90 G. H. KNIBBS. 



These last equations reduce to 

 Tl =^[o 1 .2sin(a 1 +15°) sin 15°+i a 2 . 2sin(a 2 +15°) sin 30° 

 + ias .2 sin 3 (a 3 + 15°) sin 45° + etc....] 



r 2 = - [oi. 2sin(ai.+45°) sin 15°+i a 2 .2sin (a 2 + 45°) sin 30° 

 -f ia 3 . 2 sin 3 (a, + 45°) sin 45° + etc....] 

 etc., etc., the angles increasing downwards 15°, 45° 75° etc., 

 and horizontally 15°, 30°, 45°, etc. 



The following results are now obvious : — 



a 4 sin 



+ fa 6 sin6K + 15 )] 



Q 



|Vsin4(a 4 +105 o ) 

 + fa 6 sin6(a 6 + 105°)] 



(23)...r 1 + r 7 = -[a 2 sin2(a 2 + 15°)+^a 4 sin4(a 4 +15 ) 



TV 2i 



(24)...r 4 +r 1 o=-[a 2 sin2(a a + 105 )H-^a 4 sin4(a 4 +105 



7T 2 



etc. 



It is thus evident that by making suitable combinations 

 the whole of the values Oi, a 2 , a 3 ...a 5 and a u a 2 ...a 5 can be 

 obtained, and if a 6 be assumed to be 0, then a& can also be 

 obtained. 



6v3 



(25).. .n 4- 1*4+^7+1*10= — a 4 sin 4(a 4 +15°) =L 1? say, 



For example we have from (23) and (24) 



5). ..*>! + 



similarly 



(25a)...r 2 + r 5 + r 8 +fn = — a 4 sin 4 (a 4 + 45°) = L 2 



7T 



(25b)...r 8 + r 6 + r 9 + r ia = — a 4 sin 4 (a 4 + 75°) . = L 3 . 



7T 



These last three equations may be written 



(26)... L x = ^2 a 4 (| sin 4<* 4 + -^ cos 4<* 4 ) 



7T 2 



(26a)...L 2 — — a 4 . sin 4a 4 



7T 



(26b)...L 3 = ^? a 4 (| sin 4« 4 - 4? cos 4« 4 ) 



TT 2 



from which we obtain 



