7.2 Wave Spectrum Evolution in Water with Ice Cakes
321
Due to the expression in round brackets (7.21), the frequency angular
spectrum S(w,{3,x) might seem to become infinitely large with the wave
packet arriving at the point, where w 2 = w~. However, as shown below, this
does not happen due to energy dissipation.
In order to obtain the complete spectrum (7.21) at some point {x0 ,y0 },
it is necessary to bring together all the rays there. To do this, the set of
equations (7.17), (7.18) must be solved. However, there are other ways to do
this using the following kinematic notions. It should be noted that a filtration
of wave components takes place due to dissipation as waves propagate within
the ice cake area, whereas its thickness is increased in the wave propagation
direction. The wave, which value 1(x) = 1-w 2 p3L(x)jgp2 becomes less than
or equal to zero, and ceases to exist in the area under consideration. When
the complete set of equations for the characteristics (7.17), (7.18) cannot be
solved, it is necessary to impose an additional kinematic condition onto the
solution (7.21), describing the absence of proper frequency components in
the spectrum at the considered point. This is attained by multiplying the
spectrum (7.21) by the Heaviside function B(w):
5
S(w,{3,x) = (1- ~;) -
2 So(w,f3o) exp[E(w,{3,x)] 8 [1- ~;]. (7.22)
The initial angle {30 is determined using the ratio (7.19) and can be written
as:
. ( sin({3) )
f3o = arcsm 1 _ w2 /w~ .
(7.23)
The arcsine argument value in (7.23) must be no larger than 1.0, so there
is an additional restriction on the spectrum arguments:
I
sin({3) I
2j 2 ::::: 1.
1-w wb
(7.24)
A breach of this condition means that the corresponding spectral component is absent at this point. In this case it should be taken as:
S(w,{3,x) = 0.
(7.25)
If the initial spectrum is accepted to be:
So(w, f3o) = So(w) cosn(f3o) ,
(7.26)
it can be seen that the value 1 - w 2 /w~ decreases with increasing ice thickness L and the angular distribution of wave energy is narrowed. If the angle
range is originally limited as 1!31 :::; n/2, then the range is reduced in the
presence of ice as 1!31:::; arcsin (1- w 2 /w~), and the following can be written:
321
Due to the expression in round brackets (7.21), the frequency angular
spectrum S(w,{3,x) might seem to become infinitely large with the wave
packet arriving at the point, where w 2 = w~. However, as shown below, this
does not happen due to energy dissipation.
In order to obtain the complete spectrum (7.21) at some point {x0 ,y0 },
it is necessary to bring together all the rays there. To do this, the set of
equations (7.17), (7.18) must be solved. However, there are other ways to do
this using the following kinematic notions. It should be noted that a filtration
of wave components takes place due to dissipation as waves propagate within
the ice cake area, whereas its thickness is increased in the wave propagation
direction. The wave, which value 1(x) = 1-w 2 p3L(x)jgp2 becomes less than
or equal to zero, and ceases to exist in the area under consideration. When
the complete set of equations for the characteristics (7.17), (7.18) cannot be
solved, it is necessary to impose an additional kinematic condition onto the
solution (7.21), describing the absence of proper frequency components in
the spectrum at the considered point. This is attained by multiplying the
spectrum (7.21) by the Heaviside function B(w):
5
S(w,{3,x) = (1- ~;) -
2 So(w,f3o) exp[E(w,{3,x)] 8 [1- ~;]. (7.22)
The initial angle {30 is determined using the ratio (7.19) and can be written
as:
. ( sin({3) )
f3o = arcsm 1 _ w2 /w~ .
(7.23)
The arcsine argument value in (7.23) must be no larger than 1.0, so there
is an additional restriction on the spectrum arguments:
I
sin({3) I
2j 2 ::::: 1.
1-w wb
(7.24)
A breach of this condition means that the corresponding spectral component is absent at this point. In this case it should be taken as:
S(w,{3,x) = 0.
(7.25)
If the initial spectrum is accepted to be:
So(w, f3o) = So(w) cosn(f3o) ,
(7.26)
it can be seen that the value 1 - w 2 /w~ decreases with increasing ice thickness L and the angular distribution of wave energy is narrowed. If the angle
range is originally limited as 1!31 :::; n/2, then the range is reduced in the
presence of ice as 1!31:::; arcsin (1- w 2 /w~), and the following can be written: