2d computer graphics - w3.impa.brw3.impa.br/~diego/teaching/vg/slides-4.pdf · impa 1....

2D Computer Graphics

Diego NehabSummer 2019

IMPA

1

Bézier curves

Curve modeling by splines

2

Curve modeling by splines

Thin strip of wood used in building construction

Anchored in place by lead weights called ducks

Physical process

Interpolating, smooth, energy minimizing 3

No local control 7

3

Curve modeling by splines

Thin strip of wood used in building construction

Anchored in place by lead weights called ducks

Physical process

Interpolating, smooth, energy minimizing 3

No local control 7

3

Curve modeling by splines

Thin strip of wood used in building construction

Anchored in place by lead weights called ducks

Physical process

Interpolating, smooth, energy minimizing 3

No local control 7

3

Curve modeling by splines

Thin strip of wood used in building construction

Anchored in place by lead weights called ducks

Physical process

Interpolating, smooth, energy minimizing 3

No local control 7

3

Curve modeling by splines

Thin strip of wood used in building construction

Anchored in place by lead weights called ducks

Physical process

Interpolating, smooth, energy minimizing 3

No local control 7

3

Burmester curve

4

Lagrangian interpolation

Computational process

k+ 1 vertices {p0, . . . ,pk} define a curve

γ(t) =k∑i=0

pi

∏j 6=i(t − j)∏j 6=i(i− j)

, t ∈ [0, k]

Similar issues

5

B-splines

Define a family of generating functions βn recursively

β0(t) =

1, − 12 ≤ t < 1

2

0, otherwise

βn = βn−1 ∗ β0, n ∈ N

Notation for convolutionh = f ∗ g ⇔ h(t) =

∫ ∞

−∞f (u)g(u− t)dt

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

6

B-splines

Define a family of generating functions βn recursively

β0(t) =

1, − 12 ≤ t < 1

2

0, otherwise

βn = βn−1 ∗ β0, n ∈ N

Notation for convolutionh = f ∗ g ⇔ h(t) =

∫ ∞

−∞f (u)g(u− t)dt

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

6

B-splines

Define a family of generating functions βn recursively

β0(t) =

1, − 12 ≤ t < 1

2

0, otherwise

βn = βn−1 ∗ β0, n ∈ N

Notation for convolutionh = f ∗ g ⇔ h(t) =

∫ ∞

−∞f (u)g(u− t)dt

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

6

B-splines

Define a family of generating functions βn recursively

β0(t) =

1, − 12 ≤ t < 1

2

0, otherwise

βn = βn−1 ∗ β0, n ∈ N

Notation for convolutionh = f ∗ g ⇔ h(t) =

∫ ∞

−∞f (u)g(u− t)dt

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

6

B-splines

Define a family of generating functions βn recursively

β0(t) =

1, − 12 ≤ t < 1

2

0, otherwise

βn = βn−1 ∗ β0, n ∈ N

Notation for convolutionh = f ∗ g ⇔ h(t) =

∫ ∞

−∞f (u)g(u− t)dt

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

-4 -2 2 4-0.2

0.2

0.4

0.6

0.8

1.

1.2

6

B-splines

Examples

β0(t) =

1, − 12 ≤ t < 1

2

0, otherwise

β1(t) =

1+ t, −1 ≤ t < 0

1− t, 0 ≤ t < 1

0, otherwise

β2(t) =

18(3+ 2t)2, − 3

2 ≤ t < − 12

14(3− 4t2), − 1

2 ≤ t < 12

18(9− 12t + 4t2), 1

2 ≤ t < 22

0, otherwise

7

B-splines

Examples

β0(t) =

1, − 12 ≤ t < 1

2

0, otherwise

β1(t) =

1+ t, −1 ≤ t < 0

1− t, 0 ≤ t < 1

0, otherwise

β2(t) =

18(3+ 2t)2, − 3

2 ≤ t < − 12

14(3− 4t2), − 1

2 ≤ t < 12

18(9− 12t + 4t2), 1

2 ≤ t < 22

0, otherwise

7

B-splines

Examples

β0(t) =

1, − 12 ≤ t < 1

2

0, otherwise

β1(t) =

1+ t, −1 ≤ t < 0

1− t, 0 ≤ t < 1

0, otherwise

β2(t) =

18(3+ 2t)2, − 3

2 ≤ t < − 12

14(3− 4t2), − 1

2 ≤ t < 12

18(9− 12t + 4t2), 1

2 ≤ t < 22

0, otherwise

7

B-splines

k+ 1 vertices {p0, . . . ,pk} and generating function βn define a curve

γ(t) =k∑i=0

βn(t − i)pi, t ∈ [0, k]

Local control 3

Differentiable n times everywhere 3

Non-interpolating 7

• Interpolation requires solving a banded linear system 3

Many, many interesting properties

8

B-splines

k+ 1 vertices {p0, . . . ,pk} and generating function βn define a curve

γ(t) =k∑i=0

βn(t − i)pi, t ∈ [0, k]

Local control 3

Differentiable n times everywhere 3

Non-interpolating 7

• Interpolation requires solving a banded linear system 3

Many, many interesting properties

8

B-splines

k+ 1 vertices {p0, . . . ,pk} and generating function βn define a curve

γ(t) =k∑i=0

βn(t − i)pi, t ∈ [0, k]

Local control 3

Differentiable n times everywhere 3

Non-interpolating 7

• Interpolation requires solving a banded linear system 3

Many, many interesting properties

8

B-splines

k+ 1 vertices {p0, . . . ,pk} and generating function βn define a curve

γ(t) =k∑i=0

βn(t − i)pi, t ∈ [0, k]

Local control 3

Differentiable n times everywhere 3

Non-interpolating 7

• Interpolation requires solving a banded linear system 3

Many, many interesting properties

8

B-splines

k+ 1 vertices {p0, . . . ,pk} and generating function βn define a curve

γ(t) =k∑i=0

βn(t − i)pi, t ∈ [0, k]

Local control 3

Differentiable n times everywhere 3

Non-interpolating 7

• Interpolation requires solving a banded linear system 3

Many, many interesting properties

8

Polylines

k+ 1 vertices {p0, . . . ,pk} define k segments{γ0(t), . . . , γk−1(t)

}, t ∈ [0, 1]

Each segment defined by linear interpolationγi(t) = (1− t)pi + t pi+1, i ∈ {0, . . . , k− 1}

Local control 3

Interpolates all control points 3

Not differentiable at interpolated points 7

9

Polylines

k+ 1 vertices {p0, . . . ,pk} define k segments{γ0(t), . . . , γk−1(t)

}, t ∈ [0, 1]

Each segment defined by linear interpolationγi(t) = (1− t)pi + t pi+1, i ∈ {0, . . . , k− 1}

Local control 3

Interpolates all control points 3

Not differentiable at interpolated points 7

9

Polylines

k+ 1 vertices {p0, . . . ,pk} define k segments{γ0(t), . . . , γk−1(t)

}, t ∈ [0, 1]

Each segment defined by linear interpolationγi(t) = (1− t)pi + t pi+1, i ∈ {0, . . . , k− 1}

Local control 3

Interpolates all control points 3

Not differentiable at interpolated points 7

9

Polylines

k+ 1 vertices {p0, . . . ,pk} define k segments{γ0(t), . . . , γk−1(t)

}, t ∈ [0, 1]

Each segment defined by linear interpolationγi(t) = (1− t)pi + t pi+1, i ∈ {0, . . . , k− 1}

Local control 3

Interpolates all control points 3

Not differentiable at interpolated points 7

9

Polylines

k+ 1 vertices {p0, . . . ,pk} define k segments{γ0(t), . . . , γk−1(t)

}, t ∈ [0, 1]

Each segment defined by linear interpolationγi(t) = (1− t)pi + t pi+1, i ∈ {0, . . . , k− 1}

Local control 3

Interpolates all control points 3

Not differentiable at interpolated points 7

9

Polylines

k+ 1 vertices {p0, . . . ,pk} define k segments{γ0(t), . . . , γk−1(t)

}, t ∈ [0, 1]

Each segment defined by linear interpolationγi(t) = (1− t)pi + t pi+1, i ∈ {0, . . . , k− 1}

Local control 3

Interpolates all control points 3

Not differentiable at interpolated points 7

9

Bézier curves

Generalization of linear interpolation

kn+ 1 vertices {p0, . . . ,pkn} define k segments of degree n{γn0 (t), γnn(t), . . . , γn(k−1)n(t)

}, t ∈ [0, 1]

Defined recursively, for j ∈ {0, . . . , k− 1}γ0i (t) = pi, i ∈

{nj, . . . ,n(j+ 1)

},

γmi (t) = (1− t) γm−1i (t) + t γm−1

i+1 (t), i ∈{nj, . . . ,n(j+ 1)−m

}De Casteljau algorithm

Geometric interpretation

10

Bézier curves

Generalization of linear interpolation

kn+ 1 vertices {p0, . . . ,pkn} define k segments of degree n{γn0 (t), γnn(t), . . . , γn(k−1)n(t)

}, t ∈ [0, 1]

Defined recursively, for j ∈ {0, . . . , k− 1}γ0i (t) = pi, i ∈

{nj, . . . ,n(j+ 1)

},

γmi (t) = (1− t) γm−1i (t) + t γm−1

i+1 (t), i ∈{nj, . . . ,n(j+ 1)−m

}De Casteljau algorithm

Geometric interpretation

10

Bézier curves

Generalization of linear interpolation

kn+ 1 vertices {p0, . . . ,pkn} define k segments of degree n{γn0 (t), γnn(t), . . . , γn(k−1)n(t)

}, t ∈ [0, 1]

Defined recursively, for j ∈ {0, . . . , k− 1}γ0i (t) = pi, i ∈

{nj, . . . ,n(j+ 1)

},

γmi (t) = (1− t) γm−1i (t) + t γm−1

i+1 (t), i ∈{nj, . . . ,n(j+ 1)−m

}

De Casteljau algorithm

Geometric interpretation

10

Bézier curves

Generalization of linear interpolation

kn+ 1 vertices {p0, . . . ,pkn} define k segments of degree n{γn0 (t), γnn(t), . . . , γn(k−1)n(t)

}, t ∈ [0, 1]

Defined recursively, for j ∈ {0, . . . , k− 1}γ0i (t) = pi, i ∈

{nj, . . . ,n(j+ 1)

},

γmi (t) = (1− t) γm−1i (t) + t γm−1

i+1 (t), i ∈{nj, . . . ,n(j+ 1)−m

}De Casteljau algorithm

Geometric interpretation

10

Bézier curves

Generalization of linear interpolation

kn+ 1 vertices {p0, . . . ,pkn} define k segments of degree n{γn0 (t), γnn(t), . . . , γn(k−1)n(t)

}, t ∈ [0, 1]

Defined recursively, for j ∈ {0, . . . , k− 1}γ0i (t) = pi, i ∈

{nj, . . . ,n(j+ 1)

},

γmi (t) = (1− t) γm−1i (t) + t γm−1

i+1 (t), i ∈{nj, . . . ,n(j+ 1)−m

}De Casteljau algorithm

Geometric interpretation

10

Bernstein polynomials

Algebraic interpretation

Expanding and collecting the pi terms,

γni (t) =n∑j=0

(nj)(1− t)n−j tj pi+j

Using Bernstein polynomials

γni (t) =n∑j=0

bj,n(t)pi+j with bj,n(t) =(nj)(1− t)n−j tj

Basis for the space of polynomials Pn with degree n or less (Why?)

11

Bernstein polynomials

Algebraic interpretation

Expanding and collecting the pi terms,

γni (t) =n∑j=0

(nj)(1− t)n−j tj pi+j

Using Bernstein polynomials

γni (t) =n∑j=0

bj,n(t)pi+j with bj,n(t) =(nj)(1− t)n−j tj

Basis for the space of polynomials Pn with degree n or less (Why?)

11

Bernstein polynomials

Algebraic interpretation

Expanding and collecting the pi terms,

γni (t) =n∑j=0

(nj)(1− t)n−j tj pi+j

Using Bernstein polynomials

γni (t) =n∑j=0

bj,n(t)pi+j with bj,n(t) =(nj)(1− t)n−j tj

Basis for the space of polynomials Pn with degree n or less (Why?)

11

Bernstein polynomials

Algebraic interpretation

Expanding and collecting the pi terms,

γni (t) =n∑j=0

(nj)(1− t)n−j tj pi+j

Using Bernstein polynomials

γni (t) =n∑j=0

bj,n(t)pi+j with bj,n(t) =(nj)(1− t)n−j tj

Basis for the space of polynomials Pn with degree n or less (Why?)

11

Control points and blending weights

In matrix form

γni (t) =[pni pni+1 · · · pni+n

]

︸ ︷︷ ︸Bézier control points CB

i

b0,n(t)b1,n(t)...

bn,n(t)

blending weights Bn(t)

= CBi Bn(t)

Linear invariance is quite obvious in this form

12

Control points and blending weights

In matrix form

γni (t) =[pni pni+1 · · · pni+n

]︸ ︷︷ ︸

Bézier control points CBi

b0,n(t)b1,n(t)...

bn,n(t)

blending weights Bn(t)

= CBi Bn(t)

Linear invariance is quite obvious in this form

12

Control points and blending weights

In matrix form

γni (t) =[pni pni+1 · · · pni+n

]︸ ︷︷ ︸

Bézier control points CBi

b0,n(t)b1,n(t)...

bn,n(t)

blending weights Bn(t)

= CBi Bn(t)

Linear invariance is quite obvious in this form

12

Control points and blending weights

In matrix form

γni (t) =[pni pni+1 · · · pni+n

]︸ ︷︷ ︸

Bézier control points CBi

b0,n(t)b1,n(t)...

bn,n(t)

blending weights Bn(t)

= CBi Bn(t)

Linear invariance is quite obvious in this form

12

Change of basis

Can be converted back and forth to power basis

Pn(t) =[1 t · · · tn

]TBn(t) = Bn Pn(t)

Examples

B1 =

[1 −10 1

]B2 =

1 −2 10 2 −20 0 1

B3 =

1 −3 3 −10 3 −6 30 0 3 −30 0 0 1

γni (t) = CBi Bn(t) = CBi

Change of basis︷︸︸︷Bn︸ ︷︷ ︸

Power basis control points CPi

Pn(t) = CPi Pn(t)

13

Change of basis

Can be converted back and forth to power basis

Pn(t) =[1 t · · · tn

]TBn(t) = Bn Pn(t)

Examples

B1 =

[1 −10 1

]B2 =

1 −2 10 2 −20 0 1

B3 =

1 −3 3 −10 3 −6 30 0 3 −30 0 0 1

γni (t) = CBi Bn(t) = CBi

Change of basis︷︸︸︷Bn︸ ︷︷ ︸

Power basis control points CPi

Pn(t) = CPi Pn(t)

13

Change of basis

Can be converted back and forth to power basis

Pn(t) =[1 t · · · tn

]TBn(t) = Bn Pn(t)

Examples

B1 =

[1 −10 1

]B2 =

1 −2 10 2 −20 0 1

B3 =

1 −3 3 −10 3 −6 30 0 3 −30 0 0 1

γni (t) = CBi Bn(t) = CBi

Change of basis︷︸︸︷Bn︸ ︷︷ ︸

Power basis control points CPi

Pn(t) = CPi Pn(t)

13

Bézier curves

Local control 3

Interpolates every nth point 3

Differentiable except (perhaps) at interpolation points 3

PostScript, SVG (Inkscape), RVG

Mathematica

14

Bézier curves

Local control 3

Interpolates every nth point 3

Differentiable except (perhaps) at interpolation points 3

PostScript, SVG (Inkscape), RVG

Mathematica

14

Bézier curves

Local control 3

Interpolates every nth point 3

Differentiable except (perhaps) at interpolation points 3

PostScript, SVG (Inkscape), RVG

Mathematica

14

Bézier curves

Local control 3

Interpolates every nth point 3

Differentiable except (perhaps) at interpolation points 3

PostScript, SVG (Inkscape), RVG

Mathematica

14

Bézier curves

Local control 3

Interpolates every nth point 3

Differentiable except (perhaps) at interpolation points 3

PostScript, SVG (Inkscape), RVG

Mathematica

14

Affine invariance of Bézier segments

Let T be an affine transformation and let

γn(t) =n∑j=0

bj,n(t)pj

be a Bézier curve segment.

We want to show that

T( n∑j=0

bj,n(t)pj)=

n∑j=0

bj,n(t) T(pj).

This will be true if and only if all points in the Bézier curve are affinecombinations of the control points.

15

Affine invariance of Bézier segments

Let T be an affine transformation and let

γn(t) =n∑j=0

bj,n(t)pj

be a Bézier curve segment.

We want to show that

T( n∑j=0

bj,n(t)pj)=

n∑j=0

bj,n(t) T(pj).

This will be true if and only if all points in the Bézier curve are affinecombinations of the control points.

15

Affine invariance of Bézier segments

Let T be an affine transformation and let

γn(t) =n∑j=0

bj,n(t)pj

be a Bézier curve segment.

We want to show that

T( n∑j=0

bj,n(t)pj)=

n∑j=0

bj,n(t) T(pj).

This will be true if and only if all points in the Bézier curve are affinecombinations of the control points.

15

Affine invariance of Bézier segments

Indeed,n∑j=0

bj,n(t) =n∑j=0

(nj)(1− t)n−j tj

=((1− t) + t

)n= 1n = 1.

The Bernstein polynomials therefore form a partition of unity

To apply an affine transformation to a Bézier curve, simply transformthe control points

16

Affine invariance of Bézier segments

Indeed,n∑j=0

bj,n(t) =n∑j=0

(nj)(1− t)n−j tj =

((1− t) + t

)n

= 1n = 1.

The Bernstein polynomials therefore form a partition of unity

To apply an affine transformation to a Bézier curve, simply transformthe control points

16

Affine invariance of Bézier segments

Indeed,n∑j=0

bj,n(t) =n∑j=0

(nj)(1− t)n−j tj =

((1− t) + t

)n= 1n = 1.

The Bernstein polynomials therefore form a partition of unity

To apply an affine transformation to a Bézier curve, simply transformthe control points

16

Affine invariance of Bézier segments

Indeed,n∑j=0

bj,n(t) =n∑j=0

(nj)(1− t)n−j tj =

((1− t) + t

)n= 1n = 1.

The Bernstein polynomials therefore form a partition of unity

To apply an affine transformation to a Bézier curve, simply transformthe control points

16

Affine invariance of Bézier segments

Indeed,n∑j=0

bj,n(t) =n∑j=0

(nj)(1− t)n−j tj =

((1− t) + t

)n= 1n = 1.

The Bernstein polynomials therefore form a partition of unity

To apply an affine transformation to a Bézier curve, simply transformthe control points

16

Convex hull property for Bézier segments

p =∑

i αipi is a convex combination of {pi} if∑

i αi = 1 and αi ≥ 0.

A set of points C is convex if every convex combination of points in Calso belongs to C

The convex hull of a set points S is the smallest convex set thatcontains S

If γ is a Bézier curve, then{γ(t) | t ∈ [0, 1]

}is contained in the convex

hull of its control points• From partition of unity and positivity in [0, 1]• Useful for curve intersection, quick bounding box, etc

17

Convex hull property for Bézier segments

p =∑

i αipi is a convex combination of {pi} if∑

i αi = 1 and αi ≥ 0.

A set of points C is convex if every convex combination of points in Calso belongs to C

The convex hull of a set points S is the smallest convex set thatcontains S

If γ is a Bézier curve, then{γ(t) | t ∈ [0, 1]

}is contained in the convex

hull of its control points• From partition of unity and positivity in [0, 1]• Useful for curve intersection, quick bounding box, etc

17

Convex hull property for Bézier segments

p =∑

i αipi is a convex combination of {pi} if∑

i αi = 1 and αi ≥ 0.

A set of points C is convex if every convex combination of points in Calso belongs to C

The convex hull of a set points S is the smallest convex set thatcontains S

If γ is a Bézier curve, then{γ(t) | t ∈ [0, 1]

}is contained in the convex

hull of its control points• From partition of unity and positivity in [0, 1]• Useful for curve intersection, quick bounding box, etc

17

Convex hull property for Bézier segments

p =∑

i αipi is a convex combination of {pi} if∑

i αi = 1 and αi ≥ 0.

A set of points C is convex if every convex combination of points in Calso belongs to C

The convex hull of a set points S is the smallest convex set thatcontains S

If γ is a Bézier curve, then{γ(t) | t ∈ [0, 1]

}is contained in the convex

hull of its control points

• From partition of unity and positivity in [0, 1]• Useful for curve intersection, quick bounding box, etc

17

Convex hull property for Bézier segments

p =∑

i αipi is a convex combination of {pi} if∑

i αi = 1 and αi ≥ 0.

A set of points C is convex if every convex combination of points in Calso belongs to C

The convex hull of a set points S is the smallest convex set thatcontains S

If γ is a Bézier curve, then{γ(t) | t ∈ [0, 1]

}is contained in the convex

hull of its control points• From partition of unity and positivity in [0, 1]• Useful for curve intersection, quick bounding box, etc

17

Derivative of Bézier segment

Since derivative operator is linear and γn(t) =∑n

j=0 bj,n(t)pj, all wehave to do is differentiate the Bernstein polynomials(

bj,n(t))′=((n

j)(1− t)n−j tj

)′

= j(nj)(1− t)n−j tj−1 − (n− j)

(nj)(1− t)n−1−j tj

= n(n−1j−1

)(1− t)(n−1)−(j−1) tj−1 − n

(n−1j)(1− t)n−1−j tj

= n(bj−1,n−1(t)− bj,n−1(t)

)Therefore,

(γn)′(t) =n−1∑j=0

bj,n−1(t)qj with qj = n(pj+1 − pj).

18

Derivative of Bézier segment

Since derivative operator is linear and γn(t) =∑n

j=0 bj,n(t)pj, all wehave to do is differentiate the Bernstein polynomials(

bj,n(t))′=((n

j)(1− t)n−j tj

)′= j

(nj)(1− t)n−j tj−1 − (n− j)

(nj)(1− t)n−1−j tj

= n(n−1j−1

)(1− t)(n−1)−(j−1) tj−1 − n

(n−1j)(1− t)n−1−j tj

= n(bj−1,n−1(t)− bj,n−1(t)

)Therefore,

(γn)′(t) =n−1∑j=0

bj,n−1(t)qj with qj = n(pj+1 − pj).

18

Derivative of Bézier segment

Since derivative operator is linear and γn(t) =∑n

j=0 bj,n(t)pj, all wehave to do is differentiate the Bernstein polynomials(

bj,n(t))′=((n

j)(1− t)n−j tj

)′= j

(nj)(1− t)n−j tj−1 − (n− j)

(nj)(1− t)n−1−j tj

= n(n−1j−1

)(1− t)(n−1)−(j−1) tj−1 − n

(n−1j)(1− t)n−1−j tj

= n(bj−1,n−1(t)− bj,n−1(t)

)Therefore,

(γn)′(t) =n−1∑j=0

bj,n−1(t)qj with qj = n(pj+1 − pj).

18

Derivative of Bézier segment

Since derivative operator is linear and γn(t) =∑n

j=0 bj,n(t)pj, all wehave to do is differentiate the Bernstein polynomials(

bj,n(t))′=((n

j)(1− t)n−j tj

)′= j

(nj)(1− t)n−j tj−1 − (n− j)

(nj)(1− t)n−1−j tj

= n(n−1j−1

)(1− t)(n−1)−(j−1) tj−1 − n

(n−1j)(1− t)n−1−j tj

= n(bj−1,n−1(t)− bj,n−1(t)

)

Therefore,

(γn)′(t) =n−1∑j=0

bj,n−1(t)qj with qj = n(pj+1 − pj).

18

Derivative of Bézier segment

Since derivative operator is linear and γn(t) =∑n

j=0 bj,n(t)pj, all wehave to do is differentiate the Bernstein polynomials(

bj,n(t))′=((n

j)(1− t)n−j tj

)′= j

(nj)(1− t)n−j tj−1 − (n− j)

(nj)(1− t)n−1−j tj

= n(n−1j−1

)(1− t)(n−1)−(j−1) tj−1 − n

(n−1j)(1− t)n−1−j tj

= n(bj−1,n−1(t)− bj,n−1(t)

)Therefore,

(γn)′(t) =n−1∑j=0

bj,n−1(t)qj with qj = n(pj+1 − pj).

18

Derivative of Bézier segment

What is the derivative at the endpoints?

How do we connect segments so that they are C1 continuous?

What about G1 continuity?

Show in Inkscape

19

Derivative of Bézier segment

What is the derivative at the endpoints?

How do we connect segments so that they are C1 continuous?

What about G1 continuity?

Show in Inkscape

19

Derivative of Bézier segment

What is the derivative at the endpoints?

How do we connect segments so that they are C1 continuous?

What about G1 continuity?

Show in Inkscape

19

Derivative of Bézier segment

What is the derivative at the endpoints?

How do we connect segments so that they are C1 continuous?

What about G1 continuity?

Show in Inkscape

19

Reparameterization

Given a Bézier curve segment γn(t), with control points {p0, . . . ,pn},and a reparameterization t 7→ (1− u)r + us, how can we obtain thecontrol points {q0, . . . ,qn} for the curve segment piece γn[r,s](u)?

γn[r,s](u) = γn((1− u)r + us

)= γn(a+ bu), for a = r, b = s− r

γn(a+ bu) =[p0 · · · pn

]Bn(a+ bu) = CB Bn Pn(a+ bu)

Pn(a+ bu) =

1

a+ bua2 + 2abu+ b2 u2

...(n0)an + · · ·

(ni)an−ibi ui + · · ·

(nn)bn un

20

Reparameterization

Given a Bézier curve segment γn(t), with control points {p0, . . . ,pn},and a reparameterization t 7→ (1− u)r + us, how can we obtain thecontrol points {q0, . . . ,qn} for the curve segment piece γn[r,s](u)?

γn[r,s](u) = γn((1− u)r + us

)= γn(a+ bu), for a = r, b = s− r

γn(a+ bu) =[p0 · · · pn

]Bn(a+ bu) = CB Bn Pn(a+ bu)

Pn(a+ bu) =

1

a+ bua2 + 2abu+ b2 u2

...(n0)an + · · ·

(ni)an−ibi ui + · · ·

(nn)bn un

20

Reparameterization

Given a Bézier curve segment γn(t), with control points {p0, . . . ,pn},and a reparameterization t 7→ (1− u)r + us, how can we obtain thecontrol points {q0, . . . ,qn} for the curve segment piece γn[r,s](u)?

γn[r,s](u) = γn((1− u)r + us

)= γn(a+ bu), for a = r, b = s− r

γn(a+ bu) =[p0 · · · pn

]Bn(a+ bu) = CB Bn Pn(a+ bu)

Pn(a+ bu) =

1

a+ bua2 + 2abu+ b2 u2

...(n0)an + · · ·

(ni)an−ibi ui + · · ·

(nn)bn un

20

Reparameterization

Given a Bézier curve segment γn(t), with control points {p0, . . . ,pn},and a reparameterization t 7→ (1− u)r + us, how can we obtain thecontrol points {q0, . . . ,qn} for the curve segment piece γn[r,s](u)?

γn[r,s](u) = γn((1− u)r + us

)= γn(a+ bu), for a = r, b = s− r

γn(a+ bu) =[p0 · · · pn

]Bn(a+ bu) = CB Bn Pn(a+ bu)

Pn(a+ bu) =

1

a+ bua2 + 2abu+ b2 u2

...(n0)an + · · ·

(ni)an−ibi ui + · · ·

(nn)bn un

20

Reparameterization

Pn(a+ bu) =

1a ba2 2ab b2...

...... . . .(n

0)an

(n1)an−1b

(n2)an−2b2 · · ·

(nn)bn

1uu2...un

= Ma,b Pn(u)γn[r,s](u) = γn(a+ bu)

= CB Bn Pn(a+ bu)= CB BnMa,b Pn(u)= CB BnMa,b Bn

−1 Bn(u)[q0 · · · qn

]= CB BnMa,b Bn

−1

21

Reparameterization

Pn(a+ bu) =

1a ba2 2ab b2...

...... . . .(n

0)an

(n1)an−1b

(n2)an−2b2 · · ·

(nn)bn

1uu2...un

= Ma,b Pn(u)

γn[r,s](u) = γn(a+ bu)

= CB Bn Pn(a+ bu)= CB BnMa,b Pn(u)= CB BnMa,b Bn

−1 Bn(u)[q0 · · · qn

]= CB BnMa,b Bn

−1

21

Reparameterization

Pn(a+ bu) =

1a ba2 2ab b2...

...... . . .(n

0)an

(n1)an−1b

(n2)an−2b2 · · ·

(nn)bn

1uu2...un

= Ma,b Pn(u)

γn[r,s](u) = γn(a+ bu)

= CB Bn Pn(a+ bu)= CB BnMa,b Pn(u)= CB BnMa,b Bn

−1 Bn(u)[q0 · · · qn

]= CB BnMa,b Bn

−1

21

Reparameterization

Pn(a+ bu) =

1a ba2 2ab b2...

...... . . .(n

0)an

(n1)an−1b

(n2)an−2b2 · · ·

(nn)bn

1uu2...un

= Ma,b Pn(u)

γn[r,s](u) = γn(a+ bu)

= CB Bn Pn(a+ bu)

= CB BnMa,b Pn(u)= CB BnMa,b Bn

−1 Bn(u)[q0 · · · qn

]= CB BnMa,b Bn

−1

21

Reparameterization

Pn(a+ bu) =

1a ba2 2ab b2...

...... . . .(n

0)an

(n1)an−1b

(n2)an−2b2 · · ·

(nn)bn

1uu2...un

= Ma,b Pn(u)

γn[r,s](u) = γn(a+ bu)

= CB Bn Pn(a+ bu)= CB BnMa,b Pn(u)

= CB BnMa,b Bn−1 Bn(u)[

q0 · · · qn]= CB BnMa,b Bn

−1

21

Reparameterization

Pn(a+ bu) =

1a ba2 2ab b2...

...... . . .(n

0)an

(n1)an−1b

(n2)an−2b2 · · ·

(nn)bn

1uu2...un

= Ma,b Pn(u)

γn[r,s](u) = γn(a+ bu)

= CB Bn Pn(a+ bu)= CB BnMa,b Pn(u)= CB BnMa,b Bn

−1 Bn(u)

[q0 · · · qn

]= CB BnMa,b Bn

−1

21

Reparameterization

Pn(a+ bu) =

1a ba2 2ab b2...

...... . . .(n

0)an

(n1)an−1b

(n2)an−2b2 · · ·

(nn)bn

1uu2...un

= Ma,b Pn(u)

γn[r,s](u) = γn(a+ bu)

= CB Bn Pn(a+ bu)= CB BnMa,b Pn(u)= CB BnMa,b Bn

−1 Bn(u)[q0 · · · qn

]= CB BnMa,b Bn

−1

21

Degree elevation of Bézier segment

Express a segment γn(t) as γn+1(t)? (write bj,n(t) in terms of bk,n+1(t)?)

Easy to express both bj,n+1(t) and bj+1,n+1(t) in terms of bj,n(t)bj,n+1(t)

=(n+1

j)(1− t)n+1−j tj = n+1

n+1−j (1− t)bj,n(t)

bj+1,n+1(t) =(n+1j+1

)(1− t)n−j tj = n+1

j+1 t bj,n(t)

From these,bj,n(t) = n+1−j

n+1 bj,n+1(t) + j+1n+1 bj+1,n+1(t)

Expanding and collecting terms,

γn(t) =n∑i=0

bi,n(t)pi =n+1∑j=0

bj,n+1(t)qj = γn+1(t)

with q0 = p0, qn+1 = pn, andqi = j

n+1 pi−1 + (1− jn+1)pi

22

Degree elevation of Bézier segment

Express a segment γn(t) as γn+1(t)? (write bj,n(t) in terms of bk,n+1(t)?)

Easy to express both bj,n+1(t) and bj+1,n+1(t) in terms of bj,n(t)bj,n+1(t) =

(n+1j)(1− t)n+1−j tj

= n+1n+1−j (1− t)bj,n(t)

bj+1,n+1(t) =(n+1j+1

)(1− t)n−j tj = n+1

j+1 t bj,n(t)

From these,bj,n(t) = n+1−j

n+1 bj,n+1(t) + j+1n+1 bj+1,n+1(t)

Expanding and collecting terms,

γn(t) =n∑i=0

bi,n(t)pi =n+1∑j=0

bj,n+1(t)qj = γn+1(t)

with q0 = p0, qn+1 = pn, andqi = j

n+1 pi−1 + (1− jn+1)pi

22

Degree elevation of Bézier segment

Express a segment γn(t) as γn+1(t)? (write bj,n(t) in terms of bk,n+1(t)?)

Easy to express both bj,n+1(t) and bj+1,n+1(t) in terms of bj,n(t)bj,n+1(t) =

(n+1j)(1− t)n+1−j tj = n+1

n+1−j (1− t)bj,n(t)

bj+1,n+1(t) =(n+1j+1

)(1− t)n−j tj = n+1

j+1 t bj,n(t)

From these,bj,n(t) = n+1−j

n+1 bj,n+1(t) + j+1n+1 bj+1,n+1(t)

Expanding and collecting terms,

γn(t) =n∑i=0

bi,n(t)pi =n+1∑j=0

bj,n+1(t)qj = γn+1(t)

with q0 = p0, qn+1 = pn, andqi = j

n+1 pi−1 + (1− jn+1)pi

22

Degree elevation of Bézier segment

Express a segment γn(t) as γn+1(t)? (write bj,n(t) in terms of bk,n+1(t)?)

Easy to express both bj,n+1(t) and bj+1,n+1(t) in terms of bj,n(t)bj,n+1(t) =

(n+1j)(1− t)n+1−j tj = n+1

n+1−j (1− t)bj,n(t)

bj+1,n+1(t) =(n+1j+1

)(1− t)n−j tj

= n+1j+1 t bj,n(t)

From these,bj,n(t) = n+1−j

n+1 bj,n+1(t) + j+1n+1 bj+1,n+1(t)

Expanding and collecting terms,

γn(t) =n∑i=0

bi,n(t)pi =n+1∑j=0

bj,n+1(t)qj = γn+1(t)

with q0 = p0, qn+1 = pn, andqi = j

n+1 pi−1 + (1− jn+1)pi

22

Degree elevation of Bézier segment

Express a segment γn(t) as γn+1(t)? (write bj,n(t) in terms of bk,n+1(t)?)

Easy to express both bj,n+1(t) and bj+1,n+1(t) in terms of bj,n(t)bj,n+1(t) =

(n+1j)(1− t)n+1−j tj = n+1

n+1−j (1− t)bj,n(t)

bj+1,n+1(t) =(n+1j+1

)(1− t)n−j tj = n+1

j+1 t bj,n(t)

From these,bj,n(t) = n+1−j

n+1 bj,n+1(t) + j+1n+1 bj+1,n+1(t)

Expanding and collecting terms,

γn(t) =n∑i=0

bi,n(t)pi =n+1∑j=0

bj,n+1(t)qj = γn+1(t)

with q0 = p0, qn+1 = pn, andqi = j

n+1 pi−1 + (1− jn+1)pi

22

Degree elevation of Bézier segment

Express a segment γn(t) as γn+1(t)? (write bj,n(t) in terms of bk,n+1(t)?)

Easy to express both bj,n+1(t) and bj+1,n+1(t) in terms of bj,n(t)bj,n+1(t) =

(n+1j)(1− t)n+1−j tj = n+1

n+1−j (1− t)bj,n(t)

bj+1,n+1(t) =(n+1j+1

)(1− t)n−j tj = n+1

j+1 t bj,n(t)

From these,bj,n(t) = n+1−j

n+1 bj,n+1(t) + j+1n+1 bj+1,n+1(t)

Expanding and collecting terms,

γn(t) =n∑i=0

bi,n(t)pi =n+1∑j=0

bj,n+1(t)qj = γn+1(t)

with q0 = p0, qn+1 = pn, andqi = j

n+1 pi−1 + (1− jn+1)pi

22

Degree elevation of Bézier segment

Express a segment γn(t) as γn+1(t)? (write bj,n(t) in terms of bk,n+1(t)?)

Easy to express both bj,n+1(t) and bj+1,n+1(t) in terms of bj,n(t)bj,n+1(t) =

(n+1j)(1− t)n+1−j tj = n+1

n+1−j (1− t)bj,n(t)

bj+1,n+1(t) =(n+1j+1

)(1− t)n−j tj = n+1

j+1 t bj,n(t)

From these,bj,n(t) = n+1−j

n+1 bj,n+1(t) + j+1n+1 bj+1,n+1(t)

Expanding and collecting terms,

γn(t) =n∑i=0

bi,n(t)pi =n+1∑j=0

bj,n+1(t)qj = γn+1(t)

with q0 = p0, qn+1 = pn, andqi = j

n+1 pi−1 + (1− jn+1)pi 22

Degree elevation of Bézier segment

Examples [p0 p1

]⇔

[p0 1

2(p0 + p1) p1]

[p0 p1 p2

]⇔

[p0 1

3(p0 + 2p1) 13(2p1 + p2) p2

]

23

References

G. Farin. Curves and Surfaces for CAGD, A Practical Guide, 5th edition.Morgan Kaufmann, 2002.

24