148 Long Waves in Canals and Standing Waves in Closed Basins 



- 8u F—hy 8r\ 



vh = (hy—F) — or vh = — = . 



8 * h 8t 



If we put 



t] = Zcos(o7— xx) ; u= Ucos(at—xx) and v= Vsm(at— xx) (VI. 15) 



we obtain 



V 2n yh — F 



u x it 



This means that the ratio of the maximum values of the cross-current and longitudinal current 

 is proportional to the area by which the hatched part F exceeds the rectangle yh; besides, u is pro- 

 portional to A and h. For a canal with a rectangular cross-section we have always F = yh, so that 

 everywhere v = 0; for a concave canal (as in Fig. 67) we always have F> yh until at y = \b = a 

 (half the width of the canal) F ' = \Q = ah, and there v also becomes zero. Therefore, v has always 

 the same sign as Brjjdt. If the water rises in the canal, the cross-current flows from the middle to 

 the bank; if it drops, it flows from the bank to the middle. These conclusions, of course, apply 

 only if there is no tranversal slope of the water surface in the canal. 



Airy (1842, par. 358—63) has expressed the opinion that, in a wide canal, 

 in which cross-currents are probable, the wave crests are bent because the wave 

 will travel faster in the middle of the canal than at the shallower sides. But 

 rather soon it was recognised that this idea did not seem to be conclusive. 

 Much later Proudman (1925, p. 466) has proven that for a canal with 

 a parabolic cross-section, Airy's assumption in regard to the bending of the 

 wave crests is not correct. The crest line is always a straight line perpendicular 

 to the canal, and the effect of different velocities of each part of the wave are 

 equalized. To the contrary, the wave height perpendicular to the canal does not 

 remain constant and increases towards the bank. At high water, the current 

 flows in the direction of the canal and the cross-current is zero ; with falling 

 tide the current flowing towards the middle of the canal becomes stronger and 

 attains its maximum, when the surface goes through its position of equi- 

 librium, then it decreases again and becomes again zero at low water, when 

 the current flows again in the direction of the canal, but now in the opposite 

 direction. With rising water, the cross-current flows from the middle to the 

 banks. The current, therefore, can be represented by an ellipse with a dif- 

 ferent direction of rotation at both banks; in the middle of the canal there 

 is only an alternating current. 



The equations of motion and the equation of continuity, in this case have the form 



8u 8r\ 8v 8ij 



dt~~ g dx' 8t~ 8 8y 



d 8 8n 



— (h.u)+—(h.v)+ — = 0. 



8x 8y 8t 



(VI. 16) 



Moreover, the solution must satisfy the condition that (VI. 18) 



hv = for y= ±a, (VI.17) 



