Reading. w Foley, Section 11.2 Optional

Transcription

1 Parametric Curves

2 w Foley, Section.2 Optional Reading w Bartels, Beatty, and Barsky. An Introduction to Splines for use in Computer Graphics and Geometric Modeling, 987. w Farin. Curves and Surfaces for CAGD: A Practical Guide, 4th ed.,

3 The loftsman s spline : Curves before computers w long, narrow strip of wood or metal w shaped by lead weights called ducks w gives curves with second-order continuity, usually Used for designing cars, ships, airplanes, etc. 3

4 What do we use curves for? w building models Motivation for curves w movement paths w animation 4

5 Mathematical curve representation w Explicit y=f(x) what if the curve isn t a function? w Implicit f(x,y) = 0 hard to work with x + y R = 0 w Parametric (f(u),g(u)) x(u) = cos 2πu y(u) = sin 2πu 5

6 Parametric polynomial curves We ll use parametric curves where the functions are all polynomials in the parameter. Advantages: xu ( ) yu ( ) = = n k= 0 n k= 0 w easy (and efficient) to compute w infinitely differentiable au k bu k k k 6

7 Cubic curves Fix n=3 For simplicity we define each cubic function within the range 0 t [ ] x y z 3 2 () t t t t Q () t = a t + b t + c t+ d 3 2 x x x x x 3 2 y y y y y 3 2 z() = z + z + z + z Q( t) = x( t) y( t) z( t) or Q ( t) = a t + b t + c t+ d or Q ax ay az b b b = cx cy cz dx dy dz Q t a t bt c t d 7

8 C Compact representation Place all coefficients into a matrix ax ay az b b b x y z 3 2 = T= t t t cx cy c z d d d x y z ax ay az b 3 2 x by b z Qt () = [ xt () yt () zt ()] = t t t = TC cx cy cz dx dy dz d d d d 2 Qt () = Qʹ () t = ( TC ) = TC + T C= 3t 2t 0 dt dt dt dt C 8

9 Controlling the cubic Q: How many constraints do we need to specify to fully determine the scalar cubic function Q(t)? Q: How many constraints do we need to specify to fully determine the 3-vector cubic function Q(t)? 9

10 Constraining the cubics Redefine C as a product of the basis matrix M and the 4-element column vector of constraints or geometry vector G Q C= M G 3 2 () t t t t m m m m G G G x y z m2 m22 m23 m G 24 2x G2y G 2z m3 m32 m33 m34 G3x G3y G3z m G 4 m42 m43 m44 4x G4y G4z = = TMG 0

11 Hermite Curves Determined by So w endpoints P and P 4 w tangent vectors at the endpoints R and R 4 Q() t = T Mh Gh P 4 Where G h Px Py Pz P4x P4y P 4z = Rx Ry Rz R4x R4y R4z P R R 4

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13 Computing Hermite basis matrix The constraints on Q(0) and Q() are found by direct substitution: Q(0) = M G Tangents are defined by [ ] [ ] Q() = M G Qʹ t t t M G so constraints on tangents are: 2 () = h h [ ] [ ] Qʹ (0) = M G Qʹ () = M G h h h h h h h h Q(t) P 4 P R 4 R 3

14 Computing Hermite basis matrix Collecting all constraints we get P Q(t) R P 4 R 4 Px Py Pz P P P 4x 4y 4z = Gh = Mh Gh Rx Ry Rz R4x R4y R 4z So M h = =

15 Computing a point Given two endpoints (P,P 4 ) and two endpoint tangent vectors (R, R 4 ): So Q(t) P 4 P R 4 R 2 2 P P R R4 3 2 Q() t = t t t 5

16 Blending Functions Polynomials weighting each element of the geometry vector Q() t = = 2 2 P t t t P R R4 P () P Bh t R R 4 4 B h (t) P P 4 R R 4 t 6

17 Continuity We want our curve to have continuity. There shouldn t be an abrupt change when we move from one segment to the next. There are nested degrees of continuity: C 0 : C : C 2 : C 3, C 4, : 7

18 Continuity of Splines P P 2 =P 3 R 2 P 4 R 4 C 0 : points coincide, velocities don t R R 3 P P 2 =P 3 R 2 P 4 R 4 R 3 G : points coincide, velocities have same direction P R P 2 =P 3 R 2 = R 3 P 4 R 4 C : points and velocities coincide R Q: What s C 2? 8

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20 Bézier Curves Indirectly specify the tangent vectors by specifying two intermediate points P 3 P P 4 G b P 2 Px Py Pz P P2x P2y P 2z P2 = = P3x P3y P3z P3 P4x P4y P 4z P4 R = 3( P P) 2 R = 3( P P )

21 Bézier basis matrix Establish the relation between the Hermite and Besier geometry vectors: R = 3( P P) 2 R = 3( P P ) G P P P P = = = M G 4 2 h bh b R P3 R P 4 4 2

22 Bézier basis matrix ( ) ( ) Q T M G T M M G () t = h h = h hb b = T M M G = T M G h hb b b b M = M M = b h hb Q() t = T Mb Gb 22

23 Bézier Blending Functions a.k.a. Bernstein polynomials Q( t) = t 3 t 2 t B b (t) B B 4 P P P 2 P 2 = Bb() t P3 P3 P P 4 4 B 2 B 3 t 23

24 Alternative Bézier Formulation 3 3 ( ) i Qt = Pi t ( t ) i= 0 i 3 i Q() t = t 3 t 2 t P P P P n i Q( t) P n = i t ( t) i= 0 i n i 24

25 More complex curves Suppose we want to draw a more complex curve. Why not use a high-order Bézier? Instead, we ll splice together a curve from individual segments that are cubic Béziers. Why cubic? There are three properties we d like to have in our newly constructed splines 25

26 Local control One problem with Béziers is that every control point affects every point on the curve (except the endpoints). Moving a single control point affects the whole curve! We d like our spline to have local control, that is, have each control point affect some well-defined neighborhood around that point. 26

27 Interpolation Bézier curves are approximating. The curve does not (necessarily) pass through all the control points. Each point pulls the curve toward it, but other points are pulling as well. We d like to have a spline that is interpolating, that is, that always passes through every control point. 27

28 Let s look at continuity first. Ensuring continuity Since the functions defining a Bézier curve are polynomial, all their derivatives exist and are continuous. Therefore, we only need to worry about the derivatives at the endpoints of the curve. First, we ll rewrite our equation for Q(t) in matrix form: 3 3 P P 3 3 P2 P3 3 2 Qt () = t t t 28

29 Ensuring C 2 continuity Suppose we want to join two cubic Bézier curves (V 0,V,V 2,V 3 ) and (W 0,W,W 2,W 3 ) so that there is C 2 continuity at the joint. W W 2 W 3 Qʹ (0) = 3( V V ) 0 Qʹ ( ) = 3( V V ) 3 2 Qʹ ʹ (0) = 6( V 2 V + V ) 0 2 Qʹ ʹ ( ) = 6( V 2 V + V ) 2 3 Q () = Q (0) V = W v w 3 0 Qʹ () = Q ʹ (0) V V = W W v w Qʹ ʹ () = Q ʹ ʹ ( 0) V 2V + V = W 2W v w W W = V + 4V 2 3 4V 2 + W 29 2

30 A-frames and continuity Let s try to get some geometrical intuition about what this last continuity equation means. If a and b are points, what is (2a-b)? V V2 V = W 3 0 B W W = V + 4V 4V 2 3 ( V V ) ( V V ) = W B W 2 2 V 0 W 2 W 3 30

31 Building a complex spline Instead of specifying the Bézier control points themselves, let s specify the corners of the A-frames in order to build a C 2 continuous spline. These are called B-splines. The starting set of points are called de Boor points. 3

32 B-splines Here is the completed B-spline. 2 V0 = B0 + ( B B0) + B+ ( B2 B) V = B+ ( B2 B ) 3 2 V2 = B0 + ( B B0 ) 3 V =... 3 What are the Bézier control points, in terms of the de Boor points? 32

33 Endpoints of B-splines We can see that B-splines don t interpolate the de Boor points. It would be nice if we could at least control the endpoints of the splines explicitly. There s a hack to make the spline begin and end at control points by repeating them. 33

34 B-spline basis matrix 3 2 Q() t = t t t P P P P

35 A third option If we re willing to sacrifice C 2 continuity, we can get interpolation and local control. Instead of finding the derivatives by solving a system of continuity equations, we ll just pick something arbitrary but local. If we set each derivative to be a constant multiple of the vector between the previous and next controls, we get a Catmull-Rom spline. 35

36 Catmull-Rom splines The math for Catmull-Rom splines is pretty simple: D = C C 0 0 D = ( C C ) D = ( C C ) M Dn = Cn Cn 36

37 Catmull-Rom basis matrix Q() t = t 3 t 2 t P P P P

38 C 2 interpolating splines Interpolation is a really handy property to have. How can we keep the C 2 continuity we get with B-splines but get interpolation, too? Here s the idea behind C 2 interpolating splines. Suppose we had cubic Béziers connecting our control points C 0, C, C 2,, and that we somehow knew the first derivative of the spline at each point. What are the V and W control points in terms of Cs and Ds? 38

39 Derivatives at the endpoints Qʹ (0) = 3( V V ) 0 Qʹ ( ) = 3( V V ) 3 2 Qʹ ʹ (0) = 6( V 2 V + V ) 0 2 Qʹ ʹ ( ) = 6( V 2 V + V ) 2 3 [ t ] Qʹ ʹ () t = V V 3 3 V2 V3 In general, the nth derivative at an endpoint depends only on the n+ points nearest that endpoint. 39

40 Finding the derivatives Now what we need to do is solve for the derivatives. To do this we ll use the C 2 continuity requirement. V = C 0 0 V = C + D V = C D V 2 3 = C 3 W = C 0 W = C + D 3 W = C D W = C 3 2 6( V 2 V + V ) = 6( W 2 W + W ) D + 4D + D = 3( C C )

41 Finding the derivatives, cont. Here s what we ve got so far: D + 4D + D = 3( C C ) D + 4D + D = 3( C C ) M D + 4D + D = 3( C C ) m 2 m m m m 2 How many equations is this? How many unknowns are we solving for? 4

42 Not quite done yet We have two additional degrees of freedom, which we can nail down by imposing more conditions on the curve. There are various ways to do this. We ll use the variant called natural C 2 interpolating splines, which requires the second derivative to be zero at the endpoints. This condition gives us the two additional equations we need. At the C 0 endpoint, it is: 6( V 2 V + V ) =

43 Solving for the derivatives Let s collect our m+ equations into a single linear system: D 3( C2 C0) m m m It s easier to solve than it looks. D 3( C C ) 4 D2 3( C3 C) = O M M 4 Dm 3( Cm Cm 2) 2 D 3( C C ) We can use forward elimination to zero out everything below the diagonal, then back substitution to compute each D value. 43

44 Forward elimination First, we eliminate the elements below the diagonal: 2 D0 E0 4 D E 4 D2 E2 = O M M 4 Dm- Em- 2 Dm Em 2 D0 F0 = E0 0 7/2 D = -(/ 2) F E E 4 D2 E2 = O M M 4 Dm- Em- 2 Dm Em 0 44

45 Back subsitution The resulting matrix is upper diagonal: UD = F u um D0 F0 D F D 2 F 2 = O M M M D m- Fm- u D F mm m m We can now solve for the unknowns by back substitution: u D = F mm m m u D +u D = F m-m- m- m-m m m- 45

46 C 2 interpolating spline Once we ve solved for the real D i s, we can plug them in to find our Bézier control points and draw the final spline: Have we lost anything? 46

47 What if we want a closed curve, i.e., a loop? With Catmull-Rom and B-spline curves, this is easy: 47

48 Displaying Bézier curves How could we draw one of these things? DisplayBezier( V0, V, V2, V3 ) begin end; if ( FlatEnough( V0, V, V2, V3 ) ) Line( V0, V3 ); else do something smart; It would be nice if we had an adaptive algorithm, that would take into account flatness. 48

49 Subdivide and conquer DisplayBezier( V0, V, V2, V3 ) begin end; if ( FlatEnough( V0, V, V2, V3 ) ) Line( V0, V3 ); else Subdivide(V) L, R DisplayBezier( L0, L, L2, L3 ); DisplayBezier( R0, R, R2, R3 ); 49

50 Testing for flatness Compare total length of control polygon to length of line connecting endpoints: V V + V V + V V V V 0 3 < + ε 50

51 Curves in Animator Project One of the requirements is to implement wrapping so that the animation restarts smoothly when looping back to the beginning. This is a lot like making a closed curve: the calculations for the θ - coordinate are exactly the same. The t-coordinate is a little trickier: you need to create phantom t-coordinates before and after the first and last coordinates. 5

52 Curves in Animator Project In the animator project, you will draw a curve on the screen: Q( u) = ( x( u), y( u) ) You will actually treat this curve as: θ ( u) = y( u) tu ( ) = xu ( ) Where θ is a variable you want to animate. We can think of the result as a function: θ( t) You have to apply some constraints to make sure that θ(t) actually is a function. 52

53 Summary Enforcing constraints on cubic functions The meaning of basis matrix and geometry vector General procedure for computing the basis matrix Properties of Hermite and Bézier splines The meaning of blending functions Enforcing continuity across multiple curve segments How to display Bézier curves with line segments. Meanings of C k continuities. Geometric conditions for continuity of cubic splines. Properties of C 2 interpolating splines, B-splines, and Catmull-Rom splines. 53