Interpolation to Data Points

Lizheng Lu

Oct. 24, 2007

Problem

Interpolation VS. Approximation

Interpolation Approximation

Classification

Curve

Constraint

(piecewise) Bezier curves B-spline curves Rational Bezier/B-spline curves

Outline

Some classical methods

Some recent methods on geometric interpolation

Estimate the tangent

C2k-1 Hermite Interpolation

Cubic Interpolation

C2 Cubic B-spline Interpolation

Given: A set of points and a knot sequence Find: A cubic B-spline curve, s.t.

2

3 2

4 3

5

* *

* * *

* * *

* *

s

e

p r

p s

p s

p r

Geometric Hermite Interpolation (GHI)

Given: Planar points pi, with positions, tangents and curvatures

Result: Piecewise cubic Bezier curves, having G2 continuity 6th order accuracy Convexity preservation

[de Boor et al., 1987]

Comments on GHI

Independent of parameterization High accuracy

But, it usually includes nonlinear problems Questions on the existence of solution and

efficient implement Difficult to estimate approximation order,

etc…

References on GHI

High Order Approximationof Rational Curves

Given: A rational curve , where f and g are of degree M and N, let k = M+N,

with parameters values

Find: A polynomial p of degree at most n+k-2,

and scalar values satisfying the 2n interpolation conditions:

[Floater, 2006]

Geometric Interpolation by Planar Cubic Polynomial

Curves

Comp. Aided Geom. Des. 2007, 24(2): 67-78

Jernej Kozak Marjeta Krajnc FMF&IMFM IMFM

Jadranska 19, Ljubljana, Slovenia

Problem

Given: six points

Find: a cubic polynomial parameter curve

which satisfies

An Alternative Solution:Quintic Interpolating CurvesFind a quintic curve

s.t.,

where ti are chosen to be the uniform

and chord length parameterization.

Essential of Problem

Know: t0, t5, p0, p3

Unknown: t1, t2, t3, t4 , p1, p2

Equations: P3(ti) = Ti, i = 2, 3, 4

Solution of Problem

Solved by Newton Iteration with initial values:

Know: t0, t5, p0, p3

Unknown: t1, t2, t3, t4 , p1, p2

Equations: P3(ti) = Ti, i = 2, 3, 4

Existence of Solution

Provide two sufficient conditions guaranteeing the existence

Summarize cases in a table which does not allow a solution

Comparison

cubic uniform chord length

On Geometric Interpolation by Planar Parametric Polynomial

Curves

Mathematics of Computation 76(260): 1981-1993

Problem

Given: 2n points

Find: a cubic polynomial parameter curve

which satisfies

Main Results

If the data, sampled from a convex smooth

curve, are close enough, then equations that determine the interpolating

polynomial curve are derived for general n (Theorem 4.5)

if the interpolating polynomial curve exists, the approximation order is 2n for general n (Theorem 4.6)

the interpolating polynomial curve exists for n≤ 5 (Theorem 4.7)

On Geometric Interpolation of Circle-like

Curves

Comp. Aided Geom. Des. 2007, 24(4): 241-251

What is Circle-like Curve?A circular arc of an arclength is defined by

Suppose that a convex curve is parameterized by the

same parameter as . The curve will be calledcircle-like, if it satisfies:(1)

(2)

The Result

Outline

Some classical methods

Some methods on geometric interpolation

Estimate the tangent

Tangent Estimation Methods

FMill , 1974 Circle Method Bessel

[Ackland, 1915] Akima, 1970 G. Albrecht, J.-P. Bécar, G. Farin, D. Ha

nsford, 2005, 2007

Problem

?

FMILL

Circle Method

Bessel

Parabola f (t)

Bessel

Akima’s Method

Albrecht’s Method Albrecht G., Bécar J.P.

Univ. de Valenciennes et du Hainaut–Cambrésis, France Farin G., Hansford D.

Dep. Comp. Sci., Arizona State Univ.

Détermination de tangentes par l’emploi de coniques d’approximation.

On the approximation order of tangent estimators. CAGD, in press

Main Idea Method: Estimate the tangent by using the

interpolating conic of the given five points

Solution: solved by Pascal’s theorem in projective geometry

Advantages Conic precision Less computations without computing the

implicit conic

Idea Derivation Any conic section is uniquely determined by f

ive distinct points in the plane, pi=(xi, yi).

[Farin, 2001]

2 2

2 21 1 1 1 1 12 22 2 2 2 2 22 23 3 3 3 3 32 24 4 4 4 4 42 25 5 5 5 5 5

1

1

1( , ) 0

1

1

1

x xy y x y

x x y y x y

x x y y x yf x y

x x y y x y

x x y y x y

x x y y x y

Idea Derivation[Pascal, 1640]

Projective Geometry in CAGD

Express rational forms

Implicit representation of rational forms

0 0

( ) ( ), ( ), ( ), ( )

( ) ,

( ) ( ), ( ), ( ) ( )

m n

i iji j

x y z

B

x y z

R

R

R

u u u u u

u

u = u u u u

Projective Geometry in CAGD

Express rational forms

Implicit representation of rational forms Chen, Sederberg

Conic sectionLine conics

Projective Geometry

Projective Geometry

A line in is represented by

The line joining the two points is

The intersection of two lines is

Estimate the Tangent

Estimate the Tangent

Degenerate Cases

(b)

(c)

(a)

Examples

Experimental results

Non-convex Case

Conic method Akima

Bessel Circle method

Approximation order

Theoretical Analysis

Consider a planar curve:

Theoretical Analysis

Consider a planar curve:

Take five points:

Theoretical Analysis

Consider a planar curve:

Take five points:

Let:

Theoretical Analysis

Taylor expansion:

Exact tangent:

Exact norm:

Theoretical Analysis

For a point , with the tangent:

Its corresponding tangent in the projective space is:

Compute the Approximation Order

Taylor expansion Symbolic computation: MAPLE

To solve the k in:

Numerical Result (1)

Numerical Result (2)

Summary

Obtain order four approximation for the convex case, two for the inflection point

Estimate the approximation order with theoretical justification

Estimate the direction of the tangent only, not the vector!

Thank You!