218 Mr. O. Heaviside on the Speed of Signalling 



between x=Q and x = l. The general solution is 



t> = 2Asin(y +b\6~ a i\ . . . . (1) 



where T = ckP, if v vanishes for £ = oo. Three sets of con- 

 stants, A, a, and b, have to be determined from the terminal 

 conditions for x and t. In A B and C D the current follows 

 Ohm's law. Therefore 



v 1 dv , „ 



7! = — T — when x = v, 



mkl h ax 



and 



v 1 dv , . 



iihii k ax 



for all values of £. Therefore, by (1), 



sin 6 = ??ia cos b, or tan ^»-wa = 0, 

 and 



sin (a + b) = —na cos (a + 6), 

 or 



tan (a + 5) + na = 0. 



Hence, eliminating b, 



. (m + n)a 



tan a— — — y %, ? 

 ???ncr — 1 



from which the a's can be found when m and n are given. 

 The £>'s are already known in terms of the a's, and the A's can 

 be found by integration if the potential of every part of the 

 conductor of the cable is given for £=0. Let it be that pro- 

 duced by an electromotive force E in A B, i. e. 



T(l+m + ny 

 then, by integration, 



2E cos b 



1-f 



I' + toV .1+wW 



and finally ; the potential at time t is 



1 



'=*>2E 14-mVj ;..■- i ,a { x _£' 



v = Z — — - — L-~ (sin + ma cos) -=- e t f 



i=i a % -. rn n *> 



+ T+m 2 «l + 1+nV, 



from which the arrival-curves of the current may be found by 

 making x = I. In the diagram six cases are shown. The 



