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