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

Introduction to Computer Graphics – Modeling (1) – April 13, 2017 Kenshi Takayama

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 𝑏 𝑗 𝑢 𝑗 2

Cubic Hermite curves x 𝑦 0 = 𝑦 0 • Cubic polynomial curve interpolating 𝑦 1 = 𝑦 1 derivative constraints at both ends ′ 𝑦′(0) = 𝑦 0 𝑦 ′ 1 = 𝑦 1 (Hermite interpolation) 0 1 t ′ • 4 constraints  4 DoF needed 𝑦 0 = 𝑏 0 = 𝑦 0  4 coefficients  cubic 𝑦 1 = 𝑏 0 + 𝑏 1 + 𝑏 2 + 𝑏 3 = 𝑦 1 𝑦 ′ 0 = 𝑏 1 ′ = 𝑦 0 • 𝑦 𝑢 = 𝑏 0 + 𝑏 1 𝑢 + 𝑏 2 𝑢 2 + 𝑏 3 𝑢 3 𝑦 ′ 1 = 𝑏 1 + 2 𝑏 2 + 3 𝑏 3 ′ = 𝑦 1 • 𝑦′ 𝑢 = 𝑏 1 + 2𝑏 2 𝑢 + 3𝑏 3 𝑢 2  𝑏 0 = 𝑦 0 • Coeffs determined by substituting ′ 𝑏 1 = 𝑦 0 ′ − 𝑦 1 constrained values & derivatives ′ 𝑏 2 = −3 𝑦 0 + 3 𝑦 1 − 2 𝑦 0 ′ + 𝑦 1 ′ 𝑏 3 = 2 𝑦 0 − 2 𝑦 1 + 𝑦 0 3

Bezier curves • Input: 3 control points 𝑄 0 , 𝑄 𝑄 1 , 𝑄 2 1 • Coordinates of points in arbitrary domain (2D, 3D, ...) 𝑄 𝑢 = ? Eq. of Bezier curve 𝑄 2 • Output: Curve 𝑄 𝑢 satisfying t=1 𝑄 0 = 𝑄 0 𝑄 0 𝑄 𝑢 = 1 − 𝑢 𝑄 0 + 𝑢 𝑄 2 𝑄 1 = 𝑄 2 t=0 Eq. of line segment while being “pulled” by 𝑄 1 4

𝑄 1 Bezier curves Eq. of line 𝑢 𝑄 12 𝑢 1 − 𝑢 Eq. of line 1 − 𝑢 • 𝑄 01 𝑢 = 1 − 𝑢 𝑄 0 + 𝑢 𝑄 𝑄 01 𝑢 𝑢 1 − 𝑢 1 𝑄 012 𝑢 • 𝑄 12 𝑢 = 1 − 𝑢 𝑄 1 + 𝑢 𝑄 2 𝑄 2 • 𝑄 01 0 = 𝑄 0 𝑢 • 𝑄 12 1 = 𝑄 2 𝑄 0 • Idea: ”Interpolate the interpolation” As 𝑢 changes 0 → 1 , smoothly transition from 𝑄 01 to 𝑄 12 • 𝑄 012 𝑢 = 1 − 𝑢 𝑄 01 𝑢 + 𝑢 𝑄 12 𝑢 = 1 − 𝑢 1 − 𝑢 𝑄 0 + 𝑢 𝑄 1 + 𝑢 1 − 𝑢 𝑄 1 + 𝑢 𝑄 2 = 1 − 𝑢 2 𝑄 0 + 2𝑢 1 − 𝑢 𝑄 1 + 𝑢 2 𝑄 2 Quadratic Bezier curve 5

𝑄 1 Bezier curves • 𝑄 01 𝑢 = 1 − 𝑢 𝑄 0 + 𝑢 𝑄 1 • 𝑄 12 𝑢 = 1 − 𝑢 𝑄 1 + 𝑢 𝑄 2 𝑄 2 • 𝑄 01 0 = 𝑄 0 • 𝑄 12 1 = 𝑄 2 𝑄 0 • Idea: ”Interpolate the interpolation” As 𝑢 changes 0 → 1 , smoothly transition from 𝑄 01 to 𝑄 12 • 𝑄 012 𝑢 = 1 − 𝑢 𝑄 01 𝑢 + 𝑢 𝑄 12 𝑢 = 1 − 𝑢 1 − 𝑢 𝑄 0 + 𝑢 𝑄 1 + 𝑢 1 − 𝑢 𝑄 1 + 𝑢 𝑄 2 = 1 − 𝑢 2 𝑄 0 + 2𝑢 1 − 𝑢 𝑄 1 + 𝑢 2 𝑄 2 Quadratic Bezier curve 6

Cubic Bezier curve Quad. Bezier 𝑄 123 𝑢 𝑄 2 𝑄 1 • Exact same idea applied to 4 points 𝑄 0 , 𝑄 1 , 𝑄 2 𝑄 3 : Quad. Bezier 1 − 𝑢 𝑄 012 𝑢 • As 𝑢 changes 0 → 1 , transition from 𝑄 012 to 𝑄 𝑄 0123 𝑢 123 𝑄 3 𝑢 𝑄 0 • 𝑄 0123 𝑢 = 1 − 𝑢 𝑄 012 𝑢 + 𝑢 𝑄 123 𝑢 = 1 − 𝑢 1 − 𝑢 2 𝑄 0 + 2𝑢 1 − 𝑢 𝑄 1 + 𝑢 2 𝑄 2 + 𝑢 1 − 𝑢 2 𝑄 1 + 2𝑢 1 − 𝑢 𝑄 2 + 𝑢 2 𝑄 3 1 + 3𝑢 2 1 − 𝑢 𝑄 2 + 𝑢 3 𝑄 3 = 1 − 𝑢 3 𝑄 0 + 3𝑢 1 − 𝑢 2 𝑄 Cubic Bezier curve 7

Cubic Bezier curve 𝑄 2 𝑄 1 • Exact same idea applied to 4 points 𝑄 0 , 𝑄 1 , 𝑄 2 𝑄 3 : • As 𝑢 changes 0 → 1 , transition from 𝑄 012 to 𝑄 𝑄 0123 𝑢 123 𝑄 3 𝑄 0 • 𝑄 0123 𝑢 = 1 − 𝑢 𝑄 012 𝑢 + 𝑢 𝑄 123 𝑢 = 1 − 𝑢 1 − 𝑢 2 𝑄 0 + 2𝑢 1 − 𝑢 𝑄 1 + 𝑢 2 𝑄 2 + 𝑢 1 − 𝑢 2 𝑄 1 + 2𝑢 1 − 𝑢 𝑄 2 + 𝑢 2 𝑄 3 1 + 3𝑢 2 1 − 𝑢 𝑄 2 + 𝑢 3 𝑄 3 = 1 − 𝑢 3 𝑄 0 + 3𝑢 1 − 𝑢 2 𝑄 Cubic Bezier curve • Can easily control tangent at endpoints  ubiquitously used in CG 8

n-th order Bezier curve • Input: n+1 control points 𝑄 0 , ⋯ , 𝑄 𝑜 𝑜 𝑜 C 𝑗 𝑢 𝑗 1 − 𝑢 𝑜−𝑗 𝑄 𝑗 𝑄 𝑢 = 𝑜 (𝑢) 𝑗=0 𝑐 𝑗 Bernstein basis function 1 − 𝑢 4 𝑄 0 1 − 𝑢 5 𝑄 0 + + Quartic (4 th ) Quintic (5 th ) 4𝑢 1 − 𝑢 3 𝑄 5𝑢 1 − 𝑢 4 𝑄 + + 1 1 6𝑢 2 1 − 𝑢 2 𝑄 2 + 10𝑢 2 1 − 𝑢 3 𝑄 2 + 4𝑢 3 1 − 𝑢 𝑄 3 10𝑢 3 1 − 𝑢 2 𝑄 3 + + 5𝑢 4 1 − 𝑢 𝑄 𝑢 4 𝑄 + 4 4 𝑢 5 𝑄 5 9

Cubic Bezier curves & cubic Hermite curves • Cubic Bezier curve & its derivative: 1 + 3𝑢 2 1 − 𝑢 𝑄 2 + 𝑢 3 𝑄 3 • 𝑄 𝑢 = 1 − 𝑢 3 𝑄 0 + 3𝑢 1 − 𝑢 2 𝑄 • 𝑄 ′ (𝑢) = −3 1 − 𝑢 2 𝑄 0 + 3 1 − 𝑢 2 − 2𝑢(1 − 𝑢) 𝑄 1 + 3 2𝑢 1 − 𝑢 − 𝑢 2 𝑄 2 + 3𝑢 2 𝑄 3 • Derivatives at endpoints: 𝑄′(0) 1 • 𝑄 ′ 0 = −3𝑄 0 + 3𝑄 3 𝑄 ′ 0  𝑄 1 = 𝑄 0 + 1 𝑄 2 • 𝑄 ′ 1 = −3𝑄 2 + 3𝑄 3 1 3 𝑄 ′ 1  𝑄 2 = 𝑄 3 − 𝑄 1 𝑄 3 • Different ways of looking at cubic curves, 𝑄 0 𝑄′(1) essentially the same 10

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 11

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 12

Further control: Rational Bezier curve • Another view on Bezier curve: w 0 = w 2 = 1 “Weighted average” of control points • 𝑄 012 𝑢 = 1 − 𝑢 2 𝑄 0 + 2𝑢 1 − 𝑢 𝑄 1 + 𝑢 2 𝑄 2 = 𝜇 0 𝑢 𝑄 0 + 𝜇 1 𝑢 𝑄 1 + 𝜇 2 𝑢 𝑄 2 • Important property: partition of unity 𝜇 0 𝑢 + 𝜇 1 𝑢 + 𝜇 2 𝑢 = 1 ∀𝑢 • Multiply each 𝜇 𝑗 𝑢 by arbitrary coeff 𝑥 𝑗 : 𝜊 𝑗 𝑢 = 𝑥 𝑗 𝜇 𝑗 (𝑢) • Normalize to obtain new weights: ′ 𝑢 = 𝜊 𝑗 𝑢 𝜇 𝑗 Non-polynomial curve  can represent arcs etc. 𝑘 𝜊 𝑘 𝑢 13

x 0 (t) x 1 (t) x 2 (t) x 3 (t) x 4 (t) Cubic splines x • Series of connected cubic curves • Piecewise-polynomial t 0 t 1 t 2 t 3 t 4 t 5 t • Share value & derivative at every transition of intervals (C 1 continuity) • Parameter range can be other than [0, 1] • Assumption: 𝑢 𝑙 < 𝑢 𝑙+1 • Given values as only input, we want to automatically set derivatives Curve tool in PowerPoint 14

Cubic Catmull-Rom spline • Cubic function 𝑦 𝑙 (𝑢) for range 𝑢 𝑙 ≤ 𝑢 ≤ 𝑢 𝑙+1 is defined by adjacent constrained values 𝑦 𝑙−1 , 𝑦 𝑙 , 𝑦 𝑙+1 , 𝑦 𝑙+2 𝑦 𝑦 𝑙−1 𝑦 𝑙+1 𝑦 𝑙 (𝑢) 𝑦 𝑙+2 𝑦 𝑙 𝑢 𝑙−1 𝑢 𝑙 𝑢 𝑙+1 𝑢 𝑙+2 𝑢 15

Cubic Catmull-Rom spline: Step 1 • As 𝑢 𝑙 → 𝑢 𝑙+1 , interpolate such that 𝑦 𝑙 → 𝑦 𝑙+1  Line 𝑢 − 𝑢 𝑙 𝑢 − 𝑢 𝑙 𝑚 𝑙 (𝑢) = 1 − 𝑦 𝑙 + 𝑦 𝑙+1 𝑢 𝑙+1 − 𝑢 𝑙 𝑢 𝑙+1 − 𝑢 𝑙 𝑦 𝑚 𝑙−1 (𝑢) 𝑚 𝑙 (𝑢) 𝑚 𝑙+1 (𝑢) 𝑢 𝑙−1 𝑢 𝑙 𝑢 𝑙+1 𝑢 𝑙+2 𝑢 16

Cubic Catmull-Rom spline: Step 2 • As 𝑢 𝑙−1 → 𝑢 𝑙+1 , interpolate such that 𝑚 𝑙−1 → 𝑚 𝑙  Quadratic curve 𝑢 − 𝑢 𝑙−1 𝑢 − 𝑢 𝑙−1 𝑟 𝑙 (𝑢) = 1 − 𝑚 𝑙−1 (𝑢) + 𝑚 𝑙 (𝑢) 𝑢 𝑙+1 − 𝑢 𝑙−1 𝑢 𝑙+1 − 𝑢 𝑙−1 • Passes through 3 points 𝑢 𝑙−1 , 𝑦 𝑙−1 , 𝑢 𝑙 , 𝑦 𝑙 , 𝑢 𝑙+1 , 𝑦 𝑙+1 𝑦 𝑟 𝑙 (𝑢) 𝑟 𝑙+1 (𝑢) 𝑢 𝑙−1 𝑢 𝑙 𝑢 𝑙+1 𝑢 𝑙+2 𝑢 17

Cubic Catmull-Rom spline: Step 3 • As 𝑢 𝑙 → 𝑢 𝑙+1 , interpolate such that 𝑟 𝑙 → 𝑟 𝑙+1  Cubic curve 𝑢 − 𝑢 𝑙 𝑢 − 𝑢 𝑙 𝑦 𝑙 𝑢 = 1 − 𝑟 𝑙 𝑢 + 𝑟 𝑙+1 (𝑢) 𝑢 𝑙+1 − 𝑢 𝑙 𝑢 𝑙+1 − 𝑢 𝑙 Summary: Derivative at each CP is defined by a quadratic 𝑦 • curve passing through its adjacent CPs Each interval is a cubic curve satisfying • derivative constraints at both ends 𝑦 𝑙 (𝑢) 𝑢 𝑙−1 𝑢 𝑙 𝑢 𝑙+1 𝑢 𝑙+2 𝑢 18

Evaluating cubic Catmull-Rom spline 𝑄 𝑢 3 = 𝑄 3 𝑄 𝑢 1 = 𝑄 1 𝑄 𝑢 𝑢 1 ≤ 𝑢 ≤ 𝑢 2 𝑄 𝑢 2 = 𝑄 2 𝑄 𝑢 0 = 𝑄 0 A recursive evaluation algorithm for a class of Catmull-Rom splines [Barry,Colgman,SIGGRAPH88] 19

Ways of setting parameter values 𝑢 𝑙 (aka. knot sequence) • Assume: 𝑢 0 = 0 • Uniform 𝑢 𝑙 = 𝑢 𝑙−1 + 1 • Chordal 𝑢 𝑙 = 𝑢 𝑙−1 + 𝑄 𝑙−1 − 𝑄 𝑙 • Centripetal 𝑢 𝑙 = 𝑢 𝑙−1 + 𝑄 𝑙−1 − 𝑄 𝑙 Parameterization of Catmull-Rom Curves [Yuksel,Schaefer,Keyser,CAD11] 20

Application of cubic Catmull-Rom spline: Hair modeling Parameterization of Catmull-Rom Curves [Yuksel,Schaefer,Keyser,CAD11] 21