Written Assignment #2: Appearances, Curves and Surfaces15-462 Graphics I, Fall 2003Doug JamesDue: Thursday, October 9, 2003 (before lecture)70 POINTSOctober 9, 2003• The work must be all your own.• The assignment is due before lecture on Thursday, October 9.• Be explicit, define your symbols, and explain your steps. This will make it a lot easier for us to assignpartial credit.• Use geometric intuition together with trigonometry and linear algebra.• Verify whether your answer is meaningful with a simple example.11 Angel, Chapter 6, Exercise 6.14Show that the halfway vector h is at the angle at which a surface must be oriented so that the maximumamount of reflected light reaches the viewer.If we orient the surface so it points in the direction of the halfway vector, then we have equal anglesbetween the halfway vector and the viewer, and between the halfway vector and the light source.This configuration is exactly that of a mirror which reflects all the light towards the viewer.2 Angel, Chapter 6, Exercise 6.22Generalize the shadow-generation algorithm (Section 5.10) to handle flat surfaces at arbitrary orientations.(NOTE: This is similar to your second programming assignment.)Probably the easiest approach to this problem is to transform the given plane to the plane ad-dressed for the simplified shadow algorithm, and also transform the light source and the objects inthe same way. After finding the projected shadow points, we can transform everything back.Notation: Let the quantities of interest be the light position pl= (xl, yl, zl, 1), the plane h =(n, d) = (a, b, c, d) s.t. hTp = 0, the model point to be projected p, and the projected point pp.First, let’s translate the problem so that the plane passes through the origin as in our simplealgorithm. If we translate in a direction normal to the plane, we must translate by−dnnTn.Denote this by the 4-by-4 translation matrix, T.Next, let’s rotate the plane so that the plane normal is ˆy. This is achieved by the axis-anglerotation with axis n × ˆy, and angle θ s.t. cos θ = ˆnTˆy. Denote this by the 4-by-4 rotation matrix, R.Therefore we have transformed the original problem to the simpler one with parameters:p0l= RTpl(Fictitious light point)p0= RTp (Fictitious model point)p0p= RTpp(Fictitious projected shadow point)Using the new light position p0lin our old shadow projection transformation, P0= P(p0l) (see p.263Angel), we have thatp0p= P0p0or(RTpp) = P0(RTp)so that the actual projected shadow point ispp= T−1RTP0RTp.Therefore the new shadow projection matrix isT−1RTP0RT.23 Angel, Chapter 7, Exercise 7.3How is an image produced with an environment map different from a ray-traced image of the same scene?The major problem is that the environment map is computed without the object in the scene.Thus, all global lighting calculations of which it should be a part are incorrect. These errors can bemost noticeable if there are other reflective objects, since these will not show the reflection of theremoved object. Other errors can be caused by the removed object no longer blocking light andby its shadows being missing. Other visual errors can be due to distortions in the mapping of theenvironment to a simple shape, such as a cube, and to errors in a two stage mapping.4 Angel, Chapter 10, Exercise 10.4Show that, as long as the four control points for the cubic interpolating curve are defined at unique values ofthe parameter u, the interpolating geometry matrix always exists.The matrix for interpolating points p0, p1, p2and p3isA =1 u0u20u301 u1u21u311 u2u22u321 u3u23u33wherep = Acand c are the polynomial coefficients. The interpolating geometry matrix is MI= A−1, andtherefore A must be invertible. The determinant of this A matrix has the formdet A = const(u0− u1)(u0− u2)(u0− u3)(u1− u2)(u1− u3)(u2− u3)because if any pair of the interpolating points are the same the matrix will have two identical rowsand a zero determinant. By this same reasoning, if the interpolating points are all distinct, then thedeterminant cannot be zero, and therefore the A matrix must have an inverse.35 Angel, Chapter 10, Exercise 10.6Verify the C2continuity of the cubic spline.46 Angel, Chapter 10, Exercise 10.16Find the zeros of the Hermite blending functions. Why do these zeros imply that the Hermite curve issmooth in the interval (0,1)?(Given that the blending functions are just cubic polynomials, and therefore C∞, the term “smooth-ness” really refers to how big the “wiggles” in the interpolated curve can be. However, if the blend-ing functions are guaranteed to be nonnegative, then their cubic interpolating linear superpositioncan undergo only very limited “wiggling.”)7 Angel, Chapter 10, Exercise 10.18For a 1024 × 1024 display screen, what is the maximum number of subdivisions that are needed to render acubic polynomial surface?Because 1024 = 210< 1280 < 2048 = 211, after at most 11 subdivisions, we are at the resolutionof less than a
View Full Document
Unlocking...