MR. G. I. TAYLOR ON TIDAL FRICTION IN THE IRISH SEA. 
23 
The tidal current is a maximum when x = 0 and t — 0. Its value is then 
V = (a + 6)/y/». 
(35) 
At the point a? = 0 the phases of the current and of the height of the tide are the 
same. In applying these equations to the channels of the Irish Sea, we must choose 
as origin, x = 0, the place where the height of the tide and the current have the 
same phase. As we have already seen in the South Channel, this point must be very 
close to the section AB from Arklow Bank to Bardsey Island, for the phase difference 
in that region is only 10m. of time. It is hardly worth while in the present investi¬ 
gation to take account of so small a difference as 10m. 
At the point x = 0, the range of tide, which we have called 2Hi, is evidently 
2{a—b). Now we know the values of V, g and D * we can therefore calculate 
a + h 
from (35), a and h can therefore be found separately. 
Using the values already given for the section AB, 
it will be seen that 
and 
Hi = tidal range = if feet = 145 cm. 
D = 37 fathoms = 6800 cm. 
9 ^ = 981 . 
V = 3'2 knots =163 cm. per second. 
a + h 
ICO A /3800 ,0,1 
163 A/ -= 430 cm 
V 981 
a — h = Hj = 145 cm 
(36) 
(37) 
Hence, solving (36) and (37) we get for the semi-amplitudes of the in- and out¬ 
going tidal waves, 
a = 287 cm. 
6 = 143 cm. 
and 
(38) 
It appears, therefore, that at spring tides, the tidal wave is reduced almost exactly 
to half its amplitude during its passage into and out of the Irish Sea. The wave 
which comes out of the Irish Sea therefore contains only quarter of the energy 
of the wave which goes in. 
The result just obtained does not appear to be open to any theoretical objection, 
but it is opposed to the generally accepted view that tidal friction has very little 
effect on the regime of the tides. It is worth while, therefore, to try and confirm it 
