26 Mr. 0. Heaviside on the 



is mathematically equivalent to instantaneous non -oscillatory 

 subsidence. 



The following will serve to show the relative importance of 

 R, S, K, and L in determining the amplitude of periodic 

 currents at the distant end of a long submarine cable, of fairly 

 high insulation : resistance : — 



4 ohms per kilom. makes R=40 4 , 



imicrof. „ „ S=2jj3 , 



100 megohms,, „ K = 10" 22 . 



Here, it should be remembered, K is the conductance of 

 the insulator per centim. The least possible value of L would 

 be such that LS = v~ 2 , where v = 30 10 ; this would make 

 L = 4/9 only. But it is really much greater, requiring to be 

 multiplied by the dielectric constant of the insulator in the 

 first place, making L = 2 say. It is still further increased by 

 the wire, and considerably by the sheath and by the extension 

 of the magnetic field beyond the sheath, to an extent which 

 is very difficult to estimate, especially as it is a variable quan- 

 tity; but it would seem never to become a very large number, 

 as of course an iron wire for the conductor is out of the 

 question. But leaving it unstated, we have, by (96), taking 

 R'=R, L'=L, 



,« i {(le „. +L v).(i-. + »l)' + (»o-!j--)}'. 



Now nftir is the frequency, necessarily very low on an 

 Atlantic cable. We see then that the first L 2 n 2 is quite 

 negligible in its effect upon P, even when we allow L to 

 increase greatly from the above L = 2. The high insulation 

 also makes the (RK — LSn 2 ) part negligible, making approxi- 

 mately 



P = Q = (J r n)^.10- 8 , 



P being a little greater than Q, at least when L is small. Now 

 this is equivalent to taking L = 0, K = 0, when 



P=Q=(iRS7i)i, (25b) 



reducing (196) to 



C =2V (S72/R)^{G G 1 e m + H H 1 6' m -2(G o a l H H 1 )*cos 2P/}*. (266) 



