16 Sir G. Greenhill on 



Substitute in the dynamical equation (4), and divide 

 out C, using (6), 



jake^c^rji sin (4a— h) +rj 2 cos (4a — h)~\ 



TJ 2 

 + e — c 2 ^ sin (2a— K) + rj 2 cos (2u — h)~\ — ^ sinA + »7 2 cosA 



U 2 

 + — c\_rfi cos (a — ft) — ^ 2 sin (a— A)] = 0, .... (33) 



and equating to zero the coefficients of rj 2 and rji, 



1 U 2 U 2 

 j^£e 3 c 4 cos(4a — h)4-e — c 2 cos (2a— A) + cos A c sin (a— A) = 0, (34) 



■*•*' .7 y 



1 U 2 U 2 

 .r^&eVsin (4a— h) +e — c 2 sin (2a— A) — sin A H ccos(a — A) = 0, (35) 



c/ if 



1 U 2 U 2 



i-o^V cos 4a + e — c 2 cos 2a — - csin a + 1 



-tan A-ifj 2___ 2_ , . . (36 ) 



— - n ke*c A sin 4a + e — c 2 sin 2a H c cos a 



1 U 2 . U 2 



^ ^« 3 c 4 sin 4a + e — c 2 sin 2a H c cos a 



tan /i — —z Tj2 tt2 » • • v" ' / 



:— &*? 3 c 4 cos 4a + £ — c 2 cos 2a c sin a -f 1 



12 £ ^ 



/ I U 2 U 2 \ 2 



( ttk ke z c i cos 4a + e — c 2 cos 2a c sin a-f 1 ) 



\12 9 9' 



/ 1 U 2 U 2 \ 2 



+ ( t-cs keW sin 4a + e — -c 2 sin 2a H ccosa) = 0, . . (38) 



112 g g ) K } 



1 U 2 U 2 

 r^foVcos4a+<? — c 2 cos 2a c sin a+1 = 0, . . . (39) 



1 U 2 U 2 

 -- /ta 3 c 4 sin4a + e— c 2 sin2aH ccosa = 0, (40) 



12 g g 



1 TT 2 . U 2 



~ke z (a + bi) i + e—(a + biy + i-~(a + bi) + 1 = 0. . . . (41) 



Then, with a + bi = im, 



1 TT 2 TT 2 



±-keW-e-m 2 -^-m + l = 0, (42) 



12 9 9 



1 1 7 3 3 



TT2 - + To W 



Vl = n±12 (43) 



^ 1 + em 



as 



before in (9) § 4, in deep water, with coth mft = L 



