Lecture_Chapter3_c

Lecture_Chapter3_c - Review of Last Thursday Texture...

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Review of Last Thursday 57 ECS 175 Chapter 3: Object Representation Texture mapping; Aliasing caused by sampling Triangles are suitable input, but not optimal for modeling Explicit, implicit, and parametric functions y = f ( x ) f ( x, y )=0 f ( u )=( x ( u ) ,y ( u ) ,z ( u ))
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Review of last Thursday 58 ECS 175 Chapter 3: Object Representation Parametric curves: Bézier curves b in ( u )= ° n i ± u i (1 u ) n i n ° i =0 b in ( u )=1 0 b in ( u ) 1 f ( u n ° i =0 b in ( u ) p i convex combination convex hull property
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Review of last Thursday 59 ECS 175 Chapter 3: Object Representation Subdivision – example: cubic Bézier curve p k i ( u )=(1 u ) · p k 1 i 1 + u · p k 1 i ( p 0 0 ,p 1 1 2 2 3 3 ) ( p 3 3 3 2 3 1 3 0 ) Cubic curve 1 Cubic curve 2
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Review of last Thursday 60 ECS 175 Chapter 3: Object Representation Derivatives f ( u )= n ° i =0 b in ( u ) · p i Derivatives are tangents. Lighting computations require derivatives. Example: Utah Teapot consists of bicubic Bézier patches
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Review of last Thursday 61 ECS 175 Chapter 3: Object Representation Derivatives df ( u ) du = n ° i =0 n ( b i 1 ,n 1 ( u ) b i,n 1 ( u )) · p i ( u ) du = n 1 ° i =0 b i,n 1 ( u ) · n ( p i +1 p i ) Derivative is a Bézier curve of order (n-1) with “combined” control points
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Review of last Thursday 62 ECS 175 Chapter 3: Object Representation Joining Bézier curves Discontinuous C 0 continuity: f continuous f’ discontinuous C 1 continuity: f continuous f’ continuous
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Bézier Curves 63 ECS 175 Chapter 3: Object Representation Interpolation with Bézier curves Choose parameters for “curve points” and solve for control points Example: four points = cubic Bézier curve f (0 . 33) = f 1 f (0 . 67) = f 2 f (1) = f 3 p 3 = f 3 f (0) = f 0 p 0 = f 0
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Bézier Curves 64 ECS 175 Chapter 3: Object Representation Interpolation with Bézier curves f (0 . 33) = f 1 f (0 . 67) = f 2 f (1) = f 3 p 3 = f 3 f (0) = f 0 p 0 = f 0 f 0 f 1 f 2 f 3 = b 03 (0) b 13 (0) b 23 (0) b 33 (0) b 03 (0 . 33) b 13 (0 . 33) b 23 (0 . 33) b 33 (0 . 33) b 03 (0 . 67) b 13 (0 . 67) b 23 (0 . 67) b 33 (0 . 67) b 03 (1) b 13 (1) b 23 (1) b 33 (1) p 0 p 1 p 2 p 3 p 0 p 1 p 2 p 3 =
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Lecture_Chapter3_c - Review of Last Thursday Texture...

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