MR. F. W. DYSOiSr ON THE POTENTIAL OF AN ANCHOR RING. 
1081 
Thus /Sj will be of the fifth order. 
Therefore, correctly to the fourth order, the equation of the cross-section is 
R = a {1 +/dj cos 2;)(-{-ySg cos 3^ +cos 4;)(} .... (58). 
5 30. Let oi be the molecular rotation, and the stream-line function. 
Then 
outside the ring, and 
rf-'F 
dz^ 
+ 
1 
CT dw 
d-^F d^q^ 
dz^ 
1 . o 
— -— -f- 2(770) = 0 
m dTtS 
inside the ring, while q^ and d^/dn are continuous at the surface. 
Let the vorticity be constant throughout the ring. 
Then O) = (M/2) nr, where M is a constant. 
Write q^ = aiid the above equations become 
and 
1 I 1 
dz- du5^- dvs 
dlx I _| 1 lx 
d.z~ dm~ OT dur 
" I 
• 
^„ + Mr^ = 0 i 
Therefore x cos (f) is the potential of matter of density (Mct cos (f>)f7T occupying 
the same space as the ring. 
Therefore, at any external point, ct', z, (f)', 
X cos (f)' 
]\1 ,, r r r cos ^ dm dz d^ 
Ttt ' J J J \/ — 2wct' cos (^ + ot® + (fi — 2 )"} 
Therefore 
j\I , r r r m~ cos (f> dm dz d(j) 
Ttt J J ]\/{m'^ — 2mm' COS h + m“ + (z' — z'f] 
(59), 
the integration extending all over the ring. 
(The value of q^ at an internal point may be found as the solution of 
d^^F d-^Jf 
d^ + dm"- 
1 
m dm 
Mto" = 0, 
which gives values continuous at the surface). 
Let TT = c — X. 
Then 
MDGCCXCIII.—a. 6 Y 
