48 PROCEEDINGS OF THE AMERICAN ACADEMY. 



B D = 2 + S c = 2.025,608 hyps. 



The sending-end resistance of the grounded composite line is then, 

 at D 2 , by (12), (37), (38), and (45), 



RgD = ~2 tanh 8 D ohms (47) 



= 2,000 tanh 2.025,608 = 1931.58 ohms, 



and the sending-end conductance, 



G gD = y 2 coth 8 D = \/R gD mhos (48) 



= 0.000,517,71 mho. 



The formula for finding the architrave resistance of the equivalent n 

 of the line AD is 



p" = c 2 sinh B D ■ — V^r ohms (49) 



r cosh b c v ' 



cosh 2 

 = 2,000 sinh 2.025,608 X ' 



cosh 0.525,608 



= 24,553.55 ohms 

 and y" = 1/p" = 0.407,273 X 10" 4 mho. 



Formula (49) differs from the corresponding formula (40) of the pre- 

 ceding case by the application of the ratio — ~r~Y or ^ ne ra ^° °f * ne 



cosines of the line-angles across the junction BC. 



The formula for finding the conductance of the leak at D is, as before 

 (20) and (41), 



g"n = G gD - y" = \IB gD - y" mhos (50) 



= 4.769,785 X 10 -4 mho. 



In order to complete the equivalent l~l of the line AD hyperbolically, 

 we must repeat the above process from the opposite end, by grounding 

 the end D x , as shown at A x Di (Figure 7). The line angle at C is S c = 

 1.5 hyps. Across the junction BC this angle changes suddenly to 



B B = tanh-Y C2fa " h ^ hyps (51) 



= tanh -1 1.810,296. 



This involves at first sight an impossible result ; but in all cases of a 

 hyperbolic tangent greater than unity, we may resort to the following 

 formulas: 



