Graphics Reference
In-Depth Information
Therefore,
=
2u
2
2u
1
v
2
+
2v
2v
2
+
−
2
v
2
2u
2
x
r
uv
−
−
2v
+
−
+
2u
+
and
2u
1
v
2
+
u
2
2u
2v
2v
2
+
2u
v
2
y
r
uv
=
−
2v
+
+
−
=
2u
2u
2
1
v
2
+
1
4u
2
2v
2v
2
+
2u
2u
2
v
2
z
r
uv
−
−
2v
+
+
4u
−
−
−
Simplifying the terms gives
2u
2
4u
2
v
x
r
uv
=
−
−
2u
+
4uv
+
2v
2u
2
v
2u
2
v
2
y
r
uv
=
2u
+
−
4uv
2
2u
2
2v
2
4u
2
v
4u
2
v
2
z
r
uv
=
2u
+
2v
+
4uv
−
−
−
−
+
Expressing the preceding three equations as a vector equation gives
2u
2
4u
2
v
r
=
−
−
2u
+
4uv
+
2v
i
+
+
2u
2
v
−
2u
2
v
2
j
2u
+
+
+
−
4uv
2
−
2u
2
−
2v
2
−
4u
2
v
+
4u
2
v
2
k
2u
2v
4uv
Differentiating gives
+
2
4uv
2
j
+
2
8uv
2
k
r
u
=
4v
2
4u
−
8uv
−
2
+
4v
i
+
4uv
−
+
4v
−
−
4u
−
8uv
+
v
=
−
2
i
+
2u
2
4u
2
v
j
+
2
8u
2
v
k
r
4u
2
4u
2
+
4u
+
−
+
4u
−
8uv
−
4v
−
+
The differentials at u
=
v
=
0 are
r
u
=−
2
i
+
2
j
+
2
k
r
v
=
2
i
+
2
k
The surface normal is given by
i
j k
r
u
×
r
v
=
n
=
−
222
202
n
=
4
i
+
8
j
−
4
k
(8.9)
But how can we confirm that this is correct?
Well, a Bézier surface patch has the property that the surface is tangential to the mesh of
corner control points, as shown in Fig. 8.5.
