AND THE ACTION OF TWO VORTICES IN A PERFECT FLUID. 
509 
The velocity along the radius vector 
m'd 3 a 
+ cos 6\ — 
4/a „ a . ^(F* 3 +G£ 3 +H£+K) 
4 (c v + 4ft 3 sm 3 f e. t 3 )-' ' 
{fc 3 sin e—6cv sin fe.£+2 sin 3 Je sin e.vH 2 } 
/ /o 
ma* 
where 
2 (c 3 + 4ft 3 sin 3 f ei 3 ) 
+ cos 2(9f— 
/ JO 
^(FT 3 +GT 3 +m+K') 
(c 3 + 4ft 3 sin 2 fei 3 ) 
F=r 3 sin 3 fe(2 sin fe —14 sin fe) 
G= cv 3 sin 3 fe(cos fe—13 cos fe) 
K=±c 2 v sin fe(sin fe +13 sin fe) 
K = Jc 3 (cos fe+7 cos fe) J 
(28) 
F'=2i; 3 sin 3 fe(sin f e+ sin fe) ^ 
G'=cv 2 sin 3 fe(cos fe—21 cos fe) I. 
IT=fc 3 ?; sin fe(sin f e+21 sin fe) 
K/—fc 3 (cos fe— cos fe) 
* « 
(29) 
We can now write down the differential equations giving a» and ft. We shall 
begin with that giving ft as the simplest, as a reference to equation (19) will show 
that the vortex ring C D contributes nothing to this term, so that 
/ /o 
m a A a 
1= ^^ E ^+°^+D) 
integrating and writing for brevity k instead of 4r 2 sin 3 fe we find 
Ac 3 C 
ft- 
f m'd^a ■ 
c 4 /c 3 x 
3) 
A 
A. 
5 (C 3 + /C 3 ftt 3 ( C 3 + *¥)?’ 
Br 3 
d-4 
i 
AD . B 
t 
O Q 
a c 3 
(c 3 4- /c'V 3 )t 15 (c 3 + a; 3 £ 3 ) 
'8D . 2B 
+*(- 
„6 1 rAu-’i 
t 
, 1 \ 
(c 3 + K 3 t 3 )* 
_r 2ft sin fe J 
where the arbitrary constant arising from the integration has been determined so as 
to make ft=0 when t= — oo . 
Substituting for A, B, C, D their values we find 
ft- 
1 m'a?a \ *-**»¥) 
L " ft Sill f e 
C Bin %6 
(c 3 A /c 3 t 3 j 
cos fe — Scosfe 
1 
6ft sin fe (c 3 +aA 3 ) 2 
(sin fe + sin fe)e (sin fe + sin fe)/ e 1 > 
(c 3 + K 3 t 3 f 12 c(c 3 +M ! 6c 3 l(c 3 + « 3 i 3 )^2ft sine. 
(30) 
