110 PROCEEDINGS OF THE AMERICAN ACADEMY. 



so that the voltage at the distant end of the line is : 



g ^ g - sinh(2m/+<A) =I ° r//sinh ^ V0ltS ' (44) 



Thus, in Figure 7, the voltage at the distant end B is 



sinh 0.5794 -- no - 

 100 X !Sh2» " UML 



The voltage at the nth leak is, by (29) and (30) : 



sinh[(2n-l)fl + <ft] sinh[(2n-l)g + fl , 



tn ~ €m sinh [(2 m - 1) e + 0] " ° ° cosh $ v °" 5 s j 



The current at the sending end is, by (31) : 



m " "" r tanh (X 2 a + </>) == r tanh (2 md + <j>) am P eres - ^ 



At junction n it is, by (32) : 



_ cosh(2nfl + <ft) _ ejn cosh(2nfl + </> ) „- 



in. - ^««r cosh (2m6/ + ^ - ro sinh (2m # + ^) amperes. ^) 



At the distant end, through a, it is, by (33) : 



r " sinh (2 m0 + <£) r " sinh (Z 2 a + <£) 

 I m<T cosh I m<T cosh ^ 



amperes. (48) 



cosh (2 to0 + ^) cosh (Z 2 a + 0) 



At the nth leak, the ratio of ongoing to arriving current is, by (34) 



h,n-i __ cosh [2 (n- 1)0+ fl 

 I a , n cosh (2 n6 + </>) 



(49) 



For example, the received current to ground through cr is, by (48), 

 100/6285.4 = 0.01591 ampere. 



Second Case, with a not less than r . 

 Find the hyperbolic angle of the terminal load cr from the formula : 



tanh <f>' = -. (50) 



cr 



Then treat the artificial line, actually grounded through cr, as though 

 it were freed at the far end, but with its angular length increased at all 



