62 PROCEEDINGS OF THE AMERICAN ACADEMY. 



the equivalent n of the loaded line, ground F, as at F x . Develop the 

 line-angles towards A in the usual way. No change from the corre- 

 sponding conditions of Figure 8 occurs until after we have reached S A . 

 We then have 



RgA — ~i tanh S A ohms 



= 1,000 coth 2.567,48 = 1,011.607 ohms, 



and if o- be the impedance of the terminal load at A , 



Rqaq — °" + ~i tanh 8 A ohms (103) 



= c tanhS^ ohms (104) 



= 1,111.84 ohms, 



where z is the apparent surge-impedance of the line at A ; or 



~o — z i + v c °t n &a ohms (105) 



= /^/tanh 8 A ohms (106) 



= 1,098.829 ohms. 



The architrave resistance is then, following (77), 



,, . . . cosh B c cosh8 £ , /- rt _v 



p" = z a sinh 8 A • — r-r 5 ■ — ttt onms ( 107 ) 



cosh h B cosh 6 D 



= 48,619.7 ohms. 

 The A-leak of the n, as in the case of Figure (8), is 



g" A = \/E gA0 - \lp" mhos. (108) 



To complete the n, we ground the loaded line at A, as at A F 2 , and 

 develop the line-angles towards F, commencing with 



8^ = tanh" 1 ( - J hyps (109) 



= tanh" 1 (^r-) = 0.100,336 hyp. 



\ioooy J ^ 



The architrave impedance is then 



,, • , -, coshSj coshSj, cosh , . , 



p" = z z sinhSj, • — r-£ • — r-r- • ■ — r-s- ohms (110) 

 cosh b x cosh o c cosh 6 A 



= 48,619.7 ohms. 

 The F-leak is then computed as in (82). 



