Graphics Reference
In-Depth Information
ordinates (
r
F
,s
F
,t
F
) relative to the points
A
,
B
and
C
respectively, then
r
F
=
area
∆
BCF
area
∆
ABC
And if
D
has barycentric coordinates (
r
D
,s
D
,t
D
) relative to the points
A
,
B
and
C
respectively, then
s
D
=
area
∆
CAD
area
∆
ABC
Similarly, if
E
has barycentric coordinates (
r
E
,s
E
,t
E
) relative to the points
A
,
B
and
C
respectively, then
t
E
=
area
∆
ABE
area
∆
ABC
Substituting
r
F
,s
E
and
t
D
in (11.13) we obtain
1=
r
F
+
s
D
+
t
E
+
area
∆
DEF
area
∆
ABC
From (11.12) we see that
r
F
=
2
7
s
D
=
2
7
t
E
=
2
7
therefore
1=
6
7
+
area
∆
DEF
area
∆
ABC
and
area
∆
DEF
=
1
7
×
area
∆
ABC
which is rather neat.
But just before we leave this example, let's state a general expression for
the
area
∆
DEF
for a triangle whose sides are divided in the ratio 1:
n
.Once
again, I'll leave it to the reader to prove that
area
∆
DEF
=
n
2
2
n
+1
n
2
+
n
+1
×
−
area
∆
ABC
Note that when
n
=1
,area
∆
DEF
= 0, which is correct.
[Hint: The corresponding values of
r
F
,s
D
and
t
E
are
n/
(
n
2
+
n
+1).]
11.7 Volumes
We have now seen that barycentric coordinates can be used to locate a scalar
within a 1D domain, a point within a 2D area, so it seems logical that the
description should extend to 3D volumes, which is the case.
