Chap. XVI. HYDROSTATICS.

will express the superficies of the trapezium AIKD; thus we shall have

,, 2a —26 2a

a a — 66, 66 :: ---, x-- ;

1115

whence we have — x

2 a66—6 s

If we multiply the magnitude a by a a— .66, we shall have -x

aa — 662 a66 -

6 3 + a 3 — abb

aa — 66

a a — 6*a 9 — 66

= x : whence we draw the following general rule.

To have exactly the interval from the Surface of the Water to the centre of Impression of aLock Gate. — Measure the greatest and least height of the water; cube these twoheights; subtract the lesser cube from the greater; take two-thirds of the difference;divide this quantity by the difference of the square of the greatest height of the water fromthat of the least; the quotient will give the answer required.

If the height E F were 6 feet, and the lesser, E L, 4 feet, subtracting 64 (the cube of 4)from 216 (the cube of 6), we shall have 152 for the difference, of which we must take two-thirds, 101£, which we must divide by the difference of the squares of 6 and 4, which is20; the quotient will give 5-^, that is to say, 5 feet 10 inches nearly, for the interval E O.

As the centres of impression of circular surfaces are the same as the centres of gravity ofthe solids which express these impressions, and

as we cannot have these last centres without l

knowing that of the frustum, we must examinethis solid under a different aspect than before.

For this purpose we must consider the solidityof a right cylinder, A 13 C1), as composed ofseveral surfaces E, F, G, II, of an infinitely smallthickness, which pass intersecting from the axisI K to the great surface ABC1); if we imaginethat the circle which serves as the base to thecylinder is composed of an infinity of concentriccircles, forming as many rings of an infinitelysmall thickness, each of them will serve as basefor the corresponding element of the cylinder.

Following this idea, if we cut the cylinder by aplane, M L N D O M, passing through the centreI, and through the extremity, D, of the diameterA D, this plane will detach a frustum, having for its base the semicircle M L C.

Now, as all the cylinders which pass from the axis to the surface A B C D, have been cutin the same manner as the preceding, each of them furnishing also a frustum, it followsthat the largest may be considered as being composed of an infinity of other frustums,similar to each other, all intersecting from the smallest, which answers to the centre, I, tothe largest, and similar to each other, because all the triangles I G P will mark their sec-tions through the centre ; consequently we may consider the frustum which answers to thegreat triangle I C D as composed of an infinity of portions of cylindrical, similar, concentricsurfaces of an infinitely small thickness, each of which will be equal to the rectangle com-prised by the diameter of the demicircle which serves as its base, and by the correspondingelement G P of the triangle I D C.

'io have the solidity of the frustum in this manner: let A B C be the semicircle whichserves for base: describe the semicircles FGEa °d fge , near each other; call DB or B I,a, 1) G or G P, x, Gg will be dx; thus weshall have for the differential element of thefrustum FE x G P x Gg (2 xxdx\ whose inte-

, . 2 x 5 2 a 3

gral gives--- —

Fig. 1705.

> °r -- , when x —a, for the solidityof the frustum.

To have its centre of gravity, we must mul-tiply the differential solidity 2 xxdx by theradius DG (x), which gives 2x 3 dx, whose inte-2 x 4 x* a*

gral is ~~ t or —, or — , which being divided by2 a 3 3 a

3 » tae solidity of the frustum, gives —which shows that the centre of gravity of the

frustum is distant >{ of the radius from the centre of its semicircle.

lohave the centre of gravity of the complement of the frustum, we only regard the

Fig. 1706