F = 760 feet for the 90-niile per hour windspeed, so that the equiva- 

 lent fetch is 



Fg = 760 feet . 



The vegetation does not match any of the curves in Figure 15, but falls 

 between curves B and C. Assuming that a moderate amount of brush will 

 give a friction effect about halfway between the two curves, from curve 

 B, where d = 10 feet, f f = 0.20, and from curve C, where d = 10 feet, 

 ff = 0.485. The bottom friction is then taken, in this case, as the 

 average of the two values 



For ff = 0.01, 



for ff = 0.343, 



f 0.20 ^ 0.485 , 0343 



J 2 



^f % ^^ _ 0.01 X 6 X 3,000 



d2 = 102 = ^• 



^f ^i ^^ 0.343 X 6 X 3,000 ^, ^ 

 —72 =• J32 = 61.7 



for T = 4.5 seconds and d = 10 feet, 



ilA= 2u (101 ^ 0,096 . 

 gT^ g (4.5)2 



From Figure 16, 



^/. 01 = 0-80 ^°^ f/ = 0-01 ^^ ff H^ Ax/d^ =1.8 



Kf^ = 0.105 for ff = 0.343 and ff H^ Ax/d^ =61.7 . 

 From Equation (13), 



_ ^ ~ ^fa _ 1 - 0.105 _ 0.895 _ . .„ 

 "^ " 1 _ K/.oi ~ 1-0.80" 0.20"^-'^^' 



from equation (12) , 



Fq = a^. Ax = 4.48 (3,000) = 13,440 feet 



(i.e., the wave decay over 3,000 feet of tall grass with, some brush is 

 equal to the wave decay over 13,440 feet of a sand bottom for this water 

 depth and windspeed) . 



F = Fg + F^ = 760 + 13,440 = 14,200 feet . 

 40 



