Singularity of Cubic B zier Curves and Surfaces - …enadler/presentations...Singularity of Cubic Bézier Curves and Surfaces Edmond Nadler Eastern Michigan University. joint work

Singularity of Cubic Bézier Curves andSurfaces

Edmond Nadler

Eastern Michigan University.

joint work with

Tae-wan KimMin-jae Oh

Sung-ha ParkSeoul National University

SIAM Conference on Geometric and Physical Modeling

October 24, 2011

E. Nadler, Eastern Michigan University Bézier Curve & Surface Singularity 1/35

Outline

Introduction

Singularity of a Parametric Curve

Bézier Curves

Singularity of Bézier Curves

Tangent at Bézier Curve singularity

Surfaces


Parametric Cubic Curve

C(t) = a0 + a1t + a2t2 + a3t3

Example (“twisted cubic”): C(t) =⟨

t, t2, t3⟩

-2

-1

0

1

2

x

01

23

4

y

-5

0

5

z

:

-2-1012

x

0 1 2 3 4

y

-5

0

5

z

,

01234 y-2-1012

x

-5

0

5

z

,

-2-1012

x

0

1

2

3

4

y

-505z

>


Singularity of a Parametric Curve

Singularity of a curve C(t): t∗ where C′(t∗) = ~0

Geometrically, can be a cusp

Example: C(t) =⟨

4t3 − 3t2 + 1, 4t3 − 9t2 + 6t⟩, t ∈ [0, 1]

C′(t) =⟨

12t2 − 6t, 12t2 − 18t + 6⟩

C′(12) = 〈 0, 0〉 =⇒ t∗ = 1

2

C(t∗) =⟨

34, 5

4

⟩

0.5 1. 1.5 2.x0.

0.5

1.

1.5y


Bézier Curves

A representation of parametric polynomial curves

Geometric and intuitive, facilitating creative design process

Computationally efficient and stable

At the core of Computer Aided Geometric Design (CAGD)


Bézier Curves of degree 1

C(t) = (1− t)P0 + t P1, t ∈ [0, 1]

P0

P1

C H0.25L



C(t) = (1− t)P0 + t P1, t ∈ [0, 1]

P0

P1

C H0.5L



C(t) = (1− t)P0 + t P1, t ∈ [0, 1]

P0

P1

C H0.75L



C(t) = (1− t)P0 + t P1, t ∈ [0, 1]

P0

P1

C H1.L



P01 = (1− t)P0 + t P1 ; P12 = (1− t)P1 + t P2

C(t) = (1− t)P01 + t P12, t ∈ [0, 1]

P0

P1

P2



P01 = (1− t)P0 + t P1 ; P12 = (1− t)P1 + t P2

C(t) = (1− t)P01 + t P12, t ∈ [0, 1]

P0

P1

P2

P01

P12

C H0.25L



P01 = (1− t)P0 + t P1 ; P12 = (1− t)P1 + t P2

C(t) = (1− t)P01 + t P12, t ∈ [0, 1]

P0

P1

P2

P01

P12

C H0.25L



P01 = (1− t)P0 + t P1 ; P12 = (1− t)P1 + t P2

C(t) = (1− t)P01 + t P12, t ∈ [0, 1]

P0

P1

P2P01

P12

C H0.5L



P01 = (1− t)P0 + t P1 ; P12 = (1− t)P1 + t P2

C(t) = (1− t)P01 + t P12, t ∈ [0, 1]

P0

P1

P2

P01

P12

C H0.75L


Bézier Curve – DefinitionDegree 3:

C(t) =

3∑

i=0

(3i

)(1− t)3−it i Pi

=

3∑

i=0

B3i (t) Pi , t ∈ [0, 1]

where

B3i (t) =

(3i

)(1− t)3−it i is

the ith Bernstein (basis) polynomial of degree 3, and

Pi are known as (Bézier) control points.


Bézier Curve – DefinitionDegree n:

C(t) =

n∑

i=0

(ni

)(1− t)n−it i Pi

=

n∑

i=0

Bni (t) Pi , t ∈ [0, 1]

where

Bni (t) =

(ni

)(1− t)n−it i is

the ith Bernstein (basis) polynomial of degree n, and

Pi are known as (Bézier) control points.


de Casteljau algorithmthis method of recursive convex combinations to evaluate theBézier function from its control points

C(t) = (1− t)[(1− t)[(1− t)P0 + t P1] + t [(1− t)P1 + t P2]

]

= 1- + t[(1− t)[(1− t)P1 + t P2] + t [(1− t)P2 + t P3]

]

Subdivision:C = C− ∪ C+, with domains [0, t], [t, 1].

The control point sequences P−, P+ of C−, C+ are given byP− = 〈P0, P01, P012, P0123〉,P+ = 〈P0123, P123, P23, P3 〉.



C(t) = (1− t)[(1− t)[(1− t)P0 + t P1] + t [(1− t)P1 + t P2]

]

= 1- + t[(1− t)[(1− t)P1 + t P2] + t [(1− t)P2 + t P3]

]

P0

P1

P2

P3

P01

P12 P23

P012

P123

C H0.375L





C(t) = (1− t)[(1− t)[(1− t)P0 + t P1] + t [(1− t)P1 + t P2]

]

= 1- + t[(1− t)[(1− t)P1 + t P2] + t [(1− t)P2 + t P3]

]

P0

P1

P2

P3

P01

P12 P23

P012

P123

C H0.375L





C(t) = (1− t)[(1− t)[(1− t)P0 + t P1] + t [(1− t)P1 + t P2]

]

= 1- + t[(1− t)[(1− t)P1 + t P2] + t [(1− t)P2 + t P3]

]

P0

P1

P2

P3

P01

P12 P23

P012

P123

C H0.375L




Bernstein Basis PolynomialsDegree 3:

{B3i (t)}

3i=0 = {(1− t)3, 3(1− t)2t, 3(1− t)t 2, t3}

0.25 0.5 0.75 1.t

0.25

0.5

0.75

1.

BHtL

Partition of unity:


Bernstein Basis PolynomialsDegree 3:

{B3i (t)}

3i=0 = {(1− t)3, 3(1− t)2t, 3(1− t)t 2, t3}

0.25 0.5 0.75 1.t

0.25

0.5

0.75

1.

BHtL

Partition of unity:n∑

i=0

Bni (t) =

n∑

i=0

(ni

)(1− t)n−it i

=(1− t + t

)n

= 1


Examples of Cubic Bézier Curves

from Farin & Hansford from from Far from Farin & Hansford 2000


Some Properties of Bézier Curves

Endpoint interpolation: C(0) = P0 and C(1) = Pn

Endpoint tangency to control polygon:C′(0)‖(P1 − P0) and C′(1)‖(Pn − Pn−1)

Convex Hull Property: C[P] ⊂ ConvexHull(P)which implies

(P planar=⇒ C[P] planar

)

Affine invariance: C[ΦP] = ΦC[P]


Derivative of Bézier Curve

C′(t) =

n∑

i=0

Bni′(t) Pi

=n∑

i=0

(ni

)((1− t)n−it i)′ Pi

= nn−1∑

i=0

Bn−1i (t) (Pi+1 − Pi)

That is, the Bézier control points of C′ are simplyThat is, the Bézier c 〈 n (Pi+1 − Pi) 〉 , i = 0..n − 1

Differentiate a Bézier Curve by differencing its control points


Derivative of Bézier Curve

C′(t) =

n∑

i=0

Bni′(t) Pi

=n∑

i=0

(ni

)((1− t)n−it i)′ Pi

= nn−1∑

i=0

Bn−1i (t) (Pi+1 − Pi)

That is, the Bézier control points of C′ are simplyThat is, the Bézier c 〈 n (Pi+1 − Pi) 〉 , i = 0..n − 1

Differentiate a Bézier Curve by differencing its control points


Singularity of Bézier Curve of degree 1

Recall definition of singularity of a curve C(t):t∗ where C′(t∗) = ~0

Apply this to Bézier Curve of degree 1:

C(t) = (1− t)P0 + t P1

C′(t) = P1 − P0

= ~0 ∀t iff P0 = P1

That is, the only case of singularity of a polynomial curve ofdegree 1 is the trivial case when its two endpoints agree!





C(t) = (1− t)P0 + t P1

C′(t) = P1 − P0

= ~0 ∀t iff P0 = P1






C(t) = (1− t)P0 + t P1

C′(t) = P1 − P0

= ~0 ∀t iff P0 = P1



Singularity of Bézier Curve of degree 2 – part 1

C′(t∗) = ~0 for n = 2:

C(t) = B20P0 + B2

1P1 + B22P2

C′(t) = B10(P1 − P0) + B1

1(P2 − P1) by derivative formula

= (1− t)(P1 − P0) + t (P2 − P1)

Hence, the only cases of singularity of a polynomial curveof degree 2 occur when its Bézier control points satisfy

(P1 − P0)‖(P2 − P1)

i.e., they are collinear.

Hence, by the Convex Hull Property, the curve actually lieson a line.



C′(t∗) = ~0 for n = 2:

C(t) = B20P0 + B2

1P1 + B22P2

C′(t) = B10(P1 − P0) + B1


= (1− t)(P1 − P0) + t (P2 − P1)


(P1 − P0)‖(P2 − P1)





C′(t∗) = ~0 for n = 2:

C(t) = B20P0 + B2

1P1 + B22P2

C′(t) = B10(P1 − P0) + B1


= (1− t)(P1 − P0) + t (P2 − P1)


(P1 − P0)‖(P2 − P1)





Equation for singularity:

C′(t∗) = (1− t∗)(P1 − P0) + t∗ (P2 − P1) = ~0, t∗ ∈ [0, 1]

For singularity, in addition to being collinear, must have P0, P1, P2 “outof order”, i.e.,P0 between P1 and P2: t∗ ∈ [0, 1

2]ORP2 between P0 and P1: t∗ ∈ [ 1

2, 1]

In all cases, the singularity is at P1; curve reverses direction there.

Special cases of coincident adjacent end control points:

If P0 = P1, singularity there at t = 0





C′(t∗) = (1− t∗)(P1 − P0) + t∗ (P2 − P1) = ~0, t∗ ∈ [0, 1]



2, 1]








C′(t∗) = (1− t∗)(P1 − P0) + t∗ (P2 − P1) = ~0, t∗ ∈ [0, 1]



2, 1]








C′(t∗) = (1− t∗)(P1 − P0) + t∗ (P2 − P1) = ~0, t∗ ∈ [0, 1]



2, 1]






Singularity of Bézier Curve of degree 3 – part 1Basics

C′(t∗) = ~0 for n = 3:

C(t) = B30P0 + B3

1P1 + B32P2 + B3

3P3

C′(t) = 3[ B20(P1 − P0) + B2

1(P2 − P1) + B22(P3 − P2) ]

= 3[ (1− t)2(P1 − P0) + 2(1− t)t (P2 − P1) + t2(P3 − P2) ]

Hence, the only cases of singularity of a polynomial curveof degree 3 occur when a linear combination of{(P1 − P0), (P2 − P1), (P3 − P2)} equals ~0.

Hence, for singularity, these three vectors, and hence,{P0, P1, P2, P3} themselves, must be coplanar.

Hence, for singularity, by the Convex Hull Property, thecurve must be planar.



C′(t∗) = ~0 for n = 3:

C(t) = B30P0 + B3

1P1 + B32P2 + B3

3P3

C′(t) = 3[ B20(P1 − P0) + B2

1(P2 − P1) + B22(P3 − P2) ]

= 3[ (1− t)2(P1 − P0) + 2(1− t)t (P2 − P1) + t2(P3 − P2) ]






C′(t∗) = ~0 for n = 3:

C(t) = B30P0 + B3

1P1 + B32P2 + B3

3P3

C′(t) = 3[ B20(P1 − P0) + B2

1(P2 − P1) + B22(P3 − P2) ]

= 3[ (1− t)2(P1 − P0) + 2(1− t)t (P2 − P1) + t2(P3 − P2) ]






C′(t∗) = ~0 for n = 3:

C(t) = B30P0 + B3

1P1 + B32P2 + B3

3P3

C′(t) = 3[ B20(P1 − P0) + B2

1(P2 − P1) + B22(P3 − P2) ]

= 3[ (1− t)2(P1 − P0) + 2(1− t)t (P2 − P1) + t2(P3 − P2) ]





Singularity of Bézier Curve of degree 3 – part 1C′(t) = 3[ (1− t)2(P1 − P0) + 2(1− t)t (P2 − P1) + t2(P3 − P2) ]

Notice

C′(0) = 3(P1 − P0) and C′(1) = 3(P3 − P2)

Hence, singularity at t = 0 iff P1 = P0

Hence, singularity at t = 1 iff P2 = P3.

And for t ∈ (0, 1),13C

′(t) is a strict convex combination of

∆Pi := Pi+1 − Pi, i = 0, 1, 2

With P := 〈∆P0,∆P1,∆P2〉,

C[P]′ = C[3∆P] =⇒ C[P] singular at t∗ : C[∆P](t∗) = ~0



Notice

C′(0) = 3(P1 − P0) and C′(1) = 3(P3 − P2)



And for t ∈ (0, 1),13C


∆Pi := Pi+1 − Pi, i = 0, 1, 2

With P := 〈∆P0,∆P1,∆P2〉,




Notice

C′(0) = 3(P1 − P0) and C′(1) = 3(P3 − P2)



And for t ∈ (0, 1),13C


∆Pi := Pi+1 − Pi, i = 0, 1, 2

With P := 〈∆P0,∆P1,∆P2〉,




Notice

C′(0) = 3(P1 − P0) and C′(1) = 3(P3 − P2)



And for t ∈ (0, 1),13C


∆Pi := Pi+1 − Pi, i = 0, 1, 2

With P := 〈∆P0,∆P1,∆P2〉,



Singularity of Bézier Curve of degree 3 – part 2Solution for singularity using Bézier singularity condition

Define pointsO = ℓ(P0, P3−P2)∩ ℓ(P3, P1−P0)R = ℓ(P1, P3 −P2)∩ ℓ(P2, P1 − P0)

where ℓ(P, V) is the line defined bypoint P and vector V .

P0P1

P2

P3

O

R

From the geometry above, R − O = ∆P0 − ∆P2,

∆P0 ‖ (P3 − O) and ∆P2 ‖ (P0 − O)

∆P0 = x(P3 − O), x = det(∆P2,∆P0)det(∆P2,P3−P0)

(1)

∆P2 = −y(P0 − O), y = det(∆P0,∆P2)det(∆P0,P3−P0)

(2)

with (x, y) capturing the essential shape of the control polygon.Under the Bézier singularity condition C[∆P](t∗) = ~0, (1), (2) −→

(x, y) =(

2t∗

3t∗−1,2(1−t∗)2−3t∗

), which satisfies

(x − 4

3

) (y − 4

3

)= 4

9 (∗)

Condition (∗) for singularity was found by [Su & Liu 1990] using other methods that did not make





P0P1

P2

P3

O

R


∆P0 ‖ (P3 − O) and ∆P2 ‖ (P0 − O)


(1)


(2)


(x, y) =(

2t∗

3t∗−1,2(1−t∗)2−3t∗

), which satisfies

(x − 4

3

) (y − 4

3

)= 4

9 (∗)






P0P1

P2

P3

O

R


∆P0 ‖ (P3 − O) and ∆P2 ‖ (P0 − O)


(1)


(2)


(x, y) =(

2t∗

3t∗−1,2(1−t∗)2−3t∗

), which satisfies

(x − 4

3

) (y − 4

3

)= 4

9 (∗)






P0P1

P2

P3

O

R


∆P0 ‖ (P3 − O) and ∆P2 ‖ (P0 − O)


(1)


(2)


(x, y) =(

2t∗

3t∗−1,2(1−t∗)2−3t∗

), which satisfies

(x − 4

3

) (y − 4

3

)= 4

9 (∗)






P0P1

P2

P3

O

R


∆P0 ‖ (P3 − O) and ∆P2 ‖ (P0 − O)


(1)


(2)


(x, y) =(

2t∗

3t∗−1,2(1−t∗)2−3t∗

), which satisfies

(x − 4

3

) (y − 4

3

)= 4

9 (∗)

Condition (∗) for singularity was found by [Su & Liu 1990] using other methods that did

not make essential use of the Bézier form.





P0P1

P2

P3

O

R


∆P0 ‖ (P3 − O) and ∆P2 ‖ (P0 − O)


(1)


(2)


(x, y) =(

2t∗

3t∗−1,2(1−t∗)2−3t∗

), which satisfies

(x − 4

3

) (y − 4

3

)= 4

9 (∗)







P0 P1

P2

P3O

R


∆P0 ‖ (P3 − O) and ∆P2 ‖ (P0 − O)


(1)


(2)


(x, y) =(

2t∗

3t∗−1,2(1−t∗)2−3t∗

), which satisfies

(x − 4

3

) (y − 4

3

)= 4

9 (∗)







P0 P1

P2

P3O

R


∆P0 ‖ (P3 − O) and ∆P2 ‖ (P0 − O)


(1)


(2)


(x, y) =(

2t∗

3t∗−1,2(1−t∗)2−3t∗

), which satisfies

(x − 4

3

) (y − 4

3

)= 4

9 (∗)







P0 P1

P2

P3O

R


∆P0 ‖ (P3 − O) and ∆P2 ‖ (P0 − O)


(1)


(2)


(x, y) =(

2t∗

3t∗−1,2(1−t∗)2−3t∗

), which satisfies

(x − 4

3

) (y − 4

3

)= 4

9 (∗)







P0 P1

P2

P3O

R


∆P0 ‖ (P3 − O) and ∆P2 ‖ (P0 − O)


(1)


(2)


(x, y) =(

2t∗

3t∗−1,2(1−t∗)2−3t∗

), which satisfies

(x − 4

3

) (y − 4

3

)= 4

9 (∗)







P0 P1

P2P3O R


∆P0 ‖ (P3 − O) and ∆P2 ‖ (P0 − O)


(1)


(2)


(x, y) =(

2t∗

3t∗−1,2(1−t∗)2−3t∗

), which satisfies

(x − 4

3

) (y − 4

3

)= 4

9 (∗)

Additional case: special doubly degenerate case of (x, y) = (0, 0) =⇒

P1 = P0 & P2 = P3 =⇒ singular at t = 0 & t = 1





P0P1

P2P3OR


∆P0 ‖ (P3 − O) and ∆P2 ‖ (P0 − O)


(1)


(2)


(x, y) =(

2t∗

3t∗−1,2(1−t∗)2−3t∗

), which satisfies

(x − 4

3

) (y − 4

3

)= 4

9 (∗)

Additional case: special doubly degenerate case of (x, y) = (0, 0) =⇒

P1 = P0 & P2 = P3 =⇒ singular at t = 0 & t = 1



Summary of main result

Define affine coordinates (x, y) of the control polygon of a cubicBézier curve by R − O = (P3 − O)x + (P0 − O)y

P0P1

P2

P3

O

R

The curve has a singularity at t = t∗ iff

(x, y) =(

2t∗

3t∗−1,2(1−t∗)2−3t∗

), which satisfies

(x − 4

3

) (y − 4

3

)= 4

9

or two singularities at t∗ ∈ {0, 1}, for the case (x, y) = (0, 0).



Summary of main result

Define affine coordinates (x, y) of the control polygon of a cubicBézier curve by R − O = (P3 − O)x + (P0 − O)y

P0P1

P2

P3

O

R

The curve has a singularity at t = t∗ iff

(x, y) =(

2t∗

3t∗−1,2(1−t∗)2−3t∗

), which satisfies

(x − 4

3

) (y − 4

3

)= 4

9

or two singularities at t∗ ∈ {0, 1}, for the case (x, y) = (0, 0).


Singularity of Bézier Curve of degree 3 – part 3aExamples of singular solution

Singular cubic curves with various values of t∗ ∈ [0, 1], withP0,P3, and the directions of P1 − P0 and P3 − P2 fixed.

P0 P1

P2

P3

t_sing=0.5

O

R




P0 P1

P2

P3

t_sing=0.55

O

R




P0 P1

P2

P3

t_sing=0.6

O

R




P0 P1

P3

t_sing=0.65

O




P0 P1

P3

t_sing=0.7

O




P0 P1

P2

P3

t_sing=0.75

O

R




P0 P1

P2

P3

t_sing=0.8

O

R




P0 P1

P2

P3

t_sing=0.85

O

R




P0 P1

P2P3

t_sing=0.9

OR




P0 P1

P2P3

t_sing=0.95

OR




P0 P1

P2P3

O R




P0P1 P2

P3

t_sing=0

O

R




P0P1 P2

P3

t_sing=0.05

O

R




P0P1 P2

P3

t_sing=0.1

O

R




P0P1P2

P3

t_sing=0.15

O

R




P0P1P2

P3

t_sing=0.2

O

R




P0P1P2

P3O

R




P0P2

P3O




P0

P2

P3

t_sing=0.35

O




P0 P1

P2

P3

t_sing=0.4

O

R




P0 P1

P2

P3

t_sing=0.45

O

R




P0 P1

P2

P3

t_sing=0.5

O

R


Singularity of Bézier Curve of degree 3 – part 3bExamples of cubic curves in x-y space

A "tour" of cubic curves in other regions of x-y space, again,with P0,P3, and the directions of P1 − P0 and P3 − P2 fixed.

P0 P1

P2

P3

t_sing=0.5

O

R




P0 P1

P2

P3O

R




P0 P1

P2

P3O

R




P0 P1

P2

P3O

R




P0P1

P2

P3O

R




P0P1

P2

P3O

R




P0P1

P2

P3O

R




P0P1

P2

P3O

R




P0P1 P2

P3O

R




P0P1

P2P3OR




P0P1

P2

P3O

R




P0P1

P2

P3OR




P0P1

P2P3OR




P0P1

P2

P3

O

R




P0 P1P2

P3

O

R




P0 P1

P2

P3O

R




P0 P1

P2

P3

t_sing=0.5

O

R

A complete description cubic curve shapes in x-y space is given in [Su & Liu 1990].


Singularity of Bézier Curve of degree 3 – part 4aInterval interior acts like endpoints – coincident end-controlpoints

Seen t∗ = 0 =⇒ P0 = P1 ; t∗ = 1 =⇒ P2 = P3.For t∗ ∈ (0, 1), singularity can also be regarded as coincidentend-control points (from e.g., Farin & Hansford 2000) :

C = C− ∪ C+, with domains [0, t̂] , [̂t, 1], andcontrol point sequences P−, P+,

with P−

3 = P+0 = C (̂t)

If t̂ = t∗, then, also, P−

2 = P−

3 & P+1 = P+

0 =⇒ P−

2 = P+1

By de Casteljau :P− = 〈P0, P01, P012, P0123〉,

P+ = 〈P0123, P123, P23, P3 〉.P0123 = C(t∗)

P−

2 = P+1 : P012 = P123 :

[(1− t∗)P01 + t∗ P12] = [(1− t∗)P12 + t∗ P23]

. . . =⇒ C[P]′(t∗) = ~0E. Nadler, Eastern Michigan University Bézier Curve & Surface Singularity 23/35




with P−

3 = P+0 = C (̂t)


2 = P−

3 & P+1 = P+

0 =⇒ P−

2 = P+1


P+ = 〈P0123, P123, P23, P3 〉.P0123 = C(t∗)

HP-L0

HP-L1

HP-L2

HP-L3IP+M

0

IP+M1 IP+M

2

IP+M3





with P−

3 = P+0 = C (̂t)


2 = P−

3 & P+1 = P+

0 =⇒ P−

2 = P+1


P+ = 〈P0123, P123, P23, P3 〉.P0123 = C(t∗)

HP-L0

HP-L1

HP-L2

HP-L3IP+M

0

IP+M1 IP+M

2

IP+M3





with P−

3 = P+0 = C (̂t)


2 = P−

3 & P+1 = P+

0 =⇒ P−

2 = P+1


P+ = 〈P0123, P123, P23, P3 〉.P0123 = C(t∗)

P0

P1

P2

P3

P01

P12 P23

P012

P123

C H0.375L

P−

2 = P+1 : P012 = P123 :

[(1− t∗)P01 + t∗ P12] = [(1− t∗)P12 + t∗ P23]





with P−

3 = P+0 = C (̂t)


2 = P−

3 & P+1 = P+

0 =⇒ P−

2 = P+1


P+ = 〈P0123, P123, P23, P3 〉.P0123 = C(t∗)

P0

P1

P2

P3

P01

P12 P23

P012

P123

C H0.375L

P−

2 = P+1 : P012 = P123 :

[(1− t∗)P01 + t∗ P12] = [(1− t∗)P12 + t∗ P23]





with P−

3 = P+0 = C (̂t)


2 = P−

3 & P+1 = P+

0 =⇒ P−

2 = P+1


P+ = 〈P0123, P123, P23, P3 〉.P0123 = C(t∗)

P0

P1

P2

P3

P01

P12 P23

P012

P123

C H0.375L

P−

2 = P+1 : P012 = P123 :

[(1− t∗)P01 + t∗ P12] = [(1− t∗)P12 + t∗ P23]





with P−

3 = P+0 = C (̂t)


2 = P−

3 & P+1 = P+

0 =⇒ P−

2 = P+1


P+ = 〈P0123, P123, P23, P3 〉.P0123 = C(t∗)

P0

P1

P2

P3

P01

P12 P23

P012

P123

C H0.375L

P−

2 = P+1 : P012 = P123 :

[(1− t∗)P01 + t∗ P12] = [(1− t∗)P12 + t∗ P23]


Singularity of Bézier Curve of degree 3 – part 4bExample: Interval interior acts like endpoints

Convergence of de Casteljau points P012 and P123 as t → t∗:

P0 P1

P2

P3

t_sing=0.55




P0 P1

P2

P3

P01

P12

P23

P012

P123

C H0.35L

t=0.35




P0 P1

P2

P3

P01

P12

P23P012

P123

C H0.45L

t=0.45




P0 P1

P2

P3

P01

P12

P23P012

P123

C H0.5L

t=0.5




P0 P1

P2

P3

P01

P12

P23

P012P123

C H0.525L

t=0.525




P0 P1

P2

P3

P01

P12

P23

P012P123

C H0.55L

t=0.55


Singularity of Bézier Curve of degree 3 – part 4cInterval endpoints act like interior – cusp

Singularity at ends acts like one in the interior: exhibits acusp . . .. . . if parameter interval is extended beyond [0, 1]:




P0

P1

P2

P3

t_sing=0.05




P0P1

P2

P3

t_sing=0




P0P1

P2

P3

t_sing=0




P0

P1

P2

P3

t_sing=0.05




P0

P1

P2

P3

t_sing=0.1




P0

P1

P2

P3

t_sing=0.2




P0

P1

P2

P3

t_sing=0.3




P0

P1

P2

P3

t_sing=0.4




P0

P1

P2

P3

t_sing=0.5




P0

P1

P2

P3

t_sing=0.6




P0

P1

P2 P3

t_sing=0.7




P0

P1P2 P3

t_sing=0.8




P0

P1P2 P3

t_sing=0.9




P0

P1P2 P3

t_sing=0.95




P0

P1P2P3

t_sing=1.


Geometric tangent at singularity of Bézier Curve

Recall definition of Bézier curve of degree n :

C(t) =

n∑

i=0

Bni (t) Pi

Suppose m coincident control points at endpoint P0 :

P0 = P1 = · · · = Pm−1, m ≤ n

Suppose so, as we have seen, singularity at t = 0.

Then geometric tangent at P0 is in the direction of Pm − P0

Proof: Under the reparameterization C̃(τ) = C(τ1m ),

C̃′(0) =(n

m

)(Pm − P0) .

And symmetrically for m control points coincident at other endpoint Pn.

To obtain geometric tangent at interior singularity,first apply de Casteljau Subdivision at the singularity,then reparameterize either curve segment as above.




C(t) =

n∑

i=0

Bni (t) Pi


P0 = P1 = · · · = Pm−1, m ≤ n




C̃′(0) =(n

m

)(Pm − P0) .






C(t) =

n∑

i=0

Bni (t) Pi


P0 = P1 = · · · = Pm−1, m ≤ n




C̃′(0) =(n

m

)(Pm − P0) .






C(t) =

n∑

i=0

Bni (t) Pi


P0 = P1 = · · · = Pm−1, m ≤ n




C̃′(0) =(n

m

)(Pm − P0) .




Singularity of a Parametric Surface

Local parametric surface: S(u, v) : (D ∈ R2) → R

3

Example:S(u, v) =

⟨u2 + v2, u, v

⟩: paraboloid with axis the x-axis

Normal vector N̂(u, v) is unit vector ⊥ S at (u, v), if this exists,

usually computed as N̂(u, v) =Su × Sv

‖Su × Sv‖(u, v) .

Surface is regular at any (u, v) where N̂(u, v) exists,equivalently, where Su,Sv are linearly independent.Otherwise, singular.




3

Example:S(u, v) =

⟨u2 + v2, u, v




‖Su × Sv‖(u, v) .





3

Example:S(u, v) =

⟨u2 + v2, u, v




‖Su × Sv‖(u, v) .



Example of surface with singularity

S(u, v) =⟨

u3, v3, uv⟩, u ∈ [−1, 1], v ∈ [−1, 1]

-1.0

-0.5

0.0

0.5

1.0

x

-1.0 -0.5 0.0 0.5 1.0

y

-1.0

-0.5

0.0

0.5

1.0

z

Singularity at (u, v) = (0, 0)


Example of surface with singularity

S(u, v) =⟨

u3, v3, uv⟩, u ∈ [0, 1], v ∈ [0, 1]

0.0

0.5

1.0

x

0.0 0.5 1.0

y

0.0

0.5

1.0

z

Singularity at (u, v) = (0, 0)


Rectangular Bézier Surface – DefinitionBivariate Polynomial of Degree (m, n) in (u, v),i.e., S(u, v0) degree m in u, S(u0, v) degree n in v :

S(u, v) =

m∑

i=0

n∑

j=0

ai,j ui vj

Better to represent as . . .Bézier Surface of degree (m, n) in (u, v):

S(u, v) =m∑

i=0

n∑

j=0

Bmi (u) Bn

j (v) Pi,j , u ∈ [0, 1], v ∈ [0, 1]

where, as before,

wher Bpk(t) =

(pk

)(1− t)p−kt k : Bernstein (basis) polynomial

wher Pi,j : Bézier control points.


Rectangular Bézier Surface – DefinitionBivariate Polynomial of Degree (m, n) in (u, v),i.e., S(u, v0) degree m in u, S(u0, v) degree n in v :

S(u, v) =

m∑

i=0

n∑

j=0

ai,j ui vj

Better to represent as . . .Bézier Surface of degree (m, n) in (u, v):

S(u, v) =m∑

i=0

n∑

j=0

Bmi (u) Bn

j (v) Pi,j , u ∈ [0, 1], v ∈ [0, 1]

where, as before,

wher Bpk(t) =

(pk

)(1− t)p−kt k : Bernstein (basis) polynomial

wher Pi,j : Bézier control points.


Bézier Surface

S(u, v) =

m∑

i=0

n∑

j=0

Bmi (u) Bn

j (v) Pi,j , u ∈ [0, 1], v ∈ [0, 1]

Bézier surface inherits many properties from Bézier curves, includingconvex hull property.Its boundary curves are Bézier curves, with control points from theboundary of the surface control net [Pi,j], i = 0..m, j = 0..n,e.g., curve S(u, 0) has control point sequence 〈Pi,0〉, i = 0..m

etc.E. Nadler, Eastern Michigan University Bézier Curve & Surface Singularity 30/35

Singularity of a Bézier SurfaceConsider singularity at a corner, say, (u, v) = (0, 0) and find thegeometric normal, in case one exists.

Note: Singularity at points other than the corners can be handled using de Casteljau

subdivision in each variable, in analogy with the curve case, allowing treatment of any

point in the domain as a corner of a rectangular sub-domain.

Recall

S singular at (0, 0) iff Su(0, 0),Sv(0, 0) linearly dependent

Su(0, 0) ‖ (P1,0 − P0,0) and Sv(0, 0) ‖ (P0,1 − P0,0)

Hence, S singular at (0, 0) if P0,0, P1,0, P0,1 are collinear, whichincludes the cases of P1,0 = P0,0 and P0,1 = P0,0.

Consider the tangential bilinear surface: degree (1, 1) surface withP0,0, P0,1, P1,0, P1,1 from a higher order surface.This bilinear surface shares N̂(0, 0) with the higher order surface,if both exist.






Recall


Su(0, 0) ‖ (P1,0 − P0,0) and Sv(0, 0) ‖ (P0,1 − P0,0)








Recall


Su(0, 0) ‖ (P1,0 − P0,0) and Sv(0, 0) ‖ (P0,1 − P0,0)




Cases for Tangential Bilinear SurfaceLet N̂m,n denote N̂(0, 0) of a degree (m, n) surfaceLet N̂1,1 denote N̂(0, 0) of its tangential bilinear surface

0 P0,0, P1,0, P0,1 not collinearN̂1,1 = unit ((P1,0 − P0,0) × (P0,1 − P0,0))N̂m,n = N̂1,1

1 P0,0, P1,0, P0,1 collinear, and P1,1 not on this lineTangential bilinear surface is planar.N̂1,1 = unit ((P0,1 − P1,1) × (P1,0 − P1,1))N̂m,n may not exist ; if exists, = N̂1,1.

2 P0,0, P1,0, P0,1, P1,1 all collinearN̂1,1 does not exist, as bilinear surface degenerates to aline (or even a point).N̂m,n may exist.


Example of Planar (Case 1) Tangential Bilinear SurfaceDegree (1, 3) Bézier Surface with P0,1 = P0,0 =⇒ singularity

Geometric singularity:No N, as two possible ones:using geometric tangent alongboundary curve S(0, v) vs.using planar tangential bilinearsurface

No geometric singularitywith blue control points coplanar :adjusted P0,2 to achieve


Example of Planar (Case 1) Tangential Bilinear SurfaceDegree (1, 3) Bézier Surface with P0,1 = P0,0 =⇒ singularity

Geometric singularity:No N, as two possible ones:using geometric tangent alongboundary curve S(0, v) vs.using planar tangential bilinearsurface

No geometric singularitywith blue control points coplanar :adjusted P0,2 to achieve


SummaryParametric polynomial curves & surfaces of degree 3 areuseful.

Need to understand their singularities

Use Bézier form to describe singularities of parametricpolynomial curves of degrees 1,2,3.

Unify cases of endpoint and interior Bézier curvesingularity.

Express geometric tangent of Bézier curve singularity.

Analogous approach for Bézier surface singularity, butmuch more complex; use curve results with additionalconsiderations.

Current and future related workComplete analysis of singularity of Bézier surfaces.Analyze curvatureApplication: G1 surface fitting in the presence of T-junction


For Further Reading I

G. FarinCurves and Surfaces for CAGDAcademic Press, 1990

G. Farin, J. Hoschek, M.-S. Kim, eds.Handbook of Computer Aided Geometric DesignElsevier, 2002

R.T. FaroukiPythagorean-Hodograph Curves: Algebra and Geometry Inseparable

Springer-Verlag, 2008

A. GrayModern Differential Geometry of Curves and SurfacesCRC Press, 1993

B.-Q. Su, D.-Y. LiuComputational Geometry – Curve and Surface ModelingAcademic Press, 1990


Documents

Singularity of Cubic B zier Curves and Surfaces - …enadler/presentations...Singularity of Cubic Bézier Curves and Surfaces Edmond Nadler Eastern Michigan University. joint work