Long Waves in Canals and Standing Waves in Closed Basins 227 



new state of equilibrium is at 1, in the case of the wind disturbance (W = 1), 

 it is at 1 -5. In both cases the water level is at the offset higher than the new 

 state of equilibrium. Then, by decreasing oscillations it returns to its new 

 equilibrium position. The second figure shows two kinds of action of a gust 

 of wind on the level at one end of the basin. The gust is followed by a rap- 

 id piling up of the water at the closed end of the basin, which decreas- 

 es by oscillations after the wind gust has ceased. According to Proud- 

 man, we find for different values of the depth h, the length of the basin / and 

 a coefficient of viscosity v, the unit time in the Figs. 96 and 97: 



The period of the oscillations is for the first case 4-68 h, whereas that of the 

 free period of oscillations (without friction) would be 3 97 h. 



These theoretical computations show that atmospheric disturbances of 

 any nature lead always to seiches in lakes and bays, and are always connected 

 with them. 



Of particular interest is the case where a pressure disturbance travels 

 along the length of a uniform canal, so that pressure gradients act only per- 

 pendicularly to the canal. This will cause transverse and longitudinal currents 

 in the canal, which must be accompanied by transverse oscillations. As long 

 as the rotation of the earth is neglected, it can be expected that any pressure 

 disturbance will produce transverse oscillations in the canal, whose period 

 is identical with the free oscillations in the canal. Proudman has given a solu- 

 tion for the case where the earth's rotation is considered. 



Let us assume that the canal extends in the >>- direction and that its shores are at x = and 

 x = 2 a. The depth be constant and equal to h. (Proudman deals also with the case of a canal of 

 variable depth in its cross-sections.) The atmospheric disturbance rj be also only dependent of x. 

 If the friction is neglected, equations (VI. 130) take the form 



du 8 



— -fv = -g—(>l-il), 



dt dx 



8v 



-+fit = 0, (VI. 138) 



du dri 

 dx 8t 



Apart from the equilibrium solution as given in (VI. 131), there is a stationary solution in the 

 form of the gradient current: 



gdZ 

 77 = 0, r) = Z(x), u = and v = - — , (VI. 139) 



f dx 



in which Z(x) is an arbitrary function. 

 15* 



