Problems of Denudation. 



183 



1000&, where k is thetkinematic coefficient of viscosity, prac- 

 tically 0*02 cm. 2 /l sec. Thus we must have AZ>20cm. 2 /sec. 

 A moderate rainfall would be one centimetre in an hour, or 

 about 3 x 10~ 4 cm. /sec. For the law to be correct I must 

 then be greater than 0'7 kilometre. Again, using the 

 relation fV 2 =g£ sin a, we have 



/V 3 = 0*(<?AZ sin a) 

 = 0(gAh), 



where h is the height of the highest point. Thus we have 



t=«{>m 



= 0(7x10" 

 with the above data. Taking h 



> 



(7) 



10 5 cm. and Z=10 6 cm., 



these figures corresponding to a mountainous region, we 

 have J= 0(0*7) cm.; thus a steady depth under a moderate 

 steady rainfall would be attained after a time of the order of 

 an hour. In flat regions, on the other hand, we may have 

 A = 10* cm. and Z = l0 7 cin.; then f= 0(30) cm. This is 

 evidently incorrect, for rain does not ordinarily last long 

 enough to flood the ground to this depth. It follows that 

 Vf must be less than 1000 k, and the friction is not due to 

 turbulence, but to ordinary viscosity. 



Now if u be the velocitv at distance v above the 



surface, the equation of viscous motion is 



~bv 2 



k ^r-7> = — g sin «, 



"du 



solid 



(8) 



while u — when v = Q, and =r— =0 when v = t. 



These give 



q sin ol a 



u =~Y k 0£-*)- • • 



00 



In the equation of continuity we can take \ to be the mean 

 value of u over a normal section. Then 



CJ = 3 ^" T~ ' ' * ■ (10) 



Thus V£=?f 3 sina/3£ (11) 



O(.v) denotes a " quantity of the order of x" 



