EM 1110-2-1100 (Part II)
30 Apr 02
Example Problem II-1-1 (Concluded)
By trial-and-error solution (Equation II-1-21) with d/Lo it is found that
d
' 0.05641
L
hence
3
1
d
1
' 53.2 m (174 ft) transitional depth, since
L'
<
<
0.05641
25
L
2
L
53.2
' 5.32 m/s (17.4 ft/s)
C'
'
T
10
An approximate value of L can also be found by using Equation II-1-11
gT 2
4π2 d
L.
tanh
2π
T2 g
which can be written in terms of Lo as
2πd
L L0
tanh
L0
therefore
2π(3)
L 156
tanh
156
L 156 tanh (0.1208)
L 156 0.1202 ' 54.1 m (177.5 ft)
which compares with L = 53.3 m obtained using Equations II-1-8, II-1-9, or II-1-21. The error in this case is
1.5 percent. Note that Figure II-1-5 or Plate C-1 (SPM 1984) could also have been used to determine d/L.
Water Wave Mechanics
II-1-11
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