[PPT] - Introduction to Computer Graphics Modeling (1) April 13, 2017 PowerPoint Presentation

SLIDE 1

Introduction to Computer Graphics – Modeling (1) –

April 13, 2017 Kenshi Takayama

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Parametric curves

X & Y coordinates defined by parameter t (≅ time)
Example: Cycloid

𝑦 𝑢 = 𝑢 − sin 𝑢 𝑧 𝑢 = 1 − cos 𝑢

Tangent (aka. derivative, gradient) vector: 𝑦′ 𝑢 , 𝑧′ 𝑢
Polynomial curve: 𝑦 𝑢 = 𝑗 𝑏𝑗𝑢𝑗

2

SLIDE 3

Cubic Hermite curves

Cubic polynomial curve interpolating

derivative constraints at both ends (Hermite interpolation)

4 constraints  4 DoF needed

 4 coefficients  cubic

𝑦 𝑢 = 𝑏0 + 𝑏1𝑢 + 𝑏2𝑢2 + 𝑏3𝑢3
𝑦′ 𝑢 = 𝑏1 + 2𝑏2𝑢 + 3𝑏3𝑢2
Coeffs determined by substituting

constrained values & derivatives

3

𝑦 0 = 𝑦0 𝑦 1 = 𝑦1 𝑦′(0) = 𝑦0

′

𝑦′ 1 = 𝑦1

′

𝑦 0 = 𝑏0 = 𝑦0 𝑦 1 = 𝑏0 + 𝑏1 + 𝑏2 + 𝑏3 = 𝑦1 𝑦′ 0 = 𝑏1 = 𝑦0

′

𝑦′ 1 = 𝑏1 + 2 𝑏2 + 3 𝑏3 = 𝑦1

′

 𝑏0 = 𝑦0 𝑏1 = 𝑦0

′

𝑏2 = −3 𝑦0 + 3 𝑦1 − 2 𝑦0

′ − 𝑦1 ′

𝑏3 = 2 𝑦0 − 2 𝑦1 + 𝑦0

′ + 𝑦1 ′ t x 1

SLIDE 4

Bezier curves

Input: 3 control points 𝑄0, 𝑄

1, 𝑄2

Coordinates of points in

arbitrary domain (2D, 3D, ...)

Output: Curve 𝑄 𝑢 satisfying

𝑄 0 = 𝑄0 𝑄 1 = 𝑄2 while being “pulled” by 𝑄

1

4

𝑄 𝑢 = 1 − 𝑢 𝑄0 + 𝑢 𝑄2 𝑄 𝑢 = ? 𝑄

1

𝑄2 𝑄0

t=0 t=1

Eq. of line segment
Eq. of Bezier curve

SLIDE 5

Bezier curves

𝑄01 𝑢 = 1 − 𝑢 𝑄0 + 𝑢 𝑄

1

𝑄

12 𝑢 = 1 − 𝑢 𝑄 1 + 𝑢 𝑄2

𝑄01 0 = 𝑄0
𝑄

12 1 = 𝑄2

Idea: ”Interpolate the interpolation”

As 𝑢 changes 0 → 1 , smoothly transition from 𝑄01 to 𝑄

12

𝑄012 𝑢 = 1 − 𝑢 𝑄01 𝑢 + 𝑢 𝑄

12 𝑢

= 1 − 𝑢 1 − 𝑢 𝑄0 + 𝑢 𝑄

1 + 𝑢

1 − 𝑢 𝑄

1 + 𝑢 𝑄2

= 1 − 𝑢 2𝑄0 + 2𝑢 1 − 𝑢 𝑄

1 + 𝑢2𝑄2

5

𝑄

1

𝑄2 𝑄0 𝑄

12 𝑢 𝑢 𝑢 𝑢 1 − 𝑢 1 − 𝑢

𝑄01 𝑢 𝑄012 𝑢

1 − 𝑢 Quadratic Bezier curve

Eq. of line
Eq. of line

SLIDE 6

Bezier curves

𝑄01 𝑢 = 1 − 𝑢 𝑄0 + 𝑢 𝑄

1

𝑄

12 𝑢 = 1 − 𝑢 𝑄 1 + 𝑢 𝑄2

𝑄01 0 = 𝑄0
𝑄

12 1 = 𝑄2

Idea: ”Interpolate the interpolation”

As 𝑢 changes 0 → 1 , smoothly transition from 𝑄01 to 𝑄

12

𝑄012 𝑢 = 1 − 𝑢 𝑄01 𝑢 + 𝑢 𝑄

12 𝑢

= 1 − 𝑢 1 − 𝑢 𝑄0 + 𝑢 𝑄

1 + 𝑢

1 − 𝑢 𝑄

1 + 𝑢 𝑄2

= 1 − 𝑢 2𝑄0 + 2𝑢 1 − 𝑢 𝑄

1 + 𝑢2𝑄2

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𝑄

1

𝑄2 𝑄0

Quadratic Bezier curve

SLIDE 7

Exact same idea applied to 4 points 𝑄0, 𝑄

1, 𝑄2 𝑄3:

As 𝑢 changes 0 → 1, transition from 𝑄012 to 𝑄

123

𝑄0123 𝑢 = 1 − 𝑢 𝑄012 𝑢 + 𝑢 𝑄

123 𝑢

= 1 − 𝑢

1 − 𝑢 2𝑄0 + 2𝑢 1 − 𝑢 𝑄

1 + 𝑢2𝑄2 + 𝑢

1 − 𝑢 2𝑄

1 + 2𝑢 1 − 𝑢 𝑄2 + 𝑢2𝑄 3

= 1 − 𝑢 3𝑄0 + 3𝑢 1 − 𝑢 2𝑄

1 + 3𝑢2 1 − 𝑢 𝑄2 + 𝑢3𝑄3

Cubic Bezier curve

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𝑄

123 𝑢

𝑄012 𝑢 𝑄0123 𝑢

𝑢 1 − 𝑢

𝑄

1

𝑄3 𝑄0 𝑄2

Cubic Bezier curve

Quad. Bezier
Quad. Bezier

SLIDE 8

Exact same idea applied to 4 points 𝑄0, 𝑄

1, 𝑄2 𝑄3:

As 𝑢 changes 0 → 1, transition from 𝑄012 to 𝑄

123

𝑄0123 𝑢 = 1 − 𝑢 𝑄012 𝑢 + 𝑢 𝑄

123 𝑢

= 1 − 𝑢

1 − 𝑢 2𝑄0 + 2𝑢 1 − 𝑢 𝑄

1 + 𝑢2𝑄2 + 𝑢

1 − 𝑢 2𝑄

1 + 2𝑢 1 − 𝑢 𝑄2 + 𝑢2𝑄 3

= 1 − 𝑢 3𝑄0 + 3𝑢 1 − 𝑢 2𝑄

1 + 3𝑢2 1 − 𝑢 𝑄2 + 𝑢3𝑄3

Can easily control tangent at endpoints  ubiquitously used in CG

Cubic Bezier curve

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𝑄

1

𝑄3 𝑄0 𝑄2 𝑄0123 𝑢

Cubic Bezier curve

SLIDE 9

n-th order Bezier curve

Input: n+1 control points 𝑄0, ⋯ , 𝑄

𝑜

𝑄 𝑢 =

𝑗=0 𝑜 𝑜C𝑗 𝑢𝑗 1 − 𝑢 𝑜−𝑗 𝑄𝑗

9

𝑐𝑗

𝑜(𝑢)

Bernstein basis function

1 − 𝑢 4𝑄0 + 4𝑢 1 − 𝑢 3𝑄

1

+ 6𝑢2 1 − 𝑢 2𝑄2 + 4𝑢3 1 − 𝑢 𝑄3 + 𝑢4𝑄

4

1 − 𝑢 5𝑄0 + 5𝑢 1 − 𝑢 4𝑄

1

+ 10𝑢2 1 − 𝑢 3𝑄2 + 10𝑢3 1 − 𝑢 2𝑄3 + 5𝑢4 1 − 𝑢 𝑄

4

+ 𝑢5𝑄5

Quartic (4th) Quintic (5th)

SLIDE 10

Cubic Bezier curves & cubic Hermite curves

Cubic Bezier curve & its derivative:
𝑄 𝑢 = 1 − 𝑢 3𝑄0 + 3𝑢 1 − 𝑢 2𝑄

1 + 3𝑢2 1 − 𝑢 𝑄2 + 𝑢3𝑄3

𝑄′(𝑢) = −3 1 − 𝑢 2𝑄0 + 3 1 − 𝑢 2 − 2𝑢(1 − 𝑢) 𝑄

1 + 3 2𝑢 1 − 𝑢 − 𝑢2 𝑄2 + 3𝑢2𝑄3

Derivatives at endpoints:
𝑄′ 0 = −3𝑄0 + 3𝑄

1

 𝑄

1 = 𝑄0 + 1 3 𝑄′ 0

𝑄′ 1 = −3𝑄2 + 3𝑄3

 𝑄2 = 𝑄3 −

1 3 𝑄′ 1

Different ways of looking at cubic curves,

essentially the same

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𝑄

1

𝑄3 𝑄0 𝑄2

𝑄′(0) 𝑄′(1)

SLIDE 11

Evaluating Bezier curves

Method 1: Direct evaluation of polynomials
Simple & fast, could be numerically unstable
Method 2: de Casteljau’s algorithm
Directly after the recursive definition of Bezier curves
More computation steps, numerically stable
Also useful for splitting Bezier curves

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SLIDE 12

Drawing Bezier curves

In the end, everything is drawn as polyline
Main question: How to sample paramter t?
Method 1: Uniform sampling
Simple
Potentially insufficient sampling density
Method 2: Adaptive sampling
If control points deviate too much from straight

line, split by de Casteljau’s algorithm

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SLIDE 13

Further control: Rational Bezier curve

Another view on Bezier curve:

“Weighted average” of control points

𝑄012 𝑢 = 1 − 𝑢 2𝑄0 + 2𝑢 1 − 𝑢 𝑄

1 + 𝑢2

𝑄

2

= 𝜇0 𝑢 𝑄

0 + 𝜇1 𝑢

𝑄

1+ 𝜇2 𝑢 𝑄2

Important property: partition of unity

𝜇0 𝑢 + 𝜇1 𝑢 + 𝜇2 𝑢 = 1 ∀𝑢

Multiply each 𝜇𝑗 𝑢 by arbitrary coeff 𝑥𝑗:

𝜊𝑗 𝑢 = 𝑥𝑗 𝜇𝑗(𝑢)

Normalize to obtain new weights:

𝜇𝑗

′ 𝑢 = 𝜊𝑗 𝑢 𝑘 𝜊𝑘 𝑢

13

w0 = w2 = 1

Non-polynomial curve  can represent arcs etc.

SLIDE 14

Cubic splines

Series of connected cubic curves
Piecewise-polynomial
Share value & derivative at every transition
f intervals (C1 continuity)
Parameter range can be other than [0, 1]
Assumption: 𝑢𝑙 < 𝑢𝑙+1
Given values as only input,

we want to automatically set derivatives

14

t x t0 t1 t2 t3 t4 t5 x0(t) x1(t) x2(t) x3(t) x4(t) Curve tool in PowerPoint

SLIDE 15

Cubic Catmull-Rom spline

Cubic function 𝑦𝑙(𝑢) for range 𝑢𝑙 ≤ 𝑢 ≤ 𝑢𝑙+1 is defined by

adjacent constrained values 𝑦𝑙−1, 𝑦𝑙, 𝑦𝑙+1, 𝑦𝑙+2

15

𝑢𝑙−1 𝑢𝑙 𝑢𝑙+1 𝑢𝑙+2 𝑢 𝑦 𝑦𝑙 𝑦𝑙−1 𝑦𝑙+1 𝑦𝑙+2 𝑦𝑙(𝑢)

SLIDE 16

Cubic Catmull-Rom spline: Step 1

As 𝑢𝑙 → 𝑢𝑙+1, interpolate such that 𝑦𝑙 → 𝑦𝑙+1  Line

𝑚𝑙(𝑢) = 1 − 𝑢 − 𝑢𝑙 𝑢𝑙+1 − 𝑢𝑙 𝑦𝑙 + 𝑢 − 𝑢𝑙 𝑢𝑙+1 − 𝑢𝑙 𝑦𝑙+1

16

𝑢𝑙−1 𝑢𝑙 𝑢𝑙+1 𝑢𝑙+2 𝑢 𝑦 𝑚𝑙−1(𝑢) 𝑚𝑙(𝑢) 𝑚𝑙+1(𝑢)

SLIDE 17

Cubic Catmull-Rom spline: Step 2

As 𝑢𝑙−1 → 𝑢𝑙+1, interpolate such that 𝑚𝑙−1 → 𝑚𝑙  Quadratic curve

𝑟𝑙(𝑢) = 1 − 𝑢 − 𝑢𝑙−1 𝑢𝑙+1 − 𝑢𝑙−1 𝑚𝑙−1(𝑢) + 𝑢 − 𝑢𝑙−1 𝑢𝑙+1 − 𝑢𝑙−1 𝑚𝑙(𝑢)

Passes through 3 points 𝑢𝑙−1, 𝑦𝑙−1 , 𝑢𝑙, 𝑦𝑙 , 𝑢𝑙+1, 𝑦𝑙+1

17

𝑢𝑙−1 𝑢𝑙 𝑢𝑙+1 𝑢𝑙+2 𝑢 𝑦 𝑟𝑙(𝑢) 𝑟𝑙+1(𝑢)

SLIDE 18

Cubic Catmull-Rom spline: Step 3

As 𝑢𝑙 → 𝑢𝑙+1, interpolate such that 𝑟𝑙 → 𝑟𝑙+1  Cubic curve

𝑦𝑙 𝑢 = 1 − 𝑢 − 𝑢𝑙 𝑢𝑙+1 − 𝑢𝑙 𝑟𝑙 𝑢 + 𝑢 − 𝑢𝑙 𝑢𝑙+1 − 𝑢𝑙 𝑟𝑙+1(𝑢)

18

𝑢𝑙−1 𝑢𝑙 𝑢𝑙+1 𝑢𝑙+2 𝑢 𝑦 𝑦𝑙(𝑢)

Summary:

Derivative at each CP is defined by a quadratic

curve passing through its adjacent CPs

Each interval is a cubic curve satisfying

derivative constraints at both ends

SLIDE 19

Evaluating cubic Catmull-Rom spline

19

A recursive evaluation algorithm for a class of Catmull-Rom splines [Barry,Colgman,SIGGRAPH88]

𝑄 𝑢0 = 𝑄0 𝑄 𝑢1 = 𝑄

1

𝑄 𝑢2 = 𝑄2 𝑄 𝑢3 = 𝑄3 𝑄 𝑢 𝑢1 ≤ 𝑢 ≤ 𝑢2

SLIDE 20

Ways of setting parameter values 𝑢𝑙 (aka. knot sequence)

Assume: 𝑢0 = 0
Uniform

𝑢𝑙 = 𝑢𝑙−1 + 1

Chordal

𝑢𝑙 = 𝑢𝑙−1 + 𝑄𝑙−1 − 𝑄𝑙

Centripetal

𝑢𝑙 = 𝑢𝑙−1 + 𝑄𝑙−1 − 𝑄𝑙

20

Parameterization of Catmull-Rom Curves [Yuksel,Schaefer,Keyser,CAD11]

SLIDE 21

Application of cubic Catmull-Rom spline: Hair modeling

21

Parameterization of Catmull-Rom Curves [Yuksel,Schaefer,Keyser,CAD11]

SLIDE 22

B-spline

Another way of defining polynomial spline
Represent curve as sum of basis functions
Cubic basis is the most commonly used
Deeply related to subdivision surfaces

 Next lecture

Non-Uniform Rational B-Spline
Non-Uniform = varying spacing of knots (𝑢𝑙)
Rational = arbitrary weights for CPs
(Complex stuff, not covered)
Cool Flash demo:

http://geometrie.foretnik.net/files/NURBS-en.swf

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SLIDE 23

Parametric surfaces

One parameter  Curve 𝑄(𝑢)
Two parameters  Surface 𝑄 𝑡, 𝑢
Cubic Bezier surface:
Input: 4×4=16 control points 𝑄𝑗𝑘

𝑄 𝑡, 𝑢 =

𝑗=0 3 𝑘=0 3

𝑐𝑗

3 𝑡 𝑐 𝑘 3 𝑢 𝑄𝑗𝑘

23

Bernstein basis functions 𝑐0

3 𝑢 = 1 − 𝑢 3

𝑐1

3 𝑢 = 3𝑢 1 − 𝑢 2

𝑐2

3 𝑢 = 3𝑢2(1 − 𝑢)

𝑐3

3 𝑢 = 𝑢3

SLIDE 24

3D modeling using parametric surface patches

Pros
Can compactly represent smooth surfaces
Can accurately represent spheres, cones, etc
Cons
Hard to design nice layout of patches
Hard to maintain continuity across patches
Often used for designing man-made objects

consisting of simple parts

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SLIDE 25

Pointers

http://en.wikipedia.org/wiki/Bezier_curve
http://antigrain.com/research/adaptive_bezier/
https://groups.google.com/forum/#!topic/comp.graphics.algorithms/2

FypAv29dG4

http://en.wikipedia.org/wiki/Cubic_Hermite_spline
http://en.wikipedia.org/wiki/Centripetal_Catmull%E2%80%93Rom_splin

e

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