the Problem of Columbus. 461 



Using (1), (8), (9), (10) this becomes 

 PE[(« + A cos Bye 2 + {^ sin 6{h + a cos 0) — « 3 a sin [ 2 

 + A 2 sinWJ + iA(<9 2 + ^ 2 sin*0) + i(Ja> : > 2 + ]%A cos 4= constant. 



•■*{»■+ (^^^^ 



-2^ sin 2 6>/~ + cos #) + ^ 2 ( a ^ -*■ *in*0) + ^cos 0=K, . (11) 



K being the initial value of the left-hand side of the equation. 

 Multiply (2) by a sin e and (4) by (h + a cos 0) and subtract; 



then . ... 



Cffl 3 (/t + a cos.0) — Cft) 3 a sin 00 -f aK^n sin 2 + 2aA0i/r sin cos = 0- 



Integrating and dividing by a 

 CaJ- + cos0) + A^sin 2 = \=C^/^ 4- cos O ) H- Am sin 2 o , (12) 



where o> 3 , m, O are the initial values of co s , ty, 0. 



This is the equation of angular momentum round the 

 vertical through 0. 



5. Substituting from (8) and (9) in (6) : 



yjr sin QQi + a cos 0)+^0 cos 0(A + a cos 0) — <r\^0sin 2 

 — o> 3 a sin 0— a cos00&> 3 + (a + A cos 0) 0-^ = ^ 2 



c ?3__£rom(4). 



«M sin 



.'. ^ sin 2 0(- + cos 0^ + 2^0 sin cos fl(- + cos 6} 



' • / C A 



= &> 3 ( -^ -f sin 2 Wfiij sin cos 00, 



