For T = 4.7 seconds and d = 3 meters , 



2iTd ^ 271 (3) 

 gT^ 9.8 (4.7)^ 



= 0.087 



Using Figure 3-38, for 2Trd/(gT^) = 0.087 , 



K^ ni = 0.76 for f „ = 0.01 and f . H. Ax/d^ = 2.22 

 f.Ol f f ^ 



K^ = 0.08 for f^ = 0.343 and f „ H. Ax/d^ = 76.22 

 fa f / t 



From equation (3-47), 



1 - K^ 



- Ta ^ 1 - 0.08 ^ 0.92 ^ ^ j,^ 



"r 1 - K„ Q^ 1 - 0.76 0.24 "^'^"^ 



From equation (3-48) 



F = a Ax = 3.83 (1000) = 3,830 m (12,566 ft) 

 a r 



(i.e., the wave decay over 1000 meters of tall grass with some brush is 

 equal to the wave decay over 3,830 meters of a sand bottom for indicated 

 water depth and windspeed) . 



The total fetch, using equation (3-49) is 



F = F + F = 367 + 3,830 = 4,197 m (13,770 ft) 

 e a 



Using Figure 3-21 for a windspeed of 40 meters per second and a fetch of 

 2907 meters 



gd _ 



< 



it is found that 



= 0.0184 (as previously determined) 



£F ^ 9.8 X 4,197 ^ , 



2 2 ^J.'i 

 U^ (40)^ 



■^ = 0.0059 



"a 



Solving for the equivalent wave height , 



0.0059 U^ . .__ ,..,2 

 TT A 0.0059 (40) „ o,^ f^ ,. jr.s 



H = = ^—~ — ^— = 0.963 m (3.16 ft) 



8 g y .0 



From equation (3-53), the fractional growth is 



sm 

 The decayed wave height is then given by equation (3-54) as 



1L,=H -G. (H -H )= 2.34 - 0.789 (2.34 - 1.22) = 1.46 m (4.78 ft) 

 V m 1 m sm 



3-76 



