234 Prof. Morton and Mr. Tobin on Times of Descent 



Therefore KA'B = inclination of BA' to the horizontal 

 tangent at B = 7, 



AKA'=AA / K=7r-7-(7-«). 



Fig. 6. . Fig. 7. 



Therefore in the triangle AKA 7 



2 (it — 27 + a) + (7 — «) = 7T 



or 37 — a = 7r, 



a simple relation connecting the inclinations of the two parts 

 of the path. 



(2) Secondly displace A horizontally (fig. 7), then the 

 speed at A is unchanged in the varied path, (/all this v u and 

 the speed at 0, v 2 . The time is 2cjv l -\-2a\{v l -\ : v^. 



The variation of this vanishes, so 



But 



and 



Bc/v 1 + Ba/(v l ^-v 2 )=0. 



Be — — A A' cos 7, 8a = A A' cos a, 



COS7(v 1 + f 2 ) =cosa. I*! 



r 2 2 : r 2 2 = c. sin 7 : fr sin /3 



= siny sin (/3— a) : sin /3 sin (7 — a), 

 .'. cos a sec 7= 1+ {sin /3 coaecy sin (7 — a) cosec(#— a)K' 

 If u = 7r— 37 be put into this, it gives 



\ 2 sin/3 cos7Cosec (07 — j3) \* = —2 cos 27, 

 which can be transformed into 



tan /3 = tan 7 (1 + cos 47)/ (cos 47 — 2 cos 27 + 2). 



So starting with 7 we may calculate /3 and a, and then 

 plot against the inclination /3 the corresponding values of 



