1114

THEORY AN1) PRACTICE OF ENGINEERING.

Book II.

demicircle I B Z sustains is to that which theupper 11) Z sustains, as 47 to 19

We shall see in like manner that the sum ofall the double ordinates QR by the correspond-ing elements OP of the trapezium N STMmay be expressed by a solid similar to fig. 1697.To have the pressure which the second circlesustains, we must multiply its superficies bythe mean element V X, or its equal X L, whichmarks the height of the water above the centre

It follows that having a tube A B curvedbelow to adapt thereto a species of funnelGECDFH closed by a piston, pouring thewater in this tube to the height K, the power Papplied to the piston will sustain a pressureequivalent to the weight of a column of water,which should have the circle E F of the pistonfor its base, and the line K B which marks theelevation of the level of the water above the centreI for its height, however small the diameter ofthe tube may be; thus this power will be inthe same case as if it were applied to the pistonof fig. 1702.

If the preceding surfaces, instead of beingvertical, were inclined, all that we have just saidwould still hold good, having shown that thepressure against both must be measured in thesame manner.

It follows, that having an inclined tube A B C Dfilled with water, and the bottom A D closed by apiston, the pow'er applied thereto will sustain a weightequivalent to that of a column of water having thecircle of the piston for its base, and the perpendicularE E, which marks the greatest elevation of the wateralx>ve the centre E for its height, of whatever form thetube is, without considering its thickness.

The Centre of Pressure is that point of a surface ex-posed to the action of a fluid, to which if a force equalto the whole pressure upon a surface were applied, theeffect would be the same as when that pressure is dis-tributed over the w’hole surface. The centre of pressurecoincides with the centre of percussion, which is two-thirds the height of the body.

Since the action of all the films of water against asurface A B C D may be expressed by the elements ofan isosceles triangle A E D, it is evident that there is apoint M in the perpendicular E F, where a power Pbeing applied in the opposite direction, PM, will sus-tain them all in equilibrio, and if we pay attention weshall see that this point, which is called the centre ofimpression, can only be the centre of gravity of thetriangle A ED; whence it follows that the centre ofimpression of a rectangular surface A BCD is placedat two-thirds of the line E F, which divides it intoequal parts, and marks the height of the water.

Fig. 1702.

Fig. 1703.

r?

\ c

\

>/

N \K H

/ L

""\

/ y

M \

/ /

!> \

y k i

Fig. 1704.

Regarding only the pressure which the rectangle AG H D sustains, which we may applyto a lock gate, the water being always at the same height, E E, the centre of impression ofthis surface will be the same as the centre of gravity, O, of the trapezium A I DK. Tofind it let us suppose that the point N marks that of the triangle I E K ; then calling

EF, a, EL, b, EO, *, we shall have EM = ! ^,EN=p 5111 = ?? 26

and MO

If, instead of the superficies of the similar triangles I E K and A E D, we take the squaresof their perpendiculars, E L and E F, the difference of these two squares, or a a — b b