282 Mr. 0. Heaviside on the 



when the return is outside, and 



RiWBifl + (irhdfan) 2 x -503] 



=K 1 [l + (7r%a^ 1 ny 2 x-031] 



when the return is inside. In the case of a solid wire, the 

 decimals are '083, so that whilst the correction is reduced, in 

 this a 1 /a =2 example, the reduction is far greater when the 

 return is outside than when it is inside. 



The high-speed tube formulse are readily obtained. Those 

 for the inner tube are the same as for a solid wire, and that 

 for the outer tube depends not on its bulk, but on its inner 

 radius. That is, in both cases it is the extent of surface that 

 is in question, next the dielectric, from which the current is 

 transmitted into the conductors. Let Gr (V) = (2/tt) K (#), 

 and G ] (^) = (2/7r)K 1 («^) ; then, when x is very large, 



J (x) = — G x (^) = (sin x + cos x) -f- Qirxy 1 . 



J 1 (^)= Gr (V) = (sin«2?— cos#)-=-(w#)*J 



Use these in the R/' fraction, and put in the exponential 

 form. We shall obtain 



But 



therefore 



But 



therefore 



K 1 "=( PlSl i)/(27ra 1 ). 



p 2 -=.—n 2 , 



so that, finally, 



pi- tw/ L'- Rl ' nrX 



tt] - — ~ — > ^i-— > • • • (75) 



q = n/27r = the frequency. To get R 2 ' and L 2 ', change the 

 /u. and p of course, and also % to a 2 . 



It is clear that the thinner the tube, the greater must be 

 the frequency before these formulse can be applicable. For 

 the steady-flow resistance is increased indefinitely by reducing 

 the thickness of the tube, whilst the high-speed resistance is 

 independent of the steady-flow resistance, and must be much 

 greater than it. In (75) then, q must be great enough to 

 make R' several times R, itself very large when the tube is 

 very thin. Consequently thin tubes, as is otherwise clear, 



