computer aided design - cadxfem · 2017-10-10 · computer aided design bézier curves (following)...

1

Computer Aided Design

Polynomial approximation

2


Outline

Approximation Bézier curves B-Splines We still focus on curves for the moment.

3


Bézier curves

4


Bézier curves

Bézier curves Pierre Bézier (1910-1999) Develops UNISURF –

first surface modelling software at Renault's (1971)

Publicizes the theory under his name in 1962... however, the principle was discovered in 1959 by Paul de Casteljau (at Citroën's) ! Because of the “culture of secret” at Citroën, De Casteljau will have his works recognized only in 1975.

5


Bézier curves

Use of Bézier curves : Postscript fonts (cubic Bézier) & TrueType

(quadratic Bézier)

Computer graphics In geometrical modeling and CAD, they tend to be

replaced by more general techniques (NURBS)

AaBbCc

6


Bézier curves

Modelling by interpolation is not very practical We seldom have interpolation points at our hand

Instead, we hope to define these points as the result of a modeling process instead of as an input data

Approximation gives more freedom in the design of the curve

7


Bézier curves

Elements of a Bézier curve :

n=d+1 control points

Bézier curve

Control Polygon with d=n-1 sides

(also called characteristic

polygon)

For Bézier curves, the notion of knot is trivial :

u0=0 u1=1

8


Bézier curves

Characteristics of Bézier curves More freedom than interpolation

Any degree Precise control of the curve's shape Numerical stability even with high degree (not as Lagrange !)

The are Bernstein polynomials (Sergei N. Bernstein, 1880-1968 - don't mistake for Leonard Bernstein...:) :

They form a complete polynomial basis They are a partition of the unity Sometimes called ‘blending functions’ The curve is described as one polynomial (unlike splines)

P (u)=∑i=0

d

P i Bid(u)

Bidu

9


Bézier curves

Bernstein polynomials

Bidu=di u

i1−ud−i

Binomial coefficients

10


Bézier curves

Binomial coefficients : computed with Pascal's triangle

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 11 2 3 4 5

d=0

1

2

3

4

5

di di =d !

d−i ! i !

(di )=(d−1

i )+(d−1i−1 )

i=0

11


Bézier curves

Bernstein polynomials

By design, they form a partition of unity...

Bid(u)=(di )u

i(1−u)d−i

1

Binomial coefficients

=∑i=0

d

Bid(u)

=∑i=0

d

(di )Ai Bd−i

=[(1−u)+u ]d=[A+B ]d =∑

i=0

d

(di )ui(1−u)d−i

12


Bézier curves

Some characteristics of the B. polynomials. has a root of multiplicity i for u=0 has a root of multiplicity d-i for u=1 (symmetry of the basis) If i≠0, has a unique maximum at u=i/d

Bid(u)=0 if i< 0 or i> d

Bid(0)=δi 0 and Bi

d(1)=δi d

Bidu

Bid(u)≥0 for u∈[0,1]

Bidu

Bid(1−u)=Bd−i

d(u)

Bi' d=d Bi−1

d−1u −Bi

d−1u

Bidu

Bid(i /d )=i i d−d

(d−i)(d−i )(di )

13


Bézier curves

Demonstration of Bi' d(u)=d (Bi−1

d−1(u)−Bi

d−1(u))

Bi' d(u)=(di )i u(i−1)

(1−u)d−i−(di )(d−i)ui

(1−u)d−i−1

(di )=d !

(d−i)! i !=(d−1

i−1 )di

Bi' d(u)=d (d−1

i−1 )u(i−1)(1−u)d−i

−d (d−1i )ui

(1−u)(d−1)−i

(di )=d !

(d−i)! i !=(d−1

i ) dd−i

d Bi−1d−1(u) d Bi

d−1(u)

QED

14


Bézier curves

Recurrence relations of Bernstein's basis

... but no practical interest other than demonstrating algebraic relations (cf. following)

These polynomials are usually not computed explicitly

Bidu=1−uBi

d−1uu Bi−1

d−1u

Bddu=u Bd−1

d−1u B0

du=1−uB0

d−1u

15


Bézier curves

Demonstration of the recurrence relations

Bidu=di u

i1−ud−i (di )=(

d−1i )+ (d−1

i−1 )

=(d−1i )⋅ui

(1−u)d−i+ (d−1

i−1 )⋅ui(1−u)d−i

=(1−u)⋅(d−1i )⋅ui

(1−u)(d−1)−i+ u⋅(d−1

i−1 )⋅u(i−1)(1−u)(d−1)−(i−1)

(1−u)⋅Bid−1(u) u⋅Bi−1

d−1(u)

QED

16


Bézier curves

Degree 4 No negative values

Therefore, no value above 1!

17


Bézier curves

Degree 20 No extreme values

Existence of a limit envelope

e (u)=1

√2 d πu(1−u)

e (u)

18


Bézier curves

The characteristics of Bernstein polynomials involve that the Bézier curve

: interpolates P

0 and P

d ,

is invariant by affine transformations , is contained in the convex

hull of its control points(because P(u) is a combination with positive coefficients of control points – also called convex combination) ,

P u=∑i=0

d

Pi Bidu

19


Bézier curves

(following) is variation diminishing : the curve has less inflexion

points (wiggles) than there are undulations of the characteristic polynomial (proof by the fact that a Bézier curve is obtained by recursive subdivision, see further) ,

delimits a closed convex domain if the control polygon itself is convex and closed... ,

Its length is smaller than that of the control polygon.

20


Bézier curves

Same examples as shown earlier on Lagrange interpolation Circle with an increasing number of points Perturbation of the control points

21


Bézier curves

Degree 2

22


Bézier curves

Degree 4

23


Bézier curves

Degree 10

24


Bézier curves

Degree 20

25


Bézier curves

When the number of control points increases, the curve tends to the control polygon (under the assumption that the control polygon itself converges to a smooth curve ... )

The approximation involves a substantial error between the curve and the control points However, an interpolation is not the objective here...

26


Bézier curves

Perturbation of a point We shift the indicated point

27


Bézier curves

Degree 4

28


Bézier curves

Degree 10

29


Bézier curves

Degree 20

30


Bézier curves

Editing Bézier curves Degree elevation Computation of points on the curve (De Casteljau's

algorithm and others ) Changing the range of a curve

Cutting, extension Curves defined by pieces and recursive subdivision

31


Degree elevation A curve of degree d+1 is able to represent any

curve of degree d If there aren't enough control points to design a

given shape, the degree may be increased... New control points must be determined (one more !) Forrest's equations [1972]

Q0=P0

Q i=i

d+1P i−1+(1−

id+1

)Pi for i=1,⋯ , d

Qd1=P d

Bézier curves

32


Bézier curves

Demonstration of Forrest's equations :

Let's express in function of

P (u)=∑i=0

d

P i Bid(u) Q (u)=∑

i=0

d+1

Q i Bid+1(u)

Q (u)=P (u) ∀ u∈[0,1]

Bid(u)=(di )u

i(1−u)d−i

Bid +1(u)Bi

d(u)

Bid +1(u)=(d+1

i )ui(1−u)d−i+1

=(1−u)(d+1

i )

(di )Bi

d(u)

Q i= f (P0 ... P d)

33


Bézier curves

We replace the terms of the binomial

(d+1i )=

(d+1)!(d+1−i) ! i !

=(di )d+1

d+1−i

Bi+1d +1(u)=(d+1

i+1 )ui+1(1−u)d−i

=u(d+1

i+1 )

(di )Bi

d(u)

(d+1i+1 )=

(d+1)!(d−i)!(i+1)!

=(di )d+1i+1

(di )=d !

(d−i)! i !

Bid +1(u)=(1−u)

d+1d+1−i

Bid(u) Bi+1

d +1(u)=u

d+1i+1

Bid(u)

(1− id+1 )Bi

d +1(u)=(1−u)Bi

d(u)

i+1d+1

Bi+1d +1(u)=u Bi

d(u)

34


Bézier curves

Bid(u)=(1− i

d+1 )Bid +1(u)+

i+1d+1

Bi+1d +1(u)

Q (u)=P (u)⇒∑i=0

d +1

Qi Bid+1(u)=∑

i=0

d

P i((1− id+1 )Bi

d+1(u)+

i+1d+1

Bi+1d +1(u))

=∑i=0

d

P i(1− id+1 )Bi

d+1(u)+∑

i=1

d+1

P i−1i

d+1Bi

d+1(u)

=∑i=1

d

(P i(1− id+1 )+P i−1

id+1 )Bi

d+1(u)+P0 B0

d+1(u)+Pd Bd +1

d +1(u)

Qi

Qd +1

Q0

Bid(u)=(1−u)Bi

d(u)+u Bi

d(u) We split up

Bi

d(u)

QED

Bid(u)and replace :

We renumber :

35


Bézier curves

Degree elevation in practice ...

Degree 4

36


Bézier curves

Degree 5

37


Bézier curves

Degree 6

38


Bézier curves

Degree 7

39


Bézier curves

Degree 8

40


Bézier curves

Degree 9

41


Bézier curves

Degree 21

42


Bézier curves

De Casteljau's algorithm Allows the robust construction of points on the

curve Very simple geometrical interpretation

43


Bézier curves

Principle of De Casteljau's algorithm Construction of the centroids of the control

points : We continue with ....

As far as possible, until only one control point remains,That control point is P(u).

P i0

P i1

P i1=(1−u)P i

0+u P i+1

0

P i2

P0d

44


Bézier curves

Kig

45


Bézier curves

The algorithm is :

What is its complexity ? Consists of 3d(d+1) multiplications

and 3d(d+1)/2 additions , so quadratic with respect of the the degree d.

Initialization of For j from 1 to d For i from 0 to d-j

EndForEndFor is the point we want.

P ij=1−uPi

j−1u Pi1

j−1

P0d

P i0

46


Bézier curves

Demonstration with the help of recurrence relations

By gathering the terms :

by setting

P (u)=∑i=1

d−1

Pi ((1−u)Bid−1(u)+u Bi−1

d−1(u))

+P0(1−u)B0d−1(u)+Pd uBd−1

d−1(u)

Bid(u)=(1−u)Bi

d−1(u)+u Bi−1

d−1(u)

Bdd(u)=u Bd−1

d−1(u) B0

d(u)=(1−u)B0

d−1(u)P u=∑

i=0

d

Pi Bidu

P (u)=∑i=0

d−1

[(1−u)P i+u P i+1 ] Bid−1(u)

P i1=1−u Pi

0u Pi1

0

Bid−1u

P u=∑i=0

d−1

P i1 Bi

d−1u

47


Bézier curves

We do it again with P u=∑i=0

d−1

P i1 Bi

d−1u

P(u)=∑i=1

d−2

Pi1((1−u)Bi

d−2(u)+ u Bi−1

d−2(u))

+ P01(1−u)B0

d−2(u)+ Pd−1

1 u Bd−2d−2(u)

P u=∑i=0

d−2

Pi2 Bi

d−2u

P u=P0d B0

0u , B0

0u≡1 QED

P u=∑i=0

d−2

[1−uP i1u P i1

1 ]Bid−2u

......

P (u)=(1−u)P0d−1 B0

1(u)+u P1

d−1 B11(u)

48


Bézier curves

Evaluation with Horner's method We rewrite the polynomials to a monomial form

a i=M hP i M h=(1 0 0 ⋯ 0−d d 0 ⋯ 0

d (d−1)2

−d (d−1)d (d−1)

2⋯ 0

⋮ ⋮

−1d−1 d d (d−1) ⋯ ⋯ 0

−1d−1d−1 d −1d−2 d (d−1)

2⋯ 1

)ai=di ∑j=0

i

−1i− j ij P j

P u=∑i=0

d

Pi Bidu=∑

i=0

d

ai ui

49


Bézier curves

Complexity of Horner's method : 3d multiplications + 3d additions ... without taking

into account the change of polynomial basis (to be done only once)

But : change of the internal representation Some authors have shown that numerical errors

introduced during the change of internal representation are substantial.

a i=M hP i

50


Bézier curves

Method based on Bernstein polynomials We factor the terms in

This is a monomial form that can be evaluated with Horner's method – without any change of the internal representation (points P

i )

But beware when u → 1 !!!! We'd rather factor rather the terms , That gives :

P u=∑i=0

d

Pi Bidu=1−ud∑

i=0

d

Pi di u

1−u i

1−u d−i Bidu=di u

i1−ud−i

ui

P u=∑i=0

d

Pi Bidu=ud∑

i=0

d

Pi di 1−u

u d−i

51


Bézier curves

To summarize :

P u={1−ud∑

i=0

d

P i di u

1−u i

, 0u≤12

ud∑i=0

d

P i di 1−u

u d−i

, 12u≤1

P(u)=(1−u)d (P0 P1 ⋯ P d)((d0 )

(d1)u

1−u

(d2)(u

1−u )2

⋮

(dd )(u

1−u )d )

52


Bézier curves

Algorithmic complexity of the vectorial method Requires 6d multiplications and 3d additions. No change in the internal representation

53


Bézier curves

Choice of algorithm ? Robustness

1 De Casteljau 2 Vectorial 3 Horner

Speed 1 Horner 2 Vectorial 3 De Casteljau *

De Casteljau

Vectorial

Horner

d

Tot

al f

lops

54


Bézier curves

In practice The vectorial algorithm is often used to quickly

compute points for display purposes De Casteljau's algorithm is used for increased

robustness and ... It allows to obtain derivatives of curve (see later...) It allows interesting geometrical operations (see later …) If many points are to be computed, it may actually

perform well (see later !)

55


Bézier curves

Restriction of a curve (cutting) Let us compute the intersection of two curves

We need a independent representation of each segment One wants 0<u<1 on each segment

56


Bézier curves

Let us start from De Casteljau's geometrical construction

57


Bézier curves

Let us start from De Casteljau's geometrical construction

The control polygon of the both parts is obtained from points coming from De Casteljau's algorithm !

58


Bézier curves

Curve to trim

u=0

I0

P0

0

u=1

I1

P3

0

P10

P2

0

59


Bézier curves

For a curve that we want to trim :1 – Compute the intersection point I

1 – at u=u

1 with help of De

Casteljau's algorithm – this gives the points Pij

60


Bézier curves

1 – Compute the intersection point I1 – at u=u

1 with help of De


u=0

I0

P0

0

u=1

I1

P3

0

P10

P2

0

P0

1

P1

1

P2

1

P0

2

P1

2

P0

3

61


Bézier curves

For a curve that we want to limit :1 – Compute the intersection point I

1 – at u=u

1 with help of De


2 – Among these points, consider the points P0

j : they are vertices of the characteristic polygon of the curve's restriction at the interval P

0-I

1 , and the new parametrization

is u*=u/u1

62


Bézier curves



0-I


is u*=u/u1

u=0

I0

P0

0

u=1

I1

P0

1

P0

2

P0

3

u*=0

u*=1

63


Bézier curves


1 – at u=u

1 with help of De




0-I


is u*=u/u1

3 – Calculate the intersection I0 on the new curve – at u*=u*

0 -

gives the points P*ij

64


Bézier curves


0 -


P*2

1

P*1

1

P*01 P*

03

P*0

2

P*1

2

u=0

I0

u=1

I1

P*2

0

P*3

0

u*=0

u*=1

u0*

P*1

0

P*0

0

65


Bézier curves


1 – at u=u

1 with help of De


2 – Among these points, consider the points P0j : they are

vertices of the characteristic polygon of the curve's restriction at the interval P

0-I


is u*=u/u1


0 -


4 – Consider the points P*id : vertices of the characteristic

polygon of the curve's restriction to the interval I0-I

1 : new

parametrization is u'=(u*-u*0)/(1-u*

0)

66


Bézier curves

4 – Consider the points P*id-i : vertices of the characteristic

polygon of the curve's restriction to the interval I0-I

1 : new

parametrization is u'=(u*-u*0)/(1-u*

0)

P*2

1

P*0

3

P*1

2

u=0

I0

u=1

I1

P*3

0

u*=1

u*=0

u'=1u'=0

67


Bézier curves

P'2

P'0

P'1

I0

I1

P'3

u'=1

u'=0

68


Bézier curves

With the same algorithm, we can increase the parametric domain of a curve Intersection with objects close but not touching the

curve's extremities Beware : Bézier curves are variation diminishing

and convex combinations only when 0≤u≤1...

69


Bézier curves

u=0.8u=0.1

Desired cutting points

70


Bézier curves

Cutting for u=0.8

71


Bézier curves

Cutting for u=0.8...and for u=0.1 (at u*=0.1/0.8)

72


Bézier curves

Extension for u=1.1

73


Bézier curves

Extension for u=1.1 and cutting for u=0.1 - at (new parameter) u*=0.1/1.1 ...

74


Bézier curves

Curves defined by pieces Bézier curves do have a global control If we need local control, we have to assemble

several of them We have to impose some continuity at the interface

points between curves

75


=d P0B−1d−1−B0

d−1

P1B0d−1−B1

d−1

P2B1d−1−B2

d−1

⋯

Pd Bd−1d−1−Bd

d−1

Bézier curves

Expression of the derivatives of a Bézier curve with

By factoring :

dPdu(u)=d∑

i=0

d−1

(P i+1−P i)Bid−1(u)

dPduu=∑

i=0

d

Pi Bi' du Bi

' d=d Bi−1

d−1u −Bi

d−1u

dPduu=d∑

i=0

d

Pi Bi−1d−1u−Bi

d−1u

Bid−1u

=0

=0

P i'=P i+1−P i

76


Bézier curves

First derivative

The control points P'0 and P'

d-1

are interpolated so the first derivativeat the extremities only depends

on the two first (resp. last) control points

dPduu=d∑

i=0

d−1

P i1−P iBid−1u

dPduu=d∑

i=0

d−1

Pi' Bi

d−1u

O P(u)

P2

P0

P1

P3

P'0

P'1

P'2

1d

dPduu

O

First hodograph

77


Bézier curves

Second derivative = derivative of the derivative

The control points P''0 and P''

d-2...

are interpolated, so the second derivative atthe extremities only depends on the three first (resp. last)control points

d 2 Pdu2 u=d−1d∑

i=0

d−2

Pi1'−P i

'Bi

d−2u

d 2 Pdu2 u=d−1d∑

i=0

d−2

Pi'' Bi

d−2u

P'0

P'1

P'2

1d

d Pd uu

O1

d d−1d 2 Pdu2 u

P''1

P''2

Second hodograph

78


Bézier curves

Derivative of order k

The derivative of order k at the extremities only depends on the k+1 first (resp. last) control points.

Of course, there must be enough control points ... ( k < d+1 )

d k Pduk u=d−k1⋯d−1d∑

i=0

d−k

Pi1(k-1)−Pi

(k-1)Bi

d−ku

d k P

duku =∏

l=1

k

d−l1∑i=0

d−k

P i(k) Bi

d−ku , Pi

(k)=P i1

(k-1)−Pi

(k-1)

=(Pi+ 2(k-2)−2 Pi+ 1

(k-2)+ Pi

(k-2))

=∑j=0

k

(−1) j

(dj )P j+ i(0)

79


Bézier curves

80


Bézier curves

Relation between De Casteljau's algorithm and the derivatives From control points, it is possible to build a curve

that gives directly the value of the derivative.

We can also get derivatives while computing the position by De Casteljau's algorithm.

Let us consider the penultimate iteration of De Casteljau's algorithm :

dPduu=d∑

i=0

d−1

P i1−P iBid−1u

P i1=1−uPi

0u Pi1

0

P id−1=1−uP i

d−2u P i1

d−2

⋮

81


Bézier curves

Relation that links the points at the step d -1 to control points :

P id−1=∑

j=0

d−1

Pi j0 B j

d−1u

dPduu=d∑

i=0

d−1

P i10−Pi

0Bi

d−1u

=d ∑i=0

d−1

P i10 Bi

d−1u−∑

i=0

d−1

P i0 Bi

d−1u

=d P1d−1−P0

d−1

82


Bézier curves

So d times the difference between the two penultimate points of the De Casteljau algorithm yields the derivative !

dPduu

d=3

83


Bézier curves

Derivatives of higher order, same treatment

P id−k=∑

j=0

d−k

P i j0 B j

d−ku

=∏l=1

k

d−l1∑i=0

d−k

Pi2(k-2)−2 P i1

(k-2)P i

(k-2)Bi

d−ku

=∏l=1

k

(d−l+ 1)∑i=0

d−k

(∑j=0

k

(−1) j

(kj)P j+ i(0) )Bi

d−k(u)

=∏l=1

k

(d−l+ 1)∑j=0

k

(−1) j

(kj )P jd−k

d k P

duku=∏

l=1

k

d−l1∑i=0

d−k

Pi(k) Bi

d−ku

⋮

84


Bézier curves

Example for k=2 and d=3

By using De Casteljau’s algorithm to compute P(u), one also obtains derivatives up to order d, the latter one being constant. All higher order derivatives vanish.

dk P

duk(u)=∏

i=1

k

(d−i+ 1)∑j=0

k

(−1) j

(kj)P jd−k

d 2 P

du2u =3⋅2⋅P0

1−2 P1

1P 2

1

85


Bézier curves

Connecting two curves is the same as imposing constraints on the control points on both sides of the « sticking » point We assume regular curves

G0 continuity (positions) (same as C

0 continuity)

P0

P1

P3=P*

0

P*1

P*2

P*3

P2

86


Bézier curves

G1 continuity ( C

1 continuity is more strict)

Minimum degree : 2

P0

P1

P3=P*

0

P*1 P*

2

P*3

P2

P d−1 Pd= P0* P1

* , > 0 (=1 for a C1 continuity )

87


Bézier curves

G2 continuity ( C


Continuity of the orientation of the osculatory plane

Continuity of the curvature radius

Pd−Pd−1×Pd−1−Pd−2=P2*−P1

*×P1

*−P0

*

d∥Pd−1−Pd−2∥3

(d−1)∥(Pd−Pd−1)×(P d−1−P d−2)∥=

d *∥P1*−P0

*∥3

(d *−1)∥(P2

*−P1

*)×(P1

*−P0

*)∥

R=∥dP

du∥3

∥d 2 P

du2 ×dPdu∥

88


Bézier curves

G2 continuity ( C


Continuity of the osculatory plane's orientation

Continuity of the curvature radiusPd−Pd−1×Pd−1−Pd−2=P2

*−P1

*×P1

*−P0

*

P0

P1

P3=P*

0

P*1

P*2

P*3

P2

d‖P d−P d−1‖3

(d−1)‖(P d−P d−1)×(Pd−1−Pd−2)‖=

d *‖P1*−P0

*‖3

(d *−1)‖(P2

*−P1

*)×(P1

*−P0

*)‖

89


Bézier curves

The Gk continuity with k > 2 in the general case is

complex to impose The C

k continuity is easier to impose (simple

expression of higher order derivatives) Curve should be regular ! Same as imposing the continuity of functions x(u), y(u)

and z(u) , independently of each other.

d k P

duku=∏

l=1

k

d−l1∑i=0

d−k

Pi(k) Bi

d−ku

90


Bézier curves

C0 continuity

P0

P1

P3=P*

0

P*1

P*2

P*3

P2

Pd=P0*

91


Bézier curves

C1 continuity

P0

P1

P3=P*

0

P*1

P*2

P*3

P2

P1*=PdPd−Pd−1

92


Bézier curves

C2 continuity

P0

P1

P3=P*

0

P*1

P*2

P*3

P2

P2*=Pd−24P d−Pd−1

93


Bézier curves

C3 continuity

P0

P1

P3=P*

0

P*1

P*2

P*3

P2

P3*=8 Pd−12 Pd−16 Pd−2−Pd−3

94


Bézier curves

The curve has now a unique representation of degree 3.

P0

P*3

95


Bézier curves

Recursive subdivision Allows to draw the curve quickly with the help of De

Casteljau's algorithm Idea : splitting up the curve in two parts at u=0.5, then

these sub-curves in four parts ( still for u*=0.5) and so on. The control points of the sub-curves are obtained like a

residual of the De Casteljau algorithm at each step The control points quickly converge toward the curve When the gap between the starting and ending points of

each sub-curves is lower than a factor (depends on the resolution), we join simply the points of the characteristic polygon by straight line segments.

It's a « divide and conquer » approach – a famous paradigm in software engineering.

96


Bézier curves

0 subdivision

2 subdivisions

4 subdivisions

8 subdivisions

16 subdivisions

32 subdivisions

97


Bézier curves

Cost of the recursive subdivision algorithm In for m levels of subdivision Number of generated points: For each point that is generated, the algorithm

becomes linear... Competitive in comparison with Horner It is not very accurate, nevertheless very robust.

O d 2⋅2m

d⋅2m

computer aided design - cadxfem · 2017-10-10 · computer aided design bézier curves (following)...

Documents