Lecture 23: Hermite and Bezier Curves

Transcription

1 Lecture 23: Hermite and Bezier Curves November 16, /16/17 CSU CS410 Fall 2017, Ross Beveridge & Bruce Draper 1

2 Representing Curved Objects So far we ve seen Polygonal objects (triangles) and Spheres Now, polynomial curves Hermite curves Bezier curves B-Splines NURBS Bivariate polynomial surface patches 11/16/17 Bruce Draper & J. Ross Beveridge 2

3 Beyond linear approximation Instead of approximating everything by zillions of lines and planes, it is possible to approximate shapes using higher-order curves. Advantages: More compact Reduces artifacts Use of sphere in ray tracer is an example of an implicit curve. 11/16/17 Bruce Draper & J. Ross Beveridge 3

4 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. 11/16/17 Bruce Draper & J. Ross Beveridge 4

5 Design Criteria Local control of shape Smoothness and continuity Ability to evaluate derivatives Stability Ease of rendering 11/16/17 Bruce Draper & J. Ross Beveridge 5

6 Review Forms - Explicit Explicit representation: y = f(x) Drawbacks: y = ax 3 + bx 2 + cx + d Multiple values of y for a single x impossible. Not rotationally invariant. 11/16/17 Bruce Draper & J. Ross Beveridge 6

7 Review Forms - Implicit Implicit representation: f(x, y, z) = 0 x 2 a + y2 2 b + z2 2 c 1= 0 2 Advantages: On curve & relative distance to curve tests. Drawbacks: Enumerating points on the curve is hard. Extra constraints needed half a circle? Difficult to express and test tangents. 11/16/17 Bruce Draper & J. Ross Beveridge 7

8 Parametric Representations We will represent 3D curves using a parametric representation, introducing a new variable t: Q( t) = x( t) y( t) z( t) Note that x, y and z are dependent on t alone, making it clear that there is only one free variable. Think of t as time associated with movement along the curve. 11/16/17 Bruce Draper & J. Ross Beveridge 8

9 Third Order Curves Third-order functions are the standard: Why 3? 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 Lower-order curves cannot be smoothly joined. Higher-order curves introduce wiggles. Without loss of generality: 0 <= t <= 1. 11/16/17 Bruce Draper & J. Ross Beveridge 9

10 Cubic Examples 11/16/17 Bruce Draper & J. Ross Beveridge 10

11 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 11/16/17 Bruce Draper & J. Ross Beveridge 11

12 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 The same matrix as on previous slide Again the time metaphor is useful, the tangent indicates instantaneous direction and speed. 11/16/17 Bruce Draper & J. Ross Beveridge 12

13 Hermite Curves We want curves that fit together smoothly. To accomplish this, we would like to specify a curve by providing: The endpoints The 1 st derivatives at the endpoints The result is called a Hermite Curve. 11/16/17 Bruce Draper & J. Ross Beveridge 13

14 Hermite Curves (cont.) Since Q(t) = TC, we factor C into two matrices: G (a 3x4 geometry matrix) M (a 4x4 basis matrix) such that C = G M. This step is a big deal. It makes thinking about curve geometry tractable. Note: G will hold our geometric constraints (endpoints and derivatives), while M will be constant across all Hermite curves. 11/16/17 Bruce Draper & J. Ross Beveridge 14

15 Let us concentrate on the x component: P(t) x = a x t 3 + b x t 2 + c x t + d x Remember that its derivative is: d P(t) x = 3a x t 2 + 2b x t + c dt x Therefore P(0) x P(1) x d/dt P(0) x d/dt P(1) x = a x b x c x d x 11/16/17 Bruce Draper & J. Ross Beveridge 15

16 Therefore: a x b x c x d x = And taking the inverse: a x b x c x d x = P(0) x P(1) x d/dt P(0) x d/dt P(1) x P(0) x P(1) x d/dt P(0) x d/dt P(1) x 11/16/17 Bruce Draper & J. Ross Beveridge 16

17 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 11/16/17 Bruce Draper & J. Ross Beveridge 17

18 Punchline Q( t) = x( t) y( t) z( t) = T C = T M H " $ $ $ $ $ $ # $ ( ) ( ) P 0 P 1 d P 0 dt ( ) d P 1 dt ( ) % ' ' ' ' ' ' &' Since M H and T are known, you can write down a cubic polynomial curve by inspection ending at points P(0) and P(1) with tangents d/dt P(0) and d/dt P(1). 11/16/17 Bruce Draper & J. Ross Beveridge 18

19 Expand the Math Look Inside Recall Parametric Equation: Q(t) = T M H G Where Fully Expanded (note transpose) T 11/16/17 Bruce Draper & J. Ross Beveridge 19

20 If you Prefer There are two equivalent setups The difference is solely transposition 11/16/17 Bruce Draper & J. Ross Beveridge 20

21 Examples G = G = /16/17 Bruce Draper & J. Ross Beveridge 21

22 More Examples G = G = /16/17 Bruce Draper & J. Ross Beveridge 22

23 Hermite Blending Functions Conceptual Realignment Curves are weighted averages of points/vectors. Blending functions specify the weighting. é x1 ù é x2 ù édx1ù Q = y1 ( 2 t 3-3 t 2 + 1) + y2 (- 2 t t 2 ) + dy1 ( t 3-2 t 2 + t) + ê ú ê ú ê ú ë z1 û ë z2 û ëdz1û édx2ù dy2 ê ú ëdz2û ( t 3 - t 2 ) Bh 1 = 2 t 3-3 t Bh 2 = - 2 t t 2 Bh 3 = t 3-2 t 2 + t Bh 4 = t 3 - t 2 11/16/17 Bruce Draper & J. Ross Beveridge 23

24 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. 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 24

25 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 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 25

26 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 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 26

27 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 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 27

28 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. 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 28

29 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 ) 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 29

30 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 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 30

31 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. 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 31

32 Examples 1. 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 32

33 Examples 2. 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 33

34 Example 3D 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 34

35 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 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 35

36 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. 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 36

37 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 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 37

38 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 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 38

39 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) 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 39

40 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 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 40

41 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 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 41

42 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 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 42

43 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 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 43

44 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 ) 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 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 44

45 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. 11/16/17 CSU CS410 Bruce Draper & J. Ross Beveridge 45