Chai*. XVI.

HYDROSTATICS.

1113

As for the solids, fig. 1698. and 1699., for the first we shall have the sum of all theplanes DING by bedding to the solidity of the demi-cylinder ALMHO that of thefrustum O L B M, and we shall have that of the second by taking from the cylinderEOMNIID the value of the frustum INI N IIO.

It will be easy to calculate the pressure of water against all sorts of circular surfaces; forexample, if we wish to know that which the superficies of the demicircle ABC sustainswhose diameter A C answers to the level R Z of the water, we remark that in making theright-angled isosceles triangle DBE, all of whose elements represent the heights of films ofwater, which answer to all the points of the height D B, we shall have the pressure whichacts against, the double ordinate F G, by multiplying this line by the corresponding elementI H. Now as the sum of all these products will be equal to the solidity of a frustum,which would have the demicircle ABC for its base, and the line BE equal to the radiusfor its height, this pressure will then be expressed by a volume of water equal to two-thirds the cube of the radius D B.

If the demicircle were situated in an opposite direction to the preceding, as KEM, weshall see that since we must, to have the action of all the films of water against the doubleordinates O P, multiply each of these lines by the corresponding element Q R of the triangleK L N, the pressure which this demicircle will sustain may be expressed by the comple-ment of a frustum which would have this same demicircle for its base, and the radius for

its height. Thus we shall find it by saying 14 is to 19,

2 LN

is to ^ x L N 3 , which

shows that this pressure is equal to ^ of the volume of water expressed by the cube of theradius. It follows that if the two demicircles are equal, the pressure which the first willsustain, is to that which the second will sustain as 14 to 19.

Fig. 1700.

If the two demicircles were below the level RZ, as A BCD and HRS, the lines IB and OXexpressing the greatest height of the water, making the isosceles and right-angled trianglesIBN and OXL, we must multiply the double ordinates F G and QT by the correspond-ing elements H E and P V of the trapezium NKDB and LMRX; then the sum of theproducts for the demicircle ABC being expressed by a solid similar to that of fig. 1698.,we must multiply its superficies by the line D K or DI, which marks the least height ofthe water, to have the demi-cylinder, and add thereto that of the frustum, that is to say,two-thirds the cube of the radius, we shall have the volume of water whose weight thisdemicircle will sustain.

As for the other demicircle HRS, as the sum of the products we have been speaking ofwill be expressed by a solid similar to that of fig. 1699., we see that to have the pressurewhich it sustains, we must multiply its superficies by the line X L or X O to have thesolidity of the cylinder, from which we must subtract that of the frustum, that is to say,two-thirds the cube of the radius.

Lastly, if we have two circles, one ofwhich answers to the level D L of thewater, and the other w f as lower, making theisosceles and right-angled triangles C D Band N L M, the sum of the products of thedouble ordinates E F, by the correspondingelements G H of the triangle C D B, maybe expressed by a solid similar to that of fig.1696. We must, to have the pressure whichthe first circle sustains, multiply its super-ficies by the mean element I K, which isthe radius K D, and which marks the heightof the water above the centre K.

If we recal what has been said, weshall see that the pressure which the low r er

Fig. 1701.