WITH THE TIDES OP A VISCOUS SPHEROID. 
573 
wv Y 
p— v i - l$aV 2 +(Fa 4 }, 
Where 
« = 
7(7+1) 
2.4(27 + 5)1’ 
s = »^+ 1) (2^+ 1 0i + 9) | and 
® = 
21 
4(7-1) (27+ 5)I 3 
(7+1) (27 + 3) 
•/ 7+; 
7-1/ 1 (27+1) (27 + 5)J 2.4(7-1) (27+1)1 
7(7+l) 2 
If © be reduced to the form of a single fraction, I think it probable that the 
numerator would be divisible by 27+1, but I do not think that the quotient would 
divide into factors, and therefore I leave it as it stands. 
In the case where i— 2 this formula becomes 
p= ~ 1 sin vt{1 9+-148+V + 287a 4 }, 
2 3 .3.19v 
which agrees (as will appear presently) with the same result obtained in a different 
way. 
1 shall now go on to the special case where 7=2, which will be required in the 
tidal problem. 
From (39) w r e have 
From (36) 
From (43) 
From (52) 
wva? 4 
a ° v 2 ‘ 19 3 
TT tov 1 . 
U = - • 0 r -i sm vt. 
v 2.5.19 
, 3.7 MY 2 d .. 
4a- — 0 -+- — — pr 7 — (Y r °) 
2.o / dx 0 dx ' ' 
sm vt. 
1 
y 3 2 3 .3.5.19 
1 
^S+vi< Y + 
sm vt. 
wva 
a = 
y 3 2+3.5.19 2 
( 5 . 97 ^ 7 . 4 ++++-^ 
Adding these expressions together, and adding a 0 , we get 
, wv 1 
.3.5.19 2 
c/Y 
(5.287a 4 —37.4.7aV 2 +9.19r 4 )— 
— §(2.3 7a 2 — 19r 2 )r 7 ^-(Yr“ 5 ) 
sm 
(53) 
and symmetrical expressions for /3 and y. 
• co y z 
In order to obtain the radial flow we multiply a by -, (3 by -, y by -, and add, 
and find 
nmi 1 V 
• • (M) 
wv 1 Y 
p=p 0 +— • 1Q i (287a 4 —4.37aV+19r 4 ) - sin (W + e) 
y 2 2 2 .3.19 2 
the e which was omitted in the trigonometrical term being now replaced. 
