published at Leyden in the year 1634. In this work the Hydrostatic Paradox, the discovery of which is due to Stevin, or the proposition that a fluid of any assigned weight may be so employed as to produce any assigned pressure, either upwards or downwards, is explained and exemplified. In the fifth book of this treatise, Stevin illustrates his conceptions of fluid action by the discussion of the following question, which, as having relation to the fundamental principles of the transmission of hydrostatical pressure, and as serving to exemplify the accuracy of his mechanical ideas of fluids, we will here quote at length.

“ Declarer la raison pourquoy un homme nageant au fond de l'eau, ne meurt pour la grande quantité d'eau, qui est au dessus de luy.

“Soit un homme 20 pieds de profondeur dans l'eau, le pied d'eau pesant 65 lb, et la superfice entière de son corps 10 pieds : cela estant ainsi, 13000 lb presseront contre son corps. Partant on pourroit demander, comment il est possible qu'une personne ne creve d'une si grande charge? A quoy la response sera telle.

“A. Tout pressement qui blesse le corps, pousse quelque partie du corps hors de son lieu naturel.

“0. Ce pressement causé par l'eau, ne pousse aucune partie du corps hors de son lieu naturel. “O. Ce pressement donc causé par l'eau, ne blesse nullement

l

le corps.

“La mineure est manifeste par l'experience, dont la raison est, que s'il y avoit quelque chose qui soit poussée hors de son lieu, il faudroit que

cela rentrast en un autre lieu, mais ce lieu n'est pas dehors, à cause que l'eau presse de tout costé egalement (quant à la partie de dessous, elle est un peu plus pressée que celle de dessus, ce qui n'est d'aucune estime, d'autant que telle différence ne peut pousser aucune partie hors de son lieu naturel,) ce lieu n'est pas aussi dedans le corps, car il n'y a rien de vuide non plus que dehors; d'où il s'ensuit que les parties s'entrepoussent egalement, pource que l'eau a une mesme raison à l'entour du corps. Ce lieu-là donc n'est dehors, ny dedans le corps, et par consequent en nulle part, ce qui fait que nulle partie n'est poussée hors de son lieu, et partant ne blesse nullement le corps.

“Ce que pour declarer plus apertement, soit ABCD (fig. 1) une eau, ayant au fond DC un trou, fermé d'une broche E, sur lequel fond gist un homme F, ayant son dos sur E; ce qu'estant ainsi, l'eau le pressant de tout costé, celle qui est dessus luy ne pousse aucune partie hors de son lieu.

Mais si on veut voir par effect que cecy est la cause véritable ; il ne faut qu'oster la broche E; alors il n'y aura aucun poussement contre son dos en E, comme aux autres lieux de son corps; pourtant aussi son corps patira là une compression, voire aussi forte; assavoir autant que pèse la colomne d'eau, ayant le trou E pour base, et AD hauteur; et ainsi le dessein est demonstré apertement."

1. A rectangular parallelogram is immersed in a fluid; to find the whole pressure upon it.

Let a, b, be the lengths of two unequal sides, and h the depth of its centre of gravity, that is, of its middle point, below the surface of the fluid; then, by COR. (1), the whole pressure upon it will be equal to

gp ab h.

This result shews that, so long as h remains constant, the pressure on the parallelogram will remain the same at whatever inclination it may be placed to the horizontal surface of the fluid.

2. A solid cylinder is immersed in a fluid, the depths of the centres of its circular ends being h, h, its radius r, and its length l; to determine the pressure on its surface, including its extremities.

The areas of its two ends are each equal to are, and, the depth of the centre of gravity of one being h, and of the other h', the pressures on the two ends are

gporrih, gp.mrih'. Again, the area of the convex portion of the cylinder being 2url and the depth of its centre of gravity }(h + h'), the pressure upon it will be

gp. 2trl. ž (h + h').

Hence the whole pressure on the cylinder will be equal to

agpr.{r (h + h') + ?(h + h')}

agpr (r + 1) (h + h'). This result shews that, so long as ž (h + h'), the depth of the centre of gravity of the whole cylinder, remains constant, the whole pressure on the cylinder will be invariable, however the inclination of its axis to the surface of the fluid may be varied.

We have solved this problem by means of Cor. (2). It might have been solved however somewhat more simply by means of Cor. (1). In fact, the whole surface of the cylinder being equal to

2π2 + 2πrl = 2 r (r + 1), and the depth of the centre of gravity of the surface, which is evidently the middle point of the axis, being equal to 1 (h+h'), it follows that the whole pressure on the cylinder must be equal to

agpr(r + l) (h + h').

(η + 3. An isosceles triangle is immersed vertically in a fluid with its vertex coincident with the surface of the fluid and its base horizontal; to determine how it must be divided by a line parallel to the base, that the pressures upon the upper and lower portions may be respectively in the ratio of m:n.

Let h be the altitude of the proposed triangle, and h' that of the triangle cut off by the dividing line. Let c be the length of the base of the proposed triangle ; that of the triangle cut off

K will be equal to c

. h The pressures on these two triangles will be

gp.sh. ich = šgph'c,

h'
and gp.jh'.ch' = $92

ñ

}gp

he The pressure on the lower portion of the proposed triangle will therefore be

3gp.g. (ho – h").

=

hisc

с

or, if the dividing line bisect the perpendicular altitude, the pressure on the lower portion of the triangle will be seven times as great as that on the upper.

4. To find the pressure which a diver sustains when the centre of gravity of the surface of his body is 32 feet under water.

“ The surface of a middle-sized human body is about 10 square feet. Multiply then 32, the depth of the center under water, by 10, the surface of the body, and the product, or 32 times 10 solid feet, will be a magnitude of water whose weight is equivalent to the pressure which the diver sustains. A cubick foot of water has been found by experiment to weigh 1000 averdupois ounces, therefore 32 times 10 feet, or 16 times 20 feet of water, will weigh 16 times 20000 averdupois ounces, or 20000 averdupois pounds. This therefore is the pressure of the water to which a diver at 32 feet depth is exposed.' Cotes : Hydrostatical and Pneumatical Lectures, Lect. 3.

5. To compare the pressures on the upper and lower portions of a hemispherical vessel full of fluid, the axis of the vessel being vertical, and the two portions being separated by a horizontal plane bisecting the axis.

Let O be the inclination of any radius of the bowl to its axis ; then, r representing the radius, the pressure on an annular strip of the bowl included between two consecutive horizontal planes, will be equal to

gp. 27r sin 0.rdo.r cos 0 = mgprisin 2000.

6. A plane surface, bounded by the arc of a parabola and the tangents at its extremities, is immersed in water, the point of intersection of the tangents coinciding with the surface of the fluid, and the area of the parabola being vertical: to determine , the whole pressure on the plane surface.

Let OA, OB, (fig. 2) be the two tangents to the parabola HABK, and let 04, OB, produced indefinitely, be the axes of x, y, respectively. Let a, ß, be the inclinations of OA, OB, respectively, to the horizon.

The area of an elementary parallelogram Pp, the coordinates of P, p, being x, y, and x + dx, y + dy, respectively, will be

dx dy sin (a + b), and the depth of Pp below the surface of the fluid will be equal to

x sin a + y Hence the pressure on the area AOB will be equal to

gp sin (a + B) Ssdx dy (x sin a + y sin ß). Now, if 0 A = a, OB = b, the equation to the parabola will be

sin ß.

hence, integrating with regard to y from y = 0 to y = y, the y

0 in the limit being the ordinate of a point in the curve, and then

: 0 to x = a, we have
Sbx dx dy = fx dx .y

from x =