Graphics Reference
In-Depth Information
float d21 = Dot(v2, v1);
float denom = d00 * d11 - d01 * d01;
v = (d11 * d20 - d01 * d21) / denom;
w = (d00 * d21 - d01 * d20) / denom;
u=1.0f-v-w;
}
If several points are tested against the same triangle, the terms
d00
,
d01
,
d11
, and
denom
only have to be computed once, as they are fixed for a given triangle.
The barycentric coordinates can be computed for a point with respect to a simplex
(Section 3.8) of any dimension. For instance, given a tetrahedron specified by the
vertices
A
,
B
,
C
, and
D
, the barycentric coordinates (
u
,
v
,
w
,
x
) specify a point
P
in 3D
space,
P
=
uA
+
vB
+
wC
+
xD
with
u
+
v
+
w
+
x
=
1. If 0
≤
u
,
v
,
w
,
x
≤
1, then
P
is
inside the tetrahedron.
Given the points specified as
A
=
(
a
x
,
a
y
,
a
z
),
B
=
(
b
x
,
b
y
,
b
z
),
C
=
(
c
x
,
c
y
,
c
z
),
D
(
p
x
,
p
y
,
p
z
), the barycentric coordinates can be solved for by
setting up a system of linear equations:
=
(
d
x
,
d
y
,
d
z
), and
P
=
a
x
u
+
b
x
v
+
c
x
w
+
d
x
x
=
p
x
a
y
u
+
b
y
v
+
c
y
w
+
d
y
x
=
p
y
a
z
u
+
b
z
v
+
c
z
w
+
d
z
x
=
p
z
u
+
v
+
w
+
x
=
1
Alternatively, by subtracting
A
from both sides of
P
=
uA
+
vB
+
wC
+
xD
— giving
P
−
A
=
v
(
B
−
A
)
+
w
(
C
−
A
)
+
x
(
D
−
A
)
— it follows that three of the four barycentric coordinate components can be obtained
by solving
(
b
x
−
a
x
)
v
+
(
c
x
−
a
x
)
w
+
(
d
x
−
a
x
)
x
=
p
x
−
a
x
,
+
c
y
a
y
w
+
d
y
a
y
x
−
−
−
=
−
(
b
y
a
y
)
v
p
y
a
y
, and
(
b
z
−
a
z
)
v
+
(
c
z
−
a
z
)
w
+
(
d
z
−
a
z
)
x
=
p
z
−
a
z
=
−
−
−
(with the fourth component given by
u
x
). Either system is easily solved
using Cramer's rule or Gaussian elimination. For example, in the former system the
coordinates are given by Cramer's rule as the ratios
1
v
w
=
u
d
PBCD
/
d
ABCD
,
