Lecture 20: Bezier Curves & Splines

Transcription

1 Lecture 20: Bezier Curves & Splines December 6, /6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 1

2 Review: The Pen Metaphore Think of putting a pen to paper Pen position described by time t Seeing the action of drawing is the key, so this static drawing only partly captures the point of this slide. 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 2

3 Review: Third Order Curves Third-order functions are the standard: 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 Why 3? Lower-order curves cannot be smoothly joined. Higher-order curves introduce wiggles. Without loss of generality: 0 <= t <= 1. 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 3

4 Review: Cubic Examples 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 4

5 Review: Notation T =! " t 3,t 2,t,1# $ Q( t) = T C! # # C = # # # "# a x a y a z b x b y b z c x c y c z d x d y d z $ & & & & & %& Alternatively: Q( t) T = C T T T 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 5

6 Review: Tangents to Cubic Curves The derivative of Q(t) is its tangent: d dt d d d Q(t) = [ x(t), y(t), z(t)] dt dt dt d dt Q(t) x = 3a x t 2 + 2b x t + c x d dt Q(t) = [3t2, 2t, 1, 0] C Again the time metaphor is useful, the tangent indicates instantaneous direction and speed. 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 6

7 Review: The Hermite Matrix OK, that was the x dimension. How about the others? They are, of course, the same: a x a y a z b x b y b z c x c y c z d x d y d z = M P( 0) P( 1) dp( 0) dt dp( 1) dt 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 7

8 From Hermite to Bezier What s wrong with Hermite curves? Nothing, unless you are using a point-and-click interface Bezier curves are like Hermite curves, except that the user specifies four points (p 1, p 2, p 3, p 4 ). The curve goes through p 1 & p 4. Points p 2 & p 3 specify the tangents at the endpoints. 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 8

9 More Precisely... d R 1 = P 1 = 3(P 2 - P 1 ) dt d R 4 = P 4 = 3(P 4 - P 3 ) dt Tangents at start and end are now defined by intermediate points. Q: Why 3? Why not R 1 =(P 2 -P 1 )? A: Think of 4 evenly spaced points in a line 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 9

10 Hermite Bezier The Hermite geometry matrix is related to the Bezier geometry matrix by: G H = P1 P4 R1 R4 = P1 P2 P3 P4 = M HB G B 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 10

11 Hermite Bezier For Hermite curves, Q(t) = T M H G H, where G H = [P 1, P 4, R 1, R 4 ] T, T = [t 3, t 2, t, 1] and M H = So, Q(t) =T M H M HB G B 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 11

12 The Bezier Basis Matrix Q(t) =T(M H M HB )G B M B = M H M HB = Q(t) = TM B G B See page 364 to connect with description in our textbook. 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 12

13 The Bezier Blending Functions Q(t) = P 1 (-t 3 + 3t 2-3t + 1) + P 2 (3t 3-6t 2 + 3t) + P 3 (-3t 3 + 3t 2 ) + P 4 (t 3 ) 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 13

14 Add them up. ( -t 3 + 3t 2-3t + 1) + ( 3t 3-6t 2 + 3t) + (-3t 3 + 3t 2 ) + ( t 3 ) 0t 3 +0t 2 +0t+1 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 14

15 Stay within the Convex Hull If you graph the four Bezier blending functions for t=0 to t=1, you find that they are always positive and always sum to one. Therefore, the Bezier curve stays within the convex hull defined by P 1, P 2, P 3 & P 4. 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 15

16 Examples 1. 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 16

17 Examples 2. 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 17

18 Example 3D 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 18

19 Bezier Curves are Common You have probably already used them. For example, in PowerPoint Build a shape Then select edit points Notice the control wings Enhancements Ways to introduce constraints Smooth Point Straight Point Corner Point 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 19

20 Stepping Back What should you be learning? Should you memorize M H and M B? No! That s what reference books are for. You should know what G H and G B are. You should know how to derive M H from the parametric form of the cubic equations. You should know how to derive M B from M H. If you understand these concepts, you can look up or rederive the matrices as necessary. 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 20

21 de Casteljau Curves Now for something completely different. There is another way to motivate curves. P 2 P 3 P 1 Lets say that I have four control points P 4 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 21

22 To find the midpoint of the curve corresponding to those control points: t=0.5 P 2 P 3 t=0.5 t=0.5 P 1 Connect the point between P 1 and P 2 where t = 0.5 with the point between P 3 & P 2 where t = 0.5; Do the same with the P 2 -P 3 & P 3 -P 4 lines P 4 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 22

23 Now, connect these two lines at their t = 0.5 points P 2 t=0.5 P 3 t=0.5 t=0.5 t=0.5 P 1 The t = 0.5 point on the resulting segment is P 4 the midpoint (t = 0.5 point) of some type of curve which is made up of weighted averages of the control points (we ll soon see what kind of curve) 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 23

24 We can extend this idea to any t value. To compute the t = 0.25 point, connect the 0.25 points of the original lines... P 2 t=0.25 P 3 t=0.25 t=0.25 P 1 P 4 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 24

25 Now, connect these lines at their t = 0.25 locations, and find the point where t = 0.25 of the resulting line P 2 t=0.25 P 3 t=0.25 t=0.25 P 1 In this way, you can compute the 3rd-order curve for any value of t P 4 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 25

26 Algebraic Definition The equations of the three original lines are: A 1 ( t ) = ( 1 t) P 1 + t P 2 A 2 ( t ) = ( 1 t) P 2 + t P 3 A 3 ( t ) = ( 1 t) P 3 + t P 4 The equations of the next two joining lines are: B 1 ( t ) = ( 1 t) A 1 + t A 2 B 2 ( t ) = ( 1 t) A 2 + t A 3 Finally, the line between the two joining lines is: C 1 ( t ) = ( 1 t) B 1 + t B 2 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 26

27 Begin Substitutions Substitute equations for A 1 and A 2 into B 1 B 1 ( t ) = ( 1 - t ) ( ( 1 - t ) P 1 + t P 2 ) + t ( ( 1 - t ) P 2 + t P 3 ) = = P 1-2 t P 1 + t 2 P t P 2-2 t 2 P 2 + t 2 P 3 ( t t ) P 1 + ( - 2 t t ) P 2 + t 2 P 3 Factoring the result B 1 ( t ) = ( t - 1) 2 P 1-2 t ( t - 1) P 2 + t 2 P 3 Likewise, substitute A 2 and A 3 into B 2 B 2 ( t ) = ( t 1 ) 2 P 2 2 t ( t 1 ) P 3 + t 2 P 4 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 27

28 Resulting Third Order Curve Substitute equations for B 1 and B 2 into C 1 C 1 ( t ) = ( 1 - t ) ( ( t - 1) 2 P 1-2 t ( t - 1) P 2 + t 2 P 3 ) + t ( ( t - 1) 2 P 2-2 t ( t - 1) P 3 + t 2 P 4 ) = ( - 3 t - t t 2 + 1) P ( 3 t - 6 t + 3 t ) P 2 + ( - 3 t t 2 ) P 3 + t 3 P 4 And After Factoring C 1 ( t ) = ( 1 - t ) 3 P t ( t - 1) 2 P 2-3 t 2 ( t - 1) P 3 + t 3 P 4 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 28

29 Bezier = de Casteljau But those last four functions are exactly the Bezier blending functions! The recursive line intersection algorithm can therefore be used to gain intuition about the behavior of Bezier functions Not something completely different afterall. 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 29

30 The Road Less Traveled (for good reason) In principle, we can use de Casteljau s algorithm to fit a continuous curve through an arbitrary number of points P 1 P 2 P 3 P 4 P 8 P 7 P 9 P 10 P P 11 5 P 6 One problem: if any point moves, then every point on the curve moves. There is no locality. 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 30

31 Locality of Control is Good! P 1 segment #1 segment #2 P 2 P 3 P 4 P 8 P 7 P P 11 5 P 6 segment #3 P 10 Points P 1 through P 4 define the first segment, P 2 through P 5 the second, and P N through P N+3 the Nth segment. P 9 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 31

32 Some New Requirements Note that these 3rd-order segments are neither exactly Hermite nor Bezier curves: 1) The curve from P i to P i+3 is only drawn between P i+1 and P i+2 (otherwise segments would overlap) 2) The curve is not constrained to pass through either P i+1 or P i+2. Therefore, what is their equation? 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 32

33 B-Splines 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 33 By convention, the geometry matrix and basis matrix for B-Splines are: ú ú ú ú û ù ê ê ê ê ë é = i i i i B P P P P G ú ú ú ú û ù ê ê ê ê ë é = M Bs

34 B-Spline Blending Functions 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 34 Blend = T M Bs [ ] ú ú ú ú û ù ê ê ê ê ë é = t t t Blend B o = 1 6 t 3 + 3t 2 3t +1 ( )P i = 1 6 t 1 ( ) 3 P i ( ) = i P t t B ( ) = i P t t t B ( ) = i+ P t B

35 Plot of Blending Functions 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 35

36 See Cycle in Blending Functions B 3 B 2 B 4 B 1 As the Spline Sequences from one segment to the next, control points are passed from B 4, to B 3, to B 2 and finally to B 1. Consequently, the weight exerted on the curve rises then falls as indicated by the curve above. 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 36

37 G 1 G 2 G 3 G 4 = = = = Q j ( t ) = G j B( t ) Example of B-Spline é ù ê ë ú û é ù ê ë ú û é ù ê ú ë û é ù ê ú ë û 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 37

38 More Variations We have just described uniform, non-rational B-Splines Uniform means that the control points are evenly spaced (in terms of the parameter t). It is also possible to have non-uniform B-Splines. Why? because it is easier to interpolate starting and ending points, and it is possible to reduce the continuity at specific join points. 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 38

39 Non-uniform B-Splines Every control point must have a corresponding t-value This is called a knot vector If the spacing (in t) between two control points is small, then a sharp curve will result. If the spacing (in t) is zero, the curve becomes discontinuous. 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 39

40 The Standard Knot Vector The standard knot vector begins and ends with a four-fold knot: e.g., for 5 control points T = (0, 0, 0, 0, 1, 2, 3, 4, 5, 5, 5, 5,) This means that the B-Spline will not loose the last point(s), and will behave correctly near the endpoints. 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 40

41 Rational BSplines So far we have represented curves as curves with one free parameter (t) and three dependent parameters (x, y, x): X(t) = a x t 3 + b x t 2 + c x t + d Y(t) = a y t 3 + b y t 2 + c y t + d Z(t) = a z t 3 + b z t 2 + c z t + d What else could we do? 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 41

42 Homogenous Coordinates! Why not use four dependent functions? X(t) = a x t 3 + b x t 2 + c x t + d Y(t) = a y t 3 + b y t 2 + c y t + d Z(t) = a z t 3 + b z t 2 + c z t + d W(t) = a w t 3 + b w t 2 + c w t + d The image position at t is X(t)/W(t), Y(t)/W(t) and Z(t)/W(t) 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 42

43 What does this gain us? Why are (non-rational) curves too expensive? For every value of t, we compute x, y, z Then we project x, y, z (given viewpoint) How small should the increment in t be? Depends on the projection Very small to avoid gaps in the projection But this makes curves very expensive Rational form fixes this problem (next slide) 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 43

44 Closing Argument - NURBS A remarkable property (trust me): The curve made by computing a rational BSpline for all t and then projecting it is the same curve made by projecting the control points, and then computing the 2D BSpline between the control points So Non-Uniform Rational B-Splines are drawn by projecting the control points, and then recursively computing the (minimal set of) intermediate points in 2D 12/6/16 CSU CS410 Bruce Draper & J. Ross Beveridge 44