Uses of Transformations. 2D transformations Homogeneous coordinates. Transformations. Transformations. Transformations. Transformations and matrices

Transcription

1 Uses of Transformations 2D transformations Homogeneous coordinates odeling: position and resie parts of a comple model; Viewing: define and position the virtual camera Animation: define how objects move/change with time Transformations Eamples of transformations: Transformations ore eamples: reflection with respect to the -ais translation rotation scaling shear reflection with respect to the origin Transformations Transformations and matrices Linear transformations: take straight lines to straight lines. All of the eamples are linear. Affine transformations: take paralllel lines to parallel lines. All of the eamples are affine, an eample of linear non-affine is perspective projection. Orthogonal transformations: preserve distances, move all objects as rigid bodies. rotation, translation and reflections are affine. An affine transformation can be written as a a2 b + p Ap a2 a22 b2 Images of basis vectors under affine transformations: e e 0 0 (column form of writing vectors) a a2 a a2 Ae a2 a22 0 a Ae 2 a 22

2 Transformations and matrices Problem atrices of some transformations: 0 cos α sinα 0 reflection with respect to the origin 0 0 reflection with respect to 0 -ais shear s 0 sinα cos α 0 s rotation scale b factor s Even for affine transformations we cannot write them as a single 22 matri; we need an additional vector for translations. We cannot write all linear transformations even in the form A +b where A is a 22 matri and b is a 2d vector. Eample: perspective projection [,] [, ] / equations not linear! Homogeneous coordinates Homogeneous coordinates replace 2d points with 3d points, last coordinate for a 3d point (,,w) the corresponding 2d point is (/w,/w) if w is not ero each 2d point (,) corresponds to a line in 3d; all points on this line can be written as [k,k,k] for some k. (,,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 the plane from the origin [,] line corresponding to [,] From homogeneous to 2d: [,,w] becomes [/w,/w] From 2d to homogeneous: [,] becomes [k,k,k] (can pick an nonero k!) Homogeneous Transformations Homogeneous Transformations An linear transformation can be written in matri form in homogeneous coordinates. Eample : translations [+t,+t ] [,] in hom. coords is [,,] [,] +t + t +t +t w t t 0 Eample 2: perspective projection / w [,] Can multipl all three components b the same number -- the 2D point won t change! ultipl b. [, ] w line

3 atrices of basic transformations FRV ƒ VLQ VLQ FRV V [ V \ D D D D D D rotation scaling W[ W\ V general affine transform translation skew Homogeneous line equation Implicit line equation in 2D: (n, q-p) 0, n 2D vector, p 2D point on the line. Goal: rewrite in homogeneous coordinates. n p 2D point corresponds to a 3D line through origin; 2D line corresponds to a plane through origin In other words, the 2D line is intersection of a plane through origin with the plane. Homogeneous line equation Rewrite the line equation: n,q p) n + n + (n, p) ([n,n (n,p)],[,, ]) (, (, q) where [n,n,-(n,p)] is the normal to the plane corresponding to the line, and q is the homogeneous form of q[,]: q[,,] Homogeneous form of the line equation: n (,q) 0 p Intersection of two lines Let homogeneous equations of the two lines be (,q)0, and ( 2,q)0. The two lines are intersections of planes with normals and 2 with the plane. Intersection of lines intersection point of the common line of the two planes with the plane. 2 The direction of the common line is the cross product: 2 This is also (one of possible) homogeneous forms of the intersection point! How to intersect two lines Line through two points Step : Convert equations to homogeneous form. Step 2: Compute the vector product Step 3: Convert 2 from homogeneous coordinates to 2D, dividing b the last component (ma be not!) 2 Goal: given two points in homogeneous coordinates, find the homogeneous equation of the line through these points, that is, the vector in the equation (,q) 0 Homogeneous form of a point 3D vector from the origin to the point. Line through 2 points p intersection of the plane spanned b two lines with the plane p 2 ormal to that plane (same as in the line equation): p p [p,p, ] [p,p, ] 3

4 Intersection of two lines intersection of two lines through two pairs of points all given in homogeneous form. Appl formulas for p 4 line through two points p 2 and for intersection of p 3 lines: p q i (p p 2 ) (p 3 p 4 ) q i apping quads to quads A common operation for teture mapping: find a transformation mapping a rectangular image to a quadrilateral. Goal: solve a somewhat more general problem: find a linear transform mapping one quad (A,B,C,D) to another (a,b,c,d). Assume all points given in homogeneous coordinates. Idea: to define a 33 linear transform (matri), it is sufficient to define its action on three vectors. Choose suitable points for which we know what the images should be. Transform from 3 points apping quads to quads Suppose for known vectors H, H 2, H 3 we know that the images should be h, h 2, h 3 : H h, H 2 h 2, H 3 h 3. Combine three vector equations into one matri equation: H H2 H3 h h2 h3 H H2 H3 h h2 h3 H H2 H3 h h2 h3 matri H matri h solving Hh, we get hh - Choosing the 3 points: it turns out, intersections of the sides and diagonals are convenient: H H 2 H 3 h We can compute all points using the formula for intersecting 2 lines through two pairs of points. h 2 h 3 Intersections of sides and diagonals apping rectangles to quads H H 2 H 3 H (B A) (C D) H (A D) (B C) 2 H (A C) (B D) 3 D A C B h a d h 2 h 3 c b Similar for h, h 2, h 3 To find the matri, build matrices H and h, and compute hh -. Problem: map a square to an arbitrar quadrilateral with vertices a,b,c,d using a linear map - - d a c b 4

5 Points at infinit apping rectangles to quads Idea: look at images of homogeneous points H 2 [,0,0], H [0,,0], H 3 [[0,0,] In this case the matri H is identit, and h. What are points with last component ero? [,0,0] [,0,0] [0,0,] intersection of horiontal sides (point at infinit for direction [,0] intersection of vertical sides (point at infinit for direction [0,] intersection of diagonals Using formulas for intersections of lines, we get h (b a) (c d) h2 (a d) (b c) h3 (a c) (b d) Finall, we obtain the transform: 0 0 h h2 h3 0 0 h h2 h3 0 0 h h2 h3 ote the last line is not [0,0,], this is not an affine transform! Teture apping Put image on a quadrilateral (can be thought of as rectangle viewed in perspective) ap image to the square [-,][-,] Recall scan conversion: we traverse piels along scan lines intersecting the quadrilateral; for a piel p[i,j] need to determine the piel [m,n] in the image. to determine color, need the inverse of : first, find the point q in the square which is mapped to p b, second, find the piel in the image that maps to q. Teture apping m n - (t,r,s) convert [i,j] to homogeneous form: [i,j,] p appl - : [t,r,s] - p; we get a point in the square convert back to 2D coordinates: [t/s,r/s] rescale and round to get piel coordinates: m (int)( width*(t/s+)/2.0); n (int)( height*(r/s+)/2.0); - - d a c piel [i,j] b Putting it all together Implementation Given a an image of sie width X height and four corners of a quadrilateral a, b,c,d in homogeneous coordinates compute the matri using intersections of sides h, h2 and intersection of diagonals h3 compute the inverse of scan convert the quadrilateral; for each piel to be painted, determine the piel of the image corresponding to it; use its color to paint the piel Representing matrices in C: I recommend against using 2d arras (two man pitfalls) use a d arra with 9 elements to represent 3 b 3 matri, and epressions of the tpe m[3*i+j] to address element m[i,j]; note that indices start from 0. Write a small matri/vector librar, implementing operations like add two vectors, multipl a vector b a number, multipl a matri and a vector, dot product, vector product. 5

6 Inverse matri 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