56 PROCEEDINGS OF THE AMERICAN ACADEMY. 



The sending-end resistance at A s , as before, is 



R/a — ~i coth 8 A ohms (87) 



= 1,013.04 ohms. 



The AO line branch is therefore R u — B' = 960.961 ohms. 



Repeating the process from A 4 towards F 4 , we have for the T leak 

 conductance, as in (80), 



. , coshS^ cosh8 fl , , 



9 = *° Smh 8 ' • c^hT, " co^U c mh0S - (bh) 



The sending-end resistance at F is likewise 



B fF =z 3 cothSj,. ohms (89) 



= 2.500 tanh 1.533,091 = 2,277.39 ohms, 

 from which the resistance of the line branch FO follows. 



Equivalent T. Second Method. 



The second method of arriving at the T-leak conductance is by fol- 

 lowing (83) and (84). Freeing at A 4 , we have 



, ... sinh S D sinh 8, , 



q = ?/! smh 0! • . , ^ • . , , mhos, (90) 



J J sinh b c smh b E 



and freeing at F 3 , after developing the line angles, we have 



, . , sinh8 c sinh 8^ 



/ = Va sinh a ■ g^ • Wj mhos. (91) 



Composite Line of n Sections. 



To compute the equivalent n of a composite line of n successive 

 sections, ground the line at the A end and develop the line-angles 

 towards the opposite end, following the process of (76). Find the 

 architrave impedance according to formula (80) or (83). This may be 

 regarded as formula (19) modified by the application of (n — 1) ratios 

 of cosines in (80), or of (n — 1) ratios of sines in (83). The opposite 

 end leak admittance will then be the sending-end admittance minus 

 the architrave admittance. The process must be repeated after ground- 

 ing the line at the distant end and developing line-angles towards A. 



'To compute the equivalent T, free the line at the A end and develop 

 the line-angles towards the opposite end, following the process of (85). 



