28 On the Mechanical Theory of Crookes's Force. 



we get 



Q* + 4 — K 3 Q 2 -« 2 FX 4 .K 4 ==0, 

 P 



which is a quadratic for Q 2 or a biquadratic for K. 

 Solving for Q ; we get 



Q= HWa { N /XV 2 + 4« 2 K 2 -2aK}% 



v p 



as evidently the other solutions are inadmissible. 



From this we may get an approximate value for K in terms of 

 Q ; for, unless a be very large, or the density or difference of 

 temperature very small, X 2 p is much greater than 2«K. For 

 instance, if V 1 and Y 2 correspond to a difference of 10° C, 





and consequently 



V 273' 



V 2 = 48500a /J*_, 

 V 273' 



^ (48500) 2 . 



.-. X = 9700, 



while p=g-^Q for air at atmospheric pressure ; 



.-. XV = 107600. 



And K would be large if it were 100 ; so that even if a were 

 50, 2aK would still be less than ^ of X 2 p ; and so we may 

 take approximately 



Q = W«.KX; 

 ,.K= « 



k\/ot. X 



From this we can calculate K; for & = 1*6 in most gases, and, 

 if a = 2*5, x/a = r5> and X = 9700, as above ; 



.-. Z:v / «X=22310 = 2xl0 4 ^ i ?. 



Now, at a distance of a fourth-metre in air at atmospheric 

 pressure, and with a difference of temperature of 10° C, 



Q = 10 6 « ? .p.; 

 so that in this case 



¥L = h0 q.p. y 



