CMSC427 Geometry and Vectors Review: where are we? Parametric - PowerPoint PPT Presentation

CMSC427 Geometry and Vectors

Review: where are we? • Parametric curves and Hw1? • Going beyond the course: generative art • https://www.openprocessing.org • Brandon Morse, Art Dept, ART370 • Polylines, Processing and Lab0? • Questions on Processing? • https://processing.org • Learning Processing – Dan Shiffman

How can we … • Find a line perpendicular to another? • Find symmetric reflecting vector on a mirror? • Find the distance from a point to a line? • Figure out if a facet is facing the camera?

Polyline and points P3 P5 P1 P4 P2 P0 • Polyline as sequence of locations (points)

Polyline and vectors V3 V5 V1 V4 V2 P0 • Polyline as point plus sequence of vectors • Assumption: displacements << locations, fewer bits

Polyline as vectors V3 V5 V1 V4 V2 P0 P1 = P0+V1 P2 = P1+V1 = P0+V1+V2 • Polyline as point plus sequence of vectors • Assumption: displacements << locations, fewer bits

Vector and vector operations • Objects: • Points (x,y) – represent position • Vector <x,y> – represent displacement, direction • Operations: • Vector addition, subtraction, scaling • Vector magnitude • Vector dot product • Vector cross product • Linear, affine and convex combinations • Applications: • Representations: using vectors for lines, planes, others • Metrics: angle between lines, distances between objects • Tests: are two lines perpendicular, is a facet visible?

Vectors • Direction and distance • Describes a • Difference between points • Speed, translation, axes • Notation • In bold a P1 • Angle brackets a = <x,y> a =P1-P0 • Free vectors a P0 • No anchor point • Displacement, not location

Vector scaling Multiplication by scalar s a a 0.5 a -1 a = - a

Vector addition and subtraction a b -b b c = a - b a a c = a + b

Vector addition and subtraction What is c in terms of a and b ? c a b

Vector addition and subtraction What is c in terms of a and b ? a + c = b c a b

Vector addition and subtraction What is c in terms of a and b ? a + c = b c c = b - a a a b b

Coordinate vs. coordinate-free representation Coordinate-free equation c = b - a Valid for 2D and 3D Prefer when possible a b Coordinate equation a = <3,3> b = <4,2> c = b - a = <4,2> - <3,3> = <1,-1>

Parametric line in coordinate-free vector format 𝑦 = 𝑢 𝑒𝑦 + 𝑞𝑦 𝑄1 𝑧 = 𝑢 𝑒𝑧 + 𝑞𝑧 t = 1 t = 0.75 Set 𝑄0 t = 0.5 𝑾 = 𝑄1 − 𝑄0 t = 0.25 =< 𝑒𝑦, 𝑒𝑧 > t = 0 Then 𝑄(𝑢) = 𝑢 𝑾 + 𝑄0 Good in 2D and 3D

� Vector magnitude The magnitude (length) of a vector is v 4 = 𝑤 64 + 𝑤 74 𝑤 64 + 𝑤 74 v = A vector of length 1.0 is called a unit vector To normalize a vector is to rescale it to unit length n : = v v Normal vectors represent direction and are used for: light direction, surface normals, rotation axes

Dot product • Inner product between two vectors • Defined using coordinate-free cosine rule 𝐛 • b = a b cos (𝜄) a θ b

Dot product • Inner product between two vectors • Defined using coordinate-free cosine rule 𝐛 • b = a b cos (𝜄) a θ b • Example: 𝐛 =< 1,1 > 𝐜 =< 1,0 > a 𝜄 = π/2 θ=π/2 𝐛 • b = 1.4142 ∗ 1 ∗ 0.70711 = 1 b

Dot product – coordinate version • Given 𝐜 = 𝒄 𝒚 , 𝒄 𝒛 𝐛 = 𝒃 𝒚 , 𝒃 𝒛 • Then 𝐛 • b = 𝒃 𝒚 𝒄 𝒚 + 𝒃 𝒛 𝒄 𝒛 • In n dimensions 𝒐 𝐛 • b = O 𝒃 𝒋 𝒄 𝒋 𝒋R𝟐

Dot product – coordinate version • Given 𝐜 = 𝒄 𝒚 , 𝒄 𝒛 𝐛 = 𝒃 𝒚 , 𝒃 𝒛 • Then 𝐛 • b = 𝒃 𝒚 𝒄 𝒚 + 𝒃 𝒛 𝒄 𝒛 • Example revised: 𝐛 =< 1,1 > a 𝐜 =< 1,0 > θ=π/2 𝐛 • b = 1 ∗ 1 + 1 ∗ 0 = 1 b

Dot product: computing angle (and cosine) • Given 𝐛 • b = a b cos (𝜄) • We have cos 𝜄 = 𝐛 • b a b 𝐛 • b 𝜄 = cos T𝟐 a b • Example 𝐛 =< 1,1 >, 𝐜 =< 1,0 > 𝐛 • b = 1 ∗ 1 + 1 ∗ 0 = 1 a a = 𝟐. 𝟓𝟐𝟓𝟑, b = 𝟐. 𝟏 𝐛 • b 𝟐 θ=π/4 a b = 𝟐. 𝟓𝟐𝟓𝟑 = 𝟏. 𝟖𝟏𝟖𝟐𝟐 𝐛 • b cos T𝟐 b = 𝟏. 𝟖𝟗𝟔𝟓𝟏 = 𝝆/𝟓 a b

Dot product: testing perpendicularity • Since cos 90° = cos 𝜌 2 = 𝟏 • Then a perpendicular to b gives 𝐛 ⊥ b ⇒ 𝐛 • b = 0 • Examples 𝐛 =< 1,0 >, 𝐜 =< 0,1 > 𝐛 • b = 1 ∗ 0 + 1 ∗ 0 = 0 𝐛 =< 1,1 >, 𝐜 =< −1,1 > 𝐛 • b = 1 ∗ −1 + 1 ∗ 1 = 0

Perp vector and lines • Given vector 𝐰 =< 𝑤 6 , 𝑤 7 > • The perp vector is =< −𝑤 7 , 𝑤 6 > • So 𝐰•𝐰 b = 0

Midpoint bisector • Given line segment P0 to P1, what line is perpendicular through the midpoint? P1= (5,3) P0 = (1,1)

Midpoint bisector • Given line segment P0 to P1, what line is perpendicular through the midpoint? P1= (5,3) v = P1-P0 n = v_perp/|v_perp| m = (P0+P1)/2 P0 = (1,1)

Midpoint bisector • Result: bisector is 𝑄 = 𝑢 n + 𝑛 • With n normalized perp vector, m midpoint • Appropriate range of t? P1= (5,3) v = P1-P0 n = v_perp/|v_perp| m = (P0+P1)/2 P = t n + m P0 = (1,1)

Midpoint displacement terrain (2D recursive algorithm) • Given two points P0,P1 • Compute midpoint m • Compute bisector line P = tn + m • Pick random t, generate P’ • Call recursively on segments P0-P’, P’-P1 • Tuning needed on magnitude of displacement

Midpoint displacement terrain (3D recursive algorithm) P0 P1 • Start with four points P0,P1,P2,P3 • Divide (quads or triangles?) • Compute midpoint bisector (how?) • Displace, repeat P3 P2 Hunter Loftis

Cross product a × b • Read as “a cross b” • a x b is a vector perpendicular to both a and b , in the direction defined by the right hand rule

Cross product a × b • Read as “a cross b” • a x b is a vector perpendicular to both a and b , in the direction defined by the right hand rule • Vector a and b lie in the plane of the projection screen. • Does a x b p oint towards you or away from you? a What about b x a ? b

Cross product: normal to plane P0 P1 • How compute normal vector to triangle P0, P1, P3? • Assume up is out of page P3 P2

Cross product: normal to plane P0 P1 • How compute normal vector to triangle P0, P1, P3? • Assume up is out of page • One answer: n = (P1-P3) x (P0-P1) P3 P2

Cross product: length of a x b • Parallelogram rule 𝐛 × b = a b sin (𝜄) • Area of mesh triangle? • When would cross product be zero? 𝐛 × b = 0

Cross product: length of a x b • Parallelogram rule 𝐛 × b = a b sin (𝜄) • Area of mesh triangle? • When would cross product be zero? 𝐛 × b = 0 • Either a,b parallel, or either degenerate

Cross product: computing, vector approach 𝑏 7 𝑐 h − 𝑏 h 𝑐 7 𝐛 × b = 𝑏 h 𝑐 6 − 𝑏 6 𝑐 h 𝑏 6 𝑐 7 − 𝑏 7 𝑐 6

Cross product: computing, matrix approach 𝑗 𝑘 𝑙 • Determinant of 𝑏 6 𝑏 7 𝑏 h 𝐛 × b = 𝑐 6 𝑐 7 𝑐 h • Computed with 3 lower minors 𝑏 7 𝑏 h 𝑏 6 𝑏 7 𝑐 h − 𝑘 𝑏 6 𝑏 h 𝑐 h + 𝑙 𝐛 × b = 𝑗 𝑐 7 𝑐 6 𝑐 6 𝑐 7 𝑏 𝑐 • with 𝑒 = 𝑏𝑒 − 𝑑𝑐 𝑑 • i, j, k are unit vectors in directions of x, y and z axes • i=<1,0,0> j=<0,1,0> k=<0,0,1>

Midpoint of triangle? P1 ? P2 P0

Midpoint of triangle? Answer 1: P1 𝑛 = 𝑄0 + 𝑄1 + 𝑄2 3 ? P2 P0

Midpoint of triangle? Answer 2: Double interpolation (blending) P1 First interpolation: vector line P0 to P1 𝑄 = 𝑢𝑊 + 𝑄0 m0 = 𝑢(𝑄1 − 𝑄0) + 𝑄0 P2 = 𝑢𝑄1 + (1 − 𝑢)𝑄0 Midpoint at t = 0.5 P0 𝑛0 = 0.5𝑄1 + (1 − 0.5)𝑄0 = 0.5𝑄1 + 0.5𝑄0 = 𝑄1 + 𝑄0 2

Midpoint of triangle? Answer 2: Double interpolation P1 Second interpolation: vector line m0 to P2 𝑄 = 𝑡𝑊′ + 𝑛0 m0 = 𝑡(𝑄2 − 𝑛0) + 𝑛0 P2 m = 𝑡𝑄2 + 1 − 𝑡 𝑛0 Midpoint at s = 1/3 P0 𝑛0 = 1 3 𝑄2 + 1 − 1 = 1 3 𝑄2 + 2 3 𝑛0 3 (0.5𝑄1 + 0.5𝑄0) = 1 3 𝑄2 + 2 3 𝑛0 = 𝑄2 + 𝑄1 + 𝑄0 3

� � Generalizing: convex combinations of points A convex combination of a set of points S is a linear combination such that the non-negative coefficients sum to 1 with and O 𝛽 t = 1 𝐷 = O 𝛽 t 𝑄 t 0 ≤ 𝛽 t ≤ 1 u tv w

� � Generalizing: convex combinations of points A convex combination of a set of points S is a linear combination such that the non-negative coefficients sum to 1 O 𝛽 t = 1 𝐷 = O 𝛽 t 𝑄 t 0 ≤ 𝛽 t ≤ 1 with and u tv w Is this equation for a line segment a convex combination? 𝑄 = 𝑢𝑄1 + (1 − 𝑢)𝑄0