SECT. 3] BEACH AND NEAKSHORE PROCESSES 531 



energy of the returning water will be less. So the corresponding frictional 

 energy loss, AE2, over the same unit area will be less. For the return water 

 displacement we have 



JE2 = a — — - gm2X2 cos B (tan ^ —tan j8). 



Ps 



But if the littoral profile is mature, as much sediment must be carried down 

 the slopes as is carried up. Hence 1712X2 = miX\ and 



AEi -AE2 SE 2 tan ^ 



AEi AEi tan ^ + tan ^ 



thence 



tan|8 = hE ts^n 4> 1{AE I +AE 2), 

 or, writing AE2 = c AEi, 



1 — c 

 tan^ = tan<^- (18) 



The friction coefficient tan being a constant (0 being approximately the angle 

 of repose of the material), the littoral angle /3 should be independent of the 

 linear scale, but dependent on the ratio of the energy deficit AE\ —AE2 = 8E 

 elsewhere to the loss AEi over the local area. If the deficit is large, so that c 

 approaches its limiting value of zero, 



tan /3 -^ tan ^ 



and the beach stands at the angle of repose of the material. 



If there were no deficit, i.e. if the water tranquilly returned without loss of 

 energy having flowed down a frictionless bed, the angle ^ would approach zero. 



C. Permeability and Sand Size 



The loss of energy between flow and return, which gives rise to the deficit 

 SE, may be due wholly to bed friction, in which case it may be expected to 

 decrease with decreasing grain size, for the grain size should determine not 

 only the boundary roughness but the additional energy loss arising from under- 

 water percolation tangentially through the bed under the large tangential 

 pressure gradients associated with the travelling waves. 



The permeability of the bed material may be expressed, independently of 

 wave scale, as the water discharge perpendicularly through unit area of a bed 

 of thickness equal to the perpendicularly applied head of water. The percolation 

 discharge decreases very rapidly as the grain size of the material is decreased 

 from that of pebbles to that of sand. This is because the flow within the pore 

 spaces then changes from turbulent to laminar. Thus, while sands are relatively 

 impermeable, pebbles and larger materials are very permeable and permit 

 considerable energy losses within the bed. This probably explains why the same 

 wave attack displaces a far greater mass of large-grain material than of small. 



