Graphics Reference
In-Depth Information
1
=
()
+
()
+
()
b
16
b
13
b
12
b
110
,,
020
,,
011
,,
010
,,
=
( (
+
( (
+
( (
)
=
(
)
16 343
,,
13 250
,,
12 241
,,
1361331
,
,
2
b
=
()
16
b
+
()
13
b
+
()
1
2
b
,,
100
,,
010
,,
001
000
,,
=
( (
+
( (
+
( (
)
=
(
)
1619610332
,
,
131361331
,
,
1213610312
,
,
7311356
,
,
,
2
and p(1/6,1/2) =
b
0,0,0
.
Next, let
S
be a triangular Bézier surface with parameterization p(u,v) of degree
d and domain
D
. Restricting p(u,v) to the boundary of the triangle
D
maps those points
to the boundary of the surface
S
. In terms of barycentric coordinates (a,b,c) the
boundary of
D
corresponds to the points with a = 0, b = 0, or c = 0. From this it is
easy to see that
(1) The surface interpolates the three corner points
b
d,0,0
,
b
0,d,0
, and
b
0,0,d
of its
control net.
(2) Restricting p(u,v) to an edge of the boundary of
D
is just a Bézier curve with
the points of the triangular net along that edge as its control points.
(3) The tangent plane to
S
at the three corner points of its control net is defined
by the three corner control points. For example, the tangent plane at
b
0,d,0
is
defined by
b
0,d,0
,
b
1,d-1,0
, and
b
0,d-1,1
.
The partial derivatives of a triangular Bézier surface are easily computed by a de
Casteljau-type algorithm. One uses the Bernstein representation, equation (12.51), for
the parameterization and an argument similar to the one for computing derivatives
of Bézier curves.
12.12.3.2
Theorem.
Let p(u,v) define a triangular Bézier surface of degree d and
control net {
b
i,j,k
}. Then
∂
∂
∂
∂
p
u
Â
(
)
=
d
-
1
(
)
(
)
uv
,
d
B
uv
, ,
1
-
u v
-
b
-
b
i
+
1
,,
j k
i j k
,,
+
1
ijk
,,
(
)
Œ
ijk I
,,
d
-
1
p
v
Â
(
)
=
d
-
1
(
)
(
)
uv
,
d
B
uv
, ,
1
-
u v
-
b
-
b
iijk
ij
,
+
1
,
k
,,
+
1
ijk
,,
(
ijk I
,,
)Œ
d
-
1
2
∂
∂
p
u
Â
(
)
=-
(
)
d
-
2
(
)
(
)
uv
,
dd
1
B
uv
, ,
1
--
u v
b
-
2
b
+
b
i
+
2
,,
j k
i
+
1
,,
j k
+
1
i j k
,,
+
2
ijk
,,
2
(
)
Œ
ijk I
,,
d
-
2
2
∂
∂
p
Â
(
)
=-
(
)
d
-
2
(
)
(
)
uv
,
dd
1
B
uv
,,
1
--
u v
b
-
2
b
+
b
ij
,
+
2
,
k
i
+
1
,
j
+
1
,
k
i
+
2
, ,
jk
ijk
,,
2
v
(
)
Œ
ijk I
,,
d
-
2
2
∂
∂∂
p
uv
uv
Â
(
)
=-
(
)
d
-
2
(
)
(
)
,
dd
1
B
uv
, ,
1
--
u v
b
-
b
-
b
+
b
.
i
++
1
,
j
1
,
k
i
++
1
, ,
jk
1
i j
,
++
1
,
k
1
i jk
, ,
+
2
ijk
,,
(
)
Œ
ijk I
,,
d
-
2
Proof.
The formulas follow from equation (12.51).
