From equation (6) , 



1 - Kf a 1-0.105 0.895 



a = U± — = = = 4.48 : 



1 " K f.01 1-0.80 0.20 



from equation (7), 



F a = a r 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) . 



The total fetch from equation (8) is 



F = F e + F a = 760 + 13,440 = 14,200 feet. 



For a windspeed of 90 miles per hour and a fetch of 14,200 feet (from 

 Fig. 1) 



gd 



■ a — = 0.0185 (as previously determined) 



JE = 52 - 2 x 14 > 200 = 26 24 

 U2 (132)2 Zb<Z4 



giving 



gH 



^r = 0.0071 . 



U 2 



From which the equivalent wave height, 



Hg = 0-0071 U 2 = 0-0071 (132) 2 = 3^ feet _ 

 g oZ . z 



From equation (12), the fractional growth is 



= H^ = 3_ 1 84 = 0>937 



%m 4 - X 



The decayed wave height is then given by equation (13) as 



% = F^j - Gi (H^ - H sm ) = 7.8 - 0.937 (7.8 - 4.1) = 4.33 feet . 



At the end of the fetch segment, the wave height and period are 

 approximated by 



H D = 4.33 feet 



T =4.5 seconds . 

 ************************************* 



34 



