FIGURE OF EQUILIBRIUM OF A ROTATING MASS OF LIQUID. 
27 
The general solution, apart from special values for the rotation, is such that, in my 
notation. 
2 7 Tp 
= n + e/n" = 014200 ( 1 + - 
J! 
014200 
Hence I find that Darwin’s solution ought to coincide with the solution obtained 
in my previous paper on assigning to n" a value fi/' D given by 
n" D = -0-015988. 
The solution given in the present paper, which is believed to give a true figure of 
equilibrium, is derived from the general solution by assigning to n" a value n "j given 
by 
n” j = +0-007423. 
In my previous paper, it was shown that there was, so far as second-order terms 
only were concerned, a doubly infinite series of figures of equilibrium, and it was 
found that these could be defined in terms of two independent parameters, e and £, 
where £ was the same thing as e 2 n". It accordingly appears that any figure of 
Darwin’s series differs from my figure of equilibrium having the same value of e, 
through his £ being different from mine. The excess of my £ over his will be 
D-d = ( n j n d) = + 0 023411c". 
It is readily seen that the increase in the moment of inertia over that of the 
critical Jacobian, which has been called Ak 2 , will be a linear function of f and e 2 , say 
^+ne 2 , 
Hence the excess of my angular momentum over that of Darwin, will be 
ALj-d- 
The value of n is easily obtained by allowing e 2 to vanish in the analysis given on 
p. 72 of my previous paper. The quantity m then appears as the rate of increase 
of k 2 as we pass along the Jacobian series of ellipsoids. The general value of k 2 , in the 
notation there used,* is 
k 2 = -l(a' 2 +b' 2 ) = i« 2 (l-12-71347f) + ^ 2 (1 + 9-20894^), 
whence 
fi = -7-84851. 
of 
The excess of my value of k 2 over that of Darwin ought accordingly to be 
/*£r_ D = -0-183060ej 2 . 
* I have changed the sign of which, by an oversight, had been taken with the opposite sign in my 
formuise as printed in the previous paper. 
E 2 
