sector of wave directions. If ^(.a^) is evaluated at a^ = 45° (NE or 

 SE in the example problem), it will have a value 12 percent higher than 

 the average value for FCa^) over 45° sector bisected by the NE or SE 

 directions. Thus, if the data warrant a high degree of accuracy. Equa- 

 tion 4-43 should be averaged by integrating over the sector of directions 

 involved. 



% = 



(71.50 + 131.78 + 8.35) X 103 = 212 X 103 or 212,000 cu.yd./yr. 

 (8.35 + 35.92 + 5.69) X 103 = 50 X 103 or 50,000 cu.yd./yr. 

 Qrf ~ Qfif = 212 X 103 - 50 X 103 - 162 X 103 or 162,000 cu.yd./yr. 

 Qft "*" Qfif "" 212 X 103 4- 50 X 103 = 262 X 103 or 262,000 cu.yd./yr. 



*Coast runs N-S so frequencies of waves from N and S are halved. 

 tCalculation of this number is showrn in detail in the text. 



If F(a^) is evaluated at 



(waves from the E in the example 



problem), then F(a^) = 0. Actually, a^ = is only the center of a 

 45° sector which can be expected to produce transport in both directions. 

 Therefore, F(oio) should be averaged over 0° to 22.5° and 0° to -22.5°, 

 giving F(a^) = ± 0.370 rather than 0. The + or - sign comes out of the 

 sin 2a.Q term in F(ao) (Equation 4-43), which is defined such that trans- 

 port to the right is positive, as implied by Equation 4-22. 



A further complication in direction data is that waves from the north 

 and south sectors include waves traveling in the offshore direction. It 

 is assumed that, for such sectors, frequency must be multiplied by the 

 fraction of the sector including landward traveling waves. For example 

 the frequencies from N and S in Table 4-10 are multiplied by 0.5 to 

 obtain the transport values listed in Table 4-11. 



4-106 



