EXAMPLE PROBLEM II-1-3 (cont)

EM 1110-2-1100 (Part II)

30 Apr 02

Example Problem II-1-3 (Concluded)

When z = -d,

H

3

' 1.81 m (5.92 ft)

A'

'

2 (0.8306)

2πd

2 sinh

L

and B = 0.

(b) With H0 = 3.13 m and z = -7.5 m (-24.6 ft), evaluate the exponent of e for use in Equation II-1-36,

noting that L = L0,

2πz

2π(&7.5)

' &0.302

'

L

156

thus,

e &0.302 ' 0.739

Therefore,

2πz

H0

3.13

L

e

(0.739) ' 1.16 m (3.79 ft)

A'B'

'

2

The maximum displacement or diameter of the orbit circle would be 2(1.16) = 2.32 m (7.61 ft) when

z = -7.5 m.

(c) At a depth corresponding to the half wavelength from the MWL, we have

L0

&156

z'&

' &78.0 m (255.9 ft)

'

2

2πz

2π(&78)

' &3.142

'

L

156

Therefore

e &3.142 ' 0.043

and

2πz

H0

3.13

L

A'B'

e

(0.043) ' 0.067 m (0.221 ft)

'

2

Thus, the maximum displacement of the particle is 0.067 m, which is small when compared with the deepwater

height, H0 = 3.13 m (10.45 ft).

II-1-20

Water Wave Mechanics