-it/to* - F + W sin (1) _ f + sin » _ (Q ^ 



dT/Td * ' W cos <* - R sin a <}> " cos 4, - r sin a * (9) 



Introducing the substitution Z a = tan 4>/2, (9) may be integrated (see Appendix III) 

 to give 



log t. - -2f p B ; Y%, tan" 1 Z a /p ♦ pJ ? 1/p . log 1 , 1 glffi* * 



'pH^-^Hjfc w> 



where t a = T 3 /T , T a = T f or > 0. 



Also from equations (1) and (4b), we obtain 



"^ = W cos f? t sin a * < 11 > 



Making the substitution Z a = tan 4>/2 , (11) becomes 



Z a 

 v . f 4Tn t 2 Z 2 dZ 2 (12) 



Y_ J w ( P a + z a s Ki/p a - z a a ) 



o 



We conclude the general analysis by deriving a relation between the tensions and the 

 length of the cable L between points determined by a prescribed ordinate Y. For a 

 given ordinate Y let T.., S-. for <t>< correspond to T a and S a for $ > 0. Then by (3), 



T 1 = T + FS 1 + WY and T 8 = T + FS a + WY. Hence 



T, - T\, = F(S a -S 1 )=FxL (13) 



noting by (1) that S 1 is always negative. 



Application to determination of cable resistance. Approximations. 



In the particular problem with which we shall concern ourselves, the ends of 

 the given cable are points having the same value for the ordinates. It is required 

 to find the tensions at these ends as functions of the angle <f> at the rear end of 

 the cable (<j><0); or, what is the same thing, as functions of Z... But by (13) 



T * S ft VT - T> = F X L *T^T 04) 



Hence a procedure for obtaining numerical solutions of T.. and T a as functions of Z.. 

 may be outlined as follows: 



1. Compute values of t 1 against Z 1 from (6) 



