FIND: The wave height at
length of x = 3 and y =
a point P having coordinates in units of wave-°
4. (Polar coordinates of x and y are r=5 at 53 .)
SOLUTION:
Since d8 = 15 feet, T = 8 seconds,
Lo
* — = ----- —----- = 0.0458 .
5.12T2
(5.12) (64)
Using Table C-l with
1 1
Lo ' Lo
0.0458
the corresponding value of
d
ds
- or — = 0.0899 ,
L
L
therefore,
L = i----- r = --------- = 167 feet .
H
Because 1 inch represents 133 feet on the hydrographie chart and
L = 167 feet, the wave-length is 1.26 inches on the chart.
This provides the necessary information for scaling Figure 2-36 to the
hydrographie chart being used. Thus 1.26 inches represents a radius!
ixnielength unit.
For this example, point P and those lines of equal Kz situated
nearest P are shown on a schematic overlay, Figure 2-41. This overlay is based on Figure 2-36 since the angle of wave approach is 135°.
It should be noted that Figure 2-41, being a schematic rather th'an a
true représentation of the overlay, is not drawn to the hydrographie
chart scale calculated in the problem. From Figure 2-41 it is seen
that Kz at point P is approximately 0.086. Thus the diffracted
wave height at this point is
H = KZH. = (0.086) (10) = 0.86 foot say 0.9 foot .
The above calculation indicates that a wave undergoes a substantial
height réduction in the area considered.
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2-96