Example Problem II-1-1 (Concluded)

By trial-and-error solution (Equation II-1-21) with *d/L*o it is found that

' 0.05641

hence

3

1

d

1

' 53.2 *m *(174 *ft*) transitional depth, since

<

<

0.05641

25

L

2

53.2

' 5.32 *m*/*s *(17.4 *ft*/*s*)

'

10

An approximate value of L can also be found by using Equation II-1-11

4π2 d

tanh

2π

which can be written in terms of Lo as

2π*d*

tanh

therefore

2π(3)

tanh

156

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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