THE AERODYNAMICS OF A SPINNING SHELL. 
235 
Finally equation ( 8 ) gives 
7 a 4X 2 a 2 cos 2 k 
The equation for g is obtained by substituting for q, h 2 and / 2 in (7), which becomes 
4X 2 (l — a 2 ) (a 2 .—g 2 ) {d 2 — g 2 -\- (l — a 2 ) k 2 } — a 4 (l— g 2 ) = 0 ; . . . (18) 
this is a quadratic for g 2 , whose solution may be written 
g " = sin 2 \cl ^1 - , (sin = a), .(19) 
where 0 is given by 
cot 0 = XF cot 2 \v-— (tan 2 |-a)/4X.(20) 
The ambiguity is settled in practice by the fact that g 2 must be small if equation (14) 
is to be taken as a first approximation to the solution. In practice, as we have said, 
b is small. Valuable information as to the nature of the solution is, therefore, obtainable 
by considering its limit as b -> 0 . This gives us a guide as to the actual relative order 
of all terms. 
Let us suppose then that b -> 0 , that s and q are definite constants, and let us assume 
that g is of the same order as a, which by (19) and (20) must be the case unless X -> 0. 
Then equation (17) shows that \ 2 a 2 cos 2 k/( 1 —g 2 )^0, and, therefore, in the limit, 
1 j s — l = — 4X 2 cos 2 k .(21) 
Equation ( 21 ) shows that, if s ^ 1, X -> 0 is impossible. Hence in all cases (s ^ 1) 
our assumption as to g is justified and a cos k -> 0 . This also justifies our previous 
statement concerning (15). There are now two cases according as s < 1 or s > 1 . 
Case (i).— s < 1 . We are supposing that s — 1 is fixed, so that as b -> 0 , 5—1 is large 
compared to b. To satisfy the signs of ( 21 ) we must have k > 45 degrees. This 
implies sin 2 k>|, so that (16) becomes in the limit 
4 q = 
1 —a" 
- + 
4X 2 sin 2 
a~ 
It follows that a 0 is impossible, and therefore cos k 0, k -> 90 degrees, and a and X 
tend to definite non-zero limits. The limiting forms for 1 /s, q and g 2 are easily found 
to be 
1/s — 1 = 4X 2 ,. (22) 
4 q = — sec 2 |-a + 4X 2 cosec 2 /a..(23) 
g 2 = sin 2 ^ol {1— (tan 2 Ta)/4X 2 },.(24) 
VOL. CCXXII.—A. 2 L 
