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Cubic Space Curves Consider coordinate functions that are cubic polynomialsEach is a linear combination of monomial terms
where 3 23 2
1 0 3 23 2
1 0 3 23 2
1 0 =^ +^
+^ +
=^ +^
+^ +^
0 ≤^ ≤ 1
=^ +^
+^ +
(^ ) x u^ a u (^ ) (^ )
a u^
a u^ a y u^ b u
b u^
b u^ b^
u
z u^ c u
c u^
c u^ c 3 = ƒ=0 3 =ƒ=0 3 =ƒ=
(^ ) (^ ) (^ )
iii iii iii x u^
a u y u^
b u z u^
c u
Vector/Matrix Forms for Cubic Curves For convenience, we can rewrite in vector formAnd in an even more condensed matrix form
0 T 1 2 3
»^ º …^ »…^ »^23
»^
= 1^
¬^
¼ …^ » …^ »¬^ ¼
(^ ) u^
u^ u^ u
a a p^
u Aa a
(^1) » º … » u … »= (^) u^2 … » u … »^3 u ¬ ¼
3 23 2
1 0 =^ +^3 =
+^ +
(^ ) u^ u^ i =^ ƒ ii
u^ u u p^ a^
a^ a^
a a
»^ º a^ i …^ »= b a i i …^ »…^ » c ¬^ ¼ i
3
3
3
=^
=^
=
=^
=^
ƒ^
ƒ^
(^ )^
(^ )^
(^ )
i^
i^
i
i^
i^
i
i^
i^
i
x u^
a u^ y u
b u
z u^
c u
Rewriting with Geometric Constraints Recall that a cubic is determined by 4 constraint points^ • want to rewrite spline formulas in terms of these constraints•^ not^
the coefficients of the monomial terms
P
T geometry matrix 11 12 basis matrix
13 14
1 21 22
23 24
2
2 3
31 32
33 34
3 41 42
43 44
4
»^
º »^ º
…^
» …^ »
…^
» …^ »
»^
= 1^
¬^
¼ …^
» …^ »
…^
» …^ »¬^ ¼
¬^
(^ )
m^ m^
m^ m m^ m^
m^ m
u^ u
u^ u^
m^ m^
m^ m m^ m^
m^ m
g g
p^
u MGg g
T^ T = =^
Ω^ =
(^ ) u p u A
u MG
A^
MG
Hermite Curves Specify 4 geometry constraints^ • endpoints of the curve segment• tangent vectors at each of the endpoints Easy to paste Hermite segments together^ • specify coincident endpoints and identical tangents• guarantees tangents are continuous —
(^1) C continuity
=^0 ( ) p p 0 =^1 ( ) p p 3 =^0 '( ) r p 0 =^1 '( ) r p 3
p^^0
p^3 r^^0
r^3
p » º^0 … » p^3 … »= (^) G … » r^0 … » r^3 ¬ ¼
Deriving the Hermite Basis Matrix These are the constraints that we want:We can rewrite this as:
0
0 3
1 0
2 3
3 1 0 0
»^ º^
»^ º
»^
…^ »^
…^ »
…^
…^ »^
…^ »
…^
…^ »
…^
…^ »^
…^ »
…^
¬^
¬^ ¼^
¬^ ¼
p^
a p^
a r^
a r^
a 0
0 3
3 2
1 0 0
1 3
3 1
1 =^ 0 ==^ 1 =^
+^ +^
=^ 0 ==^ 1 = 3
( )( )'( )^ + 2^ +'( )
p^ p^
a p^ p^
a^ a^
a^ a r^ p^
a r^ p^
a^ a^
a
3 23 2
1 0 =^ +^
+^ +
(^ ) u^ u
u^
u p^ a^
a^ a^
a
Deriving the Hermite Basis Matrix Starting from here, we invert the coefficient matrix …… and solve for the spline coefficient matrix
C
0
0 3
1 0
2 3
3 1 0 0
»^ º^
»^ º
»^
…^ »^
…^ »
…^
…^ »^
…^ »
…^
…^ »
…^
…^ »^
…^ »
…^
¬^
¬^ ¼^
¬^ ¼
p^
a p^
a r^
a r^
a − 0
0
0
1
3
3
2
0
0
3
3
3
»^ º^
»^ º^
»^ º
»^
º^
»^
…^ »^
…^ »^
…^ »
…^
»^
…^
…^ »^
…^ »^
…^ »
…^
»^
…^
=^
…^ »^
…^ »^
…^ »
…^
»^
…^
−3^3
−2^ −
…^ »^
…^ »^
…^ »
…^
»^
…^
2 −^
¬^
¼^
¬^
¬^ ¼^
¬^ ¼^
¬^ ¼
a^
p^
p
a^
p^
p
a^
r^
r
a^
r^
r
The Equation for a Hermite Curve We’re done! The curve, in terms of the constraints isWe can also look at it as a weighted sum of the constraints^ • each is weighted by a blending function• whose coefficients are the columns of the basis matrix
3 2
3 2
3 2
3 2
0
3
0
3
1 0 2
3 3 0
4 3 = 2^ − 3
+^
+ −2^ + 3
+^
− 2^ +^
+^ −
=^ +^
+^ +
(^ )^ (^
)^ (^
)^ (^
)^ (^
u^ u^
u^
u^ u^
u^ u^
u^ u
u
h^ h^
h^ h p^
p^
p^
r^
r
p^ p^
r^ r
0 3
2 3
0 3 1 0
»^
º …^ »
…^
…^
»^
= 1¬^
¼^
…^ »
…^
−3^3
−2^ −1^ …
…^
2 −^
¬^
¼ ¬^ ¼
(^ ) u^
u^ u^ u
p p
p^
r r
Hermite Blending Polynomials^1
(^ ) h u^ 1
(^ ) h u (^2) ( ) h u 3 ( ) h u 4
3 2 1
3 2 2 3 2 3 3 2 = 2^4
− 3^ +1= −2 + 3= − 2^ + = −
(^ ) h u^ u^ (^ ) (^ ) (^ )
u h^ u^
u^ u h^ u^ u
u^
u h^ u^ u
u
Exercise: Bézier Continuity Suppose that you’re given two cubic Bézier control polygonswhere the two curves
p^ and^ q
should be joined consecutively.
What constraints on these points are necessary to guarantee^1 C^ continuity between them?
0 1 2
3 0 1 2
3 ,^ ,^ , p p^ p^ p ,^ ,^ , q q^ q^ q
Catmull–Rom Splines Given a set of points in space, suppose we want a spline that^ • interpolates the data points
[rules out Bézier]
(^1) • with C continuity
[Hermite: lots of tweaking]
This is a common situation in animationWe start with the given set of points
define tangent ,^ ,^
(^
n^
i^ i^
i s
0
=^ −+1^ −
p^ p^
r^ p^
p
Catmull–Rom Splines Typically, we pick
s^ =^ ½ and we can derive a spline equation More generally, we can use any tension parameter
−3 −2 s
2 3
− 0 2
»^
º …^ »
…^
−1^0
…^ »
…^
»^
=^1 ¬^
¼^
…^ »
…^
2 −^
…^ »
…^
−1^3
−3^1
¬^
¼ ¬^ ¼
(^ )
i i i i
u^
u^ u^ u
p p
p^
p p^ −3 −
2 3
− 0 1
»^ º
»^
º …^ »
…^
−^0
0 …^ »
…^
»^
= 1¬^
¼^
…^ »
…^
3 − 2^
−^ …^
…^
−^ 2 −^
¬^
¼ ¬^ ¼
(^ )
i i i i s^
s
u^ u^ u^ u
s^ s
s^ s s^ s^
s^ s
p p
p^
p p
