MR. G. 1. TAYLOR ON TIDAL FRICTION IN THE IRISH SEA. 
19 
da- being an element of surface, and the integral extending over the whole surface of 
the Irish Sea included between the two sections AB and RC. 
The potential of the moon’s attraction on the sea is represented by the function 
a = in . ....... (28) 
where 
”] is the constant of gravitation. 
M is the mass of the moon. 
is the radius of the moon’s orbit. 
R is the earth’s radius. 
3- is the angle between the line joining the centre of the earth to the moon, and 
the radius of the earth which passes through the point on the earth’s surface 
which is being considered. 
If X be the latitude of the place {i.e., 52° in the case of the Irish Sea), and if 96 be 
the angle through which the earth has turned relative to the radius vector to the 
moon, since the moon was on the meridian, then by spherical trigonometry, 
cos S'= cos X cos 0. . ..(29) 
Also, if 2H be the range of tide at the place which is being considered, 
= H cos 2 (0 + 0 o)>.( 30 ) 
where is the phase of the tide at the time when the moon crosses the meridian. 
Combining (26), (28), (29), (30), it will be seen that 
m 
Tvrps r 2 Tr 
Hf“I ytW COS^ X COS 2(0 + 0o) ^ (cos^ (p) 
'P-n-Y 
D/ 
A 
D. 
■> tt-1 MR^ • 2 W 2 ^ \ 
= -|7rpHn ^p^siin 96 o(cos^x). 
( 31 ) may be written 
m = —f 7rpH cos^ X siiT (p^ ( 
RV\E/\D, 
where E is the mass of the earth. 
HE . 
—!- 1C2 
R2 
is the attraction of the earth at its surface, i.e., 
= g = 981 in C.G.S. units. 
(31) 
* See Lamb’s ‘ Hydrodynamics,’ p. 339 (1906 edition). 
D 2 
