At one point in the animation the view direction is w

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Unformatted text preview: .9 9A 1 0 0 0 1 which gives a single matrix: 0 0.9 0 0 B 0 0.9 0 B @0 0 0.9 0 0 0 0.1n 1 1 1C C 1A 1 Marks: d 4 The scene is to be viewed from a moving viewpoint specified by its position C and a left handed viewing coordinate system { u, v, w }. T B 0 0.9 0 B @0 0 0.9 0 0 0 Transformations" 0.1n 1C C 1A 1 Marks: d The scene is to be viewed from a moving viewpoint specified by its position C and a left handed viewing coordinate system { u, v, w }. At one point in the animation the view direction is w = (1, 0, 0) T , and the viewpoint is given by C = (20, 5, 10). Given that the view is in the horizontal plane (v = (0, 1, 0) T ) find the value of u. The reference coordinate system is assumed to be left-handed. The viewing coordinate system, { u, v, w }, is also left-handed so we have the cross-product identities: u⇥v = w w⇥u = v v⇥w = u 4 and a left handed viewing coordinate system { u, v, w }. At one point in the animation the view direction is w = (1, 0, 0) T , and the viewpoint is given by C = (20, 5, 10). Given that the view is in the horizontal plane (v = (0, 1, 0) T ) find the value of u. Transformations" The reference coordinate system is assumed to be left-handed. The viewing coordinate system, { u, v, w }, is also left-handed so we have the cross-product Department of Computing Examinations – 2011 - 2012 Session Confidential identities: u⇥v = w MODEL ANSWERS and MARKING SCHEME w⇥u = v First Examiner: Paul Aljabar Second Examiner: Daniel Rueckert v⇥w = u Paper: C317 - Graphics Question: 1 Page 4 of 16 The third identity can be used to find the answer: 01 01 0 1 0 1 0 u = @1A ⇥ @0A = @ 0A 0 0 1 Marks: e Hence, or otherwise, find the viewing transformation matrix. The viewing matrix has the form 4 u= 1 0 ⇥ 0 0 = 0 1 Marks: Transformations" e Hence, or otherwise, find the viewing transformation matrix. The viewing matrix has the form 0 u x uy uz B v x vy vz B @ w x wy wz 000 1 C·u C · vC C C · wA 1 u, v and w are known from the previous question. C·u = 0 10 1 20 0 @ 5A · @ 0A = 10 10 1 0 1 01 20 0 @ A @A e Hence, or otherwise, find the viewing transformation matrix. The viewing matrix has the form 0 u x uy uz B v x vy vz B @ w x wy wz 0 00 1 C·u C · vC C C · wA 1 u, v and w are known from the previous question. 0 10 1 20 0 C · u = @ 5A · @ 0A = 10 10 1 0 1 01 20 0 C · v = @ 5A · @1A = 5 10 0 0 10 1 20 1 C · w = @ 5A · @ 0A = 20 10 0 Putting these together gives the viewing matrix as 0 1 00 1 10 B01 0 5C B C @10 0 20A 00 01 Marks: 4...
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This document was uploaded on 03/26/2014.

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