EQUILIBRIUM OF ROTATING MASSES OF FLUID. 
405 
The greatest discrepancy occurs when /3 = ^ and 6 = 90°, and the difference 
between the two results is y- 0 - of either. It follows, therefore, that in drawing the 
figures it is not of much importance which results we take. But, as above remarked, 
the radius-vectors computed from the true ellipsoidal figure are the more correct. 
The formula (77) for Sr consists of a series of zonal harmonics. The pole of 
symmetry of these harmonics lies in the equator of the ellipsoid of revolution defined 
by r, and is that point of each mass which lies nearest to the other. Then, denoting 
by 6 co-latitude estimated from this pole, I find that the numerical values of Sr for 
each 15° of 6 are as follows :— 
7 
11 
o 
o 
o 
LO 
r— 1 
o 
O 
CO 
45° 
60° 
75° 
90° 
J3=}: 
dv 
«=+- iG5 ’ 
+ -141, 
+ -084, 
+ -019, 
— -031, 
— '055, 
— '056, 
+ -280, 
+ -257, 
+ 442, 
+ -024, 
- -060, 
- -094, 
- '092, 
105° 
120° 
135° 
150° 
1G5° 
180° 
- '037, 
— -004, 
+ -032, 
+ -065, 
+ '088, 
+ '096 
- -059, 
- -002, 
+ ■055 > 
+ 406, 
. + '143,. 
+ 455 . 
These have to be combined with r, so as to give the radius-vectors of the mass of 
fluid along two sections, one perpendicular to the axis of rotation (which may be 
called the equatorial section), the other through the axis and the two centres (which 
may be called the section through the prime meridian). Taking the case of /3 = we 
add the successive values of Sr to the equatorial value of r, viz., 1"043, and thus find 
the equally-spaced radius-vectors along the equatorial section. Next we add the 
successive values of Sr to the corresponding values of r, and thus find the equally- 
spaced radius-vectors along the prime meridian. The results are as follows :— 
O 
O 
11 
15° 
o 
O 
CO 
45° 
o 
O 
75° 
90° 
, O' 
Equator, = 1’208, 
1-184, 
1-126, 
1-062, 
1-012, 
•988, 
•987, 
7 ' Pr. Merid. = 1-208, 
1-174, 
1-091, 
•994, 
•915, 
‘871, 
•863, 
105° 
120° 
135° 
150° 
165° 
180° 
1-006, 
1-039, 
1-075, 
1-108, 
1-131, 
1-139 
*889, 
•942, 
1-008, 
1-072, 
1-121, 
1-139 
These results apply to the case, of (3 = y; those for /3 = j are found in the same 
way, and are given in the figures referred to below. 
When j3 = y the distance between the centres is given by c/a = 2’828. I have 
also worked out the case of /3 = although none of the numerical details are given 
here. 
In figs. 2, 3, 4, and 5 (Plates 22 and 23) are exhibited the figures which result 
from some of these computations. 
