SOLUTIONS MANUAL FOR 3d GRAPHICS FOR GAME PROGRAMMING EDITION 1 HAN

3 Chapter Compute a 2D affine transform matrix that rotates a 2D point by θ about a point (a, b), not about the origin 2D rotation in the homogeneous coordinates is defined as follows:   cosθ −sinθ  sinθ cosθ  0 It is always about the origin Therefore, rotating a point (x, y) about (a, b) is implemented by concatenating three transforms: (1) translating (x, y) by (−a, −b), (2) rotating the translated point about the origin, and (3) back-translating the rotated point by (a, b) The matrices are concatenated as follows:     −a cosθ −sinθ 10a M =  b   sinθ cosθ   −b  00  0 0 cosθ −sinθ −acosθ + bsinθ + a =  sinθ cosθ −asinθ − bcosθ + b  0 In linear algebra, it is proven that the inverse of a rotation is simply its transpose Given two non-standard orthonormal bases {a,b,c} and {d,e,f }, compute a 3×3 matrix that converts the vector defined in terms of {a,b,c} into the vector of {d,e,f } The vector defined in terms of {a,b,c} is first transformed into the vector defined in terms of the standard basis {e1 ,e2 ,e3 }, and then transformed into the vector defined in terms of {d,e,f } The first transform is implemented by the following basis-change matrix:   ax bx cx  ay by cy  az bz cz The second transform is implemented by the following basis-change matrix:   dx dy dz  ex ey ez  fx fy fz When they are  dx  ex fx combined, we obtain the following:    dy dz ax bx cx a·d b·d ey ez   ay by cy  =  a · e b · e fy fz az bz cz a·f b·f  c·d c·e c·f Let us define a matrix for scaling along orthonormal vectors, a, b, and c, which are not identical to the standard basis vectors e1 , e2 , and e3 The scaling factors are sa , sb , and sc along a, b, and c, respectively It is also observed that a × b = c where × denotes the cross product The scaling matrix is a combination of three 3×3 matrices Compute the three matrices Scaling a vector (x, y, z) along the non-standard orthonormal basis {a,b,c} is implemented by concatenating three transforms: (1) basis change from {e1 ,e2 ,e3 } to {a,b,c}, (2) scaling, and (3) basis change from {a,b,c} to {e1 ,e2 ,e3 } The matrices are concatenated as follows:     ax bx cx ax ay az sx 0  ay by cy   sy   bx by bz  0 sz cx cy cz az bz cz The view transform consists of a translation and √ a rotation We are given the following view parameters: EYE = (0, 0, − 3), AT = (0,0,0), and UP = (0,1,0) (a) Compute the translation  0  0 0 0 0   √0  3 (b) Compute the rotation EYE is located at −z axis, and looks at AT that is the origin Therefore, n is (0,0,−1) UP is (0,1,0), and is already orthogonal to n Applying the right-hand rule to UP and n finds that u is (−1, 0, 0) Applying the right-hand rule to n and u finds that v is (0, 1, 0) The rotation matrix is then as follows:   −1 0  0    0 −1  0 Section 2.4.3 derives the projection matrix assuming that the z -range of the cuboidal view volume is [−1,0] In OpenGL, the z -range is [−1,1] Derive the projection matrix for OpenGL presented in Equation (2.25) In OpenGL, the xy-ranges are the same as those in this book, [−1,1] Therefore Equation (2.33) is valid, and we have the following equation: z = −m3 − m4 z The difference is that the z -coordinates −f and −n are mapped to −1 and (not 0), respectively, by the OpenGL projection transform In other words, we have two known pairs of (z,z ): (−f ,−1) and (−n,1) Putting them into the above equation, we obtain the following: −1 = −m3 + = −m3 + m4 f m4 n Solving the above equations for m3 and m4 gives m3 = m4 = f +n f −n 2nf f −n Finally, both m3 and m4 are negated to be transformed to the lefthanded clip space Then, Equation (2.25) is obtained Chapter A viewport’s corners are located at (10,20,1) and (100,200,2) (a) Compute the reflection matrix for the viewport transform   −1  0 0  00 0  0 01 (b) Compute the scaling matrix The width of the viewport is 90, the height is 180, and the depth is Therefore, we have the following scaling matrix:   45 0  90 0     0 0 0 01 (c) Compute the translation matrix The center of the viewport’s front face is (55,110,1) Therefore, we have the following translation matrix:   0 55  110    0 1  000 FIGURE 2.1 The vertex processing and fragment processing stages (in rounded boxes) are programmable, and the rasterization and output merging stages (in rectangles) are hard-wired FIGURE 2.2 Transforms and spaces in the vertex processing stage Sections 2.1, 2.2, and 2.4 present the three transforms in order FIGURE 2.3 A vector p is rotated by θ to define a new vector p′ FIGURE 2.4 The sphere and teapot are defined in their own object spaces and are assembled into a single space, the world space FIGURE 2.5 The teapot is rotated about the y-axis by 90° and is then translated along the x-axis by seven units The teapot’s mouth is rotated from (0,2,3) to (3,2,0) and then translated to (10,2,0) The combined matrix of the rotation and the translation instantly transforms (0,2,3) to (10,2,0) FIGURE 2.6 The teapot shown in the middle is rotated CCW to define the one on the left If rotated CW, we have the result on the right FIGURE 2.6 The teapot shown in the middle is rotated CCW to define the one on the left If rotated CW, we have the result on the right FIGURE 2.7 The teapot is successively rotated using the Euler angles and acquires an orientation FIGURE 2.8 Vertex transform vs normal transform (modified from [8]) (a) The vertices and normal are transformed by a single matrix M The transformed normal is not orthogonal to the transformed triangle (b) Whereas the vertices are transformed by M, the normal is transformed by (M−1)T After the transforms, the normal remains orthogonal to the triangle FIGURE 2.9 Before transforms, n is orthogonal to the triangle 〈 p,q,r〉 Whereas 〈 p,q,r〉 is transformed by M, n is transformed by (M−1)T Then, the transformed normal n′ remains orthogonal to the transformed triangle 〈 p′,q′,r′〉 FIGURE 2.10 Given EYE, AT, and UP, the camera space is defined Its origin is EYE, and the basis is {u,v,n} In terms of the camera space, the camera is located at the origin and points in the −n direction, i.e., the view direction is −n FIGURE 2.11 The camera space {EYE, u,v,n} is superimposed onto the world space {O,e1,e2,e3} It is done by a translation followed by a rotation The two transforms are combined to produce the view transform, which converts the world-space objects into the camera space FIGURE 2.12 The first is the standard basis for R2 The second is a valid basis for R2, but is neither standard nor orthonormal The third is not the standard but an orthonormal basis FIGURE 2.13 The vertex position and normal are denoted by p and n, respectively Once the light source and camera positions are defined, l, r, and v are obtained Then, lighting can be computed per vertex FIGURE 2.14 The pyramid-like volume is named view frustum The polygons outside the view frustum (illustrated in red) are considered invisible FIGURE 2.15 If a polygon intersects the view frustum’s boundary, the part of the polygon outside the view frustum is discarded FIGURE 2.16 Projection transform (a) The view frustum is deformed into a cuboid The deformation named projection transform is in fact applied to the objects in the scene Observe how the teapot is deformed (b) Cross-section views show how the perspective projection effect (called foreshortening) is achieved through projection transform FIGURE 2.17 The z-coordinates are negated for switching from the right-handed clip space to the left-handed clip space Z-negation is equivalent to the z-axis inversion FIGURE 2.18 The last transform in the vertex processing stage, Mproj, converts the right-handed camera-space objects into the left-handed clip space This illustrates the combination of Fig 2.16-(a) and Fig 2.17 FIGURE 2.19 Computing projection matrix (a) Normalized coordinate y′ is computed (b) The aspect ratio can be defined in terms of fovx and fovy FIGURE 2.20 The projection transform converts the z-range [−f,−n] to [−1,0] ... Chapter A viewport’s corners are located at (10 ,20 ,1) and (10 0,200,2) (a) Compute the reflection matrix for the viewport transform   ? ?1  0 0  00 0  0 01 (b) Compute the scaling matrix The width... is the origin Therefore, n is (0,0,? ?1) UP is (0 ,1, 0), and is already orthogonal to n Applying the right-hand rule to UP and n finds that u is (? ?1, 0, 0) Applying the right-hand rule to n and... OpenGL, the z -range is [? ?1, 1] Derive the projection matrix for OpenGL presented in Equation (2.25) In OpenGL, the xy-ranges are the same as those in this book, [? ?1, 1] Therefore Equation (2.33) is