1110

THEORY AND PRACTICE OF ENGINEERING. Book II.

vessel from L to O, will depend on the length of the road LO, or of the sine of theangle O L R, for the less this sine is with regard to the total sine, the nearer the point Kwill approach E, and the more water there will be in the vessel, but the road will also belonger ; on the other hand, the nearer this sine approaches the total sine, the less water itwill hold, but also the length E O approaching more nearly the height O R, the power willtake less time to draw it up. Now, since the greatest quantity of water depends on the lineK II and the shortest road of the sine of the angle O I. R, we see that the product of thesetwo lines should be the greatest of all those which may be formed by the same lines.

Having made OV equal to O It, and drawn from the point V the line V Y parallel toL R, O V may be taken for the total sine, V Y for that of the angle V O R, and O Y forthat of the angle O V Y or O L R. Then calling O R or OV,«, ()Y,r, V* Y will be Vaa — xx ;as for the lines B E and E H, which we suppose equal, as the length is indifferent, we shallexpress them by unity. Let us consider that the similar triangles V O Y and B K E give

VY (Vaa-xx\ YO(x):: BE, EK ( ■ * - ) i

v v/ao-xx'

•/ aa — xx

whence we have EH — EK = KH ^ 1which being multiplied by O Y (x) gives,

x — —. of which we must take the differential, and equal it to zero; we shall have

x — ~~ 7 —, of which we must take the differential, and equal it to zero: we

V aa —xx

2 xdx x Van —xx —x 3 dx x aa — xx —\aa —xx

or, aadx — xxrfx — 2 xdx x '/aa — xx — X s dx x aa — xx-J=0;from which effacing the dx we have

xxx Vaa

V aa — xx2aax + x 3

or, a a — x x—~

V aa — xx

X s —2 aax

which being squared gives,

aa — x x

whence we at last have x 6 — |aax 4 + |a 4 x* — £ a* = 0;

and if we suppose x 8 = ay, we shall have y 3 — \ ay* + \ a*y — £ a 3 = 0, for the simplest equationto which this problem can be reduced.

Since we may suppose the line O R divided into as many equal parts as we please, taking thenumber 10 to express the value of a, we shall find, according to the ordinary rules, y =

To convince ourselves, it is only necessary to multiply the values of a and y in the same

manner as in the preceding equation: we shall finel y 3 +1 g*y =. and \ a if + ^ a 3 = ———

1000 1000

which showing that the sum of the plusses not differing much from that of the minuses, wemay regard them as equal.

x*

Having supposed x* = ay, or —=y, and a= 10, we shall have x*= 17, orx = V17 : nowa

if we multiply 17 by the square of 10000, which is 100000000, to extract the square rootmore exactly, it is 41231 : on the other hand, multiplying the value of a by 10000. weshall have a= 100000, which shows that OV (a) should be to OY(a) as 100000 is to41231, or nearly as 5 to 2, which is the proportion that may be used in practice. Thus wesee that, for the greatest effect, it is necessary that the height O R of the inclined planeshould be | of its length, LO; then we shall find that the base, L R, of the same plane is toits height, O R, as 23 to 10, or as 4jj to 2.

Taking the side O V (100000) for the total sine, OY (41231) will make that of theangle OV) = O L Ii, which answers in the tables to 24° 21', which is the angle which theinclined plane should make with the horizon.

On the fiction oj Water against circular , vertical , and inclined Surfaces .— It remains toshow the action of water against circular surfaces, and in what manner we must calculateits pressure, since it is always equivalent to the weight of a volume of water expressed bv theparts ofa cylinder cut under circumstances relative to the figure and situation of its surfaces.