180 Dr. H. Jeffreys on 



any point depending on the shape of the neighbouring ground. 

 The present problem is to find out how the water will flow. 



Consider any particular small portion of the surfaoe, of 

 area dS, and let the depth of the water be f and its density p* 

 Then the forces acting on the element of water of mass p£d$ 

 and affecting its movement over the ground are : — 



(1) The tangential component of gravity, of amount 

 gp£d$ sin a, along the line of greatest slope, where u is the 

 angle between the normal to the surface and the vertical. 



(2) The friction of the ground; when the velocity is con- 

 siderable it has the value fpV 2 d$, where / is a constant of 

 order 0*004 and V is the mean resultant velocity of the water 

 within the element. The direction is against the resultant 

 velocity. 



(3) The pressure of the surrounding water. Now the 

 pressure is zero at the upper surface of the water, and as 

 there is practically no movement perpendicular to this- 

 surface, the normal component of gravity must be almost 

 exactly balanced by the pressure. If v be the distance from 

 the surface of the ground, this makes the pressure equal to 

 gp(£—v) cos a; and therefore the difference between the 

 pressures at two points ds apart and at the same distance 



from the bottom is of order gp^r (f cos a)ds. Thus the whole 



Os 



thrust on the element dS in the direction of ds is of order 

 gp^- (? cos a) dS. 



The resultant of these three forces is the rate of change of 

 momentum of the element of water p%d$. 



Now the ratio of the third force to the first is 



~- (fcosa): sin a, and provided that the depth of the water 

 OS 



does not change rapidly in comparison with the height of 



the ground above sea-level, which is obviously a legitimate 



assumption, it appears that the pressure variation can be 



neglected. 



Next, suppose for a moment that most of the force of 



gravity is used in producing acceleration, friction never 



exceeding a certain definite fraction of it. Then the velocity 



acquired in descending through a vertical height h is given 



by V 2 = %A, less a correction for friction. This makes the 



f rictional force equal to 2fgphd$, whose ratio to the force due 



to gravity is 2fh : fsin a, which is of the order of the ratio of 



•008 of the linear dimensions of the area to the depth of the 



water, and is obviously very large in all ordinary cases^ 



