Graphics Reference
In-Depth Information
Figure 12.16.
Analyzing geometric matri-
ces for bicubic patches.
Solution.
See Figure 12.16. Let
(
)
=
(
(
)
(
)
(
)
)
puv
,
xuv yuv zuv
,
,
,
,
,
.
First of all, p(0,v) parameterizes the straight line segment [(0,-4,0),(0,-2,2)] because
the geometric data for x(u,0) is 0 and both y(u,0) and z(u,0) have the right geometric
data for a straight line. Similarly, p(1,v) parameterizes the straight line segment
[(2,0,0),(1,0,2)]. The geometric data for the curves p(u,0) and p(u,1) indicates that they
are parabolas. In fact, the data
(
)
=-
(
)
(
)
=
(
)
(
)
=
(
)
(
)
=
(
)
p
00
,
0 40
,
,
,
p
10
,
200
, ,
,
p
00
,
200
, ,
,
p
10
,
280
, ,
u
u
()
=-
(
)
(
)
=
(
)
(
)
=
(
)
(
)
= 140
(
)
p
01
,
0
, ,,
2 2
p
11
,
10 2
,,,
p
01
,
10 0
,,,
p
11
,
,,
u
u
is easily solved to give
(
(
)
)
(
(
)
)
2
2
(
)
=
()
=
pu
,
024 10
u
,
u
-
,
and
pu
,
1
u
,
2 12
u
-
,
.
Next, we use the data
(
)
=
(
)
(
)
=-
(
)
(
)
=-
(
)
(
)
=- -
(
)
p
00
,
022
, ,
,
p
10
,
102
, ,
,
p
00
,
100
, ,
,
p
10
,
1 40
,
,
v
v
uv
uv
()
=
(
)
(
)
=-
(
)
(
)
=-
(
)
()
=- -
(
)
p
01
,
022
, ,
,
p
11
,
102
, ,
,
p
01
,
100
, ,
,
p
uv
11
,
1
,
4 0
,
v
v
uv
to get
(
(
)
)
2
(
)
=
()
=-
pu
,
0
pu
,
1
u
,
2 1
-
u
,
2
.
v
v
Note how this agrees with p(u,1) - p(u,0). The point
C
= (0,0,4) is clearly the vertex
of the parabolic cone which contains this patch. We leave it as an exercise to show
that
(
(
)
)
2
(
)
=
(
)
(
)
puv
,
u
,
22 122
-
v
,
u
-
-
v
,
v
.
