Graphics Reference
In-Depth Information
the same directions. We'll sort this out now and also clean up the issue of
how to make sure the height field is only real-valued despite all the complex
numbers flying around. We now build the general real-valued solution from
a collection of real cosine waves:
h
(
x, z, t
)=
i,j
A
ij
cos(
k
·
(
x, z
)
−
ω
k
t
+
θ
ij
)
.
(13.8)
Here
A
ij
is the real-valued constant amplitude of the wave,
k
=2
π
(
i, j
)
/L
is the wave vector as before and now points in the direction of the wave's
motion,
ω
k
is the time frequency of the wave from Equation (13.7), and
θ
ij
is a constant phase shift. We'll get to picking
A
ij
and
θ
ij
lateroninthe
chapter.
Equation (13.8) can be used directly to evaluate the height field at any
point in space and time, but if a lot of waves are involved the summation
becomes expensive. However a cheaper alternative exists by way of the
Fast Fourier Transform (FFT). Let
n
=2
m
be a power of two—this isn't
essential as FFT algorithms exist for any
n
, but typically the transform is
fastest for powers of two—and restrict the wave indices to
−
n/
2+1
≤
i, j
≤
n/
2
.
(13.9)
We thus have an
n
n
grid of wave parameters. We'll additionally specify
that the constant term is zero,
A
00
= 0, as this is not a wave but the average
sea level, and to simplify life also zero out the highest positive frequency
(which doesn't have a matching negative frequency):
A
n/
2
,j
=
A
i,n/
2
=0.
The sum will then actually only be up to
i
=
n/
2
×
1.
We'll now show how to evaluate
h
(
x, z, t
) for any fixed time
t
on the
−
1and
j
=
n/
2
−
n
x, z < L
, i.e., where
x
p
=
pL/n
and
z
q
=
qL/n
for integer indices
p
and
q
.Thisgridof
h
values can then be
fed directly into a renderer. Note that this in fact gives an
L
×
n
regular grid of locations 0
≤
L
tile that
can be periodically continued in any direction: we'll talk more about that
at the end of the chapter.
The problem is to determine, for a fixed time
t
and all integer indices
×
0
≤
p, q < n
, the height values as specified by
n/
2
−
1
n/
2
−
1
A
ij
cos(
k
h
pq
=
·
(
x
p
,z
q
)
−
ω
k
t
+
θ
ij
)
.
(13.10)
i
=
−n/
2+1
j
=
−n/
2+1
This is not quite in the form that an FFT code can handle, so we will need to
manipulate it a little, first by substituting in the wave vector
k
=2
π
(
i, j
)
/L
