sector of wave directions. If F(a0) is evaluated at ao = 45° (NE or
SE in the example problem), it will hâve a value 12 percent higher than
the average value for F(ao) over 45° sector bisected by the NE or SE
directions. Thus, if the data warrant a high degree of accuracy, Equation 4-43 should be averaged by integrating over the sector of directions
involved. * is
♦Coast runs N-S so frequencies of waves from N and S are halved.
fCalculation of this number is shown in detail in the text.
If F(ao) is evaluated at a0 = 0 (waves from the E in the example
problem), then F(a0) = 0. Actually, ao = 0° is only the center of a
45 sector which can be expected to produce transport in both directions.
Therefore, F(a giving F(a^) = ± 0.370 rather than 0. The + or - sign cornes out of the
sin 2ao term in F(a<9) (Equation 4-43), which is defined such that transport 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.
Qao , Hq in cu.yd./yr. from Equation 4-42
Table 4-11. Computed Longshore Transport for East-Facing Coast of Inland Sea
Ho
N*
NE
E
SE
S*
1
1.61 X 103f
11.26 X 103
±1.52 X 103
5.63 X 103 -0.89 X 103
2
5.05 X 103
31.84 X 103
± 2.87 X 103 -12.74 X 103 -2.02 X 103
3
11.13 X 103
52.42 X 103
± 3.96 X 103 -17.55 X 103 -2.78 X 103
4
11.42 X 103
36.03 X 103
5
9.98 X 103
8
32.31 X 103
Totals 71.50 X 103
131.78 X 103
(±8.35 X 103)
16.70 X 103
-35.92 X 103 -5.69 X 103
= (71.50 + 131.78 + 8.35) X 103 = 212 X 103 or 212,000 cu.yd./yr.
Q£f = (8.35 + 35.92 + 5.69) X 103 = 50 X 103 or 50,000 cu.yd./yr.
Q„ = Qrt “ Qfr = 212 x 103 “ 50 X 103 = 162 X 103 or 162,000 cu.yd./yr.
Qj = Qrt + Qgt = 212 X 103 + 50 X 103 = 262 X 103 or 262,000 cu.yd./yr.
4-106
SE in the example problem), it will hâve a value 12 percent higher than
the average value for F(ao) over 45° sector bisected by the NE or SE
directions. Thus, if the data warrant a high degree of accuracy, Equation 4-43 should be averaged by integrating over the sector of directions
involved. * is
♦Coast runs N-S so frequencies of waves from N and S are halved.
fCalculation of this number is shown in detail in the text.
If F(ao) is evaluated at a0 = 0 (waves from the E in the example
problem), then F(a0) = 0. Actually, ao = 0° is only the center of a
45 sector which can be expected to produce transport in both directions.
Therefore, F(a giving F(a^) = ± 0.370 rather than 0. The + or - sign cornes out of the
sin 2ao term in F(a<9) (Equation 4-43), which is defined such that transport 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.
Qao , Hq in cu.yd./yr. from Equation 4-42
Table 4-11. Computed Longshore Transport for East-Facing Coast of Inland Sea
Ho
N*
NE
E
SE
S*
1
1.61 X 103f
11.26 X 103
±1.52 X 103
5.63 X 103 -0.89 X 103
2
5.05 X 103
31.84 X 103
± 2.87 X 103 -12.74 X 103 -2.02 X 103
3
11.13 X 103
52.42 X 103
± 3.96 X 103 -17.55 X 103 -2.78 X 103
4
11.42 X 103
36.03 X 103
5
9.98 X 103
8
32.31 X 103
Totals 71.50 X 103
131.78 X 103
(±8.35 X 103)
16.70 X 103
-35.92 X 103 -5.69 X 103
= (71.50 + 131.78 + 8.35) X 103 = 212 X 103 or 212,000 cu.yd./yr.
Q£f = (8.35 + 35.92 + 5.69) X 103 = 50 X 103 or 50,000 cu.yd./yr.
Q„ = Qrt “ Qfr = 212 x 103 “ 50 X 103 = 162 X 103 or 162,000 cu.yd./yr.
Qj = Qrt + Qgt = 212 X 103 + 50 X 103 = 262 X 103 or 262,000 cu.yd./yr.
4-106