CS 445 / 645 Introduction to Computer Graphics Lecture 22 Hermite Splines Lecture 22 Hermite...

CS 445 / 645Introduction to Computer Graphics

Lecture 22Lecture 22

Hermite SplinesHermite Splines

Lecture 22Lecture 22

Hermite SplinesHermite Splines

ACM Elections

Should take 15 minutesShould take 15 minutes

Taking place now in OLS 120Taking place now in OLS 120

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I’m showing movies that you can watch after class I’m showing movies that you can watch after class today (or some other day)today (or some other day)

Should take 15 minutesShould take 15 minutes

Taking place now in OLS 120Taking place now in OLS 120

Hurry BackHurry Back

I’m showing movies that you can watch after class I’m showing movies that you can watch after class today (or some other day)today (or some other day)

Splines – Old School

Representations of Curves

Problems with series of points used to model a Problems with series of points used to model a curvecurve• Piecewise linear - Does not accurately model a smooth linePiecewise linear - Does not accurately model a smooth line

• It’s tediousIt’s tedious

• Expensive to manipulate curve because all points must be Expensive to manipulate curve because all points must be repositionedrepositioned

Instead, model curve as piecewise-polynomialInstead, model curve as piecewise-polynomial• x = x(t), y = y(t), z = z(t) x = x(t), y = y(t), z = z(t)

– where x(), y(), z() are polynomials where x(), y(), z() are polynomials

Problems with series of points used to model a Problems with series of points used to model a curvecurve• Piecewise linear - Does not accurately model a smooth linePiecewise linear - Does not accurately model a smooth line

• It’s tediousIt’s tedious

• Expensive to manipulate curve because all points must be Expensive to manipulate curve because all points must be repositionedrepositioned

Instead, model curve as piecewise-polynomialInstead, model curve as piecewise-polynomial• x = x(t), y = y(t), z = z(t) x = x(t), y = y(t), z = z(t)

– where x(), y(), z() are polynomials where x(), y(), z() are polynomials

Specifying Curves (hyperlink)

Control PointsControl Points• A set of points that influence the A set of points that influence the

curve’s shapecurve’s shape

KnotsKnots• Control points that lie on the curveControl points that lie on the curve

Interpolating SplinesInterpolating Splines• Curves that pass through the control Curves that pass through the control

points (knots)points (knots)

Approximating SplinesApproximating Splines• Control points merely influence shapeControl points merely influence shape

Control PointsControl Points• A set of points that influence the A set of points that influence the

curve’s shapecurve’s shape

KnotsKnots• Control points that lie on the curveControl points that lie on the curve

Interpolating SplinesInterpolating Splines• Curves that pass through the control Curves that pass through the control

points (knots)points (knots)

Approximating SplinesApproximating Splines• Control points merely influence shapeControl points merely influence shape

Parametric Curves

Very flexible representationVery flexible representation

They are not required to be functionsThey are not required to be functions

• They can be multivalued with respect to any dimensionThey can be multivalued with respect to any dimension

Very flexible representationVery flexible representation

They are not required to be functionsThey are not required to be functions

• They can be multivalued with respect to any dimensionThey can be multivalued with respect to any dimension

Cubic Polynomials

x(t) = ax(t) = axxtt33 + b + bxxtt22 + c + cxxt + dt + dxx

• Similarly for y(t) and z(t)Similarly for y(t) and z(t)

Let t: (0 <= t <= 1)Let t: (0 <= t <= 1)

Let T = [tLet T = [t33 t t22 t 1] t 1]

Coefficient Matrix CCoefficient Matrix C

Curve: Q(t) = T*CCurve: Q(t) = T*C

x(t) = ax(t) = axxtt33 + b + bxxtt22 + c + cxxt + dt + dxx

• Similarly for y(t) and z(t)Similarly for y(t) and z(t)

Let t: (0 <= t <= 1)Let t: (0 <= t <= 1)

Let T = [tLet T = [t33 t t22 t 1] t 1]

Coefficient Matrix CCoefficient Matrix C

Curve: Q(t) = T*CCurve: Q(t) = T*C

z

z

y

y

x

x

zyx

zyx

d

c

d

c

d

c

bbb

aaa

ttt 123

Parametric Curves

How do we find the tangent to a curve?How do we find the tangent to a curve?

• If f(x) = xIf f(x) = x22 – 4 – 4

– tangent at (x=3) is f’(x) = 2 (x) – 4 = 2 (3) - 4tangent at (x=3) is f’(x) = 2 (x) – 4 = 2 (3) - 4

How do we find the tangent to a curve?How do we find the tangent to a curve?

• If f(x) = xIf f(x) = x22 – 4 – 4

– tangent at (x=3) is f’(x) = 2 (x) – 4 = 2 (3) - 4tangent at (x=3) is f’(x) = 2 (x) – 4 = 2 (3) - 4

Derivative of Q(t) is the tangent vector at t:Derivative of Q(t) is the tangent vector at t:

• d/dt Q(t) = Q’(t) = d/dt T * C = [3td/dt Q(t) = Q’(t) = d/dt T * C = [3t22 2t 1 0] * C 2t 1 0] * C

Derivative of Q(t) is the tangent vector at t:Derivative of Q(t) is the tangent vector at t:

• d/dt Q(t) = Q’(t) = d/dt T * C = [3td/dt Q(t) = Q’(t) = d/dt T * C = [3t22 2t 1 0] * C 2t 1 0] * C

Piecewise Curve Segments

One curve constructed by connecting many smaller One curve constructed by connecting many smaller segments end-to-endsegments end-to-end

Continuity describes the jointContinuity describes the joint

• CC11 is tangent continuity (velocity) is tangent continuity (velocity)

• CC22 is 2 is 2ndnd derivative continuity (acceleration) derivative continuity (acceleration)

One curve constructed by connecting many smaller One curve constructed by connecting many smaller segments end-to-endsegments end-to-end

Continuity describes the jointContinuity describes the joint

• CC11 is tangent continuity (velocity) is tangent continuity (velocity)

• CC22 is 2 is 2ndnd derivative continuity (acceleration) derivative continuity (acceleration)

Continuity of Curves

If direction (but not necessarily magnitude) of tangent If direction (but not necessarily magnitude) of tangent matchesmatches• GG11 geometric continuity geometric continuity

• The tangent value at the end of one curve is proportional to the The tangent value at the end of one curve is proportional to the tangent value of the beginning of the next curvetangent value of the beginning of the next curve

Matching direction and magnitude of dMatching direction and magnitude of dnn / dt / dtnn

– CCnn continous continous

If direction (but not necessarily magnitude) of tangent If direction (but not necessarily magnitude) of tangent matchesmatches• GG11 geometric continuity geometric continuity

• The tangent value at the end of one curve is proportional to the The tangent value at the end of one curve is proportional to the tangent value of the beginning of the next curvetangent value of the beginning of the next curve

Matching direction and magnitude of dMatching direction and magnitude of dnn / dt / dtnn

– CCnn continous continous

Parametric Cubic Curves

In order to assure CIn order to assure C22 continuity, curves must be of continuity, curves must be of

at least degree 3at least degree 3

Here is the parametric definition of a cubic Here is the parametric definition of a cubic (degree 3) spline in two dimensions(degree 3) spline in two dimensions

How do we extend it to three dimensions?How do we extend it to three dimensions?

In order to assure CIn order to assure C22 continuity, curves must be of continuity, curves must be of

at least degree 3at least degree 3

Here is the parametric definition of a cubic Here is the parametric definition of a cubic (degree 3) spline in two dimensions(degree 3) spline in two dimensions

How do we extend it to three dimensions?How do we extend it to three dimensions?

Parametric Cubic Splines

Can represent this as a matrix tooCan represent this as a matrix tooCan represent this as a matrix tooCan represent this as a matrix too

Coefficients

So how do we select the coefficients?So how do we select the coefficients?

• [a[axx b bxx c cxx d dxx] and [a] and [ayy b byy c cyy d dyy] must satisfy the constraints ] must satisfy the constraints

defined by the knots and the continuity conditionsdefined by the knots and the continuity conditions

So how do we select the coefficients?So how do we select the coefficients?

• [a[axx b bxx c cxx d dxx] and [a] and [ayy b byy c cyy d dyy] must satisfy the constraints ] must satisfy the constraints

defined by the knots and the continuity conditionsdefined by the knots and the continuity conditions

Parametric Curves

Difficult to conceptualize curve as Difficult to conceptualize curve as x(t) = ax(t) = axxtt33 + b + bxxtt22 + c + cxxt + dt + dxx

(artists don’t think in terms of coefficients of cubics)(artists don’t think in terms of coefficients of cubics)

Instead, define curve as weighted combination of 4 well-Instead, define curve as weighted combination of 4 well-defined cubic polynomialsdefined cubic polynomials(wait a second! Artists don’t think this way either!)(wait a second! Artists don’t think this way either!)

Each curve type defines different cubic polynomials and Each curve type defines different cubic polynomials and weighting schemesweighting schemes

Difficult to conceptualize curve as Difficult to conceptualize curve as x(t) = ax(t) = axxtt33 + b + bxxtt22 + c + cxxt + dt + dxx

(artists don’t think in terms of coefficients of cubics)(artists don’t think in terms of coefficients of cubics)

Instead, define curve as weighted combination of 4 well-Instead, define curve as weighted combination of 4 well-defined cubic polynomialsdefined cubic polynomials(wait a second! Artists don’t think this way either!)(wait a second! Artists don’t think this way either!)

Each curve type defines different cubic polynomials and Each curve type defines different cubic polynomials and weighting schemesweighting schemes

Parametric Curves

HermiteHermite – two endpoints and two endpoint – two endpoints and two endpoint tangent vectorstangent vectors

BezierBezier - two endpoints and two other points that - two endpoints and two other points that define the endpoint tangent vectorsdefine the endpoint tangent vectors

SplinesSplines – four control points – four control points • C1 and C2 continuity at the join pointsC1 and C2 continuity at the join points

• Come close to their control points, but not guaranteed to Come close to their control points, but not guaranteed to touch themtouch them

Examples of SplinesExamples of Splines

HermiteHermite – two endpoints and two endpoint – two endpoints and two endpoint tangent vectorstangent vectors

BezierBezier - two endpoints and two other points that - two endpoints and two other points that define the endpoint tangent vectorsdefine the endpoint tangent vectors

SplinesSplines – four control points – four control points • C1 and C2 continuity at the join pointsC1 and C2 continuity at the join points

• Come close to their control points, but not guaranteed to Come close to their control points, but not guaranteed to touch themtouch them

Examples of SplinesExamples of Splines

Hermite Cubic Splines

An example of knot and continuity constraintsAn example of knot and continuity constraintsAn example of knot and continuity constraintsAn example of knot and continuity constraints


One cubic curve for each dimensionOne cubic curve for each dimension

A curve constrained to x/y-plane has two curves:A curve constrained to x/y-plane has two curves:

One cubic curve for each dimensionOne cubic curve for each dimension

A curve constrained to x/y-plane has two curves:A curve constrained to x/y-plane has two curves:

d

c

b

a

ttt

dctbtattf x

1

)(

23

23

h

g

f

e

ttt

hgtftettf y

1

)(

23

23


A 2-D Hermite Cubic Spline is defined by eight A 2-D Hermite Cubic Spline is defined by eight parameters: a, b, c, d, e, f, g, hparameters: a, b, c, d, e, f, g, h

How do we convert the intuitive endpoint constraints into How do we convert the intuitive endpoint constraints into these (relatively) unintuitive eight parameters?these (relatively) unintuitive eight parameters?

We know:We know:• (x, y) position at t = 0, p(x, y) position at t = 0, p11

• (x, y) position at t = 1, p(x, y) position at t = 1, p22

• (x, y) derivative at t = 0, dp/dt(x, y) derivative at t = 0, dp/dt


A 2-D Hermite Cubic Spline is defined by eight A 2-D Hermite Cubic Spline is defined by eight parameters: a, b, c, d, e, f, g, hparameters: a, b, c, d, e, f, g, h

How do we convert the intuitive endpoint constraints into How do we convert the intuitive endpoint constraints into these (relatively) unintuitive eight parameters?these (relatively) unintuitive eight parameters?

We know:We know:• (x, y) position at t = 0, p(x, y) position at t = 0, p11




Hermite Cubic Spline

We know:We know:


We know:We know:


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c

b

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x

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)0(

1000

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)0(

1000

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We know:We know:


We know:We know:


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1111

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yphgfef

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So far we have four equations, but we have eight So far we have four equations, but we have eight unknownsunknowns

Use the derivativesUse the derivatives

So far we have four equations, but we have eight So far we have four equations, but we have eight unknownsunknowns

Use the derivativesUse the derivatives

d

c

b

a

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dctbtattf

x

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)(

2

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23

h

g

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tttf

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hgtftettf

y

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)(

2

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We know:We know:


We know:We know:


dtdp

cf

d

c

b

acbaf

xx

x

1

2

2

)0(

010203

0203)0(

dtdp

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)0(

010203

0203)0(


We know:We know:


We know:We know:


dtdp

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d

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Hermite Specification

Matrix equation for Hermite CurveMatrix equation for Hermite CurveMatrix equation for Hermite CurveMatrix equation for Hermite Curve

t = 0

t = 1

t = 0

t = 1

t3 t2 t1 t0

p1

p2

r p1

r p2

dtdp

dtdp

p

p

dtdp

dtdpp

p

h

g

f

e

d

c

b

a

y

y

y

y

x

x

x

x

2

1

2

1

1

1

2

1

0123

0100

1111

1000

Solve Hermite Matrix

h

g

f

e

d

c

b

a

dtdp

dtdp

p

p

dtdp

dtdpp

p

y

y

y

y

x

x

x

x

2

1

2

1

1

1

2

11

0123

0100

1111

1000

Spline and Geometry Matrices

h

g

f

e

d

c

b

a

dtdp

dtdp

p

p

dtdp

dtdpp

p

y

y

y

y

x

x

x

x

2

1

2

1

1

1

2

1

0001

0100

1233

1122

MHermite GHermite

Resulting Hermite Spline Equation

Demonstration

HermiteHermite

Sample Hermite Curves

Blending Functions

By multiplying first two matrices in lower-left By multiplying first two matrices in lower-left equation, you have four functions of ‘t’ that equation, you have four functions of ‘t’ that blend the four control parametersblend the four control parameters

These are blendingThese are blendingfunctionsfunctions

By multiplying first two matrices in lower-left By multiplying first two matrices in lower-left equation, you have four functions of ‘t’ that equation, you have four functions of ‘t’ that blend the four control parametersblend the four control parameters

These are blendingThese are blendingfunctionsfunctions

Hermite Blending Functions

If you plot the If you plot the blending blending functions on functions on the parameter the parameter ‘t’‘t’

If you plot the If you plot the blending blending functions on functions on the parameter the parameter ‘t’‘t’

Hermite Blending Functions

Remember, eachRemember, eachblending functionblending functionreflects influencereflects influenceof Pof P11, P, P22, , PP11, , PP22

on spline’s shapeon spline’s shape

Remember, eachRemember, eachblending functionblending functionreflects influencereflects influenceof Pof P11, P, P22, , PP11, , PP22

on spline’s shapeon spline’s shape

CS 445 / 645 Introduction to Computer Graphics Lecture 22 Hermite Splines Lecture 22 Hermite...

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