FIND: The water depths d^ and d^, giving the extent of the shoal zone at 

 this site. 



SOLUTION : From equation (4) , 



HsO.137 = ^1-35 + (5.6)(0.9)} = 6.4 feet. 



and from equation (3) , 



, 32.2(5.7)^ , ,r. r . , 

 Ln = { — I = 166.5 feet 



^s0.137 



= 0.038 



so that 



and Figure 3 gives 



L^ = 0.74 



Figure 4 gives 



tanh (0.74) = 0.63, 

 so that 



d^ = (^VMjtanh (?!iM = (0.74) (26.5) (0.63) = 12.4 feet - 12 feet, 



Note that equation (6) gives 



dj^ = {2[1.35 + 11(0.9)]} = 12.6 feet - 13 feet, 



From equation (8) , 



Hs5o = {1.35 - 0.3(0.9)} = 1.08 feet, 



so equation (9) becomes 



08)2 10.5 



\ L / |8(1.6)(32.2)(0, 



n , = 1.61 



00033)(5.7) 



Figure 4 gives 



■2TTd- 



;inh-l(1.61) = 1.26 



and 



t 



anh ( ^j = 0.85, 



15 



