184 WAVES IN LIQUIDS. [CHAP. VII. 



provided p n + ri, ri being the angular velocity of the moon in 

 her orbit, and a some constant. The equation (16) then becomes 



-+. ............... (17). 



The solution of this equation consists of two parts. The first part, 

 or complementary function/ is the solution of (17) with the last 

 term omitted, and expresses the free waves which could exist 

 independently of the moon s action. The second part of the solu 

 tion, or particular integral, gives the forced waves or tides pro 

 duced by the moon, and is 



c p a 

 The corresponding elevation rj is given by 



The tide is therefore semidiurnal, and is direct or inverted, i. e. 

 there is high or low water beneath the moon, according as c is 

 greater or less than pa. Now 



c 2 g h , _ or . li 



-5-j = 2^ ~ &amp;gt; nearly, = 289 - , 

 pa n 2 a a a 



in the actual case of the earth. Unless therefore the depth of the 

 canal were much greater than the actual depth of the sea, the 

 tides would be inverted. 



For the case of a circular canal parallel to the equator in lati 

 tude X, we should find 



X = fj, cos X sin 20, 



where 0=pt - + a. 



a cos X 



Substituting in (16), and solving as before, we find for the forced 

 waves 



i fjiah cos 2 X / x \ /in v 



V = 2 &quot;2 a~s ^ coszlpt - + a (19). 



c p a cos X v acosX / 



If the latitude of the canal be higher than arc cos , the tides 



pa 



will be direct. 



