Full text of "Mechanics of fluids for practical men [microform] : comprising hydrostatics, descriptive and constructive : the whole illustrated by numerous examples and appropriate diagrams"

See other formats

MECHANICS OF FLUIDS

FOR PRACTICAL MEN.

It is observed of Archimedes, by his philosophical biographer Plutarch, in the Life of Marcellus,
that " although we might labour long without success in endeavouring to demonstrate from our own
invention, the truth of his propositions ; yet so smooth and so direct is the way by which he leads us,
that when we have once travelled it, we fancy that we could readily have found it without assistance,
since either his natural genius, or his indefatigable application, has given to every thing that he
attempted the appearance of having been performed with ease."

W^fyrf/

[.MECHANICS OF FLUIDS

FOR PRACTICAL MEN,

COMPHISING

HYDROSTATICS,

DESCRIPTIVE AND CONSTRUCTIVE:

THE WHOLE ILLUSTRATED BY

NUMEROUS EXAMPLES AND APPROPRIATE DIAGRAMS.

ALEXANDER JAMIESON, LL.D. .

Ml
Author of" Elements of Algebra," fyc. fyc. $fe.

LONDON :
WILLIAM S. ORR AND CO.,

AMEN CORNER, PATERNOSTER ROW.

1837.

or THE

UNIVERSITY

PREFACE.

MECHANICS is the science which inquires into the laws of
equilibrium and the motion of bodies, whether solid or fluid.
The term originally applied only to the doctrine of EQUILIBRIUM,
and in this volume it is used in its primitive signification. The
adjunct by which this work has been designated, is meant to
convey the idea of a book that is self-instructing, and which,
in its details, may furnish those who have not had the benefit of
a regular academic education, with expeditious and practical
methods of operation, in applying the principles of hydrostatic
science to the general and every-day business of mechanics.

The volume is therefore a manual of principles combining the
twofold properties of precept and example, and exhibiting in
a comprehensive view whatever is generally and particularly
applicable to the mechanics of practical men. But the same
construction will render it available in any course of public or
private tuition, in which it may be desired to illustrate by
examples those operations which, in practical science, are go-
verned by the laws of fluid equilibrium, pressure, and support :
for it is hoped that these laws have been demonstrated and
illustrated with sufficient expansion to suit the progress of
modern discoveries, and to remove some part of that uncertainty
which has hitherto prevailed in the opinions of scientific men.

VI PREFACE.

Should the work achieve this, and contribute also to greater
precision than has hitherto been attained in the arrangements,
structures, and estimates required for works of public or private
utility, one of the objects for which it was originally undertaken
will be crowned with success.

In the spring of 1838, Gcov fleXovroc, another volume of
Mechanics of Fluids will appear, comprising Hydraulics, which,
as the term implies, will exhibit the principles of Dynamics in
the joint operation of air and water, hydraulic architecture, and
the principles of construction of various machines and engines
which belong to the mechanics of fluids.

In conclusion, we vir tvK\tiq 0avw, I beg leave to observe,
that as my own avocations did not allow me sufficient leisure
to complete such an undertaking as these volumes are, in any
reasonable time, I have availed myself of the services of
WM. TURNBULL, the author of a treatise on Cast-iron Beams
and Columns ; and it affords me unfeigned satisfaction to ac-
knowledge the extent of his abilities and the accuracy of his
calculations, in subjects connected with the mechanics of fluids.

Quam, sit uterque, libens (censebo) exerceat artem.

But having elsewhere alluded to the calculations and exam-
ples which abound in this volume, I shall only here remark,
that, were it necessary to plead authority for such exercises,
I might quote NEWTON himself, who has thus recorded his
opinion : — " In scientiis ediscendis, prosunt exempla magis quam
prtecepta." '' .r '

The wood-engravings, so well executed by Mr. G. VASEY,
are sufficiently intelligible, and possess besides the lasting
property of being destroyed only with the page in which they
appear.

A. J.

Wyke House , Sion Hill,
Isleworth, Oct. 19, 1837.

•I*

CONTENTS.

CHAPTER I.

DEFINITIONS AND OBVIOUS PROPERTIES OF WATERY FLUIDS,
WITH THE PRELIMINARY ELEMENTARY PRINCIPLES OF
HYDRODYNAMICS, FOR ESTIMATING THE PRESSURE OF
INCOMPRESSIBLE FLUIDS.

THE subject introduced, art. 1, page 1.— Fluid what, art. 2, page 2. — Fluid presses
equally in all directions, art. 3, page 2.— Lateral pressure equal to the perpendicular
pressure, art. 3, page 2. — Fluid equally pressed in all directions, art. 4, page 2. —
Fluid pressure against containing surface, perpendicular to that surface, art. 5,
page 2. — Fluid, surface of horizontal, or perpendicular to the direction of gravity,
art. 6, page 2. — The common surface of two fluids which do not mix, parallel to the
horizon, art. 7, page 2. — The surfaces of fluids continue horizontal when subjected
to the pressure of the atmosphere, art. 7, page 2. — Fluid particles at the same depth
equally pressed, art. 8, page 2. — Fluid pressure varies as the perpendicular depth,
art. 9, page 2. — Fluid pressure measured by the weight of a column whose base is
the surface pressed, and altitude the perpendicular depth of the centre of gravity,
art. 10, page 3. — Fluid pressure identified with a property of the centre of gravity,
art. 11, 12, and 13, pages 3 to 7. — Fluid pressure assignable, art. 13, page 8, corol. —
Pressure of different fluids on different plane surfaces immersed in them, as the
areas of the planes, the perpendicular depths of the centres of gravity, and the
specific gravities of the fluids, art. 14, page 8. — Pressure of the same fluid on dif-
ferent plane surfaces immersed in it, as the areas of the planes and the perpendicular
depths of the centres of gravity, art. 15, page 8. — Plane parallel to fluid's surface,
pressure on ditto varies as the perpendicular depth, art. 16, page 8. — Plane surface
inclined to the surface of the fluid, pressure on ditto varies as the perpendicular
depth of its centre of gravity, art. 17, page 8. — Planes of equal areas, whose centres
of gravity are at the same perpendicular depth, sustain equal pressures, whatever
may be their form and position, art. 18, page 8. — The centre of gravity remaining
fixed, the pressure upon a revolving plane is the same at all points of the revolu-
tion, art. 19, pages 8 and 9. — A plane surface being immersed in two fluids of
different densities, if the pressures are equal, the depths of the centres of gravity
vary inversely as the densities of the fluids, art. 20, page 9.— The same principle
demonstrated, art. 21, page 9.

VIII CONTENTS.

CHAPTER II.

OF THE PRESSURE OF NON-ELASTIC FLUIDS UPON PHYSICAL
LINES, RECTANGULAR PARALLELOGRAMS CONSIDERED AS
INDEPENDENT PLANES IMMERSED IN THE FLUIDS, AND
UPON THE SIDES AND BOTTOMS OF CUBICAL VESSELS,
WITH THE LIMIT TO THE REQUISITE THICKNESS OF FLOOD-
GATES.

Physical line when obliquely immersed, pressure on determined, art. 22, pages
10 and 11.— Practical rule for calculating ditto, art. 23, page 11. — Example to
illustrate ditto, art. 24, pages 11 and 12. — The same determined for a line when
perpendicularly immersed, art. 25, page 12. — The same determined for a line
obliquely immersed, when the upper extremity is not in contact with the surface
of the fluid, art. 26, pages 12 and 13. — Pressure determined when the perpendicular
depth of the upper extremity is given, art. 26, equation (5), page 14. — Practical
rule for ditto, art. 28, No. 1, page 14. — The same determined v hen the perpendi-
cular depth of the lower extremity is given, art. 27, equation (6), page 14. — Prac-
tical rule for ditto, art. 28, No. 2, page 14. — Example to illustrate ditto, art. 29,
pages 14 and 15. — The pressure determined when the line is perpendicularly im-
mersed, art. 30, page 15. — Two physical lines obliquely immersed, with their upper
extremities in contact with the surface of the fluid, to compare the pressures upon
them, under any angle of inclination, art. 31, pages 15 and 16. — The lines being
unequally inclined, the pressures directly as the squares of the lengths and sines of
inclination, art. 31, inf. 1, page 16. — When the lines are equally inclined, the pres-
sures are directly as the squares of the lengths, art. 31, inf. 2, page 16. — The lines
being similarly situated the pressures can be compared, art. 31, inf. 3, page 16. —
The lines being differently situated in the fluid, the pressure is determined, art. 32,
page 17. — Right angled parallelogram obliquely immersed, pressure determined,
art. 33, equation (7), page 1 9.— Practical rule for ditto, art. 34, No. 1, page 19. —
Plane perpendicularly immersed, the pressure determined, art. 33, equation (8),
page 19. — Practical rule for ditto, art. 34, No. 2, page 19. — Example for illustration
of the inclined case, art. 35, pages 19 and 20. — Example for illustrating the per-
pendicular case, art. 36, page 20. — The parallelogram obliquely immersed and
the longer side coincident with the surface of the fluid, pressure determined, art. 37,
equation (9), page 20. — Perpendicularly immersed, pressure determined, art. 37,
equation (10), page 20. — Pressures compared, art. 37,corols. 1, 2, and 3, page 21. —
Right angled parallelogram immersed as before and bisected by its diagonal, the
pressures on the triangles determined and compared, art. 38 and 39, pages 21, 22,
23, and 24. — Practical rules for the pressures, art. 40, page 25. — Example for
illustration, art. 41, page 25. — Centre of gravity of a right angled triangle deter-
mined, art. 42, pages 26 and 27. — Rectangular parallelogram bisected by a line
parallel to the horizon, the pressure on the two parts determined and compared, art.
43, pages 27, 28, and 29.— Practical rules, art. 44, Nos. land 2, page 29.— Examples
for illustration, art. 45, pages 29 and 30. — Rectangular parallelogram so divided by
a line parallel to the horizon, that the pressures on the two parts are equal, art. 46,
pages 30, 31, and 32.— The same effected geometrically, art. 47, page 32. — Practical
rule for the calculation of ditto, art. 48, page 33. — Example to illustrate ditto, art.

CONTENTS. IX

49, page 33. — The same determined when the point of division is estimated upwards,
art. 50, page 33. — Practical rule for calculating ditto, art. 51, page 34.— The solution
independent of the breadth of the parallelogram, art. .51, page 34, corol. — Rectangu-
lar parallelogram divided into two parts sustaining equal pressures, by a line drawn
parallel to the diagonal, art. 52, pages 34, 35, and 36.— Example for illustrating
ditto, art. 53, pages 36 and 37. — Centre of gravity of the upper portion determined,
art. 54, pages 38, 39, and 40. — Rules for the co-ordinates, art. 55, pages 40 and 41. —
Example for illustrating ditto, art. 56, pages 41 and 42. — Construction of the
figure, art. 57, page 42. — The problem simplified when the line of division is
parallel to the diagonal, art. 57, page 42. — Example for illustrating ditto, art. 57,
pages 42 and 43. — Remarks on ditto, art. 57, page 43. — The pressure not a neces-
sary datum, compare the concluding remark, art. 57, page 43, with equations (20
and 21), art. 54, page 40. — Rectangular parallelogram divided into two parts sustain-
ing equal pressures, by a line drawn from one of the upper angles to a point in the
lower side, art. 58, pages 43 and 44. — Practical rule for ditto, art. 59, page 45. —
Example for illustrating ditto, art. 60, page 45. — The same determined when the
line of division is drawn from one of the lower angles to a point in the immersed
length, art. 61, page 46. — Practical rule for ditto, art. 62, page 47. — Example for
illustrating ditto, art. 63, page 47. — Centre of gravity determined, art. 64, pages
48,49, and 50. — Remark on ditto, art. 66, page 51.— Rectangular parallelogram
so divided, that the pressures 011 the parts shall be in any ratio, art. 67, 68, and 69,
pages 51 and 52. — Remarks on ditto, art. 69, page 53. — Rectangular parallelogram
divided by lines drawn parallel to the horizon, into any number of parts sustaining
equal pressures, art. 70, pages 53, 54, 55, 56, and 57.— Practical rule for ditto,
art. 71, pages 57 and 58. — Example for illustrating ditto, art. 72, page 58. — Con-
struction of the figure, art. 72, page 59. — Requisite thickness of floodgates deter-
mined, art. 73, page 59. — Example for illustrating ditto, art. 73, pages 59 and 60.—
Construction of the figure, art. 74, page 60.— Remarks on ditto, art. 74, page 61. —
Pressure on the sides and bottom of a rectangular vessel compared, art. 75, pages
61, 62, and 63. — Examples for illustrating ditto, art. 76 and 77, page 64.

CHAPTER III.

ON THE PRESSURE EXERTED BY NON-ELASTIC FLUIDS UPON
PARABOLIC PLANES IMMERSED TN THOSE FLUIDS, WITH
THE METHOD OF FINDING THE CENTRE OF GRAVITY OF
THE SPACE INCLUDED BETWEEN ANY RECTANGULAR PARAL-
LELOGRAM AND ITS INSCRIBED PARABOLIC PLANE.

Pressure on a parabolic plane, compared with that on its circumscribing rectan-
gular parallelogram, art. 77, pages 65 and 66.— Practical rules for ditto, art. 78,
page 67. — Example for illustrating ditto, art. 79, page 67. — Centre of gravity of
the included space determined, art. 80, pages 67 and 68. — Remarks on ditto, art. 80,
page 69. — The same things determined when the base of the parabola is coincident
with the surface of the fluid, art. 81, pages 69 and 70, — Practical rules for ditto,
art. 82, page 71. — Example for illustrating ditto, art. 83, page 71.— Remarks on
ditto, art. 83, page 72.— The same things determined when the base of the parabola
is vertical, and just in contact with the surface of the fluid, art. 84, pages 72 and
VOL. I. b

X CONTENTS.

73. — Corollary to ditto, art. 84, page 74. — Practical rules for ditto, art. 85, page
73. — Example for illustrating ditto, art. 86, page 74. — Corollaries on ditto, art. 87,
page 75.

Pressure on a semi-parabolic plane, compared with that upon its circumscribing
parallelogram, art. 88, pages 75, 76, and 77. — Practical rule for ditto, art. 89,
page 77.— Example for illustrating ditto, art. 90, page 77.— The same determined
when the axis of the semi-parabola is horizontal, art. 91, pages 78 and 79. — Practi-
cal rules for ditto, art. 92, page 79. — Example for illustrating ditto, art. 93, page 79.
The centre of gravity of the included space determined, art. 94, pages 80, 81, and
82.— Construction of the figure, art. 94, pages 82 and 83. — Remark on ditto, art.
94, page 83.

Parabolic plane with axis vertical, and vertex in contact with the surface of the
fluid, divided by a horizontal line into two parts sustaining equal pressures, art. 95,
pages 83, 84, and 85. — Practical rule for ditto, art. 96, page 85. — Example for
illustrating ditto, art. 97, pages 85 and 86. — The same things determined generally,
art. 97, pages 86 and 87. — Example for illustrating ditto, art. 97, page 87.

CHAPTER IV.

OF THE PRESSURE OF INCOMPRESSIBLE FLUIDS ON CIRCULAR

PLANES AND ON SPHERES IMMERSED IN THOSE FLUIDS,

THE EXTREMITY OF THE DIAMETER OF THE FIGURE BEING
IN EACH CASE IN EXACT CONTACT WITH THE SURFACE OF

; THE FLUID.

Chord of maximum pressure in a circular plane determined, art. 98, pages 88
and 89. — Construction effected, corollaries 1 and 2, page 89. — Practical rules for
calculating ditto, art. 99, page 90. — Example for illustrating ditto, art. 100, page
90. — Pressures on two immersed spheres determined and compared, art. 101,
pages 90 and 91. — A globe being filled with fluid, the pressure on the interior
surface is three times the weight of the contained fluid, art. 101, corol. page 92. —
Example for illustrating ditto, art. 102, page 92. — The exterior surface of a sphere
divided by a horizontal circle into two parts sustaining equal pressures, art. 103,
pages 92, 93, and 94.— Practical rule for ditto, art. 104, page 94. — Example for
illustrating ditto, art. 105, page 94.— The solution generalized, art. 106, page 95< —
Practical rule for the general solution, art. 107, page 95.

CHAPTER V.

OF THE PRESSURE OF NON-ELASTIC OR INCOMPRESSIBLE FLUIDS
AGAINST THE INTERIOR SURFACES OF VESSELS HAVING THE
, FORMS OF TETRAHEDRONS, CYLINDERS, TRUNCATED CONES,
&C.

Pressure on the base of a tetrahedron compared with the pressure on its sides,
art. 108, pages 96, 97, and 98. — Comparison effected, corol. page 98. — Pressure on
the base of a cylinder compared with the pressure on its upright surface, art. 109,

CONTENTS-. Xi

pages 99 and 100. — When in any vessel whatever, the sides are vertical and the
base parallel to the horizon, the pressure on the base is equal to the weight of the
fluid, art. 110, pages 100 and 101. — The concave surface of a cylindrical vessel
divided into annuli, on which the pressures are respectively equal to the pressure
on the base, art. Ill, pages 101, 102, 103, and 104. — The limits of possibility
assigned, art. 112, page 104. — The equations expressed in terms of the radius,
art. 112, page 104. — The equation generalized, art. 112, page 104. — The practical
rule for any annulus, art. 113, page 105. — Example for illustrating ditto, art. 114,
page 105.

The pressure on the base of a truncated cone compared with that on its curved
surface, and also with the weight of the contained fluid, art. 115, pages 106, 107,
108, and 109. — The same principle extended to the complete cone, base downwards,
art. 115, page 108, equations (77 and 80). — Comparison completed, art. 115, corol.
page 109.

The same things determined for a truncated vessel with the sides diverging
upwards, art. 116, pages 109, 110, and 111. — For the case of the complete cone with
the base upwards, see equation (84).— Pressure on the base compared with the
weight of the contained fluid, art. 117, page 112. — The pressure on the base may
be greater or less than the weight of the contained fluid in any proportion, art. 117,
corol. 1, page 112. — The pressure on the bottom of a vessel not dependent on the
quantity of the contained fluid, art.117, corol. 2, page 113. Any quantity of fluid,
however small, balances any other quantity, however great, art. 117, corol. 3, page
113. — Pressure on the bottom of a cylindrical vessel equal to any number of times
the fluid's weight, art. 118, pages 114 and 115.— Practical rule for ditto, art. 119,
page 115. — Example for illustrating ditto, art. 120, page 115. — Remarks on ditto,
art. 120, corollaries 1 and 2, page 116. — Concluding remarks on the Hydrostatic
Press, page 116.

CHAPTER VI.

THE THEORY OF CONSTRUCTION AND SCIENTIFIC* DESCRIPTION
OF SOME HYDROSTATIC ENGINES, VIZ. THE HYDROSTATIC
PRESS, HYDROSTATIC BELLOWS, THE HYDROSTATIC WEIGHING
MACHINE, AND EXPERIMENTS PROVING THE QUA QUA VERSUS
PRESSURE OF FLUIDS.

Principle of the Hydrostatic Press developed, art. 121, pages 117 and 118. —
First brought into notice by Joseph Bramah, Esq., of Pimlico, art. 122, page 118. —
Not a new mechanical power, ib. — Known under the name of Hydrostatic Paradox
ib. — Principal element by which the power is calculated, art. 123, page 119. —
Example to illustrate ditto, art. 124, page 119. — General equation for the pressure
on the piston of the cylinder, art. 124, equation (89), page 119. — Practical rule for
reducing ditto, art. 124, page 120. — Example for illustrating ditto, art. 125, page
120. — General equation for the pressure on the piston of the forcing pump, art. 125,
equation (90), page 120.— Practical rule for ditto, art. 125, page 120. — Example for
illustrating ditto, art. 126, page 120. — General expression for the diameter of the
piston of the cylinder, equation (91), page 121. — Practical rule for ditto, art. 126, page
121. — Example to illustrate ditto, art. 127, page 121. — General expression for the

Xll CONTENTS.

diameter of piston in the forcing pump, art. 127, equation (92), page 121.— Practical
rule for ditto, art. 127, page 121. — Safety valve introduced, art. 128, pages 121 and
122. — General expressions established, art. 128, equations (93 and 94), page 122. —
Example for illustrating the weight upon the safety valve, art. 129, page 122.—
General expression for the weight upon the safety valve, art. 129, equation (95),
page 122. — Practical rule for ditto, art. 129, page 123. — Example for illustrating
ditto, art. 130, page 123. — General expression for the pressure on the piston of the
cylinder, art. 130, equation (96), page 123.— Practical rule for ditto, art. 130,
page 123. — Example for illustrating ditto, art. 131, page 123. — General expression
for the diameter of the safety valve, art. 131, equation (97), page 124. — Practical
rule for ditto, art. 131, page 124. — Example for illustrating ditto, art. 132, page 124.
— General expression for the diameter of the cylinder, art. 132, equation (98), page
124. — Practical rule for ditto, ib. — Example for illustrating ditto, art. 133, page 124.
— General expression for the weight upon the safety valve, art. 133, equation (99),
page 125. — Practical rule for ditto, ib. — Example for illustrating ditto, art. 134,
page 125. — General expression for the pressure on the piston of the forcing pump,
art. 134, equation (100), page 125. — Practical rule for ditto, ib. — Example for
illustrating ditto, art. 135, pages 125 and 126. — General expression for the diameter
of the safety valve, art. 135, equation (101), page 126. — Practical rule fpr ditto, ib.
— Example for illustrating ditto, art. 136, page 126. — General expression for the
diameter of the forcing pump, art. 136, equation (102), page 126. — Practical rule
for ditto, art. 136, page 127. — Concluding remarks, ib. — The thickness of the metal
in the cylinder determined, art. 137, pages 127, 128, 129, 130, and 131.— Example
for illustrating ditto, art. 138, page 131. — Appropriate remarks, art. 139, pages 131
and 132. — A determinate and uniform value assigned to the pressure on a square
inch of surface, art. 140, page 132. — Rules for the pressure in tons, and for the
diameter of the cylinder in inches, art. 141, page 133. — Remarks on the theory,
art. 142, page 133. — Examples for illustrating ditto, ib. and page 134.— Remarks,
art. 143, page 134. — Observations on presses previously constructed, with rules
and examples for examining them, art. 144 and 145, pages 134 and 135. — The
description of the Hydrostatic Press, with its several parts and appendages, both
in its complete and disjointed state, arts. 146, 147, and 148, pages 136 — 142.

The Hydrostatic Bellows and its uses introduced, art. 149, page 142. — Descrip-
tion and principles of ditto, art. 150, pages 142 and 143.— Practical rule for
calculating the weight of fluid in the tube, art. 151, page 144. — Example for
illustrating ditto, ib. — The solidity or capacity of the fluid in the tube determined,
art. 152, page 144.— Altitude of the tube determined, art. 152, page 145.— The
solution given generally, art. 153, page 145. — Practical rule for determining the
height of the tube, art. 154, page 146. — Illustrative example and remark, i&.—
Weight on the moveable board determined in the case of an equilibrium, art. 155,
page 146. — Practical rule and illustrative example for ditto, ib. — Concluding-
remark, art. 155, page 147. — The diameter of the equilibrating tube determined,
art. 156, page 147. — Practical rule for ditto, ib.-— Example for illustrating ditto,
art. 156, page 148. —The diameter of the Bellows Board determined, art. 157, page
148. — Practical rule and illustrative example for ditto, ib. — When the equilibrium
obtains, if more fluid be poured into the tube, it will rise equally in the tube and in
the bellows, art. 158, page 149.

The ascent of the weight on the moving board determined, art. 159, page 150. —
Practical rule and example for ditto, art. 159, page 151.

CONTENTS. Xlll

The Hydrostatic Weighing Machine introduced, described, and investigated,
art. 160, pages 151, 152, and 153.— Practical rule and example for ditto, art. 161,
page 154. — Quantity of ascent above the first level determined, practical rule and
example for ditto, art. 162, pages 154 and 155. — Concluding remarks, art. 162,
page 155.

Quaquaversus pressure of incompressible fluids illustrated by experiments, pages
155-160.

CHAPTER VII.

OF PRESSURE AS IT UNFOLDS ITSELF IN THE ACTION OF FLUIDS
OF VARIABLE DENSITY, OR SUCH AS HAVE THEIR DENSITIES
REGULATED BY CERTAIN CONDITIONS DEPENDENT UPON PAR-
TICULAR LAWS, WHETHER EXCITED BY MOTION, BY MIXTURE,
OR BY CHANGE OF TEMPERATURE.

Preliminary remarks on density, pages 161, 162, and 163. — The alteration of
pressure in consequence of a change of temperature determined, art. 163, pages
163, 164, and 165. — The chord in a semicircular plane on which the pressure is
a maximum determined, on the supposition that the diameter is in contact with the
surface of the fluid, and the density increasing as the depth, art. 164, pages 165,
166, and 167. — Practical rule for ditto, art. 165, page 167. — Example for illus-
trating ditto, art. 166, page 167. — Construction of ditto, art. 167, pages 167 and 168.
— The section of a conical vessel parallel to the base, on which the pressure is a
maximum, determined, the axis of the cone being inclined to the horizon in a given
, angle, art. 168, pages 168, 169, and 170.— Practical rule for ditto, art. 169, page 170.
— Example for illustrating ditto, art. 170, page 170. — Concluding remarks on
compressibility, art. 171, page 171. — The diameter of a globe determined, in
ascending from the bottom to the surface of the sea, on the supposition that the
globe is condensible and elastic, art. 172, pages 171 and 172. — Practical rule for
ditto, art. 173, page 173. — Example for illustrating ditto, art. 174, page 173. —
Remark on ditto, art. 175, page 173. — The depth of the sea determined, art. 176,
pages 173 and 174. — Practical rule and example for ditto, art. 177, page 174.—
Pressure on the bottoms of vessels containing fluids of different densities deter-
mined, art. 178, pages 174 and 175. — Law of induction expounded, art. 179, pages
175 and 176. — Example for illustration, art. 180, page 176. — Another example
under different conditions, art. 181, pages 176 and 177. — Pressure on the inner
surface determined and compared with that upon the bottom, art. 182, pages 177
and 178. — Altitudes of fluids of different densities, inversely as the specific gravities,
art. 183, pages 178 and 179.— Practical rule for the altitudes, art. 184, page 179. —
Example for illustrating ditto, art. 185, page 180. — A column of mercury of 2£ feet
and a column of water 33.995 equal to the pressure of the atmosphere, art. 186, page
180. — The principle of the common or sucking pump dependent on this property,
ib. — Altitudes in the tubes equal when the specific gravities are equal, art. 187,
page 180. — Surfaces of small pools near rivers on the same level as the rivers,
art. 188, page 180.

Water may be conveyed from one place to another, of the same or a less eleva-
tion, art. 189, page 180. — When the source and point of discharge are on the same
level the water is quiescent, but when the point of discharge is lower than the

XIV CONTENTS.

source, the water is in motion, art. 190, page 180. — On this principle large towns
and cities are supplied with water, art. 190, page 181. — Edinburgh thus supplied,
ib. — The principle of the proposition art. 183 generalized, art. 191, page 181. —
Remark on ditto, ib. — Two fluids of different densities, hut equal quantities, being
poured into a circular tube of uniform diameter, their positions determined when
in a state of equilibrium, art. 192, pages 181, 182, 183, and 184. — The principle
assumed to be similar to that of Mr. Barclay's Hydrostatic Quadrant, art.193, page
184. — Practical rule for ditto, ib. — Example illustrative of ditto, art. 194, pages
184 and 185.— The actual position of the fluids exhibited by a construction, art. 195,
pages 185 and 186. — Mercury preferable to water for the tubes of philosophical
instruments, art. 196, page 186. — Other fluids convenient for the purpose, art. 197,
page 186. — The result of the investigation only applicable to mercury and water,
arts. 196 and 198, page 186. — The same rendered general, art. 199, pages 186 and
187. — The general equation deducible from the particular one, art. 199, page 187. —
Practical rule for the general case, art. 200, pages 187 and 188. — Example for
illustrating ditto, art. 201, page 188.— The positions of the fluid different in the
two cases, art. 202, page 188. — Inquiry into the changes produced on the general
formulae, under certain conditions, art. 202, page 189.— Practical rule for reducing
the resulting equation, art. 203, page 189.— Concluding remark, ib.

CHAPTER VIII.

OF THE PRESSURE OF NON-ELASTIC FLUIDS UPON DYKES, EM-
BANKMENTS, OR OTHER OBSTACLES WHICH CONFINE THE
FLUIDS, WHETHER THE OPPOSING MASS SLOPE, BE PERPENDI-
CULAR, OR CURVED, AND THE STRUCTURE ITSELF BE MASONRY,
OR OF LOOSE MATERIALS, HAVING THE SIDES ONLY FACED
WITH STONE.

Introductory remarks, art. 204, page 190. — The manner explained in which a
dyke, mound of earth, or any other obstacle, may yield to fluid pressure, art. 205,
page 191. — Remarks on ditto, ib. — General investigation, art. 206—210, pages
191 — 197. — Reasons adduced for not giving practical rules on this subject, art. 210,
page 197. — Example for illustrating the reduction of the final equations, art. 211,
pages 197 and 198. — The same illustrated when the effect produced by the vertical
pressure of the fluid is omitted, art. 212, pages 198 and 199. — It is safer to calculate
by omitting the vertical pressure, art. 212, corol. page 199. — The breadth at the
foundation determined when the slopes are equal, art. 213, page 199. — Example
for illustrating ditto, art. 214, pages 199 and 200. — The same determined when the
side on which the fluid presses is vertical, art. 215, page 200. — More expensive to
erect a dyke of this form, than if both sides slope, art. 215, corol. page 201. — The
same determined when the side of the dyke opposite to that on which the fluid
presses is vertical, art. 216, page 201. — The stability in this case less than when
the vertical side is towards the water, art. 216, pages 201 and 202.

Thickness of the dyke determined when both its sides are perpendicular to the
plane of the horizon, art. 217, page 202. — Practical rule and example for this case,
art. 218, page 202. — Thickness of the wall determined when its section is in the

CONTENTS. XV

form of a right angled triangle, and first when the fluid presses on the perpendi-
cular, art. 219, page 203. — Practical rule and example for this case, art. 220, page
203. — The same determined when the fluid presses on the hypothenuse, art. 221,
pages 203 and 204. — Practical rule and example for this case, art. 222, page 204. —
The construction exhibited, art. 223, pages 204 and 205. — Concluding remarks,
art. 223, page 205. — The thickness of the dyke determined when it yields to the
pressure by sliding on its base, art. 224, pages 205 and 206. — The resistances of
adhesion and friction compared with the weight of the dyke, art. 225, page 206. —
Example for illustrating the resulting formula, art. 226, page 207. — The breadth of
the dyke at the top determined, art. 226, page 207. — The breadth of the dyke at
the bottom determined, when the side on which the water presses is perpendicular
to the horizon, art. 227, page 207. — The same determined when the opposite side is
perpendicular, ib. — The same when both sides are perpendicular, ib. — The same
determined when the section of the wall is triangular, having the side next the
fluid perpendicular, and the remote slope equal to the breadth, art. 228, page 207.
— The same takes place when the remote side is perpendicular, and the slope on
which the water presses is equal to the breadth, art. 228, page 208. — Concluding
remarks, ib.

The dimensions of the dyke determined when it is constructed of loose materials,
art. 229, pages 208, 209, and 210. — Example for illustration of ditto, art. 230, page
210. — The conditions necessary for preventing the dyke from sliding on its base
determined, art. 231, page 210. When the water presses against the vertical side
of a wall, the curve bounding the other side so that the strength may be every
where proportional to the pressure, is a cubic parabola, ib. Introductory remarks
to Chapter IX, art. 232, page 211.

CHAPTER IX.

OF FLOATATION, AND THE DETERMINATION OF THE SPECIFIC
GRAVITIES OF BODIES IMMERSED IN FLUIDS.

The buoyant force equivalent to the weight of the displaced fluid, art. 233, pages
212, 213, and 214. — The pressure downwards equal to the buoyant force, art. 233,
corol. page 214.— The height determined to which a fluid rises in a cylindrical
vessel, in consequence of the immersion of a given sphere of less specific gravity,
art. 234, pages 214 and 215. — Practical rule for ditto, art. 235, page 216. — Example
for illustrating ditto, art. 236, page 216. — The same determined when the sphere
and the fluid are of equal specific gravity, art. 237, page 216; — Practical rule for
ditto, ib. Example for illustrating ditto, art. 238, pages 216 and 217. The height
determined to which the fluid rises in a paraboloidal vessel, in consequence of the
immersion of a sphere of less specific gravity, art. 239, pages 217, 218, and 219.

Practical rule for ditto, art. 240, page 219. — Example for illustrating ditto, art.
241, pages 219 and 220. — The same determined when the sphere and the fluid are
of equal specific gravity, art. 242, page 220.— Practical rule for ditto, art. 243,
page 220.— Example for illustrating ditto, art. 244, pages 220 and 221. — Remarks
on ditto, art. 244, corol. page 221. — A homogeneous body placed in a fluid of the
same density as itself, remains at rest in all places and in all positions, art. 245,
pages 221 and 222. — The upward pressure against the base of a body immersed in

XVI CONTENTS.

a fluid, is equal to the weight of the displaced and superincumbent fluid, art. 246,
page 222. — The difference between the downward and upward pressures, is equal
to the difference between the weight of the solid and an equal bulk of the fluid,
art. 247, page 222.— Absolute and relative gravity, what, art. 248, pages 222 and
223. — By absolute gravity, fluids gravitate in their proper places, by relative
gravity they do not, ib. — A heavy heterogeneous body descending in a fluid, has the
centre of gravity preceding the centre of magnitude, art. 249, page 223. — The
reason of this, ib. — Concluding remarks, art. 250, page 223. — The force with which
a body ascends or descends in a fluid of greater or less specific gravity than itself,
is equal to the difference between its own weight and that of the fluid, art. 251,
pages 223 and 224. — The force of ascent and descent is nothing when the specific
gravities are equal, art. 251, page 224. — When a body is suspended or immersed in
a fluid, it loses the weight of an equal bulk of the fluid in which it is placed, art.
252, page 224. — When a body is suspended or immersed in a fluid of equal or
different density, it loses the whole or a part of its weight, and the fluid gains the
weight which the body loses, art. 253, page 225. — Bodies of equal magnitudes
placed in the same fluid lose equal weights, and unequal bodies lose weights pro-
portional to their magnitudes, art. 254, page 225. — The same body placed in
different fluids, loses weights proportional to the specific gravities of the fluids,
art. 255, page 225. — When bodies of unequal magnitudes are in equilibrio in any
fluid, they lose their equilibrium when transferred to any other fluid, art. 256, page
225. — When a body rises or falls in a fluid of different density, the accelerating
force, what, art. 257, page 225. — When the solid is heavier than the fluid it
descends, when lighter it ascends, art. 258, pages 225 and 226; hence relative
gravity and relative levity, ib. — Practical rule for the general formula, art. 259,
page 226.— Example for illustrating ditto, art. 260, page 226. — The distance of
descent determined, when the pressive and buoyant forces are equal, art. 261, pages
226, 227, and 228.— Practical rule for calculating ditto, art. 262, page 228. —
Example for illustrating ditto, art 263, page 2'J8.

CHAPTER X.

OF THE SPECIFIC GRAVITIES OF FLUIDS, AND THE THEORY OF
WEIGHING SOLID BODIES BY MEANS OF NON-ELASTIC FLUIDS.

Introductory remarks on specific gravity, end the principles or criteria of com-
parison, page 229.— The weight lost by a body, is to the whole weight, as the
specific gravity of the fluid is to that of the solid, art. 264, pages 229 and 230. —
The weight which the body loses in the fluid is not annihilated, but only sustained,
art. 264, page 230. — The weight of a body after immersion determinable, art. 265,
page 231. — Practical rule for ditto, ib. — Example for illustration, art. 266, page 231.
- — The weights which a body loses in different fluids, are as the specific gravities of
the fluids, art. 266, corol. page 231. — The real weight of a body determinable by
having its weight in water and in air, art. 267, pages 231 and 232. — Practical rule

for ditto, art. 268, page 232. — Example for illustration, art. 269, page 232. The

specific gravity of a body determinable from its weight, as indicated in water and
in air, art, 270, pages 233 and 234. — Practical rule for ditto, art. 271, page 234.—

CONTENTS. XVII

Example for illustration, art. 272, page 234. — The magnitude of a globular body
determinate from its real weight and density, art. 273, pages 234 and 235. — The
same determinable from its weight in air and in water, art. 274, pages 235 and 236.
— Practical rules for ditto, art. 275, page 236. — Examples for illustration, arts. 276
and 277, pages 236 and 237. — Different bodies of equal weights immersed in the
same fluid, lose weights that are inversely as their densities, or directly as their
magnitudes, art. 278, page 237. — The difference between the absolute weight of a
body and its weight in any fluid, is equal to the weight of an equal bulk of the
fluid, art. 279, page 237. — If two solid bodies of different magnitudes indicate equal
weights in the same fluid, the larger body preponderates in a rarer medium, art. 280,
page 237. — Under the same circumstances the lesser preponderates in a denser
medium, art. 281, page 238. — If solid bodies when placed in the same fluid sustain
equal diminutions of weight, their magnitudes are equal, art. 282, page 238. To
determine the equipoising weight, when two bodies equally heavy in air, are placed
in a fluid of greater density, the densities of the bodies being different, art. 283,
pages 238 and 239. — Practical rule for ditto, art. 284, page 239. — Necessary remark,
ib. — Example for illustration, art. 285, page 239. — The ratio of the quantities of
matter determinable, when two bodies of different specific gravities equiponderate in
a fluid, art. 286, page 240. — Example for illustration, art. 287, pages 240 and 241. —
Problem respecting the equiponderating of the cone and its circumscribing cylinder,
art. 288, pages 241, 242, and 243. — Practical rule for ditto, art. 289, page 243. —
Example for illustration, art. 290, page 244. — To compare the specific gravities of
a solid body, with that of the fluid in which it is immersed, art. 291, page 245. —
Example for illustration, art. 292, page 245. — To compare the specific gravities of
two solid bodies, when weighed in vacuo and in a fluid of given density, art. 293,
pages 245 and 246. — Example for illustration, art. 294, page 247. — The specific
gravities of different fluids compared, by weighing a body of a given density, art.
295, pages 247 and 248. — Example for illustration, art. 296, pages 248 and 249. —
Concluding remarks, ib. — The specific gravity of a solid body determined by weigh-
ing it in air and in water, art. 297, page 249. — The principle of solution explained,
ib. — The practical rule for ditto, art. 298, page 250. — Example for illustration, art.
299, page 250.— Concluding remarks, art. 300, page 250. — The specific gravity of
a solid body determined from that of the fluid in which it is weighed, art. 301,
page 251. — Practical rule for ditto, art. 302, page 252. — Example for illustration,
art. 303, page 252. — The specific gravity of a solid body determined, by immersing
it in a vessel of water of which the weight is known, art. 304, pages 252, 253, and
254. — Practical rule for ditto, art. 305, page 254. — Example for illustration, art. 306,
page 254. — Concluding remarks on the value of the opal, ib.

CHAPTER XL

OF THE EQUILIBRIUM OF FLOATATION.

Opening remarks on floatation, page 255. — The centre of gravity of the whole
body and that of the immersed part occur in the same vertical line, art. 307, pages
255 and 256. — The weight of the floating body and that of the displaced fluid are
equal to one another, art. 307, page 256. — Corresponding remarks, ib. — Homogeneous
plane figures divided symmetrically remain in equilibrio with their axes vertical,
art. 308, page 257. — Homogeneous solid bodies generated by the revolution of a

XV1I1 CONTENTS.

curve, when placed upon a fluid of greater specific gravity, maintain their equili-
brium with the axis vertical, art. 309, page 257. — Showing under what conditions
a prismatic hody remains in equilibrio, art. 310, page 257. — The magnitude of a
floating body to that of the immersed part, as the specific gravity of the fluid is to
that of the solid, art. 311, pages 257 and 258. — Examples for illustration, arts. 312
and 313, pages 258 and 259. — General determination, art. 314, page 259. — Practical
rule for ditto, ib. — Example for illustration, art. 315, page 260. — General investiga-
tion, ib. — Practical rule and example for ditto, pages 260 and 261. — Inferences
arising therefrom, arts. 316, 317, 318, 319, and 320, page 261. — Demonstration of a
general principle, art. 321, page 262. — To determine how far a paraboloidal solid
will sink in a fluid, art. 322, pages 262 and 263. — Practical rule for ditto, art. 323,
page 264. — Example for illustration, art. 324, page 264. — The same determined
when the vertex of the figure is downwards, ib. — Practical rule for ditto, art. 325,
page 265. — The elevation or depression of a body determined, when the equilibrium
is disturbed by the subtraction or addition of a certain given weight, art. 326, pages
265 and 266. — The same thing determined independently of fluxions, art. 327, pages
267 and 268. — Practical rule for ditto, art. 328, page 268. — Example for illustration,
art. 329, page 268. — The same determined for a body in the shape of a paraboloid,
art. 329, pages 26& and 269. — Remark on the resulting equations, art. 330, page 269.
— The descent occasioned by adding a weight determined, art. 330, pages 269 and
270. — Practical rule for ditto, art. 331, page 270. — Example for illustration, art.
332, page 270. — The ascent determined when a given weight is subtracted, art. 333,
page 270. — Practical rule for ditto, art. 334, page 271. — Concluding remark, art.
335, page 271. — The weight determined which is necessary to sink a body to the
same level with the fluid, art. 336, pages 271 and 272.— Practical rule for ditto,
art. 337, page 272.— Example for illustration, art. 338, page 272.— A solid body
being immersed in two fluids which do not mix, floats in equilibrio between them',
when the weights of the displaced fluids are together equal to the weight of the
body, art. 339, pages 272, 273, and 274.— The quantity of each fluid displaced by a
cubical body determined, art. 340, pages 274 and 275. — Practical rule for ditto,
art. 341, page 275. — Example for illustration, art. 342, page 275. — Another example
under different conditions, art. 376, page 343. — The specific gravity of a solid body
determined, so that any part of it may be immersed in the lighter of two unmixable
fluids, art. 344, page 276. — Practical rule for ditto, art. 345, page 277.— Example for
illustration, art. 346, page 277.— The same determined when equal parts of the body
are immersed in the lighter and heavier fluids, art. 347, page 277. — Practical rule
for ditto, art. 347, page 278. — Example for illustration, art. 348, page 278.— The
same determined when the lighter fluid vanishes, art. 349, page 278. — A very
curious property unfolded, art. 349,' page 279. — The ratio of the immersed parts
determined, when the body floats on water, in air, and in a vacuum, art. 350,
pages 279 and 280. — Practical rule for ditto, art. 351, page 280. — Example
for illustration, art. 352, page 280. — Remark and rule for determining the same
otherwise, art. 353, page 281. — The Hydrometer or Aerometer introduced, art.
354, page 281. — Improvements on ditto by various writers, art. 354, page 282.
— Description of the instrument according to Deparcieux, art. 355, pages 282
and 283. — The specific gravity of a fluid determined by the aerometer, art. 356,
pages 283 and 284. — Practical rule for ditto, art. 357, page 284. —Example for
illustration, art. 358, page 284. — The immersed quantity of the stem determined,
art. 359, page 285. — Practical rule for ditto, 360, page 285. — Example for illustra-

CONTENTS. XIX

tion, art. 361, page 285. — Transformation of the equations, art. 362, pages 285 and
286. — Concluding remarks, art. 362, page 286. — The change in the aerometer cor-
responding to any small variation in the density of the fluid, art. 363, page 286. —
Practical rule for ditto, art. 364, pages 286 and 287. — Example for illustration, art.
365, page 287. — Remarks, ib. — The sensibility of the instrument increased by
decreasing the diameter of the stem, and otherwise, art. 366, pages 287 and 288. — •
Practical rule for ditto, art. 367, page 288. — Example for illustration and remarks
on ditto, art. 368, pages 288 and 289. — Hydrostatic Balance introduced and de-
scribed, art. 369, pages 289 and 290.— The specific gravity of a solid body determined
by the balance, that of distilled water being given, art. 370, page 291. — Practical
rule for ditto, art. 371, page 291. — Example for illustration, art. 372, page 291 .

CHAPTER XII.

OF THE POSITIONS OF EQUILIBRIUM.

The positions of equilibrium of a triangular prism determined, art. 373, pages
292, 293, and 294. — Remarks on the form of the resulting equation, art. 374, pages
294 and 295. — Geometrical construction effected by means of two hyperbolas, art.
374, page 295. — Example for illustrating the reduction of the formula, art. 375,
pages 295 and 296. — Construction and calculation of the figure and its parts, arts.
376, 377, and 378, pages 296, 297, and 298. — More positions of equilibrium exhi-
bited by the equation, and remarks on ditto, art. 379, page 299. — The positions
determined when the section of the prism is isosceles, art. 380, pages 299 and 300.
— Practical rule for ditto, art. 381, page 300. — Example for illustration, art. 382,
pages 300 and 301.— Other positions determined, art. 383, page 301. — Remarks on
the resulting equation, art. 383, page 302. — Practical rules for reducing ditto, art.
384, pages 302 and 303. — Limits to the positions of equilibrium, art. 385, page 303.
— Example for illustration, art. 386, pages 303 and 304. — Remarks on ditto, art.
386, page 304. — The positions delineated, art. 387, page 304. — The same demon-
strated, pages 305 and 306.— Remarks, art. 388, page 306. — The positions of equi-
librium dependent upon the specific gravity, art. 389, pages 307 and 308. — The
true positions delineated according to the conditions of the problem, art. 390, page
308. — The positions of equilibrium determined when the section of the body is in
the form of an equilateral triangle, art. 391, pages 308 and 309. — The positions
calculated, art. 392, pages 309 and 310. — The positions delineated according to this
determination, page 310. — Other positions determined, art. 393, pages 310 and 311.
—The same positions delineated by construction, art. 393, page 31 1. — The positions
of equilibrium determined for a triangular prism, when two of its edges fall below
the plane of floatation, art. 394, pages 311, 312, 313, 314, and 315.-— The method of
reducing the general equation described, art. 395, page 315. — Example to show the
method of reduction, art. 396, pages 315, 316, and 317. — The position delineated
and verified, art. 397, page 317. — Another condition of equilibrium, what, art. 398,
page 317.— The same verified, page 318. — Other positions of equilibrium determin-
able, the determination left for exercise to the reader, art. 399, page 318. — The
same determined when the section of the figure is in the form of an isosceles
triangle, art. 400, page 318. — The equation reduced, art. 401, page 319. — Practical
rule for reducing the equation, art. 402, page 319.— Example for illustration, art.

or THE A
UNIVERSITY 3

XX CONTENTS.

403, pages 319 and 320. The position delineated by construction, art. 404, page
320. — Other two positions of equilibrium determinable, art. 405, pages 320, 321,
and 322.

Limits to the specific gravity defined when the body floats with two of its angles
immersed, art. 405, page 321. — Expression for the arithmetical mean between the
limits, equation (247), page 321. — Expressions for the extant sides of the triangular
section, (equations 248 and 249), pages 321 and 322. — Remarks on ditto, art. 405,
page 322. — Example for illustration, art. 406, page 322. — Construction indicating
the positions, art. 407, page 322. — The construction verified by calculation, art. 407,
page 323, and art. 408, page 324. — The positions of equilibrium determined when
the triangular section is equilateral, art. 409, pages 324 and 325. — Remarks on
ditto, art. 409, page 325. — Example for illustration, art. 410, page 325. — Construc-
tion indicating the positions, art. 410, page 325. — The same verified by calculation,
art. 410, page 326. — Two other positions assignable, ib. — The limits of the specific
gravity defined, art. 411, pages 326 and 327. — Positions of equilibrium determined,
art. 411, page 327. — Construction indicating the positions, art. 412, page 327. —
The same verified by calculation, arts. 412 and 413, pages 327 and 328. — Con-
cluding remarks on ditto, art. 414, pages 328 and 329. — The positions of equilibrium
determined for a rectangular prism with one of its edges immersed, art. 415, pages
329 to 331. — General remarks on ditto, art. 416, page 332.— The positions deter-
mined when the ends of the prism are squares, art. 417, page 332. — General
inference from ditto, art. 417, page 333. — Practical rule for ditto, art. 418, page
333. — Example for illustration, art. 419, page 333. — Construction indicating the
position, arts. 419 and 420, page 333. — Condition establishing the equilibrium,
art. 421, page 334.— The maximum limit of the specific gravity defined, art. 422,
page 334. — Two other positions determined, as they depend upon the limit of the
specific gravity, art. 423, page 334. — Conditions that limit the specific gravity,
art. 424, page 334. — Positions determined according to the limits, and repre-
sented by diagrams, art. 425, pages 334 and 335. — Positions determined from
the arithmetical mean of the limits, and represented by diagrams, pages 335 and
336. — The positions determined when three edges of the prism are immersed,
art. 426, page 336. — The positions determined when two edges are immersed, art.
427, page 336. — Remarks on the superior difficulty of this case, page 337.— Pre-
paratory construction, ib. — The conditions necessary for a state of equilibrium,
page 338. — The first condition established, equation (265), page 339. — The second
condition determinable by a separate construction, page 339. — The investigation
pursued, pages 340, 341, and 342. — The positions determined, art. 429, pages 342,
343, and 344. — The same determined when the section is a square, art. 430, page
344.— When the specific gravity of the prism is half the specific gravity of the
fluid, the body sinks to half its depth, art. 431, page 344. — This position repre-
sented by a diagram, ib. — The positions which the body would assume in the course
of revolution assigned, page 345. — Two other positions assignable, dependent on
the limits of the specific gravity, page 345, equations (273 and 274). — Construction
indicating the position according to the limits, art. 432, page 346. — The maximum
limit of the specific gravity determined, art. 433, page 346. — Remarks, accompanied
by diagrams to indicate the positions, art. 434, pages 346 and 347. — The arith-
metical mean between the limits assigned, art. 435, page 347. — The positions
assigned numerically, art. 436, pages 347 and 348. — The same indicated by con-
struction, art. 436, page 348. — The reverse positions exhibited, art. 437, page 348.

CONTENTS. XXI

— The construction verified by calculation, art. 438, page 349. — The positions of
equilibrium determined for a solid, of which the transverse section is in the form
of the common parabola, art. 439, pages 349-352. — Practical rule for ditto, art. 440,
page 352.— Example to illustrate ditto, art. 441, page 352.— Construction indicating
the position of equilibrium, with its verification, art. 442, page 353. — The same
determinable for the figure in the inclined position, art. 443, pages 353 to 356. —
Example to illustrate ditto, art. 444, pages 356 and 357. — The positions indicated
by construction, page 357. — The construction verified, page 358.

CHAPTER XIII.

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

Stability of floating bodies, the subject introduced, art. 446, page 359. — General
remarks concerning ditto, arts. 447, 448, and 449, pages 359 to 362. — Definitions,
art. 450, pages 362 and 363. — A floating body displaces a quantity of fluid equal
to its own weight, and in consequence, the specific gravity of the fluid is to that of
the solid, as the whole magnitude is to the part immersed, art. 451, pages 363 and
364. — A floating body is impelled downwards by its own weight, and upwards bj
the pressure of the fluid, and these forces act in vertical lines passing through the
centre of effort and the centre of buoyancy, art. 452, page 364. — When these lines
do not coincide, the body revolves upon an axis of motion, ib. — If a floating body be
deflected from the upright position, the stability is proportional to the length of the
equilibrating lever, or to the horizontal distance between the vertical lines, passing
through the centre of effort and the centre of buoyancy, art. 453, page 364. — When
this distance vanishes, the equilibrium is that of indifference, ib. — When it falls on
the same side of the centre of effort as the depressed parts of the solid, the equili-
brium is that of stability, art. 453, pages 364 and 365.— When it falls on the same
side as the elevated parts, the equilibrium is that of instability, art. 453, page 365.
— Concluding remarks, ib. — A general proposition belonging to the centre of gra-
vity, art. 454, page 365. — General and subsidiary remarks, ib.

The stability of floating bodies determined, art. 455, pages 366 to 370. — The
manner of generalizing the result explained, together with the constituent elements
of the equation, art. 456, page 370. — Example for illustration, art. 457, pages 370
and 371. — The steps of calculation illustrated by reference to a diagram, arts. 458,
459, 460, 461, 462, 463, and 464, pages 371 to 375. — Concluding observations, art.
464, page 375.

The principle of stability applied to ships, art. 465, page 376. — The conditions
of the data explained, art. 466, pages 376 and 377. — The longer and shorter axes,
what, art. 467, page 377. — In what respects a ship may be considered as a regular
body, art. 468, page 377. — The principal section of the water, what, art. 468, page
377. — The circumstances and conditions of calculation explained ,by reference to a
diagram, arts. 469, 470, 471, and 472, pages 378 to 382.— A numerical example for
illustration, art. 473, page 382. — Table of the measured ordinates, page 383. —
Construction of the example explained, pages 384 to 387.— Construction continued,
art. 474, pages 387 to 389. — The contents of the displaced volumes, how obtained,
art. 475, page 389. — Approximating rules for calculating the areas and solidities,
art. 476, page 390. — The construction completed, art. 477, page 390. — The practical
delineation of the vertical and horizontal planes, arts. 478 and 479, pages 391 and

XX11 CONTENTS.

392. — The manner of calculation described, art. 480, page 393. — The manner of
calculation exemplified, art. 481, pages 39S to 396. — Calculation continued, arts. 482
and 483, pages 396 and 397. — Concluding remarks, art. 483, pages 397 and 398. —
Reflections suggested by the importance of the subject, art. 484, page 398.

The principles of stability as referred to steam ships considered, art. 485, pages
398 and 399. — Reference to Tredgold's work on the Steam Engine, art. 485, page
399. — Tredgold's method of simplifying the investigation, art. 486, page 399. — His
subdivision of the inquiry, ib. — The steps of investigation not necessary to be
retraced, art. 487, page 399. — The expression for stability when the ordinates are
parallel to the depth, equation (290), art. 488, page 399. — Remarks deduced from
the form of the equation, art. 489, page 400. — The expression for stability in the
ease of a triangular section, equation (291), art. 489, page 400. — The practical rule
for reducing the equation, art. 490, page 400. — Example for illustrating ditto, art.
491, pages 400 and 401.— The expression for stability in the case when the trans-
verse section is in the form of a common parabola, equation (292), art. 492, page
401. — Remarks on its fitness for the purpose of steam navigation, as contrasted with
the triangular section preceding, art. 492, page 401. — Comparison of the results,
art. 493, pages 401 and 402.— Method of identifying the rule in the two cases, art.
494, page 402. — When the transverse section is in the form of a cubic parabola, the
stability is determined by equation (293), art. 495, page 402. — Remarks on the
superior form in this case, art. 496, page 402. — Practical rule for ditto, ib. —
Example for illustration, art. 497, pages 402 and 403. — The stability determined for
a parabolic section of the 5th order, equation (294), art. 498, page 403. — Remarks
on ditto, ib.— General remarks in reference to the limiting forms of steam ships,
art. 499, page 403. — When the ordinates of the transverse section are parallel to the
breadth, the stability is expressed by equation (295), art. 500, page 404. — When
the transverse section is in the form of a triangle, the stability is expressed by
equation (296), art. 501, page 404. — When the transverse section is in the form
of the common parabola, the stability is expressed by equation (297), art. 502,
page 404.— When the transverse section is in the form of a cubic parabola, the
stability is expressed by equation (298), art. 503, page 404. — This form superior
for stability, ib.— When the transverse section is formed by a parabola of the 5th
order, the stability is expressed by equation (299), art. 504, page 405. — Concluding
and general remarks, i&.— The stability the same at every section throughout the
length, under what conditions this will obtain, art. 505, page 405.

CHAPTER XIV.

OF THE CENTRE OF PRESSURE.

The centre of pressure, subject introduced, definition and preliminary remarks,
art. 506, page 406. — The centre of pressure determined for a plane surface, art. 507,
pages 406, 407, 408, and 409. — Formulae of condition, equation (302), page 409. —
The centre of pressure determined for a physical line, art. 508, pages 409 and 410.
— Practical rule for ditto, art. 509, page 410. — Example for illustration, art. 510,
pages 410 and 411. — The same determined when the upper extremity of the line is
in contact with the surface of the fluid, art. 511, page 411. — The same principle
applicable to a rectangle, art. 511, page 411.— The centre of pressure determined

CONTENTS. XXlll

for a rectangle when its upper side is in contact with the surface of the fluid, art.
512, pages 411 and 412.— The centre of pressure determined when the plane is a
square, art. 513, page 412. — The centre of pressure determined for a semi-paraholic
plane, art. 514, pages 412 and 413. — Example to illustrate ditto, art. 515, pages 413
and 414. — The same determined for one side of a vessel in the form of a parallelo-
pipedon, art. 517, pages 414, 415, and 416. — Practical rule and example for ditto,
page 416. — The balancing force, the centre of pressure, and the direction of its
motion determined for the side of a tetrahedron, arts. 518, 519, 520, and 521, pages
417 to 419.— Example to illustrate ditto, art. 522, page 420.

CHAPTER XV.

OF CAPILLARY ATTRACTION AND THE COHESION OF FLUIDS.

Capillary attraction and the cohesion of fluids, the subject introduced, with its
object, and remarks thereon, art. 525, pages 421 and 422. — Definition of capillary
attraction, art. 526, page 422. — Attraction of cohesion between glass and water,
art. 527, pages 422 and 423. — The particles of a fluid attract each other, art. 523,
page 423. — The particles of mercury have an intense attraction for each other, art.
529, page 423. — The attraction between glass and water only sensible at insensible
distances, art. 530, pages 423 and 424. — The manner described in which the
attractive influence is exerted, art. 531, page 424. — The forms assumed by the
summit of the elevated columns described, art. 532, pages 424 and 425. — The force
of attraction proposed to be determined, art. 533, page 425. — The parts by which
the fluid in the tube is attracted, described, art. 534, page 426. — The same as
respects the lower portion of the tube described, art. 535, page 426. — The fluid
attracted by the glass only, art. 535, pages 426 and 427, — A negative force in
the opposite direction, art. 536, page 427. — The force of attraction determined,
and expressed by equation (310), art. 536, page 427. — The conditions of the
rising and falling column explained, ib. — The expression for the force of at-
traction generalized, equation (311), page 428. — The height to which the fluid
ascends in the tube determined, art. 538, pages 428 and 429. — The mean altitude
expressed, equation (312), page 429. — The general expression modified, arts. 539
and 540, pages 429 and 430. — The constant determined, equation (315), page 430.
Practical rule for the mean height, art. 541, page 430. — Example for illustration,
art. 542, page 430. — The radius of the tube determinable, art. 543, pages 430 and
431. — The method illustrated by a numerical example, art. 544, page 431. — The
height to which the fluid rises between two parallel plates determined, arts. 545,
546, and 547, pages 431, 432, and 433. — The practical rule for ditto, art. 547, page
433. — Example for illustration, art. 548, page 433. — The distance between the
plates determinable, art. 549, page 434. — Concluding remarks, art. 550, page 434.
— When two smooth plates of glass meet in an angle, to determine the nature of the
curve which the fluid forms between them, art. 551 and 552, pages 435 and 436. —
The curve determined to be the common or Apollonian hyperbola, art. 553, page
436. — Concluding remarks, ib. — Two bodies that can be wetted with water, when
placed an inch asunder do not approximate or recede ; but if placed a few lines
apart, they approximate with an accelerated velocity, art. 554, page 436. — Two
bodies that cannot be wetted, when placed a few lines apart, approximate with an
accelerated velocity, art. 555, page 437. — When one body can be wetted and the other

XXIV CONTENTS.

not, they recede from each other, art. 556, page 437. — Remarks on the above, ib. —
General laws deducible from ditto, arts. 557, 558, and 559, pages 437 and 438. —
Hydrostatic pressure exemplified in springs and Artesian wells, arts. 560 to 569,
pages 439 to 442.

CHAPTER XVI.

MISCELLANEOUS HYDROSTATIC QUESTIONS, WITH THEIR
SOLUTIONS.

Miscellaneous hydrostatic questions, with their solutions, arts. 570 to 581, pages
443 to 448.

On a careful revision of the sheets, the following are the principal errors that
have been discovered.

/ Art. 9, page 2, for varies in its perpendicular depth, read varies as its perpendi-
cular depth.

J Page 97, line 17 from top, for p= the pressure upon one of the sides, readp— the
pressure upon the three containing sides.

1 Page 183, line 10 from bottom,/or cos.x + sin.20 sin.0, read cos.x -J- sin.20 sin.x.
_ ^ Page 215, line 16 from top, dele as.

Page 230, line 6 from top,/or the specific of the solid, read the specific gravity of
the solid ; and in line 10,/or respective gravities, read respective specific gravities.
Page 237, line 6 from top,/ar

d=\3/ Ill21l£l=3.93l3, or nearly 4 inches, read
V .5236X7

d= 4s/ W —\t/ - 14X16 =0.3939 feet, or 4.7267 inches.

V .5236S V .5236X7X1000

line 18 from bottom, far

d=.3/ (13.9975-12)16

.= /(13.9975-12)X ™=4^67 incheg.
.5 V .52361-001262.5

.
.o2o6(s-s)X62.5 V .5236(1-0012)62.

Art. 354, page 282,/or Levi read Lovi.

Page 432, line 13 from bottom,/or %bdzTr, read$d*ir.

- line 11 from bottom,/or ±bd*—%bd*ir=$bd*(2— TT), read ±bdz(l— |TT).

- line 8 from bottom,y!w m=&<#i-f£fed2(2— ?r), read m=bdh+±bdz(\—\ir).

- line 3 from bottom,/or {bdh-\-ibd\2— ir)}8g, read {bdh+ibd*(l-$7r)}3g.

- line 1 from bottom,/or d{fc+id(l— ^TT)}, read d{h+ld(l— ITT)}-

INTRODUCTION.

THE analytical table of contents, which the reader must have
perused, will have shown him that this volume is not a selection
of shreds and patches, garbled from contemporary authorities :
but a systematic treatise on Hydrostatic Science, containing a
vast mass of valuable and interesting facts, combining indeed
almost all that needs to be known on the equilibrium of fluids.
But for the convenience of reference, these Mechanics of Fluids
are distributed into a series of chapters, whose titles indicate
the several topics that receive mathematical demonstration.
The first of these contains, besides a few brief but necessary
definitions, the fundamental proposition upon which all the pro-
blems that are drawn up in Elementary Hydrostatics are in
reality founded.

The principle established in the general proposition, enables
the reader to proceed in the second chapter with the pressure of
incompressible fluids upon physical lines, rectangular parallelo-
grams considered as independent planes immersed in the fluid,
and to determine the position of the centre of gravity of the
various rectangular figures which the successive problems em-
brace, together with the pressures of fluids upon the sides and
bottoms of cubical vessels, with the limits which theory assigns
to the requisite thickness of flood-gates.

One distinguishing characteristic in this inquiry is, that every
problem is accompanied by a practical example ; and in order that
nothing be omitted which could render the subject intelligible to

VOL, i. c

XXVI INTRODUCTION.

the general reader, the most important formulae of a practical and
general nature have been thrown into rules, in words at length,
whereby all the arithmetical operations required in the solution
of the examples, can be performed without any reference to the
algebraical investigation, which is the surest way of uniting
precept with example.

After the same method, the third chapter treats of the pres-
sure exerted by non-elastic fluids upon parabolic planes immersed
in these fluids, with the method of finding the centre of gravity
of the space included between any rectangular parallelogram
and its inscribed parabolic plane. This is a valuable proposi-
tion in the practice of bridge-building, and it is very satisfactory
to find in prosecuting one branch of science, the means of ac-
complishing another ; to discover in a subject purely hydrostatic,
a method by which to find the position of the centre of gravity
of the arch, with all its balancing materials, and consequently
many important particulars respecting the weight and mechani-
cal thrust, with the thickness of the piers necessary to resist the
drift or shoot of a given arch, independently of the aid afforded
by the other arches. The method laid down in Problem
XII. for this purpose is presumed to be new; at any rate we
have not seen it noticed by any previous writer on Mechanics.
But its development belongs to Hydraulic Architecture ; the
principle here established being all that is required in Hy-
drostatics.

Chapter IV. introduces the reader to the pressure of non-
elastic fluids on circular planes, and spheres immersed in those
fluids as independent bodies, the extremity of the diameter of
the figure being in each case coincident with the surface of the
fluid. These problems could easily have been extended to
examples of elliptical planes and solids, but the investigation
would not embrace any practical result : and where that is un-
attainable, this work presumes not to advance.

The Fifth Chapter, in which are classed the tetrahedron,
cylinder, conical frustum, and indeed the frustum of any other
regular pyramid, completes this branch of fluid pressure ; but
the investigation is directed altogether to the pressure of the
fluid upon the internal surfaces of the vessels under considera-
tion. Indeed this was part of the inquiry when the sphere was

INTRODUCTION.

treated of in the fourth chapter; but in the fifth, the subject is
purely practical, and involves some of the most important prin-
ciples in the whole range of Hydrodynamics. The reader now
enters upon that remarkable and important principle,

That any quantity of fluid, however small, may be
made to balance or hold in equilibrio any other quantity,
however great ;

and is enabled thence to investigate the theory and expound
the construction of those mechanical contrivances known as
Bramah's hydrostatic press, the hydrostatic bellows, and weigh-
ing machine, which are all methods of balancing different
intensities of force, by applying the simple power of non- elastic
fluids to parts of an apparatus moving with different velocities :
and this is all the mechanical powers can effect.

The Sixth Chapter, which treats of these hydrostatic engines,
their theory of construction and scientific description, com-
mences with a distinct proposition ; the first having proved suf-
ficient to resolve every problem connected with fluid pressure
upon rectilinear and curvilinear figures considered as independent
planes immersed in the fluids, together with the pressure of
fluids upon the interior surfaces of vessels containing the fluids
and belonging to the class of regular bodies, — the second pro-
position, which the reader now enters upon, involves the prin-
ciple whereon depend the construction and appliancy of the
hydrostatic press, an engine very generally employed in practical
mechanics, and which should therefore be scientifically as it is
practically known. But the same proposition extends to the
investigation of the hydrostatic bellows, and furnishes the prin-
ciple of a particular machine by which goods may be weighed
as by the common balance. It may thence be inferred, that as
yet, science has but stepped on the threshold of fluids that are
heavy and liquid. How far this distinguishing property, the
power of transmitting pressure equally in all directions, may yet
carry mankind, it would be idle to conjecture. Enough, how-
ever, is here shown to satisfy the reader, that in expounding the
laws of the pressure and equilibrium of fluids, as well as those
of their motion and resistance, he will encounter principles of
great practical utility in the construction and use of machines,

XXVlll INTRODUCTION.

engines, apparatus, and instruments employed, not only in the
higher departments of natural philosophy, but in the every-day
concerns of society, in the arts, manufactures, and domestic
operations of civilized men. The occurrence of such principles
seems to present the legitimate time and place for classifying the
inventions to which they gave existence, and for directing genius
in its attempts to elicit new applications of collateral principles :
for though fortuitous circumstances and accidental hints may
have led to some discoveries in Hydrodynamics, the greater part
of modern improvements must be traced to patient induction,
which arrives at those coincidences whereby scientific men are
enabled to expound the theory of particular machines, whose
construction and principles of action depend upon the equi-
librium or motion of fluids. By this method, nothing is taken
for granted which can be investigated from a series of mathe-
matical truths ; for, as Mr. Whitehurst observes, " it is one thing
to assent to truths, and another to prove them to be true : the
former leaves the mind in a state of suspense, the latter in the
possession of truth." *

This chapter concludes with some experiments upon the
quaqua versus property of non-elastic fluids ; these experiments
have the lowly merit of placing that property, the power of
transmitting pressure equally in all directions, in a popular
point of view, " level to the capacity of ordinary minds."

Our labours hitherto refer exclusively to what may be termed
elementary principles in the mechanics of fluids ; we now com-
mence with PRESSURE, as it unfolds itself in the action of fluids
of variable density,^ or such as have their densities regulated by
certain conditions, dependent upon particular laws, whether ex-
cited by motion, by mixture, or by change of temperature. This
is the subject of Chapter Seventh, in which it will be found
that the investigation of the pressure of fluids of variable den-
sity is fruitful of some remarkably curious results : among these
we may notice the circumstance of a globe of condensible

* " Inquiry into the Original State and Formation of the Earth." — London, 1792.

t The word variable is perhaps taken in a too general sense : the densities are not
variable in all cases, they are only different — yet they are sometimes variable also ;
but there can be no more correct mode of writing upon this subject.

INTRODUCTION. XXIX

matter immersed in the sea to a given depth, as being likely to
suggest some easy and accurate methods of determining the depth
of the ocean, when it is so profound as to preclude the appli-
cation of the methods npw in use. The next fact claiming our
attention here, is the result we obtain by putting fluids of dif-
ferent densities into bended tubes, when the perpendicular alti-
tudes of these fluids above their common surface will vary
inversely as their specific gravity ; for we herein settle at once
the grand problem in our domestic policy — what is the best
method by which large towns and cities, or in fact any place,
can be supplied with water from a distance. But this is not all
— another result is, the construction of the hydrostatic quadrant,
for finding the altitude of the heavenly bodies, when from haze
or atmospheric obscurity, the horizon is rendered indistinct or
invisible. We trust our investigation of this beautiful principle
of the pressure of fluids of variable density, will in some mea-
sure facilitate the construction of the hydrostatic quadrant — an
instrument but as yet in its infancy.

The Eighth Chapter is one of vast utility in constructive me-
chanics, when it is necessary to investigate the pressure of fluids
on dykes and embankments, a subject interesting and im-
portant in the doctrine of Hydraulic Architecture, and peculiarly
applicable to the inland navigation and the maritime accommoda-
tion of a country situated like Great Britain, every where inter-
sected by canals, and seamed in all the sinuosities of her coast
by the tides and waves of the restless and turbulent commercial
ocean. Moreover, this subject is particularly applicable to the
great works now in progress, as rail-roads, docks, harbours, and
basins. The brevity of this chapter is compensated by the unity
it confers on separate and distinct portions of fluid pressure
and support : and the exact formulae it affords to practical men
in estimating expense, while their undertakings are executed with
systematic regard to permanent durability.

The Ninth Chapter treats of floatation, and the determina-
tion of the specific gravities of bodies immersed in fluids, com-
prehending therein some of the most interesting and important
principles of Hydrodynamic Science. There are two general
propositions embraced by this department of the philosophy of
fluids : — viz.

XXX INTRODUCTION.

1st. That when a body floats, or when it is in a state of
buoyancy on the surface of a fluid of greater specific gravity
than itself,

It is pressed uptvards by a force, whose intensity is
equivalent to the absolute weight of a quantity of the
fluid, of which the magnitude is the same as that portion
of the body below the plane of floatation, or the horizontal
surface of the fluid.

2dly. That if a solid homogeneous body be placed in a fluid
of a greater or less specific gravity than itself,

It will ascend or descend with a force which is equiva-
lent to the difference between its oivn weight and that of
an equal bulk of the fluid;

a proposition which is almost self-evident, but which leads to a
series of inferences, practically of vast importance in the me-
chanics of fluids.

Archimedes, the Sicilian philosopher, first established the fun-
damental laws of fluid equilibrium, and the specific gravity of
bodies immersed in fluids. Having determined the conditions
which are requisite to produce and measure the equilibrium of a
solid floating on a fluid, the philosopher readily perceived that

Two bodies equal in bulk, and immersed in a fluid
lighter than either of them, lose equal quantities of their
weight ;

or inversely, that when

Two bodies lose equal quantities of their weight in a
fluid, they are of equal volume ; —

this is the 7th Prob. of his first book — De Humido Insidentibus,
or Of bodies floating on a fluid. Mathematicians generally
suppose Archimedes employed this proposition to solve the well-
known problem proposed to him by Hiero, king of Syracuse,
who having employed a goldsmith to make a crown of pure
gold, and suspecting that the artist had not kept faith with him,
applied to Archimedes to discover the truth without injuring
the crown. The philosopher, it is said, laboured in vain at the

INTRODUCTION. XXXI

problem, till, going one day into the bath, he perceived that the
water rose in the bath in proportion to the bulk of his immersed
body ; it occurred to him at that moment that any other sub-
stance of equal 'size would have raised the water just as much,
though one of equal weight and of less bulk could not have
produced the same effect. He immediately felt that the solu-
tion of the king's question was within his reach, for taking two
masses, one of gold and one of silver, each equal in weight to
the crown, and, having filled a vessel very accurately with
water, he first plunged the silver mass into it, and observed the
quantity of water that flowed over ; he then did the same with
the gold, and found that a less quantity had passed over than
before. Hence he inferred that, though of equal weight, the
bulk of the silver was greater than that of the gold, and that
the quantity of water displaced was, in each experiment, equal
to the bulk of the metal. He next made a like trial with the
crown, and found it displaced more water than the gold, and
less than the silver, which led him to conclude that it was
neither pure gold nor pure silver.

This discovery by Archimedes, which after all is but the
application of the well-known axiom, that two bodies cannot
occupy the same space at the same time, has been considered
one of the most fortunate in the annals of science, for it has
led to great advances in the arts, and become the foundation of
chemical analysis ; just in the same way that his development
of the properties of floating bodies has formed the rudiments
of naval architecture, how much soever this branch of con-
structive mechanics may boast of its modern improvements.

In the Tenth Chapter, specific gravities and the methods of
weighing solid bodies in fluids are treated of; and the principle
here to be demonstrated is,

That when a solid body is immersed in a fluid of dif-
ferent specific gravity from itself, the weight which the
body loses will be to its whole weight, as the specific
gravity of the fluid is to the specific gravity of the solid.

In this chapter we have a full developement of that fine thought,
which rendered the truth of experiment an overmatch for the
craft of Hiero's goldsmith ; and the examples we have pro-

XXX11 INTRODUCTION.

duced, though not voluminous, fully show the different ways of
solving the ancient problem of Archimedes —

To find the respective weights of two known ingredients
in a given compound.

The principle enunciated above, may be popularly expounded
in the following manner. Every body placed on a surface of
water, has a tendency to sink by its own weight : it is, however,
resisted by a force equivalent to an equal bulk of the fluid, or
of as much fluid as will fill the space occupied by the body.
Should the body be heavier than the fluid, bulk for bulk, its
greater weight will cause it to descend, for the upward pressure
of the fluid will not prevent the descent. When, on the other
hand, the body is specifically, that is to say bulk for bulk,
lighter than the fluid, its pressure downwards will be less than
the upward pressure of the fluid at the same depth; conse-
quently, as the greater force necessarily overcomes the less,
and the upward pressure is the greater, the body will rise.
When the body and the fluid have the same specific gravity,
then equal masses of each are of the same weight, and the de-
scending force being equally balanced by the ascending force,
the body will float with its upper surface coincident with the
surface of the fluid, or in any other position whatever in which
it may be placed.

It is very obvious from these laws, that if, by any contrivance
or change, the specific gravity of a body can be so altered and
varied, as to be at one time greater, at another time less, and
then equal to the specific gravity of the fluid in which it is
placed, the said body will sink, or rise, or remain at rest, accord-
ing to the variations produced in its specific gravity. Lecturers
amuse their audiences with glass images, which, upon the
principle here adverted to, ascend or descend, or remain in mid-
water, at the pleasure of these philosophers.

The doctrine of the Equilibrium of Floatation, which appears
in Chapter XL, is as old as the days of Archimedes, who ex-
amines the conditions which are requisite to produce and pre-
serve the equilibrium of a solid floating in a fluid. He shows
that when a body floats in a state of equilibrium on the surface
of an incompressible fluid,

INTRODUCTION. XXX111

The centre of gravity of the whole body, and that of
the part immersed, must occur in the same vertical line, or
the line of pressure and the line of support must coincide;
and, secondly, that the magnitude of the body is to that of
the part immersed below the plane of floatation, as the
specific gravity of the fluid is to that of the floating body.

Of the truth of the doctrine which is here propounded, and,
let us hope, satisfactorily demonstrated in the sequel of our
work, we have a curious illustration afforded by an Arab ship-
builder in Java, whose task is thus described in GEORGE
EARL'S Eastern Seas: — "The largest merchant vessel in Java,
a ship about 1,000 tons burden, was built by an Arab merchant,
in a long but shallow river, which runs into the sea near Soura-
baya. As great expense is incurred by floating the timber in
rafts down the river, he determined to commence the work in
the forest itself, as he would thereby be enabled to select the
best trees for the purpose. He accordingly ascended the river,
accompanied by a sufficient number of workmen, conveying the
necessary materials, and commenced the undertaking about 80
miles from the sea. When the keel and the floor timbers were
laid, and a few of the bottom planks nailed on, he launched the
embryo vessel, and floated her gently down the river to a place
in which the water was deeper. Here the building was con-
tinued, until it became necessary to seek a deeper channel, and
in this manner the work proceeded, the vessel being floated
further down the river, whenever the water was found to be too
shallow for her to float, until at length, she was fairly launched,
half finished, into the sea, and completed in the harbour."

The operations of this ingenious orientalist proceeded upon
the truth stated in Inference 5, page 261, that if a body float
in equilibrio on the surface of a given fluid, and if the part
below the plane of floatation be increased or diminished by a
given quantity, the absolute weight of the body, (in order that
the equilibrium might still obtain,) must be increased or dimi-
nished by a weight which is equal to the weight of the portion
of the fluid that is more or less displaced, in consequence of
increasing or diminishing the immersed part of the body, or
that which falls below the plane of floatation.

XXXIV INTRODUCTION.

As his work proceeded, the Arab could calmly and skilfully
contemplate the effect of the antagonist forces directed to the
centre of gravity and the centre of buoyancy of his ship, and
survey her equilibrium as it might be permanent or instable ;
even though he knew nothing of the fine theory of M. Bouguer,
or the laborious calculations of the Swedish Admiral Chapman
or of Mr. Atwood, on the hull of the Cuffnells.

But we have other topics of equal practical importance with
the floatation of vessels in this chapter, as for example : 1st.
The consideration of a body floating in equilibrio between two
fluids which do not mix when the weights of the fluids respec-
tively displaced, are together equal to the weight of the solid
body which causes the displacement: 2dly. The construction
and application of the hydrometer, an instrument generally em-
ployed for detecting and measuring the properties and effects of
water and other fluids, such as their density, gravity, force and
velocity, which depends upon the principles explained and illus-
trated in the eighth proposition : 3dly. The hydrostatic balance,
an instrument by which we are enabled to measure the specific
gravities of bodies with great accuracy and expedition, whether
the bodies be in a fluid or a solid state.

A great many curious facts relating to the equilibrium of
floatation could have been here brought under the reader's con-
sideration; but these, as well as all popular illustrations of
natural philosophy, belong essentially to Somatology, or the
properties of matter, a subject which we could not amalgamate
with the calculations that illustrate the Mechanics of Fluids.

The Twelfth Chapter treats of the positions of equilibrium of
floating bodies, to determine which, from strict theory, is one of
the finest speculations in the whole range of natural philosophy :
to ascertain them, as we have done, by computation, involves
nothing intricate or repulsive, though the process is both laborious
and irksome. To construct them geometrically, demands a know-
ledge of principles higher than elementary. And although the
geometrical construction may truly represent the position which
the body assumes when floating in a state of equilibrium, it is the
application of numbers after all, which must determine the true
positions. The reason is this ; the specific gravities of the
solid and fluid bodies, which are always elements of the in-

INTRODUCTION. XXXV

quiry, cannot be represented by lines ; but having once obtained
by computation, the dimensions of the extant and immersed
portions of the body, the sides of which are always given in the
question, we can easily exhibit the geometrical construction.
The method of proof, by calculation, which we have applied to
this part of our work, seems to leave nothing to be added to an
elegant branch of the Mechanics of Fluids, so highly important
in the practice of naval architecture.

In the Thirteenth Chapter, we have considered the stability
of floating bodies and of ships. The subject of stability is the
same to whatever form of floating body it may be referred,
whether the body be a ship driven by wind or steam, logs of
wood, or masses of ice, and it consists entirely in resolving the

equation x = S sin. d>. The determination of the se-
ra

veral quantities of which this equation consists, depends entirely
upon calculations drawn from the particular circumstances of
the individual case under consideration ; and these circum-
stances as referred to a ship, it is impossible to assign by esti-
mation ; they must be obtained by actual measurement, and when
they have been obtained in this manner, they are to be inserted
in the above equation, to obtain the measure of stability. The
investigation of this subject is both laborious and intricate, but
from what we have done in Problems LXI. and LXIL, with
their subordinate examples, it may become intelligible to the
general reader. The mathematician who has consulted the
writings of the Swedish Admiral CHAPMAN, and the scientific
investigations of ATT WOOD, knows well that in considering the
properties of a vessel, the orderly arrangement requires that we
should treat, First of stability, or the power a vessel has of
resisting any change of position when afloat. Secondly, the
forms having stability which have the least resistance, and are
therefore best adapted for speed. Thirdly, the different methods
of propelling ships ; and Fourthly, the construction for strength.
But our inquiries are much more limited in this Treatise, and
might conveniently end with the exposition of the equation of
stability. We have, however, carried the subject a little farther,
and considered it in reference to steam navigation, in order to
point out that the stability of a ship is greatly increased, byaug-

XXXVI INTRODUCTION.

meriting the lateral dimension of the water line ; for the easiest
and most advantageous way of obtaining stability is by a large
area of floatation, and great fulness between wind and water ;
or, which is the same thing, by keeping the centre of gravity of
the displacement at the least possible distance below the water's
surface, in order to obtain the maximum of stability and the
fastest rate of sailing : and it will not differ much from the truth
to assume the cross section of the vessel, as of the form of a
parabola. In this species of figure, the stability and capacity
both increase as the ordinate becomes of a higher power ; but a
greater breadth is necessary in proportion to the vertical height
of the hull to give stability. The breadth, however, should be
every where in the same ratio to the depth, to render the sta-
bility equal throughout the length, or so that the vessel will
undergo no strain from change of position by pitching or rolling
in a boisterous sea.

The distinguishing characteristic of Chapman's works on
ship-building, is the application of the inductive method of
philosophy to the different parts of this subject, to found a
theory on experimental results, and where data failed, to con-
duct his investigations on the acknowledged principles of me-
chanics, and subject his conclusions to the test of observation
and experiment. His wrorks have never been surpassed ; and
in the treatise on ships of war, he collected and gave in detail
all the data which affected the qualities of ships, calculated
their effects under different circumstances, and determined on
theoretical principles, deduced from his experience, the dimen-
sions and forms of all ships of war, from a first-rate to the
smallest armed vessel. Their calculated elements are collected
in tables, and drawings of all the ships constructed agreeably
to these elements complete the work, which the reader will
find translated by MM. Morgan and Creuze, Naval Architects,
in the Papers on Naval Architecture, published about 1830.

Next to Chapman's, must be ranked the Treatise of Leonard
Euler, on the Construction and Properties of Vessels. The
Calculations relative to the Equipment and Displacement of
Ships and Vessels of War, by John Edi/e, show by tables and
plates, every element and material belonging to the British
navy.

INTRODUCTION. XXXVH

The Fifteenth Chapter embraces cohesion and capillary
attraction, — subjects replete with many curious speculations,
especially in our investigations of the phenomena of fluids.
Whatever may be the cause of fluidity, we know that ice
becomes water if a certain degree of he#t be applied to it, and
steam if more heat be used. Whether therefore, caloric or motion
be the cause of fluidity, we know that in the first instance of
the case we have cited, the atoms are fixed in crystals — in the
second they are thrown into intestine motion — and in the third
state they are forced asunder with an amazing expansive force.

Philosophers have usually assumed, that the particles of
fluids, since they are so easily moved among one another, are
round and smooth. This supposition will account for some
circumstances belonging to fluids, as, if the particles are round,
there must be vacant spaces between them, in the same manner
as there are vacuities between cannon balls when piled toge-
ther ; between these balls smaller shot may be placed, and be-
tween these, others still smaller, or gravel, or sand, may be
diffused. In a similar manner, a certain quantity of particles
of sugar can be taken up in a quantity of water without in-
creasing the bulk ; and when the water has dissolved the sugar,
salt may be dissolved in it, and yet the bulk will not be sensibly
augmented ; and admitting that the particles of water are
round, this is easily accounted for. Indeed the universal law of
gravitation, by which the constituent parts of all bodies mutually
attract each other, will cause all such as are fluid, and do not
revolve on their own axis, to assume spherical forms. Others
have supposed, that the cause of fluidity is the mere want of
cohesion of the particles of fluids, which in small quantities,
and under peculiar circumstances, arrange themselves in a
spherical manner, and form drops.

Fluids are subject to the same laws with solids. The parts
of a solid are so connected as to form a whole, their weight is
concentrated in a single point, called the centre of gravity : but
the atoms of a fluid gravitate independently of each other, and
press not only like solids perpendicularly downwards, but also
upwards, sideways, and in every direction. To the flexibility
and cohesion of their particles, is owing the singular property
which fluids possess of forming themselves into globules, and of

XXXVlll INTRODUCTION.

remaining heaped up above the brims of vessels ; and to their
attraction of cohesion, may be referred many phenomena in
evaporation and solution, their spontaneous ascent in capil-
lary tubes, whether natural or artificial, the motion of the
various juices through .animal bodies and vegetables, of water
through layers of ashes and sand or the rocky strata of the
earth and its ascent between plates of glass; to this attraction
may be referred solid bodies dissolving in fluids, whose first
colour or appearance is not changed, or changed without sen-
sible augmentation of the volume ; the mutual action of bodies
in contact with each other exhibiting this attraction, as when
dry salt of tartar is exposed to the air, it becomes fluid ; the
attraction of cohesion evinced in the process of evaporation, as
when the warm air of a room is crystallized on the panes of
glass during a cold night. We, however, are employed in con-
sidering the cohesion of water, which is known to be a com-
pound of hydrogen and oxygen, in the proportion of 15 parts
of the former, and 85 of the latter. Now this oxygen, which
exists in so large a proportion in water, makes exactly one-fourth
part of the atmospheric air which all animals breathe. It is
the pure part of the air, for the nitrogen or azotic gas which
exists in air, in the proportion of three-fourths, is incapable of
sustaining animal life or combustion for a single instant. The
atmosphere contains besides various supplementary matters, but
water is the most abundant, being there found in its different
states of cloud, mist, rain, dew, snow — answering a thousand
useful purposes in the great laboratory of nature, so that upon
the whole there is a perfect balancing of actions, preserving
the atmospheric mass in a uniform state, constantly fit for its
admirable purposes of animal and vegetable existence. The
sea-water, however, contains besides hydrogen and oxygen, a
solution of muriate of soda, or table salt, which probably
adapts this fluid for the purposes of animal life ; at all events
preserves the ocean from putrefaction. That the oxygen of the
water does not by cohesion or absorption swallow up the
oxygen of the atmosphere, and leave the earth to be surrounded
with a covering of deadly azotic gas, is perhaps to be accounted
for by the general laws of electrical attraction and repulsion,
which as they respect the physical constitution of these two

INTRODUCTION. XXXIX

fluids, preserve a perfect equilibrium between both and to each
its own due proportion of the life-giving gas, either as an
elastic or a non-elastic fluid. And it is, perhaps, owing to this
circumstance operating through the principle of specific gra-
vities, that the barometer — the prophet of the weather, indicates
the changes which diversify the climate of our earth. When
the atmosphere becomes surcharged with water it falls as rain,
and the weight and bulk of the mass being diminished, the
rising column of mercury presages serene and dry weather, as
previously the falling barometer had prognosticated wind and
rain. Our inquiries cease the moment we approach the limit,
which separates chemical analysis from the mechanics of fluids.

From the time of Archimedes till the age of Pascal,* the
annals of scientific discovery present no improvement in hydro-
statics. Pascal has the merit of discovering the pressure of
the atmosphere, and his treatise on the Equilibrium of Liquids
raised hydrostatics to the dignity of a science. The midnight
of barbarism, that for a thousand years had brooded over the
discoveries of the Sicilian philosopher, and had concealed the
Commentary of Sextus Julius Frontinus on the Aqueducts of
Rome, fled before the genius of Pascal and the powers of
Newton's mind ; the former, in the most perspicuous and simple
manner, demonstrating and proving by experiments the laws of
fluid equilibrium ; and the latter expounding the oscillation of
waves, a subject the most refined in Hydrodynamic science,
which, from that time, counts among its votaries the engineers and
philosophers of Italy, France, Sweden, Germany, and Britain.

It is proper here to state, that we believe in the compres-
sibility of water ; but we hold it true that for all general opera-
tions in the mechanics of fluids this compressibility is so
small as not to occasion any error in the numerous and varied
formulae, from which we have drawn practical rules for the

* Pascal gave proof of his skill in hydrostatics, by the celebrated well which he
dug at Port Royal des Champs, about six miles from Versailles. The well still exists,
in the midst of the farm yard of Les Granges ; but its machinery, by which a child
of ten years old could with ease and safety draw up water, is now no more. On
one side of the farm yard is a hovel, in which that good man studied during his
visits to Port Royal- a place rendered famous also by the name of the devout
Arnauld D'Andilli.

Xl INTRODUCTION.

solution of such questions as may engage the attention of our
readers.

LESLIE computes that air would become as dense as water
at the depth of 33 j miles ; it would even acquire the density
of quicksilver at a further depth of 163| miles ; and he hence
concludes with the probability that the ocean may rest on a
bed of compressed air. Water at the depth of 93 miles would
be compressed into half its bulk ; at the depth of 362.5 miles
it wrould acquire the ordinary density of quicksilver. Even
marble itself, subjected to its own pressure, would become twice
as dense as before, at the enormous depth of 283.6 miles. But
air, from its rapid compressibility, would sooner acquire the
same density with water, than this fluid would reach the con-
densation of marble.

For the coincidence of air and water the depth is 35.5 miles ;
for equal densities of water and marble 172.9 miles. At the
depth of 395.6 miles, or one-tenth the radius of the earth, air
would attain the density of 101960 billions ; while at the same
depth water would acquire but the density of 4.3492, and marble
only 3.8095. At the centre of the earth, the density of air would
be expressed by 764 with 166 ciphers annexed ; while water
would be condensed three millions nine thousand times its bulk
at the surface of the ocean ; and marble would acquire the
density of 119. The inference is, that if the structure of our
globe were uniform, and its mass consisted of such materials as
we are acquainted with, its mean density would far surpass the
limits assigned by Astronomy. Now both Dr. Maskelyne and
Cavendish nearly concur in representing the mean density at
only about five times greater than that of water. Leslie is
thence of opinion, that our planet must have a vast cavernous
structure, the crust of which, for aught we know to the
contrary, may cover some very diffusive medium of astonishing
elasticity, as light, which when embodied constitutes elemental
heat or fire.*

* Elements of Natural Philosophy, vol. i. pp. 447—457, second edition.

MECHANICS OF FLUIDS

CHAPTER I.

DEFINITIONS AND OBVIOUS PROPERTIES OF WATERY FLUIDS, WITH
THE PRELIMINARY ELEMENTARY PRINCIPLES OF HYDRODYNA-
MICS, FOR ESTIMATING THE PRESSURE OF INCOMPRESSIBLE
FLUIDS.

1 . THE phenomena of Hydrodynamics are those truths which explain
the peculiarities of equilibrium and motion among- fluid bodies, espe-
cially those that are heavy and liquid. As that branch of natural phi-
losophy which points out and explains the properties and affections of
fluids at rest, it comprehends the doctrine of pressure, specific gravity,
equilibrium, together with the circumstances attending the positions,
equilibrium, and stability of floating bodies, the phenomena of cohe-
sion and capillary attraction. And as that other branch of natural
philosophy which points out and explains the motions of such fluids as
have weight and are liquids, it investigates the means by which such
motions are produced, the laws by which they are regulated, the dis-
charge of fluids through orifices of various dimensions, forms, and
positions, — the motion of fluids in pipes, rivers, and canals, and the
force or effect they exert against themselves, or against solid bodies
which may oppose them. Hydrodynamics, therefore, from Greek
words signifying water and force, comprehend the entire science of
watery fluids, whether in a state of rest or of motion ; and this
science, practically considered, enables us to investigate and apply
a fruitful source of maxims and principles, upon which depend the
construction and efficiency of engines and machines employed in the
arts, manufactures, and domestic concerns of society, together with
that extensive class of mechanical combinations displayed in the more
delicate and important operations of HYDRAULIC ARCHITECTURE.

VOL. i. B *

2 ELEMENTARY PRINCIPLES OF FLUID PRESSURE.

2. A Fluid is a body so constituted, that its parts are all ready to yield
to the action of the smallest force or pressure, in whatsoever direction
it may be exerted. The following are some of the simplest and most
obvious properties of fluids.*

3. Every particle of a fluid presses equally in all directions, whether
it be upwards or downwards, laterally or obliquely; consequently, the
lateral pressure of a fluid is equal to its perpendicular pressure. The
converse of this is equally obvious, and is thus expressed.

4. Every particle of a fluid in a state of quiescence, is pressed equally
in all directions.

5. When a fluid is m a state of rest, the pressure exerted against the
surface of the vessel which contains it, is perpendicular to that surface.

6. When a mass of fluid is in a state of rest, its surface is horizontal,
or perpendicular to the direction of gravity.

7. If two fluids which do not mix, are poured into the same vessel,
and suffered to subside, their common surface is parallel to the horizon ;
consequently the surfaces of fluids continue horizontal, when sub-
jected to the pressure of the atmosphere.

8. The particles of a fluid, situated at the same perpendicular depth
below the surface, are equally pressed.

9. When a fluid is in a state of rest, the pressure upon any of its
constituent elements, wheresoever situated, is equal to the weight of
a column of fluid particles, whose length is equal to the perpendicular
depth of the particle or element pressed ; consequently, the pressure
on any particle varies ^s its perpendicular depth, and in any vessel
containing a fluid in a state of rest, the parts that are deepest sustain
the greatest pressure.

These principles, which flow immediately from the conditions of
fluidity, are too simple and obvious to require demonstration, yet
nevertheless, the writers on hydrostatical science generally accompany
them with a sort of popular proof, which may be found in almost
every treatise that has appeared on the subject. But our immediate
object being to unfold the more important elementary principles, by
the resolution of a series of examples dependent upon one general
proposition, we have thought it unnecessary to exhibit the demonstra-
tions here (Note A). The general proposition is as follows : —

* Fluids are generally divided into two sorts, compressible and incompressible, or
elastic and non-elastic ; the latter of which, or incompressible and non-elastic fluids,
such as water, mercury, wine, &c., form the subject of the present article ; the dis-
cussion of the compressible and elastic fluids, is reserved for another place. The
compressibility of water is so small, that in all practical operations in mechanics
its bulk or mass may generally be considered unalterable : for at a thousand fathoms
depth it can only be compressed one-twentieth of its bulk at the surface.

ELEMENTARY PRINCIPLES OF FLUID PRESSURE.

PROPOSITION I.

10. When an incompressible and non-elastic fluid is in a state
of equilibrium, and subjected only to the action of gravity: —

The magnitude, or the intensity of pressure exerted by the
fluid, perpendicularly to any surface immersed in it, or other-
wise exposed to its influence, is measured by the weight of a
column of the fluid, whose base is eqval to the area pressed,
and whose altitude is the same as the depth of the centre of
gravity of that area beneath the upper surface of the fluid.

This is an elegant and most important proposition in the doctrine of
fluid pressure, and in order that the principle may be the more readily
perceived, and the demonstration the more easily comprehended, it
will be proper, in the first place, to exhibit and demonstrate an analo-
gous property, in reference to the common centre of gravity of a
system of bodies, or of the particles of matter of which the system is
composed.

The property which we have alluded to above, is noticed by almost
every writer on the principles of mechanical science, and it has at
various times received most beautiful and rigorous demonstrations ; it
may therefore, at first sight, appear superfluous to introduce it here;
but in order to bring the subject more immediately before the atten-
tion of our readers, we do not hesitate to repeat the process.

PROPOSITION (A).

11. If there be any system of bodies and a plane given in position
with respect to them : —

The distance of that plane from the common centre of
gravity of the system, is equal to the aggregate of the pro-
ducts, arising from multiplying each body into its distance
from the given plane, divided by the sum of the bodies.

The proposition just enunciated, being of the greatest use in many
departments of philosophical inquiry, and of essential importance in
establishing the truth of the hydrodynamic principle above specified,
we shall therefore bestow some attention on its illustration for the
purpose of rendering it as clear as possible, by connecting the steps
with separate diagrams, -and pursuing the reasoning, until we shall
have proceeded so far that the law of induction becomes manifest, and
from thence, the truth of the principle announced in the proposition.

To accomplish this purpose, let a and b, be two very small bodies

E 2 V

4 ELEMENTARY PRINCIPLES OF FLUID PRESSURE.

or particles of matter, supposed to be col-
lected into their respective centres of gravity,
and let A B c D be a smooth rectangular
plane or surface, placed in any position
with respect to the bodies a and b.

Connect a and b by the straight line
ab, and let m be the place of their

common centre of gravity; draw the straight lines a p, m q and
br respectively perpendicular to the plane A B c D, and consequently
parallel to one another ; join pr, then because the points a, m, b are
situated in a straight line, the points p, q, r are also in a straight line,
and therefore p r will pass through the point q.

Through mt the common centre of gravity of the two bodies a and
b, draw st parallel to pr, meeting br in s, and pa produced in t ;
then the triangles ami and bms, are similar to one another; but
by the property of the lever, we have

a : b :: bm : am,
and by similar triangles, it is

bm : am : : bs : at;
therefore, by the equality of ratios, we obtain

a : b : : bs : at ;

from which, by equating the products of the extreme and mean terms,
we get

a X at — bx bs.

Now, it is manifest by the construction, that at — pt—pa, and
b s — r b— r s; therefore, by substitution, we obtain

a (p t—pa)=b (rb — rs)',

but by reason of the parallels p r and t s, the lines p t and r s are
respectively equal to m q ; hence we have

a (m q —p a) m: b (r b — m q),

and from this, by collecting the terms and transposing, we get
(a + b) mq — a X p a ' -f- b X r b,
and finally, by division, we obtain
a xa b X rb

- -

a -f- b

COROL. Here then, the truth of the proposition is manifest with
respect to a system composed of only two bodies ; that is,

The distance of the common centre of gravity from the
plane to which the bodies are referred, is equal to the sum of
the products, arising by multiplying each body into its dis-
tance from the given plane, divided by the sum of the bodies.

ELEMENTARY PRINCIPLES OF FLUID PRESSURE. 5

12. Again, let a, b and c, be a system of three very small bodies or
particles of matter, any how situated
with respect to the plane A B c D, and
connected together by the straight
lines a b, be and a c ; and suppose
the two bodies a and b to be collected
into their common centre of gravity
at the point m.

Join the points m and c by the
straight line me, and let n be the place of the common centre of
gravity of the three bodies a, b and c ; draw the lines mq,nu and
cv parallel to each other, and respectively perpendicular to the plane
A B c D; join qv, and because the points m, n and c are situated in
the straight line m c ; it follows, that the points q, u and v must also
occur in a straight line ; consequently, q v will pass through the
point u.

Through n, the common centre of gravity of the three bodies o, b
and c, draw st parallel to qv, meeting mq in t and vc produced in
s ; then are the triangles m n t and ens similar to one another ; but
by the property of the lever, and because the body at m is equal to
a 4- 6, we obtain

a ~\- b : c : : c n : m w,
and by similar triangles, we have

c n : m n : : c s : m t ;
therefore, by the equality of ratios, we get

a -|- b : c : : c s : m t ;

consequently, by equating the products of the extreme and mean
terms, we shall obtain

(a -f b) X m t zz c X c s ;
now mt — mq — t q, and c s—v s— vc; hence we get

(a -f- b) (mq — tq) — c(ys—vc).

But it is manifest by the construction, that t q and v s are each of
them equal to nu ; therefore, by substitution we have

(a -|- b (mq — n «) zr c (nu— v c) ;
therefore, by collecting the terms and transposing, we get

(a + b + c) Xnu=(a+b) Xmq + cXvc;
now, it has already been shown in the case of two bodies, that
a X^a-f b X rb

n<*= ^+T~

therefore, by substituting this value of m q in the step immediately
preceding, we shall obtain

ELEMENTARY PRINCIPLES OF FLUID PRESSURE.

(a 4-6-f- c) Xnu — aXp

and finally by division, we have
_aXpa-\-bXrb-\-cXvc
(a + b + c)

Now, n u is the distance of the common centre of gravity of the
three bodies a, b and c, from A B c D the plane to which they are
referred ; hence again, the truth of the proposition is manifest, and
if another body were added to the system, a similar investigation
would exhibit the same law, and thus we might proceed to any extent
at pleasure, the nature of the induction being fully disclosed.

COROL. If therefore, we suppose the system to be constituted of an
indefinite number of small bodies or particles of matter, it will become
assimilated to a fluid mass, and consequently, the proposition which
we have just demonstrated in reference to the centre o'f gravity, is
identified with the well-known theorem for estimating the pressure of
fluids ; to which subject we must now return.

13. Resuming therefore, the conditions specified in Proposition I
preceding, let us suppose that ABCD, de-
notes a vertical section of a reservoir full of
water, E and F representing the corresponding
sections of the walls or embankments by
which it is contained ; then, since the fluid
is supposed to be quiescent or in a state of
equilibrium, it follows, that the surface AB is parallel to the horizon.

Let b df hkm\>e the portion of the containing section or boundary,
on which the pressure exerted by the water is required to be investi-
gated, and conceive it to be constituted of an indefinite number of
minute bodies or particles of matter, placed at infinitely small dis-
tances from one another, or so near, that their aggregate or sum shall
make up the entire area which forms the subject of our investigation.

Suppose the points b, d, f, A, k and m, to be so many individual
particles of the surface pressed, and through the points thus assumed
draw the vertical lines b a, dc,fe, hg, ki and m /, which lines are
severally in the direction of gravity, and consequently perpendicular
to the surface of the fluid, indicating by their lengths, the respective
depths of the several bodies of which our immediate system is corn-

But according to Proposition I, the pressures exerted by the fluid
on the particles 6, d,f, h, k and m, are respectively represented by
the products

b X ba, dXe?c,/X fe, h X h g, k X k i and m X m I,

ELEMENTARY PRINCIPLES OF FLUID PRESSURE. 7

and the aggregate or sum of these products becomes
p — b X ba + d X dc+fxfe + h X hg -\- k X ki + mX ml,

where p denotes the sum of the computed pressures.
Now, it is manifest, from what we have demonstrated in Proposition
(A), respecting the centre of gravity of a system of bodies, that

The sum of the products, arising from multiplying each
body into its distance from a certain plane given in position,
is equal to the sum of the bodies, drawn into the distance of
their common centre of gravity from that plane.

Let therefore, the particles b, d,f, h, k and m be considered as a
system of very minute bodies, and let the surface of the fluid denote
the plane given in position, to which the system is referred ; then, if
G be the place of the common centre of gravity of that system, put
71 G == 3, and we shall obtain

8 (b+d+f+h+k+m) = b.ba+d.dc+f.fe+h . hg+k . ki+m.ml.

But we have seen above, that the sum of the products on the right
hand side of the equation, expresses the aggregate pressure on the
several points of the containing surface, to which the present step of
the inquiry refers, and that pressure we have briefly represented by the
symbol p ; therefore we have

and this expression implies, that the pressure exerted by a fluid, on
any number of points of the surface that contains it,

Is equal to the sum of the points, drawn into, the perpendi-
cular distance of their common centre of gravity below the
upper surface of the fluid.

Now, it is evident, that the same law would obtain if another point
were added to the system, and even if the number of points were to
become indefinite, or such that their aggregate or sum shall be essentially
equal to the area pressed, the law of induction would remain the
same ; consequently, if a denote the sum of the material points, or
particles of space in the surface on which the fluid presses ; then we
shall have

p = Za. (1).

This equation supposes, that the specific gravity of the fluid by
which the pressure is propagated, is represented by unity, which cir-
cumstance only holds in the case of water; therefore, let s denote the
specific gravity of any incompressible fluid whatever, and the general
form of the equation becomes

p=$as. (2).

8 ELEMENTARY PRINCIPLES OF FLUID PRESSURE.

Now, it is obvious, that the expression 5 a s indicates the weight of
a column of the fluid, the area of whose base is a, perpendicular alti-
tude S, and the specific gravity s ; hence the truth of the proposition
is manifest.

COROL. From what has been demonstrated above, it appears, that
whatever may be the form of the surface on which the fluid
presses, if its area, and the position of its centre of gravity can be
ascertained, the intensity of pressure which it sustains, is from thence
assignable.

The truth of the proposition being thus established, we shall proceed
to deduce from it a few of the most useful and obvious inferences.

14. INF. 1. If different planes be immersed perpendicularly, hori-
zontally, or obliquely, in fluids of different specific gravities : —

The pressures upon those planes perpendicularly to their
surfaces, are as their areas, the perpendicular depths of their
centres of gravity, and the specific gravities of the fluids
jointly.

15. INF. 2. If different planes be immersed perpendicularly, hori-
zontally, or obliquely in the same fluid : —

The pressures upon those planes perpendicularly to their
surfaces, are as their areas, and the perpendicular depths of
their centres of gravity.

16. INF. 3. If a plane surface of given dimensions be parallel to
the surface of the fluid in which it is immersed : —

The pressure sustained by the plane, in a direction perpen-
dicular to its surface, varies directly as its vertical depth
below the upper surface of the fluid.

17. INF. 4. If a plane surface of given dimensions be any how
inclined to the surface of the fluid in which it is immersed : —

The pressure sustained by the plane, in a direction perpen-
dicular to its surface, varies directly as the vertical depth of
its centre of gravity, below the upper surface of the fluid.

18. INF. 5. If any number of planes of equal areas be immersed
in the same fluid, and have their centres of gravity at the same vertical
depth below the surface : —

The pressures which they sustain are equal to one another,
whatever be their form, and whatever be their position with
respect to the surface of the fluid.

19. INF. 6. If any plane surface revolve about its centre of gravity,
which remains fixed in position : —

ELEMENTARY PRINCIPLES OF FLUID PRESSURE. 9

The pressure which it sustains in a direction perpendicular
to its surface, will be the same at every point of the revolution
.-, as if it remained constantly horizontal.

20. INF. 7. If the perpendicular pressures upon a given surface be
equal, when it is immersed in two fluids of different densities : —

The perpendicular depths of the centres of gravity below
the surface, will vary inversely * as the densities or specific
gravities of the fluids.

21. The above inferences are immediately deducible from the
general proposition, but it is probable that the last may require a
little illustration ; for which purpose —

Put p •=. the pressure sustained by the plane in both the fluids,
a~ the area of the plane or the surface pressed,
s — the density or specific gravity of one of the fluids,
d = the depth at which the given surface is immersed in it,
/ — the density or specific gravity of the other fluid,
and 5 = the depth of immersion.

Then, according to the principle indicated by the general equation (2),
we have, in the case of the first fluid,

p—das,
and in the case of the second fluid, it is

p :zr 5 a sf ;

but according to the conditions of the question, these expressions are
equal to one another, for the pressure is the same in both cases ; con-
sequently by comparison, we have

d a s ~ S a /,
and this, by suppressing the common factor, becomes

ds = W;

therefore, by converting this equation into an analogy or proportion,
we shall exhibit the precise conditions of the inference ; hence, we have

d : a : : / : s.

* One quantity is said to vary inversely as another, when of two quantities the
one increases as the other decreases.

CHAPTER II.

OF THE PRESSURE OF NON-ELASTIC FLUIDS UPON PHYSICAL
LINES, RECTANGULAR PARALLELOGRAMS CONSIDERED AS INDE-
PENDENT PLANES IMMERSED IN THE FLUIDS, AND UPON THE
SIDES AND BOTTOMS OF CUBICAL VESSELS, WITH THE LIMIT TO
THE REQUISITE THICKNESS OF FLOODGATES.

1. OF THE PRESSURE OF FLUIDS ON PHYSICAL LINES.

THE principle established in the general proposition enables us now
to proceed with the resolution of a numerous class of curious and
important problems, which will be found of the greatest practical
utility, in all cases in which the pressure of watery fluids is concerned.
These problems we shall accompany by examples, which will unfold
their geometrical and analytical character, and leave no truth in the
phenomena of this branch of hydrodynamics unrevealed.

PROBLEM I.

22. A physical line,* of a given length, is obliquely immersed
in an incompressible fluid in a state of equilibrium, in such a
manner that its upper extremity is just in contact with the
surface ; —

It is required to determine what pressure it sustains, the
angle of obliquity being a given quantity.

Let ABC, represent a vertical or upright section of a lake or
pool of stagnant water, confined by the walls or embankments of
which E E is a vertical section, and let A B be
the | surface of the water, supposed by the
problem to be in a state of equilibrium.

In A B take any point a, and at the point
a thus assumed, immerse the line a b of the
given length, and tending downwards at the
given inclination or angle b a A.

* A physical line is that which belongs to, or exists in nature, and is so called to
distinguish it from a mathematical line, which exists only in the imagination.

OF THE PRESSURE OF FLUIDS ON PHYSICAL LINES. 11

Bisect a 6 in m, and through the point m draw mn perpendicular
to AB, the surface of the fluid; then, because the centre of gravity
of, a straight line is at the middle of its length, m is the place of the
centre of gravity, and nm its perpendicular depth below the surface
AE; through b draw the straight line be parallel to mn, and cb is
the perpendicular depth of the lower extremity at b.

Put Z zr a 6, the length of the line whose upper extremity is at a,
d~=.nm, the perpendicular depth of the centre of gravity,
$ ~ b a c, the angle of inclination, or the given obliquity.
Then, because m is the centre of gravity of the straight line a b, we
have a m zz \ I, and by the principles of Plane Trigonometry ,we obtain

rad. : sin. (j> : : \ I : d,
and since the tabular radius is expressed by unity, we get

d— \ I sin. <j>.

Now, the whole pressure which the line sustains in a direction
perpendicular to its length, according to the second inference pre-
ceding,

Is proportional to its area, drawn into the perpendicular
depth of its centre of gravity below the upper surface of the
fluid.

But the area of a physical line is simply equal to its length ;
therefore, if the symbol p denote the pressure, and s the specific
gravity of the fluid by which it is propagated, we shall have

p = isl*sm.<t>. (3).

and this, in the case of water, where the specific gravity is expressed
by unity, becomes

p~ JZ*sin. 0.

23. This equation, as well as the more general one from which it is
derived, is sufficiently simple in its form for practical application;
but in order that nothing may be omitted, which tends to render the
subject intelligible to our readers, we shall in this, and in all the
succeeding formulae of a practical or general nature, draw up a rule,
describing the manner in which the several steps of the process are to
be performed ; pursuant to this plan, therefore, the rule for the present
case will be as follows : —

RULE. Multiply the square of the length by half the specific
gravity of the fluid, and again by the natural sine of the
angle of inclination, and the product will express the required
pressure on the line in the oblique position.

24. EXAMPLE 1. A physical line whose length is 36 feet, is im-
mersed in a cistern of water, in such a manner that the upper extremity

12 OF THE PRESSURE OF FLUIDS ON PHYSICAL LINES.

is just in contact with the surface, and the other inclining downwards
in an angle of 67° 35'; what pressure does the line sustain, supposing
the fluid in which it is placed to be in a state of equilibrium ?

Here by the question, the fluid in which the line is supposed to be
immersed is water, of which the specific is unity; consequently,
according to the rule, we have

p zz 36 X 36 X \ X sin. 67° 35' ;

but by the Trigonometrical Tables, the natural sine of 67° 35' is
.92444 ; hence we get

p=\296 X J X .92444 = 599.03712.

In this case, however, the resulting pressure is only relative, the
absolute pressure being indeterminable, upon a line where length
merely is indicated and no breadth assigned ; the existence of surface
being indispensable for the expression of a determinate measure.

25. If the line were immersed perpendicularly in the fluid, or so
as to make a right angle with its surface, the equation (3) would
become transformed into

p=%sl*sin. 90°;
but by the principles of Trigonometry, we have

sin. 90° = 1 ;
hence, by substitution, we obtain

j.= J«P; (4).

and this, in the case of water, where the specific gravity is unity,
becomes

P=ii*.

Therefore, the relative pressure for a perpendicular immersion, on
the line, as given in the above example, is

p = 36*36* 1=648.

26. If the upper extremity of the line be not in contact with the
surface of the fluid, but placed as in the

annexed diagram, then the method of solu-
tion, and consequently the form of the re-
sulting equation, will be somewhat different.

Let A B be the surface of the water or
fluid in which the line is immersed, and

A B c D a vertical section, in whose plane the line a b is situated, E E
being the corresponding section of the walls or embankments by
which the fluid is contained.

Bisect the given line ab in m, and through the point m thus deter-
mined, draw m n perpendicular to A B, the surface of the fluid ; and
through a and b the extremities of the given line, and parallel to mn,
draw ad and be, and produce ba to meet AB in A, or in any other

OF THE PRESSURE OF FLUIDS ON PHYSICAL LINES. 13

point, according to circumstances ; then is mn the depth of the centre
of gravity of the line a b, below the surface of the quiescent fluid, and
ad, be are respectively the depths of its extremities, b AC being the
angle which the direction of the given submerged line makes with the
horizontal line A B.

Put d=ad, the depth of the upper extremity of the given line,
^zrmTi, the depth of the centre of gravity,
D in b c, the depth of the lower extremity,
I zz a by the length of the proposed line,
p zz the relative pressure upon it as propagated by the fluid,
and ^ zz b A c, the angle which the given line makes with the horizon.
Through a the upper extremity of the given line, draw ae parallel
to AB the surface of the fluid; then is the angle bae equal to the
angle b A c, and by the principles of Plane Trigonometry, we have

a b : b e : : rad. : sin. <f>;

but be is manifestly equal to be — ad; that is, 6ezzD — c?, and
according to our notation, ab — l; hence, the above analogy becomes

I : (D — c?) : : rad. : sin. <p,
or by putting radius equal to unity, we get

This equation enables us to determine the obliquity of the line,
when the perpendicular depths of its two extremities are given ; but
when it is required to determine the relative pressure from the same
data, we have only to observe, that mn the perpendicular depth of
the centre of gravity, is equal to half the sum of the depths of the two
extremities ; that is,

consequently, we obtain

Again, if the angle of inclination and the perpendicular depth of
one extremity of the line are given, together with its length, the per-
pendicular depth of the other extremity can easily be found ; thus,
suppose that da is the given depth, then, by the principles of Plane
Trigonometry, we have

b e zz I sin. 0 ;

but by addition, we obtain

b c~b e -\~ e c ; that is, D zz I sin. 0 -f" d ;

consequently, the perpendicular depth of the centre of gravity, is

£z= | I sin. <f> -\- d;
and the relative pressure becomes
/>zz J I* sin. p Id.

14 OF THE PRESSURE OF FLUIDS ON PHYSICAL LINES.

But the equation, in its present form, supposes the specific gravity
of the fluid to be expressed by unity, which only takes place in the
case of water ; in order, therefore, to generalize the formula, we must
introduce the symbol which denotes the specific gravity ; hence,
we obtain

pi= |/2 s sin. 0 -\- Isd;
or by collecting the terms, we get

p — ls(\ Zsin.^>-|-c?). (5).

27. This is the general form of the equation, on the supposition
that the perpendicular depth of the upper extremity of the line is
given; it however assumes a different form, when the depth of the
lower extremity is known ; for by Plane Trigonometry, we have

as above

b erz:£sin.0,

and by subtraction, we obtain

ec — bc — be; that is, dznp — / sin. 0 ;

therefore, the perpendicular depth of the centre of gravity is

3=iD — \ I sin. cf),

and consequently, the general expression for the pressure becomes
p=ls(o — \ I sin. 0). (6).

28. Therefore, the practical rule for each of these cases, when
expressed in words at length, is as follows : —

1. When the perpendicular depth of the upper end is given (5).

RULE. To half the length of the given line drawn into the
natural sine of the angle of inclination, add the depth of the
upper extremity ; then., multiply the sum by the length of the
line, drawn into the specific gravity of the fluid, and the pro-
duct will give the pressure sought.

2. When the perpendicular depth of the lower end is given (6).

RULE. From the perpendicular depth of the lower extre-
mity, subtract half the length of the given line drawn into the
natural sine of the angle of inclination ; then, multiply the
remainder by the length of the line, drawn into the specific
gravity of the fluid, for the pressure sought.

29. EXAMPLE 2. A physical line, whose length is 56 feet, is immersed
in a cistern of water, in such a manner that its upper extremity is at
the distance of 9 feet below the surface, and its direction making with
the horizon an angle of 58 degrees ; required the relative pressure on
the line, the water being in a state of quiescence ?

The natural sine of 58 degrees, according to the Trigonometrical
Tables, is .84805 ; therefore by the rule, we have

28 X .84805 + 9 = 32.7454, the perpendicular

OF THE PRESSURE OF FLUIDS ON PHYSICAL LINES. 15

depth of the centre of gravity ; then, finally, because the specific
gravity of water is unity, we have

p — 32.7454 X 56 = 1833.7424.

Let the length of the line and its inclination to the horizon remain
as above, and suppose the depth of the lower extremity to be 56.4908
feet ; then, by the rule for the second case, we have

56.4908 - 28 x .84805= 32.7454, the depth of the
centre of gravity, the same as above, from which the relative pressure
is found to be 1833.7424, as it ought to be.

30. If the line were immersed perpendicularly, or at right angles
to the horizon, then sin. 0 is equal to unity, and the formulse for the
pressure become

p=l8($l + d), and^ = Z5(D — J I),

where it is manifest, that the parenthetical expressions are equal to
one another, each of them expressing the perpendicular depth of the
centre of gravity, or the middle point of the given line.

PROBLEM II.

31. Two physical lines of different given lengths, have their
upper extremities in contact with the surface of an incompressible
and non-elastic fluid in a state of equilibrium : —

It is required to compare the pressures which they sustain
at right angles to their lengths, supposing them to be immersed
at given inclinations to the horizon.

Let A B c D, represent a vertical section of a vessel filled with
water, or some other incompressible and non-
elastic fluid, and suppose the lines a b and cdto
be situated in the plane of the section, in such a
manner that the upper extremities a and c are
respectively in contact with A B the surface of
the fluid, while their directions make with the
horizon the angles ba A and dc B respectively.

Through the points b and d, the lower extre-
mities of the lines ab and cd, draw be and df respectively perpen-
dicular to A B the surface of the fluid ; and through m and r, the
middle points of a b and cd, draw the lines mn and rs respectively
parallel to the perpendiculars b e and df; then are mn and rs the
perpendicular depths of the centres of gravity.

Put I zz a b, the length of the line whose upper extremity is a,
I' — cd, the length of that whose upper extremity is c;
d — nm, the perpendicular depth of the centre of gravity of the
line a b ;

16 OF THE PRESSURE OF FLUIDS OK PHYSICAL LINES.

3rz s r, the perpendicular depth of the centre of gravity of the

line e d ;

^>:n b a A, the inclination of the line a b to the horizon,
0'= d c B, the inclination of c d to the horizon, or to the line A B ;
j^nrthe relative pressure upon a b,
j/zzthe relative pressure upon cd,
and 5 — the specific gravity of the fluid.

Now, because the centre of gravity of a physical straight line is at
the middle of its length, we have

a m — \ I, and c r rz J /';

therefore, by the principles of Plane Trigonometry, we obtain from
the right-angled triangle a m n

d=% lsiu.(j>,
and from the right-angled triangle c r s we get

£ = | V sin.0.

consequently, the general expressions for the relative pressures on the
lines a b and c d, according to equation (5) are

p~ \ r s sin.0, and;/— J l'*s sin. 0',
from which, by comparison, we get

p :p' : : I* sin. 0 : Z'a sin. ^'.

INF. 1. Hence it appears, that the pressures on the lines, when
their directions make different angles of inclination with the horizon,
Are directly as the squares of the lengths, and the sines of
the inclinations jointly.

2. Where $=<£', that is, when the lines are equally inclined to the
horizon, whatever may be the magnitude of the inclination, then

p:p'::l'-.l";

therefore, when the lines are perpendicularly immersed, or when they
are equally inclined to the surface of the fluid, with which their
upper extremities are supposed to be in contact,

The pressures which they sustain perpendicular to their
lengths, are directly proportional to the squares of those
lengths.

3. Consequently, if two or more lines are similarly situated in the
same fluid, the relative pressures can easily be compared ; thus, for
example : —

Suppose two physical lines, whose lengths are respectively 36 and
56 feet, to be perpendicularly immersed in the same fluid, and having
their upper extremities in contact with the surface, or equally depressed
below it; then, the pressures sustained by these lines, are to one ano-
ther as the numbers 1296 and 3136 ; that is,

pip':: 362: 56' :: 1296 : 3136.

OF THE PRESSURE OF FLUIDS ON PHYSICAL LINES. 17

But when the lines are differently situated in the fluid, the compa-
rison of their relative pressures requires a more particular exemplifi-
cation; for which purpose take the following example.

32. EXAMPLE 3. Two physical straight lines, whose lengths are
respectively 18 and 27 feet, are immersed in the same fluid, in such
a manner that their upper extremities are just in contact with its
surface, and the angles which they make with the horizon are respec-
tively equal to 42 and 29 degrees ; what is the pressure on the longer
line, supposing that on the shorter to be expressed by the number
78.54?.

If we convert the preceding analogy for the oblique lines of different
inclinations into an equation, by making the product of the mean
terms equal to the product of the extremes, we shall obtain

Now, by assimilating the several quantities in this equation to the
lines in the foregoing diagram, and according to the conditions of the
question, it appears that p' is the required quantity, all the rest being
given ; therefore, let both sides of the equation be divided by /2 sin. 0,
and we shall obtain t

, _ p /* sin. <f
/2sin.</> '

But it is a well-known principle in the arithmetic of sines, that to
divide by the sine of any arc, is equivalent to multiplying by the
cosecant of that arc ; hence we have

p'=—(p sin. 0' cosec.</>).

Let therefore the numerical values, as proposed in the example, be
substituted for the respective symbols in the above equation, and we

shall obtain

.972

»' = — (78.54 sin. 29° cosec.42°) ;
182

now, the natural sine of 29°, according to the Trigonometrical Tables,
is .48481, and the natural cosecant of 42° is 1.49447 ; therefore, by
substitution, we get

if — ?Z!x 78.54 X .48481 X 1.49447 — 128 nearly ;

182

consequently the pressures on the inclined lines, are to one another as
the numbers 78.54 and 128; but had the inclinations been equal,
the comparative pressures would have been as 78.54 to 176.72 very
nearly.

VOL. i. c

2. OF THE PRESSURE OF FLUIDS THAT ARE NON - ELASTIC UPON
RIGHT ANGLED PARALLELOGRAMS CONSIDERED AS INDEPENDENT
PLANES IMMERSED IN FLUIDS.

PROBLEM III.

33. A right angled parallelogram is immersed in a quiescent
fluid, in such a manner, that one of its sides is coincident with
the surface, and its plane inclined to the horizon in a given
angle : —

It is required to determine the pressure perpendicular to the

plane, both when it is inclined to the surface of the fluid, and

when it is perpendicular to it, the nature of the fluid, and

consequently its specific gravity, being known*

Let A B c D represent a vertical section of a volume of incompressible

fluid in a state of equilibrium, of which A B E F is the surface, and

consequently parallel to the horizon ; let a b c d be a rectangular

plane immersed in the fluid, in such a manner that the upper side a b

coincides with the surface, and the plane abed is inclined to the

horizon in a given angle.

Draw the diagonal a c, which bisect
in m, and through m the centre of gra-
vity of the parallelogram, draw mn
parallel to ad or be, meeting ab the
line of common section perpendicularly
in the point n.

In the horizontal plane A BE F, and
through the point n, draw nr also at right angles to ab, and from
m the centre of gravity of the immersed plane abed, let fall the
perpendicular m r ; then is the angle m n r the inclination of the plane
to the horizon, and rm the perpendicular depth of its centre of gravity
below the upper surface of the quiescent fluid.
Put b = ab, the horizontal breadth of the immersed parallelogram,
/ zn a d or b c, the immersed length,
d-=rm, the perpendicular depth of the centre of gravity,
0 — mn r, the inclination of the plane to the horizon,
p = the pressure on the plane perpendicularly to its surface,
and 5 — the specific gravity of the fluid.

* By the pressure upon any plane or curvilineal surface, is always understood
the aggregate of all the pressures upon every point of those surfaces, estimated in
directions perpendicular to them at each point, no part heing lost by obliquity of
direction.

OF THE PRESSURE OF FLUIDS ON RIGHT ANGLED PARALLELOGRAMS. 19

Then, because the point m is at the middle of «e, and mn parallel
to ad, it follows, that m win \ I; and by reason of the right-angled
triangle m r n, we have, from the principles of Plane Trigonometry,

rm~d—^l sin.^ ;

consequently, the entire pressure upon the plane perpendicularly to
its surface, is expressed by

p ~\ b I* s sin. (f>.

This is manifest from Problem I. (art. 22), for \l sin. § expresses
the perpendicular depth of the centre of gravity, and b I the area of
the surface pressed ; therefore, the solidity of the fluid column is

\ I sin. 0 X bl—%bl* sin. 0,

and since s denotes the specific gravity of the fluid, the weight of the
column is

| I sin. <j>XblXs — %bl*s sin. <f> ;

but the perpendicular pressure upon the plane, is equal to the weight
of the fluid column ; therefore, we obtain

p~ J6Zsssin. 0. (7).

When the plane of the immersed rectangle is perpendicular to the
surface of the fluid, we have 0 in 90°, and sin.^> zz: 1 ; consequently,
by substitution, the above equation becomes

P — iblzs. (8).

These equations are sufficiently simple in their form for practical
application, and we shall show hereafter, that they are extremely
useful in many important cases of hydrostatical construction.

34. The practical rules derived from these equations, for determining
the pressure in the particular cases, may be expressed as follows.

1. When the plane is oblique to the horizon. (Eq. 7).

RULE. Multiply the square of the immersed length of the
plane, by the horizontal breadth drawn into the specific
gravity of the fluid, and again by the natural sine of the
angle of inclination,, and half the product will give the
pressure sought.

2. When the plane is perpendicular to the horizon. (Eq. 8).

RULE. Multiply the square of the immersed length of the
plane, by the horizontal breadth drawn into the specific
gravity of the fluid, and half the product will give the
pressure sought.

35. EXAMPLE 4. A rectangular parallelogram, whose sides are re-
spectively 1 8 and 3 feet, is immersed in a quiescent body of water, in
such a manner, that its shorter side is in contact with the surface,

c 2

20 OF THE PRESSURE OF FLUIDS

and its plane inclined to the horizon in an angle of 68 degrees;
required the pressure which it sustains, both in the inclined and the
perpendicular position ?

In this example the area of the parallelogram is 1 8 X 3 == 54 square
feet, and the longer side is that which is immersed downwards in the
fluid; therefore, according to the rule for the oblique position, the
solidity of the column by which the pressure is propagated, becomes
3 X 182 X s X 1 sin. 68° = 486 X s sin. 68°.

Now, in the case of water, the specific gravity is represented by
unity, and by the Trigonometrical Tables, the natural sine of 68
degrees, is 0.92718 ; consequently, by substitution, the pressure
becomes

^ = 486x .92718 = 450.60948;

the pressure here obtained, however, is estimated in cubic feet of
water ; but in order to have it expressed in a more appropriate and
definite measure, it becomes necessary to compare it with some
weight ; now, it has been found by experiment, that the weight of a
cubic foot of water is very nearly equal to 62 1 Ibs. avoirdupois;
therefore, the absolute pressure upon the plane, is

p — 450.60948 X 62.5 = 28163.0925 Ibs.

36. Let the dimensions of the plane remain as in the preceding
case, which condition is supposed in the example ; then, the pressure
on its surface, when perpendicular to the horizon, is

p — 3 x 18 X 18 X 1 X J =486 cubic feet

of water ; but we have stated above, that the weight of one cubic foot
is equal to 62 \ Ibs. ; therefore, we have

p — 486 X 62^ zz 30375 Ibs. ;

consequently, the pressures on the plane in the two positions, are to one
another as the numbers 450.60948 and 486, when expressed in cubic
feet of water ; but when expressed in pounds avoirdupois, they are as
the numbers 28163.0925 and 30375.

37. If the longer side of the rectangular parallelogram were coin-
cident with the surface of the fluid, while its plane is obliquely inclined
to the horizon ; then, the formula for the pressure perpendicular to its
surface becomes

p=. ±b*ls sin. <f>. (9).

But if the plane of the parallelogram, instead of being inclined to
the horizon, or which is the same thing, to the surface of the fluid,
were immersed perpendicularly to it ; then, 0 zz 90°, and sin. 0 zz 1 ;
hence, the formula for the pressure becomes

p =*#**. (10).

ON RECTANGULAR PARALLELOGRAMS. 2l

Therefore, by retaining the data of the preceding example, the
absolute pressure on the plane in the oblique position, is

p=3* X 18 x 62.5 X \ X .92718 = 4693.84875 Ibs.
But when the plane is perpendicularly immersed, the absolute
pressure on its surface is

p=3* X 18 X 62.5 X J = 5062Jlbs.

COROL. 1 . Hence, the pressures on the plane in the oblique and per-
pendicular positions, are to one another as the numbers 4693.84875
and 5062 \ ; but in order to compare the pressures under the same
conditions, when the shorter and longer sides of the parallelogram
are respectively in contact with the surface of the fluid, we have as
follows, viz.

2. When the shorter side of the parallelogram is horizontal, the
absolute pressure in the inclined position is 28163.0925 Ibs. ; but
when the longer side is horizontal, the absolute pressure is 4693. 84875
Ibs. ; consequently, the absolute pressures in the two cases are to
one another as 6 to 1 .

3. Again, when the shorter side of the parallelogram is horizontal,
the pressure in the perpendicular position is 30375 Ibs. ; and when
the longer side is horizontal, the pressure is 5062 1 Ibs. ; therefore, the
pressures in these two cases are to one another as 6 to 1 , the same as
before ; from which we infer, that the quantity of inclination affects
only the magnitude of the pressures, and that in so far as it changes
the position of the centre of gravity, but it has no effect upon the
ratio ; therefore, if the plane were to vibrate round its shorter and
longer sides respectively as axes, the pressures on its surface, in the
two cases, would be to one another in a constant ratio.

3. OF THE AGGREGATE PRESSURE EXERTED BY THE FLUID ON THE
IMMERSED PARALLELOGRAM, AND ON EACH 0¥ THE CONSTITUENT
TRIANGLES FORMED BY ITS DIAGONAL.

PROBLEM IV.

38. Suppose the parallelogram to be placed under the same
circumstances as in the preceding problem, and let it be bisected
by one of its diagonals : —

It is required to determine the aggregate pressure exerted
by the fluid, in a direction perpendicular to the surface of
each triangle into which the diagonal divides the parallelo-
gram, and to compare the pressures on the two triangles.

22 OF THE AGGREGATE PRESSURE OF FLUIDS

Let A B c D represent a vertical section of a mass or collection of
quiescent fluid, contained by the walls or embankments indicated by
the shaded boundary ; and let A B E F be the horizontal surface of the
fluid, with which one side of the immersed rectangle is supposed to be
coincident.

Now, suppose abdc, to be the immersed rectangle, and draw
the diagonal b d; then are a b d and bdc
the triangles, into which the parallelogram
abed is divided by the diagonal b d, and
for which the pressures are required to be
investigated.

Draw the diagonal a c, which divide into
three equal portions in the points m and n >
then are m and n respectively the centres of gravity of the constituent
triangles a b d and bdc.

Through the points m and n, and parallel to a d or b c, the immersed
sides of the figure, draw me and nf meeting a b perpendicularly in
the points e and/; then, through the points e and /thus determined,
and in the plane of the fluid surface, draw er and/s respectively
perpendicular to ab; then are the angles mer and nfs equal to
one another, and each of them is equal to the angle which the plane
of the immersed parallelogram makes with the surface of the fluid.

From m and n, the centres of gravity of the triangles a b d and
b d c, demit the lines m r and n s respectively perpendicular to e r and
fs; then are rm and sn the perpendicular depths of the centres of
gravity.

Put b — ab, the horizontal breadth of the immersed parallelogram,
/ = ad or be, the immersed or downward length,
d = rm, the perpendicular depth of the centre of gravity of the

triangle abd,
S ~ sn, the perpendicular depth of the centre of gravity of the

triangle bdc,

D— ac or ba, the diagonal of the parallelogram,
ty~mer, or nfs, the inclination of the plane to the surface of

the fluid,

Pzz the whole pressure on the parallelogram abed,
p m the pressure on the triangle abd,
y—the pressure on the triangle bdc,
and s — the specific gravity of the fluid.

Then, because the parallelogram abed is rectangular, the triangle

ON IMMERSED RIGHT ANGLED PARALLELOGRAMS. 23

udc is right angled at d; therefore, by the property of the right-
angled triangle, we have

aczz -v/ ad*
or by employing the appropriate symbols, we have

Dizrv/r-M2.

But, according to the construction, and by the nature of the centre
of gravity, we have

am"=z -i-ac, and an='%ac,

or symbolically, we obtain

am — -ly Z2-f 62, and an zr f^/^-j-fe2.

Now, by reason of the parallel lines em, fn, and be, the triangles
a em, afn, and a be, are similar among themselves ; consequently, by
the property of similar triangles, we have

ac : be : : an :fn : : am : em;

therefore, by separating the analogies, and employing the symbols,
it is

D:Z::|V~F+&2:/w,
and again, we have

D : I :: i\/^ + & : em'>

from these analogies, therefore, we obtain fn n: |7, and em~\l\
which is otherwise manifest by drawing the dotted lines mt and nu.

Now, in the right angled triangles erm and fsn, there are given
the hypothenuses em and fn, and the equal angles mer and nfs, to
find rm and sn, the perpendicular depths of the centres of gravity;
consequently, by Plane Trigonometry, we have, from the triangle
mer,

rad. : sin. $ : : $1 : d,
and from the triangle nfs, it is

rad. : sin. <j> : : %l : £,
and since radius is equal to unity, these analogies become

d-=.\l sin. 0, and £ zz f I sin, 0.

But according to Inf. 2, Proposition (A), the pressure sustained by
each triangle, in a direction perpendicular to its surface,

Is expressed by the product of its area, drawn into the
perpendicular depth of the centre of gravity.

Now, the area of each triangle is manifestly equal to half the area
of the given parallelogram, and by the principles of mensuration, the
area of the rectangular parallelogram is equal to the product of its
two dimensions ; that is, of the length drawn into the breadth ; there-
fore, we have for the pressure on the triangle « bd,

24 OF THE AGGREGATE PRESSURE OF FLUIDS

sn. 0,
and in like manner, the pressure on the triangle bdc is

p' ~\bl* sin. <f>.

These equations, however, express the pressures simply by the
magnitude of a fluid column, whose base is the area pressed, and
whose altitude is equal to the depth of the centre of gravity below the
upper surface of the fluid. In order, therefore, to have the pressures
expressed in general terms, the specific gravity of the fluid must be
taken into the account; in which case, the pressure on the triangle
abd becomes

p=±bl*ssm.<t>, (11).

and the pressure on the triangle bdc is

p'i=i&Z2ssin.;>. (12).

COROL. Hence it appears, that the pressure perpendicular to the
plane of a triangle, when its vertex is upwards and coincident with
the surface of the fluid, is double the pressure on the same triangle,
when its base is upwards, and placed under the same circumstances.

39. If the immersed plane be perpendicular to the surface of the
fluid, then <b nz 90°, and sin. ty ml; therefore, by substitution, the
preceding equations become

p = ±bl*s, andp'=ibr~s;

here again, the pressure in the one case is double the pressure in the
other, and the same thing will obtain, whatever may be the inclination
of the plane, provided only that a b coincides with the surface of the
fluid ; for then, the perpendicular depths of the centres of gravity will
vary in a given ratio.

When the immersed plane is a square, that is, when b and I are
equal to one another, the equations for the pressures in the oblique
position become

p :z: %bss sin. 0, and p' — ±b3s sin. 0,

and when the plane is perpendicular to the surface of the fluid, we
have

p = £ bs s, and p' — | b3s*

Since the aggregate pressure upon the plane is equal to the sum
of the pressures on the constituent triangles, the expression for the
aggregate pressure in the oblique position, becomes in the case of a
rectangle

P =p +pf ; that is, P z= ±bl*s sin. 0 -f ±bl*s sin. 0 = £bl*s sin. 0.
COROL. Hence it appears, that the pressures on the constituent
triangles and that on the entire plane, are to one another as the
numbers 1 , 2 and 3 ; and the same thing obtains in the case of a
square, whatever may be the inclination of the plane.

ON IMMERSED RIGHT ANGLED PARALLELOGRAMS. 25

40. The practical rules for calculating the pressures on the triangles,
as deduced from the equations (11) and (12) are as follows.

1. When the base of the triangle is coincident with the surface.

RULE. Multiply the square of the immersed length, or the
perpendicular of the triangle, by the base drawn into the
specific gravity of the fluid, and again by the natural sine of
the plane's inclination, and one sixth part of the product will
express the whole pressure upon the triangle. (Eq. 11).

2. When the vertex of the triangle is coincident with the surface.

RULE. Multiply the square of the perpendicular of the
triangle, by the base drawn into the specific gravity of the
fluid, and again by the natural sine of the plane's inclination,
and one third of the product will express the whole pressure
on the triangle. (Eq. 12).

41. EXAMPLE 5. A rectangular parallelogram, whose sides are
respectively 26 and 14 feet, is immersed in a cistern of water, in such
a manner, that its shorter side is coincident with the horizontal sur-
face ; what will be the pressure on each of the triangles, into which
the parallelogram is divided by its diagonal, supposing its plane to be
inclined to the surface of the fluid in an angle of 56° 35' ?

Here, by the rule, we have

p= 262 X 14 X .83469 X i= 1316.58436 cubic feet
of water; but one cubic foot of water weighs 62 J Ibs. ; therefore, to
express the pressure in Ibs., we have

p= 1316.58436 X 62J = 82286.5225 Ibs.

The pressure which we have just obtained, refers to that portion of
the parallelogram which has its base coincident with the surface of
the fluid; that is, to the triangle abd, and the pressure on the other
portion, or the triangle bdc, is determined as follows.

/z=:262 X 14 X .83469 X i = 2633. 16872 cubic feet of water;

or to express the pressure in Ibs. we have

p'= 2633. 1687 X 62.5= 164573.045 Ibs.

If the plane of the immersed parallelogram were perpendicular to
the surface of the fluid, the pressures on the triangles abd and bdc
would be respectively as follows.
p=:262X 14 X 62 J X i = 98583ilbs.,andy — 26* X 14 X 62|X

i= 197 166| Ibs.

COROL. The circumstance of the aggregate pressure on the paral-
lelogram, being equal to the sum of the pressures on the constituent
triangles, furnishes a very simple and elegant method of determining

26 OF THE AGGREGATE PRESSURE OF FLUIDS

the centre of gravity; which method, in so far as respects plane
figures of particular forms, may, in many instances, be very advan-
tageously applied.

It would be foreign to our present purpose to enter into a detail of
the method alluded to in this place ; nevertheless, for the satisfaction
of our readers, we shall briefly introduce it, not being aware that it
has been suggested by any other writer in ancient or modern times.

PROBLEM B.

42. The base and perpendicular of a right angled triangle
being supposed known : —

It is required to determine the position of its centre of
gravity, or that point on which, if the surface were supported,
it would remain at rest in any position.

Let ABC be the triangle given, of which it is required to determine
the centre of gravity.

Complete the parallelogram A B c D, by drawing the
dotted lines AD and DC; then, because the entire
pressure on the parallelogram ABCD, is equal to the
sum of the pressures on the triangles ABC and ACD;
it follows, that the pressure on the triangle ABC, is
equal to the difference between the entire pressure on
the parallelogram, and that on the triangle ACD; consequently, we
have, by retaining the foregoing notation,

p = P — pr ; that is,

But it has been elsewhere demonstrated, that the pressure on any
surface, is expressed by the area of that surface, drawn into the
perpendicular depth of its centre of gravity ; consequently, the per-
pendicular depth of the centre of gravity, must be equal to the
pressure divided by the area of the surface.

Now, in the present instance the pressure is known, and since by
the problem, the base and perpendicular of the triangle are given, its
area can easily be found.

Thus, the writers on mensuration have shown, that the area of a
triangle is equal to half the product of the base drawn into the per-
pendicular altitude ; consequently, if a be put to denote the area of
the triangle ABC, we shall have

therefore, by division, the perpendicular depth of the centre of gra-
vity, is

ON DIFFERENT SECTIONS OF PARALLELOGRAMS. 27

And in like manner it may be shown, that if the side BC were
horizontal, the perpendicular depth of the centre of gravity would be

d=±b;

therefore, take T.m and BW respectively equal to one third of BC and
BA, and through the points m and n, draw mG and no parallel to BA
and BC, meeting each other in the point G ; then is G the position of
the centre of gravity.

The intelligent and attentive reader will readily perceive that the
above determination is not legitimate, since it supposes the pressure
upon the triangle ABC to be given ; now, this pressure depends
entirely upon the position of the centre of gravity, and consequently,
the problem supposes the position of the centre of gravity of the
triangle ADC to be known; the principle, however, will be more
distinctly indicated when applied to other figures, where the above
determination may be admitted, without infringing on the precepts of
scientific propriety.

4. OF THE PRESSURE OF INCOMPRESSIBLE FLUIDS ON DIFFERENT
SECTIONS OF PARALLELOGRAMS PARALLEL TO THE HORIZON.

PROBLEM V.

43. A rectangular parallelogram is obliquely immersed in an
incompressible and non-elastic fluid, in such a manner, that one
side is just coincident with the surface : —

It is required to compare the pressure on the upper and the
lower portions, supposing the parallelogram to be bisected by
a line drawn parallel to the surface of the fluid.

Let A ED represent a vessel full of water, or some other non-elastic
and incompressible fluid, of which ABEF
is the surface, and suppose one side of the
vessel to be removed, exhibiting the fluid
and the immersed rectangle as represented
by A B c D and abed.

Bisect the parallelogram abed by the
straight lines ef and gh respectively parallel
to a b and ad\ then are abfe and efcd,
the portions on which the pressures are to be compared, and gh is the
line in which the centres of gravity occur.

28 OF THE PRESSURE OF INCOMPRESSIBLE FLUIDS

Draw the diagonals d/and/a meeting the straight linear A in the
points m and n ; then are m and n the centres of gravity of the respec-
tive portions into which the parallelogram is divided by the line ef.

Through the point g and in the plane of the fluid surface ABEF,
draw gsr at right angles to a b, and the angle mgr will be the incli-
nation or obliquity of the plane ; then, through the points m and n,
draw the straight lines mr and ns respectively perpendicular to gsr,
and rm and sn will be the vertical depths of the centres of gravity
below the upper surface of the fluid.

Put b=iab, the horizontal breadth of the given parallelogram,
/ zr ad or be, the immersed length tending downwards,
d~rm, the vertical depth of the centre of gravity of the lower

portion efcd,
$ ~ sn, the vertical depth of the centre of gravity of the upper

portion abfe,

(j>=:mgr, the inclination of the plane to the surface of the fluid,
Pn= the pressure on the whole parallelogram abed,
p — the pressure on the lower portion efcd,
and p — the pressure on the upper portion abfe.

Then, because the straight line gh is bisected in 0, and each of the
portions g © and h © respectively bisected in the points n and m ;
it follows that g n zz |, and gm — % of gh; that is

g n nz J I, and g m — j / ;
consequently, by the principles of Plane Trigonometry, we have

swrzrSzz \l sin. <j>, and rm — d=%l sin. 0;

therefore, since the area of each portion of the parallelogram is
expressed by \ bl, the pressure on each portion is as below, viz.
The pressure perpendicular to the surface a bfe, is

p'zz: ±blz sin. 0,
and the pressure perpendicular to the surface efcd, is

p m %bl* sin.0 ;

consequently, by comparison, the pressures on the upper and lower
portions of the parallelogram, are to each other as the numbers 1 and
3 ; that is

pT:p::l:3.

But according to the third problem, the aggregate pressure sustained
by the plane, in a direction perpendicular to its surface, is

consequently, the pressures on the two portions and on the whole
plane, are to one another as the numbers 1, 3 and 4.

ON DIFFERENT SECTIONS OF PARALLELOGRAMS. 29

In the preceding values of the pressure, it is supposed, that the
specific gravity of the fluid in which the plane is immersed, is repre-
sented by unity, which is true only in the case of water ; therefore, in
order to render the formulae general, we must introduce the symbol
for the specific gravity, and then the above equations become,

1. For the upper half of the parallelogram,

p'=z±bl2 s sin. 0. (13).

2. For the lower half of the parallelogram,

pz=f bl*s sin. 0. (14).

When the plane is perpendicularly immersed in the fluid, or when
0 — 90°, then sin. 0 — 1, and the equations (13) and (14) become
p' — \bl^s, andprzi&/25.

In which equations the co-efficients or constant quantities remain ;
therefore, the ratio of the pressure is not varied in consequence of a
change in the angle of inclination, the variation takes place in the
magnitude of the pressures only, and not in the ratio, the magnitude
increasing from zero, where the plane is horizontal, to its maximum
where the plane is perpendicular.

44. The practical rules for calculating the pressures, as derived by
the equations (13) and (14) are as follows.

1. For the pressure on the first, or upper half of the paral-

lelogram.

RULE. Multiply the square of the immersed length, by the
breadth drawn into the specific gravity of the fluid, and again
by the natural sine of the angle of elevation ; then, one eighth
part of the product will be the pressure sought. (Eq. 13).

2. For the pressure on the second, or lower half of the paral-

lelogram.

RULE. Multiply the square of the immersed length, by the
breadth drawn into the specific gravity of the fluid, and again
by the natural sine of the angle of elevation; then, three
eighths of the product will be the pressure sought. (Eq. 14).

45. EXAMPLE 6. A rectangular parallelogram, whose sides are
respectively 20 and 30 feet, is immersed in a cistern of water, in such
a manner, that its breadth or shorter side is just coincident with the
surface ; required the pressures on the upper and lower portions of
the plane, supposing it to be bisected by a line drawn parallel to the
horizon, the inclination of the plane being 59° 38' ?

Here, by operating according to the rule, we have
p — 30* X 20 X sin. 59° 38' X i ;

30 OF RECTANGULAR PARALLELOGRAMS DIVIDED INTO

but by the Trigonometrical Tables, the natural sine of 59° 38' is
.86281 ; therefore, we have

p'=i 302 X 20 x .86281 X i= 1941.3225,

and again, by a similar process we have
p= 302 X 20 X .86281 x 1 = 5823.9675.

Now, these results are obviously expressed in cubic feet of water,
for they are respectively equal to the solidity of a fluid column, whose
base is equal to one half the given parallelogram, and whose altitude,
in the one case, is expressed by \l sin. $ rz 7.5 x .86281, and in the
other by |J sin. <j>=: 22.5 X .86281 ; but the weight of one cubic foot
of water is .equal to 62J Ibs. ; consequently, the pressures expressed
in Ibs. avoirdupois, are

pf = 1941.3225 X 62.5 = 121332.65625 Ibs.
and;? =5823.9675 X 62.5 = 363997.96875 Ibs.
When the plane is perpendicular to the surface of the fluid, the
pressure is a maximum, and in that case, the respective pressures on
the two portions of the parallelogram, are

p' = 302 x 20 X 62.5 x I = 140625 Ibs.
and jo rz302 X 20 x 62.5 x |zr 421875 Ibs.

and the sum of these, is obviously equal to the whole pressure on the
plane ; hence we get

P = 140625 + 421875 == 562500 Ibs.

COROL. If the plane, instead of being immersed in the fluid, as we
have hitherto supposed it to be, should only be in contact with it, as
we may conceive the surface of a vessel to be in contact with the fluid
which it contains ; then, the pressure will be the same ; for the quan-
tity of pressure at any given depth upon a given surface, is always the
same, whether the surface pressed be immersed in the fluid or just in
contact with it, and whether it be parallel to the horizon, or placed in
a position perpendicular or oblique to it.

5. OF RECTANGULAR PARALLELOGRAMS IMMERSED IN NON -ELASTIC
FLUIDS, AND DIVIDED INTO TWO PARTS SUCH THAT THE PRESSURES
OF THE FLUID UPON THEM SHALL BE EQUAL BETWEEN THEMSELVES.

PROBLEM VI.

46. A rectangular parallelogram is obliquely immersed in an
incompressible and non-elastic fluid, in such a manner, that one
side is just coincident with the surface : —

It is required to divide the parallelogram into two parts by
a line drawn parallel to the horizon, so that the pressures on
the two parts shall be equal to one another.

TWO PARTS SUSTAINING EQUAL PRESSURES. 31

Let A E D represent a rectangular vessel filled with water, or some
other incompressible and non-elastic fluid,
of which ABEF is the surface, and ABCD
the fluid as exhibited in the vessel, on the
supposition that one of its upright sides is
removed.

Let abed be the immersed parallelo-
gram, having its upper side a b coincident
with the surface of the fluid, and its plane
tending obliquely downwards in the given angle of inclination.
Bisect a b in g, and through g draw the v straight line gh parallel
to ad or be, the side of the given immersed rectangle, and let ef
parallel to ab or cd, denote the line of division; then, by the
problem, the pressure on the rectangle abfe, is equal to the pressure
on the rectangle efcd.

Draw the diagonals dfo.ud.fa, cutting the bisecting line gh in the
points m and n ; then are m and n respectively, the places of the
centres of gravity of the spaces efcd and abfe. Through the point
g and in the plane of the fluid surface, draw g r at right angles to a b,
and from m and n demit the straight lines mr and ns, respectively
perpendicular to the horizontal line gr ; then are sn and rm the
perpendicular depths of the centres of gravity of the rectangles a bfe
and efcd on which the pressures are equal.
Put b rz ab, the horizontal breadth of the proposed rectangular plane,

Z=n ad or be, the immersed length of ditto, or that which tends
downwards,

d = rm, the vertical depth of the centre of gravity of the lower
portion efcd,

$i=sn, the vertical depth of the centre of gravity of the upper
portion abfe,

0z= mgr, the inclination of the plane to the surface of the fluid,

P=r the pressure on the entire parallelogram,

p =z the pressure on each of the portions into which the paral-
lelogram is divided,

s rr the specific gravity of the fluid,
and x — ae, the immersed length of the upper portion abfe.

Then is ed—l — x; gn — \x, and gm — # -|- \ (I — x) ; con-
sequently, by the principles of Plane Trigonometry, we have

57i^rS= | a sin. 0, and rm — d— [x -\- %(l — #)J sin.0,
and moreover, by the principles of mensuration, the area of the upper
portion is expressed by b x, and that of the lower portion by b (I — x) ;

32 OF RECTANGULAR PARALLELOGRAMS DIVIDED INTO

consequently, the absolute pressures as referred to the respective
portions, are

but according to the conditions of the problem, these pressures are
equal to one another ; hence by comparison, we have

and this by a little farther reduction, becomes

2*2 = Z2. (15).

47. The equation in its present form, suggests a very simple geometri-
cal construction ; for since Z2 is equal to twice z2, it is manifest, that Z
is the diagonal of a square of svhich the side is x ; hence
the following process.

Draw the straight line AB of the same length as the
side of the given parallelogram, and bisect AB per-
pendicularly in c by the straight line CD; on AB as a
diameter, and about the centre c describe the semi-
circle ADB, cutting the straight line CD in the point
D; join AD, and about the point A as a centre, with
the distance A D, describe the arc DE meeting AB in E ; then is F the
point of division sought.

Upon A B and with the given horizontal breadth, describe the paral-
lelogram ABHG, and through the point E, draw the straight line EF
parallel to AG or BH ; then will EF divide the parallelogram, exactly
after the manner required in the problem. The truth of the above
construction is manifest ; for by the property of the right angled
triangle, we have

A D2 — A c2 -J- c D? ;

but AC is equal toco, these being radii of the same circle, hence
we get

AD2 =2 AC2;

but by the construction, we have

AE m AD ;
consequently, by substitution, it is

AE2Z= 2 AC2,

and doubling both sides of the equation, we get

now AC is equal to one half of AB, and it is demonstrated by the
writers on geometry, that the square of any quantity is equal to four
times the square of its half; consequently, we have

4AC2=IAB2;

therefore, by substitution, we obtain

2AE2ZZZ AB2,

TWO PARTS SUSTAINING EQUAL PRESSURES. 33

being the very same expression as that which we obtained by the
foregoing analytical process, a coincidence which verifies the pre-
ceding construction.

Returning to the equation numbered (15), and extracting the square
root of both sides, we obtain

xy2=li
and by division, we have

x = kW2. (16),

48. The practical rule for determining the point of division, as
supplied by the above equation, is extremely simple ; it may be thus
expressed :

RULE. Multiply half the length of the immersed side of
the parallelogram by the square root of 2, or by the constant
number 1.4142, and the product will express the distance
downward from the surface of the Jluid.

49. EXAMPLE 7. A rectangular parallelogram, whose sides are
respectively 14 and 28 feet, is immersed in a cistern of water, in such
a manner, that its shorter side is just coincident with the surface;
through what point in the longer side must a line be drawn parallel
to the horizon, so that the pressures on the two parts, into which the
parallelogram is divided, may be equal to one another?

Here, by operating according to the rule, we have
x— i (28 X 1.4142)= 19.7988 feet.

50. If the point through which the line of division passes, were
estimated in the contrary direction ; that is, upwards from the lower
extremity of the immersed side of the parallelogram; then, the ex-
pression for the place of the point will be very different from that
which we have given above, as will become manifest from the follow-
ing process.

Recurring to the original diagram of Problem 5, and putting
x — ed, the rest of the notation remaining, we shall have by sub-
traction,

ae=il — x;

consequently, sn the depth of the centre of gravity of the rectangle
abfe, is

3— J (/ — x) sin. 0,

and in like manner, it may be shown, that rm, the depth of the centre
of gravity of the rectangle efcd, is

c?= (/ — !#) sin. (f>.

Now, according to the writers on mensuration, the area of the
rectangle abfe is expressed by b (i — x}, and that of the rectangle
efcd by bx; consequently, the respective pressures are
VOL, i. i>

34 SECTIONS OF RECTANGULAR PARALLELOGRAMS

p— \ b(l — a;)8 s sin.0, andprrr bx(l — \ x} s sin.0,
but by the conditions of the problem, these pressures are equal ;
hence we get

J(Z — *)2 = a? (/ — I *),
and this, by reduction, becomes

** — 2/3? = — i/2;
consequently, the root of this equation is

and more elegantly, by collecting the terms, it becomes
x = l(l — \ V~)'

51. This is manifestly the same result as would arise, by subtract-
ing the value of x in equation (16), from the whole length of the
parallelogram ; and the rule for performing the operation is simply
as follows :

RULE. From unity subtract one half the square root o/2 ;
then multiply the remainder by the length of the parallelo-
gram, and the product will be the distance of the point
required from the lower extremity of the immersed dimension.

Therefore, by taking the length of the parallelogram, as proposed
in the preceding example, we shall have for the distance from its
lower extremity, through which the line of division passes,
x = 28 (1 — i ^/2) = 8.2012 feet.

COROL. It is manifest from the equations (16) and (17), that the
solution is wholly independent of the breadth of the parallelogram, its
inclination to the horizon, and the specific gravity of the fluid ; these
elements, therefore, might have been omitted in the investigation ;
but since it became necessary to express the pressure either absolutely
or relatively, we thought it better to exhibit the several quantities, of
which the measure of the pressure is constituted.

PROBLEM VII.

52. A given rectangular parallelogram is immersed in a fluid,
in such a manner, that one side is coincident with the surface,
and its plane tending obliquely downwards at a given inclination
to the horizon : —

It is required to draw a straight line parallel to one of the
diagonals, so that the pressures on the parts into which the
parallelogram is divided, may be equal to one another.

SUSTAINING EQUAL PRESSURES

Let A ED represent a cistern filled with
fluid, of which ABEF is the surface, sup-
posed to be perfectly quiescent, and con-
sequently, parallel to the horizon ; and let
ABCD be a vertical section of the cistern,
exhibiting the fluid with the immersed
rectangle abed.

Draw the diagonal ac, and in a d take
any point e ; through the point e thus
assumed, draw the straight line ef parallel to a c the diagonal of the
parallelogram; then is edf the triangle, on which the pressure is
equal to that upon the polygonal figure e a b cf.

Take dn and dt respectively equal to one third of de and df, and
through the points n and t, draw nm and tm respectively parallel to
ab and ad, the sides of the parallelogram, and meeting one another
in the point m ; then, according to problem B, m is the place of the
centre of gravity of the triangle e df.

Produce tm directly forward, meeting ab the upper side of the
parallelogram perpendicularly in s ; then, through the point s, and in
the plane of the fluid surface, draw the straight line sr also at right
angles to a b, and from m, the centre of gravity of the triangle edf,
demit the line mr perpendicularly on sr ; then is rm the perpendicular
depth of the centre of gravity of the triangle edf, and msr is the
angle of inclination of the plane to the horizon.

Put b=^ab, the horizontal breadth of the given parallelogram,
I = ad, the length of the immersed plane tending downwards,
d — rm, the perpendicular depth of the centre of gravity of the

triangle efd,

p ~ the whole pressure perpendicular to its surface,
<t> — msr, the angle which the immersed plane makes with the

horizon,

s =the specific gravity of the fluid,
and x — ed, the perpendicular of the triangle edf, of which the base

isrf/.

Then, by reason of the parallel lines ac and ef, the triangles adc
and ec?/are similar to one another, and consequently, by the property
of similar triangles, we have

ad : dc : : ed : df,
which, by restoring the symbols, becomes

I : b : : x : df,
and from this analogy we have

36 SECTIONS OF RECTANGULAR PARALLELOGRAMS

therefore, by the principles of mensuration, the area of the triangle
efd is

bx bx*

. 4*XT=2T

Now, according to the construction, dn is equal to one third of ed,
and an is equal to ad minus dn; but sn is obviously equal to an\

hence we have

sninl — \x,

and by the principles of Plane Trigonometry, it is

rm zz c?~ (I — ^ x) sin. <f> ;
consequently, the pressure on the triangle edf becomes

_ bx*s (31 — x} sin. <p
P~ ~6T

and this, by the conditions of the problem, is equal to half the pres-
sure on the entire parallelogram ; therefore, and by equation (7), we
have

bx* s (3 1 — x") sin. <£ _ b l*s sin. p e

61 ~T~

hence, by expunging the common quantities, we get

— ,

and furthermore, by separating and transposing the terms, it 'IB

2x3 — 6lx*= — 3l3,
and dividing all the terms by 2, we obtain

x8 — 3lx*=i — 1.51s. (18).

It is somewhat remarkable, that the solution of a problem appa-
rently so simple, should require the reduction of a cubic equation ;
but so it happens, and it may be proper to observe, that in the present
instance, it cannot be resolved by means of an equation of a lower
degree.

Now, in order to determine the value of x from the above equation,
we have only to substitute the numerical value of / as given in the
question, and then to resolve the equation by the rules given for that
purpose.

53. EXAMPLE 8. Suppose the immersed length of the rectangle, or
that tending downwards, to be 20 feet ; how far below the surface of
the fluid must the point be situated, through which a line drawn
parallel to the diagonal, will divide the parallelogram into two parts
sustaining equal pressures ?

SUSTAINING EQUAL PRESSURES. 37

Here the given length is 20 ; therefore, by substituting 20 and 208,
respectively for / and P in the above equation, we shall obtain

a3 — 60 x* = — 12000.

In order, therefore, to take away the term — 60 #8 and prepare the
equation for solution, we must put x = z -\- 20, and then by involu-
tion, we have

x* = z3 -f- 60 z2 -f 1200 z + 8000
— 60 xz = — 60 z* — 2400 z — 24000,

from which, by summing the terms, we get

xs — 60 *8 = z* — 1200 z — 16000 = — 12000 ;

therefore, by transposition, we obtain

z3 — 1200 z = 4000.

Now, since the equation falls under the irreducible case of cubics,
it is manifest, that its solution cannot be effected by Cardan's formula ;
we must therefore have recourse to some other method, and in the
present instance, it will be convenient to adopt the concise and
elegant theorem of the Chevalier de Borda.

For which purpose, put a zz any arc such, that cosec.3a = - \/ \m ;
then, we shall have

z = — 2 v/T™ sin.a, (19).

where m is the co-efficient of the second term, and n the absolute
number ; consequently, by substitution, we obtain

therefore, by the Trigonometrical Tables, we have

3« =14° 28' 39",
and by division, we get

a = 4° 49' 33".

But the natural sine of 4° 49' 33" to the radius unity, is 0.08412,
and }m =400 ; consequently, by equation (19), we have
z—— 40 x 0.08412 = — 3.3648;
now, we have seen above, that

arzrs+20;
therefore, by substitution, we obtain

x = — 3.3648 4- 20 = 16.6352.

COROL. Hence it appears, that if we take 3.3648 feet downwards
from the surface of the fluid, or 16.6352 feet upwards from the lower
side of the plane, and through the point thus determined in either
case, let a straight line be drawn parallel to the diagonal ; then shall
the rectangle be divided as required in the problem.

38 CENTRE OF GRAVITY OF MIXED SPACE

6. METHOD OF FINDING THE POSITION OF THE CENTRE OF GRAVITY
OF ANY MIXED SPACE OF RECTILINEAR FIGURES IMMERSED IN
NON-ELASTIC FLUIDS.

54. Since the pressure on the entire parallelogram, is equivalent to
the sum of the pressures on the two parts into which it is divided by
the line ef; it follows from thence, that the position of the centre of
gravity of the figure abcfe can be determined, as is shown in what
follows.

Let ABCFE be the figure, of which the centre of gravity is required
to be found, the angles at A, B and c being right
angles ; join the points A and F by the straight line
AF, dividing the figure into two parts, one of which is
the triangle AFE and the other the trapezoidal space
ABCF.

Now, almost every writer on mechanical science
has given the method of finding the centre of gravity
of those figures separately, from which that of the compound space
may easily be determined ; but we are not aware of any method that
has been proposed, for the direct discovery of the centre of gravity
of the mixed space ABCFE, and that is what we are now about to
investigate.

Produce the sides AE and CF till they meet in D; then, because
the angles at A, B and c are right angles, the angle at D is also a right
angle; from the point D, set off Da and D&, respectively equal to
one third of DE and DF, and through the points a and b, draw am
and bm parallel to DA and DC, which produce directly forward to d
and c ; then are cm and dm respectively, the perpendicular depths of
the centre of gravity of the triangle EDF, according as the side AB or
BC is supposed to be coincident with the surface of the fluid.

Put b m AB, the breadth of the rectangular parallelogram A BCD,
I — AD, the length of ditto,

/3= DF, the base of the right angled triangle EDF,
Z'nz DE, the corresponding perpendicular,
d — cm, the perpendicular depth of the centre of gravity of the

triangle EDF, when the side AB is horizontal,
3 — dm, the perpendicular depth of the centre of gravity m, when

the side BC is horizontal,

jazr the pressure perpendicular to the surface of the triangle EDF,
7>'z= the pressure on the irregular figure ABCFE,

OF IMMERSED RECTILINEAR FIGURES. 39

P= the pressure on the entire parallelogram A B c D,

x—rn, the perpendicular depth of the centre of gravity of the

figure ABCFE, when the side AB is horizontal,

and y — sn, the perpendicular depth, when the side BC is horizontal.
Then, because the sides AE and CF are given quantities, it follows,
that DE and DF are also given, and consequently, AC or cm, and cb
or dm are given ; therefore, the perpendicular pressure on the triangle
EDF can easily be ascertained.

Now, A a is manifestly equal to the difference between A D and a D,
and by the construction an is equal to one third of D E ; therefore, by
restoring the analytical representatives, we have

cm = d = l — il'.

Again c b is equal to the difference between c D and D b ; but D b by
the construction, is equal to one third of DF ; hence, by restoring the
analytical symbols, we shall obtain

dm — $=:b — ± ft.

But, according to the writers on mensuration, the area of the
triangle EDF is equal to half the product of the base DF by the per-
pendicular DE ; that is

JZ'X/3 = jr|3;

consequently, if we suppose the plane to be perpendicularly immersed
in the fluid, while the side AB is coincident with its surface ; then, the
pressure on the triangle EDF becomes

p = %pl's(3l-l').

Now, the pressure on the irregular figure ABCFE, is obviously
equal to the difference between the pressures on the entire paral-
lelogram A B c D, and the triangle EDF; but the pressure on the entire
parallelogram, according to equation (8), is

consequently, by subtraction, the pressure on the figure ABCFE,
becomes

but its area is also equal to the difference between that of the
parallelogram and triangle; therefore, we obtain J(2£Z — (31') for
the area of the irregular figure ABCFE; consequently, by division,
the perpendicular depth of the centre of gravity below the line AB,
becomes

and if we suppose the fluid in which the plane is immersed to be
water, the specific gravity of which i* unity, we finally obtain

40 CENTRE OF GRAVITY OF MIXED SPACE

zbi* — fii'(3i— r)

3(2ft/-00

Again, if we suppose the side BC to be horizontal, the area of the
triangle remains the same, and the pressure which it sustains in a
direction perpendicular to its surface, becomes
p = tfll's(3b— 0).

But the pressure on the whole parallelogram A BCD, on the suppo-
sition that the side BC is horizontal, according to what has been
proved in Problem 3, is

consequently, the pressure on the irregular figure ABCFE, becomes

Now, the area of the figure corresponding to the above pressure, is
obviously the same as we have previously determined it to be ; that is,
the difference between the areas of the triangle and the entire paral-
lelogram ; consequently, by division, we shall obtain
_

The equations (20) and (21) are manifestly symmetrical ; if there-
fore, we carefully attend to the conditions of the problem, from which
they are respectively derived, the position of the centre of gravity of
the figure ABCFE can easily be ascertained byresolvng the equations.

55. The practical rules for determining the co-ordinates which fix
the position of the centre of gravity, may be expressed in the follow-
ing manner :

1. When the side AE is horizontal, as indicated by equation (20).

RULE. From three times the vertical length of the given
rectangular parallelogram, subtract the perpendicular of the
triangle, and multiply the remainder by twice its area ; then^
subtract the product from three times the square of the length
of the parallelogram drawn into its breadth, and the remain-
der will be the dividend.

Divide the dividend above determined, by three times the
difference between twice the area of the parallelogram^ and
twice that of the triangle, and the quotient will give the
co-ordinate of the line A B.

2. When the side BC is horizontal, as indicated by equation (21).

RULE. From three times the vertical breadth of the paral-
lelogram, subtract the base of the triangle, and multiply the

OF IMMERSED RECTILINEAR FIGURES, 41

remainder by twice its area ; then, subtract the product from
three times the square of the breadth of the parallelogram
drawn into its length, and the remainder will be the dividend.
Divide the dividend above determined, by three times the
difference between twice the area of the parallelogram, and
twice that of the triangle, and the quotient will give the
co-ordinate of the line BC.

56. EXAMPLE 9. The sides of a rectangular parallelogram are
respectively 28 and 50 feet, and from one of the lower corners, is
separated a right angled triangle, by means of a straight line ter-
minating in the adjacent sides; it is required to determine the
position of the centre of gravity of the remaining part, the base and
perpendicular of the separated triangle, being respectively equal to
20 and 42 feet ?

Here then, by operating as directed in the first rule, we have

3 x 50— 42 = 150 — 42 = 108,
and by the principles of mensuration, twice the area of the triangle, is

42 X 20 = 840 square feet ;
therefore, by multiplication, we obtain

108 x 840 = 90720.
Again, three times the square of the length of the parallelogram, is

3x 502 = 7500,
which being multiplied by its breadth, gives

7500 x 28 = 210000;
consequently, by subtraction, the dividend is

210000 — 90720=119280.

Now, twice the area of the parallelogram, is 2 x 50 x 28 = 2800
square feet, and twice the area of the triangle, is 42 x 20 = 840
square feet ; therefore, by the second clause of the rule, we obtain

a = H928° = 20.286 feet nearly.

3(2800 — 840)

Hence it appears, that the co-ordinate of the line AB, according to
the proposed data, is very nearly 20.286 feet; and by operating as
directed in the second rule, we shall have

3 x 28 — 20 = 84 —20 = 64,
and by the principles of mensuration, twice the area of the triangle, is

42 X 20 = 840 square feet ;
therefore, by multiplication, we obtain

64 x 840 = 53760.

Again, three times the square of the breadth of the parallelogram, is
3 x 28* = 2352,

42 CENTRE OF GRAVITY OF MIXED SPACE

which being multiplied by its length, gives

2352 X 50=117600;
consequently, by subtraction, the dividend becomes

117600 — 53760 = 63840.

Now, the second clause of the second rule, being the same as the
second clause of the first rule, it follows, that the divisor must here
be the same, as we have found it to be in the preceding case ; conse-
quently, by division, we obtain

therefore, from the numerical values of the co-ordinates as we have
just determined them, the position of the centre of gravity of the
proposed figure can easily be found, in the following manner.

57. Let ABCD represent the rectangular parallelo-
gram, of which the side AB is 28 feet, and the side EC
50 feet; and let EDC be the right angled triangle,
whose perpendicular E D is 42 feet, and its base D F 20
feet, all taken from the same scale of equal parts.

From the angle B, and on the sides BC and BA, set
off BS and Br respectively equal to 20.286 and 10.857
feet; then, through the points s and r, draw the lines sn and rn,
respectively parallel to AB and BC, and the point n is the Centre of
gravity of the figure ABCFE, which remains after the right angled
triangle EDF is separated from the parallelogram ABCD.

If the line of division, or hypothenuse of the triangle EF, were
parallel to AC the diagonal of the parallelogram, as is distinctly speci-
fied in the foregoing problem, the solution would become much more
simple ; for then, in order to determine the position of the centre of
gravity, it is only necessary to reduce one of the equations, and it is
altogether a matter of indifference which of them it is, provided that
the conditions of the equation be strictly attended to.

Supposing E D the perpendicular of the triangle, to remain as above ;
then the base, when the hypothenuse is parallel to the diagonal of the
rectangle, will be found by the following analogy, viz.
50 : 28 : : 42 : 23.52.

Then, by calculating according to rule first, or equation (20), our
dividend and divisor are 103313.28 and 5436.48 respectively; con-
sequently, we get

103313.28

*= 5436.48 =19fe

OF IMMERSED RECTILINEAR FIGURES. 43

therefore, by analogy, we obtain
50 : 19 :: 28 : y = 10.64 feet.

Here, the whole process of calculating the second co-ordinate, is
replaced by the simple analogy above exhibited.

The example now before us, affords a striking instance of the
advantages to be derived from this mode of considering the centre of
gravity; in the case of the triangle illustrated under Problem (B), its
immediate utility was not so conspicuously displayed ; but we are
convinced, that in figures of more difficult and complicated forms, its
usefulness will become still more evident.

In the investigation of the formulae, we have thought it necessary
to consider the pressure on the surface whose centre of gravity is
sought ; but in the actual application of the resulting equations, the
consideration of pressure does not enter; for it is manifest, that
besides the dimensions of the figure and constant numbers, no other
element is found in the equations, and consequently, the reduction
depends upon them alone.

7. OF EQUAL FLUID PRESSURES ON THE SECTIONS OF A RECTANGULAR
PARALLELOGRAM AND THE PERPENDICULAR DEPTHS OF THE CENTRE
OF GRAVITY.

PROBLEM VIII.

58. A given rectangular parallelogram is immersed in an
incompressible and non-elastic fluid, in such a manner, that one
of its sides is coincident with the surface, and its plane tending
downwards at a given inclination to the horizon : —

It is required to draw a straight line from one of the upper
angles to the lower side, so that the pressures on the two
parts into which the parallelogram is divided, may be
equal to one another.

Let AED represent a rectangular cistern filled with water, or some
other incompressible and non-elastic fluid,
of which ABEF is the horizontal surface,
and suppose one of the upright sides, as
ABCD to be removed, exhibiting the fluid
together with the immersed rectangle abed.

In dc the lower side of the immersed
parallelogram, take any point/, and draw
af to represent the line of division ; then
the triangle adf, and the trapezoid abcf, T>

are the figures into which the parallelogram is divided, and on which
the pressures are equal.

44 PARALLELOGRAM DIVIDED TO SUSTAIN

From the angle d, set off dn and dt respectively equal to one third
of da and df, and through the points n and t, draw nm and tm
parallel to df and da the sides of the triangle, and meeting each
other in the point m ; then, according to what has been demonstrated
in Problem (B), m is the centre of gravity of the triangle adf.

Produce tm directly forward, meeting a b at right angles in the
point r, and through the point r and in the plane of the fluid surface,
draw rs also at right angles to a b, and demit ms meeting rs perpen-
dicularly in s; then is mrs the inclination of the plane to the horizon,
and sm the perpendicular depth of the centre of gravity of the triangle
ad/* below the upper surface of the fluid.

Put b — ab, the horizontal breadth of the parallelogram abed,
I — ad, the immersed length tending downwards,
p zn the pressure perpendicular to the surface of the triangle a df.
P= the pressure on the entire parallelogram abed,
0 = mrs, the angle which the immersed plane makes with the

horizon ,
d^ism, the perpendicular depth of the centre of gravity of the

triangle adf,

s =z the specific gravity of the fluid,
and x zz df, the distance between d and the point through which the

line of division passes.

Then, according to the principles of Plane Trigonometry, the per-
pendicular depth of the centre of gravity of the triangle adf, becomes

6?rr-|Zsin.0;
consequently, the pressure on its surface, is

p — %l*xs sin.0 ; J see equation (10) J .

But according to equation (7) under the 3rd problem, (art. 33), the
pressure on the entire rectangle, is

and by the conditions of the present problem, the pressure on the
triangle, is equal to one half the pressure on the entire parallelogram ;
therefore, we have

/?== !P; that is

from which, by expunging the common terms, we get

4x = 3b;

consequently, by division, we obtain
_3b

~ 4 '

EQUAL FLUID PRESSURES. 45

59. This equation is too simple in its arrangement to require any
formal directions for its resolution ; nevertheless, the following- rule
may be useful to many of our readers.

RULE. {Take three fourths of that side of the given rec-
tangular parallelogram, in which the line of division terminates ,
and the point thus discovered, is that through which the line
of division passes.

60. EXAMPLE 10. A rectangular parallelogram, whose sides are
respectively equal to 24 and 42 feet, is immersed in a cistern full of
water, in such a manner, that its shorter side is coincident with the
surface of the fluid, and its plane inclined to the horizon in an angle
of 52 degrees ; it is required to determine a point in its lower side,
to which, if a straight line be drawn from one of the upper angles, the
parallelogram shall be divided into two parts sustaining equal pres-
sures ; and moreover, if a straight line be drawn from the same point
in the lower side, to the other upper angle, it is required to assign the
pressure on the triangle thus cut off?

Here, by operating according to the rule, the point of division is

x = % X 24= 18 feet.

In the next place, to determine the pressure sustained by the
triangle bcft cut off from the parallelogram abed, by means of the
line/6 drawn from the point f to the angle at b, we have according
to equation (12), (Problem 4),

p z= i (b — x) I9 s sin.0,

where (b — #) in this equation, takes place of b in the one re-
ferred to.

The natural sine of 52 degrees according to the Trigonometrical
Tables, is .78801 ; hence, by substituting the respective data in the
above equation, we shall have

p—l (24—18) X 422 X .78801 — 2746.09928 cub. ft. of
water ; consequently, the pressure expressed in Ibs. avoirdupois, is

p1 = 2746.09928 X 62.5 = 181631.205 Ibs.

This seems to be an immense pressure, on a triangle whose surface
is only 126 square feet ; it is however but one sixth part of the pres-
sure on the entire parallelogram ; this is manifest, for the pressure on
the triangle adf, is three times the pressure on the triangle bcf, since
the base df is equal to three times the base cf, and the altitudes of
the triangles, as well as the perpendicular depths of the centres of
gravity, are the same; but the pressure on the parallelogram abed,

46 PARALLELOGRAM DIVIDED TO SUSTAIN

according to the problem, is double the pressure on the triangle adf;

hence we have

P — 2p i= 6p' = 1089787.23 Ibs.

61. If the line of division were drawn from one of the lower angles
to a point in the immersed length, after the manner represented in the
annexed diagram ; then, the equation (22), would assume a different
form, as will become manifest from the following investigation.

From the angle d on da and dc the sides of the
parallelogram, set off dn and dt, respectively equal to
one third of df and dc, and through the points n and
t thus found, draw the straight lines nm and tm \
parallel to dc and df, the base and perpendicular of
the triangle fdc, and meeting one another in m, the
place of its centre of gravity.

Produce tm directly forward, meeting a b, the hori-
zontal side of the given parallelogram perpendicularly in the point r ;
at the point r in the straight line mr, make the angle mrs equal to
the angle of the plane's inclination, and draw ms perpendicularly to
rs ; then is sm the perpendicular depth of the centre of gravity of the
triangle /We.

Let therefore, the notation of the preceding case be retained, and
put x == df; then we have

am^irmm. I' — ^x, and consequently sm — di=: (I — ^x) sin .0 ;
but the area of the triangle fdc is expressed by \b x; therefore, the
pressure perpendicular to its surface, is

now, according to the conditions of the problem, the pressure on the
separated triangle is equal to half the pressure on the entire paral-
lelogram ; consequently, we obtain

and this, by expunging the common quantities, becomes

2 x (3 / — x) = 3 l\
or dividing by 2 we get

x(3l — x} = l.5l\
and from this, by separating and transposing the terms, we have

x* — 3lx — — 1.5 Z2. (23).

If the equations (18) and (23) be compared with one another, it
will readily appear, that they are precisely similar in form, but dif-
ferent in degree ; the former being an incomplete cubic, wanting the
first power of the unknown quantity, and the latter an adfected
quadratic, having all its terms. Indeed, the diagrams from which the

EQUAL FLUID PRESSURES. 47

two equations are derived, as well as the specified conditions of the
problems, are nearly similar, the difference consisting simply in the
position of the dividing line, it being parallel to the diagonal of the
parallelogram in the one case, and oblique to it in the other.

62. Let the quantity 2JZ2be added to both sides of the preceding
equation, and we shall obtain

from which, by extracting the square root, we get

x— i$i = + \i V3";

therefore, by transposition, we have

(24).

The practical rule by which the point of division is to be determined,
may be expressed as follows :

RULE. Multiply the difference between 3 and the square
root of 3, by half the length of that side of the parallelogram
in which the line of division terminates, and the product will
'be the distance of the required point from the lower extremity
of the given length.

63. EXAMPLE 11. Let the numerical data remain precisely as in
the preceding case; from what point in the length of the paral-
lelogram, must a straight line be drawn to the opposite lower angle,
so that the parallelogram may be divided into two parts sustaining
equal pressures ; and moreover, if a straight line be drawn from the
same point, to the opposite upper angle, what will be the pressure on
the triangle thus cut off?

Here, by proceeding as directed in the above rule, we have

x — 21 (3 — V 3) = 26.628 feet.

In order to find the pressure on the triangle a bf cut off by the
line bf, we have af—l — x, and ae — rp — \(l — x)\ conse-
quently, vp^n^(l — x) sin.0, where it must be observed, that ep
and vp are respectively parallel to a b and sm.

Now, the pressure perpendicular to the surface of the triangle a bf,
is found by multiplying its area into vp, the perpendicular depth of
its centre of gravity ; hence, we have

p' — ^bs (I — #)2 sin.^. ;
but the value of x, according to equation (24), is

x = .634 I ;

consequently, by substitution, we have

p' = %b Z2 s (1 — .634)* sin.0, from which, by substituting the
several numerical values, we obtain

>' = 4 X 42s X 62.5 X .366* X .78801 = 46551. 35 Ibs.

S. PERPENDICULAR DEPTH OF THE CENTRE OF GRAVITY Of A PARAL-
LELOGRAM DIVIDED INTO TWO PARTS SUSTAINING EQUAL PRESSURES.

64. With respect to the centre of gravity of the figure a b cf, which
remains after the triangle adf, or fdc has been separated from the
parallelogram, it is in this particular instance very easily determined ;
for, since the area of trapezoids, whose parallel sides and perpendi-
cular breadths are equal each to each, are also equal ; it follows, that
the centre of gravity must occur in the straight line which bisects the
parallel sides ; it is therefore, only necessary to investigate the theorem
for calculating one of the co-ordinates, the other being determinable
from the circumstance just stated.

Let ABCF be the trapezoid, having the angles at B and c respec-
tively right angles, and of which the position of the centre of gravity
is required.

Produce the side CF directly forward to any convenient length at
pleasure, and through the point A, draw the straight line AD parallel
to BC, the longer side of the trapezoidal figure, and
meeting c F produced perpendicularly in the point D.

Then, the pressure upon the trapezoid ABCF, is
manifestly equal to the difference between the pressure
on the parallelogram A BCD, and that upon the triangle
ADF, and its area, is also equal to the difference be-
tween their areas. Bisect the parallel sides AB and CF
in the points a and b, and join a b ; then, according to
what lias been demonstrated by the writers on mechanics, the centre
of gravity of the trapezoid ABCF occurs in the straight line ai.

Suppose it to occur at m, and through the point m draw mr and
ms respectively parallel to BC and BA, meeting AB and BC perpen-
dicularly in the points a and b ; then are rm and sm the co-ordinates,
whose intersection determines the position of the point m.

Put b =z AB, the breadth of the parallelogram A BCD,
/ — AD, or BC, its corresponding length,

£ = m r the depth of the point m as referred to the line A B con-
sidered to be horizontal,
cTzu ms, the depth of the point m as referred to the line BC under

similar circumstances,
and /3~ DF, the base of the triangle ADF.

Then, by conceiving the plane to be immersed perpendicularly in a
fluid whose specific gravity is expressed by unity, the pressure upon

CENTRE OF GRAVITY OF PARALLELOGRAMS, &C. 49

the entire surface ABCD, according to equation (8) under the third
problem, becomes

P = J&/2;

and moreover, by equation (12) under the fourth problem, the general
expression for the pressure on the triangle ADF, is

p — 4/3 P s sin.0 ;

but according to the particular case now under consideration, the
above expression becomes

the terms s and sin.0, being each equal to unity, they disappear in
the equation.

Now, according to what we have stated above, the pressure on the
trapezoid ABCF, is equal to the difference between the pressure on the
entire parallelogram ABCD, and that on the triangle ADF ; that is

or, by reducing the fractions to a common denominator and collecting
the terms, we obtain

^ j/ = ^(36 — 2/3).

By the principles of mensuration, the area of the trapezoid ABCF,
is equal to the product that arises, when half the sum of the parallel
sides AB andcF, is multiplied by BC the perpendicular distance
between them ; that is,

BC X |(AB + CF)Z= ¥(^— /?),

and the perpendicular depth of the centre of gravity, is equal to the
pressure on the surface, divided by the area of the figure ; conse-
quently, we obtain

— jS) '

The form of this equation is extremely simple, but it may be arrived
at independently of the preceding investigation, by having recourse
to equation (20) under Problem 6 ; for according to the conditions of
the question, the line of division AF originates at the angle A, and
consequently, the perpendicular of the triangle and the length of the
parallelogram are equal ; therefore, by putting / instead of /' in equa-
tion (20), the above expression immediately obtains.

Now, by taking the length and breadth of the parallelogram, as
given in the preceding example, and the base of the triangle as com-
puted by equation (22), we shall obtain,

3(2 x24— 18)

65. Having thus determined the magnitude of the co-ordinate BS
or rm from the equation (25), the magnitude of the corresponding

VOL. I. E

50 CENTRE OF GRAVITY OF PARALLELOGRAMS

co-ordinate sr or sm, can very easily be found; for through the
point b the bisection of FC, draw bn parallel to BC and meeting AB
perpendicularly in n ; then, the triangles ban and mar are similar to
one another, and the sides bn, mr and an, are given to find ar, ajid
from thence the rectangular co-ordinate sr or sm\ consequently,
we have

b a : na : : mr : r a;

therefore, by subtraction, we get

BrorsrazzaB — r a.

Now, £CZZBW., is obviously equal to half the difference between
D c, the breadth of the parallelogram, and DF, the base of the triangle
ADF ; therefore, we have

B» = j(6.-0);
but an zr a B — BW ; that is, a n zz J/3, and

••• 3(24-0)-

therefore, by reducing the analogy, we get
_0(36 — 20).
' 6(26-0)'
hence, by subtraction, we obtain

36(6-0)4-0*

3(26 — 0) (26).

After the same manner that equation (25) is deducible from equa-
tion (20), by putting Z'zzZ; so also, is equation (26) deducible from
equation (21), by means of the same equality ; we might therefore have
dispensed with the preceding investigation, and derived the expression
from principles already established ; we however preferred obtaining
it as above, for the purpose of exhibiting that agreeable variety which
gives additional embellishment to scientific investigations. The method
of establishing the formulee, on the supposition that the side BC is
horizontal, is sufficiently obvious from what has been done in the
sixth problem preceding, and therefore, it need not be repeated here.

COROL. By substituting the numerical values of b and 0, as given
in the preceding example, we shall have from equation (26)

Therefore, from the point B, set off BS and sr respectively equal to
16.8 and 8.4 feet; and through the points s and r, draw sm and rm
parallel to AB and BC, the perpendicular sides of the given trapezoid,

SUSTAINING EQUAL PRESSURES. 51

and meeting one another in the point m ; then is m the required place
of the centre of gravity.

66. In computing numerically the values of the rectangular co-
ordinates mr and ms, we have supposed, that DF the base of the
applied triangle, is determiriable by the application of equation (20) ;
this supposition however is perfectly unnecessary, for the base of the
triangle is always equal to the difference between the parallel sides of
the given trapezoid ; and moreover, the equation (20), applies only to
the particular case for which it has been deduced, viz. when the
pressure on the applied triangle and that on the trapezoid to which it
is applied are equal to one another.

9. WHEN THE PARALLELOGRAM IS SO DIVIDED, THAT THE PRESSURES
ON THE TWO PARTS ARE TO ONE ANOTHER IN ANY RATIO WHATEVER.

67. In the sixth, seventh and eighth problems preceding, we have
supposed the given rectangular parallelogram to be divided into two
parts, such, that the pressures upon them shall be equal between
themselves, and the investigation has accordingly been limited to that
particular case ; but in order to render the solution general, we shall
consider the division to be so effected, that the pressures on the two
parts may be to one another in any ratio whatever, such as that of
m to n,

For which purpose then, by referring to the fifth problem, where the
given parallelogram is divided horizontally, we find, that the pressure
on the upper portion is expressed by J6a;2ssin.^, and that on the
lower portion, by { J (I — x)*-\-x(l — ,z) } 6ssin.0 ; but these ex-
pressions in their present state are equal to one another, and they
are now required to be reduced in the ratio of m to n ; consequently,
we have

1*': \ k(l — x)* + x(l — x}} ::m:n,
and this, by expanding the second term, becomes

x* : I* — a8 : : m : n;
or by equating the products of the extremes and means, we obtain

n a;2 nz m /2 — ma? \
therefore, by transposition, we get

(m -\- n} o^zr mZ%
and finally, by division and evolution, we have

52 SECTIONS SUSTAINING PRESSURES

68. Again, in the case of the sixth problem, where the given paral-
lelogram is divided by a line drawn parallel to the diagonal ; we find,
that the pressure on the triangle cut off by the line of division, is
expressed by bx*s(3l — x) sin.0 -H 6 /, and consequently, by sub-
traction, that on the remaining portion is expressed by bssm .<f>
{ 3 la — x* (3 I — #) } x ~ 6 / ; now, these expressions, by the condi-
tions of the problem, are equal to one another ; but in the present
case, they are to be reduced in the ratio of m to n ; for which purpose

we have

a* (3 / — x) : 3 /8 — x*(3 I — x) : : m : n ;

therefore, by equating the products of the extreme and mean terms,
we get

nx*(3l — x) — 3ml3 — mx*(3l — a?) ;
and from this, by transposition, we shall obtain

(m + ») (3 lx* — xs) = 3 m Is ;

therefore, by dividing and transposing the terms, we have

3ml9

(28).

Now, in order to reduce the above equation, there must be substi-
tuted the numbers which express the given ratio, together with the
length of the parallelogram, and then, the value of a; will be obtained
by any of the rules for resolving cubic equations.

69. In like manner as above, by referring to the eighth problem,
where the given parallelogram is divided by a line drawn from one of
the upper angles, and terminating in the lower side ; we find, that the
pressure on the triangle cut off by the line of division, is expressed by
^/8#ssin.0, and consequently, by subtraction, the pressure on the
remaining portion is expressed by ±Fssm.(f>(3b — 2#); and these
expressions, according to the conditions of the problem, are equal to
one another ; but in the present instance, they are to be reduced in
the ratio of m to n ; hence, we have

consequently, by equating the products of the extreme and mean
terms, we get

2nx nr 3 bm — 2 war,

from which, by transposition, we obtain

2(m + n)x = 3bm,
and finally, by division, we have

36m
•r-2(m-fn)' (29).

ANY RATIO TO ONE ANOTHER.

Hence then, the equations (27,) (28,) and (29,) express generally
the relation between the parts of division, which in the several pro-
blems is restricted to a ratio of equality ; and it is presumed, that by
paying a due attention to the examples that have been proposed and
illustrated, the diligent reader will find no difficulty in resolving any
example that may present itself under one or other of the general
forms above investigated.

In all the above cases, we have supposed the breadth, or that side
of the parallelogram which is denoted by b to be horizontal, and
coincident with the surface of the fluid ; but it is manifest, that
equations of the same form would be obtained from the other side,
having b in place of I, and / in place of b.

10. OF RECTANGULAR PARALLELOGRAMS DIVIDED INTO SECTIONS
SUSTAINING EQUAL PRESSURES; WITH THE METHOD OF DETER-
MINING A LIMIT TO THE NECESSARY THICKNESS OF FLOOD-GATES,
AND OTHER CONSTRUCTIONS OF A SIMILAR NATURE.

PROBLEM IX.

70. A given rectangular parallelogram, is immersed in an
incompressible and non-elastic fluid, in such a manner, that one
of its sides is coincident with the surface, and its plane inclined
at a given angle to the horizon : —

It is required to divide the rectangle by lines drawn
parallel to the horizon, into any number of parts, such,
that the pressures on the several parts of division shall be
equal to one another.

Let A ED represent a rectangular cistern filled with water, or some
other transparent and incompressible fluid
in a state of rest ; one side of the vessel
being removed, for the purpose of exhibit-
ing the fluid and the immersed parallelo-
gram, together with the several subordinate
lines on which the investigation depends.

Suppose e, I, g and i to be the several
points of division, and through these points
draw the lines em, If, gk and ih, respec-
tively parallel to a b or dc, the horizontal sides of the figure. Bisect
the sides a b and dc in the points G and H; join GH, and draw the
zigzag diagonals am, ml, Ik, ki and ic, cutting the bisecting line

54 HORIZONTAL SECTIONS CONTAINING EQUAL FLUID PRESSURES.

GH in the points z, q, p, o and n, which points are the respective
centres of gravity of the several parts into which the given parallelo-
gram is divided, and on which the pressures are supposed to be equal
among themselves.

Through the point G, and in the plane of the horizon, draw the
straight line or at right angles to ab, making the angle ron equal to
the given angle of the plane's inclination, and from the points n, o,p,
q and z, let fall the perpendiculars zv, qu, pt, os and nr, meeting
the line or respectively in the points v, u, t, s and r; then are the
lines zv, qu, pt, os and nr, the perpendicular depths of the centres
of gravity of the several portions into which the proposed rectangle is
divided, the points of division being estimated from the surface down-
wards.

Put£ =iab or dc, the horizontal breadth of the given parallelogram

abed,
I — ad or be, the entire immersed length, or that tending down-

wards,

^ zzrGH, the given angle of inclination,

d zn vz, the vertical depth of the centre of gravity of the part abme,
d ' zz w £, the vertical depth of - e mfl,
d" — tp, the vertical depth of—

S —so, the vertical depth of -- ykhi,
I' — rn, the vertical depth of -- iked',
n zzthe number of parts into which the parallelogram is divided,
P zzrthe entire pressure on the parallelogram abed,
p nz the pressure, common to each of the parts into which the

given parallelogram is divided,

v =n ae, the required length of the upper portion abme,
w ~el, the length of the second portion emfl,
x m Ig, the length of the third portion If kg,
y —gi, the length of the fourth portion gkhi,
z zn id, the length of the fifth portion i hcd.

Then, according to equation (7) under the third problem, the entire
pressure on the parallelogram abed, is

therefore, the pressure on each part of the divided figure, is
bl2

But because the value of $, the angle of inclination, and s the
specific gravity of the fluid, are the same for all the parts ; those

HORIZONTAL SECTIONS SUSTAINING EQUAL FLUID PRESSURES. 55

quantities may be omitted in the equation, and then the element of
comparison, or the nthpart of the total pressure becomes

_bl*
p-^~n (30).

Now, according to the principles of Plane Trigonometry, the lines
vz, uq, tp, so and rn, are respectively as below, viz.

$ nz G o sin.^>, and 3' zz G n sin.0 ;

but the lines GZ, G^, Gp, GO and GW, when expressed in terms of the
respective lengths, are as follows, viz.

G2 = |v; Gq=:v-\-%w; Gp — v -f- w -f- J#;
Gozzv-j-w-|-a;4" iy> and ftwss:* +'IIP + * + y + J«;

therefore, by substitution, the above values of the vertical depths of
the respective centres of gravity, become

d — \v sin.<£ ; d' zz: (v 4- Jw;) sin.^ ; dw zz (w •+• w 4- Jar) sin.0 ;
5 zz (v 4- w 4- a? + Jy) sin.^, and ^ zz: (v 4- w -f x 4- 2/ + |z) sin.0.

Consequently, by throwing out the common factor sin.0 and
neglecting the specific gravity of the fluid, the value of jo, or the
pressure sustained by each of the parts, may be expressed as follows,
viz.

The pressure on the part abme, is p~%bv*, (1).

- em fly isp~bw (v -\- Jw), (2).

- If kg, isp=ibx(v-{-w+^, (3).

- gkhi, isp = by(v-\-w + x+ jy), (4).

- ihcd, isp = bz (v •+• w -{- x+y+£z).(5).

Now, according to the conditions of the problem, all these expres-
sions for the value of p, are equal to one another, and each of them is
equal to the element of comparison, as given in the equation (30) ;
hence, from the first of the above equations, or values of p, we have

or by expunging the common factor, |6, we obtain

*< = -?!;
n

therefore, by extracting the square root, we have

56 HORIZONTAL SECTIONS SUSTAINING EQUAL FLUID PRESSURES.

Again, by comparing equation (30), with the second of the fore-
going expressions for the value of p, we shall have

bl*
bw(v + \w}= — -

and substituting the value of v in terms of / and n, we obtain

n w* -f- 2/-V//T7 w — P,
from which, by dividing by n, we get

w -- — . w •=. — >
n n

therefore, if this be reduced by the rule which applies to the resolution

of adfected quadratic equations, we shall obtain

I _ ___

w :rr — -( ^/2w — Vn)'
n

Proceed as above, by comparing the equation (30), with the third
of the preceding expressions for the value of p, and we shall have

= —

and if the above values of v and w, as expressed in terms of I and w,
be substituted instead of them in this equation, it will become

and dividing by n, we get

n n

therefore, by completing the square, evolving and transposing, we
obtain

By pursuing a similar mode of comparison, and reasoning in the
same manner, with respect to the fourth value of p foregoing, we
shall have

let the values of v, w and x, as determined above, be respectively
substituted in this equation, and it becomes

complete the square, and we obtain

PARALLELOGRAM DIVIDED TO SUSTAIN EQUAL FLUID PRESSURES. 57

and from this, by evolution and transposition, we get

Pursuing still the same mode of induction for the fifth value of p,
and substituting the respective values of v, w, x and y, as we have
determined them above in terms of I and n ; then we shall have

z — — { |/5» — \/4n~}.

And in like manner we may proceed for any number of divisions at
pleasure ; but what we have now done is sufficient to exhibit the law
of induction.

The formulae which we have investigated, for determining the
several sections of the given parallelogram, may now be advantage-
ously collected into one place ; for it is manifest, that by exhibiting
them in juxta-position, the law of their formation is more easily
detected, and the difference which obtains between the co-efficients of
the successive terms becomes at once assignable.

The several equations therefore, when arranged according to the
order of the corresponding sections, will stand as under, viz.

3- x = — (V^-

&C.&C.ZZ &C. &C.

71. The practical rule for determining the points of section, in
reference to their respective distances from the upper extremity of the
parallelogram, may be expressed in words, as follows, viz.

RULE. Multiply the number of parts into which the paral-
lelogram is proposed to be divided, by the number that indicates
the place of any particular section; then, multiply the square

58 PARALLELOGRAM DIVIDED TO SUSTAIN EQUAL FLUID PRESSURES.

root of the product by the length of the parallelogram, and
divide by the whole number of sections, and the quotient will
express the distance of any particular point from the upper
extremity of the divided length.
If the length of any particular section, or the distance between any

two contiguous points should be required, which is the condition

expressed by each of the above equations ; then,

Calculate for each of the points according to the preceding
rule, and the difference of the results will give the length of
the required section.

72. EXAMPLE 12. A rectangular parallelogram whose length is 25
feet, is perpendicularly immersed in a fluid, in such a manner, that
its breadth or upper side is just in contact with the surface ; now, if
it be proposed to divide the parallelogram by lines drawn parallel to
the horizon, into five parts sustaining equal pressures ; it is required
to determine the distance of each point of section from the surface of
the fluid, and the respective distances between the several points ?

Here then, we have given I — 25 feet, and n =r 5 an abstract
number ; therefore, by proceeding according to the rule, we shall have,
for the distance of the first point,

25V5--- 5 = 11. 18034 feet.
For the distance of the second point, we get

25V2 x 5 + 5 = 15.81139 feet.
For the distance of the third point, we obtain

25 /3~x^5 H- 5 = 19.36492 feet,
and for the distance of the fourth point, it is

25 V 4~X~5 ~ 5 = 22.36068 feet.

The preceding is all that is necessary to be calculated, for the
distance of the fifth point is manifestly equal to the whole length of
the parallelogram, and consequently, by the question, it is a given
quantity.

Having thus determined the distances of the several points of
section below the upper surface of the fluid, the respective distances
between them, or the breadths of the several sections can easily be
ascertained, since they are merely the consecutive differences of the
quantities above calculated ; hence, we have

Distances 11.18034, 15.81139, 19.36492, 22.36068, 25.

Differences 4.63105, 3.55353, 2.99576, 2.63932 ;

LIMIT TO THE REQUISITE THICKNESS OF FLOOD-GATES. 59

therefore, the breadths of the respective sections, estimated in order
from the surface of the fluid, are

11.18034, 4.63155, 3.55352, 2.99576, 2.63932 feet.
Let A BCD be the rectangular parallelogram, whose length AD or
B c is equal to 25 feet, and the breadth A B or DC
of any convenient magnitude at pleasure. Upon
the length A D, set off the distances AE, A a, A b
and AC, respectively equal to the preceding num-
bers, taken in the order of their arrangement,
and through the points E, a, b and c, draw the
straight lines EF, af, be and cd, respectively
parallel to AB or DC the horizontal sides of the parallelogram ; then
are the rectangular spaces AF, ae, bd and cc, the respective portions
into which the given parallelogram ABCD is divided, and on which,
according to the conditions of the problem, the perpendicular pres-
sures are equal among themselves.

OF THE REQUISITE THICKNESS OF FLOOD-GATES, &C.

73. The problem which we have just resolved is a veiy important
one; by it we can determine a limit to the requisite thickness of
flood-gates and other constructions of a similar nature, and also the
form which the section ought to assume, in order that the strength in
every part may be proportional to the pressure sustained.

For according to the preceding notation, and by equation (7) under
the third problem, the pressure on the rectangle ABFE, is

p — J6v2ssin.^,
and the pressure on the whole rectangle ABCD, is

Pzz: ^bFssin.ty;
consequently, by analogy and comparison, we get

P:p:i?i v\

And in like manner it may be shown, that the same relation
obtains in respect of any other rectangle AB/a, when compared
with the entire figure ABCD; consequently, the pressures are uni-
versally as the squares of the depths, the breadth being constant ;
therefore, the thickness should be as the square of the depth, being
greatest at the bottom and decreasing upwards to the surface of the
fluid.

Thus for example, let the flood-gate be of the same depth as the
rectangle in the foregoing question, and let the thickness at the
bottom be equal to one foot or twelve inches ; then, the corresponding
thicknesses for the several feet of ascent estimated upwards, will be as
follows.

LIMIT TO THE REQUISITE THICKNESS OF FLOODGATES.

At 25 feet, we

have 25s : 252

OC2 . 0x14

: 12 : 12 inches.
• ln • 11 060 •

058 . 032

• 12 : 10.156 • -

r»o

, OC2 . 00«

• 1° • 9 °9°

OAJ2 . 012

• 1° • 8 4fi8

052 . OQ2

• 1° • 7 680

. °52 • 192

• 12 • 6 93°

1 9

, 052 . 1 g2

. JO . Q OOQ

.. OA2 . 172

• 1° • 5 'i/IR

Ifi

OC2 . Ifi2

. 10 . xi qcf?

1 ^

052 . ^52

• 1° • 1 3°0 •

\*\

OC2 . 1/12

• 1° • S 7fi^l

0 « . 1 02

. 10 . 0 0^^ , ., ..

, 0« . 102

. 10 . o 7«4

0 £2 . 112

. 10 . o qo^ ,

052 . 1Q2

• 1° • 1 9°0

052 . os

• 1° • 1 556 •

. 0 C2 . Q«

. 10 . 1 000

o« . 72

. 10 . 0 C]A() ,

fi .

0« . f!2

. 10 . o fiQ2

052 . y,

• 1° • 0 160

OC2 . /|2

• 1° • 0 308 •

'} j

052 . 02

• 10 . 017°

OC2 . 02

. 10 • n 07fi -

9/5* : I8

• 19 • 0.090 .. -

74. Having calculated the numerical values of the thicknesses or
ordinates, corresponding to each foot in length, estimated from the
bottom where the pressure is a maximum, upwards to the summit
where it vanishes ; we shall now proceed to construct the section, in
order to exhibit the particular form which the preceding theory assigns.

Let the straight line A B represent the perpen-
dicular depth of the flood-gate supporting the
fluid F, and whose vertical section is denoted by
the figure ABC, the exterior boundary of which is
the curve line ABC, and the greatest thickness,
or that at the bottom, equal to BC.

Divide the depth AB into twenty five equal
parts, having an interval of one foot for each ;
then, through the several points of division, and parallel to the horizon,
draw straight lines, beginning at the bottom and proceeding upwards,
making these lines respectively equal to 12, 11.06, 10.156, 9.292, &c.

OF FLUID PRESSURE ON THE SIDES AND BASE OF CUBICAL VESSELS. 61

inches, according to the numbers in the foregoing tablet, and through
the remote extremities of the several ordinates, trace the curve line
ADC, which will mark the exterior boundary of the section.

The intelligent reader will readily perceive, that in the actual con-
struction of the above figure, it has been found impossible to preserve
the proper proportion between the several abscissse and their corre-
sponding ordinates ; if this had been attempted, the figure must either
have been enlarged to an inconvenient size, or the ordinates would
have been so small as to render the general appearance of the section
very indistinct.

We have therefore thought it preferable to preserve the line of the
abscissee within moderate bounds, and to enlarge the ordinates in a
given constant ratio ; by this means the form of the curve is correct,
and the whole diagram is sufficiently distinct for practical illustration.

In all that has hitherto been done respecting the rectangular
parallelogram, we have constantly considered it as being an inde*
pendent plane immersed in the fluid, and having its upper side
coincident with the surface ; but we must now observe, that whatever
relations have been shown to exist on such a supposition, the same
will hold, if the plane be considered as the side of a vessel filled with
the fluid by which the pressure is propagated.

We have already alluded to this principle, at the conclusion of our
illustration of the fourth problem ; it therefore only remains to deter-
mine by it, the pressure on the bottom and sides of a vessel filled with
a fluid of uniform density, on the supposition that the bottom and the
sides are respectively rectangular planes.

11. METHOD OF COMPARING THE PRESSURE ON THE PERPENDICULAR
SIDES AND ON THE BOTTOM OF ANY RECTANGULAR CISTERN, BASIN,
OR CANAL LOCK.

PROBLEM X.

75. Suppose that a vessel in form of a rectangular paral-
lelopipedon, is filled with fluid of uniform density, and placed
with its sides perpendicular to the horizon : —

It is required to compare the pressure on the upright sides
with that upon the bottom, both when the sides are all equal,
and when the opposite sides only are equal.

62 OF FLUID PRESSURE ON THE SIDES AND BASE OF CUBICAL VESSELS.

In the solution of the present problem, it will be unnecessary to
exhibit the construction of a separate diagram ; because, the bound-
aries when considered individually, being rectangular parallelograms,
the investigation for each would be similar to that required in Pro-
blem 4, and the resulting formulae would coincide in form with that
exhibited in equation (8).

Therefore, put b = the horizontal breadth of the greater opposite

sides,

/3 n: the horizontal breadth of the lesser ditto ditto,
I z= the perpendicular depth of the fluid, whose den-

sity is uniform,
P= the aggregate, or total pressure on the upright

surface, and
p zz the pressure on the bottom.

Then, according to equation (8) under Problem 3, the pressure on
each of the narrower sides is expressed by |/3/2s, and that on each of
the broader sides by \bl*s\ consequently, the entire pressure on the
upright surface, is

P±=l'«04..&)<

and by the third inference to Proposition (A), the pressure sustained
by the bottom of the vessel, is

consequently, by analogy, we obtain
P:p::l(P + b):pb.

Therefore, when the opposite sides of the rectangular vessel only,
are equal to one another,

The total pressure on the upright surface, is to the pressure
on the bottom, as half the area of the former is to the area of
the latter.

If b in /3, or if all the sides of the vessel are equal to one another ;
then, the entire pressure on the upright sides, is

and that on the bottom, is

p=b*ls;

therefore, by analogy, we obtain
P :p :: 2/ : 6.

OF FLUID PRESSURE ON THE SIDES AND BASE OF CUBICAL VESSELS. 63

Consequently, when all the four sides of the rectangular vessel are
equal to one another,

The total pressure on the upright surface, is to the
pressure on the bottom, as twice the length of the side is to
its breadth.

Again, when all the sides of the vessel have the same breadth, and
the length I equal to the breadth b ; then, the vessel becomes a cube,
and the total pressure on the upright surface, is

P=263s,

and that on the bottom, is
p = b9 s ;

therefore, by analogy, we obtain

P :p :: 2 : 1.

Hence it appears, that when the vessel is a cube, that is, when the
bottom and the four upright sides are equal to one another,

The total pressure upon the four sides, is to the pressure on
the bottom, in the ratio of 2 to 1 .

Since the pressure on the upright surface of a cubical vessel, is
double the pressure on the base ; it follows, that the entire pressure
which the vessel sustains, is equal to three times the pressure upon its
bottom ; that is,

P+p=:'3b*s. (32).

But the expression b*s is manifestly equal to the weight of the
fluid ; consequently, the total pressure upon the sides and base of
the vessel,

Is equal to three times the weight of the fluid which it
contains.

Now, in the case of water, where the specific gravity is represented
by unity, the equation marked (32) becomes

P+p = 3b*;

but when the dimensions of the vessel are estimated in feet, and
the pressure expressed in pounds avoirdupois, of which 62| are equal
to the weight of one cubic foot of water ; then, the above equation is
transformed into

p' = 187.56s. (33).

COROL. This equation in its present form implies, that if the solid
content of the vessel in cubic feet, be multiplied by the constant

64 OF FLUID PRESSURE ON THE SIDES AND BASE OF CUBICAL VESSELS.

co-efficient 187.5, the product will express the number of Ibs. to
which the pressure on the bottom and the four upright sides is
equivalent.

76. EXAMPLE 13. Suppose the length of the side of a cubical
cistern to be 35 feet ; what is the pressure sustained by it, when it is
completely filled with water ?

Here we have given, b zz 35 feet ; therefore, by proceeding accord-
ing to the composition of the foregoing equation, we shall obtain

p' = 35 X 35 x 35 X 187.5= 8039062.5 Ibs.
Hence it appears, that the aggregate pressure upon the bottom
and the upright surface of a cubical vessel whose side is 35 feet,
is 8039062.5 Ibs. or

8039062.5 -f- 2240 = 3588.867 tons,

while the absolute weight of the contained fluid, is only one third
of that quantity, or

35 X 35 x 35 X62.5 -; 2240 = 1196.289 tons.

77. EXAMPLE 14. Let the dimensions of the vessel remain as in the
preceding example, and suppose it to be filled with wine, of which the
specific gravity is .96, instead of water, whose specific gravity is unity,
what pressure does it then sustain ?

For the weight of a cubic foot of wine, we have

1 : 62.5 : : .96 : 60 Ibs.

and the pressure on the bottom and sides of a vessel containing a
cubic foot is 60 X 3 zz: 180 Ibs. ; consequently, the pressure on the
bottom and sides of a vessel whose side is 35 feet, is

p' = 35 X 35 X 35 X 180 = 7717500 Ibs.,
and this, by reducing it to tons, is

= 3445-3125 tons' ; :"

while the absolute weight of the contained fluid, is only one-third of
that quantity, or

35 X 35 X 35 x 60 4- 2240 = 1 148.4375 tons.

CHAPTER III.

t)N THE PRESSURE EXERTED BY NON-ELASTIC FLUIDS

PARABOLIC PLANES IMMERSED IN THOSE FLUIDS, WITH THE
METHOD OF FINDING THE CENTRE OF GRAVITY OF THE SPACE
INCLUDED BETWEEN ANY RECTANGULAR PARALLELOGRAM AND
ITS INSCRIBED PARABOLIC PLANE.

1. WHEN THE AXIS OF THE PARABOLIC PLANE IS PERPENDICULAR TO
THE HORIZON, AND ITS VERTEX COINCIDENT WITH THE SURFACE
OF THE FLUID.

PROBLEM XL

77. If a parabolic plane be just perpendicularly immersed
beneath the surface of an incompressible fluid : —

It is required to compare the pressure upon it, with that
upon its circumscribing rectangular parallelogram, and to
determine the intensity of pressure, according as the vertex
or the base of the parabola is in contact with the surface
of the fluid.

First, when the vertex of the parabola is uppermost, and just in
contact with the surface of the fluid ; let AC B
be the parabolic plane, of which AB is the
base or double ordiriate parallel to the hori-
zon, and CD the vertical axis just covered by
the fluid whose surface is EF, and let ABFE
be a rectangular parallelogram circumscribing
the parabola.

Now, it is demonstrated by the writers on
mechanics, that the centre of gravity of a parabolic plane is situated
in the vertical axis, and the point where it occurs, is at the distance
of three fifths of that axis from the summit of the figure.

Therefore, if the axis CD be divided at m, into two parts such, that
cm is to Dm as 3 is to 2,* then is m the centre of gravity of the

VOL. I. F

66 OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA.

parabolic space ACB ; and if the axis CD be bisected in the point n ;
n is the centre of gravity of the circumscribing rectangular parallelo-
gram ABEF.

Put b zz A B, the base of the parabola, or the horizontal breadth of its
circumscribing rectangular parallelogram,

/—CD, the vertical axis of the parabola, or the depth of its sur-
rounding rectangle,

dm en, the depth of the centre of gravity of the parallelogram
ABFE below the upper surface of the fluid,

S zz cm, the, depth of the centre of gravity of the parabola ACB,

Pzz the pressure on the circumscribing rectangle,

p zz the pressure on the parabolic plane,

A zz the area of the parallelogram,

a zz the area of the parabola, and

s — the specific gravity of the fluid.

Then, according to the writers on mensuration, the area of the cir-
cumscribing rectangular parallelogram, is

A = bl',

but the area of a parabola, is equal to two thirds of the area of its
circumscribing rectangle ; therefore, we have

a zz %bl.

Now, 5 zz %l by the construction, and we have shown in Proposition
(A), that the pressure upon any surface,

Is expressed by the area of that surface, drawn into the
perpendicular depth of its centre of gravity, and also into the
specific gravity of the fluid.

Consequently, the pressure perpendicular to the surface of the
parabolic plane, is

/>==#/ X#X * = lbP*. (34).

But in order to compare the pressure on the parabola, as repre-
sented by, or implied in the above equation, with that upon its cir-
cumscribing parallelogram, we have only to recollect, thatc?zz|/,
and consequently, the pressure on the rectangle, is

consequently, by omitting the common factors, and rendering the
fractions similar, we have

p : P : : 4 : 5.

78. Now, the practical rule for determining the pressure on the
parabolic plane, as deduced from the equation (34), or from the pre-
ceding analogy, may be expressed in words, as follows.

OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA. 67

RULE. Multiply two fifths of the base of the parabola, by
the square of the length of its axis drawn into the specific
gravity of the fluid, and the product will express the pressure
sustained by the parabolic plane, in a direction perpendicular
to its surface. Or thus :

Find the pressure on the circumscribing rectangular paral-
lelogram, according to the second case of the rule under the
third problem, and four fifths of the pressure so determined,
will express the pressure perpendicular to the parabolic
surface.

79. EXAMPLE 14. A parabolic plane, whose base and vertical axis
are respectively equal to 28 and 42 feet, is perpendicularly immersed
in a reservoir of water, so that its vertex is just in contact with the
surface ; what weight is equivalent to the pressure on the plane, the
weight of a cubic foot of water being equal to 62 J Ibs. ?

Here, according to the rule, we have
p = f x 28 X 422 X 62J = 1234800 Ibs.,

or by the second clause of the rule, it is
p — \ X 28 x 42* X 62J X * = 1234800 Ibs.
Either of these methods is sufficiently simple for every practical
purpose ; but it will be found of essential advantage, to bear in mind
the relation between the pressure on the parabola and that on its
circumscribing rectangle ; for which reason, the latter method may
probably claim the preference.

2. METHOD OF FINDING THE CENTRE OF GRAVITY OF THE SPACE
INCLUDED BETWEEN ANY RECTANGULAR PARALLELOGRAM AND
ITS INSCRIBED PARABOLA.

80. It is a principle almost self-evident, that the centre of gravity,
and the centre of magnitude of a rectangular parallelogram, exist in
one and the same point ; consequently, admitting the position of the
centre of gravity of the rectangle to be known or determinable a priori,
the position of the centre of gravity of the inscribed parabola can
very readily be found.

For by knowing the position of the centre of gravity of a rectangu-
lar surface, the pressure upon it can easily be ascertained, and we
have shown above, that the pressure upon a parabolic plane, and that
upon the surface of its circumscribing parallelogram, are to one
another in the ratio of 4 to 5 ; hence, when the pressure on the

68 OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA.

rectangular parallelogram is known, the pressure on the inscribed
parabola is also known, being equal to four fifths of that upon the
parallelogram.

Now, it has already been demonstrated, that the pressure upon any
surface, whatever may be its form, is always equal to its area, drawn
into the perpendicular depth of the centre of gravity, below the upper
surface of the fluid ; therefore, conversely, the perpendicular depth
of the centre of gravity of any surface, is equal to the pressure which
it sustains, divided by the area.

But from what, has been demonstrated above, it is manifest that
the area of the parabola and the pressure upon it, are respectively
expressed by

f&Z, and f&/2s;

consequently, by division, we obtain

and when s is expressed by unity, as in the case of water, we get

a=#

Now, because the parabola ACB is symmetrically divided by the
axis CD, it follows, that the centre of gravity occurs in that line, and
we have just shown, that it occurs at the distance of three fifths of its
length from the vertex ; hence, the position is determined, and that
independently of computing the corresponding horizontal rectangular
co-ordinate, whose intersection with the axis fixes the place of the
centre sought.

The aggregate pressure upon the two equal and similar spaces A E c
and BFC, is obviously equal to the difference between the pressures
on the rectangular parallelogram ABFE, and that on the inscribed
parabola A c D ; that is,

where p' denotes the pressure on the spaces A EC and BFC.

Again, the area of the spaces A EC and BFC, is equal to the dif-
ference between the area of the parallelogram ABFE, and that of the
inscribed parabola ACB; therefore, if a1 denote the area of the trian-
gular spaces, we have

a' — A — a=zbl — %bl = %bl.

But the depth of the centre of gravity of any surface, is equal to
the pressure upon that surface divided by its area ; consequently, the
depth of the centre of gravity of the figure A EC FB, which is composed
of the two triangular spaces AEC and BFC, is

OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA.

hence, if the specific gravity of the fluid be expressed by unity, we get

V = Thl. (35).

COROL. It therefore appears, that the centre of gravity of the space,
included between any rectangular parallelogram and its inscribed
parabola, is situated in the axis, at the distance of three tenths of its
length from the vertex.

The preceding investigation determines the place of the centre of
gravity to be at three tenths of the axis below a tangent line passing
through the vertex of the parabola, and that it is situated in the axis
is manifest; for the spaces A EC and BFC are equal, and they are
similarly and symmetrically placed with respect to the axis c D ; there
can therefore be no reason, why the centre of gravity should occur at
a point, which is nearer to the one than it is to the other ; it conse-
quently occurs at a point which is equally distant from both, and that
point must obviously be found in the axis of the figure.

The above is a valuable proposition in the practice of bridge build-
ing ; for by it, we can readily assign the position of the centre of gravity
of the arch with all its balancing materials, and consequently, many im-
portant particulars respecting the weight and mechanical thrust, may
be determined and examined with the greatest facility : all of which will
be investigated and applied in our treatise on Hydraulic Architecture.

3. WHEN THE PARABOLIC PLANE PERPENDICULARLY IMMERSED HAS
ITS BASE COINCIDENT WITH THE SURFACE OF THE FLUID.

81 . What has hitherto been done under the present problem, applies
only to the case, in which the parabola is perpendicularly immersed in
the fluid, and having its vertex coincident with the surface ; but when
the parabolic plane is perpendicularly immersed, and having its base
coincident with the fluid surface, the circumstances will be something
different, as will become manifest from the following investigation.

Let ABFE be a rectangular parallelogram immersed in a fluid, with
its plane perpendicular to the plane of the
horizon, and having its upper side coincident
with the surface of the fluid ; and let A c B be
a parabola described upon the rectangular
plane, in such a manner, that its vertex may
be downwards, its axis vertical, and its base
in contact with the surface of the fluid in which
it is placed.

JV C

70 OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA.

Draw the diagonal AF intersecting CD, the vertical axis of the
parabola in the point n ; then is n the centre of gravity of the rectan-
gular parallelogram ABFE ; and because, as we have stated in the
construction of the preceding case, the centre of gravity of a parabolic
plane is situated in the axis, at the distance of three fifths of its length
from the vertex ; it follows, that if the axis DC be divided in the point
m, into two parts such, that Dm is to me in the ratio of 2 to 3 ; then
shall m be the centre of gravity of the parabola ACB.

Put 6 = AB, the horizontal breadth of the rectangular parallelogram

ABF,E, or the base of its inscribed parabola ACB,
/ — DC or AE, the vertical axis of the parabola, or the depth of

its circumscribing rectangle,

d — D n , the depth of the point n, below A B the surface of the fluid ,
£ nr DW, the depth of the point m as referred to AB,
P — the pressure on the surface of the rectangle ABFE,
j9=:the pressure on the parabolic surface ACB,
Az=the area of the rectangular parallelogram,
a — the area of its inscribed parabola, and
s zn the specific gravity of the fluid.

Now, it is manifest from the principles of mensuration, that the
area of a rectangular parallelogram, is equal to the product that
arises when the two dimensions of length and breadth are multiplied
into one another ; that is,

A=bl,

and according to the writers on conic sections, the area of a parabola
is equal to two thirds of the area of its circumscribing rectangle ;

therefore, we have

oz=f6Z.

Referring to the construction of the figure, we find that the axis
DC is divided at m, into the two parts Dm and me, having to one
another the ratio of 2 to 3 ; it therefore follows, that

Dmzi: £ — -§:/;
consequently, the pressure on the parabolic surface ACB, is

p — ^blX^lXs — ^b l*s. (36).

Again, since AF the diagonal of the parallelogram, bisects DC the
axis of the parabola in the point n ; it follows, that

Dw = d—|/;

therefore, the pressure perpendicular to the surface of the rectangular
parallelogram ABFE, becomes

P — blX i/X s=

OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA. 71

hence, by analogy, we obtain

p : P ::T\bl*s : \bl*s,
and this, by omitting the common quantities, b, I1 and s, becomes

P : P : : rV : 1,

and finally, by reducing the fractions in this analogy to a common
denominator, we shall obtain

p : P : : 8 : 15.

82. Having thus established the formula for determining the pres-
sure on the parabolic plane, and also the ratio which compares the
said pressure with that upon the circumscribing rectangular paral-
lelogram ; we shall in the next place deduce the rules, by which the
numerical process is to be performed, when the principle is applied to
the actual determination of the pressure in reference to cases of prac-
tice. The rule, as derived from the equation numbered (36), may be
expressed in the following manner.

RULE. Multiply the base of the parabola by the square of
its vertical axis, and again by the specific gravity of the
fluid ; then, four fifteenths of the product will express the
pressure perpendicular to the surface of the parabolic plane.

The rule for determining the pressure on the surface of a parabola,
as deduced from the analogy of comparison investigated above, may
be expressed in words in the following manner.

Find the pressure perpendicular to the surface of the
circumscribed rectangular parallelogram, after the manner
described in the second case of the rule under the third
problem; then, eight fifteenths of the pressure so determined,
will express the pressure on the parabolic plane.

83. EXAMPLE 15. The data remaining as in the preceding case ;
what will be the pressure on the parabolic plane, its axis being verti-
cal, and its base in contact with the surface of the fluid ?

28 X 42 X 42 X 62J = 3087000,

and four fifteenths of this, is
3087000 X 4 -r 15 zz 823200 Ibs.

This result is derived from the first of the above rules, or that which
corresponds to the equation (36) ; and the process as performed by the
second rule, or that obtained from the ratio of comparison, will stand
as below.

} X 28 X 42 X 42 X 62£ X TV = 823200 Ibs.

72 OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA.

Hence, the pressure on the plane in this case, is only two thirds of
what we found it to be in the foregoing case, where the vertex is in
contact with the surface of the fluid.

With respect to the position of the centre of gravity in this case, it
is manifest, that the mode of discovering it, is similar to that which
we employed in the case immediately preceding, where the axis of the
parabola was supposed lo be vertical, and its s-ummit in contact with
the surface of the fluid ; it is therefore unnecessary to repeat the
investigation, but there is another condition of the figure remaining
to be considered, in which a knowledge of the position of the centre
of gravity becomes of more importance, as will readily appear from
the circumstances which present themselves in the solution of the
following problem.

4. WHEN THE BASE OF THE PARABOLIC PLANE IS PERPENDICULAR TO
THE HORIZON, ITS AXIS HORIZONTAL, AND THE PRESSURE UPON
IT IS TO BE DETERMINED AS COMPARED WITH THAT UPON ITS
CIRCUMSCRIBING RECTANGULAR PARALLELOGRAM.

PROBLEM XII.

84. If a parabolic plane be perpendicularly immersed in an
incompressible fluid, in such a manner, that its base may be
vertical, and just in contact with the surface : —

It is required to determine the pressure upon it, and to
compare it with that upon its circumscribing rectangular
parallelogram.

Let ABEF be a rectangular parallelogram immersed in a fluid, with
its plane perpendicular to the plane of the
horizon, and its upper side AB coincident
with the surface of the fluid in which it is
immersed.

Bisect AF and BE in the points D and c;
join DC, and upon AF as a base, with the
corresponding axis DC, describe the parabola
ACF, touching AB the surface of the fluid in
the point A ; then is AC F the surface for which the pressure is required
to be investigated.

Join BF, intersecting DC the axis of the parabola in the point n-r
then is n the centre of gravity of the rectangular parallelogram ABEF.

OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA. 73

Divide the axis DC into two parts Dm and cm such, that Dm is to
cm, in the ratio of 2 to 3; and the point m thus determined, is the
place of the centre of gravity of the parabola ACF. Through the
points m and n, draw the straight lines mr and ns respectively per-
pendicular to the axis DC; then are mr and ns7 the perpendicular
depths of the points m and n below AB, the surface of the fluid, and
which in the present case are equal to one another.

Put &ZZIAB, the horizontal breadth of the rectangular parallelogram
ABEF, or the axis of its inscribed parabola ACF,

I — AF, the length of the circumscribing rectangle, or the base
of the inscribed parabola,

d — rm or sn, the vertical depths of the centres of gravity, below
AB the surface of the fluid,

Pmthe pressure on the rectangle ABEF,

/> = that on the inscribed parabola ACF,

A = the area of the circumscribing rectangular parallelogram,

a — the area of the parabola, and

s —the specific gravity of the fluid in which they are immersed.

Then, according to the principles of mensuration, the area of the
rectangular parallelogram ABEF, is equal to the product of the
breadth AB drawn into the depth AF ; that is,

A = bl;
and by the property of the parabola, its area is

a = \b I.

But the pressure perpendicular to the surface of the rectangular
parallelogram, is, as we have already frequently stated, expressed by
the area drawn into the perpendicular depth of the centre of gravity ;
and this being the case, whatever may be the form of the surface
pressed, it follows, that the pressure on the rectangle ABEF, is

P — bl X d X s — bdls'y
and that on the parabola ACF is

Now, it is manifest from the nature of the figure, and from the
principles upon which it is constructed, that rm and sn are each of
them equal to JAF; that is, d=L\l\ therefore, let \l be substituted
for d in each of the above equations, and we shall obtain
For the rectangle ABEF,

and for the parabola ACF, it is

(37).

74 OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA.

Consequently, by analogy, the comparative pressures on the para-
bola and its circumscribing rectangle, are as follows :

p: P ::ibl*s:%bl*s;
and from this, by casting out the common quantities and assimilating

the fractions, we get

p : P : : 2 : 3.

COROL. Hence it appears, that when the axis of the parabola is
horizontal, and its base perpendicular to the horizon ; the pressure
perpendicular to its surface, when compared with that on its circum-
scribing parallelogram, bears precisely the same relation, that its area
bears to the area of the rectangle by which it is circumscribed.

85. The practical rule for determining the pressure on the parabolic
plane, when placed in the position specified in the problem, may be
expressed in words at length in the following manner.

RULE. Multiply the horizontal axis, by the square of the
vertical base or double ordinate, and again by the specific
gravity of the fluid ; then, take one third of the product for
the pressure perpendicular to the surface of the parabolic
plane.

Or thus, Calculate the pressure on the circumscribing
rectangle, and take two thirds of the result for the pressure
on the parabola.

86. EXAMPLE 16. The data remaining as in the example to the
foregoing problem, it is required to determine the pressure on the
parabolic plane, supposing its axis to be horizontal, its base or double
ordinate vertical, and the upper extremity of the base in contact with
the surface of the fluid, which, according to the conditions of the
previous question, is water, whose specific gravity is expressed by
unity, and the weight of one cubic foot of which is equal to 62£ Ibs.
avoirdupois ?

Referring the numerical data to the same parts of the figure, as in

the preceding cases, we have given 6 = 42 feet; /iz:28 feet, and

s zz: 62 j Ibs. ; therefore, by proceeding according to the rule, we have

42 X 28 X 28 X 62 J = 2058000,

which being divided by 3, gives

p = 2058000 -r 3 = 686000 Ibs.

Hence it appears, that the total pressures perpendicular to the
parabolic surface, according to the several positions in which we have
placed it, are to one another respectively as the numbers
3087, 2058 and 1715;

OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA. 75

the first two of which, by reason of the parts of the figure being the
same in each, are obviously dependent upon one another; but the
third, in which the parts of the figure are reversed, is wholly inde-
pendent and distinct from the other two.

87. COROL. 1. Admitting therefore, that the pressure upon the
parabolic surface, under the three circumstances of position in which
we have considered it, is represented by the equations (35), 36) and
(37) ; it follows, that the situation of the centre of gravity can easily
be ascertained ; for the pressure in each case, as we have elsewhere
shown, is represented by the area of the figure, drawn into the per-
pendicular depth of its centre of gravity ; consequently by reversing
the process, the depth of the centre of gravity will become known, if
the pressure be divided by the area of the surface on which the fluid
presses.

COROL. 2. Since the parabola is a figure symmetrically situated with
respect to its axis, it is obvious, that the centre of gravity of its surface
must occur at the same point, in whatsoever position it may be placed;
but when the place of its centre is referred to the surface of the fluid
in which it is immersed, the distance varies for each particular case:
thus,

In the first instance, the perpendicular distance, is in f ths of the axis,

second, izrfths

third, m ^ the base.

But as we have just stated, the centre of gravity of the parabolic sur-
face as referred to its vertex, or any other fixed point, in all these cases,
remains unaltered, in whatever position the figure itself may be placed.

5. THE METHOD OF DETERMINING THE PRESSURE OF THE FLUID UPON
A SEMI-PARABOLIC PLANE AS COMPARED WITH THAT ON THE
CIRCUMSCRIBING RECTANGULAR PARALLELOGRAM.

88. When the semi-parabola only is considered, the determina-
tion of its centre of gravity, and consequently, of the pressure on
its surface becomes more difficult ; for, since the figure is not sym-
metrical with respect to its axis, we are under the necessity of com-
puting the two rectangular co-ordinates, whose in-
tersection determines the place of the required centre.

Let CBD be a semi-parabola, perpendicularly
immersed in a fluid, so that its axis CD is vertical,
and the vertex in contact with CF the surface of the
fluid, and let CFBD be the circumscribing rectan-
gular parallelogram.

76 OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA.

Now, since by Problem XI, it has been proved that the pressure on
the entire parabola with its axis vertical, is equal to four fifths of that
upon its circumscribing parallelogram ; it follows, that the pressure
on the semi-parabola in the same position, is also equal to four fifths
of that upon its circumscribing parallelogram, it being manifestly
equal to half the pressure on the whole parabola.

Divide the axis CD into two parts, such, that Dm and cm shall be
to one another in the ratio of 2 to 3 ; and in like manner, let the
ordinate or base BD be divided into two parts, such, that DW and vn
shall be to one another in the ratio of 3 to 5* ; then, through the
points m and n, and respectively parallel to DB and DC, draw the
straight lines WIG and no, meeting each other in G, the centre of
gravity of the semi-parabola DC B.

Put &IZICF or DB, the horizontal breadth of the rectangle CFED, or

the base of the semi-parabola c B D,
I =: CD or FB, the vertical depth of the rectangle, or axis of its

inscribed semi-parabola,

cZzz cm or EG, the perpendicular depth of the centre of gravity,
Pzzthe pressure on the rectangular parallelogram CFBD,
p zz the pressure on its inscribed semi-parabola,
A zz the area of the parallelogram,
a zz the area of the semi-parabola, and

s zz the specific gravity of the fluid in which they are immersed.
Then, according to the principles of mensuration, the area of a
rectangle is expressed by the product of its two dimensions ; that is,
by its length drawn into its breadth ; therefore, we have

A = bl,

and by Proposition (A), the pressure exerted by a fluid, perpendicu-
larly to any surface immersed in it, or otherwise exposed to its
influence,

Is equal to the area of the surface pressed, drawn into the

perpendicular depth of its centre of gravity, and again into

the specific gravity of the fluid.

Consequently, the pressure on the circumscribing rectangular paral-
lelogram CFBD, becomes

* It is demonstrated by the writers on mechanics, that the centre of gravity of a
semi-parabola is situated in its plane, at the distance of three eighths of the ordinate
from the axis, and two fifths of the axis from the ordinate, or three fifths of the axis
from its vertex.

OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA. 77

Now, since the pressure on the semi-parabola is equal to four fifths
of the pressure on its circumscribing parallelogram, we shall obtain

f P = $ X Ibl^s — ^b^s—p. (38).

The expression which we have here determined for the pressure on
the surface of the semi-parabola D c B, is precisely the same as that which
we have given in equation (35) for the entire figure ; only in the pre*
sent instance, the value of 6, the horizontal breadth of the parallelogram,
is but one half the value as applied to the parabola, when placed
under the conditions specified in the eleventh problem foregoing.

89. If the symbol b retain its former value, that is, if it be referred
to the base of the entire parabola, or to the breadth of the parallelo-
gram circumscribing the entire parabola, then, the pressure on the
semi-parabola, becomes '

p — \bl^s. (39).

Consequently, the practical rule for determining the pressure on the
semi-parabola as deduced from this equation, may be expressed as
follows.

RULE. Multiply one fifth of the base, or double ordinate of
the whole parabola, by the square of the length of its axis,
and again by the specific gravity of the fluid, and the product
will express the pressure on the semi-parabola in a direction
perpendicular to its surface.

But if the symbol b refer to the ordinate, or base of the semi-
parabola, then, the rule as deduced from the equation (38), will be
precisely the same as that which we have given under the equation
numbered (35) in Problem XI, to which place the reader is referred
for the purpose of avoiding a direct repetition.

90. EXAMPLE 17. A plane in the form of a semi-parabola whose
base or ordinate is 16 feet, and its axis 40 feet, is perpendicularly
immersed in a cistern of water, in such a manner, that its axis is
vertical, and its vertex in contact with the surface of the fluid ; what
pressure does it sustain, the weight of a cubic foot of water being
equal to 62 Jibs.?

The equation in its present state, supposes the ordinate, or base of
the semi-parabola to be given, and therefore, the pressure is deter-
mined by the rule to the equation (35) or (38), in the following
manner :

p—l x 16 X402 X 621= 640000 Ibs.

But in order to determine the pressure by the rule immediately
preceding, we must suppose the breadth or base of the figure to be

78 OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA.

doubled, in which case it will have reference to the whole parabola,
and the pressure will be reduced to that upon its half, by employing
the constant \ instead of f- according to the rule, thus,
p = i x 32 x 402 X 62J = 640000 Ibs.

91. If the axis of the semi-parabola were horizontal and its ordinate
vertical, as in the annexed diagram ; then,
the area of the semi-parabolic figure, as well
as that of its circumscribing parallelogram,
will remain the same, but the pressures
perpendicular to their respective surfaces
will be very different.

Divide BD in n, in such a manner, that En and DW may be to one
another in the ratio of 5 to 3 ; and in like manner, divide CD in m,
so that cm and Dm shall be to each other in the ratio of 3 to 2.

Through the points n and m, and parallel respectively to B F and
BD, draw the straight lines no, and ma, intersecting one another in
the point G, arid produce m G to E ; then , by the note to the pre-
ceding case, the point G is the centre of gravity of the semi-parabola
DCB, and EG is its perpendicular depth below BF, the horizontal
surface of the fluid.

Therefore, let the preceding notation remain, and let the several
symbols refer to the same parts of the figure as in the preceding case,
disregarding the change of position which has taken place ; then, as
before, the area of the parallelogram BFCD, is

A = 6/,

and the pressure perpendicular to its surface, is

For draw the diagonals B c and F D intersecting each other in the
point r, and through r draw rs parallel to BD or Em; then, r is the
centre of gravity of the rectangle BFCD, and sr its perpendicular
depth below BF the surface of the fluid; but according to our notation
srzr \b, and we have seen above, that A zz: bl; now, the pressure on
any surface, whatever may be its form,

Is equal to the product that arises, when the area of the
surface pressed, is drawn into the perpendicular depth of its
centre of gravity, and again, into the specific gravity of the
fluid.

Consequently, the pressure perpendicular to the surface of the
rectangular parallelogram BFCD, is

OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA. 79

Again, the area of the semi-parabola BCD, is equal to two thirds of
its circumscribing rectangular parallelogram BFCD ; therefore we have

a = l X bl=$bl,
and the pressure perpendicular to its surface, is

p = &Vl8.

This is manifest, for according to the construction and the nature
of the figure of the parabola, BW or EG is equal to five eighths of BD ;
therefore, we have

p — ^blXibX s = -frb*ls', (40).

consequently, by analogy, we obtain

p : P : : &&ls : ±b*ls.

Therefore, by suppressing the common factors, and rendering the
fractions T\ and | similar, we shall get

p:P::5: 6;

hence it appears, that when the ordinate of the semi-parabola is
vertical, and its upper extremity in contact with the surface of the
fluid :—

The pressure upon the semi-parabola, is to that upon its
circumscribing rectangular parallelogram, as 5 is to 6, or as
1 is to 1.2.

92. Consequently, the practical rule for determining the pressure
in the present instance, as deduced from the equation marked (40),
or from the above analogy, may be expressed as fdllows.

RULE. Multiply the square of the given ordinate by the
axis of the semi-parabola, and again by the specific gravity
of the fluid ; then, Jive twelfths of the result will give the
pressure sought. Or thus,

Find the pressure on the circumscribing parallelogram, and
take five sixths of the pressure thus found, for the pressure on
the semi-parabola.

93. EXAMPLE 18. Let the numerical values of the axis and ordi-
nate remain as in the preceding example ; what will be the pressure
on the surface of the semi-parabola, supposing the axis to be hori-
zontal, the ordinate vertical, and its remote extremity in contact with
the surface of the fluid ?

If the operation be performed according to the rule deduced from
equation (40), we shall obtain

p= 162 X 40 X 62£ X TV = 2666661- Ibs.

80 OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA.

but if the operation be performed according to the rule derived from
the analogy of comparison, we shall have

p = f P ; that is, p — \ x 16* X 40 X 62 1 X £ = 266666f Ibs.

6. THE METHOD OF DETERMINING THE POSITION OF THE CENTRE
OF GRAVITY OF THE SPACE COMPREHENDED BETWEEN THE
PARABOLIC CURVE AND ITS CIRCUMSCRIBING PARALLELOGRAM.

94. By means of the pressure on the semi-parabola, as we have
investigated it in the two foregoing cases, we are enabled to determine
the pressure on the space CFB, and from thence, the position of its
centre of gravity.

This is an important inquiry in the practice of bridge building, for,
in determining the thickness of piers necessary to resist the drift or
shoot of a given arch, independently of the aid afforded by the other
arches, it becomes requisite to find the centre of gravity of the span-
drel or space BFC, which is used for the purpose of balancing the arch
and filling up the haunches or flanks.

Now, the method which has generally been employed for the deter-
mination of this centre is extremely operose, and in many cases it
involves considerable difficulty, requiring the calculations of solids
and planes, which are by no means easy ; but the method which we
are about to employ, requires no such tedious and prolix operations,
as will become manifest from the following investigation, which refers
to the space comprehended between a semi-parabola and its circum-
scribing rectangle.

Let BCD be a semi-parabola, having the axis DC vertical while the
corresponding ordinate DB is horizontal, and
let B DC F be the circumscribing rectangular
parallelogram.

Suppose the point D to remain fixed, and
conceive the semi-parabola BCD to revolve
about the point D until it comes into the
position A ED, where the axis DE is horizon-
tal, and the corresponding ordinate DA vertical; then it is manifest,
that the circumscribing rectangular parallelogram ADEH in this
latter position, is equal to BDCF in the former, and consequently,
the space AHE comprehended between the sides of the rectangle
AH, HE and the curve AE, is equal to the space BFC similarly con-
stituted.

OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA. 81

It is further manifest, that while the semi-parabola revolves about
the point D, from the position BCD to that of A ED, the points B and c,
the extremities of the ordinate and axis, describe respectively, the
circular quadrants BA and CE, while the point F describes another
quadrant, whose containing radii are the diagonals DF and DH.

Put 6:zr BD or AD, the ordinate of the semi-parabola in either posi-

tion,

I n= CD or ED, the corresponding axis,
d — nG, the depth of the centre of gravity of the space BFC,

when the axis is vertical,
3 rz m G, or A ft, the depth of the centre of gravity of the space

AHE or BFC, when the axis is horizontal,
Pzz: the pressure on the circumscribing rectangular parallelogram

BDCF, Or AH ED,

p — the pressure on the inscribed semi-parabola, and

p — the pressure on the space comprehended between the semi-

parabola and its circumscribing rectangular parallelo-

gram.

Then, according to equation (8) under the third problem, the
pressure on the circumscribing rectangular parallelogram when the
length is vertical, is

and agreeably to equation (38) under the eleventh problem, the
pressure on the inscribed semi-parabola with the axis vertical, is

consequently, by subtraction, the pressure upon the space BFC, com-
prehended between the sides of the parallelogram BF, FC and the
curve of the parabola BC, is

hence, by suppressing the symbol for the specific gravity, we get
p' — bl^(l— f)rrTV^2. (41).

Now, according to the writers on mensuration, the area of the
semi-parabola is equal to two thirds of the area of the circumscribing
parallelogram ; it therefore follows, that the area of the space BFC, is
equal to one third of the rectangle BDCF; that is,
bi — $bl=;$bl.

But it has been demonstrated, that the pressure upon any surface,
is equal to the area of that surface, drawn into the perpendicular
depth of the centre of gravity ; consequently, we have

VOL. i. G

82 OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA.

and we have shown above in equation (41), that when the axis of the
semi-parabola is vertical, the pressure on the space BFC is

consequently, by comparison, we obtain
#*!=&*+

and finally, dividing by ^bl, we shall have

Again, when the axis of the semi-parabola is horizontal, as indicated
by A ED, the pressure on the circumscribing rectangle, according to
equation (10) under the third problem, is

and the pressure upon the inscribed parabola, according to equation
(40) under the eleventh problem, is

therefore, by subtraction, the pressure upon the space comprehended
between the rectangular parallelogram and its inscribed semi-parabola,

p' — P — p = J6* /* — TM>2 ' «,
and by suppressing the symbol for the specific gravity, we have

Now, the area of the inscribed semi-parabola is, as we have seen
above, equal to two thirds of its bounding rectangle, and the area of
the space comprehended between the rectangle and the semi-parabola,
is therefore, equal to one third of the same quantity ; that is,

bl—lbl = *bl;
consequently, the pressure on the irregular space A HE, is

p'=ibtl;
hence, by comparison, we shall have

^=TV^;

therefore, by division, we obtain

Having thus determined the values of the rectangular co-ordinates,
as represented by the equations (42) and (43), the position of the
centre of gravity can easily be found ; for, from the point F, set off
rm equal to three tenths of the axis CD, and jfn equal to one fourth
of the ordinate BD, or its equal FC ; then, through the points m and n,
and parallel respectively to the ordinate BD and axis CD, draw the

OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA. 813

straight lines ma and HG, intersecting each other in the point G ; and
the point G thus determined, is the position of the centre of gravity
of the space comprehended between the semi-parabola and its circum-
scribing rectangular parallelogram.

This method of determining the position of the centre of gravity of
the space comprehended between the curve and its circumscribing
parallelogram, will be illustrated and applied in all its generality,
when we come to treat on the subject of Hydraulic Architecture, to
which it more properly belongs ; and for this reason, we shall take
no further notice of it in this place, but proceed with our inquiry
respecting pressure, which is more immediately the object of our
research.

7. METHOD OF DIVIDING A PARABOLIC PLANE PARALLEL TO ITS BASE,
SO THAT THE FLUID PRESSURES ON EACH PART MAY BE EQUAL
TO ONE ANOTHER.

PROBLEM XIII.

95. If a parabolic plane be immersed perpendicularly in an
incompressible fluid, in such a manner, that its vertex is j ust in
contact with the surface : —

It is required to determine at what distance from the
vertex, a straight line must be drawn parallel to the base, so
that the figure may be divided into two parts, on which the
pressures are equal to one another.

Let ACB be the given parabola, immersed in the fluid after the
manner specified in the problem, and let aAEb be the circumscribing
rectangular parallelogram.

Take cm for the distance from the vertex
through which the line of division passes, and
draw EF parallel to the base AB; then are
the spaces ABFE and ECF, the parts into
which the parabola is divided, and on which,
by the conditions of the problem, the pres-
sures are equal.

Through the points E and F, the extremities of the line of division,
draw EC and FG? respectively parallel to CD the axis of the figure;
then is CEFC?, the rectangular parallelogram circumscribing the para-
bola ECF.

84 OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA.

Put 26 = AB or ab, the base of the parabola ACB, or the horizontal

breadth of the circumscribing parallelogram OAB&,
I zz CD or a A, the vertical axis of the parabola ACB, or the

depth of the rectangle by which it is encompassed,
2y zz EF or cd, the base of the parabola ECF, or the breadth of

the rectangle c E F d,
x zz cm or CE, the axis of the parabola ECF, or the depth of its

circumscribing rectangle CEFC?,
P zzthe pressure on the rectangular parallelogram CEAB&, cir-

cumscribing the parabola ACB,
p zz the pressure on the inscribed parabola,
P' zz the pressure on the rectangular parallelogram CEF^, cir-

cumscribing the parabola ECF,
p' zz the pressure on the inscribed parabola, and
* zz the specific gravity of the fluid.

Then since ABzz26 and EFZz2y, it follows, that AD — 6 and
EWI zz y ; therefore, by the property of the parabola, we have

and consequently, by equating the products of the extreme and mean
terms, we shall obtain

y=p*,

and by division, it is

b*x

y=T>
and this, by extracting the square root, becomes

and finally, multiplying by 2, we obtain
EFZZ ^ — ^

Now, the pressure perpendicular to the surface of the rectangular
parallelogram aAB#, according to equation (8) under the third
problem, is

but we have seen elsewhere, that the pressure on the surface of a
parabola, is equal to four fifths of that upon its circumscribing paral-
lelogram ; consequently, the pressure on the parabola ACB, is

(44).

OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA. 85

Again, the pressure perpendicular to the surface of the rectangular
parallelogram CEF<?, is

and the pressure upon the inscribed parabola ECF, is four fifths of
the pressure on the circumscribing rectangle ; that is,

I ' (45).

but according to the conditions of the problem,
p' •=. \p ; hence we have

and by suppressing the common factors, we get

from which, by squaring both sides, we obtain

o:5-|*5;
consequently by extracting the fifth root of both sides, we get

x = l#1i
but according to the arithmetic of surd quantities

^7=1^87

therefore, by substitution, we shall have

now, the sursolid, or fifth root of 8, is 1.51571 ; hence we get

a: = .75785*. (46).

96. The practical rule supplied by this equation is extremely
simple; it may be expressed in words at length in the following
manner.

RULE. Multiply the axis of the given parabola by the
constant number .75785, and the product will give the distance
from the vertex through which the line of division passes.

97. EXAMPLE 19. The axis of a parabola is 29 feet, and its plane
is perpendicularly immersed in a cistern of water, in such a manner,
that its vertex is just in contact with the surface ; through what point

86 OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA.

in the axis must a line be drawn parallel to the base, so that the pres-
sures on the two parts into which the parabola is divided, may be
equal to one another ?

By operating according to the preceding rule derived from equation
(46), we shall have for the distance from the vertex

a,— .75785 X 29 = 21.97765 feet.

In the case which we have investigated above, the parabola is
divided into two parts on which the pressures are equal ; but in order
to render the solution general, we must so arrange it, that the parts
of division may bear any ratio to one another, as denoted by the
symbols m and n ; that is,

P '• P — P' '• ' m : n.

Now, we have seen in equation (44), that the pressure on the entire
parabola ACB, is

and according to equation (45), the pressure on the parabola
ECF, is

but the pressure upon the space AEFB, is manifestly equal to the
difference of these ; that is,

consequently, we obtain

: : m : n;

and from this, by equating the products of the extreme and mean
terms, we shall obtain

which, by transposing and collecting the terms, becomes
(m -f w) x*^/ -^- = ml\

from which, by division, we shall get
/~~x _ ml*

XV T—^r~n;

OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA. 87

therefore, by involving or squaring both sides of this equation, we
shall obtain

and by extracting the sursolid root, we get

= '1/0

»)*' (47).

If m and n be equal to one another as in the preceding case, then
it is obvious that the equation becomes

and if any other numerical ratio be proposed, such as 4 to 5; then
we shall have

Thus for example; let the axis of the parabola remain as in the
foregoing question, and let it be required to find a point, through
which a line must be drawn parallel to the base, so that the pressure
on the part above the dividing line, may be to that below it in the
ratio of 4 to 5 ?

Here we have

= 20.967 feet.

Hence it appears, that if a point be taken in the axis of the given
parabola at the distance of 20.967 feet from the vertex, and if through
that point, a line be drawn parallel to the base, the parabola will be
divided into two parts, on which the pressures are to one another
as 4 to 5.

CHAPTER IV.

OF THE PRESSURE OF INCOMPRESSIBLE FLUIDS ON CIRCULAR

PLANES AND ON SPHERES IMMERSED IN THOSE FLUIDS, THE

EXTREMITY OF THE DIAMETER OF THE FIGURE BEING IN EACH
CASE IN EXACT CONTACT WITH THE SURFACE OF THE FLUID.

PROBLEM XIV.

98. Suppose a circular plane to be immersed perpendicularly
in an incompressible fluid, in such a manner, that the extremity
of the diameter is just in contact with the surface : —

It is required to draw from the lowest point of the circular
plane, that chord on which the pressure shall be a maximum.

Let ABC be the circular plane immersed in the fluid according to
the conditions of the problem ; draw the ver-
tical diameter BC touching the surface of the
fluid in the point B, and let CA be the chord
required.

Bisect the chord CA in the point m, and
through the point m thus determined, draw
mn parallel to BC the vertical diameter, meeting the surface of the
fluid in n j then is n m the perpendicular depth of the centre of gravity
of the chord AC, below the surface of the fluid in which it is immersed ;
draw also AE and mD respectively perpendicular to the diameter BC.

Now, we have already demonstrated in the first Problem, that the
pressure upon a physical line, is equal to the product of its length by
the perpendicular depth of its centre of gravity, and again by the
specific gravity of the fluid ; consequently, we have

p = AC x nm X s.
Put d~ BC, the diameter of the immersed circular plane,

I •=. mn, the perpendicular depth of the centre of gravity of the
chord AC,

Of FLUID PRESSURE ON CIRCULAR PLANES AND ON SPHERES. 89

/ = AC, the length of the chord on which the pressure is a

maximum,

p in the pressure on the chord AC,
x =. c D, the perpendicular height of the centre of gravity of the

chord A c above the lower extremity of the diameter,
and s ~ the specific gravity of the fluid.

Then we have o = c? — x; CE=z:2a:, and by the property of the
circle, the length of the chord becomes

consequently, the pressure upon it is

but this, according to the conditions of the problem, is to be a maxi-
mum ; therefore, by putting the fluxion of the expression equal to
nothing, we obtain

MX (d* — 4dx + 3z2) = 0 ;

therefore, by omitting the common factors 2dx and transposing, we
shall have

and this quadratic equation being reduced, we get

x — ^d. (48).

COROL. 1. Consequently, to determine the chord by construction,
make BE equal to one third of the vertical diameter BC, and through
the point E, draw the straight line EA at right angles to BC, and
meeting the circumference in the point A ; then from c, the lower
extremity of the diameter, inflect the straight line CA, and the thing
is done. Or thus :

2. Find 0 an angle such, that tan. 0 — J^/2"—. 707 11, which
happens when 0zn 35° 15' 51" ; therefore, at c the lower extremity
of the diameter, make the angle BCA equal to 35° 15' 51", and the
straight line CA will be the chord required.

It has been shown above, that according to the property of the
circle, the length of the chord is £== ^2dx; if, therefore, the value
of x as determined in equation (48), be substituted instead of it, in
the foregoing value of /, we shall have

l=ldi/6. (49).

99. This is the proper form of the expression when adapted for
numerical operation, and the practical rule which it supplies, may be
expressed as follows.

90 OF FLUID PRESSURE ON CIRCULAR PLANES AND ON SPHERES.

RULE. Multiply one third of the diameter of the given
plane by the square root of 6, and the product will be the
length of the chord required. Or thus :

Multiply the given diameter by the constant number
.81647, and the product will be the length of the chord
required.

100. EXAMPLE 20. A circular plane whose diameter is equal to
36 feet, is perpendicularly immersed in a fluid, so that the upper
extremity of the vertical diameter is in contact with the horizontal
surface ; what is the length of a chord, which being inflected from
the lower extremity of the diameter, sustains a greater pressure than
any other chord which can be drawn from the same point ?

Here by the rule, we have
f = i X 36 x VQ = 29.3929 feet,
and by the second part of the rule, we have
1=36 X .81647 —29.3929 feet.

PROBLEM XV.

101. If two spheres or globes of different diameters, be
immersed in a fluid, in such a manner, that the uppermost point
on their surface is just in contact with the horizontal surface
of the fluid :—

It is required to determine the pressure on each of the
spheres, and to compare the pressures with one another.

Let ABD and abd be the two spheres, whose diameters AB and ab,
have their upper extremities A and a in
contact with EF, the horizontal surface of
the fluid.

Bisect the diameters AB and a b, respec-
tively in the points c and c ; then are the
points c and c, the centres of magnitude of

the respective spheres ; but in a sphere, the centre of magnitude and
the centre of gravity occur in the same point ; therefore, c and c are
the centres of gravity of the spheres ABD and abd, and AC and
ac are the perpendicular depths below the surface of the fluid.

OF FLUID PRESSURE ON CIRCULAR PLANES AND ON SPHERES. 91

Put D zz AB, the diameter of the greater sphere ABD,

d n= ab, the diameter of the lesser sphere abd,
'|D = AC, the radius of the greater sphere, or the perpendicular

depth of its centre of gravity,
\d zz a c, the radius of the lesser sphere, or the perpendicular

depth of its centre of gravity,
•S zz the surface of the greater sphere ABD,
P z= the pressure perpendicular to its surface,
>S = the surface of the lesser sphere a b d,
p — the pressure perpendicular to its surface,

and TT nr 3.1416, the circumference of a circle whose diameter is
expressed by unity.

Then, according to the principles of mensuration, the surface of a
sphere or globe —

Is equal to four times the area of one of its great circles,
or that whose plane passes through the centre of the sphere.

Consequently, the convex surface of the greater sphere ABD, is
expressed as follows.

and that of the lesser sphere abd, is

S =: 3.1416 eP.

But the pressure perpendicular to any surface, is equal to the area
of that surface multiplied by the perpendicular depth of the centre of
gravity, and again by the specific gravity of the fluid ; consequently,
when the specific gravity of the fluid is denoted by unity, we have for
the pressure on the surface of the greater sphere,

P=r3.1416Da X iD=1.5708D8. (50).

and for the pressure on the lesser sphere, it is

hence, by comparison, we shall have

P : p : : D3 : d3.

Consequently, if two spheres of different diameters be placed in a
fluid under similar circumstances, the pressures perpendicular to their
surfaces, are to one another as the cubes of their diameters.

By the principles of mensuration, the solid content of a sphere or
globe, is equal to the cube of the diameter multiplied by the constant
number .5236 ; therefore, if c denote the solid content, we have

cin.5236D3,

92 OF FLUID PRESSURE ON CIRCULAR PLANES AND ON SPHERES.

or multiplying both sides of the equation by 3, we get

3czzl.5708D3;
consequently, by equation (50), we have

or if s denote the specific gravity of the fluid, we shall obtain

COROL. Hence we infer, that if a hollow sphere or globe be filled
with an incompressible and non-elastic fluid : —

The whole pressure sustained by the internal surface of the
sphere is equal to three times the weight of the fluid which it
contains.

102. EXAMPLE 21. A hollow spherical shell or vessel, whose inte-
rior diameter is equal to 30 feet, is completely filled with water ; what
weight is equivalent to the pressure sustained by its internal surface ?

Here, by operating according to the process indicated in equation
(50), we have

P=: 1.5708 X 303zr42411.6cub.ft.

Now, since the fluid with which the vessel is filled, is water, giving a
weight of 62 \ Ibs. to a cubic foot, we have

P •=. 4241 1.6 X 62 J =: 2650725 Ibs. ;

but 2240 Ibs. are equal to one ton; therefore, the pressure on the
internal surface of a hollow spherical vessel whose diameter is 30 feet,
when completely filled with water, is

P = 2650725 H- 2240 — 1 183 Hi tons.

PROBLEM XVI.

103. Suppose a sphere or globe to be immersed in an incom-
pressible and non-elastic fluid, in such a manner, that the upper
extremity of the vertical diameter is just in contact with its
surface : —

It is required to determine through what point of the axis
a horizontal plane must pass, so to divide the sphere, that the
pressure on the convex surface of the lower segment, may be
equal to the pressure on the convex surface of the upper.

Let AD BE represent the sphere in question, so placed, that A the
upper extremity of the vertical diameter, is just in contact with FG
the surface of the fluid.

OF FLUID PRESSURE ON CIRCULAR PLANES AND ON SPHERES. 93

Suppose that P is the point in the vertical
diameter through which the plane of division
passes, separating the sphere into the seg-
ments DAE and DBE, sustaining equal pres-
sures on their convex surfaces.

Bisect A p and B P in the points m and n ;
then are m and ny the points thus determined,

respectively the centres of gravity of the surfaces of the spheric seg-
ments DAE and DBE,* and Am, A.n are their perpendicular depths
below FG the horizontal surface of the fluid.

Put D — AB, the vertical diameter of the sphere or globe A DBE,

d=:Am, the depth of the centre of gravity of the surface of the

upper segment DAE,
§ zz A n, the depth of the centre of gravity of the surface of the

lower segment DBE,
S zz the surface of the upper segment,
P zn the pressure upon it,
&'zz the surface of the lower segment,
p — the pressure upon it,
s zz the specific gravity of the fluid,
x zz AP, the perpendicular depth of the point through which the

plane of division passes, and
7TZZ3.1416, the circumference of the circle whose diameter is

unity.

Then, according to the principles of mensuration, the convex sur-
face of a spheric segment : —

Is equal to the circumference of the sphere, drawn into the
versed sine or height of the segment ivhose surface is sought.

And moreover, the circumference of a sphere, or the circumference
of any of its great circles : —

Is equal to the diameter multiplied by the constant quantity
TT, or the number 3.1416.

consequently, the convex surface of the upper segment DAE, is

and that of the lower segment DBE, is
S zz D TT (D — x) zz 3.1416 D (D — x}.

* It is demonstrated by the writers on mechanics, that the centre of gravity of
the surface of a spheric segment, is at the middle of its versed sine or height.

94 OF FLUID PRESSURE ON CIRCULAR PLANES AND ON SPHERES.

But the pressure perpendicular to any surface, whatever may be its
form, as we have already sufficiently demonstrated : —

Is equal to the area of the surface multiplied by the
perpendicular depth of the centre of gravity, and again by
the specific gravity of the fluid.

Therefore, the pressure perpendicular to the convex surface of the
upper segment DAE, is

P = 3.1416Da: X i* X * = 1.5708 DSX\ (51).

and the pressure perpendicular to the convex surface of the lower
segment DBE, is

(52).

Now these two expressions, according to the conditions of the
problem, are equal to one another; consequently, by comparison,
we get

1 .5708 D s x* — 3.1416 D s {(D — x) x -f g (D — a;)8},
and from this, by suppressing the common quantities, we have

*a:=:2{(D-aO*+i(D-.af};
therefore, by expanding the terms, we obtain

2x* — DS ;
consequently, by division and evolution, we get

ar=zJDV2. (53).

104. The ultimate form of this equation is extremely simple, and
the practical rule which it supplies, may be expressed as follows.

RULE. Multiply the radius, or half the diameter of the
sphere by the square root of 2, and the product will give
the point in the vertical diameter through which the plane of
division passes, estimated downwards from the surface of the
fluid.

105. EXAMPLE 22. A sphere or globe, whose diameter is 18 inches,
is immersed in a fluid, in such a manner, that the upper extremity is
just in contact with the surface ; through what point of the diameter
must a horizontal plane be made to pass, so to divide the sphere, that
the pressures on the curve surface of the upper and lower segments
may be equal to one another ?

The square root of 2, is 1.4142, and half the given diameter is 9
inches ; consequently, by the rule we have

x =1.4142 X 9 — 12. 7278 inches,

OF FLUID PRESSURE ON CIRCULAR PLANES AND ON SPHERES. 95

106. The preceding investigation applies to the particular case, in
which the pressures on the curve surfaces of the segments are equal
to one another ; but in order to render the solution general, we must
investigate a formula to indicate the point of division, when the
pressures are to one another in any ratio whatever ; for instance, that
of m to n.

By expunging the common factors from the equations (51) and
(52), we obtain

a» : 2 { (D — a) x + } (D — *)2 } : : m : n ;

therefore, by equating the products of the extreme and mean terms,
we get

2m{(D—x)x + J(D — xY} = nx*,
which, by expanding the bracketted expression, becomes

nxt — m^ — x*},
or by transposition, we obtain

(m -\- n) x* nz m D2,
and finally, by dividing and extracting the square root, we have

rm
m+n (54).

107. The general equation just investigated, is sufficiently simple
in its form for every practical purpose that is likely to occur ; it may
therefore appear superfluous to reduce it to a rule, yet nevertheless,
that nothing may be wanting for the general accommodation of our
readers, we think proper to draw up the following enunciation.

RULE. Divide the first term of the ratio by the sum of the
termSj and multiply the square root of the quotient by the
diameter of the sphere ; then, the product thus arising, will
express the distance below the surface of the fluid, of that
point through which the plane of division passes.

It is unnecessary to propose an example for the purpose of illus-
trating the above rule ; that which we have already given, where the
values of m and n are equal to one another, being quite sufficient.

CHAPTER V.

OF THE PRESSURE OF NON-ELASTIC OR INCOMPRESSIBLE FLUIDS
AGAINST THE INTERIOR SURFACES OF VESSELS HAVING THE

; FORMS OF TETRAHEDRONS, CYLINDERS, TRUNCATED CONES,
&C.

1. WHEN THE VESSEL IS IN THE FORM OF A TETRAHEDRON.

PROBLEM XVII.

108. Suppose a vessel in the form of a tetrahedron, or equi-
lateral triangular pyramid, to be filled with an incompressible
and non-elastic fluid : —

It is required to compare the pressure on the base with that
upon the sides, and also with the weight of the fluid; the base
of the vessel being parallel to the horizon.

Let ABCD be the tetrahedron filled with fluid, of which ABC is the
base parallel to the horizon, and ABD,
c B D and ADC the sides or equal contain-
ing planes.

From D the vertex of the figure, let fall
the perpendicular DP, upon the base or
opposite side ABC; then will DP be the
vertical depth of the centre of gravity of
the base ABC, below the horizontal plane passing through D, the
summit of the figure, or highest particle of the fluid.

Bisect AD and AB, two of the adjacent edges of the figure, in the
points m and n\ draw the straight lines BWI and DW, intersecting
each other in the point r ; then is r the centre of gravity of the
triangular plane ABD.

Through the point r draw rs perpendicular to DP, the altitude of
the vessel or pyramid; then is DS, the perpendicular depth of the
centre of gravity of the triangular plane ADB, below the vertex D, or
the uppermost particle of the fluid.

OF FLUID PRESSURE UPON THE INTERIOR OF A TETRAHEDRON. 97

By the nature of the figure, the three containing planes A D E, ADC
and BDC, are equal to one another, and they are also equally inclined
to, or similarly situated with respect to the base ABC; consequently
DS, the perpendicular depth of the centre of gravity, is common to
them all.

Now, by the property of the centre of gravity, we know, that or is
equal to two thirds of D n ; therefore, by reason of the parallel lines
np and rs, DS is also equal to two thirds of DP.

Put a zn the area of the base and each of the other containing planes,
I ~ the length of the side of each triangular plane, or the edges

of the figure,
d zz: DP, the perpendicular depth of the centre of gravity of the

base ABC,
3 zr DS, the perpendicular depth of the centre of gravity of the

side ADB ;

P zz the pressure upon the b*ase,
i '

p zz the pressure upon

wzzthe weight of the fluid contained in the vessel, and
s zz the specific gravity.

Then, by the principles of Plane Trigonometry and the property of
the right angled triangle, we have

DP — d~ — \/4 — sec.* 30°,

but by the arithmetic of sines, we know, that

secJ30°zzli-;
consequently, by substitution, we have

. i? • d=-LVe. ' $ v; :•:§":•:•";

Now, according to the construction of the figure and the property
of the centre of gravity, it follows, that DS is equal to two thirds of
DP ; hence we get

* 2J _
DszzrSz:: — ^6.

By the nature of the figure, it is manifest, that the area of each of
the triangular sides is equal to the area of the base ; and by the prin-
ciples of mensuration : —

The area of an equilateral triangle, is equal to one fourth
the square of the side, multiplied by the square root. *f
three.

X-oTST-^

/ or THE

98 OF FLUID PRESSURE UPON THE INTERIOR OF A TETRAHEDRON.

Consequently, the area of the base, and each of the containing
sides of the vessel, is expressed by

therefore, the area of the three containing equilateral triangular
planes, becomes

hence, for the pressure upon the base, we have

P = 4JV3"X iV6 X s- iP5V% (55).

and the pressure upon the three containing planes, is

j> = JJV3 X JV6 X s=il*s^2; (56).

consequently, by analogy, we shall have

P :p ::±Psi/iTi JJ»«t/2;

and this, by suppressing the common factors, becomes
P : p : : 1 : 2.

If the two equations marked (55) and (56) be added together, the
sum will express the aggregate pressure upon the vessel ; therefore
we have

P + P=J>'=(i + 1) l*sV2 = $l*s^. (57).

According to the principles of mensuration, the solid content of a
tetrahedronal vessel, is equal to the area of its base, multiplied by one
third of its perpendicular altitude ; therefore, we have

IJViFx $1^/6 x i=ZTW2;

now, the weight of the contained fluid, is manifestly equal to its
magnitude multiplied by the specific gravity ; consequently, we
obtain

w = -frPsV2l (58).

hence, by analogy, we get

P : w : : %lss j$ : ^s ^2,
and this, by suppressing the common factors, gives

P : w : : 3 : 1.

COROL. It therefore appears, that the pressure upon the base, is to
the pressure on the three sides, in the ratio of 1 to 2, and to the
weight of the contained fluid in the ratio of 3 to 1 ; consequently, the
weight of the fluid, the pressure on the base, and the pressure on the
sides, are to one another as the numbers 1, 3 and 6.

2. WHEN THE VESSEL IS IN THE FORM OF A CYLINDER.

PROBLEM XVIII.

109. If a cylindrical vessel be completely filled with an
incompressible and non-elastic fluid, and so placed, that its
bottom may be parallel to the horizon: —

It is required to compare the pressure against its bottom,
with that against its upright surface, and also with the
weight of the fluid which it contains.

Let ABCD be a vertical section of a cylindrical vessel, filled with
an incompressible and non-elastic fluid, whose
surface AB is horizontal, and let it be required
to compare the pressure exerted by the fluid on
the bottom DC, with that upon the whole upright
surface.

Draw the diagonals AC and BD intersecting
one another in the point p, and through the point
p, draw the vertical line mn, meeting DC the
bottom of the vessel in the point m, and A B the surface of the fluid in
n; then is m the position of the centre of gravity of the bottom, and
nm its perpendicular depth below the surface AB.

Bisect AD in E, and through E draw EP parallel to A B or DC, and
meeting the vertical line mn in p ; then is p the position of the centre
of gravity of the upright surface, and n p its perpendicular depth below
AB the surface of the fluid.

Put D nz AB or DC, the diameter of the cylindrical vessel proposed,
d ~ nm, the perpendicular depth of the centre of gravity of the

bottom DC, below AB the surface of the fluid,
3 ~ nv, the perpendicular depth of the centre of gravity of the

upright surface,

Anr the area of the base or bottom of the cylinder,
P — the pressure upon it,
« nz the area of the curved or upright surface,
p — the pressure upon it,
w zzz the weight,
and s z= the specific gravity of the fluid.

Then, by the principles of mensuration, the area of the base or
bottom of the cylindric vessel, is

A=r.7854D2,
H 2

100 OF FLUID PRESSURE UPON TTTE INTERIOR OF A CYLINDER.

and that of the upright surface, is

a — 3.1416DG?;
consequently, the pressure on the bottom becomes

P = .r7S54v*ds, (59).

and for the pressure upon the upright surface, we have

p=:3.Ul6vdSs; (60).

therefore, by analogy, we obtain
P :p : : .7854 D*ds : 3.1416D^s;
from which, by omitting the common factors, we get

P :p :: D : 43;

now, by the construction of the figure, we have 3 z= \d ; therefore
4£ zz 2d, and by substitution, we obtain

P : p : : D : 2d : : |D : d.

Hence it appears, that the pressure upon the bottom of a cylindrical
vessel, is to the pressure upon its upright surface, as the radius of the
base is to the perpendicular altitude.

Since the entire pressure sustained by a cylindrical vessel, is equal
to the sum of the pressures on the bottom and the upright sides, it
follows, that

P -f p = p'= .7854 (D + 43) vds,
or substituting \d for 3, we shall obtain

p' = .7854 (D + 2d) vds. (61).

It is demonstrated by the writers on mensuration, that the solid
content of a cylinder, is equal to the area of its base, drawn into its
perpendicular altitude ; therefore, we have

where C denotes the solid content of the cylinder.

Now, it is manifest, that the weight of an incompressible and non-
elastic fluid, is equal to its magnitude drawn into its specific gravity;
hence we have

but this is precisely the expression which we have given in equation
(59), for the pressure perpendicular to the bottom of the vessel ; con-
sequently, the weight of the fluid, and the pressure on the bottom of
the vessel, are equal to one another ; hence, the following inference.

110. When the sides of a vessel of any form whatever, are perpen-
dicular, and its base parallel to the horizon : —

OF FLUID PRESSURE UPON THE ANNULI OF A CYLINDER. 101

The pressure perpendicular to the base of the vessel, is
equal to the whole weight of the fluid which it contains.

This is manifest, for the whole pressure of the fluid is sustained by
the base and the sides together, and the sides being in the direction
of gravity, sustain no part of the pressure which is exerted perpendi-
cularly downwards ; consequently, the whole weight of the fluid is
sustained by the base.

3. WHEN THE PRESSURE UPON THE ANNULI OF A CYLINDER IS TO BE

DETERMINED.

PROBLEM XIX.

111. If a cylindrical vessel whose bottom is parallel, and
sides perpendicular to the horizon, be filled with an incom-
pressible and non-elastic fluid: —

It is required to divide the concave surface, into any
number n of horizontal annuli, in such a manner, that the
pressure on each annulus shall be equal to the pressure on
the bottom of the vessel.

Let ABCD be a vertical section, passing along the axis of the
cylinder, or vessel containing the fluid, whose surface
is AB ; draw the diagonals AC and BD intersecting
one another in the point p ; then is p the centre of
gravity of the cylindrical surface.

Through p the point of intersection, draw the ver-
tical line mn parallel to AD or BC, and let a, b and c
be the points, which with the extremities A and D of
the side AD, terminate the several annuli : then, through the points
a, b and c, and parallel to AB or DC, draw the straight lines af,
be and cd, cutting BC the opposite side of the section in the points
ft e and d.

Draw the zigzag diagonals Af, fb, bd and dv, intersecting the
vertical line mn in the points k, i, h and g ; then are the points thus
determined, respectively the centres of gravity of the several annuli,
into which the concave surface of the vessel is supposed to be divided ;
and n k, ni, nh and ng, are the respective depths below the surface
of the fluid, WP being the depth of the centre of gravity of the whole
upright surface of the cylindrical vessel, and nm the vertical depth
of the bottom.

' - i ^ o *»y

• ' ' ' : • '

« &pO e *'

102 OF FLUID PRESSURE UPON THE ANNULT OF A CYLINDER.

Put D = AB or DC, the diameter of the proposed cylindrical vessel,
A •=. the area of its bottom,
d =nm, the whole perpendicular depth, or altitude of the

cylinder,

x ~Aa, the breadth of the first annulus,
x' z=ab, the breadth of the second,
x" — £c, the breadth of the third,
x'"— CD, the breadth of the fourth, and so on, to any number

of annuli n,
P — the pressure on the concave surface of the cylinder, or the

sum of the pressures on the several annuli into which

it is divided,
p — the pressure on the bottom of the vessel, and each of the

several annuli,
TT =z 3.1416 the circumference of a circle whose diameter is

unity, and
s zr the specific gravity of the fluid.

Then we have, nk — \x ; ni in x -\- \x ; n h — x -f- x' -|- \x",
and ng — x -\- x1 -|- x" + kx>" » an(* by the principles of mensuration,
the area of the bottom of the vessel, is

AZrjTTD*,

and by Proposition (1), the pressure upon it, is

p — l7n>*ds= .7854 i>M$, (62).

Again, by the principles of mensuration, the concave surfaces of the
respective annuli are as follows, viz.

For the first annulus, we have TT D x — 3.1416 D x, the surface,

second TTDO;' ~ 3.1416 D x',

third TTDX" — 3.1416 DX",

fourth TT DB"' = 3.1416 Da?1",

&c. &c. — &c.

And by Proposition (1), the pressures perpendicular to these sur-
faces, are respectively as below, viz.
For the first annulus, the pressure isp=zl.5708Da:a s,

second j0=3.1416D*' s(x+ J#'),

third p— 3.1416Dx"s(aj+x/+Ja:1'),

fourth - -P=i3.U\6»x'"s(x+x'+x"+%xf"),

&c. &c. &c.

Now each of these pressures, according to the conditions of the
problem, is equal to the pressure upon the bottom, exhibited in the
equation (62) ; consequently, by comparison, we have

OF FLUID RRESSURE UPON THE ANNULI OF A CYLINDER. 103

1. 5708 vx*s :=.
and casting out the common factors, we get

2a??= vd;
therefore, by division and evolution, we have

x=%^2nd'. (63).

By proceeding in a similar manner for the breadth of the second
annulus, we shall obtain

and this, by expunging the common terms, becomes

4*'(* + J*') = Drf;

therefore, by substituting for x, its value as expressed in equation (63),
we shall get

complete the square, and we have

a'2 -\-^2iTdx + %vd=: vd;
and finally, extracting the square root and transposing, we obtain

*' = 4(2—^2) J~d. (64).

Again, by performing a similar process for the breadth of the third
annulus, we shall have

3.1416D3"s (x -f- x' + i«") = .7854 D«d«,
from which, by casting out the common quantities, we get

4x"(x + x'+%x")=:i>d;

therefore, by substituting for x and x', their values as expressed in the
equations marked (63) and (64), and we shall obtain

4x" {J Jtod+ J(2— V2) V~^~d+ ix"} = Dd,
and this, by a little reduction and proper arrangement, gives

complete the square, and we obtain

consequently, by extracting the square root and transposing, we get

a/;-=JV6— 2)^25- (65>

Pursuing a similar train of reasoning for the breadth of the fourth
annulus, we shall obtain

3.1416D*"'s (x + x' + x" -f iO = .7854D«rf«,
and by suppressing the common factors, we have

4*"' (x + x' + x" 4- 1*'") = »d ;

104 OF FLUID PRESSURE UPON THE ANNULI OF A CYLINDER.

sustitute in this equation, the values of x, x and x" as represented in
the equations marked (63), (64), and (65), and we get

(2 — -t/2) VD^-KV6 — 2) V
and this, by a little further reduction, becomes

therefore, by completing the square, we obtain

a;"'* 4- i/fiDd.x"1 -f i Drf— 2od;

and finally, by extracting the square root and transposing, we have
x1" = J (2 V~2 — V 6) V^ (66).

112. And thus we may proceed to any extent at pleasure ; that is,
to any number of annuli within the limit of possibility ; for it is mani-
fest, from the nature of the problem, that impossible cases may be
proposed, but the limit can easily be ascertained in the following
manner.

It is obvious, that the sum of the breadths of the several annuli, is
equal to the whole depth of the vessel ; and that the sum of the
pressures is equal to the pressure on the concave surface ; but in the
problem immediately preceding, we have demonstrated that the pres-
sure on the bottom of a cylindrical vessel, is to that upon its upright
surface, as the radius of the base is to the perpendicular altitude.

Now, according to the conditions of the question, the pressure on
each annulus is equal to that upon the base ; consequently, in order
that the problem may be possible, the depth of the vessel must be
equal to the radius of the base, drawn into the number of annuli.

If instead of D the diameter of the cylindrical vessel, we substitute
2R its equivalent in terms of the radius, the preceding equations (63),
(64), (65), and (66), will become transformed into

x = (/f — v'O) y^, (67).

*' — (V2— /f) Va3, (68).

~ " (69).

(70).

From these equations the law of induction becomes manifest, and
the general expression for the breadth of the n^ annulus, is

(71).

OF FLUID PRESSURE UPON THE ANNULI OF A CYLINDER. 105

113. And from the above general form of the equation, the follow-
ing practical rule may be derived, for calculating the breadth of any
proposed annulus, independently of the breadths of those which pre-
cede it.

RULE. From the square root of the number which expresses
the place of the required annulus, subtract the square root of
that number minus unity ; then, multiply the remainder by
the geometric mean between the altitude of the vessel and the
radius of its base, and the product will give the breadth of the
required annulus.

114. EXAMPLE. A cylindrical vessel has the radius of its base, and
its perpendicular depth, respectively equal to 4 and 24 feet; now,
supposing the concave surface to be divided into 6 horizontal annuli,
such, that the pressure upon each shall be equal to the pressure upon
the base ; required the breadth of the fourth annulus ?

By performing the operation as directed in the preceding rule, we
shall obtain

x'" —(^— V3)X v/4 X 24 z=2.625 feet nearly.

The annulus which we have just determined, corresponds to the
fourth of the preceding class of equations, or that marked (69), and
the distance of its centre of gravity below the surface of the fluid, or
its position with respect to the bottom or top of the vessel, can easily
be ascertained.

The area of the cylinder's base, is

A — 3.1416B9;

the pressure which it sustains, is

p = 3. 1416 R2d= 1206.3744,

and this is equal to the pressure on the annulus.

Now, according to the writers on mensuration, the area of the

annulus, or the curved surface of a cylinder, whose radius is 4 feet

and its perpendicular altitude 2.625 feet, is expressed as follows, viz.

6.2832 X 4 X 2.625 = 65.9736.

If therefore, we divide the pressure on the base of the vessel, by the
area of the annulus, the depth of its centre of gravity will become
known ; thus, we have

106

4. WHEN THE VESSEL ASSUMES THE FORM OF A TRUNCATED CONE,
THE BASE OF WHICH IS ALSO THE BOTTOM OF THE VESSEL, AND
ITS AXIS PERPENDICULAR TO THE HORIZON.

PROBLEM XX.

115. If a vessel in the form of the frustum of a cone, be
filled with an incompressible and non-elastic fluid, and have its
axis perpendicular to the horizon : —

It is required to compare the pressure on the bottom of the
vessel with that upon its curved surface, and also with the
weight of the fluid which it contains, both when the sides of
the vessel converge, and when they diverge from the extremities
of the bottom.

Let ABCD represent a vertical section of a vessel in the form of the
frustum of a cone, and filled with an incom-
pressible and non-elastic fluid whose horizontal
surface is AB ; produce AB both ways, to any
convenient distance, and through D and c the
extremities of the bottom diameter, draw Da
and c b respectively perpendicular to D c, and
meeting AB produced in the points a and b;
then is abcv the vertical section, passing
along the axis of the cylinder which circumscribes the conic frustum.

Bisect AB and DC respectively in the points m and n, and draw
the straight line mn\ then, because the figure ABCD is symmetrical
with respect to the axis mn, it follows, that mn bisects the figure
or trapezoid ABCD, and consequently passes through its centre of
gravity.

Draw the diagonal AC, dividing the figure ABCD into the two
triangles ABC and ADC ; then it is manifest, that the common centre
of gravity of the two triangles, and that of the trapezoid constituted
by their sum, must occur in one and the same point ; therefore, bisect
the diagonal A c in the point t, and draw A n and D t intersecting each
other in the point r, and c m, B t intersecting in s ; then are r and s
the centres of gravity of the triangles ADC and ABC; draw rs inter-
secting mn in G, and G will be the centre of gravity of the trapezoid
ABCD.

Now, it is demonstrated by the writers on mechanics, that the
centre of gravity of the surface of a conic frustum :

OF FLUID PRESSURE UPON THE INTERIOR OF CONICAL VESSELS. 107

Is situated in the axis, and at the same distance from its
extremities, as is the centre of gravity of the trapezoid, which
is a vertical section passing along the axis of the solid.

Therefore, since by the construction, the point G has been shown to
be the centre of gravity of the trapezoid, it is also the centre of gravity
of the surface of the conic frustum, and TWO is its perpendicular depth
below the surface of the fluid.

Put AZZ: the area of the base or bottom of the vessel, whose diameter

is DC,
v=mn, the perpendicular depth of its centre of gravity, or the

length of the axis of the vessel,
a — the curve surface of the conic frustum,
d^nmo, the perpendicular depth of the centre of gravity,
/3 zz: DC, the diameter of the base or bottom of the vessel,
S zz: AB, the diameter of the top,
Pzz: the pressure on the bottom,
p HZ the pressure on the curve surface,
wzz: the weight, and s the specific gravity of the fluid.

Then, according to the principles of mensuration, the area of the
lower base of the conic frustum, or the bottom of the vessel on which
the fluid presses, becomes

A = . 7854/3%
and consequently, the pressure which it sustains, is

Pzz:. 7854 /32DS. (72).

In the next place, the area of the curved surface of the conic
frustum, or the sides of the vessel containing the fluid, is

a= 1.5708 08 + 3)X V** + 1 (ft — W I
and therefore, the pressure which it sustains, is

p = 1.5708 ()8 4- 5) ds VD* 4- 403— S)2. (73).

Now, according to the writers on mechanics, the depth of the centre
of gravity of the trapezoid ABCD, below the horizontal line AB, is
obtained in the following manner :

3 (/3 4 3) : D : : 2/3 4 3 : d,
and by equating the products of the extremes and means, we get

therefore, dividing by 3 (/3 4- 3), we obtain
_
=

108 OF FLUID PRESSURE UPON THE INTERIOR OF CONICAL VESSELS.

Let this value of d be substituted instead of it, in the equation
marked (74), and we shall have for the pressure on the curved surface
of the vessel

p — .5236 D s (2/3 4- 3) V^-4- J03 — 3)* ; (75).

consequently, by comparing the equations (72) and (74), we get
Pip:: .7854/3*DS : .5236Ds(2/3 + 3) J D2

and this, by suppressing the common factors, becomes

P : p : : 3/3* : 2 (2/3 + 3) V»* + i (P - *)«- (76).

If J, the upper diameter of the frustum vanishes, the figure becomes
a complete cone, and consequently, the pressure upon the base, is to
that upon the curve surface, as three times the diameter of the cone,
is to four times its slant height ; that is

P:p:: 3/3: 4^+1^. (77).

According to the principles of solid geometry, the capacity of the
conic frustum, or the quantity of fluid which the vessel contains, is

where c denotes the solid content of the vessel.
But the weight of any quantity or mass of fluid, varies directly as
the magnitude and specific gravity conjointly; consequently, the
weight of fluid in the vessel, is expressed by

w — .2618 D s (/3s + /3 J -f a2). (78).

Hence, if the equations marked (72) and (78), be compared with
each other, we shall obtain

p^i-.s/^os'-h/sa+a2), (79).

and when 3 vanishes, the vessel becomes a complete cone, and conse-
quently, we get

P : w : : 3 : 1, (80).

It therefore appears, that the pressure against the bottom of a
conical vessel, when filled with an incompressible and non-elastic
fluid, (the bottom being downwards) :

Is equivalent to three times the weight of the fluid which it
contains.

The solidity of the cylinder circumscribing the conic frustum, of
which abci> is a vertical section, is

c' =r.7854 f? D,
where c' denotes the capacity of the cylinder circumscribing the vessel;

OF FLUID PRESSURE UPON THE INTERIOR OF CONICAL VESSELS. 109

and because the weight of any quantity or mass of fluid, is propor-
tional to the magnitude and specific gravity jointly ; it follows, that if
w' denote the weight of the circumscribing column of fluid, we obtain
w/z=.7854/32Ds. (81).

COROL. Now this expression is precisely the same, as that which we
obtained for the pressure on the bottom, indicated by the equation
marked (72) ; hence it appears, that when the sides of the vessel
converge from the extremities of the diameter of its base towards each
other : —

The pressure on the base or bottom of the vessel, is equal to
the weight of a column of the fluid, of the same magnitude
as the cylinder circumscribing the conic frustum, or the vessel
by which the fluid is contained.

But the circumscribing cylinder is manifestly greater than the conic
frustum ; consequently, the pressure upon the base or bottom of the
vessel, is greater than the weight of the fluid which it contains ; and
it is obvious, that the additional pressure arises from the re-action of
the converging sides.

5. WHEN THE VESSEL REPRESENTS AN INVERTED TRUNCATED CONE,
WITH ITS AXIS PERPENDICULAR TO THE HORIZON.

116. If the sides of the vessel diverge from the extremities of the
base, as represented in the subjoined diagram ; then, it may be
shown, that the weight of the fluid which the vessel contains, exceeds
the pressure upon its" base.

Let ABCD be a vertical section, passing
along the axis of a vessel in the form of a
conic frustum, and which is filled with an
incompressible fluid whose horizontal surface
is AB ; the greater base of the frustum being
uppermost, or which is the same thing, the
sides diverging from the extremities of the lower diameter.

Bisect the diameters A B and c D respectively in the points m and n ;
draw mn, and through the points D and c, the extremities of DC, draw
the straight lines Da and cb respectively parallel to mn, and meeting
AB in the points a and b; then is abcn a vertical section passing
along the axis of the inscribed cylinder.

Draw the diagonal AC, dividing the trapezoid ABCD, into the two
triangles ABC and ADC; bisect the diagonal AC in the point t, and

110 OF FLUID PRESSURE UPON THE INTERIOR OF CONICAL VESSELS.

draw B t and D t, which will be intersected by the straight lines, c m
and \n iu the points r and s; then are r and s respectively the
centres of gravity of the triangles ABC and ADC.

Join the points r and s, by the straight line rs, intersecting mn in
the point G ; then it is obvious, that the common centre of gravity of
the triangles ABC and ADC, (which coincides with that of the trapezoid
ABCD), must occur in the line rs, which joins their respective centres.

Now, because the trapezoid A B c D is symmetrically situated with
respect to the axis mn, it follows, that its centre of gravity must occur
in that line ; but we have shown above, that it also occurs in the line
rs, it consequently must be situated in the point G, where these lines
intersect one another ; hence, the centre of gravity of the surface of
the conic frustum occurs at the point G, and m G is its perpendicular
depth below AB the upper surface of the fluid.

Put A •=. the area of the lower base or bottom of the vessel, whose

diameter is DC,
P — the pressure perpendicular to its surface, or the weight of a

quantity of fluid equal to the inscribed cylinder,
Dzzmw, the axis of the frustum, or the perpendicular depth of

the centre of gravity of the bottom,

a zz: the area of the curve surface of the vessel or conic frustum,
p zz: the pressure perpendicular to the curve surface,
d zz: m&, the perpendicular depth of its centre of gravity,
3 zz: DC, the diameter of the lower base or bottom of the vessel,
ft zz: AB, the diameter of the top or upper base,
w— the weight, and s the specific gravity of the contained fluid.

Then, by the principles of mensuration, the area of the lower base
of the conic frustum, or the bottom of the vessel on which the fluid

presses, is

Azzr.785422,

and consequently, the pressure upon it, is

Pzz:.7854j2Ds. (82).

This equation, having 52 instead of /32, is the same as that which we
obtained for the pressure on the bottom in the preceding case, when
the greater base of the vessel was downwards ; it therefore follows,
since our notation is adapted to the same parts of the vessel, that not-
withstanding the inversion, the pressure on the curved surface of the
conic frustum, will still be expressed as in the equation marked (73) ;
consequently, we have

P :p: :.7854$2ns:: 1.5708 (/3-f 2)ds VD'

OF FLUID PRESSURE UPON THE INTERIOR OF CONICAL VESSELS. Ill

or by expunging the common quantities, we get

P :p : : 3*D : 2(/3 + 3) dji>* + J(/3 — 3)'.

But the writers on mechanics have demonstrated, that the depth of
the centre of gravity of the trapezoid A BCD, below the horizontal line
AB, is expressed as follows, viz.

' (83).

let therefore, this value of d be substituted instead of it in the above
analogy, and we shall obtain

P : p : : 33* : 2 (ft + 23) V »* + I (ft — V-

If £, the diameter of the bottom or lower base should vanish, the
vessel becomes a complete cone with its vertex downwards, in which
case, the value of d as expressed in the equation marked (83), is

Let this value of d be substituted instead of it, in the equation
marked (73), and suppose 3 to vanish ; then, the pressure on the con-
cave surface of a conical vessel with its vertex downwards, becomes

p = .5236 ft D s V D9 + J r/P. (84).

The solid content of the inscribed cylinder, of which the vertical
section passing along the axis is a6co, becomes

c'=r.785432D,

and as we have already stated, its weight is proportioned to its mag-
nitude drawn into the specific gravity ; hence we have

«/ = . 7854 3*Ds;

but this is the same expression which indicates the pressure on the
bottom, as exhibited in the equation marked (82) ; hence it follows,
that the pressure on the bottom or lower base of the conic frustum,
when the sides diverge from the extremities of its diameter,

Is equal to the weight of a column of the fluid, of the same
magnitude as the cylinder inscribed in the conic frustum.

But the solid content of the inscribed cylinder, and consequently
its weight, is manifestly less than the content of the vessel ; hence we
infer, that when the sides of the vessel diverge from the extremities of
the diameter of its bottom, the pressure on the bottom is less than
the weight of the fluid which it contains, the remaining weight being
supported by the resistance of the diverging sides.

112 OF FLUID PRESSURE UPON THE INTERIOR OF CONICAL VESSELS.

117. In order to compare the pressure on the bottom of the vessel,
with the weight of the fluid which it contains, we must again have
recourse to the principles of solid geometry ; from which we learn,
that the solid content of a conic frustum, whose diameters are denoted
by (3 and 5, and its perpendicular altitude by D, is

and consequently, its weight becomes

to = .2618 DS (0* + ft 3 + S2) ; (85).

therefore, by comparing this equation with that marked (82), we get

. P : w : : 33* : (/32 + /3 3 + a2). (86).

Again, if we compare the equations marked (73 and (85) with one
another, we shall have

p : w :: 1.5708 (0 + 3) dsv'D* -*•%(&— 3) : .2618 Ds(/32 + /33 + J') ;
if therefore, we expunge the common quantities, from the third and
fourth terms of the above analogy, and in the third term, substitute the
value of d as it is expressed in the equation (83), then we shall obtain

p : w : : 2(0 + 2 3) <J D* -f i (ft — *> : (/32 + ft 3 + *).
When 5 vanishes, or when the vessel becomes a complete cone with
its vertex downwards, the preceding analogy gives

p : w : : 2 ^ D* + |/32 : /3.

In complying with the conditions of the 20th problem, the fore-
going investigation has been conducted on the supposition, that the
vessel in question is in the form of the frustum of a cone ; but the
attentive reader will readily perceive, that the same mode of procedure
will apply to the frustum of any other regular pyramid, and the result-
ing formulae will partake of similar forms and combinations, differing
only in so far as depends upon the constant numbers which express
their respective areas and solidities ; it is therefore unnecessary to
pursue the inquiry further, taking it for granted, that by a careful
perusal of what has been done above, no difficulty will be met with
in applying the same principles to any other case of form or condition
that is likely to occur.

COUOL. 1. By the preceding investigation, then, and the formulae

arising from it, we learn, that by causing the sides of a vessel, which

is filled with an incompressible and non-elastic fluid, to converge or

diverge from the extremities of the base, supposed to be horizontal : —

The pressure on the base, may be greater or less than the

weight of the fluid which the vessel contains, in any propor-

tion whatever.

OF FLUID PRESSURE USED AS A MECHANICAL POWER. 113

2. Upon these principles therefore, and others of a similar nature,
which we have mentioned at the outset, is explained the paradoxical
property of non-elastic fluids : —

That the pressure on the bottom of a vessel Jilted with
fluid, does not depend upon its quantity, but solely upon the
perpendicular altitude of its highest particles above the
bottom of the vessel or the surface by which the pressure is
sustained.

3. And from the property here propounded, is deduced the remark-
able and important principle : —

That any quantity of fluid however small, may be made to
balance, or hold in equilibrio, any other quantity, however
great.

Let the upright or vertical section of a vessel containing an incom-
pressible and non-elastic fluid, be such as is repre-
sented by A BCD in the annexed diagram, and let
cdbe the corresponding section of a small pipe or
tube inserted into its upper surface at the point d.

Then, supposing the vessel and the tube to be
filled with fluid as far as the point c ; it is manifest
from the first case of the preceding problem, that
the pressure upon DC the bottom of the vessel, is precisely the same
as if it were entirely filled to the height acb\ for the pressure upon
the bottom DC, is equal to the weight of a fluid column, the diameter
of whose base is DC and its perpendicular altitude an or be; but this
is evidently greater than the weight of the fluid in the vessel, and by
increasing the height of fluid in the tube, the pressure on the bottom
will be increased in the same proportion, while the actual increase
of weight is very small, being only in proportion to the increase of
pressure, as the area of a section of the tube is to the area of the
bo torn.

PROBLEM XXI.

118. Having given the diameter and perpendicular height of
a cylindrical vessel, together with the diameter of a tube fixed
vertically into the top of it :—

VOL. I. I

114 OF FLUID PRESSURE USED AS A MECHANICAL POWER.

It is required to find the length of the tube, such, that
when it and the vessel are filled with an incompressible fluid,
the pressure on the bottom of the vessel may be equal to any
number of times the fluid's weight.

In resolving this problem, it will be sufficient to refer to the pre-
ceding diagram, because a separate construction would exhibit no
variety ; for this purpose then,

Put D zz DC, the diameter of the base of the cylindrical vessel,
A zz the area of its base,

h zz A D, the altitude, or perpendicular depth of the vessel,
C — the capacity, or solid content,
P — the pressure on the bottom,

d — ed, the diameter of the tube inserted in the top of the vessel,
a zz the area of its horizontal section,
c zz the capacity, or solid content of the tube,
w ~~ the weight of the fluid in the vessel, and w' the weight of

that in the tube ;

s zz the specific gravity of the fluid,

n zz the number of times the pressure exceeds the weight,
and x zz the required length of the tube.

Then, according to the principles of mensuration, the area of the
bottom of the vessel becomes

AZZ.7854D2,
and that of a horizontal section of a tube, is

azz.7854d2;
and again, by the geometry of solids, the capacity of the vessel is

Czz.7854D2A,
and for the capacity of the tube, we have

the respective weights being

wzz.7854D2As, and w1 zz .7854d2a-s;

consequently, the whole weight of the fluid in the vessel and tube, is

w + w' = .7854s (D2A, -f d*x).

Now, we have shown above, that the pressure on the bottom of the
vessel is equal to the weight of a fluid cylinder, whose diameter is
DC and altitude <ZD; consequently, the pressure on the bottom is
expressed by

OF FLUID PRESSURE USED AS A MECHANICAL POWER. 115

and this, according to the conditions of the problem, is equal to n
times the entire weight ; hence we have

.7854 D2s (h + x) = .7854 ns (tfh + d*x),
therefore, by casting out the common factors, we get

or by separating the terms and transposing, we get

( D2 — n (T ) x = D2 h ( n — 1 ) ,
from which, by division, we obtain
_D8A(n--l)
1 D'^-nd- ' (87).

119. It may be perhaps proper to illustrate this case by an example;
but in the first place, it becomes necessary to give the rule by which
the operation is to be performed.

RULE. From the number of times which the pressure on the
bottom of the vessel, is proposed to exceed the weight of the
fluid, subtract unity ; multiply the remainder by the square
of the vessel 's diameter, drawn into its depth or perpendicular
altitude, and the result will be the dividend.

Then, from the square of the vessel's diameter, subtract
n times the square of the diameter of the tube, and divide the
above dividend by the remainder for the length of the tube
required.

120. EXAMPLE. If the perpendicular height of a cylindric vessel be
18 inches, its diameter 5 inches; the diameter of a tube fixed to the
top of the vessel one inch ; and if the vessel and tube be filled with an
incompressible and non- elastic fluid, till the pressure on the bottom
of the vessel is equal to twelve times the entire weight of the fluid;
what is the length of the tube into which the fluid is poured ?

Here, by proceeding according to the rule, it is

n— 1=12— 1~11,

D«A=z5X5x 18=450;

therefore, by multiplication, we obtain

«2/* (n — 1) = 450X11 == 4950 dividend.

Again, to determine the divisor, we have

D2 — Bd»=z5X5 — 12=13;

consequently, by division, we obtain

^ = 49504- 13 = 380|f inches.

i 2

116 OF FLUID PRESSURE USED AS A MECHANICAL POWER.

COROL. 1. If the tube, instead of being fixed perpendicularly to the
top of the vessel, were inserted obliquely into any part of its sides and
inclined upwards, the principle above exemplified would still obtain ;
and the pressure in the narrow tube may be produced, not merely by
the addition of a little fluid, but by the application of any kind of
force, such as the working of a piston and the like.

2. If the bottom, or the cover of the cylindric vessel be made move-
able, the pressure on either may be brought to bear on any one point
of an external body, and may then produce an inconceivable com-
pression, as is very successfully done in the Hydrostatic Press, an
instrument, which, on account of the simplicity of its application, its
expeditious performance, and the almost unlimited extent of its power,
is altogether without a parallel in the annals of mechanical invention,
and the numerous purposes to which it is applied, entitle it to no
small share of popular approbation.

This machine is not only used for pressing bodies together, with a
view of diminishing their bulk, in order to render them the more easily
stowable ; but it is equally applicable to the operation of drawing and
lifting great loads, and overcoming immense resistances, however
opposed to its action ; even piles, which have been driven to a great
depth for the purpose of forming coffer-dams, can be drawn by it with
the greatest facility, and moreover, trees of the greatest size and most
tenacious growth, offer but a feeble resistance to its energy ; and in
addition, iron bolts and cables, capable of holding the largest ships in
the British navy, are totally incompetent to resist its influence.

An instrument possessing such immense power in combination with
so many other advantages, such as cheapness of construction, porta-
bility, and simplicity of application, certainly merits the greatest
attention, and too many attempts cannot be made to simplify the
theory, and render its operations easily understood ; we shall there-
fore, in the following pages, endeavour to unfold the principles, and
to describe its construction and mode of operation.

CHAPTER VI.

THE THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION
OF SOME HYDROSTATIC ENGINES.

1. OF THE HYDROSTATIC OR BRAMAH PRESS.

PROPOSITION II.

121. IF there be any number of pistons of different magni-
tudes, any how applied to apertures in a cylindrical vessel filled
with an incompressible and non-elastic fluid : —

The forces acting on the piston to maintain an equilibrium,
will be to one another as the areas of the respective apertures,
or the squares of the diameters of the pistons.

Let ABC D represent a section passing along the axis of a cylindrical
vessel filled with an incompressible and non-elastic
fluid, and let E,F be two pistons of different magni-
tudes, connected with the cylinder and closely fitted
to their respective apertures or orifices ; the piston
F being applied to the aperture in the side of the
vessel, and the piston E occupying an entire section
of the cylinder or vessel, by which the fluid is contained.

Then, because by the nature of fluidity, the pressures on every part
of the pistons E and F, are mutually transmitted to each other through
the medium of the intervening fluid ; it follows, that these pressures
will be in a state of equilibrium when they are equal among them-
selves.

Now, it is manifest, that the sum of the pressures propagated by
the piston E, is proportional to the area of a transverse section of the
cylinder ; and in like manner, the sum of the pressures propagated by
the piston F, is proportional to the area of the aperture which it
occupies; consequently, an equilibrium must obtain between these
pressures : —

118 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION

When the forces on the pistons, are to one another, respec-
tively as the areas of the apertures or spaces which they
occupy.

And it is obvious, that the same thing will take place, whatever
may be the number of the pistons pressed.

Hence it appears, that by taking the areas of the pistons E and F,
in a proper ratio to one another, we can, by means of an incompres-
sible fluid, produce an enormous compression, and that too by the
application of a very small force.

Put P zz the force or pressure on the piston E,
A=z the area of the orifice which it occupies,
p zz the pressure on the piston F, and
a zz the area of the orifice or space to which it is fitted.

Then, according to the principle announced in the foregoing pro-
position and demonstrated above, we shall obtain

a : A ::p : P.

But because, by the principles of mensuration, the areas of different
circles are to one another as the squares of their diameters ; if there-
fore, we substitute d? and D2 respectively for a and A in the above
analogy, we shall have

d* : D9 : : p : P,

and from this, by making the product of the mean terms equal to the
product of the extremes, we get

p^ — ^d\ (88).

122. This is the principle upon which depends the construction and
use of that very powerful instrument, the Hydrostatic Press, first
brought into notice about the year 1796, by Joseph Bramah, Esq.,
of Pimlico, London ; who announced it to the world as the discovery
of a new mechanical power.

In this however, he was mistaken, for although the principle upon
which it depends may be said to constitute a seventh mechanical
power, yet the principle announced in Proposition II. was not new
to philosophers at the time when Mr. Bramah applied it to the
construction of his presses, it having long been familiarly known
under the designation of the Hydrostatic Paradox ; and besides,
the celebrated Pascal obscurely hinted at its application to mecha-
nical purposes, but did not pursue the idea far enough to produce
any thing useful, or to entitle him to the full merit of the discovery.

The improvement introduced by Mr. Bramah, consisted in the
application of the common forcing pump to the injection of water,

OF THE HYDROSTATIC PRESS. 1 19

or some other incompressible and non-elastic fluid, into a strong
metallic cylinder, truly bored and furnished with a moveable piston,
made perfectly water-tight by means of leather collars or packing,
neatly fitted into the cylinder.

123. The proportion which subsists between the diameter of this
piston, and that of the plunger in the forcing pump, constitutes the
principal element by which the power of the instrument is calculated ;
for, by reason of the equal distribution of pressure in the fluid, it is
evident, that whatever force is applied, that force must operate alike
on the piston in the cylinder, and on the plunger in the forcing
pump, and consequently,

In proportion as the area of the transverse section of the
one, exceeds the area of a similar section of the other, so must
the pressure sustained by the one, exceed that sustained by the
other.

Therefore, if the piston F in the preceding diagram, be assimilated
to the plunger in the barrel of a forcing pump, and the piston E to
that in the cylinder of the hydrostatic press ; then, the equation
marked (88), notwithstanding the very simple and concise form in
which it appears, involves every particular respecting the power and
effects of the engine, of which a detailed description with illustrative
drawings will be given a little further on.

This being premised, we shall now proceed to exhibit the use and
application of the formula, by the resolution of the following practical
examples.

124. EXAMPLE 1. If the diameter of the cylinder is 5 inches, and
that of the forcing pump one inch ; what is the pressure on the piston
in the cylinder, supposing the force applied on the plunger or smaller
piston, to be equivalent to 750 Ibs. ?

Here we have given D =r 5 inches ; d — 1 inch, and p — 750 Ibs. ;
therefore, by substitution, equation (88) becomes

5?X750 = PX12; that is, Pz= 18750 Ibs.

Or the equation for the value of P, may be expressed in general
terms, as follows.

P_/>D!

P-^r- (89).

And from the equation in its present form, we deduce the following
practical rule.

120 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION

RULE. Multiply the square of the diameter of the cylinder
by the magnitude of the power applied, and divide the product
by the square of the diameter of the forcing pump, and the
quotient will express the intensity of the pressure on the piston
of the cylinder.

125. EXAMPLE 2. If the diameter of the cylinder is 5 inches, and
that of the forcing pump one inch ; what is the magnitude of the
power applied, supposing the entire pressure on the piston of the
cylinder to be 18750 Ibs. ?

Here we have given D = 5 inches; d= 1 inch, and P= 18750
Ibs. ; therefore, by substitution, equation (88) becomes

5a Xp— 18750 X 1*; or p — 750 Ibs.

If both sides of the fundamental equation (88) be divided by D*, the
general expression for the value of p, is

_Pd2

p~ D2 ' (90).

And the practical rule which this equation supplies, may be
expressed in words at length in the following manner.

RULE. Multiply the given pressure on the piston of the
cylinder, by the square of the diameter of the forcing pump,
and divide the product by the square of the diameter of the
cylinder for the power required.

126. EXAMPLE 3. The diameter of the forcing pump is one inch,
and the power with which the plunger descends is equivalent to 750
Ibs. ; what must be the diameter of the cylinder, to admit a pressure
of 18750 Ibs. on the piston ?

Here we have given c?n= 1 inch; ^ = 750 Ibs., and P n= 18750 Ibs. ;
consequently, by substitution, the equation marked (88) becomes

750 D2=l 8750 x I2;
hence, by division, we obtain

consequently, by evolution, we have
D — ^ 25 — 5 inches.

If both sides of the equation (88) be divided by p, and the square
root of the quotient extracted, the general expression for the diameter
of the piston, is

OF THE HYDROSTATIC PRESS. 121

And the practical rule for the determination of D, may be expressed
in words as follows.

RULE. Multiply the pressure on the piston of the cylinder,
by the square of the diameter of the forcing pump, and divide
the product by the force with which the plunger descends ;
then, the square root of the quotient will be the diameter of
the cylinder sought.

127. EXAMPLE 4. The diameter of the cylinder is 5 inches, and
the force with which the plunger descends, is equivalent to 750 Ibs. ;
what must be the diameter of the forcing pump, in order to transmit
a pressure of 18750 Ibs. to the piston of the cylinder ?

Here we have given D r= 5 inches ; p — 750 Ibs., and P zz 18750
Ibs. ; consequently, by substitution, equation (88) becomes

1 8750 d* = 750 X 52,
and by division, we shall have
750 X 25

therefore, by extracting the square root, we get

d— /f zzl inch.

If both sides of the original equation marked (88), be divided by
P, and the square root extracted, the entire pressure on the piston,
the general expression for the value of d becomes

= /l/ "P"' (92).

And the practical rule which this equation supplies, may be
expressed in words in the following manner.

RULE. Multiply the force with which the plunger descends,
by the square of the diameter of the cylinder, and divide the
product by the entire pressure on the piston; then, extract
the square root of the quotient for the diameter of the forcing
pump.

128. The foregoing is the theory of the Hydrostatic Press, as
restricted to the consideration of the diameters of the cylinder and
forcing pump, and the respective pressures on the piston and plunger;
but since the instrument is generally furnished with an indicator or

122 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION

safety valve for measuring the intensity of pressure, the theory would
be incomplete without considering it in connection with the diameters
of the pump and cylinder. For which purpose

Put 3 zz the diameter of the safety valve, expressed in inches or

parts,
and w — the weight thereon, or the force that prevents its rising.

Then, according to the principle announced in Proposition II., we
obtain the following analogies, viz.

D2 : ^ : : P : w,
d* : ^ : : p : w ;

and from these analogies, by making the products of the extreme
terms equal to the products of the means, we get

D'wzz^P, (93).

zudd*w = tfp. (94).

Now, in order to pursue the expansion of these equations, we shall
suppose the value of & to be one fourth of an inch, while the numerical
values of the other letters remain the same as supposed for the several
examples under equation (88) ; then, to determine the corresponding
value of w, or the power which prevents the safety valve from rising,
when all the parts of the instrument, or the several powers and pres-
sures are in a state of equilibrium, we have the following examples to
resolve according to the proposed conditions.

129. EXAMPLE 5. The diameter of the cylinder is 5 inches, that of
the indicator or safety valve J of an inch, and the entire pressure
upon the piston of the cylinder 18750 Ibs. ; what is the corresponding
force preventing the ascent of the safety valve, on the supposition of a
perfect equilibrium ?

Here we have given D zz 5 inches ; 3 zz J of an inch, and P zz 1 8750
Ibs. ; consequently, by substitution, the equation (93) becomes

52wzz.252 X 18750;
from which, by division, we get

.0625 X 18750

wzz — zz46.875 Ibs.

But the general expression for the value of iv, as derived from the
equation (93), becomes

_J2P
- V' (95).

From which we derive the following rule.

OF THE HYDROSTATIC PRESS. 123

RULE. Multiply the entire pressure on the piston of the
cylinder, by the square of the diameter of the indicator or
safety valve, and divide the product by the square of the
diameter of the cylinder for the weight required.

130. EXAMPLE 6. The diameter of the safety valve is J of an inch,
that of the cylinder 5 inches, and the weight on the safety valve
46.875 Ibs. ; what is the corresponding pressure on the piston of the
cylinder ?

Here we have given 3 = £ of an inch ; D zz: 5 inches, and
w zz:46.875 Ibs ; therefore, by substitution, equation (93) becomes

.252P = 52x 46.875,
and by division, we obtain

The general expression for the value of P, as derived from the
equation marked (93), becomes

*
-

(96).

And the practical rule supplied by this equation, may be expressed
in words as follows.

RULE. Multiply the weight on the safety valve, by the
square of the diameter of the cylinder, and divide the product
by the square of the diameter of the safety valve, and the
quotient will give the entire pressure on the piston of the
cylinder.

131. EXAMPLE 7. The diameter of the cylinder is 5 inches, the
entire pressure of the piston is 18750 Ibs., and the weight on the
safety valve is 46.875 Ibs. ; what is its diameter ?

Here we have given ozz:5 inches ; P zz: 18750 Ibs., and wzr46.875
Ibs. ; therefore, by substitution, equation (93) becomes

18750a2z=52X 46.875,
and from this, by division, we get

and by extracting the square root, we obtain

3= -V/ .0625 zz: .25, or £ of an inch.

The general expression for the value of 3, as derived from the
equation (93), is as follows, viz.

124 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION

*=y -p-- (97).

And the practical rule which this equation affords, may be expressed
in words in the following manner.

RULE. Multiply the load on the safety valve by the square
of the diameter of the cylinder ; divide the product by the
entire pressure on the piston, and the square root of the
quotient will give the diameter of the safety valve required.

132. EXAMPLE 8. The diameter of the safety valve is ^ of an inch,
the load upon it 46.875 Ibs., and the entire pressure on the piston
of the cylinder is 18750 Ibs. ; what is its diameter?

Here we have given £= J of an inch, w= 46.875 Ibs., and Pzr 18750
Ibs. ; consequently, by substitution, we have

46.875 D*=. 25* X 18750,
from which, by division, we shall obtain
.25aX 18750

46.875
and finally, by extracting the square root, we get

D = -v/ 25 — 5 inches.

If both sides of the equation marked (93), be divided by w the
weight on the safety valve, we get

and by extracting the square root, the general expression for the value
of D the diameter of the cylinder, becomes

-y ^' (98).

And from this equation we derive the following rule.

RULE. Multiply the entire pressure on the piston of the
cylinder by the square of the diameter of the safety valve,
divide the product by the weight upon the safety valve, and
extract the square root of the quotient for the diameter of the
cylinder sought.

133. EXAMPLE 9. The diameter of the forcing pump is one inch,
that of the safety valve is one fourth of an inch, and the power or
force with which the plunger descends, is equivalent to 750 Ibs. ; what
is the corresponding weight on the safety valve ?

OF THE HYDROSTATIC PRESS. 125

Here we have given d— 1 inch; Ji=^ of an inch, and p — 750
Ibs. ; consequently, by substitution, the equation (94) becomes

P X w = .25* X 750 ; that is, w = 46.875 Ibs,, the
very same value as we derived from the fifth example.

If both sides of the equation marked (94) be divided by d*, the
general expression for the value of w becomes

_3>

~d2* (99).

And the practical rule supplied by this equation, may be expressed
in words at length in the following manner.

RULE. Multiply the force with which the plunger descends
by the square of the diameter of the safety valve, and divide
the product by the square of the diameter of the plunger ;
then the quotient will express the load upon the safety valve.

134. EXAMPLE 10. The diameter of the safety valve is J of an
inch, that of the forcing pump is one inch, and the load upon the
safety valve is 46.875 Ibs. ; what is the power applied, or the force
with which the plunger in the forcing pump descends ?

Here we have given 3=J of an inch, d=:l inch, and w=r46.875
Ibs. ; consequently, by substitution, equation (94) becomes

.252j9=46.875 X I2,
and from this, by division, we obtain

' '

The general expression for the value of p, as obtained from the
equation marked (94), becomes

_d*w

:~F* (100).

from which we derive the following rule.

RULE. Multiply the load on the safety valve by the square
of the diameter of the forcing pump ; then, divide the product
by the square of the diameter of the safety valve, and the
quotient will give the force with which the piston descends.

135. EXAMPLE 11. The diameter of the plunger or the piston of
the forcing pump is one inch, the force with which it descends is
equivalent to 750 Ibs., and the load on the safety valve is 46.875 Ibs.;
what is its diameter ?

126 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION

Here we have given dnz 1 inch, p •=. 750 Ibs., and w zz 46.875 Ibs. ;
consequently, by substitution, we have

750 3"= 1s X 46.875,
and from this, by division, we obtain

and finally, by evolution, we have

a = V.0625 = .25 of an inch. ,

Let both sides of the equation marked (94) be divided by p, the
power or force with which the piston of the forcing pump descends,
and we shall have

and by extracting the square root, we get

(101).
Hence, the following practical rule.

RULE. Multiply the weight or load upon the safety valve,
by the square of the diameter of the forcing pump, and divide
the product by the force with which the plunger or piston
of the forcing pump descends ; then, the square root of the
quotient will be the diameter of the safety valve.

136. EXAMPLE 12. The diameter of the safety valve is one fourth
of an inch, the weight upon it is 46.875 Ibs., and the power applied,
or the force with which the plunger descends, is 750 Ibs ; what is the
diameter of the forcing pump ?

Here we have given £ — J of an inch, w> — 46.875 Ibs., and /?=z750
Ibs. ; consequently, by substitution, the equation marked (94) becomes

46.875d2=.25*x750;

therefore, by division, we obtain

.25^X750

: 46.875 Z

and finally, by extracting the square root, we get

<frz 1 inch.

The general expression for the value of the diameter of the forcing
pump, as derived from the equation (94), is

d=^ (102).

OF THE HYDROSTATIC PRESS. 127

And from this, we obtain the following practical rule.

RULE. Multiply the force with which the piston of the
forcing pump descends, by the square of the diameter of the
safety valve ; divide the product by the load on the safety
valve, and extract the square root of the quotient for the
diameter of the forcing pump.

The foregoing twelve examples exhibit all the varieties of cases that
can arise, from the combination of the six data which we have em-
ployed in our theory, viz. the diameters of the cylinder, the forcing
pump and the safety valve ; together with the entire pressure on the
piston of the cylinder, the power applied to the plunger of the forcing
pump, and the weight upon the safety valve.

We have determined each of the quantities, composing the several
fundamental equations, in terms of the others, and have drawn up
rules from the general expressions, merely for the assistance of those
who are not accustomed to algebraic reductions; those who are,
will prefer finding each quantity directly from the general equation
expressing its value.

137. It is manifest from the principles of mensuration, that the area
of a transverse section of the cylinder, or the base of the piston, is
expressed by .7854 D2 ; and we have shown, equations (89) and (96),
that the entire pressure upon the base of the piston in the case of
equilibrium, is

consequently, if n denotes the pressure in pounds avoirdupois on one
square inch of the piston, then we have

P p w

n=7imtf'r' ~-7JS54d*al -T7854F- (104).

Now, from principles investigated by Professor Barlow, of the Royal
Military Academy at Woolwich, it appears, that if c denote the cohe-
sive force of the material employed in the construction of the cylinder,
t its thickness, and r the interior radius ; then, in order that the strain
produced by the pressure, shall not exceed the elastic power of the
material ; it is necessary that

In order to demonstrate this, let ABD be a transverse section of the
cylinder, perpendicular to the axis passing through c ; then, sup-

128 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION

posing a certain uniform pressure to be exerted all round the interior
boundary; it will readily appear, from the
theory of resistance, that each successive
circular lamina, estimated from the interior
towards the exterior circumference, offers a
less and less resistance to the straining force.
But it is obvious from the very nature of
the subject, that by reason of the internal
pressure or strain, the metal must undergo a
certain degree of extension, and since the resistance of the outer
boundary is less than that of the inner one, it follows, that the exten-
sion must also be less ; this is manifest, for the resistance which any
body offers to the force by which it is strained, is proportional to the
extension which it undergoes, divided by its length ; now, since the
resistances of the several laminee, decrease as they recede from the
interior boundary towards the exterior, while at the same time, the
corresponding circumferences increase; it is manifest, that the exten-
sion for the several laminse decreases to the last or exterior boundary,
where it is the least of all : — It is therefore the law of the decreasing
resistance, that the present enquiry is instituted to determine.

Put d ~ ab, the interior diameter of the cylinder before the pressure

is applied,

e m the increase of d occasioned by the pressure,
d'— A B, the exterior diameter in its original state,
e — the increase induced by pressure.

Then (d -\- e) and d' -f e'), are respectively the interior and exterior
diameters of the cylinder as affected by extension.

By the principles of mensuration, the area of the annulus, or cir-
cular ring contained between the interior and exterior boundaries : —

Is equal to the difference of the squares of the diameters,
drawn into the constant fraction 0.7854 ; or it is proportional
to the sum of the diameters, drawn into their difference.

But according to the nature of the present enquiry, the area of the
ring is the same, both before and after the extension takes place ;
consequently, we have

(d' + c')f — (<* + *)* = <** — d*;

therefore, by expanding the terms on the left hand side, we get
e'* — d"' — 2de — e? — d3 — d* ;

OF THE HYDROSTATIC PRESS. 129

or by transposing and expunging the common terms, it is

and this equation being converted into an analogy, gives
2^4-e' : 2d-\-e : : e : e'.

Now, the quantity of extension that the material will allow before
rupture being very small, especially as compared with the quantities
2d' and 2d ; it therefore follows, that the quantities e' and e, in the
first and second terms, may be conceived to vanish, and the above
analogy becomes

d' : d : : e : e'.

From this it appears, that the extensions of the respective circum-
ferences, are inversely as the corresponding diameters ; but we have
stated above, that the resistance is as the extension divided by the
length; therefore, we have

d d'

d'''d'

or which amounts to the same thing,

d2 : d'1 ;

hence this law, that the magnitude of the resistance offered by each
successive circular lamina : —

Is inversely as the square of its diameter, or, which is the
same thing, inversely as the square of its distance from the
common centre to which they are referred.

From the general law thus established, the actual resistance due to
any point in the annulus, or to any thickness of metal, can very easily
be ascertained.

Put r iz: C«, the interior radius of the cylinder, of which the annexed

diagram is a section,
£r±«A, the entire thickness of the

metal,
#:zi an, any variable thickness esti- A

mated from a, the interior

surface,
n — the pressure on a square inch of

the inner surface in pounds avoirdupois,
/m the measure of the straining force, or the resistance sustained

by the first or interior lamina, and
c =i the cohesive force of the material.
VOL. i. K

130 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION

Then, agreeably to the law of the resistances which we have
established above, we have

::^: (r + x)2'

this result expresses the strain at the point x, or the resistance of the
material whose thickness is an ; and the fluxion of this quantity as
referred to the variable thickness x, is

consequently, the fluent, or the sum of all the strains, is

/fr*x
— - - 4- C, and this when x — t becomes
(r + xf

Therefore, if the strain or resistance /, were to act uniformly on the
thickness expressed by ^— , it would produce the same effect, as if

all the variable strains were to act on the whole thickness t.

The above law being admitted, let us suppose that the interior
radius of the cylinder, and the pressure per square inch on the surface
are given, and let it be required to determine the thickness such, that
the strain and resistance may be in equilibrio.

Here it is manifest, that the greatest strain the thickness — —

CT t

can resist, is — - , and the strain to which it is actually exposed, is
nr; consequently, when these are equal, we have

crt

***%?

from which, by expunging the common factor r, we get

-7+? (105).

If this value of n be compared with its respective values, as indi-
cated in the equations (104) preceding, we shall have the following
expressions, for the thickness of metal in the cylinder to resist any
pressure, while the elastic power of the material remains perfect, viz.

Pr _ pr . _ wr

— .7854CD9 — P .7854cd8 — ' .78540? — w

OF THE HYDROSTATIC PRESS. 131

Therefore, if for c in each of the preceding expressions, we substitute
its value as determined by experiment, and which for cast iron, accord-
ing to Dr. Robison, is 16648 pounds avoirdupois upon a square inch ;
then we shall have

,_ Pr *

- 13076 Da — P' (106).

A107)-

-130763s — w (108).

The following example will illustrate the use of these equations, the
value of t the thickness of the metal coming out the same by each.

138. EXAMPLE 13. What must be the thickness of metal in the
cylinder of a Hydrostatic Press, to resist a pressure of 30000 Ibs. ;
the diameter of the cylinder being 5 inches, that of the forcing pump
one inch, and of the safety valve one fourth of an inch ; being the
same dimensions which we have employed in the preceding examples?

Here we have given P ±= 30000 Ibs. ; D = 5 inches ; and conse-
quently, r— 2J inches; therefore, by substitution, equation (106)
gives

30000

n . ,

'= 13076 X 5*- 30000 = '^ "^ bem*
thing more than one fourth of an inch.

In order that the entire pressure on the piston of the cylinder, may
be equal to 30000 Ibs. according to the conditions of the question ;
the force with which the plunger of the forcing pump descends, must
be equal to 1200 Ibs. ; therefore, by equation (107), we have

before.

Again, in order that the entire pressure may be equal to 30000 Ibs.
the weight upon the safety valve must be 75 Ibs. ; hence, from equa-
tion (108), we obtain

the two cases foregoing.

139. It may not be improper here to remark, that although the

requisite thickness of metal is alike assignable from either of the

above equations, when the respective pressure and diameters are

* Where the constant number 13076 = 16648 X .7854.
K2

132 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION

known; yet it is the first of the class only, or that marked (106),
which becomes available in practice, and for this reason, that the
power of the press, or the aggregate pressure which it is capable of
exciting, is known d priori, or immediately assignable from the con-
ditions of construction, while the load upon the safety valve, and the
force with which the plunger descends, have each to be determined
by calculations founded on circumstances connected with the aggre-
gate or ultimate pressure.

140. Referring to equation (105), which has been purposely inves-
tigated, for expressing the intensity of pressure on a square inch of
surface, and multiplying both sides by r -\- t the denominator of the
fraction, we shall have

nr -f- nt~ct,

from which, by transposing and collecting the terms, we get

(c — n) t — n r ;

then by division, the value of #, or the thickness of metal in the
cylinder to withstand the pressure, becomes

.,nr
-^T* (109).

From which it appears, that if a constant value adapted to practical
purposes, can be assigned to n, the rule for" calculating the thickness
of metal in the cylinder will become exceedingly simple.

Now, it has been remarked by several eminent practical engineers,
as well as by the most approved and intelligent manufacturers, that
the extreme pressure on a square inch of the piston,* should never
exceed half the cohesive power of the material ; but according to Dr.
Robison, the cohesive power of cast iron of a medium quality is equal
to 16648 Ibs. ; hence we have

therefore, if 8324 Ibs. be adopted as the limit of pressure upon a
square inch of surface, the foregoing value of t becomes
- 8324 r
"16648 — 8324"

consequently, in order that the strain produced by the pressure may
not exceed the elastic power of the material ; —

* There is no occasion to limit the pressure to the piston only, since every square
inch of surface in contact with the fluid sustains the same pressure. This limitation
has frequently caused a misapprehension respecting the mode of ascertaining the
pressure on an inch of surface.

OF THE HYDROSTATIC PRESS. 133

The thickness of metal ought never to be less than the
interior radius of the cylinder.

By the first equation of class (104), it has been shown, that the
pressure on a square inch of the piston in Ibs. avoirdupois, is

~.7854D8'
or by substituting the foregoing value of n, it is

from which, by multiplication, we obtain

8324 X .7854D2rz:P;

but in order to express the pressure in tons, it is
6537.6696D*

2240 (110).

141. Therefore, when the diameter of the cylinder is given, the
entire pressure in tons is determined by the following very simple

rule.

RULE. Multiply the square of the diameter in inches, by

the constant number 2.9186, and the product will be the
pressure in tons.

And again, when the pressure in tons is given, the diameter of the
cylinder may be determined by reversing the process, or by the fol-
lowing rule.

RULE. Divide the given pressure in tons by the constant
number 2.9186, and extract the square root of the quotient,
for the diameter of the cylinder in inches.

142. The preceding theory, as we have developed it, unfolds every
particular connected with the Hydrostatic Press, and by paying
proper attention to the equations, rules, and examples, as we have
delivered them, every difficulty attending the construction of the
instrument will be removed ; to practical persons, however, that part
of the theory exhibited in the equations marked (109) and (110)
will be found the most valuable, as they do the more immediately
contain the particulars which direct their operations. The following
examples will prove the truth of these remarks.

EXAMPLE 14. The diameter of the cylinder in a Hydrostatic Press,
is 10 inches ; what is its power, or what pressure does it transmit?
Here by the first rule above, we have
PzzlO2 X 2.9186zz291. 86 tons.

134 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION

EXAMPLE 15. What is the diameter, and what the thickness of
metal, in a press of 300 tons power ?

By the second rule above, we have
D2 = 300 ~ 2.9186 — 102.81 nearly ;
therefore, by extracting the square root, we obtain

D= V 102.81 = 10.13 inches;

consequently, according to the remark under the equation (109), the
thickness of metal is

t = 10.13 4- 2 = 5.065 inches.

143. The rules by which the preceding examples have been resolved,
are very nearly, but not precisely the same as those employed by
Messrs. Bramah in the construction of their excellent presses ; the only
difference, however, consists in their assuming a higher number as
the limit of pressure, the standard which they employ being 8556 Ibs.
upon a square inch of the piston, thereby indicating, that they reckon
on a higher cohesive power in the material, than that which we have
adopted as the basis of our theory.

Now, 8556 Ibs. on a square inch, is equivalent to 6619.8824 Ibs.
upon a circular inch ; whereas the constant which we have chosen is
only 6537.6696 Ibs., being a difference of 82.2128 Ibs. upon the
circular inch, a difference that need not be regarded in practice, as
the error will always fall on the side of safety, giving a smaller power
to the press than what it really possesses.

144. It sometimes, indeed it very frequently happens, that presses
are constructed, without any attention being paid, to the relation which
subsists between the strength of the parts, and the strain which they
have to resist ; in all such cases, therefore, it may be interesting to
possess a rule, by which the merits or demerits of a press so con-
structed can be ascertained, for in this way a failure in the instrument
may be prevented, and a remedy applied to any defect that may
exist.

Now, according to the first equation of class (104), the pressure
upon a square inch is

-.~7854D~"

and according to equation (105), it is.
ct

OF THE HYDROSTATIC PRESS. 135

therefore, by comparison, we have
P ct

consequently, by multiplying and substituting the cohesive power of
cast iron, we have

(*4-r)P:=13076D2f. (111).

Let 4r2, be substituted in this equation, instead of D2 its equivalent,
and we shall obtain

consequently, the pressure in tons, is

52304 r*t _ 23.35 r*t
~2240(*+r)~- (f+r) ' (112).

From which it appears, that by knowing the interior radius of the
cylinder and the thickness of the metal, the power of the press can
easily be ascertained ; the following is the rule for that purpose.

RULE. Multiply 23.35 times the thickness of metal by the
square of the radius of the cylinder, and divide the product
by the radius plus the thickness of metal, and the quotient
will give the power of the press in tons.

145. EXAMPLE 16. A Hydrostatic Press is so constructed, as to
have the interior radius of its cylinder equal to 3 inches, and the
thickness of metal 4 inches ; now this press is designed for packing
flax, and is estimated to stand a pressure of 180 tons; query if its
power is not overrated ?

According to the above rule, it is

consequently, the power of the press is overrated by about 60 tons,
being one third less than the estimated pressure according to the
question.

The thickness of metal necessary to resist a pressure of 180 tons or
403200 Ibs. is equal to 17.9 inches nearly, and the proposed thickness
is only 4 inches, being less than one fourth of the thickness which is
really necessary to resist the strain ; hence we infer that the press in
its present state, is entirely unfitted for its intended purpose, and
altogether inconsistent with safety and precision of operation. Here
follows the description of a press when completely furnished in all its
parts and fit for immediate action.

136 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION

146. The Hydrostatic Press,, in its present high and refined state of
improvement, is a machine that is capable of generating and trans-
mitting a 'greater degree of force, for the purpose of overcoming
immense resistances, and raising enormous loads to a small height,
than any other instrument or engine with which we are acquainted ;
it is therefore of the highest importance th,at the principles of its
construction and the mode of operation should be rightly understood,
and in order to render the subject as clear and intelligible as possible,
we think proper to lay before our readers the following detailed

description.

Fig. 1.

The wood cut before us, fig. 1, exhibits an elevation of the press
in its complete state, accompanied by the forcing pump and all its
appurtenances as fitted up for immediate action : F is a strong metal-
lic cylinder of cast iron, or some other material of sufficient density
to prevent the fluid from issuing through its pores, and of sufficient
strength to preclude the possibility of rupture, by reason of the im-
mense pressure which it is destined to withstand.

The cylinder F is bored and polished with the most scrupulous
precision, and fitted with the moveable piston D, which is rendered
perfectly water-tight, by means of leather collars constructed for the
purpose, and fixed in the cylinder by a simple but ingenious con-
trivance to be described hereafter.

OF THE HYDROSTATIC PRESS. 137

Into the side or base of the cylinder F, the end of a small tube bbb
is inserted, and by this tube the water is conveyed or forced into the
cylinder; the other end of the tube is attached to the forcing pump,
as represented in the diagram ; but this will be more particularly
explained in another place.

A A are two very strong upright bars, generally made of wrought iron,
and of any form whatever, corresponding to the notches in the sides of
the flat table E, which is fixed upon the end of the piston r>, and by
workmen, is usually denominated the ' Follower or ' Pressing Table.'

B is the top of the frame into which the upright bars A A are fixed,
and cc is the bottom thereof, both of which are made of cast, in
preference to wrought iron, being both cheaper and more easily
moulded into the intended form.

The bottom of the frame cc, is furnished with four projections or
lobes, with circular perforations, for the purpose of fastening it by iron
bolts to the massive blocks of wood, whose transverse sections are
indicated by the lighter shades at GG. The top B has two similar
perforations, through which are passed the upper extremities of the
vertical bars A A, and there made fast, by screwing down the cup-nuts
represented at a and a.

Fiy. 2 represents the plan of the top, or as it is more frequently
termed, the head of the frame ; the lower side Fig. 2.

or surface of which is made perfectly smooth,
in order to correspond with, and apply to the
upper surface of the pressing table E in^. 1 ;
this correspondence of surfaces becomes ne-
cessary on certain occasions, such as the copy-
ing of prints, taking fac-similes of letters and

the like ; in all such cases, it is manifest, that smooth and coincident
surfaces are indispensable for the purpose of obtaining true impres-
sions.

The figure before us represents the upper side of the block, where
it is evident, that the middle part B, (through whose rounded extremi-
ties a and a, the circular perforations are made for receiving the
upright bars or rods AA,^. 1), is considerably thicker than the parts
on each side of it ; this augmentation of thickness, is necessary to
resist the immense strain that comes upon it in that part ; for although
the pressure may be equally distributed throughout the entire surface,
yet it is obvious, that the mechanical resistance to fracture, must prin-
cipally arise from that part, which is subjected to the re-action of the
upright bars.

138

THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION

Fig. 3 represents the plan of the base or bottom of the frame ; it is
generally made of uniform thickness, Fig. 3.

and of sufficient strength to withstand
the pressure, for be it understood, that
all the parts of the machine are sub-
jected to the same quantity of strain,
although it is exerted in different ways.*

The circular perforations cc correspond to a a in the top of the
frame, and receive the upright bars in the same manner ; the perfora-
tions dddd, receive the screw bolts which fix the frame to the beams
of timber represented at GG, fig. \ ; the large perforation r receives
the cylinder, the upper extremity of which is furnished with a flanch,
for the purpose of fitting the circular swell around the perforation,
and preventing it from moving backwards during the operation of
the instrument.

When the several parts which we have now described are fitted
together, they will present us with that portion of the drawing in
fig. 1 denominated " Elevation of the Press."

A side view of the engine as thus com-
pleted, is represented in fig. 4, where, as is
usual in all such descriptions, the same letters
of the alphabet refer to the same parts of the
structure.

F is the cylinder into which the fluid is
injected ; D the piston, on whose summit is
the pressing table E ; A one of the upright
rods or bars of malleable iron ; B the head of
the press, fixed to the upright bar A by means
of the cup-nut a; c the bottom, in which the
upright bar is similarly fixed ; and G a beam
of timber supporting the frame with all its
appendages.

147. But the Hydrostatic Press as here
.described and constructed, must not be con-
sidered as fit for immediate action ; for it is
manifestly impossible to bore the interior of
the cylinder so truly, and to turn the piston

Fig. 4.

* The upright bars, cylinders, and connecting tubes, resist by tension, the pistons
by compression, and the pressing table, together with the top and bottom of the
frame, resist transversely.

OF THE HYDROSTATIC PRESS. 139

with so much precision, as to prevent the escape of water between
their surfaces, without increasing the friction to such a degree, that it
would require a very great force to counterbalance it.

In order, therefore, to render the piston water-tight, and to prevent
as much as possible the increase of friction, recourse must be had to
other principles, which we now proceed to explain.

The piston D is surrounded by a collar of pump leather oo, repre-
sented in Jig. 5, which collar being doubled up, so Fig. 5.
as in some measure to resemble a lesser cup placed
within a greater, it is fitted into a cell made for its
reception in the interior of the cylinder ; and when

there, the two parts are prevented from coming toge- ^^m^
ther, by means of the copper ring pp, represented in FtS- 6-

Jig. 6, being inserted between the folds, and retained in its place, by a
lodgement made for that purpose on the interior of the cylinder.

The leather collar is kept down by means of a brass or bell-metal
ring mm, Jig. 7, which ring is received into a Fig. 7.

recess formed round the interior of the cylinder,
and the circular aperture is fitted to admit the
piston D to pass through it, without materially
increasing the effects of friction, which ought to be
avoided as much as possible.

The leather is thus confined in a cell, with the edge of the inner
fold applied to the piston D, while the edge of the outer fold is in
contact with the cylinder all around its interior circumference ; in
this situation, the pressure of the water acting between the folds of
the leather, forces the edges into close contact with both the cylinder
and piston, and renders the whole water-tight ; for if the leather be
properly constructed and rightly fitted into its place, it is almost
impossible that any of the fluid can escape; for the greater the
pressure, the closer will the leather be applied to both the piston and
the cylinder.

The metal ring mm is truly turned in a lathe, and the cavity in
which it is placed is formed with the same geometrical accuracy ; but
in order to fix it in its cell, it is cut into five pieces by a very fine
saw, as represented by the lines in the diagram, which are drawn
across the surface of the ring. The four segments which radiate to
the centre are put in first, then the segment formed by the parallel
kerfs, (the copper ring pp and the leather collar oo being previously
introduced), and lastly, the piston which carries the pressing table.

That part of the cylinder above the ring mm, where the inner

140 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION

surface is not in contact with the piston, is filled with tow, or some
other soft material of a similar nature ; the material thus inserted has
a twofold use ; in the first place, when saturated with sweet oil, it
diminishes the friction that necessarily arises, when the piston is
forced through the ring mm; and in the second place, it prevents
the admission of any extraneous substance, which might increase the
friction or injure the surface of the piston, and otherwise lessen the
effects of the machine.

The packing here alluded to, is confined by a thin metallic annulus,
neatly fitted and fixed on the top of the cylinder, the circular orifice
being- of sufficient diameter, to admit of a free and easy motion to the
piston .

If a cylinder thus furnished with its several appendages be placed
in the frame, and the whole firmly screwed together, and connected
with the forcing pump, as represented in fig. 1, the press is completed
and ready for immediate use ; but in order to render the construction
still more explicit and intelligible, and to show the method of con-
necting the press to the forcing pump, let jig. 8 represent a section
of the cylinder with all its furniture, jr^. 3.

and a small portion of the tube im-
mediately adjoining, by which the
connexion is effected.

Then is FF the cylinder; D the
piston; the unshaded parts oo the
leather collar, in the folds of which
is placed the copper ring pp, dis-
tinctly seen but not marked in the
figure; mm is the metal ring by
which the leather collar is retained

in its place ; nn the thin plate of copper or other metal fitted to the
top of the cylinder, between which and the plate m m is seen the soft
packing of tow, which we have described above, as performing the
double capacity of oiling the piston and preventing its derange-
ment.

The combination at wx, represents the method of connecting the
injecting tube to the cylinder : it may be readily understood by in-
specting the figure; but in order to remove all causes of obscurity, it
may be explained in the following manner.

The end of the pipe or tube, which is generally made of copper^
has a projecting piece or socket flanch soldered or screwed upon it^
which fits into a perforation in the side or base of the cylinder, accord-

OF THE HYDROSTATIC PRESS. 141

ing to the fancy of the projector, but in the figure before us the per-
foration is in the side.

The tube thus furnished, is forcibly pressed into its seat by a hollow
screw w, called an union screw, which fits into another screw of equal
thread made in the cavity of the cylinder ; the joint is made water-
tight, by means of a collar of leather, interposed between the end of
the tube and the bottom of the cavity.

A similar mode of connection is employed in fastening the tube to
the forcing pump, the description of which, although it constitutes an
important portion of the apparatus, does not properly belong to this
place ; the principles of its construction and mode of action, must
therefore be supposed as known, until we come to treat of the con-
struction and operation of pumps in general.

Admitting therefore, that the action of the forcing pump is under-
stood, it only now remains to explain the nature of its operation in
connection with the Hydrostatic Press, the construction of which we
have so copiously exemplified.

148. In order to understand the operation of the press, we must
conceive the piston D Jig. 1, as being at its lowest possible position in
the cylinder, and the body or substance to be pressed, placed upon the
crown or pressing table E ; then it is manifest, that if water be forced
along the tube b b b by means of the forcing pump, it will enter the
chamber of the cylinder F immediately beneath the piston D, and cause
it to rise a distance proportioned to the quantity of fluid that has been
injected, and with a force, determinable by the ratio between the square
of the diameter of the cylinder and that of the forcing pump. The
piston thus ascending, carries its crown, and consequently, the load
along with it, and by repeating the operation, more water is injected,
and the piston continues to ascend, till the body comes into contact
with the head of the frame B, when the pressure begins ; thus it is
manifest, that by continuing the process, the pressure may be carried
to any extent at pleasure ; but we have already stated, in developing
the theory, that there are limits, beyond which, with a given bore
and a given thickness of metal, it would be unsafe to continue the
strain.

When the press has performed its office, and it becomes necessary
to relieve the action, the discharging valve, placed in the furniture
of the forcing pump, must be opened, which will admit the water to
escape out of the cylinder and return to the cistern, while the table
and piston, by means of their own weight, return to their original
position.

142 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION

The method of calculating the power of the press, as well as every
other particular respecting it, has been fully exemplified in the fore-
going theory ; it is hence unnecessary to dwell longer on the subject :
we shall therefore conclude our description of the press, and proceed
with that of the Hydrostatic Bellows, which depends upon the same
principle, viz. the quaqua versum pressure of non-elastic fluids.

2. THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION OF THE
HYDROSTATIC BELLOWS.

149. In the preceding pages we have developed the theory and
exemplified the application of the hydrostatic press ; and furthermore,
in order to render the subject as complete as possible, we have given
a minute and comprehensive description of its several parts, and for
the purpose of guiding the practical mechanic in its erection, the
instrument is exhibited in its complete and finished state, accompanied
by the forcing pump and all its requisite appendages.

The next subject, therefore, that claims our attention, is the
Hydrostatic Bellows, an instrument of very frequent occurrence in
philosophical experiments ; it is chiefly employed in illustrating the
upward pressure of non-elastic fluids and the hydrostatic paradox,
and consequently, it depends upon the same principle as the hydro-
static press, admitting of a similar, but a more concise mode of dis-
cussion and illustration.

150. The Hydrostatic Bellows consists of a tube or pipe FEI, of
very small diameter, and of any convenient length

at pleasure, connected by means of the elbow at ^
i, with a cylindrical vessel whose vertical section
is CDGH, and whose sides are made of leather
like a common bellows, represented by the waving
lines AmD and BWK; the upper and the lower
surfaces AB and DC, being formed of circular
boards corresponding to the cylindrical form of
the vessel.

When the bellows is empty, it is manifest that the boards A B and
DC are very nearly in contact, and would be completely so, but for
the leather sides forming into folds and preventing a coincidence:
in this state, when water or any other incompressible and non-elastic
fluid, is poured into the tube, it flows into the bellows and separates
the boards ; a heavy weight as w is then placed upon the upper

OF THE HYDROSTATIC BELLOWS. 143

board, and by pouring more fluid into the tube, the moveable plane
A B and its incumbent load w, will be raised and kept in equilibrio
by the column of fluid in the tube ; and when the equilibrium obtains,
we infer, that : —

The iveight of the supporting column of fluid in the tube, is
to the weight upon the moveable plane, as the area of a section
of the tube, is to the area of the plane.

This is manifest, for the fluid at i, the lowest point of the vertical
tube FEI, is pressed by a force varying as the altitude LI, and by the
nature of fluidity, this pressure is communicated horizontally to all
the particles in DC, and thence transmitted throughout the whole mass
of fluid in the bellows ; consequently, the pressure upwards on the
board AB, is equal to the weight of a column of the fluid, the diameter
of whose base is DC, and altitude LI or GD ; but the actual weight of
the fluid supported, is that of a column whose diameter is DC, and
altitude EI or AD.

Hence, the weight which maintains the equilibrium, will be that of
a cylinder of fluid, whose base is A B and altitude A G ; consequently,
the weight w, placed upon the moveable plane of the bellows, since it
balances the column of fluid L E, is equivalent to the weight of a fluid
cylinder, whose section along the axis is ABHG.

Put D rz AB or DC, the diameter of the cylindrical vessel or bellows,
d zz LM, the diameter of the vertical tube,
w ~ the weight upon the moveable plane, and
w'-=. the weight or pressure of the fluid in the column LE.

Then, because by the principles of mensuration, the areas of circles
are to one another as the squares of their diameters ; the foregoing
inference gives

w' : w : : d* : D2,

and this, by equating the products of the extreme and mean terms,
becomes

tfw'=:d*w. (113).

Let both sides of this equation be divided by the quantity D2, which
is found in combination with the weight or pressure of the fluid in the
tube, and we shall obtain

, d*w
W=^' (114).

Here again, that singular property of non-elastic and incompressible
fluids becomes manifest, viz.

144 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION

That any quantity however small, may be made to balance
any other quantity however great.

151. If the diameter of the tube, the diameter of the cylinder or
bellows, and the weight upon the moveable board AB be given, the
weight of the fluid in the tube, or its perpendicular altitude to main-
tain the equilibrium, can easily be determined by means of the equa-
tion (114), which affords the following practical rule.

RULE. Multiply the square of the diameter of the tube by
the load upon the moveable board, and divide the product by
the square of the diameter of the bellows or cylinder ; then,
the quotient will give the weight of the fluid by which the
equilibrium is maintained.

EXAMPLE. The diameter of the bellows or cylindrical vessel is 18
inches, that of the tube or pipe, through which the fluid is conveyed
into the vessel, is one fourth of an inch, and the weight upon the
moveable board is 5760 Ibs. ; what weight of water must be poured
into the vertical tube, so that the whole may remain at rest ?

In this example there are given, Dm 18 inches; c?=:| of an inch,
and w = 5760 Ibs. ; therefore, by performing as directed in the rule,
we shall have

.25* X 5760 360

Here it appears, that a quantity of water weighing l£lbs., disposed
in a tube of £ of an inch in diameter, is capable of balancing another
quantity of 5760 Ibs., disposed in a cylinder of 18 inches diameter;
it is therefore manifest, that the height of the one column must far
exceed the height of the other, and the excess of altitude may be
determined in the following manner.

152. It has been abundantly proved by experiment, that a cubic
foot of distilled water, at the temperature of about 39° of Fahrenheit's
Thermometer, weighs very nearly 1000 avoirdupois ounces, or 62 Jibs. ;
consequently, the number of cubic inches in the column whose weight
is l^lbs., is found by the following analogy, viz.

62£ : 1728 :: l£ : 30±f inches;

hence, the solidity of a column which maintains the equilibrium is
30£f- inches, and according to the conditions of the question, the
diameter of its base or section, is one fourth of an inch, and con-
sequently, the area of the base or section, is

.25* X. 7854 = . 0490875 square inches.

OF THE HYDROSTATIC BELLOWS. 145

Now, according to the principles of mensuration, the solidity of a
cylinder is determined, by multiplying the area of its base into its
perpendicular altitude ; consequently, if h denote the perpendicular
height of the column, we have

.0490875 h = 30.72;
therefore, by division, we shall obtain

OA (TO

153. The solution which we have here given, applies to the parti-
cular example preceding, in which the data are assigned ; but in order
to accommodate the theory to every case, it becomes necessary to
draw up the solution in general terms ; for which purpose, we must
recur to equation (114), where the weight of the equilibrating column
has already been found ; then, according to the above analogy, we
have

62i : 1728 : : £f , .,

where s denotes the solidity of the column.

If in the above analogy, we make the product of the jnean terms
equal to the product of the extremes, we shall have

and from this, by division, we get

tf (115).

Therefore, if the solidity of the equilibrating column be divided by
the area of its base, viz. the quantity .7854d2, the quotient will fur-
nish the perpendicular altitude ; hence we have

_ 35.2024 w

D* (116).

154. From this it appears, that in order to determine the altitude
of the equilibrating column, it is not necessary that its diameter
should be known, for the equation is wholly independent of that
element, the diameter of the bellows, and the weight upon tne
moveable board only, entering into its composition. The following
practical rule will therefore determine the altitude of the column by
which the equilibrium is maintained.

VOL. I. L

146 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION

RULE. Divide 35.2024 times the load to be sustained upon
the moveable board, by the square of the diameter of the
bellows, and the quotient will be the altitude of the equi-
librating column.

We shall determine the perpendicular altitude by this rule, on the
supposition that the diameter of the bellows and the weight upon the
moving plane, are the same as in the foregoing example ; therefore
we have

. 35.2024X5760 _,Q10. ,
^— — nz 625.8 19 inches.

The equation (114) for the weight of the equilibrating column, was
deduced from the equation (113), by simple division only, without the
enunciation of any problem ; but in order to render the subject a
little more systematic, we shall determine the other elements of the
general equation, severally from the resolution of their respective and
appropriate problems.

PROBLEM XXII.

155. In a hydro-statical bellows of a cylindrical form, there
are given, the diameters of the bellows and of the equilibrating
tube, together with the weight of the fluid by which the equili-
brium is maintained : —

It is required to determine the weight upon the moveable
plane, at the instant when the equilibrium obtains.

Let both sides of the general equation (113), be divided by d* the
square of the diameter of the balancing tube, and we shall obtain

_pV

~-~d^' (117).

And this equation affords the following practical rule.

RULE. Multiply the weight of the equilibrating fluid, by
the square of the diameter of the bellows, and divide the
product by the square of the diameter of the tube, for the
weight upon the moveable plane.

EXAMPLE. The diameter of a cylindrical bellows is 24 inches, the
diameter of the balancing tube is one fourth of an inch, and the
weight of the fluid in the tube is 2 J Ibs. ; what weight will this coun-
terpoise on the moving board of the bellows ?

OF THE HYDROSTATIC BELLOWS. 147

Here, by proceeding as directed in the rule, we obtain

This is something more than 10 tons and a quarter, which is mani-
festly a great load to be suspended by 2 Jibs.; but the altitude of the
suspending column must be proportionably great, which circumstance,
without the aid of some artificial force, would render the instrument
very inconvenient for any practical purpose ; it was, no doubt, by
viewing the matter in'this light, that Mr. Bratnah, senior, was led to
apply the forcing pump, and thereby to produce that very powerful
engine, which formed the subject of our last article.

PROBLEM XXIII.

156. In a hydrostatical bellows of a circular form, there are
given, the diameter of the bellows, the load suspended, and the
weight of the suspending fluid : —

It is required to determine the diameter of the equilibrating
tube, so that the instrument may be just in a state of equi-
librium.

Let both sides of the general equation (113), be divided by w the
weight upon the bellows, and we shall obtain

D2M/

w
and from this, by extracting the square root, we get

D2^'

w ' (118).

_
And the practical rule which this equation supplies, may be

expressed in words at length in the following manner.

RULE. Multiply the square of the diameter of the bellows,
by the weight of the fluid which maintains the equilibrium,
and divide the product by the weight upon the bellows, then,
the square root of the quotient will be the diameter of the
equilibrating tube.

* This equation for the diameter of the tube may be otherwise expressed ; thus *

i 2

148 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION

EXAMPLE. The diameter of the bellows or cylindrical vessel, is 24
inches, the weight of the suspending fluid is 2 Ibs., and the weight
suspended on the bellows 8000 Ibs. ; what is the diameter of the

tube?

Performing according to the rule, we have

and from this, by extracting the square root, we obtain
6?=V°-144 = -38 of an i

PROBLEM XXIV.

157. In a hydrostatic bellows of a cylindrical form, the
diameter of the tube, the weight suspended, and the weight of
the suspending fluid, are given : —

It is required to determine the diameter of the bellows, so
that the whole may be in a state of equilibrium.
Let both sides of the general equation (113), be divided by w' the
weight of the suspending fluid, and we shall have

__
w'

from which, by extracting the square root, we get

w1 ' (119).

And from this equation, we obtain the following practical rule.

RULE. Multiply the square of the diameter of the suspend-
ing tube, by the weight suspended, and divide the product by
the weight of the fluid which maintains the equilibrium; then,
the square root of the quotient will be the diameter of the
cylinder sought.

EXAMPLE. The diameter of the suspending tube in a cylindrical
hydrostatic bellows, is half an inch, the weight of the suspending fluid
is 2 Ibs., and the weight suspended on the bellows board is 12000 Ibs. ;
what is the diameter of the bellows ?

Here, by proceeding as directed in the foregoing rule, we get

.5X. 5X12000
D*~ — 1500,

* This equation for the diameter of the bellows may be otherwise expressed ; thus :

OF THE HYDROSTATIC BELLOWS. 149

and by extracting the square root, we have

:rr 38.73 inches.

158. The foregoing problems and rules, unfold every particular
respecting the calculation of the hydrostatic bellows, and from them
we may infer, that in the case of an equilibrium, if more fluid be
added : —

It will ascend equally in the suspending tube, and in the
cylindrical vessel composing the bellows, whatever may be
their relative magnitudes.

The demonstration of this is very simple, for let A BCD be a vertical
section, passing along the axis of the cylindrical
vessel, and also along the axis of the suspending
tube KI ; and suppose that F and c are the points
to which the fluid rises in the vessel and the tube,

when the bellows is in a state of equilibrium.

Take ic equal to Da, and through the points a D C I
and c let a horizontal plane be drawn, intersecting
the vertical plane A BCD in the line ab ; then it is manifest, that the
weight w in the position EF, is equivalent to the weight of the fluid
column a&FE. Let more fluid be poured into the tube at K; the
equilibrium will then be destroyed, and the weight w will ascend,
until by discontinuing the supply, the equilibrium is restored, and the
fluid in the vessel and the tube becomes again quiescent at the points
n and K.

Take IK equal to DA, and through the points A and K, let a hori-
zontal plane be drawn, cutting the vertical plane A BCD in the line
AB; then as before, the weight w in the position mn, is equivalent
to the weight of the fluid cylinder, of which A&nm is a vertical
section.

Now, the weight w is not altered in consequence of the change of
position from EF to mn\ therefore, because EF is equal to mn, it
follows, that E« is equal to mA; consequently, by taking away the
common space ma, the remainders Em and a A are equal to one
another; but by reason of the parallel lines ac and AK, the spaces
a A and CK are equal to one another; therefore CK is equal to
Em.

From the principle here demonstrated, the resolution of the follow-
ing problem may readily be derived.

150 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION

PROBLEM XXV.

159. If a hydrostatic bellows of a cylindrical form, have a
given quantity of fluid poured into the equilibrating or suspend-
ing tube : —

It is required to determine through what space the weight
on the moving board will ascend in consequence of the supply.

Before we proceed to the resolution of this problem, it may be
proper, as in the foregoing cases, to exhibit an appropriate notation ;
for which purpose,

Put D= AB or DC, the diameter of the cylindrical vessel or bellows,
c?zz the diameter of the equilibrating or suspending tube,
q HZ the quantity of fluid poured into the tube, and
a; — Em, the space through which the weight ascends by reason
of the supply.

Then, according to the principles of mensuration, the area of a
transverse section of the cylindrical vessel or bellows, is

and the area of the corresponding section of the tube, is

where the symbols a and a' denote the respective areas.
But by the property demonstrated above, the fluid rises equally in
the bellows and in the tube ; therefore, the quantity of fluid which
flows into the bellows in consequence of the supply, is

and the quantity which remains in the tube, is

where the symbols s and s' denote the solidities of the cylinders,
whose diameters are D and d, and their common altitude x.

Now, the sum of these quantities, is manifestly equal to the quantity
of fluid poured into the tube ; hence we have

and by division, we obtain

- 9

(120).

OF THE HYDROSTATIC WEIGHING MACHINE. 151

It therefore appears, that the space through which the weight
ascends by reason of the supply : —

Is equal to the quantity of fluid which is poured into the
tube, divided by the sum of the areas of a cross section of the
tube and the cylindrical vessel or bellows.

The practical rule, or method of applying the equation, may there-
fore be expressed in words at length in the following manner.

RULE. Divide the quantity of fluid which is poured into
the tube, by .7854 times the sum of the squares of the dia-
meters, and the quotient will give the quantity of ascent, or
the space through which the weight is raised in consequence
of the supply.

EXAMPLE. The diameter of a cylindrical vessel is 20 inches, and
that of the suspending tube is one inch ; now, suppose that an incom-
pressible fluid is poured into the tube, until its weight sustain in
equilibrio, a load of 8760 Ibs. upon the moveable bellows board ;
then, how much higher will the load be raised, when 150 cubic inches
of the fluid are superadded ?

Here then, we have given D=r20 inches, d— 1 inch, and <?— 150
cubic inches ; consequently, by the rule, we obtain

* Hi * = .7854(450°0+1) ='476 °f a" inch' • r: ' I'

From this it appears, that if a machine of the given dimensions be
brought into a state of equilibrium, the addition of 150 cubic inches
will raise the load .476, or very nearly half an inch higher ; in which
case the equilibrium will still obtain, for the altitude of the fluid in
the suspending tube, is increased exactly as much as the load has
been raised, while the magnitude of the load, and consequently, the
height of the equilibrating column, remain the same.

3. THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION OF THE
HYDROSTATIC WEIGHING MACHINE.

160. The preceding principles may also be applied to the con-
struction of a very simple and convenient weighing machine ; for, if
into the side of an open cylindrical or other vessel, a bent tube be
inserted, and if on the surface of the fluid, a moveable cover exactly

152 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION

fitting the vessel be placed with a weight upon it, and the tube
graduated : —

Then, any additional weight placed upon the cover, may be
determined by knowing the height to which the fluid rises in
the tube ; and conversely : —

If the additional weight be known, the height to which the
fluid rises in the tube may be found.

Let ABCD represent a vertical section of a cylindrical vessel, or of
any other vessel, whose sides are perpendicular
to the horizon ; and let K i c be the corres-
ponding section of the equilibrating tube.

Let both the vessel and the communicating
tube be open at the upper parts AB and de,
and conceive the vessel to be filled with fluid
to the line EF or altitude DE; then, on the
surface of the fluid at EF, let there be placed
a moveable cover exactly fitting the vessel, so
that the whole may be water-tight.

Produce EF to b, then is the point b at the same level in the tube
IK, as the surface of the fluid in the vessel whose level is EF : upon
the cover EF let the weight w be placed, and suppose a to be the
point in the tube, to which the fluid will rise by the action of the
cover, together with the weight w which is placed upon it; in this
case, the machine is in a state of equilibrium.

If some additional weight w' be placed upon the cover, then the
original equilibrium will be destroyed, and can only be restored, by
the fluid ascending in the tube to a sufficient height to balance the
additional weight.

Put DZZ AB or DC, the diameter of the cylindrical vessel, of which

ABCD is a section,

d zz de, the diameter of the communicating tube KIC,
h zz ba, the height of the original equilibrating column,
wzz the weight supported by the column b a,
w/zzthe additional weight, whose quantity is required,
A'zzaK, the increased altitude of the supporting column,
$ zz Em, the descent of the cover occasioned by the additional

load w', and
$ zz the specific gravity of the fluid,

OF THE HYDROSTATIC WEIGHING MACHINE. 153

Then it is manifest, that when the equilibrium originally obtains ;
that is, when the surface of the fluid in the tube is at a, and that in
the vessel at EF, the pressure of the fluid in the tube exerted at b, is

where the symbol p denotes the pressure at b ;

but this is manifestly in equilibrio with the pressure of the column
wEFxt or the weight w ; consequently, we have

.7854rf2As : w : : .7854tf : .7854D2;
or by suppressing the common factors, we obtain

hs : w:: 1 : .7854 D8;
and by equating the products of the extremes and means, it is

wn=.7854£sD2. (121).

Again, by means of the additional weight w', whose magnitude is
required, the cover EF is supposed to descend to the position mn;
while, in order to regain the equilibrium, the fluid rises in the tube as
far as the point K, in which case, the altitude of the equilibrating
column CK becomes (h -f- ti -f- 5) and consequently, the pressure at
c, is

;>'=.7854eZ2s (& + &' + 3),

and this is in equilibrio with the pressure of the column ymnz, or
the weight (w + w') ; consequently, we have

.7854d2s (h + h' -f 3) : w -f w' : : .7854d2 : .7854D* ;
or by suppressing the common factors, we have

s(k-\-h' -M) : w + w' : : 1 : .7854o2;

therefore, by equating the products of the extremes and means, we get
w -\- w/zz .7854D2s(/z-r-A'-r-S). (122).

But we have seen above, equation (121), that wnr.7854/i5D2;
consequently, by substituting and separating the terms, we obtain

W/i=.7854D2s(A/ + a). (123).

Now, it is manifest, that the descent of the cover in the vessel, and
the rise of the fluid in the tube, must be to one another, inversely as
the squares of the respective diameters ; therefore, we have

Ztf^h'tf,

* or by division, we get
h'd*

S = l^'

and finally, by substitution, we obtain

w' = .7854 h's (D2 -f d3). (124).

154 SCIENTIFIC DESCRIPTION OF THE HYDROSTATIC WEIGHING MACHINE.

161. If the fluid be water, whose specific gravity is represented by
unity, the equation becomes somewhat simpler ; for in that case, we
have

w' = .7854 &' (D« -f d2). (125).

From this equation the magnitude of the additional weight, or the
measure by which it is expressed, can very easily be ascertained ;
and the practical rule by which it is discovered, is as follows.

RULE. Multiply the sum of the squares of the diameters,
by .7854 times the rise of the fluid in the tube, or the eleva-
tion above the first level, and the product will express the
magnitude of the additional weight.

EXAMPLE. The diameter of a cylindrical vessel is 16 inches, and
that of the communicating tube one inch ; now, supposing the machine
in the first instance, to be in a state of equilibrium, and that by the
addition of a certain weight on the moveable cover, the water in the
tube rises 6 inches above the original equilibrating level ; how much
weight has been added ?

By proceeding according to the rule, we have
D2 -|- d*= 162 + P i=256 + 1 = 257,

and by multiplication, we obtain
w' = .7854X6X257 zz 1211.0868 avoirdupois Ibs.*

162. If the additional weight, by which the water is made to rise
in the tube be given, the distance above the first level to which it will
rise, can easily be found ; for let both sides of the equation (125), be
divided by the quantity .7854 (D* -\- dz), and we shall obtain

,,_

~.7854(D*4-d2)'

And from this equation, we deduce the following rule.

RULE. Divide the additional weight, by the sum of the
areas of the moveable cover and the cross section of the
communicating tube, and the quotient will give the height to
which the fluid will rise above the first level.

* It is manifest from the form of the equation which supplies the rule, that
without paying particular attention to the nature of the load which produces the
equilibrium in the first place, the value of vf is ambiguous, and may be read in
ounces, Ibs., cwts., or tons ; and indeed, in any denomination of weight whatever ;
but it must always be read in the same name as that by which the equilibrium
is produced.

EXPERIMENTS ON THE QUAQUAVERSUS PRESSURE OF FLUIDS. 155

EXAMPLE. The diameter of the moveable cover is 16 inches, and
that of the communicating tube one inch ; then, supposing that the
machine in the first instance is brought to a state of equilibrium,
and that a load of 1211 Ibs. is applied on the cover, in addition to
that which produces the equipoise ; to what height above the first level
will the water ascend in the communicating tube?

Proceeding according to the rule, we obtain

.7854 (D8 + <P) = .7854 (16* + I2) = 201 .8478 divisor ;

consequently, by division it is

And exactly after the manner of these two examples, may any
other case be calculated ; but in applying the principles to the
determination of weights, mercury ought to be employed in preference
to water, as it exerts an equal influence in less space, and besides, it
is not subject to a change of density by putrefaction and the like.

4. EXPERIMENTS ILLUSTRATING THE QUAQUAVERSUS PRESSURE OF
INCOMPRESSIBLE FLUIDS.

Before we conclude our inquiries on fluid pressure, it may be both
interesting and instructive to the readers of this work, to describe a
few select experiments, by which the equal distribution of pressure,
among the particles of an incompressible fluid is beautifully and
rigorously demonstrated, and its equal propagation in all directions,
placed beyond the possibility of the smallest doubt.

Allied to the preceding subject, is the following, by which is exhi-
bited a very surprising effect of the equilibrium of incompressible
fluids, but which, for the sake of convenience, we shall suppose to be
water, since that is more easily obtained in small or large quantities,
than any other fluid whatever.

EXPERIMENT 1. Let A BCD represent an upright section of a square
or cylindrical vessel, closed at top with a cover of
which A D is a section ; make a hole in the top at E,
and fix a tube FE therein of any convenient dia-
meter at pleasure, but small in comparison of the
diameter of the vessel. Let the tube be closely
fixed in the cover with pitch, or some other glutin-
ous matter, so as to be rendered air and water-tight
all round the orifice, and suppose its length or
height to be twelve or fifteen inches according to
circumstances ; then, fill the vessel with water by some holes made

156 THE QU AQUA VERSUS PRESSURE OF FLUIDS

in the top and afterwards stopped up, or it may be filled through
the tube alone.

Now, if a load of about seven or eight hundred Ibs., be laid upon
the cover of the vessel, it will be depressed into a concavity repre-
sented by the dotted line AMD, the displaced water ascending in the
tube, in proportion as the cover is bent by the pressure of the super-
incumbent load ; but if we pour water into the tube FE, the cover of
the vessel, together with its incumbent load, will not only be raised
to the original situation, but will even assume a convex form, as
represented by the dotted line AND, rising in the middle as much
above the point E, as it was formerly depressed below it, the quantity
of elevation being measured by the index or ruler IL, which is fixed
in an adjoining support in such a manner, as to remain immoveable,
the point H which is marked on the tube, ascending or descending
with the cover of the vessel.

If the tube be increased in length and more water added, it will be
found that the cover of the vessel, together with its load, will rise
higher and higher until a rupture takes place by overstretching the
fibres of the material ; this however, is a case not admitted in the
experiment, and consequently, we may conclude, that the small
column of water in the tube : —

Exerts the same force in raising the cover of the vessel,
together with the load upon it, as if the tube and the vessel
were of equal diameters, and the incumbent column equal to
AGHD, instead of that contained in the tube.

Now, this is precisely the property of the weighing machine for-
merly exemplified ; for the water in the small tube F E, will raise the
cover of the vessel, (supposing it to be moveable and water-tight),
together with its load, even although it were a thousand times greater:
this is manifest, because the velocity with which the water descends
in the tube, is to the velocity with which it ascends in the vessel, as
the area of a section of the vessel, is to the area of a corresponding
section of the tube ; for instance, if the vessel is 30 inches in diameter,
while the supplying tube is only one ; then, we know by the prin-
ciples of mensuration, that the area of the top of the vessel, is to that
of a section of the tube, in the ratio of 900 to unity ; consequently,
when the water in the tube has descended one inch, the top of the
vessel, and the load upon it, has ascended by a one nine-hundredth
part of an inch ; therefore, if the water in the tube weighs one lb., it
will be in equilibrio with 900 Ibs. in the vessel, or which is the same

ILLUSTRATED BY EXPERIMENTS.

157

thing, one Ib. of water in the tube, will suspend the top of the vessel,
together with the load upon it, supposing them to weigh conjointly
900 Ibs.

EXPERIMENT 2. Let mno represent a vertical section of a spherical
vessel filled with water, or some other incompressible and non-elastic
fluid, and let AB be a common
tumbler glass, held vertically with
its mouth exactly in contact with
the fluid's surface ; then it is mani-
fest, that in this state the glass is
completely rilled with air of the
natural density ; that is, with air of
the same density as atmospheric air
on the surface of the earth.

If the tumbler be still held in a
vertical position, but a little depressed below the surface of the fluid,
as represented by c D, then it is obvious, that a small quantity of the
fluid has entered, and the rest of the glass is filled with air in a state
of slight condensation, corresponding to the pressure of the super-
incumbent column of water represented by T>d. And moreover, if the
glass be still farther depressed, the fluid will ascend higher and higher,
and the air will be compressed into a less and less space.

Again, if the glass be inclined in any degree from the vertical
position, as represented by EF and GH, taking care to have its mouth
wholly immersed in the water, then it is evident, that the greater the
degree of inclination, the greater is the quantity of fluid which enters,
and the greater also is the condensation of the included air; but when
the quantity of fluid which enters the glass is the same, both in the
vertical and the inclined position, the density of the air is also the
same, being compressed by the same force ; consequently, the water
or fluid in which the glass is placed, exerts the same pressure in
whatever direction it is propagated. One sees this experiment verified
daily by empty casks having only one end, thrown into water.

EXPERIMENTS. If the several tubes A, B, c, D A;B
and E, bent at various angles, be .inserted in an
empty vessel, or if they be held in the hand, and
mercury be introduced at their lower extremities,
in such a manner, as to come close to the ori-
fices ; then let water be poured into the vessel,
and it will be seen, that during the time of its
filling, the mercury is pressed gradually from the

158 THE QUAQUAVERSUS PRESSURE OF FLUIDS

lower towards the higher extremities of the tubes, which are supposed
to rise to a height considerably above the surface of the water.

Now, since the lower extremities of the tubes may be conceived to
point in every possible direction, it follows, that the pressure of the
superincumbent fluid is also propagated in every direction. But when
it is required that the lower orifice should point directly downwards,
in order to show the upward pressure of fluids, a straight tube must
be employed, and the mercury which is introduced must be kept in
by the finger, until the height of the water above the lower surface, is
about fourteen times the height of the mercurial column ; for if the
finger be removed before the water has attained that height, the mer-
cury will fall out of the tube, since its weight is fourteen times greater
than the weight of an equal bulk of water. If the finger be continued
upon the orifice, until the height of the water be equal to fourteen
times the height of the mercury, then, on removing the finger, and
pouring in more water, the mercury will be seen to ascend in the
tube, and will continue to rise higher and higher, according to the
quantity of water poured in, thereby showing the upward pressure of
the water.

EXPERIMENT 4. The pressure of fluids at different points of their
depths, may be very simply illustrated
in the following manner : let K be a bag
of leather, or some other tough and
flexible material, filled with mercury,
and attached to the extremity of a
glass tube zi, in such a manner, that
the mercury may just enter the tube
when the bag is held in air.

Then, if the bag be immersed in
water, it is manifest that the pressure
of the fluid will cause it to collapse, and the mercury will ascend in
the tube to a certain height, corresponding to the pressure exerted by
the water, at the depth where the bag is placed. If the bag continue
to be lowered in the water, it will become more and more collapsed in
consequence of the increased pressure, and the mercury will ascend
higher and higher in the tube, and the heights to which it rises, will
indicate the magnitude of pressure at different depths.

EXPERIMENT 5. There is a very simple and amusing experiment,
by which the propagation of pressure through fluids is illustrated,
called the " Cartesian Devil" from M. Descartes, the celebrated
French philosopher, by whom it was discovered ; it is as follows.

ILLUSTRATED BY EXPERIMENTS. 159

Let the little figure in the inverted jar AB represent the " Cartesian
Devil," surmounted by a bag-like crown of great size in proportion to
his body, filled with some very light substance, such as air, and we
shall therefore suppose that air is the body which it contains. The imp
himself must be constructed of glass or enamel, so as to possess the
same specific gravity as water, and therefore to remain suspended in
the fluid. A

At the bottom of the vessel or jar, is
placed a diaphragm or bladder, that can be
pressed upwards by applying the finger to
the extremity of a lever eo, moving round o
as its fulcrum or centre of motion. The pres-
sure applied at a is communicated through the
water to the bag of air at m, which is thus
compressed, and consequently, the specific
gravity of the figure is increased, by which
it sinks to the bottom of the jar.

By removing the pressure on the dia-
phragm at a, the figure will again ascend,
so that it may be made to oscillate, or rise
upwards and sink downwards alternately, and to dance about in the
jar, without any visible cause for its movements.

Other figures, such as fishes made of glass, are sometimes employed
in this experiment, but the principle is nevertheless the same, and
when a common jar is used, the pressure is applied to the upper
surface of it, as at A.

EXPERIMENT 6. The pressure of fluids at very great depths, is
beautifully illustrated by an experiment which has often been made
at sea, where the water is sufficiently deep to admit of the principle
being accurately put to the trial.

The experiment is this : an empty bottle well corked is made to
descend to a great depth, on which the pressure of the fluid becomes
so great as to drive in the cork, and the bottle when brought up is
always filled with water. Several methods have been employed to
prevent the cork from being driven inwards, but although this has
been effected, yet the bottle on being brought to the surface, is con-
stantly filled with the fluid in which it has been sunk.

The following experiments of this sort, are detailed by Mr. Campbell,
the author of " Travels in the South of Africa," published at London
in the year 1815; the experiments were tried on his voyage home-
wards from the Cape of Good Hope. He drove very tightly into an

160 THE QUAQUAVERSUS PRESSURE OF FLUIDS.

empty bottle, a cork of such a large size, that one half of it remained
above the neck : a cord was then tied round the cork and fastened to
the neck of the bottle, and a coating of pitch was put over the whole.

When the bottle was let down to the depth of about fifty fathoms,
he perceived by the additional weight, that it had instantly filled ;
and on drawing it up, the cork was found in the inside of the bottle,
which of course was filled with water.

Another bottle was prepared in a similar manner ; but in order to
secure the cork, and to prevent it from being pressed within the bottle,
a sail needle was passed through it, so as just to rest on the margin of
the glass, and the whole was carefully covered with a coating of pitch.

When the bottle had descended to the depth of about fifty fathoms,
as in the former case, it was again perceived to have been filled with
water; and on bringing it to the surface, the cork and needle were
found in the same position, and no part of the pitch appeared to be
broken, although the bottle was completely filled with water. Here
the water must have insinuated itself through the pores of the pitch
and the cork, and not as the experimentalist supposes, through the
pores of the glass.

The equality of fluid pressure in every direction, is very easily
demonstrated in the following manner.

EXPERIMENT 7. If a piece of very soft wax, as GUI, and the egg
E, be placed in a bladder, or some other flexible vessel filled with
water, and if the bladder be put into a brass box, and a moveable
cover laid upon the bladder so as to be wholly supported by it.

Then, if one hundred, or one hun-
dred and fifty Ibs. be laid upon this
cover, so as to press upon the bladder
and its contained fluid ; this enor-
mous force, although propagated
throughout the fluid, and acting
upon the soft wax and the egg, will
produce no effect, the wax will not
change its form, and the egg will not be broken. And in like manner,
if a living fish should be put into the cylinder of a hydrostatic press, when
under a very high degree of pressure, it will not suffer the least incon-
venience ; from which it is obvious that every particle of the fluid is
equally pressed, and presses equally in all directions.

Numerous other examples might be adduced for proving the same
thing, but since the principle is manifest, it is needless to dwell longer
on the subject.

CHAPTER VII.

IN the former part of this treatise, we have displayed the nature of
pressure as it occurs in the action of non-elastic fluids of uniform
density, and in addition, we have investigated the theory and exem-
plified the application of the Hydrostatic Press, the Hydrostatic
Bellows, and the Hydrostatic Balance or weighing machine ; instru-
ments whose operations depend upon the quaqua-versus principle of
non-elastic and incompressible fluids: — We come therefore in the
next place, to consider pressure as it unfolds itself in the action of
fluids of variable density, or such as have their densities regulated by
certain conditions, dependent upon particular laws, whether excited
by motion, by mixture, or by change of temperature.

In mechanical science density is used as a term of comparison,
expressing the proportion of the number of equal moleculee in the
same bulk of another body; density, therefore, is directly as the
quantity of matter ; and inversely as the magnitude of the body*

We cannot by means of our senses discover the figure and magnitude of the ele*
mentary particles of matter. Mechanical inventions have wonderfully magnified
objects invisible to the unassisted eye; but no microscopical assistance has yet en-
abled us to assume that we have seen an elementary particle of matter. A number of
elementary particles uniting bv the power of cohesion form greater particles, and
these again uniting, by the same power, form still greater; and we may consider the
aggregate of many such formations to become at length an atom of a sensible bulk.
All bodies seem to be composed of these derivative corpuscles, which, formed of
more or fewer of these repeated unions, compose bodies more or less dense. These
derivative corpuscles are sometimes similar, as the coloured rays of a beam of light,

VOL. I. M

162 OF THE PRESSURE OF FLUIDS OF VARIABLE DENSITY

separated by the prism ; mercury, when squeezed through the pores of leather, or
raised in fume and received upon clean glass, which exhibits globules similar and
undistinguishable. In short, every mass of matter is divisible into particles, which
we designate by the Greek term atom, or that which is so exceedingly minute that
it cannot be further cut or divided, and which therefore, as far as sense is concerned,
is the ultimate resisting particle. It must be obvious, that the density or quantity
of atoms which exist in a given space is very different in different substances.
Hence, if it be asked why bodies are called dense ? the answer is, Because they
contain more atoms than others of the same size. There are more atoms in a cubic
inch of lead than in a cubic inch of cork : the former is forty times heavier than the
latter. A cubic foot of rain water weighs 62£ Ibs. ; but an equal volume of mercury,
which is fourteen times heavier than water, weighs (62^x14) =875 Ibs.

Density must depend on three circumstances, to which we should carefully attend
in all our disquisitions : first, the size or weight of the individual atom ; secondly, on
porosity, or the arrangement of the atoms by cohesion, or mechanical and physical
arrangement ; thirdly, the proximity of the atoms determined by the substance of
which they are constituent particles, possessing tenacity and incompressibility.
Thus, heat dilates some bodies and contracts others. A pound of tin and a pound of
copper melted together form bronze ; but this new mass occupies less space by one
fifteenth than the two masses did when separate j proving that the atoms of the one
are partially received into what were empty spaces of the other. In other words,
the affinity of cohesion is one fifteenth greater in the bronze than in the tin and
copper separately. Two pounds of brine are made out of a pound of salt and a
pound of water ; but the mass is of less bulk than the aggregate of the ingredients
apart.

Water, we have seen, resists compression very powerfully, but at the depth
of 1000 fathoms yielding a very small part of its bulk at the surface, shows the
particles not to be in contact, and that the fluid may acquire density in propor-
tion to its depth. Wood swims in water, because the water has more atoms in the
same bulk than the wood, and therefore more weight or central force than the wood ;
consequently, the water falls first and leaves the wood behind; in other words, the
wood floats upon the water — the wood is borne on the surface of the water with a force
exactly proportional to the difference between its weight and that of an equal bulk
of water. The pressures which the fluid exerts in supporting the wood are together
equivalent to a force directed upwards through the centre of gravity of the fluid
displaced, and equal to the weight of a quantity of the fluid so displaced by the
immersed part of the body. But it is not necessary here to dwell further on this
topic, the density of water. We therefore pass on to another character it possesses,
viz. gravity or weight : and it is, in fact, by comparing the weight of a body with
the force which holds it up in the fluid, that the comparative weights or specific
gravities are found, as of metals compared with water, and of admixtures of metals
for the purpose of ascertaining at once the proportion of each in the compound
mass.

Water is the common standard with which all other substances are compared,
whose weight we would fix and record in tables of specific gravities. When
we say, therefore, that gold is of the specific gravity of 19, and copper of 9, and
cork of one seventh, we mean that these substances are just so much heavier or
lighter than their bulk of pure water in its densest form, viz. at the temperature of
40 degrees of Fahrenheit's thermometer. It appears, therefore, that the terms

0fTHe ^\
( UNIVERSITY J

VSi^WX

UPON THE SIDES AND BOTTOM OF A CYLINDRICAL VESSEL. 163

density and specific gravity express the same thing" under different aspects ; the
former being' more accurately restrained to the greater or less vicinity of particles,
the latter to a greater or less weight in a given volume ; hence, as weight depends
upon the closeness of the particles, the density varies as the specific gravity, and
the terms may in most cases be indiscriminately used.

The specific gravities of fluids are usually considered without any regard to the
empty spaces between the particles, though if the particles of fluids are spherical,
the vacuities make at least one fourth of the whole bulk. But it is sufficient that
we know precisely in what sense the specific gravities of fluids are understood.

PROBLEM XXV.

163. A cylindrical vessel whose sides are perpendicular to
the horizon, has a certain quantity of fluid in it; which fluid,
by reason of a sudden change of temperature, has its magnitude
or bulk increased by a certain part of itself : —

It is therefore required to determine what will be the
alteration of pressure on the sides and bottom of the vessel.

Let ABCD, and abed respectively, represent vertical sections of
the cylindrical vessel, of which the sides are perpendicular and the
base parallel to the horizon ; then in
the first instance, let E F be the height
to which the vessel is filled, and ef
the height to which the fluid rises,
by reason of the change that takes
place in the temperature.

Draw the diagonals EC, FD and
ec, fd intersecting respectively in

the points G and g, and through the points G, g, draw the Vertical lines
MN and mn ; then are MG and mg, the respective depths of the centres
of gravity of the cylindric surfaces, in contact with the fluid before
and after the expansion, and MN, mn, are the depths of the centres
of gravity of the bases or bottoms D c and dc.

Through the points G and g, draw the straight lines GT and gs,
parallel to the horizon and to one another; then is GS or rg the
height which the centre of gravity of the cylindric surface is elevated,
by reason of the expansion of the fluid.

Put d = DC or dc, the diameter of the cylindric vessel,

h = MN, the height to which the vessel is originally filled,
h'—mn, the height at which the fluid stands in the vessel after
expansion,

M 2

164 OF THE PRESSURE OF FLUIDS OF VARIABLE DENSITY

Put P zz: the pressure on the bottom DC, by the fluid in its original

state,

p zz: the corresponding pressure on the cylindric surface,
P'zz: the pressure on the bottom dc, by the fluid after expansion,
p' zz: the corresponding pressure on the cylindric surface,
s zz: the specific gravity of the fluid before expansion,
s' z= the specific gravity of the fluid after expansion,
a zz: the area of the base or bottom of the vessel in both cases ;

<j> and $' the cylindric surfaces, and
£zz: the part of its bulk by which the fluid is increased.

Then, since d denotes the diameter of the bottom, the area accord-
ing to the principles of mensuration, becomes

and the pressure exerted by the fluid in its original state, is

(126).

Again, according to the principles of mensuration, the cylindric
surface in contact with the fluid before expansion, is

and consequently, the pressure upon it, is
/;zz:3.1416d/iXjAXszz:1.5708dA2s. (127).

Now, it is manifest, that since the diameter of the vessel is the
same both before and after the expansion of the fluid, the capacity
and the altitude must vary directly as each other; consequently,
because the capacity or bulk is increased by ^th part of itself, it
follows, that the altitude is increased in the same proportion ; there-
fore we have

but when the weight of the fluid remains the same, the density, and
consequently the specific gravity, varies inversely as the magnitude.

The specific gravity of the fluid, after it has expanded by reason
of an increase of temperature, is therefore,

*J>± ') .ft ........-»'

n n-j-l '

hence, the pressure on the bottom of the vessel, after the fluid has
increased by expansion, becomes

P' = .7864d«AV; that is,

'Vv

ON SEMICIRCULAR PLANES IMMERSED VERTICALLY. 165

P' — .7854ef X h - X --7 = .7854d« h s. (128).

\ n ' n -f- 1

The cylindric surface in contact with the fluid after expansion, may
be expressed as follows, viz.

but it has been shown above, that

0 — 3.1416eM;
therefore, by substitution, we obtain

n
and consequently, the pressure becomes

If therefore, the equations (126) and (128) be compared with one
another, it will be found that the pressure is the same, and equal to
the weight of the fluid in both cases; but if the equations (127) and
(129) be compared, the pressure in the one case, is to that in the other,
in the ratio of n : n + 1 ; that is,

p : p' : : n : n -\- I .

PROBLEM XXVI.

164. A semi-circular plane is vertically immersed in a fluid
whose density increases as the depth, and in such a manner,
that the horizontal diameter coincides with the upper surface of
the fluid :—

It is required to determine, on which chord parallel to the
horizon, the pressure is a maximum, or greater than the
pressure on any other chord.

Let A BCD represent a vertical section of a mass of fluid, of which
AB is the surface, and whose density
varies directly as its depth ; and let jmr
aGFHB be the semi-circular plane im-
mersed in it, in such a manner, that
the horizontal diameter a b, coincides
with AB the upper surface.

Let m be the point in the vertical

166 OF THE PRESSURE OF FLUIDS OF VARIABLE DENSITY

radius EF, through which the chord of maximum pressure is supposed
to pass; draw the chord GH, and the radius EG; then, because the
chord GII is parallel to the diameter aby it follows, that GH is bisected
in m by the vertical radius EF; consequently, m is the place of the
centre of gravity of the chord GH, and Em is its perpendicular depth
below AB, the upper surface of the fluid.

Put r IZTEG, the radius of the semi-circular plane,

0 = GF, half the arc subtended by the required chord GH,
and x — Em, the distance of the chord below the surface of the fluid.

Then, because by the conditions of the problem, the density of the
fluid varies directly as its depth ; it follows, that the pressure on the
chord GH varies directly as Gra drawn into Em2; that is,

where p denotes the pressure upon the chord ; but this, by the condi-
tions of the problem, is to be a maximum ; therefore, we have

x* \/ r2 — #2 n: a maximum,
from which, by equating the fluxion with zero, we get

or by transposing and expunging the common factors, we obtain

3z2 = 2r2;
therefore, by division, we have

z'zn-fr2,
and finally, by evolution, it becomes

* = rV?. (130).

The same result, however, may be otherwise determined ; for by
the arithmetic of sines, we have, to radius unity

Gmzzsin.0, and Emzncos.0 ;

but in order to accommodate these quantities to the radius r, it is

Gmmr sin.0, and Em2zziiK2z=:r2cos.2^;

consequently, by multiplication, we obtain

and this, by the conditions of the problem, is to be a maximum ;
hence we get

r*sin.0 cos.2^ = a maximum,
which being thrown into fluxions, becomes
0 — r3(0- cos.3<£ — 20- sin.2^ cos.^) ;

ON SEMICIRCULAR PLANES. THE CHORD OF MAXIMUM PRESSURE. 167

therefore, by transposing and casting out the common terms, it is *

cos.20nz2sin.20.

But according to the principles of Plane Trigonometry, we have
2sin.20 = 2 — 2cos.2<£; consequently, by substitution, the above
expression becomes

cos.20 zz 2 — 2 cos.20 ;

therefore, by transposition and division, we obtain

cos.20=rf,
and by extracting the square root, we get

cos.0rz:-v/-| ;

finally, let both sides of this expression be multiplied by r, for the
purpose of adapting it to the proper radius, and we shall have

# = rcos.0:zrr|/!-, the same as above.

165. The practical rule for reducing this equation, may be expressed
in words at length in the following manner.

RULE. Multiply one third of the radius of the given semi-
circular plane by the square root of 6, or by the constant
number 2.44947, and the product will give the distance of
the point on the vertical axis below the surface of the fluid,
through which the chord of maximum pressure passes.

166. EXAMPLE. The radius of a semi-circular plane, immersed in a
fluid agreeably to the conditions of the problem, is 27 inches ; at
what distance below the surface of the fluid must a horizontal chord
be drawn, so that the pressure which it sustains may be greater than
the pressure sustained by any other chord drawn parallel to it?

By operating according to the rule, we shall obtain

x — 9 ^6*= 9 X2.44947 = 22.04523 inches.

167. The same example admits of a very simple and elegant geo-
metrical construction, which may be effected in the following manner.

Let ACS be the semi-circular plane, of
which the diameter AB is parallel, and the
radius D c perpendicular to the horizon ; draw
the chord BC, and from c as a centre, with
the radius CD, describe the circular arc DKH
cutting BC produced in the point H.

Through the point c draw the tangent c K,
and let fall the perpendicular H i meeting c K
in i; join DI to intersect the arc A EC in E, and through the point E

168 OF THE PRESSURE OF FLUIDS OF VARIABLE DENSITY.

draw the chord EGF parallel to AB the diameter of the semicircle;
then is EGF the chord, on which the pressure is a maximum.

That the line DG corresponds with x in the equation marked (168),
may be thus demonstrated.

By reason of the parallel lines AB and KC, the angles ABII and
KCH are equal to one another ; but the angle ABH is manifestly equal
to half a right angle or 45 degrees, therefore, the angles KCH and
CHI, are each of them equal to half a right angle, and the lines ci
and HI are equal, being respectively the sine and cosine of 45 degrees
to the radius CH or CD.

Now, according to the principles of Plane Trigonometry, the sine
and cosine of 45 degrees to the radius unity, are respectively expressed
by \ V% > hence we have

ci^JvC

and by the property of the rightangled triangle, it is

Dizr-v/ci*-4-DC2=:r-v/|-,
and by similar triangles, we have

r V i : r '• :'r '• D G -=- r V $••

The length of the chord line EF is very easily found, for by reason
of the right angled triangle EDG, of which the two sides DE and DG
are known, it is

E G2 ~ D E? — D G2 ;

but by the elements of geometry, the square of a line is equal to four
times the square of its half, therefore, we have

EF2H=4(DE2 — DG2);

hence, by extracting the square root, we get

E F Zr 2 ^ D E2 D G2 J

now D L2 1=1 r*, and DG*I=: f r9 ; therefore it is

EF=z|rV~3. (131).

Wherefore, if we take the radius of the semi-circle equal to 27
inches, as in the preceding example, the whole length of the chord
will be 18X1. 7321=31. 176inches.

PROBLEM XXVII.

168. If a given conical vessel be filled with fluid, and sup-
ported with its axis inclined to the horizon at a given angle : —

It is required to determine, on vjhat section parallel to the
base the pressure is a maximum.

MAXIMUM PRESSURE ON A SECTION OF A CONICAL VESSEL. 169

Let ABC be a section passing along the axis of the conical vessel,
of which c is the vertex, and AB the diameter of its base.

Conceive AI to be horizontal, and produce
the axis CD to meet the horizontal line AI in
the point i ; then is A ic the angle of inclina-
tion between the axis and the horizontal line
AI.

Let G be the point in the axis through which
the plane of the required section is supposed
to pass, arid through G draw the straight line

EF parallel to AB, and GH perpendicular to AI; then is EF the dia-
meter of the section, and GH the perpendicular depth of its centre of
gravity below A, the highest particle of the fluid.

Put Rizr AD, the radius of the base of the conical vessel,
H — c i), the axis or height,
r ZZTEG, the radius of the section on which the pressure is a

maximum,

a ~ the area of the section,

d ~ GH, the perpendicular depth of its centre of gravity,
p — the pressure perpendicular to its surface,
0 zr: A ic, the angle of inclination between the axis of the cone

and the horizon,
x zr CG, the distance between the section and the vertex of the

cone,
and s z= the specific gravity of the fluid.

Then, because of the rightangled triangle ADI, and from the prin-
ciples of Plane Trigonometry, we have

R : DI : : tan.0 : 1,
and from this, we obtain

consequently, by adding the axis, we get

cizz R cot.<j> -f-H,
and again by subtraction, it is

G i ~ R cot.p 4" H — x.

But the triangle GHI, is by construction right angled at H ; there-
fore, by Plane Trigonometry, we have

GHmc?zz:{Rcot.^ -j~ H — a?}sin.0.

Again, the triangles CD A and CGE are similar to one another;
therefore, by the property of similar triangles, we have

170 OF THE PRESSURE OF FLUIDS OF VARIABLE DENSITY

H2 : R2 : : x* : r2 ;
or from this, by equating and dividing, we get

r*-^.

~ H2 '

consequently, by the principles of mensuration, the area of the section,
on which the pressure is proposed to be a maximum, becomes

3.1416R2*2
-rf -- ;

therefore, the pressure upon its surface is
3.1416Ra*2s

Now, because the whole of the quantities which enter this equa-
tion are constant, excepting x, and the bracketted expression
{Rcot.^-f H — x} which is affected by it; it follows, that the value
of p varies as rc2{RCOt.0-|- H — x}, and consequently, is a maximum,
when the quantity which limits its variation is a maximum ; hence
we have

a?2 {R cot.^ -|- H — x} zz a maximum.

Let the above expression for the maximum be thrown into fluxions,
and we shall obtain

2 (R cot.0 -f- H) xx — 3a?ai •=. 0 ;

therefore, by transposing and expunging the common quantities, we
get

3a?i=2(RCot.0+ H),
and finally, by division, we obtain

a? = |-(RCOt.^ + H). (132).

169. The practical rule for reducing this equation, may be expressed
in words at length in the following manner.

RULE. Multiply the natural cotangent of the angle which
the axis of the cone makes with the horizon, by the radius of
the vessel's base, and to the product add the altitude or axis
of the cone ; then, two thirds of the sum will give the distance
of the section, on which the pressure is a maximum, from the
vertex of the cone.

170. EXAMPLE. A conical vessel whose altitude is 20 inches, and
the radius of its base 8 inches, is filled with fluid and so inclined, that
its axis makes with the horizontal line passing through the extremity
of the diameter of its base, an angle of 48 degrees ; on what section,
parallel to the base is the pressure a maximum ?

ON A GLOBULAR BODY OF CONDENSIBLE AND ELASTIC MATTER. 171

Here we have given, RZH 8 inches, H m 20 inches, and 0m 48
degrees, of which the natural cotangent is 0.9004 very nearly; con-
sequently, by the rule, we have

x = $ {8X0.9004 + 20} =18.1355 inches.

If therefore, 18.1355 inches be set off from the vertex, a straight
line drawn through that point parallel to the base, will be the diameter
of the section on which the pressure is a maximum.

171. In any fluid the particles towards its base support those that
are immediately above ; these again bear the load above them, and
so on to the surface, where the whole mass supports the super-
incumbent atmosphere. There is therefore a pressure among the
successive strata of an homogeneous fluid increasing in exact propor-
tion to the perpendicular depth. Hence a bubble of air or of steam,
set at liberty far below the surface of water, is small at first, and
gradually enlarges as it rises. This phenomenon shows that the
compressive power of the fluid slackens by ascent. Experiment and
calculation most readily demonstrate the compressibility of water :
and the next problem exhibits the striking effects from the increase of
pressure at great depths of the sea.

PROBLEM XXVIII.

172. A globular body of condensible and elastic matter, is
suffered to ascend vertically from the bottom to the surface
of the sea : —

It is required to determine its diameter at the surface, the
depth of the sea, and the diameter at the bottom being given.

Let AB be the surface of the sea, ab its bottom, and ecf, ECF, two
positions of the globular body in its ascent
from the bottom to the surface, and A sea,
B vfb the curves described by the extremities
of the diameter.

Through G and g, the centre of the globe
in the two positions, draw the vertical line
cc, which is manifestly the abscisse to the
curve, the radii ge and GE, as well as ca and
CA, being ordinates.

Produce the abscisse cc to D, and make CD equal to the height of
a column of sea water, which would be in equilibrio with the pressure

172 OF THE PRESSURE OF FLUIDS OF VARIABLE DENSITY

of the atmosphere ; through the point D draw the straight line 11 1
parallel to the surface of the sea, and HI will manifestly be the
asymptote to the curve described by the extremities of the diameter.

Put r=ac, the radius of the globe at the bottom of the water,
d — cc, the depth of the water at the place of immersion,
7i — CD, the height of a column of water equal to the weight of

the atmosphere,
x rr CG, any abscissa,
y nz G E, the corresponding ordinate, or radius of the globe.

Then, because the magnitude of the globular body, is inversely as
the density, (the weight and the quantity of matter remaining the
same,) and the density is directly as the pressure ; it follows, that the
magnitude of the body at different points of its ascent, is inversely as
the pressure at those points, and the pressure is directly as the depth ;
therefore, we have

DC : DG : : GES : ca8;
but according to the foregoing notation, we have

d+h : h + x:: y* : r3 ;

from which, by equating the products of the extremes and means,
we get

hence, by division, we obtain

(*+*)

and by extracting the cube root, it is
,/(d + h)

ry (T+T)' (133).

The equation in its present form, exhibits the nature of the curve
described by the diameter of the body during its ascent; or it ex-
presses generally, the value of the ordinate or radius corresponding
to any depth ; but in order to determine the radius at the surface,
which is the primary demand of the problem, we must suppose the
quantity x to vanish, in which case, the above equation becomes

d+h

h (134).

173. The practical rule supplied by, or derived from this equation,
may be expressed in words at length in the following manner.

ON A GLOBULAR BODY OF CONDENSIBLE AND ELASTIC MATTER. 173

RULE. To the given depth of the sea, add the height of a
column of sea water, which is equal to the weight or pressure
of the atmosphere ; divide the sum by the height of the atmo-
spheric column, and multiply the radius of the body at the
bottom of the sea by the cube root of the quotient, and the
product will give the radius at the surface.

174. EXAMPLE. The radius of a globe of elastic arid condensible
matter, when placed at the depth of 75 fathoms in sea water, is equal
to 4 inches ; what will be the radius on ascending to the surface, the
atmospheric column being equal to 33 feet ?

Here we have given d~ 75 fathoms, or 75 x 6 m 450 feet ; h zz 33
feet; fizz 4 inches; consequently, by the rule, we have

-— izz 9. 785 inches nearly.

175. From this it appears, that if a globe of condensible matter,
whose radius is 9.785 inches, be immersed in the sea to the depth of
450 feet, its radius will be decreased to 4 inches ; this circumstance
may suggest some easy and accurate methods of determining the depth
of the ocean, when it is so great as to preclude the application of
other methods.

176. In order, however, to adapt our equation to the determination
of the depth, we must consider the radii at the surface and at the
bottom, together with the height of the atmospheric column, to be
accurately known at the time of trial ; then, by a very obvious trans-
formation, the depth of descent may be ascertained ; for let R be
substituted instead of y in the foregoing equation, to denote the radius
at the surface, and we shall have

R— rV ~T~~'

in which equation, d is the unknown quantity.
Let both sides of the equation be divided by r, the radius of the
globe at the bottom of the sea, and we shall obtain

(d+h),
~T~~'
and cubing both sides, it becomes

multiply by h, and we obtain

174 OF THE PRESSURE OF UNMIXABLE FLUIDS OF DIFFERENT DENSITIES,

**=- + *.

and finally, by transposition, we have

r3 (135).

177. The practical rule for reducing the above equation, may be
expressed in words at length in the following manner.

RULE. Multiply the difference of the cubes of the radii)
by the height of the atmospheric column, and divide the
product by the cube of the lesser radius for the depth
required.

EXAMPLE. The radius of a globe of condensible matter is 10 inches
before immersion, and it is suffered to descend so far as to have its
radius diminished to 3 inches ; required the depth of descent, the
atmospheric column at the time of the experiment being equivalent
to 33 feet.

Here we have given R — 10 inches, r — 3 inches, and h — 33 feet ;
therefore, by proceeding according to the rule, we have
Rs_ r8__ 10QO _27 — 973 ;

consequently, multiplying by 33 feet, we obtain

973x33 — 32109,
therefore, by division, it is

PROBLEM XXIX.

178. Let a vessel of any form whatever, whose base is hori-
zontal, be filled with fluids of different densities which do not

mix : —

It is required to determine the pressure on the bottom of the
vessel, supposing the fluids to succeed each other in the order
of their densities.

Let ABGH represent a vertical section of the
vessel, containing fluids of different densities or
specific gravities, as indicated by the shading of
the several strata AC, DF and EG; and for the
sake of simplicity of investigation, let the bottom
HG be parallel, and the sides AH, BG perpendicular

ON THE BOTTOM OF ANY VESSEL. 175

to the horizon. Then are AB, DC and EF, the respective surfaces of
the several fluids, as mercury, water, and olive oil, also parallel to
the horizon ; for, as we have elsewhere stated : —

The common surface of two fluids which do not mix, is
parallel to the horizon.

Now, it is manifest, (since the sides A H and B G are perpendicular
to the base HG), that the pressure upon the base HG, is equal to the
pressures or weights of the several fluids contained in the vessel ;
therefore

Put d zz EH, the perpendicular depth of the lowest stratum EG,
d' •=. DE, the perpendicular depth of the middle stratum DF,
d"=: AD, the perpendicular depth of the upper stratum AC,
p zz the pressure of the stratum E G upon the line H G,
jt/zz the pressure of the stratum DF upon the line EF,
p"— the pressure of the stratum A c upon the line D c ; and let s,

s' and s" denote the specific gravities of the respective

fluids.

Then, since the pressure upon any surface, is equal to the area of
that surface, drawn into the perpendicular depth of its centre of
gravity; it follows, that the pressure upon HG, occasioned by the
fluid in EG, is

^ZZTHGX^S,
and in like manner, the pressure upon EF, is

j/ = EFXdV,
and lastly, the pressure upon D c, is

But the total pressure upon H G, is manifestly equal to the sum of
these pressures ; therefore, if P denote the entire pressure on the line
H G, we have
P—

but the lines HG, EF and DC, are equal among themselves, therefore
we get

P = HG (ds + d's' -f- d"8u). (136).

179. In the preceding investigation, we have considered three fluids
of different densities to be contained in the vessel; but the same
mode of procedure will extend to any number whatever, and what we
have done respecting three fluids is sufficient to discover the law of
induction for any other number. It is this : —

176 OF THE PRESSURE OF UNMIXABLE FLUIDS OF DIFFERENT DENSITIES.

The perpendicular pressure upon the horizontal base of a vessel
containing any number of fluids of different densities, which do not
mix in the vessel : —

Is equal to the area of the base, multiplied by the sum of
the products of the specific gravities drawn into the altitudes
of the several fluids .

But the pressure upon the base, will manifestly be the same, if we
suppose the vessel to be filled with a fluid of uniform density, arising
from the composition of the densities of the several fluids according to
their magnitudes ; or if the magnitudes are equal, the uniform density
will be a medium between the several given densities.

180. EXAMPLE. A cylindrical vessel, whose diameter is 6 and alti-
tude 24 inches, is filled with mercury, water and olive oil, in the
following proportions, viz. mercury 7, water 8, and olive oil 9 inches;
what is the pressure on the bottom of the vessel, the specific gravities
being 13598, 1000 and 915 respectively?

Here, by the principles of mensuration, the area of the bottom of
the vessel containing the fluids, is

36 X .7854 — 28.2744 square inches ;
consequently, the pressure produced by the mercury, is

;? = 28.2744x7X13598 = 2691327.0384,
and in like manner, the pressure of the water, is

p1 = 28.2744 X8 X 1000 = 226195.2,
and lastly, the pressure produced by the oil, is

p"~ 28.2744x9x915 = 232839.684 ;

and the sum of these is manifestly the whole pressure ; hence we get
P =. 2691327.0384 + 226195.2 + 232839.684 z=z 3150361 .9224.
If the pressure as here expressed be divided by 1728, the number
of solid inches in a cubic foot, we shall have

181. Again, suppose the dimensions of the vessel to remain as
above, and let it be filled with the same fluids in equal quantities ;
that is, 8 inches of mercury, 8 of water, and 8 of olive oil ; what then
is the pressure upon the bottom ?

Here, by proceeding as above, we have for mercury,
p = 28.2744 X 8 X 1 3598 = 3075802,3296 ;

ON THE CONCAVE SURFACE OF A VESSEL. 177

for water, it is
p' zz 28.2744X 8 X 1000 — 226195.2,

and for olive oil, it is

p" zz 28.2744 X 8 X915 zz 206968. 608 ;

hence by summation, the entire pressure on the bottom, is

P zz 3075802.3296 -f 226195.2 4. 206968.608 zz 3508966.1376,

and lastly, dividing by 1728, we obtain

172o

The pressure which we have found in this last instance, is the very
same as that which would arise, if the vessel were filled with fluid of
a medium density ; for we have

i (13598 4- 1000 + 915) zz 5171 medium density ;

hence, the entire pressure on the bottom, is

P — 28.2744 X 24X5171— 3508966.1376;

which by division, gives

3508966.1376
P— - r^rr - zz 2030.6517 ounces, the same as before.

182. Let the conditions of the problem remain as above, and let it
be required to determine the pressure on the concave surface of the
vessel, and to compare it with that upon the bottom.

Let ABGH, as in the preceding case, represent an upright section
of the vessel, of which the base HG is parallel, and
the sides AH, BG perpendicular to the horizon;
and suppose the fluids of different densities to be
contained in the strata AC, DF and EG.

Bisect the surface and base A B and HG, in the ™
points m and n, and join mn ; then do the centres
of gravity of the several cylindric surfaces occur in
that line. Draw the diagonals AC, DF and EG,
cutting the vertical line mn in the points c, b and a, which mark
the places of the respective centres of gravity.

Put D zz AB or HG, the diameter of the vessel containing the fluids,
\d zz ea, the depth of the centre of gravity of the lower cylindric

surface,

\d' zz db, the depth of the middle cylindric surface,
£<i"zz me, the depth of the upper cylindric surface ; each of these

being referred to the surface of the respective fluid.
VOL. i. N

178 OF THE PRESSURE OF UNMIXABLE FLUIDS OF DIFFERENT DENSITIES

Then, the pressures and the specific gravities being denoted as

before, by the letters p,p',p", and s, s', s", we shall obtain as follows.

According to the principles of mensuration, the circumference of

the vessel is expressed by 3. 1416 D; consequently the several surfaces,

estimated in order upwards, are

3.1416Drf; 3.1416Dd', and 3.1416D<f ;

and the corresponding pressures, are
j?=1.5708Dd2*; p' = 1.5708 D<T s1, and p" =1 1.5708 D d"2 s" ;

and the total pressure is

P' z= (p 4-p' 4-p") = 1 .5708 D (d*s 4- d'*s' 4- dns"). (137).
But the area of the base is expressed by .7854 D8; consequently,
the equation numbered (136) becomes

P = .7854 D9 (d s 4- d's' 4- d"s"),

and the pressures on the base, and on the upright surface of the vessel,
are to one another as D (ds 4- d's' 4- d" s") : 2 (d*s 4- d'*s'+ d"*s".

PROPOSITION III.

183. If two fluids of different densities or specific gravities,
communicate with one another through a bent tube or otherwise,
and remain in a state of equilibrium : —

The perpendicular altitudes of these fluids above their com-
mon surface, will vary inversely as their specific gravities.

Let ABCD be a tube, through the bent arms of which, two fluids of
different specific gravities, communicate
with one another in the common surface
a b, and suppose the horizontal plane E F
to pass through the surface of communi-
cation.

Take G the centre of gravity of the
plane EF, and through o draw the vertical
line GW, meeting Am and DTI respectively

in the points m and n ; then, because the lines Am and DTI are parallel
to the horizon, TKG and no are the perpendicular depths of the plane
EF, below the surfaces of the fluids at A and D.

Now it is manifest, that since the part of the plane ab, which is
contiguous to the common surface of the fluids, is sustained in its
place by the downward pressure of the lighter fluid in A b, and by the

ON THE MERCURIAL COLUMN EQUILIBRATING ONE OF WATER. 179

upward pressure of the heavier fluid in DCB; it follows, that these
pressures are equal to one another. But because the plane EF is
parallel to the horizon, the pressure of the fluid in £B, is equal and
opposite to the pressure in cc, the fluid in de serving no other pur-
pose than for mutually transmitting the opposing pressures ; conse-
quently, the pressure of the lighter fluid in A by is counterpoised by
the pressure of the heavier fluid in DC, the fluid in CCB& serving only
as a medium of communication.

Put dinwG, the perpendicular depth of the plane EF, below the
surface of the lighter fluid at A,

3 in w G, the perpendicular depth of the plane EF, below the
surface of the heavier fluid at D,

p in the pressure on the plane, occasioned by the lighter fluid
in A 5,

7/zz: the pressure on the plane, occasioned by the heavier fluid in
DC ; and let s and s' represent the corresponding spe-
cific gravities of the lighter and the heavier fluids.

Then, since the pressures on the plane, occasioned by the actions
of the two fluids, are respectively as the depths of the centre of
gravity, and the specific gravity of the fluids jointly ; it follows, that

p : p' : : ds : Ss' ;

but according to the conditions of the proposition, these pressures are
equal to one another ; hence we have

ds = Ss'; (138).

and by converting this equation into an analogy, it becomes

d : $ : : sf : s •

hence, the truth of the proposition is rendered manifest.
Let both sides of the equation numbered (138), be divided by s, the
specific gravity of the lighter fluid, and we shall obtain

,7-!fl

• s * (139).

184. Hence, in order to determine the perpendicular depth, or
altitude of a column of the lighter fluid, that will balance or keep in
equilibrio a given column of the heavier ; we must observe the follow-
ing practical rule.

RULE. Multiply the altitude of the heavier fluid by its
specific gravity, and divide the product by the specific gravity
of the lighter fluid, for the altitude sought.

180 OF THE PRESSURE OF UNMIX ABLE FLUIDS OF DIFFERENT DENSITIES

185. EXAMPLE. The height of a column of mercury is 30 inches,
or 2 1 feet, and its specific gravity 13598 ounces per cubic foot; what
is the height of the equilibrating column of water, its specific gravity
being 1000 ounces per cubic foot?

The operation performed according to the rule, is as below.

186. A column of mercury 30 inches or 2| feet in perpendicular
height, is found to equiponderate with the atmosphere in a medium
state of temperature; consequently, a column of water 33.995 or 34
feet nearly in perpendicular height, will produce the same effect ; it
is therefore manifest, that water will ascend in a vacuum tube, to the
height of about 34 feet, by means of the pressure of the atmosphere ;
and on this principle depends the operation of the sucking pump,
to which we shall have occasion to advert in another place.

187. If in the equation numbered (139), the specific gravities are
equal to one another, that is, if szr s', then £/ — o, or the perpendi-
cular altitudes of two fluids whose specific gravities are equal, are also
equal, the fluids being supposed to communicate with one another in
the arms of a bent tube, whatever may be the shape or position of the
arms through which the communication takes place.

188. This explains the reason why the surfaces of small pools
or collections of water near rivers, are always on a level with the
surfaces of the rivers, when there is any subterraneous communication
between them.

189. It is on this principle also, that water may be conveyed from
any one place, to any other place of the same or a less elevation ; for
by means of pipes, a communication can be opened between the places,
arid whatever may be the number of elevations and depressions, or
deviations from the same vertical plane, and whatever may be the
distance from the source to the point of discharge, the water will
continue to flow along the communicating vessels, provided always,
that none of the intervening elevations exceeds the level of the stag-
nant fluid, or the source from which the water flows.

190. When the point to which the water is conveyed, is of the
same altitude as that from which it proceeds, the surface will be in a
state of quiescence ; but if the point of discharge be lower than the
point of supply, the fluid, by endeavouring to rise to the same level,
will cause a stream to flow.

It is by this property of fluids endeavouring to rise to the same

ORIGINATING THE CONSTRUCTION OF A HYDROSTATIC QUADRANT. 181

level, that large towns and cities are supplied with water from a
distance; the city of Edinburgh, in Scotland, is supplied in this way;
but the successful execution of all such complicated and elaborate
undertakings, requires an immense outlay of capital, directed by the
skill and judgment of the most eminent engineers.*

191. The principle which we have demonstrated in the foregoing
proposition, with respect, to two fluids of different specific gravities,
may in like manner, be shown to obtain with any number of fluids
whatever ; a separate demonstration, however, would here be out of
place, we shall therefore content ourselves with the general enuncia-
tion ; but the reader may, for his own satisfaction and improvement,
supply the demonstration ; the enunciation is as follows.

If any number of fluids of different specific gravities, communicate
with one another through the arms of a bent tube, and remain in
equilibrio : —

The sums of the products of their perpendicular heights and
specific gravities, in each branch of the communicating tube,
shall be equal to one another.

Various interesting and important problems might be proposed, on
the principle of fluids of different densities, communicating with one
another in the opposite branches of a bent tube ; but our limits will
only admit of the following, which on account of its elegance, is
worthy of a place in our present inquiry.

PROBLEM XXX.

192. The ratio of the specific gravities of two fluids being
given, if equal quantities of the fluids be poured into a circular
tube of uniform diameter : —

It is required to determine their position when in a state of
equilibrium.

* Baths and aqueducts contributed largely, in the Roman empire, to the wealth
and comfort of the meanest citizen, whether in the solitudes of Asia and Africa, or
the well watered provinces of the west. The splendour, the wealth, even the
existence of those numerous and populous cities which are now no more, was
derived from such artificial supplies of a perennial stream of fresh water. The
boldness of the enterprise and the solidity of the workmanship may be judged of by
those of Spoleto, Metz, Segovia, &c. The aqueduct of Troas, constructed partly at
the expense of the generous Atticus, is but a solitary instance of the spirit of
those PATRICIANS who were not afraid of displaying to the world that they had the
wisdom to conceive and wealth to accomplish the noblest undertakings. — See
Gibbon's Decline and Fall, vol. i. chap. II.

182 OF THE PRESSURE OF UNMIXABLE FLUIDS OF DIFFERENT DENSITIES

Let AB be the interior diameter of the circular tube, and ci, IE the
spaces occupied by the fluids when they have
attained the state of quiescence, GI being the
common surface, or the plane in which the
communication occurs.

Through the common surface IG, the surface
of the heavier fluid at c, and the surface of the
lighter at E, draw the lines GH, CD and EF
respectively parallel to the horizon, and meeting
the diameter AB at right angles in the points H, D and F ; then is DH
the vertical altitude of the heavier fluid, above the common surface IG,
and FH is the vertical altitude of the lighter fluid, as referred to the
same plane.

Put d z=: FH, the perpendicular altitude of the lighter fluid,
S nr DH, the perpendicular altitude of the heavier fluid,
s =i the specific gravity of the lighter fluid,
s' •=: the specific gravity of the heavier fluid,
0 — c G or G E, the portion of the circular tube which is occupied

by each of the fluids,
x zr c B, the number of degrees between the highest and lowest

points of the heavier fluid. •*

Then, because by the preceding proposition, the perpendicular
altitudes of two fluids of different densities, which communicate with
one another through the branches of a bent tube, are inversely as the
densities or specific gravities ; it follows, that

d : 5 : : s' : s,

and from this analogy, by making the product of the mean terms
equal to the product of the extremes, we obtain

c?snr£s' ;

the very same result as equation (138), and if both sides be divided
by 5, we shall obtain

Now, if the specific gravity of the lighter fluid be expressed by
unity, while that of the heavier is denoted by m ; then the above
equation becomes

d = md. (140).

By referring to the diagram, it will readily appear, that the space
occupied by both the fluids, when in a state of equilibrium, is repre-
sented by

ORIGINATING THE CONSTRUCTION OF A HYDROSTATIC QUADRANT. 183

CGEHI20,

and according to our notation, it is shown that

c B zr x ;
consequently, by subtraction, we obtain

B G ~ 0 — x, and B G E rz: 20 — x.

But by the nature of the circle and the principles of Plane Trigo-
nometry, it is manifest, that BD is the versed sine of the arc BC ;
B H the versed sine of the arc B G, and B F the versed sine of the arc B G E ;
therefore, by re-establishing the respective symbols, we shall obtain
F H =r d •=. vers. (20 — x) — vers. (0 — a1),
and by a similar subtraction, we get

D H zz I =: vers.z — vers. (0 — x).

Let both sides of this equation be multiplied by m, the co-efficient
of 8 in equation (140), and we shall have

mS — m {vers. a; — vers. (0 — x)} ;
consequently, by comparison, we obtain
vers. (20 — x} — vers<(0 — x) ~ m{ vers.x — vers. (0 — x)}.
Now, in order to simplify the reduction of this equation, it will be
proper to substitute for the several versed sines of which it is com-
posed, their corresponding values in terms of the radius and cosines ;
for, by such a substitution we obtain

cos. (0 — x) — cos. (20 — x) = m {cos. (0 — x) — cos. a-} ;

but by the arithmetic of sines, we have
cos. (0 — #)i=:cos.0cos.a7 -f- sin. 0 sin. a:, and cos. (20 — ^)zzcos.20

cos. x -)-. sin. 20 sinfjgj

consequently, by substitution, we get

cos. 0 cos. x -f- sin.0sin.a: — cos. 20 cos. x — sin. 20 sin. z:zzwi{cos.0

cos. a? -j- sin.0 sin. a: — cos. a;} ;

and from this, by transposition, we have

mcos.x — cos.20cos.a:-j-sin.20sm.,r-}-(7W — l)(cos.0cos.o;-4-sin.0sin.a:).

Let all the terms of this equation be divided by cos. x, and we shall
obtain

w=icos.20 -f- sin. 20 tan .x -}- (m — l)(cos.0-|- sin.0 tan. a:) ; *
therefore, by separating and transposing the terms, we get

* It is demonstrated by the writers cm Analytical Trigonometry, that the sine
divided by the cosine to the same radius, is equal to the tangent; hence we have

sin.r

tan. a zr .

cos..r

184 OF THE PRESSURE OF UNMTXABLE FLUIDS OF DIFFERENT DENSITIES

— sin. 20 tan. re — (m — 1) sin. 0 tan. x = cos.20 -\-(m — 1) cos.0 — m,
and finally, by division, we obtain

cos.2$ -j- (m — 1) cos.0 — m

~l)sin. (141).

193. We believe that Mr. Barclay's Hydrostatic Quadrant, for
finding- the altitude of the heavenly bodies when the horizon is
obscure, is founded on principles similar to those propounded in this
problem, and expressed in the above equation ; but it would be
improper in this place to attempt a delineation of this instrument ;
it will therefore suffice, to illustrate the reduction of the above
formula, by a numerical example performed according to the direc-
tions contained in the following practical rule.

RULE. From the specific gravity of the heavier fluid, sub-
tract unity ; multiply the remainder by the natural cosine of
the circular space occupied by each fluid; to the product add
the natural cosine of the circular space occupied by both
fluids ; then, from the sum subtract the greater specific gravity,
and the remainder will be the dividend.

Again. From the specific gravity of the heavier fluid, sub-
tract unity ; multiply the remainder by the natural sine of
the circular space occupied by each fluid ; then, to the pro-
duct add the natural sine of the circular space occupied by
both fluids, and the sum will be the divisor.

Lastly. Divide the dividend by the divisor, and the quotient
will give the natural tangent of a circular arc, which being
found in the tables, enables us to assign the position of the
fluids when in a state of equilibrium.

194. EXAMPLE. Suppose that 100 degrees of the inner circum-
ference of a circular tube, exhibits equal quantities of mercury and
water, whose specific gravities are to one another, very nearly, as
14 to 1 ; it is required to assign the position of the fluids, with respect
to the vertical diameter of the tube, when they are in a state of equili-
brium with each other ; that is, when they excite equal pressures on
the plane passing through their common surface ?

Here we have given mm 14 :0z=:500, its natural sine and cosine
.76604 and .64279 respectively; 20zr 100°, its natural sine .98481,
and its cosine — .17365; therefore, by the rule, we have

For the dividend,
— cos.2(H-(w-l)cos.0-?w—— .17365+13 X. 64279— 14r=— 5.81738,

ORIGINATING THE CONSTRUCTION OF A HYDROSTATIC QUADRANT 185

For the divisor,

— sin.20— (w— l)sin.20= — .98481— 13 X.76604z= — 10.94333;
consequently, by division, we obtain

= -53159 = nat.tan. 27° 59' 41".

195. Having discovered the value of x by the preceding operation,
the actual position of the fluids, with respect to the vertical diameter
of the tube, may from thence be very easily exhibited.

Let AC BE represent a circular tube of glass, or some other trans-
parent matter, partly filled with mercury and
water, in such quantities, that when the tube
is retained in a vertical plane, and the fluids
in equilibrio, a space equivalent to fifty de^
grees of the inner surface comes in contact
with each ; it is therefore required to assign
the actual~position of the fluids.

Draw the vertical diameter AB, and from the point B where it meets
the inner circumference of the tube, set offsc from a scale of chords,
equal to 27° 59' 41" ; then, take 50° in the compasses, and setting one
foot on c extend the other to G, thereby marking off the space occu-
pied by the mercury, including the lowest portion of the tube ; then,
with the same extent of the compasses, set off G E the space occupied
by the water, and the position of the fluids is from thence determined.

Through the points c, G and E draw the straight lines CD, GII and
EF respectively parallel to the horizon, and meeting AB the vertical
diameter perpendicularly in the points D, H and F ; then are DH and
FH the perpendicular altitudes of the mercury and the water, as
referred to the plane passing through their common surface at G ; and
BD, BF are the respective altitudes, as referred to the vertical dia-
meter AB.

Now, according to the question, the specific gravity of the mercury,
is fourteen times greater than that of the water ; and by the third
proposition preceding, the perpendicular altitudes are inversely as the
specific gravities; consequently, FH must be fourteen times greater
than DH; when the positions of the fluids are properly determined ;
let us therefore inquire if this be the case.

It has been found above, that BC is equal to 27° 59' 41", and by
construction CG and GE are each equal to 50 degrees; consequently,
BG — 50° — 27° 59' 41"— 22° 0' 19", and B E = 50° + 22° 0' 19" —
72° 0' 19" ; hence we have

186 OF THE PRESSURE OF UNMIXABLE FLUIDS OF DIFFERENT DENSITIES

BEz=72°0' 19" - - - nat. vers. — .69106,
BC — 27° 59' 41" - - - nat. vers. — .11700,
B G = 22° 0' 19" - - - nat. vers. = .07285 ;

consequently, by subtraction, we obtain
FH = BF — BH ; that is, FIX = .69106 — .07285 — .61821 ,

and after the same manner, we get

DH = BD — BH; that is, DH — .11700 — .07285 = .04415 ;

therefore, according to the proposition, we have

.04415 : .61821 : : 1 : 14 very nearly.

196. The above result has been obtained on the supposition, that
the fluids enclosed in the tube are mercury and water ; mercury, on
account of ,its great density and high degree of purity, is very fre-
quently enclosed in tubes and applied to permanent purposes ; but
water, by reason of its liability to become putrid, is not so well
adapted for the occasion, and consequently, is seldom or never em-
ployed in the construction of philosophical instruments.

197. There are however, several other fluids that do not partake
of the putrescent nature of water, and whose specific gravities may be
either greater or less according to the required circumstances ; some
of these, on account of the colouring matter which they contain, are
very convenient, and from the length of time that they retain their
spirit and purity, are generally employed in preference to others, which
do not possess these very requisite and important characteristics.

198. Now, the result of our investigation, as we have already
observed, is only applicable in the case of mercury and water ; at
least, equation (141) implies, that the specific gravity of one of the
fluids is expressed by unity, and this, according to our present
standard, can only obtain when water becomes the subject of
reference.

199. It is therefore necessary, in order that our formula may apply
to fluids without distinction, to bring it into a general form, and this is
very easily done; for we have shown in the preceding investigation, that

but it has also been shown, that
d — vers. (20 — x) — vers. (0 — a:) ;

consequently, by comparison, we obtain

$s'

— — vers. (20 — a:) — vers.(0 — a-) ;

ORIGINATING THE CONSTRUCTION OF A HYDROSTATIC QUADRANT. 187

and moreover, it has been further proved, 'that
5 =r vers.a; — vers.(<£ — a;) ;

therefore, multiplying both sides by—, we shall get
— m— {vers.a? — vers.(<£ — x}} ;

S 5

let these two values of — be compared with one another, and we
shall obtain

vers.(20 — a?) — vers.(0 — #) m — {vers.z — vers.(</> — x)} ;

or multiplying by s, we have

s vers.(2^> — ar) — s vers.(^> — x} zz: s' vers.a? — s' vers.(0 — x).
Now, by substituting for the several versed sines, their values in
terms of the cosines and radius, we shall obtain

s{cos.(0 — x} — cos. (2^ — a:)} — s'{cos.(0 — x) — cos. a- }

from which, according to the arithmetic of sines, we get
s {cos.0 cos.a: -f- sin.0 sin.a; — cos.2^ cos. x — sin.20 sin. a;} zz: s' {cos.<£

cos. a; -j- sin.0 sin. a? — cos. a;}.

Let all the terms of this equation be divided by cos. a;, and it
becomes transformed into
s cos.0 -{- s sin.0 tan. a; — s cos.20 — s sin.2^> tan. a: zz: sr cos.0 -f~ s/ X

sin. ^ tan. x — 5';

therefore, by bringing to one side, all the terms that involve tan. a:,
we shall have *

— sin.^(s' — s)tan.a; — 5 sin. 20 tan. x n: scos.2^ -|- (*' — s)cos.^> — s' ;
hence, by division, we shall obtain

s cos.2<6 4- (s' — s) cos.0 — 5'

tan x "~~ — -

s sin. 20 + (s> — s) sin.0 (142).

If the equation which we have just obtained, be compared with that
numbered (141), it will readily appear, that the one might have been
deduced immediately from the other, by simply substituting s' for m,
and s for unity, in the several terms of the numerator and denominator;
but in order to render the formation of the formula more intelligible,
we have thought proper to trace the steps throughout.

200. The practical rule for reducing the above equation, will require
a different mode of expression from that which we have given in the
rule to equation (141), but it will not be more operose ; the rule is
as follows.

188 OF THE PRESSURE OF UNMIXABLE FLUIDS OF DIFFERENT DENSITIES

RULE. Multiply the difference of the yiven specific gravities,
by the natural cosine of the circular space in contact with one
of the fluids ; to the product, add the natural cosine of the
whole circular space drawn into the less specific gravity , and
from the sum subtract the greater specific gravity for a
dividend.

Again. Multiply the difference between the specific gravi-
ties, by the natural sine of the circular space in contact^with
one of the fluids, and to the product, add the natural sine of
the whole circular space drawn into the less specific gravity,
and the sum ivill be the divisor.

Lastly. Divide the dividend by the divisor, and the quotient
will give the natural tangent of a circular arc, which being
found in the tables, enables us to assign the actual position of
the fluids when in a state of equilibrium.

201. EXAMPLE. On the inner surface of a circular tube containing
mercury and rectified alcohol, it is observed, that when the tube is
held in a vertical plane, and the fluids in a state of equilibrium, a
space of 75 degrees of the circumference, is occupied by, or in contact
with each fluid ; it is required to determine the position of the fluids
at the instant of observation, their specific gravities being 14000 and
829 respectively ?

In this example there are given /= 14000; s — 829; ^ — 7£°7
its natural sine and cosine equal to .96593 and .25882; 2^>rr: 150°,
its natural sine being .50000, and its cosine — .86603 ; consequently,
by proceeding according to the directions contained in the foregoing
rule, we shall obtain

For the dividend — scos.2^ -f (s' — s)cos.<£ — s'= — 829 X .86603 -|-

(14000 — 829) x .25882 — 14000 = — 11 309 .02065 ;

For the divisor — s sin .2<£ — (V — s) sm.^> — — 829 X .50000 — (14000

— 829) X .96593 = — 1 3136.76403 ;

consequently, by division, we obtain

ta"-* = = -86086 = °at- tan- 40- 43' 25'.

202. The positions of the fluids in this example, are manifestly
very different from what they are in the preceding, the point F in the
vertical diameter falling on the other side of the centre; but in this
case, we shall leave the construction for the reader's amusement, and
proceed to inquire what changes the general formula will undergo, in

ORIGINATING THE CONSTRUCTION OF A HYDROSTATIC QUADRANT. 189

consequence of certain assumed spaces of the inner surface, being in
contact with each of the contained fluids.

If 0:zzhalf a right angle, that is, if each fluid cover a space of 45
degrees; then 20 zz 90°, and consequently, sin.20 zr 1 and cos.20:zrO,
while sin. 0:=! ^/ 2, and cos.0:=r | ^2 ; therefore, by substitution,
equation (142) becomes

l(s'

-J(s'

Now, let the fluids be mercury and rectified alcohol, as in the pre-
ceding example, then we shall have

829 V2-

i (14000—829)^2 -f 829

which answers to the natural tangent of 28° 55' nearly.
Again, if 0zza right angle, that is, if each fluid cover a space of
90 degrees on the inner surface of the tube; then, 20zzl80°, of
which the sine and cosine are respectively 0 and — 1, while the sine
and cosine of 0, are respectively 1 and 0 ; consequently, by substitu-
tion, equation (142) becomes

s' -4- s

tan.zzz-7-- — . /IAA\

s1 — s (!44).

203. This is a very neat and obvious expression, and the practical
rule derived from it, may be enunciated in the following manner.

RULE. Divide the sum of the specific gravities of the two
fluids by their difference, and the quotient will give the
natural tangent of an arc, which being estimated from the
lowest point of the tube, will indicate the highest point of the
heavier fluid.

If therefore, the contained fluids be mercury and rectified alcohol,
as in the preceding cases, we shall have
14000 -|- 829

ta"-*=14000-829=U2587-
which answers to the natural tangent of 48° 23' 19''.
We might assume other particular values of the spaces in contact
with the fluids, and thereby deduce corresponding forms of the equa-
tion ; but what we have already done on this subject is quite
sufficient.

CHAPTER VIII.

OF THE PRESSURE OF NON-ELASTIC FLUIDS UPON DYKES, EM-
BANKMENTS, OR OTHER OBSTACLES WHICH CONFINE THEM,
WHETHER THE OPPOSING MASS BE SLOPING, PERPENDICULAR
OR CURVED, AND THE STRUCTURE ITSELF BE MASONRY
OR OF LOOSE MATERIALS, HAVING THE SIDES ONLY FACED
WITH STONE.

1. OF FLUID PRESSURE AGAINST MASONIC STRUCTURES.

204. BEFORE we proceed to develope the theory of Floatation, and
to explain the method of weighing solid bodies by immersing them in,
or otherwise comparing them with liquids ; it is presumed that it will
not be considered out of place, to take a brief survey of the circum-
stances attending the pressure of non-elastic fluids, when exerted
against dykes or other obstacles, that may be opposed to the efforts
which they make to spread themselves.

This is an interesting and important subject in the doctrine of
Hydraulic Architecture, and since the principles upon which it is
founded, depend in a great measure on Hydrostatic pressure, it cannot
properly be omitted in unfolding the elementary departments of the
Mechanics of Fluids, which come so directly before our view in what
is called level cutting in the practice of canal making. Every one
knows that in cutting a canal, no further excavation is required than
that which will hold the water at a given depth and breadth ; when
a bank is made on both sides with the earth excavated, the 'level sur-
face of the canal may be elevated above the natural surface of the
adjacent land, and in this case great part of the cost of excavation
will be saved. But when the canal is to be carried along wholly
within embankments, too much attention cannot be paid to the prin-
ciples of fluid pressure, if we would avoid unnecessary expense, and
at the same time complete the work with systematic regard to its
permanent durability ; this therefore is the object of the present

OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS. 191

section, intended as a preliminary article to our Inland Navigation,
which will consequently form a part of Hydraulic Architecture.

205. When an incompressible and non-elastic fluid presses against
a dyke, mound of earth, or any other obstacle that it endeavours to
displace, there are two ways in which the obstacle thus opposed may
yield to the effort of the fluid.

1 . It may yield by turning upon the remote extremity of
its base.

2. It may yield by sliding along the horizontal plane on
which it stands.

In either case, the effort to overcome the obstacle, arises from the
force which the fluid exerts in a horizontal direction ; and the stability
of the obstacle, or the resistance which it opposes to being overcome
or displaced, arises from its own weight, combined with the vertical
pressure of the fluid upon its sloping surface.

206. When the vertical pressure of the fluid is considered, the
investigation, as well as the resulting formulae, a're necessarily tedious
and prolix ; but when the effect of the vertical pressure is omitted, the
subject becomes more easy, and the computed dimensions are better
adapted for an effectual resistance ; but in order to render the inves-
tigation general, it becomes necessary to include its effects.

Now, it is manifest from the nature of the inquiry, that when an
equilibrium obtains between the opposing forces, the momentum of
the horizontal pressure must be equal to the momentum of the vertical
pressure, together with the weight of the body on which the pressure
is exerted ; and for the purpose of showing when this condition takes
place, let A BCD represent a vertical section
of the dyke, whose resistance is opposed to
the pressure of the stagnant fluid, of which
the surface is ME and the perpendicular
depth EF.

Let AB and DC be parallel to the hori-
zon, and consequently parallel to one another; and from the points
E, A and B, demit the straight lines EF, AK, and BL, respectively
perpendicular to DC the base of the section.

Take EG any small portion of the sloping side AD, and through
the point G, draw the lines GH and 01, respectively parallel and
perpendicular to the horizon, constituting the similar triangles EHG,
EFD and GID.

The figure being thus prepared, it only remains to establish the
proper symbols of reference, before proceeding with the investigation.

192 OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKM EN TS.

Put b zz: DC, the breadth of the section's base, or the thickness of the
dyke at the foundation,

D zz AK or BL, the perpendicular altitude or height of the
section,

d =z EF, the perpendicular depth of the fluid whose surface is
at EM,

I zz: DF, the distance between the near extremity of the base at
D, and the perpendicular E F,

c zz: DK, the measure of the slope AD, or the distance between
the near extremity of the base at D, and the perpendi-
cular from the extremity of the opposite side at A,

c zz: CL, the distance between the remote extremity of the base
at c, and the perpendicular from the extremity of the
opposite side at B, or the measure of the slope BC,

a zz: ABCD, the area of a vertical section of the obstacle to be
displaced,

p — the horizontal pressure of the fluid on the increment of EG,

/ zz: the force with which the horizontal pressure operates to
overcome the resistance of the dyke,

m zz: the momentum of that force,

p' = the vertical pressure of the fluid on the increment of EG,

/' zr the force with which the vertical pressure resists the dis-
placement of the obstacle,

tfi'nr the momentum of that force,

w zz: the symbol which denotes the weight of the dyke or obstacle
of resistance,

F =z the force with which it opposes the horizontal pressure of
the fluid,

M zz the momentum of that force,

s zz- the specific gravity of the fluid,

s' zz: the specific gravity of the dyke, or opposing body,

z zz: EG, any small portion of the sloping side AD on which the
fluid presses,

z zz: the increment or fluxion of that portion,

y zz: EH, the perpendicular depth of the point G,

y z= the increment or fluxion of y,

x zz: G H, the ordinate or horizontal distance,

and x z= the increment or fluxion of the horizontal ordinate or dis-
tance G H.

Then, since the pressure upon any line or surface, is equal to, or
expressed by the magnitude of that line or surface, multiplied by the

OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS. 193

perpendicular depth of its centre of gravity, and again by the specific
gravity of the fluid ; it follows, that the horizontal pressure on the
increment of EG, is

but by the principles of mechanics, the aggregate or accumulated force,
with which the horizontal pressure operates to overturn or remove the
dyke, is

and by taking the fluent of this, it is

/= %sy*.

But the perpendicular distance from E, at which this force must be
applied, is manifestly equal to %y ; for the centre of gravity of the
triangle EHG, occurs in the horizontal line passing through that point;
therefore, the length of the lever on which the force operates to over-
turn the dyke is

consequently, for the momentum of the force, we have

and when y becomes equal to d, the whole height of the fluid, it is

m — ±sd\ (145).

Again, the vertical pressure exerted by the fluid on the increment of
EG, is obviously equal to the weight of the incumbent column ; that is

and this pressure expresses the force, with which the fluid operates
vertically to retain the obstacle in its position, or to prevent it from
rising to turn about the point c ; consequently,

p'=f'=syx.

Now, the length of the lever on which this force acts, is evidently
equal to ic, the distance between the fulcrum c, and the point i, where
the perpendicular passing through G cuts the base DC ; but ic accord-
ing to the figure, is equal to DC — DF -f i F ; that is

ic = b — £ + *;
consequently, the momentum of the force /', is

m' = syx(b — 3 -far),
or taken collectively, the momentum on EG, is

VOL. I.

194 OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS.

but by reason of the similar triangles EHG and EFD, we have the
following proportion, viz.

d : 3 : : y : x,

from which we obtain

and because the fluxions of equal quantities are equal, it is

Let these values of x and x, be substituted instead of them in the
preceding value of m', and we shall obtain

consequently, by taking the fluent, it becomes

there being no correction, since the whole expression becomes equal to
nothing when y is equal to nothing.

When y becomes equal to d the whole perpendicular height of the
fluid, then the foregoing value of m' becomes

wf = «8d(J6— >8). (146).

The foregoing equations (145) and (146), exhibit the horizontal
and vertical momenta of the pressure exerted by the fluid on the
sloping side of the obstacle ; and it is manifest from the nature of
their action, that they operate in opposition to one another ; the
horizontal pressure, endeavouring to turn the body round the point c
as a fulcrum or centre of motion, and the vertical pressure tending to
turn it the contrary way round the same point, or otherwise to render
it more stable and firm on its foundation.

208. But the stability of the dyke is farther augmented by means
of its own weight, which being conceived to be collected into its
centre of gravity, opposes the horizontal pressure of the fluid with a
force, which is equivalent to its own weight drawn into a lever, whose
length is equal to the perpendicular distance between the centre of
motion, and a vertical line passing through the centre of gravity of
the section ABCD.

Now, it is manifest from the principles of mensuration, that if the
transverse section of the dyke be uniform throughout, the weight is
proportional to the area of the section, multiplied into the specific

OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS. 195

gravity of the material of which it is composed, and again into ijs
length ; but the length of the dyke is the same as the length of the
fluid which it supports ; consequently, the weight is very properly
represented by the area of the section and the specific gravity of the
material ; thus we have

w — as'. (147).

But according to the writers on mensuration, the area of the trape-
zoid A BCD, is equal to the sum of the parallel sides AB and DC, drawn
into half the perpendicular distance AK or BL ; hence we have

a=(AB -f- DC)XjAK,

but by the foregoing notation, it is

A B — b — (c -{- e) ;
consequently, by addition, we have

A B -f DC = 26 — (c + e) ;
therefore, the area of the section is

a = ID (26 — c — e);

let this value of a be substituted instead of it in the equation marked
(147), and it becomes

w •=. |DS' (26 — c — e) ;

but the weight of the dyke is equivalent to the force whose momentum,
combined with that of the vertical pressure of the fluid, counterpoises
the momentum of the horizontal pressure, which force we have repre-
sented by F ; hence we have

F— IDS' (26 — c — e\
and the momentum of this force, is

IF — M zz JD Is' (26 — c — e),

where / denotes the lever whose length is equal to the distance
between the fulcrum, or centre of motion at c, and the vertical line
passing through the centre of gravity of the section A BCD ; conse-
quently, in the case of an equilibrium, we have

mi^.m' -j- M,
and this, by restoring the analytical values, becomes

Lsd* = s3d(tib — £3) -{- ID/*' (26 — c — e). (148).

209. This is the general equation which includes all the cases of
rectilinear sloping embankments, but it has not yet obtained its
ultimate form ; for the value of I has still to be expressed in terms
of the sectional dimensions, and in order to this, a separate investi-
gation becomes necessary.

196 OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS.

Thus, let A B c D be a vertical section of
the dyke as before, and bisect the parallel /Hn\

sides AB and DC in the points m and w,
and join mn\ then, the straight line mn
will pass through the centre of gravity of
the figure ABCD.

Take g a point such, that mg is to ng, as 2n c -j- A B is to BC -}- 2AD,
and g will be the centre of gravity sought ; through the points m and
g, draw the straight lines mr and gs, respectively perpendicular to
DC the base of the section, then is sc the length of the lever by which
the weight of the dyke or embankment opposes the horizontal pressure
of the fluid.

From the points A and B, draw the straight lines AK and BL, re-
spectively perpendicular to D c ; then it is manifest from the principles
of geometry, that

rw= £(CL — DK),

and this, by restoring the symbols for CL and DK, becomes
rn~^(e — c).

But mrzrD; consequently, by the property of the right angled
triangle, we have

m n* m m i* -f- n r3 ;
or by restoring the analytical values, it is

raw2=iD94- \(e — cf\
therefore, by extracting the square root, we have

m n — J -Y/ 4o2 -|- (e -=— c)".

By the property of the centre of gravity, and according to the
foregoing construction, the point g is determined in the following
manner.

2DC-{- AB = 36 — c — e
— 36 — 2c — 2e

66 — 3c — 3e : i^/4D24-(e — c)8: : 36 — 2c— 2e : gn,
from which, by reducing the analogy, we get

(36 — 2c — 2e) v/4o2-f- (e — c)s
9n-~ 6(26-c-e)

and by the property of similar triangles, it is

— 2c —

OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS. 197

wherefore, by reducing the analogy, we obtain
_(e — c)(3b — 2c — 2e)
6(26 — c — e) '

But by referring to the diagram, it will readily appear that scur
sn -\- nc ; therefore, by addition, we obtain

.36(26 — c — e) + (e — c)(36 — 2c — 2e)

6(26 — c — e)

Let this value of I be substituted instead of it in the equation
marked (148), and we shall obtain

and this being reduced to its simplest general form, becomes
sd8~das(36 — S} + 3bvJ(b — c)-hDs'(c8 — e2). (149).

210. The general equation in the form which it has now assumed,
is very prolix and complicated ; but its complication and prolixity, as
we have before observed, are much increased by the introduction of the
vertical pressure; if that element be omitted, the equation becomes

sd3 = 3bDs'(b — c) + DS'(C* — e8). (150).

An expression sufficiently simple for every practical purpose ; but
it must be observed, that if e2 be greater than c2, the term in which it
occurs will be subtractive.

We shall not attempt to express these equations in words, or to
give practical rules for their reduction ; the combinations are too
complex, to admit of this being done in a neat and intelligible
manner; it is necessary, however, to illustrate the subject by proper
numerical examples, for which purpose, the following are proposed in
this place.

211. EXAMPLE 1. The water in a reservoir is 24 feet deep, and the
wall which supports it is 30 feet in perpendicular height, the slope of
the side next the water being one foot, and that of the opposite side
one foot and a half; it is required to determine the transverse section
of the wall or dyke, supposing it to be built of materials whose mean
specific gravity is 2 J, that of water being unity ?

By contemplating the conditions of the question as here proposed,
it will readily be observed, that the breadth of the section at the
base, is the first thing to be determined from the equation ; for since
the quantity of the slopes, as well as the perpendicular height are
given, the breadth of the dyke at top can easily be found, when the
breadth at the foundation is known.

198 OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS.

In the first place then, let us take into consideration the effect
produced by means of the vertical pressure of the fluid ; this will
refer us to equation (149), but previously to the substitution of the
several numerical quantities, it becomes necessary to assign the nu-
merical value of S, which is not expressed in the question, but is
determinable from the perpendicular altitudes of the wall and the
fluid, together with the slope of that side on which the fluid presses :
thus,
£•'-' 30 : 24 : : 1 : 8 = ± of a foot.

Let therefore, the several given numbers replace their representa-
tives in equation (149), and we shall have

248=24X£X2K3&— i)4-3X30x2J(6 — 1)6 — 30X2|(1F— I2),

in which expression, b is the unknown quantity.
If the several terms be expanded, collected, and arranged, accord-
ing to the dimensions of the unknown quantity, we shall have

2256s — 816=13956.15;

complete the square, and we get

2025006* — 729006 X 8 12 = 12567096,

and extracting the square root, it is

4506— 81= |/1 2567096 = 3545 nearly;

therefore, by transposition and division, we get

b — 8.05 feet.

Consequently, if from the breadth of the foundation as above
determined, we subtract the sum of the slopes, the remainder will be
the breadth of the dyke at the top ; hence, the section can be deli-
neated.

212. The above is the method of performing the operation, when
the effect produced by the vertical pressure of the fluid is taken into
consideration ; but when that effect is omitted, the process is consi-
derably shortened ; for in the first place, there is no occasion to cal-
culate the value of S, that term not occurring in equation (150), and
in the next place, there are fewer quantities to be substituted for ;
this greatly abbreviates the labour of reduction ; but the equation is
still of the same degree, and consequently, it must be resolved in the
same manner.

Let the several given quantities remain as in the preceding case,
and let them be respectively substituted in the equation (150), and
we shall obtain

2256* — 2256= 13917.75;

OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS. 199

if all the terms of this equation be divided by 225, the co-efficient of
b*, we shall get

b* — 6 = 61.856,
and this, by completing the square, becomes

therefore, by extracting the square root and transposing, we have
b rz 8 . 1 4 feet nearly.

COROL. It therefore appears, that under the same circumstances,
the computed breadth of the foundation differs very little, when the
vertical pressure of the fluid is considered, from what it is when the
pressure is omitted ; and what is very remarkable, the difference,
whatever it may amount to, leans to the side of safety and conve-
nience in the case of the omission ; it will therefore be sufficient in
all cases of practice, to employ equation (150), but under certain
circumstances of the data, it will admit of particular modifications.

213. When the slopes c and e are equal ; that is, when the vertical
transverse section of the dyke or embankment,

is in the form of the frustum of an isosceles
triangle, as represented by ABC D in the annexed
diagram ; then, the general equation (149), be-
comes transformed into

sd*=idSs(3b — S) + 36D/(6 — c). (151).

If the perpendicular height of the dyke, and

the depth of the fluid, are equal to one another ; that is, if the water
is on a level with the top of the wall ; then, rfzz D and 3 = c, and the
above equation becomes

sd* = cs(3b — c) + 3bs'(b — c). (152).

Again, if we neglect the effect of vertical pressure, and express the
specific gravity of water by unity, we get

3s'(tf — cb) = d\ (153).

And finally, if both sides of the equation be divided by the quantity
3/, we shall obtain

b*-cb = 37- (154).

The method of applying this equation is manifest, for we have only
to substitute the given numerical values of c, d and /, and the value
of b will become known by reducing the equation.

214. EXAMPLE 2. The dyke or embankment which supports the
water in a reservoir, is 20 feet in perpendicular height, and it slopes
equally on both sides to the distance of 2 feet ; what is the breadth of

200 OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS.

the base, supposing the water to be on a level with the top of the
wall, the specific gravity of the materials of which it is built being If,
that of water being unity ?

Let these numbers be substituted for the respective symbols in the
above equation, and we get

b* — 26 — 61.619,
complete the square, and it becomes

&* — 26+ 1—62.619,
from which, by evolution and transposition, we get

b — 8.9 13 feet nearly.

Here then, the transverse section of the dyke is 8.913 feet across at
the bottom, and consequently it is 8.913 — 4 — 4.9 13 feet broad at
the top ; hence the delineation is very easily effected.

215. If the slope c should vanish; that is, if
the side of the dyke on which the fluid presses A.JB

be vertical, as represented by A BCD in the an-
nexed diagram ; then S vanishes also, and the
equation marked (149) becomes

(155).

where it is manifest there is no vertical pressure
on the dyke, the whole effect of the fluid being exerted in the hori-
zontal direction, tending to turn the wall about the remote extremity
of its base.

When the perpendicular altitude of the wall or dyke, and the depth
of the water are equal ; then dizzo, and admitting that the value of
5, or the specific gravity of water is represented by unity, we obtain

and this, by transposition and division, becomes

*-=^'

and lastly, by extracting the square root, we get

'— y —

(156).

Let the slope of the dyke be two feet, its perpendicular altitude, or
the depth of the fluid 20 feet, and the specific gravity of the material
If, as in the preceding example; then, by substitution, we obtain

6=i/400+" = 8.804 feet.
V 3X1.75

OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS. 201

COROL. The breadth at the base, as determined by this and the
preceding equation, exhibits but a small difference, being in excess in
the former case, by a quantity equal to 0.109 of a foot; but the
breadth at the top in the latter case, exceeds that in the former, by a
quantity equal to 1.891 feet; and the difference in the area of the
section, is 17.82 feet: it is consequently more expensive to erect a
dyke or embankment, with the side next the fluid perpendicular, than
it is to erect one of equal stability with both sides inclined or sloping
outwards.

216. If the slope e should vanish; that is, if the side of the dyke,
opposite to that on which the fluid presses, be
perpendicular to the horizon, as represented by
A BCD in the annexed diagram, then, the equa-
tion (149) becomes
sd*=d$s(3b— £)-}-3Ds'(6?— c£)4-DcY. (157).

But when the effect of the vertical pressure of
the fluid is omitted, we obtain

sds=i SDS' (b* — c&) -f- DC9 s', (158).

and by supposing the altitude of the dyke, and the depth of the fluid
to be equal (the specific gravity of the fluid being expressed by unity) ;
then we have d — D, and the foregoing equation becomes

d* = 3s'(b* — c&)4-cV;
consequently, by transposition and division, we get

A* I — *— CV

~~37~ (159).

from which equation, the value of b is easily determined.
Let the slope of that side of the dyke on which the fluid presses, be
equal to 2 feet, and the perpendicular altitude of the dyke, or the
depth of the fluid 20 feet, the specific gravity of the material being
If as before ; then by substitution, the foregoing equation becomes

,-„==£$!!=««,,

by completing the square, we obtain
p — 2b + 1 = 74.8571 4- 1—75.8571 ;

consequently, by extracting the square root and transposing, we get

b = 9.709 feet.

In this case, the dyke has less stability than it has when the perpen-
dicular side is towards the water, as is manifest from its requiring a
greater section, and consequently, a greater quantity of materials to
resist the effort of the pressure which tends to overturn it.

202 OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS.

The sectional area in the one case, is 7. 804 X 20 =: 156.08 square
feet, and in the other, it is 8.709x20=: 174.18 square feet, being a
difference of 18.1 square feet in favour of magnitude in the latter
form, where the sloping side is adjacent to the fluid ; and this being
multiplied by the length of the dyke, will give the extra quantity of
materials necessary for obtaining the same degree of stability.

217. If both the slopes c and e become evanescent; that is, if the
section of the dyke be rectangular, having both its
sides perpendicular to the horizon, as represented
by ABC D in the annexed diagram ; then, the general
equation (149), becomes transformed into

sda = 3vs'b\ (160).

Then, by supposing the depth of the fluid, and
the perpendicular altitude of the dyke to become
equal, (the specific gravity of water being expressed by unity,) we
have

3s'&2zrd8;

and this by division becomes

-I-

consequently, if the square root of both sides of this equation be
extracted, we shall have

'- (161).

218. This is indeed a very simple form of the, equation, applicable
to the very important case of rectangular walls ; it is however accu-
rate, and corresponds in form with that investigated by other writers
for the same purpose, and by different methods ; the mode of its
reduction is simply as follows.

RULE. Divide the specific gravity of the fluid to be sup-
ported, by three times the specific gravity of the dyke or
embankment, and multiply the square root of the quotient
by the perpendicular altitude of the dyke, for the required
thickness.

Let the perpendicular depth of the water, or the altitude of the
dyke be equal to 20 feet, and the specific gravity of the materials of
which it is built If, as in the foregoing cases; then, by proceeding
according to the rule, we have

OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS. 203

219. There is still another case of very frequent occurrence that
remains to be considered, viz. that in which the section is in the form
of a right angled triangle, having its vertex on the same level with
the surface of the fluid.

This case will also admit of two varieties, according as the perpen-
dicular side of the dyke is, or is not in contact with the fluid ; when
it is in contact with it c vanishes, and since the section is in the form
of a triangle, the breadth of the base b is equal to the remote slope e,
and the vertical pressure of the fluid on the dyke is evanescent; con-
sequently, the equation marked (149) becomes

sd* = 3Ds'P — DsV; (162).

but by the nature of the problem e2 is equal to 62, and by the hypo-
thesis of equal altitudes c?=z= D ; therefore, in the case of water, whose
specific gravity is expressed by unity, we obtain

Zs'tfi^d*;
and from this, by division, we get

and finally, by extracting the square root, it is

b — d\/ _L

V 2s'' (163).

220. This is also a very simple expression for the base of the section,
and the rule for its reduction is simply as follows.

RULE. Divide the specific gravity of the incumbent fluid,
by twice the specific gravity of the dyke or embankment, and
multiply the perpendicular depth of the fluid by the square
root of the quotient, for the required thickness of the dyke.

Let the perpendicular altitude and the specific gravity of the wall,
be 20 feet and 1 j respectively, as in the foregoing cases, and we shall
have

6 = 2<V2-^5= 10.68 feet.

221. Lastly, if the fluid come in contact with, or press upon the
hypothenuse of the triangle ; then the slope e vanishes, and b and c
are equal ; consequently, equation (149) becomes

and if the vertical pressure of the fluid be omitted, the first term on
the right hand side of the equation vanishes, and consequently, we get

sd*=vs'b*;

204 OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS.

but C?:=D, and s=z»l ; therefore, we shall have
and finally, by division and evolution, we obtain

t=i^

222. This equation is of a still simpler form than that which arises
when the perpendicular side is towards the pressure, and the rule for
its reduction is as follows.

RULE. Divide the specific gravity of the fluid, by that of
the dyke or embankment, and multiply the perpendicular alti-
tude by the square root of the quotient, for the breadth of
the base.

Therefore, by taking the altitude and specific gravities hitherto
employed, the rule will give

6 = 20A/ — =15.11 feet.
V 1.75

223. This gives a thickness for the base of the section, exceeding
the thickness in the former case by 4.33 feet; which seems to be a
very great difference, when it is considered that both the form and
the perpendicular altitude of the wall are the same in both cases ; but
the reason of the difference will become manifest from the following
construction.

Let ABC and a be be two right angled triangles, equal to one
another in every respect, but having
their perpendiculars opposed in such a
manner, that the water pressing in the
same horizontal direction, is resisted by
the perpendicular AB in the one case,
and by the hypothenuse ac in the other.

Bisect the sides AB, AC in the point
M and N, and a b, ac in the points m

and n respectively; draw the lines CM and BN intersecting in G, and
cm and bn intersecting in g; then are G and g, the centres of gravity
of the respective triangles ABC and a be.

Demit the straight lines GH and gh, perpendicularly to BC and be;
then are HC and h b the levers, by which the weights of the sections,
supposed to be concentrated in their respective centres of gravity,
resist the horizontal pressure of the fluid which tends to turn them
round the points c and b.

OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS. 205

Now, according to the property of the centre of gravity, HC is
equal to two thirds of BC, while h b is only one third of be ; but the
horizontal pressure of the water is the same in both cases ; it will
therefore require the same mechanical energy to resist it; and since,
by the conditions of the problem, the altitudes A B and a b are equal,
it follows, that in order to produce an equilibrium, the product of the
base of the triangle a be, into the length of the lever h b, must be
increased in such a manner, that

BCXHC = 6cX^&, (166).

and by converting this equation into an analogy, it becomes
BC : be : : hb : HC.

We have seen, that by the construction and the property of the
centre of gravity, the lever HC is equal to two thirds of BC, and h b
equal to one third of be; let therefore, fsc and %bc, be substituted
for HC and h b, in the equation marked (166), and we shall obtain

or dividing both terms by -^ , it becomes

2BC2 = £C2,

and finally, by extracting the square root, it is

(167).

Hence the reason for, and the nature of the increased breadth
become obvious, the one being the side, and the other the diagonal
of a square.

Now, we have found that the breadth of the dyke at the base, is
equal to 10.68 feet, when the perpendicular side is in contact with
the fluid ; consequently, when the pressure is exerted on the hypo-
thenuse, we have

6=10.68X1.4142 = 15.11 feet,

being the very same result as that which we obtained from the reduc-
tion of the equation (165).

224. What we have hitherto done, has reference to the case in
which the obstacle yields to the pressure of the fluid, by turning upon
the remote extremity of its base ; we have therefore, in the next place,
to investigate the conditions of equilibrium, when the obstacle is sup-
posed to yield, by sliding along the horizontal plane on which it is
erected.

Since the base of the dyke or wall is horizontal, it is manifest that
the mass which it sustains, resists the horizontal pressure of the fluid,
only by its adhesion to the base, and the resistance occasioned by
friction.

206 OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS.

Suppose therefore, that the resistances of adhesion and friction, are
equal to n times the weight of the dyke, which we have represented
by w ; then we have

nw — \d*s\
but we have shown, in the investigation of the preceding case, that

w = jD/(26 — c — e}\
consequently, by substitution, we obtain

d2s = i>nJ(2b — c — e). (168).

This is the equation of equilibrium, or that in which the resistance
of the dyke is counterpoised by the horizontal pressure of the fluid,
the effect of the vertical pressure not being considered ; but in order
to express the breadth of the base in terms of the other quantities, let
both sides of the equation be divided by ons', and it becomes

D7ZS

consequently, by transposition and division, we obtain

d*s

6 = 2^7+i(c + e): (169).

and finally, if the perpendicular depth of the fluid and the height of
the dyke are equal, we shall have

- - . (170).

225. In order therefore, to illustrate the reduction of the above
equation;1by means of a numerical example, we must assume a value
to the letter n, having some relation to the nature of the materials of
which the resisting obstacle is constructed ; now, it has been found
by numerous experiments, that when rough and uneven bodies rub
upon one another, or when a heavy body composed of hard and
rough materials, is urged along a horizontal plane, the effect of the
friction is equivalent to about one third of the weight of the body
moved ; or in other words, it requires about one third part of the force
applied to overcome the effects of the friction ; and moreover, in the
case of a wall built of masonry, there is, in addition to the friction,
the adhesion of the materials to the plane on which the wall is built.

If therefore, we consider the effect of adhesion to be equivalent to
the effect of friction, it is manifest, that their conjoint effects will
destroy about two thirds of the force applied ; consequently, in the
case of masonry, we may suppose that the value of w, is very nearly
equal to 1J, but for other materials it will vary according to the
specific gravity or weight.

OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS. 207

Having thus assigned a particular value to the letter n, we shall
next proceed to illustrate the reduction of the equation ; for which
purpose, take the following example.

226. EXAMPLE 3. The vertical transverse section of the wall which
supports the water in a reservoir, is 24 feet in perpendicular height ;
what is the thickness at the base of the wall, supposing the section to
be in the form of the frustum of an isosceles triangle, the slope or
inclination on each side, being equal to 2 feet, and the specific gravity
of the material If, that of water being expressed by unity ?

Let the several numerical values here specified, be substituted
instead of the respective symbols in the equation (170), and we shall
obtain

_24>O_ ___ nearly.

The breadth of the section, or the thickness of the dyke at the
bottom, being thus determined, the breadth or thickness at the top
can easily be found, for we have

6.571 — 4 = 2.571 feet.

127. If the slope c should vanish; that is, if the side of the dyke
on which the water presses be perpendicular to the horizon ; then, the
equation (170), becomes

• (171).

And if the opposite slope e becomes evanescent, while the slope c
remains ; then we have

* = 2^7+ic< (172).

But if both slopes vanish, or the section of the wall becomes
rectangular; then, the equation (170) is

(173).

If therefore, the perpendicular altitude of the section, and the
specific gravity of the materials of which the dyke is composed, remain
as in the preceding example ; then we shall have

b= 24X1 =4.571 feet.
2XfX£

228. When the section of the wall assumes the form of a right
angled triangle ; that is, when the slope c vanishes, and e becomes
equal to the whole breadth b ; then we have

208 OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS.

-W' (174).

And exactly the same equation would arise, if the slope e remote
from the fluid were to vanish, and the slope c adjacent to the fluid,
become equal to b the whole breadth of the section; consequently,
the thickness of a dyke in the case of a triangular section, whether
the water presses on the perpendicular or hypothenuse of the tri-
angle, is

In all the preceding cases, it is supposed that the section of the
dyke or embankment is of such dimensions, as to oppose an equipois-
ing resistance to the pressure of the fluid which it supports ; but in
the actual construction of all works of this nature, it becomes neces-
sary, for the sake of safety, to enlarge the dimensions considerably
beyond what theory assigns to them ; but it does not belong to this
place to determine the limits of the enlargement.

2. OF THE PRESSURE OF FLUIDS AGAINST EMBANKMENTS OF LOOSE

MATERIALS.

229. The theory which we have established above, supposes that a
perfect connection obtains between all the parts of the dyke or
embankment which is opposed to the pressure of the fluid, so that any
one portion of it cannot be displaced or overthrown, unless the whole
be overthrown at the same time; the formulge thence arising, are
therefore, only applicable to dykes or embankments that are con-
structed of masonry ; in those which are constructed of earth or other
loose materials, and having the sides faced or fortified with stone, the
same connection between the component portions of the wall does not
exist, and consequently, although the several equations apply when
the whole perpendicular height of the dyke is considered, yet the dyke
will not resist equally at every part of the height, but is liable to
be separated into horizontal sections.

In order therefore, to adapt our prin-
ciples to this case also, it becomes neces-
sary to trace out the steps of another
investigation ; for which purpose,

Let ACD represent a vertical section of
the dyke or embankment, whose summit
at A is on a level with the surface of the fluid AE.

OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS. 209

Take any point G in the line AGD, and through the point o thus
assumed, draw the horizontal ordinate GB, cutting the vertical axis
AC in the point B : now, it is required to determine the nature of the
curve AGD such, that each portion of the dyke, or of its section, as
AGE, estimated from the vertex, may be equally capable of resisting
the horizontal pressure of the fluid exerted against AG ; or which is
the same thing, that each portion may retain its stability and remain
in equilibrio on its base GB; not separating from the lower portion
GBCD, either by turning about the point c as a centre of motion, or
by sliding in a horizontal direction along the base GB.

Put x — AE, the abscissa of the curve estimated from the vertex

at A,
y = EG, the horizontal ordinate corresponding to the ab-

scissa x,
s z= the specific gravity of the fluid, which endeavours to

displace the dyke by pushing it along the lin« BG,
/ zn the specific gravity of the materials of which the dyke

is constituted,

m m the momentum of the horizontal pressure,
m'— the momentum of the resistance offered by the dyke, and
n :=z the number of times that the adhesion and friction of the

dyke are equal to its weight.

Then we have already seen, equation (145), that the momentum of
the horizontal pressure of the fluid as referred to the point c, is

from which, by substituting xs instead of d3, we obtain m —
which equation indicates the momentum of pressure at the point D.

But the momentum of the resistance offered by the wall, that is,
the momentum of the portion of the section represented by ABG, is

m' = is'fy*x;

and these momenta in the case of an equilibrium must be equal to
one another ; hence we have

from which, by taking the fluxion, we shall obtain

or by suppressing the common factors, it becomes sx^ — s'y*;
by extracting the square root of both terms, we get x\/s — y\/s'.
Now, when x becomes equal to d, the whole perpendicular depth
of the fluid, or the altitude of the section ; then y becomes equal to
b, the thickness of the dyke, or the greatest breadth of the section ;
VOL. i. p

210 OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS.

consequently, if d and b be respectively substituted for x and y in
the preceding equation, we shall have

d^/7—b^S (175).

This equation involves the conditions necessary for preventing the
dyke from turning about the point B, and if the equation be resolved
into an analogy, we shall have b : d : : \/s : ^/sf.

COROL. From which we infer, that the section is in the form of a
rectilinear triangle, whose base is to the perpendicular height, as the
square root of the specific gravity of the fluid, is to the square root of
the specific gravity of the wall or dyke.

230. EXAMPLE. The perpendicular altitude of an embankment of
earth is 20 feet ; what must be the breadth of its base, so that each
portion of it estimated from the vertex, shall resist the effort of the
fluid, to turn it round the remote extremity of the base, with equal
intensity ; the water and the dyke having equal altitudes, and their
specific gravities being 1 and 1 .5 respectively ?

Here we have given rfrz 20 feet, szr 1, and sr zz 1 .5 ; consequently,
by the preceding analogy, we have ^ 1.5 : ^ 1 : : 20 : 16.33 feet.

231. The conditions necessary for preventing the portion of the
section ABG from sliding on its base, may be thus determined.

We have seen (art. 220), that the momentum of the horizontal
pressure, to urge the section along its base, is m— Jse?%

consequently, by substituting x9 for d2, we have m — Jsa:2,

but the momentum of the section opposed to this, is m' zz ns'fyx;

therefore in the case of an equilibrium, we have isx* = ns'fyx,

from which, by taking the fluxion, we obtain sxx~ ns'yx,
and by casting out the common factor, we get sx^nns'y. (176).
From this equation, when converted into an analogy, we shall obtain
x : y : : ntf : s.

Which also indicates a rectilinear triangle, whose altitude is to the
base, as n times the specific gravity of the embankment, is to the
specific gravity of the fluid.

If the water presses against the perpendicular side of the wall, the
curve bounding the other side, so that the strength of the wall may
be every where proportional to the pressure which it sustains, must be
a semi-cubical parabola, whose vertex is at the surface of the fluid,
and convex towards the pressure.

OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS. 211

232. We are now arrived at that particular division of our subject,
which comprehends some of the most interesting and important
departments of hydrodynamical science ; it unfolds the principles of
floatation, explains the method of weighing solid bodies in fluids,
determines the relations of their specific gravities ; and moreover, it
investigates the laws of equilibrium, and assigns the conditions neces-
sary for a state of perfect or imperfect stability. Every term in this
enumeration conveys the idea of mechanical action.

Floating bodies, those which swim on the surface of a fluid, which is bulk for
bulk heavier than the body afloat, are pressed downward by their own weight in a
vertical line passing through their centre of gravity : and they are supported by
the upward pressure of the fluid, which acts in a vertical line passing through the
centre of gravity of the part which is under the water. When these lines are coin-
cident, the equilibrium of floatation will be permanent. In the present instance we
have merely to consider the principles of floatation as fluids exhibit the properties
of the mechanical powers, as the lever or balance, the screw, &c. The pulley, in
lowering a great weight or in lifting it up again, does no more than the ocean tide
when it silently recedes and leaves dry, or majestically advances and without effort
floats a stupendous ship. The lever or balance does no more than a canal lock
effects, when it transfers from one level to another a heavy barge or vessel laden
with ponderous commodities. And we behold too the ocean, like a vast screw or
press, forcing down to its dark recesses vast masses, which in shipwrecks are sub-
merged in its bosom, and which yet might be fashioned to be bulk for bulk much
lighter than the devouring flood which has swallowed them up in its insatiable
womb. The eternal and immutable laws of Nature, in all these cases, are most
satisfactorily accounted for in the doctrine of fluid pressure and support j but this
doctrine, like all the rudiments of human skill applied to natural phenomena, must
depend on matters of fact, which can only be learned from observation and experi-
ment, and which can generally and successfully be applied by the help of mathema-
tical and philosophical investigations. This is the only scientific view we ought
to take of all those truths that are denominated the phenomena of fluids, whose
affections, from a series of concurring experiments, we undertake to expound j or
assuming these as established pi-inciples that operate generally in the pressure and
elasticity of fluids, we demonstrate them to be adequate to the production, not only
of the particular effects adduced to prove their existence and power, but of all
similar phenomena. This is the only method by which to make the results of
practical men available in scientific discussions, and on the other hand render these
discussions the handmaids of genius in constructive mechanics. This is the province
of the mathematician; and we shall in the sequel follow it very closely, in expound-
ing the doctrine of floatation and the specific gravities of bodies, the laws of equili-
brium, and the conditions necessary for a state of perfect or imperfect stability, &c.

P 2

CHAPTER IX.

OF FLOATATION, AND THE DETERMINATION OF THE SPECIFIC
GRAVITIES OF BODIES IMMERSED IN FLUIDS.

OF the several particulars with which we concluded the last
chapter, we shall speak in order, beginning with the theory of float-
ation and the determination of the specific gravity of bodies, the
leading principles of which are contained in the following proposition.

PROPOSITION III.

233. When a body floats, or when it is in a state of buoyancy
on the surface of a fluid of greater specific gravity than itself: —

It is pressed upwards by a force, whose intensity is equi-
valent to the absolute weight of a quantity of the fluid, of
which the magnitude is the same as that portion of the body
below the plane of floatation.*

Let ABC represent a vertical section of a solid body floating on a
fluid, whose horizontal surface is DE, mn being the plane of floata-
tion, and men the immersed portion of the floating body.

Take any two points G and H on
the surface of the solid, indefinitely
near to each other, and through the
points G and H thus arbitrarily as-
sumed, draw the straight lines G F
and HI, respectively parallel to DE

the surface of the fluid, and meeting the opposite sides of the solid in
the points r and i, so that each point in either of the intercepted
portions GH and FI, may be considered as being at the same perpen-
dicular depth h& or ZF below the horizontal surface of the fluid.

At H and i erect the perpendiculars H r and i s, which produce to
t and w, and through the points G and F, draw the straight lines G b
and F/, respectively perpendicular to the surface of the solid in the

* The Plane of Floatation is the imaginary plane, in which the floating solid is
supposed to be intersected by the horizontal surface of the fluid.

OF FLOATATION AND THE SPECIFIC GRAVITY OF BODIES. 213

points G and F ; make ob and F/each equal to /IG or ZF, the perpen-
dicular depth of the points G and F below the surface DE; then,
according to the principles which we have propounded and demon-
strated in the first proposition and its subordinate inferences, the
perpendicular pressures upon the indefinitely small portions of the
body GH and FI, may be expressed as follows, viz.

/> zr s XG H X G &, and p' m s X F i X F/,

where s denotes the specific gravity of the fluid, and p, j/ the respec-
tive pressures exerted by it perpendicularly to GH and FI, any
indefinitely small portions of the floating body.

But it is manifest from the resolution of forces, that the pressures
of the fluid in the directions bo and y*F, may each be decomposed
into two other pressures, the one vertical and the other horizontal ;
for by completing the rectangular parallelograms oabc and Tdfe, it
is obvious that the pressures in the directions «G, CG and dr, ev are,
when taken two and two, respectively equivalent to the pressures in
the directions bo andy*F.

Now, the horizontal pressures CD and CF, by construction are equal
to one another, and they operate in contrary directions; consequently
they destroy'each other's effects, and the upward vertical pressures on
the solid at the points G and F, are respectively indicated by the
straight lines «G and C?F drawn into the specific gravity of the fluid;
therefore, the whole vertical pressures on the indefinitely small por-
tions GH and FI, are as follows, viz.

j9 = 5XGHX«G, and jy'zzrsXFiX^F,

where p and p', instead of indicating the perpendicular pressures as
formerly, are now considered in reference to the vertical pressures.

Since the parallel straight lines GF and HI are indefinitely near to
one another, the lines G n and FI may be assumed as nearly straight,
and consequently, the elementary triangles Giir and FIS are respec-
tively similar to the triangles GBO and F/W; therefore, by the pro-
perty of similar triangles, we have

G 6 : G a : : G H : G r, and $f: F d : : F i : F s ;

and from these analogies, by equating the products of the extreme
and mean^terms, we obtain

G&XGnziGaXGH, an(j Fy XFSZHFC? XFI.

Let therefore, the products G&Xor and F/XFS be substituted
instead of GiiXctG and FiXc?r in the above values of p and //, and
we shall have

/; — s X G b X G r, and p m s X F/ X F s.

214 OF FLOATATION AND THE SPECIFIC GRAVITY OF BODIES.

Now, these pressures are manifestly equal to the weights of the
columns Gt and vu considered as fluid, and since the same may be
demonstrated with respect to every other portion of the immersed
surface, we therefore conclude, that the whole pressure upwards, is
equal to the sum of the weights of all the columns Gt, FU, &c. ; that
is, to the weight of a quantity of the fluid equal in magnitude to the
immersed part of the body ; hence the truth of the proposition is
manifest.

COROL. From the principles demonstrated above, it follows, that
when a solid body floating on the surface of a fluid is in a state of
quiescence : —

The pressure downwards is equal to the buoyant effort ;

that is, the weight of the floating body, is equal to the weight

of a quantity of the fluid, whose magnitude is the same as

that portion of the solid, which falls below the plane of

floatation.

PROBLEM XXXI.

234. A cylindrical vessel of a given diameter, is filled to a
certain height with a fluid of known specific gravity, and a
spherical body of a given magnitude and substance is placed
in it : —

It is required to determine how high the fluid will rise in
consequence of the immersion of the spherical segment which
falls below the plane of floatation.

Let ABCD represent a vertical section passing along the axis of a
cylindrical vessel, filled with an incompressible and
non-elastic fluid to the height ED, EF being the sur-
face of the fluid before the sphere whose diameter is
mn, is placed in it, and ab the surface after the
immersion of the segment tnu, the liquid rising to the
height A D.

Then it is manifest from the nature of the problem,
that the spherical segment tvnwu, together with the quantity of fluid
in the vessel, must be equal to the capacity of the cylinder whose
diameter is DC, and perpendicular altitude «D; for the fluid rises in
consequence of the immersion of the segment, and fills the spaces
atvE and buw^s all around the vessel ; we have therefore to calculate

OF FLOATATION AND THE SPECIFIC GRAVITY OF BODIES. 215

the spherical segment tvmuu, and the cylinders EFCD and a ben,
for which purpose,

Put d zz ED, the height to which the vessel is originally filled with

the fluid,

$ == DC, the diameter of the cylindrical vessel,
r zz ct, or cv, the radius of the sphere,
s = the specific grav;ty of the fluid in the vessel,
s' nr the specific gravity of the floating body, and
x — «D, the height to which the fluid rises on the immersion of

the spheric segment.

Then, since by the principles of mensuration, the solid content or
capacity of a sphere, is equal to two thirds of that of its circumscribing
cylinder, it follows, that the capacity of the sphere mvnw, is ex-
pressed by

3.1416r2x2rXf=: 4.1888^;

but^we have elsewhere demonstrated, that the magnitudes of bodies
are inversely as their specific gravities ; consequently, the magnitude of
the part immersed, is determined by the following analogy, viz.

.:.':-. 4.1888^:

Now, as we have already observed, the quantity of fluid in the

vessel at first, is

.7854X^X^—7854^,

and the capacity of the cylinder formed by the fluid and the spherical

segment, is

consequently, by addition, we shall have

.7854 tf x = .7854^-f H^^i .

and therefore, if all the terms of this equation be divided by the
quantity .78543% we shall obtain

16rV
'

(177).

Or if the height to which the vessel is originally filled, be subtracted
from both sides of the above expression, the increase of height in con-
sequence of the immersion of the spheric segment, becomes

, 16rV

— d=x— 3?7'
where &•'=: a£ the increase of height.

216 OF FLOATATION AND THE SPECIFIC GRAVITY OF BODIES.

235. Either of these equations will resolve the problem, but the
latter form is the most convenient for a verbal enunciation, and the
practical rule which it supplies is as follows.

RULE. Multiply sixteen times the specific gravity of the
sphere, by the cube or third power of its radius ; then, divide
the product by three times the specific gravity of the fluid,
drawn into the square of the cylinder's diameter, and the
quotient will give the increase of height, in consequence of the
immersion of the spheric segment.

236. EXAMPLE. A cylindrical vessel whose diameter is 8 inches, is
filled with water to the height of 10 inches; how much higher will
the water rise, and what will be its whole weight, when a globe of
alder of 6 inches diameter is dropped into the vessel ; the specific
gravity of alder being equal to .8, when that of water is expressed by
unity ?

Here, by operating according to the above rule, we get

16rV=16x3x3x3x. 8 z= 345.6,

and in like manner we have

33^ = 3x8x8x1 — 192;

consequently, by division, we obtain

ar'zr— -— — z= z= 1.8 inches, and the whole height is 11.8 inches.

v oo s ly.2

237. If the specific gravity of the globe, and that of the fluid in which
it is placed, are equal to one another, then equation (178) becomes

, 33* ' (179).

In this case it is manifest, that the sphere is wholly immersed in
the fluid ; consequently, the increase of height wilt be equal to the
altitude of a cylinder, whose diameter is 3, and whose capacity is
equal to that of the immersed body ; hence, the method of computa-
tion is obvious ; but the practical rule deduced from the equation for
this purpose, may be expressed in the following manner.

RULE. Divide sixteen times the cube or third power of the
radius of the sphere, by three times the square of the cylin-
der s diameter, and the quotient will give the increased height
of the fluid.

238. EXAMPLE. A cylindrical vessel whose diameter is 12 inches,
is filled with fluid to the height of 6 inches; to what height will

OF FLOATATION AND THE SPECIFIC GRAVITY OF BODIES. 217

the fluid ascend when a sphere of 4 inches diameter is placed in it,
the specific gravities of the fluid and the sphere being equal to one
another ?

lii this example there are given 5 ~ 6 inches and r zz 2 inches ;
therefore, by operating as directed in the rule, we shall have

16r3=:16X23— 128, the dividend,
and in like manner for the divisor, we get

3S2 = 3 X 122 — 432, the divisor ;
consequently, by division, we obtain

xf — Hi= 0.2962 of an inch.

Hence it appears, that the height of the fluid in the vessel, is
increased by a quantity equal to 0.2962 of an inch, in consequence
of the immersion, and the whole height to which it rises, is 6.2962
inches.

PROBLEM XXXII.

239. A vessel in the form of a paraboloid, is placed with its
vertex downwards and its base parallel to the horizon; now,
supposing the vessel to be filled to the wth part of its capacity
with a fluid of known specific gravity, and let a spherical body
of a given size and substance be placed in it : —

It is required to ascertain the height to which the fluid
will rise, in consequence of the immersion of the spherical
segment.

Let ABC represent a vertical section passing along the axis of the
vessel, whose form is that of a paraboloid,
generated by the revolution of the common
parabola ; and suppose the vessel to be filled
with an incompressible and non-elastic fluid
to the height sc, DE being its horizontal
surface when in a state of quiescence, before
the sphere whose diameter is mn is placed in
it; then will a b be the surface or the plane
of floatation after the immersion of the segment rnw, the fluid rising
to the height ts all around the spherical body.

Now, it is obvious from the nature of the problem, that, the solidity
of the spherical segment rnw, together with the quantity of fluid in
the vessel, is equal to the magnitude of the paraboloid acb, whose

218 OF FLOATATION AND THE SPECIFIC GRAVITY OF BODIES.

base is ab and axis cf ; therefore, in order to calculate the solidity
of the segment, and that of the paraboloids DCE and acb,

Put d n: me, the whole axis or height of the paraboloid,
p zn the parameter or latus rectum of the axis,
r zz: cr, cm or cw, the radius of the sphere,
s — the specific gravity of the fluid in the vessel,
/ zz the specific gravity of the floating body, and
x zz tc, the whole height to which the fluid ascends, n being
the part originally filled.

By the principles of solid mensuration, the capacity or solidity of a
sphere, is equivalent to two thirds of that of its circumscribing cylin-
der ; consequently, the capacity of the floating sphere, is

now, we have demonstrated in another place, that the magnitudes of
bodies, are inversely as their respective gravities ; hence we have for
that portion actually immersed,

*:^: 4.1888,* :

Again, by the principles of mensuration, the solidity of a paraboloid
is equal to one half the solidity of its circumscribing cylinder, and by
the property of the parabola, we have

niA*~pd;
therefore, the capacity of the paraboloidal vessel, is

3.1416X^^X^—1-5708^^,
and consequently, the quantity of fluid in it is expressed by

But the capacity, or the solid content of the paraboloid a c b, whose
axis is tc, becomes

consequently, by addition and comparison, we have

4.1888rV \.570Spd*
1 .57(% X* zz -- -f - £— ,
s n

and dividing all the terms by 1.5708, we get

8rV pd*
**- = — + -

OF FLOATATION AND THE SPECIFIC GRAVITY OF BODIES. 219

and again, if all the terms be divided by p the parameter of the
parabola, and the square root be extracted from both sides of the
equation, we shall have

But because the parameter of a parabola is a third proportional to
any abscissa and its ordinate ; it follows, that if b denote the base A B
of the paraboloid, of which the axis is d, we shall have

let this value of the parameter be substituted instead of it in the above
equation, and we shall obtain

(180).

240. The following practical rule supplied by this equation, will
serve to direct the reader to the method of its reduction.

RULE. Multiply thirty -two times the axis of the paraboloid,
by the cube or third power of the radius of the sphere drawn
into its specific gravity ; then, divide the product by three
times the square of the base of the vessel multiplied by the
specific gravity of the Jluid, and to the quotient, add the
square of the axis or depth of the vessel, divided by the
number, which expresses what part of it is occupied by the
Jluid ; then, the square root of the sum, will give the height
to which the Jluid rises after the immersion of the spheric
segment.

241. EXAMPLE. The axis of a vessel in the form of a paraboloid is
27 inches, and the diameter of its mouth is 18 inches ; now, supposing
that the vessel is one fifth full of water, into which is dropped a sphere
of hazel whose diameter is 8 inches ; to what point of the axis will
the fluid ascend, the specific gravity of hazel being 0.6, when that of
water is expressed by unity ?

By proceeding according to the rule, we get

32X27X4X4X4X.6 = 33177.6, the dividend,

and in like manner, for the divisor, we have

3X18X18X1 = 972, the divisor ;
consequently, by division, we obtain

220 OF FLOATATION AND THE SPECIFIC GRAVITY OF BODIES.

This is the value of the first term under the radical sign, in the
expression for x equation (180), and the value of the second term, is

278

- = 145.8 ;

therefore, by addition and evolution, we obtain

a;=^/179.93 = 13.414 inches nearly.

242. If the specific gravity of the ball, and that of the fluid in
which it is placed, be equal to one another, then equation (180)
becomes

„_ /32^ ^
'V -W~ + n'

and by reducing the fractions under the radical sign or vinculum to a
common denominator, we obtain

a? =

243. The practical method of reducing the above equation, is
expressed in words at full length in the following rule.

RULE. Multiply the cube or third power of the- radius of
the sphere, by thirty two times the number which indicates
what part of the vessel is occupied by the fluid, and to the
product add three times the axis of the vessel drawn into the
square of its diameter ; then, divide the sum by three times
the square of the vessel's diameter, drawn into the number
which denotes what part of it the fluid occupies ; multiply
the quotient by the axis of the vessel, and extract the square
root of the product , for the height to which the fluid rises.

244. EXAMPLE. Let the dimensions of the vessel and the immersed
body, remain as in the preceding example, the vessel containing also
the same quantity of fluid ; to what height on the axis will the fluid
ascend, supposing its specific gravity to be the same as that of the
immersed body ?

Here, by operating as directed by the rule, we get
32wr3= 32X5X4X4X4 = 320X32 = 10240, and

3b*d= 3X 18 X 18 X27 = 972 X27 = 26244 ;
consequently, the sum of the parenthetical terms is

32»r3 4- 3b*d = 10240 + 26244 = 36484,

and for the denominator of the fraction, we have

367w = 3X18X18X5 = 324X15 = 4860;

OF FLOATATION AND THE SPECIFIC GRAVITY OF BODIES. 221

consequently, by division, we obtain
32«r»+3ffd_ 36484

36s n 4860 ~

therefore, by multiplication we shall have

7.507X27 = 202.689,

and finally, by evolution, it is

x= y 202.689= 14.23 inches.

COROL. Hence it appears, that when the specific gravities of the
fluid and the immersed body, are equal to one another, the fluid rises
in the vessel to the height of 14.23 inches ; but when the specific
gravities are to each other as 1 : 0.6, it rises only to 13.414; the
reason of the difference, however, is manifest, for in the case of equal
specific gravities, the spherical body is wholly immersed ; but when
the specific gravities are unequal, only a part of the body falls below
the plane of floatation. From the above we deduce the following
inferences.

245. INFERENCE 1. If a homogeneous body be immersed in a fluid
of the same density with itself: —

It will remain at rest> or in a state of quiescence, in all
places and in all positions.

Let ABCD represent a vessel, filled with an incompressible and
non-elastic fluid to the height «D, and let G be a
homogeneous body, of the same density or specific
gravity as the fluid.

Now, it is manifest, that when the body G is put
into the vessel and left to itself, it will by reason of
its own weight, sink below ab the original surface,
and raise the fluid to the height E D, where the body
will be entirely under the fluid, and the whole mass
in a state of equilibrium with the surface at EF.

Then it is evident, that the body being of the same density as the
fluid in which it is placed, it will press the fluid under it, just as much as
the same quantity of the fluid would do if put in its stead, and conse-
quently, the pressure exerted by the solid, together with that of the
superincumbent fluid, presses downwards with the same energy, as if
it were a column of fluid of equal depth.

Therefore, the pressure of the body against the fluid at H, is equal
to the pressure of the fluid against the body there; consequently,

222 OF FLOATATION AND THE SPECIFIC GRAVITY OF BODIES.

these two pressures are equal and opposite to one another, and must
therefore be in a state of equilibrium, in which case, the body will
remain at rest.

Hence, the truth of the inference is manifest with respect to a
vertical pressure ; but it is equally true in reference to a motion
horizontally and obliquely; for the horizontal pressures are obviously
equal to one another, and they are in opposite directions; therefore,
they are in equilibrio with one another, and no motion can take place.

And again, with regard to the oblique pressure, it is evidently
compounded of a vertical and horizontal one ; but we have just
demonstrated that these are equal and opposite; consequently, the
body can have no oblique motion, it must therefore remain at rest in
any place and in any position.

If the specific gravity of the immersed body be greater than that of
the fluid, the pressure downwards will exceed the pressure upwards ;
consequently, the weight of the body will overcome the resistance of
the fluid under it, and it will therefore sink to the bottom.

But if the specific gravity of the body be less than that of the fluid,
the pressure upwards will exceed the pressure downwards ; therefore,
the buoyant principle will overcome the weight of the solid, and it
will rise to the surface of the fluid.

246. INF. 2. If a solid body be immersed in a fluid, and the whole
mass be in a state of equilibrium : —

The pressure upwards against the base of the body, is
equal to the weight of a quantity of fluid of equal magnitude,
together with the weight of the superincumbent fluid.

247. INF. 3. If a solid body be placed in a fluid of greater or less
specific gravity than itself : —

The difference between the pressures downwards and up-
ivards is equal to the difference between the weight of the
solid and that of an equal bulk of the fluid.

248. INF. 4. Heavy bodies when placed in fluids have a twofold
gravity, the one true and absolute, the other apparent or relative.

Absolute gravity is the force with which bodies tend
downwards.

By reason of this force, all sorts of fluid bodies gravitate in their
proper places, and their several weights, when taken conjointly,
compose the weight of the whole ; for the whole is possessed of
weight, as may be experienced in vessels full of liquor.

OF FLOATATION AND THE SPECIFIC GRAVITY OF BODIES. 223

Apparent or relative gravity, is the excess of the gravity
of the body above that of the fluid in which it is placed.

By this sort of gravity, fluids do not gravitate in their proper places ;
that is, they do not preponderate ; but opposing one another's descent,
they retain their positions as if they were possessed of no weight.

249. INF. 5. If a heavy irregular heterogeneous body descends in
a fluid, or if it moves in any direction, and a straight line be drawn,
connecting the centres of magnitude and gravity of the body : —

It will so dispose itself as Jo move in that line, the centre
of gravity preceding the centre of magnitude .

This is a manifest and a beautiful fact ; for the centre of gravity
being surrounded by more matter and less surface than the centre of
magnitude, it will meet with less resistance from the fluid ; conse-
quently, the body will so arrange itself, as to move in the line of
direction with its centre of gravity foremost.

250. What has been here adverted to, in regard to bodies of greater
density or specific gravity sinking in a fluid, must only be understood
to apply to such as are solid ; for if a body be hollow, it may swim in
a fluid of less specific gravity than that which is due to the substance
of which the body is composed ; but if the hollows or cavities are
filled with the fluid, the body will then descend to the bottom.

Again, if bodies of greater specific gravity than the fluid in which
they are placed, be reduced to extremely small particles, they may
also be suspended in the fluid ; but the principle or force by which
this is effected, does not belong to hydrodynamics.

PROPOSITION IV.

251. If a solid homogeneous body, be placed in a fluid of
greater or less specific gravity than itself: —

It will ascend or descend with a force, which is equivalent
to the difference between its own weight, and that of an equal
bulk of the fluid.

The principle announced in this proposition is almost self-evident,
yet nevertheless, it may be demonstrated in the following manner.

Put m — the common magnitude of the body and the fluid,
w' — the weight of the solid body,
s' — its specific gravity.

224 OF FLOATATION AND THE SPECIFIC GRAVITY OF BODIES.

w zz the weight of an equal quantity of the fluid,
s zz: its specific gravity, and

/ zz the force with which the body ascends or descends in
the fluid.

Then, because as we have elsewhere demonstrated, the absolute
weights of bodies, are as their magnitudes and specific gravities ; it
follows, that

w zz: ms, and w' zz ms' ;

but according to the third inference preceding, the difference between
the pressures downwards and upwards : —

Is equal to the difference between the weight of the solid
body, and that of an equal bulk of the fluid.

But the difference between the upward and downward pressures, is
equivalent to the force of ascent and descent ; consequently, we have

/zz: w^w'-mms^ms',
and this, by collecting the terms, becomes

/z=m(sv-s'). (182).

If, therefore, the specific gravity of the solid be less than that of
the fluid, the force of ascent will be

/z=m(s-/);

but when the specific gravity of the solid exceeds that of the fluid, the
force of descent becomes

/zz:m<X-s),

and when the specific gravities are equal to one another, the force of
ascent and descent vanishes, in which case, the body will remain at
rest, in whatsoever position it may be placed ; this agrees with what
we have already stated in the first inference to Problem 32.

From the above proposition and its subordinate formulae, the fol-
lowing inferences may be deduced.

252. INF. 1. When a solid body is immersed, or suspended in a
fluid of equal or of different specific gravity : —

It loses the weight of an equal magnitude of the fluid in
which it is placed.

This is obvious, for when the specific gravities are equal, the body
loses the whole of its weight ; and therefore, it neither endeavours to
ascend nor descend ; but when the specific gravities are unequal, the
body only endeavours to ascend or descend, by the difference between

OF FLOATATION AND THE SPECIFIC GRAVITY OF BODIES. 225

its own weight and that of an equal bulk of the fluid, and has there-
fore lost the weight of as much fluid.

253. INF. 2, When a solid body is immersed, or suspended in a
fluid of the same, or of different specific gravity : —

It loses the whole or a part of its weight, according as it
is totally or partially immersed, and the fluid gains the weight
which the body loses.

This is manifest, for the sum of the weights of the body and the
fluid must be the same, both before and after the immersion.

254. INF. 3. If bodies of equal magnitude are placed in the same
fluid, whatever may be their specific gravities : —

They lose equal weight, and unequal bodies lose weights
that are proportional to their magnitudes.

255. INF. 4. If the same body be immersed, or suspended in fluids
of different specific gravities: —

The weights lost by the body are as the densities or specific
gravities of the fluids.

256. INF. 5. When two bodies of unequal magnitude are in equilibrio
with one and the same fluid : —

They will lose their equilibrium, if they be transferred to
another fluid of different density.

257. INF. 6. When a body ascends or descends, in a fluid of greater
or less specific gravity than itself: —

The force which accelerates its ascent or descent, is equal
to the quotient that arises, when the difference between the
weight of the body, and that of an equal bulk of the fluid, is
divided by the common magnitude.

It consequently follows, that when the solid is entirely immersed in
the fluid, the force which urges its ascent or descent is constant; in
which case, the motion upwards or downwards must be uniformly
accelerated, if it be not disturbed by the resistance of the medium in
which it moves.

258. If the solid is specifically heavier than the fluid, it will tend
downwards, and press the bottom of the vessel, with a force which is
equivalent to the excess of its weight above an equal bulk of the
fluid ; and this is what we understand by the relative giavity of the
body in the fluid.

But if the body be specifically lighter than the fluid in which it is
placed, it seems to lose a greater weight than it actually possesses,

VOL. I. Q

226 OF FLOATATION AND THE SPECIFIC GRAVITY OF BODIES,

and consequently, it tends upwards, with a force equal to the dif-
ference between its own weight and that of an equal bulk of the
fluid ; and continuing to ascend, it will attain a position in which its
weight is equal to that of a quantity of the fluid of the same magni-
tude as the part immersed ; and this is what we understand by the
relative levity of the body in the fluid.

259. These inferences being admitted, we shall now proceed to
exemplify the general formula resulting from our proposition, viz.
that in which we have

The practical rule for reducing this equation, may be expressed in
general terms in the following manner.

RULE. Multiply the common magnitude of the body and
the fluid, by the difference of their specific gravities, and
the product will be the force of ascent or descent, according
as the specific gravity of the body is less or greater than that
of the fluid in which it is placed.

260. EXAMPLE. A mass of dry oak, whose magnitude is equal to
7 cubic feet, and specific gravity equal to 0.8 (that of water being
unity), is plunged into a vessel of fluid, whose specific gravity is
0.932 ; with what force will it ascend ?

Here, according to the rule, we have

/=: m (s — s'} =: 7(.932 — .8) — .924 of a cubic foot of the
body whose specific gravity is 0.932; consequently, for the force in
Ibs. avoirdupois, we have 1 : 62.5 : : 924 : 57.75 Ibs.

PROBLEM XXXIII.

261. In a vessel filled with an incompressible and non-elastic
fluid, is placed a hollow cylinder, which we shall consider as
being perfectly void of gravity or weight; to the bottom of
which, a cylindrical body of a given magnitude, and whose
specific gravity is greater than that of the fluid, is so closely
fitted that no fluid can enter : —

It is required to determine, how far below the surface of
the fluid the body will descend, before the tendency down-
wards, and the pressure upwards, are in equilibria with one
another.

Let A BCD in the annexed diagram, be a vertical section of the

OF FLOATATION AND THE SPECIFIC GRAVITY OF BODIES. 227

vessel containing the fluid, abed a corresponding
section of the hollow cylinder, and EFGH that of
the attached or cylindrical body.

Now, it is manifest, that in consequence of the
connection between the hollow cylinder and the
attached body, the downward pressure of the fluid
can have no effect upon that portion of the upper
surface of the body whose diameter is dc ; and be-
cause the hollow cylinder abed, is supposed to be
without weight, it can have no influence on the
downward tendency of the body EFGH : an equilibrium will therefore
obtain, when the downward pressure on the surface EeZ, CF, together
with the weight of the body, is equal to the upward pressure on the
bottom HG.

Put d — dc, the diameter of the hollow cylinder destitute of weight,
5 zz: EF, the diameter of the attached body,
I zz: EH, its perpendicular length, or vertical altitude,
a zz: the area of the end of the hollow cylinder,
A z± that of the attached cylindrical body,
s zz: its specific gravity, greater than that of the fluid,
s' zzz the specific gravity of the fluid »
w zz: the weight of the body,
p zz: the pressure on its upper surface,
p zz: the pressure on its base, and
x zz: e d, the distance below AB, the upper surface of the fluid.

Then, by the mensuration of surfaces, the area of the lower extre-
mity of the hollow cylinder abed, becomes «:z± .7854J2,

and that of the base of the attached body, is A zz: .78543* ;
and the difference of these, or the quantity of the upper surface of
the body, which is exposed to the downward pressure of the fluid, is

A — «z=.7854(32 — <Z»);

consequently, the downward pressure becomes p zz: .7854 (32 — c?2) s' x \
but the absolute weight of the body is expressed by its magnitude or
solidity, drawn into its specific gravity ; consequently the expression
for the weight of the attached body becomes tuzz: A/szzr.785432Js;
therefore, we have for the whole tendency downwards,

p 4- w zz: .7854 (32 — d*} s'x + .78543* /s,
and from this, by collecting the terms, we obtain
p 4- w = .7854 {(32— «P) s' x + tf Is},
Q 2

228 OF FLOATATION AND THE SPECIFIC GRAVITY OF BODIES.

Now, the pressure upwards on the bottom of the attached cylindri-
cal body, according to the principle of the first proposition, is

//=.78543V(J + a?) ;

but in the case of an equilibrium, or when the body has attained a
state of quiescence, the pressure upwards is exactly equal to the
downward tendency ; consequently, by comparison, we have

.78543V (l + x) = .7854 { (3* — d*) s' x + Vis} ;
therefore, by suppressing the common factor .7854, and transposing,

we get

3V (I + x} — (a2 — d2) s' x = V Is, and this,

by expanding and collecting the terms, becomes d?s'x=i&l(s — s'),
and finally, by division, we obtain x zz: — ~-f — . (183)

262. The following practical rule, drawn out in words at length,
will serve for the reduction of the equation.

RULE. Multiply the difference between the greater and
less specific gravities, by the square of the diameter of the
attached body drawn into its length ; then divide the product
by the square of the diameter of the hollow cylinder, drawn
into the specific gravity of the fluid, and the quotient will
be the distance below the surface of the fluid at which the
body rests.

263. EXAMPLE. A cylinder of lignum vitse, whose diameter is
8 inches, length 36 inches, and specific gravity 1.327, is attached to
the lower end of a hollow tube, whose diameter is 3 inches, in such a
manner that no fluid can enter ; now, supposing the body, and the
hollow cylinder to which it is attached, to be placed in a vessel full of
water, it is required to determine, at what distance below the surface
of the fluid the body will become quiescent ?

Here, by operating according to the rule, we obtain

82X36(1.327 — 1.000)
,= - ( - -' = 83.707 inches,

Hence it appears, that a cylinder of lignum vitae of the proposed
dimensions, will sink to the depth of 83.707 inches, or very nearly
7 feet below the surface of the water, before the upward pressure
becomes an equipoise for its downward tendency ; and this being
added to the three feet which it is in length, gives 10 feet for the
depth of the body of water, necessary for admitting an equilibrium,
under the specified conditions of magnitude and attachment.

CHAPTER X.

OF THE SPECIFIC GRAVITIES OF FLUIDS, AND THE METHOD OF
WEIGHING SOLID BODIES BY MEANS OF NON-ELASTIC FLUIDS.

THE specific gravity of a body is its weight compared with that of another body
of the same magnitude. The magnitude may be expressed by a number denoting its
relation to some standard generally used, and as a criterion of comparison, similar
to itself, as a cubical inch, a foot, &c. ; and if the criteria be different, as in solids
and fluids, the magnitudes of bodies are to each other as the criteria multiplied
into the numbers expressing these magnitudes. The clear and scientific expositions
which in the last chapter were given of absolute and relative weight, must have
well prepared the reader for entering upon the doctrine of specific gravities, which
distinguishes different species of matter from each other, in one of their most
obvious properties, namely, the weight of matter contained in a given space. The
weight of any portion of matter is easily ascertained ; but it is not always easy to
measure the space occupied by a body, or its magnitude; and in some instances
this cannot well be effected without artificial means. We employ for this purpose
distilled water, the specific gravity of which, or weight of a given bulk, is nearly
at all times the same. Adopting, therefore, this pure homogeneous substance as
our criterion or unit of measure, by comparing it with other substances the ratio of
their specific gravities may be easily discovered ; and denoting the specific gravity
of water by any number taken at pleasure, the numbers expressing the specific
gravities of other bodies are hence given, or, at least, assignable.

PROPOSITION V.

264. When a solid body is immersed in a fluid of different
specific gravity from itself: —

The weight which the body loses, will be to its whole
weight, as the specific gravity of the fluid is to the specific
gravity of the solid.

This is a very important proposition in hydrostatics, but its demon-
stration does not require the assistance of a diagram ; we must there-
fore endeavour to establish its validity by the application of symbolical
arithmetic ; for which purpose —

230 OF FLOATATION AND THE SPECIFIC GRAVITY OF BODIES.

Put m nr the magnitude of the immersed solid,
w ~ its weight,

w"~ the weight lost by the pressure of the fluid,
w' — the weight of a quantity of the fluid, of the same bulk as

the solid tody, -

s — the specif of* the solid, and „
s' in the specific gravity of the fluid.

Then, because, as we have already stated in a former part of this
work, the wejghts are directly as their magnitudes drawn into their
respective* gravities ; it follows, that

A j / /

w=.ms, and w ~ms ;

consequently, by analogy, we obtain
w' : w : : ms' : ms.

Now, when a body is immersed in a fluid of less specific gravity
than itself, it obviously must descend, in consequence of its superior
gravity, and the force of descent is equal to the difference between
the weight of the solid body and the weight which it loses by the
action of the fluid ; but it has been shown in the preceding propo-
sition, that the force of descent is equal to the difference between the
weight of the solid, and that of an equal bulk of the fluid ; conse-
quently, if f denote the force of descent, then in the one case, we have

f—w — w'9
and in the other case, it is

f—w — w";
therefore, by comparison, we obtain

w — w' ~ w — 10" ;
hence, by expunging w, it becomes

w' — w".

If, therefore, we substitute w" instead of w' , in the first term of the
above analogy, we shall get

w" : w : : ms* : ms : : s' : s ;
hence the truth of the proposition is manifest.

The part of the weight which the body loses by descending in the
fluid, is not annihilated, it is only sustained by the upward pressure
of the fluid opposing the descent of the body ; and this is the reason
why the weight of a vessel full of water is not perceptible while it is
beneath the surface.

OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES. 231

By equating the products of the extreme and mean terms in the
preceding analogy, we obtain

sw" — s'w,

and dividing by s, we have
s'w

but according to our notation, w" denotes the weight which the body
loses ; consequently, the weight which it retains in the fluid, becomes

s'w w (s — s')

W W'-^HW m - . /1Q/|\

s . s (184)-

265. If, therefore, the weight of the body, together with its specific
gravity, be known, before it is immersed in a fluid of a given specific
gravity ; its weight after immersion can easily be ascertained by the
following practical rule.

RULE. From the specific gravity of the body, subtract that
of the fluid in which it is immersed ; multiply the remainder
by the weight of the body, and divide the product by its
specific gravity for the weight which it retains after im-
mersion.

266. EXAMPLE. A piece of cast iron which weighs 14 Ibs. is
plunged into a cistern of water; what force will be required to
sustain the iron at rest in any point, its specific gravity being to that
of water as 7 to 1 ?

Here, by operating according to the rule, we have

14(7—1)

— - ~ 12 Ibs. avoirdupois.

From which it appears, that 14 Ibs. of cast iron being suspended
in water, loses 2 Ibs. of its weight ; or which is the same thing, the
upward pressure of the water exceeds its downward pressure, by a
force which is equivalent- to 2 Ibs.

COROL. We may also infer from the above, that the weights which
the same body loses by being immersed in different fluids : —
Are as the specific gravities of the fluids.

PROBLEM XXXIV.

267. If a body be weighed in air and in water respectively,
and the weights be exactly ascertained : —

It is required, from the weights thus exhibited, to deter-
mine the real weight.

232 OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES.

Here again, we are under the necessity of dispensing with the
assistance of a diagram, the investigation being wholly analytical ; in
order therefore to proceed,

Put w = the weight of the body when weighed in water,
w'~ the weight when weighed in air,

s — the specific gravity of water, generally expressed by unity,
s* nz the specific gravity of air, and
x rzz the real weight of the body required.

Then we have x — w' ', and x — w> for the weights which the body
loses in air and in water ; but we have deduced it as an inference
from Proposition V., that when the same body is weighed in different
fluids, it loses weights in proportion to the specific gravities of the
fluids in which it is weighed ; consequently, we have

x — w' : x — w : : s' is;

therefore, by making the product of the mean terms equal to the
product of the extremes, we have

s (x — w') — s' (x — w),
and from this, by separating the terms, and transposing, we get

(s — s') x ~ s w' — s'w;

consequently, by division, we obtain

__sw' — s' w

s — s' (185).

268. The equation in its present form, supplies us with the follow-
ing practical rule for its reduction.

RULE. Divide the difference between the products of the
alternate weights and specific gravities, by the difference of
the specific gravities, and the quotient will be the real weight
of the body.

269. EXAMPLE. A certain body when weighed in water and in air,
is found to equiponderate 12 and 13.9975 Ibs. respectively; what is
its real weight, the specific gravities of air and water being as 1 to
.0012?

Here, by operating as directed in the rule, we have
1X13.9975 — .0012X12

x —

— .0012

— 14 Ibs.

From which it appears, that a body of 14 Ibs. avoirdupois, will
completely fulfil the conditions of the question.

OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES, 233

PROBLEM XXXV.

270. If the weights which a body indicates, when weighed
in air and in water, are exactly ascertained : —

It is required from thence to determine the specific gravity
of the body, the specific gravities of air and water being
known.

Here also, as in the case of the preceding Problem XXXIV., and
Proposition V., the aid of a diagram is not required ; for it would be
totally inconsistent with scientific precision, to denote the specific
gravities of bodies by geometrical magnitudes,

Put W i= the real weight of the solid body,
w =z the weight when weighed in water,
w' zz the weight when weighed in atmospheric air,
s =. the specific gravity of water, expressed by unity,
s' •=. the specific gravity of air, and
S = the required specific gravity of the solid body.

Then, according to the principle announced and demonstrated in
the 5th proposition, we have

W — w : W ::s : S ;

where it is manifest, that W— w; expresses the weight which the body
loses by being weighed in water ; therefore, we have
W — (W— -w) : W : : S—s : S;
or by equating the products of the extremes and means, we get

Sw = (S—s)Vf;

and by proceeding in a similar manner when the body is weighed in
air, we obtain

Now, from the first of these equations, we have

and from the second, it is

W = (S=7);

hence by comparison, we obtain
Sw Sw'

234 OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES,

and by taking away the denominators, we get

Sw — s'w~Sw' — s uf,
from which, by transposing and collecting the terms, we obtain

(10* — w)S-=.sw' — s'w,
and finally, by division, we have

sw' — s' w

'' w' — w ' (186).

271. The practical rule or method of reducing this equation, may
be expressed in words in the following manner.

• - RULE. Divide the difference between the products of the
alternate weights and specific gravities, ly the difference of
the weights when weighed in air and in water, and the
quotient will express the specific gravity of the body.

272. EXAMPLE. A certain body when weighed in water indicates
exactly 12 Ibs. avoirdupois ; but when the same body is weighed in
air, it indicates 13.9975 Ibs.; required the specific gravity of the
body, the specific gravities of water and air being as in the preceding
problem, or as 1 to .0012 ?

The process performed according to the directions given in the
rule, or after the manner indicated in equation (186), will stand as
follows.

1X13.9975— .0012X12

13.9975 — 12

Therefore, a body whose specific gravity is seven times the specific
gravity of water, will fulfil the conditions of the question. -

PROBLEM XXXVI.

273. If the weights which a solid body indicates, when weighed
in air and in water, together with its specific gravity and real
weight, are exactly ascertained : —

It is required from thence, to determine the magnitude of
the body, on the supposition that it is globular.

If the specific gravity of the body and its real weight were unknown,
the solution of the present problem would include that of the two pre-
ceding ones ; but in order to abbreviate the investigation, we have
supposed the specific gravity and the real weight of the body to be
given ; the process of the solution is therefore as follows.

OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES. 235

Put W zz the real weight of the globular body,
S zz its specific gravity,
w zz the weight which the body indicates when weighed in

water,

s zz the specific gravity of water,

vf zz the weight which the body indicates when weighed in air,
s' zz the specific gravity of air, and
d zz the required diameter of the solid body.

Then, according to the principles of mensuration, the solidity of a
globe is expressed by the cube or third power of its diameter, multi-
plied by the constant decimal .5236 ; therefore, we have
dx dxd* .5236 zz .5236d8 ;

but it has been stated in a former part of this work, that the absolute
weight of any body, is expressed by its magnitude drawn into the
specific gravity ; hence we have

Wz=.5236Sd3;
consequently, by division, we obtain

and from this, by extracting the cube root, we get

.52365* (187).

274. The equation in its present form, expresses the diameter of
the body in terms of its absolute weight and specific gravity ; this is
certainly the simplest and only mode of determining the magnitude
of any body or quantity of matter, when the weight and specific
gravity are known a priori ; but when this is not the case, we must
have recourse to other methods ; and a very elegant and simple one,
consists in weighing the body in water and in air, as implied in the
problem, and then proceeding as follows.

By equation (185), Problem XXXIV., it appears, that the real or
absolute weight of the solid, expressed in terms of its relative weights,
and the specific gravities «of the fluids in which it is weighed, viz. water
and air, is

!j£/-V^
s— s1 '

and by equation (186), Problem XXXV., the specific gravity of the
solid expressed in terms of the same quantities, is

236 OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES.

_ sw* — s'w

w' — w

But the real or absolute weight of any body, is expressed by its
magnitude drawn into the specific gravity ; consequently, we have

s w> — s/ w

W - W

let this value of the real weight be compared with that above, and we
shall have

.5236 d3 (s w' — s'w) _ sw' — s' w

w — w s — s'

If the expression (sw' — s'w) be suppressed on both sides of the
above equation, we shall obtain

.5236rf»_ 1
w' — w s — s' '
and again, by suppressing the denominators, we get

.5236(5 — t')d* = u/ — w;
therefore, dividing by . 5236 (s — 5'), we have

".5236(5 — 57
and finally, by extracting the cube root, we obtain

.5236(5 — 5') (188).

275. Now, the methods of reducing the equations (187) and (188),
or the practical rules derived from them, may be expressed as follows.

1. When the absolute weight and specific gravity are given.

RULE. Divide the absolute weight of the body, by .5236
times the specific gravity, and the cube root of the quotient
will be the diameter of the solid sought.

2. When the weights indicated by the body in water and in air

are given.

RULE. Divide the difference between the weights, as obtained
from weighing the body in air and in water, by .5236 times
the difference between the specific gravities of water and air ;
then, the cube root of the quotient will be the diameter of the
solid sought.

276. EXAMPLE 1. The absolute weight of a globular body is 14 Ibs.*
and its specific gravity 7 ; what is its diameter ?

OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES. 237

This example corresponds to equation (187), and must therefore
be resolved by the first case of the foregoing rule, observing to bring
the numerator into the same denomination with the denominator, that
is, reducing Ibs. avoirdupois into ounces; or thus, 14Xl6:zrW, the
absolute weight, from which we get

Hence it appears, that a globular body, whose specific gravity is
seven times greater than that of water, will weigh 14 Ibs. when its
diameter is 3.9313 inches, which corresponds very nearly with a globe
of cast iron.

277. EXAMPLE 2. A globular body whose specific gravity and
absolute weight are unknown, indicates 12 Ibs. avoirdupois when
weighed in water, and 13.9975 Ibs. when weighed in air; what is its
magnitude, the specific gravity of water and air being to one another
as the numbers 1 and .0012 ?

This second example corresponds to the conditions represented in
equation (188), and must therefore be resolved by the second case of
the foregoing rule, the numerator being brought into the same deno-
mination with the denominator, or the Ibs. avoirdupois being turned
into ounces, as (13.9975 — 12) 16, from which we obtain

l-3-^-7^=r=^^4^^§9i»eg nearly 4 inches, the same $**,
.5236(1— .0012)
as above.

From the principles established in the foregoing Proposition (V), and
the problems derived from it, we deduce the following inferences.

278. INF. 1. When bodies of equal weights, but of different magni-
tudes, are immersed in the same fluid : —

The weights which they lose, are reciprocally proportional
to their specific gravities, or directly proportional to their

279. INF. 2. When a solid body is weighed in air, or in any other
fluid whatever :

The difference between its absolute weight, and the weight
exhibited in the fluid, is the same as the weight of an equal
bulk of the fluid.

280. INF. 3. If two solid bodies of different magnitudes, when
weighed in the same fluid indicate equal weights : —

The greater body will preponderate when they are brought
into a rarer medium.

238 OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES.

281. INF. 4. If two solid bodies of different magnitudes, indicate
equal weights when weighed in the same fluid : —

The lesser body will preponderate when they are placed in
a denser medium.

282. INF. 5. If two or more solid bodies, when placed in the same
fluid, sustain equal diminutions of weight : —

The magnitudes of the several bodies are equal among
themselves.

This is manifest, for the losses of weights are as the weights of the
quantities of fluid displaced ; and these are as the magnitudes of the
bodies which displace them.

PROBLEM XXXVII.

283. If two bodies of equal weights, but different specific
gravities, be exactly equipoised in air, and then immersed in a
fluid of greater specific gravity, the smaller body will prevail : — ••

It is therefore required to determine, what weight must
be added on the part of the greater body, to restore the
equilibrium. • •

Put s zz: the specific gravity of the fluid, in which the bodies are

immersed, after being equipoised in air,
/ zz: the specific gravity of the greater body,
s" zz: the specific gravity of the smaller body,
w — the common weight of each,
w' zz: the weight lost by the greater body, by reason of the

immersion,

w" zz: the weight lost by the lesser body, and
x zz: the weight which must be added to restore the equi-
librium.

Then, because by the preceding proposition, when a body is im-
mersed in a fluid, the weight which it loses, is to its whole weight, as
the specific gravity of the fluid is to that of the body ; it follows,

that

sr : s : : w : w' ',

and this, by reducing the proportion, gives w' zz:— —•>

Again, for the weight lost by the lesser body, we have

s" : s : : w : w" :

OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES. 239

which by reduction gives
ws

»=-F-

Now, it is manifest, that the weight required to restore the equili-
brium, must be equal to the difference between the results of the
above analogies ; therefore, we obtain

H __ w s ws

~~7 7r '

which, by a little farther reduction, becomes

X = 77' ' (189).

284. The practical rule which this equation affords, may be ex-
pressed in words at length in the following manner.

RULE. Multiply the difference between the specific gravities
of the bodies, by their common weight in air, drawn into the
specific gravity of the fluid in which they are immersed, and
divide the result by the product of the specific gravities of the
bodies, for the weight to be added in order to restore the

It may be proper here to observe, that the weight determined by this
rule must not be immersed in the fluid, it must only be attached to
that side of the balance on which the greatest weight is lost.

285. EXAMPLE. Suppose that 84 Ibs. of brass, whose specific gravity
is 8.1 times greater than that of water, is equipoised in air by a piece
of copper, whose specific gravity is 9 times greater than that of water ;
how much weight must be applied to the ascending arm of the balance
to restore the equilibrium, the same being destroyed by immersing the
bodies in water, of which the specific gravity is expressed by unity ?
Here, by attending to the directions in the rule, we get

84X1X(9 — 8.1) ..

*= 8.1X9 =l-0371bB.

Hence it appears, that if a mass of brass and of copper, each equal
to 84 Ibs. when weighed in air, be immersed in a vessel of water, the
copper will preponderate, in consequence of its greater specific gravity;
and in order that the equilibrium may be again restored, a weight of
1 .037 Ibs. must be attached to the ascending arm of the balance, or
that from which the brass is suspended .

240 OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES*

PROBLEM XXXVIII.

286. If two bodies of different, but known specific gravities,
equiponderate in a fluid of given density : —

It is required to determine the ratio of the quantities of
matter which they contain.

Put 5 — the specific gravity of the fluid, in which the bodies are
found to equiponderate,

m zz the magnitude of the greater body,

s' zz its specific gravity,

m'zz the magnitude of the lesser body,

s" zz its specific gravity,

w zz the weight of the greater body in the fluid, and

w/zz the weight of the lesser body under the same circum-
stances.

Then, by Proposition V., when a solid body is immersed in a fluid
of different specific gravity, the weight which it loses, is to its whole
weight, as the specific gravity of the fluid, is to the specific gravity of
the solid ; it therefore follows, that

/ : s : : m s' : m s zz the weight lost by the greater body ;
but the weight of the body in the fluid, is manifestly equal to the
difference between its absolute weight, and that which it loses in
consequence of the immersion ; hence we have

w zz m s' — m s zz m (s' — s) ;
and by a similar mode of procedure, we obtain
s" : s : : m1 s" : m' s zz the weight lost by the lesser body;
consequently, the weight which it possesses in the fluid, is

w' zz m1 s" — m' s zz m' (s" — s).

Now, according to the conditions of the problem, these are in
equilibrio with one another; therefore by comparison, we have

m (s' — s) zz m'(s" — s),
and by converting this equation into an analogy, it is

m : m' : : (s"— s) : (s'— s);

and finally, if we multiply the first and third terms by s', and the
second and fourth by s", we shall have

ms' : m's11 : : s'(s"— s} : s"(s'—s).

287. EXAMPLE. Twenty ounces of brass, whose specific gravity is
eight times greater than that of water, and a piece of copper whose

OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES. 241

specific gravity is nine times greater, are in equilibrio with one another
in a fluid whose specific gravity is unity ; required the weight of the
copper ?

Here we have given m s' = 20 ounces ; / = 8 ; s" — 9, and s — \ ;
consequently, by substitution, the above analogy becomes

20 : m's" : : 8(9—1) : 9(8—1);
and by equating the products of the extremes and means, we get

64mVzzl260,
and dividing by 64, we have

m1 s" in =19-f^- ounces.

It therefore appears, that 20 ounces of brass and 19|^ ounces of
copper, are in equilibrio with each other, when immersed in a fluid
whose specific gravity is unity ; but if put into a fluid of greater
•density, the copper will prevail.

PROBLEM XXXIX,

288. Suppose a cylinder and cone, of the same altitude, base,
and specific gravity, to balance each other at the extremities of
a straight lever, when immersed in a fluid of given density ; the
cone being suspended at the vertex, and the cylinder at the
extremity of the axis. Now, suppose a cone equal to the one
proposed, to be abstracted from the cylinder, and its place sup-
plied by another of the same magnitude and half the specific
gravity ; it is manifest that in this state, the cone will prepon-
derate : —

It is therefore required to determine, how much must be

taken from the cone, in order that the equilibrium may be

again restored.

Let AB be a straight inflexible lever, supported upon and easily
moveable about the fulcrum F, and
let the cone CDE and the cylinder
GHIK, (equal in altitude, base, and
specific gravity,) be suspended from
the extremities at A and B.

Then it is manifest, that in conse-
quence of the equality of the bases ,y.
and altitudes, the magnitude of the cylinder is equal to three times
the magnitude of the cone ; and since the specific gravities of the

VOL. i. B

242 OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLED BODIES.

bodies are the same, it follows also, that the weight of the cylinder is
equal to three times the weight of the cone ; consequently, by the
principles of the lever, the length of the arm AF, is three times the
length of the arm BF; for it is a well known property in the doctrine
of mechanics, that when two bodies of different weights are in equi-
librio on the opposite arms of a straight lever : —

The lengths of the arms are to each other, reciprocally as
the weights of the suspended bodies.

Now, suppose the cone LKI, which is obviously equal in magnitude
to CDE, to be abstracted from the cylinder, and to have its place sup-
plied by another cone of half the specific gravity as the former ; then
it is evident, that if the cone CDE is suffered to retain its magnitude,
it will preponderate and cause the cylinder to ascend ; it is therefore
necessary, in order that the equilibrium shall'not be disturbed, to
diminish the magnitude, and consequently the weight of the equili-
brating cone; and for the purpose of assigning the quantity of
diminution,

Put m m the magnitude of the conical body CDE,

m'~ the magnitude of the cylindrical body GHIK,

m"-=. the magnitude of the remaining portion cab,

w 1=1 the weight which the cone loses in the fluid,

w' •=. the weight lost by the cylinder,

w"— the weight lost by the remaining cone cab,

s =n the specific gravity of the fluid, and

s' =i the specific gravity of the cone and cylinder.

Then, since the weight which a body loses by being immersed in a
fluid, is to its whole weight, as the specific gravity of the fluid is to
the specific gravity of the body, we have

s' : s : : m s' : w ;

therefore, by equating the products of the extremes and means, it is
wmmsi=. the weight lost by the cone ; but according to the
principles of mensuration, the magnitude of a cylinder is equal to
three times the magnitude of a cone of the same base and altitude ;
consequently, we have

m' nr 3m,

and for the weight lost by the cylinder, we get

s' : s : : 3ms1 : w' ;

Horn which, by equating the product of the extremes and means, we
obtain

w/zr 3ms~m's,

OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES. 243

and in like manner, the weight lost by the cone cab, is found to be

w"—m"s.

But the weights which the several bodies possess in the fluid, are
manifestly equal to the difference between the absolute weights and
the weights lost ; and the absolute weights are equal to the magni-
tudes drawn into the specific gravities ; therefore, we have

ms' — ms zz m (s' — 5) zz the weight of the cone in the fluid,

3ms' — 3ms—3m(s' — s)zzthe weight of the cylinder,

m"s' — w" szzm" (sr — s) zz the weight of the remaining cone.

Now, if from the weight which the cylinder possesses in the fluid,
we subtract the corresponding weight of the cone, and to the remainder
add the weight of another cone of equal magnitude and half the
specific gravity ; then, the reduced weight of the cylinder in the fluid
becomes

2ws' 4- J*w*' — 3mszzwi(2Js' — 3s).

But according to the conditions of the problem, this weight is to
be in equilibrio with the weight of the remaining cone; therefore, by
the property of the lever, we have

ro(2js' — 3s) : m"(s' — s) : : 3 : 1 ;

and from this, by equating the products of the extremes and means,
we get

3m" (s' — s)=m (2 Is' — 3s),

in which equation ra" is unknown ; in order ^therefore to determine its
value, divide both sides of the equation by 3(s' — 5), and it becomes

6(s' — s) ' (190),

But this that we have determined, is the magnitude of the part
which remains, whereas the problem requires the magnitude of the
part to be cut off; now, the magnitude of the whole cone is m ; con-
sequently, by subtraction, we have

«__ m(5s' — 65) ms'

- 6(s--s) :- 6(7=0' 09')-

289. The practical rule for reducing this equation is very simple,
it may be expressed in the following manner.

RULE. Multiply the magnitude of the cone by its specific
gravity, and divide the product by six times the difference
between the specific gravity of the cone and cylinder, and that
of the fluid, and the quotient will give the magnitude of the
part to be cut off, in order to restore the equilibrium^

R 2

244 OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES,

290. EXAMPLE. Suppose that a cone and cylinder of copper, whose
specific gravity is nine times greater than that of water, are immersed
in a fluid, whose specific gravity is 1.85, and placed under the con-
ditions specified in the problem ; how much must be cut from the
lower part of the cone to restore the equilibrium, the diameter of the
bases and the altitude of the cone and cylinder being respectively
2 and 5 inches ?

Here, by operating according to the rule, we shall have

5.236* X 9 47.124
= 6(9=L85)=:1^

the part to be cut off from the cone, in order that the remainder may
equipoise the cylinder; or we may calculate the magnitude of the
equipoising cone by equation (190), in the following manner.

5.236(5X9 — 6X1.85)

m"— ^— — ~ i—4.135 cubic inches;

o(y — l.oo)

which, by subtraction, gives 1.098 cubic inches for the part to be
cut off.

Put d •=. the diameter of the base of the cone and cylinder, and
h ~ the common height or altitude.

Then, the equations (190) and (191) will become transformed, in
terms of the dimensions, into those that follow, viz.
,,__.2618<fA(5s'— 6s)

~6(s'-5) (192).

This equation expresses the magnitude of the cone which restores
the equilibrium, and the following one expresses the magnitude of the
frustum which has to be deducted, in order that the equilibrium may
obtain ; that is,

.2618^5'

6(sf — s)' (193).

The above is a better mode of expressing the quantities, than that
exhibited in equations (190) and (191) ; since it is not probable, that
the magnitudes or solid contents of the bodies will be proposed, with-
out having previously stated their linear dimensions.

It would be superfluous to propose an example, for the purpose
of illustrating the reduction of the equations in their modified form ;
for since the expression .261 8e?2 A, is equivalent to the magnitude of

* The number 5.236 is that which expresses the magnitude of the cone, for
22X. 7854X5X$=5.236.

OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES. 245

the cone, the rule in words would be the same in both cases, and
there fore, 4t need not be repeated.

PROBLEM XL.

291. If a solid body be weighed in vacuo and in a fluid, and
the different weights correctly noted : —

It is required from thence, to compare the specific gravities
of the solid, and the fluid in which it is immersed.

The solution of this problem is extremely easy, for the difference
between the weight of the body in vacuo and in the fluid, gives the
weight lost ; therefore,

Put w zr the weight indicated by the body when weighed in vacuo,
w'izr the weight when weighed in the fluid,
s — the specific gravity of the fluid in which the body is

weighed, and
s' zz the specific gravity of the body.

Then we have w — w' == the weight which the body loses by being
weighed in the fluid ; therefore, by the fifth proposition, we obtain
w — w' : w : : s : s' ; that is
w — w' s

~^r -T' (194).

consequently, since the one ratio is given, the latter can be found.

292. EXAMPLE. Suppose a piece of metal to indicate 40 ounces
when weighed in vacuo, and 35 ounces when weighed in water ; what
is the specific gravity of the metal ?

Here, by substituting the given numbers in equation (194), we get
4Q_ 35__1 _s_^
40 " 8 "~ s ;

hence, the specific gravity of the solid, is eight times greater than the
specific gravity of water.

PROBLEM XLI.

293. If two solid bodies be weighed in vacuo and in a fluid,
and the different weights correctly noted : —

It is required from thence, to compare the specific gravities
of the bodies.

246 OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES.

The intelligent reader will readily perceive, that the present
problem is only an extension of that which immediately pj^cedes it,
and is proposed with the design of detecting the law, by which the
specific gravities of different bodies are compared ; for which purpose,

Put W = the weight of the heavier body when weighed in vacuo,
Wizr its weight when weighed in the fluid,
w — the weight of the lighter body when weighed in vacuo,
wf zn its weight when weighed in the fluid,
s = the specific gravity of the fluid in which the bodies are

weighed,

s' zz: the specific gravity of the heavier body, and
s" zz: the specific gravity of the lighter.

Then, the weights which the bodies lose by being weighed in the
fluid, are respectively W — W, and w — w', and because the weight
lost, is to the whole weight, as the specific gravity of the fluid is to
the specific gravity of the solid ; it follows in the case of the heavier
body, that

W — W : W ::s : s'f

and in the case of the lighter body, it is

w — w : w : : s : s" ;

therefore, by equating the products of the extreme and mean terms in
each of these analogies, we have

s' ( W— W) = s W, and *" (w — w') zz: s w ;

consequently, by division, we obtain

sW sw

-, and s zz:

• W— W " w — w'

hence, by analogy, we have

' sW " sw

W— W : w — w"

and finally, by suppressing s, it becomes

W w

f * • •»'/. _______

W— W ' w — w''

Hence it appears, that the specific gravities of the two bodies, are
to one another, as their absolute weights divided by the weights which
they lose in the fluid; and it is manifest, that the same law will
extend to any number of bodies whatever ; therefore, the method of
comparison has been determined.

OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES. 247

294. EXAMPLE. A solid body whose absolute weight is 23 ounces,
when weighed in a certain fluid, loses 3 ounces of its weight; and
another body of 800 ounces, when weighed in the same fluid, loses
1 02 ounces ; what is the ratio of their respective gravities ?

Here, since the loss of the one is 3 ounces, and that of the other
102 ounces, it follows, that the specific gravities of the bodies, are to
one another as the numbers 391 and 400 ; for we have

And in like manner, if three or more bodies be weighed in the
same fluid, their specific gravities may be compared with one another,
and also with that of the fluid in which they are weighed.

PROBLEM XLII.

295. If a solid body of known specific gravity, be weighed in
several different fluids, and the weights correctly indicated : —

It is required from thence, to determine the ratio of their
respective gravities.

This problem, it will readily be perceived, is exactly the reverse of
the preceding one, and therefore, the method of its solution may easily
be discovered ; it is, however, of equal utility in philosophical inqui-
ries, for which reason we have proposed it in this place.
Put Wzz the weight of the solid when weighed in vacuo,
s zz the specific gravity of the solid,
w zz the weight which it indicates in a fluid whose specific

gravity is s', and

w' zz the weight which it indicates in a fluid whose specific
gravity is s\

Then the weights which the body loses, by being weighed in the
two fluids, are respectively W — w and W — w' ; but the weight lost,
is to the whole weight, as the specific gravity of the fluid is to the
specific gravity of the solid ; hence, for the first fluid, we have

W — w: W ::s':s;

from which, by equating the products of the extremes and means,
we get

s'Wzzs(W — w);

therefore, by division, we obtain

_*(W—

(195).

248 or SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES.

Again, for the second fluid, we have

W— w' : W ::«" : s,

and from this, by equating the products of the extremes and means,
we get

f*W=r*(W — «**),
and this by division becomes

W-tiQ

~W~ (196).

hence, by analogy, we obtain

w v

and finally, by suppressing the common quantities in the second and
fourth terms, we get

,': W — w :: 5" : W — w/;

that is, the specific gravities of different fluids, are as the weights,
which the body loses.

296. EXAMPLE. A mass of brick whose absolute weight is 64
ounces, and its specific gravity equal to twice the specific gravity of
water; when weighed in one fluid indicates 37 ounces, and when
weighed in another, it indicates only 30 ounces ; it is required from
thence, to determine the ratio of the respective gravities of the fluids,
and also the specific gravity of each ?

Here it is manifest, that the weight lost by the solid when weighed
in one fluid, is

W — w — 64 — 37 z= 27 ounces,

and on being weighed in another fluid, it loses

W — w' = 64 — 30 — 34 ounces.

Now, we have seen above, that the specific gravities of the fluid
in which the solid is weighed, are to one another, respectively as the
weights which the solid loses in them ; consequently, we have

s' : s" : : 27 : 34.

This is the ratio of the specific gravities ; but it appears from equa-
tions (195) and (196), that when the specific gravity of the solid is
known, the specific gravity of the fluid in which it is weighed can
easily be ascertained.

If, therefore, we employ the specific gravity of the body as given in
the question, the specific gravity of the first fluid, by equation (195),
becomes

OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES. 249

and for the specific gravity of the second fluid, it is
,.=*(M-3°) = 1.062 nearly.

Hence, the absolute specific gravities of the fluids, are respectively
equal to 0.844 and 1.062, that of water being unity ; and if we refer
to a table of the specific gravities of known fluids, we will find these
numbers to correspond with alcohol and acetic acid, the two fluids in
which the brick is weighed.

It is manifest, that what has been done above in respect of two
fluids, may be extended to any number of fluids whatever, the law,
and the method of determining the specific gravity, being the same
in all.

PROBLEM XLIII.

297. If a body be weighed in air, and again in a vessel filled
with water, the weight of the vessel and water being known : —

It is required from thence, to determine the specific gravity
of the body.

The principle upon which the solution of this problem depends is
not so evident; it may, nevertheless, be briefly expounded in the
following manner.

The body is first weighed in air ; then being put into a vessel filled
with water, the weight of which is known, it will expel a quantity of
the fluid equal to its own bulk, and because the specific gravity of the
body is supposed to be greater than that of water, it is obvious, that
the vessel and its contents will now be heavier than it was before the
body was put into it, by a quantity, which is equal to the difference
between the weight of the body, and that of an equal bulk of the
water ; but the body loses as much of its own weight in the fluid, as
is equal to that of the water displaced ; hence, the determination of
its specific gravity becomes easy, for which purpose,

Put w zz the weight of the body when weighed in air,

w' zz .the weight of the vessel and the water before the body

is put into it,

w" zz the weight of the vessel, the water, and the solid, and
5 zz the specific gravity of the solid.

250 OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES.

Then the weight gained by the vessel, by reason of the immersion
of the solid body, is w" — wf, and this expresses the weight of the
body in the fluid ; consequently, the weight which the body loses, is
w—(w"—w').

But by the 5th Proposition preceding, the weight lost, is to the
whole weight, as the specific gravity of the fluid is to the specific
gravity of the body ; therefore, because the specific gravity of water
is expressed by unity, we have

w — (w" — w') : w : : I : s ; that is

-w_.(w»_-f£,')- (197).

298. The practical rule for reducing the above equation, may be
expressed in words at length in the following manner.

RULE. Divide the weight of the body in air, by the difference
between that weight, and what is gained by the vessel in con-
sequence of the immersion, and the quotient will express the
specific gravity of the solid.

299. EXAMPLE. A solid body when weighed in air, indicates a
weight of 16 ounces; and when put into a vessel filled with water,
the vessel, the solid and the water together, indicate a weight of 36
ounces ; whereas the vessel when filled with water alone weighs only
32 ounces ; required the specific gravity of the body, that of water
being expressed by unity ?

Here, by following the directions of the rule, we have
_ 16 _

""16 — (36 — 32) -

being a measure of specific gravity, which corresponds very nearly
with American ebony, a very suitable material for hydrostatical expe-
riments.

300. We have hitherto been considering the nature of bodies that
are specifically heavier than the fluids in which they are weighed, and
consequently, such as would sink to the bottom, if they were left to
the free action of their own gravity ; we have therefore, in the next
place, to consider such bodies as are specifically lighter than the fluids
on which they are placed, and consequently, such as would float on
the surface, if left to the free exercise of their own buoyancy.

This is a very abstruse, but interesting and important department of
Hydro-Dynamical science ; for on it depends the principles by which
we determine the conditions of equilibrium, and the stability of floating
bodies.

OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES. 251

PROBLEM XLIV.

301. If a solid body is weighed in a fluid of greater specific
gravity than itself, and of which the specific gravity is given : —

It is required from thence, to determine the specific gravity
of the solid.

The usual method of resolving this problem, is, to attach the body
whose specific gravity is required, to another body specifically heavier
than the fluid, and of a sufficient magnitude to cause the compound
mass to sink ; then, by observing the weights indicated by the sub-
sidiary body, and by the compound mass, when they are separately
placed in the fluid, the specific gravity of the lighter body will become
known.

Put rv zz the weight of the lighter body when weighed in vacuo,
TV zz the difference between the weight of the compound mass,
and that of the heavier body, when weighed separately
in the fluid,

s zz: the specific gravity of the fluid, and
s' zz the specific gravity of the solid required.

Now, it is manifest, that the effort of buoyancy, or the force of
ascent of the lighter body, is equal to the difference between the
weight of the compound mass in the fluid, and that of the heavier
body in the fluid ; therefore,

w' zz: the force of ascent of the lighter body.

But, the force of ascent of the lighter body, or, as it is more ele-
gantly denominated, the effort of buoyancy, is evidently equivalent to
the excess of the weight of a quantity of the fluid, equal in magnitude
to the lighter body, above the weight of the lighter body when weighed
in vacuo ; consequently, the weight of a quantity of the fluid, equal
in bulk to the lighter body, is expressed by rv -\- w' ; hence, we have

from which, by reducing the proportion, we get
. sw

rV-\-n'

Or put W zz the weight of the compound mass when weighed in the
fluid, and W zz: the corresponding weight of the heavier body when
weighed separately ; then we have

/ -\\r \\rt .

rv — w — vv ,

252 OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES.

let this value of the force of ascent, be substituted instead of it, in the
above value of s', and we shall obtain

-w_j_W— W" (198).

302. The practical rule for reducing this equation, may be expressed
in words at length in the following manner.

RULE. Multiply the absolute weight of the body, of which
the specific gravity is required, by the specific gravity of the
fluid; then, divide the product by the absolute weight of the
body, increased by its force of ascent, and the quotient will be
the specific gravity sought.

303. EXAMPLE. A piece of wood which weighs in vacuo 22 ounces,
is attached to a piece of metal of such a magnitude as to weigh 12
ounces in water ; now, supposing that when the compound mass is
placed in water, it is found to weigh 20 ounces ; what is the specific
gravity of the wood, that of water being expressed by unity?

Here, by proceeding according to the rule, we get

and this, by referring to a table of specific gravity, is found to cor-
respond very nearly with the medium species of citron wood.

PROBLEM XLV.

304. Having given the weight of a vessel full of water, both
before and after a body of a given weight in air is immersed
in it, together with the specific gravity of the air at the time of
observation : —

It is required to determine the specific gravity of the
immersed solid, the weight of the air being taken into
consideration .

This problem is perhaps more curious than useful ; but since it
tends to excite the reader's attention, and to render him familiar with
the resources of analysis, we have thought proper to introduce it in
this place ; and in order to its resolution,

Put w i= the weight of the vessel when full of water,

TV =n the weight of the solid body when weighed in air,
w":n the weight of the vessel with the solid in it, when filled
up with water,

OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES. 253

Put m iz: the magnitude of the solid,

s ~ its specific gravity, or the quantity which the problem

demands, and
s' zr the specific gravity of air at the time of the experiment.

Then, because the weight of any body, is expressed by the product
of its magnitude drawn into its specific gravity ; it follows, that when
the weight of the air is disregarded, the weight of the solid is

rv' zn m s ;

but the weight of a quantity of air equal in magnitude to the body,
is ms', and it evidently loses as much weight as that of the fluid which
it displaces ; consequently, when the weight of the air is considered,
the weight of the body when weighed in air, becomes

rv' m m (s — s'),

therefore, by division, the magnitude of the solid is
w'

consequently, the weight in vacuo, is
snf

s — s '

Then, because rv" denotes the weight of the vessel with the solid in
it, when filled up with water, and rv the weight of the vessel when full
of water ; then w" — rv expresses the weight which the vessel has gained
by the immersion of the solid, and this is manifestly equal to the dif-
ference between the weight of the solid, and that of an equal bulk of
the fluid ; therefore, the weight of a quantity of water, equal in bulk
to the solid, is

sw s(rv -\-w' — n")-\-s'(w" — rv)

-. — rv -4- rv — 7 •

s — / s — /

Then, as the specific gravity of the body, is to the specific gravity
of water, so is the weight of the body, to the weight lost ; that is,

sw s(w-\- rv' — rv") + s' (w" — w)
:7=^: s — s'

and this, by suppressing the denominator in the homologous terms,
becomes

s : 1 : : sw : s(rv-\-rv' — rv")-\-s'(w" — rv),

and by equating the products of the extreme and mean terms, we
obtain

s(w 4- rv' — rv") -\- s' (rv" — rv) — rv' ;

254 OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES.

therefore, by transposition, we get

s(rv + nf — rv"} rz: n> — s' (TV" — w),

and finally, by division, we shall obtain

__ ?v' — s'(w" — TV)

~ w + rv'-w" ' (199).

305. The practical rule by which the reduction of the above equa-
tion is effected, may be expressed in words at length in the following,
manner.

RULE. From the weight of the vessel with the solid in it,
when filled up with water -, subtract the weight of the vessel
when full of water only ; then multiply the remainder by the
specific gravity of the air at the time of observation, and
subtract the product from the iveight of the solid in air for a
first number.

To the vjeight of the vessel when full of water, add the
weight of the solid when weighed in air, and from the sum,
subtract the weight of the vessel with the solid in it, when
filled up with water, and the remainder mill be a second
number.

Divide the first number by the second, and the quotient
will give the specific gravity of the solid.

306. EXAMPLE. The weight of a vessel when full of water is 68 Ibs.
avoirdupois, and the weight of a solid body when weighed in air of a
medium temperature, is 34 Ibs. ; now, when the solid is placed in the
vessel, its bulk of water is expelled, and the vessel being then weighed,
is found to indicate 86 Ibs. ; required the specific gravity of the solid
body?

When the air is of a medium temperature, its specific gravity is very
nearly expressed by the fraction 0.0012, that of water being unity;
therefore, by proceeding according to the rule, we have

_ 34 -.0012(86- 68) _

68 + 34—86

which, by referring to a table of specific gravities, is found to correspond
very nearly with the opal stone, a silicious material of very great value,
for the senator Nonius preferred banishment to parting with his
favourite opal, which was coveted by Antony.

CHAPTER XL

OF THE EQUILIBRIUM OF FLOATATION.

BY the equilibrium of floatation is generally meant the position of a floating body^
when its centre of gravity is in the same vertical line with the centre of gravity of
the displaced fluid. When the lower surface of the floating body is spherical or
cylindrical, the centre will coincide with the centre of the figure ; as, in all circum-
stances, the height of this point, as well as the form of the volume of fluid displaced,
must remain invariable. In the next proposition, we shall prove, that the place of
the centre is determined by the doctrine of forces combined with the elementary
principles of hydrostatics, by considering the form and extent of the surface of the
displaced portion of the fluid, compared with its bulk, and with the situation of its
centre of gravity. Our inquiry will be found to embrace also rectangular figures ;
solids in the form of paraboloids and cylinders, together with the equilibrium of
floatation of solids immersed in fluids of different specific gravities ; the theory of
construction of the aerometer ; the hydrostatic balance, and the method of weighing
solid bodies in fluids. The reader will therefore consider this syllabus as the
paraphrase of a definition for the term equilibrium of floatation.

PROPOSITION VI.

307. When a body floats in a state of equilibrium on the
surface of an incompressible and non-elastic fluid : —

The centre of gravity of the whole body, and that of the
part immersed ; or which is the same thing, that of the fluid
displaced, must occur in the same vertical line*

Let ADBH be a vertical section passing through G and g, the centres
of gravity of the whole body AHBD; and of the immersed part A DB,
which falls below EF the plane of floatation.

* The vertical line, which passes through the centres of gravity of the whole
body, and the part immersed beneath the plane of floatation, is generally denomi-
nated the line of support.

256 OF THE EQUILIBRIUM OF FLOATATION,

For since g is the centre of gravity
of the immersed part of the body, or of
the fluid ADB, it is also the centre of all
the forces or weights of the particles of
fluid in ADB, tending downwards; but
because the body is at rest, the same
point g is also the centre of all the

pressures of the fluid tending upwards, by which the weight of the
body ADB H is sustained in a state of equilibrium.

Now, it is manifest, that the sum of all the downward forces, is
equal and opposite to the sum of all the upward forces, otherwise the
body could not be in a state of rest ; but the direction in which the
weight of the body tends downwards from c, is perpendicular to the
horizon; consequently, the line CD which passes through G and g,
the centres of gravity of the whole body and of the part immersed,
must also be perpendicular to the horizon ; for if it is not, the body
must have a rotatory motion, but according to the hypothesis, the body
is at rest; therefore, the line CD is perpendicular to the horizon.

From Propositions III. and VI., it is obvious, that for a floating
body to remain at rest, or in a state of equilibrium, two conditions
must obtain, and these are,

The weight of the floating body, and that of the fluid dis-
placed, must be equal to one another.

This is manifest from the inference under the third Proposition, and
the second condition to which we have alluded forms the substance of
Proposition VI., viz.

The centre of gravity of the whole body and that of the
part immersed, or of the fluid displaced, occur in the same
vertical line.

It is extremely obvious, from the nature of the subject, that both the
above conditions must have place ; for if the first do not obtain, the
body will ascend or descend in a direction which is perpendicular to
the horizon ; and if the second fail, the body will turn about its centre
of gravity, until the centre and that of the fluid displaced occur in
the same vertical line ; and if both these conditions fail together, the
body will partake of a progressive and a rotatory motion at one and
the same time.

From the Proposition which we have demonstrated above, two or
three very useful inferences may be deduced, as follows.

OF THE EQUILIBRIUM OF FLOATATION. 257

308. INF. 1. If any homogeneous plane figure be divided symme-
trically by its vertical axis, and placed in a fluid of greater specific
gravity than itself : —

It will remain in equilibria with its bisecting axis vertical.

309. INF. 2. If any homogeneous solid, generated by the revolu-
tion of a curve round its vertical axis, be placed in a fluid of greater
specific gravity than itself: —

It will remain in equilibria in that position, that is, with its
axis vertical.

310. INF. 3. If in any homogeneous prismatic body, whose axis is
horizontal, the centre of gravity of the section made through its middle
parallel to its base, be in the same vertical line with the centre of
gravity of that part of the solid which falls below the plane of float-
ation : —

The body will remain in equilibria in that position, if placed
in a fluid of greater specific gravity than itself.

This is manifest, for the centres of gravity of the whole prism, and
of the part immersed, may be conceived to lie in those points, and
consequently, the prismatic body is in a state of equilibrium.

PROPOSITION VII.

311. When a solid body floats upon a fluid of greater specific
gravity than itself, and has attained a state of equilibrium : —

The magnitude of the body, is to that of the part immersed
below the plane of floatation, as the specific gravity of the
fluid is to that of the floating body.

For by the inference to the third proposition, when the body floats
in a state of equilibrium": —

The weight of the floating body, is equal to the weight of a
quantity of the fluid, whose magnitude is the same as that
portion of the solid which falls below the plane of floatation.

And according to this principle, the truth of the above Proposition
is demonstrated ; for,

Put m = the magnitude of the whole floating body,
m'— the magnitude of the part immersed,
s := the specific gravity of the floating body,
w — its weight,

VOL. I. S

258 OF THE EQUILIBRIUM OF FLOATATION.

s' n: the specific gravity of the fluid, and
w/iz: the weight of a quantity of the fluid, of the same magni-
tude as that part of the body which falls below the
plane of floatation ; then, according to the above in-
ference just stated, we get

w~w.

But because the weight of any body is expressed by the product of
its magnitude drawn into its specific gravity ; it follows, that

w — ms, and w' — m's',
consequently, by comparison, we have

ms = m's'. (200).

Therefore, if this equation be converted into an analogy, the truth
of the Proposition will become manifest ; for

m : m' : : s' : s,

being precisely the principle which the Proposition implies.
From the principle demonstrated above, various curious and inte-
resting questions may be resolved, and by selecting a few which point
directly to practical subjects, the information afforded by their reso-
lution will sufficiently repay the labour of an attentive perusal.

312. EXAMPLE I. A cubical block of fir, whose specific gravity is
0.55, floats in equilibrio on the surface of a fluid whose specific gravity
is 1.026 ; how much of the block is above, and how much below the
plane of floatation, the entire magnitude being equal to 512 cubic
inches ?

Here, by the Proposition, we have

512 : m' : : 1.026 : 0.55,

and from this, by equating the products of the extreme and mean
terms, we get

1.026m' zz 281.6,
and finally, dividing by 1.026, we obtain

m'zz ' m 274. 464 cubic inches.

It therefore appears, that the quantity of the solid immersed below
the plane of floatation, is 274.464 cubic inches ; consequently, the
part extant is 512 — 274.464 = 237.536 cubic inches, being less than
half the magnitude of the body, by 18.464 cubic inches.

313. EXAMPLE 2. Let the specific gravities and the magnitude of
the body remain as in the last example ; what weight must be added

OF THE EQUILIBRIUM OF FLOATATION. 259

to the body, in order that its upper surface may be made to coincide
with that of the fluid ?

Put x nz the weight which must be added to the solid, in order that
it may sink to a level with the surface of the water ; then, we have

m: m + x :: 0.55 : 1.026,
and by equating the products of the extremes and means, we get

0.55 (m + x) = 1.026m;
therefore, by transposition, we obtain

0.55* = 0.476m;
but according to the question, mzr:512 cubic inches, hence we get

0.55*:= 243.712,
and finally, by division, we have

243 712
x zn — -—— - — — 443.1 13 cubic inches; but a

cubic inch of fir of the given specific gravity, weighs 0.0198 Ibs.
avoirdupois, very nearly ; consequently, the weight to be added for
the purpose of making the solid sink to the same level as the surface
of the fluid, is

.0198X443.113 — 8.774 Ibs. nearly.

314. But to determine generally, the magnitude which must be
added to the original solid, in order that its surface maybe coincident
with that of the fluid :-— Let zziithe weight to be added; then, by
the Proposition, we have

m -{- x : m : : s : s,

from which, by equating the products of the extremes and means,
we get

s (m -\- x) — s' m,

and by separating the terms, it becomes

^m -\- sx^ns'm,

and finally, by transposition and division, we obtain
_m(s' — s)

s (201).

Therefore, the practical method of reducing this equation, may be
expressed in words in the following manner.

RULE. From the specific gravity of the fluid, subtract that
of the solid ; then, multiply the remainder by the magnitude
of the solid, and divide the product by its specific gravity for
the weight to be added.

260 OF THE EQUILIBRIUM OF FLOATATION.

315. EXAMPLE 3. A cubic mass of oak, whose specific "gravity is
0.872, is placed in a cistern of water, and when it has attained a state
of equilibrium with its sides vertical, it stands 20 inches above the
surface of the fluid; what is the magnitude of the solid, the specific
gravity of the water being represented by unity ?

In order to resolve this example, we shall first investigate a general
formula, which will apply to every case of a similar nature, when the
specific gravity of the fluid and that of the solid are given ; for which
purpose,

Put x nr the length of one side of the solid ; then is

x — a*— the length of that portion which is below the plane of
floatation.

But by the principles of mensuration, the magnitude of the whole
body is

razz a?8,

and that of the part immersed, is

m'= (x — a) Xs'zz a:8 — ax* ;

therefore, by the Proposition, we obtain

m ' m! \ '. s' i s ;
and this, by substitution, becomes

a;8 : x* — a a?2 : : s' : s ;

from which by equating the products of the extremes and means,
we get

sz'zz s'O8 — as2);
and by separating the terms, we obtain

sx9 zz 5' x3 — as' x9;
consequently, by transposition and division, it is

as'

~7^7)' (202).

And the practical rule supplied by this equation, may be expressed
in words at length in the following manner.

RULE. Multiply the specific gravity of the fluid by the
height of the body above its surface, and divide the product
by the difference between the specific gravity of the fluid
and that of the solid, and the quotient will give the side
of the cube required.

* The quantity a, is here put to denote the height of the body abovelhe fluid.

OF THE EQUILIBRIUM OF FLOATATION. 261

Taking, therefore, the data as proposed in the foregoing example,
and we shall obtain

20X1
#~ , Q S7, =r 156.29 inches; consequently,

the magnitude, or cubical contents of the body, is

156.29X 156.29X 156.29 = 3815627.7 inches.
In addition to the foregoing examples, which might very appro-
priately have been ranked under the head of problems, the seventh
proposition affords the following inferences.

316. INF. 1. If two bodies floating on the same fluid, be in a state
of equilibrium : —

The specific gravities of those bodies, will be to one another
directly as the parts below the plane of floatation , and in-
versely as the whole magnitudes of the bodies.

317. lNF.^2. If the same body float upon two different fluids, and
be in a state of equilibrium on each : —

The specific gravities of the fluids, will be to one another,
inversely as the parts of the body below the plane of floatation.

318. INF. 3. If different bodies float in equilibrio on the surfaces
of different fluids, and if the parts below the planes of floatation be
equal among themselves : —

The specific gravities of the fluids, will be to one another
directly as the weights of the bodies, or directly as the magni-
tudes of the bodies drawn into their specific gravities.

319. INF. 4. If any number of bodies of the same weight, but of
different specific gravities, float in equilibrio on the surface of the
same fluid : —

The magnitudes of the parts below the plane of floatation,
are equal to one another.

320. INF. 5. If a body float in equilibrio on the surface of a given
fluid, and if the part below the plane of floatation be increased or
diminished by a given quantity :

The absolute weight of the body, (in order that the equili-
brium may still obtain,) must be increased or decreased by a
weight, which is equal to the weight of the portion of the fluid
that is more or less displaced, in consequence of increasing or
diminishing the immersed part of the body, or that which falls
below the plane of floatation.

262 OF THE EQUILIBRIUM OF FLOATATION.

321. The principle announced in the last inference, may be demon-
strated in the following manner.

Put ra — the magnitude of the body at first, when in a state of
equilibrium,

ra':=r the part originally below the plane of floatation,

m"=: the part by which it is increased or diminished,

5 zn the specific gravity of the body,

s' — the specific gravity of the fluid, and

w ~ the weight by which the body is increased or diminished,
in consequence of the increase or decrease of the im-
mersed part.

Then, because the quantity of fluid displaced, is equal to the mag-
nitude of the body which displaces it, it follows, that the weight of
the displaced fluid is expressed by (m'-f- w")s', and the weight of the
whole solid with which it is in equilibrio, is (ms + w} ; consequently,
we have

(rn1 -|- mH)s' i=.ms-\-w.
Now it is manifest, that in the case of the first equilibrium,

m V rr m s ;

it therefore follows, that
tm"s' = w.

That is, the weight by which that of the body must be increased or
diminished, to restore the equilibrium : —

Is equal to the weight of that quantity of the fluid which
is more or less displaced, in consequence of the increase or
decrease of the part below the plane of floatation.

PROBLEM XLVI.

322. If a solid body in the form of a paraboloid, be in a state
of quiescence on the surface of a fluid, whose specific gravity
bears any relation to that of the body : —

It is required to determine how much of the solid will sink
beneath the plane of floatation.

Let GACBH be a vertical section passing along the axis of the solid,
and cutting the plane of floatation in the line AB ; CD being the axis

OF THE EQUILIBRIUM OF FLOATATION. 263

of the solid, GH the diameter of its base,
and ABHG the portion which falls below
EF the surface of the fluid.

Put a :=: CD, the axis of the paraboloid,
d •=. G H, the diameter of its base,
m — the magnitude of the entire

solid GACBH,

w'mthe magnitude of the immersed part,
s —the specific gravity of the body,
s' =z the specific gravity of the fluid, and
x nrci, the axis of that portion of the body, which in a state
of equilibrium, remains above the plane of floatation.

Consequently, by the seventh proposition, we have

m : m' : : s' : s.

But by the principles of mensuration, the solidity of a paraboloid is
equal to one half the solidity of its circumscribing cylinder ; there-
fore, we get

m — .3927ad2,

and similarly, we obtain

solidity ACB = 1.5708^ a;2,

where p is the parameter of the generating parabola.
Now, the writers on conic sections have demonstrated, that accord-
ing to the property of the generating curve,

4ap z= d* ;

let this value of d? be substituted instead of it in the preceding value
of m, and we shall obtain

consequently, by subtraction, we get

therefore, by substituting these values of m and m' in the above
analogy, it is

a2 : a2 — re2 : : / : s ;

and from this, by equating the products of the extreme and mean
terms, we obtain

s'a? — s'#2n=$a2,
and by transposition we have

s'x1 z=a2(s' — s) ;
therefore, by division and evolution, we obtain

264

OF THE EQUILIBRIUM OF FLOATATION.

IHCl A/ f _

V '

and finally, by subtraction, we have

(203).

323. The practical rule for effecting the reduction of the above
equation, may be expressed in words at length in the following
manner.

RULE. Divide the difference between the specific gravity
of the fluid and that of the solid, by the specific gravity of
the fluid; then, from unity subtract the square root of the
quotient, and multiply the remainder by the axis of the
parabola, and the result will give the height of the frustum
that falls below the plane of floatation.

324. EXAMPLE. The axis of a paraboloid which floats in equilibrio
on the surface of a fluid, is 29 inches ; what part of the axis is im-
mersed below the plane of floatation, supposing the body *to be of
oak, whose specific gravity is 0.76, that of water being unity ?

Here, by proceeding as directed in the rule, we get

1 — 0.76

) zz 14.79 inches very nearly.

If the vertex of the figure be downwards, as in the annexed dia-
gram, then the part of the axis which
falls below the plane of floatation will
be greater than it is in the preceding
case ; for it is manifest, that since the
same magnitude or part of the body
must be immersed in both cases, it will
require a greater portion of the axis,
towards the vertex of the figure, to
constitute that magnitude, than it would require towards the base.

Therefore, by retaining the foregoing notation, we have, by the
principles of mensuration,

GACBII — mzz 1.5708pa% and ACBZZIW'ZZ \.57Q8px?;
consequently, by the seventh proposition, we obtain

1.5708pa2 : 1.5708;?*2 : : s' : s ;

therefore, by suppressing the common factors 1.5708p, and equating
the products of the extreme and mean terms, we get

s' x" m s a2,

OF THE EQUILIBRIUM OF FLOATATION. 265

and this, by division, becomes

*=%>

and finally, by extracting the square root, we get

(204).

325. The practical rule deduced from this equation is very simple ;
it may be expressed in words at length in the following manner.

RULE. Divide the specific gravity of the solid, by that of
the fluid on which it floats ; then, multiply the square root
of the quotient by the axis of the body, and the product will
give the height of the part below the plane of floatation.

Therefore, by retaining the data of the foregoing example, we shall
obtain as under,

c i — x = 29 V0.76 = 25.251 inches;

being a difference of 10.46 inches, in the depths of immersion, for the
two cases of the question.

PROBLEM XLVII.

326. When a body floats in equilibrio on the surface of a
homogeneous fluid, it is a necessary condition, that the centre
of gravity of the body, and that of the fluid displaced, shall be
in the same vertical line. Supposing, therefore, that the equili-
brium is disturbed by the addition or subtraction of a certain
weight : —

It is required to determine, how far the body will be
depressed or elevated in consequence of the extraneous weight?

The above problem will obviously admit of two cases, for a weight
may be added to a body, and it may be subtracted from it ; in the
one case the body will descend, and in the other it will ascend ; the
following general solution, however, wiH
answer both cases.

Let AFBC represent a vertical section
of the solid body, floating in a state of
equilibrium on a fluid of which the hori-
zontal surface is K L ; and suppose, that
when the body is acted on by its own
weight only, the straight line D E is coin-

266 OF THE EQUILIBRIUM OF FLOATATION.

cident with the surface of the fluid ; but in consequence of the addi-
tional weight abed, the body descends through the space GH, where
it again attains a state of quiescence, and the plane of floatation
mounts to AB.

Now, it is manifest, that when the body is acted on by means of its
own weight only, (in which case, DE is coincident with the surface of
the fluid,) the weight of the whole body is equivalent to that of a
quantity of the fluid, whose magnitude is DCE; but when the weight
abed is applied, the compound weight is equivalent to that of a
quantity of the fluid, whose magnitude is ACB; consequently, the
subsidiary weight abed, and the weight of a quantity of the fluid,
whose magnitude is ABED, are equal to one another.

Draw the -straight line ef parallel and indefinitely near to DE;
then is DE/e, the small elementary increase of the immersed portion
of the body, corresponding to any indefinitely small increase of the
weight abed.

Put a •=. the area of the horizontal section passing through AB,
determinable from the position and the figure of the
body, before the weight abed is applied,
n :z: the weight abed,

rv nr the fluxion or small elementary variation of w ;
x — G H, the distance through which the body passes in con-

sequence of the weight n being applied.
x zz the fluxion or elementary variation of x, corresponding

to w, and
s =: the specific gravity of the fluid.

Then, because the line ef is supposed to be indefinitely near to D E,
it follows, that the portion of the body whose section is D E/e, may be
considered as uniform in area between its bases, and consequently, its
magnitude is expressed by ax ; but DE/C, is equal to the quantity of
fluid displaced by reason of the elementary weight TV, and it is a well
attested principle in hydrostatics, that the weight of the quantity of
fluid displaced, and that of the body which displaces it, are equal to
one another ; therefore we have

and the aggregate of the small elementary weights, or the whole weight
added, is

?y = /as*. (205).

This is the general form of the equation of equilibrium ; but it
admits of various modifications, according to the conditions of the

OF THE EQUILIBRIUM OF FLOATATION. 267

question and the nature of the body. For instance, if the body be a
solid of revolution, and r the radius of the section coincident with the
plane of floatation ; then, the above equation becomes

w^nrsfax, (206).

where the symbol TT denotes the number 3. 1416, or four times the area
of a circle whose diameter is expressed by unity.

327. The solution of the problem, however, may be effected inde-
pendently of the fluxional analysis, especially in all cases where the
floating body is symmetrical with respect to its axis ; for if it be in
the form of a right cylinder with its axis vertical, as in the annexed
diagram; then, the solution becomes an object of the greatest
simplicity ; for since the area of the horizontal section is constant,
the space through which the body moves will be the same, whether
the weight be added to it or subtracted from it.

Let A B c D be a vertical section of a
cylinder, floating in equilibrio on a fluid
whose surface is GH, the axis mn being
perpendicular to the horizon, and sup-
pose the weight n to be placed on the
upper end of the cylinder ; it is obvious
that the equilibrium will then be de-
stroyed, and the body will continue to
descend, until it has displaced a quan-
tity of the fluid, whose weight is equal

to that of the compound mass, consisting of the cylinder, together with
the applied body whose weight is TV; or it will continue to descend, until
the weight of the fluid displaced by the space IKFE is equal to n the
weight of the applied body ; in which case, the equilibrium will again
obtain, and the plane of floatation, which originally cut the cylinder
in E F, will now be transferred to i K.

Again, on the other hand, if a,weight rv be subtracted from the
cylinder, supposed to be in a state of equilibrium with the plane of
floatation passing through EF, the body will then ascend, until the
weight of the fluid which rushes into its place becomes equal to the
weight subtracted, in which case the solid will again be in a state of
quiescence with the plane of floatation passing through a b.

Put r ~ id or nd, the radius of the horizontal section,
m — the magnitude of the space IKFE or EF#a,
x — de or ec, the space through which the body is depressed
or elevated in consequence of the extraneous weight,

268 OF THE EQUILIBRIUM OF FLOATATION.

w zr the weight which is added to or subtracted from the

cylinder, and

5 z= the specific gravity of the fluid.

Then, by the principles of mensuration, the solidity of the cylinder,
of which the section is IKFE or EF&O, becomes

and the weight of an equal magnitude of the fluid, is

m s = 3.l4l6r*sx',

but this, by the nature of equilibrium, is equal to the disturbing
weight ; hence we have

w=3.Ul6r*sx,
and from this, by division, we obtain

~3.1416rV (207).

328. The practical rule for reducing this equation is very simple ;
it may be expressed in words at length in the following manner.

RULE. Divide the given disturbing weight, whether it be
added to or subtracted from the cylinder, by the area of the
horizontal section, drawn into the speci/ic gravity of the fluid,
and the quotient mill express the quantity of descent or ascent
accordingly.

329. EXAMPLE. A cylinder of wood, whose diameter is 24 inches,
floats in equilibrio with its axis vertical, on the surface of a fluid
whose specific gravity is expressed by unity ; now, supposing the
equilibrium to be destroyed by the addition or subtraction of another
body, of which the weight is 56 Ibs. ; through what space will the
body move before the equilibrium be restored ?

Here, by proceeding as directed in the rule, we have

_ 56

a'~~3.1416xl22X.03617*~~3'42

Again, if the body should be in the
form of a paraboloid, floating in equi-
librio on the surface of a fluid with its ~~
vertex downwards, as represented in the xj
annexed diagram, where ACB is a ver-
tical section passing along the axis CD,
and GH the surface of the fluid on

* The decimal fraction 0.03617 expresses the weight in Ibs. of a cubic inch of water.

OF THE EQUILIBRIUM OF FLOATATION. 269

which the body floats, with the plane of floatation originally passing
through EF, but which, on the addition or subtraction of the weight rv,
ascends to i K or descends to a b.

Put p zr the parameter of the generating curve,

S ~ ec, the distance of the vertex below the surface of the

fluid at first,
m ±r the magnitude or solidity of the paraboloidal frustum, of

which IKFE is a section,

#/:=r the magnitude of the frustum whose section is EF^a ;
x rn de, the descent of the body in consequence of the addi-

tion of the weight rv,

x' rr ec, the corresponding ascent in the case of subtraction,
and s rzi the specific gravity of the fluid.

Then we have crf = 5 -j- x, and c c rr 8 — otf, and according to the
writers on mensuration, we have

and in a similar manner, we obtain

m' = 1.5708p(28*f — *'2) ;

and since the weight of any body is expressed by its magnitude drawn
into its specific gravity, it follows, that the weight of a quantity of
fluid equal respectively to m and m1 , are

THSIZ: 1.5708;?s(23;r + x2), andm's— 1.5708ps(fcc'— x'3).
Now, these according to the conditions of the problem, are respec-
tively equal to the disturbing weight ; hence we have in the case of
addition,

m = 1 .57Q8ps(Z$x + *2), (208).

and in the case of subtraction, it is

tv= 1.5708^5(2^'— *'8). (209).

330. The equations which we have just obtained, are precisely the
same as would arise, by taking the fluent of the expression in equation
(206) ; it therefore appears, that although the fluxional notation is the
most convenient for expressing the general result, yet in point of sim-
plicity as regards symmetrical bodies, there is little advantage to be
derived from its adoption.

Suppose that in the first instance, the equilibrium is destroyed by
the addition of the weight rv, and let it be required to determine how
far the body will descend in consequence of the addition.

Equation (208) involves this condition ; consequently, if both sides
be divided by the expression 1 .5708^5, we shall obtain

270 OF THE EQUILIBRIUM OF FLOATATION.

which being reduced, gives

x — 1/8"H — 8- /oim

Y 1.5708j9s (210).

331. And the practical rule for reducing this equation, may be
expressed in words at length, as follows.

RULE. Divide the weight which disturbs the equilibrium,
by 1.5708 times the parameter of the generating parabola,
drawn into the specific gravity of the fluid, and to the quotient
add the square of the distance between the vertex of the figure
and the plane of floatation in the first position of equilibrium ;
then, from the square root of the sum, subtract the said dis-
tance, and the remainder will express the quantity of descent.

332. EXAMPLE. A solid body in the form of a paraboloid, floats on
a vessel of water in a state of equilibrium with its vertex downwards,
when 12 inches of the axis are immersed below the plane of floata-
tion ; how much farther will the body sink, supposing a weight of
28 Ibs. to be laid on its base, the parameter of the generating parabola
being 16 inches?

Here, by pursuing the directions of the rule, we get,

122H — — 12 =z 1.22 inches.

1.5708X16X. 03617

333. If the weight w should be subtracted from the paraboloid
instead of being added to it, the quantity of ascent will then be deter-
mined by equation (209), where we have

divide both sides of this equation by the quantity 1.5708p5, and we
shall obtain

1.5708psJ
which, by transposing the terms, becomes

By completing the square, we get

Or THE EQUILIBRIUM OF FLOATATION. 271

and finally, by extracting the square root and transposing, it is

' = a V a'- (211).

334. And the practical rule by which this equation is reduced, may
be expressed in words at length, in the following manner.

RULE. Divide the weight which is proposed to be subtracted
from the paraboloid, by 1.5708 times the parameter of the
generating parabola drawn into the specific gravity of the
fluid, and subtract the quotient from the square of the dis-
tance of the vertex below the plane of floatation in theflrst
position of equilibrium ; then, from the said distance , subtract
the square root of the remainder for the quantity of ascent
required.

335. If we take the data of the foregoing example, and proceed
according to the directions of the rule, we shall obtain

x1 = 12 — 4 / 144 __ _ ?? _ —1.35 inches,
y 1.5708X16X.03617

exceeding the descent in the former example, by 0.13 of an inch.

Numerous other examples akin to the above, respecting bodies of
various forms, and placed in different positions, might here be pro-
posed ; but since they are all resolvable by the general formula,
equation (206), we have thought proper to omit them.

PROBLEM XLVIII.

336. If a body which is symmetrical with respect to its verti-
cal axis, floats upon a fluid in a state of equilibrium : —

It is required to determine what weight must be placed
upon the body, so that it shall sink to a level with the surface
of the fluid, the specific gravities of the solid and the fluid,
together with the magnitude of the solid, being given.

In the second example to the seventh proposition, we have advanced
principles of nearly a similar import to those required for the solution
of the present problem, yet nevertheless, we think a separate solution
in the present instance is necessary, since it can be somewhat differ-
ently represented ; for which purpose,

Put m ~ the entire magnitude of the floating body,

w — its weight before the external body or weight is super-
added,

272

OF THE EQUILIBRIUM OF FLOATATION.

rv' rr the added weight,
n>" — the weight of the fluid displaced,
s =z the specific gravity of the fluid, and
s' = the specific gravity of the floating solid.

Then, because the absolute weight of any body, is expressed by its
magnitude drawn into its specific gravity ; it follows, that the weight
of the floating solid, is

w — ms',

and in like manner, the weight of the displaced fluid, is
n" — m s ;

now, it is manifest, from the nature of the problem, that the weight
of the displaced fluid is equal to the weight of the floating body,
together with the superadded weight; consequently, we have

«/ -J- rv ~w' -f- w/™wis ;
from which, by transposition, we obtain

w'=ffi(* — «'). .. (212).

337. The practical rule for the reduction of this equation is very
simple : it may be expressed as follows.

RULE. Multiply the difference between the specific gravities
of the fluid and the floating solid, by the whole magnitude of
the floating bodyt and the product will express the value of
the added iveight.

338. EXAMPLE. A mass of oak, whose specific gravity is .872, that
of water being unity, floats in equilibrio on the surface of a fluid
whose specific gravity is 1.038 ; what weight applied externally to the
floating body, will depress it to the level of the fluid surface, sup-
posing the magnitude of the body to be equal to 8 cubic feet ?

Here, by operating as the rule directs, we shall have
w'— 8(1.038 — .872) — 1.328 cubic feet of water ;

but it is a well known fact, that one cubic foot of water weighs 62 £
Ibs. avoirdupois, very nearly ; consequently, we have

wf=1.328X62J = 831bs.

PROPOSITION VIII.

339. If a solid body, which is specifically heavier than one of
two fluids which do not mix, and specifically lighter than the
other, be immersed in the fluids : —

/ UNIVERSITY

V Of .y±S

OF TIIR EQUILIBRIUM OF FLOATATION. 273

It will float in equilibria between them, when the weight
of the fiuids respectively displaced, are together equal to the
weight of the solid body which causes the displacement ; the
specific gravities of the fluid being supposed known.

Let A BCD be a vertical section passing through the centre of gravity
of the floating body, and suppose that
IK is the common surface of the two
fluids, in which the solid is quiescent,
GH being the surface of the lighter fluid.

Now, it is manifest, that since the
body ABCD is specifically heavier than
one of the fluids, and specifically lighter
than the other, it cannot be entirely at

rest in either, but must rest between them in such a position, that the
sum of the weights of the fluids displaced shall be equal to the whole
weight of the solid.

Let EFD be perpendicular to i K, the common surface of the fluids
in which the body floats ; then it is evident, that the pressure down-
ward on any point of the base D, is equal to the weight of the incum-
bent line of solid particles, whose altitude is BD the thickness of the
body, together with the weight of EB the superincumbent column of
trie lighter fluid ; and again, the pressure upwards on the same point
D, is equal to the weight of a column of the heavier fluid whose alti-
tude FD, together with the weight of a column of the lighter fluid,
whose altitude is EF.

Put d — EB, the depth of the body below the upper surface of the

lighter fluid,
d' — EF, the whole depth of the lighter fluid, or the depth of

the common surface,
•3 — FD, the depth of the body below the common surface, or

the surface1 of the heavier fluid,
$ = BD, the whole thickness of the solid body,
5 = the specific gravity of the lighter fluid,
s' zz the specific gravity of the heavier fluid,
s" = the specific gravity of the solid body,
p rzr the downward pressure, and
p' ~ the upward pressure.

Then, because the weight of any body, whether it be fluid or solid,
is expressed by the product of its magnitude drawn into its specific
gravity, it follows that the downward pressure on the point D, is

VOL. I. T

274 OF THE EQUILIBRIUM OF FLOATATION.

p i= 2' s" -|- ds,

and in like manner, the pressure upwards, is
p' = Ss' -\-d's.

But when the body floats in a state of equilibrium, the upward and
the downward pressures are equal to one another ; hence we have

from which, by transposing and collecting the terms, we get

SY'zzas'-Kd'— d)s.

Now, it is obvious, that what we have demonstrated above with
respect to the point D, may also be demonstrated to hold with
respect to every other point of the surface whose section is ADC;
consequently, by taking the aggregate of the upward and downward
pressures, we obtain

&c.)«"=(3 + S + 3+ &c.)s' + {(d' + ef+ d'-f&c.)

Put w zz (o; -f- S' •+• 2' -j- &c.) to infinity, equal to the magnitude

of the entire floating body,
m — (3 + I 4- 3 4- &C0 to infinity, equal to the part immersed

in the heavier fluid, and
m"= {(d' 4- d' 4- d' 4- &c.) — (d -f d 4- d + &c.)} to infinity,

equal to the part immersed in the lighter fluid ; conse-

quently, by substitution, we get

ms" = m's' + m"s. (213).

This equation involves the principle announced in the Proposition,
and its application to practical cases will be exemplified in the reso-
lution of the following Problems.

PROBLEM XLIX.

340. Suppose that a solid body in the form of a regular cube,
is observed to float in equilibrio between two unmixable fluids
of different specific gravities : —

It is required to determine, how much of each fluid is dis-
placed by the body, the specific gravities of the body and the
fluid being given.

Let A BCD be a cubical body, floating in equilibrio between the
two unmixable fluids, whose upper horizontal surfaces are respec-

OF THE EQUILIBRIUM OF FLOATATION. 275

tively IK and GH, GH being the surface
of contact of the two fluids.

Put m =: A BCD, the magnitude of the

whole body,

x m ABCFE, the magnitude of the
part immersed in the lighter
fluid, and
m — a; rr EFC D, the magnitude of the part immersed in the heavier

fluid.

Therefore, if the specific gravities of the body and the fluids be
respectively denoted by s", s and s', as in the Proposition; then, we
shall have

ms"=:xs-\- (m — x)s',

from which, by transposition, we obtain

x (s' — 5) zz m(s' — s"),
and finally, by division, it becomes

341. An equation of an extremely simple and convenient form, from
which we deduce the following practical rule.

RULE. Multiply the magnitude of the immersed solid by
the difference between its specific gravity and that of the
heavier Jluid; then divide the product by the difference between
the specific gravities of the fluids, and the quotient will give
the magnitude of the part immersed in the lighter Jluid.

342. EXAMPLE. A cubical piece of oak containing 2000 inches,
and whose specific gravity is 0,872, that of water being unity, floats
in equilibrio between two fluids, whose specific gravities are respec-
tively 1.24 and 0.716 ; what portion of the solid is immersed in each
of the fluids, supposing them to be altogether unmixable ?

The operation being performed according to the rule, will stand as
below.

2000(1.24 — 0.872)

(1.^4-0.716)- 14°4'58 °ublC mches'

This result expresses the solid contents of that part of the body
which is immersed in the lighter fluid ; consequently, the part which
is immersed in the heavier fluid, is

2000 — 1404.58 = 595,42 cubic inches.
T 2

276 OF THE EQUILIBRIUM OF FLOATATION.

343. In this example, a greater portion of the body is immersed in
the lighter fluid than what is immersed in the heavier ; but this cir-
cumstance manifestly depends upon the nature of the immersed body,
and the relation of the specific gravities ; for an instance may readily
be adduced, in which exactly the reverse conditions will obtain : —
Thus, let the magnitude of the body and the specific gravities of the
fluids remain as above, and suppose the specific gravity of the body
to be 1.17 instead of 0.872 ; what then are the parts immersed in the
respective fluids ?

The numerical process is represented as below.

2000(1.24—1.17)
* = - (1.24- 0.716) = 267.175 cub.c.nches.

Here then, we have 267.175 cubic inches for the portion which is
immersed in the lighter fluid, while that immersed in the heavier is
2000-267.175~1732.825 cubic inches; this agrees with the case
represented in the diagram, for there the body displaces a greater
quantity of the heavier than it does of the lighter fluid.

PROBLEM L.

344. Having given the specific gravities of two unmixable
fluids, and the magnitude of a solid body which floats in equili-
brio between them : —

It is required to determine the specific gravity of the solid,
so that any proposed part of it may be immersed in the
lighter fluid.

Put m zr the magnitude of the immersed solid, the same as in the

preceding Problem,

s ~ the specific gravity of the lighter fluid,
s' — the specific gravity of the heavier fluid,
x = the specific gravity of the solid body, being the required

quantity, and
n n= the denominator of the fraction which expresses the part

of the body immersed in the lighter fluid.

Then, according to the principle demonstrated in Proposition VIII.,
we shall obtain

ms , (m — m)s'

mx= \-- — ,

n n

and from this, by a little farther reduction, we get
mnxmmns' — m(sr — s) ;

OF THE EQUILIBRIUM OF FLOATATION. 277

consequently, by division, we obtain

_mns' — m(s' — s) *

mn (215).

345. And from this equation we deduce the following rule.

RULE. Multiply together, the magnitude of the body, the
number which expresses what part of it is immersed in the
lighter, and the specific gravity of the heavier fluid ; then,
from the product subtract the difference between the specific
gravities of the fluid drawn into the magnitude of the solid
body, and divide the remainder by the magnitude of the bodyt
multiplied by the number which expresses what part of it is
immersed in the lighter fluid ; then shall the quotient express
the specific gravity of the body.

346. EXAMPLE. The specific gravities of two unrnixable fluids are
respectively 1.24 and 0.716, that of water being unity ; now, sup-
posing that when these fluids are poured into the same vessel, a body
of 2000 cubic inches which is in equilibrio between them, has one
seventh part of its magnitude immersed in the lighter fluid ; what is
the specific gravity of the body ?

Here, by proceeding according to the rule, we have

mns/ = 2000x7Xl.24 — 17360
— 7M(5'—s):= 2000 (1.24 — 0.716) zz — 1048 subtract

difference zr 163 12;
consequently, by division, we shall obtain

mns' — m(s' — s) 16312

x — s = r—-— — - zz: 1.17 nearly.

mn 2000X7 J

347. From what has been done above, it is easy to ascertain what
will be the specific gravity of the body, when equal portions of it are
immersed in the lighter and in the heavier fluids ; for in that case, we
have n equal to 2, which being substituted in equation (215), gives

*~s'— J(*'— s). (216).

And the practical rule for reducing this equation may be expressed
in words at length, in the following manner.

* This equation is susceptible of a simpler form, for by casting out the common
factor m, it is

278 OF THE EQUILIBRIUM OF FLOATATION.

RULE. From the specific gravity of the heavier fluid, subtract
half the difference between the given specific gravities, and the
remainder will be the specific gravity of the solid body.

348. EXAMPLE. Let the specific gravities of the fluids remain as
above ; what must be the specific gravity of the body, so, that when
it is in a state of equilibrium, one half of it may be immersed in each
solid ?

One half the difference of the given specific gravities, is

£(1.24 — 0.716) =0.262;
consequently, by subtraction, we have

a- =1.24 — 0.262 = 0.978, and with this specific
gravity, a body, whatever may be its magnitude, will be equally
immersed in the two unmixable fluids.

349. If the specific gravity of the lighter fluid vanish, or become
equal to nothing; then equation (215) becomes

and by converting this equation into an analogy, we get
m : m' : : s1 : s".

This analogy expresses the identical principle, which is announced
and demonstrated in Proposition VII. preceding ; it is therefore pre-
sumed, that the examples already given will be found sufficient to
illustrate the application of this very elegant and important property.
Since the magnitude of the whole floating body is equal to the sum
of its constituent parts, it follows, that according to our notation,

m = m' 4~ m" '*
consequently, by substitution, equation (215) becomes

(m 4- m") s" = m' s' 4- m" s,
or by transposing and collecting the terms, we get

m« («"-«) = »»'(*-*"),
and by converting this equation into an analogy, we obtain

m : m" ::(s" — s) : (s'—s»).

By comparing the terms of the proportion as they now stand, it
will readily appear, that if the specific gravity of the lighter fluid be
increased, the term (s" — s) is diminished, while (s' — s") remains the
same ; consequently, the first term ni will be diminished with respect
to the second term m" ; which implies, that the part of the body in
the lighter fluid will be increased ; hence arises the following very
curious property, that

OF THE EQUILIBRIUM OF FLOATATION. 279

If any body float upon the surface of a fluid in vacuo, and
air be admitted, the body will ascend higher above the surface,
and consequently, the proportion of the immersed part to the
whole will be diminished.

PROBLEM LI.

350. Suppose a solid body to float in equilibrio on the surface
of water, both in air and in a vacuum : —

It is required to determine the ratio of the parts immersed
in the water in both cases.

Put m zz: the magnitude of the whole floating body,

m' zr the magnitude of the part immersed below the surface

of the water, when the incumbent fluid is air,
m"— the portion immersed when the body floats on water in

vacuo,

s zz: the specific gravity of air,
s' zz: the specific gravity of water, and
s" zz: the specific gravity of the floating body.
Then we have m — m', for the part above the surface of the water,
when the incumbent fluid is air, and m — m" for the extant part when
the body floats in vacuo ; consequently, by equation (213), we have,
when the body floats in air,

ms" zz: m' s' -\-(m — ra') s,
from which, by a little reduction, we obtain

m' (s' — s) zz m (s" — s),
and finally, by division, it becomes

,_*»(*"-*)

- (s'-s) ' (217).

and again, when the body floats in vacuo, we have

but in this case, s vanishes, hence we get

ms" = m"s,
and by division, it is
_ms"

(218).

Let the equations (217) and (218) be compared with one another in
the terms of an analogy, and we shall have

' - m(s''—s) . . a . ms" .

' (S' — 5) S'

280

OF THE EQUILIBRIUM OF FLOATATION.

therefore, by equating the products of the extreme and mean terms
and casting out the common quantity m, we obtain
m"(s"—s) _ m's"

s' — sY~ :~7"'
by clearing the equation of fractions, we get

m"s'(s" — s) = nf8l!(S — s),
and finally, by division, we have

«'(«" — «) (219).

351. Now, it is manifest, that in order to determine from this
equation, what part of the body is immersed in the water when it
floats in vacuo, it is necessary in the first place, to ascertain how
much of it is immersed when the floatation occurs in air : — Equation
(217) determines this, and the practical rule deduced from the equa-
tions (217) and (219) may be expressed in words at length in the
following manner.

RULE. From the specific gravity of the floating body, sub-
tract the specific gravity of air ; multiply the remainder by
the magnitude of the body, and divide the product by the
difference between the specific gravities of water and air, for
the part which is immersed in water, when the incumbent fluid
is air.

Again. Multiply the difference between the specific gravities
of water and air by the specific gravity of the floating body ;
divide the product by the difference between the specific gra-
vities of the solid body and air, drawn into the specific gravity
of water ; then, multiply the quotient by the magnitude of the
part immersed in water when the body floats in air, and the
product will be the magnitude of the part immersed in water,
when the body floats in vacuo.

352. EXAMPLE. A mass of oak whose specific gravity is 0.925, con-
tains 185 cubic inches; what quantity of it exists below the plane of
floatation, supposing it to float on water in vacuo, the specific gravity
of the air being 0.0012 at the instant of observation ?

By operating according to the directions given in the first clause of
the rule, the quantity below the plane of floatation when the incumbent
fluid is air, becomes

185(0.925 — 0.0012)

m •=. : -±£ 171.108 cubic inches;

(1—0.0012)

OF THE EQUILIBRIUM OF FLOATATION. 281

therefore, according to the second clause of the rule, the part immersed
when the body floats in vacuo, becomes

0.925(1 — 0.0012)
= X171-108:

353. If we refer to the equation (218) preceding, it will readily
appear, that the above result may be determined with much less labour
and greater simplicity ; for the magnitude of the immersed part, when
the body floats in vacuo, is there expressed in terms of the weight of
the body and the specific gravity of water, and the practical rule for
reducing the equation, may be expressed in words at length, as follows.

RULE. Multiply the magnitude of the body by its specific
gravity ; then divide the product by the specific gravity of
water, and the quotient will express the magnitude of the
immersed part when the body floats in vacuo.

Therefore, by taking the data as proposed in the above example,
the magnitude of the immersed part becomes

185X0.925
m'~ - - - ^ = 171.125 cubic inches; being precisely

the same quantity as we obtained by the foregoing prolix operation.

If we compare the computed values of m' and m" with one another,
we- shall find that the latter exceeds the former by a very small quan-
tity, that is,

171.125— 171. 108 =z 0.017,

which verifies the concluding inference under Problem L.

354. On the principles which we have explained and illustrated in the
foregoing problems, depends the construction and application of the
Hydrometer, an instrument which is generally employed for detect-
ing and measuring the properties and effects of water and other fluids,
such as their density, gravity, force, and velocity.

When the hydrometer is employed to determine the specific gravity
of water, it is sometimes denominated an aerometer or water-poise ;
and being an instrument of very general utility in numerous philoso-
phical experiments, we think it will not be amiss in this place, to
discuss its nature and properties a little in detail ; and we may here
observe, that the following problems and remarks are quite sufficient
to establish and exemplify its most important applications.

The hydrometer, or aerometer, in general consists of a long cylin-
drical stem of glass, or other metal, connected with two hollow balls,

282

OF THE EQUILIBRIUM OF FLOATATION.

into the lower of which is introduced a small quantity of mercury or
leaden shot, for the purpose of preventing the instrument from over-
turning, and causing it to float steadily in a vertical position, or per-
pendicularly to the surface of the fluid in which it is immersed.

Numerous schemes have been promulgated by different ingenious
and experienced philosophers for the improvement of this instrument ;
but however much the forms which have been recommended may differ
among themselves, yet the general principle is the same in all.

The following is a list of the principal writers who have registered
their improvements in the annals of science, viz.

Desaguliers, Guyton, Nicholson, Speer,

Adie, Charles,

Atkins, Clark, Dicas,

Brewster, Deparcieux, Fahrenheit,

Jone

Quin,
Sikes,

and
Wilson.

355. It would be quite superfluous to detail the various alterations
and improvements suggested by these authors; suffice it to say, that in
all there is something different and in all there is something common ;
but that which merits the greatest share of our attention, by reason
of the extreme delicacy of its indications -and the simplicity of its
construction, is the hydrometer of Deparcieux, which was presented
to the Academy of Sciences in the year 1766.

This instrument, which was intended by its inventor to measure the
specific gravities of different kinds of water, is
represented in the annexed figure, where AC is a
glass phial about seven or eight inches in length,
loaded with mercury or leaden shot, to prevent it
from overturning; and in order that no air may
lodge below it, when it is immersed in the fluid, the
lower part is rounded off into the form of a spheric
segment.

In the cork of the phial at A, is fixed a brass wire
of one twelfth of an inch in diameter, and from thirty to thirty six
inches long, or of any other length which may be found convenient
for the purpose, but such, that when the phial is loaded and immersed
in spring water of a medium temperature, the entire phial and about
one inch of the wire should be below the graduated scale DH, which
is fixed upon the side of the tin vessel DEFG ; to the other end, or
summit of the wire, is attached a small box B, intended for containing
the minute weights which it may be found necessary to apply, in order
to cause the instrument to sink to a certain fixed point in the

OF THE EQUILIBRIUM OF FLOATATION. 283

different kinds of water whose specific gravities are required to be
found.*

The white iron vessel DEFG, is used for holding the fluid on which
the experiment is to be performed ; it is generally about three feet in
length, and from three to four inches in diameter, according to the
circumstances under which it may happen to be employed. The small
scale DH, is attached for the purpose of measuring the different depths
to which the instrument sinks when differently loaded, or when it is
immersed in fluids of different specific gravities.

The indications of this instrument are so extremely delicate, that if
a small quantity of alcohol, or a little common salt, be added to the
fluid, the phial will ascend or descend through a very sensible dis-
tance, which circumstance greatly enhances the value of the aerometer;
for in proportion to its sensibility and the delicacy of its indications,
are its importance and utility to be appreciated.

We come now to consider the theory of this instrument, and we
shall just remark in passing, that the same principles, under very
slight and obvious modifications, will apply to any other hydrometric
instrument, of a similar, or nearly similar nature and construction, to
that which forms the subject of our present discussion.

PROBLEM LII.

356. Having given the capacity or volume of the phial, toge-
ther with the dimensions of the immersed wire, and the entire
weight of the aerometer : —

It is required to determine the specific gravity of the fluid,
in which the instrument settles in a state of equilibrium.

Now, because the weight of any body when floating in equilibrio,
whatever may be its form and the substance of which it is composed,
is equal to the weight of the fluid which it displaces ; it follows, that if
we put c =: the capacity or volume of the phial immersed in the fluid,
I zz the length of the immersed wire,
r nr the radius of its transverse section,

TTZZ: 3.1416, the number which expresses the circumference of
a circle whose diameter is equal to unity,

* To the bottom of the box B we have affixed the arm a b, from one extremity of
which is suspended the wire cd carrying the index i, the whole being truly balanced
by the small ball b attached to the other extremity of the horizontal arm ab. In all
other respects the instrument is that of Deparcieux.

284 OF THE EQUILIBRIUM OF FLOATATION.

s m the specific gravity of the fluid sought, and

w=. the entire weight of the aerometer, always known.

Then, it is manifest, that the capacity or volume of the phial,
together with the magnitude of the immersed wire, is equal to the
quantity of fluid displaced ; and the weight of this quantity of fluid
is equal to the weight of the aerometer ; but by the principles of men-
suration, the magnitude of the immersed wire is expressed by Trr*l;
consequently, the quantity of fluid displaced is c-f-7rr2/, and the
magnitude of any body multiplied by its specific gravity is equal to
its weight ; hence we have

t0 = (c + *•»*/)*; (220).

therefore, by division, we obtain
w

(221).

357. Here follows the practical rule for reducing the equation.

RULE. Divide the entire given weight of the aerometer, by
the capacity or volume of the phial, increased by the quantity
of wire immersed, and the quotient will give the specific gra-
vity of the fluid.

358. EXAMPLE. The whole weight of an aerometer, when so loaded
as to have the attached wire depressed 15 inches below the surface of
the fluid, is 23 ounces; required the specific gravity of the fluid,
supposing the diameter of the wire to be one twelfth of an inch, and
the capacity of the phial 40 inches ?

Here, by the mensuration of solids, the magnitude of the wire is

IGX^TXTT^0-082 of a cubic inch, very nearly;
therefore, the whole quantity of fluid displaced, is
40 4- 0.082 zz: 40.082 cubic inches ;
therefore, by the rule, we obtain

The number 0.5738, which we have obtained from the above calcu-
lation, expresses the weight of one cubic inch of the fluid in ounces ;
but since it is customary to express the specific gravity of bodies in
ounces per cubic foot, it becomes necessary, for the sake of compari-
son, to reduce the above result to that standard ; hence we have

s zn 0.5738 X 1728 z= 991 .5264 ounces per cubic foot for
the specific gravity of the fluid on which the experiment was tried.

OF THE EQUILIBRIUM OF FLOATATION. 285

PROBLEM LIII.

359. Having given the capacity or volume of the phial, the
whole weight of the aerometer, the specific gravity of the fluid,
and the radius of the wire : —

It is from thence required to determine, how much of the
stem or wire is immersed below the surface of the fluid when
the instrument rests in a state of equilibrium.

By recurring to the equation marked (220), and separating the
terms, we obtain

Trr*sl— w — cs;

from which, by division, we get

w — cs

T7T' (222).

360. The practical rule for reducing this equation, may be expressed
in words at length, in the following manner.

RULE. From the entire weight of the hydrometer, subtract
the capacity of the phial drawn into the specific gravity of
the fluid ; then, divide the remainder by the area of a trans-
verse section of the wire, drawn into the specific gravity of
the fluid, and the quotient will express how far the wire is
immersed below the upper surface of the fluid, when the
instrument floats in a state of equilibrium.

361. EXAMPLE. The entire weight of an aerometer, when so adjusted
as to remain at rest in a fluid whose specific gravity is 0.5738*nr23
ounces ; what length of the stem or upright wire falls below the sur-
face of the fluid, supposing its diameter to be one twelfth of an inch,
and the magnitude of the immersed phial 40 inches ?

Here, by the foregoing rule, we have
23 — 40x0.5738
'= 3^1416X^7x07673-8 = 15'33 mcheS ™**

362. If the entire weight of the aerometer be multiplied by 1728,
the number of cubic inches in one cubic foot, the formulas (221) and
(222) become transformed into

* The number 0.5738, by which the specific gravity is here expressed, is the
weight in ounces of one cubic inch, which being reduced to the standard of one
cubic foot, gives srr 0.57 38x1728 =98 1.5624 oz.

286 OF THE EQUILIBRIUM OF FLOATATION.

d,__1728?<; — cs
c

from the first of which the standard specific gravity is obtained, and
in the second, the specific gravity as calculated from the first must be
employed.

By comparing the quantities in equation (222) with each other, it
will readily be perceived, that a very small variation in w the weight
of the instrument, or in s the specific gravity of the fluid, will produce
a very considerable variation in Z, the immersed portion of the stem
or wire ; for it is manifest, that the numerator of the fraction w — cs,
expresses the weight of the fluid displaced by the wire or upright
stem of the instrument, and consequently, since r the radius of the
stem is a very small quantity, it follows, that the weight of the fluid
which it displaces must also be very small.

PROBLEM LIV.

363. Suppose that a small variation takes place in the density
of the fluid in which the instrument is immersed : —

It is required to determine the corresponding variation that
takes place in the depth to which it sinks before the equili-
brium is restored.

Let the notation for the first position of equilibrium remain as in
Problem LII., and let I' denote the immersed length of the stem or
wire, corresponding to the specific gravity s' ; then, by equation (222),

we have

_ w — cs'

7T?'V '

consequently, by subtraction, the variation in length becomes
w — cs w — cs'

and this, by a little farther reduction, gives

(223).

364. The practical rule for reducing this equation, may be expressed
in words as follows.

RULE. Multiply the whole weight of the aerometer by the
variation in the specific gravity ; then, divide the product by
the area of the transverse section of the upright stem or wire,

OF THE EQUILIBRIUM OF FLOATATION. 287

drawn into the greater and lesser specific gravities of the fluid,
and the product will express the required variation in the
position of the instrument.

365. EXAMPLE. Suppose the specific gravity of the fluid to vary
from 0.5738 to 0.5926 ounces per cubic inch during the time of the
experiment, what is the corresponding variation in the depth of the
instrument, its whole weight being 23 ounces, and the diameter of the
upright stem equal to one twelfth of an inch ?

Here, by attending to the directions in the rule, we obtain
. ., 23(0.5926 — 0.5738)

Hence it appears, that by a difference of 0.0325 in the absolute
specific gravity of the fluid, there arises a difference of 233 inches in
the position of the instrument ; this seems a very great difference, and
is in reality far beyond the bounds prescribed for the whole apparatus
to occupy ; it serves, however, to exemplify the extreme delicacy of
the principle, and when the changes in the specific gravity are very
minute, the corresponding changes in depth will nevertheless be suffi-
ciently distinct to admit of an accurate measurement.

366. By diminishing the diameter of the upright stem, or increasing
the entire weight of the instrument, which is equivalent to an increase
in the weight of the fluid displaced, the sensibility of the aerometer
may be greatly increased. This is manifest, for by inference 5,
equation (202), it will readily appear, that if the specific gravity
remains the same, the quantity by which the instrument sinks in the
fluid on the addition of a small weight wr, varies directly as the mag-
nitude of the weight added, and inversely as the square of the radius
of the upright stem.

Let us suppose, that by the addition of the small weight w', the
length of the part of the stem Z, which is originally immersed, becomes
equal to /' ; then, by the principles of mensuration, the increased
magnitude of the immersed stem is 7rr*(l' — I); but the weight of a
body is equal to its magnitude multiplied by its specific gravity;
hence we have

*r\l' — l)s = w';

and this, by division, becomes

r-r=+-.

irr^s

Now it is obvious, that by the supposition of a constant specific
gravity, the quantity ITS is also constant : it therefore follows, that

288 OF THE EQUILIBRIUM OF FLOATATION.

V — I varies as — .
r*

In the above investigation, we have supposed the specific gravity of
the fluid to remain constant ; but admitting it to vary, so that s may
become equal to s' ; then, in order that the upright stem may rest at
the same depth of immersion, w must become equal to (w -f- w') ; if,
therefore, we substitute s and (w -)- w'), for s and w in equation (223),
we shall obtain

_ w -\-w' — cs'

I — „ , ,

TrrV

an equation from which we find the value of s' to be
w — w'

and by a similar reduction, equation (223) gives

consequently, by analogy, and suppressing the common denominator,
we get

s' : s : : w -}- w' : w.

From this analogy, the difference between the specific gravities in
the two cases can very easily be ascertained, for by the division of
ratios, we have

s' — s : s : : w -\- w' — w : w,

which, by reduction, becomes
__ w' s
~ 17* (224),

367. This is a very simple equation for expressing the difference
of the specific gravities ; it may be reduced by the following practical
rule.

RULE. Multiply the added weight by the lesser specific
gravity ; then, divide the product by the lesser weighty and
the quotient will be the difference between the specific gravities
sought,

368. EXAMPLE. An aerometer, whose absolute weight is equal to
23 ounces, is quiescent in a fluid whose specific gravity is 0.5738
ounces, as referred to a cubic inch ; but on being put into a denser
fluid, it requires the addition of 0.7536 of an ounce, to cause the
instrument to sink to the same depth ; what is the specific gravity
of the denser fluid?

OF THE EQUILIBRIUM OF FLOATATION. 289

Here then we have given w' = 0.7536 of an ounce, and s = 0.5738 ;
consequently, by the above rule, we have

consequently, the specific gravity of the heavier fluid, is

s' = 0.5738 -f 0.0188 = 0.5926 ;

and this, when reduced to the standard of one cubic foot, becomes
0.5926x1728 = 1024.0128, which, on being referred to a table of
specific gravities, will be found to correspond with sea water at a
medium temperature.

In the above operation, we have taken the specific gravity as re-
ferred to one cubic inch of the fluid only, but the well informed reader
will readily perceive, that the same result would obtain if the specific
gravity should be estimated by the cubic foot ; for in that case, we
should have w1 — 0.7536 of an ounce, and 5 = 991.5264, conse-
quently, by the rule, we have

therefore by transposition, the specific gravity of the denser fluid, is
s' = 991.5264 4- 32.4864 = 1024.0128, being precisely the same
result as that which we obtained on the former supposition.

369. The diagram which we have employed to illustrate the general
principle of the aerometer, is at the best but a very rude and imper-
fect representation, and in its present state, it is altogether unfitted
for ascertaining the specific gravities of fluids with any degree of pre-
cision ; it is therefore requisite, in cases where extreme accuracy is
required, to have recourse to some other method of indicating the
precise measure of density, and for this purpose, the hydrometer or
aerometer, is very advantageously replaced by the

HYDROSTATIC BALANCE,

an instrument which determines the specific gravities of bodies with
the greatest correctness, and which, on account of its simplicity and
cheapness, is rendered available for almost every purpose in which
the specific gravity of bodies forms the subject of inquiry.

The Hydrostatical Balance, so called, is nothing more than a
common balance, furnished with some additional apparatus for ena-
bling it to measure the specific gravities of bodies with accuracy and
expedition, whether the bodies be in a solid or a fluid state. The
description of the instrument is as follows.

VOL. i. u

290

OF THE EQUILIBRIUM OF FLOATATION.

Let AB be the beam of a balance very nicely equipoised upon its
centre of motion at c, and suspended
from the fixed object represented at F,
the centering being so delicately exe-
cuted, that the equilibrium of the instru-
ment is disturbed by the smallest por-
tion of a grain being added to or sub-
tracted from either arm of the beam.

D and E are two scales, which, together
with their appendages are also balanced
with the greatest exactness ; one of them
as E having a hook in the middle of its
bottom surface, to which the weight w
is suspended by means of a horse hair,

or any other flexible substance of such extreme levity, as to have no
sensible effect upon the equilibrium.

p is an upright pillar placed directly under the centre of motion,
and carrying the circular arc mmm, which serves to prevent a too
great vibration on either side, and also, by means of the index i, which
is fixed on the beam immediately under the fulcrum, it indicates the
exact position of equilibrium ; for it is manifest, that when the beam
is horizontal, the pointer must be directly over the middle of the arc.
The pieces in the scale D, denote the weight of the body when
weighed in air ; but when the body is immersed in water, as repre-
sented by the figure abed, the scale D with its accompanying weights,
must evidently preponderate, and for the purpose of restoring the
equilibrium, small weights must be placed in the opposite scale at E ;
and since the weights thus added, indicate the weight of a quantity of
water of equal bulk with the immersed body, it follows, that the specific
gravity of the body can from thence be determined.

The hydrostatical balance, like the hydrometer or aerometer pre-
viously explained, has undergone various alterations and improve-
ments, according to the ideas of the different individuals who have
had occasion to apply it in their inquiries respecting the specific
gravities of bodies ; but since the general principle is the same in all,
under whatever form the instrument may appear, it would lead to
nothing useful to enter into a detailed description of the various
improvements which it has received, and the numerous changes that
have been made upon it ; we shall therefore refrain from farther dis-
cussion on the nature of its construction, and proceed to exemplify the
manner in which it is applied to the determination of specific gravities.

OF THE EQUILIBRIUM OF FLOATATION. 291

PROBLEM LV.

370. Having given the specific gravity of distilled water,
equal to 1000 ounces per cubic foot : —

It is required to determine the specific gravity of a solid
body that is wholly immersed in it.

It is manifestly implied by the total immersion of the body, that its
specific gravity exceeds the specific gravity of the fluid in which it is
immersed ; therefore, attach the body to the hook in the bottom of
the scale E by a very fine and light thread, and balance it exactly by
weights put into the other scale at D ; then, immerse the body in the
water, and find what weight is required to restore the equilibrium, the
weight thus required will measure the specific gravity of the body.
Put w z= the weight of the body when weighed in water,
w' ~ the weight when weighed in atmospheric air,
s zn the specific gravity of water, and
s' — the specific gravity of the body sought.

Then is w — w' equal to the weight which must be put into the scale
E to restore the equilibrium; consequently, by the fifth proposition,
we have . w' — w : w' : : s : s' ;

from which, by reduction, we get

,'-_^i-

~w' — w (225).

371. The following is the practical rule in words at length for
reducing the above equation.

RULE. Multiply the weight of the body when weighed in
air, by the specific gravity of the fluid, and divide the product
by the weight which it loses in water for the specific gravity
of the body.

This rule determines the specific gravity of the body when it
exceeds that of the fluid in which it is weighed ; but when the body
is specifically lighter than the fluid, the method of finding its specific
gravity is shown in Problem XLIV., it is therefore unnecessary to
repeat it here.

372. EXAMPLE. If a piece of stone weighs 20 Ibs. in air, but in water
only 13J Ibs.; required its specific gravity, that of water being 1000 ?

Here, by the rule, w=: 13J, w'= 20, s=z 1000,

20X1000 20000
therefore s =— T^T— ~ 3076.923 rz the specific gravity

Z(j —~ Aug O.O

of the mass when it is wholly immersed in water.

u 2

CHAPTER XII.

OF THE POSITIONS OF EQUILIBRIUM.

PROBLEM LVI.

373. Suppose that a solid homogeneous triangular prism,
floats upon the surface of a fluid of greater specific gravity
than itself, with only one of its edges immersed : —

It is required to determine in what position it will rest,
when it has attained a state of perfect equilibrium.

Let ABC be a vertical transverse section, at right angles to the axis
of the homogeneous prism, floating A
in a state of equilibrium on the fluid
whose horizontal surface is IK.

Bisect AB, BC the sides of the tri-
angle in the points r and n, and D E,
D c in the points H and m ; draw the
straight lines CF and AH, intersecting
one another in the point G, and CH,
EWI intersecting in g ; then is G the
centre of gravity of the whole triangle
ABC, and g the centre of gravity of
the triangle DEC, which falls below
DE the plane of floatation.

Join the points G, g by the straight
line G # ; then, according to the prin-
ciple announced and demonstrated in the sixth proposition, the straight
line gc, is perpendicular to DE the surface of the fluid.

Draw FH, and because CF and CH the sides of the triangle CFH,
are cut proportionally in the points G and g, it follows from the prin-
ciples of geometry, that the straight lines Gg and F H are parallel to
one another ; but we have shown that gG is perpendicular to the hori-

OF THE POSITIONS OF EQUILIBRIUM. 293

zontal surface of the fluid, or the plane of floatation passing through
DE; consequently, F H is also perpendicular to DE, and FD, FE are
equal to one another.

Put a — A B, the unimmersed side of the triangular section,
b nz BC, one of the sides which penetrate the fluid,
c zz AC, the other penetrating side,
d =2 CF, the distance between the vertex of the section, and

the middle of the extant side,
0 zz the angle ACF, contained between the side AC and the

line CF,
^' zz the angle BCF, contained between the line CF and the

side B,

s zz the specific gravity of the solid body,
s' zz the specific gravity of the fluid on which it floats,
x zz CD, the immersed portion of the side AC, and
y zz CE, the immersed portion of the side BC.

Then, according to the principles of geometry, since the line CF is
drawn from the vertex of the triangle at c, to the middle of the base
or opposite side at F, it follows, that

AC2 -f BC2ZZ2(AF2-f CF2),

or by taking the symbolical representatives, we shall obtain

from which, by reduction, we get

dzziV2(^4-c2) — a*. (226).

Since all straight lines drawn parallel to the axis of the prism are
equal among themselves ; it follows, that the weight of the whole solid
ABC, and that of the portion DEC below the plane of floatation, which
corresponds to the magnitude of the fluid displaced, are very appro-
priately represented by the areas drawn into the respective specific
gravities of the solid and the fluid on which it floats.

Now, the writers on the principles of mensuration have demon-
strated, that the area of any right lined triangle : —

Is equal to the product of any two of its sides, drawn into
half the natural sine of their contained angle.

Therefore, if we put a' and a" to represent the areas of the triangles
ABC and DEC respectively, we shall have for the area of the triangle
ABC,

294 OF THE POSITIONS OF EQUILIBRIUM.

and for the area of the triangle DEC, it is

But according to the principle demonstrated in the third proposition
preceding, the weight of a floating body : —

Is equal to the weight of the quantity of fluid displaced.

Consequently, the weight of the solid prism whose section is ABC,
is equal to the weight of the fluid prism, whose section is D EC ; that is,

\bcs sin.ty 4- 4>') = \xys' sin.ty -f 0'),
and from this, by suppressing the common quantities, we get

bcs — xys. (227).

By the principles of Plane Trigonometry, it is
F D2 zz d* + x* — 2c?a- cos.<£, and F E2 =z d* -|- y* — 2dy cos.^' ;
but these by construction are equal ; hence we have

Let the value of d as expressed in equation (226), be substituted
instead of it in the above equation, and we shall obtain

a?*— x cos.(j> V 2(c2 -f 62) — a2 — ?/2— y cos. f V 2(c2-f 62) — a2. (228).
Recurring to equation (227), by division, we have

bcs , r , . , . 6Vs2

?/ zn . , the square of which is w2 zn ;

xs x*s 2

substitute these values of y and y9 in equation (228), and it is

bcscos.Q' .

and multiplying by a?2 we obtain,
£rcV bcs cos.0'

5 • s

from which, by transposition, we get

a-4— cos. / 2^+c2— a2 X x*-\ -- — - 262 c2 — a2 X x=

(229).

374. The equation as we have now exhibited it, involves the several
circumstances that accompany the equilibrium of a floating body, and
its root determines the position in which the equilibrium obtains ; the
general form of the expression, is however exceedingly complex, and
rising as it does to the fourth order or degree, the resolution is neces-
sarily attended with considerable difficulty, especially when the sides

OF THE POSITIONS OF EQUILIBRIUM. 295

of the transverse section are represented by large numbers ; in parti-
cular cases, the ultimate form will admit of being modified, and may
in consequence, be rendered somewhat more simple ; but it must
nevertheless be understood, that whenever the position of equilibrium
is required by computation, we must inevitably perform a very irksome
and laborious process.

A geometrical construction may also be effected by the intersection
of two hyperbolas ; but since this implies a knowledge of principles
higher than elementary, we think proper to pass it over, and proceed
to illustrate the application of the above equation by the resolution of
a numerical example.

375. EXAMPLE. Suppose a triangular prism of Mar Forest fir,
the sides of whose transverse section are respectively equal to 28, 26,
and 18 inches, to float in equilibrio in a cistern or reservoir of water,
having only one angle immersed ; it is required to determine the posi-
tion of equilibrium, on the supposition that the two longest sides of
the section penetrate the fluid, the specific gravity of the prism being
to that of water as 686 to 1000 ?

By recurring to equation (229), and comparing its several consti-
tuent quantities with the parts of the diagram to which they respec-
tively refer, it will readily appear, that x, cos.0 and cos.0' are the only
terms whose values require to be calculated ; of which cos.0 and cos.0'
are to be determined from the nature of the figure, and x from the
resolution of the biquadratic equation in which its values are involved.

The length of the straight line CF, which is drawn from the vertex
of the section at c, to the middle of the opposite side at F, is accord-
ing to equation (226), expressed by

consequently, by substituting the numerical values of the sides, we
obtain _

d= 1^2(28* + 262) — 182 = 25.4754784 inches.
Hence, in the triangles ACF and BCF respectively, we have given
the three sides AC, AF, FC and BC, BF, FC to find COS.ACF and cos.
BCF; for which purpose, we have the following equations as deduced
from the elements of Plane Trigonometry, viz. In the triangle ACF,
it is

4c« d« — a2

and in the triangle BCF, it is

296 OF THE POSITIONS OF EQUILIBRIUM.

Now, by substituting the numerical values of a, b and c, as given
in the question, and the value of d as deduced from calculation, the
absolute values of cos.0 and cos.0' will stand as below.

Thus, for the absolute numerical value of cos.^>, we have
4(784-f649) — 324

and for the absolute numerical value of cos.^', it is
649) — 324

Let the numerical values of cos.^> and cos.^' as determined by the
above computation, together with the numerical values of a, b, c, s,
and s', as given in the question, be respectively substituted in equa-
tion (229), and we shall obtain

x4 — 48.2S59*3 4- 23894.7* =: 249408 ;

but in order to simplify the resolution of this equation, it will suffice
to take the co-efficients to the nearest integer, for the error thence
arising will be of very little consequence in cases of practice, and the
modification will very much abridge the labour of reduction ; the
equation thus altered, will stand as below.

x4—4Sxs -f- 23895* = 249408.

Therefore, if this equation be reduced by the method of approxima-
tion, or otherwise, the value of or will come out a very small quantity
less than 22 inches ; but taking it equal to 22, the result of the equa-
tion is

224— 48 X223 4- 23895X22 = 248842.

By substituting the given values of b, c, s and s', with the com-
puted value of x, in equation (227), we shall have

22000^ = 499408,
from which, by division, we obtain
499408

22000 '

Consequently, from these computed dimensions, together with the
sides of the section given in the question, the prism may be exhi-
bited in the position which it assumes when floating in a state of
equilibrium.

376. Construct the triangle ABC to represent the transverse section of
the floating prism, and such that the sides AC, BC, and AB are respec-
tively equal to 28, 26, and 18 inches; make CD and CE respectively

OP THE POSITIONS OF EQUILIBRIUM. 297

equal to 22 and 22.7 inches, and through the points D and E, draw

the straight line IK, which will coincide

with the plane of floatation, or the sur- A

face of the fluid on which the body floats. \ Jx^xT; — -3

Bisect A B, the extant side of the sec- T \^ ^^1

tion in the point F, and join F D and F E ;
then, the conditions of equilibrium ma-
nifestly are, that the lines FD and FE
are equal to one another, and that the
area of the immersed triangle DCE, is
to the area of the whole triangle ACB,
as the specific gravity of the solid is to
the specific gravity of the fluid.

That the lines FD and FE are equal to one another, appears from
an inspection and measurement of the figure ; but the following proof
by calculation will be more satisfactory, inasmuch as numbers can be
more correctly estimated than measured lines, which depend for their
accuracy upon the delicacy of the instruments and the address of the
operator.

In the plane triangle DCF, we have given the two sides DC and CF,
respectively equal to 22 and 25.4754784 inches, and the natural cosine
of the contained angle DCF equal to 0.94769 ; consequently, the third
side D F can easily be found ; for by the principles of Plane Trigono-
metry, we know that

DF2=:DC24- FC2 2DC.FCCOS.DCF;

therefore, by substituting the respective numerical values, we obtain

DF2 = 484 + 649 — 2X22X25.4754784X0.94769 = 70.72;

consequently, by extracting the square root, it is

DFZZI V70.72 = 8.4 inches.

Again, in the plane triangle ECF, we have given the two sides
EC and CF, respectively equal to 22.7 and 25.4754784 inches, and
the natural cosine of the contained angle ECF equal to 0.93906;
consequently, by Plane Trigonometry, we have

EF2zrEC9-|-CF2 2EC.CFCOS.ECFJ

and substituting the respective numerical values, we obtain
EF2 =: 515.29 -f- 649 — 2 X22.7 X25.4754784 X 0.93906 =: 77.895 ;
therefore, by extracting the square root, we shall have
E F zz: V 777895 — 8 .82 inches.

298 OF THE POSITIONS OF EQUILIBRIUM.

Hence/the length of the line DF is 8.4 inches, and the length of EF
is 8.82 inches, giving a difference of 0.42, or something less than half
an inch ; being as small a difference as could be expected, from the
manner in which the co-efficients of the equation that furnished the
value of x were modified, and also from the circumstance of x being
determined only to the nearest integer, without considering the frac-
tions with which it might be affected.

377. Upon the whole then, the position of equilibrium is sufficiently
manifest, from the condition of equality between the straight lines D F
and EF ; we shall therefore proceed to inquire if it be equally apparent,
from the proportionality between the triangles ACB and DCE.

Since cos. </> = 0.94769, and cos. f — 0.93906, it follows that
<j> = l8° 37' and </>':=: 20° 6'; consequently, by addition, the whole
angle ACB becomes

04-f = 18° 37'4-20° 6' = 38° 43';

therefore, in each of the triangles ACB and DCE, we have given the
two sides AC, BC and DC, EC with the contained angle ACB common
to both, to find the respective areas.

Now, the writers on mensuration have demonstrated, that when
two sides of a plane triangle, together with the contained angle are
given : —

The area of the triangle is equal to the product of the two
sides drawn into half the natural sine of their included angles

378. This is a principle which we have already stated in the investi-
gation, and expressed analytically in deducing equation (227) ; we
shall now employ it in determining the areas of the triangles according
to the magnitudes of the sides and the contained angle, as given in
the example and derived from computation.

The natural sine of 38° 43' is 0.62547, and the sides AC and BC are
respectively 28 and 26 inches ; consequently, by the above principle,
we have

a' = 1(28X26X0.62547) — 227.671 square inches.
The natural sine of the contained angle remaining as above, the
sides DC and EC as derived from computation, are equal respectively
to 22 and 22.7 inches ; hence, from the same principle, we have

a"= 1(22X22.7X0.62547) = 156.1798 square inches.
Now, according to the conditions of the question, the specific
gravity of the fluid is 1000, and that of the floating body is 686;
consequently, we obtain

1000 : 227.671 : : 686 : 156.1823.

OF THE POSITIONS OF EQUILIBRIUM.

299

In this case the errorx is extremely small, amounting only to
156.1823 — 156.1798 = 0.0025 of a square inch; hence we con-
clude, that the position of equilibrium under the given conditions, is
very nearly the same as we have found it to be from the resolution of
the equations (227) and (229).

379. The preceding solution, however, indicates only one position
of equilibrium ; but it is manifest from the nature of the equation
(229), that there may be more, for by transposition, we have

bcscos.ti'

/ a i T

-a2 X x—

.
= 0,

-,/E

and it is demonstrated by the writers on algebra, that in every equa-
tion of an even number of dimensions, having its last term negative,
there are at least two real roots, the one positive and the other nega-
tive ; consequently, the above equation has two of its roots real and
determinable ; but the equation being of four dimensions has also four
roots, hence, the other two roots may also
be real, and in that case, there will be three
values of x positive and the fourth negative;
but for every positive value of x there may
be a position of equilibrium, that is, there
may be three positions, in which the body
may float in equilibrio with the angle ACB
downwards ; but there cannot be more.

380. If the sides b and c are equal to one
another, as represented in the annexed dia-
gram, then cos.<£ and cos.^' are also equal,
and the general equation becomes

— «*Xas-f.

?2— a* X x — -JT = 0. (^30) .

Now, it is manifest from the relation of the terms in this equation,
that it is resolveable into the two quadratic factors

b*s bzs

a? — -~r, and x* — cos.<W462 — cfXx-}-— r, each of which
« s

is equal to nothing ; consequently, the four roots of the equation
(230), are the same as the roots of the two quadratic equations

o;2iz:— r> and x2 — cc
s

and the positions of equilibrium are indicated by the number of real
positive roots which these equations contain.

300 OF THE POSITIONS OF EQUILIBRIUM.

By extracting the square root of both sides of the equation or*zz _ »
we shall obtain

This expression exhibits two roots of the original equation (230), one
positive and the other negative ; but the positive root only becomes
available in determining the position of equilibrium, the negative one
referring to a case whjch does not exist.

It has already been shown in equation (227), that when a solid
body floats in equilibrio on a fluid of greater specific gravity than

itself; then we have

xys' — bcs,

but according to the supposition, b and c are equal to one another ;
hence we get

from which, by division, we obtain

b*s

y = ^'
or, by substituting the above value of x, it becomes

. (232,

Hence it appears, that the values of x and y are each of them
expressed by the same quantity; consequently, the triangle DCE is
isosceles, and AB the extant side of the section, is parallel to DE the
base of the immersed portion, both of them being parallel to the plane
of floatation or the horizontal surface of the fluid.

381. The practical rule for the reduction of the equation (231) or
(232), may be expressed in words at length, in the following manner.

RULE. Divide the specific gravity of the solid body, by the
specific gravity of the fluid on which it floats ; then, multiply
the square root of the quotient, by the length of one of the
equal sides of the section, and the product will give the portion
of that side which is immersed below the plane of floatation, or
that which is intercepted between the vertex of the section
and the horizontal surface of the fluid.

382. EXAMPLE. A prism of wood, the sides of whose transverse
section are respectively equal to 20, 28 and 28 inches, is placed
with its vertex downwards in a cistern or reservoir of water whose

OF THE POSITIONS OF EQUILIBRIUM. 301

surface is horizontal ; it is required to determine, what position the
solid will assume when in a state of equilibrium, its specific gravity
being to that of water as 686 to 1000 ?

Here, by the rule, we have

from which, by extracting the square root, we get
^0.686—0.8282;

and finally, by multiplication, we obtain
x = 28 X 0.8282 = 23.1896 inches.

But according to equation (232), y possesses the very same value ;
consequently, if 23.1896 inches be set off from the vertex of the
section upwards on each of its equal sides, the straight line which
joins these points will coincide with the plane of floatation, or the
horizontal surface of the fluid on which the body floats.

383. This is the most natural and obvious position of equilibrium,
and such as must always obtain when the body is homogeneous, and
symmetrical with respect to a vertical plane passing through the axis
and bisecting the base ; but there may be other situations in which
the body may float in a state of quiescence, and the circumstances
under which they occur must be determined by the resolution of the
following equation, viz.

b*s

Complete the square, and we shall have

2— o8 X * -f ( 1
and by extracting the square root, we get

x— COS.0V— a = : JCOS2^(463 — a2)— -j»

hence, by transposition, we shall obtain, (233).

/ ^

— a*^y J cos2.^(4^— a2) — -,

The corresponding values of y are (234).

y = \ cos.<£ V462 — a?±2\/

302 OF THE POSITIONS OF EQUILIBRIUM.

Expressions of this form, arising from the reduction of an adfected
quadratic equation, are in general rather troublesome and difficult to
render intelligible in words, and even when intelligibly expressed, they
are to say the least of them, but very dull and uninviting guides, from
which a tasteful reader turns with disgust ; we are therefore unwilling
to crowd our pages with long and formal directions for the purpose of
reducing equations, when it is probable after all, that nine out of every
ten of our readers will pass them over, and proceed immediately to
discover their object by the direct resolution of the original equation.

384. It is however necessary, in conformity to the plan of our
work, to express the most important final equations in words at length,
and since the preceding forms are of considerable utility in the doctrine
of floatation, it would be a direct violation of systematic arrangement,
to omit the verbal description, and leave the subject open only to
algebraists ; we shall therefore, in order to render both parts of the
operation intelligible, endeavour to express the method of reduction
in as brief and comprehensive a manner as the nature of the subject
will admit.

1 . To determine the value of x.

RULE. From four times the square of one of the equal
sides of the section, subtract the square of the base, or side
opposite to the vertical angle ; multiply the square root of the
remainder by one half the natural cosine of half the vertical
angle, and call the product ?n.

From four times the square of one of the equal sides of the
section, subtract the square of the base, or side opposite to the
vertical angle, and multiply the remainder by one fourth of
the square of the natural cosine of half the vertical angle,
or that which is immersed in the fluid ; then, from the product,
subtract the quotient that arises, when the specific gravity of
the solid, drawn into the square of one of the equal sides
of the section, is divided by the specific gravity of the fluid,
and call the square root of the remainder n.

Finally, to and from the quantity denoted by m, add and
subtract the quantity denoted by n ; then, the sum in the one
case, and the difference in the other, will give the two values
of*.

2. To determine the corresponding values of y.

RULE. Calculate the values of m and n, precisely after the
manner described above; then, from and to the quantity

OF THE POSITIONS OF EQUILIBRIUM. 303

denoted by m, subtract and add the quantity denoted by n,
and the difference in the one case, and the sum in the other,
will give the values ofy corresponding to above values of x.

385. These are the rules by which the other positions of equilibrium
are to be determined ; but it is necessary to remark, that beyond cer-
tain limits no equilibrium can obtain. In the first place, in order that
the body may float with only one of its angles immersed, it is mani-
festly requisite, that the equal sides of the section should each be
greater than m -j- n ; and in the second place, in order that x and y
may be real positive quantities, the expression J cos2.^>(462 — a2) must

. tfs s , cos2.d>(4£2— a )

exceed — — , or — must be less than - r. ,, - .
' 2

reason of these limitations is obvious from the nature of the
quadratic formula (233) and (234), but it will be more satisfactory to
show, that unless the data of the question are so constituted as to
fulfil these conditions, the rules will fail in determining the positions
of equilibrium ; or in other words, there is no other position in which
the body will float at rest, but that which is indicated by the equa-
tions (231) and (232).

386. EXAMPLE. The data remaining as in the preceding example,
let it be required to determine from thence, whether under the pro-
posed conditions, the body can float at rest in any other position than
that which we have already assigned for it, by the reduction of the
equations (231) and (232), in which the extant side or base of the
figure is parallel to the horizon.

By the principles of Plane Trigonometry, we have

\ cos.0 n= TV v7 28 + 1 0) (28 — 1 0) = !(0.93406) = \ cos.20° 55' 29" ;

consequently, by proceeding according to the rule, we'get
m~\ cos.</>(462— as)*zz0.46703 ^4 X282— 20*=: 24.429 very nearly.

Again, to determine the value of w, it is

i cos.80(462 — a2) •=. 0.467032(4 X282 — 202) = 596.768,

and for the value of the term, involving the specific gravities, we have

consequently, by subtraction, we get

596.768 — 537.824 = 58.944.

It therefore appears from the last result, that both the values of x
and y are real positive quantities ; consequently, one of the limiting

304 OF THE POSITIONS OF EQUILIBRIUM.

conditions is answered, and we shall shortly see, whether or not the
data are sufficient to satisfy or fulfil the other condition.

By extracting the square root of 58.944, we get

w=-V/58.944 = 7.677 nearly;
therefore, by addition and subtraction, the values of x, are

a: = m + 72 = 24.429 +7.677 = 32. 106, and x — m — n — 24.429—
7.677 zz 16.752 inches.

Now, we have seen by equation (234), that the corresponding
values of y are expressed in the same terms, having the signs of the
second member reversed ; hence we have

y = 16.752, and y = 32.106 inches.

But here we have m -\- n zn 32.106 inches, being greater than b the
downward side of the transverse section, which by the question is only
28 inches ; it therefore follows, that with the proposed data and under
the specified circumstances, there is only one position in which the
body can float in a state of rest, and it is that which we have already
determined, where the base of the section, or the extant side of the
body, is parallel to the surface of the fluid.

But we may here observe, that notwithstanding the values of a: and
y, as we have just assigned them, do not satisfy the conditions of the
question, yet they are not to be considered as being useless ; for they
actually serve, with a slight modification of the body, to furnish posi-
tions in which it will float at rest, although those positions do not
agree with the case, in which only one angle of the figure falls below
the plane of floatation.

387. The positions of equilibrium corresponding to the preceding
values of a; and y, are
as represented in the an-
nexed diagrams, where
E D is the horizontal sur-
face of the fluid, ABC
being the position which
the body assumes when
x is equal to 32.106 and
y equal to 16.752 inches,
and abc the correspond-
ing position when y is equal to 32.106 and x equal to 16.752 inches ;
these being the respective values as obtained by the above numerical
process.

OF THE POSITIONS OF EQUILIBRIUM. 305

That the positions here exhibited are those of equilibrium, is very
easy to demonstrate, for produce the sides CA and cb to meet the
surface of the fluid in the points E and D, and bisect AB and a b in
the points F and/; then, if the straight lines FE, FH and/D, /i be
drawn, they will be equal among themselves.

This is one of the conditions of equilibrium, as we have already
demonstrated in the construction of the original diagram, and the
other condition is, that the areas of the immersed figures ECH and
DCI, are respectively to the whole areas ABC and a be, as the specific
gravity of the solid, is to the specific gravity of the fluid which sup-
ports it.

Now, if the first of these conditions obtain, that is, if the straight
line FE be equal to FH, and/D equal to/i, then, by the principles of
Plane Trigonometry, we shall have

EC2-j-CF2 — 2EC.CFCOS.ECFZTHC*-}- c^ — 2HC.CF COS.FCH ;

but the angles ECF and FCH are equal to one another, and each of
them equal to 0 ; consequently, by substituting the literal representa-
tives, we have

xa -f- d* — 2rf x cos.^> — y3- -)- d* — 2dy cos.0,
or by expunging the common term e?4, we get

x* — 2dx cos.0 z= ?/* — ^dy cos.0,
and this, by transposing and collecting the terms, becomes

#2 — ?/2 zz: 2e? cos.^(o? — y) ;

therefore, jf both sides of this equation be divided by the factor (x — y),
we shall obtain

x -f- y zr: 2rf cos.^>.

Now, by a previous calculation we found x to be equal to 32.106
inches, y 16.752 inches, d equal to \/288 — 10% and cos. 0 equal to
0.93406 ; consequently, by substitution, we have

32.106 + 16.752 = 2x0.93406X6/19";
hence the equality of the lines FE and FH is manifest.
What we have shown above with respect to the triangle ABC, may
also be shown to obtain in the triangle a be, the one being equal and
subcontrary to the other ; this being the case, it is needless to repeat
the process; but we have yet to prove, that the area CEH, is to the
whole area ABC, as the specific gravity of the floating body, is to that
of the fluid on which it floats.

therefore, by the principles of Plane Trigonometry, we get

AC : AF : : rad. : sin.ACF,
VOL. i. x

306 OF THE POSITIONS OF EQUILIBRIUM

or numerically, we shall obtain

28 : 10 : : 1 : sin.«/>= 0.35714,

and we have already found that

cos.^ =: V (28 -f 10) (28 — 10) -f- 28 = 0.93406 ;

but according to the arithmetic of sines, it is

sin.20 — 2 sin.0 cos.0,
and by substituting the above numerical values, we get

\ sin.20 = 0.35714 X 0.93406 = 0.33359.

Then in the triangle ECH, there are given the two sides EC and HC,
respectively equal to 32.106 inches and 16.752 inches, together with
half the natural sine of the contained angle ; to find the area of the
triangle.

Now, by the principles of mensuration, the area of any plane tri-
angle is expressed by half the product of any two of its sides, drawn
into the natural sine of the contained angle, hence we get

32. 106X 16.752 X0.33359 = 179.417 square inches.
Again, in the isosceles triangle ABC, there are given the sides AC
and BC, respectively equal to 28 inches, and half the natural sine of
the contained angle ACB, equal to 0.33359 ; to find the area.
Here, by the principles of mensuration, we have
28X28X0.33359 zz 261.53456 square inches;

then, by the property of floatation, it is
1000 : 686 : : 261.53456 : 179.413 square inches.

388. Since this result agrees so very nearly with that derived from
a direct computation of the triangular area, we may reasonably con-
clude, that the positions exhibited in the diagram are those of equili-
brium ; it is however necessary to remark, that since the weight of the
body remains unaltered in what position soever it may be situated, it
does not readily appear in what manner the adequate quantity of fluid
is displaced, unless we conceive some physical plane, of sufficient
breadth and totally destitute of weight, to be fixed on that edge of the
solid which becomes immersed by reason of the change of position
that the body is supposed to undergo.

This plane, during the oscillation of the prism, will dislodge the
fluid which occupies the space EAW or vbn, and the weight of this
quantity of fluid added to that which is displaced by the quadrilateral
figure cj mn or cbni, will be equal to the whole weight of the float-
ing body

OF THE POSITIONS OF EQUILIBRIUM. 307

389. The above modification, however, does not strictly accord with
the conditions of the problem ; we must therefore inquire, whether the
just principles of equilibrium do not depend upon some other element,
such as the specific gravity. Now, we have already stated, that in
order to have the values of x and y real positive quantities, it is neces-
sary that

s , coss.0(46a — -a9)

— should be less than - prs - »
s 4kb9

and for a similar reason

s .

— must be greater than

And if the specific gravity of the fluid be denoted by unity, as is
the case with water, then the specific gravity of the floating body must
lie between the limits

cos9^(4&g — aa) and cos.ft V 46* — a9
4&2 b

The specific gravity of the floating body, as we have proposed it in
the question, is 686, that of water being denoted by 1000; conse-
quently, when the specific gravity of water is expressed by unity, that
of the solid is 0.686; let us therefore try if this number lies between
the above limits ; for which purpose, we must substitute 28 for b, 20
for a, and 0.93406 for cos.0 ; then we shall have as follows.

0.934062(4X28? — 202)
For the greater limit we have s = - ;. OQa - - — 0.761

4X ~°

nearly.

It is therefore manifest, that the specific gravity of the floating
body, as we have employed it, is less than the greater limit, and con-
sequently properly chosen with regard to it, and we have next to
inquire if it exceeds the lesser limit ; for which purpose, it is

Here then it is obvious, that the lesser limit exceeds the given
specific gravity ; and from this we infer, that without the modification
specified above, the body will not fulfil the conditions of the problem
in any other position than that in which its base is parallel to the
surface of the fluid ; but if the specific gravity of the floating body
fall between the numbers 0.761 and 0.745, all other things remaining,
then the prism, besides the situation of equilibrium in which its base
is parallel to the surface of the fluid, may have two others, in both of

308

OF THE POSITIONS OF EQUILIBRIUM.

which the conditions of the question will be truly satisfied, for only
one angle of the figure will fall below the plane of floatation.

In order therefore to exhibit those positions, we shall suppose the
specific gravity of the floating prism to be expressed by 0.753, which
is the arithmetical mean between the limits above assigned ; then, by
operating according to the rules under equations (233) and (234), we
shall obtain _

— tf — 0.46703 ^4x28'— 202rz 24.429 as for-

merly computed ;

and after a similar manner, we have

— a9) — =

0.46703S(4X289 — 20*)

1UUO

=z2.528;

consequently, by addition and subtraction, we shall get
x = m -f n — 24.429 -f 2.528 = 26.957 inches, and x = m — n =

24.429 — 2.528 — 21.901 inches;
and the corresponding values oft/, are
21.901 and 26.957 inches respectively.

390. The positions of equilibrium corresponding to the above values
of a: and y, are as represented
in the annexed diagrams, where
it may be shown that the
straight lines F E, F H and /D,
fi are equal to one another,
and also that the areas of the
immersed spaces E c H and D ci
are respectively to the whole
areas ABC and abc, as the
specific gravity of the solid, is to that of the fluid on which it floats,
or as 0.753 to unity in the case of water.

These conditions being satisfied, the body will float in equilibrio in
the positions here exhibited; and it from hence appears, that the
problem admits of a complete solution, by retaining the specific gra-
vity of the solid within determinate limits.

391 . When the transverse section of the floating prism, is in the
form of an equilateral triangle ; then a and b are equal to one another,
and equation (230) becomes

— _ b s cos id) \/ 3 b^ s*

r $' 5/9 >

OF THE POSITIONS OF EQUILIBRIUM. 309

and if the value of s' be expressed by unity, as in the case of water,

then we have

— 6V = 0. (235).
Now, it is manifest that this equation is composed of the two
quadratic factors x* — b* srz:0, and .r'^-^cos.^V^Xa; 4- b*s=:&,
whose roots give the positions of equilibrium.

Since the sides a and b are equal to one another, and s' equal to
unity ; then, the limits between which the value of s must be retained,
are

Jcos*.0 and cos.<j>^~3—l ;

but in the case of the equilateral triangle, ^ zr 30° ; consequently,
cos.^> — J ^/ 3, and cos2.0 — j ; therefore, by substitution, the above
limits become

Tgs= 0.5625, and $• — lrz:0.5,

the arithmetical mean of which, is

£(0.5625 + 0.5) = 0.53125.

Let this value of s be substituted instead of it, in each of the con-
stituent quadratic factors, and the equations whose roots determine
the positions of equilibrium, become respectively

** — 0.531256% and x* — bcos.(j>^/3^x = — .531256*;
but by the property of the equilateral triangle,

0 =n 30°, and consequently cos.0 — \<J 3 ;
hence, the above adfected quadratic equation becomes,

x%— 1.56ar = — .5312563.

392. If b the side of the triangle be equal to 28 inches, as we have
hitherto supposed it to be; then, the preceding equations become

a2 = 416.5, and a2 — 42a = — 416.5.

Now, it is manifest, that the first of these equations has one positive
and one negative root, each of them being expressed by the same
numerical quantity, viz. the square root of 416.5 ; for by extracting
the square root of both sides of the equation, we have

tf = d=V/416.5=:±: 20.4083 inches.

But according to equation (227), we have xys'~bcs, where by
the present supposition, b and c are equal to one another, and s' is
equal to unity ; therefore, it is

x y — b^s =416.5;
hence, by division, we shall get

310 OF THE POSITIONS OF EQUILIBRIUM

* ~ T = ±^83 = -20'4083

Then, by taking the positive values of x and y respectively, the
position of equilibrium indicated by the
above results, is represented in the an-
nexed diagram, where CE and CH are
respectively equal to 20.4083 inches,
and consequently, AB the base of the
section is parallel to L K the surface of
the fluid.

Bisect the base AB in the point F, and
draw the straight lines er, FE and FII ; then because CE is equal to
CH, and the angle ECF equal to the angle HCF, it follows, that the
line F E is equal to the line F H ; this is one condition that must be
satisfied, when the body floats in a state of quiescence ; and another
is, that the area of the immersed triangle ECH, is to the area of the
whole section ACB, as the fraction 0.53125 is to unity.

Now, by the property of the equilateral triangle, the area of the
section ACB, is expressed by the product of one fourth of the square
of its side, drawn into the square root of the number 3, and the same
property holding with respect to the area of the triangle ECH; it
follows, that in the case of an equilibrium,

J«V3> : i#V3~: : 0.53125 : 1*
or by suppressing the common quantity J\/3, we have

x2 : 62 : : 0.53125 : 1 ;

but xz~ 416.5, and £2~ 784; therefore, by substitution, we obtain
416.5 :784 : : 0.53125 : 1.

It is therefore evident, that by the above results, both the condi-
tions of equilibrium are satisfied, and consequently, the body floats in
a state of equilibrium when placed as represented in the diagram ;
that is, with 20.4083 inches of its side immersed, and its base parallel
to the plane of floatation.

393. The adfected quadratic equation a2 — 42z:zz — 416.5, has
obviously two positive roots, each of them less than b the side of the
section ; from which we infer, that besides the position of equilibrium
above exhibited, the body may have other two, and these will be
determined by the resolution of the equation, as follows.

Complete the square, and we obtain
^ _ 42a; + 212 = — 416 5 + 441 = 24.5,

OF THE POSITIONS OF EQUILIBRIUM. 311

extract the square root of both sides, and we get

x — 21 = dby7 2475 = =t 4.95 nearly;
consequently, by transposition, we obtain

x :r±21 -f 4.95 == 25.95 inches, and x = 2l — 4.95 — 16.05 inches;
and the corresponding values of y are

Now, the positions of equilibrium supplied by the above values of
a; and y, are as exhi-
bited in the subjoined
diagrams, where LK is
the surface of the water,
ABC the position of the
body corresponding to
x equal 25.95 inches,
and y equal 16.05
inches; abc being the
position which the solid

assumes, when the values of x and y reverse each other ; that is, when
y equal 16.05 inches and y equal 25.95 inches.

Bisect A B and a b in the points F and /, and draw the straight lines
FE, FH andyi,yD to meet the surface of the water in the points E, H
and i, D, the points in which the plane of floatation intersects the im-
mersed sides of the solid ; then are the lines FE, FH andyi,yD equal
among themselves, and the areas ECH, ICD, are respectively to the
whole areas ABC, a b c as the number 0.53125 to unity.

PROBLEM LVII.

394. Suppose that a solid homogeneous body, in the form of
a triangular prism, floats upon the surface of a fluid of greater
specific gravity than itself, in such a manner, that two of its
edges shall fall below the plane of floatation : —

It is required to determine its position, when it has attained
a state of perfect quiescence.

Let ABC represent a section perpendicular to the axis of a solid
homogeneous triangular prism, floating in a state of quiescence on a
fluid whose horizontal surface is IK; A DEB and DCE being respec-
tively the immersed and extant portions.

312

OF THE POSITIONS OF EQUILIBRIUM,

Now, it is manifest, that since the
whole section ABC, is divided by DE
the line of floatation, into the two
parts A DEB and DCE; it follows,
that the centre of gravity of the
section ABC, and the common centre
of gravity of the two parts into which
it is divided occur in the same point ;
consequently, the centres of gravity
of the triangular areas ABC and DEC,
with that of the quadrilateral space
A DEB, are situated in the same
straight line.

But by the principles of floatation we know^ that when the solid is
in a state of quiescence, the centre of gravity of the whole section
ABC, and that of the immersed portion A DEE occur in the same verti-
cal line; that is, the vertical line passing through their centres of
gravity is perpendicular to the horizontal surface of the fluid ; and for
this reason, the vertical line passing through the centre of gravity of
the whole section ABC, and that of the extant portion DEC, is also
perpendicular to the horizon.

Bisect the sides AB, BC and D E, EC in the points F, n and H, m and
draw the straight lines CF, ATI and CH, Din intersecting two and two
in the points G and g, which points are respectively the centres of
gravity of the triangular spaces ABC and DEC.

Draw the straight lines Gg and FH, and because CF and CH the
sides of the triangle CFH, are cut proportionally in the points o and
g, it follows, that Gg and FH are parallel to one another; but it has
been demonstrated, that Gg is perpendicular to TK the horizontal
surface of the fluid ; therefore, FH is perpendicular to DE the line of
floatation; and since DE is bisected in H, it follows that FD and JE
are equal to one another.

Put a — AB, the immersed side of the triangular section ABC,

b ~ AC, one of the sides of the triangular section which

penetrate the fluid;

c :n BC, the other penetrating side of the figure,
d m CF, the distance between the vertex of the section and

the middle of the immersed side ;
0 :rz ACF, the angle contained between the side AC and the

bisecting line c F*

OF THE POSITIONS OF EQUILIBRIUM. 313

ty' — BCF, the angle contained between the bisecting line CF

and the side B c,

s zn the specific gravity of the floating solid,
s' :zr the specific gravity of the supporting fluid,
x :zz CD, the extant portion of the side AC, and
y — CE, the corresponding portion of the side BC.

Then, since the area of any plane triangle, is expressed by the
product of any two of its sides, drawn into half the natural sine of
their included angle, it follows, that the area of the entire section
ABC, is expressed as under, viz.

and for the area of the extant triangle DEC, we have

where the symbols a and a", denote the areas of the whole section
and the extant portion respectively ; consequently, by subtraction,
the area of the immersed part ADEB, is

(a1 — a") = i3in.(0 + f)(bc — xy).

But by the principles of floatation, the area of the whole section
ABC, is to the area of the immersed portion ADEB, as the specific
gravity of the supporting fluid, is to the specific gravity of the floating
solid ; that is

!&csin.(4>4-<£') : £sin.(</> -\- 0') (be — xy) : : s' : s ;
from which, by casting out the common terms, we get

be : (be — xy) : : s' : s,
and equating the products of the extremes and means, it is

bcs — bcs' — xys' ;
therefore, by transposing and collecting the terms, we obtain

xys' = bc(s' — s). (236).

Since the line CF is drawn from the vertex of the triangle ABC, to
the middle of the opposite side or base AB, it follows from the prin-
ciples of geometry, that

AC?4-BC8:=2(AF2-f CF3)*

and this, by substituting the literal representatives, becomes

62 + c2=:2(K-f d2);
therefore, by transposition, we have

4d2zz2(62 + c2)-a2,
and finally, by dividing and extracting the square root, we get

d— 4^2(^4- <r)~-a2.' (237),

314 OF THE POSITIONS OF EQUILIBRIUM.

This is the very same expression for the value of d as that which we
obtained in equation (226), as it manifestly ought to be, since the
same letters refer to the same parts of the figure ; but we have thought
proper to repeat the investigation, in preference to directing the
reader's attention to the former result ; for by this means, our per-
formance is rendered more systematic, and the several steps of the
operation are more readily traced and applied.

Now, in the plane triangle DFC, there are given the two sides CD
and CF, with the contained angle DCF ; to find the side FD.

Therefore, by the principles of Plane Trigonometry, it is

a? 4- d2 — 2dx cos.0 zz F o2 ;

and in the triangle EFC, there are given the two sides CE and CF, with
the contained angle ECF ; to find the side FE.

Consequently, as above, we have

2/2 -f d2 — 2dy cos.f zz F E2 ;

but we have demonstrated, that according to the principles of floata-
tion, the lines FD and FE are equal to one another; therefore, their
squares must also be equal ; hence, by comparison, we have

xz — %dx cos ,(j> zz 2/2 — Zdy cos.0' ;
or by substituting the value of d, equation (237), we get

x*— cos.0 V 2(63+c2)— a3 X zzrz/9— cos.f V 2(^4-c2)— a2 X y . (238).
If both sides of equation (236) be divided by the expression #/, we
shall obtain

bc(s' — s)

y=—^—-

consequently, by involution, we have

y^ ° j>s~s -

Let these values of y and y* be substituted instead of them in equa-
tion (238), and we shall have

^bc(s'— s)cos.<j>'
xs

and multiplying all the terms by x , we get

6V(s' — s)2 bc(s —
-- — --

and finally, by transposition, we have

OF THE POSITIONS OF EQUILIBRIUM. 315

) — a2

_62cV-5)2

"I72"" (239).

395. The above is the general equation, whose roots give the several
positions in which the solid may float in a state of equilibrium ; it is
similar to equation (229), having (5' — s) instead of s, and (s' — sf
instead of s2 ; the body may therefore have three positions of equili-
brium, but it cannot have more, the very same as in the case, where
it floated with only one of its edges below the surface of the fluid.

The method of applying the general equation to the determination
Of the positions of equilibrium, is to calculate the value of d, cos.0
and cos.^>' from the given dimensions of the section, and to substitute
the several given and computed numbers instead of their symbolical
equivalents ; this will give a numeral equation of the fourth degree,
which may be reduced either by approximation or otherwise, accord-
ing to the fancy of the operator.

396. EXAMPLE. Suppose a solid homogeneous triangular prism,
the sides of whose transverse section are respectively equal to 28, 23
and 18 inches, to float in equilibrio on a cistern of water with two of
its edges immersed ; it is required to determine the positions of equi-
librium, on the supposition that the two longest sides of the section
include the extant angle, the specific gravity of the prism being to
that of water, as 565 to 1000 ?

In order to resolve this question, we must first of all determine the
length of the line cr, which is drawn from the extant angle at c to the
middle of the opposite side AB ; for which purpose, let the dimensions
of the section be respectively substituted according to the combination
exhibited in equation (237), and we shall have

d = \ V2(282 + 232) — 1 82 — \ v/ 2302 = 23.99 inches nearly.

Consequently, in the triangles ACF and BCF respectively, we have
given the three sides AC, AF, FC and BC, BF, FC to find cos. ACF and
cos. BCF; for which purpose, the elements of Plane Trigonometry
supply us with the following equations, viz.

In the triangle ACF, it is

and in the triangle BCF, it is

316 OF THE POSITIONS OF EQUILIBRIUM.

Therefore, by substituting the respective values of a, b and c, as
given in the question, and the value of d as computed above, we shall
have the following values of cos.0 and cos.^'.

Thus, for the absolute numerical value of cos.0, it is

4(28* + 23.99*)— 182_A

COS'* = -- 8X28X23.99 ~ °'95166'

and for the corresponding value of cos. 0', we have

4(232-4-23.992 — 182
COS-» = 8X23X23.99 =°-92747'

Having ascertained the numerical values of cos. 0 and cos.^>', let the
respective quantities be substituted in equation (239), and it becomes

28X23X435X0.92747

4 — 0.

1000
282X232X4352

10002
from which, by computing the several terms, we get

*4 — 45.66*3 -f 12466* = 78478.36.

The root of this equation will be most easily discovered by approxi-
mation, and for this purpose, we shall adopt the method of trial and
error, which Dr. Hutton has so successfully applied to the resolution
of every form and order of equations, however complicated may be
their arrangement.

By a few simple trials, indeed it is almost self evident, that the
value of x will be found between 15 and 16; consequently, by sub-
stitution we obtain

154 — 45.66X158 4- 12466X15 — 78478.36 = 15113.64 too little.
164 — 45.66X163 + 12466X16 — 78478.36 = 509.72 too great.
Here it is manifest that the errors are of different affections, the one
being in defect and the other in excess ; hence we have

15113.64 + 509.72 : 16 — 15 :: 509.72 : 0.032;
consequently, for the first approximation, we get

x = 16 — 0.032 = 15.97 very nearly.

Supposing therefore, that x lies between 15.9 and 16 ; by repeating
the process, we shall have

164 — 45.66X 163 -f 12466X 16 —78478.36 = 509.72 too great.
15.94 __ 45.66 X15.93 4- 12466 X 15.9— 78478.36 zz!05.393too little.
Here again, the e/rors are of different affections, the one being in
excess and the other in defect; consequently, we have

OF THE POSITIONS OF EQUILIBRIUM. 317

509.72-1-105.393 : 16—15.9 : : 105.393 : 0.017 nearly;
therefore, the second approximate value of x, is

x = 15.9 + 0.017 = 15.917 inches.

By again repeating the process, a nearer approximation to the true
value of x would be obtained, but the above is sufficiently accurate
for our present purpose ; therefore, let this value of x, together with
the numerical values of b, c, s and s', be substituted in equation (236),
and we shall obtain

159172/m 280140,
and from this, by division, we get
280140

397. And the position of equilibrium corresponding to the above
values of x and z/, is represented in the
annexed diagram, where IK is the hori-
zontal surface of the fluid, ABED the im-
mersed part of the section, and DCE the
extant part.

Bisect AB in F, and draw the straight
lines CF, FD and FE; then, as we have
previously demonstrated, when the body
floats in a state of equilibrium, the lines
FD and FE are equal to one another.

Now, in order to determine if this equality obtains, we must have
recourse to equation (238), where we have

c*) — «'X x = t/2— cos.
then, let the computed values of or, y, cos.0, cos.^', and the given
values of a, b and c, be substituted instead of them in the above equa-
tion, and we shall obtain

15.9172 — 0.95166 v/2302 X 15.917 — 17.62— 0.92747 /2302X 17.6,
and this, by transposition and reduction, gives

— 726.77 = 56.43 — 783.20.

398. Another condition of equilibrium is, that the area of the im-
mersed part ABED, is to the area of the whole section ABC, as the
specific gravity of the solid is to that of the supporting fluid. This is
a more necessary condition than the equality of the lines FD, FE; for
such an equality may exist when no equilibrium obtains ; but it may
be considered as a universal fact, that whenever the two conditions are
satisfied at the same time, the body floats in a state of quiescence.

318 OF THE POSITIONS OF EQUILIBRIUM.

We have already found that cos. 4>zz 0.95 166, and cos.0'= 0.92747 ;
consequently, ^=17° 53', and <j>' = 21° 57' ; hence we have (<£ -f- 0')
zr39° 50', and by the principles of mensuration, we get

a' = 1(28 X 23) sin.39° 50' = 322 X 0.64056 = 206.26032,

and the area of the extant part, is

a" = J(15-917 X 17.6) sin.39° 50' = 140.0696 X 0.64056 = 89.72298 ;

therefore, by subtraction, the area of the immersed part becomes

(a' — a") = 206.26032 — 89.72298 = 1 16.53734 ;

consequently, by the principle of floatation, it is

206.26032 : 116.53734 : : 1000 : 565 nearly;

from which it appears, that both the conditions of equilibrium are
satisfied, and therefore the body as exhibited in the figure indicates a
state of quiescence.

399. By finding the other roots of the equation, other situations of
equilibrium may be assigned ; but since the one above given is that
which would be adopted in practice, we consider that it would be a
waste of both labour and time to search after the others ; we therefore
leave the reduction of the resulting cubic equation for exercise to the
reader, presuming that he will find his trouble and attention amply
repaid, by the satisfaction to be derived from the confirmation of the
principles by an actual construction.

400. When the triangle ABC becomes isosceles; that is, when the
sides b and c are equal to one another ; then cos.0 and cos.0' are
also equal, and the general equation (239), becomes transformed into

b\s' — s)cos.d> — - . b\s'— s)2

2 2

—

or by transposing the absolute given quantity - - ^ - , we get

(240).

Now, by carefully examining the nature of this equation, it will
immediately appear to be composed of the two quadratic factors

b\s'—s) _ W— s}

*2 -- r— • =0, and*2— cosV4£2— a2X*4--^— " '

where it is manifest, that each of these expressions involve two roots
of the original equation, and the number of the real positive roots,
indicates the number of positions in which the body may float in a

OF THE POSITIONS OF EQUILIBRIUM. 319

state of quiescence, while the absolute values of the roots determine
the positions themselves.

401. Let each of the above quadratic factors be transformed into

an equation, by transposing the given term --^ — -, and we shall

obtain

<_yy—<) 2 b\s'— s)

and when the value of s', or the specific gravity of the supporting fluid
is expressed by unity, as is the case with water ; then, we have for
the pure quadratic,

xz— £2(1— s). (241).

and for the adfected quadratic, it is
^2 ___ cos^ ^ 4#! _ a2 x x __ „ #(i_s). (242).

Let the square root of both sides of equation (241) be extracted,
and we shall obtain

x = b^/T^7; (243).

but from equation (236), we have

xy = b\l— s),

and this, by substituting the above value of x, becomes
b<jT^~sXy=ib\\ — s);
hence, by division, we get
_61 — *

Here then it is manifest, that the values of x and y are each
expressed by the same quantity ; from which we infer, that the solid
floats in a state of equilibrium, when the base of the section is parallel
to the surface of the fluid ; that is, when the extant portion of the
section is also isosceles, having its base coincident with the plane of
floatation.

402. The practical rule for computing the equation (243) or (244),
may be expressed in words at length, as follows.

RULE. From unity, or the specific gravity of the fluid,
subtract the specific gravity of the floating solid, and multiply
the square root of the difference by one of the equal sides of
the section, and the product will express the value of x and y.

403. EXAMPLE. Suppose the two equal sides of the section to be
respectively equal to 28 inches, the base 18 inches, and the specific

320

OF THE POSITIONS OP EQUILIBRIUM.

gravity of the solid 0.565 as in the preceding example ; how much of
the equal sides is immersed in the fluid, and how much is extant, the
body being in a state of quiescence ?

Here, by the rule, we have

l—s= 1 — 0.565 = 0.435,

the square root of which is

v/ 0.435 zz 0.659;

consequently, by multiplication, we have
*or2/zz28x0.659:z: 18.452 inches.

404. And the position of equilibrium indicated by the above value
of x and y, is as represented in the subjoined

diagram, where IK is the horizontal surface
of the fluid, ABED the immersed portion of
the section, and DCE the extant portion, DE
being the water line or plane of floatation.

Since CD and CE are each equal to 18.452
inches, it follows, that AD and BE are each
equal to 28 — 18.452 zz 9.548 inches, the
extant part of the equal sides being nearly
double of the immersed part.

Bisect A B in the point F, and draw the straight lines F D and F E ;
then shall FD and FE be equal to one another; this is manifest, for
AF, BF and AD, BE are equal, and the angle DAF is equal to the
angle EBF ; therefore, FD is equal to FE.

By examining the nature of the equation (243) or (244), it is
manifest that the values of x and y depend entirely on the value of s,
or the specific gravity of the floating body ; now, since this may admit
of all magnitudes between zero and unity, which is the specific gravity
of water, it follows, that x and y may be of all magnitudes between
zero and 28 inches ; but whatever may be the magnitude of the extant
sides, the position in which the body floats will be the same ; viz. that
in which the base of the section is parallel to the surface of the fluid.

405. Admitting the specific gravity of the solid to fall within the
limits of possibility, the formula equation (242), when reduced, will
supply us with other two positions of equilibrium, in which the body
may float with two of its angles immersed ; here follows the reduction
of the equation.

Complete the square, and we obtain

2— a2)— b\\— s),

OF THE POSITIONS OF EQUILIBRIUM* 321

and by extracting the square root, it is

«2) — b\l— s),
and finally, by transposition, we get

2 — a) — #2(1 — *); (245).
the corresponding values of y being

• — a?)—b\l—s}. (246).
In order to satisfy the conditions implied in the foregoing equations,
it is requisite that the value of s, the specific gravity of the floating
solid, should fall between the limits indicated by the following expres-
sions, viz.

— -rjz ' ana — — , j

now, by the principles of Plane Trigonometry, we have

b : g \/462 — a* : : rad. : cos.0,

which being reduced, gives

and by involution, we obtain

Let these values of cos.<£ and cos2.^ be substituted in the above
expressions for the limits of 5, and we shall get for the greater limit,

and for the lesser limit, it is

__(862 — a*)X«8

1664

and the arithmetical mean of these two limits, is
_

Then, if this value of s be substituted instead of it in the equations
(245 and 246), we shall obtain for the values of x, as follows,

2 /4/2 2\ j.2/1 (,100 — ft )a V /048)

VOL. I. Y

322

OF THE POSITIONS OF EQUILIBRIUM.

and similarly, the corresponding values of y are

The practical rule for reducing the above equations must be omitted
in this case, the forms being too complex to admit of a clear and com-
prehensive description ; it is however presumed, that the attentive
reader will be enabled to understand the method of solution, by care-
fully tracing the several steps of the process as exhibited in the fol-
lowing arrangement.

406. EXAMPLE. If the dimensions of the section be the same as in
the preceding example ; then, the value of s as computed from equa-
tion (247), becomes

__(16X282— 18a)Xl8*__ 3959280

— 0.2013 nearly.

32X284 19668992

Therefore, let the mean value of s as thus determined, together
with the numerical values of a2, b*, cos.0 and cos2.^, be substituted in
equation (248), and we shall have, in the case of the positive sign,

*i=0.47347.v/4X28*— 188-f ^0.22417(4 X282— 18*)— 0.7989X28*
— 27.152 inches,

and in the case of the negative sign, it is

a: =: 0.47347 -v/4X28*—182—V0.22417(4X282—182)— 0.7989x28*

HZ 23.06 inches,

and the corresponding values of y, are
t/— 23.06, and y — 27.152 inches.

407. Now, the positions of equilibrium corresponding to the above
values of x and ?/, are as
denoted in the subjoined
diagram ; where IK is the
horizontal surface of the
fluid, ABC the position cor-
responding to x = 27. 152
and y zz: 23.06 inches, and
abc the position indicated
by x z= 23.06 "and y =
27.152 inches.

Bisect AB and ab in the points F andy, and draw the straight
lines FD, FE and fd, fe intersecting the plane of floatation in the
points D, E, and d, e ; then shall the lines so drawn be equal among

OF THE POSITIONS OF EQUILIBRIUM. 323

themselves. This is very easily verified, for by the principles of Plane
Trigonometry, we know, that when two sides of a plane triangle are
given together with the contained angle ; then, the square of the side
opposite to the given angle, is expressed as follows, viz.
In the triangle DCF, it is

D F2 IZ: D C2 -|- C F2 - 2DC.CFCOS.DCF,

and in the triangle ECF, it is

EF2HrEC2 -f- CF2 - 2EC.CFCOS.ECF.

But by the construction, the angles DCF and ECF are equal to one
another, and consequently, cos. DCF = cos. ECF; therefore, by sub-
stituting the analytical values of the several quantities, the above
expressions become

D F* =z x* -\-d? — 2dx cos.0, and E F2 zz yz -f- dz — 2dy cos.0 ;
but when the body floats in a state of equilibrium, these are equal,
hence we have

x* — 2dx cos.0 zz y* — %dy cos.<£,
and from this, by transposition, we obtain

2t? cos.(p(x — r/) zz a;2 — y* ;
therefore, by division, we have

Now, we have seen by the preceding solution, that a; zz 27. 152, and
y zz23.06 inches ; consequently, by substitution we get

2d cos.0 zz 27. 152 -f 23.06 zz 50.212 ;
but by the property of the right angled triangle, it is

and we have already seen, that

Let the numerical values of a2, b and 62 be substituted in each of
these expressions, and we shall have

d= i V/4X282— 182=z>/703~26.514,
and similarly, for cos.^>, we get

cos.0 = — - V 4 X 28* — 1 82 r= 0.94693 ;
oo

consequently, by substitution, we obtain
zz 26.514X0.94693X2 ~ 50.212.
Y 2

324 OP THE POSITIONS OF EQUILIBRIUM.

408. The expression for the area of the immersed figure ABED, is
^sin.2^>(62 — xy}, and the expression for the area of the whole section
ABC, is \tf sin. 2^ ; and by the principles of floatation, these are to one
another, as the specific gravity of the floating solid, is to that of the
fluid on which it floats ; hence we have

— xy) : i&2sin.20 : : 0.2013 : 1,
and by suppressing the common term ^sin.2^, we get
{tf — xy} : 62:: 0.2013 : 1,

and from this, by putting the product of the extreme terms, equal to
the product of the means, we obtain

and finally, by substituting the numerical values, we have
27. 152X23.06 = 0.7987 X282 very nearly,

which satisfies the other condition of equilibrium ; hence we infer, that
the subcontrary positions represented above, are those which the body
assumes when floating in a state of quiescence with two of its angles
below the plane of floatation.

409. When a, b and c are equal to one another ; that is, when the
triangular section is equilateral; then, the general equation (239),
becomes

b\s'— *)*
and from this equation, by transposing the given term -- pj -- , we

get

8'-

Now, it is manifest, that the equation in its present form, is com-
posed of the two quadratic factors

—

and these factors, by transposing the given term - —, -- in each,
become transformed into the following quadratic equations, viz.

OF THE POSITIONS OF EQUILIBRIUM.

325

and supposing the specific gravity of the fluid, or the value of s' to be
expressed by unity ; then, these equations become

'Resolving these equations by the rules which the writers on algebra
have laid down for that purpose, we shall have for the pure quadratic,

and ?/zz:6\/(l — s};
and again, for the adfected form, it is

(251).

and y— b {0.866 cos.^v/ 0.75 cos8.^ — (1 — r^ f ^
In the equations, (251), it is obvious that the values of a; and y are
assignable, whatever may be the value of s, provided that it is less than
unity ; and since x and y are each expressed by the same quantity, it
follows that they are equal to one another, and consequently the body
will float in equilibrio, when the immersed side or base of the section
is parallel to the surface of the fluid.

410. EXAMPLE. If the floating prism be of fir from the forest of
Mar, of which the specific gravity is 0.686, that of water being unity;
then we have

and if the value of b, or the side of the equilateral triangle be 28 inches,
we get xi=y — 0.56X28 = 15.69 inches;

and the position of equilibrium corresponding to this common value
of a; and y, is exhibited in the
annexed diagram, where IK is
the horizontal surface of the fluid,
DCE being the extant portion of
the floating body, and ABED the
part immersed below the plane of
floatation ; c D and c E being re-
spectively equal to 15.69 inches.
Bisect AB in F, and draw the
straight lines FD and FE to inter-
sect the surface of the fluid in the points D and E ; then, because the
triangle ABC is equilateral, and CD equal to CE by the construction,
it follows, that FD and FE are equal to one another; this satisfies one
of the conditions of equilibrium, and we have now to inquire if the area
of the immersed portion ABED, is to the area of the whole section ABC,
as the fraction 0.686 is to unity.

\|/ i==~=-

326 OF THE POSITIONS OF EQUILIBRIUM.

Now, by the principles of mensuration, we know that the area of a
plane triangle, of which the three sides are equal, is expressed by one
fourth of the square of the side, drawn into the square root of the
number 3 ; consequently, the area of the whole section ABC, is

and the area of the extant part DEC, is

therefore, the area of the immersed part ABED, is

(a' — a") = JftV 3"— i*V~3 = 0-433 (fi» — x2) ;

hence, by the principles of floatation, we get

0.433 (&• — *') : 0.43362 : : 0.686 : 1,
and by equating the products of the extremes and means, it is

x* = b*(\ — 0.686) — 0.31462.

But b is 28 and x 15.69 inches ; therefore, if these values of b and
x be substituted instead of them in the preceding equation, we shall

have

15.69* nr0.314x282=r 246.176.

In this case also, one of the conditions of equilibrium is satisfied ;
hence we conclude, that the position which we have represented above
is the true one, since both the conditions upon which the equilibrium
depends, have been fulfilled by the results as obtained from the reduc-
tion of the formula.

The value of x and y, as exhibited in equations (252), will indi-
cate two other positions of equilibrium, subcontrary to each other ;
but in order that those positions may be coiisistent with the conditions
of the problem, it becomes necessary to assign the limits of s, or the
specific gravity of the floating body ; for it is manifest, that beyond
certain limits, the conditions specified in the problem cannot obtain.

411. Now, in the case of the isosceles triangle, it has been shown,
that the greater limit of the specific gravity, is

and consequently, when the triangle is equilateral,
s=i~ -ft = 0.5;

and moreover, it has also been shown, that when the triangle is
isosceles, the lesser limit of the specific gravity, is

OF THE POSITIONS OF EQUILIBRIUM.

which, when the triangle is equilateral, becomes

s = —=0.4375,
16

1 and the arithmetical mean of these, from equation (247), is

327

Let therefore this value of s be substituted instead of it in the
expressions, class (252), and we shall obtain

x = 25.95, and x = 16.05 inches,

the corresponding values of y being

y =z 16.05, and y =. 25.95 inches.

412. The positions of equilibrium, as indicated by these values of
x and?/, are as represented in the annexed diagrams, where IK is
the horizontal surface of the fluid, ABED, abed the immersed, and
DEC, dec the extant portions of the section corresponding to the
positions ABC and abc, in which CD and ce are each equal to 25.95
inches, and CE, cd equal to 16.05 inches, being the respective values
of x and y, as determined from equation (252).

Bisect A B and a b in the points F and f, and draw the straight lines
FD, FE and fd, fe intersecting the horizontal surface of the fluid in
the points D, E and d, e ; then, when the body floats in a state of equi-
librium, the lines FD, FE,/C? and/e are equal among themselves.

This is very easily proved, for since the triangle ABC is equilateral,
the angle ACB is equal to sixty degrees, and consequently its half, or
the angles ACF and BCF are each of them equal to thirty degrees;
therefore, by the principles of Plane Trigonometry, we have

DF8=CD2-4-CF? 2CD.CFCOS.30°,

and similarly, by the same principles, we get

F E2 HZ C E4 -4- C F2 — 2C E.C F COS. 30° ;

328 OF THE POSITIONS OF EQUILIBRIUM.

but according to the conditions of equilibrium, these are equal, hence
we have

CD2 — 2C D.C F COS.300 = C E2 — 2CE.CF COS.300 ;

therefore, by substituting the analytical expressions, and transposing,
we get

x3 — z/2 = 2d cos.30°(a?— y\

and dividing both sides by (x — y), we shall have
2dcos.30° = a:-f y.

By Plane Trigonometry cos. 30° nr sin. 60°, and by the property of
the equilateral triangle, we have e? = b sin. 60° ; consequently, by sub-
stitution, we get

2&sin2.60°~ x + y;

or numerically, we obtain

2X28x1 = 25.954-16.05 — 42.

413. Hence it appears, that in so far as the equilibrium of floatation
depends upon the equality of the lines FD and FE, the condition is
completely satisfied, and the same may be said respecting the lines

fd and/e ; but it is manifest, that another condition must be fulfilled
before the body attains a state of perfect quiescence, and that is, that
the area of the immersed part ABED, is to the area of the whole section
ABC, as the specific gravity of the solid body, is to that of the fluid on
which it floats, or as 0.46875 to unity : now, this condition is evidently
satisfied, when

x y = P(l— 0.46875),

therefore, numerically we obtain

25.95X16.05 — 282X0.53125i=41.65.

Here then, both the conditions of equilibrium are satisfied, and from
this we infer, that the positions exhibited in the diagram are the true
ones, the downward pressure of the body in that state, being perfectly
equipoised by the upward pressure of the fluid.

414. What we have hitherto done respecting the positions of equi-
librium, has reference only to a solid homogeneous triangular prism,
floating on the surface of a fluid with its axis of motion * horizontal ;

* When a solid homogeneous body, in a state of equilibrium on the surface of a
fluid is disturbed by the application of an external force, it will endeavour to restore
itself by turning round a horizontal line passing through its centre of gravity, and
th^s line on which the body revolves, is called the axis of motion.

OF THE POSITIONS OF EQUILIBRIUM.

329

and

but there are various other forms, which are not less frequent in the
practice of naval architecture, nor less important as subjects of theo-
retical inquiry : some of these we now proceed to investigate.

PROBLEM LVIII.

415. Suppose that a solid homogeneous body in the form of
a rectangular prism, floats upon the surface of a fluid of greater
specific gravity than itself, in such a manner, that only one of
its edges falls below the plane of floatation : —

It is required to determine what position the body assumes,
when it has attained a state of perfect quiescence.

Let ABCD be a vertical section, at right angles to the horizontal
axis passing through the centre of
gravity of the rectangular prism,
and let I K be the surface of the
fluid, on which the body floats
in a state of equilibrium,
being the extant portion
mvn the part which falls below
the plane of floatation.

Bisect mn in F and vn in H,
and draw the straight lines DF
and mil, intersecting each other
in g the centre of gravity of the
immersed triangle mvn. Join the

points A, c and B, D by the diagonals AC and BD, intersecting in o
the centre of gravity of the rectangular section ABCD, and draw og.

Then, because the body floats upon the surface of the fluid in a
state of equilibrium according to the conditions of the problem ; it
follows from the laws of floatation, that the straight line Gy is perpen-
dicular to IK. Through F the point of bisection of mn the base of
the immersed triangle, and parallel to go, draw FP meeting the
diagonal BD in the point p, and join PTW, PW; therefore, because the
straight line ga is perpendicular to mn the plane of floatation, it is
evident that FP is also perpendicular to mn, and consequently, pm
and PW are equal to one another.

This is a condition of equilibrium which holds universally, and
another is, that the area of the immersed triangle m D n, is to the area
of the whole section ABCD, as the specific gravity of the solid, is to

330 OF THE POSITIONS OF EQUILIBRIUM.

the specific gravity of the fluid on which it floats ; when both these
conditions obtain, the body will float permanently in a state of equi-
librium.

Put a zz: A D or BC, one of the sides of the section that contain the

immersed angle,

b zz: DC or AB, the other containing side;
s zz the specific gravity of the floating solid,
s' zz the specific gravity of the supporting fluid, or that on

which the body floats,
x zz Dm, the part of the AD which is immersed under mn the

plane of floatation,

y zz DW, the corresponding portion of the side DC ;
a' zz the area of the whole rectangular section A BCD, and
o"zz the area of the immersed portion mvn.

Then, since the section of the solid is considered to be uniform,
with respect to the axis of motion, throughout the whole of its length,
we have

a"s' = a's. (253).

But by the principles of mensuration, the area of the whole rectan-
gular section A BCD, is expressed by the product of its two sides;
that is,

a' zz a b,

and the area of the immersed triangle mow, is

Let these values of a' and a" be substituted instead of them in the
equation (253), and we shall have

abs=\xys'. (254).

By the property of the right angled triangle, it is

B D2 zz A D* 4- A B?>
or by putting d to denote the diagonal BD, we get

from which, by extracting the square root, we obtain

and by the principles of Plane Trigonometry, it is

\/a8 + 62 : a : : rad. : COS.ADB ;

and similarly, by Trigonometry, we have

1^/0? -j- &2 : b : : rad. : cos. BDC ;

OF THE POSITIONS OF EQUILIBRIUM. 331

therefore, by working out the above analogies, and putting radius
equal to unity, we shall have

COS.ADBIZZ — — , and cos. BD cur — .

Since gG and FP are parallel, and g? equal to one third of DF ; it
follows, that GP is equal to one third of DP; or which is the same
thing, DP is equal to three fourths of BD ; that is
DPz=|Va2-f b\

When two sides of a plane triangle are given, together with the
angle of their inclination, as is the case in the triangles WIDP and
WDP; then, the writers on Trigonometry have demonstrated, that

7WP2ZZDW2-f- DP2 - 2D97Z.DPCOS.ADB, and WP2 — DW2-|- DP2 -
2DW.DPCOS.BDC ;

and these, by the principles of floatation, are equal, hence we get

Dm2 - SDTW.DPCOS.ADBUZDW2 - 2DW.D P COS.B DC.

Let the analytical expressions of the several quantities Dm, on, DP,
cos. ADB and COS.B DC, be substituted in the above equation, and we
shall obtain

*-!£:=,•-»» (255).

If both sides of the equation (254), be divided by the expression
Jars', we shall obtain as follows, viz.

the square of which, is

• y ~ :

Now, if these values of?/ and z/2, be respectively substituted instead
of them in equation (255), we shall obtain

3ab*s

2 __

": :

and finally, by reduction and transposition, we get

(256).

And if we consider the value of s', or the specific gravity of the
fluid, to be expressed by unity, as is the case with water ; then the
above general equation becomes

332 OF THE POSITIONS OF EQUILIBRIUM.

416. When the specific gravity of the solid body is so related to
that of the fluid, as to fulfil the conditions of the problem, the roots
of the above equation will determine the positions of equilibrium ;
but since there cannot be more than three real positive values of x in
the equation, it follows, that there cannot be more than three positions
in which the prism will float in a state of rest, with only one of its
edges below the surface of the fluid.

417. If a and b are equal to one another ; that is, if the transverse
section of the floating body be a square at right angles to the axis of
motion ; then, equation (257) becomes

O j

x4' --

and from this, by transposition, we obtain

a;4— — X*3-f 3b*sx — 46V = 0. (258).

Now, it is obvious, that this equation is composed of the two fol-
lowing quadratic factors,

O I

a?2— 262s, and x9 — X a; -f 2&2*;

which being converted into equations, gives

x^ — Ws, (259).

and similarly, from the adfected factor, we obtain

x9— ^Xx=— 262s. (260).

Since these two quadratic equations are deduced from the factors
which constitute the particular biquadratic (258), it follows, that the
real positive roots which they contain, must indicate the positions of
equilibrium according to their number.

If we extract the square root of both sides of the equation (259),
we shall obtain

x = b^; (261).

but by equation (254), we have

lxy = bzs\
consequently, by division, we get

OA2 «

(262).

OF THE POSITIONS OF EQUILIBRIUM.

333

Here then it is manifest, that the values of x and y are each expressed
by the same quantity ; hence we infer, that the body floats with one
diagonal of its vertical section perpendicular to the surface of the
fluid, and the other parallel to it.

418. The practical rule afforded by the equations (261 and 262), may
be expressed in words at length as follows.

RULE. Multiply the square root of twice the specific gravity
of the solid, by the side of the square section, and the product
will give the length of the immersed part , when the body is in
a state of rest.

419. EXAMPLE. Suppose a square parallelopipedon, whose side is
equal to 18 inches, to be placed upon a fluid with one of its angles
immersed, and one of its diagonals vertical ; how much of the body
will fall below the plane of floatation, supposing its specific gravity to
be 0.326, that of the supporting fluid being equal to unity?

Here, by operating according to the rule, we get

ar= 18V 2 X0.326 zz 14.526 inches,
and for the corresponding value of y, we have

, = ^ = ^6^,

Consequently, the position of equilibrium thus indicated, is as
represented in the annexed
diagram; where IK is the
surface of the fluid, AC the
horizontal and B D the ver-
tical diagonal ; Dm and i>n
being respectively equal to
14.526 inches, as deter-
mined by the foregoing
arithmetical process.

420. Take DP equal to
three fourths of BD, and
draw pm and pn meeting

the surface of the fluid in the points m and n ; then are pm and pn
equal to one another ; this is one of the conditions necessary to a state
of equilibrium, when neither of the diagonals is vertical ; but in the
present instance, the condition of equality will obtain wherever the
point P may be taken, and consequently, the equilibrium is not in-
fluenced by the position of that point.

334 OF THE POSITIONS OF EQUILIBRIUM.

421. The only condition, therefore, which establishes the equilibrium
in this case, is, that the area of the immersed triangle mi>n, is to the
area of the whole section A BCD, as the specific gravity of the solid is
to that of the supporting fluid.

422. If the specific gravity of the solid be equal to one half that of
the fluid on which it floats; then, AC will coincide with IK, and in
this state the specific gravity attains its maximum value; for if it
exceeds this limit, more than one angle of the solid will become
immersed, and this is contrary to the conditions of the problem.

423. When the specific gravity of the floating solid is properly
limited, the equation (260), has two real positive roots; hence we
infer, that there are two other positions in which the body may float
in a state of equilibrium, and these will be determined by the resolu-
tion of the equation.

Therefore, complete the square, and we get

and by extracting the square root, it is

3b b

x — •— — =4= -V (9 — 32s) ;

consequently, by transposition, we have
b C

4 - (263).

and the corresponding values of y, are

(264).

424. Now, by attentively examining these equations, it will appear,
that in order to have the values of x and y real quantities, the value
of sy or the specific gravity of the solid body, must be such, that thirty
two times that quantity shall not exceed the number 9 ; and moreover,
in order that the greatest value of x and y may be less than b the side
of the square section, it is necessary that thirty two times the specific
gravity of the solid shall not be less than the number 8.

425. When the value of s is taken such, that 32s zz 9; then we
have \/9 — 325 ~ 0 ; in which case the values of x and y are each of
them equal to three fourths of b ; but when the value of s is such, that
325 — 8 ; then we have <\/9 — 32s = =t 1 , and consequently, the two
values of x are b and \b respectively, the corresponding values of y
being \b and b; and the positions of equilibrium corresponding to

OF THE POSITIONS OF EQUILIBRIUM. 335

the above values of x and y, are as represented in the annexed

diagrams, where i K is the horizontal surface of the fluid ; No. 1 the
position corresponding to x and y, when they are respectively equal
to three fourths of b ; No. 2 the position indicated by x zn b and
y=\b, and No. 3 that which corresponds to the reverse values of
x and y, viz. when x is equal to \b, and y equal to b ; in both of
which cases, one angle of the figure is under the plane of floatation,
and another coincident with it; but this is scarcely consistent with
the conditions of the problem, which distinctly intimates, that only
one edge or angle of the floating body shall be immersed in the fluid,
and this implies, that all the other edges or angles shall be wholly
extant, or in other words, that the greatest values of x and y, shall be
less than the side of the square section.

In order, therefore, that this condition may obtain, the specific
gravity of the body must be less than T9T, which gives the position in
No. 1 ; and greater than 3-8T, which gives the positions in Nos. 2 and 3 ;
consequently, by taking the arithmetical mean between these limits,
we shall have s— 0.265625, and the equations (263 and 264) become

the corresponding values of y, being

But the square root of 9 — 8.5 is 0.7071 very nearly ; therefore, if
b be equal to 18 inches, as in the preceding example, we shall have

18X3.7071

the corresponding values of y being
y = 10.318, and y = 16.682 inches.

Now, it is manifest, that none of the positions represented above,
resemble that which is indicated by the values of x and y just deter-
mined ; but the true positions which these values furnish, are such as
correspond to a state of equilibrium, and they are exhibited in the

336

OF THE POSITIONS OF EQUILIBRIUM,

subjoined figures, whereas in all the previous cases, IK is the horizontal
surface of the fluid ; men and mo A Bra being the areas of the immersed
and extant portions of the body, corresponding to #— 16.682 inches,

and 2/z=:10.318 inches ; the subcontrary figures odp and oabcp being
the respective areas when #=: 10.318 inches, and y zn 16.682 inches.

Bisect mn in F, and through the point F draw FP at right angles to
mn, meeting the diagonal AC in the point p, and join pm and PW;
then it is manifest, that the straight lines pm and PW are equal to one
another, as ought to be the case when the solid floats in a state of
equilibrium; and moreover, the area of the immersed portions men,
and odp, are to the area of the entire sections A BCD and abed, as the
specific gravity of the floating solid, is to that of the supporting fluid.

426. If the conditions of the problem should be reversed, that is,
if three angles of the figure be immersed beneath the plane of floata-
tion, and one extant above it; then, by a similar mode of investiga-
tion, it may be shown, that

\xy s' — «&(«' — s)
and furthermore, that

.. __

~ '

2 ~ 2

Now, these being similar equations to those which correspond to the
case of one angle being immersed beneath the surface of the fluid ; it
follows, that all the other steps of the investigation would also be
similar, and consequently they need not be repeated.

PROBLEM LIX.

427. Suppose that a solid homogeneous prismatic figure,
whose transverse section is rectangular, is found to float in a
state of equilibrium on the surface of a fluid with its two
edges immersed : —

OF THE POSITIONS OF EQUILIBRIUM.

337

It is required to determine the positions assumed by the
solid, when it is in a state of quiescence.

The solution of this problem is attended with greater difficulty than
either of the preceding ones respecting the positions of equilibrium ;
the superior difficulty in this case, arises from the situation of the
lines, whose equality constitutes the second condition of equilibrium ;
in the foregoing cases, this equality was determined by the resolution
of the simple problem in Plane Trigonometry, where two sides and
the contained angle are given, and it is required to find the third side,
or that which subtends the given angle ; in the present instance, how-
ever, this mode of comparison does not take place, and the equality
of the lines alluded to, or rather the condition of equilibrium depend-
ing on such an equality, can only be established by a series of com-
plicated analogies, arising from the similarity of triangles determined
by the construction.

Let A BCD represent a transverse section perpendicular to the axis
of a homogeneous rectangular
prism, which floats in equilibrio
on the surface of a fluid of greater
specific gravity than itself, and in
such a manner, that two of its
angles are wholly immersed be-
neath the plane of floatation re-
presented by HE; IK being the
horizontal surface of the fluid,
HECD the immersed portion of
the section, and ABEH the extant
portion.

Let G and g be the centres of gravity of the whole section ABCD,
and the immersed part HECD; join G#, then, if the position which
the body has assumed be that of equilibrium, the line Gg is perpen-
dicular to HE the plane of floatation, and the area of the immersed
part HECD, is to the area of the whole section ABCD, as the specific
gravity of the solid is to that of the supporting fluid.

Through the point c, the most elevated of the immersed angles of
the figure, draw cb perpendicular to IK, and through the points o
and g draw GC and ge perpendicular to cb; then, if the position
which the body has assumed be that of equilibrium, the straight lines
GC and g e are equal to one another. The conditions under which the
body floats in a state of quiescence, therefore are,

VOL. I. 7.

338 OF THE POSITIONS OF EQUILIBRIUM.

1. That the area of the immersed part, and that of the
whole section, are to one another as the specific gravities of
the solid and the fluid.

2. That the horizontal lines, intercepted between the centres
of gravity, and the vertical line passing through the most
elevated of the immersed angles, are equal to one another.

Through the points o and g, draw the straight lines oa and gv
perpendicular to BC the side of the section; and through the points
a and v, draw ab and vd parallel to the horizon, and am,vn perpen-
diculars to GC and ge\ and finally, through E and g, and parallel to
CD and CB, draw the straight lines E£ and sr, and the construction is
finished.

Then it is manifest, that by means of the parallel and perpendicular
lines employed in the construction, we can form a series of similar
triangles, which will lead us by separate and independent analogies,
to the comparison of the lines GC and ge, on whose equality the equi-
librium of floatation depends.

Put a =z AD or b c, the longest side of the transverse section,
b =: AB or DC, the shortest side,
d ~ gv, the perpendicular distance between the centre of

gravity of the immersed part, and the side of the

section B c ;

a' — the area of the whole section ABC D,
a"~ the area of the immersed part HECD ;
x zz DH, the distance between the lowest immersed angle, and

the corresponding extremity of the line of floatation,
y — CE, the distance between the highest immersed angle and

the other extremity ;

and as heretofore, let s denote the specific gravity of the solid body,
and s' the specific gravity of the fluid on which it floats ; then, by the
principles of floatation, we have

a" :a'::s: s',

and from this, by equating the products of the extreme and mean
terms, we get

a's = a"sf.

Now, by the principles of mensuration, the area of the rectangular
section ABCD, is expressed by the product of its two containing side*
AB and BC; hence we have

OF THE POSITIONS OF EQUILIBRIUM. 339

and moreover, the area of the immersed part HE CD, is expressed by
half the sum of the parallel sides drawn into the perpendicular dis-
tance between them ; consequently, we obtain

abs= \bs\x + y),
and finally, by multiplication and division, it is

2«s — s'O + y)- (265).

The equation which we have just investigated, involves one of the
conditions of equilibrium, viz. that in which the area of the immersed
part, and that of the whole section, are to each other, as the specific
gravity of the solid is to that of the fluid ; but in order to discover the
equation which involves the other condition, we must have recourse to
a separate construction, as follows.

Let IK be the surface of the fluid, and HECD the immersed portion
of the section, as in the general dia-
gram preceding. Bisect D H and c E, ~ \'- "
the parallel sides of the figure, in the
points t and z, and draw tz;* then,
by the principles of mechanics, the l =^^-^
straight line tz passes through the
centre of gravity of the figure HECD,
and divides it into two parts such, that
gz : gt : : SDH-J-CE : DH + 2cE.

Through the point E draw E£* parallel to DC, and bisect DC and
HE in the points i and h; draw ih, bisecting E£ the side of the
triangle HE* in the point q, and join th and uq, intersecting one
another in the point o ; then is the point o thus determined, the centre
of gravity of the triangular space H E t.

Draw the diagonal D E, bisecting q i in p the centre of gravity of
the rectangular space ECD t, and join po intersecting tz in g ; then is
g the centre of gravity of the quadrilateral space HECD which falls
below the plane of floatation. Through the point g thus determined,
draw the straight line gv parallel to DC, the lowest immersed side of
the section, and meeting CE in the point v; then is gv the quantity
to be assigned by the construction.

From the point z and parallel to CD, draw zf meeting DH perpen-

* It is a circumstance entirely accidental, that the lines tz and <E terminate in
the same point t, and consequently, has nothing to do with the conditions of the
problem j it only happens when D H is double of c E.

z 2

340 OF THE POSITIONS OF EQUILIBRIUM.

dicularly in f\ then are the triangles tzf and zgv similar to one
another, and tf is half the difference of the sides D H and CE ; that is,

tf-\(x-y}.

Consequently, by the property of the right angled triangle, that the
square of the hypothenuse is equal to the sum of the squares of the
base and perpendicular, we shall have

and by extracting the square root, it is

But by the property of the centre of gravity alluded to in the con-
struction of the diagram, it follows, that

* + (x-y? : gz : : 3(* + y) : 2
or by equating the products of the extremes and means, we get

and from this, by division, we shall obtain
_(<2

then, because of the similarity of the triangles tzf and zgv, it is

tz : zf: : gz : gv,
or by substituting the analytical expressions, it becomes

and finally, by working out the analogy, we obtain

d-W* + y\

~ 3(x + y)

Referring now to the original diagram, or that on which the prin-
cipal part of the investigation depends, it will readily appear, that
since sr passes through g, the centre of gravity of the quadrilateral
figure HECD, it follows, that sg is equal to gr; but by the construc-
tion cv is equal to sg, and consequently equal to |sr; now sr is
manifestly equal to sw and wr taken conjointly ; therefore we have
cvnr \(sw -\- wr).

Since by the construction, the lines DH and sr are parallel to one
another, the triangles HE* and EW are similar; therefore, by the
property of similar triangles, we have

E£ : tii : : EM; : wr.

OF THE POSITIONS OF EQUILIBRIUM. 341

or by substituting the analytical expressions, it becomes

from which, by working out the proportion, we get

tt.r-(*— y)(2* + y).

**±9]
consequently, by adding and dividing by 2, we obtain

6(0: + y)

Again, it is obvious by the construction, that the triangles HE* and
vcd are similar to one another ; hence we have

HE : ut : : cv : vd;
but by the property of the right angled triangle, it is

HE — ^/b*+(x — yf;
consequently, by substitution, we obtain

hence, by reducing the proportion, we get

It is furthermore manifest, that the triangles HE* and vgn are
similar to one another ; consequently, we have

HE : E * : : gv : gn\
or by substituting the respective values, we get

from which, by reduction, we obtain
)

but ge = yn 4. en ; therefore, by addition, we have

(* + y)

(266).

428. This is one side of the equation which involves the second
condition of equilibrium, and in order to determine the other side, we
must have recourse to the triangles acb and a cm, which together

342 OF THE POSITIONS OF EQUILIBRIUM.

with the triangle HE*, are similar among themselves; consequently,
we have first, from the triangles list and ac£, as follows.

H E : ii t : : a c : a b ;
which by substitution becomes

y)»:(* — y):: Ja:a6
therefore, by reduction, we obtain

a(x — y)
- v yj -

Again, from the triangles HE£ and aowt, we get

HE : E£ : : ao : om;
consequently, by substitution, it is

from which, by reduction, we obtain
but GC — Gw-f-^c; therefore, by addition, we have

GC.

429. Here then, we have discovered the other side of the equation
which involves the second condition of equilibrium, and consequently,
we are now prepared to determine the positions which the body assumes
when floating in a state of rest ; for which purpose, let the equations
(266 and 267) be compared with each other, and we shall have

Here it is manifest that the equation involves two unknown quan-
tities ; in order therefore to render it capable of solution, one of those
quantities must be eliminated, and this can very easily be done by
means of the equation (265), where we have

therefore, by transposition and division, we get
2as

y— 7 — x>

or by supposing s' equal to unity, as is the case with water, we have
t/:n2as — x.

OF THE POSITIONS OF EQUILIBRIUM. 343

Let this value of y be substituted instead of it, wherever it occurs in
the above equation, and we shall obtain

which being reduced and thrown into a simpler form, becomes
2x8 — 6asx2-f (12aV — 6a2s -f 62)a;:=: 8aV-|- a£2s — 6aV. (268).

Now, according to the nature of the generation of equations, it is
manifest that the above expression is composed of one simple and one
quadratic factor; but as — #zzO, is obviously one of the members
from which the equation is derived, for in that case, the whole vanishes,
or which is the same thing, when all the terms of the equation are
arranged on one side with their proper signs, the sum total is equal
to nothing.

Granting therefore, that as — a:~0, is one of the constituent
factors, then we shall have

x~ as,
and by referring to equation (265), we shall obtain

therefore, by transposition, it is

Consequently, the position of equilibrium assumed by the solid in
this instance, is when x and y are equal to one another ; that is, when
the side of the body is parallel to the horizon, the depth to which it
sinks being determined by the measure of its specific gravity.

Let all the terms of the equation (268) be transposed to one side,
and let their aggregate be divided by (as — #), and there will arise

2x2 — 4asx+$a2s* — 6a2s-j-62 — 0,

and from this, by transposition and division, we obtain

x2 — 2a sx = a*s (3 — 4*) — j#.

From this equation it may be inferred, that if the roots or values of
x be both real and positive quantities, and each of them less than a
the upward side of the section ; then the body may have two other
positions of equilibrium, which will be determined by reducing the
equation.

Complete the square, and we obtain

V — 3aV(l — *) — J# ;

344

OF THE POSITIONS OF EQUILIBRIUM.

therefore, by evolution, it becomes

x — asi=

3azs(l — s) — i
and by transposition, we have

x = as+^/3a?s(l— s) — %b* ;
and the corresponding values of yf are

(269).

3a8s(l— s) — \tf. (270).

It would be superfluous in this place, to give a numerical example
to illustrate the reduction of equations (269 and 270) ; we shall
therefore drop the discussion of the oblong rectangular section, and
proceed to inquire, what are the circumstances which combine to
establish the equilibrium in a square.

430. Therefore, when a and b are equal to one another, that is,
when the transverse section is a square ; then the general equation
(268), becomes

2x* — 6£sa;2-f 68(12s8 — 6s -f 1 ) x = ^(Ss8 — 6s*-fs); (271).

but one of the constituent factors of this equation is,

bs — x — 0 ;
consequently, by transposition and division, the other factor becomes

2a8— 4bsx 4- &X8s8— 65 -f 1) = 0 ;
and from this, by transposing and dividing by 2, we shall get

a?8— 2bsx — b\3s — 4s8— J). (272).

Now, it is manifest, that when the section is a square, as we have

assumed it to be in the present instance, the factor bs — arzzO, gives

x = bs,

and from equation (265), we obtain
y ' — 2bs — x ~%bs — b$~ bs.

Hence it appears, that the body will float in a state of quiescence,
when any of its sides is horizontal, and in this case, the problem is
reduced to the determination of the depth to which the body sinks,
and this is entirely dependent on the A

measure of its specific gravity.

431. If the specific gravity of the
floating solid, be to that of the fluid j=
on which it floats in the ratio of I to 2 ; j
then the body will sink to one half its f
depth, as represented in the annexed
diagram, where IK is the horizontal

OF THE POSITIONS OF EQUILIBRIUM. 345

surface of the fluid; EF the water line, or line of floatation; EFCD
the immersed portion of the section, and ABFE the part that is extant,
and these in the present case, are equal to one another, since the
specific gravity of the fluid is double the specific gravity of the float-
ing body.

It is also obvious from the figure in this case, that if the body were
to revolve about its axis of motion, till one of the diagonals assumed
a vertical position, it would then float in equilibrio with one half the
section immersed, the horizontal diagonal in that case coinciding with
the surface of the fluid.

If we resolve the quadratic equation (272), we shall obtain twa
other positions, in which the body will float in equilibrio, provided
that the specific gravity be retained within proper limits ; for it is on
this limitation solely, that the equilibrium of floatation depends.

Complete the square, and we shall have

a*— %bsx + 6V= b\3s — 3s2— j),
extract the square root, and we obtain

x — bs—=^b^3s(l— s) — ,
therefore, by transposition, we have

x = b(s+)/3s(l — s) — J. (273).

But by equation (265), we have 2bs = x -\-y ; consequently, by
transposition, we obtain

y — 2bs — x;
therefore, by substitution, we get

y — b(s =pV 35(1— *) — J. (274).

Now, with regard to the limits of the specific gravity, it is easy to
perceive, that if the quantity ^3s(\ — s) — £ be greater than s9
the least values of x and y will be negative ; and if the expression
* -|--V/3s(l — s) — J be greater than unity, the greatest values of x
and y will be greater than b ; consequently, neither of them satisfies
the conditions of the problem. But it is further manifest, that in
order to have the values of x and y real quantities, the expression
3s(l — s) must exceed the fraction ^ ; now the least value of s that
will fulfil this condition, is s^z j, in which case we have

35(1— S)ZZ3X|XJ = TQ6,

from which subtracting | or T8F, we get

346

OF THE POSITIONS OF EQUILIBRIUM.

the square root of which is J, hence it is

432. The position of equilibrium indicated by these values of a; and
y, is represented in the annexed

diagram, where IK is the hori-
zontal surface of the fluid; AE
the line of floatation ; A E c D the
immersed, and ABE the extant por-
tion of the section.

Here it is obvious, that since the
plane of floatation passes through
the angle A, and bisects the oppo-
site side in the point E; the immersed part AECD, is equal to three
fourths of the entire section ABCD, as it ought to be, in consequence
of the specific gravity of the body, being assumed equal to three
fourths of the specific gravity of the fluid.

It may also be readily shown, that the centre of gravity of the
whole section, and that of the immersed part occur in the same
vertical line ; but this is not necessary in the present instance, as we
are only endeavouring to discover the limits of the specific gravity.

433. The position of equilibrium corresponding to the value of
x—\b, and y zn b, is similar and subcontrary to the position represented
in the preceding diagram, and this being the case, it is unnecessary to
exhibit it ; we shall therefore proceed to determine the greatest limit
of the specific gravity that will fulfil the conditions of the problem ;
for which purpose, we have

&(1 -.) = !,
from which, by separating the terms, we get

3S — 3s2— J;
therefore, by transposition and division, it becomes

Complete the square, and we obtain

s2— s + imi — 1 = ^,
hence, by extracting the square root, we get

(275).

and finally, by transposition, we have

434. From what has been done above, it is manifest, that the least
limit of the specific gravity is f , and the greatest is £(3 -f- ^ 3) ; the

OF THE POSITIONS OF EQUILIBRIUM.

347

former giving the position represented in the preceding diagram, and

the latter that which is exhibited in

the marginal figure ; where the body

floats with one of its flat surfaces

horizontal, IK being the surface of

the fluid ; E F the water line, or line

of floatation; EFCD being the im-

mersed part of the section, and

A B F E the part which is extant, the

immersed part being to the whole

section, as 0.788675 to 1 ; that is

ED: AD:: i(3 4-^/3) •!•

Since £(3 — y/'S is also a root of the equation (275), it follows, that
the body will float in equilibrio with
one of its flat surfaces horizontal, as
in the annexed figure, when the spe-
cific gravity is equal to the above
quantity ; for in that case the radical
expression ^ 3s(l — s) — J in equa-
tions (273 and 274) vanishes, and
x and y become each equal to

, and the immersed part of the section is to the whole, as
0.211 to 1 ; that is

ED : AD : : i(3 — V*3) : 1.

435. Having established the limits between which the solid floats
in equilibrio with a flat surface upwards, but inclined to the horizon
in various angles depending on the specific gravity ; we must now
return to the equations (273 and 274), in which the conditions are
indicated, that have enabled us to assign the above limits to the
relative weight of the floating body.

Taking the arithmetical mean between the limits above determined,
we shall have

s = |(0.75 + 0.788675) = 0.7693375 ;

consequently, if the side of the square section be equal to 20 inches,
the values of x and y will be determined by the following operation.

436. Let the mean calculated value of s the specific gravity of the
floating body, and the given value of b the side of its square section,
be respectively substituted in equation (273), and we shall have, for
the greatest value of a-,

x =20(0.7693375+ V 3x0.7693375 x 0.2306625-0.5) =18.96inches,

348

OF THE POSITIONS OF EQUILIBRIUM.

and for the least value of x, it is
a:zz20(0.7693375— v/ 3 X 0.7693375 X 0.2306625— 0.5)zzl 1 .8 inches,

the corresponding values of y as found from equation (274), are
y — 11.8 inches, and y — 18.96 inches.

Now, the positions of equilibrium corresponding to the above values
of x and ?/, are
as represented
in the annexed
diagrams, where 1
IK is the hori-
zontal surface
of the fluid ; E F
and ef the lines
of floatation

ABCD the position corresponding to am 18.96, and y rz 11.8 inches ;
the position abed, being that arising from the reversed values of x
and?/; that is, a;— 11.8 and y— 18.96 inches; EFCD being the
immersed part of the section in the one case, and efcd in the other.

437. If the specific gravity be taken equal to the complement of
the above mean, we shall obtain two other positions of equilibrium in
which the body will float, corresponding precisely to the above figures
inverted, af-
ter the man- A^ ^^o
ner exhibited
in the mar-
ginal dia-
gram ; where
ABCD is the
position cor-
responding

to £ — 20 — 18.96zz 1.04 inches, and abed the position correspond-
ing to 20 — 11.8:^:8.2 inches; the immersed portions EFCD and
efcd in the one case, being equal to ABFE and abfe, the extant
portions in the other.

It is moreover manifest, that the centres of gravity of the immersed
and extant portions of the section are situated in the same vertical ;
for they are connected by a straight line which passes through the
centre of gravity of the entire figure ABCD; but in the case of an
equilibrium, the centres of gravity of the whole, and the immersed
part, are situated in the same vertical line ; therefore also, the centres

OF THE POSITIONS OF EQUILIBRIUM. 349

of gravity of the immersed and extant parts occur in the same
vertical.

438, We have now to inquire if the second condition of equilibrium
be satisfied ; that is, if the area of the whole section and that of the
immersed part, are to one another as the specific gravity of the fluid,
is to that of the solid.

Now, the area of the whole section, is 20x20 = 400, and that of
the immersed portion, is (1 8.96 -f 11.8)10 = 307.6, or 92.4; and the
mean specific gravity is, 0.7693375 or 0.2306625 ; therefore we have

400 : 307.6 : : 1000 : 769, and 400 : 92.4 : : 1000 : 231 ;
consequently, the positions of equilibrium are as exhibited above.

PROBLEM LX.

439. A solid homogeneous body, having the section which
cuts the axis of motion perpendicularly, in the form of a common
or Apollonian parabola, is supposed to float upon a fluid of
greater specific gravity than itself: —

It is required to determine the position it assumes when in
a state of equilibrium, supposing its base or extreme ordinate
to be entirely above the surface of the fluid.

In the resolution of this problem, we shall have occasion to advert
to several properties of the common parabola, a curve which, by
reason of its easy construction, and the simplicity of its equation, has
been very extensively introduced into mechanical science; and from
the frequency of its occurrence, it is presumed, that its chief properties
are familiar to and clearly understood by the greatest part of our
readers ; so that in tracing the positions of equilibrium, it will not be
requisite to demonstrate any of the properties to which we refer, and
which, by the nature of the investigation, we are constrained to
employ. A

Let AD B be a common para-
bola, representing a transverse
section of a solid uniform body>
floating at rest upon a fluid
whose surface is IK, and let T___ J[j
DC be the axis, AB the base or
extreme ordinate, (which, by "1
the conditions of the problem,

350 OF THE POSITIONS OF EQUILIBRIUM.

is entirely above the surface of the fluid), and FII the line of
floatation.

Bisect FH in w, and through n draw nm parallel to DC the axis
of the parabola, and meeting the curve in the point m ; then is mn
when produced to r a diameter of the curve, whose vertex is in the
point m.

Through the point m, draw mt parallel to AB the base of the
parabola, and meeting the axis DC in the point K ; produce CD to E,
making BE equal to DK, and join EW, then by the property of the
parabola, Em is a tangent to the curve in the point m, and it is
parallel to FH the line of floatation, or the double ordinate to the
diameter mr.

Let P be the place of the focus; join ?WP, and through H the
extreme point of the line of floatation, draw uv parallel to AB the
base of the figure, and meeting the diameter mr perpendicularly in
the point v; then are the triangles KETW and vnii similar to one
another.

Take DG equal to three fifths of DC, and mg equal to three fifths
of mn; then are G and g respectively the centres of gravity of the
whole parabola ADB and of the part FDH; join og, then by the
principles of floatation, the straight lines Gg and FH are perpendicular
to one another, and consequently, the triangles KEW and wng are
similar.

Put a m DC, the axis of the parabola or section of the floating

body,
26zz AB, the base or double ordinate corresponding to the

axis DC,

a' zz the area of the whole parabola ADB,
a"z= the area of the immersed part FDH,
x ~ mn, an abscissa of the diameter mr,
y zz: H n, the corresponding ordinate,

z zz Km, the ordinate passing through m the point of contact,
p zz the parameter or latus rectum to the axis,
s zz the specific gravity of the floating solid, and
s' zz the specific gravity of the fluid on which it floats.

Now, supposing that all the sections which are perpendicular to the
axis of motion, are equal to one another; then, according to the
principles of floatation, we have

a'szza'Y. (276).

OF THE POSITIONS OF EQUILIBRIUM. 351

But the writers on mensuration have demonstrated, that the area of

the common parabola, is equal to two thirds of its circumscribing

rectangle, or equal to four thirds of the rectangle described upon the

axis and the ordinate ; according to this principle therefore, we have

a'zzi^DcXAC, and a"~%vnXmn.

By the equation to the curve, it is

therefore, by division, we obtain

&
DKrr — ;

P
but according to the construction, we have

and by the property of the right angled triangle, it is

K E2-f K m>— E m3 ; that is, E w2= —
therefore, by extracting the square root, we shall have

and from the similar triangles KETW. and vrm, we get

Ts.m : Km : : nn : VH;

and this, by substituting the analytical equivalents, becomes
z

: z : : y : v H ;
P '
consequently, by working out the analogy, we have

VH±T

Hence then, by substituting the respective literal representatives,
for the quantities DC, AC and VH, mn, the preceding values of d and
a", become

a'±z 4« b, and a"=

Therefore, let these values of a' and a" be substituted instead of
them in the equation (276), and we shall obtain

If we suppose the axis of the parabola to be vertical, and its base
or double ordinate horizontal ; then the points m and D coincide with

352 OF THE POSITIONS OF EQUILIBRIUM.

one another, and KTWZHZ vanishes; consequently, in that case, equa-
tion (277), becomes

a b s — x y s' ;
but by the property of the parabola, we have

y— ^/px;
and similarly, we obtain

b zz ^p a ;
therefore, by substitution, we get

a s \/pa zz x s \/p x ;
by squaring both sides, it is

s'V — 5s a8, (278).

and finally, by division and evolution, we have

440. The practical rule supplied by this equation, may be expressed
in words at length in the following manner.

RULE. Divide the square of the specific gravity of the
floating solid, by the square of the specific gravity of the fluid
on which it floats, then multiply the cube root of the quotient
by the axis of the parabola, and the product will give the
portion of the axis, which falls below the plane of floatation,
or the surface of the fluid.

441. EXAMPLE. A solid body whose transverse section is in the
form of a parabola, floats in equilibrio on the surface of a fluid with
its vertex downwards, and its base or double ordinate horizontal ; it is
required to determine how deep the body sinks, supposing the vertical
axis to be equal to 40 inches, the specific gravity of the body and that
of its supporting fluid, being to one another, as 686 to 1000.

Here, by operating as directed in the rule, we shall have

s2=6862z=: 470596;

s'2=10002=: 1000000;

from which, by division, we obtain

the cube root of which, is
0.470596 = 0.7778 ;

OF THE POSITIONS OF EQUILIBRIUM.

353

consequently, by multiplication, we finally obtain
a; =: 40X0.7778 = 31. 112 inches.

442. Therefore, the position of equilibrium corresponding to the
above value of a?, is as represented

in the annexed diagram, where
AB is the base or double ordinate
of the parabolic section, DC its
axis; FH the water line, or double
ordinate of the immersed portion
FDII, DE the corresponding ab-
scissa, and IK the horizontal sur-
face of the fluid.

That the condition is satisfied, in which the centres of gravity of
the whole and the immersed part are situated in the same vertical, is
manifest from the circumstances of the case ; and that the other con-
dition is satisfied, in which the areas of the whole and the immersed
part, are to each other, as the specific gravities of the fluid and the
solid, will appear from the following calculation.

Since the parabolas ADB and FDII, are similar to one another,
having the same parameter and being situated about the same axis ;
it follows, that

a^ a : x \/ x : : a : a" ;

but by the question, a is equal to 40 inches, and by the foregoing
computation, we have found that & = 3 1.1 12 inches; therefore, we get

SOv/To": 31.112t/3nT2 : : 1000 : 686,

which satisfies the other condition of equilibrium, from which we infer,
that if the specific gravity of the solid be taken within proper limits,
the preceding diagram exhibits a position of floating.

443. The equation (278) was obtained on the supposition, that the
axis of the parabola is vertical and the points D and m coincident, in
which case the quantity z vanishes entirely from the figure ; but the
same result will obtain whether we consider the points D and m to be
coincident or not, as will appear from what follows.

By the property of the parabola, that the distance of any point of
the curve from the focus, is equal to the perpendicular distance
between that point and the directrix, it follows, that mp (see Jig.
art. 439) is equal to the sum of DK and DP taken jointly ; that is,

VOL. I.

354 OF THE POSITIONS OF EQUILIBRIUM.

now, we have already seen that D K is expressed by — , and by the
nature of the curve, we have DP =r \p ; therefore, by addition, it is

But since by the property of the parabola, the parameter of any
diameter, is equal to four times the distance between the vertex of
that diameter and the focus, we have

P
and by the equation to the curve, it is

consequently, by extracting the square root, we obtain

Let this value of y be substituted instead of it in equation (277),
and we shall obtain

VP
but by the equation to the curve, we have

therefore, by substitution, we shall get

VP
and multiplying both sides by ^/p, we obtain

from which, by squaring both sides, we get

which is the identical expression, obtained on the supposition of a
coincidence between the points D and m ; consequently, the value of
x must be the same in both cases, and the position of floating depend-
ing upon the specific gravity must also be the same.

Now, by the construction we have seen, that the triangles KEW and
wng are similar to one another; hence we get
Em : EK : : gn : wn,

OF THE POSITIONS OF EQUILIBRIUM. 355

or by taking the analytical equivalents, it becomes

- — —,

and from this, by working out the proportion, we get
4xz

hence, by subtraction, we have ,.,'< j
ws — ns — w;w;

but because Emus is a parallelogram, it follows, that ws— E?n;
therefore, it is

Again, the triangles wgn and WGS are similar to one another; but
we have shown above, that KEW is similar to wng; therefore, KEW
is similar to WGS, and we have

EK : EWZ : : ws : SG ;
taking therefore the analytical values, we obtain

2£: ~

P P

consequently, by reduction, we have

2p 5

But by the nature of the figure, SG is manifestly equal to EG — ES ;

z2
now ESZZ mn^nx, and EG ^ZDG -J- DE; that is, EG — fa-j

consequently, we have

s G — f- a -| # ;

let these two values of SG be compared with each other, and we get

~~Tp "If—If"1'"*' (279).

and finally, by reduction, we obtain

~~ 10
Now, the value of a;, as we have determined it from equation (278), is

2A2

356 OF THE POSITIONS OF EQUILIBRIUM.

therefore, by substitution, we have

(280).

Here then, we have obtained a pure quadratic equation, which gives
two subcontrary positions of equilibrium, provided that the specific
gravity be taken within proper limits.

Extract the square root of both sides of equation (280), and we
shall obtain

and if the specific gravity of the fluid be expressed by unity, as is the
case when the fluid is water, then we shall have

(281).

But in order to have the value of z a real positive quantity, it is
necessary that 6a should be greater than 5p -f- 6a^s2 ; in order there-
fore, to find the greatest value of * that will satisfy this condition, we
must put these two quantities equal to one another, arid in that case
we shall obtain

transpose, and we obtain

6a$72=6a— 5p;
divide by 6a, and it becomes
5p

therefore, by involution, we get

'=<'-£>'•.. '"'..,..

and finally, by evolution, it is

,1, •=(>-!•)*•

Here then it is manifest, that in order that the positions determined
by the equation may be those of equilibrium, it is necessary that the

specific gravity of the floating body shall be less than ( 1 — — ^

> 6a/

444. EXAMPLE. A solid body whose transverse section is in the
form of a parabola, is placed in a cistern of water with its vertex

OF THE POSITIONS OF EQUILIBRIUM. 357

downwards, in such a manner, that its base or extreme ordinate is
entirely above the surface ; it is required to determine the position of
the body when in a state of equilibrium, the parameter of the parabolic
section being 16 inches, the axis 40 inches, and the specific gravity of
the floating solid, to that of the supporting fluid as 1 to 2 ?

In this example there are given, p — 16 inches, am 40 inches, and
s — 0.5, the specific gravity of water being unity ; therefore, by sub-
stitution, we get from equation (281)

* —^ 1.6(6X40 — 5X16 — 6x40/0125) — 3.75 inches.

And the positions of equilibrium corresponding to this value of g,
are as represented in the subjoined diagrams, and the following is the
method of construction.

With the parameter or latus rectum equal to 16 inches, and the
subcontrary axes DC and dc each equal to 40 inches, describe the
parabolas ADB and adb ; from c the middle of the base and towards
the depressed part of the figure, set off cr equal to 3.75 inches, the
computed value of z ; through the point r, draw rm parallel to the
axis CD, and meeting the curve in the point m ; draw the tangent WE,
and on the diameter mr, set off mn equal to 25.19 inches, the value
of x as obtained by the reduction of equation (278) ; then through the
point n, and parallel to the tangent ?WE, draw the straight line IK,
which will coincide with the surface of the fluid, and cut the parabolas
ADB and adb in F, H and/", h the extremities of the lines of floata-
tion, corresponding to the positions of equilibrium which we have
exhibited in the diagrams.

We must now endeavour to prove, that the positions in which we
have represented the body are those of equilibrium; and for this
purpose, we must inquire if the equation (279) is satisfied by the sub-
stitution of the computed values of x and z ; for when that is the
case, the line which joins the centres of gravity of the whole section
and the immersed part of it, is perpendicular to the surface of the fluid.

358 OF THE POSITIONS OF EQUILIBRIUM.

Now, the values of x and z as we have determined them by calcu-
lation, are respectively equal to 25.19 and 3.75 inches ; therefore, by
substitution, equation (279) becomes

162-j- 4X3.75* 2X25.19 __ 3X40 3.752
2X16 ~5~~ ~~5~~ ~16"

from which, by transposition, we have
162-f 4X3.75* 2X25.19 3x40 3.752

2X16

445. Here then, it is manifest, that one of the conditions of equi-
librium is satisfied, viz. that in which the line which passes through the
centres of gravity of the whole section and the immersed part of it, is
perpendicular to the surface of the fluid; we have therefore in the
next place, to inquire if the areas of the whole section and the im-
mersed part, are to one another, as the specific gravity of the fluid is
to that of the solid. Now, we have seen, equation (277), that

——

vy+422

but by the nature of the parabola, b zn \/ ap ; hence it is

and we have elsewhere seen, that the value of yy is

consequently, by substitution, we get

therefore, by expunging the common term \/p, and converting to an
analogy, we get a^a : x^/x : : s' : s,

and this, by substituting the given value of a, and the computed value
of x, is

80/10~: 25.19V2539 : : 1 : 0.5.

From this it appears, that the second condition of equilibrium is
also satisfied ; we may therefore conclude, that the positions in which
we have represented the body are the true ones ; but we may further
observe, that by altering the specific gravity of the body, other posi-

tions may be exhibited, provided that the expression — shall never
exceed fl — - — j* in the common or Apollonian parabola.

CHAPTER XIII.

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

446. IN the preceding pages, we have investigated and exemplified
the method of determining the positions of equilibrium in a few of the
most important cases, where the forms of the floating bodies are such
as to render them of very frequent occurrence in practical construc-
tions ; we shall therefore, in the next place, proceed to investigate and
exemplify the conditions of Stability, or that power, by which, when
the equilibrium of a floating body has been disturbed, it endeavours,
in consequence of its own weight and the upward pressure of the
fluid, either to regain its primitive settlement, or to recede farther
from it, by revolving on an axis passing through its centre of gravity
parallel to the horizon, until it arrives at some other position of
equilibrium, in which the principles of quiescent floatation are again
displayed.

1. DEFINITIONS AND PROPOSITIONS OF STABILITY IN FLOATING BODIES.

447. It is familiar to every person's experience, that when bodies of
certain forms and dimensions, placed under particular circumstances
on the surface of a fluid, have their equilibrium deranged by the
action of some external force, they return to their original position
after a few movements or oscillations backwards and forwards, in a
direction determined by that of the disturbing impulse.

It is equally obvious with regard to other bodies, that however
small may be the quantity of their deviation from the original state of
quiescence, they have no tendency whatever to return to it, but con-
tinue to recede farther and farther from it, by revolving about a
horizontal axis, until the deviating effort obtains a maximum depend-
ing upon the angle of deflexion ; after which the deflecting energy

360

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

continues to decrease until it vanishes, in which case the body settles
in another situation, which also satisfies the conditions of equilibrium.

Again, a solid body may be so constituted with respect to shape
and dimensions, that in every position which can be given to it on
the surface of a fluid, it will rest in a state of equilibrium ; for in all
situations and under every condition, the centre of gravity of the
whole body and that of the immersed part, will occur in the same
vertical line ; this being the case, it is manifest that in such a body,
the equilibrium cannot be disturbed, because the external force, how-
ever it may be applied, can only operate to turn the solid round an
axis, passing through the centre of gravity in a direction parallel to
the horizon.

Homogeneous spheres are bodies of this sort, so also are homo-
geneous cylinders floating with the axis horizontal; these have no
tendency to solicit one situation in preference to another, and con-
sequently, in whatsoever position they are placed, with reference to
the axis of revolution, they are still in a state to satisfy the conditions
of equilibrium, for the centre of gravity of the whole body and that
of the immersed part, are always situated in the same vertical line.

In the first case then, where the body has a tendency to restore
itself to the original position, the equilibrium is said to be stable ; in
the second case, where the body deviates farther and farther from the
original state, the equilibrium is unstable ; and lastly, in the case where
the body has no tendency to remain in, or to solicit one position in
preference to another, the equilibrium is said to be insensible.

448. The conditions of equilibrium as here stated, are in themselves
sufficiently simple and explicit, but in order that none of our readers
may enter upon this important and difficult subject, with incorrect
notions respecting the different species of equilibrium, and the various
conditions or circumstances of floating under which a body may be
placed, we have thought it expedient to subjoin the following expo-

sitions and illustra-
tions.

Let the dotted line
inj^. 1 represent a
transverse section of
any uniform prismatic
body, placed verti-
cally on the surface
of a fluid, and sup-
pose the specific

Ffe.i.

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

361

gravity of the body to be such, when compared with that of
the fluid on which it floats, as to sink it to the depth mn. The
body floats in equilibrio in the upright position ; suppose therefore
that by the application of some extraneous agent, it is deflected into
the position abed, where it is conceived to revolve about a horizontal
axis, passing through G its centre of gravity, at right angles to the
plane abed. If therefore, the body when thus inclined, requires the
force /'to retain it in that state, or to prevent it from returning to the
upright position ; then the equilibrium in which the body is originally
placed, is what we understand by the equilibrium of stability.

Again, let the dotted Fig. 2.

line in fig. 2 represent
a vertical section of
any uniform homogene-
ous prismatic body , float-
ing upright and quies-
cent on the surface of a
fluid, and let the specific
gravity of the solid be
such as to sink it in the
fluid to the depth mn\
suppose now, that by the action of some external force, the body is
deflected from the vertical position into that represented, by a b cd ; it
is obvious, that the revolution is made about the horizontal axis
passing through G the centre of gravity, at right angles to the plane
abed.

Hence, if the body when thus inclined, requires the application of
the force/ to retain it in that state, or to prevent it from inclining
further ; then the equilibrium in which the body is originally placed,
is what we comprehend by the equilibrium of instability .

Finally, let abed, fig. 3, be the vertical -*»£• 3-

section, and let the specific gravity of the
body be such, as to sink it to the depth md;
the solid floats in equilibrio in the upright
position, and in such a manner, that an
evanescent force will either retain it in that
state, or deflect it from it ; this is the insen-
sible equilibrium, or the equilibrium of indif-
ference, and the solid is said to overset.

449. Of these three species of equilibrium, bodies floating on the
surface of a fluid are manifestly susceptible; but they admit of a

362 OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

more perspicuous and comprehensive definition, which may be scien-
tifically read in the following manner.

1 . The Equilibrium of Stability, is that property in floating bodies,
by which on being slightly inclined to either side, they endeavour to
redress themselves and to recover their original position.

2. The Equilibrium of Instability, is that property in floating
bodies, by which on being slightly inclined from the upright position,
they tumble over in the fluid and assume a new situation, in which
the conditions of floating again occur.

3. The Equilibrium of Indifference, is that property in floating
bodies, by which they are enabled to retain whatever position they
are placed in, without exhibiting the smallest tendency, either to
regain the original position, or to deviate farther from it.

In addition to the different species of equilibrium described above,
there are several other terms of very frequent occurrence in the
doctrine of floatation, which it will be proper to explain before we
proceed to develop the laws that regulate the conditions of stability.
The most common and the most important of the terms here alluded
to, are the following.

450. DEFINITION 1. The Centre of Effort, is the same with the centre
of gravity of the entire floating body ; it is that point through which
the horizontal axis passes, and about which the body is supposed to
revolve.

DEFINITION 2. The Centre of Floatation, or the Centre of Buoy-
ancy, is the same with the centre of gravity of the immersed part of
the floating body, or it is the same as the centre of gravity of the fluid
displaced in consequence of the floatation.

DEFINITION 3. The Line of Pressure, is the vertical line passing
through the centre of effort, in the direction of which, the body is
impelled downwards by means of its own weight.

DEFINITION 4. The Line of Support, is the vertical line passing
through the centre of buoyancy ; it is either parallel to, or coincident
with the line of pressure, and is that in whose direction the body is
propelled upwards by the pressure of the fluid.

DEFINITION 5. The Axis of Motion, as we have already observed
in treating of the positions of equilibrium, is the horizontal line passing
through the centre of effort, and about which the body revolves on
being deflected from its original position.

DEFINITION 6. The Transverse Section of the Solid, is that indi-
cated by a vertical plane at right angles to the axis of motion, and
separating the body into any two parts : all the transverse sections

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS. 363

are parallel to one another, and the principal transverse section is that
which passes through the centre of effort.

DEFINITION 7. The Axis of the Section, is the straight line which
passes through its centre of gravity, dividing it into two parts, which
in the case of a regular body are equal and similar to one another.
When this axis is vertical, it either coincides with, or is parallel to
the line of pressure.

DEFINITION 8. The Line of Floatation, or The Water Line, is the
horizontal line in which the surface of the fluid meets a vertical trans-
verse section of the floating body.

DEFINITION 9. The Plane of Floatation, is the horizontal plane
coincident with the surface of the fluid, and which passes through
the water line, dividing the body into the immersed and extant
portions.

DEFINITION 10. The Equilibrating Lever, is a straight line equal
to the horizontal distance between the verticals passing through the
centre of effort and the centre of buoyancy ; or it is the horizontal
distance between the line of pressure and the line of support.

DEFINITION 11. The Stability of Floating, or the Measure of
Stability, is that force by which a body floating on the surface of a
fluid, endeavours to restore itself, when it has been slightly inclined
from a position of equilibrium by the action of some external agent ;
or it is a force precisely equal to the fluid's pressure, or to the entire
weight of the floating body acting on the equilibrating lever. (See
Proposition (XI.) following).

DEFINITION ]2. The Metacentre, is that point in which the axis
of the section and the line of support intersect each other ; it limits
the elevation of the centre of effort.

Upon these definitions, therefore, in combination with the following
simple and obvious propositions, depends the whole doctrine of the
stability of floating bodies.

PROPOSITION IX.

451. It has already been admitted as a principle in the theory
of hydrostatics, that every body, whatever may be its form and
dimensions, if it floats upon the surface of a fluid of greater
specific gravity than itself, displaces a quantity of the fluid on
which it floats equal to its own weight, and consequently : —

364 OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

The specific gravity of the supporting fluid, is to that of
the floating body, as the whole magnitude of the solid is to
that of the part immersed. (See Proposition VII.)

PROPOSITION X.

452. If a solid body, of whatever form or dimensions, floats
upon the surface of a fluid of greater specific gravity than itself: —

It is impelled downwards by its own weight acting in the
direction of a vertical line passing through the centre of
effort ; and it is propelled upwards by the pressure of the
fluid which supports it acting in the direction of a vertical
line passing through the centre of buoyancy. (See Proposi-
tion VI.)

Therefore, if these two lines are not coincident, the floating body
thus impelled must revolve upon an axis of motion, until it attains a
position in which the centre of effort and the centre of buoyancy are
in the same vertical line.

PROPOSITION XL

453. If a solid body of any particular form and dimensions,
floating on the surface of a fluid of greater specific gravity than
itself, be deflected from the upright position through a given
angle : —

The stability of the body is proportional to the length of
the equilibrating lever, or to the horizontal distance between
the vertical lines passing through the centre of effort and the
centre of buoyancy. (See Problem LXI. following.)

When the horizontal distance here alluded to is equal to nothing ;
that is, when the centre of effort and the centre of buoyancy are
situated in the same vertical line; then the stability, or the force
which urges the body round its axis of motion vanishes, and the equi-
librium is that of indifference ; for in this case, the metacentre coin-
cides with the centre of effort.

If the floating body be any how inclined from the upright position,
and if, in consequence of the inclination, the line of support falls on
the same side of the centre of effort as the depressed parts of the
solid, then the length of the equilibrating lever is accounted positive,

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS. 365

and the pressure of the fluid operates to restore the equilibrium ; in
this case, therefore, the equilibrium is that of stability.

But when the line of support falls on the same side of the centre of
effort as the parts of the solid which are elevated in consequence of
the inclination ; then the length of the equilibrating lever is accounted
negative, and the equilibrium is that of instability.

Hence it appears, that the stability of a floating body is positive,
nothing or negative, according as the metacentre is above, coincident
with, or below the centre of effort: these consequences, however,
will be more readily and more legitimately deduced from the general
formula which indicates the conditions of stability, and this formula
we shall shortly proceed to investigate.

PROPOSITION XII.

454. The common centre of gravity of any system of bodies
being given in position, if any one of these bodies be moved from
one part of the system to another, it is manifest, from the principles
of mechanics, that : —

The motion of the common centre of gravity, estimated in
any given direction, is to the motion of the body moved,
estimated in the same direction, as the weight of the said
body, is to the weight of the entire system.

Therefore, by means of these propositions and the definitions that
precede them, the whole doctrine of the stability of floating bodies,
with the train of consequences which immediately flow from it, may
be easily and expeditiously deduced ; but in proceeding to develope
the laws on which the stability of floating depends, it will be con-
venient for the sake of simplicity, to consider the body as some
regular homogeneous solid, of uniform shape and dimensions through-
out the whole of its length ; for in that case, all the vertical transverse
sections will be figures precisely equal and similar to each other ; and
if the body be divided by a vertical plane passing along the axis of
motion, the two parts into which it is separated will be symmetri-
cally placed with respect to the dividing plane.

This being premised, the principles upon which the stability of
floatation depends, will be determined by the resolution of the follow-
ing problem, in which all the transverse sections are trapezoids.

366

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

2. PRINCIPLES OF THE STABILITY OF FLOATING BODIES.

PROBLEM LXI.

455. A solid homogeneous body of uniform shape and dimen-
sions throughout the whole of its length, is placed upon a fluid
of greater specific gravity than itself, in such a manner, that the
centre of effort and the centre of buoyancy are in the same
vertical line.

It is required to determine the stability, when by the
application of some external force, the body is deflected from
the upright position, or from a position of equilibrium through
u given angle.

Let the solid to which our investigation refers be such, that the
vertical transverse sections perpendicular to the axis of motion, are
equal and similar trapezoids, as indicated by ABCD and abed in the
annexed diagrams. The solid floats upon the surface of the fluid IK,

and ABCD is its position when in a state of equilibrium ; ABFE being
the extant portion of the vertical section, and EFCD the part immersed
beneath the fluid's surface. The point G is the centre of effort, or the
centre of gravity of the whole section, the plane of which is supposed
to pass through the centre of gravity of the body, and g is the centre
of buoyancy, or the centre of gravity of the part immersed below the
surface of the fluid ; then since the body floats in a state of equili-
brium, it follows from Proposition X., that PQ the axis of the section,
which passes through the points G and g, is perpendicular to EF the
line of floatation.

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS. 367

We are now to suppose, that by the application of some external
force, the solid revolves about its axis of motion until it comes unto
the position represented by abed, in which state the equilibrium does
not obtain.

Here it is manifest that PQ, the axis of the section which was
vertical in the first instance, is transferred, in consequence of the
inclination, into the position pq ; and in like manner, the line EF, which
before was horizontal, is transferred into the oblique position ef, and hi
is now the line of floatation, or as it is otherwise called, the water line.

Since the absolute weight of the body remains unaltered, whatever
may be the position of floating, the area of that portion of the section
which is immersed below the surface of the fluid, must also be inva-
riable; it therefore follows, that the areas hied and EFCD are equal
to one another ; but the space efcd is equal to EFC D, hence the spaces
hied and efcd are each of them equal to EFCD; they are therefore
equal to one another, and consequently, the extant triangle hke is
equal to the immersed triangle fh i.

On pq the axis of the section, set off GH equal to Gg, the distance
between the centre of effort and centre of buoyancy in the original
position of equilibrium ; then it is manifest, that in consequence of the
inclination, the point g, which is the centre of gravity of the space
EFCD, will be transferred to the point n, which is the centre of gravity
of the equal space efcd; and the pressure of the fluid would act upon
the body in the direction of a vertical line passing through 71, if efcd
were the portion of the section immersed under the fluid's surface ;
but this is not the case, for in consequence of the inclination, the
triangle fki, which was before above the fluid's surface, is now
depressed under it, and in like manner the triangle hke, which was
previously under the surface, is now elevated above it.

It is therefore obvious from Proposition XII, that by transferring the
triangle hke into the position fki, the point n, which is the centre
of gravity of the space efcd, must partake of a corresponding motion
and in the same direction ; that is, the point n must move towards
those parts of the body that have become more immersed in conse-
quence of the inclination, until it settles in g the centre of gravity of
the immersed volume hied.

Through # the centre of gravity of the immersed part hied, draw
ym perpendicular to hi the line of floatation, and meeting pq the
axis of the section in the point m ; then is m the metacentre, and the
pressure of the fluid will act in the direction of the vertical line g'm,
with a force precisely equal to the body's weight ; and according to

368 OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

the principles of mechanics, it will act with the same energy at what-
soever point of the line gm it may be applied.

Through the point n and parallel to hi the line of floatation, draw
nz cutting the vertical line gm in the point z, and through G the
centre of gravity of the whole space abed, draw or perpendicular
and GS parallel to nz, and let k be the point in which the lines ef
and hi intersect one another; then, as we have stated above, the
pressure of the fluid will have the same effect to turn the body round
its axis, whether it be applied at the point g or the point s ; we shall
therefore suppose it to be applied at the point s, in which case GS
will represent the point of the lever, at whose extremity the pressure
of the fluid acts to restore the body to its original state of equilibrium,
or to urge it farther from it.

Since the effect of the fluid's pressure, acting in the direction of
the vertical line which passes through g the centre of buoyancy, has
no dependence on the absolute position of that point, but on the
horizontal distance between the vertical lines rG and gm ; it follows,
that in the actual determination of the positions which bodies assume
on the surface of a fluid, and their stability of floating, the situation
of the centre of buoyancy in the inclined position is not required, for
the horizontal distance between the vertical lines which pass through
that centre and the centre of effort, is sufficient for obtaining every
particular in the doctrine of floatation.

Bisect the sides of the triangles h ke and/Az in the points u, v and
w, a:, and draw the straight lines ku, ev and kw, ix intersecting two
and two in the points / and o ; then are / and o the points thus deter-
mined, respectively the centres of gravity of the triangles hke and

fki.

Through the points I and o draw the straight lines ly and of,
respectively perpendicular to hi the line of floatation, corresponding
to the inclined position of the body ; then is yt the horizontal distance
through which the centre of gravity of the triangle hke has moved in
consequence of the inclination ; therefore, by the principle announced
in Proposition XII., we obtain

area efcd : area hke : : yt : nz.

It is easy to comprehend in what manner the proposition cited above
applies to the case in question ; for we may assume the area efcd as
a system of bodies, of which the common centre of gravity is n. One
of the bodies composing this system, viz. the triangular area hke,
conceived to be concentrated in the point /, is transferred, in conse-
quence of the inclination from the point I to the point o, in which the

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS. 369

equal volume fki is similarly concentrated ; this will have the effect
of moving the common centre of gravity of the system from n to z,
in a direction parallel to yt, and the distance nz, through which the
common centre is moved, is what the proposition determines.

Let the position of the point z be supposed known ; then, if a vertical
line be drawn through that point perpendicular to the line of floatation,
the centre of gravity of the immersed space hied, will occur in some
point of that line, as for example at g ; but we have already observed,
that it is not necessary to determine the absolute position of the point
in question, the horizontal distance GS or rz between the verticals
Gr and mz, being all that is required.

Put a zz: hied or efcd, the area of the immersed space,

d z= hke or fki, the area of the triangle which has been
assumed as constituting an individual body of the
system ;

d zzr yt, the horizontal distance through which the centre of
gravity of the triangle hke has moved, in shifting to the
position o in the triangle fki,

S zz: Gg or Gn, the distance between the centre of effort and
the centre of buoyancy, when the axis of the section is
vertical ;

b zz: A B or a b, the length of the greater parallel side of the
trapezoidal section,

/3 zz: DC or dc, the length of the lesser parallel side ;

D zz: PQ or pq, the perpendicular distance between the paral-
lels AB and DC, or ab and dc,

c zz: EF or ef, the water line or line of floatation in the
upright position,

I zz: the axis of motion, or the whole length of the floating
body, passing through o the centre of effort ;

s zz: the specific gravity of the floating body,

s' zz: the specific gravity of the supporting fluid, which in the
case of water, is expressed by unity ;

5 zz: the stability of the body, or the momentum of the redress-
ing force ;

<£ zzr/Az, or nGr, the angle of deflexion, and

x zz: GS or rz, the length of the equilibrating lever.

Then, by substituting the literal representatives of the several quan-
tities in the foregoing analogy, we shall obtain

a : a' : : d: nz, f*\* ™>

VOL. i. 2 B

UN1VERS!

370 OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

from which, by equating the products of the extreme and mean terms,
we get

aXnz~ a'd;

therefore, by division, we have

a'd

nz — .

But by the principles of Plane Trigonometry, it is

rad. : 3 : : sin.0 : nr,
which being reduced, gives

wrzzd sin.^,

and according to the construction of the figure, it is manifest that rz
or G s, is equal to the difference between nz and nr, the two quantities
whose values have just been determined ; consequently, by subtrac-
tion, we have

dd . .
X~~a asm*0' (282).

456. The equation which we have just investigated has reference
only to a particular case of the general problem, viz. that in which the
vertical transverse sections, throughout the whole length of the body,
are equal and similar figures ; this condition, although it is a restric-
tion upon the general applicability of our result, yet it allows an im-
mense latitude, for the figures of bodies whose parallel transverse
sections are equal and similar areas are very numerous ; and if we
substitute the magnitude of the whole immersed volume, and that
of the volume which becomes immersed in consequence of the inclina-
tion, instead of the areas of the respective sections, the above
equation becomes general, because its form and the manner of
combining the terms admit of no change.

The expression consists of five members on one side, one of which,
that is, the angle of deflexion, must always be a given datum, or it
must be directly assignable from the circumstances of the case, and
the others must all be determined by means of the given dimensions,
and other particulars dependent upon the figure of the section ; but
the method of applying the formula, and the whole operation neces-
sary for its reduction, will be sufficiently exemplified by the resolution
of the following example.

457. EXAMPLE. A solid homogeneous body, of which the trans-
verse parallel sections at right angles to the axis of motion, are
equal and similar trapezoids, is placed upon the surface of a fluid in
such a manner, that its broadest side is upwards and parallel to the

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

371

horizon ; the body floats in equilibrio in this position ; but suppose
that some external force is so applied to it, as to deflect it from the
upright and quiescent state through an angle of 15 degrees; it is re-
quired to determine the stability or the momentum of restoration, the
parallel sides of the section being respectively 40 and 30 inches, the
perpendicular distance between them 20 inches, and the whole length
of the body 14 feet, its specific gravity when compared with that of
the fluid being as 270 to 1000, or as 0.27 to 1 ?

458. For the purpose of rendering the several steps of the operation
perfectly clear and comprehensible, we shall refer to the annexed
diagram, which re-
presents a trans-
verse section of the
body in the inclined
position ; ef being
the line in which
it is intersected by
the water's surface
when it is upright,
and hi the corres-
ponding line when
it is deflected
through the angle fki. PQ is the perpendicular distance between
ab and cd the parallel sides of the section, QH the depth to which
the body sinks in the fluid as induced by the specific gravity ; o and
I are the centres of gravity of the triangles fki and hke; mt the
projected distance between them on the line hi, and Gg the distance
between the centre of effort and the centre of buoyancy, or the dis-
tance between the centre of gravity of the whole body and that of the
immersed part, when the body is upright.

Now, the several parts which have to be calculated in order to
resolve the question, are the areas efcd, fki, and the distances mt
and og ; for which purpose, we have given a b, dc, PQ, the angle fk i
and the specific gravity of the solid.

Therefore, according to the principles of mensuration, the area of
the whole section is expressed by

20
|(6 -f- /3) X D =(40 -f 30) X — = 700 square inches,

and by the nature of floatation, we have

2fi2

372 OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

or by putting s' equal to unity, and substituting the respective num-
bers, we obtain

area efcd = a = 700 X 0.27 — 189 square inches. (283).

459. Consequently, by having the area of the trapezoid efcd,

and one of its parallel sides dc given, the other parallel side ef and

the perpendicular depth Q H can easily be found ; for by the nature

of the figure and the property of the right angled triangle, we have

D C / - 7 20

«» = b=piV *

900+ 189 — 3o|— 6 inches.

therefore, by the property of the trapezoid, we have

3(30 + c)=r 189,

or by separating the terms and transposing, we get

3c = 189 — 90=99,

and by division, it is

99
ef= c=--=33 inches.

460. We must next endeavour to discover the point k, in which
the primary and secondary water lines intersect each other, and for
this purpose,

\)utfk nr y, then by subtraction, we have e k = 33 — y ;
but by the rules of mensuration, it is

and by restoring the above values of fk and e k, it becomes

yXki = (33 — y)Xkh. (284).

Through the point c and parallel to PQ, draw en meeting ef per-
pendicularly in n\ then it is manifest, from the principles of Plane
Trigonometry, that

*"•*•=£="=«•

which corresponds to the natural tangent of 75° 57' 49".
But by the principles of Geometry, the exterior angle cfn is equal
to both the interior and opposite angles fki and fik ; consequently,
by subtraction, we have

angle fik = 75° 57' 49" — 15° = 60° 57' 49",

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS. 373

and by Plane Trigonometry we get
sin.60°57'49" : sin. 75° 57' 49" : : y : ki = 1.10967,
and by proceeding in a similar manner with the triangle hke, we
shall have

sin.90°57'49" : sin.75° 57' 49" : : 33 — y : M zz 0.9703(33 -- y) ;
therefore, by substituting these values of ki and kh in equation (284),
we get

1.1096/nz 0.9703(33 — y)*,
from which, by reciprocating the terms, we obtain

and extracting the square root, it is

— = V 0.874434 — 0.9351 ;
66 — y

therefore, finally by reduction, we have

fk zz y zz: 15.94 inches nearly.

461. Having thus determined the value of fk, the value of ki can
very easily be found; for we have seen above, that ki = l.lQ96y ;
consequently, by substitution, we have

^—1.1096X15.94 = 17.687 inches;

therefore, by the principles of Mensuration, we get

area/A zzzo'zrjx 15.94 X 17.687 X 0.25882=36.47 squ. inches. (285).

Let the values of a and a' as determined in equations (283) and (285),
be respectively substituted in equation (282), and we shall obtain

(286).

but in this equation the values of d and 3 are still unknown ; in
order therefore to assign their values, we must have recourse to other
principles.

462. Now, since the line kw which passes through o, the centre of
gravity of the triangle fki, bisects the side fi in w, we know from
the principles of Geometry, that

from which, by extracting the square root, we get

2 k w =

and dividing by 2, it becomes

kw =

374 OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

But we have already found that/A zz 15. 94 inches, and k zzz 17.687
inches; consequently, their squares are 15. 942rz 254.0836, and
17.6872zz312.83 respectively; therefore, we have

kw — \ <J 1 133.8272 — /z2,

and the value offi2 is found by the following logarithmic operation,
angle fik =60° 57' 49" - - log. cosec. 0.058334
angle fki — 15 00-- log. sin. - 9.412996
side-fk —15.94 inches - log. 1.202488

Sum of the logs. = 0.67381 8

/t* = 22.2657 nat. number - - twice the sum zz 1.347636;
therefore, by substitution and reduction, we obtain

k w = W 1 133.8272 — 22.2657 zz 16.68 inches nearly.

But by the property of the centre of gravity as referred to the plane
triangle, we know that k o zz -f & w ; hence we have
Aozzf-of 16.68 ml 1.12 inches,

and by a well known theorem in the doctrine of Plane Trigono-
metry, we have

from which, by substituting the numerical values, we get
312.83 + 277.89 — 5.5664

2X16.68X17.687
consequently, by multiplication, we get
kt = 11. 12X0.99231 = 11.0345 inches.

463. Returning to the triangle hke, we find that £e = 33 —
15.94 zz 17.06 inches, and kh — 0.9703x17.06 zz 16.55 inches;
therefore, by Plane Trigonometry,

sin.75 57' 49" : sin. 15° : : 16.55 : he,
which being actually reduced, gives

Aezz4.415 inches.

Therefore, by pursuing a train of reasoning, similar to that by
which we discovered the value of k t, we shall obtain

— he9

from which, by substituting the numerical values, we get

291.0436 + 821.7075—19.4923
m= - - — - zz 11.01 inches.

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS. 375

Let this value of km be added to that of kt already determined,
and the aggregate will give the value of mt; therefore, we have

mt = d= 11.0345 + 11.01 =22.0445 inches. (287).

464. With respect to the value of the remaining quantity S, we
have only to observe, that from the nature of the trapezoid, the posi-
tions of the points G and g can easily be ascertained, and the distance
between them, is therefore expressed by the following equation, viz.

g_D .

Now, according to the conditions of the question, we have 6 = 40
inches, ft nr 30 inches, and D zrz 20 inches ; and moreover, by com-
putation, we have found that ^=33 inches, and QHzrG inches;
consequently, by substitution, the above equation gives

* on C 20(40 + 60) 6(66 + 30)7 - .0 - ,
d =i 20 — ) ; ~ - L j — i - IL - L J. z= 7.43 inches very nearly.
I 3(40 + 30) T 3(33 + 30) } (288).

Let therefore the values of d and S, as obtained in the equations
(287 and 288), be substituted in equation (286), and we shall obtain

which being reduced, gives
x = 4.25 — 1 .923 =. 2.327 inches.

This is the length of the equilibrating lever, but the whole weight
of the body in cubic inches of water, is

(40 + 30) 10 X 12 X 14x 0.27 = 41752 cubic inches,

which being reduced to Ibs. gives
41752X62.5-^1728=: 1510.125 Ibs. ;

hence the momentum of stability is
S= 1510.125X2.327 z= 3514.054 Ibs.

Such is the method of calculating the measure of stability, when
the transverse sections are all equal and similar figures ; but when
this happens not to be so, as in the case of ships and other vessels
designed for the purposes of navigation, the difficulty of calculation.
is greatly increased, for the several terms of which the equation is
constituted, must have their values separately determined by intricate
forms of approximation, the nature of which can only be known from
the circumstances which regulate the particular constructions, to which
the investigations are referred.

376 OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

3. PRINCIPLES OF THE STABILITY OF SHIPS.

465. We have seen from the formula (282), that the measure of
stability, when the body is inclined through any angle from the per-
pendicular, is

a'd

in which equation, the symbol x expresses the horizontal distance
between the two vertical lines, one of which passes through the centre
of effort, and the other through the centre of buoyancy.

The same principle has now to be applied in estimating the stability
of ships, and this object will be attained, if either by calculation or
geometrical construction, we find the value of rr, which in the inclined
position of the diagram to Problem LXI. is represented by GS or rz ;
then, if we suppose the whole weight of the ship or floating mass to
be denoted by w, it is manifest, that the momentum of stability will
be expressed by the weight of the vessel drawn into the horizontal
distance between the vertical lines above described ; that is,

771 — WXj

where 772 denotes the momentum of stability, or the effort by which
the vessel endeavours to regain the upright position, from which it is
deflected by the action of the wind, or some other equivalent force
similarly applied.

If, therefore, we put v to denote the whole volume of fluid displaced
by the immersed part of the vessel, and v for the volume which is
depressed below the plane of floatation, in consequence of the vessel
heeling from the upright position through an angle equal to0; then,
the general form of the equation for the momentum of stability becomes

dv >

- aSm'^W;- (282-).

466. Now, in applying this expression to any particular case in
practice, it is understood, that the position of the centre of gravity of
the entire ship, and also the position of the centre of gravity of the
immersed volume when the ship is upright and quiescent, are both
known, and consequently, the distance between those centres, which
is represented by the line Gg or GW, is a given or assignable quan-
tity ; and moreover, the total displacement occasioned by the floating
body, is supposed to have been determined by previous admeasure-
ments, and hence, the weight of a quantity of water, which is equal

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS. 377

in magnitude to the displacement, will likewise be equal to the whole
weight of the vessel.

The quantity 0 is necessarily given from the circumstances of the
case, and may be of any magnitude whatever, and therefore, the
only quantities required to be ascertained, for the purpose of dis-
covering the momentum of the ship's stability, are v and d in the
numerator of the fractional term, the one denoting the magnitude of
the volume which becomes immersed in consequence of the inclina-
tion, and the other, the distance through which the centre of gravity
of that volume is moved in a horizontal direction, during the deflexion
of the ship from the upright and quiescent position. In order, there-
fore, to facilitate the determination of those quantities, the following
observations are necessary.

467. If a straight line be conceived to pass through the centre of
gravity of the ship, in the direction of its length and parallel to the
horizon, traversing from the head to the stern of the vessel ; then, such
a line is called the longer axis of the vessel ; it is the same with the
axis of motion described in the fifth definition preceding, and is so
called, for the purpose of distinguishing it from another line also hori-
zontal, which passes through the centre of gravity at right angles to
the former, and is called the shorter or transverse axis of the vessel ;
it is on this axis that the vessel turns in the process of pitching, a
motion which is easily understood by considering an alternate eleva-
tion and depression of the head and stern.

468. A vertical plane drawn through the longer axis, when the
vessel floats in an upright and quiescent position, divides it into two
parts which are perfectly similar and equal to one another, and in this
respect at least, the figures of vessels may be considered regular,
although that their forms are not otherwise restrained to any uniform
or particular proportions.

From the similarity and equality of these two divisions, it necessa-
rily follows, that when a vessel floats in a state of upright quiescence,
the similar parts on the opposite sides of the plane of division will
be equally elevated above the water's surface. A ship thus floating
in a position of equilibrium, may be conceived to be divided into two
parts by the horizontal plane which is coincident with the water's
surface, and the section formed by this plane passing through the
body of the vessel, is called the principal section of the water ; it
corresponds with the plane of floatation in the particular case where
the vessel is upright and quiescent, as will readily be perceived by a
reference to the ninth definition preceding.

378

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

469. Let A B c K D represent a tranverse section of the hulk of a ship,
perpendicular to the longer axis, and passing through G its centre of
gravity, and suppose the vessel as it floats upon the surface of the
water to be upright and quiescent; then LK the axis of the section,
according to the seventh definition, is perpendicular to the horizon,
and in this state
the principal sec-
tion of the water
passes through the
line D c throughout
the whole length
of the vessel ;
which may
bly be better un-
derstood, if the
principal water
section be viewed
endways, with the eye at a great distance, it will appear as if it were
projected into the straight line DC.

While the vessel retains its upright position and remains in a state
of rest, the transverse or shorter axis, is that which is represented by
the dotted line a b, and the place of the centre of gravity of the
immersed portion DKC, is somewhere in the line passing through g in
a direction parallel to the horizon ; for g is the place of the centre of
gravity of the section DKC, which falls below the principal section of
the water passing through DC.

When the ship is caused to heel or to revolve about the longer axis
passing through G, until it moves through an angle equal to FPC;
then it is manifest, that the principal section of the water, or the plane
in the ship which passes through the line DC, will be transferred into
the position EF ; but the section of the water will intersect the sides
of the vessel, in the direction of a plane passing through DC, which is
inclined to the former plane passing through EF in an angle equal to
the angle FPC. The plane which passes through the line DC in a
direction parallel to the plane of the horizon, may therefore be termed
the secondary section of the water, merely to distinguish it from that
which formerly passed through EF, and which we denominated the
principal section.

The principal and secondary sections of the water must therefore
intersect one another in the line denoted by the point P, or rather in
the line which being viewed endways, is projected into the point P,

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS. 379

and stands at right angles to the plane A KB. Consequently, since
the vessel is supposed to be inclined around the longer axis, it
follows, that the intersection of the planes which we have supposed to
be projected into the point P, will be parallel to the axis round which
the vessel is supposed to revolve in passing from one position to
another.

But by the laws of hydrostatics, since the whole weight of the
vessel is considered to be precisely the same, however much it may
be deflected from the upright and quiescent position ; it follows from
hence, that the volume which becomes immersed below the water's
surface in consequence of the inclination, is equal in magnitude to
that which is elevated above it by the same cause, and consequently,
the position of the line which is represented by the point P, will depend
entirely upon the form of the sides DE and CF.

Now, in a ship, of which the breadth is continually altering from
the head to the stern, and in no regular proportion expressible by
geometrical laws, it is manifest, that the place of the point P, repre-
senting the line in which the water's surface intersects the vessel in
the upright and inclined positions, must be practically determined by
some method of approximation, dependent upon the ordinates in the
vertical and horizontal sections into which the ship is supposed to be
divided.

By similar modes of approximation, the other quantities necessary
for the solution may also be ascertained ; but in ships of war and of
burden, constructed after the forms which they generally assume at
sea, the calculations necessary for the purpose are unavoidably prolix
and troublesome ; and after all, they must depend for their accuracy
entirely upon the skill and address of the persons by whom the requi-
site ordinates are measured and registered, according to the different
parts of the vessel to which they particularly belong ; for if a very
nice and accurate arrangement be not preserved with regard to the
magnitudes and places of the several ordinates, it is easy to be per-
ceived, that the results may come out very wide of the truth, and must
therefore necessarily vitiate the whole process.

In our diagram, the lines DC and EF, through which the principal
and secondary sections of the water pass, are supposed to bisect each
other, and consequently, the point P must occur at the middle of them
both ; in which case its position is known ; but the careful and atten-
tive reader will easily perceive, that this can very seldom happen,
unless the extreme sides of the zone which limits the angle of the ship's
inclination, are equally inclined to, and similarly situated in respect
of the extremities of the intersecting lines DC and EF.

380 OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

When the curves D E and c F are dissimilar between themselves, and
dissimilarly situated in respect of the intersecting lines DC and EF ;
then it is manifest, that the point p cannot fall in the middle of either,
but must occur to the right or to the left, according as it is influenced
by the nature of the curves, which define the exterior contour or bound-
ary of the vessel.

Suppose therefore, that the intersection takes place at the point p,
a little to the right of the place where the two water lines are sup-
posed to bisect one another ; and through the pointy, let the straight
line <?/be drawn parallel to EF, and meeting the sides of the vessel in
the points e and/; then is e/the line through which the secondary
section of the water passes, on the supposition that it intersects the
principal section in the straight line passing through p, parallel to the
longer axis of the vessel.

470. Now, in order to determine the position of the point p, it will
be expedient to conceive the volumes which become immersed under,
and elevated above the fluid's surface, in consequence of the inclination,
and of which fpc and epv are transverse sections, to be divided into
segments by vertical and parallel planes cutting the longer axis of the
vessel at right angles, and at the distance of a few feet from each
other, the distances being regulated by the dimensions of the vessel,
and the nature of the curves by which it is bounded ; they may in
general, however, be from 3 to 5 feet in large vessels, and from 2 to
3 feet in smaller ones ; but in all cases, they must be chosen accord-
ing to circumstances.

i 7P

Each of these segments will be of a wedge-like form, contained
between two vertical and parallel planes fpc, f'p'c'; two inclined
planes cpp'c', fpp'f, making with each other an angle fpc or
f'p'c' equal to the angle of the vessel's inclination, and the portion of
the ship's side which is represented by fee1 'fr.

The horizontal distance between the planes fpc and f'p'c', is the
line pp', which being produced both ways to the head and stern of
the vessel, forms the line in which the two sections of the water cut

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS. 381

each other, and is therefore coincident with the water's surface, and
parallel to the longer axis of the ship.

471. Since the dimensions of the vessel are supposed to be known,
the lines D c and ef will be known ; and from these data, the lines pf
and pc are to be assumed by estimation ; but the angle fpc through
which the ship is deflected from the upright position, is given by the
nature of the particular conditions from which the inclination or de-
flexion arises, and consequently, by the rules of Trigonometry, the
area of the triangle fpc becomes equal to J/p X jpcsin. 0.

If therefore, the area of the small circular space fnc be determined
by any of the methods of approximation, and added to the area of the
triangle fpc, the sum will be the area of the mixed space fpcn, and
by proceeding in a similar manner, the area of f'p'c'm will become
known ; then, if a mean of these two areas be multiplied by the
perpendicular distance pp' , the product will be a near approximation
to the solidity of the wedge contained between the planes fpp'f
and cppc*.

And exactly in the same manner, the solid contents of the opposite
segment which is elevated by the inclination is to be obtained, and if
the aggregate of all the elevated segments be equal to the aggregate
of all the depressed ones ; that is, if the entire volume which becomes
immersed by the inclination, is equal to that which becomes elevated
by the same cause, the point p has been properly determined ; but if
they are not equal, the operation must be repeated until they exactly
agree, and when this agreement has been obtained, the value of v in
equation (282°) becomes known.

472. Now, in order to determine the momentum of stability eli-
cited by the ship under the proposed inclination, it is requisite that the
product dv in the numerator of the fraction should be completely de-
termined; and for this purpose, the area of the space fpcnf, and
the position of its centre of gravity have to be found by approxima-
tion, and also, the area of the space f'p'c.'mfy with the position of
its centre of gravity. Let the points o and t respectively, denote the
positions of those centres, and let the lines or and ts be drawn at
right angles to pc andjo'c'; then are pr and p' s the respective dis-
tances of the points o and t from the horizontal line pp'.

Take the arithmetical mean of the two distances pr and p's9 for
the distance between the horizontal line pp', and the centre of gravity
of the solid wedge or segment fpcf'p'c'. Find similar distances for
all the segments between the head and stern of the vessel, for those
which are elevated by the inclination, as well as those which are

382 OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

depressed by it ; then, if the solidity of each segment is multiplied
into the distance of its centre of gravity from the horizontal line
passing along pp', and produced both ways to the head and stern of
the vessel ; the aggregate of the products thus arising, will constitute
the value of the numerator dv of the fractional term in equation
(282a), where the momentum of the vessel's stability is

<dv 7

,=:< o sm.d>>?tf.

L \ ry

Consequently, since the several quantities w, v, S and 0, are either
given d priori, or determinable from the circumstances of the case, it
follows, that the momentum of stability for any angle of inclination,
and for any form of body, can be found by the above formula ; but
the labour and intricacy of the calculation, increases with the irregu-
larity of the body to which such calculations are referred, and in
particular cases, the labour required to accomplish the purpose is
immensely great.

PROBLEM LXII.

473. The vertical transverse sections of a ship, taken at the
distance of five feet from each other along the principal longi-
tudinal axis, are thirty-four in number, and are bounded by
curves approaching to a parabola of a very high order ; cor-
responding to these are twelve horizontal sections between the
keel and the plane of floatation, taken at intervals of two feet
on the vertical axis, the first section occurring at the distance
of nine inches from the upper surface of the keel : —

It is required to determine the measure of stability, when
by the action of the wind, or some other equivalent external
force, the vessel is deflected from the upright position through
an angle of thirty degrees; the ordinates corresponding to
the several sections, being as registered in the following
table.

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

383

TABLE SHOWING THE ORDINATES CORRESPONDING TO THE
SEVERAL SECTIONS.

Horizontal sections, intervals on the vertical axis 2 feet.

1
b

h
ft

1
1

No.

ft.

4
ft.

ft.

7
ft.

8
ft.

9
ft.

10
ft.

11
ft.

12
ft.

2
3
4
5

7
8
9
10

2.55
3.09

3.50
6.85
9.78

1.80
6.35
10.45
13.50

4.09
9.00
13.20
15.09

6.20
11.30
15.30
17.55

1.70
8.20
13.50
16.85
18.64

3.30
10.00
15.10
17.84
19.30

20.12
20.65
20.90
21.14

21.28

21.51
21.51
21.51
21.35
21.35

4.90
11.73
16.24
18.55
19.73

6.60
13.25
17.08
19.10
20.08

8.10
14.40
17.70
19.45
20.30

9.60
15.35
18.10
19.70
20.50

20.92
21.21
21.44
21.55
21.60

21.60
21.60
21.60
21.60
21.60

10.78
16.00
18.40
19.85
20.54

5.50
6.75
8.00
9.10
9.80

12.15
13.60
14.55
15.20
15.50

15.60
16.80
17.45
17.88
18.15

18.30
18.30
1830
18.30
18.30

17.70
18.65
19.05
19.35
19.53

19.60
19.75
19.75
19.75
19.75

18.91
19.60
20.05
20.25
20.40

20.50
20.60
20.60
20.60
20.60

19.65
20.32
20.62
20.78
20.93

20.93
21.10
21.10
21.10
21.10

21.05
21.00
20.90
20.81
20.67

20.45
20.90
21.20
21.34
21.47

20.65
21.05
21.34
21.45
21.50

20.80
21.15
21.38
21.52
21.60

20.94
21.20
21.38
21.48
21.50

21.56
21.58
21.58
21.56
21.56

21.55
21.53
21.51
21.48
21.32

11
12
13
14
15

10.50
10.50
10.50
10.50
10.50

15.90
15.90
15.90
15.90
15.90

21.51
21.51
21.51
21.51
21.51

21.59
21.59
21.59
21.59
21.59

21.60
21.60
21.60
21.60
21.60

16
17

18
19
20

10.30
9.80
9.20
8.50
8.00

15.70
15.50
15.35
15.00
14.60

18.20
18.05
17.95
17.75
17.52

19.65
19.55
19.45
19.30
19.15

20.52
20.45
20.35
20.25
20.10

21.32
21.30
21.20
21.10
21.00

21.51
21.50
21.40
21.30
21.20

21.59
21.55
21.52
21.44
21.30

21.60
21.60
21.55
21.50
21.35

21.25
21.05
20.83
20.61
20.40

21.60
21.60
21.52
21.50
21.35

22
23
24
25

7.20
6.40
5.90
5.10
4.20

14.20
14.62
12.90
11.90
10.60

17.25
16.90
16.40
15.70
14.80

18.90
18.62
18.28
17.75
17.10

19.90
19.70
19.30
19.00
18.46

•20.50
20.32
20.05
19.75
19.30

21.85
20.68
20.40
20.15
19.85

21.05
20.89
20.65
20.40
20.10

21.18
21.00
20.75
20.55
20.30

21.24
21.08
20.82
20.61
20.43

21.22
21.05
20.82
20.61
20.44

26
27
28
29
30

3.35
2.50
1.80
1.40
1.11

9.20
7.20
5.38
3-55
2.40

13.40
11.60
9.35
6.65
4.25

16.15
14.80
12.85
10.10
7.05

17.88
16.60
15.45
13.10
10.10

6.12
2.90
1.32
0.75

18.80
18.05
17.11
15.35
12.90

19.35
18.80
18.10
16.85
15.05

19.73
19.30
18.80
17.90
16.80

20.00
19.59
19.22
18.55
17.82

20.10
19.75
19.44
19.00
18.40

20.12
19.80
19.52
19.15
18.62

20.15
19.85
19.58
19.25
18.77

31
32
33
34

0.90
0.80
0.62
0.60

1.45
0.98
0.75
0.63

2.30
1.25
0.80
0.65

3.75
1.90
1.00
0.70

9.10
4.55
1.85
0.90

12.00
7.10
2.70
1.05

14.60
10.35
4.31
1.35

16.30
13.40
7.45
1.95

17.40
15.65
14.50
3.40

17.90
16.90
14.90
7.00

18.21
17.52
1650
12.95

From these data, combined with others that are either assumed or
determined by the circumstances of the case, the stability of the
vessel or the momentum of the redressing force is to be found by
calculation ; it will, however, be an improvement on the mode of

384 OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

procedure, if in the first place, we take a brief survey of the principles
of construction; for this purpose, let act represent any transverse

section of the vessel, at right angles to the principal longitudinal axis
passing through the centre of effort ; then is ef the breadth of this
section at the water line when the .ship is loaded and the plane of the
masts vertical, and hi becomes the water line, coincident with the
surface of the fluid, when the vessel is deflected from the upright
position through the given angle fki.

It is however manifest from the ordinates in the foregoing table,
that in this case, the vertical sections are all different, both in form
and in magnitude, and consequently, the primary and secondary
water lines do not intersect one another in the point k which bisects
ef\ let p be the point of intersection, and through the point/?, draw
the straight line mn parallel to hi, and making with e/the angle fpn
equal to the given angle of inclination.

Now, by considering the conditions of the problem, it will readily
appear, that the position of the point p in any of the sections
parallel to acz, cannot be determined on the same principles by
which the place of that point was fixed according to the foregoing
solutions, viz. by equating the areas of the triangular spaces mpe and
fpn; for it is evident, that the volume which becomes immersed
below the fluid's surface in consequence of the inclination, and that
which emerges above by the same cause, will not now be proportional
to those areas, in the same manner as they are, on the supposition of
all the vertical sections being equal and similar figures.

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS. 385

In the present instance, the vertical sections being all different,
both in form and magnitude, the water's surface intersecting the
vessel in the plane passing through the line mn when the vessel is
inclined, will so divide the areas of the several sections, that although
the space fpn may not be equal to mpe in any one of them, yet the
immersed volume corresponding to all the spaces fpn, estimated
from the head to the stern of the ship, shall be equal to the volume
corresponding to all the emerged spaces mpe estimated in the same
manner.

Let ef, the breadth of the section at the water line, be bisected in
the point k by the vertical line dc, and suppose a plane to pass
through dc from head to stern of the vessel, such a plane will divide
the vessel into two parts that are equal and symmetrical, and it will
pass through the point k in all the parallel vertical sections made
throughout the whole length.

But it is easily shown, that at whatever distance kp from the
middle point k, the plane of floatation in the inclined position, inter-
sects the primary line ef in one of the vertical sections, it will
intersect the corresponding line in all the other sections at the same
distance from the middle point; that is, the distance kp will be the
same in all the parallel sections, (the same lines and letters of
reference being understood to belong to each ;) for according to the
conditions of the problem, the revolution of the vessel is supposed to
be made about the principal longitudinal axis, and consequently, the
intersection of the two planes passing through the lines ef and mn,
must be parallel to the axis of motion, and therefore parallel to the
line drawn through the point k in all the sections, estimated from
head to stern of the ship.

We have now to determine the distance kp at which the inter-
section takes place ; and for this purpose we must consider, that
according to the given conditions, whatever may be the position of
the point p in all the sections, if lines mn are drawn through those
points, making with ef, an angle equal to the given angle of inclina-
tion; then it is manifest, that the same plane will pass through all
the lines mn that occur betwixt the head and stern of the vessel.

It is therefore required to determine, at what distance kp from the
middle points k, the plane of floatation corresponding to the inclined
position of the vessel must pass, so as to cut off a volume on the
depressed side fpn equal to that which rises above the water on the
side mpe.

In each of the parallel vertical sections, let the common line hi
VOL. i. 2 c

386 OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

be drawn through k, the middle point of ef, and inclined to ef at an
angle equal to that of the vessel's deflexion ; then, from what we have
stated above, all the lines hi will lie in the same plane; that is, the
same plane will pass through the line hi in all the sections. If there-
fore, the areas of the spaces fk i and h k e in each of the vertical
sections, be determined by some mode of mensuration adapted to the
particular case, it is easy from these equidistant areas, to ascertain the
solidity of the volumes contained between the planes passing through
the lines kf, ki and he, kh.

Put m — the magnitude or solid contents of the volume, bounded

by the side of the vessel and the planes passing through

A/and ki,
m'— the magnitude or solid contents of the volume, bounded

by the side of the vessel and the planes passing through

ke and k A,
A — the area or superficial contents of the plane passing

through the line hki in all the sections, estimated from

head to stern of the vessel, which area is determined

by having given all the lines hi;
(j) ~fki, the angle through which the vessel is inclined from

the upright and quiescent position, and
e nr m — m', the difference of the volumes or solidities,

denoted by the symbols m and m'.

Then, if upon the line kf, which coincides with the line of floata-
tion when the vessel is upright and quiescent, there be set off in each
of the parallel sections, the line

kp— f_

AXsin.<^'

and through all the points p thus found, let lines mpn be drawn
parallel to hi, and consequently, cutting ef in the points p, at an
angle equal to that of the vessel's inclination; then, if a plane be
drawn through all the lines mpn, it will so divide the vessel, that the
solidity of the volume contained between the planes passing through
the lines fp and np, will approximate to an equality with the volume
contained between the planes passing through the lines ep and mp.

Therefore, since the surface of the water coincides with the plane
passing through ef when the vessel is upright, it will also coincide
with the plane passing through all the lines mpn, when the vessel is
deflected through the angle fpn, whose magnitude is given.

It is very easy to show, that by setting off the distance kp in all the

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS. 387

sections, as determined by the preceding equation, and thereby draw-
ing a plane through all the lines mpn, the plane thus drawn is
coincident with the water's surface, and is situated very nearly in
its true position. For through the point /<:, draw the line kr meet-
ing mn perpendicularly in the point r ; then is krp a right angled
triangle in which the angle kpr is given, and by the principles of
mensuration, it is manifest, that the solid contained between the
planes passing through the lines hki and mpn from head to stern of
the vessel, is very nearly equal to the area of the plane passing
through hkij drawn into kr the perpendicular thickness of the solid.
Now, the solid of which hinm is a section, is obviously equal to
the difference of the solids of which fki and hke are sections; hence
we have

m — m —e — AXkr nearly ;
consequently, by division, we obtain

and by the principles of Plane Trigonometry, we get

kr : kp : : sin. kpr : rad.,
or by restoring the analytical values, it is

e
— : kp : : sm.^> : rad. ;

and from this, by reducing the proportion and putting radius equal to
unity, we obtain

*„ — _ e_

-AXsin.0' (289).

An equation which is very nearly true for small inclinations, and
this being the case, it fully establishes the propriety of the above
construction ; if the areas of the planes passing through the lines hki
and mpn are equal to one another, the construction as thus effected
would be rigorously correct.

474. In pursuing the construction, it will be necessary, in order to
avoid confusion in the lines and letters of reference, to redraw that part
of the section which includes the angle of the vessel's inclination, viz.
the space contained between the sides of the vessel me, nf and the
dotted lines en, inf. We shall not, however, attempt to preserve the
due proportion between the several parts of the figure ; this indeed
would be troublesome and altogether unnecessary, since it is the prin-
ciples of construction only that we mean to illustrate, and not the
actual solution of any particular example.

2c2

388

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

Let enfm be the space in question, including the angle of the
vessel's inclina-
tion ; draw the
lines me and nf»
cutting off the
curvilinear areas
mre and nsf;
bisect the sides
me, pe in the
points u and TT,
and draw the
lines pu and mir

intersecting each other in l\ /is the centre of gravity of the triangular
space mpe. Suppose z to be the centre of gravity of the curvilinear
segment mre, and through the points I and z, draw lq and zy respec-
tively perpendicular to mn, the line of floatation in the inclined posi-
tion of the vessel.

Again, bisect the sides w/and/p in the points w, $, and draw pw
and n(f> intersecting each other in the point o\ o is the centre of
gravity of the triangular space npf. Let v be the centre of gravity
of the curvilinear area n sf, and through the points o and v, draw the
straight lines ot and vx respectively perpendicular to the water line
mpn; then, in the line tx intercepted by the perpendiculars ot and
vx, take tc such, that it shall be to £0: in the same proportion, as the
curvilinear space nsf, is to the compound space pnsf, and by the
property of the centre of gravity, c will be the point in mn, where it
is intersected by the perpendicular through the common centre of the
triangular and curvilinear spaces npf and n sf.

Through the point p in all the sections, let a line PQ be drawn at
right angles to mn ; then, the same plane will pass through all these
lines, and cp will be the perpendicular distance of this plane, from
the centre of gravity of the mixed space pnsf. Therefore, if the
products arising from multiplying each area, into the distance pc of
its centre of gravity from the plane passing through PQ, be truly cal-
culated in all the sections contained between the head and stern of
the vessel ; then, by the principle announced and demonstrated in
Proposition (A), Chapter I, the distance of the centre of gravity of
the volume, whose sections are represented by all the areas pnsf,
from the vertical plane passing through PQ can easily be ascertained.

Let pR be that distance, and by a similar mode of computation,
suppose that pE is found to be the corresponding distance of the

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS. 389

centre of gravity of the volume whose sections are the areas pmre ;
then is B R the horizontal distance between the centres of gravity of
the volumes that are respectively immersed and emerged, below and
above the water's surface, in consequence of the vessel being deflected
from the upright position, through an angle of which the magnitude
is known.

475. The solid content of the entire volume immersed, or the quan-
tity of water displaced by the immersed part of the vessel, is to be
obtained from the areas of the several horizontal sections ; for the
ordinates drawn in the several sections being arranged in regular order,
after the manner which we have adopted in the preceding table, the area
of any section can readily be assigned, by methods of approximation
adapted to the particular case, and from these areas the solidity of
the immersed volume is to be inferred ; making allowance for the
irregularities of the vessel towards the head and stern, if it be at all
necessary to take those parts into the account ; in all practical cases,
however, they may safely be omitted.

That part of the immersed volume, comprehended between the keel
and the lowest horizontal section, is obtained, by first finding the areas
of the several vertical planes, between the keel and the nearest ordi-
nates, and from these areas, by means of some appropriate mode of
approximation, the magnitude of the part cut off by the lowermost
horizontal plane will be determined ; which being added to the solidity
of the part contained between the extreme planes, will give the mag-
nitude of the immersed volume, or the quantity of fluid displaced.

476. Referring to the original diagram, it will be observed, that from
the areas of the several horizontal sections, made between the keel of
the vessel and the plane of floatation, the distance kg, that is, the
distance between the water line ef and the centre of buoyancy, or
the centre of gravity of the immersed volume, can also be determined
by the application of particular approximating rules, and the best
with which we are acquainted for this purpose, are those given by
Stirling in his " Methodus Differ enlialis" and by Simpson in his
" Essays ;" these rules may be expressed in general terms as
follow.

RULE 1. Jyx=i(?—\S}Xr,

where x is the fluxion of the abscissa, y the perpendicular ordinate,
expressing a general term or function of x ; r the common distance
between the ordinates ; S the sum of the first and last ordinates, and
p the sum of the whole series.

390 OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

RULE 2. y#* = (S-f4p-|-2Q)Xir,

where x, y, r and 5 denote as in rule 1st; p the sum of the 2nd,
4th, 6th, 8th, &c. ordinates, and Q the sum of the 3rd, 5th, 7th, 9th,
&c. (the last of the series excepted).

RULES. /X=(S + 2p + 3g)X|r,

here again, x, y, r and S denote as in the preceding cases ; P the sum
of the 4th, 7th, 10th, 13th, &c. ordinates (the last excepted), and Q
the sum of the 2nd, 3rd, 5th, 6th, 8th, 9th, &c.

With respect to the applicability of the above rules, it may be
observed, that the first approximates to the fluent, whatever may be
the number of the given ordinates, and the second only requires that
the number of ordinates shall be odd. But in order to apply the
third rule, it is a necessary condition, that the number of given
ordinates shall be some number in the series 4, 7, 10, 13, 16, &c. ;
that is, the number of ordinates must be some multiple of 3 increased
by unity. In every case, however, the approximate fluent can be
obtained, either from the second or third rule considered separately,
or from both taken conjointly.

477. But to return from this short digression, we may remark, that
the position of the point G, which marks the centre of effort, or the
centre of gravity of the whole vessel, depends partly on the equipment
and construction, and partly upon the distribution of the loading and
ballast; which circumstances, therefore, determine G#, the distance
between the centre of effort and the centre of buoyancy when the
vessel is upright.

These several conditions having been determined, the remaining
part of the construction, limiting the measure of the vessel's stability,
may be effected as follows.

Through g the centre of buoyancy, or the centre of gravity of the
immersed volume, draw gt parallel to mn, and make gt to BR (see
the subsidiary figure), as the volume immersed in consequence of the
inclination, is to the whole immersed volume induced by the weight
of the vessel ; through G the centre of effort, draw GZ parallel and 05
perpendicular to gt, and through t draw tm parallel to GS, and meet-
ing the axis cd in M ; then is M the metacentre, and GZ the measure
of the vessel's stability when inclined from the upright position through
the angle fpn or gGs.

The principles of the preceding construction are general, and can
be applied in all cases, whatever may be the figure of the vessel, or

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS. 391

the nature of its bounding curves ; but the arithmetical operations, as
applied to any particular case, are unavoidably tedious, and necessa-
rily extend to considerable length ; they are, however, very far from
being difficult, as the ensuing process will fully testify.

478. By referring to the table of ordinates, it will appear, that the
greatest, or principal transverse section, intersects the longer axis, at
about the distance of 60 feet, or 12 intervals from the section nearest
to the head of the ship ; we shall therefore delineate that section, and
in order that nothing may be wanting to the proper understanding of
the subject, we shall also delineate the plane of floatation, which cor-
responds to the twelfth horizontal section in the preceding table of
ordinates.

The ordinate in the table opposite the twelfth vertical and under
the twelfth horizontal section, is 21.58 feet, and the whole vertical
distance between the keel and the plane of floatation, is 22.75 feet;
therefore, draw
the horizontal
line a a which
make equal to
43.16 feet, and
bisect a a per-
pendicularly by
AK equal to
22.75 feet.

Divide the
vertical axis A K

into twelve parts, eleven of which are 2 feet each, and the first or
lowermost only three fourths of a foot, or nine inches ; then, through
the several points of division 1, 2, 3, 4, 5, 6, &c. and parallel to aa,
draw the several ordinates, taken from the twelfth horizontal row in
the preceding table, which set off both ways, and through the extre-
mities of the several ordinates, let the curve line ana be drawn, which
will represent the boundary of the principal lateral section, so far as
it is immersed below the fluid's surface.

And exactly in the same manner may the whole of the 34 vertical
sections, into which the longer axis is divided, be delineated ; but the
above being sufficient for illustration, we shall next proceed to describe
the twelfth horizontal section which is coincident with the water's
surface, and of which the greatest ordinate is 21.58 feet, correspond-
ing to a A in the above vertical section.

479. Since the body of the vessel is divided at intervals of 5 feet into
34 vertical sections, it follows, that between the first section adjacent

392

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

to the head, and the 34th section adjacent to the stern, there must be
33 intervals, or 33x5 = 165 feet; therefore, draw the horizontal line
mn to represent the longer
axis of the plane of floata-
tion, and make be equal
to 165 feet, which divide
into 33 equal intervals of
5 feet each; then at right
angles to mn and through
the several points of divi-
sion 1,2,3,4, 5, &c. to 34,
draw the ordinates dd, ee,
//, a a, gg, kh, &c. to iit
and from a scale of equal
parts, of the same dimen-
sions as that from which
be is taken, set off both
ways, (beginning at the
1st division adjacent to the
head of the vessel), the
numbers contained in the
twelfth column of the pre-
ceding table; then, through
the extremities of the seve-
ral ordinates, let the curve
man a be drawn, and it
will represent the plane of
floatation when the vessel
is upright, according to the
foregoing tabulated mea-
surements. It is manifest,
that by a similar mode of
procedure, the eleven remaining horizontal sections, situated between
the keel and the plane of floatation, might also be delineated; it is,
however, unnecessary to pursue the subject of construction farther,
since what has already been done, is quite sufficient to show the
reader, the method and nature of the delineation when pursued
throughout the entire vessel.*

* It may be proper to remark, that the scales from which the vertical and
horizontal sections have been constructed, are to one another as 2 to 1 ; the one for
the vertical section being l-20th of an inch to a foot, and the other l-10th.

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS. 393

480. It now only remains to calculate the measure of stability, when
the vessel is deflected from the upright position through an angle of 30
degrees, and for this purpose we must again refer to the original
diagram, where, on the supposition that it is correctly constructed,
the lines ki and kh are to be carefully measured in each of the sec-
tions, on the same scale with the original dimensions; then, if the
lines fi and he be drawn, the areas of the triangular spaces fki and
like, can easily be determined from the two sides and the included
angle ; and if a series of perpendicular ordinates be measured on the
lines fn and he, the areas of the curvilinear spaces/#i and hye may
from thence be found, which being added to the triangles fki and
hke, the sums will be the whole areas of the compound spaces fki x
and hkey.

Pursuing a similar process throughout the 34 vertical sections, we
shall at last arrive at the magnitudes of the volumes which are con-
tained between the planes passing through kf, ki and kh, he, a
knowledge of these volumes being necessary to determine the posi-
tion of the point p.

It is presumed that it will be sufficient for the exemplification of the
rules, to exhibit the calculation of one of the spaces fk ix ; to perform
the operation for the whole series, would be a very tedious and at the
same time a superfluous proceeding; and for this reason, that the
constructions and calculations founded on them, for inferring the
results in any one of the sections, are similar to those required for
obtaining the corresponding results in any other section ; and this
being the case, the representation of one process will suffice for all
the rest.

481. But to proceed, the line ki being taken in the compasses, and
applied to an accurate scale of the proper dimensions, it is found to
indicate 22.6 feet, and according to the table of ordinates, the line fk
is 21.58 feet; and moreover, according to the conditions of the
problem, the angle of inclination, or that contained between the lines
kf and ki, is equal to 30 degrees, of which the natural sine is j, radius
being unity; consequently, if a denote the area of the triangle fki,
we have by the principles of mensuration,

a=|(22.58x21.6)=: 121.927 square feet.

Now, according to the principles of Plane Trigonometry, the line/i
is expressed by the equation

; — V/22.6H-21.582— 2x22.6X21.58xcos.30°zz: 11.55 feet.

394 OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

consequently, if this line be divided into six equal parts of 1.925 feet
each, and perpendicular ordinates be erected thereon, they will be
found to measure as follows, viz.

No. of ordinates
Feet - - -

0.15

3
0.30

4
0.43

5
0.38

6
0.23

Therefore, if a' denote the area of the curvilinear space fxi ; then
by the second of the preceding approximating rules, we have

•S— 0 4- 0 — 0, the sum of the extreme ordinates,

4p = 4(.15 4- .43 4- .23) = 3.24, the second term of the series,

2Q = 2(.30 4- .38) — 1.36, the third and last term ;

hence, by addition, we get
S _|_ 4? 4- 2Q = 0 4- 3.24 + 1 .36 = 4.6 ;

but one third of the common interval is 0.642 of a foot nearly ; con-
quently, by multiplication, the area of the curvilinear space fxi,
becomes

a1 = .642 X 4.6 = 2.9532 square feet,

which being added to the area of the triangle above determined, the
area of the compound space fk i x becomes

a 4- a' = 121.927 4- 2.953= 124.88 square feet.

Again, if we put b to denote the area of the triangle hke, and b'
the area of the curvilinear space hye; then, by proceeding in a
manner similar to the above, the area of the mixed space hkey
becomes

b 4- b' = 1 33.68 square feet.

Now, if in this way, the values of a 4- a' and b 4- b' be calculated
for each of the 34 vertical sections, the several results will be as exhi-
bited in the following table.

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

395

Table of Areas for the thirty -four Vertical Sections.

No. of

Values of

No. of

Values of

sections.

a -\- a'

& + &'

sections.

a + a'

6 + 6'

feet.

42.86

23.61

124.62

132.53

81.53

58.92

123.91

131.05

100.80

86.80

123.21

129.57

114.16

105.27

121.06

127.48

121.56

115.70

118.91

125.40

121.75

120.90

117.50

122.66

123.47

125.36

116.10

119.93

125.20

129.82

114.01

116.88

124.87

131.04

111.91

113.83

124.54

132.27

108.96

109.81

124.69

132.97

106.01

105.80

12*

124.88

133.68

101.82

98.92

124.87

133.68

97.24

91.71

124.87

133.68

92.41

79.95

124.82

133.42

86.31

66.06

124.78

133.17

81.60

48.20

124.20

132.85

68.35

17.92

Sums

1953.85

1963.14

Sums

1813.93

1737.70

Therefore, the sum of all the (a -f- a')"zz 3767.78 ; and the sum of
all the (b -f i'yzz 3700.84, and by the conditions of the problem, the
vertical sections intersect the principal longitudinal axis at intervals
of 5 feet ; therefore, by applying the third of the preceding approxi-
mative rules, the solid contents of the volume contained between the
planes passing through fk and ki, will be found as follows.

S= 42.86 + 68.35 zz 1 1 1.21, the sum of the extreme ordinates,

2pzz(611.82 -f 555.25) X 2= 2334. 14, the 2nd term of the series,
or twice the sum of the 4th, 7th, 10th, 13th, &c. ordinates,

3Qzz(1299.17 -\- 1 190. 33) X 3 zz 7468. 5, the third term of the series,
or the sum of the 2nd, 3rd, 5th, 6th, 8th, 9th, &c. ordinates ;

consequently, by addition, we have
S -f 2r -f SQ zz 1 1 1.21 -f 2334. 14 -f 7468.5 zz 9913.85 ;

* This is the vertical section for which we have exhibited the process of com-
putation.

396 OF THE STABILITY OF FLOATING BODIES AND OF SHIPS.

and finally, by multiplying by three eighths of the common interval,
the magnitude of the volume becomes

m = (S4-2p-f3Q)X5Xf z= 9913.85 x V = 18588.47 cubic feet

very nearly.

Proceeding exactly in the same manner with the areas (b -\- £>'), the
solidity of the space comprehended between the planes passing through
the lines kh and ke, and the intercepted side of the vessel, becomes

m'=. (5-f • 2p 4- 3Q)X y = 18433.47 cubic feet;

therefore, by subtraction, we obtain
m — m'= e = 18588.47 — 1 8433.47 = 155.

482. In the next place, we have to determine the area of the plane
passing through all the lines hki in the several vertical sections ; this
is effected by measuring all the ordinates in that plane, taken at the
common interval of 5 feet along the axis passing through k from head
to stern of the vessel.

When this operation is performed in a dexterous manner, the area
of the plane will be found to be 7106 square feet very nearly ; that is,

A = 7106 square feet;
consequently, by equation (412), we have

Hence it appears, that the distance of the pointy from the middle
point k, is too small to cause any material error in the result, we shall
therefore suppose that the plane of floatation corresponding to both
positions of the vessel, intersect each other in the axis passing through
k from head to stern of the vessel. Taking, therefore, the mean
between the two foregoing solidities, we shall have

$(ro 4- m') — |(18588.47 -f 18433.47) = 18510.97 cubic feet.

This, therefore, is the solidity of the volume which becomes immersed
in consequence of the inclination ; and by pursuing a similar mode of
procedure with respect to the areas of the twelve horizontal sections,
the solidity of the whole volume immersed, will be found to be 119384
cubic feet very nearly ; and moreover, by referring to the subsidiary
figure employed in the construction, and introducing the principles by
which the distance BR is ascertained, we shall have
BRzr27.32 feet;

consequently, by Proposition XII., Chapter XIII., the distance gt in
the original figure is thus found,

OF THE STABILITY OF FLOATING BODIES AND OF SHIPS. 397

119384 : 27.32 : : 18510.97 : 4.25 feet,

483. But in order to infer the stability of the vessel from the value
of gt just discovered, it is necessary to have given the distance eg, or
the distance between the centre of effort and the centre of buoyancy ;
now it is obvious, that the position of this latter point is regulated
entirely by the form and dimensions of the immersed portion of the
vessel, and consequently, it may be considered as absolutely fixed
with respect to the plane of floatation ; but since the position of the
centre of effort is regulated partly by the construction and equipment,
and partly by the distribution of the loading and ballast, it can only
be assumed on the ground of supposition, unless in cases where the
position of that point has been actually ascertained by accurate
mensuration.

In several instances, the distance Gg has been measured, and found
to be equal to about one eighth of the greatest breadth at the plane
of floatation ; therefore, by assuming this to be the case generally, we
have

eg == J- of 43.16 = 5.396 feet,

therefore, by Plane Trigonometry, it is

rad. : 5.396 : : sin. 30° : gs,
from which, by reducing the proportion, we obtain

#5 = 2.698 feet,
which being subtracted from g t, the remainder is

S£=:GZ = 4.25 — 2.698=:1.552 feet, the measure of
the vessel's stability, or the length of the equilibrating lever. But the
whole weight of the vessel, as found from the solidity of the immersed
part, is

119384

w nr — 341 1 tons very nearly ; 35 cubic feet

of sea water being equal to one ton weight ; consequently, the
momentum of the redressing force, or the power which the pressure
of the water exerts to restore the vessel to the upright position, is
equal to 3411 tons acting on a lever whose length is 1.552 feet; or
it is equivalent to a force or pressure of 245 tons acting at the dis-
tance of half the greatest breadth of the vessel from the axis ; for by
the principles of the lever, it is

21.58 : 3411 : : 1.552 : 245 tons nearly.

The preceding is the method of determining the stability of a ship,
on the supposition that the data are all assignable ; the process con-

398 OF THE STABILITY OF STEAM SHIPS.

sidered in its full extent is unavoidably tedious and prolix, we have
merely pointed out the method of conducting the calculations ; but
when it is necessary to determine the stability of a ship in actual
practice, every individual process must be separately performed, and
the result obtained as above, will approximate very nearly to the
truth.

484. Those who have ever witnessed the spectacle of a ship tossed in
a tempest, or have read any of the brilliant accounts which maritime
tales afford, will appreciate the subject we have just investigated.
They may have seen, moreover, the vast bulwark slide from her cradle
into the calm water, on which she first swung round and heeled till
she regained her stability of equilibrium ; giving the imagination a
contrast of the stormy element on which she was soon to ride in awful
grandeur. But seamen will best appreciate our labours, especially
those who in the days of battle and the nights of danger have had to
manage the noblest work of art and skill ; and who in their country's
cause have encountered all weathers and every clime, traversing the
wide expanse of ocean's bosom, visiting all the ends of the earth, and
identifying themselves as part of the stupendous ship which figuratively
has to do and to suffer for her country, and which in peace or in war, in
sunshine or in storm, carries with her the benediction of mankind
pronounced as on a living being, when she was first launched in pre-
sence often thousand enthusiastic spectators, one and all sympathizing
in the national solemnity.

4. PRINCIPLES OF THE STABILITY OF STEAM SHIPS.

485. When a ship is set afloat upon the surface of the waters, and
impelled by some power acting in the direction of its length, as is the
case with steam vessels, now so extensively employed, the subject of
stability becomes of very great importance. This remark does not
strictly apply to vesels navigating still waters, or rivers where the
tides produce but small effects ; but it is well known that the natural
motion of the sea, even in its calmest state, causes a considerable
lateral motion in a vessel placed upon its surface, and in consequence
of this motion, the paddles are made to dip unequally in the water,
by which means some part of the impelling power is lost.

It is with the view of avoiding this waste of power, that the subject
of stability acquires such vast importance when referred to steam
vessels ; and it is easy to perceive, that the best method of attaining
this object, is to adapt the form and capacity of the vessel to the

OF THE STABILITY OF STEAM SHIPS. 399

several circumstances by which the floatation is regulated, and on
which the mode of action depends.

The late Thomas Tredgold has considered this subject in his work
on the STEAM ENGINE, and his views in this case, as in all others
where the powers of his comprehensive and refined mind have been
called into action, are concise, elegant, and original ; and we cannot
close this chapter to greater advantage than by adopting his theory,
which however we shall modify to suit the plan and arrangement of
the present work.

486. In order to simplify the investigation of stability, Tredgold
considers the vessel to be a solid homogeneous body of the same
density as water, with vertical or circular surfaces at the water lines
when the vessel is in a state of quiescence. Now, it is obvious, that
with regard to a ship which is designed to carry burdens at sea, the
first of these conditions cannot obtain ; this however is of no conse-
quence as respects the result of the inquiry, for in reality it refers to
a mass of water equal in bulk to the immersed portion of the floating
body. As another means of simplification, he supposes the transverse
sections of the ship at right angles to the axis of motion to be in the
form of a parabola, of which the equation is px=iyn, and for the
purpose of contrasting the extremes of form, he branches the subject
into the two following varieties, viz.

1 . When the ordinates are parallel to the depth, and

2. When the ordinates are parallel to the breadth.

For each of these cases a general equation is deduced, involving
the sine of the angle of inclination, the breadth and depth of the
vessel, and the index or exponent by which the order of the parabola
is expressed.

487. Having already investigated an expression by which the
stability of a floating body is indicated, we do not consider it neces-
sary to trace the steps of inquiry in the present instance, for the
intelligent reader will at once perceive, that although the form of the
equation is somewhat different, by reason of its involving different
parts and different data, yet the principles upon which the investiga-
tion proceeds, are, and necessarily must be, the same, or nearly the
same as before.

488. When the ordinates of the parabola are parallel to the depth,
the general equation by which the stability is indicated, becomes

(290).

400 OF THE STABILITY OF STEAM SHIPS.

where £> — DC is the breadth of the vessel at the water line when
upright and quiescent, C?=LK the corresponding depth, 0 — FPC
the angle of inclination from the upright position, n the exponent
denoting the order of the parabolic section, and S the stability.

489. If we examine the structure of the above equation, it will

readily appear, that while b* is greater than - ' — , the stability is

n — |— A>

positive, and the vessel endeavours to regain the upright position ; if
these two quantities are equal to one another there is no stability ;
and if the latter exceeds the former, the stability is negative, and the
vessel oversets. Hence it appears, that between the breadth and
depth of the vessel, a certain relation must obtain to render it fit and
sufficiently stable for the purposes of navigation; and it is further
manifest, that the stability increases directly as the exponent of the
ordinate, so does the area of the transverse section ; but in order to
give the proper degree of stability, the breadth must increase more
rapidly than the depth.

By giving different values to the symbol n in the preceding
general equation, we shall obtain expressions to indicate the stability
for sections of different forms ; thus for instance, if n zz 1 the section
is a triangle, and the expression for the stability becomes

(291).

490. This equation is very simple, and can easily be illustrated by an
example ; the practical rule for its reduction may be expressed in the
following terms.

RULE. From the square of the breadth of the water line
when the vessel is upright, subtract twice the square of the
corresponding depth ; multiply the remainder by the breadth
drawn into the natural sine of the angle of inclination, and
one twelfth of the product will express the stability.

491. EXAMPLE. A floating body in the form of a triangular prism,
has its breadth at the water line equal to 28 feet, the corresponding
depth under the water equal to 19| feet, and its density equal to the
density of water; now, suppose the body to be in a state of equili-
brium when the axis is vertical ; what will be its stability, or what is
the relative value of the force by which it would endeavour to regain
the upright position, on the supposition that it has been deflected
from it through an angle of 15 degrees ?

This is obviously a case that is not likely to occur in the practice

OF THE STABILITY OF STEAM SHIPS.

401

of steam navigation, because the form is altogether unsuitable for
vessels of that description, and our only object forgiving it here is to
show the method of reducing the equation ; this being the more
necessary for the sake of system, as it forms a particular case of the
general problem, and is deducible from it by merely assuming a
particular value for the exponent of the parabolic ordinate.

By the rule, we have (b* — 2e?2) =r 784— - 380.25X2 = 23.5 ;
therefore, by multiplication and division, we obtain

b sin. rf>

—^(^—2rfa)zz28x0.25882X23.5-rl2z=5— 14.19 very nearly.

492. Returning to the general
equation, if we suppose wzr2,
then the section is in the form of
the common or Apollonian para-
bola, as represented in the an-
nexed diagram, wherein AB is the
base or double ordinate of the
parabolic section, DC its axis, FII
the water-line, or double ordinate
of the immersed portion FDH, DE the corresponding abscissa, and IK
the horizontal surface of the fluid. Then, with ram 2, the expression
for the stability becomes

(292).

The form of the vessel of which the stability is expressed by the
above equation, is much better adapted for the purposes of steam
navigation, than the triangular form already discussed; but it is obvious
from the relation of the parenthetical terms, that it requires a much
greater breadth at the water line under the same depth and inclina-
tion, to give an equal degree of stability ; and the breadth necessary
for this purpose maybe determined by reversing the expression, which
will then assume the form of a cubic equation, wanting the second
term, and whose reduction will give the necessary breadth.

493. Now, by the preceding calculation we have found the stability
to be 14.19 very nearly, while the depth is 19| feet, and the inclina-
tion from the upright position, 15 degrees, of which the natural sine is
0.25882; consequently, by substitution we obtain

0.0215763— 24.66=14.19,

and if this equation be reduced, we shall find the value of b or the
breadth of the vessel at the water line, to be a very small quantity in

VOL. i. 2 D

402 OF THE STABILITY OF STEAM SHIPS.

excess of 34 feet ; but taking it at 34, the value of the stability for
a vessel in the form of a common parabola becomes

s_ 34X0.25882 (1156_ 1140t75)= 1M84;

hence it appears, that the breadth at the water line, in the case of the
parabola, requires an increase of more than 6 feet, to give the same
stability as the triangle under the same depth and deflexion.

494. If the equation for the stability in the case of the parabola,
be compared with that for the triangle, it will be seen that 3d2 occurs
in the one case, instead of 2d2 in the other ; consequently, the practical
rule as given for the triangle, will also apply to the parabola, if the
phrase " thrice the square of the corresponding depth" be substituted
for " twice the square," as it is now expressed ; the repetition of the
rule is therefore unnecessary.

495. Again, if we put wnr3, then the transverse section of the
vessel is in the form of a cubic parabola, and the general equation
for the value of the stability becomes

496. This form is greatly superior to the preceding one for a steam
vessel, as it gives the surfaces in contact with the water a less degree
of curvature ; but it requires a greater increase of the breadth at the
water line in proportion to the depth to obtain the same degree of
stability, which is manifest from the increase of the negative co-
efficient, the form of the equation being in every other respect the
same as before.

The practical rule for this form, may be expressed in words at
length as follows.

RULE. From the square of the breadth at the water line
when the vessel is upright, subtract 3.6 times the square
of the corresponding depth ; multiply the remainder by the
breadth drawn into the natural sine of the angle of inclina-
tionr and one twelfth of the product will express the stability.

497. EXAMPLE. Let the breadth of the water line be 38 feet, and
let the depth and the deflexion, as well as the density of the vessel,
be the same as before ; what then will be the value of the stability ?

Here by the rule we have

(62— 3.6d9)i=382— 3.6 X19.52=: 75.1 ;

consequently, by multiplication and division, we have

OF THE STABILITY OF STEAM SHIPS.

403

12 v- ;= 38x0.02157x75.1 n: 6 1.6 nearly.

498. In order to pursue the inquiry a step further, let us suppose
that 7zz=5; then, by substituting this value of n in the general
equation for the value of the stability, we shall get

(294).

an equation which differs in nothing from those that precede it, but in
the value of the constant co-efficient of the negative term within the
parenthesis, a quantity which indicates the increase of breadth at the
water line, necessary to give the vessel the same degree of stability,
under the same depth and deflexion, which it possesses when bounded
by curves of the lower orders.

499. If the curves which we have just considered were delineated
from a fixed scale, according to the relation that subsists between the
ordinates and the corresponding abscissas, it would be seen, that the
breadths towards the vertex become greater and greater as the exponent
of the ordinate increases; the figure therefore approaches continually
to the form of a rectangular parallelogram, and essentially coincides

fc'

with it, when the value of n
becomes infinite, as in the
parabola A KB, wherein DC is
the breadth, and LK the depth
of the vessel ; E F the water
line, and k'k the line of sup-
port in the inclined position ;
y the ordinate parallel to the
depth, and x the abscissa;
DPE the immersed triangle, I
and F PC the extant triangle. This extreme case has a manifest rela-
tion to the subject of stability ; for whatever may be the effect of
giving to the sides of vessels the forms of the higher orders of
parabolas, it is evident, that as the exponent of the ordinate is
increased, the stability will approach to that which would obtain if
the sides were made parallel to the plane of the masts.

Now, it may easily be shown, that when the sides of the vessel are
made to coincide with the form of a conic parabola, (fig. art. 492,)
the stability is the same as when the sides are parallel planes ; hence
it is inferred, that if the sides of a vessel be formed to coincide with
a parabola of the lowest order, and another to coincide with one of
the highest, all other circumstances being the same, the stabilities
will be equal in these two cases.

2 D 2

404 OF THE STABILITY OF STEAM SHIPS.

500. But we must now proceed to consider the second variety, in
which the ordinates are parallel to the breadth of the vessel at the
water line when the vessel is placed in an upright and quiescent
position ; and in this case, the general equation expressing the value
of the stability, is

^sin^x .12nd2 \

"ITA6 ~n*+3rc + 2/' (295).

where the several letters which enter the equation indicate precisely
the same quantities, and refer to the same parts of the vessel as before ;
and by giving particular values to the quantity n, we shall obtain
another series of equations, indicating the stability according to the
order of the parabolic curve by which the vessel is bounded.

501. If we put w:nl, then the transverse section of the vessel
becomes a triangle, and the equation expressing the value of the
stability in that case, is

(296).

which is manifestly the same expression as that which we obtained for
the triangle in the first variety, where the ordinates were supposed to
be parallel to the depth; hence, the value of the stability when
estimated in numbers will also be the same.

502. Again, if we suppose the bounding curve of a cross section to
be the same as the common parabola, then 11=12, and this being sub-
stituted in the general equation, the expression for the stability in this
case, is

(297).

the very same as for the triangle ; hence it appears, that when the
ordinates are parallel to the breadth, the stability for a triangular
section is the same as it is for a section in the form of the common or
Apollonian parabola.

503. But when the boundary of the section is in the form of a
cubic parabola, then n =: 3, which being substituted in the general
equation, the expression for the value of the stability in this case, is

s= »§* (*.!.«•>. .'.'•: r. (298)

If this equation be compared with the corresponding one for the
cubic parabola, in the case when the ordinates are parallel to the
depth, it will be seen that the present form is superior in point of
stability, since it requires a less breadth in proportion to the depth to
offer an equal resistance. This inference is drawn from a comparison

OF THE STABILITY OF STEAM SHIPS. 405

of the constants belonging to the negative term, for in the one case
it is double of what it is in the other, and consequently, in the latter
case, a less breadth is necessary to give a positive result.

504. Lastly, if n zr 5, then the parabola which bounds the trans-
verse section A KB of the vessel, is defined by the equation px=iys,
as in the annexed figure, in
which the several letters indi-
cate the parts already men-
tioned, viz. DC the breadth,
LK the depth of the vessel,
E F the water line, k' k the line
of support, y the ordinate
parallel to the breadth, x the
abscissa ; then, the expression
for the stability in this case, becomes

.,/ (299,

from which it appears, that the higher the order of the parabola, the
less increase of breadth is necessary with the same depth to obtain
an equal degree of resistance ; but in the case when the ordinates are
parallel to the depth, as in the first variety, the contrary takes place,
a greater increase of breadth being necessary for the same purpose.
Hence we conclude, that the higher the order of the parabola, the
greater is the degree of stability ; but the form in which the ordinates
are parallel to the breadth, is preferable to that in which they are
parallel to the depth; and, as Mr. Tredgold justly remarks, " this
species of figure may be easily traced through all the varieties of
form, and it has obviously a decided advantage in point of stability,
and it is so easy to compute its capacity and describe it by ordinates,
that it is much to be preferred to the elliptical figures which foreign
writers have chosen for calculation."

505. In order that the stability may be the same at every section
throughout the whole length of the vessel, this being a necessary
condition in the most advantageous cases, the breadth should be
every where in the same ratio to the depth ; for when this is the case,
the vessel will suffer no lateral strain from a change of position.
The preceding determinations relate to the vessel's stability when the
inclination is made about the longer axis ; but the position of the
shorter axis, round which the ship revolves in pitching, in all cases of
practical inquiry, must also be considered ; but since the investigation
sf the several conditions would be similar to that which refers to the
longer axis, we deem it unnecessary to extend the inquiry any further.

CHAPTER XIV.

OF THE CENTRE OF PRESSURE.

506. THE subject of the present chapter might have been placed in
juxtaposition with the doctrine of pressure on plane surfaces ; but we
chose to reserve it for the conclusion of fluid equilibrium, and in as
brief a manner as possible we shall now view the centre of pressure,
by illustrating a few select examples dependent upon its principles.

The Centre of Pressure of a plane surface immersed in a fluid, or
sustaining a fluid pressing against it, is that point, to which, if a force
be applied equal and contrary to the whole pressure exerted by the
fluid, the plane will remain at rest, having no tendency to incline to
either side.

It is manifest from this definition, that if a plane surface immersed
in a fluid, or otherwise exposed to its influence, be parallel to the
horizon ; then, the centre of pressure and the centre of gravity occur
in the same point, and the same is true with respect to every plane on
which the pressure is uniform; but when the plane on which the
pressure is exerted, is any how inclined to the horizon, or to the
surface of the fluid whose pressure it sustains ; then, in order to
determine the centre of pressure, we must have recourse to the
resolution of the following problem.

PROBLEM LXIII.

507. Having given the dimensions and position of a plane sur-
face immersed in a fluid, or otherwise exposed to its influence : —

It is required to determine the position of the centre of
pressure, or that point, to which, if a force be applied equal
and opposite to the pressure of the fluid, the plane shall
remain in a state of quiescence, having no tendency to incline
to either side.

01' THE- CENTRE OF PRESSURE.

407

Let ABC be a cistern filled with an incompressible and non-elastic
fluid, and let abed be a rectangular
plane immersed in it at a given
angle of inclination to its surface ;
produce the sides da and cb di-
rectly forward to meet the surface
of the fluid in the points e and /;
join ef, and through the points e
and /, draw es and/r respectively
perpendicular to the plane produced,
and coinciding with the surface of
the fluid in ef; draw also ds and cr,

meeting es andyY at right angles in the points s and r\ then is des
or c/r, the angle of the plane's inclination, and ds, cr are the per-
pendicular depths of the points d and c.

Let P be the position of the centre of pressure, and through p draw
pm and PW, respectively perpendicular to cb and cd the sides of the
rectangular plane ; then are cb and cd the axes of rectangular co-ordi-
nates originating at c, and pm, PW are the corresponding co-ordinates,
passing through p the centre of pressure, supposed to be situated in
that point.

Now, it is manifest from the nature of fluid pressure demonstrated
in the first chapter, that the force of the fluid against d : —

Is equal to the weight of a column of the fluid, whose base
is the point d, and altitude the perpendicular depth of that
point below the upper surface of the fluid. ^

Consequently, the force against the point d, varies as dx ds ; but
by the principles of Plane Trigonometry, we have
rad. : ed : : sin. des : ds;
hence by reduction, we get

ds=:ed$m.des;
therefore the pressure on the point d varies as

d XedXsin.des,

and the effort or momentum of this force, to turn the plane about the
ordinate pm, manifestly varies as

d Xee?Xsin.c?esXPw,

where p n is the length of the lever on Avhich the force acts.
But by subtraction, p n -=.ed — /m, for ed—fc; therefore by sub-
stitution, the force to turn the plane about the ordinate pm, varies as

408 OF THE CENTRE OF PRESSURE.

dXed* Xsm.de s — e?Xec?X/wXsi
therefore, the accumulated effect of all the forces, to turn the plane
round rw, must be proportional to the sum of

{dXed*}Xs\n.des — /mXsum of {dXed}Xsin-des,
and this by the definition is equal to nothing; hence we get
yVwXsum of {c?Xec?}sin.rfeszzsum of {dXed*} sin.e?es ;
therefore, by division, we obtain
_ sum o
~ su

But the sum of {dXed}, is obviously the same as the body, or
sum of all the constituent particles, multiplied into the distance of
the common centre of gravity ; and therefore, by the principles of
mechanics, fm is also the distance of the centre of percussion, if ef,
the common intersection of the plane with the fluid, be considered as
the axis of suspension, the plane being supposed to vibrate flat-ways.
By reasoning in the same manner as above, it will readily appear,
that the effect or momentum of the pressure on c?, to turn the plane
about the ordinate PW, varies as

dXe dXdnX$in.de s;

but by subtraction, it is dn — cd — en ; therefore by substitution, the
force on d to turn the plane round PW, varies as

and consequently, the effect of all the forces to turn the plane around
pn, must be proportional to the

sum of {dxdeycd} siu.des — crcXsum of {dxed} s'm.des ;
but by the definition, the sum of these forces is equal to nothing; for
the plane has no tendency to incline to either side, being sustained in
a state of quiescence by means of the fluid, and the equivalent oppos-
ing force applied at the centre of pressure ; hence we get
cwXsum of [dXed}~ sum of {dXdeXcd} ;
therefore, by division, we shall have
__sum of {dXdeXcd}

sumof{dXde} ' (301).

which expression also indicates the distance of the centre of percus-
sion ; from which it is manifest, that the centres of pressure and
percussion coincide, when the line of common section between the
plane or the plane produced, and the surface of the fluid is made the
axis of suspension. This being the case, it is evident, that the formulse

OF THE CENTRE OF PRESSURE.

409

which are employed to determine the centre of percussion, may also,
and with equal propriety, be employed to determine the centre of
pressure.

Now, the writers on the general principles of mechanical science
have demonstrated, that if .

xmed, the side of the plane extending downwards, and
?/:zicc?, the horizontal side parallel tofe;

t\\Gnfm and pm, the respective distances of the point P, fromfe and
fc the sides of the plane, are generally represented by the following
fluxional equations, viz.

/•

fm — •- - , andpwzn

Jxyx ZJyxx

(302).

From these two equations, therefore, the centre of pressure cor-
responding to any particular case, can easily be found, as will become
manifest, by carefully tracing the several steps in the resolution of the
following problems.

PROBLEM LXIV.

508. A physical line of a given length, is vertically immersed
in a fluid : —

It is required to ascertain at what distance below the
surface of the fluid the centre of pressure occurs.

Let be be a physical line, perpendicularly
immersed in a fluid of which the surface is
AB, and produce cb to f, so that the point
f may be considered as the centre of suspen-
sion, and let m be the centre of pressure, or
the centre of percussion ; then

by equation (302), we shall obtain

-^
f'

in which equation y is constant ; therefore,

Put d —fc, the distance of the lower extremity below AB,
8 — /£, the distance of the higher extremity, and
I ~bc, the whole length of the line.

410 OF THE CENTRE OF PRESSURE.

Therefore, by addition we have d= I -f- £, and by taking the fluent
of the above equation, we get

fm._

J™ -

3(a*- ar

and when a; me?, we shall obtain

(303).

The equation as it now stands, is general in reference to a line
of which the extremities are both situated below the surface ; but
when the upper extremity is coincident with it, then 8 vanishes, in
which case c?— Z, and our equation becomes

fm = \L* (304>-

509. This last form of the expression is too simple to require any
illustration ; but the form which it assumes in equation (303), may
be expressed in words at length in the following manner.

RULE. Divide the difference of the cubes of the depths of
the extremities of the given line below the surface of the
fluid, by the difference of their squares, and two thirds of the
quotient will give the distance of the centre of pressure below
the surface ; from which, subtract the depth of the upper
extremity, and the remainder will show the point in the line
where the centre of pressure is situated.

510. EXAMPLE. Required the position of the centre of pressure in
a line of 4 feet in length, when immersed vertically in a fluid, the

* Now what is here true of a physical line is true also of a plane, which, if it
reach the surface of the fluid whose pressure it sustains, will have its centre of
pressure at a distance equal to two thirds of its breadth or depth from the upper
extremity ; and this holds true also, whatever may be its inclination, its centre of
pressure will he distant from the upper edge by two thirds of its surface or breadth.
A single force, therefore, applied at that distance, and exactly in the middle of the
length of the plane, would hold it at rest. And the same would manifestly be the
case, if the rod, in place of being applied longitudinally at a single point, were
placed across the plane over the point which indicates the position of the centre of
pressure. All that is required in order to procure the equilibrium is, that a
sufficient balancing force be applied to that centre ; thus, a sluice or floodgate may
be held in its place by the pressure of a single force, applied at one third of its
length from its base, and at two thirds of its length below the surface of the fluid.
And this suggests the practical importance of placing the beams and hinges of
flood and lock-gates at equal distances above and below the centre of pressure,
which is at two thirds the depth of the gate. See Problem LXVI. p. 416.

OF THE CENTRE OF PRESSURE. 411

upper extremity being 2 feet below the surface, and the lower extremity
6 feet ?

Here we have d3 — £3:z: 6s — 23— 208,

and (F— c~ — 62— 2'zz: 32 ;
consequently, by division, we obtain

208

-55= 6.5 feet,

and by taking two thirds of this, we get

fm i= -| of 6.5 rz 44 ft.
and finally, by subtracting the depth of the upper end, we obtain

511. If the upper extremity of the line had been in contact with
the surface of the fluid, then would

imzzfof 4z=2tft.

This is manifest from equation (304), and if a rectangle be described
upon the vertical line, the distance of its centre of pressure below the
surface of the fluid will be expressed by the equation (303 or 304),
according as the upper extremity is situated below, or in contact with
the surface, and this distance will obviously be measured in the line
by which the rectangle is bisected.

512. If the upper side of the rectangle coincides with the surface
of the fluid, as in the annexed diagram, where

adfe is the rectangular parallelogram, having T_ _
the upper side ad in contact with IK ; then,
according to equation (304), bm the distance
of the centre of pressure, is equal to two
thirds of be, the whole length of the paral-
lelogram, and consequently, by subtraction,
we have mc—^bc; therefore, the tendency
of the plane to turn about the base ef, is

equal to the pressure which it sustains, drawn into the length of the
lever mc — \l, where I denotes the whole length of the plane.

Put b nr ad or ef, the horizontal breadth of the plane,
p =: the entire pressure which it sustains, and
s iz: the specific gravity of the fluid in which it is immersed.

Then, by equation (8), Problem III. Chapter II. the whole pressure
sustained by the immersed plane, is expressed by

412

OF THE CENTRE OF PRESSURE.

and this pressure being applied at m, operates on the lever me to turn
the plane about ef, with a force which is equal to

Through the point m, draw mn parallel to ef, the base of the
immersed plane ; then, the tendency to turn round the vertical side
a e, is equal to the whole pressure upon the plane, drawn into the
length of the lever mn ; but mn=i \b, and we have seen above, that
the pressure is expressed by %bl*s; consequently, the tendency to
turn round ae, is

let these two momenta be compared, and we shall have

513. This is obvious, for by casting out the common factors, and
assimilating the fractions, the ratio becomes

21 : 3b ;

and when I and b are equal to one another, or when the immersed
plane is a square ; then the ratio is simply as 2:3; that is, the
tendency of the plane to turn round the lower horizontal side, is to its
tendency to turn round a vertical side, as 1 : 1 J, or as 2 : 3.

PROBLEM LXV.

514. A semi-parabolic plane is immersed vertically in a fluid, in
such a manner, that the extreme ordinate is just in contact with
the surface : —

It is required to determine the position of the centre of
pressure, both with respect to the axis and the extreme
ordinate , which is coincident with the surface of the fluid.

Let IK be the surface of the fluid, and acd the semi-parabola
vertically immersed in it, in such
a manner, that ad the extreme
ordinate coincides with i K, while
the axis ac is perpendicular to it.

Let m be the point at which the
centre of pressure is supposed to
be situated, and through m draw
mb and mn, respectively perpen-
dicular toad and ac, and suppose
the axes of co-ordinates to originate at a ; then, if we

OF THE CENTRE OF PRESSURE.

413

Put I — ac, the axis of the semi-parabola,

b zr ad, the extreme ordinate, which is in contact with IK,
x =z any abscissa estimated from the vertex at c, and
y zz the corresponding ordinate,

then is I — x the distance between the ordinate and the origin of
the axes, corresponding to x in the general investigation. Problem
LXIII. ; but by the property of the parabola, we have

l:bl::x:y*;
and from this, by reduction, we get

Therefore, if l — x and by |, be respectively substituted for
and y in the equations of condition numbered (302), we shall have

and for the corresponding co-ordinate, it is

mn~

But by the writers on the fluxional analysis, the complete fluents
of these expressions are respectively as follows, viz.

— - ,

bm=: - ^ - ST- - ,and?wn—
351 —

the correction in both cases being equal to nothing; but when arrr/,
we get

imzz|/, and mn — Tsjb, (305).

and from these values of the co-ordinates, is the centre of pressure to
be found.

515. EXAMPLE. A plane in the form of a semi-parabola, is immersed
perpendicularly in a fluid, in such a manner, that the extreme ordinate
coincides with the surface; whereabouts is the centre of pressure
situated, the axis being 9 and the ordinate 6 inches ?

414 OF THE CENTRE OF PRESSURE.

Here then we have given I — 9 inches, and b zz: 6 inches ; conse-
quently, by the equations numbered (305), we have

bm — $ of 9 zn 5j- inches, and mn^n T\ of 6 zz l£ inches.

Therefore, with the abscissa ac:=9 inches, and the ordinate
ad —6 inches, construct the semi-parabola adc,
by means of points or otherwise, as directed by the
writers on conic sections ; then, on the axis ac
and the ordinate ad, set off an and a b, respectively
equal to 5-f and If inches, as obtained by the pre-
ceding calculation ; and through the points n and
b as thus determined, draw nm and bm, respec-
tively parallel to ad and ac, intersecting each other
in m; then is m the place where the centre of pressure occurs, as was
required by the question.

516. It would be easy to multiply cases and examples, respecting
the parabola and other curves of a kindred nature, considering them
either entire or in part, and situated in different positions, as referred
to the surface of the fluid ; but since the resolution in every instance,
depends upon the integration of the general fluxional equations num-
bered (302), when accommodated to the particular figure, we think
it quite unnecessary to dwell longer on this part of the inquiry ; we
therefore proceed to resolve a problem or two that depend upon
similar principles, and consequently, are well adapted for illustrating
the manner in which the inquiry is to be extended.

PROBLEM LXVI.

517. A vessel in the form of a parallelopipedon with the sides
vertical, has one side loose revolving on a hinge at the bottom,
and is kept in its position by a certain power applied at a
given point: —

It is required to determine how high the vessel must be
filled withjtuid, before the revolving side is forced open.

Let ABC represent the vessel in question, and let avdc be the
loose side moveable about the hinges at e andy*; bisect c d in c, and
draw cb perpendicular to cd, and let n be the point at which the
given power is applied ; then, because the side axdc is just sustained
by means of the power acting at n, it follows, that the whole force
of the fluid acting at the centre of pressure must produce the equipoise.

OF THE CENTRE OF PRESSURE.

415

Suppose m to be the centre of pres-
sure, and make mb equal to twice
me; then by equation (304), the
point b must coincide with the sur-
face of the fluid.

Through the point b and parallel
to «B or cc?, draw the straight line
rs, which marks the height to which
the vessel must be filled with fluid,
before the side is forced open.

Put b = au, the breadth of the loose side of the vessel,

S zz en, the distance from the bottom at which the force is

applied,

yzz: the magnitude of the force applied at the point n,
s zr the specific gravity of the fluid contained in the vessel,
p HZ the pressure of the fluid against its side, and
z = cb, the height to which the fluid rises.

Then, by the principle indicated in equation (8), Problem III.
Chapter II. we have

and this takes place at the centre of pressure, which, according to
equation (304), is situated at two thirds of the depth below the
surface, and consequently, its effect to turn the side andc on the
hinges e and /*, is, according to the principle of the lever, expressed by

Now, the effect of the force applied at w, to prevent the side from
being thrust open by the pressure of the fluid, is expressed by the
magnitude of the given force, drawn into en the length of the lever
on which it acts, and is precisely equal to the effect of the fluid
acting at the centre of pressure ; hence we get

/8 =#*»,;
and by division this becomes

Z-

Z - -7— *

bs
from which, by extracting the cube root, we obtain

416 OF THE CENTRE OF PRESSURE.

This is the general form of the equation, corresponding to a fluid oi
any density whatever denoted by s; but when the fluid is water, of
which the specific gravity is unity, the above equation becomes

(306).

The method of reducing this equation, may be very simply expressed
in words as follows.

RULE. Divide six times the momentum of the given force,*
by the horizontal breadth of the side to which it is applied,
and the cube root of the quotient will be the height to which
the vessel must be filled.

EXAMPLE. The horizontal breadth of one side of an oblong pris-
matic vessel, is 30 inches ; now, supposing this side to be loose and
moveable about a Hinge at the bottom ; how high must the vessel be
filled with water, in order that the pressure of the water, and a force
of 400 Ibs. applied at the distance of 12 inches from the bottom, may
exactly balance each other ?

Here we have given

b = 30 inches ; 3 zr 12 inches ; and /zz 400 Ibs ;
consequently, we. obtain

z=f/ 6x4QQXl2
30

—9.865 inches.

But the centre of pressure, at which the weight of the fluid is sup-
posed to be applied in opposition to the given force, is situated at
one third of the above distance from the bottom of the vessel ; hence
we have

cm zzi of 9.865 — 3.288 inches.

Therefore, the pressure of the fluid acting on a lever of 3.288 inches,
must be equal to a force of 400 Ibs. acting on a lever of 12 inches ;
that is

400 X 12 = 30 X 9.865s-:- 6.

* The momentum of the given force, is equivalent to its magnitude, drawn into
the distance above the bottom of the point at which it is applied.

OF THE CENTRE OF PRESSURE.

417

PROBLEM LXVII.

518. A vessel in the form of a tetrahedron is entirely filled with
water, and has one of its planes bisected by a line drawn from
the vertex to the middle of the opposite side ; now supposing
one half of the bisected plane to be loose, and moveable about a
hinge at its lower extremity : —

It is required to determine the magnitude of the force, the
point of application, and the direction in which it acts with
respect to the horizon, when the moveable half of the contain-
ing plane is just retained in a state of quiescence.

Let ACB be one side of the vessel, divided by the line CD into two
parts, which are equal and similar to
one another; and let the part BCD
be moveable about the hinges at e
and/.

Suppose the centre of pressure of
the loose part B c D to be at the point
m, and through m draw the straight
lines mb and mn, respectively paral-
lel to CD and AB; in CD take any
other point E, and through E draw
EF perpendicular to CD, or parallel
to DB, making the triangles CDB and CEF similar to one another.

Put I =z AB, BC or AC, the side of the containing plane, or the

edge of the tetrahedron,
d r= CD, the length of its perpendicular,
5 =z the perpendicular depth of its centre of gravity below the

vertex at c,

p = the pressure of the water on the loose triangle CDB,
ar^: CE, any variable distance,
y zz EF, the corresponding co-ordinate, and
(f> z= the angle which the direction of the retaining force makes

with the horizon.

Then it is manifest from the nature of the .problem, that in the case
of an equilibrium, the magnitude of the retaining force must be equal
to the whole pressure of the water upon the moveable triangle CDB ;

VOL. i. 2 E

418 OF THE CENTRE OF PRESSURE.

but by Problem XVII. Chapter V. equation (56), the whole pressure
upon the side ACB is expressed by £/8\/2 '» consequently, the pressure
upon the triangle CDB, becomes

This is manifest, for by Proposition I. Chapter I. the pressure upon
the triangle ACB, is equal to its area drawn into the perpendicular
depth of the centre of gravity, the specific gravity or density of the
fluid being expressed by unity ; but by Problem XVII. Chapter V.
the perpendicular depth of the centre of gravity of the side of a
tetrahedron below the vertex, is

and according to the writers on mensuration, the area of an equilateral
triangle, is expressed by one fourth of the square of the side, drawn
into the square root of 3 ; therefore, we have

where a denotes the area of the triangle ACB; consequently, by
multiplication we obtain

from which, by division, we get

p-=L -rW's. (307).

519. This equation satisfies the first demand of the problem, and the
second manifestly requires the determination of the centre of pressure ;
for by the definition, the point of application of a force, equal in
intensity to the pressure of the water, must occur at the centre of
pressure of the plane CDB, on which that pressure is exerted.

Now, by the principles of Plane Trigonometry, the length of the
perpendicular c D is thus determined,

rad. : I : : sin. 60° : CD,
from which, by reduction, we get

but sin.60°=r %\/3, and consequently, by substitution, we get

Therefore, since the triangles CDB and CEF are similar, by the
property of similar triangles, we have

JJ/3 : ¥ ' • x : y,
and by reduction, we get

OF THE CENTRE OF PRESSURE. 419

Let this value of y be substituted in the first of the equations of
condition (302), and we shall have, for the value of the distance en,
as follows, viz.

> ,.

Jxyx fx*x

the correction being equal to nothing ; and when a: zr c D or d, we have

en = 1^*3. (308).

520. Again, for the horizontal co-ordinate nm, by substituting the
above value of y in the second of the equations of condition, we obtain

here again the correction is nothing, and in the limit when a? = CD or
dj we have

T3GL (309)-

8\/3

521. It is shown by the writers on mensuration, that the planes com-
posing a tetrahedron, are inclined to each other in an angle whose
sine is equal to §V2> and by the principles of mechanics, the direction
of the force applied at m, must be perpendicular to the plane ; itjs
therefore inclined to the horizon at an angle whose cosine is $^/2 ;
but by the principles of Trigonometry, we have

sin.^> — -v/ 1 — cos2.0,
or by substituting as above, we get

sin.^ i= v/l — f = $ = .33333,
corresponding to the natural sine of 19° 28' 15".

Hence it appears, that at whatever point in the plane the retaining
force may be applied, its direction will be inclined to the horizon, at
an angle of 19° 28' 15" ; the third demand of the problem is therefore
satisfied, and we have seen that equation (307) fulfils the first, while
the second requires the application of equations (308) and (309), and
the method of reduction will become manifest from the resolution of
the following example.

2E2

420 OF THE CENTRE OF PRESSURE.

522. EXAMPLE. A vessel, in the form of a tetrahedron has the length
of its edges equal to 15 inches; now, supposing it to be filled with
water, and placed upon its bottom with the axis vertical ; conceive
one of its sides or containing planes to be bisected by a line drawn
from the vertex to the middle of one of the bottom edges, and let one
half of this plane be considered as loose, and moveable about a hinge
at the bottom ; what must be the magnitude of a force that will just
retain it in its place, and at what point must it be applied ?

By equation (307), the magnitude of the required force is precisely
equal to the pressure of the fluid upon the moveable plane ; therefore,
by substituting the datum of the above example, we have

p z= TVX 15s X -/2 — 397.74 cubic inches of water ;

which being reduced to Ibs. avoirdupois, becomes

p = 397.74 X62.5 -4-1728 — 14.38 Ibs.

523. Again, for the point at which this force must be applied, in order
to counteract the pressure of the water, we have by equation (308),

cw~|7^/3"— £ of 15 X -v/3 — 9| inches nearly.
And in like manner, for the corresponding co-ordinate nm, we have
by equation (309),

nm — ^l — -?^ of 15 — 2.8125 inches,
from which the position of the point m can easily be ascertained.

524. If the vessel should be placed with the bottom upwards and
parallel to the horizon, then we shall have

j9 1=7.19 Ibs. ; DW — 6.495 inches, and nm~ 1.875 inches.

A great variety of useful and interesting problems similar to the
preceding, might be proposed in this place ; but as we have already
overleaped the limits of this subject in the present volume, we must
defer their consideration till another opportunity.

CHAPTER XV.

OF CAPILLARY ATTRACTION AND THE COHESION OF FLUIDS.

525. THE subject of Capillary Attraction, and the Cohesion of
Fluids, considered merely as a branch of philosophical inquiry, is
exceedingly seductive and interesting ; but when viewed in the light
of a demonstrative and practical department of physical science, its
application is necessarily very circumscribed, and its character is
unimportant as an analytical theory.

It has, however, been very extensively studied, both in this and in
foreign countries, and numerous philosophers of the greatest eminence,
possessed of the loftiest conceptions and the most profound mathema-
tical attainments, have deemed it a topic worthy of their most atten-
tive consideration, and have ascribed to its influence, a numerous
class of phenomena, in reference to the operations of nature on the
various objects of our sublunary world. To this it is owing, that the
rains which fall on the higher elevations, do not immediately descend
and run to the sea with an increasing velocity, but are retained by
the soil, and being slowly filtered down through it, are cleansed from
their impurities, and delivered in springs and fountains at the foot of
the hills, so as to afford a constant and nearly uniform supply of
moisture to the lower levels.

By capillary attraction, does the oil or melted tallow rise slowly
through the wick of a lamp or candle, where it is converted into
vapour by the heat of the surrounding flame, and rushing out in every
direction, is ignited when it comes in contact with the circumambient
air. By capillary attraction, the juices of the earth are absorbed by
plants, and carried through their numerous ramifications to the
remotest leaf, where they are again partly discharged by evaporation,
after a similar manner to that in which the oil is dissipated from the
wick of a lamp, or the melted tallow from the wick of a candle.

422 OF CAPILLARY ATTRACTION AND THE COHESION OF FLUIDS.

It is also by the principles of capillary attraction, that the lymph
and other fluids are taken up, and transferred through the ramifying
vessels to every part of the animal frame ; other causes dependent on
the organical structure both of plants and animals, may assist in
producing this effect ; but it is abundantly proved by observations,
that by far the greatest part of it is produced by capillary attraction
alone. It is solely owing to it, that a piece of dry wood absorbs a
considerable quantity of moisture, and in consequence of this absorp-
tion it swells with a force almost irresistible, thereby splitting rocks
and other bodies of inconceivable hardness and tenacity.

Consequently, since the principles of capillary attraction are found
to exercise such extensive influence in the operations of nature, philo-
sophers are justified in attempting to acquire a more precise and
comprehensive knowledge of the manner in which it acts, and of the
laws by which that action is regulated during the period of its opera-
tion on natural bodies.

526. DEFINITION. Capillary Attraction is that principle in nature,
by which water and other liquids are made to ascend in slender tubes,
to heights considerably above the level of the fluid in the containing
vessel ; it is so called, because its influence is only sensible in tubes
whose bore is extremely small, in general very little exceeding the
diameter of a hair, but never greater than one tenth of an inch. The
tube thus limited, and in which the fluid is found to ascend, is called
a capillary tube, from the Latin word capillus, a hair. The principles
of capillary attraction, and the theory which it unfolds, together
with its application to tubes of various forms and diameters, we shall
very briefly consider in the present chapter ; the chief and most im-
portant properties peculiar to this subject are detailed in the following
experiments.

527. EXPERIMENT 1. There is an attraction of cohesion between the
constituent particles of glass and water.

This is manifest, for if a very smooth plate of glass be brought into
contact with water, and then gently removed from it, it will be found
that a small portion of the fluid adheres to the glass, and remains
suspended from the lower surface when placed in a horizontal posi-
tion ; hence the existence of an attraction is inferred, and its intensity
must be such, as to balance and sustain the gravitating power of the
water.

And again, if a smooth plate of glass be suspended horizontally
from one arm of a lever, and kept in equilibrio by a weight applied
at the other arm ; then, if the glass be brought into contact with the

OF CAPILLARY ATTRACTION AND THE COHESION OF FLUIDS. 423

surface of the water, it will be found that an additional weight must
be applied to the opposite arm to effect a separation ; and the magni-
tude of this additional weight, is a precise measure of the force of
cohesion.

If the water and the glass be placed in a vacuum, and then brought
into contact, the same effect will be found to obtain, and conse-
quently, the. cohesion is not produced by the pressure of the atmos-
phere ; hence an attraction must exist between the particles of the
water and the glass.

528. EXPERIMENT 2. The constituent particles of a mass of fluid
are mutually attracted ; that is, they have an attraction towards
each other.

According to the preceding principle, when a smooth plate of glass
is brought into contact with the water and gently withdrawn from it,
a thin stratum of fluid adheres to its lower surface ; now, if this
stratum of fluid be carefully weighed, it will be found that its weight
is much less than that which is required to detach it ; consequently,
an attraction necessarily exists, which would keep the stratum united
to the fluid in the vessel independently of its weight, and hence it is
inferred, that the particles are mutually attracted ; that is, they have
an attraction towards each other.

529. EXPERIMENT 3. The constituent particles of a mass of mer-
cury have an intense mutual attraction ; that is, they are strongly
attracted towards each other.

This becomes manifest from the circumstance of the smallest quan-
tity constantly assuming a globular form, and from the resistance
which it opposes to the separation of its parts.

Another circumstance which proves the attractive principle in the
particles of mercury, is, that if a quantity of it be separated into a
great number of parts, they will all of them be spherical ; and if any
two of them be brought into contact, they will instantly unite, and
constitute a single drop of the same form which they separately
assumed.

530. EXPERIMENT 4. The attractive power which is evolved between
the particles of glass and water, is sensible only at insensible dis-
tances ; that is, 'the attraction between the particles is imperceptible,
unless the distance between them be very small.

This is inferred from the following circumstance, viz. whatever may
be the thickness of the plate of glass which is brought in contact with
the water, the force required to detach it is always the same. This
indicates that any new laminte of matter that may be added to the

424 OF CAPILLARY ATTRACTION AND THE COHESION OF FLUIDS.

plate, have no influence whatever upon the fluid with which it is
brought in contact ; whence it follows, that the indefinitely thin lamina
of fluid which attaches to the surface of the plate, interposes between
it and the rest of the fluid in the vessel, a sufficient distance to prevent
any sensible effect from their mutual attraction ; and furthermore, it
appears that the force which is requisite to detach all the equal laminae
of the fluid is the same, being that which is required to separate an
individual film of the fluid from the rest.

Again, it is manifest from observation, that water of the same tem-
perature, rises to the same height in capillary tubes of the same bore,
whatever may be the thickness of the glass of which they are con-
stituted; from this we infer, that the laminse of the glass tube,
however small their distance from the interior surface, have no
influence in promoting the ascent of the fluid.

If the inner surface of a capillary tube be covered with a very thin
coating of tallow, or some other unctuous substance, the water will
not ascend, for in that case the capillary attraction is destroyed ;
hence we conclude, that the action of gravity and capillary action are
different in their nature, for if they were similar, the capillary force,
like the force of gravity, would act through media of all kinds, and
consequently, would cause the fluid to rise in the tube, notwithstand-
ing its inner surface being coated with grease.

531 . From the preceding experiments, and others of a kindred charac-
ter, it is inferred, that the force of attraction in a capillary tube, when it
exceeds the mutual attraction of the fluid particles, extends its influence
no farther than to the fluid immediately in contact with it, which it
raises; and the water thus raised, by forming an interior tube, in virtue
of its own attraction, raises that which is immediately in contact with
itself, and this again, by extending its influence to the lower particles,
continues the operation to the axis of the tube.

The direction of the first elements of the fluid, depends entirely
upon the respective natures of the fluid, and the solid with which it
comes in contact ; if these are the same in all cases, the direction is
invariable, whatever may be the form of the attracting surface, whether
it be fashioned into a tube, or retains the simple form of a plane ; but
the direction of the other elements, or those which are situated out of
the sphere of sensible activity of the attracting surface, depends solely
on the mutual effect of the fluid particles, and the form which the
surface of the fluid assumes, is also regulated by the same cause.

532. From numerous experiments and careful micrometrical observa-
tions^ has been ascertained, that when water moves freely in a capillary

OF CAPILLARY ATTRACTION AND THE COHESION OP FLUIDS. 425

tube, the surface forms itself into a hemisphere, with its vertex down-
wards and its base horizontal, in which position it nearly touches the
interior surface of the tube ; but when the fluid rises between two
planes, the surface assumes a circular form, having for tangents the
planes by which it is attracted.

These experiments and particulars being premised, we shall now
proceed to develope the theory of calculation ; and in order that it
may be invested with all the interest of which it is susceptible, we
deem it advisable to adopt the method of M. le Comte La Place,
one of the most profound and sagacious philosophers that have existed
in this, or in any preceding age.

PROBLEM LXVIII.

533. A cylindrical tube of glass, whose diameter is exceedingly
small, has its lower extremity immersed in a vessel of water,
and its axis vertical : —

It is required to determine with what force the water rises
in the tube, by means of the attractive influence of its surface.

Let abed represent a vertical section of a vessel, filled with water
to the height BC, and let AB be the
corresponding section of a small cylin-
drical tube immersed in it at the lower
extremity, and having its axis perpen-
dicular to the surface.

The fluid rises in the tube above its
natural le^vel, a thin film being first
raised by the attraction of the inner
surface of the glass ; this first film of
fluid raises a second, and the second a
third, and so on, until the weight of the

elevated fluid exactly balances all the forces by which it is actuated,
viz. the attractive influence of the glass, and the mutual adherence of
its own particles.

Let us now suppose that the inner surface of the tube is produced to
E, then carried horizontally to D and vertically to c, and let the sides
of this extended tube be conceived to be so extremely thin, as to have
no action whatever on the contained fluid, and not to prevent the
reciprocal attraction which obtains between the real tube AB and the
particles of the fluid ; that is, let the portion BEDC of the tube be so

426 OF CAPILLARY ATTRACTION AND THE COHESION OF FLUIDS.

circumstanced, as neither to attract nor repel the fluid particles, and
consequently, the circumstances of the problem will not be at all
affected by supposing the tube to assume the form represented in the
diagram.

Now, since the fluid in the tube A E is in equilibrio with that in the
tube CD, it is manifest, that the excess of pressure in AE, arising from
the superior height of the column, is destroyed by the vertical attrac-
tion of the tube, together with the mutual attraction of the fluid
particles in the tube AB ; in order therefore to analyze these different
attractions, we shall first consider those that take place under the
tube AB, in which the fluid rises above its natural level.

534. In the first place then, it is evident, that the fluid in the
imaginary tube BE, is attracted,

1. By the reciprocal action of its own particles,

2. By the exterior fluid surrounding the tube BE,

3. By the vertical attraction of the fluid in AB, and

4. By the attraction of the glass in the tube AB.

Now, the first and second of these attractions, are obviously
destroyed by the equal and similar attractions experienced by the
fluid in the opposite branch DC; consequently, their effects may be
entirely disregarded. But the vertical attraction of the fluid in the
tube AB, is also destroyed by an equal and opposite attraction
exerted by the fluid in BE, so that these balanced effects may likewise
be neglected, and there remains the attraction of the glass in AB,
which operates to destroy the excess of pressure exerted by the
elevated column BF.

535. Again, the fluid in the lower portion of the cylindrical tube
AB, is attracted,

1. By the reciprocal action of its own particles,

2. By the fluid in the imaginary tube BE,

3. By the attraction of the glass in the tube by which it is

contained.

But since the reciprocal attractions of the particles of a body, do
not communicate to it any motion if it is solid, we may, without
altering the circumstances of the problem, imagine the fluid in AB to
be frozen ; then, since the fluid in the lower part of AB, and that in the
imaginary tube BE, are acted on by equal and opposite attractions, these
attractions destroy each other, and consequently, their effects may be
neglected ; hence, the only effective force which remains to actuate
the fluid in AB, is the attraction of the glass containing it. Let this

OF CAPILLARY ATTRACTION AND THE COHESION OF FLUIDS. 427

force be denoted by /, which obtains equally in both the cases above
stated ; therefore, if F denote the intensity of the vertical attractive
force, we shall have

* = 2f.

536. But there is a negative force acting- in the opposite direction, by
which this value of F is influenced, and which arises from the attrac-
tion exerted by the fluid surrounding the imaginary tube, on the lower
particles in the column BE, and the result of this attraction is a
vertical force acting downwards, in opposition to the force 2/"; let
this antagonist force be denoted by/', and we shall obtain

FZZ2/-/'.

Put m zz the magnitude, or solid contents of the column B F,
I zz the density or specific gravity of the fluid, and
g zz the power of gravity.

Then by multiplying these quantities together, the weight of the
elevated column is expressed by m%g ; but in the case of an equili-
brium between this weight and the attractive forces by which it is
elevated, it is manifest that they are equal ; hence we have

mlg—lf—f. (310).

If the force 2/ be less than — /', the value of m or the magnitude
of the attracted column will be negative, and the fluid will sink in the
tube ; but whenever the force 2/ exceeds — /', the value of m will be
positive, and the fluid will rise above its natural level.

537. Since the attractive forces, both of the glass and the fluid, are
insensible at sensible distances, the surface of the tube A B will have
a sensible effect on the column of fluid immediately in contact with
it; this being the case, we may neglect the consideration of curvature,
and conceive the inner surface to be developed upon a plane ; the
force f will therefore be proportional to the width of this plane, or
which is the same thing, to the inner circumference of the tube.

Put d zz the inner diameter of the tube,

TT zz the ratio of the circumference to the diameter,

0 zz a constant quantity, representing the intensity of the

attraction of the tube upon the fluid, and
9' zz another constant, representing the intensity of attraction

which the fluid exerts upon itself.

Then, by the principles of mensuration, we have tin equal to the
inner circumference of the tube, and also to the exterior circum-
ference of a column of fluid of the same diameter ; therefore, it is

428 OF CAPILLARY ATTRACTION AND THE COHESION OF FLUIDS.

f=i dirty,
which being substituted for /and/' in equation (310), gives

m$g = dir('2<l> — f). (311).

This is the general equation that expresses the force by which the
water is raised in a cylindrical tube, and its application to particular
cases will be exemplified by the resolution of the following problems.

PROBLEM LXIX.

538. In a cylindrical capillary tube of a given diameter, the
top of the elevated column is terminated by a hemisphere : —

It is therefore required to determine the height to which
the1 fluid ascends above its natural level.

Let abed be a section passing along the axis of a very small
cylindrical tube, of whioh the diameter is
ab; let the tube be vertically immersed in
the fluid whose surface is IK, and suppose
that in consequence of the immersion, the
fluid rises in the tube to e on a level with
the surface IK, and from thence it is at-
tracted by the glass in the tube, together
with the mutual action of its own particles,
until it arrives at ab, where it forms the
spherical meniscus abfg, and in which

position, the weight of the elevated column is in equilibrio with the
attractive forces.

Now, the problem demands the height to which the fluid rises in
the tube in consequence of the attraction, and on the supposition
that its diameter is very small.

Put r = am, the radius of the interior surface of the tube,

h =n en, the height of the uniform column, or the distance

between the surface of the fluid and the lowest point

of the spherical meniscus,
A'n: ev, the mean altitude, or the height at which the fluid

would stand, if the meniscus were to fall down and

form a level surface,

TT — the ratio of the circumference to the diameter, and
m i= the magnitude of the whole elevated column.

OF CAPILLARY ATTRACTION AND THE COHESION OF FLUIDS. 429

Then, by the principles of mensuration, it is manifest that the
inner circumference of the tube is 2r7r, and the solidity of the
uniform column whose height is en, becomes r*/nr; now, the solidity
of the meniscus ganbf, is obviously equal to the difference between
the cylinder abfg and its inscribed hemisphere an b.

But by the rules for the mensuration of solids, we know that tl>e
solidity of the cylinder abfg is r87r, and that of the inscribed hemi-
sphere is fr37r ; consequently, the solidity of the meniscus is

rs7r — fr37r = £r87r,
which being added to the solidity of the uniform column, gives

wzrr2/i7r -f- T^TJ
from which, by collecting the terms, we get

Now this is equivalent to the solidity of a cylinder, whose radius is
r and altitude evrz h' ; consequently, we have
m = r*K (h + £r) zz rV h' ;
whence it appears, that

h' = h + ir. (312).

539. Instead of din the equation (3 11), let its equal 2r be substituted,
and instead of m in the same equation, let its equivalent h'r^ir be
introduced, and we shall obtain

h' r9 TT $g = 2r TT (20 — 0'),
and from this, by casting out the common factors, we get

A' r 30 = 2(20— A
and dividing by 5#, it becomes

Now, since the symbols <j>, 0', 3 and g are constant for the same
fluid and material, it follows that the whole expression is constant ;
hence, the height to which the fluid rises, varies inversely as the
radius of the tube.

540. Instead of h' in the equation (313), let its equivalent (h -f- £r)
in equation (312) be substituted, and we shall obtain

" '

2(2<6— 0')
Hence it is manifest, that the constant quantity -'—- — — , is

equal to the mean altitude of the fluid multiplied by the radius of the
tube ; and it has been shown in equation (312), that the mean altitude

430 OF CAPILLARY ATTRACTION AND THE COHESION OF FLUIDS.

is equal to the observed altitude of the lowest point of the meniscus,
increased by one third of the radius of the tube, or which is the same
thing, by one sixth of the diameter ; the value of the constant quantity,
can therefore only be determined by experiment, and accordingly we
find, that various accurate observations have been made for the purpose
of assigning the value of this element; the mean of which, according
to M. Weitbrecht, gives

hence, finally, we obtain

#r=.0214. (315).

541 . The equation (315), it may be remarked, is general for cylindrical
tubes, if the elevated column of fluid is terminated by a hemispherical
meniscus, and the practical rule which it supplies, is simply as follows.

RULE. Divide the constant fraction .0214 by the radius of
the capillary tube, and the quotient will express the mean
altitude to which the Jluid rises above its natural level.
If it be required to determine the highest point to which the fluid
particles ascend, it will be discovered, by adding to the mean altitude
two thirds of the radius of the tube, or one sixth of the diameter.

542. EXAMPLE. The diameter of a cylindrical tube of glass, is .06 of
an English inch ; now, supposing it to be placed in a vertical position,
with its lower extremity immersed in a vessel of water ; what is the
mean altitude to which the fluid will ascend, and what is the altitude
of the highest particles ?

Since, according to the question, the diameter of the tube is .06
of an inch, the radius is .03 or half the diameter ; consequently, by
the rule, the mean altitude to which the water rises, is
h' =.. 0214 -~ . 03 — 0.713 of an inch,
and therefore, the point of highest ascent, is
0.713 4- .02 =z 0.733 of an inch.

543. If the mean altitude of the fluid is given, the radius of the tube
can easily be found from equation (315), for it only requires the con-
stant number .0214 to be divided by the given altitude ; but when the
observed altitude, or the distance between the surface of the fluid in
the vessel, and the lowest point of the meniscus is given, the radius
can only be determined by the resolution of an adfected quadratic
equation ; for by equation (314), we have

i**4- Ar-zz.0214,

OF CAPILLARY ATTRACTION AND THE COHESION OF FLUIDS. 431

which being multiplied by 3, becomes

7-2-f3£r=.0642. (316).

544. Suppose now, that the observed altitude is 0.703 of an inch;
then, .by substituting 0.703 instead of h in equation (315), we obtain

consequently, by completing the square, we get

r»+ 2.11r -4- 1.0552— 1.177225,
from which, by extracting the square root, we obtain

r + 1. 055 = 1.085;

hence, by subtraction, we have

r— 1.085 — 1.055 = . 03 of an inch.

Now, if one third of the radius just found, be added to the observed
altitude, the sum thence arising will express the mean altitude ; and
if the whole radius be added to the observed altitude, the sum will
express the greatest height to which the fluid rises in the tube.

PROBLEM LXX.

545. Two parallel planes of glass or other materials, are placed
in a vertical position, with the lower sides immersed in a fluid : —

It is required to determine how high the fluid rises between
them, their distance asunder being very small in comparison
to their surfaces.

Let ad and be represent the ends or sections of two plates of glass,
placed in a position of vertical parallelism,
and having their lower edges d and c im-
mersed in a fluid of which the surface is IK.
Suppose now that a b or cd, the distance
between the plates, is very small in com-
parison to their extent of surface ; then it is
obvious, that the fluid will rise between them
as high as e in consequence of the immersion,
and from thence it moves upwards by the
attractive influence of the glass, and the

mutual action of its own particles, until it arrives at rs, where it forms
a semi-cylindrical meniscus rsfg, whose diameter is equal to the dis-
tance between the planes, and its length the same as their horizontal
breadth.

432 OF CAPILLARY ATTRACTION AND THE COHESION OF FLUIDS.

In this position, the whole weight of the elevated fluid, and the
united efforts of the attractive forces, are in equilibrio among them-
selves, and the problem requires the height to which the fluid rises,
when the powers of gravitation and attraction become equal to one
another ; for this purpose,

Put b zr the horizontal breadth of the planes by whose attraction

the fluid is elevated,

d •=. ab, the perpendicular distance between the planes,
li zz e n, the distance between the lowest point of the meniscus

and the surface of the fluid,

h' zz ev, the mean altitude of the fluid, or the height at which
it would stand, if the meniscus were to fall down and
form a level surface,
TT zz the ratio of the circumference of a circle to its diameter,

and

m zz the magnitude of the volume of fluid raised.
Then, if the constants <£, 0', S and g denote as before, the magnitude
of the elevated volume will be found as follows.

546. By the principles of mensuration, the solidity of the fluid
parallelopipedon, whose breadth is b, thickness d, and height h, is
expressed by bdh; and the solidity of the fluid meniscus whose
section is grnsf, is equal to the difference between a semi-cylinder
and its circumscribing parallelopipedon, the length being equal to
bj and the diameter equal to d, the distance between the attracting
planes.

Now, the solidity of the circumscribing parallelopipedon is |6d8,
and the solidity of the semi-cylinder is J6eP?r; consequently, the
solidity of the meniscus, is

to which if we add the solidity of the uniform solid, the whole magni-
uid becomes
m — bdh -\- \bd\1 — TT).

tude of the elevated fluid becomes

But the periphery of the fluid which is elevated between the planes,
is manifestly equal to 2(£ + d) ; consequently, by substituting this
value of the periphery for dir in equation (311); and for m, let its
value as determined above be substituted, and we shall obtain
{bdh+lbd\1-Tr}}Zg=i1(b+d)(<l<t> — f),
and dividing both sides by b$g, it becomes

OF CAPILLARY ATTRACTION AND THE COHESION OF FLUIDS. 433

but since d is conceived to be very small in comparison with b, the

—

horizontal breadth of the plates, the fraction — may be considered as

evanescent, and then we get

The solidity of the fluid parallelopipedon corresponding to the
mean altitude, is expressed by bdh'i but this is equal to the whole
quantity of fluid raised ; therefore we have

bdh'=bd{h + Jd(l— J*0},
from which, by casting out the common terms, we get

A'= A +Jrf(l — £TT); (318).

Now, (1 — |TT) is a constant quantity ; hence it appears, that the
mean altitude varies inversely as the distance between the planes.

547. Let the symbol for the mean altitude, be substituted in
equation (317), instead of its analytical value as expressed in equation
(318), and we shall obtain

where the value of the constant quantity is the same as before ; hence
we have

rfA'=.0214. (319).

The practical rule which this equation supplies, may be expressed
in words, in the following manner.

RULE. Divide the constant number 0. 021 4 by the perpen-
dicular distance between the planes, and the product will
give the mean altitude to which the fluid rises.

548. EXAMPLE. The parallel distance between two very smooth plates
of glass, is 0.06 of an inch ; now, supposing the lower edges of the
plates to be immersed in a vessel of water ; what is the mean altitude
to which the fluid ascends ?

Here, by operating as the rule directs, we have
h' = 0.0214 -f. 0.06 — 0.356 of an inch.

In this example, the distance between the planes is the same as the
diameter of the tube in the preceding case, but the mean altitude of
the fluid is only one half of its former quantity ; hence it appears,
that if the tube and the planes are of the same nature and substance,
and the radius of the one the same as the distance between the

VOL. i. 2 F

434 OF CAPILLARY ATTRACTION AND THE COHESION OF FLUID9.

other, the fluid will rise lo the same height in them both, if they are
placed under the same or similar circumstances.

549. Having given the mean altitude to which the fluid rises, the
distance between the plates can easily be ascertained ; for we have
only to divide the constant number 0.0214 by the given altitude,
and the quotient will give the distance sought ; but if the observed
altitude, or the distance between the lowest point of the meniscus
and the surface of the fluid in the vessel be given, the operation is
more difficult, since it requires the reduction of an adfected quadratic
equation.

By recurring to equation (317), it appears, that

but we have shown, equations (315 and 319), that the constant

quantity —L~ — — , has, from the comparison of experiments, been

*9
assumed zz 0.0214 ; hence it is

d{ h + \d (1 — |TT) = 0.0214 ;

now, the value of the parenthetical quantity (1 — JTT) is also known,
being equal to 1 — . 7854 — . 2146 ; consequently, by substitution,
we have

0.1073d2+hd= 0.0214,

and dividing both sides by 0.1073, it becomes

d2-f- 9. 32hd= 0.1994.

Let us therefore suppose, that the observed altitude of the fluid, or
the value of h is equal to 0.2913 parts of an inch, and on this
supposition, the above equation will become

rf*+ 2.71473^:= 0.1994,

and this equation being reduced according to the rules for quadratics,
we finally obtain

d = 1.429 — 1.357 zz 0.072 of an inch.

550. The preceding theory has reference to the phenomena of capil-
lary attraction, as they are displayed in cylindrical tubes and parallel
plates of glass ; it would however, be no difficult matter to extend
the inquiry to figures of other forms, and placed under various cir-
cumstances ; but being aware that an extended inquiry would elicit
no new principle, we have thought proper to omit it ; the property
disclosed in the following problem, is however, of too curious and
interesting a character to be passed over without notice, we shall
therefore endeavour to draw up the solution in the most concise and
intelligible manner which the nature of the subject will permit.

OF CAPILLARY ATTRACTION AND THE COHESION OF FLUIDS. 435

PROBLEM LXXI.

551. If two smooth plates of glass be inclined to each other at
a very small angle, having their lower sides brought in contact
with a fluid, to the surface of which the coincident edges are
vertical : —

It is required to determine the nature of the curve which
the fluid forms upon the plates, by rising up in virtue of the
attraction.

Let ABEF and CDEF be the smooth plates of glass, having their
edges coinciding in the line EF, and
their planes inclined to each other
in the angle BED; and suppose the
edges BE and DE to be coincident
with the fluid, while EF the line in
which the plates are brought toge-
ther, is perpendicular to its surface,
which is represented by the plane
BED; then shall FTWOB and FW^D,
be curves described by the particles
of the fluid upon the surface of the
plates.

Take any two points t and r in

the line DE, and in the plane CDEF, draw tn and rp perpendicular
to DE, and meeting the curve FW^D in the points n and p ; the lines
tn and rp are therefore parallel to EF the line of coincidence, and
perpendicular to BED the surface of the fluid.

Again, from the same points t and r, and in the plane BED co-
incident with the fluid's surface, draw the straight lines ts and rq
perpendicular to DE, and consequently, parallel to each other; then,
from the points s and q, in which the lines ts and rq meet BE the
lower edge of the plane ABEF, draw sm and qv respectively parallel
to tn and rp, and meeting the curve FWOB in the points m and o;
these lines are consequently parallel to EF and perpendicular to the
plane BED.

552. Since by the supposition, the angle BED which measures the
inclination of the planes, is very small, the fluid in each section may
be conceived to be elevated by the attraction of parallel planes, and
consequently, by an inference under the preceding problem, the
altitude of the fluid at any two points, will vary inversely as the
distances between the planes at those points ; therefore, we have

2r2

436 OF CAPILLARY ATTRACTION AND THE COHESION OF FLUIDS.

in : rp : : rq : ts ;
but by the property of similar triangles, it is

Er : E£ : : rq : ts;
consequently, by the equality of ratios, we obtain

Er : E£ : : tn : rp,
and by equating the products of the extreme and mean terms, it is

vrXrp = EtXtn. (320).

Now, according to the principles of conic sections, we have it, that
in the common or Apollonian hyperbola, if the abscissae be estimated
from the centre along the asymptote, the corresponding ordinates are
to one another inversely as the abscissae ; hence it is manifest, that
the curve which the surface of the elevated fluid traces on the plates,
is the curve of a hyperbola, whose properties are indicated by equation
(320).

553. Such then is the theory of capillary attraction, in so far as it
is necessary to pursue it; but we shall just remark in passing, that
other fluids, such as alcohol, spirit of turpentine, oil of tartar, spirit
of nitre, oil of olives, and the like, are elevated in the same manner
as water, but to a less degree ; thereby showing that the affinity of
glass to water, is greater than its affinity to any other liquid.

Again, on the other hand, some fluids are depressed by the action
of the capillary force, such as mercury, melted lead, and indeed all
the metals in a state of fusion, are more or less depressed, according
to their density or specific gravity ; but an inquiry into the quantity
of depression in this place, would lead to nothing new or interesting,
and as a subject of practical utility, it is altogether unimportant ; we
therefore pass it" over, and hasten to lay before our readers a detail
of the experiments performed by the celebrated M. Monge, on the
approximation and recession of bodies floating near each other on the
surface of a fluid.

The following are a few of the principal experiments that have been
made on this subject.

554. EXPERIMENT 1. If two light bodies, capable of being wetted
with water, are placed one inch asunder on its surface, in a state of
perfect quiescence, they will float at rest, and experience no motion
but what is derived from the agitation of the air ; but if they are
placed apart only a few lines, they will approach each other with an
accelerated velocity.

Also if the vessel be of glass, or such as is capable of being wetted with
water, and if the floating body is placed within a few lines of the edge
of the vessel, it will approach to the edge with an accelerated velocity.

OF CAPILLARY ATTRACTION AND THE COHESION OF FLUIDS. 437

555. EXPERIMENT 2. If the two floating bodies are not capable of
being wetted with the fluid, such as two balls of iron in a vessel of
mercury, and if they are placed at the distance of a few lines, they will
move towards each other with an accelerated velocity; and if the vessel
is made of glass, in which the surface of the mercury is always convex,
the bodies will move towards the side when they are placed within a
few lines of it.

556. EXPERIMENT 3. If one of the bodies is susceptible of being
wetted with water, and the other not, such as two globules of cork,
one of which has been carbonized with the flame of a taper ; then, if
we attempt, by means of a wire or any other small stylus, to make the
bodies approach, they will fly or recede from each other as if they
were mutually repelled; and if the vessel is of glass, having the
carbonized ball of cork placed in it, it will be found impossible to
bring the cork in contact with the sides of the vessel.

In these experiments it is manifest, that the approximation and
recession of the floating bodies, are not produced by any attraction
or repulsion between them ; for if the bodies, instead of floating on
the fluid, are suspended by slender threads, it will be observed that
they have not the slightest tendency either to approach or recede,
when they are brought extremely near to each other.

From an attentive consideration of the phenomena exhibited in
these experiments, we may deduce the following laws.

557. (1.) If two bodies, capable
of being wetted by a fluid, are
placed upon its surface and
brought near to each other, they
will approach as if they were
mutually attracted.

For if two plates of glass AB,
CD are brought so near each
other, that the point H, where
the two curves of elevated fluid
meet, is on a level with the rest
of the mass, they will remain in
a state of perfect equilibrium.

If, however, they are brought
nearer together, the water will
rise between them to the point
G ; the water thus raised, attracts
the sides of the glass plates, and

438 OF CAPILLARY ATTRACTION AND THE COHESION OF FLUIDS.

causes them to approximate in a horizontal direction, the mass of fluid
having always the same effect as a heavy chain attached to the
plates.

The same thing is true of two floating bodies, when they come
within such a distance that the fluid is elevated between them ; for
it is obvious that the bodies A and
B, being placed at a capillary
distance asunder, have the fluid
elevated between them, and are
therefore brought together by the
attractive influence of the fluid
upon the sides of the globules.

558. (2.) If two bodies are not susceptible of being wetted by the
fluid, they will still approach each other when brought nearly into
contact, as if they were mutually attracted.

For if the two floating bodies A and B, are not capable of being
wetted by the liquid, it will be

depressed between them as at H, A. B

below its natural level, when they
are placed at a capillary distance ;
hence it appears, that the two
bodies are more pressed inwards

by the fluid which surrounds them, than they are pressed outwards
by the fluid between them, and in virtue of the difference between
these pressures, they mutually approach each other.

559. (3.) If one of the two bodies is susceptible of being wetted by
the fluid, and the other not, they will recede from each other as if
they were mutually repelled.

For if one of the bodies as A, is capable of being wetted, while the
other as B is not, the fluid will rise
round A and be depressed round B;
hence, the depression round B will
not be uniform, and therefore, the
body B, being placed as it were upon
an inclined plane, its equilibrium is destroyed, and it will move
towards that side where the pressure is least.

These laws, deduced from experiment by M. Monge, have been
completely verified by the theory of capillary attraction as developed
by La Place ; from his theory it follows, that whatever be the nature
of the substances of which the floating bodies are made, the tendency
of each of them to a coincidence, is equal to the weight of a prism of

HYDROSTATIC PRESSURE EXEMPLIFIED IN SPRINGS, &C. 439

the fluid, whose height is the elevation of the fluid between the bodies,
measured to the extreme points of contact of the interior fluid, and
minus the elevation of the fluid on the exterior sides. The elevation,
however, must be reckoned negative when it changes into a depression,
as is the case with mercury and other metals in a state of fusion, as
has been observed elsewhere.

HYDROSTATIC PRESSURE EXEMPLIFIED IN SPRINGS AND
ARTESIAN WELLS.

560. The atmosphere is the uninterrupted source of communication
between the sea and the earth ; it is the capillary conductor of water
from the ocean to the land. Water ascends in the form of vapour,
and descends as dew or rain upon the earth, which however it pene-
trates but a small depth, except by fissures and permeable strata, which
conduct it to subterranean reservoirs, whence it again issues as in the
discharge of springs; or when the earth is bored through, it rises as
in wells. Some wells are fed by land springs — springs of shallow
depth ; others are fed from the percolation of water through strata
that act as conduits, conveying a current of the fluid through their
permeable texture from one high land to another. Hence it is, that
in valleys and champaign districts, very deep wells are dug, in order
to arrive at those great feeders, where the hydrostatic pressure sends
the water up with amazing force. In some cases we can trace the
source of springs ; and with the help of FATHER KIRCHER'S Mundus
Subterraneus, a man of a fanciful wit might present the public with a
very learned treatise on Natural Hydraulics and Artesian Wells.*

561. As regards rock springs, we know of none that surpass the
sources of the Scamander, in Asia Minor, an account of which will be
found in some notes accompanying Poems of the Rev. Mr. Carlyle, who
saw the stream of the Menderi issuing from a cave surrounded with
trees, and tumbling down the crags in a foaming cascade ; for there
the cavern that " broods the flood divine," discharges its sacred stores
by two large openings in the rock, which leads into the cavern. Upon
entering the recess, two other openings, nearly answering to the out-
ward ones, like arches in a cloister, present themselves to the sight;
and through one of them, in a basin below, the traveller perceives the

* From Artois (the ancient Artesium of Gaul), where perpetually flowing artifi-
cial fountains are obtained, by boring a small hole through strata destitute of water,
into lower strata loaded with subterraneous sheets of this important fluid, which
ascends by pipes let down to conduct it to the surface.

440 HYDROSTATIC PRESSURE EXEMPLIFIED

just emerged Scamander. The channel that conducts the stream into
the basin is a cleft in the rock towards the right, only about four feet
wide and nearly twenty in height; it winds inwards in a curve, and
is soon lost in darkness ; at its bottom glides the current, which for a
few moments seems to repose in the basin beside, and then, by another
subterraneous channel, rushes to the mouth, from whence it issues to
the day : it here bursts from the precipice, and forms a noble water-
fall between forty and fifty feet high, broken, and furnished with
every accompaniment that the admirer of picturesque beauty could
require : its sides are fringed with pine and brushwood ; below, it is
almost hidden from the view by immense fragments of rock that have
fallen from the precipice ; and above it hang crags of from two to
three hundred feet in height, that jut over the bases in large angular
prominences. Such is the spring which flows through the sweet
vale of Menderi in many a winding turn. Far above this is the
summit of Ida — the snowy head of Khasdag, the seat of the immortals
— from whence the bard of yore could view Mysia, the Propontis, the
Hellespont, the JEgean sea, Lydia, Bythynia, and Macedonia.

562. If sea water, which is nauseous to taste, and of perceptible
smell, be the constituent condition of the fluid we call water, then rain
water, which is without smell and taste, is salt water distilled by the
atmosphere; and this is the common quality of rain, river, and
spring water, except where accidental varieties of this last occur,
distinguished by the physical qualities of taste, odour, colour, and
temperature.

563. Two constructions in the physical constitution of the earth
contribute to originate springs, which from the same circumstances
never cease to flow : one is the adaptation of the atmosphere to
transport water from the sea to high lands ; and the other is, the
porous beds of sand, and stone, and clay, which exert a capillary
influence in conveying the fluids they may be charged with from one
elevation to another. Those beds or strata of sand and stone,
resemble sponge, paper, or pipes, as conductors of fluids that are
heavy and incompressible, as water; clay strata, which are impe-
netrable by water, form the great reservoirs or basins in which the
treasure of the skies lies hid. Dislocations in the general mass,
resulting from fractures, intersect the strata and facilitate the dis-
charge from the reservoirs formed by the clay stratum.

564. The water-bearing strata are at various depths, from 50 to 500
feet below the surface, and a sheet of impure, or mineral water, may
be perforated till the operation conducts to a stratum containing pure

IN SPRINGS AND ARTESIAN WEILS. 441

water; for the pipe let down into the lower stratum will not allow
the impure water from above to mix with the pure ascending from
below. Water from two different strata may thus be brought to the
surface by one borehole of a sufficient size to contain a double pipe,
viz. a smaller pipe included within a larger one, with an interval
between them for the passage of the water. The smaller pipe may
thus discharge the water of the lower, and the larger pipe that of the
upper stratum ; for in either case the fluid is but endeavouring to
regain the level at its feeding source on the surface of the earth.
Fountains of this sort — Artesian wells — are very well known on the
eastern coast of Lincolnshire by the name of blow wells. This district
is low, covered by clay between the wolds of chalk near Louth and
the sea shore ; and by boring through the clay to the subjacent chalk,
a spring is found that yields a perpetual jet, rising several feet above
the surface. But wells of this kind are common in many parts of the
world ; in the neighbourhood of London ; in Artois, Perpignan, Tours,
Roussillon, and Alsace, in France ; in some parts of Germany ; in the
duchy of Modena ; in Holland, China, and North America.

565. But whence come those vast issues of fresh water that sometimes
rise up in the sea, as in the Mediterranean near Genoa, and in the
Persian Gulf, where the ascending volume is so vast as to allow
mariners in the one case, and divers in the other, to water ships ?
Springs such as these are the issues of subterranean rivers, all of
which consist of meteoric water, or that which the atmosphere had
transferred to itself from the ocean, distilled and discharged upon the
undulating surface of the earth.

566. The annual fall of rain between the tropics is about ten feet
in depth ; and estimating this in other countries as nearly propor-
tional to the cosine of the latitude, the quantity of moisture ex-
haled in a year, over the surface of the globe of our earth, would
form a sheet of water five feet deep ; therefore the number of cubic
feet of water turned into vapour, and dispersed through the mass of
the atmosphere every minute, would be 5x10,424,000,000, or fifty-
two thousand one hundred and twenty millions. But this enormous
mass Leslie further multiplies by 18,000, the mean height* of the
atmosphere in feet, and again by 62 J, the weight in pounds avoir-

* In taking 18,000 feet as the mean height of the atmosphere, we have followed
Leslie ; but the mean height is 27,800 feet in round numbers, for air is to water as
1§ to 1000; therefore we have 1§ : 1000 : : 34 : 27,818 feet for the height of the
cloud sustaining atmosphere j that is to say, there are no clouds carried higher than
five miles.

442 HYDROSTATIC PRESSURE EXEMPLIFIED IN SPRINGS, &C.

dupoise of a cubic, foot of water ; and the final measure of effect he
therefore takes leave to express by 58,635,000,000 million Ibs. and
equal to the labour of 80,000,000 millions of men. Now the whole
population of the globe being reckoned at 800 millions, of which
only half, or less than that, is incapable of labour, it follows, that
the power exerted by Nature, in the mere formation of clouds, to
produce rain and make rivers and springs, exceeds by two hundred
thousand times the whole accumulated toil of mortals, who, if all
employed in carrying the water of the ocean to the mountain tops,
for streams, and watering the fields, meadows, and woods, could not
rival Nature in her simple process of evaporation, absorption, and
distribution.

567. Such is the enormous power exerted in the great laboratory
of Nature above the earth. Let us now contemplate her exertions
beneath its crust, in the grand hydraulic apparatus of permeable
strata — the casual introduction of faults and dislocations in imper-
vious strata, causing natural vents of water — the interposition of
syphons, cavities, thermal springs, mineral waters — all resulting from
the sea co-operating with the atmosphere to irrigate, to fertilize, to
bless the habitable earth.

568. The surface of our own island contains 67,243 square miles,
which are watered annually by a pool of water about 36 inches deep,
of which, if one-sixth flow to the sea, there is still 2| feet depth left
to fertilize the land, to feed the permeable strata, and afford to each
individual the most abundant supply of this inestimable blessing.

569. If a vertical section of Hertfordshire, Middlesex, Kent, and
Surrey, be taken, we shall have a pretty fair type of the sources of
Artesian or any other wells. Below the London clay we have plastic
clay, then chalk, then fire-stone, then gault clay, and below that
woborn sand. It is sufficient to bore through the tenderest plastic clay
into the chalk, to obtain the finest fresh water in the world. Kent and
Surrey abound in chalk, which dips deeply below the plastic clay
stratum, and makes its appearance at St. Alban's and Dunstable.
The woborn sand met with at Sevenoaks sinks below the fire-stone
and gault clay, and re-appears at Leighton Buzzard. Any one may
for himself sketch a perpendicular section of these districts, and a
few perpendiculars let fall through the London clay, to penetrate
into the chalk, by passing entirely through the plastic clay, will
exhibit the exact position of the borer in searching for water ; or the
reader will find it done to his hand in Dr. Buckland's Geology.

CHAPTER XVI.

MISCELLANEOUS HYDROSTATIC QUESTIONS, WITH THEIR
SOLUTIONS.

570. QUESTION 1. How deep will a cube of oak sink in fresh water,
each side of the cube being 15 inches, and its specific gravity 0.925,
that of the water in which it is immersed being expressed by unity ?

The solution of this question is extremely simple, for by art. 311,
page 257, it is announced as an established hydrostatical principle,
that the magnitude of the whole body is to the magnitude of the
immersed part, as the specific gravity of the fluid is to the specific
gravity of the solid. But since the base of the whole solid and that
of the immersed portion are the same, it follows from the principles of
mensuration, that the magnitudes are as the altitudes, and conse-
quently, the altitudes are as the specific gravities ; hence we have

1 : 0.925 : : 15 : 13.875 inches, the depth required.

571. QUESTION 2. If a cube of wood floating in fresh water, have
three inches of it dry, or standing above the surface of the fluid, and
3J-I-3 inches dry when in sea water; it is required to determine the
magnitude of the cube, and what sort of wood it is made of?

This question may be resolved on the same principles as the last ;
for if we put a; ~ the side of the cube in inches, and s the specific
gravity of the wood ; then, by art. 311, page 257, we have

1000 : s : : x : TT™» the part immersed in fresh water,

and 1026 : s : : x : SX 0 the part immersed in sea water;

but the part immersed and the part extant, together make up the
whole altitude or side of the cube ; hence we have

444 MISCELLANEOUS HYDROSTATIC QUESTIONS, WITH THEIR SOLUTIONS.

-f- 3 :rz x, in the case of fresh water,

1000

O -V*

and - + 3fH = * in the case of sea

therefore, if one of these equations be subtracted from the other, we
shall have

533520

26xzz ' , or arm 40 inches, the side of the cube required ;
51 o

hence, the altitude of the immersed part, as referred to fresh water,
is 40 — 3 = 37 inches ; and the altitude as referred to sea water, is
36 yVy inches ; and from either of these, the specific gravity of the wood
is found by the proposition referred to above ; for we have

40 : 37 : : 1000 : siz:925; indicating the specific gravity
of oak, when that of fresh water is expressed by 1000.

572. QUESTION 3. If a cube of wood floating in sea water be | below
the plane of floatation, and it sinks VV of an inch deeper in fresh
water ; what is its magnitude, and what is its specific gravity ?

This question at first sight would appear to be the same as the last ;
it may indeed be resolved by the same principles ; but since the im-
mersed parts are given in this instance, instead of the extant parts, as
was the case in the preceding question, this circumstance suggests a
simpler and a better mode of solution ; for by the inference in art. 317,
page 261, it appears that the parts immersed below the surface of the
different fluids, are to each other inversely as the specific gravities of
the fluids ; hence, if x denote the side of the cube in inches, then by

the question, — is the altitude of the part immersed below the sur-

3ar 3 3Qx -4- 12
face of sea water, and — — 4- TZ ~ IT; is the altitude of the

4 ' 10 40

part immersed below the surface of fresh water ; consequently, by the
inference above cited, we obtain

1000 : 1026 : : — : 30* ."*" 12 ;
4 40

and from this, by equating the products of the extreme and mean
terms, we get

78x1= 1200, or arzr 15TST inches, the side of the cube required.

Having thus determined the side of the cube, the specific gravity
of the material will be found as in the last question, for we have

15TT : IX 15TT : : 1026 : 769, the specific gravity sought.

VHlVERSiTY

MISCELLANEOUS HYDROSTATIC QUESTIONS, WITH THEIR SOLUTIONS. 445

Mr. Dalby makes the side of the cube equal to 13| inches, and the
specific gravity 772 ; but this only shows that he has employed a
higher number for the specific gravity of sea water; 1030 brings out
his results.

573. QUESTION 4. How deep will a globe of oak sink in fresh
water, the diameter being 12 inches and the specific gravity 925, that
of water being 1000?

By the rules for the mensuration of solids, the solidity of the globe
is expressed by the cube of its diameter multiplied by the decimal
.5236 ; consequently, we have 1728 X .5236 = 904.7808 cubic inches
for the solidity of the globe; therefore, according to art. 311, page
257, we get

1000 : 925 : : 904.7808 : 836.923 cubic inches, the solidity of the
immersed segment. Now, according to the principles of mensuration,
as applied to the segment of a sphere, if x be put to denote the height
of the segment, then its solidity is expressed by .5236 (36ar — 2zs),
and this must be equal to the solidity of the segment found by the
above analogy ; hence we get

1 So;2— x3= 799.2.

In order to reduce this equation, let the signs of all the terms be
changed, and put a;zr z -f- 6 ; then, by substitution, we have

xa= zs -f 18z2+ 108z + 216,
and— 18**=:*— 18s2— 216z — 648;
hence, by summation, we obtain

38—108zr= — 367.2,

and from this equation, the value of z is found to be 3.9867 very
nearly ; but by the supposition, x — z -f- 6, and consequently, it is

x •=. 3.9867 4- 6 = 9.9867 inches very nearly, for
the height of the segment, or the depth to which a globe of oak
descends in fresh water, the diameter being 12 inches, and the specific
gravity 925. This result agrees with that obtained by Dr. Hutton, in
the second volume of his Course of Mathematics.

574. QUESTION 5. If a sphere of wood 9 inches in diameter, sinks
by means of its own gravity, to the depth of 6 inches in fresh water ;
what is its weight, and also its specific gravity ?

By the corollary to the third proposition, art. 233, page 212 — 214,
it is manifest, that the weight of the body is the same as the weight
of the fluid displaced by its immersion ; that is, the weight of the
entire sphere, is equal to the weight of as much fluid as is represented
by the solidity of the immersed segment ; but by the principles of

446 MISCELLANEOUS HYDROSTATIC QUESTIONS, WITH THEIR SOLUTIONS.

mensuration, the solidity of the segment is (9x3 — 12 X 2) x 36 x
.5236 zn 282.744 cubic inches ; consequently, the whole weight of the
body, is 282.744X0.03617 — 10.226 Ibs., the decimal fraction 0.3617
being the number of Ibs. in a cubic inch of fresh water. (See note to
art. 329, page 268.)

Having thus determined the weight of the globe, the specific gravity
of the material may be found in various ways ; but we shall here
determine it by the principle of Proposition VII. art. 311, page 257;
from which we have the following process, viz.

98 : (27 — 12)X36 : : 1000 : 740|f, the specific gravity sought.

575. QUESTION 6. An irregular piece of lead ore, weighs in air 12
ounces, but in water only 7 ; and another piece of the same material,
•weighs in air 14| ounces, but in water only 9 : it is required to
compare their densities or specific gravities ?

This question may be very simply resolved, by the principle stated
in Proposition V. art. 264, page 229 ; which is the same as the
principle employed by Dr. Hutton for the same purpose; from it
we have •

12 — 7 : 12 : : 1000 : 2400, the specific gravity of the
lightest fragment ; and again, we have

14.5 — 9 : 14.5 : : 1000 : 2636.36, the specific gravity
of the heavier piece.

The specific gravities are therefore to one another, as the numbers
2400 and 2636.36. Dr. Hutton makes the ratio as 145 to 132. (See
question 52, page 298, vol. ii. 10th ed. Course, 1831 ;) his formulae
will be found in arts. 250 or 251.

The above solution however, is not correct, for the weight of the
body in air is not its real weight, as it would be exhibited in vacuo ;
the correct specific gravity will therefore be obtained by equation
(186), art. 270, page 234; and the operation is as follows, the
specific gravity of air being 1|-, that of water being 1000.

of the lighter fragment ; and for the specific gravity of the heavier,
we have

14.5X1000-9X1* .d

14.5 _ 9 34-J6-

By the results of our solution, it appears that the heavier fragment
is also the densest; by Dr. Hutton's solution, exactly the reverse is
the case.

MISCELLANEOUS HYDROSTATIC QUESTIONS, WITH THEIR SOLUTIONS. 447

576. QUESTION 7. An irregular fragment of glass, weighs in air 171
grains, but in water it weighs only 120 grains; what is its real
weight ; that is, what would it weigh in vacuo ?

The answer to this question is obtained by equation (185), art. 267,
page 232 ; and the operation as there indicated is simply as follows.
171X1000 — 120XH lry

the real weight of the glass in vacuo.

577. QUESTION 8. A fragment of magnet weighs 102 grains in air,
and in water it weighs only 79 grains ; what is its real weight, or what
does it weigh in vacuo ?

The solution of this question is effected exactly in the same manner
as the preceding, the conditions from which the data are obtained
being precisely the same; that is, the body is weighed in air and
in water ; consequently, the operation is as under.

102X1000 — 79X1|-
'•

the real weight of the magnet in vacuo.

From the real weights of these materials, as determined in the above
examples, the absolute specific gravities can be found by the principle
of Proposition V. page 229 ; for the weights lost, are to the whole
weights, as the specific gravity of water, is to the specific gravities of
the substances in question ; hence we have

— 120 : 171-gVVs- : : 1000 : 3350^, the specific gravity
of the glass.

— 79 : 102¥y¥V : : 100° : 443°Hf» the specific gravity of
the magnet.

Dr. Hutton makes the specific gravities of the glass and the
magnet, respectively equal to 3933 and 5202, and says that the
ratio is very nearly as 10 to 13; our own numbers give the same
ratio.

578. QUESTION 9. Taking the specific gravity of glass equal to 3350,
suppose that a globe is found to weigh 10 Ibs. avoirdupoise ; what is
its diameter ?

The cubic inch of glass of the given specific gravity weighs
0.1211695 of alb.; therefore, according to the equation 187, page
235, we have

/ Z-T

in 5.402 inches nearly.

.5236X0.1211695

448 MISCELLANEOUS HYDROSTATIC QUESTIONS, WITH THEIIl SOLUTIONS.

579. QUESTION 10. Supposing the same piece of glass to weigh 9.996
Ibs. in air, but in water only 7.015 Ibs. ; what is its diameter?

The specific gravity of water, when reduced to pounds per cubic
inch, is 0.03617, and that of air is 0.000045 ; therefore, by equation
(188), page 236, we have

9.996 — 7.015 .— 5.402 inches, the same as before.

.5236 (0.03617 — 0.000045)

580. QUESTION 1 1 . The lock of a canal is 40 feet wide, and the lock
gates being rectangular planes, stand 16 feet above the sill, with their
upper edges on a level with the surface of the water ; now, supposing
that the gates are found to meet each other in an angle of 141° 35';
what is the amount of pressure which they sustain, a cubic foot of
water weighing 62 \ Ibs. avoirdu poise ?

Here the lock gates meet each other in an angle of 141° 35' ; which,
according to Barlow, is the situation in which, with a given section
of timber, they obtain the greatest strength. But by the principles
of Plane Trigonometry, the length of each gate is

20Xsec.l9° 25'=21.206 feet;

and by the question, the depth is 16 feet ; therefore, the whole surface
exposed to the pressure of the water, is 2 1. 206 X 32 = 678. 592 square
feet. Now the centre of gravity of each gate is 8 feet below the
surface of the water, the specific gravity of which is unity ; conse-
quently, by equation (8), page 19, the entire pressure upon the gates, is

p=: 8X678.592 — 5428.736 cubic feet of water;

or when reduced to Ibs. it is
5428.736X62J z= 339296 Ibs., or 151 tons 9 cwt. 0 qrs. 48 Ibs.

581. QUESTION 12. If the diameter of a cylindrical vessel be 20
inches; required its depth, so that when filled with a fluid, the pres-
sure on the bottom and sides may be equal to each other ?

This question is resolved by the equations (59 and 60), page 100,
where it is manifest, from the construction of the equations, that
Bzz: \d, and consequently, by substitution, we obtain

.7854D2dzz 3.1416Ddx Jd, and this expression is equivalent to

.7854D2d— 1.5708D<f,

and by casting out the common factors, we have 2</~D, or by
division, c?m 10 inches ; hence when the depth of the vessel is equal
to half the diameter of the base, the concave surface and the bottom
of the vessel sustain equal pressures.

NOTES.

NOTE A.— CHAPTER I.

ARTICLE 3. Every particle of a non-elastic fluid presses equally in every
direction.

The truth of the principle enunciated in this proposition, is abundantly illus-
trated by the experiments introduced at the end of the sixth chapter, and conse-
quently, it needs no further confirmation here ; but from it we may infer, that

The lateral pressure of a fluid is equal to its perpendicular pressure,

Art. 4. Every particle of fluid in a state of quiescence, is pressed equally in all
directions.

This is obvious ; for if possible, let any particle receive a greater pressure in one
direction than in another ; then, since by art. 2, the particles of a fluid yield to the
smallest force or pressure, and are easily moveable amongst themselves, it follows,
that motion will take place in that direction in which the pressure is greatest ; but
by the proposition, the fluid is in a state of quiescence ; that is, there is no motion
taking place among its particles ; they are therefore equally pressed in all direc-
tions.

Art. 5. When a fluid is in a state of rest, the pressure exerted against the
surface of the vessel which contains it, is perpendicular to that surface.

This also is manifest ; for if the pressure be not perpendicular to the containing
surface, the re-action of that surface cannot destroy it ; let the pressure therefore
be resolved into two, the one perpendicular and the other parallel to the surface ;
then it is manifest, that the former will be destroyed by the re-action, and the
latter continuing to act on the particles of the fluid, will be transmitted in every
direction, and consequently, motion will take place ; but this is contrary to the
supposition, for the fluid is stated to be at rest ; therefore, the pressure must be
perpendicular to the surface.

Art. 6. When a mass of fluid is in a state of rest, its
surface is horizontal, or perpendicular to the direction of
gravity.

For let A B D c represent a vessel of fluid, such as water,
and conceive the right line A B to be parallel to the hori-
zon. Suppose the surface of the fluid to be in the position
PB, any how inclined to the horizontal line AB; then,
since by art. 2, the particles of the fluid are easily
moveable among themselves, it follows, that the higher
VOL. i. 2 G

450

NOTES.

P Q

particles at E, will, in consequence of their gravity, continually descend towards
the lower parts at F.

Again, the greater pressure which obtains among the particles under E, and the
lesser under F, will obviously cause the particles at E to descend, and those at F to
ascend ; and thus the higher parts of the fluid at E, descending and spreading
themselves over the lower parts at F, which at the same time are ascending ; it is
obvious, that the surface will at last be reduced to the horizontal position A B ; and
having attained that position, it must continually remain in it, for then there is no
part higher than another, and consequently, there is no tendency to descend in
one part more than in another, and therefore the fluid must rest in a horizontal
position.

Art. 7. If two fluids that do not mix, are poured into the same vessel, and suf-
fered to subside, their common surface is parallel to the horizon.

Let A B D c be the vessel containing the two fluids which do not mix, and let E F
denote the common surface, or that in which
the fluids come in contact. A a L ft

The upper surface A B of the lighter fluid is
horizontal by art. 6; therefore, let P and Q
be two contiguous particles of the heavier •]
fluid, equally distant from a horizontal plane,
and consequently, equally distant from A B ; if
they are not also equally distant from E F the
common surface, the vertical pressures upon
them will be unequal, for this pressure is made

up of the weights of two columns, containing different quantities of fluid matter,
viz. PC, qd of the heavier fluid, and ca, db of the lighter; consequently, the
pressures in opposite directions will be unequal, and motion must take place, which
is contrary to the supposition.

The particles P and Q are therefore equally distant from E F the common surface
of the fluids; and the same being true for every other two contiguous particles in
the same horizontal plane, it follows, that E F must also be horizontal.

Art. 8. The particles of fluid situated at the same perpendicular depth below the
surface, are equally pressed.

This is almost self-evident, but nevertheless it may be thus demonstrated ; for
let the plane passing through E F, be parallel to the surface
AB; then, since the height of the fluid is the same at all the
points of E F, it is manifest that the weights of the fluid
columns standing upon any equal parts of it, must also be
equal, and consequently, the pressure on all the points of the
plane passing through EF is the same, since they are all
situated at equal depths below the surface AB.

Art. 9. When a fluid is in a state of rest, the pressure upon
any of its constituent elements, wheresoever situated, varies
as the perpendicular depth of the particle or element pressed.
The demonstration of this principle is evident from that to article 8; for the
pressure depends upon the weight of the superincumbent column, and the weight
of this column manifestly varies directly as its height ; hence, the pressure upon
any particle varies as its perpendicular depth below the surface of the fluid.

NOTES. 451

NOTE B.— PROPOSITION I.— CHAPTER I.

In this proposition, and the several laws and consequences deduced from it, the
effect of the atmospheric pressure is entirely disregarded. It may however be
proper to remark, that in numerous delicate hydrostatical inquiries, the pressure
thus excited must be taken into the account: it is equal to the pressure of a
column of water 34 feet in perpendicular height,

NOTE C.— CHAPTER VI.

Experiment 7. Page 160. — Since these experiments were selected and inserted
in this work, a living eel has been killed in the cylinder of the hydrostatic press,
in which also an egg has been broken. But the eel would have been killed by
suffocation, if no pressure had been applied to the fluid, and the fracture of
the egg was due to the air it contained between the white and the shell, or
to the different densities of the shell, the white, the yolk, and the water.
Thus we can easily conceive, as Mr. Tredgold remarks, that the trial of an
experiment may be the means of condemning a very useful principle, merely
through inattention to the proportions and the mode of action. We may still
affirm, that fishes will endure a very high degree of fluid pressure, provided they
be allowed to breathe ; indeed it is recorded, that a whale in the arctic seas, being
struck by a harpoon, descended perpendicularly by the line about 900 fathoms,
before it returned to the surface to respire ; it was then under a pressure of nearly
164 atmospheres, or 2,460 Ibs. upon a square inch of its surface; now if the living
animal could sustain this natural pressure without inconvenience, we are at liberty
to conclude that it could sustain an equal degree of artificial pressure. It is manifest,
that fishes which do not come to the surface, breathe the air with which the water
is impregnated, at whatever depth they may be found. Moreover, if an eel were
killed by pressure, we suppose it would be crushed, or burst asunder. In short, we
require evidence of the death by pressure, to remove our belief in death by suffo-
cation. Air, which is invisible, by squeezing the heat out of it by strong pressure,
may be compressed into water ; but the contraction which water suffers at every
increase of pressure, exceeds not the twenty thousandth part of what air would
undergo in like circumstances ; and fishes are at their ease in a depth of water,
where the pressure around will instantly break or burst inwards almost the strongest
empty vessel that can be let down.

We are perfectly aware of the experiments of Mr. Canton, in 1760, which
established incontestably the compression of water. Indeed the theory of com-
pression extends to all bodies : Dr. Young says that steel would be compressed
into one-fourth, and stone into one-eighth of its bulk at the earth's centre ; but a
density so extreme is not borne out by astronomical observation. And the late Sir
John Leslie, who suggests the idea that the ocean may rest upon a subaqueous
bed of compressed air,* says that water at the depth of 93 miles would be com-
pressed into half its bulk at the surface of the earth ; and at the depth of 362.5
miles it would acquire the ordinary density of quicksilver.t Practical men, m
reply to all this physical science, may justly reply, " We are seldom called upon

* Article " Meteorology," in the Supplement to the Encyclopaedia Britannica.
t See Leslie's Elements of Natural Philosophy, vol. i.

2G 2

452 NOTES.

to execute undertakings much below the level of low water, and those investigations
suit us best, which are confined to the depth of a few fathoms, where we know that
water is, to all intents and purposes in our business, wholly unaltered by com-
pression."

NOTE D.— CHAPTER X.

The principle of fluid support, and the doctrine of specific gravity, which we are
now considering, explain many curious facts that daily pass unobserved. Thus, a
stone which two men on land can h ardly lift, may be borne along by one man in water ;
and in diving, a dog will bring to the surface a human body, which the strongest of
his species could not lift on land : hence the ease also with which a bucket is lifted
from the bottom of a well to the surface of the water. And as the human body in
an ordinary healthy state, with the chest full of air, is lighter than its equal bulk
of water, a man naturally floats with about half the head extant ; " having,"
as Dr. Arnott says, " then no more tendency to sink than a log of fir." When a
swimmer floats on his back, with merely his face above water, in which position
he can breathe freely, he exhibits the true position of floatation, in which the human
body is lighter than water, for its specific gravity is one ninth less than that of water,
being about 0.891. In some cases however, the bodies of men are heavier: thus, a
person who weighs 135 Ibs. would be 12 Ibs. heavier than two cubic feet of river
water, and would require a float of cork equal to 4 Ibs. to keep him from sinking ;
for 123 + 4#— 135 -\-x, where x represents the weight of cork; consequently,
123 + 3a;=135; therefore, 3x= 135— 123=12; whence x =4=.

When a solid specifically heavier than a fluid, is immersed to a depth which is to
its thickness, as the specific gravity of the solid to that of the fluid, and the pressure
of the fluid from above is removed, the body will be sustained in the fluid ; for the
pressure from above being removed, the body is in the same state with respect to
the contrary pressure, as if the same weight filled the whole space to the surface of
the fluid; which means, as if its specific gravity and that of the fluid icere equal.

The principle here enunciated helps the philosophers in their explanation of the
common experiment of making lead to swim, in consequence of being fitted to the
bottom of a glass tube.

In the case cited above, of solid bodies being lighter in water than in air— that
is to say, being more easily moved in the water than on dry land — the meaning of
the proposition is, that all bodies, when immersed in a fluid, lose the weight of an
equal volume of that fluid. Thus, in raising a bucket of water from the bottom of
a well, so long as the bucket is under the water, we do not perceive it to have any
additional weight beyond the wood it is made of; but the moment we raise the
bucket to the surface, and suspend it in air, then we feel the additional weight of
the water, which if equal to 6£ gallons, or to one cubic foot, will add nearly 62£
pounds, or 1000 ounces avoirdupois weight to the bucket. Now all this weight
existed in the bucket when under the surface of the water, being supported by an
equal bulk, or 62J pounds. The weights thus gained or lost by immersing the
same body in different fluids, are as the specific gravities of the fluids ; hence
we affirm that all bodies of equal weight, but of different volume, lose in the
same fluid, weights which are reciprocally as the specific gravities of the bodies,
or directly as their volumes. In the salt sea it will be one thirty-fifth lighter than
i n fresh water.

NOTES. 453

NOTE E.— CHAPTER XII.

It will not, however, be out of place to remark, that the weight of the whole
solid, and that of the portion immersed below the plane of floatation— which
corresponds to the magnitude of the fluid displaced — are very appropriately
represented by the areas drawn into the respective specific gravities of the solid
and the fluid on which it floats. But the most cursory observation shows, that a
solid may be immersed in a fluid in numberless different ways, so that the part
immersed, shall be to the whole magnitude in the given proportion of the specific
gravities, and yet the solid shall not rest permanently in any of these positions. The
reason is obvious : the floating body is forced down by its own weight, and borne
up by the pressure of the fluid ; it descends in the direction of a vertical line
passing through its centre of gravity ; it is pushed up in the direction of a vertical
line passing through the centre of gravity of the part immersed, or the displaced
fluid. Unless therefore, these two lines are coincident, or that the two centres of
gravity shall be in the same vertical line, it is evident that the solid thus impelled,
must revolve on an axis until it finds a position in which the equilibrium of floating
will be permanent.

To ascertain therefore, the positions in which the solid floats permanently, we must
have given the specific gravity of the floating body, in order to fix the proportion of
the part immersed to the whole; and then, by geometrical or analytical methods,
determine in what positions the solid can be placed on the surface of the fluid, so
that the centre of gravity of the floating body, and that of the part immersed may
be situated in the same vertical line, while a given proportion of the whole volume
is immersed beneath the surface of the fluid.

The incumbent weight may be considered as collected in the centre of gravity of
the floating body, and the sustaining efforts as united in the centre of buoyancy,
which, as we have already said, is the same as the centre of gravity of the water
displaced, or of the immersed portion of the uniform solid. To these two points
therefore, the antagonist forces are directed, and the line which joins them, called
the line of support, will have constantly a vertical position in the case of equilibrium.

The centre of gravity of the whole mass, about which it turns in the water, must
evidently continue invariable ; * but the centre of buoyancy will change its relative
place, according to the situation of the immersed portion of the solid. If these
two centres should coincide, the body will float indifferently in any position of
stability. It will therefore float, as often as a vertical line, drawn from the centre
of buoyancy, shall pass through the centre of gravity. But this will obtain when-
ever the line of support becomes perpendicular to the horizon. The equilibrium
may, however, be either permanent or instable. It is permanent, if on pulling the
body a little aside it has a tendency to redress itself, or to recover its original
position ; it is instable, when the body, on being slightly inclined, tumbles over in
the liquid and assumes a new situation. These opposite conditions will occur in a
body of irregular form, when the centre of gravity occupies the highest or the lowest
possible position, (when the centre of gravity is the lowest possible, the situation is
that of maximum stability) for though the volume of immersion remains the same,
the solid will evidently be less or more depressed in the fluid medium, according

* This is not strictly true, but it causes no difference in the theory that it is otherwise.

454 NOTES.

to the width of its section or water lines. We have a curious proof of this in the
construction of the French ship of the line, of 74 guns, called Le Scipion, fitted for
sea at Rochfort, in 1779; but she wanted stability, which, after various fruitless
attempts, was achieved by applying a bandage or sheathing of light wood to the
exterior sides of the vessel. This cushion, bandage, or sheathing, was from one
foot to four inches in thickness, extending throughout the whole length of the
water line, and ten feet beneath that line. We are left to infer Le Scipion then
floated with permanent stability.

If the centre of buoyancy stand higher than the centre of gravity, the floating
body will, in every declination maintain its stability, and regain its perpendicular
position ; for though made to lean towards either side, the vertical pressure exerted
against that variable point will soon bring it back again into the line of support.
But the elevation of the centre of buoyancy above that of gravity, is by no means
an essential requisite to the stability of floatation; on the contrary, it falls in most
cases considerably below the centre of gravity about which the body rolls. The
buoyant efforts may be considered as acting upon any point in the vertical line,
and consequently, as united in the point where the line crosses the axis of the
floating body. If the point of concourse, thus assigned, should stand above the
centre of gravity, the body will float firmly, and will right itself after any small
derangement. If it coincide with the centre of gravity of the homogeneous body,
this will continue indifferent with regard to position; but if the vertical line
should meet the axis below the centre of gravity, the body will be pushed forwards,
its declination always increasing till it finally oversets.

Thus, a sphere of uniform consistence floated in water, will sink till the weight
of the fluid displaced by the immersed portion shall be equal to its own load. The
centre of gravity of this body is the centre of the sphere itself; but the centre of
buoyaney must be the centre of gravity of the volume of immersion, which there-
fore lies below the centre of gravity of the body, in an axis perpendicular to the
water line, or line of floatation. The ball is hence pressed down by its own
weight collected in its centre of gravity, and pushed up in the opposite direction by
an equal force combined in the centre of buoyancy ; both of the forces, however,
concurring in the centre of gravity of the immersed sphere. Wherefore, being
always held in equilibrium by those antagonist forces, it will remain still in any
position which it may happen to occupy. But this indifference to floating will
obtain only when the sphere is perfectly homogeneous, and its centre of gravity
coincides with the centre of magnitude, for otherwise, the former descending as
low as possible, would always assume a determinate position.

A cylinder will, according to its density, and the proportion of its diameter and
altitude, exhibit the three features of a floating body, in indifference, instability, or
permanence of equilibrium. For example, a cylinder, the specific gravity of which
is to that of the fluid in which it floats, as 3 to 4, its axis being to the diameter of
the base as 2 to 1, if placed on the fluid with its axis vertical^ will sink to a depth
equal to a diameter and a half of the base ; and as long as the axis is sustained in
a vertical position by external force,, the centre of gravity of the solid and the
centre of the immersed part will be situated in the same vertical line ; but the solid
will not float permanently in that position, for as soon as the external force is
removed, it will overset and float with its axis horizontal. But a cylinder whose
axis is one half, instead of twice the diameter of the base, being placed in a fluid
with its axis vertical, will sink to the depth of three fourths of a diameter^ and will

NOTES. 455

float' permanently in that position. Incline it as you may, on being left to itself it
will ultimately settle permanently, with its axis perpendicular to the horizon. The
differences of the phenomena in this case, arise from the change which takes place
in the position of the line of support; and what is true of the cylinder is true also
of other figures ; for when a solid changes its position, by revolving on an axis on
the surface of a fluid, any position of equilibrium is always succeeded by a position
of equilibrium which is of a contrary description.

A segment of a sphere floating in water, will have its centre of gravity below
the centre of the sphere, when the segment floats with its vertex downwards,
arid in an axis at right angles to its base ; but the centre of buoyancy, or the
centre of gravity of the immersed segment, must, in every situation of the
floating mass, occur in a perpendicular bisecting the water line, and conse-
quently passing through the centre of the sphere. In the case of equilibrium this
perpendicular must have a vertical position, or the involved base of the segment
must form a horizontal plane. If this body be now drawn aside, into a position
which shall incline ,jts base in any angle with the water line, it will be pressed
down by its own weight, collected in the centre of gravity of this body, and pushed
upwards by an equal buoyant power exerted at the centre of buoyancy. This force
may now be conceived to act upon any point in the line connecting the centres of
gravity and buoyancy, and therefore at the concourse of the axis in the case of equi-
librium, and of the vertical line when the body is drawn aside. The buoyancy trans-
mitted to this point pushes the axis of inclination obliquely, the greater part of it
bearing the point of concourse in the direction of the axis of permanent floatation,
while another small part of this force, pressing perpendicular to the axis of inclina-
tion, makes the body turn about its centre of gravity, from the higher or lower
point of inclination of its upper surface, till it ultimately coincides with the water
line. Every derangement is thus corrected by a restoring energy which maintains
a permanent equilibrium.

An oblate homogeneous spheroid will sink in a manner similar to the segment of
the sphere, and carry the centre of buoyancy in a like position. The declination of
its axis, by drawing the body aside from the position of permanent equilibrium, is
restored to its vertical position by the effort of buoyancy exerted at a point above
the centre of gravity of the spheroid, which tends to redress the floating body and
secure its stable equilibrium.

On the other hand, a prolate spheroid will have its centre of buoyancy and plane
of floatation, each the same height as in a sphere described on the longer axis of
the spheroid. But the shifting of its centre of buoyancy will be diminished in
proportion to the narrowness of the spheroid. The vertical will meet the principal
axis below the centre of gravity of the solid, and will push it aside more and more
till the spheroid falls, and extends its longer diameter in a horizontal position. It
may then roll indifferently upon that line, as the sphere turns about its diameter.

A solid of any form, not abruptly irregular, set to float in water, will be divided
into correspondent equal portions by its principal axis, which will cross the plane
of floatation at right angles. If the body be inclined, its centre of buoyancy will
shift its place as the inclination varies, until the antagonist forces meet in a point
in the axis, where the effort of the body to redress itself remains unaltered, like the
centre of gravity itself. That characteristic point standing always above the centre
of gravity of the mass, and limiting its greatest elevation in the case of permanent
stability, has been called the metacentre.

If the floating body be a homogeneous parallelepiped placed vertically in the

456 NOTES.

fluid, it will evidently sink till the immersed part shall be to its whole height, as its
density is to that of the fluid. The centre of buoyancy will be below the centre of
gravity, but both will be in the axis of the solid ; the former midway between the
base of the parallelepiped and the water line ; the latter halfway between the base
and summit of the body. If the solid be inclined to one side, its water line will
shift its position on the body; the centre of buoyancy will make a corresponding
change, describing a small arc of a circle, till it be raised in relation to the altitude
of the centre of gravity of the extant triangle, as the area of the adjacent rectangular
figure is to that of the triangle, while the moveable centre of buoyancy is carried
laterally in the same ratio. And when the metacentre coincides with the centre
of gravity, the solid floats passively and indifferent to its position. If the paral-
lelopiped become a cube, then its breadth and length being equal, the two densities
of indifferent floatation are expressed in the numbers ^ and |§. Between these
limits there can be no stability, but above and below them the floating body
acquires permanence.

Both experiment and calculation prove, that a parallelepiped of half the density
of water, and having 9 inches for its altitude, and 11 inches for the side of its
square base, will float indifferently ; but it will gain stability if its density be either
increased or diminished. With a density two thirds that of water, the metacentre
will stand ^ parts of an inch above the centre of gravity ; and $ parts of an inch
above it if the density be reduced to one third. With such proportions, a paral-
lelopiped might therefore in every case continue erect ; and copper or sheet-iron
tanks, with such proportions, would float safely as pontoons for flying bridges.

But this is not all : we can prove, that if the parallelepiped be set upon water,
with one of its solid angles uppermost, the stability will be limited within the
densities of 395 and ||. jn a Word, let the specific gravity be greater than ^ or less
than |§, the solid would permanently float in that position : but were the specific
gravity either less than the former, or greater than the latter, the body would
overset. Were the parallelepiped thus set on water, with one of its diagonals
immersed and the other vertical, its equal side being 18 inches, then it would sink
about 14^ inches on the side ; 9% inches of the diagonal would be immersed, and
nearly 16 extant; supposing the specific gravity of the solid to be 0.326, that of the
fluid being equal to unity.

In short, the determination of the positions of equilibrium of a solid body, floating
on a fluid of a given density greater than itself, is reducible to a problem of pure
geometry, which may be better expressed as follows : —

To cut any proposed "body by a plane, so that the volume of one of the
segments may be to that of the whole body in a given ratio ; and such that
the centre of gravity of the whole body, and that of one of its segments, may
be both found in a line perpendicular to the cutting line.

In order to the complete solution of this problem, it is necessary in each parti-
cular case, to express the two conditions of equilibrium by means of equations, the
solutions of which will make known all the directions that can be given to the cut-
ting plane, and whence necessarily result all the positions of equilibrium of the body.

This is precisely the plan we have pursued, and all our investigations proceed to
ascertain these two conditions of equilibrium; and from the resulting or final
equations, to draw up a geometrical construction of the positions so determined ;
for calculation is here an instrument of necessity, and not a vain exhibition of
analytical formulae, difficult to follow and still more difficult to apply.

NOTES. 457

NOTE F.— CHAPTER XIII.

The investigations pursued in this and the previous chapter, explain the cause
of the oversetting of the large icebergs which sometimes float within the limits of
the temperate zone. These enormous blocks of frozen fresh water assume various
forms i some are columnar, others approach the parallelepiped in their outline,
others again resemble mis-shapen cylinders; but all evidently different in form
below the plane of floatation to what they exhibit in their extant volume. The
action of the atmosphere as the summer advances, slowly thaws the upper surface ;
the under side likewise melts at first, but becomes soon protected by a pool of fresh
water of the same temperature, consisting of the dissolved portion of the ice which
is upheld by the superior density of the surrounding medium. The principal waste
of the icy mass taking place along its immersed sides, the current of melted water
continually rises upwards, and leaves a new surface to the attack of a warmer
current. Whenever therefore, the breadth of the vast column becomes so reduced
that it approaches to three fourths of its altitude, the icy parallelepiped will
overset, and present a new position of equilibrium. Thus, if the whole height
of the mass were 1000 feet, 890 feet would be submerged in the ocean, and 110 feet
would be extant, towering amidst the waves. In this case, the elevation of the
centre of gravity beyond that of buoyancy would be 35 feet, which is the limit of
the metacentre after the base of the column has been reduced to a breadth of
766 feet.

An iceberg of a cylindrical form 1000 feet high, would sink 889 feet in the
ocean ; but when the diameter of its base was reduced to nearly the same dimen-
sions, say 885, it would overset and take a new position. The instability of the
cylinder takes place earlier than that of the parallelepiped, or when the width below
becomes eight ninths, instead of three fourths of the whole height. Since then the
extant portion wastes more slowly than the immersed portion, the greater the
extension of the summit, the more it will hasten the change of position by over-
whelming the icy mass.

And if the block be wasted and rounded below into the shape of a parabolic
conoid, it will suffer a total inversion the moment its base is reduced to Its depth
in the ratio of about 11 to 20, and its lowest point will become the summit of the
extant mass. This form of a body of ice would therefore suffer a greater previous
waste ; but its balance is in the end more effectually destroyed. In every case,
stability becomes precarious after the breadth of the block is inferior to its depth.

NOTE G.— CHAPTER XIV.

Upon the pressure, cohesion, and capillary attraction of fluids that are heavy,
depends their transmission through fissures of the earth and between its strata,
which are pervious to the percolation of water. We can penetrate but a small
distance, say 500 fathoms, in digging for coal ; a less depth suffices for some ores,
and water is found at all depths, from a few feet to three hundred, as in the neigh-
bourhood of London. In the great coal area of Britain, extending lengthwise 260
miles, and in breadth about 150 miles, in a diagonal line from Hull to Bristol, in
England, and from the river Tay to the Clyde, in Scotland, we find a great variety

458 NOTES.

of rocks or strata, piled up at a small angle with the horizon, though in some
instances, like the primitive, nearly vertical. These strata consist of sand-stone,
clay-slate, bituminous slate, indurated argillaceous earth, or fireclay, argillaceous
ironstone, and greenstone or blue whinstone. And to possess the valuable treasures
concealed among these rocks, we employ a vast capital in money, and tax all the
ability of the human mind in the science of engineering.

To bring the subject matter of capillary attraction, as regards Artesian wells,
springs, mountainous marsh lands, or bogs, fairly before the reader in a very brief
manner, we shall avail ourselves of a vertical section of the strata in Derbyshire,
selecting our materials from the valuable work of Mr. Whitehurst, " On the original
State and Formation of the Earth."

If the reader conceive the alluvial covering to be removed, the strata will at
once appear on the upper surface, as in the external contour of the country between
.Grange Mill and Darley Moor, in Derbyshire. Let now the numbers 1, 2, 3, 4,
&c. represent the strata in their vertical position, bassetting towards 8, with the
river Derwent running over a fissure filled with rubble in the centre.

Then, the upper stratum, or No. 1, at Darley Moor, is Millstone Grit, a rough
sandstone, 120 yards deep, composed of granulated quartz and quartz pebbles,
without any trace of the animal or vegetable kingdoms.

The next stratum, called No. 2, which is found on both sides the Derwent, is a
bed of Shale, or Shiver, 120 yards deep, being a black laminated clay, much indu-
rated, without either animal or vegetable impressions. It contains ironstone in
nodules, and the springs issuing from it are chalybeate, as that at Buxton Bridge,
or that at Quarndon, and another near Matlock Bridge, towards Chatsworth.

Next in- succession we have No. 3, Limestone, 50 yards thick, productive of lead
ore, the ore of zinc, calamine, pyrites, spar, fluor, cauk, and chert. This stratum
is full of marine debris, as anomince bivalves, not known to exist in the British seas ;
also coralloids, entrochi or screw stones; and amphibious animals of the Saurian,
or lizard and crocodile tribe, some of which, in a fossil state, are of enormous size.

Following this we have No. 4, a bed of Toadstone, 16 yards thick, but in some
instances varying in depth from 6 feet to 600 feet. It is a blackish substance,
resembling lava, very hard, with bladder holes, like the scoria of metals or Iceland
lava. This stratum is known by different names in different parts of Derbyshire.
At Matlock and Winster it is loadstone and blackstone ; at Moneyash and Tidswell
it is called channel} at Castleton, cat-dirt; and at Ashover, black-clay. " This
toadstone, channel, cat-dirt, and black-clay, is actually lava, and flowed originally
from a volcano, whose funnel or shaft did not approach the open air, but which
disgorged its contents between the adjacent strata in all directions," at a period
when the limestone strata and the incumbent beds of millstone-grit, shale, argilla-

NOTES. 459

ceous stone, clay, and coal, had an uniform arrangement concentric to the centre
of the earth.

Beneath all these we have No. 5, a Limestone formation, 50 yards thick, and
similar to No. 3; that is to say, laminated, containing minerals and figured stones.
It is productive of marble ; it abounds with entrochi and marine exuviae : it was
thence at one time the bed of a primaeval ocean.

No. 6 is Toadstone, 40 yards deep, and similar to No. 4, but yet more solid,
showing that the fluid metal was much more intensely heated and combined than
No. 4.

No. 7, Limestone, very white, 60 yards deep ; laminated like No. 3 and 5, and
like them it contains minerals and figured stones, and was either a continuation of
Nos. 3 and 5, the entire mass having been split at different depths by the expansive
power of the boiling lara.

No. 8, is Toadstone, 22 yards deep, similar to No. 6, but yet more solid.

No. 9, Limestone, resembling Nos. 3, 5, and 7.

To this enumeration of the Derbyshire strata we must now add six other strata f
too minute to be expressed in the same scale, but which are in fact the capillary
strata, which we may liken to the glass plates referred to hi Problem LXXI.
Miners call these minute parallel strata, clays, OT way-boards : in general they are not
more than four, five, or six feet thick, and in some instances not more than one foot.
They are the channels for water, and all the springs flowing from them are warmy
like those at Buxton and Matlock Bath. The first stratum of clay separates Nos. 3
and 4 ; the second, Nos. 4 and 5; the third, Nos. 5 and 6; the fourth, Nos. 6 and
7 ; the fifth, Nos. 7 and 8 ; the sixth, Nos. 8 and 9 : and what is very remarkable,
by these clays the thickness of the other strata may be ascertained, which would
otherwise be difficult, as the limestone beds consist of various lamina.

There are several circumstances illustrative of this capillary attraction, which
receive illustration from the diagram before us ; to these we shall now address-
ourselves ; and, first, it is observable that all the parallel strata basset or shoot
towards the surface, occasioning thereby a diversity of soil ; and as the beds or
layers of rock, £c. contain fossil remains, we may expect to meet with shells, corals,,
bones, plants, trees, &c. on or near the surface. All these rocks ranged in beds or
layers, whether perfectly horizontal or shooting up at any angle, are called strati-
fied; while abrupt masses of granite,* having none of this masonic appearance, are
said to be unstratified. It is obvious, from what has been observed above, that the
stratified parts of the globe are those in which we must look for capillary veins
and sheets of water.

In the diagram before us all the strata are distinctly marked with their various
dislocations and fissures. The river Derwent is supposed to flow over a vast fissure,
R ; the letters A, A, A indicate lesser fissures ; G, G, G do the same, and all these
fissures are hi the limestone strata. Hence it appears that the toadstone or lava

* Granite consists of distinct aggregations of quartz, felspar, mica, and hornblende, each in a
crystalline form. Felspar is of a whitish, sometimes of a reddish colour, quite opaque, aad occasion-
ally crystallized in a rhomboidal form ; quartz is less abundant, somewhat transparent, and of a
glassy appearance ; mica is dispersed throughout in small glistening plates, the colour is dark and
the appearance metallic ; hornblende imparts a deep green colour to rocks called greenstone and
basalt.

460 NOTES.

strata are attended with many peculiar circumstances, very different from their
associates, 3, 5, 7, 9. These peculiarities are :

1. Toadstone is similar to Iceland Lava both in its appearance and chemical
qualities. 2. It is extremely variable in thickness. 3. It is not universal. 4. It
has no fissures corresponding to those in limestone. 5. It frequently fills up the
fissures in the stratum underneath it, as at H, and the bottom of the shaft s, which
enters a fissure of toadstone that in a liquid state has flowed into the limestone
stratum, numbered 9. Throughout the limestone strata of Derbyshire the fissures
we have marked correspond ; and in these fissures, and between their lamina, the
minerals are found. The mines in the fissures are called rdke-wor ks ; the mines in
the laminae are called pipe-works. Thus in the stratum No. 3, we find Yatestoop
mine ; the Portaway and Placket mines are in No. 5. ; in No. 7. we have Mosey-
meer, and in No. 9 Gorseydale mines ; Hangworm mine is on Bonsai Moor. The
stratum No. 5 bassets and forms the surface of the earth at Foolow and Bonsai
Moor. No. 8 bassets and becomes the base of the land called Grange Mill. No.
3 again bassets and becomes the districts Trogues Pasture to the right, and
Wensley to the left of the great shaft sunk at o and trending below ground to the
fissure G in No. 5. Here we have a beautiful illustration of the genius of geological
engineering. A spring occurs at I in the fissure G, No. 3, too powerful to be
overcome, or too expensive to be kept under ; accordingly a shaft is sunk at o
higher up the acclivity. The miners pioneer to a, descend to the fissure G by
driving a gallery or gate, as they term this tunnel, and this is a common practice,
and never fails in producing dry work in the stratum No. 5, for the close texture
of the toadstone will not allow the water in the seam between 3 and 4 to percolate
its impervious mass, although the pool may accumulate from 10 to 15 fathoms in
No. 3. If the water in 3 rise not to the horizontal level L L, it can never incom-
mode the shaft o a. The grand geological fact elicited here is, as regards capillary
attraction, that toadstone turns water, is free from fissures, nay, sometimes fills
up fissures, as at s and H, which the miners call troughing. In the Slack and
Salterivay mines on Bonsai Moor, some forty years ago, these cros.-ralte fissures
were noticed by Mr. Whitehurst. Their occurrence in other mines need not
astonish geological engineers.

In other districts in Britain, we find that the coal formations sometimes repeat,
in precisely the same order, arid in nearly the same thickness, the following earths
and minerals : sandstone, bituminous shale, slate clay, clay iron, stone, coal ; or
the coal is covered with slate, trap, or limestone, or rests upon these rocks. The
strata generally follows every irregularity of the fundamental rock on which they
rest j but in some instances their directions appear independent, both of the surface
of the rock, and of the cavity or hollow in which they are contained, and in general
take a waved outline, seldom rising greatly above the level of the sea.

We have now, however, merely represented the general arrangement of the
strata ; not all the particular circumstances accompanying them, with respect to
their several fractures, dislocations, &c. ; but it will enable us to reason upon the
chemical effects of water upon limestone and gypsum rocks, where we meet with
caverns, caves, and extensive fissures, that reach sometimes to the surface, some-
times dip to a greater or less distance, and afford channels for great springs and
subterranean rivers. These caves in the gypsum and chalk formations vary in
magnitude from a few yards to many fathoms in extent, forming upon the surface
of the ground, when their superincumbent roofs give way, those funnel-shaped

NOTES. 461

h6llows of such frequent occurrence in gypsum districts. The limestone strata,
besides being " loaded with the exuviae of innumerable generations of organic
beings," says Dr. Buckland, " afford strong proofs of the lapse of long periods of
time, wherein the animals from which they have been derived, lived, and multiplied
and died, at the bottom of seas which once occupied the site of our present con-
tinents and islands." * With how much reason then may we not suppose those
formations to have held large beds of rock salt, which the percolation of water, in
the lapse of ages, removed, and left the chambers empty, or the receptacles of
meteoric water. The percolation of water through felspar rocks, must of necessity
wash away the alkaline ingredient, which combining with iron will form hydrate,
or by its decomposition oxidate the metallic substance. Hence result chalybeate,
acidulous, sulphureous, and saline springs, all the result of capillary attraction in
the strata of the earth, and the disintegration by water of the various ingredients
which the universal solvent holds in a state of fluidity.

Supposing these cavities, to which we have just referred, to have been freed from
their original salt deposits, by water percolating the fissures leading to and from
the masses of salt, we trace the operation of salt springs. For in all cases in which
water holds any mineral in solution, it acts by combination, but where it simply
destroys the mineral aggregation, the mineral falls into small pieces with an audible
noise, as is observed in bole ; or it falls without noise into small pieces, which are
soon diffused through the fluid, without either dissolving in it or becoming plastic,
as in Fuller's earth, and some minerals, as unctuous clay ; it renders plastic other
minerals, absorbs water in greater or less quantity, by which their transparency,
and also their colour, are changed.

The toadstone, which intersects mineral veins, totally cuts off all communication
between the upper and lower fissures, and by the closeness of its texture permits
not the water in the clay strata, or way-boards, to filtrate. Hence toadstone is
said to be capable of turning water, as we have shown in the shaft and gallery,
o a G G. Sandstone strata, of an open porous texture, becomes a great feeder
of water. Several of the sandstones are, however, impervious to water, and
almost all the beds of light-coloured argillaceous schistus, or fine clays, are
particularly so, being very close in their texture. But the percolation of water
at the beds or partings of two strata is an occurrence so general, that our
wonder ceases when examining parts of the country where the strata basset or
shoot to the surface in an acute angle, to find the alluvial covering in places
swampy, marshy, and overrun with puddles, springs, and all that species of soil,
which, being damp and cold, subjects its inhabitants to rheumatism, agues, and a
train of diseases, unknown in regions that are not incumbent on the extremities of
way-boards and capillary strata. The source or feeder of these subterranean
capillaries receiving a constant supply, keeps up the train of human ills from one
generation to another, while local interests or associations bind the natives to their
hereditary doom.

Capillary attraction and cohesion, besides expounding the phenomena of fluid
ascent in strata of earth, direct us in penetrating those troublesome quicksands and
beds of mud, which in the winnings of collieries are met with in mining, and where

* Dr. Buckland's Bridgewater Treatise, pp. 112-116, 1st ed. vol. i.

462 NOTES.

cast-iron tubbing is employed to support the sand or mud bed, and carry the water
down to the bottom of the pit.

Water stands higher in narrow than in wide glass tubes, but quicksilver mounts
higher if the inside of the tube be lined with bees-wax or tallow. We can easily
conceive that the lateral action may yet cause the perpendicular ascent j for it is
a fundamental property in fluids, that any force impressed in one direction may be
propagated equally in every other direction. Hence the affinity of the fluid to the
internal surface producing the vertical ascent. A drop of water let fall on a clean
plate of glass spreads over the whole surface, in as far as there is liquid to cover
the glass, the remoter particles extending the film, yet adhering with the closest
union. The adhesiveness of fluids is still more clearly shown in their projection
through the pores of minerals, plants, animals, gravel, earth, and sand. Water
rises through successive strata of gravel, coarse sand, fine sand, loam, and even
clay : and hence, on the sea-coast, those quicksands, which have engulfed armies
and ships, the pressure and elevation of the ocean at flood tide sending its
advanced column up in the sand to a level with its surface far out at sea. Gravel
divided into spaces of the hundredth part of an inch, will allow water to ascend
above four inches ; it would mount up through a bed of sixteen inches of this
material, supposing sea gravel to be the 500th part of an inch. Fine sand, in
which the interstices are the 2,500th part of an inch, allow the humidity to
ascend seven feet through a new stratum; and if the pores of the loam were
only the 10,000th part of an inch, it would gain the further height of 25f feet
through the soft mass ; thence originate v astce syrtes. The clay would retain the
moisture at a greater altitude ; but the extreme subdivisions of the clay, which
enable it to carry water to almost any elevation, yet make it the most efficient
material in puddling or choking up the interstices of masonry.

The ascent of water in a glass tube is due chiefly, we think, to the excess of the
attractive power of the glass above the cohesive power of the fluid mass over itself.
Were the attractive and cohesive forces equal, the fluid would remain balanced at
a common level. Mercury hence sinks, by reason of the strong cohesive power of
its own particles. Hence we account for mercury closing over a ball of crude
platinum, which nevertheless, being gently laid on the mercury, will float, although
its specific gravity is above that of mercury.

It is however the province of chemistry, rather than of mechanics, to measure
the cohesive power possessed by different fluids, or by the same fluid under different
degrees of temperature.

The suspension of water in any stratum through which it can percolate, must
depend entirely upon the smallness of the upper orifice, or superficial extent of
the deflection with which the stratum slopes off horizontally above ground, and
upon the relative elevation of the extremities of the impervious stratum. Thus,
suppose a and b to be two extremities of a stratum pervious to water ; the central
column of water at c is pressed with the whole weight of the space be, and this
pressure upon c a pushes the fluid out at a
by the excess of force in be above that in
«a; and therefore, while the ground or
land at b is generally dry, that at a is per-
haps boggy; at all events it will exhibit
springs at its surface, be cold, damp, and
its inhabitants subject to rheumatism or

NOTES. 463

agues. A column of water of this description may occupy a space of many miles
extent between b and a ; and c may be many hundred feet deep below the hori-
zontal level of a. In digging for water at d, we should find it at e.

The cohesion of the particles of water, and its extreme facility to obey any
impression, fit it admirably for percolating through fissures of the earth, when in
the tenderest filaments it is detached from the general fluid mass, and penetrates
only by the laws of capillary attraction from one point to another in an extensive
stratum of clay, precisely as if it flowed through a pipe in passing from one hill to
another. Hence the certainty with which we meet with water in boring to a
proper depth in the earth, and hence also the origin of Artesian wells, which finely
expound the varied phenomena of a retreating and subsiding column towards the
body of the fluid, as if an equal and opposite pressure from the sides of a capillary
tube had come into action. We may hence infer, that in strata pervious to water,
the capillary ascension, however much it may be accelerated or retarded by the
parallel sides of the stratum and the material of which it is composed, is governed
by these three principles which we have fully discussed, pressure from above,
cohesion subsisting among the particles of the liquid, and attraction of the parallel
sides of the stratum. Were this attraction equal to the antagonist cohesion, the
fluid would remain at rest, balanced at a common level, till overcome by the weight
of the contents in the longer branch of the fluid column forcing the contents of the
shorter column out at the discharging orifice. All the springs which are below the
London clay, at the depth of 150, 200, 250, or 300 feet, are fed by sources con-
siderably elevated above the Hampstead level. With what ease then might the
metropolis be provided in every street with spring water from an Artesian Well !

Any of our readers who may be desirous of acquiring a practical and thorough
knowledge of geology, must chiefly prosecute his studies by laborious researches in
the great field of nature, and must there explore for himself the various phenomena
presented to his view. His first step must be to understand by reading the leading
facts and principles of the science ; he must learn to recognise at once the principal
simple minerals, entering into the composition of rocks, and also the various
metallic ores and other minerals which usually occur in veins. He must likewise
be acquainted, the more minutely the better, with at least the more common forms
of fossil organization, and with the general mode of their distribution in the rocky
masses constituting the crust of the globe. Some preliminary knowledge of
chemistry, although not perhaps essential, will form a very desirable addition to
the qualifications already named. Thus provided with the knowledge requisite
to decipher the instructive pages on which nature has recorded, in her own
language, the history and revolutions of our planet, the student may now com-
mence the most valuable, but far the most laborious part of his career. He must
visit the deep recesses of our mines, which, although too much neglected, afford
the finest examples of many of the most important facts on which the science of
geology is built. He must observe the strata as laid open in our quarries, and as
displayed in the deep cuttings of our roads, railways, and canals. Every excavation
will indeed present something worthy of notice to Ms view; but not contented
with observing merely these spots, where the labour of man has penetrated into
the interior of the earth, he must wander around the base of the lofty cliffs which
overhang the ocean, and observe the grand and instructive sections which nature
herself presents, and of which our own islands afford such numerous and admirable
examples. He must pursue the course of rivers into the interior, and observe the

464 NOTES.

strata laid open by the excavations of their currents; but his most instructive
studies will be found, when he has arrived far inland at the mountains, where they
take their rise. Here he will find that nature has revealed the structure of the
globe on the grandest scale ; here the marks of ancient revolutions will be found
imprinted in characters not to be mistaken, and the truth both of facts and
theories, before known only by description, will at once be impressed on his mind.
By researches of this kind, extended over considerable tracts of countries, so as to
embrace all the great series of geological formations, and by careful study and
comparison of all the phenomena presented to his view, both as regards the
mineral structure of the globe, the forms of organized bodies peculiar to each
species of rocks, and the physical changes now taking place on the earth's surface,
the student wifl at length become a practical geologist, and be enabled by his own
observations to improve and advance the science he has been studying. But the
course which has here been pointed out, although essential to a practical and
thorough knowledge of the subject, can only be pursued by few; and a general idea
of its most important facts, and the practical consequences arising from them, is
of comparatively easy attainment. The great principles of geology have been most
ably brought together in various publications ; and where only a general knowledge
is required, geological maps and sections may be made in some measure to supply
the place of travelling and observation. A few words then on these important
documents, which are the medium of expressing some of the most important
practical results of the labours of geologists in the field, may not be misplaced. A
map which combines with the geographical and physical features of a country,
a view of its internal structure, supposes all wood and vegetation to be absent,
and that every species of superficial soil and covering removed, so that the
actual rocks and strata which compose the solid crust of the globe beneath shall be
perfectly exposed and laid open to our view. The space occupied at the surface by
these rocks and strata is then distinctly shown by different tints of colour, in the
same manner as territorial divisions are indicated on ordinary geographical maps.
But although we thus obtain a perfect view of the surface distribution of the solid
materials of the globe, it is evidently essential to know in what manner they are
arranged below, and what relations they bear to each other in the internal parts of
the elobe. This object is accomplished by means of geological sections, the nature
of which will acquire but little explanation. A geological section supposes, that on
any given line the internal structure of the earth is laid open in the direction of a
vertical plane, as in our section between Barley Moor and Grange Mill in Derby-
shire. It therefore merely represents, although generally on a much more
extended plane, the same thing which we see in many artificial excavations, and
which nature herself exhibits to our view in cliffs and precipices. Geological
sections are indeed merely a combination of sections of this kind, in which they
bear the same relation as the map of a large country would do, to the smaller
plans and sketches from which it was compiled, especially connected with the
Mechanics of Fluids. Such a map is that of Messrs. J. and C. WALKER.

It appears that the present annual value of the mineral produce of Great
Britain, may be estimated at somewhere about 20,000,000?. sterling; independent
of any subsequent process of manufacture, and not including the cost of carriage
on coal. — Burr's " STUDY OF GEOLOGY," London, 1836.

A TABLE

THE SPECIFIC GRAVITIES OF DIFFERENT BODIES.

In consulting this Table of Specific Gravities, it must be borne in mind that
water is taken as the unit of measure for solids and liquids ; and atmospheric air
as the unit of measure for the different gases. Water at the common temperature
is 1,000, and mercury 13.568 ; whence we conclude that mercury is 13£ times
heavier than water. We mean that a cubic foot of water weighs 1000 ounces ;
therefore a foot of mercury weighs 1 3,568 ounces, and a cubic foot of bar iron
7788 ounces; a cubic foot of vermilion 4230, of Portland stone 2496, of indigo
0,769, and of cork 0,240 ounces.

METALS.

Antimony, crude ' ^ .. .- .,. 4064

glass of . ,-.:-. 4946

molten . . . 6702

Arsenic, glass of, natural . . 3594

molten vi>?;i;i, • • 5763

native orpiment . . 5452

Bismuth, molten ... 9823

native . . . 9020

ore of, in plumes . 437 1

Brass, cast, not hammered . 8396

ditto, wire-drawn . . 8544

cast, common . . 7824

Cobalt, molten . . . 7812

blue, glass of ,.,^ ,' . 2441

Copper, not hammered . . 7788

the same wire-drawn . 8878

ore of soft copper, or

natural verdigris &ggjy. • 3572
Gold, pure, of 24 carats, melted,

but not hammered . . 19258

the same hammered . . 19362

Parisian Standard, 22 car.

not hammered . 17486
VOL. I.

Gold, hammered . ,,/••••'•;./•!• 17589

guineaofGeo.il. . . 17150

guinea of Geo. III. . .17629

Spanish gold coin . . 17655

Holland ducats . . 19352

trinket standard, 20 carats

not hammered . . 15709

the same hammered . . 15775

Iron, cast ;[,';•,!;:••,• • 7207

cast at Carron . „ ' , . 7248

ditto at Rotherham . .7157

bar, either hardened or not 7788

Steel, neither tempered nor hard-
ened .... 7833

hardened,butnot tempered 7840

tempered and hardened . 7818

. ditto, not hardened . . 7816

Iron, ore prismatic . . . 7355

ditto specular . . . 5218

ditto, lenticular . . .5012

Lead, molten .... 11352

ore of, cubic . . . 7587

ditto horned . . . 6072

ore of black lead v . 6745

2 H

466 A TABLE OF THE SPECIFIC GRAVITIES OF D1FFEUENT BODIES.

Lead, ore of white lead

ditto ditto vitreous .

ditto red lead .

ditto saturnite .

Manganese striated . •» ,.
Molybdena • •*'> v „..
Mercury, solid or congealed

fluent . .!•

natural calyx of .

precipitate, per se

precipitate, red .

brown cinabar .

red cinabar - • ,

Nickel, molten . • . -••-•* ,•>

ore of, called Kupfer-

nickel of Saxe

Kupfer-nickel of Bohemia 6007

Platina, crude, in grains .
purified, not ham-
mered

ditto hammered .

•• ditto wire-drawn .

ditto rolled .

Silver, virgin, 12 deniers, fine,

not hammered

ditto, hammered .

Paris standard

shilling of Geo. II. .

shilling of Geo. III.

French coin .

Tin, pure Cornish, melted, and
not hardened .

the same hardened .

of Malacca, not hardened .

the same hardened .

ore of, red

ore of, black .

- ore of, white . • " *. '• ;: ' : r i '
Tungsten ; . • .. V . ij •"",-
Uranium ....
Wolfram ....
Zinc, molten ....

PRECIOUS STONES.

Beryl, or aqua-marine oriental .

ditto, occidental

Chrysolite of the jewellers

4059

Chrysolite of Brazil .

2692

6558

Crystal, pure rock of Madagascar

2653

6027

of Brazil . ...'<*/

2653

5925

European

2655

4756

rose-coloured . ,

2670

4738

yellow ....

2654

15632

violet, or amethyst ,-, *

2654

13568

white amethyst

2651

9230

Carthaginian - . . • v

2657

10871

hlirlr

Qflf>4

1UO 1 I

8399

Diamond, white oriental .

£\JiJi±

3521

10218

rose-coloured oriental

3531

6902

orange ditto

3550

7807

, , 1-4.4.

3524

blue ditto . • * v **»*'

3525

6648

Brazilian .

3444

L 6607

-^,.11 nrr.

3519

yeiiow ...

15602

Emerald of Peru

2775

Garnet of Bohemia . >• «r

4189

19500

of Syria

4000

20337

dodecaedral .

4063

21042

volcanic, 24 faces .

2468

22069

Girasol

4000

Hyacinth, common . . "•'."

3687

10474

Jargon of Ceylon

4416

10511

Quartz, crystallized .

2655

10175

in the mass . ^"v^W

2647

10000

• brown, crystallized

2647

10534

^Fftflfflfe

2640

10408

milky ....

2652

fat, or greasy

2646

7291

Ruby, oriental ....

4283

7299

fivtivhAH

3760

7296

Ballas ....

3646

7307

Brazilian ....

3531

6935

Sapphire, oriental

3994

6901

sJi4-4-jt v'liifn

3991

6008

of Puys

4077

6066

Brazilian .

3131

6440

Spar, white sparkling

2595

7119

red ditto ....

2438

7191

green ditto

2704

blue sparkling .

2693

green and white ditto

3105

transparent ditto

2564

adamantine

3873

3549

Topaz, oriental ....

4011

2723

pistachio ditto .

4061

2782

Brazilian l;- .

3536

A TABLE OF THE SPECIFIC GRAVITIES OF DIFFERENT BODIES. 467

Topaz of Saxe . . x .
white ditto

3564
3554
4230

2590
2638
2625
2607
2667
2632
2616
2664
2606
2665
2587
2615
2630
2612
2623
2623
2591
2613
2594
2582
2612
2565
2950
2966
2983
2359
2681
2661
2691
2710
2711
2735
2696
2816
2750
2628
2114
2684
2664
2654
2609
2612

Pebble, stained ....
Prasium .....

Sardonyx, pure . . .

speckled .
veined . .

blackish .
Schorl, black prism, hexaedral .
octaedral
tourmalin of Ceylon

2587
2581
2603
2606
2622
2595
2595
2628
3364
3226
3054
2923
3156
3286
2416
2111
2143
2484

&c.

2730
2762
2699
2744
2691
2693
2699
2713
2638
2876
1078
926
909
2313
2578
3073
2864
1104
2000
2790
2727
2784
2168
2306
2274
2311
2312

SILICIOUS STONES.

Agate, oriental . •>•(:& . • .. ..._.„

cloudy . •, ;.. ifutt •-.*>},
speckled
veined . . %>»;<.;•',:-

Chalcedony, common »:.-.* ,'>t;*'->

Stone, paving . . . •>« tfai
cutlers' . :f?',;HfiJifJ-jr~— *— •
grind . . . . ~»~
_ -mm

veined . . H-JK'
reddish .

rj hliifch

onyx <^d»f<<K3vf •»;?$
Cornelian, pale . .• • * ;j .y Tt'i n • < : • : • • •
speckled . jt;»*i*nt<>»
- veined , ^ •.,.>;•

stalactite . ft&dftt/l

VARIOUS STONES, EARTHS,

Alabaster, oriental white . id •-*-
do. semi-transparent .

Flint, white . ? -.. •->... ' awittr. -
black . ..'.-ray Ira*;'
veined . Ltf ttvf'J lu jt >A <
Egyptian . i i^i'l .huMlfo
Jade, white . *>O Mfiiit^iiltfb
green . . U-*iteV.
olive . . -tifM -,t$
Jasper, clear green . * j .*&
brownish green . ••< .-us

brown . . . V> '*•>

stained brown . -V v*
veined . ) "V-^'w •$•-
— of Piedmont . -3 -.—
of Malta . o&9o8-tf-

• of Valencia *I*tfoIv~r-
• of Malaga . . •>! -.*--
Amber, yellow transparent '*.-.*•
Ambergris . . ,/T^ -.—
Amianthus, long . . sstfiv * -.—

Asbestos, ripe ....
starry.
Basaltes from Giants' Causeway
Bitumen of Judea
Brick .. ., 'astral..
Chalk, Spanish . . .-•'.'••««»

violet ....
cloudy ....
• - veined . . ..'•-.*&
onyx . . . ^hJ«
red and yellow

Opal

British ....
Gypsum, opaque . . - .
— semi-transparent

Pearl, virgin oriental
Pebble, onyx ....

English .. '.:«*a!*

rhomboidal

468 A TABLE OF THE SPECIFIC GRAVITIES OF DIFFERENT BODIES.

Gypsum, cuneiform crystallized
Glass, green . .

2306
2642
2892
2733
3189
3329
2654
2876
2894
3054
4360
5000
2500
2270
3179
3156
3182
2742
2724
2717
2838
2726
2695
2650
2700
2710
2705
2678
2755
2708
2858
2728
2718
2668
2714
2649
2348
1329
1714
2146
2341
2385
2765
2676
2793
2754
2728
4954
3900

Pyrites, ferruginous, round
ditto, of St. Domingo .
Serpentine, opaque, green Italian
ditto, veined black and
olive

4101
3440
2430

2594
2627
2586
3000
2669
2672
2854
2186
2766
2324
2478
915
2722
2415
2945
2771
2520
2510
2049
2496
2470
1981
2460
2122
2357
2274
2378
2034
2201
2033
1991
2792
2089
2246
2655
2900
2704

1841
2125
1271
1580

bottle ....
. Leith crystal . . --*-

- fluid

Granite, red Egyptian . > ~v
Hone, white razor .- 'A I1. •« i - -r
Lapis nephriticus
.— Ti7ii1i

ditto, fibrous
ditto, from Dauphiny
Slate, common ....

Judaicus ....
Manati . ."rr,»Vf-t-- ;.-> i-
Limestone •;*••- * • ~W&fj:;'t
white fluor . VsuJu*

Marble, green, Campanian Ife
red . . > . .:'•'.
white Carrara . i*

flesh polished .- Iv^-.-
Stalactite, transparent
opaque . . - •-; —
Stone, pumice . fcOBtoy -»rS
prismatic basaltes . - — .—

~ tOUCh . . i"--54 -«-;—

Siberian blue . £$»**'• -«nH
oriental ditto . -sin * ,s-s*>-

- Bristol

- Portland ff/Ji** - -»—»

Castilian . i-ralfcv- — .--
Valencian . infctfa — *~
white Grenadan . -r-dfr
Siennien . i5I io — *-
Roman violet & '!•.•• — » -
African . . — *-

rotten . . . '•''., i a _
hard paving
— rockofChatillon .
clicard, from Brachet
ditto, from Ouchain v.r"

Norwegian . !£!?<> -nr-

St. Maur . . . 9?&*

St Cloud : "5 VEV"> 5*

green Egyptian .
Switzerland .
French ....
Obsidian stone . .. » >•
Peat, hard . . rvm^' —
Phosphorus . ...» • .
Porcelaine, Sevres . :*!s-i 'i- » ew

Sulphur, native ....
molten
Talc, of Muscovy . :'. rfi*-
black crayon
ditto German . **«lbi7: *-
yellow . • • -'-. •
black ....
white

LIQUORS, OILS, &c.

Acid, sulphuric ....
ditto, highly concentrated
nitric ....
ditto, highly concentrated .

Porphyry, red ....
green
.- red, from Dauphiny .
.- red, from Cordoue
- - green, from ditto
Pyrites, coppery . -
ferruginous cubic .

A TABLE OF THE SPECIFIC GRAVITIES OF DIFFERENT BODIES. 469

Ac-id, muriatic .

. 1104

Oil, of walnuts ....

0923

red, acetous . jih •>;.'*

. 1025

of whale ....

0923

white acetous .

. 1014

of hernpseed

0926

distilled ditto .

. 1010

of poppies .

0924

fluoric

. 1500

• rapeseed

0919

acetic . .

. 1063

Spirit of wine. See Alcohol.

0837

phosphoric . . '-. »>

. 1558

Turpentine, liquid . . • — .-

0991

formic . . i . ;

. 0994

Urine, human .•

1011

Alcohol, commercial .

. 0837

nooq

Water, rain c

Atr.-i.ivi^A

1000

T Ai"lA

mixed with water

uo^y

• sea (average) . . ;

lUUU

1026

15-16ths alcohol

. 0853

of Dead sea .

1240

14-1 6ths ditto

. 0867

Wine, Burgundy . . fi:u;';j

0992

13-16ths ditto

. 0882

Bourdeaux . . .

0994

12-16ths ditto

. 0895

Madeira ....

1038

ll-16ths ditto

. 0908

Port . . . - \t

0997

10-lGths ditto

. 0920

.T*A«ftAW**

1033

9-16ths ditto

. 0932

8-16ths ditto

. 0943

7-16ths ditto

. 0952

RESINS, GUMS, AND ANIMAL

6-16ths ditto

. 0960

SUBSTANCES, &c.

5-16ths ditto

. 0967

4-16ths ditto

. 0973

Aloes, socotrine . . ^-Hji'-I

1380

3-l6ths ditto

. 0979

hepatic . . .

1359

2-16ths ditto

0985

Asafcetida . i.

1328

l-16th ditto

. 0992

Bees-wax, yellow

0965

Ammoniac, liquid . < o >;

. 0897

white

0969

Beer, pale . .

. 1023

Bone of an ox . . . '.

1656

_ — - brown • • •

. 1034

Butter

0942

Cider .,,,..-.

. 1018

Calculus humanus

1700

Ether, sulphuric

. 0739

flQftQ

ditto ....

l)<|44>.n

1240

T Af\A

muriatic . ,,« ;r'5 «„ Li

uyuy
. 0730

Camphor .....

14d4

0989

acetic

. 0866

Copal, opaque . . . .'. »*i

1149

Milk, woman's . .

. 1020

Madagascar .

1060

cow's

. 1032

Chinese . . . •.;....

1063

ass's ' .r.. ,' V , ( • ,,

. 1036

Crassamentum, human blood

1126

ewe's ....

. 1041

Dragon's blood ....

1205

goat's .. ".". ' . ". '

. 1035

Elemi

1018

mare's . . . .!&

. 1034

Fat, beef

0923

cow's clarified . fc:<

. 1019

hog's wj--.

0937

Oil, essential, of turpentine

. 0870

mutton . . . . ..

0924

psscTiti3/l of* IsvcndGr

0894

0934

, , (lit to of cloves

1036

1212

j - - - (Jit/to of cin.n£Linoii

1044

1222

of olives

. 0915

Gum, ammoniac . . , . ^

1207

of sweet almonds

. 0917

Arabic ....

1452

of filberts . ' ~. V7^

. 0916

Euphorbia

1124

linseed

. 0940

seraphic . -

1201

470 A TABLE OF THE SPECIFIC GRAVITIES OF DIFFERENT BODIES.

Gum, tragacanth

. 1316

Cedar, Indian . . . ':&.'*

. 1315

bdellium .

. 1372

American i^^fAt^

. 0561

Scammony of Smyrna

. 1274

Citron . . ;.---• -

. 0726

ditto of Aleppo w0 , , j,

. 1235

Cocoa-wood

. 1040

Gunpowder, shaken . fi&&

. 0932

Cherry-tree . . . .**

. . 0715

in i lnn~p TIPIH

. 0836

Cork . . - . - «« , Srt

. 0240

nstlS'l

. 1745

Cypress, Spanish - . *. -.1 ->**«r

. 0644

Honey ,(,_,., r. , nstiu

. 1450

Ebony, American . i H«

. 1331

Indigo . . . . in

. 0769

Indian . ^:|i|;n.-«rs»r"««

. 1209

Ivory . .,. . ... !i::»>

. 1826

Elder-tree . . . ••• '. ; -

. 0695

Juice of liquorice ' *.:"•• v;i.) i

. 1723

Elm, trunk of . • iJ «w hv.*»

. 0671

— — of Acacia . -...•- 't »i'f

. 1515

Filbert-tree .t-«U «Uti; l-«.I.

. 0600

Labdanum ,;+. . ..'^fcao*;1

. 1186

Fir, male . . >-.;'.M-t-».

. 0550

Lard , . . .', s ; h<>.

0948

female

0498

Mastic . . . i-iKj

1074

Hazel . -' :.

0600

Myrrh . . . .,

. 1360

Jasmine, Spanish . ' -1 '..

. 0770

Opium . . . -fftfii

. 1336

Juniper-tree . . •# ' .

. 0556

Scammony. See Gum.

Lemon-tree . • »«(i<JI^! .

. 0703

Serum of human blood

. 1030

Lignum-vitae . . -^ .

. 1333

Spermaceti . . ) <i >

. 0943

Linden-tree . . • - V .

. 0604

Storax ....

. 1110

Logwood. See Campechy.

. 0913

Tallow . . . *

0942

Mastick-tree . . -^ .

. 0849

Terra Japonica . .... a« »(??••.

. 1398

Mahogany iMti& «&£ i -f- .

. 1063

Tragacanth. See Gum. .

. 1316

Maple . '-'.U -')•:*«'' .

. 0750

Wax. See Bees-wax.

. 0965

Medlar . *«*$ if . T--v .

0944

shoemakers' . >ria '.

. 0897

Mulberry, Spanish . -I .

. 0897

Oak, heart of, 60 years old

. 1170

Olive-tree ... . .

. 0927

WOODS.

Orange-tree . • . • *;

. 0705

Pear-tree ..<.*• ., . .'.-.

. 0661

Alder ....,; ... ^.,- ^

. 0800

Pomegranate-tree . -•»*';;•:•

. 1354

Apple-tree ..,.,.. <-v^

. 0793

Poplar . -:A . • . .a'

. 0383

Ash, the trunk . » . . .

. 0845

white Spanish '. - H

. 0529

Bay-tree . ,„ , -..•*, '<'*;•**

. 0822

Plum-tree . .- . v. •##

. 0785

Beech . .

. 0852

Quince-tree . . - s'iiu

. 0705

Box, French . . .

. 0912

0482

Dutch . ;f4:.-i--

••Htfj. 1328

Vine ....

. 1327

Brazilian red . ^ V;

**•#• 1031

Walnut . » , . • . *

. 0671

Campechy-wood

. 0913

waiow . . ... ;•¥

. 0585

Cedar, wild . . ..,," . •

: •*.;• 0596

Yew, Dutch

. 0788

Palestine , . .. . . .

- -,y; 0613

Spanish

. 0807

471

WEIGHT AND SPECIFIC GRAVITY OF DIFFERENT
GASES.

Fahrenheit's Thermom. 55° Barometer 30 inches.

Spec. Gray. Wt. Cub. Foot.

Atmospheric air . 1.2 . 525.0 grs.

Hydrogen gas . 0.1 43.75

Oxygen gas . ; 1.435 . 627.812

Azotic gas . . 1.182 . 517.125

Nitrous gas . v<fj 1.4544 . 636.333

Ammoniac gas. ""*.-.. .7311 . 319.832

Sulphureous acid gas . 2.7611 . 1207.978

In this table the weights and specific gravities of the principal gases are given,
as they correspond to a state of the barometer and thermometer which may be
chosen for a medium. The specific gravity of any one gas to that of another will
not exactly conform to the same ratio under different degrees of heat and other
pressures of the atmosphere.

And if common air, the standard, be taken at unity (1); chlorine oxy muriatic
acid will be 2.500 ; and hydrogen 0.069 ; whence we conclude that chlorine is 2£
times heavier than hydrogen, and this last is 14 times lighter than common air.
For, to arrive at the absolute weight of the gases, we have only to assume 100 cubic
inches of atmospheric air to weigh 30.5 grains, and as there are 1728 cubical inches
in a cubic foot, the simple proportion

100 : 30.5 grains : : 1728 : 527.04 grains,
the weight of a cubic foot of common air.

And for any other gas, it is only necessary to observe its specific gravity in
relation to that of common ah-; for example, chlorine has a specific gravity of 2.5;
hence a cubic foot of chlorine will weigh 2£ times as much as a cubic foot of
common air ; for

527 .04 x2£ =1317.6 grains,
the weight of a cubic foot of chlorine.

To determine the weight of any gas lighter than common air, we also compare
their specific gravities : thus, the specific gravity of ammoniacal gas is 0.5, and
that of atmospheric air being =1, we have 1 : 0.5 : : 1728 : 864.0, or simply
1728-J-2=864 grains, for the weight of a cubic foot of ammoniacal gas; and so on
for all the other gaseous bodies, as they are arranged in the following table.

472

WEIGHT AND SPECIFIC GRAVITY OF DIFFERENT GASES.

Tf atmospheric air be taken at unity (1), then the various gases will stand as
under : —

Atmospheric air . .
Ammoniacal gas
Carbonic acid .
Carbonic oxide .
Carburetted hydrogen
Chlorine .
Chlorecarbonous acid
Chloroprussic acid ,.
Cyanogon
Euchlorine
Fluoboric acid .
Fluosilicie acid .
Hydriodie acid .

1.000
0.500
1.527
0.972
0.972
2.500
3.472
2.152
1.805
2.440
2.371
3.632
4.346

Hydrogen .... 0-69

Muriatic acid .... 1.284

Nitric oxide .... 1.041

Nitrogen 0.972

Nitrous acid .... 2.638

Nitrous oxide .... 1 .527

Oxygen 1.111

Phosphuretted hydrogen . . 0.902

Prussic acid .... 0.937

Subcarburetted hydrogen . . 0.555

Subphosphuretted ditto . . 0.972
Sulphuretted ditto . . .1.180

Sulphureous acid ... . 2.222

CONCLUSION.

THE reader will have seen in this volume how the road to abstract
science may be smoothed ; but he may rest assured that any popular
version of Hydrostatics is quite illusory, for no portion of sound know-
ledge was ever acquired without some corresponding exertion of mind.
It is one of the improvements to be made in our systems of education
for the various professions, and in books written to retrieve the de-
clining taste for science, that students in Mechanics should devote
themselves methodically to the profitable but toilsome drudgery of
computation ; and, in their geometrical constructions, be as clever
with their hands as ingenious with their heads. Science and know-
ledge are subject, in their extension and increase, to this law of
progression : the further we advance, instead of anticipating the ex-
haustion of their treasures, the larger the field becomes — the greater
power it bestows upon its cultivators to add new measures to its
rapidly-expanding dominions. It is the science of calculation which
has grasped the mighty masses of the universe, and reduced their
wanderings to fixed laws ; which prepares its fetters to chain the
flood, to bind the ethereal fluid ; and which must ultimately govern
the whole application of Hydrostatics to the Arts of Life.

London : J. Rider, Printer, 14, Bartholomew Close.

0^ CALIFORNIA

NGV 5 1947

One dott«on-Sev

REC'D LD

JUL Ibl

e«h day overdue.

RECEIVED

NOV27'67-1QP

ewe

28Ju1'59AJ

-10Om.l2,'46(A2012sl6)4120

UNIVERSITY OF CALIFORNIA LIBRARY

Internet Archive Audio

Featured

Top

Images

Featured

Top

Software

Featured

Top

Books

Featured

Top

Video

Featured

Top

Mobile Apps

Browser Extensions

Archive-It Subscription

Save Page Now

Full text of "Mechanics of fluids for practical men [microform] : comprising hydrostatics, descriptive and constructive : the whole illustrated by numerous examples and appropriate diagrams"

See other formats