112 PROCEEDINGS OF THE AMERICAN ACADEMY. 



The current strength at the sending end or junction m is : 



Im ° = To COth (Za + <}>') = T COth (2 luB + </>') ^^^ ^ 



amperes. (61) 



At junction n it is, by (16) : 



sinh (2 n6 + <£') 

 inv ~ m,r sinh(2m0 + <£') 



At the receiving end, or junction 0, it is: 



7 °* = l ™ sinh (2 m0 + <£') = r '" cosh (2^0 + 0') am P eres - ( 62 ) 



At the nth. leak the ratio of ongoing to arriving line current is, by (17) : 



I n -i, a _ sinh [2 (n- l)0 + 4>'] 



In* sinh (2 n$ + <£') 



The current escaping at the nth leak is, by (18) 



(63) 



,_ , cosh[(2n- 1)0 + <£'] / cosh[(2 n- 1)9 + <f> '] 



W - * M g - e *g cQsh {6 + 0/) r - t m „9 cosh[(2m-l)d + <l>'] 



, cosh [(2 n -1)6+ <}>'] 

 = «""* cosh 6 cosh (2m* + *') am P eres ' (64) 



As an example, let <x = 3750 ohms. Then <£' = tanh - * — — - — = 

 1 3750 



0.403535 hyp. The sending-end resistance at junction 1 is, by (52), 



1436.1 coth 0.755255 = 2250 ohms, which by Figure 8 is evidently 



correct. Again, the received current strength for the same case with 



ei = 10 volts, at junction 1, will be,by (62), I 0<T = at- cog 



3464.1 cosh 0./oo26 



= 0.00222 amperes, which is also easily seen to be correct, from 

 Figure 8. 



Third Case a- = r . Exponential Case. 



In the particular and intermediate case in which a = r , either of 

 the preceding sets of formulas applies under limit conditions. We 

 have cf> = <$>' = x, by (36) and (50). Consequently, the sending-end 

 resistance becomes at any junction: 



Rgro = r ° ohms. (65) 



The resistance at any leak, excluding the same, is : 



# Vo = r o «* ohms. (66) 



e being the Napierian base. 



