11/6/ :55 Graphics II Introduction to Parametric Curves and Surfaces Session 2

21/6/ :55 Computer Graphics Conceptual Model Application Model Application Program Graphics System Output Devices Input Devices API Function Calls or Protocol Data

31/6/ :55

4 Gouraud Shading: Use Mean Normal at Each Vertex

51/6/ :55 The Utah Teapot: 32 Bezier Patches

61/6/ :55 Explict Representation of Curve: Two Dimensions Dependent Variable Independent Variable General case: Neither variable is a single-valued function of the other. Example: a circle centered at the origin

71/6/ :55 Explicit Representation of a Curve: Three Dimensions c y x z Plane x = c

81/6/ :55 Implicit Representation Two Dimensions: Should be thought of as a “membership” or “testing” function. Divides space into points on the curve and not on the curve. Circle: Describes a curve Line :

91/6/ :55 Implicit Representation: Three Dimensions Defines a surface in three dimensions. Example: a sphere of radius r at origin There is no easy way to represent a curve implicitly in three dimensions. Algebraic surfaces are those in which f is polynomial in x, y, z. Quadric surfaces are algebraic surfaces where the polynomial is of degree at most 2.

101/6/ :55 Problems with Implicit Representation 0 Difficult to evaluate for rendering because identifying points on the curve requires explicit solution for (x,y). 0 Limited variety of curves that can be obtained.

111/6/ :55 Parametric Representation of Curves and Surfaces Express each spatial variable for points on the curve as a function of a non-spatial independent variable. Parametric surfaces require two parameters: Where u is defined over some closed range, e.g. [0, 1]

121/6/ :55 What does a Parametric Representation Really Mean?

131/6/ :55 Advantages of Parametric Representation 0 Solves problem of choice of independent variable 0 Easy computation of derivatives 0 Provides mechanism for “tracing” a curve or surface 0 Facilitates joining of multiple curves, surfaces 0 Generates a rich variety of curves, surfaces 0 Ease of rendering

141/6/ :55 Parametric Polynomial Curves Defines points on a parametric curve. A polynomial parametric curve of degree n is defined:

151/6/ :55 Parametric Polynomial Surfaces u=1 v=1 v=0 u=0 y z x

161/6/ :55 Continuity Considerations C 0 continuity: = C 1 continuity: G 1 continuity:

171/6/ :55 What degree do we want? High order polynomials provide more control over shape of curve. Degree n provides 3(n+1) degrees of freedom, in the choice of c k. The higher the order of a polynomial, the less “smooth” it is. A polynomial of degre n can change directions n-1 times. Consider polynomial of order 5.

181/6/ :55 Parametric Cubic Polynomial Curves Matrix Notation

191/6/ :55 Parametric Curves: How Do I Control the Shape?

201/6/ :55 Interpolation: Four control points on curve This constraint can be expressed by the equations:

211/6/ :55 Computing the coefficients Expressing the constraints in matrix form: Where:

221/6/ :55 Computing the coefficients

231/6/ :55 Blending Functions Substituting fromgives: which can be written as: whereis a column vector of blending functions.

241/6/ :55 Blending Functions

251/6/ :55 Expressing a point on the curve in terms of the blending polynomials Control

261/6/ :55 What the blending functions look like:

271/6/ :55 0 Contains scalars and polynomials (vectors) 0 Addition, zero polynomial defined 0 Can express a basis, e.g. 0 Representation of a polynomial in terms of this basis can be expressed as a column vector of scalar coefficients. Let’s look at polynomials as a vector space.

281/6/ :55 Conversion of representation from one basis to another Representation: Basis:

291/6/ :55 Conversion of representation from one basis to another Representation : Basis : The central issue in parametric curves (and surfaces) is the selection of an appropriate basis for the control point representation.

301/6/ :55 How do we extend this to 3 dimensions? Interpolating Patch u = constant v = constant

311/6/ :55 Bicubic interpolating surface patch 16 coefficients allow the interpolation of 16 points.

321/6/ :55 Solving for Coefficients Substituting the values at 16 points: Gives the equation:Where as before

331/6/ :55 Solving for Coefficients Note: separability of blending polynomial in u, v.

341/6/ :55 Lines of constant u and v are Interpolating curves u = constant v = constant