1. Splines I – Curves and Properties based on: Michael Gleicher Curves, chapter 15 in Fundamentals of Computer Graphics, 3 rd ed. (Shirley & Marschner) Slides by Marc van Kreveld 1

2. What is a curve? Intuitively: what you can draw with a pen without lifting it up Mathematically: image of a continuous mapping of an interval into the plane (or higher-d space) f : [0,1]  R 2 [] 2

3. Types of curves Normal (two endpoints) Closed loops (no endpoints) Infinite curves (no endpoints) 3

4. More curves Curves may self-intersect, have sharp corners, trace back on themselves, … Well-known curves: line segment, line, circle, ellipse, square (boundary), parabola, circular arc, … General curves are called free-form curves 4

5. How to specify a curve (3 ways) Implicit curve Parametric curve Procedurally defined curve 5

6. Implicit curves Given by a function in 2 variables (for curves in the plane) and an outcome value of that function – Circle with radius 2 centered at origin: x 2 + y 2 = 4 the function is f(x,y)= x 2 + y 2 – Some weird curve: log x – x 2 + y 3 = 0 (if this equation yields a connected set???) We can assume the given value is 0 Not every function+value gives a connected set, implying that it cannot be the image of a continuous mapping 6

7. Parametric curves Use a parameter, e.g. t, and use two functions f(t) and g(t) to describe corresponding x and y coordinates – f(t) = cos tfor t in [0, 2  ); this is a circle g(t) = sin t – f(t) = t 2 g(t) = t – f(t) = t 4 g(t) = t 2 – f(t) = t 6 g(t) = t 3 7

8. Parametric curves Representations are not unique The functions f and g can be written together as a vector-valued function f, we then write (x, y) = f (t) as a shorthand for (x, y) = (f(t), g(t)) 8

9. Procedural curves Defined by a process or a grammar, e.g. Koch snowflake (a fractal curve) 9

10. Parametric curves Most common in graphics Easy to work with (we can draw by evaluating at the parameter) 10

11. About the parameterization The value of the parameter tells you where you are on the curve The parameter determines “speed” along the curve f (t) = (t, t) (0,1) (0,0) (1,0) t=0 t=1 t=½ f (t) = (t 2, t 2 ) t=0 t=1 t=½ t=0 t=1 t=½ f (t) = (t 1/2, t 1/2 ) 11

12. About the parameterization Speed is the derivative (magnitude): | d f(t)/d t | f (t) = (t, t) (0,1) (0,0) (1,0) t=0 t=1 t=½ f (t) = (t 2, t 2 ) t=0 t=1 t=½ t=0 t=1 t=½ f (t) = (t 1/2, t 1/2 ) speed is  2 everywhere speed increases from 0 to 2  2 speed decreases from  to ½  2 12

13. Arc-length parameterizations Informally, constant-speed parameterizations Formally, when s is used as parameter when it is an arc-length parameterization f(t) = (t, t) 13

14. Arc-length parameterizations Informally, constant-speed parameterizations Formally, when s is used as parameter when it is an arc-length parameterization f(t) = (t 2, t 2 ) 14

15. Tangent vector along a curve Has a direction and magnitude; the magnitude is speed f(s) = sin s s  [0,  /2] g(s) = cos s f(t) = t t  [0,1] g(t) =  ((1 – t) 2 ) The direction of tangent depends purely on the position on the curve, whereas the magnitude depends on the parameterization as well 15

16. Length of a parameterized curve You get the length of a curve by integrating over the magnitude of the tangent vector of the curve The arc length from parameter value a to b, so the arc length between f (a) and f (b), is: (x,y) = f (a) (x,y) = f (b) 16

17. Piecewise curves Defined by a sequence of functions Should line up head-to-tail to get a continuous overall mapping f1f1 f2f2 f3f3 f4f4 A piecewise curve is a curve by definition; its description just happens to be done using pieces 17

18. Piecewise curves Example – f 1 (t) = (t, 0)for t  [0,1) – f 2 (t) = (t, t – 1)for t  [1,3) – f 3 (t) = (t 2 – 2t, t – 1)for t  [3,4] 18

19. Piecewise curves Pieces must connect  continuous curve Tangent vector at both sides is the same  differential curve 19

20. Types of continuity C 0 is the name/notation for normal continuity Parametric continuity C 1 : The tangent vectors are the same “just before and just after every point on the curve”, in particular for breakpoints Geometric continuity G 1 : The tangent vector directions are the same just before and just after Parametric continuity depends on the parameterization, while geometric continuity depends only on the shape 20

21. Types of continuity You cannot “see” parametric continuity, you need the parameterization Parametric continuity implies geometric continuity Re-parameterization may make a geometrically continuous curve parametrically continuous C n and G n means that all derivatives up to (and including) the n-th are continuous Practice: C 1 / G 1 continuous curves look good, C 2 / G 2 continuous curves look slightly better, more is not needed 21

22. Types of continuity Consider the piecewise curve – f 1 (t) = (t, t)for t  [0, ½) – f 2 (t) = (2t – ½, 2t – ½)for t  [½, 1] 22 1 1½ ½

23. Types of continuity Consider the piecewise curve – f 1 (t) = (t 2, 1)for t  [0, 1) – f 2 (t) = (1 + sin(t-1), 2 – cos(t-1))for t  [1, 1+  /2] 23 1 1 2 2 Question: What is the speed just before and just after (1,1)? Is it C 1, G 1, both, neither?

24. Polynomial pieces Both functions used for the f are polynomials in the parameter Simplest example: piecewise linear (polygonal) curve, consisting of straight line segments; only C 0 – Each piece can be expressed using linear interpolation between two points: f(u) = (1 – u) p 0 + u p 1, u  [0,1] – Also : f(u) = a 0 + u a 1 (where a 0 = p 0 and a 1 = p 1 - p 0 ) p0p0 p1p1 u is used as parameter notation when it ranges over the unit interval [0,1] 24

25. Polynomial pieces Both functions used for the f are polynomials in the parameter Simplest example: piecewise linear (polygonal) curve, consisting of straight line segments; only C 0 – Each piece can be expressed using linear interpolation between two points: f(u) = (1 – u) p 0 + u p 1, u  [0,1] – Also : f(u) = a 0 + u a 1 (where a 0 = p 0 and a 1 = p 1 - p 0 ) a0a0 a1a1 u is used as parameter notation when it ranges over the unit interval [0,1] 25

26. Polynomial pieces Higher-degree polynomials can be written in canonical form: f(u) = a 0 + u a 1 + u 2 a 2 (quadratic), or f(u) = a 0 + u a 1 + u 2 a 2 + … + u n a n (n-th degree) In vector/matrix notation: these have an x- and a y-coordinate 26

27. Quadratic pieces, quadrics The (vector) coefficients a 0, a 1 and a 2 of the canonical form are not convenient for making curves It is more natural to describe quadrics using 3 points, one for f(0), one for f(0.5), and one for f(1) p 0 = f(0) = a 0 + 0 a 1 + 0 2 a 2 p 1 = f(0.5) = a 0 + 0.5 a 1 + 0.5 2 a 2 p 2 = f(1) = a 0 + 1 a 1 + 1 2 a 2 With linear algebra, we can convert a 0, a 1 and a 2 into p 0, p 1 and p 2 and vice versa 27

28. Quadrics p2p2 p1p1 p0p0 p2p2 p1p1 p0p0 28

29. Quadratic curves in the real world 29

30. Quadrics p 0 = f(0) = a 0 + 0 a 1 + 0 2 a 2 p 1 = f(0.5) = a 0 + 0.5 a 1 + 0.5 2 a 2 p 2 = f(1) = a 0 + 1 a 1 + 1 2 a 2 constraint matrix 30

31. Quadrics p 0 = f(0) = a 0 + 0 a 1 + 0 2 a 2 p 1 = f(0.5) = a 0 + 0.5 a 1 + 0.5 2 a 2 p 2 = f(1) = a 0 + 1 a 1 + 1 2 a 2 base matrix If you have the three points p 0, p 1 and p 2, you can get the coefficients of the polynomial in canonical form 31

32. Quadrics Example 1: Given the points p 0 = (0,1), p 1 = (2,0) and p 2 = (4,4), compute the coefficients of the polynomial in canonical form for which f( 0 ) = p 0, f( 0.5 ) = p 1, and f( 1 ) = p 2 Solution: simply apply the base matrix B to ( p 0 p 1 p 2 ) T 32

33. Quadrics Example 2: Given the points p 0 = (0,1), p 1 = (2,0) and p 2 = (4,4), compute the coefficients of the polynomial in canonical form for which f( 0 ) = p 0, f( 0.3 ) = p 1, and f( 1 ) = p 2 Solution: determine the base matrix B for this case and apply it to ( p 0 p 1 p 2 ) T 33

34. Quadrics p 0 = f(0) = a 0 + 0 a 1 + 0 2 a 2 p 1 = f(0.3) = a 0 + 0.3 a 1 + 0.3 2 a 2 p 2 = f(1) = a 0 + 1 a 1 + 1 2 a 2 Now compute B = C – 1 and apply it to ( p 0 p 1 p 2 ) T 34

35. Quadrics We can specify the quadric using points and derivatives too – two points and one derivate at a point – one point, first derivative at a point and second derivative at a point p1p1 p0p0 35

36. Quadrics f(u) = a 0 + a 1 u + a 2 u 2 f’(u) = a 1 + 2 a 2 u f’’(u) = 2 a 2 By specifying anything like p 0 = f(0), p 1 = f’(0.6), p 2 = f’’(0.7) we can determine a (and therefore the quadric in canonical form) by setting up the constraint matrix, doing matrix inversion, and applying it to ( p 0 p 1 p 2 ) T 36

37. Quadrics f(u) = a 0 + a 1 u + a 2 u 2 f’(u) = a 1 + 2 a 2 u f’’(u) = 2 a 2 For example: p 0 = f(0.1) = a 0 + 0.1 a 1 + 0.1 2 a 2 p 1 = f’(0.6) = a 1 + 2  0.6 a 2 p 2 = f’’(0.7) = 2 a 2 Note: p 1 and p 2 are not points, but specifications of the first and second derivatives 37

38. Cubics Cubics are the most popular polynomial pieces – They are determined by 4 coefficients so we can for instance specify the two endpoints and the two first derivatives at these endpoints – This is important for building splines – f(u) = a 0 + a 1 u + a 2 u 2 + a 3 u 3 p2p2 p0p0 p3p3 p1p1 38

39. Cubics p 0 = f(0) = a 0 +0 a 1 + 0 2 a 2 + 0 3 a 3 p 1 = f’(0) = a 1 + 0a 2 + 3  0 2 a 3 p 2 = f(1) = a 0 +1 a 1 + 1 2 a 2 + 1 3 a 3 p 3 = f’(1) = a 1 + 2  1 a 2 + 3  1 2 a 3 This is the so-called Hermite form p2p2 p0p0 p3p3 p1p1 39

40. Cubics For example: p 0 = (0,0), p 1 = (0,2), p 2 = (3,0), p 3 = (2,0) p 0 = (0,0) p 1 = (0,2) p 2 = (3,0) p 3 = (2,0) point tangent vector 40

41. Summary There are three main ways to specify curves, parametric, implicit, and procedural There are different degrees of continuity of curves, important when designing piecewise defined curves We can build (parametric) linear curves from two specifications, quadric curves from 3 specifications and cubic curves from 4 specifications (through point, tangent at point, …) The constraint matrix and base matrix allow us to convert specifications into canonical form and v.v. 41

42. Questions 1.Given the piecewise, parameterized curve f(t) = (3t, t 2 ) for t  [0, 2) and f(t) = (4+t, 2t) for t  [2, 4]. Is the curve G 0 ? Is it G 1 ? Is the parameterization C 0 ? Is it C 1 ? Explain 2.Suppose we make a cubic curve by specifying f(0), f’(0), f(0.5) and f(1). Give the constraint matrix and the base matrix Suppose f(0) = (1,1), f’(0) = (0,1), f(0.5) = (3,1) and f(1) = (4,0). Put the curve in canonical form 3.Analyze the curve given by f(0) = (1,1), f’(0) = (1,0), f(1) = (3,1) and f’(1) = (-2,0). 42