182 
MR. -J. H. JEANS ON THE CONFIGURATIONS 
28. The material for numerical computations has already been given in § 18. 
Using the exact solution there given, I find, by direct computation from equation (90), 
4e, = 4e 2 = -0T3404 ; 4e 3 = 0-06808.(112) 
From equation (I ll), and the companion to (108), I now find 
=£ = -0-016037; ^ = 0-056337,.(113) 
d c 
and equation (108) now gives 
A 2 
A n= - 7 = -0*04400.(114) 
2irp 0 ClOC 
These quantities are all comparatively small; they would all have vanished under 
Approximation B. 
Second Order Solution. 
29. We now pass to the solution of higher order still. On equating terms in e 2 in 
equation (50), we obtain 
7r< 
abc 
(<SV,.( 1 )-AE ,) + Sn(x‘+ V ’) 
= -e 
«3 0 -(y-2)P 0 (2^)+i(y-2)(y-8VS^ 
a- 
X 
2 \ 3 
a 
30. From equation (28), taking terms as far as first powers of e only, 
E, + AE ( - = -wabc 
in which the lower limit q is given by 
d\ 
~K 
+ ——r + ——7 +^> (?) = 7' 
« s + A If + A tf + A 
As far as e, equation (12) gives 
A?) = UP-i/DP+A/w], 
• (ns) 
(116) 
• (117) 
while from S 8, 
2 _ x 2 
y 2 
qJ - + + ?Ta ~ q ' ' • 
. (118) 
• (H9) 
D 
9 
<r 
a ” + (k- 
v « 2 a 2 + X/dg 2 \b 2 b 2 + xj Cyf \c 
1 \ ^2 
I \ 0 
+ 1 ~ ~ „2 
o+x; ci 
0 2 
( 120 ) 
