KENNELLY. — ARTIFICIAL LINES FOR CONTINUOUS CURRENTS. 109 



Then treat the artificial line as grounded directly, but with the angles 

 of all its leaks and junctions increased by </>. Formulas (19) to (35) 

 will then apply, except where the strength of the current to ground 

 enters into consideration, as in (26), (28), (30), and (33). The surge- 

 resistance r must then be replaced by a new surge-resistance 



r " = — %- ohms. (37) 



cosh 4> 



Thus, the sending-end resistance becomes, by (19) : 



R g <r = T tanh (L°a + <j>) = r tanh (2 md + <f>) ohms. (38) 

 The resistance at and excluding the nth leak becomes, by (20) : 



*---»' Sg::g;:s ° hms - (39) 



The resistance at and including the nth leak becomes, by (21) : 



.sinh [(2 w - 1) 6 + <£] 

 fr*"-*' cosh (2 n$ + «) ° hmS ' (40) 



The ratio of local maximum resistance just before a leak to the local 

 minimum just after the preceding leak is : 



R'n+i.a* sinh[(2tt+l)0 + fl 

 R' gtna ' sinh [(2 n- 1)6 + *] * K * l) 



For example, the sending-end resistance of the line in Figure 7, with 



750 



o" = 750 ohms, whose hyperbolic angle is tanh -1 ■= 0.57941 



Jtf 6 1436.1 



hyp., becomes by (38), 1436.1 X tanh 2.338 = 1409.6 ohms. 



The receiving-end resistance is, by (26) and (37) : 



Ri* = r " sinh (L 2 a + <f>) = To" sinh (2 m6 + <j>) 



= r sinh 2 mO + a cosh 2 md ohms. (42) 



Thus, in Figure 7, r " = 1224.7 ohms by (37), and 



R l<r = 1224.7 sinh 2.338 = 6285.4 ohms. 



The voltage at the nth junction is, by (27) and (28) : 



