and the nature of the parabola is given by its equation y = p.xv. 727. If x denote the distance of any element of the orifice ydx (parallel to the horizon) from the surface of the fluid, and Q the quantity of fluid issuing through the whole orifice in a second, we have x √ a. dQ = 2ydx × √2gx :: Q = 2fydx √2gx integrated between x = 0 and x = x, the altitude of the fluid. A M. Let now the given area of the orifice be A, and that of the descending surface, which is also given, be M; then the velocity of the descending surface is x Q" by the question :. Sydx √2gx = 1 nMC and y = 2A√29 or the required curve is a parabola. A 2g M 2A 728. The time of emptying any depth x of the fluid is (696) Xdx √x But here b2 a2 a and b being the semiaxes of the generating ellipse. xf(2a xdx) X = xy2 = π 729. t= = HYDRODYNAMICS. Hence πb2 a2 A√ 2g a2 A 2g Make xa, and then x = sults gives the time required, viz., πb2 √ a A2g By 696 1 t = = X = T X 28/217 60 A√2g But since, as it may be easily shewn, the section of an oblate to its major axis at the distance from the vertex, is an ellipse, whose semiaxes are spheroid made by a plane b √2ax a 2πb aA√2g 2πb for the whole spheroid. a a and b being the semiaxes of the spheroid, we have b (2ax-x) a A√ 2g 167ba3 x2, and √2ax x2 15 Ag Xdx , and the difference of the re < HYDRODYNAMICS IN GENERAL. 1 a2y 731. LET be the angular velocity round the axis; then when the water and cylinder are relatively at rest, the surface of the water will be that of a certain solid of revolution, whose ordinate, normal, and subnormal, referred to the axis of rotation, will represent the centrifugal force, the re-action of the fluid, and the force of gravity, respectively. Hence if y denote the ordinate, and the linear velocity of any particle of the fluid in the surface, the centrifugal force is 0 v2 a2y 2 y y and the subnormal is i. e. to = a2y × gy = 요 a2 or the generating curve of the volume of the part evacuated by the fluid is the common parabola, whose parameter is 2g Hence, if r be the radius of the base of the cylinder, the quantity of water thrown out by the revolution of the cylinder is equal in volume to the paraboloid, whose base is r3, and altitude a2x2 2g aspe 24. 2g and this taken from the given volume of the whole cylinder, will leave the quantity of water required. παι 4g Let be the angle required; then (Vince's Hydroscos. = max. and by 732. tatics, Prop. XXVIII.) the effect c sin. the rule for maxima and minima, we get sin. 0 = √2/3 733. If p be the parameter of the given paraboloid, and r the radius of its base; then the angular velocity necessary to empty the vessel is (731) 734. Let a be the required angular velocity; then the volume of fluid evacuated is that of a paraboloid, whose base is wr2 (r being the radius of the base of the cylinder,) and altitude (see 731,) ars Hence, and by the question, 2g 735. The descending force is w, and the resistance is sx v2 (Vince's Hydrostatics, Prop. XXV.) s being the spe xd2 4 cific gravity of the fluid, and v the velocity... the force which accelerates the whole apparatus downwards is F=w #d3sv2 |