1. CS 376 Introduction to Computer Graphics 04 / 23 / 2007 Instructor: Michael Eckmann

2. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 Today’s Topics Questions? Piecewise Cubic Parametric curves –Hermite –Bezier –Spline

3. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 Curves Some motivations for having mathematical equations define curves and surfaces 1) polylines and polygons are 1 st degree piecewise linear approximations to curves and surfaces (they are approximations unless they are the exact description of the outline or shape) –if they are approximations, then it takes large numbers of points to describe them for more accurate representations == BAD –if interactively building these surfaces (or edges/outlines) then the more points, the more tedious it is to build We want to be able to get more accurate representations of curved surfaces (or edges/outlines) than polygons or polylines give us AND we want to be able to represent them with less points. This will benefit us by having less storage and be more easily built interactively.

4. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 Curves These smooth curves are also used in graphics for other purposes like describing a motion path for an animation sequence.

5. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 Curves Let's consider 2d dimensional curves which can be later generalized to 3d surfaces. Instead of using piecewise linear functions, we use piecewise smooth curves. These are of a higher degree than linear functions. We have several choices –a) we can use functions like y = f(x) problem that for each value of x there is one and only one y value –can't represent circles/ellipses etc. problem that you can't represent vertical lines and is problematic to represent curves with a tangent/slope of infinity –b) we can use implicit equations like f(x,y) = 0 (e.g. a circle of radius 1 like: x 2 + y 2 – 1 = 0) may have more solutions than we want --- e.g. to model half a circle you'd need constraints on the values of x or y or both. joining 2 implicitly defined curves is hard to make the tangents at the join point agree

6. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 Curves We have several choices –c) we can use parametric equations of the curve x = f(t) and y = g(t) –where f(t) and g(t) are typically cubic polynomials the problems of the other 2 choices are solved as we shall see –we typically use cubic polynomials because 1) quadratic or linear polynomials do not allow a curve segment to be described with 2 endpoints and specific slopes at the 2 endpoints (examples on the board) 2) higher degree (than 3) polynomials are more compute intensive and allow (possibly) unwanted artefacts in the curve A cubic polynomial to define a finite curve segment has 4 coefficients per parametric equation x(t) = a x t 3 + b x t 2 + c x t + d x y(t) = a y t 3 + b y t 2 + c y t + d y z(t) = a z t 3 + b z t 2 + c z t + d z where 0 <= t <= 1

7. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 Curves x(t) = a x t 3 + b x t 2 + c x t + d x y(t) = a y t 3 + b y t 2 + c y t + d y z(t) = a z t 3 + b z t 2 + c z t + d z where 0 <= t <= 1 can be written in matrix form as: Q(t) = [x(t) y(t) z(t)] = T C where T = [t 3 t 2 t 1] and [a x a y a z ] C = [b x b y b z ] [c x c y c z ] [d x d y d z ]

10. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 Curves Continuity - when joining two curve segments together (piecewise curves). There are several different kinds of continuity to consider Geometric continuity –G 0 geometric continuity = the curve segments are joined together –G 1 geometric continuity = the curve segments are joined together and the directions of tangent vectors are equal at the point where they are joined Parametric continuity –C 0 parametric continuity = same as G 0 geometric continuity –C 1 parametric continuity = the curve segments are joined together and the directions AND magnitudes of tangent vectors are equal at the point where they are joined –C k parametric continuity = the curve segments are joined together and the directions AND magnitudes of all the derivatives up to the k th are equal at the point where they are joined (that is, the 1 st through k th derivatives exist and are continuous themselves)

11. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 Curves Let me draw on the board some example joined curve segments and point out their continuity.

12. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 Curves There are several families of cubic polynomial curve segments that are determined by different things –Hermite determined by the 2 endpoints and the tangent vector at each of the 2 endpoints –Bezier determined by the 2 endpoints P 1 & P 4 and 2 other intermediate points P 2 & P 3 not on the curve –the tangent vectors at the end points are the vectors P 1 P 2 and P 3 P 4 –which are [P 2 – P 1 ] and [P 4 – P 3 ] –Spline determined by 4 specific points Note: there can be higher degree polynomials of these families of curves (see a few slides ahead for a 4 th degree polynomial Bezier curve segment)

13. Hermite Curve segment examples (on the board)

14. Bezier Curve segment examples

15. more Bezier Curve segment examples

16. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 Recall x(t) = a x t 3 + b x t 2 + c x t + d x y(t) = a y t 3 + b y t 2 + c y t + d y z(t) = a z t 3 + b z t 2 + c z t + d z where 0 <= t <= 1 can be written in matrix form as: Q(t) = [x(t) y(t) z(t)] = T C where T = [t 3 t 2 t 1] and [a x a y a z ] C = [b x b y b z ] [c x c y c z ] [d x d y d z ]

17. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 Basis Matrix and Geometry Matrix Q(t) = [x(t) y(t) z(t)] = T C where T = [t 3 t 2 t 1] and [a x a y a z ] C = [b x b y b z ] [c x c y c z ] [d x d y d z ] can be rewritten so that C = M G where –M is a 4x4 matrix called the Basis Matrix and –G is a 4x1 column vector called the Geometry Matrix so Q(t) = T M G

18. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 Recall Q(t) = [x(t) y(t) z(t)] = T M G where T = [t 3 t 2 t 1] and [m 11 m 12 m 13 m 14 ] M = [m 21 m 22 m 23 m 24 ] [m 31 m 32 m 33 m 34 ] [m 41 m 42 m 43 m 44 ] [G 1 ] G = [G 2 ] [G 3 ] [G 4 ] Note: the values of G i are the conditions that define the curve --- such as endpoints and tangent vectors at those end points

19. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 Blending functions look at x(t) x(t) = (t 3 m 11 + t 2 m 21 + t m 31 + m 41 )g 1x + (t 3 m 12 + t 2 m 22 + t m 32 + m 42 )g 2x + (t 3 m 13 + t 2 m 23 + t m 33 + m 43 )g 3x + (t 3 m 14 + t 2 m 24 + t m 34 + m 44 )g 4x where g 1x is the x coordinate of G 1 similar equations for y(t) and z(t) the curve is a weighted sum of the elements of the Geometry Matrix where the weights are cubic polynomials of t which are called the Blending Functions the Blending Functions B are given by B = T M so Q(t) = B G

20. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 Example with lines Recall the parametric equation of a line: x(t) = x 0 + t ( x end – x 0 ) y(t) = y 0 + t ( y end – y 0 ) z(t) = z 0 + t ( z end – z 0 ) what are the endpoints of the line segment described above? can be rewritten as: x(t) = a x t + b x y(t) = a y t + b y y(t) = a z t + b z

21. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 x(t) = a x t + b x y(t) = a y t + b y y(t) = a z t + b z To get this in the form of Q(t) = T C = T M G T = [ t 1 ] and C = [ a x a y a z ] [ b x b y b z ] C = M G where M = [ m 11 m 12 ] and G = [ g 1x g 1y g 1z ] [ m 21 m 22 ] [ g 2x g 2y g 2z ] Example with lines

22. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 The geometry matrix is simply the two endpoints G = [ G 1 ] = [ g 1x g 1y g 1z ] = [ x 0 y 0 z 0 ] = [ P 0 ] [ G 2 ] [ g 2x g 2y g 2z ] [ x end y end z end ] [ P 1 ] we need to find out the Basis Matrix M M = [ m 11 m 12 ] [ m 21 m 22 ] Q(t) = T M G = [ t 1 ] M G Let's do the rest on the board to figure out M Example with lines

23. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 Q(t) = T M G = [ t 1 ] M G Recall that T M are the blending functions. What then, are the blending functions of lines? Example with lines

24. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 Q(t) = T C = T M G = [ t 3 t 2 t 1 ] M G M is the 4x4 Basis matrix and G is the Geometry matrix For Hermite curve segments, recall that they are defined with 2 endpoints and the tangents at those endpoints. So, the Geometry matrix is made up of the 2 endpoints and the 2 tangent vectors. The tangent vectors are determined by the derivative of the curve with respect to t at each of the end points (when t=0 and when t=1). x(t) = a x t 3 + b x t 2 + c x t + d x y(t) = a y t 3 + b y t 2 + c y t + d y z(t) = a z t 3 + b z t 2 + c z t + d z where 0 <= t <= 1 x'(t) = 3a x t 2 + 2b x t + c x y'(t) = 3a y t 2 + 2b y t + c y z'(t) = 3a z t 2 + 2b z t + c z Hermite Basis Matrix and Blending Functions

25. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 x'(t) = 3a x t 2 + 2b x t + c x y'(t) = 3a y t 2 + 2b y t + c y z'(t) = 3a z t 2 + 2b z t + c z So, Q'(t) = [ 3t 2 2t 1 0 ] C agreed? Let's call the 2 endpoints in the Geometry Matrix P 1 and P 4 and the 2 tangent vectors at those 2 endpoints R 1 and R 4. [ P 1 ] G = [ P 4 ] [ R 1 ] [ R 4 ] Hermite Basis Matrix and Blending Functions

26. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 Let's just work with the x coordinates (the y and z coordinates will be handled similarly) So, x(t) = a x t 3 + b x t 2 + c x t + d x = T C x = [ t 3 t 2 t 1 ] M G x and x'(t) = 3a x t 2 + 2b x t + c x = T C x = [ 3t 2 2t 1 0 ] M G x when t = 0, x(t) gets us the first endpoint and when t = 1, x(t) gets us the other endpoint. x(0) = P 1 = [ 0 0 0 1 ] M G x x(1) = P 4 = [ 1 1 1 1 ] M G x when t = 0, x'(t) gets us the tangent at the first endpoint and when t = 1, x'(t) gets us the tangent at the other endpoint. x'(0) = R 1 = [ 0 0 1 0 ] M G x x'(1) = R 4 = [ 3 2 1 0 ] M G x Hermite Basis Matrix and Blending Functions

27. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 [P 1 ] [ 0 0 0 1 ] [P 4 ] = G x = [ 1 1 1 1 ] M G x [R 1 ] [ 0 0 1 0 ] [R 4 ] [ 3 2 1 0 ] just like when we solved for M (the basis matrix) with the line equations, we do the same thing here because clearly [ 0 0 0 1 ] [ 1 1 1 1 ] M = Identity [ 0 0 1 0 ] [ 3 2 1 0 ] so, that matrix on the left is equal to M -1 Hermite Basis Matrix and Blending Functions

28. Michael Eckmann - Skidmore College - CS 376 - Spring 2007 There are techniques to get the inverse of a matrix which we won't go into here --- if you've had linear algebra, you should know how. The inverse of [ 0 0 0 1 ] [ 1 1 1 1 ] [ 0 0 1 0 ] [ 3 2 1 0 ] is [ 2 -2 1 1 ] [ -3 3 -2 -1 ] [ 0 0 1 0 ] [ 1 0 0 0 ] This is the Basis Matrix for Hermite curve segments. Let's verify that it is indeed the inverse of the matrix above. Hermite Basis Matrix and Blending Functions