Introduction to Computer Graphics Modeling (1) April 18, 2019 - PowerPoint PPT Presentation

Introduction to Computer Graphics – Modeling (1) – April 18, 2019 Kenshi Takayama

Some additional notes on quaternions 2

Another explanation for quaternions (overview) 1. Any rotation can be decomposed into even number of reflections 2. Quaternions can concisely describe reflections in 3D 𝑦 = − Ԧ 𝑦 Ԧ 𝑔 −1 𝑆 Ԧ 𝑔 Ԧ 𝑔 Ԧ 3. Combining two reflections equivalent to the rotation leads to the formula = cos 𝜄 2 + 𝜕 sin 𝜄 𝑦 cos 𝜄 2 − 𝜕 sin 𝜄 𝑆 𝑕 𝑆 Ԧ 𝑔 Ԧ 𝑦 Ԧ 2 2 3

Any rotation can be decomposed into even number of reflections • Mathematically proven 𝜄 • Valid for any dimensions 3𝜄 4 𝜄 𝜄 • Not unique (of course!) 2 2 𝜄 4 R One way 𝜄 𝜄 R 𝜄 2 Another way R 4

Quaternions recap • Complex number: real + imaginary 𝑏 + 𝑐 𝐣 • Quaternion: scalar + vector 𝑏 + Ԧ 𝑤 • Definition of quaternion multiplication: Scalar part Vector part 𝑏 1 + 𝑤 1 𝑏 2 + 𝑤 2 ≔ 𝑏 1 𝑏 2 − 𝑤 1 ⋅ 𝑤 2 + 𝑏 1 𝑤 2 + 𝑏 2 𝑤 1 + 𝑤 1 × 𝑤 2 • Pure vectors can take multiplication by interpreting them as quaternions: 𝑤 1 𝑤 2 = −𝑤 1 ⋅ 𝑤 2 + 𝑤 1 × 𝑤 2 • Notable properties: 𝑤 Ԧ 𝑤 −1 = − 𝑤 2 If Ԧ 𝑤 ⋅ 𝑥 = 0 , then Ԧ 𝑤 Ԧ Ԧ 𝑤 = − Ԧ Ԧ 𝑤 𝑥 = −𝑥 Ԧ 𝑤 • (Relevant later) 𝑤 2 Ԧ 𝑤 is always zero Multiplying Ԧ 𝑤 to rhs produces 1 𝑤 × Ԧ Ԧ 𝑤 𝑥 = Ԧ Ԧ 𝑤 × 𝑥 = −𝑥 × Ԧ 𝑤 = −𝑥 Ԧ 𝑤 5

Describing reflections using quaternions • Reflection of a point Ԧ 𝑦 across a plane orthogonal to Ԧ 𝑔 : 𝑦 Ԧ Ԧ 𝑔 𝑦 ≔ − Ԧ 𝑦 Ԧ 𝑔 −1 𝑆 Ԧ 𝑔 Ԧ 𝑔 Ԧ • Holds essential properties of reflections: 𝑃 • Linearity : 𝑆 Ԧ 𝑔 Ԧ 𝑦 𝑆 Ԧ 𝑔 𝑏 Ԧ 𝑦 + 𝑐 Ԧ 𝑧 = 𝑏 𝑆 Ԧ 𝑔 Ԧ 𝑦 + 𝑐 𝑆 Ԧ 𝑔 Ԧ 𝑧 𝑔 gets mapped to − Ԧ • Ԧ 𝑔 : 𝑔 −1 = − Ԧ 𝑔 Ԧ 𝑔 = − Ԧ 𝑔 Ԧ 𝑔 Ԧ 𝑆 Ԧ 𝑔 • If a point Ԧ 𝑦 satisfies Ԧ 𝑔 = 0 (i.e. on the plane), Ԧ 𝑦 doesn’t move: 𝑦 ⋅ Ԧ 𝑔 −1 = − − Ԧ 𝑔 −1 = Ԧ 𝑦 = − Ԧ 𝑦 Ԧ 𝑦 Ԧ Ԧ Because if Ԧ 𝑔 = 0 , then Ԧ 𝑆 Ԧ 𝑔 Ԧ 𝑔 Ԧ 𝑔 𝑦 𝑦 ⋅ Ԧ 𝑦 Ԧ 𝑔 Ԧ 𝑦 = − Ԧ 𝑔 6 https://math.stackexchange.com/a/7263

Setup for rotation around arbitrary axis 𝑧 Ԧ • Rotation axis (unit vector) ： 𝜕 𝜄 • Rotation angle ： 𝜄 𝑦 Ԧ • Point before rotation ： Ԧ 𝑦 • Point after rotation ： Ԧ 𝑧 ≔ 𝑆 𝜕, 𝜄 Ԧ 𝑦 • Think of local 2D coordinate system ： 𝜕 ⋅ Ԧ 𝑦 𝜕 • “Right” vector ： 𝑣 ≔ Ԧ 𝑦 − 𝜕 ⋅ Ԧ 𝑦 𝜕 𝑤 Ԧ • “Up” vector ： Ԧ 𝑤 ≔ 𝜕 × Ԧ 𝑦 • Note that 𝑣 = 𝑤 Ԧ 𝑣 • (Let’s call it 𝑀 ) 𝑃 𝑀 7 https://thimbleprojects.org/kenshi84/50110/demo/quaternion-schematic.html

Decompose rotation into two reflections 1 st reflection ： 2 nd reflection ： 𝑕 ≔ − sin 𝜄 2 𝑣 + cos 𝜄 Ԧ Ԧ 2 Ԧ 𝑤 𝑔 ≔ Ԧ 𝑤 Top view Ԧ 𝜄 𝑔 𝑕 Ԧ 𝜄 2 8 https://thimbleprojects.org/kenshi84/50110/demo/quaternion-schematic.html

Combining two reflections 𝑔 −1 = − Ԧ 𝑔 −1 Ԧ 𝑕 −1 = 𝑔 −1 Ԧ • Formula ： 𝑆 𝑕 𝑆 Ԧ = 𝑆 𝑕 − Ԧ 𝑦 Ԧ 𝑕 − Ԧ 𝑦 Ԧ 𝑕 Ԧ Ԧ 𝑕 −1 𝑔 Ԧ 𝑦 𝑔 Ԧ 𝑔 Ԧ Ԧ 𝑔 𝑦 Ԧ • Substitute to the above Ԧ 𝑕 ≔ − sin 𝜄 2 𝑣 + cos 𝜄 𝑔 ≔ Ԧ 𝑤, 2 Ԧ 𝑤 • For the left part Ԧ 𝑕 Ԧ 𝑔 ： (because 𝑣 ⋅ Ԧ 𝑤 = 0 ) 𝑀 2 cos 𝜄 𝑕 ⋅ Ԧ 𝑔 = − sin 𝜄 2 𝑣 + cos 𝜄 Ԧ 2 Ԧ 𝑤 ⋅ Ԧ 𝑤 = (because 𝑣 × Ԧ 𝑤 = 𝑀 2 𝜕 ) 2 −𝑀 2 sin 𝜄 𝑕 × Ԧ 𝑔 = − sin 𝜄 2 𝑣 + cos 𝜄 Ԧ 2 Ԧ 𝑤 × Ԧ 𝑤 = 2 𝜕 Therefore, −𝑀 2 cos 𝜄 𝑕 Ԧ 𝑕 ⋅ Ԧ 𝑕 × Ԧ 2 + 𝜕 sin 𝜄 Ԧ 𝑔 = − Ԧ 𝑔 + Ԧ 𝑔 = 2 𝑔 −1 Ԧ 𝑕 −1 = 𝑔 𝑕 𝑔 −1 Ԧ 𝑕 −1 = −𝑀 −2 cos 𝜄 • The right part Ԧ 𝑀4 is analogous ： Ԧ 2 − 𝜕 sin 𝜄 2 • Finally, we get the formula ： = −𝑀 2 cos 𝜄 −𝑀 −2 cos 𝜄 2 + 𝜕 sin 𝜄 2 − 𝜕 sin 𝜄 𝑆 𝜕, 𝜄 Ԧ 𝑦 = 𝑆 𝑕 𝑆 Ԧ 𝑔 Ԧ 𝑦 𝑦 Ԧ 2 2 cos 𝜄 2 + 𝜕 sin 𝜄 𝑦 cos 𝜄 2 − 𝜕 sin 𝜄 = Ԧ 2 2 9

Representing and blending poses using quaternions ℝ 4 • Any rotations (poses) can be 𝑞 = cos 𝜄 2 + 𝜕 sin 𝜄 represented as unit quaternions 𝑟 2 • Points on hypersphere of 4D space • Fix 𝜕 and vary 𝜄  unit circle in 4D space −𝑟 𝜄 =0 (1, 0,0,0) • A pose after rotating 360 ° about a certain axis is represented as another quaternion 𝜄 = 𝜌 𝜄 =3 𝜌 (0, 𝜕 x , 𝜕 y , 𝜕 z ) (0, - 𝜕 x ,- 𝜕 y ,- 𝜕 z ) • One pose corresponds to two quaternions (double cover) 𝜄 =2 𝜌 • A geodesic between two points 𝑞, 𝑟 on the (-1, 0,0,0) hypersphere represents interpolation of these poses • Should pick either 𝑟 or −𝑟 which is closer to 𝑞 (i.e. 4D dot product is positive) 10

Normalize quaternions or not? • Any quaternions can be written as scaling of unit quaternions 2 , 𝑟 −1 = 𝑠 −1 cos 𝜄 𝑟 = 𝑠 cos 𝜄 2 + 𝜕 sin 𝜄 2 − 𝜕 sin 𝜄 2 • In the rotation formula, the scaling part is cancelled 𝑦 𝑟 −1 = 𝑠 cos 𝜄 𝑦 𝑠 −1 cos 𝜄 2 + 𝜕 sin 𝜄 2 − 𝜕 sin 𝜄 = cos 𝜄 2 + 𝜕 sin 𝜄 𝑦 cos 𝜄 2 − 𝜕 sin 𝜄 𝑟 Ԧ Ԧ Ԧ 2 2 2 2 𝑟 2  so, normalization isn’t needed ? • In practice, don’t use quaternion mults for computing coordinate transformation (because inefficient) • Just do explicit vector calc using axis & angle 𝑟 1 +𝑟 2 𝑦 − 𝜕 ⋅ Ԧ Ԧ 𝑦 𝜕 cos 𝜄 + 𝜕 × Ԧ 𝑦 sin 𝜄 + 𝜕 ⋅ Ԧ 𝑦 𝜕 2 𝑠 = 1 • Can get axis & angle only after normalization • Un-normalized can cause artifact when interpolated 𝑟 1 11

Modeling curves 12

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 = σ 𝑗 𝑏 𝑗 𝑢 𝑗 13

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 14

Bezier curves • Input: 3 control points (CPs) 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 15

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 16

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 17

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 18