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CS 354 Computer Graphics; April 5, 2012; University of Texas at Austin

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- 1. CS 354Bézier CurvesMark KilgardUniversity of TexasApril 5, 2012
- 2. CS 354 2 Today’s material In-class quiz On procedural methods lecture Lecture topic Project 2 Bézier curves
- 3. CS 354 3 My Office Hours Tuesday, before class Painter (PAI) 5.35 8:45 a.m. to 9:15 Thursday, after class ACE 6.302, just for today, in ENS basement 11:00 a.m. to 12 Randy’s office hours Monday & Wednesday 11 a.m. to 12:00 Painter (PAI) 5.33
- 4. CS 354 4 Last time, this time Last lecture, we discussed Project 2 on Programmable Shaders Procedural Methods L-Systems Particle systems Perlin Noise This lecture Project 2 discussion Bézier curves Project 2 due is due Friday
- 5. CS 354 5 On a sheet of paper Daily Quiz • Write your EID, name, and date • Write #1, #2, #3, followed by its answer Multiple choice: A True or False: Perlin’s stochastic L-system noise function sums up multiple versions of a) repeats the same rule turbulence. forever List three forces that b) varies the rules a particle system randomly could model c) uses biology to make mountains
- 6. CS 354 6 Your Mission So Far You should now have these first two shaders tasks implemented Task 1: apply decal Task 0: roll torus
- 7. CS 354 7 Procedurally Generating a Torus from a 2D Grid 2D grid over (s,t)∈[0,1] Tessellated torus
- 8. CS 354 8 GLSL Standard Library Routines You’ll Need for Project 2 texture2D—accesses a 2D texture through a sampler2D and a 2-component texture coordinate set (s,t) textureCube—access a cube map with a samplerCube and a 3-component texture coordinate set (s,t,r) normalize—normalizes a vector cross—computes a cross product of 2 vectors dot—computes a dot (inner) product of 2 vectors max—compute the maximum of two values reflect—compute a reflection vector given an incident vector and a normal vector vec3—constructor for 3-component vector from scalars mat3—constructor for 3x3 matrix from column vectors *—matrix-by-vector or vector-by-matrix multiplication sin—sine trigonometric function cos—cosine trigonometric function pow—raise a number to a power, exponentiation (hint: specular)
- 9. CS 354 9 Normal Maps Visualized texas_- longhorn2 normal map construction bumps_in normal map construction Hint: dump your normal map computeNormal calls in NormalMap::load stbi_write_tga(buffer, width, height, 3, normal_image);
- 10. CS 354 10 Other Normal Maps mosaic geforce_etch stripes texas_longhorn bumps_out brick geforce_cell
- 11. CS 354 11 Coordinate Spaces for Project 2 Parametric space 2D space [0..1]x[0..1] for 2D patch Object space Transform the patch’s parametric space into a 3D space containing a torus Has modeling transformation from object- to world-space World space Environment map is oriented in this space gluLookAt’s coordinates are in this space Surface space (0,0,1) is always surface normal direction Mapping from object space to surface space varies along torus Perturbed normal from normal map overrides (0,0,1) geometric normal Eye space gluLookAt transforms world space to eye space
- 12. CS 354 12 Making Curves Spline weights used to create curve without computers
- 13. CS 354 13 Moving Between Two Points Given 2 or more points, how can we move between them? Easy answer: in a straight line Linear interpolation p(t) = p0 + t (p1-p0) Jagged! Can we make something smoother?
- 14. CS 354 14 Types of curves Variety of curve formulations Interpolating Hermite Bézier B-spline Explore their characteristics 14
- 15. CS 354 15 Matrix-Vector Form of Cubic 3 p(u ) = ∑ c k u k k =0 c 0 1 u define c= c1 u = 2 c 2 u 3 c3 u then p(u ) = u c = c u T T 15
- 16. CS 354 16 Interpolating Curve p1 p3 p0 p2 Given four data (control) points p0 , p1 ,p2 , p3 determine cubic p(u) which passes through them Must find c0 ,c1 ,c2 , c3 16
- 17. CS 354 17 Interpolation Equations apply the interpolating conditions at u=0, 1/3, 2/3, 1 p0=p(0)=c0 p1=p(1/3)=c0+(1/3)c1+(1/3)2c2+(1/3)3c2 p2=p(2/3)=c0+(2/3)c1+(2/3)2c2+(2/3)3c2 p3=p(1)=c0+c1+c2+c2 or in matrix form with p = [p0 p1 p2 p3]T 1 0 0 0 1 1 2 1 3 1 3 3 3 p=Ac A= 2 2 2 2 3 1 3 3 3 1 1 1 1 17
- 18. CS 354 18 Interpolation Matrix Solving for c we find the interpolation matrix 1 0 0 0 − 5.5 9 − 4.5 1 M I = A = 9 − 22.5 18 − 4.5 −1 − 4.5 13.5 − 13.5 4.5 c=MIp Note that MI does not depend on input data and can be used for each segment in x, y, and z 18
- 19. CS 354 19 Interpolating Multiple Segments use p = [p0 p1 p2 p3] T use p = [p3 p4 p5 p6]T Get continuity at join points but not continuity of derivatives 19
- 20. CS 354 20 Blending Functions Rewriting the equation for p(u) p(u)=uTc=uTMIp = b(u)Tp where b(u) = [b0(u) b1(u) b2(u) b3(u)]T is an array of blending polynomials such that p(u) = b0(u)p0+ b1(u)p1+ b2(u)p2+ b3(u)p3 b0(u) = -4.5(u-1/3)(u-2/3)(u-1) b1(u) = 13.5u (u-2/3)(u-1) b2(u) = -13.5u (u-1/3)(u-1) b3(u) = 4.5u (u-1/3)(u-2/3) 20
- 21. CS 354 21 Blending Functions These functions are not smooth Hence the interpolation polynomial is not smooth 21
- 22. CS 354 22 Interpolating Patch 3 3 p(u , v) = ∑ ∑ cij i u vj i =o j =0 Need 16 conditions to determine the 16 coefficients cij Choose at u,v = 0, 1/3, 2/3, 1 22
- 23. CS 354 23 Matrix Form Define v = [1 v v2 v3]T C = [cij] P = [pij] p(u,v) = uTCv If we observe that for constant u (v), we obtain interpolating curve in v (u), we can show C=MIPMI p(u,v) = uTMIPMITv 23
- 24. CS 354 24 Blending Patches 3 3 p(u , v) = ∑ ∑ b (u ) b i j (v ) pij i =o j =0 Each bi(u)bj(v) is a blending patch Shows that we can build and analyze surfaces from our knowledge of curves 24
- 25. CS 354 25 Other Types of Curves and Surfaces How can we get around the limitations of the interpolating form Lack of smoothness Discontinuous derivatives at join points We have four conditions (for cubics) that we can apply to each segment Use them other than for interpolation Need only come close to the data 25
- 26. CS 354 26 Hermite Form p’(0) p’(1) p(0) p(1) Use two interpolating conditions and two derivative conditions per segment Ensures continuity and first derivative continuity between segments 26
- 27. CS 354 27 Equations Interpolating conditions are the same at ends p(0) = p0 = c0 p(1) = p3 = c0+c1+c2+c3 Differentiating we find p’(u) = c1+2uc2+3u2c3 Evaluating at end points p’(0) = p’0 = c1 p’(1) = p’3 = c1+2c2+3c3 27
- 28. CS 354 28 Matrix Form p 0 1 0 0 0 p 1 1 1 1 q = 3 = c p0 0 1 0 0 p3 0 1 2 3 Solving, we find c=MHq where MH is the Hermite matrix 1 0 0 0 0 0 1 0 M = H − 3 3 − 2 − 1 2 −2 1 1 28
- 29. CS 354 29 Blending Polynomials p(u) = b(u)Tq 2 u 3 − 3 u 2 + 1 − 2 u3 + 3 u 2 b(u ) = 3 u − 2 u2 + u u −u 3 2 Although these functions are smooth, the Hermite form is not used directly in Computer Graphics and CAD because we usually have control points but not derivatives However, the Hermite form is the basis of the Bézier form 29
- 30. CS 354 30 Parametric and Geometric Continuity We can require the derivatives of x, y, and z to each be continuous at join points (parametric continuity) Alternately, we can only require that the tangents of the resulting curve be continuous (geometry continuity) The latter gives more flexibility as we have need satisfy only two conditions rather than three at each join point 30
- 31. CS 354 31 Example Here the p and q have the same tangents at the ends of the segment but different derivatives Generate different Hermite curves This techniques is used in drawing applications 31
- 32. CS 354 32 Higher Dimensional Approximations The techniques for both interpolating and Hermite curves can be used with higher dimensional parametric polynomials For interpolating form, the resulting matrix becomes increasingly more ill-conditioned and the resulting curves less smooth and more prone to numerical errors In both cases, there is more math operations in rendering the resulting polynomial curves and surfaces 32
- 33. CS 354 33 Pierre Bézier French engineer at Renault Popularized Bézier curves and surfaces For computer-aided design Winner: ACM Steven Anson Coons Award for Outstanding Creative Contributions to Computer Graphics 2nd winner, after 1st winner Ivan Sutherland of SketchPad fame
- 34. CS 354 34 Bézier’s Idea In graphics and CAD, we do not usually have derivative data Bézier suggested using the 4 data points as with the cubic interpolating curve to approximate the derivatives in the Hermite form 34
- 35. CS 354 35 Approximating Derivatives p1 p2 p1 − p0 p3 − p 2 p (0) ≈ p (1) ≈ 1/ 3 1/ 3 slope p’(0) slope p’(1) p0 p3 u 35
- 36. CS 354 36 Equations Interpolating conditions are the same p(0) = p0 = c0 p(1) = p3 = c0+c1+c2+c3 Approximating derivative conditions p’(0) = 3(p1- p0) = c0 p’(1) = 3(p3- p2) = c1+2c2+3c3 Solve four linear equations for c=MBp 36
- 37. CS 354 37 Bézier Matrix 1 0 0 0 − 3 3 0 0 MB = 3 − 6 3 0 −1 3 − 3 1 p(u) = uTMBp = b(u)Tp blending functions 37
- 38. CS 354 38 Blending Functions (1− u)3 2 b(u) = 3u (1− u) 3 u2 (1− u) u 3 Note that all zeros are at 0 and 1 which forces the functions to be smooth over (0,1) 38
- 39. CS 354 39 Bernstein Polynomials The blending functions are a special case of the Bernstein polynomials d! d −k bkd (u ) = u (1 − u ) k k!(d − k )! These polynomials give the blending polynomials for any degree Bézier form All zeros at 0 and 1 For any degree they all sum to 1 They are all between 0 and 1 inside (0,1) 39
- 40. CS 354 40 Convex Hull Property The properties of the Bernstein polynomials ensure that all Bézier curves lie in the convex hull of their control points Hence, even though we do not interpolate all the data, we cannot be too far away p1 p2 convex hull Bézier curve p0 p3 40
- 41. CS 354 41 Bézier Patches Using same data array P=[pij] as with interpolating form 3 3 p (u , v) = ∑∑ bi (u ) b j (v) pij = uT M B P MT v B i =0 j =0 Patch lies in convex hull 41
- 42. CS 354 42 Analysis Although the Bézier form is much better than the interpolating form, we have the derivatives are not continuous at join points Can we do better? Go to higher order Bézier More work Derivative continuity still only approximate Apply different conditions Tricky without letting order increase 42
- 43. CS 354 43 Evaluating Polynomials Simplest method to render a polynomial curve is to evaluate the polynomial at many points and form an approximating polyline For surfaces we can form an approximating mesh of triangles or quadrilaterals Use Horner’s method to evaluate polynomials p(u)=c0+u(c1+u(c2+uc3)) 3 multiplications/evaluation for cubic 43
- 44. CS 354 44 Finite Differences For equally spaced {uk} we define finite differences ∆p k=( k () 0 ( ) p ) u u ∆(k=(k1 pk p ) p +− u () 1 u u ) ( ) (+ ∆ pk ∆ (k1 ∆ (k m u = p +− p ) 1 ()) u ) u () m () m For a polynomial of degree n, the nth finite difference is constant 44
- 45. CS 354 45 Building a Finite Difference Table p(u)=1+3u+2u2+u3 45
- 46. CS 354 46 deCasteljau Recursion We can use the convex hull property of Bézier curves to obtain an efficient recursive method that does not require any function evaluations Uses only the values at the control points Based on the idea that “any polynomial and any part of a polynomial is a Bézier polynomial for properly chosen control data” 46
- 47. CS 354 47 Splitting a Cubic Bézier p0, p1 , p2 , p3 determine a cubic Bézier polynomial and its convex hull Consider left half l(u) and right half r(u) 47
- 48. CS 354 48 l(u) and r(u) Since l(u) and r(u) are Bézier curves, we should be able to find two sets of control points {l0, l1, l2, l3} and {r0, r1, r2, r3} that determine them 48
- 49. CS 354 49 Convex Hulls {l0, l1, l2, l3} and {r0, r1, r2, r3}each have a convex hull that that is closer to p(u) than the convex hull of {p0, p1, p2, p3} This is known as the variation diminishing property. The polyline from l0 to l3 (= r0) to r3 is an approximation to p(u). Repeating recursively we get better approximations. 49
- 50. CS 354 50 Equations Start with Bézier equations p(u)=uTMBp l(u) must interpolate p(0) and p(1/2) l(0) = l0 = p0 l(1) = l3 = p(1/2) = 1/8( p0 +3 p1 +3 p2 + p3 ) Matching slopes, taking into account that l(u) and r(u) only go over half the distance as p(u) l’(0) = 3(l1 - l0) = p’(0) = 3/2(p1 - p0 ) l’(1) = 3(l3 – l2) = p’(1/2) = 3/8(- p0 - p1+ p2 + p3) Symmetric equations hold for r(u) 50
- 51. CS 354 51 Efficient Form l0 = p0 r3 = p3 l1 = ½(p0 + p1) r1 = ½(p2 + p3) l2 = ½(l1 + ½( p1 + p2)) r1 = ½(r2 + ½( p1 + p2)) l3 = r0 = ½(l2 + r1) Requires only shifts and adds! 51
- 52. CS 354 52 Every Polynomial is a Bézier Curve We can render a given polynomial using the recursive method if we find control points for its representation as a Bézier curve Suppose that p(u) is given as an interpolating curve with control points q p(u)=uTMIq There exist Bézier control points p such that p(u)=uTMBp Equating and solving, we find p=MB-1MI 52
- 53. CS 354 53 Conversion Matrix 1 0 0 0 5 3 1 − 6 3 − 2 3 MB MI = 1 −1 Interpolating to Bézier 3 5 − 3 − 3 2 6 0 0 0 1
- 54. CS 354 54 Example These two curves were all generated from the same original data using Bézier recursion by converting all control point data to Bézier control points Bézier Interpolating 54
- 55. CS 354 55 Surfaces Can apply the recursive method to surfaces if we recall that for a Bézier patch curves of constant u (or v) are Bézier curves in u (or v) First subdivide in u Process creates new points Some of the original points are discarded original and discarded original and kept new 55
- 56. CS 354 56 Second Subdivision 16 final points for 1 of 4 patches created 56
- 57. CS 354 57 Normals For rendering we need the normals if we want to shade Can compute from parametric equations ∂p(u , v) ∂p(u , v) n= × ∂u ∂v Can use vertices of corner points to determine OpenGL can compute automatically 57
- 58. CS 354 58 Utah Teapot Most famous data set in computer graphics Widely available as a list of 306 3D vertices and the indices that define 32 Bézier patches 58
- 59. CS 354 59 Quadrics Any quadric can be written as the quadratic form pTAp+bTp+c=0 where p=[x, y, z]T with A, b and c giving the coefficients Render by ray casting Intersect with parametric ray p(α)=p0+αd that passes through a pixel Yields a scalar quadratic equation No solution: ray misses quadric One solution: ray tangent to quadric Two solutions: entry and exit points 59
- 60. CS 354 60 Next Class Next lecture Vector graphics and path rendering Resolution independent 2D graphics Project 3 to be assigned Animate a virtual person using motion capture data Reading Procedural methods: Chapter 9, 465-499 Curves: Chapter 10, 503-522 Remember Project 2 Shading and lighting with GLSL Due Friday, April 6th
- 61. CS 354 61 Thanks • E. Angel and D. Shreiner: Interactive Computer Graphics 6E © Addison-Wesley 2012

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