108 PROCEEDINGS OF THE AMERICAN ACADEMY. 



in terms of the current to ground and of the surge-resistances, corrected 

 and nominal. 



Consequently, the voltages at successive ascending leaks are pro- 

 portional to the hyperbolic sines of the angles of those leaks. 



Current. Far End grounded. 

 The current at the sending end is : 



r tanh L 2 a r tanh 2m6 ^ 



The current at junction n is :' 



T T cosh 2 nO e m cosh 2 n$ 



m cosh 2 m6 r sinh2m# ^ 



The current at junction 0, or the grounded end, is : 



I - €m = 6m = Im = Im 



r sinh 2 md r sinh L 2 a cosh 2 m0 cosh L 2 a. /oo\ 



At the nth leak, the ratio of ongoing to arriving line current is : 



Jn-i _ cosh 2 (n - 1) 6 

 I n cosh 2nd 



The current escaping at the nth leak is: 



, , sinh(2n— 1)0 n T .,-.,,« <s/ , 



t„ = e n <7 = € m9 . , /r> rr^ = 2 1 sinh ^ sinh (2 n — 1)0 



sinh (2 m -1)0 amperes. (35) 



By comparing formulas (6) and (19), (13) and (27), (16) and (32), it 

 will be seen that with the far end free, the sending-end resistances 

 follow the cotangents, voltages the cosines, and currents the sines, of 

 the hyp. angles of the junctions ; but that with the far end grounded, 

 the sending-end resistances follow the tangents, voltages the sines, 

 and currents the cosines, of said angles. 



Line grounded at Far End through a Resistance a. (Figure 7.) 



First Case. Let a be not greater than r . 

 Find the hyperbolic angle <f> of the terminal load <r from 



tanh <f> = — . (36) 



To 



(34) 



