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61. not encounter any atmospheric resistance until it is level with the upper limit of [[underline]] s [[/underline]] [[subscript]] 6 [[/subscript]] (125,000 ft.). This condition will not, of course, be that which would actually obtain in practice, for a continually increasingly resistance will be experienced as the apparatus descends; but if a sufficient breaking action can be shown to exist in the present example, the parachute device will [[underline]] afortiori [[/underline]] be satisfactory in practice. The velocity acquired by the apparatus in falling freely under the influence of gravity between the two levels is -------------- \/64 x 1,103,000 = 8,400 ft/sec. Now the air resistance in poundals per square inch of section at atmospheric presure for this velocity is, from the plot of Mallock's formula, 360 x 32 poundals per square inch, making the value of [[underline]] R [[/underline]] for the area of the parachute R = 1,653,000 poundals/in [[superscript]]2[[/superscript]]. But the actual resistance is [[underline]] R [[/underline]], multiplied by the density at 125,000 ft. which is approximately 0.001, giving for the resistance, F = 1,653 poundals/in [[superscript]] 2 [[/superscript]]. A retarding acceleration must therefore act upon the apparatus, of amount given by a = F/M = 1,653 ft/sec [[superscript]] 2 [[/superscript]]. [[start individual word underline]] Hence it is safe to say that, long before the apparatus had fallen to the 125,000 ft. level, the velocity would have been reduced to, and maintained at, a safe value, with the employment of even a small parachute. [[/end individual word underline]] This case, it should be noticed, is entirely different from that of a falling meteor; in that the apparatus under discussion falls from rest, at the highest point