20 
ME. G. I. TAYLOR ON TIDAL FRICTION IN THE IRISH SEA. 
^, the ratio of the masses of the moon and the earth, is , the ratio of the 
E 
radius of the earth to the radius of the moon’s orhit, is (ftr- 
cos^ X = 0'38, 
P = 1-03, 
R = 6-4 X 10® cm. 
Hence 
m = —f X TT X 1'03 X 0’38 x 981 x x (^)®x 6‘4 x 10® x H sin^ 
= — 6‘6 X lO^H sin^ ergs. . (32) 
The mean rate at which energy is communicated by lunar attraction is found by 
dividing this by 87 x lOf the number of seconds in two semi-diurnal periods, i.e., in 
24h. 50m. 
Hence the mean rate at which work is done by the moon’s attraction on each 
square centimetre of the Irish Sea, is 
-6-6 X 10^ 
87x 10" 
(average value of H sitd <^ver the Irish Sea). 
Now the mean value of 2H, the rise and fall of tide in the Irish Sea, is about 
14 feet or 420 cm. Hence H may be taken as 210 cm. The average time of H.W. 
is about l^h. before the moon’s meridian passage. Hence = -t-22|-° and 
sin^ (pa — 07. 
A rough approximation to the average value of H siid cpa is therefore 
210x07 = 150 cm.; hence 
~ xl50 = —110 ergs per square centimetre per second. . (33) 
It will be noticed that since it is H.W. shortly before the moon’s meridian passage, 
work is done by the tides in the Irish iSea on the moon. Tins is indicated by the 
negative sign in (33). 
Dissipation of Energi/ in the Irish Sea. 
We have now seen that the rate at which energy flows into the Irish Sea through 
the North and South Channels is 6'4x Df' ergs per second. 
The area of the Irish Sea between the sections AB and RC is 11,600 square nautical 
miles = 3'Ox 10"" sq. cm. Hence energy enters the Irish Sea through the Channels 
6-4x10"" 
at a rate of —-—^ = 1640 ergs per second per square centimetre of its area. Of 
o y X i u 
this energy we have just seen (see equation 33) that 110 ergs per square centimetre 
