VII. EXAMPLE PROBLEMS 
The following example problems illustrate the methods of predicting reflec- 
tion coefficients presented in this report. 
kok kk kk Kk KOK OK OK O&K * & EXAMPLE PROBLEM 1% * * * XK KK KKAKKKAKE 
GIVEN: A smooth impermeable revetment (nonovertopped) has a toe water depth, 
d, = 7.62 meters, a slope cot® = 2.0, and the offshore slope is m= 0.02. 
FIND: The wave reflection coefficient and fraction of wave energy dissipated 
for a wave with H; = 3.05 meters and T = 10 seconds. Illustrate the influ- 
ence of wave height and period on K, and show the effect of reducing the 
slope to cot@ = 5. 
SOLUTION: From equation (7), 
d 
Hy -8OMl7 be {1.0 - exp E 4.712 Tal opal 1.33)}} 
Oo 
= 0.17 (1.56 x 10? ) {2 - exp Le AID soa fee 2(a plo\(ORO2)) as -32)]\ = 5.85 m 
From equation (17) 
tané 
g = eRe = O2_ = 3.58 
E 3.05 
lly 156 
and from equation (15) 
= .807 (3.58) 
Ky = a WSUS = 0.56 
E2 + B Glss) 2 ses 
The energy dissipation coefficient for this example is K4 = 0.69, or 69 per- 
cent of the incident wave energy is dissipated (from Fig. 3). Other reflection 
coefficient calculations for 5-, 10-, and 20-second periods for wave heights 
between 0.3 and 4.4 meters are summarized in Figure 12. Predictions are also 
shown for a structure with cot6§ = 5. Figure 12 illustrates the influence of 
wave height, period, and structure slope on Ky 
ZS) 
