From equation (3) the ponding level is 



P = 7.8(0.6)e- ( -°- 7 + i- / 7 - 8 ) 2 = 3 ' 36 feet 



From equation (7") 



V64.4 3.36 [0.8(125) 12(4 - 1) + 2(1.0) 1,000] 

 9.53(400) 4 



or 



K = 5.4 



The corresponding dimensionless velocity is 



= 0.18 for K = 5.4 from Figure 5 



c d V 2 § P 



and 



V = 0.18(0.8) V64.6 3.36 =2.11 feet per second. 



*************** EXAMPLE PROBLEM 2*************** 



GIVEN : The same breakwater system as in example 1. 



FIND ; V for irregular wave conditions with a significant wave height, 

 H s = 8.0 feet (2.44 meters). 



SOLUTION : The computer solution is illustrated in this example with the 

 input shown in Table 1. Resulting computer output is given in Table 2. 

 The predicted velocity is V =_1.33 feet (0.14 meter) per second. This 

 velocity is lower than the V obtained for monochromatic waves in example 1 

 (2.11 feet per second), because irregular waves of a given significant wave 

 height have less overtopping than monochromatic waves. 



Table 1. 



Computer program input for example problem 2. 



1 4 4C0. 



125. 7.77 8. 7. 12. 13. 



.667 .019 



.053 .8 -1.0 iOOt'. 



00000000900000 000000000 0000508000000000000 90000000 oooooooooooooooooooooooo 



1 2 1 4 5 1 J s « to n o u m u it u ■ u .'0 



nn2JJ*75M2J7injO 11 37033* to 16 37 3> M 49 41 ii 4144 4i«4l U49 »}i MS3W »« V '£ WHI (1 I7C3 M 61 Ktjl C* W TO fl 12 (J II ft TT II II n K 



111111111 111 1 11 1111 



1 1 1 ii 11 n 1 1 1 1 1 1 u 1 1: ii 111111111 1 1 1 1 1 1 1 1 1 1 ii i n 1 1 ; 1 1 i 1 1 1 1 n 



