521493S Computer Graphics Exercise 2 Solution (Chapters 4-5)
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1 5493S Computer Graphics Exercise Solution (Chapters 4-5). Given two nonparallel, three-dimensional vectors u and v, how can we form an orthogonal coordinate system in which u is one of the basis vectors? Create basis vectors using u = [ 3 T and v = [ 3 4 T. The vector a = u v is orthogonal to u and v. The vector b = u a is orthogonal to u and a. Hence, u, a and b form an orthogonal coordinate system. Note that v may not be orthogonal to u and therefore can t be part of coordinate system. For the second part we need to actually calculate the cross product between two vectors: i j k a = u v = u u u 3 = i u u 3 v v v v v + j u 3 u 3 v 3 v + k u u v v 3 = (u v 3 u 3 v )i + (u 3 v u v 3 )j + (u v u v )k u v 3 u 3 v = [ u 3 v u v 3 u v u v With u = [ 3 T and v = [ 3 4 T, we can solve second basis vector a We can then solve b using u and a a = u v = [ 3 4 = [ 3 ( ) 3 8 b = u a = [ 3 ( ) ( ) = [ ( ) 4 We can check the result by making sure that the dot products u a = u b = a b = : u a = ( ) ( ) = = (ok!) u b = ( 8) + ( ) = = (ok!) a b = ( 8) + ( ) 4 = = (ok!). We can specify an affine transformation by considering the location of a small number of points both before and after these points have been transformed. In three-dimension (3D) cases, how many points must we consider to specify the transformation uniquely? How does the required number of points change when we work in two-dimensional (D) cases?
2 There are degrees of freedom in the three-dimensional affine transformation. m m m 3 m 4 m M = [ m 3 m m 3 m 3 m 33 m 4 m 34 Suppose we consider a point p = [x, y, z, T that is transformed to p = [x, y, z, T by the matrix M. Hence we have the relationship p = Mp where M has unknown coefficients but p and p are known. Thus we have 3 equations in unknowns (the fourth equation is simply the identity = ). x = m x + m y + m 3 z + m 4 y = m x + m y + m 3 z + m 4 z = m 3 x + m 3 y + m 33 z + m 34 = x + y + z + If we have 4 such pairs of points we will have equations in unknowns which could be solved for the elements of M. Thus if we know how a quadrilateral is transformed we can determine the affine transformation. In two dimensions, there are 6 degrees of freedom in M but p and p have only x and y components. m m m 3 M = [ m m m 3 Hence if we know 3 points both before and after transformation, we will have 6 equations in 6 unknowns and thus in two dimensions if we know how a triangle is transformed we can determine the affine transformation. 3. Start with a cube centered at the origin and aligned with the coordinate axes. Find a rotation matrix that will rotate the cube first 45 around y-axis and then 45 around z-axis. Transformation order is from right to left. So rotation around y-axis is on the right side of the rotation matrix. cos(φ) sin(φ) R y (φ) = [ sin(φ) cos(φ) cos (φ) sin (φ) R z (φ) sin (φ) cos (φ) = [
3 We are using 45 degree angles to rotate and cos(45 ) = sin(45 ) = R = R z (45 )R y (45 ) = [ [ = [ = [ 4. Not all projections are planar geometric projections. Give an example of a projection in which the projection surface is not a plane, and another in which the projections are not lines. Eclipses (both solar and lunar) are good examples of the projection of an object (the moon or the earth) on to a nonplanar surface. All the maps in an atlas are examples of the use of curved projectors. If the projectors were not curved we could not project the entire surface of a spherical object (the earth) onto a rectangle. 5. If we were to use orthogonal projections to draw the coordinate axes, the x and y axes would be like in the plane of the paper, but the z axis would point out of the page. Instead, we can draw the x and y axes as meeting at a 9-degree angle, which the z axis going off at -35 degrees from the x axis. Find the matrix that projects the original orthogonal-coordinate axes to this view. We are projecting point p = [x, y, z, T using projection matrix M to point p = [x, y, z, T. p = Mp Z-axis is projected to the same plane with x and y axes with a -35 degree angle from positive x axis: x = x + α cos( 35 ) z y = y + α sin( 35 ) z z = = Here α is the length of the vector as it has not been given. We can select α cos( 35 ) = α sin( 35 ) = to obtain a simple looking projection matrix. 3
4 M = [ 6. Find the projection of a point onto the plane ax + by + cz + d = from a light source located at infinity shining to the direction (d x, d y, d z ). What if the light source L = (L x, L y, L z ) is not at infinity? 6. For light source at infinity All the points on the projection of the point (x, y, z) in the direction (d x, d y, d z ) are of the form (x + αd x, y + αd y, z + αd z ). Thus the shadow of the point (x, y, z) is found by determining the α for which the line intersects the plane, that is Substituting and solving, we find ax s + by s + cz s + d = α = d ax by cz. However, what we want is a projection matrix. Using this value of α we find x s = x + αd x = (bd y + cd z )x + ( bd x )y + ( cd x )z dd x y s = y + αd y = ( ad y)x + (ad x + cd z )y + ( cd y )z dd y z s = z + αd z = ( ad z)x + ( bd z )y + (ad x + bd y )z dd z These results can be computed by multiplying the homogenous coordinate point (x, y, z, ) by the projection matrix. M = bd y + cd z bd x cd x dd x ad y ad x + cd z cd y dd y ad z bd z ad x + bd y dd z [ 6. For light source not at infinity In that case shadow vector d x, d y, d z is no longer constant and that will change the result a bit. 4
5 From our previous result, plane equation, the shadow ray directions and their length before intersection with the plane are still valid: ax + by + cz + d = x s = x + αd x y s = y + αd y α = z s = z + αd z d ax by cz But now shadow vector direction depends on both the coordinates of the light source and the point casting the shadow: d x = x L x d y = y L y d z = z L z Let s solve x-component of shadow s intersection point on the plane s surface x s : = x + x s = x + αd x d ax by cz d ad x + bd y + cd x z = x( ) + ( d ax by cz)d x Now let s substitute d x = x L x, d y = y L y and d z = z L z : = x (a(x L x) + b(y L y ) + c(z L z )) + ( d ax by cz)(x L x ) a(x L x ) + b(y L y ) + c(z L z ) = ax al x x + bxy bl y x + cxz cl z x dx + dl x ax + al x x bxy + bl x y cxz + cl x z ax al x + by bl y + cz cl z All x and cross-terms (xy, xz) cancel out as well as many others: ax al x x + bxy bl y x + cxz cl z x dx + dl x ax + al x x bxy + bl x y cxz + cl x z ax al x + by bl y + cz cl z Then we need to collect coefficients for x, y, z and constant terms like in original E.6. However, at this time we need to do that for denominator as well as it is no longer constant: = ( bl y cl z d)x + (bl x )y + (cl x )z + dl x (a)x + (b)y + (c)z + ( al x bl y cl z ) 5
6 Finishing y s and z s are left as an exercise to the reader. The final projection matrix will look something like the following: M = bl y cl z d bl x cl x dl x???????? [ a b c ( al x bl y cl z ) The denominator should end up being the same for all x s, y s and z s cases. In this case, denominator is not constant and it s no longer possible to just divide numerator (and therefore upper three rows of the matrix) with the constant at the bottom row of the matrix. This division must be handled by perspective division operation ( x w, y w, z w, w w )after matrix multiplications have been performed. This kind of matrix should be able to project points (and polygons) of objects onto a plane. That could be useful functionality for example in airplane simulators and other 3-D scenarios with a large flat surface at the bottom. There are better techniques for creating shadows with OpenGL in general, though. So what happens if the light source is between object and the plane? In that case, shadow rays will still intersect with the plane but distance from light source term α is going to be negative for that to be possible 7. Show that the following sequences commute: a. A rotation and a uniform scaling b. Two rotations about the same axis c. Two translations a. If the scaling matrix is uniform then RS = RS(α, α, α) = αr = SR. b. Consider R x (θ), if we multiply and use the standard trigonometric identities for the sine and cosine of the sum of two angles, we find c. By simply multiplying the matrices we find R x (θ)r x (φ) = R x (θ + φ). T(x, y, z )T(x, y, z ) = T(x + x, y + y, z + z ). 6
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