2D transformations Homogeneous coordinates. Uses of Transformations

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1 2D transformations omogeneous coordinates Uses of Transformations Modeling: position and resize parts of a complex model; Viewing: define and position te virtual camera Animation: define ow objects move/cange wit time

2 Transformations Examples of transformations: translation rotation scaling sear Transformations More examples: reflection wit respect to te -axis reflection wit respect to te origin 2

3 Transformations Linear transformations: take straigt lines to straigt lines. All of te examples are linear. Affine transformations: take paralllel lines to parallel lines. All of te examples are affine, an example of linear non-affine is perspective projection. Ortogonal transformations: preserve distances, move all objects as rigid bodies. rotation, translation and reflections are affine. Transformations and matrices An affine transformation can be written as x a = a 2 a a 2 22 x b + b 2 p =Ap Images of basis vectors under affine transformations: e x e = 0 0 = (column form of writing vectors) a Ae x = a 2 a a 2 22 a = 0 a 2 a Ae = a

4 Transformations and matrices Matrices of some transformations: 0 cosα sinα sear sinα cos α s 0 0 s rotation scale b factor s 0 reflection wit respect to te origin 0 0 reflection wit respect to 0 -axis Problem Even for affine transformations we cannot write tem as a single 2x2 matrix; we need an additional vector for translations. We cannot write all linear transformations even in te form Ax +b were A is a 2x2 matrix and b is a 2d vector. Example: perspective projection [x,] [x, ] x = = /x x= equations not linear! 4

5 omogeneous coordinates replace 2d points wit 3d points, last coordinate for a 3d point (x,,w) te corresponding 2d point is (x/w,/w) if w is not zero eac 2d point (x,) corresponds to a line in 3d; all points on tis line can be written as [kx,k,k] for some k. (x,,0) does not correspond to a 2d point, corresponds to a direction (will discuss later) Geometric construction: 3d points are mapped to 2d points b projection to te plane z = from te origin omogeneous coordinates z z= [x,] line corresponding to [x,] From omogeneous to 2d: [x,,w] becomes [x/w,/w] From 2d to omogeneous: [x,] becomes [kx,k,k] (can pick an nonzero k!) x 5

6 omogeneous Transformations An linear transformation can be written in matrix form in omogeneous coordinates. Example : translations z z= [x+t x,+t ] [x,] in om. coords is [x,,] [x,] x x = x+t x =x+ t x = +t = +t w = x = t x x t omogeneous Transformations Example 2: perspective projection z z= x = = /x w = [x,] Can multipl all tree components b te same number -- te 2D point won t cange! Multipl b x. x = x = [x, ] w = x line x= x x = x 0 0 6

7 Matrices of basic transformations FRV ƒ VLQ VLQ FRV rotation V [ V \ scaling W [ W \ translation V skew D D D D D D general affine transform omogeneous line equation Implicit line equation in 2D: (n, q-p) = 0, n = 2D vector, p = 2D point on te line. Goal: rewrite in omogeneous coordinates. z= N z n p x 2D point corresponds to a 3D line troug origin; 2D line corresponds to a plane troug origin In oter words, te 2D line is intersection of a plane troug origin wit te plane z=. 7

8 omogeneous line equation Rewrite te line equation: n,q p) = n x + n + (n, p) = ([n,n (n,p)],[x,, ]) z z= N ( x x, = n x (N, q) were N=[n x,n,-(n,p)] is te normal to te plane corresponding to te line, and q is te omogeneous form of q=[x,]: q=[x,,] p omogeneous form of te line equation: (N,q) = 0 Intersection of two lines z z= N Let omogeneous equations of te two lines be (N,q)=0, and (N 2,q)=0. Te two lines are intersections of planes wit normals N and N 2 wit te plane z=. Intersection of lines = intersection point of te common line of te two planes wit te plane z=. N 2 x Te direction of te common line is te cross product: N N 2 Tis is also (one of possible) omogeneous forms of te intersection point! 8

9 ow to intersect two lines Step : Convert equations to omogeneous form. Step 2: Compute te vector product N N 2 Step 3: Convert N N 2 from omogeneous coordinates to 2D, dividing b te last component (ma be not!) Line troug two points Goal: given two points in omogeneous coordinates, find te omogeneous equation of te line troug tese points, tat is, te vector N in te equation (N,q) = 0 omogeneous form of a point = 3D vector from te origin to te point. Line troug 2 points = p intersection of te plane z spanned b two lines wit te plane z= z= p 2 Normal to tat plane (same as N in te line equation): N x N = p p = [p,p, ] [p,p, ] x x 9

10 Intersection of two lines intersection of two lines troug two pairs of points all given in omogeneous form. Appl formulas for p 4 line troug two points p 2 and for intersection of p 3 lines: p q i q i = (p p 2 ) (p 3 p 4 ) Mapping quads to quads A common operation for texture mapping: find a transformation mapping a rectangular image to a quadrilateral. Goal: solve a somewat more general problem: find a linear transform mapping one quad (A,B,C,D) to anoter (a,b,c,d). Assume all points given in omogeneous coordinates. Idea: to define a 3x3 linear transform (=matrix) M, it is sufficient to define its action on tree vectors. Coose suitable points for wic we know wat te images sould be. 0

11 Transform from 3 points Suppose for known vectors, 2, 3 we know tat te images sould be, 2, 3 : M =, M 2 = 2, M 3 = 3. Combine tree vector equations into one matrix equation: M x z 2x 2 2z 3x 3 3z = x z 2x 2 2z 3x 3 3z matrix matrix solving M=, we get M = - Mapping quads to quads Coosing te 3 points: it turns out, intersections of te sides and diagonals are convenient: We can compute all points using te formula for intersecting 2 lines troug two pairs of points.

12 Intersections of sides and diagonals 2 2 D C 3 a d 3 c A B b 2 3 = (B A) (C D) = (A D) (B C) = (A C) (B D) Similar for, 2, 3 To find te matrix M, build matrices and, and compute -. Mapping rectangles to quads Problem: map a square to an arbitrar quadrilateral wit vertices a,b,c,d using a linear map M - M d c - a b 2

13 Points at infinit Idea: look at images of omogeneous points 2 =[,0,0], =[0,,0], 3 =[[0,0,] In tis case te matrix is identit, and M =. Wat are points wit last component zero? [,0,0] [,0,0] [0,0,] intersection of orizontal sides (point at infinit for direction [,0] intersection of vertical sides (point at infinit for direction [0,] intersection of diagonals Mapping rectangles to quads Using formulas for intersections of lines, we get 2 3 = (b a) (c d) = (a d) (b c) = (a c) (b d) Finall, we obtain te transform: 0 0 x 2x 3x M = = 0 0 M z 2z 3z Note te last line is not [0,0,], tis is not an affine transform! 3

14 Texture Mapping Put image on a quadrilateral (can be tougt of as rectangle viewed in perspective) Map image to te square [-,]x[-,] Recall scan conversion: we traverse pixels along scan lines intersecting te quadrilateral; for a pixel p=[i,j] need to determine te pixel [m,n] in te image. to determine color, need te inverse of M : first, find te point q in te square wic is mapped to p b M, second, find te pixel in te image tat maps to q. Texture Mapping m n - (t,r,s) - M M - d a c pixel [i,j] b convert [i,j] to omogeneous form: [i,j,] = p appl M - : [t,r,s] = M - p; we get a point in te square convert back to 2D coordinates: [t/s,r/s] rescale and round to get pixel coordinates: m = (int)( widt*(t/s+)/2.0); n = (int)( eigt*(r/s+)/2.0); 4

15 Putting it all togeter Given a an image of size widt X eigt and four corners of a quadrilateral a, b,c,d in omogeneous coordinates compute te matrix M using intersections of sides, 2 and intersection of diagonals 3 compute te inverse of M scan convert te quadrilateral; for eac pixel to be painted, determine te pixel of te image corresponding to it; use its color to paint te pixel Implementation Representing matrices in C: I recommend against using 2d arras (two man pitfalls) use a d arra wit 9 elements to represent 3 b 3 matrix, and expressions of te tpe m[3*i+j] to address element m[i,j]; note tat indices start from 0. Write a small matrix/vector librar, implementing operations like add two vectors, multipl a vector b a number, multipl a matrix and a vector, dot product, vector product. 5

16 Inverse matrix Code for inverse: D = m[0]*m[4]*m[8] - m[0]*m[5]*m[7] - m[3]*m[]*m[8] + m[3]*m[2]*m[7] + m[6]*m[]*m[5] - m[6]*m[2]*m[4]; minv[0] = (m[4]*m[8]-m[5]*m[7])/d; minv[] = -(m[]*m[8]-m[2]*m[7])/d; minv[2] = (m[]*m[5]-m[2]*m[4])/d; minv[3] = -(m[3]*m[8]-m[5]*m[6])/d; minv[4] = (m[0]*m[8]-m[2]*m[6])/d; minv[5] = -(m[0]*m[5]-m[2]*m[3])/d; minv[6] = (m[3]*m[7]-m[4]*m[6])/d; minv[7] = -(m[0]*m[7]-m[]*m[6])/d; minv[8] = (m[0]*m[4]-m[]*m[3])/d; 6