Graphics Reference
In-Depth Information
Here we've assumed the existence of a matrix-vector product procedure and a
pseudoinverse procedure, as provided by most numerical packages.
The two procedures above can be combined, of course, to produce the function
value at a point
P
that's specified in
xyz
-coordinates.
Input:
• A triangle mesh in the form of an
n
×
3 table,
vtable
, of vertices
•A
k
3 table,
ftable
, of triangles, where each row of
ftable
contains
three indices into
vtable
•An
n
×
1 table,
fntable
, of function values at the vertices of the mesh
• A point
P
of the mesh, expressed by giving the index,
t
, of the triangle in
which the point lies, and its coordinates in 3-space
×
Output:
• The value of the function defined by the function table at the point
P
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double meshinterp2(double[,] vtable, int[,] ftable, double[] fntable,
int t, double p[3])
{
double[] barycentricCoords =
barycentricCoordinates(vtable, ftable, t, p);
return meshinterp2(vtable, ftable, fntable, t,
barycentricCoords[0], barycentricCoords[1], barycentricCoords[2]);
}
Of course, the same idea can be applied to the ray-intersect-triangle code of
Section 7.9.2, where we first computed the barycentric coordinates
(
)
of
the ray-triangle intersection point
Q
with respect to the triangle
ABC
, and then
computed the point
Q
itself as the barycentric weighted average of
A
,
B
, and
C
.
If instead we had function values
f
A
,
f
B
, and
f
C
at those points, we could have
computed the value at
Q
as
f
Q
=
α
,
β
,
γ
f
C
. Notice that this means we can
compute the interpolated function value
f
Q
at the intersection point without ever
computing the intersection point itself!
Computing barycentric coordinates
(
α
f
A
+
β
f
B
+
γ
α
,
β
,
γ
)
for a point
P
of a triangle
ABC
,
R
2
, is somewhat simpler than the corresponding problem in
3-space (see Figure 9.3). We know that the lines of constant
where
A
,
B
,
C
∈
α
are parallel to
BC
.Ifwelet
n
=(
C
−
B
)
⊥
, then the function defined by
f
(
P
)=(
P
−
B
)
·
n
is
also
constant on lines parallel to
B
C
. Scaling this down by
f
(
A
)
gives us the
function we need: It's zero on line
BC
, and it's one at
A
.Sowelet
−
(
P
−
B
)
·
n
g
:
R
2
→
R
:
P
→
n
,
(9.11)
(
A
−
B
)
·
and the value of
g
(
P
)
is just
α
. A similar computation works for
β
and
γ
.
The resultant code looks like this:
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double[3] barycenter2D(Point P, Point A, Point B, Point C)
C[2])
{
double[] result = new double[3];
result[0] = helper(P, A, B, C);
result[1] = helper(P, B, C, A);
result[2] = helper(P, C, A, B);
return result;
