Ex. 2. Given W = 20 stone, and the other elements as before, to determine the magnitude of the balloon necessary just to lift that weight from the ground. Here r = 0. Consequently 7803 65 X 4450 = 4044 cubic feet. 72 Hence if r be the radius of a spherical balloon of this capacity, we have 49p3 - 4044 3 and p3 = 1000 nearly. ::1 = 10 feet nearly. Even such a vehicle, it appears, would be bulky and expensive. Its exterior would require upwards of 1200 square feet, or 133 square yards of silk. Professor Playfair, (p. 254, vol. 1, of Outlines of Natural Philosophy,) has erroneously treated this subject. The weight of the gas is there wrongly computed. 754. Since the altitudes of fluids in a tube are inversely as their specific gravities, if a be altitude of the water sustained by the air, we have 1 1 : 302 :: • a 13568 1000 i. a = 413.824 inches. and the required length is :: I= = 2a = 827.648 inches. sin. 30° a 755. Generally, we have, by 749, at. where a is the depth of air in the tube before inversion, h the standard altitude, and x the defect from that altitude, caused by the elasticity of the air. But by the question, 2 33 + 28) = 4 inches 3 756. Let E denote the elasticity of the fluid when occupying a given space wrh, r being the radius of the base of the tube; then its elasticity, when occupying the space жr2 x Hence if W denote the weight of the piston, the moving force and the force which accelerates its descent is E h Х If the fluid be atmospheric air, its elasticity in a natural state, or when the barometer stands at 30 inches, is about 15lb. upon the square inch, and we have E = p X 15 lbs.; and if in addition to the piston there is pressing upon it the whole atmospherical column of air, for W we must put W + r X 15, and we then get 1572 h h-S v = 2. 1. + s} 1571? + W h dx Eh l.x - x)" W W dx dt = 757. In the condenser, the density of the air in the receiver after t turns is (Vince, p. 112) d = D x (r+tb) D being the density at first, and r and b the capacities of the receiver and barrel. Again, in the air-pump, after t turns the density, is (Vince, 107,) N =D x (6+r)! (26+r)" Let b = t = 3; then we have D+ 10 1 3D 353 or the density has decreased -ths. 758. Let R be the radius of the body of the pump, r that of the suction-tube. Also let H be the height of a column of water whose weight = weight of a column, having the same base, of the atmosphere, s the length above the water of the suction-pipe, p the play of the piston, u the height of a column of water equivalent to the pressure of the air below the piston, and x the elevation due to the first stroke. Then x +u =H .. (1) Also XS psH Rop+r' (s-x) Ro . u Xн R? р Н Again, let x, be the elevation of the water in the tube at the end of the second, and u, the elastic force of the air within the pump at that period; then as before X, + Ug = H and Ug =uX pol. (s-x) Rʻp + gol (sx) Hence ’ – s 77 { 2H - AF VA’-B – 4(H +s s– )} 2 and so on to an and Up Since x and u must be both less than H, it follows that the lower sign of the root is alone applicable. 759. If D be the density of the air at first, and a that after m turns of the air-pump, we have (Vince, p. 107,) de DX (6+r) - D x (#2)". . (1) . Again, let x be the required depth of air in' the tube at first, and y the standard altitude; then (Vince, p. 92,) y : y a ::1-a : x :. xy = (1 - a) (y - a) (2) i being the length of the tube. Also if ý denote the standard altitude when under the receiver, we have, by the same rule y' : • - b::1-6 : x d Therefore But y' = y which gives D.(l – b) + ad (1 – a) (a - b) (n + 1)" (a - b) Hence (1 – a) (y – a) y (-6)(n+2)* +all+b)(n+1)* 8(2-6)(n+2)" +all-a)(n+1)" 760. Let D, d, be the densities at first of the two airs, and 8 that of the atmosphere when its elasticity is E. Also let D', d', denote the densities at any other time t. |