for d = 16.5 feet. Hi = 3.17 feet, f; = 0.108, Ax = 3,000 feet, and 

 T = 3.31 seconds (from Fig. 16) 



2l$= 0.294 



Ky-_oi = 0.9 88 for ff= 0.01 and f^ H; ..x/d^ = 0.349 

 K^^ = 0.88 for ff = 0.108 and fz- \U Ax/d^ = 3.77 .. 

 Using equation (10), a = 0.1 and 



Fq = ex Ax = 0.1 (3,000) = 300 feet 

 F = Fg + Fa = 5,400 + 300 = 5,700 feet . 

 For d = 16.5 feet (from Figs. 1 and 2) 



Hf = 3.27 feet and T = 3.41 seconds . 



The remaining 2,000 feet of the fetch can then be treated as a third 

 segment. The average depth, d = 14 feet, and the average friction 

 factor, ff = 0.13. 



For d = 14 feet and Hi = 3.27 feet (from Fig. 1), 



Fg = 7,200 feet; ; 



for d = 14 feet, H^ = 3.2 7 feet, f ^ = 0,13 (from Fig. 16) 



Ax = 2,000 feet, T = 3.41 seconds, and 2TTd/(gt2) = 0.235 , 

 K/.Ol = 0-98 for ff = 0.01 and ff Hi Ax/d^ = 0.334 

 Kfa = 0.80 for f / = 0.13 and ff Hi Ax/d^ = 4.34 . 



Using equation (10), a = 0.1 and 



F^ = a Ax = Ool (2,000) = 200 feet 



F = Fg + F^ = 7,200 + 200 = 7,400 feet . 



For d = 14 feet, U = 70 miles per hour, and F = 7,400 feet (from Figs. 

 1 and 2) 



Hf = 3.34 feet and T = 3.51 seconds . 



Note. --For a sandy bottom ff= 0.01, the wave would have increased 

 to a height of approximately 4.26 feet, a 42-percent increase from 



35 



