Graphics Reference
In-Depth Information
y
2
z
3
(
rx
1
+
sx
2
+
tx
3
)+
x
2
y
3
(
rz
1
+
sz
2
+
tz
3
)+
x
3
z
2
(
ry
1
+
sy
2
+
ty
3
)
−y
3
z
2
(
rx
1
+
sx
2
+
tx
3
)
− x
3
y
2
(
rz
1
+
sz
2
+
tz
3
)
− x
2
z
3
(
ry
1
+
sy
2
+
ty
3
)
1
6
V
=
⎡
⎤
r
(
x
1
y
2
z
3
+
x
2
y
3
z
1
+
x
3
y
1
z
2
−
x
1
y
3
z
2
−
x
3
y
2
z
1
−
x
2
y
1
z
3
)+
=
1
6
⎣
⎦
s
(
x
2
y
2
z
3
+
x
2
y
3
z
2
+
x
3
y
1
z
2
−
x
2
y
3
z
2
−
x
3
y
1
z
2
−
x
2
y
2
z
3
)+
t
(
x
3
y
2
z
3
+
x
2
y
3
z
3
+
x
3
y
3
z
2
−
x
3
y
3
z
2
−
x
3
y
2
z
3
−
x
2
y
3
z
3
)
and simplifies to
x
1
y
1
z
1
V
=
1
6
r
x
2
y
2
z
2
x
3
y
3
z
3
This states that the volume of the smaller tetrahedron is
r
times the volume
of the larger tetrahedron
V
T
,where
r
is the barycentric coordinate modifying
the vertex not included in the volume. By a similar process we can develop
volumes for the other tetrahedra:
V
(
P, P
2
,P
4
,P
3
)=
rV
T
V
(
P, P
1
,P
3
,P
4
)=
sV
T
V
(
P, P
1
,P
2
,P
4
)=
tV
T
V
(
P, P
1
,P
2
,P
3
)=
uV
T
where
r
+
s
+
t
+
u
=1.
Similarly, the barycentric coordinates of a point inside the volume sum to
unity.
Let's test the above statements with an example.
Figure 11.25 shows a tetrahedron and a point
P
3
,
3
,
3
located within
its interior.
The volume of the tetrahedron
V
T
is
001
100
010
V
T
=
1
6
=
1
6
Y
P
3
1
P
P
4
P
2
P
1
1
1
X
Z
Fig. 11.25.
A tetrahedron.
