Q( t) = T C T =! " t 3,t 2,t,1# Q( t) T = C T T T. Announcements. Bezier Curves and Splines. Review: Third Order Curves. Review: Cubic Examples

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1 Bezier Curves an Splines December 1, 2015 Announcements PA4 ue one week from toay Submit your most fun test cases, too! Infinitely thin planes with parallel sies μ oesn t matter Term Paper ue one week from Thursay Instructions still on web site Single space (in case you were wonering) ACM Toay: last meeting. Winwar Coe wars teaming Dec. 11 th, 11:00-12:00, Bozeman MT start-up 2 Review: The Pen Metaphore Think of putting a pen to paper Pen position escribe by time t Seeing the action of rawing is the key, so this static rawing only partly captures the point of this slie. Review: Thir Orer Curves Thir-orer functions are the stanar: x t Why? ( ) a x t + b x t 2 + c x t + x ( ) a y t + b y t 2 + c y t + y ( ) a z t + b z t 2 + c z t + z y t z t Lower-orer curves cannot be smoothly joine. Higher-orer curves introuce wiggles. Without loss of generality: 0 < t < 1. 4 Review: Cubic Examples Review: Notation T! " t,t 2,t,1# $ Q( t) T C! a x a y a $ # z & # b x b y b z & C # & # c x c y c z & # & x y "# z %& Alternatively: Q( t) T C T T T 5 6

2 Review: Tangents to Cubic Curves 7 The erivative of Q(t) is its tangent: t Q(t) [ x(t), y(t), t t t z(t)] t Q(t) x a x t 2 + 2b x t + c x t Q(t) [t2, 2t, 1, 0] C Again the time metaphor is useful, the tangent inicates instantaneous irection an spee. Review: The Hermite Matrix 8 OK, that was the x imension. 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 x y z M P( 0) P( 1) P( 0) t P( 1) t Ross Beverige & Bruce Draper 2015 From Hermite to Bezier What s wrong with Hermite curves? Nothing, unless you are using a point-an-click interface Bezier curves are like Hermite curves, except that the user specifies four points (p 1, p 2, p, p 4 ). The curve goes through p 1 & p 4. Points p 2 & p specify the tangents at the enpoints. More Precisely... R 1 ( - ) t R 4 ( - P ) t Q: Why? Why not R 1(-)? A: Think of 4 evenly space points in a line Tangents at start an en are now efine by intermeiate points Ross Beverige & Bruce Draper 2015 Hermite Bezier The Hermite geometry matrix is relate to the Bezier geometry matrix by: Hermite Bezier For Hermite curves, Q(t) T M H G H, where G H [,, R 1, R 4 ] T, T [t, t 2, t, 1] G H P1 P4 R1 R P1 P2 P P4 an M H M HB G B So, Q(t) T M H M HB G B 11 12

3 Ro s s Bev erige & Bruc e Draper 2014 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 64 to connect with escription in our textbook. Ro s s Bev erige & Bruc e Draper 2014 The Bezier Blening Functions Q(t) (-t + t 2 -t + 1) + (t - 6t 2 + t) + P (-t + t 2 ) + (t ) 1 14 A them up. ( -t + t 2 - t + 1) + ( t - 6t 2 + t) + (-t + t 2 ) + ( t ) 0t +0t 2 +0t+1 Ro s s Bev erige & Bruc e Draper 2014 Stay within the Convex Hull If you graph the four Bezier blening functions for t0 to t1, you fin that they are always positive an always sum to one. Therefore, the Bezier curve stays within the convex hull efine by,, P & Examples 1. Ro s s Bev erige & Bruc e Draper 2014 Examples

4 Stepping Back What shoul you be learning? Shoul you memorize M H an M B? No! That s what reference books are for. Yo u s h o u l kn ow wh at G H an G B are. You shoul know how to erive M H from the parametric form of the cubic equations. You shoul know how to erive M B from M H. If you unerstan these concepts, you can look up or reerive the matrices as necessary. e Casteljau Curves Now for something completely ifferent. There is another way to motivate curves. P Lets say that I have four control points To fin the mipoint of the curve corresponing to those control points: P Now, connect these two lines at their t 0.5 points P Connect the point between an where t 0.5 with the point between P & where t 0.5; Do the same with the -P & P - lines The t 0.5 point on the resulting segment is the mipoint (t 0.5 point) of some type of curve which is mae up of weighte averages of the control points (we ll soon see what kin of curve) We can ex t en t h is i ea to any t val ue. To co mp u t e t he t 0.25 point, connect the 0.25 points of the original lines... P Now, connect these lines at their t 0.25 locations, an fin the point where t 0.25 of the resulting line In this way, you can compute the r-orer curve for any value of t P 2 24

5 Algebraic Definition The equations of the three original lines are: A 1 ( t ) ( 1 t) + t A 2 ( t ) ( 1 t) + t P A ( t ) ( 1 t) P + t 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 Finally, the line between the two joining lines is: C 1 ( t ) ( 1 t) B 1 + t B 2 Begin Substitutions Substitute equations for A 1 an A 2 into B 1 B 1 ( t ) ( 1 t ) (( 1 t) + t ) + t (( 1 t) + t P ) 2 t + t t 2 t 2 + t 2 P ( t t) + ( 2 t t) + t 2 P Factoring the result B 1 ( t ) ( t 1) 2 2 t ( t 1) + t 2 P Likewise, substitute A 2 an A into B 2 B 2 ( t ) ( t 1) 2 2 t ( t 1 ) P + t Resulting Thir Orer Curve Substitute equations for B 1 an B 2 into C 1 C 1 ( t ) ( 1 t ) (( t 1) 2 2 t ( t 1) + t 2 P ) + t (( t 1) 2 2 t ( t 1) P + t 2 ) ( t t + t 2 + 1) 2 + ( t 6 t + t) + ( t + t 2 ) P + t An After Factoring C 1 ( t ) ( 1 t) + t ( t 1) 2 t 2 ( t 1) P + t Ro s s Bev erige & Bruc e Draper 2014 Bezier e Casteljau But those last four functions are exactly the Bezier blening functions! The recursive line intersection algorithm can therefore be use to gain intuition about the behavior of Bezier functions Not something completely ifferent afterall The Roa Less Travele (for goo reason) In principle, we can use e Casteljau s algorithm to fit a continuous curve through an arbitrary number of points P P 8 P 7 1 P 5 P 6 One problem: if any point moves, then every point on the curve moves. There is no locality. P 9 0 segment #1 segment #2 Locality of Control is Goo! P P 8 P 7 segment # 0 1 P 5 P 6 Points through efine the first segment, through P 5 the secon, an P N through P N+ the Nth segment. P

6 Some New Requirements Note that these r-orer segments are neither exactly Hermite nor Bezier curves: 1) The curve from P i to P i+ is only rawn between P i+1 an P i+2 (otherwise segments woul overlap) 2) The curve is not constraine to pass through either P i+1 or P i+2. Therefore, what is their equation? B-Splines G B By convention, the geometry matrix an basis matrix for B-Splines are: Pi Pi + Pi + Pi M Bs B-Spline Blening Functions Blen T M Bs [ ] 6 Blen t t t Plot of Blening Functions B o 1 t + t 2 t +1 6 ( )P i 1 t 1 6 ( ) P i B 1 ( t 6t + 4) Pi + 1 B2 1 ( t + t + t + 1) Pi B 1 ( t ) P 6 i+ 4 See Cycle in Blening Functions B 4 B B 2 As the Spline Sequences from one segment to the next, control points are passe from B 4, to B, to B 2 an finally to B 1. Consequently, the weight exerte on the curve rises then falls as inicate by the curve above. B 1 Example of B-Spline G 1 G 2 G G Q j ( t) G j B( t) 5 6

7 Ross Beverige & Bruce Draper 2014 More Variations We have just escribe uniform, non-rational B-Splines Uniform means that the control points are evenly space (in terms of the parameter t). It is also possible to have non-uniform B-Splines. Why? because it is easier to interpolate starting an ening points, an it is possible to reuce the continuity at specific join points. Non-uniform B-Splines Every control point must have a corresponing t-value This is calle 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 iscontinuous. 7 8 The Stanar Knot Vector The stanar knot vector begins an ens with a four-fol knot: e.g., for 5 control points T (0, 0, 0, 0, 1, 2,, 4, 5, 5, 5, 5,) This means that the B-Spline will not loose the last point(s), an will behave correctly near the enpoints. Rational BSplines So far we have represente curves as curves with one free parameter (t) an three epenent parameters (x, y, x): X(t) a xt + b xt 2 + c xt + Y(t) a yt + b yt 2 + c yt + Z(t) a zt + b zt 2 + c zt + What else coul we o? 9 40 Homogenous Coorinates! Why not use four epenent functions? X(t) a xt + b xt 2 + c xt + Y(t) a yt + b yt 2 + c yt + Z(t) a zt + b zt 2 + c zt + W(t) a wt + b wt 2 + c wt + The image position at t is X(t)/W(t), Y(t)/W(t) an Z(t)/W(t) What oes 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 shoul the increment in t be? Depens on the projection Very small to avoi gaps in the projection But this makes curves very expensive 41 42

8 Closing Argument - NURBS A remarkable property (trust me): The curve mae by computing a rational BSpline for all t an then projecting it is the same curve mae by projecting the control points, an then computing the 2D BSpline between the control points So Non-Uniform Rational B-Splines are rawn by projecting the control points, an then recursively computing the (minimal set of) intermeiate points in 2D 4