Graphics Reference
In-Depth Information
(
)
()
+
(
)
+-
()
(
)
2
12
uu v
3 12
,
vuuuu v
+
,
+
.
1
2
1
2
1
2
We now think of u
1
and u
2
as fixed and find the blossoms of the polynomials in v to
get
(
)
=
()
+
(
(
)
+
()
+
(
)
-
()
+
(
)
(
)
)
Pu u v v
,
,
,
12
u
u
12
v
v
3 14
,
u
u
v
+
v
,
uu
+
vv
.
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
We would have gotten the same function P if we had first found the blossom for v and
then u.
The Triangular Surface Blossom.
Here we look for a more direct generalization
of a blossom as defined in Section 11.5.2 and treat a point (u,v) in
R
2
as a single entity.
In other words, the blossom of the polynomial function p in equation (12.47) should
be a symmetric multilinear function of the form
()
Æ
d
m
P
:
RR
,
(12.49)
where d is some appropriate “degree” of p, and it should satisfy P((u,v), ...,(u,v)) =
p(u,v).
12.12.2.2 Theorem.
If d is the total degree of p, then a unique such function P
exists and is called the
triangular polar form
or
triangular blossom
of p. If p(u,v) is a
monomial cu
h
v
k
, h + k = m £ d, then
(
(
)
)
˜
Ê
ˆ
˜
c
hk d
!!
-+
h k
!
Ê
Á
ˆ
Â
'
'
(
(
)
(
)
)
=
Pu v
,
,...,
u v
,
u
v
.
(12.50)
Á
11
dd
i
j
d
!
»Õ
{
}
«=
IJ
1
,...,
dIJ
,
f
,
IhJk
=
,
=
iI
Œ
jJ
Œ
The blossom for an arbitrary polynomial p(u,v) is obtained by adding up the blossoms
of all the monomial terms in p(u,v) using equation (12.50).
Proof.
See [Gall00].
To find the triangular blossom P of p(u,v) = (u + v - 3,uv,u
2
+ v
2
).
12.12.2.3
Example.
Solution.
The polynomial has total degree two and so using equation (12.50) on all
the monomials we get
(
(
) (
)
)
=
()
+
(
(
)
+
()
+
(
)
-
()
(
)
)
Pu v
,
,
u v
,
12
u
u
12
v
v
3 12
,
uv
+
uv uu
,
+
vv
.
1 1
2 2
1
2
1
2
12
21
12
12
The justification for the adjective “triangular” in the name of the blossom P is
that this approach leads to surfaces defined on triangular patches because one
wants to use barycentric coordinates to describe points in the plane
R
2
(as we did
for points in
R
) and for that one needs a triangle. The basic domain for p, which
was the unit interval [0,1] for curves, is now the triangle with vertices (0,0), (1,0), and
(0,1).
