1132

THEORY AND PRACTICE OF ENGINEERING.

Rook H.

VE 0,(x), multiplied by the height H F, (c), we have a —b = c VxyVtKxiAx being

squared gives | a 3 — £x2a6v' a6 + |5 s = ccx,* + 6 3 — 2 ab a b

a 9 + b 9 — 2 ab Va b

= x; whence we have

9, 4

x, or, 9 is to 4, as the sum of the cubes of the greatest and least

height of the water relatively to the orifice, minus twice the square root of the product ofthese two cubes, the difference divided by the square of the height of the orifice, is to themean required.

If the height E F (n) be 8 feet, the height E II (5) 6, the height II F (c) will be 2 feet;we shall then have 512 for the first cube, and 216 for the second, the product of which gives110592, from which extracting the square root we have 332^, twice which is 665$, whichbeing subtracted from 728, the sum of the two cubes, there remains 62$, which divided by 4,the square of the height of the orifice, gives 15^§ for the third term of the proportion : thenas 9 to 4, so is 15^ to the height required, which will be G feet 11 inches 10‘? lines.

Another and more simple method of finding the mean velocity of water from a rectan-gular orifice is to seek the velocities which answer to the greatest and least height of thewater ; multiply each of these velocities by its fall, subtract the second product from thefirst, take 3 the difference, and divide this quantity by the height of the orifice; the quotientwill give the required velocity.

When the orifices are tra-peziums whose sides II C, AD,are parallel, having their sum-mits at the level of the water,the expenditure of the first willbe obtained by multiplyingtwo-thirds of the rectangleEBC F, plus two-fifths thesum of the triangles A E B,

F D C, by the greatest velocityof the water ; and the expen-diture of the second by multiplying two-thirds the rectangle B E F C, plus four-fifths thesum of the triangles A E B, F I) C, by the greatest velocity.

If the two triangles C E A in figures 1719. and 1720. do not answer to the

level of the wa^r, the summit of the first and the base of the second being below

the summit B of the parabola, multiply the elements of these triangles by the cor-responding ordinates of the parabolic segment A E F D. Calling the perimeter ofthe parabola p , A D, o, the base C A or C E, by B E, c, E A, A, B A, n, E F, 9, Eli, x,

II !/> w c shall have, on account of the similar triangles, h, b : x, ^ = GII, which

c h

Fig. 1719.

being multiplied by ydx gives

bxydx

for the differential of the solid ; and as we have

pc + px=yy, or x —

yy-

-E£ t whose differential is dx = ~—^y

bxydx ... . 2by* dy — 2 pbcyydy

— — will give ---, whose integral is

•- pph 0 5pph

putting the values of x and dx2 by 9 2 bey 3

3 ph

, and putting in

place of y s and y 9 their value will

. 2 b -

give -j x c x

V p

c + px —

2 he

3 A ’

*s/P

c -\-p x =

25 * / 2b c - /

x (c + x) / pc + px — x c + x pc +px ; and supposing x=0, there remains/ 2bcc / Abccq

P c —34 '%' P c= - YiTF* k e * n S added with the contrary sign gives

2 bcc5 h

2b

5h

2 b c

X c + x S A / pc+px- 3 ~xc +

-**/>

p c px

4bccq15 h

"When x becomes equal to A, we shall have c + x = n, and Vpc-\-px = a ; therefore, by substi-tuting these values, we shall, after reducing the terms to the same denomination, have -~r

15 A

y. 6 ann + 4 ccq +10 acriy for the most simple expression of the solid ; so that to have theexpenditure of a triangular orifice whose summit is below the level of the water, we mustfirst multiply the greater velocity, A D, by six times the square of the height 13 A of thewater; secondly, multiply the least velocity, E F, by four times the square of the height, B E