EM 1110-2-1100 (Part II)
30 Apr 02
Example Problem II-1-5 (Concluded)
Also from Figure II-1-11 or II-1-12,
(yt & d)
% 1 ' 0.865
H
thus,
yt ' (0.865 & 1)(1) % 3 ' 2.865 m (9.40 ft)
(d) The dimensionless wave profile is given in Figures II-1-11 and II-1-12 for k2 = 1 - 10-5. The results
obtained in (c) above can also be checked by using Figures II-1-11 and II-1-12. For the wave profile obtained
with k2 = 1 - 10-5, the SWL is approximately 0.14H above the wave trough or 0.86H below the wave crest.
The results for the wave celerity determined under (b) above can now be checked with the aid of
Figure II-1-16. Calculate
H
(1)
' 0.349
'
yt
2.865
Entering Figure II-1-16 with
L 2H
(1)
' 0.349
'
2.865
3
d
and
H
' 0.349
yt
it is found that
C
' 1.126
g yt
Therefore,
C ' 1.126 (9.8)(2.865) ' 5.97 m/s (19.57 ft/s)
The differences between this number and the 5.90 m /sec (18.38 ft/s) calculated under (b) above is the result of
small errors in reading the curves.
Water Wave Mechanics
II-1-47
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