due to this column = Axw, and it acts at G. The 
 resultant pressure (at 1) is of course the sum of these forces; 
 and if jo is the depth of I below AS, we have, by the 
 theorem of plane-moments, 
 
48 Hydrostatic? 
 
 the point / dividing 7 6 T so that 
 
 VOL. I 
 
 IG 
 
 W 
 
 (4) To find the position of the centre of pressure on a 
 plane triangle occupying any position in a liquid. 
 
 Let ABC, Fig. 22, be the triangle ; let A be its area, 
 and a, /3, y the depths of its vertices below the free surface 
 of the liquid. 
 
 Take any point, P, in the area ABC, and let x, y, z be 
 
 the lengths of the perpen- 
 diculars from P on the sides 
 BC, CA, AB of the triangle. 
 The point P is the centre 
 of gravity of three masses 
 proportional to the areas 
 BPC, CPA,&nd APB placed, 
 respectively, at the vertices 
 
 Fig. 22. 
 
 //, B, C ; for if the line AP produced meets BC in m, we 
 have by Euclid VI. i, 
 
 Bm : mC = area BAm : area mAC ; 
 and also m :m C = BPiu : mPC ; 
 
 therefore Bm:mC = area BAP : area APC. 
 
 Hence the centre of gravity of the three masses above 
 named lies somewhere on AP ; similarly it lies somewhere 
 on HP, and it is, therefore, the point P itself. If the sides 
 BC, CA, AB are denoted by a, 6, c, the masses are pro- 
 portional to ax, by, cz. Now let f be the length of the 
 perpendicular from P on the free surface of the liquid, and 
 use the theorem of plane-moments (p. 1 9) thus : 
 
CH. in Liquid Pressure on Plane Surfaces 49 
 
 masses 
 
 distances from 
 free surface 
 
 products 
 
 ax 
 
 a 
 
 O.O.X 
 
 by 
 
 & 
 
 b/3y 
 
 cz 
 
 y 
 
 cyz 
 
 where A is the area of the triangle ABC, since the sum of 
 the first column is zA. 
 
 Now the intensity of pressure at P is w ; and we see 
 that this is equal to the sum of the three terms 
 
 aa b3 cy 
 
 .ivx. 7 ivy, .wz. 
 lA 2,A ' 2,A 
 
 in which the multipliers of x, y, z are the same for all points 
 P in the area. 
 
 The term wx would be the intensity of pressure at P 
 if the triangle ABC were placed vertically with the side BC 
 in the free surface of a liquid whose specific weight is w ; 
 the term -. wy would be the intensity of pressure at P if 
 the triangle ABC were placed vertically with the side CA 
 in the free surface of a liquid whose specific weight is -. w ; 
 
 C "V 
 
 and the term -. wz would be the intensity of pressure at P 
 if the triangle were placed vertically with AB in the free 
 
 C'V 
 
 surface of a liquid of specific weight ^ w. Hence we can 
 regard the actual distribution of pressure on the area ABC 
 
5o Hydrostatics VOL. i 
 
 (Fig. 23) as produced by a superposition of these three 
 fictitious pressure systems. 
 
 But if BC were placed in the surface of a liquid of 
 
 specific weight -w, the centre of pressure would be at i l 
 (Fig. 23) which is the middle point of the bisector, A A', of 
 
 the side BC ; and if p is the length of the perpendicular 
 from A on BC, the total amount of the pressure acting at i l 
 would be (Art. 1 2) 
 
 p aa . A 
 
 A--* .w, or Aaw. 
 
 3 2^ 
 
 Similarly, the total forces at i., and i z would be ? A ft to 
 and \Ayw. The actual pressure, then, on ABC in Fig. 23 
 is in magnitude and line of action the resultant of these 
 three forces, supposed acting at i 1} i z , and i 3 . We thus get 
 the result that the centre of pressure coincides with the centre 
 of gravity of three particles placed at the middle points of the 
 bisectors of the sides, their masses being proportional to the 
 depths of the corresponding vertices. 
 
 This is not quite the most convenient representation. 
 The force A aw at ^ can be replaced by ^ Aaw at B' and 
 Aaw at C' ; that at i 2 can* be replaced by ^ Afiw at C 1 
 uud | A ft ic at A'; and that at ? 3 by \Ayw at A' and 
 
CH. in Liquid Pressure on Plane Surfaces 51 
 
 ^ Ayw at B'. Thus the forces are now transferred to the 
 middle points, A', B', 6", of the sides, and their magnitudes 
 are A (/3 + y] w, &c. ; so that if we denote the depths of 
 A', B' t (7 below the free surface by , rj, the forces at A f , 
 Bf, C f whose centre coincides with the centre of pressure are 
 
 Hence we get one of the most useful rules in Hydrostatics, 
 which we shall call the Particle Rule 
 
 the centre of pressure on a triangular area which occupies any 
 
 position in a homogeneous liquid coincides with the centre of 
 
 gravity of three particles placed at the middle points of the 
 
 sides, their masses being proportional to their depths beloiv the 
 
 free surface. 
 
 If a plane area consists of two or more triangles, the 
 
 Fig. 24. 
 
 particles to be placed at the middle points of their sides are 
 to have masses proportional to the products of the several 
 areas and the depths of the middle points. Thus, for 
 example, suppose the plane area to be a trapezium ABCD 
 (Fig. 24) whose parallel sides are AB = 42, CD = 30, with 
 a perpendicular distance 18 between them, the side AB 
 being horizontal and at a depth 10 below the free surface, 
 LM. 
 
Hydrostatics 
 
 VOL. I 
 
 Break the area into two triangles, ABC and ACD. The 
 areas of these are proportional to 7 and 6, so that the particles 
 at the mid points of ABC can be taken as having masses 
 70, 133, 133 ; those at the mid points of ACD are 114, 
 1 14, 1 6 8. There is a double particle equal to 133 + 114 at 
 the middle point of AC. 
 
 Now since the trapezium can be broken up into infinitely 
 narrow horizontal strips the resultant pressure on each of 
 which acts at its middle point, the centre of pressure must 
 lie somewhere on the line joining the middle points of AB 
 and CD ; so that we have merely to find its distance from 
 some horizontal plane. The simplest plane to take is the 
 horizontal plane through PQ w T hich contains three of the 
 particles. Using, then, the principle of plane-moments, 
 and taking distances below PQ as positive, we have the 
 scheme : 
 
 % masses 
 
 distances 
 frotnPQ. 
 
 products 
 
 70 
 
 11+ 
 
 -19 
 o 
 
 -70 x 90 
 o 
 
 2+7 
 
 
 
 
 
 133 
 
 o 
 
 o 
 
 tbS 
 
 '9 
 
 tbfi x 19 
 
 732 
 
 
 
 9& x 19 
 
 Hence the distance of the centre of pressure below PQ is 
 98 x 19 
 
 73* 
 
 or 2-54. 
 
 The centre of pressure is, then, the point of intersection of 
 a horizontal line 2-54 below PQ and the line joining the 
 mid points of AJB, CD. This example serves as a type of 
 the calculation for all plane areas that can be broken up 
 into triangles. 
 
CH. in Liquid Pressure on Plane Surfaces 53 
 
 The proof of the Particle Rule given above was com- 
 municated to the author by the late W. S. M c Cay, Fellow 
 of Trinity College, Dublin ; the rule itself had been given 
 previously in a paper by the late Dr. E. J. Routh. 
 
 15. Pressure on Circular Area. Let a circular area, 
 AEBF, be immersed vertically in a liquid with its highest 
 point, A (Fig. 25), in the surface. It is required to find the 
 centre of pressure, /, on this area. This point lies on the 
 
 Fig. 25. Fig. 26. 
 
 vertical diameter, AB. Imagine a hemisphere to be con- 
 structed on the given area, and consider the separate 
 equilibrium of the water contained within this hemisphere. 
 The forces keeping this water in equilibrium are (i) its 
 weight acting vertically through its centre of gravity ; 
 
 (2) the pressure of the outside water acting over the 
 plane base AEF and producing its resultant at 7; and 
 
 (3) the pressure of the outside water acting all over the 
 curved surface of the hemisphere. This last produces 
 a resultant acting through the centre 0, since at each 
 element of the curved surface the pressure is normal to the 
 surface. 
 
 Let Fig. 26 represent a plane section of the hemisphere of 
 Fig. 25 made by a plane through AB perpendicular to the 
 plane of the paper. The circular base AEBF is represented 
 by the line CD. Now the centre of gravity, G, of a homo- 
 
54 
 
 Hydrostatics 
 
 VOL. I 
 
 geneous hemisphere of radius r is f ? from the centre 
 (Statics, vol. i, Art. 173), and the weight of the water 
 acting through G is 7rr :3 w; also the magnitude of the 
 pressure at / on the plane base is (Art. 12) -nr 3 w\ and the 
 resultant of these two forces must act through 0, because it 
 is equal and opposite to the resultant action on the curved 
 surface. Hence 
 
 TT f* 8 w nl 
 
 ~'W 
 
 where n is the point of meeting of the two forces. Now 
 nl= OG = r 
 
 i.e. the centre of pressure is one-quarter of the radius below 
 the centre. From this by the principle of Art. 13 we obtain 
 the position of the centre of pressure when the circular 
 area is at any depth. 
 
 16. Semicircular Area. In precisely the same way we 
 can find the centre of pressure on a semicircular area whose 
 
 bounding diameter is in the free surface. Let EFbe the 
 bounding diameter (Fig. 27), and consider the separate 
 equilibrium of the liquid contained within the quarter- 
 sphere having the semicircle EFB for base. Let Fig. 28 
 represent, as before, a section of this quarter-sphere through 
 OB perpendicular to the plane of the paper. The centre 
 of gravity, <?, lies on the radius which is inclined at 45 to 
 
CH.'IH Liquid Pressure on Plane Surfaces 55 
 
 OD t and its perpendicular distance, gO, from the base is the 
 same as for a hemisphere, i. e. 
 
 9 = lr=gG. 
 The weight of liquid in the quarter-sphere is j -rrr 3 w, and 
 
 the pressure on the base EBF is ^isr* . --w^iace (Statics, 
 
 Art. 165) the distance of the centroid of a semicircular area 
 
 4 V 
 from the centre is -- Also the liquid pressures all over 
 
 the curved surface pass through ; hence 
 r*w nl ? 
 
 16 
 
 EXAMPLES. 
 
 1. A triangular area whose height is 12 feet has its base 
 horizontal and vertex uppermost in water; find the depth to 
 which its vertex must be sunk so that the difference of level 
 between the centre of area and the centre of pressure shall be 
 8 inches. 
 
 Four feet. (Use the principle in p. 31 .) 
 
 2. Find the depth of the centre of pressure on a trapezium 
 having one of the parallel sides in the surface of the liquid. 
 
 If the side a is in the surface and 6 below, h being the 
 height of the trapezium, the depth of the centre of pressure 
 
 is 
 
 2 2 + a 
 
 and it lies, of course, on the line joining the middle points of a 
 and b. 
 
 3. In the last example find the position of the centre of 
 pressure by geometrical construction. 
 
 (Break up the area into a parallelogram and a triangle, or 
 two triangles.) 
 
56 Hydrostatics VOL. i 
 
 4. The plane of a trapezium being vertical, and its parallel 
 sides horizontal, to what depth must the upper side be sunk in 
 a liquid so that the centre of pressure shall be at the middle 
 point of the area ? 
 
 Ans. The parallel sides being a and 6, of which the upper, 
 
 a, must be the greater, the required depth = h, where h = 
 
 CL 
 
 height of the trapezium. 
 
 5. A rectangular area of height h is immersed vertically in a 
 liquid with a side in the surface ; show how to draw a horizontal 
 line across the area so that the centres of pressure of the parts of 
 the area above and below this line shall be equally distant 
 from it. 
 
 The line must be drawn at a depth (A/S r) - 
 
 2 
 
 6. A BCD is a trapezium whose parallel sides, AB and CD, 
 are 16 and 32 feet long, respectively, their perpendicular distance 
 being 12 feet. AB is horizontal, at a depth of 20 feet below the 
 free surface of water in which the area is immersed vertically, 
 CD being below AB. Find the position of the centre of 
 pressure. 
 
 The centre is i-i feet below the mid horizontal line of the 
 figure. 
 
 7. Given that when a circular area of radius r is immersed 
 vertically in water with its highest point in the surface the 
 
 V 
 
 centre of pressure is - below the centre, find the position of the 
 
 4 
 
 centre of pressure on a vertical circular area of radius 2 feet, its 
 centre being at a depth of 6 feet below the free surface. 
 
 2 inches below the centre. 
 
 8. The parallel sides of a trapezium are 40 and 30 feet long, 
 and the height is 24 feet. If the figure is immersed vertically 
 in water with the parallel sides horizontal and the longer 
 uppermost, what must be the depth of this side below the 
 free surface so that the centre of pressure shall lie on the 
 horizontal centre line ? 
 
 Ans. 72 feet. 
 
CH. in Liquid Pressure on Plane Surfaces 57 
 
 9. What must be the depth of the upper side of this trapezium 
 so that the centre of pressure shall be 
 
 (a) 8 inches above the horizontal centre line 1 
 
 (b) $ feet above this line 1 
 
 Ans. The first position is impossible at any depth; the 
 second requires an infinite depth. 
 
 10. ABCD is a trapezium whose parallel sides, AB, CD, are 
 30 and 1 6 inches long, respectively; the sides EG and DA 
 are 15 and 13 ; and the figure is placed in water with A in the 
 surface and AB vertical. Find the position of the centre of 
 pressure. 
 
 Its depth is 17-54, and its distance from the vertical centre 
 line is -75, towards AB. 
 
 11. A triangular area ACB is placed vertically in water with 
 the point C in the surface ; the area being turned in its plane 
 round C, prove that, so long as the area remains completely 
 immersed, the centre of pressure describes a right line in the 
 area, this line being parallel to AB at a distance equal to 
 i (height of triangle). 
 
 Apply the particle rule. If in any position x and y are 
 the masses of particles at the mid points of BC and CA, the 
 particle at the mid point of AB is x + y. 
 
 12. ABC is an isosceles triangle, CA = CB' t it is placed 
 vertically in water with A in the surface and AB vertical ; 
 prove that the c. p. is vertically below the centre of gravity at 
 
 A' 4. AB 
 
 a distance 
 
 12 
 
 Use the particle rule. 
 
 13. ABCD is a parallelogram with AB in the surface and 
 CD down. Lines are drawn from D across to points, P, on BC. 
 Find the locus of the centre of pressure of the triangle DPC as 
 P varies. 
 
 14. A plane area in the form of a regular hexagon is placed 
 vertically in water with one side in the surface ; find the 
 position of the centre of pressure by an application of the 
 particle rule. 
 
 Its depth is ^ of the height of the hexagon. 
 
5 ft Hydrostatics VOL. i 
 
 1 5. A rectangular vessel of height h contains liquid of specific 
 weight w' to a height f, the remainder being filled with a liquid 
 of specific weight w ; prove that the distance of the centre of 
 pressure on one of the vertical faces from the base is 
 
 / w) 
 
 ' w) ' 
 
 16. The water at one side of a rectangular dock gate of 
 breadth b stands at a height a, and the water at the other side 
 at a height c; find the magnitude and line of action of the 
 resultant water pressure on the gate. 
 
 If w is the weight of a unit volume of water, the 
 
 magnitude is \w (a 2 c 2 )6, and the line of action is at a 
 
 a _ f 
 
 height 4 - s above the bottom of the gate. 
 3 a' c 18 
 
 17. The gates of a canal lock are each 12^ feet wide, and the 
 breadth of the lock is 24 feet; the water at one side is 18 feet 
 high and at the other side 12 ; prove that the magnitude of the 
 thrust between the gates is about 56 tons' weight. 
 
 18. A plane area of any form is immersed vertically in water 
 with its highest point in the free surface ; and in this position 
 h is the depth of the centre of gravity and h + c that of the 
 centre of pressure. If the area is lowered into the water 
 without rotation and with uniform velocity v, prove that the 
 vertical velocity of the centre of pressure at the time t, reckoned 
 from the initial position, is 
 
 ch 
 
 19. A circular area of radius r is immersed vertically in 
 a liquid, the depth of the centre of the area below the surface 
 of the liquid being h ; show from the consideration of the 
 separate equilibrium of the hemisphere of liquid having the 
 given area for base that the depth of the centre of pressure on 
 
 the area, below its centre, is 
 
 r 
 
 20. A triangular area ABC of height h is immersed vertically 
 in'water with C in the surface and AB horizontal; the area is 
 divided by a horizontal line PQ into two parts on which the 
 
CH. in Liquid Pressure on Plane Surface* 59 
 
 pressures are equal. Prove that the depth of the centre of 
 pressure on the lower part is 
 
 21. A circular area of radius r is immersed vertically in 
 water, its centre being at a depth h ; is the lowest point of 
 the area. Show how to draw a horizontal chord, PQ, of the 
 circle so that the depth of the centre of pressure on the triangle 
 OPQ is a minimum. 
 
 Result. The depth is ( -- i V^ + r). If, however, this 
 
 -1 i 2 v/5 
 
 is<A r, that is, if h> - r, the minimum will correspond 
 
 5 
 
 to a chord PQ of infinitesimal length at the highest point of 
 the circle. The maximum depth corresponds, of course, to an 
 infinitely short chord touching at 0. 
 
 17. Self-acting Sluice. If an aperture of any shape 
 (rectangular, circular, &c.) is made in a vertical wall or a 
 lock-gate at one side of which there is a mass of water 
 rising to any height, and the aperture is closed by a rigid 
 plane surface movable about a horizontal axis in its plane, 
 this axis can be so fixed that when the level of the water 
 rises to any assigned height above the top of the sluice the 
 sluice will open, let out the water, and prevent any increase 
 in the height of the water. 
 
 It is obvious that the axis must 
 be fixed at the centre of pressure 
 on the sluice corresponding to the 
 assigned level of the water. 
 
 Thus if the sluice is the rectangle 
 ABCD whose sides are horizontal 
 and vertical, and if AD = 6 feet, 
 while the sluice is to open when the 
 level of the water is 6 feet above 
 AJ3, we can find the depth of the centre of pressure below 
 
 m n 
 
60 Hydrostatics VOL. i 
 
 the line mn which bisects AD and BC by breaking- the area 
 into two triangles whose areas may be taken as i , I ; and 
 the depths of middle points as 2, 3, 3, 3, 4, so that the 
 
 depth of c. p. below mn is - f '* - , i. e. i of a foot. 
 
 ' 
 
 EXAMPLES. 
 
 1. If a rectangular sluice 3 feet high is to open when the 
 level of the water rises i feet above its top, where must 
 the axis of the sluice be fixed? 
 
 Ans. 3 inches below the centre of the rectangle. 
 
 2. If the height of the sluice is h, and if it is to open when 
 the level of the water rises to a height x above its top, where 
 must the axis be fixed ? 
 
 ?2 
 
 Ans. i - below the centre. 
 
 ' 2X+k 
 
 3. Supposing a rectangular vessel whose base is horizontal to 
 be divided into two water-tight compartments by means of a rigid 
 diaphragm movable round a horizontal axis lying in the base of 
 the vessel ; if water is poured into the compartments to different 
 heights, find the horizontal force which, applied to the middle 
 point of the upper edge of the diaphragm, will keep this dia- 
 phragm vertical, and find the pressure on the axis. 
 
 Let a be the length of the axis, c the height of the 
 vessel, h and h' the heights of the water in the compartments 
 
 (h > h') ; then the required force is (h 3 h' 3 ) w, and the pressure 
 
 O C 
 
 on the axis is ^a(h 2 h' 2 )w (h 3 h' 3 )w. 
 
 O C 
 
 IS. Lines of Resistance. Supposing Fig. 30 to re- 
 present a vertical transverse section, ABCD, of an embank- 
 ment which is pressed by water on the side AB (assumed 
 vertical), if we take any horizontal section, PQ, of the 
 embankment and consider the equilibrium of the portion, 
 QPAD, above this section, we see that it is acted upon by 
 
OH. in Liquid Pressure on Plane Surfaces 61 
 
 its weight and also by the water pressure which is a 
 horizontal force acting at a point two-thirds of the way 
 down AP. Taking the resultant of these two forces, its 
 line of action, RS, intersects the section PQ in a point R, 
 the calculation of the position of which is of great impor- 
 tance in the construction of reservoirs. 
 
 As the section PQ varies in position, the point R 
 describes a curve which is called a line of resistance ; and 
 it is considered essential to the stability of the wall AB 
 that this curve must cut each horizontal section, PQ, in 
 
 Fig. 30. 
 
 a point, R, which lies somewhere within the middle third of 
 that section ; that is, if we divide PQ, into three equal 
 parts, R must lie somewhere within the middle part. 
 
 Our figure represents a comparatively simple case that 
 in which the vertical cross-section of the embankment is 
 a trapezium ABCD with the side next the water vertical. 
 If we make AD = o, we have the case of a triangular 
 section. 
 
 Let AB h, AD = a, BC = a + c ; draw Dpq vertical, 
 and consider the area APQD as composed of a rectangle, 
 APjiD, and a triangle j)QD ; let AP = y ; let w = specific 
 
62 Hydrostatics VOL. i 
 
 weight of water, w' = sp. weight of the masonry ; let be 
 the mid point of AD, and take as axes of reference horizontal 
 and vertical lines through 0. Let the figure represent 
 a portion of the embankment one unit of" length thick 
 the thickness being measured perpendicularly to the plane 
 
 m wiP 
 
 of the figure. Now the water pressure against AP is - 
 
 v 
 acting at the centre of pressure /, which is - above P ; the 
 
 \J 
 
 weight oAPpD is w'ay acting through its middle point, 
 
 w ct/^ 
 G; the weight of jpQD is j- acting through the centre 
 
 /' 
 
 of gravity G' and cutting PQ in a point r such that 
 
 The resultant of these forces must be counteracted by the 
 stress exerted over the section PQ on the portion APQ.D 
 by the lower portion, PCQ ; so that the resultant of this 
 stress must be a force acting through R and opposite to the 
 resultant of the above three forces. 
 
 The most simple way to find the position of R is to 
 express the fact that the algebraic sum of the moments of 
 these three forces about R is zero. Let tR = x ; then by 
 moments we have 
 
 wy z y w'cij 2 / cu a \ f 
 
 - - + -j- (-*- + # ) - w ay . x = o. 
 
 2 3 2 /t V 3 h 2 
 
 If we put w' nw, where n will usually be between 
 2 and 3, this equation is 
 
 and this gives the locus of li as the section PQ, varies. 
 We see that the locus is a hyperbola passing through 0, 
 having one asymptote horizontal and the other parallel to 
 the Hue indicated between the brackets. 
 
CH. in Liquid Pressure on Plane Surfaces 63 
 
 In the special case in which AD = o, the locus reduces 
 to the right line 
 
 If BC AD, that is if the vertical cross- section is a 
 rectangle, c o, and the locus is the parabola 
 
 f = 6an.x, ...... (3) 
 
 whose vertex is at 0. 
 
 We have assumed that the reservoir is full. 
 
 To find where the curve (i) cuts the base BC, put y = //, 
 and we have 
 
 and for safety this must be < ^ (a + 4 c). 
 
 Thus for a rectangular cross-section the condition for 
 
 stability is h < a ( 
 
 EXAMPLES OF LINES OF RESISTANCE. 
 
 1. The vertical cross-section of an embankment is a rectangle, 
 and the water reaches to a distance c from the top ; show that 
 the equation of the line of resistance referred to horizontal 
 and vertical lines through the mid point of the top side 
 of the rectangle is 
 
 / xs 6w ' 
 
 (y cf = axy, 
 
 where a is the breadth of the rectangle, w and w' being the 
 specific weights of water and the masonry. 
 
 2. The vertical cross-section of an embankment is a rectangle 
 of height h and breadth a ; find the greatest height to which 
 the reservoir can be filled so that the embankment shall be safe 
 from failure by tilting over the outer edge. 
 
 Ans. ( a?h ) 
 
 v 10 / 
 
64 Hydrostatics VOL. i 
 
 3. The vertical cross-section of an embankment is the 
 trapezium in Fig. 24, p. 51. Given AD = 6 feet, BC = 20, 
 AB = 1 6, mass of i cubic foot of masonry = 140 pounds, find 
 the point in which the line of resistance cuts the base. 
 
 At a distance of 8-56 feet from B. 
 
 4. If in the same figure AD = 6, AB = 18, / = 140 lb., 
 find the least length of BC so that the line of resistance may cut 
 the base within the middle third. 
 
 BC= 10-8. 
 
 5. If in the same figure the straight line AC ia replaced by 
 any curve, x = </>(?/), show that the equation of the line of 
 resistance is 
 
 6. If the curve AC is a parabola, so is the line of resistance. 
 
 10. Stress in thin pipes. When a hollow tube, or pipe, 
 is filled with water derived from an elevated reservoir, the 
 very considerable water pressure in the tube produces 
 a transverse tension in the tube which tends to split it. 
 Suppose that Fig. 31 represents a transverse section of the 
 tube, and consider the separate 
 equilibrium of a small part of 
 the tube made by two close 
 radial sections OP and OQ in- 
 clined at the angle POQ or 
 d, and contained between two 
 close planes perpendicular to 
 the axis of the tube i. e. by 
 Fi 3I the plane of the figure and one 
 
 above it at the small height 
 
 c. In reality we imagine the figure to lie midway be- 
 tween these close planes. 
 
 Now the normal pressure N causes the portions of the 
 
CH. in Liquid Pressure on Plane Surfaces 65 
 
 material outside P and Q to exert a tearing force, T, at P 
 and at Q. Also N acts along O;/, the bisector of POQ. 
 Resolving along On for equilibrium, we have 
 
 N = 2 Tew- , or N=TB, nearly . . . (i) 
 
 2 
 
 Now if jo is the intensity of the water pressure, 
 jV = j0.r0.e, where r is the inner radius, since the area 
 pressed is rO . e. Also if t is the intensity of the tearing 
 force i. e. the magnitude of this force per unit area 
 T = t . e . r, where T is the (small) thickness of the tube. 
 Hence we have from (i) 
 
 tr =jpr ....... (2) 
 
 The bursting of the pipe depends on the magnitude of t, 
 which is tabulated for pipes of various materials in tons' 
 weight per square inch or other appropriate units. 
 
 If instead of a thin cylinder we have a thin spherical 
 surface subject to internal pressure, 
 p, the intensity of tearing stress is 
 only half of that for the cylinder of 
 same radius. 
 
 Let Fig. 32 represent a small 
 patch of the spherical surface in- 
 cluded by four great circles ; then, 
 considering its separate equilibrium, 
 we shall have four T's instead of Fi 2 
 
 two, similarly related to the normal 
 
 pressure N, so that by resolving along the normal we have 
 now evidently 
 
 which gives 
 
 t.r = kpr .... (3) 
 
 Thus in the case of a cylindrical boiler closed at both 
 ends by spherical caps, if the thickness is the same through- 
 
66 Hydrostatics VOL. i 
 
 out, the tendency to burst at the side is twice as great as 
 that at the ends. 
 
 In the case of thick pipes the tension T (or ' hoop stress ') 
 varies throughout the thickness, and the preceding results 
 do not tell us anything about the magnitude of the stress 
 at a definite point in the substance. 
 
 In (2) and (3) it must be understood that p means the 
 excess of internal over external intensity of pressure : the 
 external pressure will often be that of the atmosphere. 
 
 EXAMPLES. 
 
 1. Assuming that the greatest intensity of stress permissible 
 in a metal pipe is 5,000 Ib. weight per square inch, find the 
 greatest intensity of pressure allowable inside a pipe i foot in 
 diameter and ^ inch thick. 
 
 Result. 138-8 Ib. weight per square inch. 
 
 2. Water is conveyed in pipes from a height of 400 feet ; the 
 diameter of the pipes is i foot, and the maximum allowable 
 stress is 2,800 Ib. weight per square inch; what is the least 
 thickness of pipe necessary ? 
 
 Result. -37 inch. 
 
 3. A thin elastic spherical envelope, of radius r, contains air 
 at atmospheric pressure ; if n times the mass of the original air 
 is forced into it, find its new radius, assuming that the increase 
 of surface is proportional to the intensity of tension. 
 
 Result. If / is the new radius, and r' 2 r 2 = kT, r' is 
 found from the equation 
 
 2 / 2 (r' 2 - v- 2 ) = kp [(n + i ) r 3 - r' 3 ]. 
 
CHAPTER IV 
 
 PRESSUEE ON CURVED SURFACES: PRINCIPLE 
 OF BUOYANCY 
 
 20. Principle of Buoyancy. If any curved closed 
 surface, M (Fig. 5, p. 10), be traced out in imagination in 
 a fluid acted upon by gravity, the pressures exerted on all 
 the elements of this surface by the surrounding fluid have 
 a single resultant, which is equal and opposite to the 
 weight of the fluid enclosed by M. 
 
 This is evident, because the fluid inside M is in equili- 
 brium under its own weight and the pressure exerted on 
 its surface by the surrounding fluid ; hence this pressure 
 must reduce to a vertical upward force equal to the weight 
 of the fluid inside M and acting through the centre of 
 gravity of this fluid. 
 
 This is obviously true whatever be the nature of the 
 fluid liquid or gaseous, homogeneous or heterogeneous. 
 
 If the curved surface M is not one merely traced out in 
 imagination in the fluid, but the surface of a solid body 
 displacing fluid, the result is the same 
 
 tfie resultant pressure of a heavy fluid on the surface of any 
 solid body M is a vertical upward force equal to the weight of 
 the fluid which could statically replace M, and this force 
 acts through the centre of gravity of this replacing fluid. 
 
 v 2 
 
68 Hydrostatics VOL. i 
 
 Let Fig. 33 represent the solid body, which we may 
 imagine to be a mass of iron, of solid rock, or any other 
 substance, the surrounding fluid being water, air, or any 
 fluid acted upon by gravity. The body is represented as 
 held in its position by cords attached to fixed points, 
 C,D, ..., and the arrows represent pressures exerted on its 
 surface by the fluid at various points. 
 
 Now it is quite clear that if the body were replaced by 
 any other one having exactly the same surface and 
 occupying exactly the same position, the pressure on each 
 element of its surface would be identically the same as 
 before, of whatever substance the new body may be. If 
 
 the new body were of 
 wood, or instead of 
 being solid were a thin 
 hollow shell, it might 
 be necessary to keep it 
 in the position repre- 
 sented by means which 
 prevent its rising up 
 33- out of the fluid ; but 
 
 we are not at all con- 
 cerned with the forces which keep the body M in this 
 position ; our object is merely to ascertain the resultant, 
 if any, of the fluid pressures exerted in the given position 
 on its surface. 
 
 In general, a number of forces acting in various lines 
 which do not lie in one plane have no single resultant : 
 their simplest reduction is to two forces whose lines of 
 action do not meet (Staticg, vol. ii, chap. xiii). But it 
 is remarkable that the pressures exerted on the various 
 elements of any closed surface by a heavy fluid lave a single 
 resultant ; and the truth of this we see by imagining the 
 place occupied by M to be occupied by a portion of the fluid 
 
CH. iv Pressure on Curved Surfaces 69 
 
 itself, placed in the vacancy without disturbing any of the 
 surrounding fluid. 
 
 With regard to this replacing fluid, observe two things : 
 firstly, it is in equilibrium ; secondly, it is kept so by its 
 own weight and the very same system of pressures as that 
 which acted on the body M, since this body and the re- 
 placing fluid present identically the same surface to the 
 surrounding fluid. Hence, then 
 
 the system of pressures has a single resultant which is a 
 vertical upivard force equal to the weight of the statically 
 replacing fluid and acting through the centre of gravity of this 
 fluid. 
 
 The centre of gravity, H, of the replacing fluid is called 
 the centre of buoyancy ; and, so far as the general principle 
 of buoyancy is concerned, there is no relation between H 
 and the centre of gravity, G, of the body ; nor is there 
 any relation between the weight, W, of this body and the 
 weight, L, of the displaced fluid. 
 
 If the fluid is water, or any homogeneous liquid, the 
 resultant pressure is the weight of the liquid which would 
 flow into the vacant space if M were removed ; but if the 
 density of the fluid is different in different layers, we must 
 not imagine the replacing fluid to be that which would 
 flow in when M is removed, but rather to be a continuation 
 of the surrounding fluid placed in the vacancy without any 
 disturbance of the external fluid, and having the same 
 surfaces of equal density as this fluid. The distiibution of 
 this replacing fluid is unique and determinate, as will be 
 subsequently proved. 
 
 COR, i. Pressure of uniform intensity exerted over any 
 closed surface produces no resultant. 
 
 For, imagine the closed surface to be one traced out in 
 a perfectly weightless fluid or a very light gas whose 
 surface is subject to any pressure. 
 
yo Hydrostatics VOL. i 
 
 The intensity of pressure will be uniform throughout the 
 whole fluid, and therefore over the given surface ; and by 
 what has just been said, the resultant pressure over this 
 surface is equal to the weight of the enclosed fluid that is 
 to say, zero. 
 
 The result of this Corollary may also be thus stated : 
 given any closed curve, plane or tortuous, in space ; if 
 a surface of any size and shape be described having this 
 curve for a bounding edge, and if pressure of uniform 
 intensity be distributed over one side of this surface, the 
 resultant of this pressure is the same whatever the size and 
 shape of the surface. 
 
 Hence if the given bounding curve is plane, the resultant 
 pressure on any surface having it for a bounding edge is 
 the same as the resultant pressure on the plane area of the 
 curve. 
 
 COR. 2. The principle of Archimedes. 
 The pai'ticular case in which the solid body M which 
 displaces fluid is in equilibrium solely under the action of its 
 own weight and the fluid pressure over its surface furnishes 
 the Principle of Archimedes. 
 
 The resultant of the 
 system of fluid pressures 
 must then be exactly 
 equal and opposite to the 
 weight of the displacing 
 body. 
 
 Thus, let Fig. 34 repre- 
 sent a heavy body whose 
 Fig. 34. centre of gravity is G } 
 
 floating in equilibrium in 
 
 a heavy fluid. The surface over which the fluid pressure is 
 exerted is ADB, which is not a closed surface ; but, as there 
 is no pressure due to the fluid exerted over the free surface, 
 
CH. iv Pressure on Curved Surfaces 71 
 
 , of the fluid, we can suppose the immersed surface ADB 
 to be closed by the section of the body made by the horizontal 
 plane AB. Hence the resultant of the pressures is the 
 weight of the fluid that would fill the space ADB ; and if 
 H is the centre of gravity of this fluid, the resultant 
 pressure acts up through //, so that G and H must be in 
 the same vertical line. Hence there are two distinct con- 
 ditions of equilibrium of a body floating freely in a heavy 
 fluid, viz. 
 
 (1) the weight of the body must be equal to the weight of the 
 fluifl ichich it displaces ; and 
 
 (2) the centre of gravity of the body and the centre of gravity 
 of the flriid that would statically Jill its place (centre of 
 buoyancy] must be in the same vertical line. 
 
 The student must observe that the principle of Archi- 
 medes applies to a closed surface, which is completely 
 surrounded by liquid or, as pointed out above, to an 
 unclosed surface, such as that of the immersed portion of a 
 ship, which may be supposed to be closed by a surface of 
 zero pressure. It does not apply to such a case as the 
 following : 
 
 ABC (Fig. 35) is a solid body with a perfectly flat base, 
 BC, resting on the base of a vessel into 
 which water is poured, the fit of the 
 bases being so accurate that no water 
 flows in between them. The water 
 pressure on the body, so far from an 
 
 -D 
 
 upward force, is a downward one, the 
 
 reason being that the surface pressed is 
 not closed and subject all over to water pressure, and 
 that it cannot be assumed to be closed by a surface of 
 zero pressure, as the intensity of the pressure at the level 
 of BC is not zero. We might as well expect the elevated 
 curved portion of the bottom of a champagne bottle to 
 
72 Hydrostatics VOL. I 
 
 be raised by the liquid as expect the body in Fig. 35 to 
 be urged upwards. It is important also to bear in mind 
 that the principle of Archimedes is not identical with the 
 general principle of buoyancy, but only an application of 
 the latter to the particular case of a body floating freely in 
 a liquid under the influence of no forces but its own weight 
 and the pressure of the surrounding liquid. 
 
 We have hitherto supposed that the only fluid displaced 
 by the body is that represented in the vessel below the 
 surface LM ; but if above this there is air, whose weight is 
 considered, there is also displaced a volume of air repre- 
 sented by ACB, and the resultant effect of the air is to 
 produce an upward vertical force, even though (as in the 
 figure) the air pressure exerted by the air actually in contact 
 with the displacing body should be a downward force ; for. 
 we must remember that the surface LM of the lower fluid 
 is all subject to air pressure which is (by Pascal's Principle) 
 transmitted undiminished all through this fluid, so that the 
 lower part, ADB t of the surface of the body is really acted 
 upon all over by air pressure of constant intensity. Now 
 by Cor. I, the resultant of this system of air pressures on 
 the curved surface ADB is the same as if the pressure was 
 applied over the lower side of the plane area AB in which 
 the surface LM cuts the body. The resultant air pressure 
 is, therefore, an upward force equal to the weight of the air 
 that would statically fill the space ACB, and it acts through 
 the centre of gravity of this air. 
 
 The case of a balloon floating in the air is also an instance 
 of the principle of Archimedes ; the force of buoyancy is the 
 weight of the air that could statically replace all the solid 
 portions of the balloon and the gas which it contains. It 
 must not be supposed that, since the balloon is a com- 
 paratively small body, the intensity of the air pressure is 
 constant all over its surface a not unnatural error ; for. if 
 
CH. IV 
 
 Pressure on Curved Surfaces 
 
 73 
 
 this air pressure were of constant intensity all over the 
 surface, its resultant would be absolutely zero, as we have 
 already seen, and there would be no force of buoyancy. 
 If the medium surrounding 1 a body is ever so slightly acted 
 upon by gravitation, its intensity of pressure cannot be 
 constant, and hence the densities of the air at the top and 
 at the bottom of the balloon tire not the same. 
 
 MISLEADING STATEMENT OF THE PRINCIPLE OF ARCHIMEDES. 
 
 A desire for conciseness of expression leads very often to 
 misleading and erroneous statements of scientific facts. 
 Thus, a very common reply to the question ' what is the 
 principle of Archimedes ? ' is simply this ' the weight of 
 a floating body is equal to the weight of the fluid displaced.' 
 To show the misleading nature of this reply, take the 
 following example : 
 
 A cylinder contains a small quantity of water whose 
 level is AB (Fig. 36) ; a cylindrical 
 block of wood, mnpq, whose diameter is 
 nearly equal to that of the vessel is 
 lowered into the water and allowed to 
 float, if it can do so. Suppose that it 
 does float without touching the bottom 
 of the vessel, and occupies the position 
 m'n'p'q'. There will remain a thin layer 
 of water at the bottom between the 
 vessel and the wood, and the remainder 
 of the water will be forced up the sides and attain the 
 level A'B'. 
 
 Now the weight of the floating block may be very many 
 times as great as that of the water actually displaced and, 
 in fact, greater than that of all the water present. The 
 force of buoyancy is not at all equal to the weight of the 
 water displaced, but to the very much greater weight of the 
 
 171, Tt, 
 
 
 
 w; . 
 
 I J 
 
 A B : 
 
 A B 
 
 t' | 
 
 
 
 Fig. 36. 
 
74 Hydrostatics VOL. i 
 
 volume, A'B'j/q', of water that would occupy the place of 
 the immersed portion of the body without disturbing- the 
 surrounding water. 
 
 As a numerical example, suppose that the radius of the 
 vessel is 3 cm., the radius of the cylindrical body mnpq 
 2-9 cm., the mass of this body 240 grammes, and the height 
 ofAB above the base i cm. In this case we find that the body 
 will float with a layer of water about -41 cm. below it, and 
 that the water will rise to a height of about 9-5 cm. above 
 the base of the vessel. 
 
 Here the fluid actually displaced is only a column of 
 radius 2-9 and height equal to the difference of level of AB 
 and f/p', that is about -6 cm. ; and the weight of the water 
 displaced is the weight of about 15-85 grammes, whereas the 
 force of buoyancy is the weight of the volume A'B'p'q', 
 and is, of course, the weight of 240 grammes. 
 
 21. Introduction of Fictitious Forces. In the case in 
 which a body is partially immersed in a fluid, or a part of 
 the body is in one fluid and the remainder in another, it is 
 often very convenient to introduce fictitious forces of buoy- 
 ancy in one part of the calculation and to take them away 
 in another. 
 
 Thus, suppose Fig. 34 to represent a body of which the 
 portion ADB is immersed in water while the portion ACB 
 is in vacuo. Then the actual force of buoyancy is due to 
 the volume ADB of water ; but we can complete the volume 
 of the displaced water by supposing the portion ACB to be 
 also surrounded by water, and then supposing that there is 
 a downward force, in addition, due to the action of this por- 
 tion ACB of water taken negatively. Thus the actual force 
 of buoyancy viz. an upward force at H equal to the weight 
 of the volume ADB of water can be replaced by an upward 
 force equal to the weight of the whole volume ADBC of 
 water acting at the centre of gravity of the homogeneously 
 
CH. iv Pressure on Curved Surfaces 75 
 
 filled volume ADBC(i\o{, G, the c. g.of the body, unless the 
 body is itself a homogeneous solid), together with a down- 
 ward force equal to the weight of the volume ACB of water 
 acting at the centre of gravity of the homogeneously filled 
 volume ACB. 
 
 In the same way, if the portion ACB is in a liquid of 
 specific weight w l5 and ATJB in one of specific weight w 2 , 
 we may regard the force of buoyancy as consisting of an 
 upward force equal to the weight of the whole volume 
 ADBC of the liquid w z together with a downward force 
 equal to the weight of a fictitious liquid of specific weight 
 w z ~ w i ac ting at the centre of gravity of the homogene- 
 ously filled volume ACB. 
 
 EXAMPLES. 
 
 1. A solid homogeneous right cone floats in a given homo- 
 geneous liquid ; find the position of equilibrium, firstly, when 
 the vertex is down and base up ; and, secondly, when the base is 
 down and the vertex up. 
 
 Let w f , V, h be the specific weight, volume, and height of the 
 cone ; let w be the specific weight of the liquid, and x the length 
 of the axis immersed when the vertex is down. Then since the 
 volumes of similar solids are proportional to the cubes of their 
 corresponding linear dimensions, the volume of the displaced 
 
 X s 
 liquid = j-g F. Hence, equating the force of buoyancy to the 
 
 weight of the cone, 
 
 *L Vw = Ft*/, 
 
 In the second case, if x is the length of the axis above the 
 
 3,3 
 liquid, the volume of the displaced liquid = (i 'r-J V, and we 
 
 I ^ ll ' 
 
 have , i 
 
 if w V 
 x = h I i ) 
 
 V />/ 
 
76 Hydrostatics VOL. i 
 
 2. A solid homogeneous isosceles triangular prism floats in 
 a given homogeneous liquid ; find the position of equilibrium in 
 each of the two previous cases. 
 
 If x is the depth of its edge below the surface, h the height of 
 the isosceles triangle which is the section of the prism by a 
 plane perpendicular to the edge, and A the area of this section, 
 
 since the areas of similar figures are as the squares of their 
 
 3.2 
 corresponding linear dimensions, A is the area of the face of 
 
 fif 
 
 the immersed prism in the first case, and if I = length of 
 edge, the volume of the prism is IA , so that the volume of the 
 
 y 
 
 immersed prism is V. Hence 
 
 li 
 
 , / 
 
 . . x = h ( 
 Vtw 
 
 In the second case, 
 
 .-. . = *(,-=.) 
 
 V W' 
 
 3. A uniform rod, AB, of small normal section and weight- W 
 has a mass of metal of small volume 
 
 and weight - W attached to one ex- 
 
 n 
 tremity, B ; find the condition that 
 
 the rod shall float at all inclinations 
 in a given homogeneous liquid. 
 
 Let AB = 2 a, let m be the middle 
 point of AS, Fig. 37, G the centre of 
 Fig. 37. gravity of the rod and the metal, w' 
 
 the specific weight of the rod, w that 
 of the liquid, and s the area of the normal section of the rod. 
 
 Then TF= zasw' and BG = . Also G must be the 
 
 n+i 
 
 centre of buoyancy if the rod floats in the oblique position repre- 
 
CH. iv Pressure on Curved Surfaces 77 
 
 sented, and the length, SO, of the displaced column of liquid 
 
 2n 
 = - a, so that if the weight of this column == (i + -\ jf 
 
 n' 
 
 both conditions of equilibrium will be satisfied, whatever be the 
 inclination of the rod. Equating the weight of the body to the 
 force of buoyancy, 
 
 , 2n 
 2 asw = asw ; 
 
 71+ I 
 
 which is the relation required between the specific weights. . 
 
 4. A solid homogeneous cylinder floats, with its axis vertical, 
 partly in a homogeneous liquid of specific weight w, and partly 
 in one of specific weight w z , 
 the former resting on the 
 latter; find the position of 
 equilibrium. 
 
 Let h be the height of the ~~ 
 cylinder, A the area of its 
 base, w its specific weight, 
 and c the thickness of tlie 
 upper liquid column. 
 
 Then if we assume the top, 
 
 "i - 
 
 _,. 
 
 A, of the cylinder to project 
 
 a distance x above the upper 
 
 surface of the upper liquid, as 
 
 in Fig. 38, and equate the weight of the cylinder to the sum of the 
 
 forces of buoyancy due to the displacements of the liquids, we 
 
 have 
 
 c x)u\ 2 , ..... (i) 
 
 T / M>,\ , / W \ 
 
 It c( i -- ) > hi i -- I. we must write 
 \ w' \ to 
 
 and it would appear that the position of equilibrium is one in 
 
78 Hydrostatics VOL. I 
 
 which, as in Fig. 39, A is below the upper surface of the upper 
 liquid by the distance 
 
 ( w i\ if w \ 
 c ( i J ^ i J > \4/ 
 
 in virtue of the usual interpretation of a negative co-ordinate in 
 algebra. 
 
 To take a numerical case, suppose c = ^h, and 
 
 w:w 1 :w 2 = 5:2:6; 
 
 then x = ^h, and it would appear that A is ^h below the 
 upper surface of the upper fluid. 
 
 itfow if we had originally assumed A to be, as in Fig. 39, at 
 an unknown distance, x, below the surface, our equation would 
 have been hw = (c-x)w l + (hc + x}w 2 , .... (5) 
 
 i w i w ,,-\ 
 
 .-. x=.ch , (6) 
 
 w z w i 
 
 which disagrees with (4), and which in the particular numerical 
 case gives x = %h, instead of x = ^h, which we had been led to 
 expect by interpretation of the negative value (3). 
 
 Why the disagreement ? Because the continuity of the values 
 of variables in algebra and algebraic geometry finds no corre- 
 sponding characteristic in the hydrostatical conditions. In fact, 
 the supposition that the negative value (3) harmonizes with the 
 physical assumptions leading to the first solution is untrue ; for, 
 in this solution we assume that, whatever be the unknown posi- 
 tion of equilibrium of the body, the whole column of the upper 
 liquid is operative in producing its force of buoyancy, as is 
 evident from the first term, cw l , at the right-hand side of (i) ; 
 whereas the supposition that A is below the upper surface of 
 this liquid is an explicit assumption that the whole column of 
 the liquid may not be so operative. Hence we ought not to 
 expect the two solutions to agree. 
 
 In the case, therefore, in which the value of x in (2) is nega- 
 tive, the correct result is (6) and not (3). 
 
 5. A heavy uniform bar, AB, of small cross-section is freely 
 movable round a horizontal axis fixed at one extremity, A, at a 
 given height above the surface of a homogeneous liquid in which 
 the rod partly rests ; find the position of equilibrium and the 
 pressure on the axis. 
 
CH. iv Pressure on Curved Surfaces 79 
 
 Let AB = 2 a ; let h be the height of A above the liquicl ; let 
 a = area of cross-section of the rod ; let it/ and w be the specific 
 weights of the rod and the liquid ; 
 and let = the angle between AB 
 and the vertical. 
 
 Then if EC is the part immersed, 
 the centre of buoyancy, H, is the 
 middle point of EC. If W-= weight 
 of rod, W 2 asw' ; also 
 
 BC = 2 ah sec 0, ^- 
 
 . . the force, L, of buoyancy . 
 
 = (2 ah sec 0) sw. 
 
 The rod is in equilibrium under the action of L, W, and the 
 pressure at A, which last must be vertical and = W L. 
 Taking moments about A for equilibrium, 
 
 W. AG sin 6-L.AH sin 0, (i) 
 
 and if we reject the factor sin 0, i. e. omit the consideration that 
 sin = o gives one position of equilibrium (the vertical one), we. 
 
 have 4 V/ = (4a 2 -A 2 sec 2 0)w, (2) 
 
 h / w \k , , 
 
 cos = I ,1 (3) 
 
 The oblique position requires w to be greater than w' and also 
 w > ^ p w' ; 
 
 so that, for example, if the bar were of metal and the liquid 
 water, the only position of equilibrium would be the vertical one. 
 
 6. A uniform pole, AB, the linear dimensions of whose cross- 
 section are small compared with its length, is supported by 
 a cord attached to A and floats, partly immersed, in water; 
 find its position of equilibrium. 
 
 Result. If 2 x is the length of the immersed portion, w and 
 w' the specific weights of the liquid and pole, 
 
80 Hydrostatics VOL. i 
 
 Hence the length of the immersed part is independent of the 
 inclination; but if w'>w, the pole must float vertically. The 
 suspending cord always assumes a vertical position. 
 
 7. If the pole is 1 2 feet long, and its specific gravity f , and 
 if the end A is fixed by a horizontal axis at a height exceeding 
 8 feet above the water, the only position of equilibrium is a 
 vertical one. If the height of A is 6 feet, prove that the 
 inclination of the pole to the vertical is cos" 1 -75. 
 
 8. Two uniform straight rods, of lengths 2 a, 26, and specific 
 gravities s, /, respectively, are joined together to form a single 
 straight rod; find the ratio a : b so that there may be an 
 inclined position of equilibrium when the system floats freely 
 in water. 
 
 Result. The ratio is given by the equation 
 
 s(i )a 2 +2s'(i s)ab + s'(i -s')^ = o, 
 or the same with 23(1 /) for the coefficient of ab. 
 
 9. The specific gravity of ice is -918 and that of sea- water 
 1-026; prove that the volume of the submerged portion of an 
 iceberg is 8^ times that of the portion above water. 
 
 10. A hollow closed cone of metal whose specific gravity is 8 
 is to be made of such uniform thickness that it will float in all 
 positions wholly submerged in water ; show that the thickness 
 must be -0109 x h, where h is the height of the cone. 
 
 1 1. If the specific weight of the metal is /, and the cone floats 
 in all positions, wholly submerged, in a liquid of specific weight 
 w, show that the semi-vertical angle of the cone must be sin" 1 , 
 and that the thickness must be 
 
 h 
 
 4 
 
 12. A solid homogeneous cone is floating in water with its axis 
 vertical and vertex downwards ; to cause it to sink until f of its 
 axis is immersed requires a load of 50 grammes on its base ; and 
 to cause to be immersed requires 96 grammes ; show that the 
 sp. gr. of the cone is -324 nearly. 
 
 13. A cylindrical block of wood the area of whose section 
 perpendicular to its axis is 50 square inches floats, with the 
 
CH. iv Pressure on Curved Surfaces Si 
 
 axis vertical, in water in a vessel whose horizontal cross- 
 section is unifoim and equal to 80 square inches; the specific 
 gravity of the wood is f and the height of the blcck is 2 feet ; 
 how much does the level of the water rise above the position 
 which it occupies when the block is withdrawn, and how much 
 of the axis is under water ? 
 
 Ans. 10 inches; 16 inches. 
 
 14. A block of wood of height h, cross-section B, and specific 
 weight w' floats in a liquid of specific weight w contained in 
 a vessel of cross-section A ; how much does the level of the 
 liquid rise after the immersion of the block, and how much of 
 the block is immersed ? 
 
 Ans. B w' w' 
 
 . h ; h. 
 
 A w ID 
 
 15. A body of weight JFis allowed to float in a tank of cross- 
 section A containing a liquid of specific weight w ; how much 
 does the level of the liquid rise ? 
 
 Ans. W 
 Aw 
 
 16. A block of wood, of sp. gr. ^, in the shape of a prism 
 whose section perpendicular to its edge is an isosceles triangle 
 whose base is 8 inches long and height 10 inches, the length 
 of the edge being 27 inches, is placed with its edge submerged 
 in a tank i foot broad and 3 feet long ; how high does the level 
 of the water rise, and what is the depth to which the edge of the 
 block is submerged 1 
 
 Ans. i^ inches; 6 inches. 
 
 17. The section of an isosceles prism of wcod perpendicular to 
 its parallel edges is a trapezium whose parallel sides are 8 and 
 6 feet long, the perpendicular distance between them being 
 4 feet ; the sp. gr. of the wood is -|f , and the prism is floating 
 in water with the longer of the parallel sides uppermost ; find 
 the position of equilibrium. 
 
 The thickness of the immersed part is 2 feet. 
 
 [Whatever the distance between the parallel sides may be, if 
 the other data remain unaltered, half the thickness will be 
 immersed.] 
 
 14U Q 
 
8 a Hydrostatics VOL. I 
 
 18. ABCD is a uniform rectangular board of specific weight 
 w* ; it is to float in a liquid of specific weight w with the 
 diagonal AC submerged and horizontal when a particle of 
 negligible volume whose mass is n times that of the board is 
 attached at the corner B; find the condition and position of 
 equilibrium. 
 
 If the side AD cuts the surface of the liquid at P, 
 
 and if we put DP = . DA. m is determined by the equation 
 m 
 
 3 m 3 ( i ) 3 m + i = o, while n and m must satisfy the 
 > w ' 
 
 .. W f I \ 
 
 equation n = -.[ i ---- ) i. 
 r V 2 m 2 ' 
 
 19. A solid homogeneous hemisphere is movable round a 
 horizontal axis which is a tangent to its rim and rests partially 
 immersed in a liquid ' find its position of equilibrium. 
 
 Result. Let h be the height of the axis above the liquid, 
 a the radius and w' the specific weight of the solid, w being 
 that of the liquid ; then the plane of the base is inclined to the 
 horizon at an angle given by the equation 
 
 (a h + asin Qf(2a + h a sin 6)w = 2a 3 (i f tan 6}w'. 
 
 20. A solid homogeneous body of any shape is movable round 
 a fixed horizontal axis and rests partially immersed in a liquid 
 contained in a trough, the section of floatation being a curve 
 AB marked on the surface of the body; liquid is poured into 
 the trough, the body rising by rotation round the axis which 
 becomes immersed at increasing depths ; it is observed that in 
 the course of revolution the same curve AB becomes again the 
 curve of floatation. Show that the specific gravity of the body 
 must be one-half of that of the liquid. 
 
 22. Resultant Pressure on an unclosed curved surface. 
 
 Suppose BCD A, Fig. 41, to represent any unclosed surface 
 in a heavy fluid, and suppose its bounding- edge to be a 
 plane curve so that the surface can be closed by a plane base, 
 
CH. IV 
 
 Pressure on Curved Surface* 
 
 represented by AB. It is required to find the resultant of 
 the fluid pressures exerted on one side of the unclosed 
 surface. 
 
 Closing- the surface by means of the plane base AB, 
 the resultant of the pressures all over the outside of the 
 completely closed surface is the vertical upward force, L, 
 represented by the line HL drawn through the centre of 
 gravity, H, of the fluid which would 
 fill the volume. But if P is the 
 resultant fluid pressure on the plane 
 base A, acting at the centre of 
 pressure,/, the force L is the resultant 
 of P and the resultant pressure over 
 the unclosed part. This latter force, 
 Q, is therefore found by producing 
 the lines of action of L and P to 
 
 Fig. 41. 
 
 meet at 0, suppose and drawing On and Om to repre- 
 sent L and P, respectively ; then the required force Q is 
 represented by the line OQ which is equal and parallel 
 to mn. 
 
 If the fluid is a homogeneous liquid of specific weight w^ 
 if A is the area of the plane base AB, z the depth of the 
 centre of area of AB below the free surface, and V is the 
 volume of the closed surface, 
 
 P Azw, and L = Vw. 
 
 Hence if 6 is the inclination of the plane base AB to the 
 
 horizon. 
 
 Q = w V r*+2,f'Az cos 6 + A* z* ; 
 
 horizontal component of Q, = Azw sin 9, 
 vertical component of Q = ( Az cos V) w. 
 
 G 2 
 
8 4 
 
 Hydrostatics 
 EXAMPLES. 
 
 VOL. I 
 
 1. Suppose a right cone whose axis is vertical and vertex 
 downwards to be filled with a liquid; find the resultant 
 pressure on one-half of the curved surface determined by any 
 plane containing the axis. 
 
 Let ACB, Fig. 42, be the vertical plane of section, and 
 ACDB the half of the curved surface on 
 which we desire to find the resultant 
 liquid pressure. 
 
 Consider the separate equilibrium of 
 the fluid contained between this curved 
 surface and the triangle ACS. It is 
 kept at rest by its weight, the pressure 
 of the remaining fluid on the area ACB 
 acting at 7, the centre of pressure on 
 this triangle, and by the pressure of the 
 curved surface ACDB. The weight acts 
 through G, the centre of gravity of the 
 semi-cone ; and if on the diameter, OD, 
 which is perpendicular to AS, we take the point n such that 
 
 On = , where r is the radius of the base, this point n is the 
 
 3^ 
 
 centre of area of the semicircle ADS, so that G lies on nC and 
 Gn = \Cn (Statics, vol. i, Art. 163). The point / is half-way 
 down OC (Art. 14). If P is the pressure on the triangle ACB, 
 h = height of cone, 
 
 Fig. 42. 
 
 P = 
 
 and W = 
 
 where W = weight of liquid. The lines of action of P and W 
 meet in a point c, whose position is thus completely known; 
 and by drawing cP and c W to represent P and W on any scale, 
 the diagonal through c of the rectangle thus determined will 
 represent Q, the resultant pressure of the curved surface on the 
 fluid in the semi-cone. The line cQ is drawn to represent this 
 pressure, and this force reversed is the pressure of the fluid on 
 the surface. 
 
 2. If the cone is closed by a base, and the axis is held hori- 
 zontal, find the resultant pressure on the lower half of the 
 curved surface. 
 
CH. iv Pressure on Curved Surfaces 85 
 
 Result. If X and Y are the horizontal and vertical com- 
 ponents of the resultant pressure, 
 
 and the line of action of the pressure passes through a point 
 whose distances from the base and the axis of the cone are 
 
 h 7T + 8 , r 37r+ 16 
 
 and - 
 
 4 7T + 6 4 377 + 4 
 
 3. If a hollow cylinder is filled with liquid and held with its 
 axis vertical, determine the magnitude and line of action of the 
 resultant pressure on one half of the curved surface cut off by a 
 vertical plane through the axis. 
 
 It is a horizontal force equal to r^hw acting in a line 
 
 - from the base. 
 3 
 
 . 4. If the cylinder is closed at both ends and held with its 
 axis horizontal, find the resultant pressure on the lower half of 
 the curved surface. 
 
 A vertical force ^ ( 2 H ) 
 
 5. In example 2 find the magnitude and line ot action 
 of the resultant pressure on the upper half of the curved 
 surface. 
 
 If X and Y are the horizontal and vertical components 
 of the pressure, 
 
 and the line of action of the resultant passes through a point 
 whose distances from the base and the axis of the cone are 
 
 h S TT r 16 377 
 
 - and ^ 
 
 4 6 TT 4 3714 
 
86 Hydrostatics VOL. i 
 
 6. A spherical shell is filled with liquid ; find the magni- 
 tude and line of action of the resultant pressure on the curved 
 surface of either hemisphere cut off by any vertical central 
 plane. 
 
 The line of action passes through the centre of the sphere ; 
 the horizontal component is TII^W, and the vertical ^-nr'w. 
 
 7. A spherical shell is filled with liquid ; find the magnitude 
 and line of action of the resultant pressure on each of the 
 hemispheres into which the sphere is divided by any diametral 
 plane. 
 
 If 6 is the inclination of the plane section to the hori- 
 zon, the pressure on one hemisphere is the resultant of two 
 forces Ttr 5 w and Trr s w, respectively perpendicular to the plane 
 section and vertical, the lines of action of these forces including 
 an angle 9, while the pressure on the other curved surface is the 
 resultant of the same forces including an angle TT ; and both 
 pass through the centre. 
 
 8. If a hole is made in the top of the shell and fitted with a 
 funnel, find the height to which the funnel must be filled with 
 the liquid in order that the resultant pressure on one of the 
 hemispheres shall be a horizontal force, 
 
 The height = r (f sec 0- i ). 
 
 When the unclosed curved surface which is exposed to 
 water pressure cannot be closed by a plane base i. e. when 
 its bounding edge is not a plane curve the total vertical 
 component of pressure on one side of the surface is easily 
 found. Thus, in Fig. 43, suppose AS to represent an 
 unclosed surface, and consider the pressures (represented 
 by the arrows) exerted on one side by the water. At the 
 various points of the bounding edge draw vertical lines 
 terminated by the free surface, PQ, of the water, and con- 
 sider the separate equilibrium of the cylindrical column, 
 PASQ, of the liquid. This column is kept in equilibrium 
 by (i) its weight, (a) the horizontal pressures exerted all 
 
CH. IV 
 
 Pressure on Curved Surfaces 
 
 round its vertical surface, and (3) the pressures on its under 
 side which are represented by the arrows in the figure 
 reversed in direction. 
 
 Resolving- forces vertically, we see that the vertical com- 
 ponent of these latter forces is equal to the weight of the 
 column standing on AB commonly called the weight of 
 the superincumbent fluid. 
 
 The curved surface may be such that some of the pressures 
 on one and the same side of it have a downward and others 
 an upward component, as in Fig. 44. In this case a portion 
 of the above cylinder will be formed by lines, such as BQ, 
 
 Fig- 43- 
 
 which are tangents to the given surface A EC ; and on the 
 corresponding part of the surface, AS, the pressures have a 
 downward vertical component equal to the weight of the 
 column ABQ ; while the pressures on the remainder, EC. f 
 have an upward component equat(by the same reasoning) to 
 the weight of the column RCJ3Q. The resultant upward 
 component is equal to 
 
 weight of DBC -weight of PADR 
 
88 Hydrostatic* VOL. i 
 
 If in Fig. 44 the point C either coincides with A or is 
 vertically under A i. e. if the surface ABC is either a 
 perfectly closed surface or one which can be closed by a 
 vertical plane base the column PADR vanishes, and the 
 upward thrust is equal to the weight of the liquid enclosed 
 by the surface ABC. Thus we are brought back to the 
 principle of buoyancy. Art. 20. 
 
 In both Fig. 43 and Fig. 44 the pressures have also a 
 horizontal component, but this cannot be obtained in the 
 same simple manner ; for, the term vertical is perfectly 
 definite, while the term horizontal is not so : vertical means 
 parallel to a line, whereas horizontal means merely parallel 
 to a plane only one vertical line can be drawn through 
 a given point, but an injinite number of horizontal lines can 
 be drawn through it. 
 
 To find the total horizontal component of the pressures on 
 one side of an unclosed curved surface along a given hori- 
 zontal line, project the bounding edge of the surface on 
 a vertical plane perpendicular to the given horizontal direc- 
 tion, by horizontal lines drawn through the points of the 
 bounding edge ; then find the magnitude of the pressure 
 exerted on the (plane) area thus obtained. This pressure, 
 calculated by Art. 12, is the required horizontal component. 
 
 23. Moulds. The vertical thrust of a fluid is well 
 illustrated in the case of a mould for producing a metal 
 casting. 
 
 Suppose that it is, desired to make a thin hollow cone of 
 metal. Take a solid cone, ABC, of clay, plaster of Paris, or 
 other substance, fastened to a horizontal base ; and over it 
 place a mould, PQRS, containing a hollow, similar, and 
 slightly larger conical cavity, EFGH, with a narrow vertical 
 aperture, DEH, the mould being fastened in position by 
 weights or bolts. If the molten metal is poured in at D, it 
 will fill the space between the cone and the mould, and 
 
CH. IV 
 
 Pressure on Curved Surfaces 
 
 8 9 
 
 solidify into a thin cone of metal. Now, while the metal 
 is in the liquid state, the mould will be acted upon by pres- 
 sure all over the conical shoulder FENG, and the resultant 
 vertical upward thrust of the liquid is equal to the weight 
 of the column of the liquid 
 standing on FENG with 
 vertical sides reaching up 
 to the level, PQ, of the 
 free surface. This column 
 is cylindrical in shape, and 
 is represented in section by 
 mFEHGn, deducting the 
 volume of the aperture 
 DEH. 
 
 If H is the height, QR, 
 of the mould, li the height 
 of the cone FEHG, r the 
 
 radius of the base of this cone, and w the specific weight 
 of the metal, the vertical upward thrust tending to lift 
 the mould off its base is, approximately 
 
 Fig. 45- 
 
 In the same way a hemispherical bowl may be cast if the 
 solid JJACis replaced by a hemispherical body, and the space 
 FEHG is also hemispherical. 
 
 EXAMPLES. 
 
 1. A hemispherical bowl of metal 2 feet in radius is cast 
 with a mould whose height in 3 feet ; the mass of the metal per 
 cubic foot is 480 pounds ; prove that the lifting force on the 
 mould is about 4^ tons' weight. 
 
 2. Two hollow cones, of the same vertical angle, are joined 
 together at their vertices so that their axes are in the same 
 
QO Hydrostatics VOL. i 
 
 vertical line, forming a figure like an hour-glass (Fig. 46). 
 
 The vessel is closed at one end with 
 a plane base and rests on a hori- 
 zontal plane with the axis vertical. 
 Find the height to which it must be 
 filled with water so that the total 
 vertical water thrust on the curved 
 surface shall be zero. / 
 
 Result. If h is the height of the 
 _ lower cone and H the height of the 
 upper cone of water, 
 
 which gives H = zh. 
 
 3. A hollow frustum of a cone, 6 feet high, is placed on the 
 ground with its base (8 feet in diameter) cemented, water-tight, 
 to the ground ; the upper extremity of the frustum has a 
 diameter of 2 feet and is connected with a vertical tube i o feet 
 high which is rigidly joined all round to the frustum. If the 
 compound vessel is filled to the top with water, what is the 
 force tending to tear the vessel from the ground ? 
 
 Ans. The weight of 20477 cubic feet of water, i. e. 17-88 tons' 
 weight. 
 
 4. A thin spherical shell of radius r and negligible weight has 
 a small hole into which a tube is fitted ; the tube is placed 
 vertically and filled with water to a height h above the centre 
 of the sphere, the system being kept in equilibrium in any way. 
 Show that the whole force with which the upper hemisphere of 
 the shell tends to separate from the lower is 
 
 A very striking illustration of the principle of buoyancy 
 is furnished by the following experiment. Take a straight 
 glass tube ; insert into it at A (Fig. 47) another glass tube, 
 Av, at right angles ; bend the first tube at m and n at 
 opposite sides of Av, making two constrictions at m and 11. 
 the branches of the tube being mB and nC ; through the 
 open ends , C insert a ball c of cork soaked in paraffin and 
 a solid marble, y, of glass or metal ; close the ends and C 
 
CH. IV 
 
 Pressure on Curved Surfaces 
 
 of the branches. If the system is held with Av vertical the 
 cork and the marble will fall to m and n and be stopped by 
 the constrictions. Now pour in water through u, filling 
 both branches. The cork will rise to B and the marble will 
 remain at n. Let a clamp be fixed at A to the tube, and 
 attach this clamp to a whirling- apparatus so that the whole 
 can be rotated rapidly about the vertical line Av. As the 
 rotation increases, the cork will come down from B and try 
 to reach A, while the marble will ascend towards C. 
 
 Let <o be the angular velocity of the tube, and suppose the 
 cork to be at c ; let r be the distance of c from the vertical 
 axis of rotation, and let W^ be 
 the weight of the cork. 
 
 Suppose the cork to be tied 
 to the point IB by a thread, 
 whose tension is T, while the 
 system is rotating about Av 
 with angular velocity o> ; then 
 the forces acting on the cork 
 are T, W lt N (the normal reac- 
 tion of the tube) and the pres- 
 
 sure of the surrounding liquid. This last is exactly the same 
 as it would be on a sphere of the liquid itself which might 
 replace the cork. Let H and V be the horizontal and 
 vertical components of this pressure, and let W be the weight 
 of the liquid that would occupy the place of the cork. Now 
 since the sphere of liquid goes round in a horizontal circle 
 without the aid of a thread but with the aid of a pressure 
 N 1 from the tube (unless the sphere is so small that it 
 is not in contact with the tube), its equation of horizontal 
 motion is M v r 
 
 Fig. 47- 
 
 H= 
 
 (i) 
 
92 Hydrostatics VOL. i 
 
 and also for vertical equilibrium, 
 
 Y- H+N'cosa ...... (2) 
 
 The equation of horizontal motion of the cork is 
 
 r,^= H-Nsina-Tcoaa, 
 
 9 
 
 = W -(N- N') sin a- T cos a, 
 9 
 
 by (i) ; while vertical equilibrium gives 
 
 T = W 1 + NCOS aT sin a 
 
 = JT+^'cosa. 
 Hence we have 
 
 o 
 
 (r-r ] )-p = (JV--JV 7 ') s ina + 7'cosa, . . (3) 
 
 W-W l =(N-N')cosa-Tsina.. . (4) 
 Eliminating N N', we have 
 
 . . . (5) 
 This equation shows that T is negative so long as 
 
 co 2 < - tan a, 
 
 r 
 
 and therefore the cork would require a push towards m 
 to keep it in position ; but when o> 2 = - tan a, no force along 
 
 the tube is required ; and when or increases beyond this 
 value, T becomes positive that is, the cork tries to go 
 down the tube. 
 
 The case of the marble is similarly discussed ; let W z 
 be the weight of the marble, then for the tension of the 
 constraining cord we have 
 
 T= _r- 
 
 2 
 
 = (r 2 -r)(sina- -cos a), ... (6) 
 
CH. IV 
 
 Pressure on Curved Surface* 
 
 which shows that when o> increases beyond the above-named 
 value, T becomes negative that is, a push downwards along 
 Cn is required, or, in other words, the marble ascends the 
 tube. 
 
 24. Work of Immersion. When a body displaces liquid, 
 a certain amount of work is done against the force of 
 buoyancy. Suppose, for example, that a solid cylinder 
 is allowed to sink into water until it finds its position 
 of equilibrium. Then the principle of work and energy 
 shows that the work done on the cylinder by gravity 
 is equal to the work which the cylinder does against 
 buoyancy from the first 
 to the second position. 
 
 Suppose Fig. 48 to 
 represent a body of 
 any form displacing a 
 liquid contained in any 
 given vessel. Let A B 
 be the level of the 
 liquid before the body 
 was immersed, and let 
 PQ be the level to Fig. 48. 
 
 which the liquid rises 
 
 when the body is immersed. [We are not assuming that the 
 body \sfloating in equilibrium : we assume merely that by some 
 means it is now in the position represented.] Now it is 
 quite clear that the volume LCN of the liquid is taken up 
 and spread into the layer whose section in the plane of the 
 figure is represented by LAPR and NJ3QS ; so that the 
 work done against buoyancy is the work of raising the 
 volume LCN into the position of the layer. This work is 
 the same as that done by raising the volume ELCNS into 
 the position of the volume PRSQ,BNLA,smcQ by this process 
 the volume RSNL would be undisturbed. 
 
94 
 
 Hydrostatics 
 
 VOL. I 
 
 Let (f be the centre of gravity of the volume RLCNS, 
 supposed filled with the liquid, and let g be the centre 
 of gravity of the layer PASQ ; then the work done against 
 buoyancy is the weight of the layer PAJ3Q multiplied by 
 the difference of level of the points g and g' . 
 
 For clearness, let us take a numerical case. 
 
 To find the work which must be done to submerge com- 
 pletely a cylinder of weight W> 1- eight /<, and specific gravity 
 | in a cylindrical vessel of water, the area of the cross-section 
 of this vessel being twice that of the body. 
 
 Let Fig. (i) represent the body just touching the surface 
 of the water ; Fig. (a) the body in its equilibrium position 
 
 A - 
 
 G' 
 
 
 
 
 
 
 
 
 r 
 
 : 
 
 
 
 
 P 
 
 B A 
 
 = 
 
 f I 
 <?a 
 
 
 R 
 Q 
 
 S-A 
 
 = 
 
 .-& - 
 
 = 
 
 R 
 
 ^ = -r^= 
 
 ^3 
 
 
 i 
 
 m MJ 
 
 
 
 ? 
 
 22 * 
 
 
 
 1 
 
 
 
 
 z 
 
 Fig. 49. 
 
 
 
 
 3 
 
 
 
 when left to itself ; Fig. (3) the body completely submerged 
 by means of a downward pressure. 
 
 Since the force of buoyancy in Fig. (2) is the weight of the 
 volume pmuq of liquid, or of the layer PAQJ3, i.e. of the volume 
 S x AP, where S = area of cross-section of vessel, we have 
 
 ,* . AP = h, and pm = h. 
 
 Now from position (i) to position (2) the work against 
 buoyancy is the work of raising the volume pmnq into the 
 position PAQ ; and us the centre of gravity of the first of 
 
CH. iv Pressure on Curved Surface* 95 
 
 these volumes is on the plane AJ3, and the centre of gravity 
 of the second is -? above this plane, the weight of the volume 
 
 pmnq being 1 equal to //' , the required work against buoyancy 
 from (i) to (2) is -| Wli. Denote this work by 12 ; also 
 denote the work against buoyancy from (i) to (3) by B ly 
 Now the volumes rmns and RABS are equal ; . . AR = % h, 
 and the centre of gravity, 6r' 3 , of rmns is on the plane AB ; 
 also the weight of this volume of liquid = f W\ hence 
 
 .RO = I JT.- = | JF. k. 
 
 4 
 
 If J9 23 denotes the work against buoyancy from (2) to (3), 
 we have = S S ; 
 
 This, however, is more than the work which we should have 
 to do to sink the body from (2) to (3), because we are assisted 
 by the work done on the body by gravity from (2) to (3). 
 This work is W x diff. of level of G 2 and G 3 ; that is | Wk ; 
 hence, deducting this from S 2S) we have 
 
 hW.h 
 
 for the work which, in addition to that contributed by 
 gravity, is required to sink the cylinder. 
 
 As a further exercise on this same question, take this 
 problem : the cylinder leing held ly the hand in the position (i) 
 and allowed to drop into the liquid, find how far it will sink 
 before coming for an instant to rest. 
 
 Applying the principle of work and energy to the body, 
 since the kinetic energy in the first position = o, and also 
 that in the second position = o, the work done by gravity 
 on the body must be numerically equal to the work done by 
 buoyancy on it. Suppose the body to sink until AP = x ; 
 
g6 Hydrostatics VOL. I 
 
 then, since vol. pmnq vol. PAHQ, we see that;;? = 2#, and 
 
 n/t 
 
 the work against buoyancy from (i) to (2) is \ S. ix . w . - , 
 
 "2f 
 
 since the centre of gravity of pmnq is on the plane AB ; 
 this work is J Sro . x*. To get the work done on the body 
 by gravity, observe that, measuring all distances from the- 
 fixed plane AB, the centre of gravity, G, of the body is, in 
 the first position, at a height of ^ h above AB y and in the 
 second at a height \h~x above AB\ and, the weight of 
 the body being \ S/nv, the work done by gravity from (i) to 
 (2) is | Shw (% h \ h + x), or ^ Shxw. Equating this to the 
 work against buoyancy, we have x = f /$, a result which 
 shows that the cylinder disappears below the liquid ; but 
 the result cannot be relied upon as expressing the final 
 position, since we have assumed in the work that p is not 
 above the top of the cylinder. (See a similar case, p. 77.) 
 We must make a fresh calculation assuming that the top of 
 the cylinder disappears to a distance z above the plane AB. 
 The work of gravity will be W(Ji z), and the work against 
 buoyancy \ Shw (| li z\ or f W (f k z). Equating these we 
 
 have _ i 7 
 
 z ? "> 
 
 in other words, three-fourths of the cylinder goes below the 
 original level of the liquid. 
 
 If the vessel, or tank, in which the body floats is of un- 
 limited cross-sectional area a lake or a river as represented 
 in Fig. 50, the work of buoyancy is calculated very simply. 
 Thus, let the body be immersed so that pqmn is under the 
 liquid ; then the work done against buoyancy in reaching 
 this position is simply the work of raising the volume pqmn 
 of liquid and spreading it out along the unlimited surface 
 AB of the liquid as an infinitely thin horizontal layer. If 
 pm = x and A = area of cross-section of the cylinder, the 
 volume raised is Ax, and the height through which its 
 
CH. IV 
 
 Pressure on Curved Surfaces 
 
 97 
 
 centre of gravity is raised is-|#; so that the work done 
 against buoyancy is \ Ax 2 w ; and the work done against 
 buoyancy in submerging it to the position in which pm = x' 
 is \Ax'*w ; therefore the work done between these two 
 positions is \A (x' 2 
 
 w. 
 
 ~m 
 
 Fig. 5- 
 
 G 
 
 -B 
 
 G \U=T 
 
 M 
 
 Let the first position be the freely floating one, and the 
 second the position of complete submergence ; then if w' 
 
 w 
 
 is the specific weight of the cylinder, as = h , and x' = h ; 
 
 therefore the work against buoyancy from the one to the 
 other is 
 
 The weight of the body itself has assisted in the sub- 
 mergence and done the amount Akw' (of x), or 
 
 so that the work required to be done by an external agent 
 to submerge the body from the floating position is 
 
98 Hydrostatics VOL. i 
 
 W W W' 
 
 where W\& the weight of the cylinder. 
 
 EXAMPLES. 
 
 1 . A solid homogeneous cylinder of height A, cross-section B, 
 and specific weight w' floats with its axis vertical in a liquid 
 of specific weight w contained in a tank of cross-section A ; 
 find the work required to submerge it. 
 
 2 A W 
 
 2. A solid body of weight W, specific gravity s, and given 
 figure rests on the bottom of a river of depth h ; find the work 
 necessary to raise it just clear of the water. 
 
 Let G be the centre of gravity of the body, Fig. 50, and let 
 the height of G above the base be k ; then the work of buoyancy, 
 which assists in the raising is that done by the falling of a 
 superficial layer of water into the position of the body. The 
 
 W 
 
 volume of this layer = vol. of body = , where w = sp. 
 
 w 
 
 W 
 
 weight of body, .-. the weight of the layer = , and the 
 
 S 
 
 original and final heights of the centre of gravity of the layer 
 above the bottom, LM, are h and k ; therefore the work of 
 
 W 
 
 buoyancy = (h k) the centre of gravity, G, of the body is 
 
 8 
 
 raised through h', .'. the required work is 
 
 Wh - -(h-k). 
 s x 
 
 3. Calculate the work required to submerge a homogeneous 
 sphere of specific weight \w in a tank of given cross-section 
 containing a liquid of specific weight w. 
 
 If V = vol. of sphere, r = radius, A = area of section of 
 tank, the work is 
 
CH. iv Pressure on Curved Surfaces 99 
 
 [Taking the three positions of Fig. 49, 
 V V 
 
 and we have 
 
 = i fw (l-r - 
 
 Deduct the work done on the sphere by gravity from (2) to 
 (3), which is Vw (r ), and we get the result.] 
 
 4. If this sphere is held in the position (i) and allowed to drop 
 into the liquid, how far will it descend before being brought 
 for an instant to rest ? 
 
 Ans. It will just reach the position (3) of complete sub- 
 mergence. 
 
 5. A solid homogeneous cone of weight W, height h, and 
 specific gravity -fa floats in water of indefinite horizontal 
 extent ; find the amount of work required to submerge it. 
 
 I^JF.A. [Work against buoyancy = ff W. h; work of 
 gravity = ^ W. h.] 
 
 6. A solid homogeneous cone of weight W, height h, and 
 specific weight w floats, vertex down, in an indefinite mass of 
 liquid of specific weight w ; find the work required to sub- 
 merge it. 
 
 Work against buoyancy = 1 W. h \j ( Y* ( 
 |_ ( w \ tv ' y 
 
 work of gravity = W , h < i ( ) ( 
 
 7. If this cone is held with its axis vertical and its vertex 
 just touching the liquid, and is allowed to drop into the liquid, 
 
 H 1 
 
ioo Hydrostatics VOL. i 
 
 how much of it will be submerged in the position of tem- 
 porary rest ? 
 
 Ans. Four times the volume submerged when the cone is 
 floating freely in equilibrium provided that this volume does 
 not exceed the whole volume of the cone. If, however, w'> ^w, 
 let the base of the cone disappear to a depth x below the 
 surface; then the work done against buoyancy 
 
 and the work done by gravity on the cone 
 
 = V 
 equating these, we have 
 
 tp-frM, 
 
 fc*/ ^~ ft-. 
 
 to 10 
 
 8. A solid sphere of radius r and specific gravity s lies on the 
 bottom of a cylindrical vessel of radius a and height h which is 
 filled to the top with water ; prove that the work required to 
 raise the sphere just clear of the water is 
 
 f 4-r* i / 2r 
 
 I* r ; 
 
 3 a 2 s V sa 
 
 TF 
 
 (When the sphere is just clear, the height, x, of the water is 
 
 41* 
 h ; Now take the mass-moment of the system in the 
 
 3 
 
 first and in the second position with respect to the base. For 
 the first consider the sphere to consist of a body of sp. gr. 
 equal to i and a sphere of sp. gr. equal to s i. Hence if 
 w = weight, per unit volume, of water, the first mass-moment 
 is TTW [^ 2 A* + f (s i ) "*], and the mass-moment in the second is 
 nw[%a i x' > + s(x + r) r 3 ]. The excess of the second over the 
 first is the work required.) 
 
 9. A solid cone of volume V, height h, and specific weight w 
 floats with its vertex down in a liquid of specific weight w con- 
 tained in a tank of uniform horizontal section A ; find the 
 amount of work necessary to submerge the cone. 
 
 *Vhw ( > / w \ /w'\5 ) V 2 (w w') z 
 
 Result. ^ *(-/ x ) + ( ) if - 
 
 4 ( v r / ^w' ) 2 Aw 
 
 (Take mass-moments about the surface of the liquid in the 
 freely floating position.) 
 
CH. iv Pressure on Curved Surfaces 101 
 
 10. A solid homogeneous body of weight W, volume F, and 
 specific gravity 3 rests on the bottom of a tank of uniform cross- 
 section A, the height of the centre of gravity of the body above 
 the base being z ; the tank contains water to a height h when 
 the body is immersed. Prove that the work required to raise 
 the body clear of the water is 
 
 A 
 
CHAPTER V 
 
 GASES 
 
 25. Definition of a Perfect Gas. A gas may be 
 
 described roughly as a fluid which can be easily compressed. 
 A more precise mathematical definition will be given 
 subsequently ; for the present, we shall define a perfect gas 
 as a fluid which obeys the law of Boyle and Mariotte. 
 This law is as follows : the temperature remaining constant, 
 the volume of a given mass of gas varies 
 inversely as its intensity of pressure. 
 
 This law is proved experimentally as 
 follows. Let HABK (Fig. 51) be a bent 
 glass tube of uniform section at least in 
 the leg AH which is closed at the top. 
 Let the gas to be experimented upon be 
 enclosed in the branch AH by means of a 
 column, ALB, of mercury, the branch LBK 
 of the tube being open to the atmosphere. 
 Suppose matters arranged so that when the 
 gas in AH is in equilibrium of temperature 
 with the surrounding air, after the pouring 
 in of the mercury has ceased, the surfaces A and B of the 
 mercury are at the same level in both branches. Then the 
 intensity of pressure at any point in the surface A is equal 
 to that at any point in B ; so that if JJ Q is the atmospheric 
 intensity of pressure, p is also the intensity of pressure of 
 the gas in AH. 
 
CH. v Gases 103 
 
 For simplicity denote p by the height of the barometer 
 at the time of the experiment. 
 
 Let thia height be h (inches or millimetres), and let 
 t' (cubic inches or cubic millimetres) be the volume of the 
 gas AH. If 10 is the weight of a unit volume (cubic inch 
 or cubic millimetre) of the mercury, we have j9 = w .h. 
 
 Let us now, by pouring mercury slowly into the open 
 branch at K, reduce the volume of the gas in AH to half 
 its value. If CH = AH, the mercury is to be poured in 
 until its level in the closed branch stands at C after all 
 disturbance and heating effect due to the pouring in of the 
 mercury have subsided. If we now read the difference 
 of level between C and the surface of the mercury in the 
 branch LK, we shall find it exactly equal to //, the height 
 of the barometer. Equating the intensity of pressure at C 
 due to the imprisoned gas to the intensity of pressure due 
 to the mercury and the atmosphere, we see that the former 
 must be equal to j) + wh, i. e. the new intensity of pressure 
 = 2/? , while the new volume is ^r . 
 
 Again, let EH = % AH, and let us pour mercury in at K 
 until the volume of the imprisoned gas is EH, i. e. ^ v . 
 We shall then find that the difference of level between E 
 and the surface, F, of the mercury (not represented in the 
 figure) in the open branch is 3 times the height of the 
 barometer, i. e. 3 h, so that the intensity of pressure of the 
 gas in EH is p Q + 3 w/i, or 4p . 
 
 Hence we have the following succession of volumes and 
 pressure intensities for the gas, its temperature being 
 the same all through, 
 
 If, in the same way, the volume is reduced to -v , the 
 
 difference of level of the mercury in the two branches is 
 found to be (n i) ft, so that the new intensity of pressure 
 
104 
 
 Hydrostatics 
 
 VOL. I 
 
 is npQ ; and from these results we see that in each case the 
 volume of the gas is inversely proportional to its intensity 
 of pressure, as stated in the law. 
 
 The law of Boyle and Mariotte may also be verified in 
 the following- simple manner by means of a single straight 
 tube, about 2 mm. in diameter. 
 
 Let AD be a tube of uniform section closed at the end A 
 and open at D ; let a portion, AB, of the tube be filled with 
 air or other gas, and let a thread of mercury, C, of length 
 
 I, separate this gas 
 D A" from the external air. 
 When the tube is 
 held horizontal and all 
 disturbance has sub- 
 B" sided, let the volume, 
 t' , of the gas AB be 
 read ; its intensity of 
 pressure is the same 
 as that at C, i.e. p 0t 
 the atmospheric in- 
 tensity. Now let the tube be held in a vertical position 
 with the closed end A' downwards and let the gas occupy 
 the volume A'B', or v'. Its intensity of pressure is now 
 equal to that at B' due to everything above B', i. e. p Q + wl, 
 where w = weight of unit volume of mercury. If li is the 
 height of the barometer during the experiment, p = wh, 
 and if p' is the intensity of pressure in A'B ', 
 
 p' = to(h + l). 
 
 Finally, let the tube be held vertically with the closed 
 end A" uppermost, and let the volume of the gas be A"B" ', 
 or v". If its intensity of pressure is p", the intensity 
 at (? is p" + wl due to everything above C" ; but p is 
 also the intensity of pressure at C" since that is a point in 
 the external air. Hence p" = w (h l). 
 
 53. 
 
CH. V 
 
 Gazes 
 
 105 
 
 Hence, as regards volume and intensity of pressure, we have 
 the succession of states 
 
 and we find on trial that 
 
 according to the requirements of Boyle's law. 
 
 This law is not accurately obeyed by any known gas, but 
 the approximation is very close in the case of all gases 
 when they are not near the state in which, either by 
 increase of pressure or by diminution of temperature, lique- 
 faction begins. When any gas is near the state of 
 liquefaction, its volume decreases more rapidly with in- 
 creased pressure than it would if it followed Boyle's law. 
 ' When it is actually at the point of condensation, the 
 slightest increase of pressure condenses the whole of it into 
 a liquid.' (Clerk Maxwell's Theory of Heat, chap, i.) 
 
 26. Graphic Representa- 
 tion. We can graphically 
 represent its various states as 
 expressed in the fundamental 
 equation (/8), thus : draw any 
 two rectangular axes, Ov, Op, 
 and let the volumes assumed 
 by the gas be measured, on 
 any scale, along Or, while 
 the intensities of pressure are 
 measured on any scale along 
 OP. 
 
 If, on these scales, OM and 
 0j\ r represent respectively any 
 volume and the correspond- 
 
 ing intensity of pressure, the point, P, whose co-ordinates 
 are OM and ON will graphically represent the state of 
 
 o NI 
 
 Fig- 53- 
 
io6 Hydrostatics VOL. i 
 
 the gas ; and all points, such as P, whose co-ordinates 
 satisfy () will be found on a rectangular hyperbola 
 passing through P and having the axes Ov and Op for 
 asymptotes. 
 
 Thus, then, the curve of transformation of a given mass of 
 gas at constant temperature is a rectangular hyperbola. Such 
 transformation is called an isothermal transformation. 
 
 The figure exhibits the fact that when the intensity of 
 pressure is infinitely increased the volume of the gas 
 becomes infinitely small, and that when the intensity 
 of pressure is infinitely reduced, the volume becomes 
 infinitely great. 
 
 The first result would be strictly true for a substance 
 whose transformations . strictly follow the law for all 
 values of p ; but it will be readily understood that there 
 exists no gas for which Boyle's law holds indefinitely, and 
 that when the intensity of pressure is greatly increased 
 the gas may approximate to, and actually become, a liquid. 
 
 27. Law of Dalton and Gay-Lussac. The volume of 
 a given mass of gas may be altered by heat as well as 
 by pressure. The law relating to this change was dis- 
 covered, independently by Dalton in 1801 and by Gay- 
 Lussac in 1802 ; and, apparently, it was discovered fifteen 
 years previously by M. Charles, although not published 
 by him. It is this 
 
 The intensity of pressure being constant, the volume of a 
 
 given mass of yets, when its temperature is raised from Hie 
 
 freezing to the boiling point of water, increases ly a fraction of 
 
 the volume at the first temperature, which fraction is the same 
 
 foi' all gases. 
 
 In short, the law is that all gases have the same coeffi- 
 cient of expansion, and that this is independent of the 
 magnitude of the (constant) intensity of pressure under 
 which they expand. 
 
CH. v Gases 107 
 
 The fraction in question is, with certain reservations 
 to be mentioned presently, 
 
 3 66 5> 
 
 so that if we measure degrees of heat by the Centigrade 
 thermometer, this is the fractional increase of volume for 
 1 00. The rate of expansion per degree is also found 
 to be uniform, and is -003665, which we shall use in 
 the form 
 
 Hence if V Q denotes the volume of a given mass of any 
 gas at o C., and v its volume at t C., we have 
 
 v=sv (i+~] (i) 
 
 273'' 
 
 whatever be the intensity of pressure (supposed constant) ; 
 and if v' is its volume at if, we have 
 
 v v' , . 
 
 273 + 1 ~ 273 + 1' ' 
 
 If the point from which the temperature is reckoned on 
 the Centigrade thermometer is removed 273 below the 
 ordinary zero, i. e. the point at which water freezes when 
 its surface intensity of pressure is that due to a standard 
 atmosphere (indicated by a mercurial column 760 mm. in 
 height), the expression 273 + 1 indicates the newly measured 
 temperature, and is always denoted by T, and called the 
 absolute temperature of the substance, the new point of 
 reckoning being called the absolute zero of temperature. 
 
 If it were possible to have T = o, that is t = 273, for 
 the gas supposing the substance to remain a gas at 
 
 * The fraction is more accurately r- , but the above is usually taken 
 
 272-85 
 
 for simplicity. Clerk Maxwell {Theory of Heat} gives various values; 
 
 thus, and , the latter deduced from experiments of Thomson 
 
 273s 273-7 
 
 and Joule. 
 
io8 Hydrostatics VOL. i 
 
 all temperatures with constant coefficient of expansion, 
 -- T equation (i) would give 
 
 i. e. the gas would be reduced to zero volume. As the 
 substance does not satisfy the above supposition, but alters 
 its state in the process of lowering the temperature, the 
 consequence is not realized, and it would thus appear that 
 the notion of an absolute zero of temperature at 273 C. is 
 a gratuitous error. Indeed, if the conception of absolute 
 temperature rested on no other foundation, we might simi- 
 larly argue from the coefficient of expansion of platinum, 
 
 for instance, that since for this body v = v n (i -I 
 
 ^ 
 nearly, where V Q is its volume at zero and v its volume 
 
 at , if we make t = 37699 we shall arrive at the absolute 
 zero of temperature. 
 
 The truth is that the measure of absolute temperature 
 rests on quite another basis, that it is intimately connected 
 with the coefficient of expansion of a perfect gas, and that 
 273 + 1 is properly to be regarded as measuring the absolute 
 temperature of a body whose temperature indicated by 
 a Centigrade thermometer is f. 
 
 Adopting absolute temperature, then, equation (2) gives 
 
 Of course in the expression of the law of Dalton and 
 Gay-Lussac it is not necessary to signalize the particular 
 temperature corresponding to the freezing of water as pos- 
 sessing any special reference to the expansion of gases. 
 
 The law may be stated thus : all gases expand, per degree, 
 by the same fraction of their volumes at any common tempera- 
 ture. This is obvious because their volumes at any tem- 
 perature, T, will all be the same multiple of their volumes at 
 
CH. v Gases 109 
 
 o, and a constant fraction of the latter will give a constant 
 fraction of the former. 
 
 In symbols, for any gas let u be the volume at r, 
 v that at t, v that at zero, and a the coefficient of expansion 
 with reference to the volume at zero ; then 
 
 v = V (l + at) ; n = V Q ( l + ar) ; 
 
 , , , . l+at 
 and therefore v u = u 
 
 l + a T I + ar 
 
 I +ar^ 
 
 where /3 = , so that /3 is obviously the rate of expan- 
 sion of the gas reckoned as a fraction of the volume u ; and 
 if a is the same for all gases, so is /3. 
 
 It is remarkable that a is the same for all gases when far 
 removed from their condensing points, i. e. from the liquid 
 states, and that it is independent of the intensity of 
 pressure under which the expansion takes place. 
 
 Clerk Maxwell (Theory of Heat) points out that if the law 
 of Dalton and Gay-Lussac u true for any one intensity of 
 pressure, and if the law of Boyle holds, it follows that the 
 former law holds for all intensities of pressure. 
 
 This is very easily proved thus. Let v be the volume 
 of a given mass of gas at (o, p), i. e. p is its intensity 
 of pressure, and let the law of Dalton hold for this pressure 
 intensity ; then if v is its volume at (t, p) 
 v = v (l +at). 
 
 Now, keeping t constant, alter p to p' ; then by Boyle's 
 law the new volume, u, is given by the equation 
 
 --,' 
 
no Hydrostatics VOL. i 
 
 But if v at (o,p) were altered by keeping its temperature 
 zero and changing its intensity of pressure to />', its value, 
 
 n 
 
 M O , would be V Q . , by Boyle's law ; so that the last equa- 
 tion gives 
 
 u = u Q (i + at), 
 
 and therefore Dalton's law holds for p' if it holds for p. 
 
 With regard to the accuracy of the law of Dalton and 
 Gay-Lussac, Mi Regnault has found that, a being the 
 coefficient of expansion per degree Centigrade, 
 
 for Carbonic acid gas a = -003710, 
 Protoxide of Nitrogen = -0037 1 9, 
 ,, Sulphurous acid gas ,, = -003903, 
 Cyanogen = -003877 ; 
 
 the last two of which are notably greater than the coeffi- 
 cient of expansion of air ; but these are precisely the gases 
 that can be most easily liquefied, while it is found that for 
 all gases which can be liquefied only with great difficulty, 
 a has very nearly the same small value, -003665, that it has 
 for air. Hence M. Regnault modifies the law of Dalton 
 and Gay-Lussac by saying that the coefficients of expansion 
 of all gases approach more nearly to equality as their 
 intensities of pressure become more feeble ; so that it is 
 only when gases are in a state of great tenuity that they 
 have the same coefficient of expansion. 
 
 28. General Equation for the Transformation of a 
 Gas. Given the volume, v, of a mass of gas at the 
 temperature t, and pressure intensity p, find its volume 
 at t' and p'. 
 
 First let the temperature be altered from t to t', the 
 pressure intensity remaining p ; then the volume v becomes 
 u, where 273 + // 
 
 V 273 + t' 
 by equation (2) of last Art. 
 
CH. v Gases in 
 
 Now keep the temperature constantly equal to t' and alter 
 p toy ; then u becomes v', where 
 
 by Boyle's law. Hence we have 
 
 V = V 
 
 p_ 
 
 p 
 
 v'.p' v.p 
 
 " 
 
 or 
 
 273 + rf 273 + t 
 
 v'. p' v .n 
 or - - 
 
 <]}' ijl ' W 
 
 where T and T are the absolute temperatures of the gas. 
 
 Hence, whatever changes of pressure and temperature 
 may be made in a given mass of gas, we have the result 
 
 ~- = constant (/3) 
 
 between its volume, pressure intensity, and absolute tem- 
 perature. 
 
 This most important result is the general equation for 
 the transformation of a given mass of gas. 
 
 29. Formula in English Measures. Since the freezing 
 point of water is marked 32 on Fahrenheit's thermometer, 
 and the boiling point 212, the fractional expansion of gas 
 
 per degree Fahrenheit is --rr- or about - > of the 
 
 180 
 
 volume at 32. This fraction is usually taken as 7 | 5 ; and 
 this, as will presently be seen, would place the absolute zero 
 of temperature 460 Fahrenheit degrees below the zero of 
 the Fahrenheit scale. The experiments of Joule and 
 Thomson indicate 460-66 as the position of the absolute 
 zero ; but for all practical purposes we can take 460. 
 If a given mass of gas has a volume u at 32 F., and its 
 
ii2 Hydrostatics VOL. i 
 
 temperatnre is raised to t, we have, if the intensity of 
 pressure is unaltered, 
 
 / t 32s 
 
 v u ( H ' } . 
 
 492 t 
 
 Hence v = u ; and if v' is its volume at t', we 
 
 492 
 
 have ,/ 
 
 460 + 1 460 -t- 1' ' 
 
 If the intensity of pressure, estimated in any way, alters 
 from p to jo', 
 
 VP _ v'p' / x 
 
 460 + 2! 460 + ;!'' 
 
 It thus appears that if the temperature of the gas were 
 reduced to 460 R, its volume would vanish, supposing 
 that it obeys the kws of a gas during the whole process. 
 
 If we denote by T the absolute temperature, 460 + ^, 
 of the gas, we have the general equation of transformation 
 of a given mass 
 
 y = ^p- = constant (/3) 
 
 30. Law of Avogadro. One of the fundamental laws of 
 gases is known as the Law of Avogadro. It is the follow- 
 ing : equal volumes of all substances when in the state of 
 perfect gas, and at the same temperature and intensity of 
 pressure, contain the same number of molecules. 
 
 This law enables us to find the relative molecular weights 
 of all substances by converting these substances into vapours, 
 and then measuring the weights of known volumes of the 
 vapours at known temperatures and intensities of pressure. 
 Thus, it is found that a cubic foot of oxygen weighs 16 
 times as much as a cubic foot of hydrogen under like 
 conditions of temperature and pressure ; hence we conclude 
 
CH. v Gase* 113 
 
 that the mass of each molecule of oxygen is 16 times that 
 of a molecule of hydrogen. 
 
 31. Air Thermometer. A long- capillary glass tube, 
 AC, terminating in a bulb, , is filled with air, and a short 
 thread, m, of mercury is inserted into it, the end of the tube 
 (beyond C) being open. In order to fill the bulb and 
 portion of the tube to the left of m with air deprived of 
 moisture, the tube and bulb are first filled with mercury 
 which is boiled in the bulb. The open end is then inserted 
 into a cork fitting into the neck of a tube, J), filled with 
 chloride of calcium, which has the property of absorbing 
 aqueous vapour from air, a fine platinum wire having been 
 inserted into the stem CA through the tube D. If the 
 instrument is supported in a position slightly inclined to 
 
 *9?1 
 
 LL 
 
 Fig- 54- 
 
 the horizon on two stands and the platinum wire is agitated, 
 air enters through the chloride of calcium, and gradually 
 displaces the mercury from the bulb and stem, the process 
 being stopped when only a very short thread of mercury is 
 left. 
 
 The air in the instrument may now be considered to be 
 dry. 
 
 Detach the stem from the drying tube D, and place it in 
 a vertical position with the bulb B in a vessel filled with 
 melting ice. Suppose the barometer to stand at 760 mm., 
 thus indicating the standard atmospheric pressure. Then 
 when the air has assumed the temperature of the melting 
 ice. mark o on the stem AC at the under limit of the 
 mercury index m. 
 
1 14 Hydrostatics VOL. i 
 
 If then the instrument is placed vertical with the bulb 
 B surrounded by the steam of boiling water close to the 
 surface of the water, but not in it the index m will move 
 up towards C, and at its lower limit let 100 be marked on 
 the stem. Thus the two standard Centigrade temperatures 
 are marked on the stem, and the intervening space is to be 
 divided into 100 equal parts if the stem has previously been 
 ascertained to be of uniform bore. The graduations may be 
 carried then below zero and beyond 100. 
 
 If the tube of the air thermometer is made cylindrical all 
 through so that the bulb B is simply a uniform continua- 
 tion of the stem and we continue the graduations to 273 
 parts below the zero, we shall here reach the bottom, , of 
 the tube. 
 
 Hence the definition of the absolute temperature of a body, 
 which we are so far justified in giving, is simply, in the 
 words of Clerk Maxwell, its temperature reckoned from the 
 bottom of the tube of the air thermometer. 
 
 The upper end of the stem of an air thermometer neces- 
 sarily remains open to the atmosphere, otherwise the index, 
 m, would not move or would scarcely move at all : if the 
 end were closed and the air uniformly heated, m would not 
 move. 
 
 Hence the air thermometer cannot be used to indicate 
 temperature except in conjunction with the barometer. 
 If the latter stands at p instead of p Q , the standard height 
 (which we have above supposed to be 760 mm.) and the 
 temperature indicated by the index m is t, the real reading 
 is not t but that at which the index would stand if the 
 intensity of pressure were altered to p Q . To find the point 
 at which the index would stand in this case, let # be the 
 area of the cross-section of the tube, c the length of the tube 
 between two successive degrees, and B the volume of the 
 bulb and tube up to the zero mark. Then when the index 
 
CH. V 
 
 m stands at the mark t', the volume of the gas is B + cftf . 
 But since at the absolute zero the volume of the gas would 
 vanish, B = 273 cs ; hence 
 
 v = (273 + 1') cs, 
 
 and this is at the intensity of pressure 7;, its true tempera- 
 ture being t. If p were altered to p Q without any change 
 in the true temperature, the index would stand at t and the 
 volume would be (273 + ^)0?. Now since these volumes 
 are inversely as the intensities of pressure, we have 
 
 2 73 + / = (273 + 0-7' 
 
 1 o 
 which gives the true reading. 
 
 EXAMPLES. 
 
 1. A circular cone, hollow but of great weight, is lowered 
 into the sea by a rope attached to its vertex ; find the volume 
 of the compressed air in the cone 
 when the vertex is at a given depth 
 below the surface. 
 
 Let Fig. 55 represent a section 
 of the cone ; let c be the depth of 
 the vertex below the surface, LN, 
 of the water, h = height of cone, 
 V = its volume, t = the tempera- 
 ture of the air at the surface, t' = 
 temperature of the water, and there- 
 fore of the air in the cone; let P 
 be the surface of the water within 
 the cone, and let k be the height 
 of a column of sea-water in a water 
 barometer. 
 
 L 
 
 N 
 
 
 
 -^^ - i 
 
 
 
 =^=^=^^=^~ -S 
 
 = =rt=-=^-_^^^= = 
 
 
 ~^=^=-==- 
 
 Fig. 55- 
 
 If these quantities are in English measure, we may regard the 
 lengths as measured in feet, and the temperature as Fahrenheit ; 
 then k will be about 33 feet. 
 
 Now if x is the depth of P below A, the volume of the air in 
 
 Fee 3 
 the cone is -77-. The intensity of pressure of this air measured 
 
 fl 
 
 I 2 
 
u6 Hydrostatics VOL. i 
 
 by a column of water is k+ c + a-. Hence the following diagrams 
 represent the histoiy of this mass of air as regards volume, 
 temperature, and intensity of pressure : 
 
 in which T and T' are absolute temperatures. 
 From Art. 28 or Art. 29 we have, then, 
 
 h 3 T' 
 
 T' 
 
 =0, 
 
 from which x can be found. 
 
 A vessel used in this manner is called a diving-bell. The 
 above is a conical diving-bell. 
 
 2. If in the above position of the cone it is desired to free 
 the interior of water completely by pumping the air above the 
 surface into the cone, find the volume of this surface air that 
 will be required. 
 
 Let U be the volume required, and h the height of the cone ; 
 then suppose the cone to be wholly filled with air of the tem- 
 perature t' of the surrounding water, and write clown the 
 history of this air, thus : 
 
 V, 
 
 T' 
 
 T ~~ T' 
 
CH. v Gases 117 
 
 (It is iiot improbable that the student will fall into the error 
 of supposing that U can be calculated as the volume of the 
 surface air which is required to occupy the lower portion of the 
 cone in Fig. 55, i.e. the portion occupied by water.) 
 
 Of course the result is the same whether the vessel is conical 
 or of any other figure. 
 
 3. If a conical diving-bell of height h feet contains a mercurial 
 barometer the column of which stands at p inches when the 
 bell is above the surface of the water, and at a height p when 
 below, infer the depth of the top of the bell below the surface. 
 
 4. Deduce the depth for a cylindrical or prismatic bell. 
 Rml , 3|96 <,-)- ^. 
 
 5. A cylindrical diving-bell 1 2 feet high is lowered into water 
 until the depth of the top of the bell is 60 feet ; the height of 
 a mercury barometer at the surface is 30 inches and the tem- 
 perature of the air 76 F. ; the temperature below is 48; find 
 the height to which the water rises in the bell. 
 
 Result. 8-1 feet. 
 
 What volume of surface air must be forced down into the 
 bell to expel all the water 1 
 
 Result. 2-29 times the volume of the bell. 
 
 6. If the bell is 10 feet high and is lowered until the depth 
 of the top is 40 feet, find the same things when the surface 
 temperature is 60 and the temperature below is 50, the height 
 of a water barometer being 34 feet. 
 
 Result. 4- 26 feet; 1-52 x vol. of bell. 
 
 7. Find the tension of the suspending chain in a diving-bell 
 which occupies any position in water. 
 
 Result. The weight of the bell and its appurtenances 
 
n8 Hydrostatics VOL. I 
 
 diminished by the weight of the water which is displaced from 
 all causes. 
 
 (The water is displaced by the chain, the thickness of the bell, 
 and the air within the bell ; the weight of this water is the 
 force of buoyancy. In strictness, the weight of the contained 
 air should be added to that of the bell.) 
 
 8. If at the bottom of a river 40 feet deep, when the tem- 
 perature is 40 F., a bubble of air has the volume ^ of a cubic 
 
 inch, what will be its volume on reaching the surface where the 
 temperature is 50 F.. and the height of a water barometer is 
 34 feet? 
 
 Ans. - cubic inches. 
 
 io 5 
 
 9. If an open vessel (such as a tumbler) made of a substance 
 whose specific gravity is greater than that of water is forced, 
 mouth downwards, into water, show that its equilibrium becomes 
 unstable after a certain depth has been reached. 
 
 (If the volume of the solid substance of the vessel is v, and 
 in any position of the vessel if X is the volume of its compressed 
 air, the downward force, P, required to hold it in equilibrium 
 is given by the equation 
 
 P = Xv: v(w w\ 
 
 where w = specific weight of water, w f = specific weight of 
 substance of vessel. 
 
 Hence when X is so far diminished by forcing the vessel 
 down that Xw = v(w'w), the pressure P vanishes, and after 
 this an upward pull would be required.) 
 
 10. If v is small (i.e., if the thickness of the vessel is small), 
 and if V is the volume of the interior of the vessel, prove that 
 when the position of instability is reached, the depth of the top 
 of the vessel below the surface of the water is approximately 
 
 Vw 
 
 I 
 
 r 
 
 where k is the height of a water barometer at the surface. 
 
CH. v Gases 119 
 
 11. In a mercury barometer tube the height of the mercury 
 column is 30 inches, the area of the cross-section of the tube is 
 i sq. inch, and the vacant space is 4 inches long. If % of 
 a cubic inch of the external air is passed up into the tube, what 
 depression of the column will it produce ? 
 
 Ans. i inch. 
 
 12. If v cubic inches of the external air at the absolute tem- 
 perature T are inserted into the Torricellian vacuum of a 
 uniform cylindrical barometer tube, calculate the depression 
 produced in the column of mercury if the absolute temperature 
 changes to T f . 
 
 Let h inches be the height of the barometer at first, 
 a = length of Torricellian vacuum, s square inches = area 
 of cross-section of tube, x = length of tube finally occupied 
 by the air; then 
 
 r vh 
 
 x(x-a) = T >~ 
 
 13. A diving-bell of any shape occupies a given position 
 below the surface of water ; the bell has a platform inside ; if a 
 large block of wood falls from the platform into the water, prove 
 that the water will rise inside the bell, but that the bell now 
 contains less water than before. 
 
 Let the depth of the top be c, let h be the height of a water 
 barometer at the surface, put k = c + h, let B = volume of the 
 block of wood, w f its specific weight, w = specific weight of 
 water, V= whole volume of the interior of the bell, let * be the 
 depth of the water in the bell below the top of the bell, and let 
 X be the volume of the interior of the bell above this surface. 
 
 Then (X-B)(x + k) = Vh ....... (i) 
 
 "When the wood falls the volume of it which remains above 
 
 the surface is B ( i -- ) . Let x' be the new depth of the water 
 V w' 
 
 surface in the bell below the top of the bell, and X' the volume 
 of the interior of the bell above this new surface. Then 
 
 F/, . (2) 
 
 Now since X obviously increases with x, we must have x' <x, 
 
120 Hydrostatics .. VOL. I 
 
 since in the opposite case each of the factors at the left-hand 
 side of (2) would be greater than the corresponding factor 
 in (i), and the equations would be inconsistent. 
 
 Hence the surface of the water rises. 
 
 Again, if 1 = the first volume of water in the bell, and H' 
 the second 
 
 IZ = r A. , 
 
 &' = V-X'-B -; 
 w 
 
 ,,,./ 
 
 ... fl-ii' = *--(jr-A") ..... (3) 
 
 Now from (i) and (2) 
 
 / Vh(x-x') 
 
 n n/_ y 
 which shows that 1' is less than il. 
 
 32. "Weight of Gas. It is obvious that the weight of 
 a cubic foot of air, or any other gas, is not the same when 
 its temperature is 20 or 100, as when it is o, supposing 
 the intensity of pressure the same. In other words, the 
 weight of a cubic foot of air depends on the temperature 
 and pressure intensity at which it is taken. 
 
 Taking the units of the Metric System, let us inquire 
 what is the weight of v litres (i. e. cubic decimetres) of dry 
 air when its temperature is f C. and its intensity of pressure 
 denoted by a column of mercury j) millimetres high. 
 
 Supposing that we knew the weight of i litre of air 
 when its temperature is o and its intensity of pressure that 
 of a standard atmosphere, denoted by a column of mercury 
 760 mm. high, we could answer the question by finding 
 the number of litres which would be occupied by the given 
 v litres if its state were changed from (/;, t) to (760, o). 
 
CH. v Gases 121 
 
 But by (i) or (2) of Art. 28, if we put t' = o, /= 760, 
 -have r _2 73 .^ 
 
 7 -^6o ~T' 
 
 Now M. Regnault found that the mass of 
 i litre of dry air at (760, o) = 1-293187 grammes . (2) 
 
 Hence the mass of v litres at (/;, f) is v multiplied by 
 this number. Denoting the mass in grammes by IP, we 
 have then 
 
 , ...... (a) 
 
 in which, be it remembered, T is the absolute Centigrade 
 temperature of the air, p its pressure intensity estimated in 
 millimetres of mercury, v its volume in litres, and W its 
 mass in grammes. 
 
 For any other gas, if its specific gravity at (760, o) is 
 denoted by *, the mass of a litre of it in this state is 
 1-293187 xsgrammes, and evidently if W \$> the mass. of v 
 at (p, t), we have simply 
 
 ^r* ...... 03) 
 
 The specific gravity of a gas is above assumed to be the 
 ratio of the weight of any volume of the gas to the weight 
 of an equal volume of dry air at (760,0) ; but it is easy to 
 see that we get exactly the same result by taking the ratio 
 of the weight of a volume of the gas at (p, f) to the weight 
 of an equal volume of air also at (p, t), whatever the pressure 
 intensity, p, and the temperature, t, may be, if it be true 
 that all gases have the same coefficient of expansion ; for, 
 equal volumes, v, of the two gases at (/?, f) will become 
 equal volumes, ?; , at (760, o) since 
 
 v .p i 
 ~~ 
 
 ~~ I + at 760 
 and a is the same for both gases. 
 
122 Hydrostatics , VOL. I 
 
 33. The equation p = kp. From (ft) we see that if for 
 
 W 
 
 we write p, where p is the mass, in grammes, of the gas 
 
 per litre, we have 
 
 p being measured in millimetres of mercury. 
 
 Let p be measured in grammes' weight per square 
 centimetre, and let p be the mass, in grammes, per cubic 
 centimetre ; then, if we have I cubic cm. of the gas at 
 (p, ), and this becomes x cubic cm. at zero and an intensity 
 of pressure of 76 x 13-596 grammes' weight per sq. cm., 
 
 we nave x x 76x1^-596 _ ix/> 
 
 273 T (2) 
 
 Now the mass of I cubic cm. at the latter temperature 
 
 and pressure being - * grammes, the mass of x cubic 
 
 cm. is obtained by multiplying this by the value of x in 
 (2), and this mass is p, the density of the gas at (t,p), i.e. 
 the mass of i cubic cm. Hence we have 
 
 T 
 p = 2926-9 -p, ...... (3) 
 
 o 
 
 where p is in grammes' weight per sq. cm. and p in 
 grammes per cub. cm. Hence if we write the relation 
 between p and p in the form p = kp, we see that 
 
 T 
 
 k = 2926-9- . 
 7 * 
 
 If /; is measured in dynes per sq. cm., we must multiply 
 this value of k by the value of g in dynes, i.e. by 981 
 (about). In this case, then, 
 
 gT 
 P = 29rf'9 P ...... (4) 
 
CH. v Gases 123 
 
 (Observe that here * is the specific gravity of the gas 
 referred to air.) 
 
 If v is the volume of the gas at (p, T 7 ), and w its mass, we 
 have, by multiplying both sides of (3) by t\ 
 
 iff 
 pv = 2926-9 -.T. (5) 
 
 o 
 
 It is sometimes useful to express p in kilogrammes' 
 weight per square decimetre, v in cubic decimetres, and tv 
 in kilogrammes ; in which case (5) becomes 
 
 w 
 pv = 292-69 - . T. .... (6) 
 
 It is usual to write (5) or (6) in the form 
 
 pv = RwT, (7) 
 
 O Q2 6* O 
 
 where R stands for the constant ~ in the first case. 
 
 s 
 
 34. Formulae in English Measures. The equation 
 connecting the volumes, &c., of a given mass of gas in 
 English measures is 
 
 _vp_ _vy_ 
 
 460 + 1 460 + V ' 
 
 To obtain the formula for the mass of air, analogous 
 to (a), Art. 32, we may either convert the metric formula 
 into English measures, or deduce a formula from special 
 observations on the mass of a given volume of air under 
 standard conditions. Dr. Prout found that the mass of 100 
 cubic inches of dry air at the temperature 60 P. at an 
 intensity of pressure indicated by 30 inches of mercury in 
 a barometer tube is 31-01 17 grains ; in other words, 
 the mass of I cubic foot at (60, 30") is -0765546 pounds, (a) 
 Now if we have v cubic feet of dry air at (t,p], where p 
 is in inches of mercury, this would, by (i), become 
 520 vjj 
 30 460 + ^ 
 
124 Hydrostatics VOL. i 
 
 cubic feet at (60, 30), and multiplying this by the number 
 (a), we have 
 
 W= 1.327!^ (2) 
 
 ' 460 + f 
 
 for the mass, in pounds, of the given v cubic feet at (t,p), 
 the intensity of pressure, p, being supposed taken in inches 
 of mercury. 
 
 For a gas of specific gravity # (referred to air), 
 
 * ' 460 + 1 ' 
 
 If jo is estimated in pounds' weight per square foot, and 
 460 + 1 is denoted by T, we have, with sufficient accuracy, 
 
 TIT _ I Vpg 
 
 = 53'3 T' 
 T 
 
 where p is the density of the gas in pounds per cubic foot, 
 p its intensity of pressure in pounds' weight per square 
 foot, &c. 
 
 To obtain the analogue of (6), Art. 33, multiply both 
 sides by v ; then 
 
 where w is the mass of the gas in pounds ; or 
 pv = RwT, 
 
 53-3 
 
 where R stands for 
 
 EXAMPLES. 
 
 1. If a cubic inch of water is converted into steam at 212 F., 
 find the volume of the steam. 
 
 Result. 1696 cubic inches. Hence it is roughly true that 
 a cubic inch of water yields a cubic foot of steam. 
 
CH. v Gaset* 
 
 2. Calculate the mass of air in a room whose dimensions are 
 1 8, 1 8, and 10 feet, the temperature being 6o c F. and the 
 barometer standing at 30 inches. 
 
 Result. 248 pounds. 
 
 3. If I pound of water is converted into steam at 212 F. at 
 an intensity of pressure of 15 pounds' weight per square inch, 
 find the volume of steam in cubic feet. 
 
 Result. 26-66. 
 
 4. If any volume of water is converted into steam at t F. 
 under a pressure of p pounds' weight per square inch, show 
 that the ratio of the volume of steam formed to that of the 
 water is about 
 
 460 + 1 
 
 "" 
 
 (This is called the relative volume of steam at the tem- 
 perature t.) 
 
 35. Barometric Formula. We are now in a position to 
 deduce a formula for the height of a mountain, by assuming 
 the temperature of the 
 air to be constant 
 within these limits. 
 The latter assumption 
 would often be far 
 from the truth, but 
 we shall presently see 
 how it can be cor- 
 rected. 
 
 Let A, Fig. 56, be Fig. 56. 
 
 a point at the base of 
 
 the mountain where the height of the barometer is jo mm. ; 
 let P be a point at a height of z decimetres above A, and 
 at P let the barometric height be p mm. ; let Q be a point 
 close to P at a height dz above P ; let t be the temperature 
 of the air at P. 
 
ia6 Hydrostatics VOL. i 
 
 Imagine a horizontal area of I square decimetre at P ; 
 then the atmospheric pressure on this area is the weight of 
 the column of air standing on it and extending to the upper 
 limit of the atmosphere. Hence the difference of the 
 pressures on this area at P and Q is the weight of the 
 vertical column of air of height dz standing on i square 
 decimetre. But if the height of the barometer at Q is 
 p + dp mm., the pressure on this area is also the weight of 
 a column of mercury dp millimetres high. Hence 
 weight of 10 dp c.c. of mercury = weight of dz litres of air; 
 that is (p. 131) 
 
 pdz 
 
 -i 35-96 ^p = .4645 --_, . . . (i) 
 
 which gives the relation between dp and dz. The integral 
 gives, if t is constant, 
 
 z = 673-962 (273 + *)lo& 10 ^, 
 
 z being in decimetres. If z is taken in metres, this becomes 
 with sufficient accuracy 
 
 ,= 18400(1+^)10^. . . . (a) 
 
 If the variation of gravity is taken into account, let p be 
 measured in grammes' weight per square centimetre, since 
 it matters not in what units p and p are measured in (2). 
 Then if r denotes the radius of the Earth, 
 
 dp _ r 2 
 
 ~dz~ ~~ 
 
 p = 2926-9 Tp, 
 
CH. v Gases 127 
 
 in which again we can put 
 
 P > + 
 
 where h Q and h are the observed barometric heights at the 
 
 z 2 
 two stations. Neglecting ^ , and taking z in metres, this 
 
 equation becomes 
 
 z-z, = 18400(1+ -L)i oglo (i + a} 
 
 Assuming r = 637 x io 4 metres, and observing that 
 
 approximately, this equation gives 
 ^-^ = 18400(1 + ^l)|i +-0025 
 
 The barometric readings /? , p in these equations require 
 slight corrections for temperature ; for we have assumed 
 that the mass of a cubic centimetre is 13-596 grammes, 
 which is true only if the temperature is zero. 
 
 Since the coefficient of expansion of mercury per degree 
 
 C. is - , a volume which was I c.c. at o would become 
 5550 
 
 c.c. at t while its mass remains 13-596; hence at 
 
 555 
 the temperature t the mass of I c.c. would be 
 
 ) grammes, 
 
 nearly, and the mass of 
 
 io dp c.c. = 135-96 (l - -;) dp ; 
 
 so that we must regard the height p observed at the tem- 
 perature t as corrected, or reduced, to (i )ff, and hence 
 
128 Hydrostatics VOL. i 
 
 at the base, where the temperature is / and the observed 
 
 barometric height is ^ , we must replace p Q by ( i - - )/? . 
 
 v 
 
 and at the top similarly use ( i -- )P instead of p. 
 
 ooo 
 These are called the corrected barometric heiglits. In the term 
 
 i H -- which occurs in (2) it is usual to take f as the 
 
 273 
 mean of the temperatures at the top and bottom. 
 
 If the Fahrenheit thermometer is used, the coefficient of 
 expansion of mercury may be taken as - , and the 
 
 IOOOO 
 
 corrected height is f I - ) . 
 
 V JOOOO'^ 
 
 The formula in English measures which corresponds to 
 (2) is found in the same way to be 
 
 r(feet) = 122-73 (460 + *) Iog 10 -^, . . (3) 
 where t is mean Fahrenheit temperature. 
 
 EXAMPLES. 
 
 1. The height of the barometer on the ground being 30 inches 
 and the temperature 32 F., what height corresponds to a fall of 
 i inch in the barometric height t 
 
 Ans. About 890 feet. 
 
 2. The temperature of the air being assumed at o C. and the 
 barometric height at the base of a mountain 76 cm., if the 
 mountain is 4600 metres high, what will be the barometric 
 height at the top ? 
 
 Arw. 42-74 cm. 
 
 3. At the foot of a mountain the temperature of the air is 
 66 F., and the height of the barometer 29-35 inches ; at the top 
 the temperature is 50 F., and the barometric height 24-81 ; find 
 the height of the mountain, assuming the coefficient of expansion 
 of mercury to be r^^ per degree F. 
 
 About 4590 feet. 
 
CHAPTEK VI 
 
 HYDRAULIC AND PNEUMATIC MACHINES 
 
 36. Water Pumps. The Common Pump. Let DB, Fig. 
 57, represent a vertical section of an iron cylinder termin- 
 ating in a much narrower cylinder 
 or pipe, BA, which dips into a well 
 from which water is to be raised. 
 In the cylinder DB, or barrel, works 
 a piston having a valve, v, which 
 opens upwards, the piston rod, r, 
 being connected at c with a lever, 
 fH, working about the fulcrum f. 
 
 At the top, , of the suction pipe 
 is a valve which also opens upwards. 
 
 Let the piston be at the bottom 
 of the barrel, the level of the water 
 in the well being A, and the pipe 
 and barrel both completely filled 
 with air. 
 
 When the piston is raised by 
 means of the handle, H, of the lever, 
 the valve v remains closed, and the 
 pressure of the air in the pipe opens Pi g< ,^ 
 
 the valve at B, the air in AB rising 
 
 and partly filling the barrel, its intensity of pressure also 
 diminishing. As a result of this diminution of pressure 
 
130 Hydrostatics VOL. i 
 
 below its original (atmospheric) value, the atmospheric 
 pressure on the water forces some of the liquid into the 
 pipe. Let P be the level of the water in the pipe at the end 
 of the upward stroke of the piston. On the downwai'd 
 stroke of the piston the valve B closes and v opens allowing 
 the air in the barrel which the downward motion of 
 the piston tends to compress to escape through the piston 
 into the atmosphere, until the piston again reaches the 
 bottom of the barrel. On again raising the piston, the 
 valve v closes and that at B opens, thus allowing the air in 
 BP to expand and to diminish its intensity of pressure ; 
 and, in consequence, more water is forced up into the pipe, 
 and perhaps into the barrel. This process being continued, 
 the water ultimately reaches the level of the spout and 
 is thus driven out. 
 
 To find the height to which the water is raised in the pipe 
 by the first stroke of the piston, let A = area of cross-section 
 of barrel, a = area of cross-section of pipe, I = length of stroke 
 of piston, c = length BA, h = height of a water barometer, and 
 AP = x. The volume of air in the suction-pipe before the 
 stroke is ac, and its intensity of pressure is represented by h. At 
 the end of the stroke the volume of this mass is a(c x) + lA, 
 and its intensity of pressure is represented by hx. Hence 
 
 (h x) [a(c x) + lA} = ach, 
 
 = o, 
 
 which determines x. 
 
 When the water is flowing out of the spout, there will 
 be a tension in the piston rod on an upward stroke, which 
 is found thus. Let z be the vertical height of the piston above 
 the level A, and let z f be the height of the column of water 
 (which we may suppose to reach to any point, D, of the barrel) 
 above the piston. The valve at B being open, there is a contin- 
 uous water communication between A and the bottom of the 
 piston. Hence if w is the weight of a unit volume of water, 
 the intensity of the upward pressure exerted by the water on 
 the bottom of the piston is w (h z) ; and the intensity of down- 
 
CH. vi Hydraulic and Pneumatic Machines 131 
 
 ward pressure exerted on the top of the piston is w(h + z'); 
 therefore the total downward pressure on the piston is w(z + z')A. 
 This is equal to T, the tension of the rod, if we neglect any 
 acceleration of the piston. Hence, approximately, 
 
 T = w.A.DA, 
 
 which shows that the tension of the rod is the weight of a 
 vertical column of water having the piston for base, and for 
 height the difference of level of the water in the barrel and that 
 in the well. 
 
 On the downward stroke there is a pressure in the rod, 
 which is approximately equal to the weight of the column 
 of water above the piston. 
 
 When the water is flowing out, the force required at // 
 
 fc 
 to work the piston on the upward stroke is T x jj\> where T 
 
 has the above value. 
 
 It is obvious that, for the working of the pump, the length 
 AB of the suction-pipe above the well must be less than 
 the height of a water barometer, i.e., about 34 feet ; and, owing 
 to imperfect fittings, AB must be considerably less than this 
 say about 25 feet. 
 
 In the middle ages a curious modification of the common 
 pump, called the bellows jmmp, was employed in Europe. Instead 
 of a piston worked by a lever, fff(Fig. 57), a large bellows was 
 attached firmly to the top of the barrel, and the nozzle of 
 the bellows was the spout through which the water was forced. 
 
 The top of the barrel fitted into the interior of the bellows 
 through a hole in the lower board of the bellows ; there was no 
 valve in the top board, but there was one opening outwards fixed 
 in the nozzle. The action was, of course, the same as that in 
 our modern pump. 
 
 The Forcing Pump. This is an instrument for raising 
 water to a great height. It differs from the previous pump 
 in having a completely solid piston. 
 
 To the barrel of the pump is attached the pipe through 
 which the water is to be raised. This delivery pipe is pro- 
 vided with a valve at D, Fig. 58, opening upwards, and 
 this pipe is represented as being provided with a spout and 
 
 K 2 
 
133 
 
 Hydrostatics 
 
 VOL. I 
 
 stop-cock by means of which the machine can be made to 
 act as a Common pump. 
 
 The action is the same as 
 in the previous. case. 
 
 On the downward stroke 
 of the piston, the valve at 
 B closes and that at D 
 opens, and through this 
 latter the water is forced 
 out of the barrel into the 
 delivery pipe, DV. 
 
 There is then a pressure 
 in the piston-rod, r, equal 
 to the weight of a column 
 of water having the piston 
 for base, and for height the 
 difference of level between 
 the piston and the water, 
 F, in the delivery tube. 
 
 On the upward stroke 
 Fig. 58. there is a tension in the 
 
 rod, whose value is the same 
 as in the previous pump. 
 
 The Fire Engine. This is simply a double forcing pump. 
 The figure (Fig. 59) represents the two barrels, P and Q, 
 as immersed in a tank, DEI, full of water ; and from this 
 tank the pumps, which are both worked by the lever AB, 
 force the water through a hose connected with the chamber 
 C at Ji. The water is forced through this hose to the place 
 where the fire is to be extinguished. The action of the 
 valves is obvious in the figure. Such would be the arrange- 
 ment in a small fire engine, the tank DEI being filled 
 by buckets of water brought by hand. 
 
 When large fires have to be dealt with, a supply thus 
 
CH. vi Hydraulic and Pneumsdic Machines 133 
 
 derived would be of no use, and the water which is pumped 
 through the hose must be derived from a well or other 
 reservoir by means of a suction pipe. The figure represents 
 at c the place where such a pipe can be attached to the 
 engine. 
 
 The chamber C is partly filled with water and partly 
 with air, and is called an air chamber. Such a chamber 
 may be, and often is, fitted to forcing and other pumps, 
 the object being to render the stream of water from the 
 
 delivery pipe at h continuous instead of intermittent ; 
 and this result is evidently secured by means of the 
 compressed air at the top of the chamber ; for, since 
 this air was originally at the atmospheric pressure (when 
 it filled the whole of the chamber), its intensity of pressure 
 after the chamber is partly filled with water is greater than 
 this value. This increased pressure is therefore continuously 
 exerted on the top of the water in the chamber and helps 
 to drive the stream through the hose. 
 
 The Hydraulic Screw. This is one of the most ancient 
 machines for raising water,, and is still employed in some 
 
134 Hydrostatics VOL. i 
 
 countries. It is often called the Screw of Archimedes, be- 
 cause its invention is supposed to be due to the philosopher 
 of Syracuse. There is, however, reason to think that it 
 was first employed in Egypt. 
 
 As represented in Fig. 60, it consists of a pipe wound 
 spirally on an axis which is fixed in a position inclined to 
 the vertical, its extremities fitting into two solid supports, 
 
 Fig. 60. 
 
 A, B, the axis (and, with it, the screw) being able to revolve 
 freely. Both ends of the spiral pipe are open, and the 
 lower is immersed in the water which is to be raised to the 
 level 2). 
 
 The revolution of the screw can be effected in various 
 ways : in the figure it is represented as produced by the 
 revolution of a shaft C fitted with a toothed wheel which 
 gears with another fitted to the top of the axis of the screw. 
 
CH. vi Hydraulic and Pneumatic Machines 135 
 
 When the screw is made to revolve so that the lower 
 end comes towards us in the figure, water entering- at this 
 end continually drops to the lowest point of each part 
 of the spiral, and is thus carried continuously up to the top 
 where it is discharged. 
 
 There is a certain condition that must be satisfied by the 
 inclination of the axis of the screw to the vertical and the angle 
 of the spiral in order that the machine may be able to raise the 
 water. The condition is this : the inclination of the axis of the 
 screw to the horizon must be less than the inclination of the tangent 
 line of the spiral to the axis of the screw. To prove this, we may 
 put the matter thus : the axis of the screw must be so much 
 inclined to tJie vertical that it is possible to draw a horizontal 
 tangent to the spiral. This is obvious, because if we imagine a 
 single particle (suppose a small marble) entering the lower end 
 of the pipe, it would not drop down farther through the opening 
 unless there were in the pipe a place in which the particle 
 could rest under the action of its own weight and the reaction of 
 the pipe on it ; and at such a place the tangent to the spiral 
 must be horizontal. 
 
 The Hydraulic Screw is capable of a differential form. Suppose 
 the screw in Fig. 60 not to dip into water at its lower extremity, 
 but to receive into the upper end of the pipe at A a stream of 
 water from any source. Then the screw, being fixed exactly as 
 represented, would be driven by this stream in the direction 
 opposite to that in which it was caused to rotate under the 
 previous circumstances. 
 
 Now suppose that it is, as before, desired to raise water from 
 a well at the extremity B to a position D, and suppose that 
 there is available a stream of water at some lower level, repre- 
 sented by the arrow L. Let a second pipe be spirally wound 
 round the axis BA outside the tube represented in the figure and 
 in tJie reverse sense, its upper end terminating at the level L. If 
 the stream L is led into the upper end of this second pipe, it will 
 cause the whole machine to rotate in the sense required to raise 
 the water from the well by means of the internal spiral pipe. 
 
 The idea of the Differential Hydraulic Screw appears to 
 be due to the ingenious Marquis of Worcester, who published 
 his notions on this machine and on many others in a work called 
 A Century of Inventions, in the reign of Charles II. 
 
136 Hydrostatics VOL. i 
 
 The Hydraulic Ham. This is a machine in which the 
 momentum of a stream of water is suddenly stopped, with 
 the result that a portion of the water is forced up to 
 a considerable height. 
 
 Let AB, Fig. 61, be a stout iron chamber to which 
 
 is attached a pipe P which admits a flow of water from 
 a stream or reservoir the level of which may be only 
 slightly higher than that of AB. The vessel AB is fitted 
 with a support, E, for a valve, v, which can move freely up 
 and down. When this valve falls, there is a free communi- 
 cation between the interior of AB and the atmosphere, and 
 
CH. vi Hydraulic and Pneumatic Machines 137 
 
 if AB is full of water, some of this water flows away through 
 the opening at v, and is wasted. 
 
 To the top of the vessel AB is screwed a chamber, C, 
 which has a valve c opening upwards, and also a valve 
 opening inwards. This latter valve is attached to a rather 
 weak spring fixed to a side pipe opening into the chamber 
 C. Finally, a supply pipe, S, is attached to the chamber C. 
 
 Imagine the whole machine free of water, so that 
 the valves c and v are down ; and then let the stream flow 
 in at P. At first the water will rush through the opening 
 at v ; but soon the rush of water will close this valve, and 
 at this instant, the water being suddenly checked, some 
 will be forced through the opening at c. This valve will 
 then close and v will drop, allowing an outflow again from 
 the vessel AB. The same process will be again and again 
 repeated until the water forced into C rises in the pipe 
 S. The upper part of the chamber C contains imprisoned 
 air, the pressure of which serves to keep the flow up the 
 pipe S continuous. The valve v falling and rising thus 
 regularly, the machine is self-acting. After a long time 
 the air in the air chamber C would be absorbed by the 
 water, and thus the advantage of an air chamber C would 
 be lost. The object of the snifting valve s is to prevent this, 
 and the air is renewed as follows. When the valve v drops, 
 the intensity of pressure immediately under it is that of the 
 atmosphere, and therefore the intensity of pressure of the 
 water at s is less than that of the outside air, so that 
 (as the spring which closes s is a weak one) this valve s is 
 forced inwards, thus allowing some air to enter the water ; 
 and this air when the valve c is next opened will rise to 
 the top of C and replace the air absorbed by the water. 
 
 (It must not be supposed that the forcing of water 
 by this self-acting machine to a height vastly greater than 
 that of the source whence the water was derived involves 
 
i 3 8 
 
 Hydrostatics 
 
 VOL. I 
 
 any contradiction of the principle of Work and Energy ; 
 for, it is by means of the kinetic energy generated by 
 gravity in the very large mass of water which flows into 
 and out of the ram that the comparatively small mass 
 of water is raised in the pipe.) 
 
 The Hydraulic Ram was invented in 1772 by Whitehurst 
 of Derby, his machine, however, not being self-acting. 
 With him, instead of the self-acting valve v, there was a 
 stop-cock through which the water flowed ; and it was on 
 the sudden closing of this stop-cock that the water was 
 forced through the valve c. 
 
 37. Air Pumps. The Common Air Pump. In Fig. 62, 
 
 B is a cylinder or barrel in 
 which works a piston with 
 a valve, c, opening upwards. 
 The barrel is screwed, or 
 otherwise firmly attached, to 
 a plate, D, through which 
 
 x^ =:: ^\ runs a groove which com- 
 
 II \\ 1 B municates with the interior 
 
 of the barrel through an 
 opening which can be closed 
 by a valve a ; the other end, 
 n, of this groove opens up 
 through a large plate PQ, 
 the upper surface of which 
 is perfectly flat. On this 
 plate is placed a large glass 
 vessel, A, called the receiver, 
 
 the mouth of which rests oc the plate PQ, ; the rim of the 
 receiver is ground, and it fits the plate so accurately that 
 the junction is air-tight, especially as a layer of grease is 
 rubbed on the rim before it is placed on the plate. 
 
 The object is to remove the air, partially or completely. 
 
 Fig. 62. 
 
CH. VI Hydraulic and Pneumatic Machines 139 
 
 from the receiver, and therefore from any body or vessel 
 that may be placed under it. 
 
 The manner in which this is effected is obvious from 
 the figure. Suppose the piston in the lowest position in 
 the barrel ; then when it is raised, a vacuum tends to form 
 above a, so that the air of the receiver raises this valve 
 and fills the barrel at the end of the stroke. On the 
 descent of the piston, a closes and c opens, and the air in 
 the barrel is thus expelled into the atmosphere. This pro- 
 cess is repeated many times, with the result that the air in 
 A is greatly diminished in mass. 
 
 To calculate the degree of exhaustion after n strokes of 
 the piston, let the volumes of the receiver and barrel be A 
 and B ; let the original intensity of pressure of the air in A 
 be^? , and Iet'j) 1 ,p 2 , p z ... be the intensities after i, 2, 3, ... 
 strokes. 
 
 Then after the piston has been raised the first time the 
 mass of air whose volume and intensity of pressure were 
 (A, p ) becomes (A + B, p^ ; hence 
 
 (A + B) Pl = Ap ...... (i) 
 
 When the down stroke is ended there is a different mass 
 of air in A, and it is denoted by (A,pj) ; and this same 
 mass of air is denoted by (A + ,j)^) at the end of the 
 second upward stroke ; hence 
 
 (A + 3)pt = Ap l ...... (2) 
 
 Similarly 
 
 (A + B)p 3 =Jp tt .... (3) 
 
 Hence, by multiplication, 
 
140 
 
 Hydrostatics 
 
 VOL. I 
 
 which gives the final intensity of pressure. If W n is the 
 weight of the air finally left, and W^ the original weight, 
 we have from (a) of Art. 32, 
 
 and a similar relation between the final density, p n , and the 
 original p . 
 
 For the very high exhaustions required in the globes of 
 incandescent electric lamps, and in the interior of vacuum 
 tubes, an air pump of this kind would be quite insufficient, 
 because, after the exhaustion has reached a certain limit, 
 the pressure of the gas is insufficient to raise the valve a. 
 
 Condensiny Air Pump. When 
 it is desired to fill a vessel, A, 
 Fig. 63, with air or any other 
 gas at a given intensity of 
 pressure, a condensing pump 
 is employed. This machine 
 consists of a cylinder or barrel, 
 fitted with a solid piston, and 
 having a valve, c, opening 
 downwards. At the side of 
 the barrel there is attached a 
 pipe having a valve, a, opening 
 inwards. When any other gas 
 than air is to be forced into 
 A, the vessel supplying this 
 gas is attached to the pipe at 
 a. The valve at a opens while 
 the piston is raised, and the 
 p. 6 barrel is filled with the gas. 
 
 When the piston is lowered, 
 
 a closes, c is forced open, and if the stop-cock, s, fitted to 
 the vessel A is properly turned, the gas enters A. On the 
 
CH. vi Hydraulic and Pneumatic Machines 141 
 
 upward stroke of the piston, c closes, a opens, and the 
 process is repeated. 
 
 Let A and B denote the volumes of the chamber A and 
 the barrel, and consider the gas which fills A after n strokes 
 of the piston. Let p be its intensity of pressure, and let p Q 
 be the intensity of pressure of the gas which fills the 
 barrel : if A is being filled with atmospheric air,^ is the 
 atmospheric intensity. 
 
 Then the gas whose volume and intensity of pressure are 
 (A, p) was once represented by 
 
 (A + n 13, p ), 
 supposing that A contained the gas originally. Then 
 
 The Geissler Pump. When very high exhaustions are 
 required, air pumps with valves and pistons are replaced by 
 pumps in which a column of mercury plays the part of a 
 piston. Of the latter kind is that represented in Fig. 64. 
 
 To a certain extent, all air pumps are identical in 
 principle. In each of them a given mass of gas occupying 
 a volume V is made to occupy a larger volume, F+ U, and 
 then the portion occupying U is mechanically expelled. If 
 this process could be repeated indefinitely, an exhaustion of 
 any degree desired could be obtained ; but we have seen 
 that the raising of valves in the common air pumps puts a 
 limit to the process. 
 
 The mercury pumps of Geissler and Sprengel are free 
 from this drawback. 
 
 AB is a glass tube of greater length than the height of 
 the mercury barometer, having part of the Torricellian 
 vacuum enlarged into a chamber, A, of large capacity. 
 
142 
 
 Hydrostatics 
 
 VOL. I 
 
 Above A is inserted a two-way stop-cock, , which 
 in the position represented establishes a communication 
 between the chamber A and a side tube, /, fitted to the 
 
 Fig. 64. 
 
 tube BAd. This tube/ has a stop-cock, c, which in the 
 
 position represented closes / ; but if c is turned through 
 
 a right angle it will establish a communication between 
 
 /and the external atmosphere at v. If a is turned through 
 
CH. vi Hydraulic and Pneumatic Machines 143 
 
 a right angle from its present position, it will close the 
 communication of A with y, and open one between A and 
 a vessel J to which the portion d is joined as represented. 
 The vessel / contains some sulphuric acid the object of 
 which is to remove aqueous vapour from air which may pass 
 over it ; and, by means of a stop-cock, t, J communicates 
 with a very stout indiarubber tube, np, which is connected 
 with the vessel G which is to be exhausted. 
 
 To the vessel / is connected a truncated manometer, 
 that is, a bent glass tube, mr, containing mercury which 
 when the air in / is at atmospheric pressure quite tills the 
 leg m. If the air in / is completely removed, the columns 
 of mercury in the legs m and r will assume exactly the 
 same level. 
 
 To the end B of the tube BA is attached a stout flexible 
 tube T, which is also fixed to a large reservoir, C, of 
 mercury. 
 
 Suppose now that c is turned so as to establish com- 
 munication between f and the atmosphere at v, and that a 
 is in the position represented (i. e., closing communication 
 of A with d) ; and let C be raised until the surface of the 
 mercury in BA reaches the stop-cock a, thus expelling all 
 the air from A through fv. Then turn a and t so as to 
 admit air from G through / and f/, and lower C to its 
 original position. The air of G now occupies the volume 
 G -f A together with the volumes of the communicating 
 tubes. 
 
 Turn t so as to break communication with the rarefied 
 air in G, and then turn a so as to establish communication 
 between A and the atmosphere at v. Again, raise C until 
 the mercury in BA drives out the gas from A into the 
 atmosphere ; and repeat the process of establishing com- 
 munication with G, &c. In this way, by repeated operations, 
 the air in G is exhausted almost completely. By this 
 
J44 
 
 Hydrostatics 
 
 VOL. I 
 
 pump the air left in G can be reduced to an intensity 
 of pressure represented by only ^ of a millimetre of 
 mercury. 
 
 The Sprengel Pump. Fig. 65 represents this pump, 
 
 in which, as in the Geissler, 
 the vessel, G, to be ex- 
 hausted is made part of the 
 Torricellian vacuum of a 
 barometer tube, H. 
 
 A funnel, F, prolonged 
 into a narrow tube fitted 
 with a stop-cock, f, is 
 supported in a vertical 
 position (support not repre- 
 sented in figure) and dips 
 into a wide tube, B, also 
 supported. is connected 
 by indiarubber tubing with 
 a narrow vertical tube, C, 
 above which is a large 
 chamber, A, open at the 
 top, and fitted with a 
 stopper, s, the tube D 
 being, like C, connected 
 with the chamber. D is 
 connected by indiarubber 
 tubing with the vertical 
 tube /, which communi- 
 F} 6 cates freely with the very 
 
 narrow tube H, the top of 
 
 which is connected with the vessel G, and the bottom 
 of which, curved up a little, dips into a vessel V full of 
 mercury. There is an overflow from V to a trough T, as 
 represented ; and there is a clamp, c, by means of which 
 
CH. vi Hydraulic and Pneumatic Machines 145 
 
 communication between D and / can be established or 
 broken. The vessel G is provided with a stop-cock, y. 
 
 The order of operations is as follows. The tubes being- 
 all completely occupied by air, fix the clamp c, remove the 
 stopper *, lower the system of tubes D, C, and pour mercury 
 into F and through its tube into B, until this mercury 
 completely fills the tubes C, D and the chamber A. Close 
 A with the stopper s, and raise the system C, D to its 
 original position ; open the clamp c, and let the mercury 
 run from B, C, D into / up to the top, expelling air from 11 
 through the mercury in V. The mercury coming over at a 
 into H will occupy a certain portion of H. Now let the 
 stop-cock, ff, be turned so that G is connected with the upper 
 part of //. Mercury may be poured into F to keep up the 
 flow from a through H, and the rate of supply of this mercury 
 can be regulated by turning the stop-cock /"more or less. 
 
 Now as each drop of mercury falls down through 77, it 
 forces the air in front of it down through the end of H ; and 
 hence the gas of G which keeps flowing into the upper part 
 of H is perpetually driven down and out by the successive 
 drops of mercury which fall over from #. If the mercury 
 in A has fallen down through D into /, //, and F, the 
 chamber A is a vacuum. 
 
 When the exhaustion of G has not been carried very far, 
 the successive threads of mercury falling down H (and 
 represented in the figure) succeed each other comparatively 
 slowly, and they can be seen forcing the gas which reacts 
 against their fall ; but when the exhaustion is nearly com- 
 plete, these drops fall much more freely through the now 
 exhausted space ; and when there is only a very small 
 quantity of gas left in , the drops falling from a on the 
 top of the mercury surface, H, produce a sharp metallic 
 sound, like that of a water hammer. This sound is an 
 indication of a high degree of exhaustion. 
 
146 Hydrostatics VOL. i 
 
 When the exhaustion is complete the surface of the 
 mercury in H will be at the barometric height above the 
 level in V, and the difference of the level of the mercury in 
 B and C will also be the barometric height. 
 
 The object of allowing the tube from F to dip into a 
 much wider tube, B t is, partly, to let any air that may be 
 carried down with the mercury from F escape into the 
 atmosphere through the mercury in the wide tube, and 
 partly to avoid filling F a very great number of times ; 
 this incessant filling will hot be necessary if the tube B is 
 very much wider than the other tubes. 
 
 The object of turning up the end of the tube H in the 
 mercury in V is to allow the gas (whatever it may be) that 
 is expelled from G through this end to be collected in 
 another vessel, a tube from which is led to the point at the 
 end of H. 
 
 The object of having the tubes C, D and the chamber A 
 is (when A is vacant of mercury) to catch in this chamber 
 any air that may have been carried over by the mercury 
 from F, so that the exhaustion in the tube H shall be 
 performed by mercury devoid of air. Hence the chamber 
 A is called an air trap. So far as the principle of the 
 Sprengel pump is concerned, we might dispense with the 
 tubes C, D and the air trap, A, and connect B directly with 
 / and the Sprengel pump is, in fact, usually so represented. 
 
 38. Manometers. A manometer is an instrument for 
 measuring the intensity of pressure of a condensed or an 
 exhausted gas. The instrument has various forms, and the 
 principle of all will be easily understood from that repre- 
 sented in Fig. 66. 
 
 Let HFED be a vertical bent glass tube, having a portion, 
 ED, of one leg enlarged into a capacious reservoir, and let 
 two necks C, D, project from this reservoir so that vessels 
 may be connected with the reservoir by means of india- 
 
CH. vi Hydraulic and Pneumatic Machines 147 
 
 rubber tubes fitting on C and D, The leg- II F is closed at 
 the top. 
 
 Suppose the cross-section of this leg- to be uniform, as 
 also that of the reservoir except near its ends. Let mercury 
 be poured into the instrument, and when the air thus 
 imprisoned in HF assumes the temperature and pressure of 
 the surrounding atmosphere (which enters at C and D] let 
 AB be the level of the mercury in both legs, and let the 
 number i be marked on the leg HF where the surface of 
 the mercury stands, this point being, of course, in the 
 prolongation of BA. The number indicates that when the 
 air in HF occupies the length ///, at a 
 given temperature, its intensity of pressure 
 is i atmosphere. Let c be the length ///. 
 
 Now suppose that it is desired to fill a 
 globe, or other vessel, with air at a great 
 pressure and to measure the intensity of 
 the pressure. Let the vessel be connected 
 with the neck (?, and let D be connected 
 with a condensing pump. When this pump 
 forces air in through D, and therefore 
 through C into the vessel, the surface of 
 the mercury falls in the reservoir ED and 
 rises in the tube HF. After the pump has 
 been working for any time, let n atmo- 
 spheres be the intensity of pressure of the air in the 
 reservoir (and therefore in the vessel which was to be 
 filled) ; let PQ be the depressed surface in the reservoir, 
 and let the surface of the mercury in HF be at x, which is 
 as inches or millimetres above the level AB. If the 
 intensity of pressure of the air in the reservoir is n atmo- 
 spheres, the number n is to be marked at the level x on the 
 tube HF. Let the depth of PQ, below AB be y inches or 
 millimetres ; a = area of cross-section of HF, A = that of 
 
 L a 
 
 Fig. 66. 
 
148 
 
 Hydrostatics 
 
 VOL. I 
 
 the reservoir ; and let h inches or millimetres represent 
 an intensity of pressure of I atmosphere. Then the dif- 
 ference of level between the point x and the surface PQ 
 being- x +y, and the volume of the air in HF being- now 
 a (c x) with an intensity of pressure denoted by nh xy, 
 we have by Boyle's law 
 
 a(c x] (nk xy] ac/i. 
 But evidently Ay = ax ; hence 
 
 a(c x] j nil (i +-j)# \ ach = o, 
 
 which determines the number, n, to be marked on any 
 part of the tube. 
 
 39. Hydrometers. These are instruments for the deter- 
 mination of the specific gravities of 
 liquids and solids. We shall describe 
 two only. 
 
 The Common Hydrometer, Fig-. 67, is 
 used for finding the specific gravities of 
 liquids. It consists of a glass bulb, or 
 cylinder, A, terminating at one end in 
 a long- narrow graduated stem, and at 
 the other end in a small bulb, , which 
 contains a little mercury, the object of 
 which is to keep the instrument vertical 
 when it is immersed in a liquid. 
 If when immersed in one liquid it floats with a volume v 
 immersed, and in another with a volume v', the specific 
 weights of these liquids being w and w', respectively ; and 
 if W is the weight of the instrument itself, we have 
 
 v . w = W ; v f . w' = W, 
 
 
 f i 
 
 p----^ 
 
 
 
 
 
 
 EEErEI 
 
 A 
 
 ~~~ 
 
 zrr _ 
 
 
 -=~- 
 
 Fig. 67. 
 
 w 
 
CH. vi Hydraulic and Pneumatic Machines 149 
 
 so that the specific weights arc inversely as the volumes 
 immersed. 
 
 The volume of the portion Aft irrespective of the stem 
 can be found by graduating- the stem (supposed of uniform 
 bore) into any number of equal parts, and then observing 
 the weight, W, of the instrument. Let masses ji, p' be 
 successively attached to the top of the stem, and with 
 these let the instrument float in water up to the nth and 
 w'th division, respectively. Then if V is the volume A ft, 
 and a the volume per division of the stem, 
 
 ( F+ no) w = W+p ; (V+ na} = W'+p', 
 
 which determine V and a. 
 
 If when the hydrometer is immersed in two different 
 liquids the readings on the stem are n and n', we have 
 from (i) 
 
 w V + n'a 
 
 which shows that if a is very small, n and n must be very 
 widely different, i. e., the instrument is exceedingly sen- 
 sitive to small differences of specific weight. 
 
 Sikes's Hydrometer is a form of the above in which 
 the stem is a very thin flat strip of metal, for which, 
 of course, a is very small. 
 
 Nicholsons Hydrometer. This hydrometer is employed to 
 measure the specific gravities of both solids and liquids. 
 It consists of a hollow metallic cylinder, A, Fig. 68, having 
 a very fine stem on which there is a fixed mark, P ; the 
 lower end of the cylinder is connected by a wire with a 
 closed cone, D, which is heavy enough to keep the stem 
 vertical ; the base, C, of this cone serves as a platform on 
 which a solid body can be placed ; the stem terminates in a 
 cup, ft, in which solids can be placed. 
 
Hydrostatics 
 
 VOL. I 
 
 To find the specific gravity of a solid, place masses in B 
 until the mark P is just sunk to the surface of the water ; 
 then place the given body in B : this will cause P to sink 
 lower ; remove weight from B until P again reaches the 
 surface ; if the weight removed is W, 
 then W is the weight of the given 
 body. Now remove the body from 
 B to the platform C, and add weight, 
 W, to B until P sinks to the surface ; 
 then W is the weight of a volume 
 of water equal to the volume of the 
 
 W 
 
 body ; and =-, is the required specific 
 
 gravity. 
 
 To find the specific gravity of a 
 liquid, let H be the weight of the 
 hydrometer itself; let the instrument 
 be immersed in the given liquid ; add 
 weight, p, to B until P sinks to the 
 the weight which must be added 
 to sink to the surface when the 
 
 Fig. 68. 
 
 surface ; let p Q 
 to B to cause 
 
 be 
 P 
 
 instrument is immersed in water ; 
 specific gravity of the given liquid is 
 
 then evidently the 
 H+p 
 
CHAPTER VII 
 STEADY MOTION UNDER THE ACTION OF GRAVITY 
 
 40. Steady Motion. "When a fluid is in motion and 
 we confine our attention to any point, P, in the space 
 through which the fluid moves, it will be readily under- 
 stood that the magnitude and direction of the velocity 
 of the molecule which is passing through P at any instant 
 may not be the same as the magnitude and direction of the 
 velocity of the molecule which is passing through P at any 
 other instant. If these should be the same at all instants, 
 and if a like state of affairs prevails at all other points, 
 the motion is said to be steady. 
 
 It is obvious, for example, that if a vessel is filled from 
 a large reservoir of water, so that it is kept constantly full, 
 while the liquid is allowed to flow out from an aperture 
 made anywhere in the vessel, the motion at any fixed point 
 in the vessel will be the same at all times. 
 
 41. Methods of Euler and Lagrange. It is at once 
 obvious that the problem of the motion of a fluid acted 
 upon by given forces may be attacked by two different 
 methods. For, firstly, we may make it our aim to discover 
 the condition of things i.e., the magnitude and direction 
 of the resultant velocity, and the pressure intensity at 
 each point, P, in space at all times, and thus, as it were, to 
 obtain a map of the whole region or a series of maps, if the 
 motion is not steady exhibiting the circumstances at each 
 point as regards velocity and pressure. 
 
152 
 
 Hydrostatics 
 
 VOL. I 
 
 Or, secondly, we may make it our aim to trace the 
 path, and other circumstances, of each individual molecule 
 throughout its whole motion. 
 
 The second object is much more difficult of attainment 
 than the first, and, moreover, is not generally so desirable. 
 
 The first method is sometimes called the statistical, or the 
 method of Euler ; the second the historical, or the method 
 of Lag-range. 
 
 42. Flow through a tube ; work of gravity. Suppose 
 a column of water to occupy at any instant a length AB of 
 
 Bl= 
 
 Fig. 69. 
 
 a straight vertical tube of uniform cross-section, and let the 
 end B of the tube be open. 
 
 If in a small element, At, of time a mass, Am, of water 
 flows out, what is the work done by gravity on the water 
 during this interval ? 
 
 Divide the tube by a series of very close horizontal planes, 
 P, Q,R,... into sections such that the mass included between 
 each adjacent pair is A m. If A $ is the distance between the 
 middle points of successive layers, PQ, QR, &c., while Am 
 flows out the middle point of each layer will fall through 
 the height A s, and the work done by the weight of this 
 layer will be 
 
 Am . As or g Am . As, 
 
CH.VII Steady Motion under Action of Gravity 153 
 
 according as we use gravitation or absolute measure of force. 
 (If mass is measured in pounds and length in feet, the first 
 expression gives the work in foot-pounds' weight, the second 
 in foot-poundals.) If we use the first, the work is 2 A m . A ?, 
 which can be written in either of the forms 
 
 Am (As + A/ + A*-" + ...), 
 
 (Am + Am' + Am" + ...) A*, 
 
 in which, of course, the successive distances A*, A*', A *".. 
 are all equal, and the successive weights AM, Am', Am",... 
 are also all equal. 
 Hence the work is 
 
 A m x AS, 
 
 where J/ is the weight of the whole column. 
 
 The first expression shows that the work done is the same 
 as if the mass Am which flows out at B fell through the 
 height AB of the column. 
 
 Precisely the same result holds if the shape of the tube 
 is that represented in the right-hand figure. Let it be 
 divided by close horizontal planes in the same manner 
 as before. If now Az is the vertical distance between the 
 middle point of the layer, PQ, and the middle point of 
 the next layer, QR, the work done in the descent of the 
 first layer into the position of the second is Am . Az, so 
 that the work done by gravity on the whole tube of liquid 
 while the quantity Am flows out at D is 
 
 Amx. z 
 
 in gravitation units, where z is the difference of level of the 
 ends C and D. t 
 
 This is again the same as the work of carrying Am from 
 Cto 1). 
 
154 
 
 Hydrostatics 
 
 VOL. I 
 
 N H 
 
 43. Stream Lines. The actual path of a particle of 
 a moving fluid is called a stream line. If at any point, A, 
 Fig. 70, we describe a very small closed curve and at each 
 
 point on the contour of this 
 curve we draw the stream line, 
 such as AP, and produce it in- 
 definitely, we obtain a stream 
 tube. When the fluid is a liquid, 
 the mass contained between the 
 normal sections of a tube at any 
 two points, A, P, must always 
 be the same ; and therefore the 
 same mass of fluid crosses every 
 normal section of the tube per 
 
 Fig. 70. 
 
 unit of time. Hence if v is the resultant velocity of the 
 liquid at P and a- the area of the cross-section of the tube, 
 the product V(r 
 
 is constant all along the tube. 
 
 44. Theorem of Daniel Bernoulli. Consider at any 
 instant the liquid contained in the stream tube between the 
 normal sections at A and P, and suppose this liquid to 
 occupy the volume A'P at the end of an infinitesimal 
 element of time ; let V Q , p , <T O be the velocity, pressure 
 intensity, and cross-section of the tube at A ; let v, j), or be 
 the same things at P ; let Z Q and z be the depths of A and P 
 below any fixed horizontal plane; let A^ be the distance 
 between the cross-sections at A and A',, A being that 
 between those at P- and P' ; and let w weight per unit 
 volume of the liquid. 
 
 Now apply the equation of work and kinetic energy to 
 the mass of liquid between A and P in the tube. The gain 
 of kinetic energy in the small motion considered is 
 
 kinetic energy of A'P' kinetic energy of AP, 
 
CH. vii Steady Motion under Action of Gravity 155 
 
 in which, as the motion is steady, the kinetic energy of 
 the portion A'P is common to the two terms, and therefore 
 disappears. Hence the gain of kinetic energy is 
 
 thatofPP'-thatofJ//', 
 
 v*~ v 2 
 
 or Am - > . . (i) 
 
 iff 
 
 where A m = weight of PP* = weight of A A ' . 
 
 The external forces doing work on the column of liquid 
 considered are 
 
 gravity, the pressure at A, the pressure at P. 
 The work of gravity is 
 
 Am.(z-z Q }, ...... (2) 
 
 by Art. 42. 
 
 The pressure at A is j) <T O , and its work = j? <r . As ; the 
 pressure at P is p <r, and its work = pa- . A s. Hence the 
 work of the pressure is 
 
 Am . . . . 
 
 -'(JP-M ...... (3) 
 
 since o- . A* = <r . A# , i. e., the volume PP = the volume 
 
 w 
 
 Equating (i) to the sum of (2) and (3), we have 
 
 + _, = 5 + 5 _, . ... ( 4 ) 
 
 2,ff w ig w 
 
 in other words, since A and P are any two points along 
 the stream line, 
 
 + -* = C ...... (5) 
 
 at every point ,of the stream line, C being a constant for 
 the stream line chosen ; but this constant may have 
 different values as we pass from one stream line to another. 
 
156 Hydrostatics VOL. i 
 
 If, however, the liquid has a plane surface, such as DC 
 (Fig. 70), at each point of which v is practically zero and p 
 is constant, the constant C is the same for all stream lines. 
 The equation (5) holds all along a stream line even if in its 
 course the liquid flows along any number of fixed smooth 
 surfaces ; but if it meets surfaces which it sets in motion 
 (as in a turbine) to the left-hand side must be added a 
 term depending, on the energy which it communicates to 
 them at the point to which v belongs. 
 
 This result is the theorem of D. Bernoulli. 
 
 If at P we draw a vertical line, PQ, of such length that 
 
 p=w.PQ, 
 
 the height PQ is called the pressure head at P. If also QR 
 is drawn vertically of such length that 
 
 v* = 2 ff . QR, 
 
 QR is called the velocity head at P. Let AB be the pressure 
 head and BN the velocity head at A. Then (4) gives 
 PR = 
 
 where AL z Z Q and is the perpendicular from A on the 
 horizontal plane through P. 
 
 Since PL is horizontal, it follows that RN is horizontal. 
 Hence the theorem of Bernoulli may be expressed in these 
 words : if at each point along a stream line there le drawn a 
 vertical line whose length = the pressure head + the velocity 
 head at the point, the extremities of all these vertical lines lie 
 in the same horizontal plane. 
 
 If the liquid has a horizontal surface, CD, at rest at 
 all points of which the intensity of pressure is constant 
 (e. g., that of the atmosphere), the extremities of these lines 
 drawn at all points of the liquid, and not merely along the 
 same stream line, will all lie in the same horizontal plane. 
 If CH is the pressure head on the surface CD (about 34 feet 
 
CH. vii Steady Motion under Action of Gravity 157 
 
 if CD supports the atmospheric pressure), the extremities 
 R, . . . of the vertical lines drawn at all points P, . . . lie in the 
 horizontal plane through //. 
 
 For a liquid in equilibrium, Q coincides with JR, since 
 QR = o, and it has already been shown that the extremities 
 of all vertical lines representing- pressure heads lie in the 
 same horizontal plane. The theorem of Bernoulli is the 
 generalization of this result for a liquid in steady motion. 
 
 An approximate method of indicating the value of p, the 
 pressure intensity at any point P in a moving liquid 
 consists in inserting a vertical glass tube, open at both 
 ends, into the liquid, one extremity of the tube being placed 
 at P. The liquid will rise to a certain height in this tube 
 and remain at rest. Thus, if the tube is so long that the 
 upper end is above the free surface CD, the liquid would 
 rise in it to the height PQ, the remainder of the tube being 
 occupied by air. Such a tube is called a pressure gauge ; 
 but it is evident that it does not strictly measure the pres- 
 sure, since the glass must, to some extent, alter the motion 
 of the liquid. 
 
 45. Flow through a small 
 orifice. Let Fig. 71 repre- 
 sent a vessel containing a 
 liquid whose level is LM 
 which flows out through a 
 small aperture made any- 
 where in the side of the 
 vessel, and let the thickness 
 of the side be so small that 
 the liquid touches the inner 
 edge, AB, of the orifice and Fig. 71. 
 
 thence passes out without 
 
 touching the outer edge or any intervening part of the 
 aperture. 
 
158 Hydrostatics VOL. i 
 
 The curved lines in the figure represent the stream lines, 
 or paths of particles, the forms and positions of which cannot, 
 however, be determined mathematically. 
 
 We are obliged to have recourse to experiment for certain 
 facts concerning the issuing jet. Firstly, it is found that 
 after leaving the orifice AB, this jet contracts to a minimum 
 cross-section, CD, beyond which, of course, the jet widens 
 out again. This minimum cross-section is called the vena 
 contracta. 
 
 The ratio of the area of the vena contracta to that of the 
 orifice AB is called the coefficient of contraction. 
 
 For a circular orifice whose diameter is AB, if CD is 
 the diameter of the vena contracta, it has been found 
 experimentally that 
 
 CD 
 
 IB = '79' 
 
 so that if S is the area of the orifice and <r that of the 
 vena contracta, 
 
 It has also been found that the distance, 10, between the 
 orifice and the vein is somewhere between -39 x AB and 
 5 x AB, where, as before, AB is the diameter of the orifice ; 
 the uncertainty arising from the fact that in the neighbour- 
 hood of the minimum section the diameter of the jet varies 
 very little. All the streams which pass through the vena 
 contracta cut its plane perpendicularly. By consideration 
 of the general equations of motion, it will follow from this 
 fact that the intensity of pressure is the same at all points 
 in the vena contracta. At all points on the outer surface, 
 ACE, BDF, of the jet the pressure intensity is, of course, 
 the same as that of the atmosphere, if the jet flows into the 
 atmosphere ; also the velocities at all points of the vein are 
 
CH. viz Steady Motion under Action of Gravity 159 
 
 equal ; but in the interior of the jet this pressure intensity 
 does not exist, except at points in the plane of the vena 
 contracta. 
 
 Of course at the orifice AB the directions of motion are 
 not all perpendicular to the cross-section of the jet, neither 
 are the velocities all the same at points in this section. 
 
 46. Theorem of Torricelli. In the case of a jet escaping 
 into the air, the velocities of particles in the vena contracta 
 are expressed by a very simple formula. 
 
 In (4) of Art. 44, let p and z refer to a point, 0, in the 
 vena contracta while p ot z refer to the point, TV, of the 
 stream line through which is on the free surface of 
 the liquid. Then p =j , as we have said above, and as 
 the velocities at the surface LNM are all very small, we 
 may consider v = o. 
 
 Hence 2 
 
 = W&, ........ (a) 
 
 where A, or z z , is the vertical depth of the vena contracta 
 below the free surface LNM. 
 
 Hence when the particles reach the vena contracta, they 
 have the same velocity as if they fell directly from the free 
 surface. This is known as Torricelli's Theorem. Obviously 
 it holds with considerable exactness in the case of a small 
 orifice only. 
 
 If a liquid devoid of friction escapes from a small orifice 
 in a vessel in which the free surface is maintained at a con- 
 stant level, the velocity in the vena contracta is theoretically 
 given by the equation (a). 
 
 In the case of water, however, it is found that the velocity 
 is not quite equal to this amount, but is very nearly a con- 
 stant fraction, //, of the value given by (a). The fraction 
 p. is nearly equal to unity (about -97). We may therefore 
 
 P ut 
 
i6o 
 
 Hydrostatics 
 
 VOL. I 
 
 Again, if S is the area of the aperture, and c denotes the 
 coefficient of contraction (Art. 45), the area of the cross- 
 section of the vena contracta is cS ; so that the volume of 
 water issuing- from the vessel per unit of time is 
 
 cfj.S V iff/i. 
 
 If the unit of length is a foot and the unit of time 
 a second, this is the discharge in cubic feet per second, and 
 multiplying it by w, the mass per unit volume (in the case 
 of water 62 1 lb.), we obtain the mass discharged per second. 
 The product c^ may be taken as -62. 
 To find the time in which a vessel of any form, ACS, 
 filled originally to a level AS, 
 Fig. 72, with water will be emptied 
 through a small orifice at a point V, 
 let PQ be the level at any time /, 
 let z be the vertical distance between 
 PQ, and AB, let x be the depth of 
 V below PQ, S = area of orifice, 
 and A area of the section PQ of 
 the vessel. 
 
 Then in the time dt the volume 
 of the liquid discharged is that 
 between PQ and FQ', i.e., Mz\ 
 but in this time the volume discharged at V is 
 
 62, S </ zgx . dt, 
 
 if we can regard the velocity at each point of PQ as 
 practically zero. Also dz = dx, so that 
 
 62 S Vigx . dt = Adx. 
 
 Now A is known in terms of # from the shape of the vessel ; 
 hence if the depth of V below AS is A, 
 
 62 V 2-ff 
 
 [ h Adx 
 
 _____ , i 
 
 2,0 Jo ->/x 
 
CH. vii Steady Motion under Action of Gravity 161 
 
 EXAMPLES. 
 
 1. A cylindrical barrel, the area of whose cross-section is 12 
 square feet, and whose axis is vertical, is filled to a height of 
 4 feet with water ; in what time will it be emptied through a 
 hole whose area is half a square inch placed in the bottom of 
 the barrel t (Take g = 32.) 
 
 Result. 46 27". 
 
 2. In what time will a cylinder of radius a feet be emptied 
 through a hole of radius r inches in the bottom, if the cylinder 
 was filled to a height h feet 1 
 
 7 1800 a 2 I 
 Result. - h 2 seconds. 
 3i ^ 
 
 3. Show that if the level of the water had been kept for this 
 time at its original height, twice as much water would have 
 been discharged. 
 
 4. Find the time in which a vertical hollow cone of volume 
 V filled with water to a height h will be emptied through a 
 small orifice of area or at its vertex. 
 
 6 V 
 Result. - - -- seconds. 
 
 3*x X<r . 
 
 5. If the level had been kept constant for this time, how 
 much water would have been discharged ? 
 
 Result, f F. 
 
 6. Find the time taken by a hollow sphere of radius r filled 
 with water to empty itself through a small orifice at the lowest 
 point. 
 
 Result. If V = volume of sphere, <r = area of orifice, 
 
 4F 
 
 3-1 xcrzgrr 
 
 47. Plow through a large orifice. The determination 
 of the discharge through a large orifice cannot be satis- 
 factorily accomplished by theory. 
 
 1424 M 
 
162 Hydrostatics VOL. i 
 
 Suppose, for example, that the oriHce is a rectangle, 
 ABCD, with vertical and horizontal sides, and that LM 
 (Fig. 73) represents the level of the free surface in the 
 vessel, the flow being supposed to take place through the 
 orifice towards us as we look at the 
 t - % - figure. 
 
 Divide the area of the aperture 
 into an indefinitely great number 
 of narrow horizontal strips, of which 
 that between the horizontal lines 
 m and n is the type. 
 Fig. 73- Let the depths below LM of 
 
 the lines AD and BC be h^ and 
 
 ^ 2 , respectively, those of the lines m and n being z and 
 z + dz. Let AD = d ; then, supposing that the aperture 
 between m and n alone existed, the volume of the discharge 
 would be given by (3) of last Article, in which S = bdz. 
 Denoting the product cp. by k, and by dQ the mass dis- 
 charged per unit time through the strip, we have 
 
 dQ = klw */ igz .dz ...... (i) 
 
 Now the assumption that k is constant for all the strips 
 enables us to find Q, the total discharge ; but clearly this 
 assumption cannot be strictly correct, for each strip does 
 not discharge as if it alone existed as an aperture. 
 
 Assuming k to be constant, we integrate (i) from 
 z = h^ to z ^ 2 , and obtain 
 
 q = kbw Jw(k$-k$ ..... (2) 
 
 To calculate the energy per second which flows through 
 the orifice, if v is the velocity of the portion clQ, its 
 kinetic energy is 
 
 v 2 
 
 . dQ, i.e. zdQ. 
 
CH. vii Steady Motion under Action of Gravity 163 
 
 Energy per unit time is called Power. Hence if dP 
 is the power of this flow, 
 
 dP = kbw 
 
 -Afi. . . . (3) 
 
 If in this expression length is measured in feet, time in 
 seconds, and w in pounds, since the unit of power called 
 a Horse-Power is 550 foot-pounds' weight per second, we 
 get the Horse-Power of the discharge equal to the right- 
 hand side of (3) divided by 550. 
 
 If AB is small compared with the depth /t l} and if 
 h is the depth of the centre of area of the orifice, we can 
 easily find from (2) that 
 
 Q = kwS </~2gb, (a) 
 
 where S = the area of the orifice. For if AB = 2 a, we 
 have / 2 = h ( i + j) > /^ = h(ij}, and expanding in 
 
 powers of j , we see that the term ^ disappears, and (a) is 
 
 a z 
 true if we neglect the small fraction r? 
 
 It/ 
 
 As another example of the same kind, suppose the orifice 
 
 T 
 
 to be circular (Fig. 73), and that powers of j beyond the 
 
 second are negligible. Then it will be found that 
 
 r 2 
 q = Tiki* w Vzgh ( x - r^p) > 
 
 where h is the depth of the centre O. 
 
 EXAMPLES. 
 
 1. The Syphon. We now take a few common practical illus- 
 trations of the application of equation (4), Art. 44, which 
 applies to the motion of a liquid acted upon by gravity. The 
 
 si 2 
 
164 Hydrostatics VOL. i 
 
 first of the simple examples is furnished by the common syphon 
 which is employed for the purpose of raising a liquid out of 
 a vessel and lowering it into another vessel. The operation 
 might, of course, in many cases be directly performed by taking 
 the first vessel in the hand and pouring out the liquid bodily 
 into the second ; but if, as often happens, the liquid is 
 Sulphuric or Nitric Acid, which it would be most undesirable 
 to pour out with splashing, this method would not answer, and 
 a syphon is used. The syphon is a bent tube (usually of glass) 
 open at both ends, and with unequal branches. 
 
 Suppose M (Fig. 74) to be the vessel which it is desired to 
 empty into another (not represented in the figure), and suppose 
 the liquid to be water. 
 
 A bent tube, DABC, (the syphon) whose branch BC is longer 
 than the branch BD is first filled with water, and the apertures 
 at D and C held closely by the fingers. 
 The end D is then inserted into the liquid 
 in the vessel M, the fingers removed from 
 D and C, and the tube held in the hand. 
 The result will be a flow of the liquid 
 through C until, if D is kept close to the 
 bottom of the vessel, nearly all the liquid 
 is removed. 
 
 Let p Q be the atmospheric intensity of 
 'g- 74* pressure, which exists on the surface of 
 
 the liquid at A and also at C\ v , the 
 
 velocity of liquid on the surface A ,is nearly zero; if v = velocity of 
 efflux at C, z = depth of C below A, equation (4) of Art. 44 
 gives, since A DBG may be taken as a simple stream line. 
 
 2<7 W W 
 
 .. v = A/ 2gz, 
 
 so that the flow will continue all through if C is at a lower 
 level than D, Of course there will be a small residue of liquid 
 in M, because when nearly all has flowed out, air will enter the 
 syphon at D. 
 
 If the liquid to be removed is an acid, as sulphuric or nitric, 
 the syphon must be filled with it at the beginning by first 
 inserting the end D into the vessel and then sucking the air 
 out through C until the liquid rises in the syphon and falls in 
 
CH. vii Steady Motion under Action of Gravity 165 
 
 the leg BC to a lower level than A ; and this suction may 
 be effected by joining another tube to the end C by means of a 
 short piece of indiarubber tubing which can be subsequently 
 removed. 
 
 2. Hero's Fountain. Hero of Alexandria (120 B.C.) con- 
 structed a fountain, which is represented in Fig. 75. It consists 
 of two glass globes, M and N, and a dish, DD. Each globe is 
 partly filled with water and partly with air at the atmospheric 
 pressure. The globes are fitted with 
 necks and are held together by two glass 
 tubes, A, B, each open at both ends, ////''"*$''"" 
 which pass through necks fitted to the 
 globes. The extremities of A are in the 
 air in the globes ; the lower extremity 
 of B dips nearly to the bottom of the 
 liquid in M, while its upper end barely 
 projects into the dish DD. A third tube, 
 C, open at both ends, passes through the 
 neck of N, its lower end dipping nearly 
 to the bottom of the liquid in N, while 
 its upper end projects beyond the upper 
 surface of the dish. 
 
 In this state of affairs the water is at 
 rest in both vessels, the intensity of 
 pressure on both water surfaces being 
 2> Q , that of the atmosphere. 
 
 If now water is poured into the dish, 
 it will fall through B into M and drive 
 some of the air into N where the surface 
 pressure on the water becomes greater 
 than p , and as a result the water from 
 N is forced up through the tube C into 
 the air. 
 
 To calculate the height to which it rises, let z be the 
 difference of level between the water in M and that in N ; let 
 c = difference of level between that in D and that in iV; and 
 let h = the height of the top of the jet above the water in D. 
 
 Then, since the pressure intensity of the air in N = that in 
 M = w (z + c)+p ; since the velocity of the water at the 
 surface in N is nearly zero, and is also zero at H, where the 
 pressure intensity isp Q , we have from (4) of Art. 44 
 
 Fig. 75- 
 
1 66 
 
 Hydrostatics 
 
 VOL. I 
 
 M 
 
 to 
 
 .. h = z, 
 
 i.e., the height of the jet above the water in D is equal to the 
 difference of level in M and N. 
 
 3. Mariettas Bottle. It is sometimes desired to produce 
 a narrow jet of water flowing for a considerable time with 
 constant velocity. Of course a very large reservoir with a very 
 small aperture made in the side would produce the result ; but 
 such a reservoir is not always at hand. The result can also be 
 produced by means of a broad flask fitted with a stop-cock near 
 the bottom. Fig. 76 represents the flask. 
 The stop-cock (not represented) is fitted 
 at C, and the aperture is supposed to 
 be very small compared with the cross- 
 section of the flask. The flask is first 
 quite filled with water, the stop-cock 
 being closed. In the top of the flask 
 there is a neck fitted with a cork, and 
 into this is inserted a tube, II D, open at 
 both ends, the tube also being quite filled 
 with water. 
 
 Now let the stop-cock be opened, and 
 Fig. 76. water will flow out, because the atmo- 
 
 sphere presses at H and at the outside of 
 
 (7, and between C and H there is a column of water. The water 
 that first flows out comes from the tube HD alone, the flask re- 
 maining filled to its upper surface ; and, moreover, the velocity 
 of efflux will be variable as the level sinks in HD. But when 
 the tube is emptied of water, some air will be forced through D 
 by superior atmospheric pressure, and it will rise to the upper 
 part of the flask, and will begin to force down the water of 
 the flask. 
 
 This being the case, the intensity of pressure at D in the 
 water is^? , the atmospheric intensity, and we may assume that 
 p t is also the intensity all over the horizontal plane, LM, 
 through D, because the motion of particles in this plane is 
 very slow. The air at the top aided by the water above LM 
 will keep the pressure intensity approximately equal to p {) at 
 points in LM other than D. 
 
CH. vii Steady Motion under Action of Gravity 167 
 
 Now let z = CM = vertical distance of orifice below D, the 
 lower extremity of the tube ; let V Q , p Q in (4) of Art. 86 refer 
 to D, while v is the velocity at C. Then, since p is also the 
 pressure intensity at C, 
 
 zg 
 
 which shows that the velocity is constant whatever the position 
 of the upper surface, AB, of the water in the flask. 
 
 The tube must, of course, have such a position that D is 
 above the aperture. If the water, instead of escaping into the 
 atmosphere, escapes into a medium (gaseous or liquid) in which 
 the pressure intensity at G is p, we shall have 
 
 *-*(+*?) 
 
 This vessel is known as Mariotte's Bottle. 
 
 4. Thomsons Jet Pump. If water flows with steady motion 
 through a horizontal pipe of variable cross-section (Fig. 77), the 
 
 Fig. 77. 
 
 pressure intensity at a narrow part of the pipe is less than it is 
 at a broad part. Let A be a part at which the section is broad 
 and B a part at which it is narrow ; let v and p be velocity 
 and pressure intensity at A, and i/, p' those at B ; let S and tf 
 be the cross-sections at A and B ; in a small interval of time, 
 At, let the water at A come to A' and that at B come to B'. 
 Employing the equation of work and energy to the mass of 
 water in the pipe between A and B in occupying the space 
 between A' and B', the work done on it is pSAx p'&Ax', 
 where A a? is the distance between the sections at A and A', and 
 Ax' the distance between those at B and B' . Also, as at p. 155, 
 
 v' 2 v 2 
 
 the gain of kinetic energy is wtfAaf, wS Ax . 
 
1 68 Hydrostatics VOL. i 
 
 Hence 
 
 ^2 2 
 
 pS&x-p'ff&af = w&'Ax' . - -- wSAx . . 
 
 2 <7 iff 
 
 Now, since the liquid is incompressible, S' &x r = SAx; hence 
 
 But Aa; = v Af, and Arc' = v'A<, .. ? = V, .*. r'>v, 
 and (a) shows that p <p- 
 
 On this principle is founded Professor James Thomson's jet 
 pump which can be employed for draining marshy land. 
 ABC (Fig-. 78) is a horizontal pipe with a narrow 
 
 cross-section at 7?, into 
 which is fitted a ver- 
 tical pipe, D, which 
 is plunged into the 
 marshy soil. This pipe 
 D ends inside the pipe 
 ABC in a nozzle situ- 
 ? 8 - ated near B. A steady 
 
 flow of water from a 
 
 reservoir enters at A under atmospheric, or nearly atmo- 
 spheric, pressure. At B the pressure is much less ; and 
 if D dips into water at atmospheric pressure, this water 
 will rise and, passing- through the nozzle (if the pipe D is 
 not very long), will be carried through B and discharged 
 through the mouth C wherever desired. 
 
 The difference between the water pressures at A and B 
 (Fig. 78) can be exhibited by inserting vertical glass tubes 
 into the pipe at A and B. The water will then ascend 
 from the pipe into these tubes and will stand at a higher 
 level in the tube at A than in the tube at B. If the water 
 in the tube inserted at A reaches to a point 77 in the tube 
 
CH. viz Steady Motion under Action of Gravity 169 
 and the water in the tube at B reaches to H' t and if we 
 
 V L 
 
 prolong AH to K so that HK = , v being the velocity 
 
 iff v '2 
 
 at A, and similarly prolong BH' to A' so that H'K' = > 
 
 v' being the velocity at B, the points K and A'' will stand 
 at the same level, which will be that of the water in the 
 reservoir from which the pipe AB is supplied. This follows 
 from equation (a). 
 
 48. Theory of Turbines. In connexion with the 
 motion of water along surfaces which are themselves in 
 motion, the following result in elementary dynamics is 
 useful:, if a particle moves under the influence of force from 
 one position to another, the change in its moment of momentum 
 about any fixed axis is equal to the time-integral of the moment 
 of the force about the axis from the one position to the other. 
 
 Let a particle of mass m (pounds, 
 suppose) at the point P, Fig. 79, have 0' 
 
 at the time t a velocity v represented 
 by PQ ; at the end of an extremely 
 small interval dt let its velocity be 
 represented by the vector PR ; then the 
 vector QR, or PS which is equal and 
 parallel to QR, is a . dt, where a denotes 
 the resultant acceleration of the particle 
 at P. Then by Newton's Second Axiom PS is the direction 
 of the resultant force acting on the particle, and this 
 
 force, in gravitation units (pounds' weight) is - 
 
 / 
 
 Now by the principle of moments, since the vector PR is 
 the resultant of PQ and PS, the moment of PR about any 
 axis, represented by 0, is equal to the sum of the moments 
 of PQ and PS about that axis; 
 
 .. moment of PR moment of PQ = moment of PS. 
 
170 Hydrostatics VOL. I 
 
 Now m x PR and w x PQ are the momenta of m at the 
 beginning and end of the interval dt ; therefore the change 
 of moment of momentum = moment of madt = moment 
 of ff. Pdt, if P is the resultant force acting on the particle. 
 
 Hence if the initial and final moments of momentum of 
 the particle are M 2 and M lt and L denotes at any instant 
 the moment of the resultant force about the axis, we have 
 
 the integral being taken throughout the whole time interval. 
 Let AB, Fig. 80, be a smooth plane curve rotating 
 with angular velocity <o about an axis 
 at perpendicular to its plane, and 
 let a particle P be moving along the 
 rotating curve. Then if N is the re- 
 action of the curve at P, acting along PN, 
 and if at the end of the time dt the curve 
 comes to A'B' and the particle to P', the 
 work which the particle does on the 
 curve may be calculated by supposing 
 the particle not to move along the curve, 
 but to be carried with it to Q, such that 
 PQ = (ardt, and to move along the curve, 
 considered fixed, from Q to P '. In this latter displacement 
 N does no work, and in the former N does on the curve 
 the work N. u>rdt .smty, where $ = OPN, i.e., u>.Nvrdt, 
 where -or is the perpendicular from or N. The work 
 which JV does on the particle is, of course, the same with 
 a negative sign. 
 
 Let AB (Fig. 81) be a vane forming a part of a wheel 
 which is rotating about a fixed axis 0, so that u is the 
 velocity of the end A, this velocity being perpendicular 
 to OA \ and suppose a particle of water moving with 
 absolute velocity V on reaching A to enter along the vane 
 
CH. vii Steady Motion under Action of Gravity 171 
 
 without shock on the vane. Then if AT is the direction of 
 
 the tangent at A to the vane, this must be the direction of 
 
 the relative velocity of the water and the vane at A ; let 
 
 AC represent u and AD 
 
 represent V \ then applying 
 
 u reversed to the particle of 
 
 water and combining V with 
 
 u reversed, we obtain the 
 
 relative velocity. Draw AC' 
 
 equal and opposite to AC; 
 
 then the resultant of AD and 
 
 AC must be along AT. If a 
 
 is the angle TAG and the 
 
 angle DAC, 
 
 V sin (a 6) 21 sin a. 
 
 Fie. 81. 
 
 Since there is no relative velocity normal to the vane, 
 there will be no sudden impulse or blow received by the 
 water on the vane, and therefore no sudden loss of kinetic 
 energy. 
 
 The velocity V of the water can be resolved into a 
 tangential component, F t , along AC and a radial com- 
 ponent, T ri along AO. 
 
 Now suppose a channel A to be carried by a wheel D 
 (Fig. 82) which rotates round with angular velocity <a ; 
 suppose a continuous supply of water from the outside 
 to be given to the wheel and the water to be discharged 
 from the channel at its other end into a space C surrounding 
 the axis and to flow away continuously out of the machine 
 in a direction perpendicular to the plane of the figure. 
 Further, suppose the angular velocity of the wheel to be 
 always the same by making the wheel do some work, 
 against external resistance, at a constant rate. Then the 
 state of affairs inside the channel will be the same at all 
 
172 
 
 Hydrostatics 
 
 VOL. i 
 
 times, so that, for example, the moment of momentum about 
 
 of the water which fills the channel will always be 
 the same. 
 
 Imagine the channel to be divided into a very great 
 number of sections i, 2, 3, 4, .... all of the same volume. 
 Let this channel revolve into the position A' in the time cU. 
 We have represented a particle m of water just outside A 
 which is ready to enter the channel, and which actually 
 enters it when the channel reaches the position A'. This 
 particle has the velocity requisite for avoiding shock, i. e., 
 
 it has tangential 
 velocity 7 1 and ra- 
 dial F r , as above. 
 Now let the con- 
 struction of the 
 vanes be such that 
 the water ejected 
 into the space OC 
 has its absolute ve- 
 locity at exit di- 
 rected towards ; 
 and consider the 
 change of moment 
 of momentum, about 0, of the water in A. The section 
 
 1 moves into the compartment i' in A'; the section 2 
 moves to 2', and so on; the last section (5 in the figure) 
 is ejected radially and has no moment of momentum about 0. 
 
 Now let M be the moment of momentum of the water 
 i, 2, 3, 4, . ... ; also M is the moment of momentum of 
 m', i', 2', 3', 4', .... ; and the new moment of momentum of 
 the water i, 2, 3, 4, is that of i', 2', 3', 4', . ... which is 
 M minus the moment of momentum of m'. If dw is the 
 weight of m, the moment of momentum of m is dw.aV ty 
 where a is the outer radius of the wheel, i. e., the distance 
 
 Fig. 82. 
 
CH. vii Steady Motion umler Action of Gravity 173 
 
 of m' from 0. Hence the change produced in time dt in the 
 moment of momentum of the water in A is 
 
 M dw . a 7\ M, or dw .aV t \ 
 
 and this is, as above proved, equal to dt multiplied by the 
 moment about of the normal reactions of the vane on the 
 elements of water I, 2, 3, 4, .... 
 
 If N is the normal reaction per unit length of the vane at 
 any point P of the channel A, the total moment of force is 
 
 the integral being taken along the whole length of the 
 channel A. Hence 
 
 dw . a V t 
 
 
 dw 
 
 = dtf N-nds. 
 
 If we denote -j- by W, the weight of water entering the 
 (It 
 
 channel per unit time, 
 
 WaV t 
 
 - l - 
 
 But we have proved above that the work, dE, done in 
 time dt on the vane is wdtfN'erd*; hence 
 
 WaV t _ i dE 
 g M dt 
 
 and since o> a is u, the circumferential velocity of the wheel 
 at its outer rim, we have 
 
 t _ dE 
 
 which gives the energy communicated per unit time by 
 the water to the wheel. 
 
 If the wheel has a large number of channels such as A, 
 then W is the total weight of water entering the chambers 
 
174 
 
 Hydrostatics 
 
 VOL. I 
 
 per unit of time. If, as is often the case, the direction of 
 the tangent AT (Fig. 81) to the vane is radial, V t = u, and 
 
 Wu 2 
 the energy per second given to the wheel is - . 
 
 / 
 
 The case just described is that of the Reaction Turbine, 
 so called because it is worked by the continuous pressure of 
 
 Fig. 83- 
 
 water flowing along curved blades, forming curved spokes 
 of a revolving wheel. When the water enters the wheel at 
 its outer rim and is ejected through the hollow axis of the 
 wheel the machine is called an imvardflow turbine. 
 
 Fig. 83 represents the main features of the inward flow 
 reaction turbine invented by Professor James Thomson. 
 
CH. VII Steady Motion under Action, of Gravity 175 
 
 Water eaters a large pipe at P, flows round a chamber 
 QjRS, which surrounds another fixed chamber ABC. In 
 this latter are fixed certain curved blades, f, f,f,f, called 
 guide blades, the chamber ABC in which they are fixed 
 being- called the guide chamber. These fixed blades guide 
 the water to apertures on the outer rim of a wheel which 
 can rotate round a fixed central axis, and the spokes of this 
 wheel are a large number of curved blades or vanes such as 
 that already described in Fig. 81. The water from the 
 guide chamber ABC enters the wheel, of course, at apertures 
 all over its rim, although the ends of the guide blades reach 
 the rim of the wheel at only four of the apertures in the 
 rim. We can assume that the whole of the water flowing 
 into the rim at the left-hand side of any guide blade f 
 strikes the spoke vanes of the wheel at an angle not greatly 
 different from that at which the water at the end of the 
 blade f strikes the wheel there. Although the guide blades 
 /"have been said to be fixed, they are really capable of some 
 adjustment, each blade /"having a pivot fixed in it near the 
 end which is close to the wheel, and the rotation of these 
 blades is easily effected from the outside of the machine. 
 
 Now in the equation p. 171, since a is constructed once for 
 all and cannot be varied, we see that if u is to be constant 
 that is, if the wheel is always to be run at constant speed 
 while from any cause 7 varies, must be capable of being 
 altered ; and this is effected by slightly rotating the guide 
 blades/ round their pivots. 
 
 The chamber QJtS, from which the water enters the 
 guide chamber, is called the vortex chamber. The space 
 between these two chambers narrows very much up to a 
 point A close to the supply pipe P ; and here the water 
 enters the guide chamber through an aperture at A, so 
 that the velocity with which it enters this chamber is 
 greatly increased. 
 
176 Hydrostatics VOL. i 
 
 Fig. 83 represents a section, perpendicular to the axis of 
 the revolving- wheel, of the wheel itself and its surrounding- 
 chambers. The whole is, of course, boxed up in a casing 
 through which the axis of the wheel projects, so that none 
 of the interior of the machine (shown in the figure) is 
 visible from the outside. 
 
 To the revolving axis of the wheel is fixed on the outside 
 a pulley, round which is passed a belt which can be used to 
 drive any revolving machine, such as a dynamo, which is 
 to be set in motion by means of the turbine. 
 
 The efficiency of the turbine is measured by the ratio 
 of the useful work which it does in any time to the energy 
 of the water which enters it in that time. Assuming that 
 there is no loss by friction, the energy imparted to the 
 
 u . V 
 wheel by a weight W of water is W - - (p. 173), where 
 
 / 
 
 V t and u are the absolute circumferential velocities of 
 the water at entrance to the wheel and of the wheel itself. 
 If the total head of the water before entering the machine 
 is h, the possible available energy per second is W .h, and 
 
 uV 
 the efficiency is ,'. 
 
 If W pounds of water are supplied per minute to the 
 turbine, V t and u being in feet per second, the horse-power 
 developed by the machine is 
 
 WuV t ^ 
 33000,7 ' 
 
 where g = 32, approximately. 
 
 We see the reason for arranging the blades so that the 
 absolute velocity of the water at exit from the wheel is 
 radial ; for if the terminal section 5' (Fig. 82) had moment 
 of momentum about 0, the change of moment of momentum 
 of the water in the channel A would not be quite so great 
 
CH. vii Steady Motion under Action of Gravity 177 
 
 as dw.aV t , and the energy given to the wheel would be 
 diminished. 
 
 If the height of the supply water above the turbine is ^, 
 and j is the pressure (atmospheric) on its surface, we have 
 by Bernoulli's theorem 
 
 , 
 
 W 2ff W 
 
 where V and /; are the velocity and pressure at entrance to 
 the wheel, and if E is the work which the water does per 
 unit weight on the wheel and 7 r ' is the velocity of the 
 water at exit, each side of the above equation is equal to 
 
 ^v 
 
 If we neglect F/, we get y 
 
 an equation which is only approximately true. 
 
 If the blades are radial at the outer circumference of the 
 wheel, u = V t , and then we have 
 
 7? = ^, 
 
 which shows that the circumferential velocity must be that 
 due to half the total head of water. 
 
 There is a simple form of turbine, sometimes used when 
 only a small power is required, in which the guide chamber 
 is placed above the wheel chamber, and the water enters 
 through the common axis of these chambers. This is called 
 an axial flow turbine. 
 
 In all forms of turbines there are two conditions to be 
 satisfied for high efficiency, namely 
 
 i. The water must enter the wheel without normal 
 velocity on the blades, because normal velocity entails 
 shock, and this entails sudden loss of energy ; 
 
178 Hydrostatics VOL. i 
 
 2. The water must leave the wheel with an absolute 
 velocity which is wholly radial. 
 
 Turbines are usually divided into two classes reaction 
 turbines and impulse turbines. These names are not accu- 
 rately expressive. By reaction turbines are meant those in 
 which the water when it reaches the wheel blades is not 
 only in motion but also under some pressure exceeding the 
 pressure of the atmosphere. If the water was originally 
 derived from a source having a height H above the level at 
 which it enters the machine, its total head at entrance 
 
 Fig. 84. 
 
 would be A + If, where ^ is the head due to the atmosphere. 
 When it leaves the wheel chamber its head is fi Q) and when it 
 is driving the wheel it has a head intermediate to li Q + //and 
 h>Q ; in addition to this it has its kinetic energy when driving 
 the wheel. If, however, the turbine is such that when the 
 water begins to drive the wheel its head is simply JI Q , i.e., 
 if the excess head, H, is wholly turned into kinetic energy 
 before striking the blades of the wheel, the machine is 
 called an impulse turbine. An example of this latter kind 
 is the Pelton Wheel, represented roughly in Fig. 84. 
 
CH. viz Steady Motion under Action of Gravity 179 
 
 To the rim of the wheel are attached curved buckets, 
 .#, C, J), ... all round, only four of which are shown. 
 
 The water from a reservoir or from a stream in which a 
 dam is fixed issues through a sluice or through a pipe P 
 fitted with a nozzle, and impinges with considerable 
 velocity on the bucket , running up into the interior of 
 the bucket. As the wheel revolves, the water falls out of 
 the bucket, and the bucket has a lip at its lower end over 
 which all the water is emptied shortly after the bucket has 
 left its lowest position, as at D. The buckets are not 
 spherical cups; their interior surfaces are more nearly 
 cylindrical, and each cup is divided by a partition into two 
 compartments separated by a diaphragm represented in 
 Fig. 85 by fg, which is in the direction of the radius of 
 the wheel. 
 
 Now, reverting to the case of a particle projected along a 
 smooth curve (Fig. 80) which 
 is moving in its own plane, let 
 v x and Vy be the components of 
 the absolute velocity of the 
 particle at P, while a and /3 
 are the components of the abso- 
 lute velocity of the point I of 
 the curve itself, parallel to two . 
 
 fixed axes of x and y. Then if 
 
 6 is the angle made with the axis of x by the tangent at P, 
 and w is the weight of the moving particle, the equations 
 of motion of the particle are 
 
 ~^T 
 
 dt , r 
 - = NCOS 0, 
 
 N 2 
 
180 Hydrostatics VOL. i 
 
 Now the components of the relative velocity of the 
 particle and tube at P are v x a and v v fS; and since the 
 resultant relative velocity is along the tube 
 
 _ A 
 
 = tan0. 
 
 v-a 
 
 x 
 Hence the above equation becomes 
 
 K ~ ) f jj fer) + K ~ /3) ^ (* ) = - 
 
 So that if a and /3 do not vary with the time, we have 
 
 (tk-qja + ^-p)* = 2 _ Constant, 
 
 or, in other words, the relative velocity is constant in 
 magnitude. 
 
 Now in the case of the Pelton wheel the water entering 
 the bucket strikes the partition^, separating the two halves 
 of the bucket, and is directed sideways, as represented in 
 Fig. 86, which is a section of the bucket perpendicular 
 
 Each such section of the bucket can be considered as 
 
 a smooth blade 
 v-" . -,, 
 
 moving with a 
 
 velocity u in the 
 direction of the 
 water jet, and if 
 
 V is the velocity 
 
 \\ 
 
 ^ of the jet, V u 
 
 jj is the relative 
 
 *' velocity of the 
 
 water and bucket 
 both at entrance 
 
 and exit. The absolute velocity of the water at exit in 
 the sense of the jet isu (V u), i.e., 2,n7; so that if 
 u = \7, the water will at exit drop out of the bucket 
 
CH. vii Steady Motion under Action of Gravity 181 
 
 with no velocity, and the whole of its original kinetic 
 
 F 2 
 energy, W , will be abstracted from it by the wheel. 
 
 If, then, the wheel is made (by doing work in lifting, or 
 in driving machinery) to run with a circumferential velocity 
 equal to half that of the impinging water, the efficiency 
 would be perfect, on the supposition that no friction is 
 encountered in the machine. 
 
 The efficiency of the wheel for any value of u is thus 
 found : if W is the weight of water supplied to the wheel 
 in any time, the loss of energy of the water is 
 
 WV 2 - W^u-V}* t 4Wu(V-u] ^ 
 
 2ff ' 2ff iff 
 
 and if this is wholly given to the wheel, the ratio of the 
 kinetic energy of the wheel to that of the water is the 
 ratio of this to 
 
 This is unity when u = \ V. 
 
 The water exerts a mean tangential pressure P on the 
 buckets which can be thus found : let s be the area of the 
 cross-section of the jet ; then, measuring backwards from 
 the wheel along the jet a length of column equal to (Fu) r, 
 the volume (Vu) T$ of water will strike the wheel in time r; 
 the absolute velocity of this water is changed from V to 
 iuV, .'. its change of momentum is, if w is the weight 
 per unit volume, 2togr( V- u) z , 
 
 and by the equation of impulse and momentum, 
 
 ) 2 _. 
 = JTT. 
 
 9 
 
 _ 
 
 .1 := 
 
 9 
 
 The Centrifugal Pump. Imagine the hollow central axis 
 
1 82 Hydrostatics VOL. i 
 
 of the turbine wheel shown in Fig. 83 to be connected with 
 a vertical pipe dipping into a well at a depth of about 20 
 feet, or less, below the centre of the wheel, and that the 
 pipe P is connected with a vertical pipe bent at its upper 
 end over a tank. Suppose also that the whole of the pump 
 chambers and the pipe dipping into the well are completely 
 full of water. If the atmosphere is excluded from the pump 
 chambers, the pressure at the centre of the wheel is less 
 than atmospheric by the pressure due to the height of the 
 centre of the wheel above the well. 
 
 Now suppose the wheel to be driven by some engine 
 coupled with the driving pulley of the wheel in the sense 
 opposite to that represented in the figure. Then the water 
 in the chambers is set in motion, and consequently pressure 
 diminishes within it, and water ascends from the well. 
 Thus the whirlpool chamber and the pipe leading to the 
 elevated tank get continuous supplies of water raised from 
 the well, and the machine acts as a force pump for filling 
 the tank. A turbine driven backward in this way is called 
 a centrifugal pump. Such pumps are frequently used in 
 pumping water out of the boiler compartments of large war 
 ships, and also for drainage and for emptying docks. 
 
 49. Pressure of a jet on 
 B. a plane. Let AS, Fig. 87, 
 
 represent a column of fluid 
 striking a fixed plane at A, 
 and let v be the velocity of the 
 particles in the neighbourhood 
 of A. Then if the stream 
 
 vV; flows continuously, the effect 
 
 A on the plane is a continuous 
 
 Fig. 87. pressure. The magnitude of 
 
 this we propose to find. 
 Measure backwards along the stream a length AQ equal 
 
CH. vii Steady Motion under Action of Gravity 183 
 
 to v.&t, where A is any infinitesimally small time ; then 
 all the particles in the column between A and Q will strike 
 the plane in the time A t. If 8 is the area of the cross- 
 section of the stream and 10 the weight per unit volume of 
 the fluid, the weight of the column AQ is wSv . A, and its 
 momentum is wSv^.Af. This momentum is destroyed by 
 the plane in the time A^ ; and if P is the force, in gravita- 
 tion units, capable of destroying the momentum, 
 
 wSv* 
 -- A/, 
 
 . 
 
 and -~ is the force, or pressure, per unit area on the plane. 
 
 If the velocity v is such as might be acquired in falling 
 
 p. 
 through a height k, and -~ is denoted by p, we have 
 
 p = zw/t, 
 
 which is the intensity of pressure that would be produced by 
 a statical column of the liquid of height 2/t. If the stream 
 is produced by a waterfall of height k, then a statical column 
 of double this height would produce the same effect on the 
 plane. 
 
 This result holds, with some modification, for the intensity 
 of pressure produced on the wall of a vessel containing a 
 gas. At each point inside the vessel the gas particles are 
 moving in all possible directions with very various velocities. 
 Now these velocities give a certain mean square of their 
 values, which mean square remains constant at all points of 
 the gas so long as the temperature remains constant. Let 
 v 2 be the value of the mean of the squares of the velocities 
 of the gas particles at a point P, velocities being taken in 
 all directions round P. Now supposing that we fix on a 
 
184 Hydrostatics VOL. i 
 
 single fixed direction, Px, at P, and consider only com- 
 ponents of velocities measured in that direction, what will 
 be the mean square of these at any instant ? Evidently 
 much less than v 2 , since in finding this the whole of each 
 velocity is taken and not merely a component of it. It is 
 proved in works on the kinetic theory of gases that the 
 mean square of components in a single direction Px is just 
 one-third of the total mean square, so that if we denote the 
 former by f B 2 , f^ 2 i $2, 
 
 The reasoning which applied to the unidirectional stream 
 A, Fig. 87, can now be applied to the gas, and we have 
 
 1 wv 2 
 
 P---3' g 
 
 for the intensity of pressure exerted on the wall of the vessel. 
 In this expression w is the weight of a unit volume of 
 the gas at the temperature of the gas. Suppose that we 
 use the C. G. S. system, and employ the formula for the 
 weight given in Chap. V. If k is a numerical constant, 
 
 we know that w = k-^ , where * and T are the specific 
 
 gravity and absolute temperature of the gas. Hence the 
 above equation gives 
 
 where /tx is a numerical constant, and v is the square root of 
 the mean square of the velocities of particles. This v is 
 called the ' velocity of mean square '. With the numbers 
 for the metric system, if velocity is measured in metres per 
 second and * is the specific gravity of the gas referred to 
 hydrogen (about 14-4 times that referred to air), we find 
 
 v = in ' 
 
 Vf- 
 
CH. vii Steady Motion under Action of Gravity 185 
 
 EXAMPLES. 
 
 1. Water enters a horizontal pipe at an end where the area of 
 the cross-section is A square feet ; the pipe contracts to a cross- 
 section B, and after this changes to a cross-section C at the 
 other end, where the water flows into the atmosphere, K cubic 
 feet being delivered per second ; find the pressure intensities ut 
 A and B. 
 
 Result. If w = weight of a cubic foot of water, the pressure 
 
 wK 2 / i i \ ., , . wK" 1 / i i\ 
 
 at A 1SA + (^- I ,); thatatlM 8J + -_(_-_), 
 
 where p (per square foot) is the atmospheric pressure. 
 
 2. A turbine whose blades are radial at the outer rim receives 
 40,000 Ibs. of water per minute ; the outer radius of the wheel 
 is i foot, and the turbine is to have a horse-power of 50 ; how 
 many revolutions per minute must the wheel make 1 
 
 Result. Ifv feet per sec. is the circumferential velocity of the 
 
 40000 v* i 
 
 wheel, we have ; . - = 50, . . v = 36-3 f/s, and 
 
 00 3 2 55 060 
 
 the number of revolutions per minute = - - x 60 = 347. 
 
 2 7T 
 
 3. A turbine whose blades are radial at the outer rim receives 
 W Ibs. of water per minute ; the outer radius of the wheel is 
 a feet, and the machine is to develop m horse-power ; how many 
 revolutions must the wheel make 1 
 
 Result. 
 
 *5 Im, 
 a V W 
 
 4. The inner and outer radii of a turbine wheel are b and a ; 
 the velocity of flow through the inner circumference is i of that 
 due to the head of water, and the outer circumferential velocity 
 is due to half the head ; find the inclination of the vanes to the 
 inner circumference. 
 
 Result, tan" 1 
 86 
 
 5. The blades of a turbine at its outer circumference are 
 radial ; 500 Ibs. of water per second are supplied, with a head 
 
1 86 Hydrostatics VOL. i 
 
 of 28 feet; find the total area of the inlet surface of the wheel 
 if the radial velocity at inlet is ^ of that due to the head, 
 taking g = 32. 
 
 Result. ^\/7~ square feet. 
 
 6. If the velocity of the ahove turbine wheel at its outer 
 circumference is 40 f/s, the head of water 49 feet, the radial 
 velocity at inlet |- of that due to the head, find the magnitude 
 and direction of the absolute velocity of the water at inlet, 
 
 Result. \/40- + 6*, tan" 1 (o- 1 5). 
 
 7. The outer radius of the wheel of a centrifugal pump is 
 a feet ; a turning-moment of L foot-pounds' weight is applied 
 to the wheel spindle, and the pump raises W Ib. per second ; 
 find the tangential velocity of discharge of the water. 
 
 Result. ^ - 
 Wa 
 
 50. Fluid revolving about vertical axis. If a vessel, 
 represented in a vertical section by ACJ3, Fig-. 88, and 
 containing a fluid, is set rotating- round 
 a vertical axis, Cz, after a short time 
 the fluid, owing to friction between its 
 particles and against the surface of the 
 vessel, will rotate like a rigid body with 
 the angular velocity &> ; each particle, 
 P, will describe a horizontal circle with 
 this angular velocity, so that if PN is 
 the perpendicular from P on the axis of 
 Fig. 88. rotation, the resultant acceleration of 
 
 the particle is directed along PN from 
 P towards N, and is o> 2 . NP in magnitude. Denote NP 
 by r, and consider an indefinitely small particle of mass dm 
 at P ; then the resultant mass-acceleration of this particle is 
 
 coPr.dm, ....... (i) 
 
 and this vector is directed from P towards N. [The reversed 
 mass- acceleration, u?rdm t is called the force of inertia of 
 
CH. vii Steady Motion under Action of Gravity 187 
 
 the particle, or its resistance to acceleration. It is most 
 important to understand that this force of inertia is not 
 a force acting- on the particle, but one exerted ly it on the 
 surrounding- medium, or, generally, on the agent or agents 
 accelerating its motion. Thus, then, if a is the vector 
 representing the resultant acceleration of a particle, dm, 
 a force completely represented by 
 
 a. dm in absolute units, or 
 . dm in gravitation units, 
 
 t/ 
 
 is the resultant force exerted ly the particle on the agents 
 acting upon it. 
 
 Now the fundamental principle of all Dynamics is this : 
 for each particle of any material system (whether rigid body, 
 natural solid, liquid, or gas) the resultant mass-acceleration 
 is in magnitude and direction the exact resultant of all the 
 forces acting upon the particle. These forces will, in 
 general, consist partly of pressures from the surrounding 
 particles, partly of attractions from these particles, and 
 partly of attractions from bodies outside the system. If we 
 consider the equivalence of the resultant mass-acceleration, 
 adm, to these forces under three separate heads, we deduce 
 three great principles of Dynamics. Thus, if we consider 
 that a dm has 
 
 1. The same virtual work for any imagined displace- 
 ment of the particle, 
 
 2. The same moment about any axis, 
 
 3. The same component along any line, 
 
 as the whole system of forces acting on the particle, and 
 that this is true for every particle of the system, we have at 
 once the principles of 
 
 i. Kinetic Energy and Work, 
 
1 88 Hydrostatics VOL. I 
 
 2. Time-rate of change of Moment of Momentum, 
 
 3. Motion of Centre of Mass, 
 
 for every material system. If the forces acting are 
 measured in gravitation units, their complete equivalent is 
 
 - dm.~\ 
 
 ff 
 
 Suppose that at any point, P, Fig. 89, we take as the 
 
 element dm a very short and thin cylinder, abed, of the fluid 
 having its axis along the tangent at P to 
 any curve AB. Let the length be = ds ; let 
 <r be the area of the cross-section, ab, of the 
 cylinder ; let F be the external force per unit 
 mass exerted on the fluid at P, and there- 
 fore Fdm the force on the cylinder, not in- 
 cluding the pressure exerted on its surface by 
 the surrounding fluid ; let a be the resultant 
 
 acceleration of the particle ; and let p be the intensity of 
 
 /*/73 
 
 pressure on the face al, and therefore p + -jr&* that on cd. 
 
 Then dm = tvcrds, if w is the mass of the fluid per unit 
 volume at P ; and if we resolve forces in the direction of 
 
 the tangent at P to the curve, we see that w<r^#.-has the 
 
 same component along this tangent, in the sense PB, as 
 
 di) 
 F.Wffds and -j-ds.v, the length of the arc s being 
 
 measured from A. Hence if a s is the component of a along 
 the tangent at P and S is the component of F, we have 
 
 ' = *-I^. (2) 
 
 g w ds 
 
 This equation connects the acceleration in any direction 
 with the force intensity and the rate of change of pressure 
 intensity in that direction. 
 
CH. vii Steady Motion under Action of Gravity 189 
 
 Now suppose the external force to be gravity. Taking 
 the right line NP (Fig. 88) as the direction of *, (2) 
 
 becomes toV dp 
 
 w. = -f- ; ...... (3) 
 
 ff dr 
 
 and again, taking the vertical downward direction at P 
 as that of #, (2) becomes 
 
 dp 
 
 = fp -,7P ....... (4) 
 
 where z is the depth of P below any fixed horizontal 
 plane. 
 
 Now p is a function of r and z only, so that 
 
 dp . dp , 
 dp = ~dr + ~(h 
 dr dz 
 
 = *>(^<lr + tty ..... (5) 
 
 */ 
 /CD 2 ?' 2 \ ~ 
 
 where C is a constant, which may be determined from a 
 knowledge of p at some one point. If p is the value of p 
 at 0, the point in which the free surface cuts the axis 
 of rotation, and if is taken as origin, since r = z o at 
 0. we have C = j) ; hence 
 
 2 .2 
 
 P=Po + w(^- +*) ...... (6) 
 
 At all points on the free surface p = p^ , therefore the 
 equation of this surface is 
 
 showing that the z of every point on it is negative, i.e., all 
 these points are higher than 0. This equation denotes 
 
 nfi 
 
 a parabola whose latus rectum is r > , and the free surface is 
 
 * 
 
190 Hydrostatics VOL. i 
 
 therefore a paraboloid generated by the revolution of this 
 surface round Oz. 
 
 If the vertical upward line Oz is taken as axis of x, and 
 a tangent at as axis of y, the equation of the parabola 
 is, in its usual form, 
 
 (8) 
 
 If the fluid contained in the vessel is a gas, equations 
 (2), (3), (4) still hold, and, in addition, p = Jew (Art. 33) ; 
 hence (5) becomes 
 
 dp i /c 
 
 x 
 
 ), ..... (9) 
 
 P g 
 
 A , x 
 
 .-. p = Ae ....... (10) 
 
 where A is a constant to be determined either from a 
 knowledge of p at some point or from the given mass of 
 the fluid. Equation (10) shows that for a gas the free 
 surface and the surfaces of constant pressure intensity are 
 still paraboloids. 
 
 In the same way, if the vessel contains two fluids that 
 do not mix, their surface of separation is a paraboloid of 
 revolution. For if w, to' are their specific weights, we have 
 (if they are liquids) 
 
 w 2 ?- 2 
 p = w (^ - + z) + C for one, 
 
 o * 
 
 p' = w' (~ -- 1- z] + C" for the other, 
 v 2 ff 
 
 and since at all points on the surface of separation p = p', 
 we have the equation of a paraboloid of revolution, as 
 before. 
 
CH. vii Steady Motion under Action of Gravity 191 
 
 EXAMPLES. 
 
 1. A cylinder contains a given quantity of water. If it is 
 rotated round its axis (held vertical), find the angular velocity 
 at which the water begins to overflow. 
 
 Let A OB represent the surface of the rotating liquid, the 
 points A and B being at the top of the cylinder ; let r and h be 
 the radius and height of the cylinder, and c the height to which 
 the cylinder, when at rest, is filled. 
 
 Then since B is on the parabola, if is the depth of below 
 AB, 
 
 But the volume of the water remains unchanged, and it is the 
 volume of the cylinder minus the volume of the paraboloid A OB. 
 
 Tins latter is ^?. 2 . Hence 
 
 taT 
 
 ^^c, ........ (2) 
 
 -12S& 3 ( 3 ) 
 
 The above is on the supposition that the water begins to over- 
 flow before the vertex, 0, of the parabola reaches the base, C, of 
 the cylinder. In this case, with any angular velocity, co, if PQ 
 is the level to which the water rises, and LM is the level at 
 which the water stands when at rest, it is easily proved that 
 
 depth of below LM = height of PQ above LM. . (4) 
 
 Take now the case in which c is so small in comparison with 
 /* that reaches C (or the base begins to get dry) before the 
 water begins to overflow. The angular velocity at which 
 
 reaches C is . Let o> = (i +n) ; then if 0' (below 
 
 0) is the vertex of the parabola, we have 
 
 C0'=2n(i+n)c (5) 
 
 the height of PQ (the water level) above the base is 20(1 + n) ; 
 and if the free surface cuts the base in R, we have 
 
 RC = 
 
192 
 
 Hydrostatics 
 
 VOL. I 
 
 which is the radius of the dry circle on the base. The water 
 will begin to overflow when the height of PQ above the base is 
 A; i.e., 
 
 = 7i A, 
 
 r \l c' 
 
 (7) 
 
 which is quite different in form from (3) ; so that if c is in- 
 finitely small, i. e., if there is only an infinitely thin layer of 
 water put originally into the cylinder, it will not begin to over- 
 flow until co is infinitely great; and in this case CR = r, as it 
 should be. 
 
 2. A heavy cylinder floats with its axis vertical in a liquid 
 contained in a vessel which rotates uniformly round a vertical 
 axis ; find the length of the portion of the cylinder immersed. 
 , , Let PQ, Fig. 90, be the level of 
 
 the liquid round the cylinder, and 
 PEDQ the immersed portion, the 
 free surface being APOQB, and 
 the vertex of the parabola. 
 
 Now, by the same reasoning as 
 that in p. 91, it is obvious that 
 the resultant action of the liquid 
 on the cylinder is the same as that 
 which the liquid would exert on the 
 liquid which would fill the volume 
 POQDEP; hence the weight, W, 
 of the cylinder must be equal to 
 -p- the weight of this volume of the 
 
 liquid. 
 Let r be the radius of the cylinder and Om the perpendicular 
 
 from on PQ ; then Om = , and the volume of the displaced 
 
 liquid 
 
 = Trr 2 . PE-vol POQ = itr* 
 hence if w = specific weight of liquid, 
 
 4 f J 
 
 PE = 
 
 W 
 
 T 
 
 4 y 
 
CH. vii Steady Motion under Action of Gravity 193 
 
 3. A vessel of given form containing water is set rotating 
 round a vertical axis, the vessel and the liquid being in relative 
 equilibrium; find the greatest angular velocity of the vessel 
 which will allow all the water to escape through a small orifice 
 at the lowest point of the vessel. 
 
 Let the vessel be ACS, Fig. 88, C being its lowest point ; 
 assume the free surface to pass through C, the latus rectum of 
 
 %9 
 the parabola being ^ ; then, taking C as origin, the tangent at 
 
 C as axis of y and the vertical upward line as axis of x, express 
 
 2 O 
 
 the condition that the parabola if = ~ x intersects the curve 
 
 ACB in no other point than C. 
 
 Thus, if the vessel is a sphere (with another small hole at 
 the top) of radius a, the parabola will be completely outside the 
 
 c 9 
 sphere if or = - 
 a 
 
 4. A cylinder whose axis is vertical is filled with a given 
 mass of gas and set rotating round its axis ; if the gas is 
 assumed to move in relative equilibrium with the cylinder, find 
 the intensity of pressure at any point. 
 
 We have, measuring z from the top of the cylinder 
 
 , /w 2 r 2 \ 
 dp = wd [ -- \- z ) , 
 v 29 ' 
 
 have 
 
 .. 
 
 k 
 
 Also if W is the weight of the gas put into the cylinder, we 
 
 where dl = element of volume at any point, P (Fig. 88). Now 
 if is the angle which the plane of P and the axis, Oz, of rota- 
 tion makes with any fixed vertical plane, 
 
194 Hydrostatics VOL. i 
 
 Integrating with respect to r, the limits of r are o and a, 
 where a is the radius of the cylinder, so that the integrations in 
 r and z may he performed independently, the limits of z being o 
 and h. We easily find 
 
 which determines A. 
 
 5. If the cylinder is replaced by a spherical shell rotating 
 about a vertical diameter, solve the previous problem. 
 
 6. A hemispherical bowl containing a given quantity of water 
 is set rotating about a vertical diameter, find the angular velo- 
 city at which the water begins to overflow. 
 
 Result. If F is the volume of the water, a the radius of the 
 bowl, 
 
 7. If in the last case the angular velocity is increased beyond 
 
 the value f ) find how much of the bowl is dry. 
 v a ' 
 
 2 (1 
 
 Result. It is dry to a vertical height a above the 
 
 CO 2 
 
 lowest point. 
 
 8. If a hollow open cone with its axis vertical and vertex 
 downwards containing a given quantity of water is made to 
 revolve round a vertical axis, discuss the question as to the 
 possibility of emptying the cone by increasing the angular 
 velocity. 
 
 9. A narrow horizontal tube, BC, has two open vertical 
 branches BA and CD, water being poured into the continuous 
 tube, thus formed, to a given height. If this tube is set rotating 
 round a vertical axis through a point in BC, find the position 
 of the liquid in its state of relative equilibrium. 
 
 Result. If BO m, OC = n, the difference of level in the 
 
 2 
 
 two vei'tical branches is (m? n*). 
 
 v 
 
CH. vii Steady Motion under Action of Gravity 195 
 
 10. A straight tube, AB, containing water is held inclined at 
 an angle to the vertical line at A , and the tube is set in rota- 
 tion with given angular velocity, co, about this vertical line ; 
 find the greatest length of column of liquid so that there shall 
 be no separation of the column anywhere. 
 
 Let x be the length of column ; let P be a point in it such 
 that r and z are the distances of P from the axis of rotation and 
 the horizontal line through B; then 
 
 At A we have p = wh, where h is the height of a water 
 
 barometer, and if I is used for -4, 
 
 co 
 
 2l 
 
 p = r 2 2 Ir cot + 2 1 (h + x cos 6) x 2 sin 2 9. 
 w 
 
 Now if P is zero anywhere in the column, the column breaks 
 there. Make it impossible for p to vanish for any value of r, 
 and we have 
 
 2 I (h + x cos 0) a: 2 sin 2 G > l z cot 2 6. 
 
 If we make these two sides equal, we have the limiting value 
 of x, viz. 
 
 I cos + V 2 hi sin 
 sin 2 
 
 11. If in Q. 9 the lengths of BA and CD are each c and 
 the liquid reaches to their tops, and if is the middle point of 
 BC, prove that no liquid will overflow until the angular velocity 
 exceeds 
 
 V 2 y (h + c) 
 
 where h is the height of a barometer formed of the liquid, and 
 BC= 20,. 
 
 Taking as origin, z being measured upwards from OC, 
 
196 Hydrostatics VOL. i 
 
 and since at D we have p = wh, we have 
 
 p w\h - (tf-r^-z + c . 
 
 2 J 
 
 Hence at 
 
 fi to? \ 
 
 p = w [ li - a- + c } , 
 2<7 
 
 and when the liquid begins to overflow, the horizontal column 
 splits at 0, .'. p = o at 0. Hence the result. 
 
 12. In the last if EC alone is filled with liquid, find the 
 angular velocity which will cause the liquid to rise to a height k 
 in the vertical branches. 
 
 n 7, 
 
 Kesult. or 
 
 
 1 3. If in the last the upper ends of BA and CD are closed 
 before the motion begins, these branches containing air at 
 atmospheric pressure, find the speed necessary to raise the 
 liquid to a height k in the vertical branches. 
 
 ch 
 
 *~k 
 
 Kesult. w 2 = 2y 
 
 a 2 AT 
 
INDEX TO VOL. I 
 
 Absolute temperature, 114. 
 Air pumps, 138. 
 Air thermometer, 113. 
 Archimedes, principle of, 70. 
 
 erroneous statement of, 73. 
 
 - screw of, 134. 
 Avogadro's law, 112. 
 
 Barometric formula, 125. 
 Bellows, hydrostatic, 14. 
 Bernoulli's theorem, 154. 
 Boyle's law for gases, 105. 
 ]'> nu nali press, n. 
 Buoyancy, principle of, 67. 
 
 illustration of by revolving 
 tube, 91. 
 
 Centre of pressure, 42. 
 
 general theorem of, 43. 
 
 particle rule for triangle, 48. 
 
 on circular area, 53. 
 
 on hexagonal area, 57. 
 
 special cases of, 44. 
 
 Centrifugal pump, 181. 
 
 Condensing air pump, 140. 
 
 Cone, emptied through small ori- 
 fice, 161. 
 
 Contracted vein, 158. 
 
 Cylinder, emptied through small 
 orifice, 161. 
 
 Dalton and Gay Lussac's law, 106. 
 Density, definition of, 45. 
 Diving-bell, 115. 
 
 Dynamics, fundamental princi- 
 ples of, 187. 
 
 Efficiency of turbine, 176. 
 
 conditions of in turbine, 177. 
 Equality of pressure round a 
 
 point, 7. 
 
 Fictitious forces, introduction of, 
 
 r, 74 ' 
 
 Fire engine, 132. 
 
 Floating body, apparent anomaly 
 connected with, 78. 
 
 in revolving fluid, 192. 
 
 Flow through a tube, 152. 
 
 small orifice, 157. 
 
 - large orifice, 161. 
 Forcing pump, 131. 
 Free surface of a liquid, 28. 
 
 Gases, 102. 
 
 weight of, 120. 
 Geissler pump, 141. 
 Gravity, specific, 14. 
 
 Hero's fountain, 165. 
 Hydraulic press, 10. 
 
 ram, 136. 
 
 screw, 131. 
 
 differential, 135. 
 
 Hydrometers, 148. 
 Hydrostatic bellows, 14. 
 
 Immersion, work done in, 93. 
 Impulse turbine, 178. 
 Intensity of pressure, 2. 
 Isothermal transformation, 106. 
 
 Jet, pressure of, on plane, 182. 
 Jet pump, Thomson's, 167. 
 
 Lines of resistance, 60. 
 
 Manometers, 146. 
 Mariotte's bottle, 166. 
 Mass-moments, 20. 
 Moulds, 88. 
 Mountain, height of, 125. 
 
198 
 
 Index to Vol. I 
 
 Nicholson's hydrometer, 149. 
 
 Orifice, large, flow through, 161. 
 
 small, flow through, 157. 
 
 Particle rule for triangular area, 
 
 49, 51- 
 
 Pascal's principle, 3, 31. 
 Pelton wheel, 180. 
 Perfect fluid, definition of, 3, 5. 
 Pipe, thin, stress in, 64. 
 Plane-moments, theorem of, 19. 
 Pressure, equal transmission of, 
 
 3i- 
 
 centre of, 42. 
 
 on plane area, 33. 
 
 due to gravity, 26. 
 
 on circular area, 53. 
 
 on unclosed curved surface, 82. 
 
 head, 156. 
 
 - gauge, 157. 
 
 - in gas due to motion, 184. 
 Pump, water, 129. 
 
 air, 138. 
 
 centrifugal, 181. 
 
 Reaction turbine, 174. 
 
 Relative volume of steam, 125. 
 Resistance, lines of, 60. 
 Revolving fluid, 186. 
 
 Separate equilibrium, principle 
 
 of, 9. 
 
 Sluice, self-acting, 59. 
 Specific gravity, 14. 
 Sphere, emptied through small 
 
 orifice, 161. 
 
 Spherical surface, stress in, 65. 
 Sprengel pump, 144. 
 Steam, relative volume of, 125. 
 Stream lines, 154. 
 Superposed liquids, pressure due 
 
 to, 29. 
 Syphon, 163. 
 
 Thomson's jet pump, 167. 
 Torricelli, theorem of, 159. 
 Transformation, general equation 
 
 of, for gas, 1 10. 
 Turbines, 169. 
 
 Velocity head, 156. 
 
 of mean square, 184. 
 
 Vena contracta, 158. 
 
OXFORD : HORACE HART, M.A. 
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