PNEUMATICS. 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 x 0. Consequently Hence if r be the radius of a spherical balloon of this capacity, we have 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 PNEUMATICS. where a is the depth of air in the tube before inversion, h the standard altitude, and the defect from that altitude, caused by the elasticity of the air. 756. Let E denote the elasticity of the fluid when occupying a given space #rh, r being the radius of the base of the tube; then its elasticity, when occupying the space Hence if W denote the weight of the piston, the moving force upon it is 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 Ex 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 + r2 x 15, and we then get If the time of acquiring this velocity be required, we have 757. In the condenser, the density of the air in the re ceiver after t turns is (Vince, p. 112,) d = D x (r+tb) r (1) 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,) = D' x (b+r) 758. = Let R be the radius of the body of the pump, 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 a the elevation due to the first stroke. Then R$ (H+ p + s) x = R2 R R2 PH R2 {H+p+s± √(H+ — * p+s) − 4 — p H A±√A -B 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 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 that after m turns of the air-pump, we have (Vince, p. 107,) Again, let a be the required depth of air in the tube at first, and y the standard altitude; then (Vince, p. 92,) :: xy=(la) (ya) 7 being the length of the tube. (2) Also if y denote the standard altitude when under the receiver, we have, by the same rule 760. Let D, d, be the densities at first of the two airs, and that of the atmosphere when its elasticity is E. Also let D', d', denote the densities at any other time t. |