Ray tracing c a point in the square triangle is given

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Unformatted text preview: direction vector d. A right-angled triangle is given by three points P1 , P2 and P3 (the right angle is at point P1 ). Show in detail how you can calculate the intersection between the ray and the face defined by the triangle. Ray tracing:" c)! A point in the square triangle is given by P = w1 u 1 + w2 u 2 + P 1 where u1 = P2 P1 , u2 = P3 P1 and 0 w1 1 0 w2 1 Department of Computing Examinations –+ w 2013 Session 0 w 2012 - 1 1 2 Confidential MODEL ANSWERS and MARKING SCHEME We first determine the intersection of the ray with the plane defined by the First Examiner: Daniel Rueckertplane is given by Second Examiner: Abhijeet Ghosh triangle. The normal of the Paper: C317 - Graphics Question: 2 Page 5 of 12 n = ± ( u1 ⇥ u2 ) / ( | u1 ⇥ u2 | ) The equation of the ray can be written as V + ad. The position of the intersection q is given by P1 + P = V + a d P=V Since P · n = 0: P1 + a d The equation of the ray can be written as V + ad. The position of the intersection q is given by P1 + P = V + a d Ray tracing" P=V P1 + a d Since P · n = 0: 0 = (V a= (V P1 )n + adn P1 )n/(dn) Now, we need to test whether the intersection point P is inside the triangle. Since u1 and u2 are orthogonal, taking the dot product with u1 and u2 yields w1 = (Pu1 P1 u1 ) / ( u1 u1 ) w2 = (Pu2 P1 u2 ) / ( u2 u2 ) For a valid intersection the following conditions must be met: 0 w1 1 0 w2 1 0 w1 + w2 1 Marks: 6 0 w1 + w2 1 Marks: d 6 Ray tracing" a ray starts at V = (7, 7, 0) and has a direction vector In a concrete example, d = (0, 0, 1). The points of the triangle are given as P1 = (6, 6, 10), P2 = d)! 6, 10), and P3 = (6, 10, 5). Calculate whether the ray intersects the (12, face defined by the triangle. First, we calculate the normal of the plane defined by the triangle: u1 = (6, 0, 0) u2 = (0, 4, 5) n = (0, 30, 24)or (0, 0.78, 0.62) Calculating the intersection of the ray with the plane yields a= and ((7, 7, 0) (6, 6, 10))(0, 0.78, 0.62)/0.62 = (0.78 6.2)/0.62 = 8.74 Department of Computing Examinations – 2012 - 2013 Session Confidential MODEL ANSWERS and MARKING SCHEME First Examiner: Daniel Rueckert Second Examiner: Abhijeet Ghosh Ray tracing" Paper: C317 - Graphics Question: 2 Page 6 of 12 P = (7, 7, 8.74) Calculating the barycentric coordinates yields: w1 = ((7, 7, 8.74)(6, 0, 0) (6, 6, 10)(6, 0, 0))/36 = (42 w2 = ((7, 7, 8.74)(0, 4, 5) 36)/36 = 0.16 (6, 6, 10)(0, 4, 5))/41 = 0.25 0 w1 1 0 w2 1 0 w1 + w2 1 So the intersection point is inside the triangle. Marks: The four parts carry, respectively, 30%, 20%, 30%, and 20% of the marks. 4 Splines" 2 Splines A cubic spline patch is defined by the parametric equation: P ( µ ) = a3 µ3 + a2 µ2 + a1 µ + a0 where P(µ) is a point that traces the locus of the curve, with 0 < µ < 1 and a3 , a2 , a1 and a0 are vector constants that define the shape of the curve. a Given that the patch is to be drawn between two points Pi and Pi+1 and the 0 0 gradients at the ends are to be Pi and Pi+1 respectively, write down four 00 equations connecting Pi Pi+1 Pi Pi+1 and a3 a2 a1 a0 . b Solve the above equations to find the values in matrix A in the equation: 01...
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