Rational curves interpolated by polynomial curves
Reporter Lian ZhouSep. 21 2006
Introduction
De Boor et al.,1987 Dokken et al.,1990 Floater,1997 Goladapp,1991 Garndine and Hogan,2004
Introduction
Jaklic et al.,Preprint Lyche and Mørken 1994 Morken and Scherer 1997 Schaback 1998 Floater 2006
Introduction
Non-vanishing curvature of the curve De Boor et al.,1987
Circle Dekken et al.,1990;Goldapp,1991; Lyche and Mørken 1994
Introduction
Conic section Fang, 1999;Floater, 1997
Introduction de Boor et al., 1987 where a 6th-order accurate cubic inter
polation scheme for planar curves was constructed.
Introduction Lian Fang 1999
point.-mid parametric at the continuity-G and
points end at thesection conic with thecontinuity-G has curve
polynomial quintic dconstructe curves.The polynomial quintic
using sections conic ingapproximatfor method a presents
1
3
Introduction
property. preserving-shape
better anderror smaller much have shown to is others theof one
and 1997) (Floater,in proposedt interpolan Hermite geometric
theis themof One .continuity geometric mentioned thesatisfying
curves polynomial quintic eexist thre theresection, conicany for
thatfound isIt
The Hermite interpolant We will approximate the rational quadratic Bézier curve
,(t)B (t)wB (t)B
(t)pB (t)wpB (t)pB r(t)
210
221l00
The Hermite interpolant ellipse when w < l parabola when w = 1 hyperbola when w >
1;
The Hermite interpolant
,(t)pJ (t)pJ (t)pJ r(t)
form useful more in ther Express
221100
,)(1
)(,)(1
)1(2)(,
(t)-1
t)-(1 (t)J
where2
21
2
0 t
ttJ
t
tttJ
t)t- )(1 - 2(1 (t)
The Hermite interpolant
.1
t t (t)K
, 1 t)- t(12 (t) K
, 1
t)-(1 t)- (1 (t)K
Let
1-m
0i
i22
m1-m
0i
i 1
m1-m
0i
i20
m
2/)1( nm
The Hermite interpolant
r. t tointerpolan Hermite geometric a is
,(t)pK (t)p K (t)pK q(t)
n, degree of q curve polynomial that the
22l100
The Hermite interpolant Lemma 1
t.allfor 0 f(r(t))
equation thesatisfiesr curve Then the
4 - y) f(x,
as defined be R R : fLet
2022
1
2
The Hermite interpolant Lemma 2
2.1 -n l -n l-n2
21-n
2m22
2
1) -(2t t) t -(11) - ()1(
2
(t)1) -(2t ) (1
-f(q(t))
as factorized becan f(q(t)) polynomial The
The Hermite interpolant Theorem 1 The curve q has a total number of 2n c
ontacts with r since the equation f(q(t)) = 0 has 2n roots inside [0, 1].
The Hermite interpolant
2l01-n
1n
1-n
2
2
H p 2p -p)n
1 -1 (
n
1
2
1) - (
)1(
)max(l,r)(q,d
then3 0 lf
2 Theorem
The Hermite interpolant
itself.ion approximat theasorder same
theof is which ),O(h is above bounderror that thefollowsIt ).O(h also is
,differenceorder second ay essentiall ,P 2P -Pquantity theshown that
further It was ).O(h is 1 - aquantity that thesubcurve,each gnormalisin and
midpoint at the gsubdividiny alternatel of consisting r,for schemen subdivisio
recursive particular a using shown, it was 1995) (Floater,paper previous aIn
.P 2P -P and 1) -(w terms theare bound in the featuressalient twoThe
2n2
2l0
2
2l01-n
The Hermite interpolant Approximate the rational tensor-prod
uct biquadratic Bézier surface
Disadvantage For general m, little seems to be kno
wn about the existence of such interpolants apart from the two families of interpolants of odd degree m to circles and conic sections found in (Lyche and Mørken, 1994) and (Floater, 1997),
each having a total of 2m contacts.
High order approximation of rational curves by polynomial curves
Michael S. Floater
Computer Aided Geometric Design 23 (2006) 621–628
Method Let be the rational curve r(t)=f(t)/g(t).
2d , Rb] [a, :r d
(1) n. , . . . 2, 1, i ), (tr ) (tp ), r(t ) p(t
conditionsion interpolat2n the
satisfying , uesscalar val and 2-k nmost at
degree of p polynomial a find web t t t a
valuesparameter of sampleeach For N. Mk let and
N and Mmost at degrees of spolynomial are g and f where
iiiii
n1
n21
Two assumptions
complex).or (real roots double no has g (A2)
b], [a,in roots no has g (A1)
Basic idea
.determined be topolynomial a is and
), t-(t )t-)(t t-(t (t)
where
(2) ),(t r(t))(r(t) p(t)
Let
n21n
nt
Basic idea
). (t)(t 1 with satisfied is (1)condition that showing
(3) ), (tr))(t)(t (1 ) (tp
thatmeanswhich
,r r r r p
gives p atingdifferenti and
), r(t ) p(t have we0, ) (t Since
inii
iinii
n
iiin
nn
Basic idea
.determined be toX polynomial somefor g(t)X(t), (t)
lettingby f oft coefficien for the arranged becan This
)()(
)()()(
)(
)()((t)-g(t)p(t)
Now
2n
tf
tg
tttf
tg
tgt n
Basic idea
g. polynomial by the divides gX-1
polynomial thesuch that X polynomial a find toremainsit
),()()()()(
)()(X(t)-1p(t)
Then
n
tfttXtf
tg
tgtn
Basic idea
(1). satisfying polynomial a is
(5) (t)f(t)X(t) Y(t)f(t) p(t)
then
(4) 1, g(t)Y(t) (t)X(t)g(t)
such that Y and X spolynomial twofind To
n
n
Basic idea
p. polynomial a of degree thedenotes d(p) wherely,respective
1, - )d(a and 1- )d(amost at degrees of Y and X
solutions polynomial real unique find toused becan
rithmg.c.d.algo Euclids then common,in roots no have
i.e., prime, relatively are a and a if that algebra fromknown isIt
g. a ,g a
where
(6) 1, (t) (t)Ya (t)X(t)a
as written becan which (4), Eq.hen Consider t
01
10
10
10
n
Theorem 3 There are unique polynomials X and Y
of degrees at most N − 1 and n + N − 2, respectively, that solve (4). With these X and Y , p in (5) is a polynomial of degree at most n+k −2 that solves (1).
Euclid’s g.c.d.gorithm Now describe how Euclid’s algorithm can
be used to find the solutions X and Y .
Euclid’s g.c.d.gorithm
).d(a )d(a and )d(a )d(a )d(q where
(7) ,a aq a
in a and
q spolynomial thedefines which remainder, the
find and aby a divide we., . . 2, 1, 0, k each For
1k2k1kkk
21kk
2k
k
1kk
kk
Euclid’s g.c.d.gorithm
k. r let point weat which ,polynomial
constant a is aremainder when thestops algorithm The 2k
zero,-non be willaconstant the
though,(A2) and (A1) sassumptionUnder
2r
Euclid’s g.c.d.gorithm
(10) 1.r , . . . 3, 2, j ,bq- b b
where
(9) ,ab ab a
withup end we way,in this Continuing .bq- b b where
, ab ab ) aq - (ab ab a
gives 1-r k with (7)Then . -q b and 1 b where
(8) ,ab ab a
1-j1j-r2-jj
11r0r2r
11-r02
r21-r1r1-r1-r1r02r
r10
1r1r02r
Euclid’s g.c.d.gorithm
.(t)b
Y(t) ,(t) b
X(t)
solutions thehas (6) that shows thisand
,)(
a
b1
get toaby (9) dividecan we
constant, zero-non a is a since Finally,
2
1r
2
r
12
10
2r
r
2r
2r
rr
r
r
aa
aa
tba
Euclid’s g.c.d.gorithm
unique. are Y and X
).d(a d(Y) and )d(a d(X) Thus
).d(a )d(a-)d(a )d(b
),d(a )d(a- )d(a ) d(q)d(q ) d(b
01
01r01r
11r1r1r
Approximation order
Algebraic form of circle or conic section
Dokken et al., 1990; Goldapp, 1991; Lyche and Mørken, 1994; Floater, 1997;
Approximation order New method
).(t (t) t (t) n
Approximation order Theorem 4
,)())((max
such that n and b, a, r,
on only depending 0 C and 0 h constants are There
2
0
1
n
tttChtptr
n
0hhfor
Interpolating higher order derivatives
. t tallow i.e., coalesce, tob] [a,
in t, . . . , tvaluesparameter theof some Allow
n1
n1
Interpolating higher order derivatives
(14) 1.-2 i 0 ),(ts )(tp
and
(13) 1,- i 0 ),(tr )(tp
ty thenmultiplici has point t theIf
5. Theorem
(i)(i)
(i)(i)
Circle case
,t1
2t) ,t- (1
rational, quadratic theisorigin at the
centred circleunit theoftion representa A typical
2
2
)15(
Circle case Add the vector (1, 0) to (15) Then
(16) , t 1
2t) (2,
g(t)
f(t) r(t)
2
Circle case
1.n most at is p of degree theand
(18) 1, )Y(t)t 1( (t)X(t)2t
n to and 1most at degrees of solutions unique theare Y and X where
(17) 2), (t)(0,X(t) 2t) Y(t)(2, p(t)
is (1) topsolution a (16),in circle theisr If
R.in valuesincreasingarbitrary be t Let t
6. Theorem
2n
n
n1
Circle case Restrict n to be odd and place the para
meter values symmetrically around t = 0.
ntt ,,1
Circle case
)u-(u )u-2u(u (u)A
where
(20) ,)t (-1)(1A
(-1)A -)(tA-Y(t) ,
(-1)A
1 X(t)
andn degree has (17)in pThen . u u0 valuessomefor
(19) ) u , . . . ,u0, ,,-u . . . ,(-u ) t, . . . ,(t
thatand 0 s somefor 1 2s n Suppose
7. Theorem
2s
210
20
02
0
0
s1
s11sn1
Circle case
0. at t contacts2n having (1994)Mrken and Lycheby
foundion approximat like-Taylor theis This
).)()(-t2t ,)(-t(2 p(t)
,)(Y(t) /2,-(-1)X(t)
obtains, one 0, u ulet weIf
1-s
0i
22s
0i
i2
0
2s
s1
si
s
i
s
tt
t
Circle case
(16). arccircular the toapplied 1997) (Floater,
oft interpolan Hermite theis p that n,calculatiolengthy aafter
finds, one (17), into dsubstitute are e when thesand
, )v-(-1
)v-(t t1 Y(t) ,
)v-2(-1
1- X(t)
gives which 0, v u u is case limitingAnother s
1ii2
1-i222
s2
s1
Example
;012.8 e 0.5), 0.375, 0.25, (0.125, u 9, n (f)
;103.6 e 0.5), (0.25, u 5, n (e)
;10 1.6 e 0.5), (0.5, u 5, n (d)
;10 4.9 e 0.0), (0.0, u 5, n (c)
;10 7.4 e(0.5), u 3, n (b)
;10 7.8 e (0.0), u 3, n (a)
)u,,(uu
10
6-
5-
4-
4-
3-
s1
Thank you Thank you