13 



Appendix III 

 A. We shall now solve the differential equation encountered in eq. (5). We had 

 dT/T = cosV^^sin^ d ♦' Put Z =- tan 6/2. Then 



tan A = - r ^s , sin A = - r |2_ y , cos A = f^-|J , d A = - r | d | s 



Hence 



f - 2Z 



w/T - _ 1 + Z a 2dZ „ 1 - 2Z/f + Z 2 ,„ 



V ~ 1 - Z a , . . 4^~ x " TTZ* " " 2f 1 + 4rZ g - Z^ dZ - 



1 + Z 5 (1 + z 2 ) 3 



But 1 + 4rZ 2 - Z 4 = 1 + 4r 2 - (2r - Z 2 ) 3 = (p a - Z 2 ) (1/p 2 + Z 2 ) 



where p 2 = V 4l " a + 1 + 2r. Hence we have 



ht/t - 9-p 1 - 2Z/f + Z a , 7 _ 2f f l - 1/p 2 - 2Z/p . 



dT/T " " 2f (p 8 - Z a )(l/p s + Z 8 ) dZ " - P a + Vp 8 L 1/p^Z 8 P + 



1 * f + 2p/f + 1 ± f - 2 p/f I 

 2p(p + Z> + 2p(p - z5 J dZ 



as may be readily verified. Integrating, we obtain 



log T/To = - pg 2f l/p a [p(1 - 1/p 2 ) tan" 1 pZ - 1/f log (1 + p 2 Z a ) + 



4p^ lQ g f^ + -r lQ g ( 1 - za /? a ) ] 



where T Q is the tension for Z = 0. This last is seen to be equivalent to equation 

 (6). 



B. Treating equation (9) in a similar fashion, we have 



d T/T = *** ^..fni,, • Put Z = tan 4/2. Then 

 cos 4 — r sin 9 



tan 4 = -j _ z a » sin 4 = ^ ^ z a , cos 4 = ^ ~ z a , d 4 = 1 + z a 



