EM 1110-2-1100 (Part II)
30 Apr 02
Example Problem II-1-2 (Concluded)
From the above known information, we find
2πd
' 0.7691
L
and values of hyperbolic functions become
cosh(0.7691) ' 1.3106
and
sinh(0.7691) ' 0.8472
Therefore, fluid particle velocities are
u ' 1.515(1.1306)(0.500) ' 0.99 m/s (3.26 ft/s)
w ' 1.515(0.8472)(0.866) ' 1.11 m/s (3.65 ft/s)
and fluid particle accelerations are
αx ' 1.190(1.3106)(0.866) ' 1.35 m/s 2 (4.43 ft/s 2)
αz ' &1.190(0.8472)(0.500) '&0.50 m/s 2 (1.65 ft/s 2)
(c) Thus,
2π(z%d)
cosh
H
L
sin θ
ξ'&
(II-1-29)
2
2πd
sinh
L
2π(z%d)
sinh
H
L
ζ'%
(II-1-30)
cos θ
2
2πd
sinh
L
II-1-16
Water Wave Mechanics
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