ACOUSTICS AND GRAVITATION. 119 



Notwithstanding the steadiness of the new results, one-minute periods, as 

 shown in figure 152, are again impossible, while the three -minute periods follow- 

 ing succeed at once. In figure 153, the highest mean exhaustion available dur- 

 ing thirty minutes was applied. The value of the triplets has not practically 

 changed size: 



Vacuum, = 30 cm. A;y = o.79 



= 8 0.83 



the smaller values being referable to slightly greater temperatures within the 

 case. Figure 153, moreover, is appreciably sinusoidal, all the branches being 

 first accelerated and finally retarded. There is no apparent inertia effect. The 

 double inflection might be associated with the elastic resilience of the fiber. It 

 is much more probably, however, the result of a repulsive radiant force which 

 requires a minute or so for development. Thus, when the weight is turned the 

 counter radiant force acts for a time with the new pull of the weight. Hence 

 the acceleration. The radiant force gradually vanishes in turn, passing through 

 zero, then to become negative and constituting an increasing resistance to the 

 new pull. Thus the retardation ensues. In other words, the radiant force 

 needing time to develop, is in phase behind the gravitational pull, which acts 

 instantaneously. The two observations between the turning-points should 

 therefore be without appreciable radiant force and the data of 95 for the rates 

 of uniform motion bear this out very fully. 



94. Tentative estimate. The resistance experienced by a sphere of radius 

 r, moving in a viscous fluid (77) with a velocity v = lu, is well known to be 

 6irrjrv. I have not found the corresponding expression for a cylinder (figure 

 139, m m') of radius r, semi-length /, and with hemispherical ends, moving with 

 constant speed, broad-sides on. To get at an order of values, however, we may 

 postulate that for equal frontal areas, 7rr 2 = 2rA/, the resistances are alike. 

 Thus the element of resistance is 



dF = 6irrjrv = 6V VTJ/WV/ Vr 2 = 6v / 7n?a>/v / 2?'.A/ 



and we may meet the further conditions by integrating for the length 2 1 of the 

 needle. To carry out the integration* put l = nX2r, where n is a serial number. 

 The equation becomes 



and the problem reduces itself to the summation of a series of cubes, 



2\/2n s = n(n+i), the length being 2/, 



Hence, finally, for two masses, M, m, at a distance R apart, disregarding 

 corrections, 



The constants of the second apparatus were : M = 1 6o2g, 1*1 = 0.563% 

 R = 5.i cm., 2r = o.4 cm., 2^ = 22.8 cm., 17 = 0.00019, 11 = 28.5 

 *A more acceptable upper limit would be y=*(B?/Mm)yr t Qr*<*n(n+'l). 



