20 MR- 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 D m 



radius of the earth to the radius of the moon's orbit, is ^ 



cos 2 X = 0'38, 

 P = 1'03, 



R = 6'4xl0 8 cm. 

 Hence 



m = -f XTTX! '03 xO'38x 981x^4; x( B 1 (T ) 3 x6'4x 10 8 xHsin 2 



= -6'6xl0 4 Hsin 2 ergs (32) 



The mean rate at which energy is communicated by lunar attraction is found by 

 dividing this by 87 x 10 4 , the number of seconds in two semi-diurnal periods, i.e., in 

 24h. 50m. 



Hence 2%, the mean rate at which work is done by the moon's attraction on each 

 square centimetre of the Irish Sea, is 



WM ~TTT^ ' j~ x (average value of H sin 2 over 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 ifh. before the moon's meridian passage. Hence </> = +22^- and 

 sin 2 = 07. 



A rough approximation to the average value of H sin 2 </>,, is therefore 

 210x07= 150cm.; hence 



/> / 



W M = Q. 7 x!50 = 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 Sea on the moon. This is indicated by the 

 negative sign in (33). 



Dissipation of Energy 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 10 17 ergs per second. 



The area of the Irish Sea between the sections AB and RC is 11,600 square nautical 

 miles = 3'9x 10" sq. cm. Hence energy enters the Irish Sea through the Channels 



6'4 x 10 17 



at a rate of - = 1640. ergs per second per square centimetre of its area. Of 



o y x j_ \j 



this energy we have just seen (see equation 33) that 110 ergs per square centimetre 



