Lecture_Chapter3_b

Ecs 175 chapter 3 object representation 51 bzier

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ints i=0 cubic Bernstein basis polynomials ￿￿ ni bin (u) = u (1 − u)n−i i ECS 175 Chapter 3: Object Representation 50 Bézier Curves •  Convex hull property ￿￿ ni bin (u) = u (1 − u)n−i i n ￿ bin (u) = 1 partition of unity i=0 0 ≤ bin (u) ≤ 1 Bézier polynomial is convex combination of control points; stays Within convex hull of control polygon. ECS 175 Chapter 3: Object Representation 51 Bézier Curves •  Subdivision – example: cubic Bézier curve −1 pk (u) = (1 − u) · pk−1 + u · pk−1 i i i f ( u ) = p3 ( u ) 3 cubic Bézier curve (p0 , p1 , p2 , p3 ) 0123 (p3 , p3 , p3 , p3 ) 3210 ECS 175 Chapter 3: Object Representation Cubic curve 1 Cubic curve 2 52 Bézier Curves •  Derivatives f ( u) = n ￿ i=0 bin (u) · pi Derivatives are tangents. ECS 175 Chapter 3: Object Representation Lighting computations require derivatives. Example: Utah Teapot consists of bicubic Bézier patches 53 Bézier Curves •  Derivatives f ( u) = n ￿ i=0 bin (u) · pi n n ￿ dbin (u) df (u) d￿ = bin (u) · pi = · pi du du i=0 du i=0 dbin (u) = n(bi−1,n−1 (u) − bi,n−1 (u)) du n df (u) ￿ = n(bi−1,n−1 (u) − bi,n−1 (u)) · pi du i=0 ECS 175 Chapter 3: Object Representation 54 Bézier Curves •  Derivatives n df (u) ￿ = n(bi−1,n−1 (u) − bi,n−1 (u)) · pi du i=0 n− 1 df (u) ￿ = bi,n−1 (u) · n(pi+1 − pi ) du i=0 Derivative is a Bézier curve of order (n-1) with “combined” control points ECS 175 Chapter 3: Object Representation 55 Bézier Curves •  Joining Bézier curves Discontinuous C0 f continuous f’ discontinuous C1 f continuous f’ continuous ECS 175 Chapter 3: Object Representation 56...
View Full Document

Ask a homework question - tutors are online