Computer Graphics (CS 543) Lecture 3 (Part 1): Linear Algebra for - PowerPoint PPT Presentation

Computer Graphics (CS 543) Lecture 3 (Part 1): Linear Algebra for Graphics (Points, Scalars, Vectors) Prof Emmanuel Agu Computer Science Dept. Worcester Polytechnic Institute (WPI)

Points, Scalars and Vectors  Points, vectors defined relative to a coordinate system  Point: Location in coordinate system  Example: Point (5,4)  Cannot add or scale points y 5 (5,4) 4 (0,0) x

Vectors  Magnitude  Direction  NO position  Can be added, scaled, rotated Length  CG vectors: 2, 3 or 4 dimensions Angle

Vector-Point Relationship  Subtract 2 points = vector v = Q – P P v  point + vector = point Q P + v = Q

Vector Operations  Define vectors Then vector addition:  a ( , ) a a a 1 , 2 3  b     a ( , )  a b a b a b b ( , ) b b b 1 1 , 2 2 3 3 1 , 2 3 a a+b b

Vector Operations  Define scalar, s Note vector subtraction:  Scaling vector by a scalar  b s  a a ( , , ) a s a s a s 1 2 3        ( ( ), ( ), ( )) a b a b a b 1 1 2 2 3 3 a a-b a 2.5a b

Vector Operations: Examples  Scaling vector by a scalar • Vector addition: s   b     a ( , , ) a ( , ) a s a s a s a b a b a b 1 2 3 1 1 , 2 2 3 3  For example, if a =(2,5,6) and b = (-2,7,1) and s =6, then       a b ( , ) ( 0 , 12 , 7 ) a b a b a b 1 1 , 2 2 3 3   a ( , , ) ( 12 , 30 , 36 ) s a s a s a s 1 2 3

Affine Combination  Given a vector  a ( , ,..., ) a a a a 1 , 2 3 n    ......... 1 a a a 1 2 n  Affine combination: Sum of all components = 1  Convex affine = affine + no negative component i.e   , ,......... a a a non negative 1 2 n

Magnitude of a Vector  Magnitude of a    2 2 2 | a | .......... a a a 1 2 n  Normalizing a vector (unit vector) a vector  a  ˆ a magnitude  Note magnitude of normalized vector = 1. i.e    2 2 2 .......... 1 a a a 1 2 n

Magnitude of a Vector  Example: if a = (2, 5, 6)     2 2 2  Magnitude of a | a | 2 5 6 65  Normalizing a   2 5 6  ˆ   a , ,   65 65 65

Convex Hull  Smallest convex object containing P 1 , P 2 ,…..P n  Formed by “shrink wrapping” points 11

Dot Product (Scalar product)  Dot product,         a b ........ d a b a b a b 1 1 2 2 3 3  For example, if a =(2,3,1) and b= (0,4,-1) then  b        ( 2 0 ) ( 3 4 ) ( 1 1 ) a     0 12 1 11

Properties of Dot Products  Symmetry (or commutative):    a b b a  Linearity:  )      ( a c b a b c b  Homogeneity:    ( a ) b ( a b ) s s   2  And b b b

Angle Between Two Vectors y   b    c b cos , b sin b b   c    c cos , c sin  c c b     b c b c cos c  b x b b b Sign of b.c : c c c b.c > 0 b.c < 0 b.c = 0

Angle Between Two Vectors  Problem: Find angle b/w vectors b = (3,4) and c = (5,2)  Step 1: Find magnitudes of vectors b and c     2 2 | b | 3 4 25 5    2 2 | c | 5 2 29  Step 2: Normalize vectors b and c     3 4 5 2 ˆ   ˆ     c , b ,     5 5 29 29

Angle Between Two Vectors ˆ  b ˆ c  Step 3: Find angle as dot product     3 4 5 2 ˆ      ˆ   b c , ,     5 5 29 29 15 8 23 ˆ  c     ˆ b 0 . 85422 5 29 5 29 5 29  Step 4: Find angle as inverse cosine     cos( 0 . 85422 ) 31 . 326

Standard Unit Vectors y Define    i 1 , 0 , 0 i k    j 0 , 1 , 0   0 j  k 0 , 0 , 1 z x So that any vector,       v , , i j k a b c a b c

Cross Product (Vector product) If       a , , a a a b , , b b b x y z x y z Then        a b ( ) i ( ) j ( ) k a b a b a b a b a b a b y z z y x z z x x y y x Remember using determinant i j k a a a x y z b b b x y z Note: a x b is perpendicular to a and b

Cross Product Note: a x b is perpendicular to both a and b a x b 0 a b

Cross Product (Vector product) Calculate a x b if a = (3,0,2) and b = (4,1,8)       b 4 , 1 , 8 a 3 , 0 , 2 Using determinant i j k 3 0 2 4 1 8 Then        a b ( 0 2 ) i ( 24 8 ) j ( 3 0 ) k     2 i 16 j 3 k

Normal for Triangle using Cross Product Method n p 2 n · ( p - p 0 ) = 0 plane n = ( p 2 - p 0 ) × ( p 1 - p 0 ) p p 1 normalize n  n/ |n| p 0 Note that right-hand rule determines outward face

Newell Method for Normal Vectors  Problems with cross product method:  calculation difficult by hand, tedious  If 2 vectors almost parallel, cross product is small  Numerical inaccuracy may result p 1 p 2 p 0  Proposed by Martin Newell at Utah (teapot guy )  Uses formulae, suitable for computer  Compute during mesh generation  Robust!

Newell Method Example  Example: Find normal of polygon with vertices P0 = (6,1,4), P1=(7,0,9) and P2 = (1,1,2)  Using simple cross product: ((7,0,9)-(6,1,4)) X ((1,1,2)-(6,1,4)) = (2,-23,-5) P1 - P0 P2 P2 - P0 (1,1,2) PO (6,1,4) P1 (7,0,9)

Newell Method for Normal Vectors  Formulae: Normal N = (mx, my, mz)     1 N     m y y z z ( ) ( ) x i next i i next i  i 0     1 N     m z z x x ( ) ( ) y i next i i next i  0 i     1 N     m x x y y ( ) ( ) z i next i i next i  0 i

Newell Method for Normal Vectors  Calculate x component of normal     x y z 1 N     m y y z z ( ) ( ) x i next i i next i 6 1 4  P0 0 i P1 7 0 9     ( 1 )( 13 ) ( 1 )( 11 ) ( 0 )( 6 ) m x 1 1 2 P2    13 11 0 m x 6 1 4  P0 2 m x

Newell Method for Normal Vectors  Calculate y component of normal x y z     1 N     m z z x x y i next ( i ) i next ( i ) 6 1 4 P0  0 i P1 7 0 9      ( 5 )( 13 ) ( 7 )( 8 ) ( 2 )( 7 ) m y 1 1 2 P2     65 56 14 m y 6 1 4 P0   23 m y

Newell Method for Normal Vectors  Calculate z component of normal x y z     1 N     m x x y y 6 1 4 ( ) ( ) P0 z i next i i next i  0 i P1 7 0 9      ( 1 )( 1 ) ( 6 )( 1 ) ( 5 )( 2 ) m z 1 1 2 P2     1 6 10 m z 6 1 4   P0 5 m z Note: Using Newell method yields same result as Cross product method (2,-23,-5)

Finding Vector Reflected From a Surface a = original vector  n = normal vector  r = reflected vector  m = projection of a along n  e = projection of a orthogonal to n  n Note: Θ 1 = Θ 2 a r   e a m m -m   Θ 1 r e m Θ 2    r a 2 m e e

Forms of Equation of a Line  Two-dimensional forms of a line  Explicit: y = mx +h  Implicit: ax + by +c =0 α P 1  Parametric: x( a ) = a x 0 + (1- a )x 1 1 - α P α y( a ) = a y 0 + (1- a )y 1 P o  Parametric form of line  More robust and general than other forms  Extends to curves and surfaces

Convexity  An object is convex iff for any two points in the object all points on the line segment between these points are also in the object P P Q Q not convex convex

References  Angel and Shreiner, Interactive Computer Graphics, 6 th edition, Chapter 3  Hill and Kelley, Computer Graphics using OpenGL, 3 rd edition, Sections 4.2 - 4.4