Re-entering Figures 1 to 12 with the fetch length, F, the windspeed, 

 U, and water depth, d, the final wave height at the end of the fetch 

 segment, H^, is determined. This is shown schematically in Figure 17 

 and in the following example problem. 



*************** EXAMPLE PROBLEM i************ 



GIVEN: A wave passes into shallow water over a flooded coastal area. 

 The water depth, d^, at the seaward edge of the area is 23 feet (7 

 meters) , and at the landward edge of the area the depth is 13 feet (4 

 meters). The distance across the area in the direction of wave motion 

 is 10,000 feet (3,050 meters). The wave height, H-i, at the seaward 

 edge of the area is limited by large sandbars seaward of the area being 

 considered and is 3 feet (0.91 meter); the wave period is 3.2 seconds. 

 The windspeed is 70 miles per hour (102.7 feet or 31.3 meters per second) 

 The flooded area is covered with thick stands of tall grass. 



FIND : The height and period of the significant wave at the landward edge 

 of the segment. 



SOLUTION: 



0.25 d^ = 0.25 (23) = 5.75 feet 



Ad = 23 - 13 = 10 feet > 0.25 d^; . 



Since this does not meet the condition of equation (5) , the area should 

 be divided into two fetch segments. Assuming a uniform variation in 

 depth, take the first segment as a distance Ax = 5,000 feet with a 

 depth variation from 23 to 18 feet. Then 



Ad = 23 - 18 = 5 feet < 0.25 d^ . 

 At the 23-foot depth (from Fig. 15, curve B) , 



ff = 0.080 



and at the 18- foot depth (curve B) , 



f^ = 0.095 



Afy = 0.095 - 0.080 = 0.015 



0.25 fj^^'= 0.25 (0.080) = 0.020 



Af^= 0.25 fj.- . 



Equations (5) and (6) are satisfied, so the fetch segment chosen is used. 

 For a uniformly varying depth, the average depth can be taken as the 

 average of the depths at the beginning and the end of the segment; i.e.. 



31 



