20 THE TIDES. 



Then, as in fig. 5, 



77" 



velocity = j- sin 2X cos 2(45 - m). 



(T being as in p. 12 the angular velocity relatively to the 

 moon). 



Now, this increase in the height of the water depends on 

 the difference in velocity at two points, of which the lati- 

 tude is X and X + a where a is very small. In fact the 

 difference in the amount of the water entering the section 

 and leaving it is equal to the area of the section multiplied 

 by the difference of velocity, and the decrease or increase 

 of height is equal to this difference of amount divided by 

 the area of the surface, i.e. 



. . . , depth x increase of velocity 



Decrease of height = 



ra 



(the height increasing when the velocity is diminishing, 

 and vice versa). 



But a being small, 



sin 2(X + a) - sin 2X = 2a cos 2X ; 



7) TT 



.'. decrease of height = - cos 2X sin 2m (D being depth), 



DHcoB2\ 



and total rise or fall = - - cos 2m. 



r 4r 2 



This is = cos 2X x half the total rise or fall in the equa- 

 torial canal with the moon in the equator. 



After passing 45 latitude, the decrease in the circles of 

 latitude becomes important. If we assume our meridional 

 canal to be of uniform width, then the canals will gradually 

 overlap, the tide thus diminishing until at the pole, as is 

 obvious, there will be no tide. 



