Computer Graphics. Curves and Surfaces. Hermite/Bezier Curves, (B-)Splines, and NURBS. By Ulf Assarsson

Transcription

1 Compter Graphcs Crves and Srfaces Hermte/Bezer Crves, (B-)Splnes, and NURBS By Ulf Assarsson Most of the materal s orgnally made by Edward Angel and s adapted to ths corse by Ulf Assarsson. Some materal s made by Magns Bondesson

2 Utah Teapot Most famos data set n compter graphcs Wdely avalable as a lst of 06 D vertces and the ndces that defne 2 Bezer patches A Bezer patch

3 Objectves Introdce types of crves and srfaces Explct Implct Parametrc

4 Modelng wth Crves data ponts approxmatng crve nterpolatng data pont

5 What Makes a Good Representaton? There are many ways to represent crves and srfaces Want a representaton that s Stable Smooth Easy to evalate Mst we nterpolate or can we jst come close to data? Do we need dervatves?

6 Explct Representaton Most famlar form of crve n 2D y=f(x) Cannot represent all crves Vertcal lnes Crcles Extenson to D y x y y=f(x), z=g(x) gves a crve The form y = f(x,z) defnes a srface z x

7 Implct Representaton Two dmensonal crve(s) g(x,y)=0 Mch more robst All lnes ax+by+c=0 Crcles x 2 +y 2 -r 2 =0 Three dmensons g(x,y,z)=0 defnes a srface (we cold ntersect two srfaces to get a crve)

8 Parametrc Crves Separate eqaton for each spatal varable x=x() y=y() z=z() For max mn we trace ot a crve n two or three dmensons p() p()=[x(), y(), z()] T p( mn ) p( max )

9 Selectng Fnctons Usally we can select good fnctons not nqe for a gven spatal crve Approxmate or nterpolate known data Want fnctons whch are easy to evalate Want fnctons whch are easy to dfferentate Comptaton of normals Connectng peces (segments) Want fnctons whch are smooth

10 Parametrc Lnes We can let be over the nterval (0,1) Lne connectng two ponts p 0 and p 1 p(1)= p 1 p()=(1-)p 0 +p 1 Ray from p 0 n the drecton d p()=p 0 +d p(0) = p 0 p(0) = p 0 d p(1)= p 0 +d

11 Parametrc Srfaces Srfaces reqre 2 parameters x=x(,v) y=y(,v) z=z(,v) p(,v) = [x(,v), y(,v), z(,v)] T Want same propertes as crves: Smoothness Dfferentablty Ease of evalaton y p(0,v) z p(,0) p(,1) x p(1,v)

12 Normals We can dfferentate wth respect to and v to obtan the normal at any pont p = v v v v ) /, z( ) /, y( ) /, x( ), p( = v v v v v v v v ) /, z( ) /, y( ) /, x( ), p( v v v = ), ( ), p( p n v

13 Parametrc Planes n pont-vector form p(,v) p(,v) v p(,v)=p 0 +q+vr n = q x r p 0 r q n (three-pont form p 2 q = p 1 p 0 r = p 2 p 0 ) p 1 p 0

14 Crve Segments After normalzng, each crve s wrtten p()=[x(), y(), z()] T, 1 0 In classcal nmercal methods, we desgn a sngle global crve In compter graphcs and CAD, t s better to desgn small connected crve segments p() jon pont p(1) = q(0) p(0) q() q(1) How shold we descrbe crve segments?

15 We choose Polynomals Easy to evalate Contnos and dfferentable everywhere Mst worry abot contnty at jon ponts ncldng contnty of dervatves p() q() jon pont p(1) = q(0) bt p (1) q (0) Let s worry abot that later. Frst let s scrtnze the polynomals!

16 Parametrc Polynomal Crves N j x( ) = cx y( ) = cyj z( ) = czk = 0 M j= 0 L k = 0 k Cbc polynomals gves N=M=L= Remember: x=x() y=y() z=z() p() p( mn ) p( max ) Notng that the crves for x, y and z are ndependent, we can defne each ndependently n an dentcal manner We wll se the form p( ) = where p can be any of x, y, z Let s assme cbc polynomals! L k = 0 c k k

17 Cbc Parametrc Polynomals Cbc polynomals gve balance between ease of evalaton and flexblty n desgn p( ) = k = 0 c k For coeffcents to determne for each of x, y and z Seek for ndependent condtons for varos vales of resltng n 4 eqatons n 4 nknowns for each of x, y and z Condtons are a mxtre of contnty reqrements at the jon ponts and condtons for fttng the data k

18 Objectves p 1 Introdce the types of crves Interpolatng Blendng polynomals for nterpolaton of 4 control ponts (ft crve to 4 control ponts) Hermte ft crve to 2 control ponts + 2 dervatves (tangents) Bezer 2 nterpolatng control ponts + 2 ntermedate ponts to defne the tangents B-splne To get C 1 and C 2 contnty NURBS Dfferent weghts of the control ponts Analyze them p 0 p 2 p

19 Matrx-Vector Form c k k k = = 0 ) p( = c c c c c = 2 1 defne c c T T = = ) p( then

20 Interpolatng Crve p 1 p p 0 p 2 Gven for data (control) ponts p 0, p 1,p 2, p determne cbc p() whch passes throgh them Mst fnd c 0,c 1,c 2, c Let s create an eqaton system!

21 p p 0 p 2 p 1 p p() = c 0 + c 1 + c c 0 1/ 2/ 1 Interpolaton Eqatons apply the nterpolatng condtons at =0, 1/, 2/, 1 p 0 =p(0) = c 0 p 1 =p(1/)= c 0 +(1/)c 1 +(1/) 2 c 2 +(1/) c p 2 =p(2/)= c 0 +(2/)c 1 +(2/) 2 c 2 +(2/) c p =p(1) = c 0 +c 1 +c 2 +c or n matrx form wth p = [p 0 p 1 p 2 p ] T p=ac p = I.e., c=a -1 p! # # # # # " p 0 p 1 p 2 p! $ # & # & # & = Ac = # & # & # % # "# ' 1 ) * ( +, ' 1 ) * 2 ( +, ' 2 * ), ( + ' 2 * ), ( + 2 ' 1 ) * ( +, ' 2 * ), ( $ &! &# &# &# &# &# &" %& c 0 c 1 c 2 c $ & & & & & %

22 Interpolaton Matrx Solvng for c we fnd the nterpolaton matrx MI = 1 A = c=m I p Note that M I does not depend on npt data and can be sed for each segment n x, y, and z p() = c 0 + c 1 + c c p 0 p 1 p 2 p

23 Interpolaton Matrx p() = c 0 + c 1 + c c means: x=x()=c x0 + c x1 + c x2 2 + c x y=y()=c y0 + c y1 + c y2 2 + c y z=z()=c z0 + c z1 + c z2 2 + c z p1 p2 p where c x = M I p x c y = M I p y c z = M I p z p0 p x are the x coordnates of p 0 p p y are the y coordnates of p 0 p p z are the z coordnates of p 0 p

24 Interpolatng Mltple Segments se p = [p 0 p 1 p 2 p ] T se p = [p p 4 p 5 p 6 ] T Get contnty at jon ponts bt not contnty of dervatves

25 Blendng Fnctons Rewrtng the eqaton for p() p 0 p 1 p 2 p p()= T c= T M I p = b() T p where b() = [b 0 () b 1 () b 2 () b ()] T s an array of blendng polynomals sch that p() = b 0 ()p 0 + b 1 ()p 1 + b 2 ()p 2 + b ()p b 0 () = -4.5(-1/)(-2/)(-1) b 1 () = 1.5 (-2/)(-1) b 2 () = -1.5 (-1/)(-1) b () = 4.5 (-1/)(-2/)

26 Blendng Fnctons p 1 p p 0 p 2 p() = b 0 ()p 0 + b 1 ()p 1 + b 2 ()p 2 + b ()p

27 Blendng Patches Crve: p()= T c= T M I p = b() T p Patch: p(, v) = = o j= 0 c j v j p(,v) = =o b ( ) j =0 b j (v ) p j = T MI P T M I v Each b ()b j (v) s a blendng patch Shows that we can bld and analyze srfaces from or knowledge of crves

28 Hermte Crves and Srfaces How can we get arond the lmtatons of the nterpolatng form Lack of smoothness Dscontnos dervatves at jon ponts We have for condtons (for cbcs) that we can apply to each segment Use them other than for nterpolaton Need only come close to the data

29 Hermte Form p (0) p (1) p(0) p(1) Use two nterpolatng condtons and two dervatve condtons per segment Ensres contnty and frst dervatve contnty between segments

30 Eqatons p (0) p(0) p (1) p(1) Interpolatng condtons are the same at ends p() = c 0 +c c 2 + c p(0) = p 0 = c 0 p(1) = p 1 = c 0 +c 1 +c 2 +c Dfferentatng we fnd p () = c 1 +2c c Evalatng at end ponts p (0) = p 0 = c 1 p (1) = p 1 = c 1 +2c 2 +c q = " p 0 % $ ' $ p 1 ' $ p' 0 ' $ ' # p' 1 & = " % $ ' $ ' c $ ' $ ' # &

31 Matrx Form p (0) p(0) p (1) p(1) q = " p 0 % $ ' $ p 1 ' = $ p' 0 ' $ ' # p' 1 & " % $ ' $ ' c $ ' $ ' # & Solvng, we fnd c=m H q where M H s the Hermte matrx MH = p() = T c => p() = T M H q

32 Blendng Polynomals p() = b() T q = ) b( Althogh these fnctons are smooth, the Hermte form s not sed drectly n Compter Graphcs and CAD becase we sally have control ponts bt not dervatves However, the Hermte form s the bass of the Bezer form = MH p() = T M H q =>

33 Contnty (a) (b) (c) (d) A) Non-contnos B) C 0 -contnos C) G 1 -contnos D) C 1 -contnos (C 2 -contnos) See page n Real-tme Renderng, rd ed.

34 Example Here the p and q have the same tangents at the ends of the segment bt dfferent dervatves Generate dfferent Hermte crves Ths technqes s sed n drawng applcatons

35 Reflectons shold be at least C1

36 Bezer Crves In graphcs and CAD, we do not sally have dervatve data Bezer sggested sng the same 4 data ponts as wth the cbc nterpolatng crve to approxmate the dervatves n the Hermte form

37 Approxmatng Dervatves p 1 located at =1/ p 1 p 2 p 2 located at =2/ dp( = 0) d p p p p = p'(0) 1 0 '(1) 1/ 1/ 2 p slope p (0) slope p (1) p 0 p

38 p 1 p 2 Eqatons p p '(0) 1/ p p '(1) 1/ 1 0 p p 2 Interpolatng condtons are the same p(0) = p 0 = c 0 p(1) = p = c 0 +c 1 +c 2 +c Approxmatng dervatve condtons p (0) = (p 1 - p 0 ) = c 1 p (1) = (p - p 2 ) = c 1 +2c 2 +c Solve for lnear eqatons for c=m B p p 0 p p() = c 0 +c c 2 + c p () = c 1 +2c c Bp=Ac c=a -1 Bp

39 Bezer Matrx MB = p() = T M B p = b() T p blendng fnctons

40 p 1 p 2 Blendng Fnctons p 0 p b( ) = (1 ) 2 (1 ) 2 2 (1 ) Note that all zeros are at 0 and 1 whch forces the fnctons to be smoother over (0,1) Smoother becase the crve stays nsde the convex hll, and therefore does not have room to flctate so mch.

41 Convex Hll Property All weghts wthn [0,1] and sm of all weghts = 1 (at gven ) ensres that all Bezer crves le n the convex hll of ther control ponts Hence, even thogh we do not nterpolate all the data, we cannot be too far away Bezer crve p 1 p 2 convex hll p 0 p

42 Bezer Patches Usng same data array P=[p j ] as wth nterpolatng form v p v b b v p T B B T j j j M P M = = = = ) ( ) ( ), ( 0 0 Patch les n convex hll v c v p j j j o = = = 0 ), (

43 Analyss Althogh the Bezer form s mch better than the nterpolatng form, the dervatves are not contnos at jon ponts p 1 p 2 p 0 p What shall we do to solve ths?

44 B-Splnes Bass splnes: se the data at p=[p -2 p -1 p p +1 ] T to defne crve only between p -1 and p Allows s to apply more contnty condtons to each segment For cbcs, we can have contnty of the fncton and frst and second dervatves at the jon ponts

45 Cbc B-splne p() = T M S p = b() T p M = 1 S

46 Blendng Fnctons = ) (1 6 1 ) b( convex hll property p() = T M S p = b() T p => p() = b 0 ()p 0 + b 1 ()p 1 + b 2 ()p 2 + b ()p T S M =

47 B-Splne Patches p(,v) = b() j =0 j =0 b (v) p j = T MS P T M S v defned over only 1/9 of regon

48 Splnes and Bass If we examne the cbc B-splne from the perspectve of each control (data) pont, each nteror pont contrbtes (throgh the blendng fnctons) to for segments We can rewrte p() n terms of all the data ponts along the crve as p( ) = B ( ) defnng the bass fnctons {B ()} p

49 Bass Fnctons ) ( ) ( 1) ( 2) ( 0 ) ( < + + < < < < + + = b b b b B In terms of the blendng polynomals p 0 p 1 p 2 p p p() = B () p = 0 B () 0 p +... n 1 B () n 1 p p 0 p 1 p 2 p p 4 Weghts for each pont along the crve

50 One more example 1 p 4 p 0 p 1 p 2 p p 4 p 0 0 p 1 p 2 = 2.7 = 2.7 p 0 p p() = B 0 ()p 0 + B 1 ()p 1 + B 2 ()p 2 + B ()p + B 4 ()p 4 I.e.,: p( ) = B ( ) p p 4

51 SUMMARY B-Splnes These are or control ponts, p 0 -p 8, to whch we want to approxmate a crve p 0 p 1 p 2 p p 4 p 6 p7 p 8 p 5 = % Illstraton of how the control ponts are evenly (nformly) dstrbted along the parametersaton of the crve p(). In each pont p() of the crve, for a gven, the pont s defned as a weghted sm of the closest 4 srrondng control ponts. Below are shown the weghts for each control pont along =0 8 p 0 p 1 p 2 p p 4 p 5 p 6 p 7 p 8

52 SUMMARY B-Splnes 100 % In each pont p() of the crve, for a gven, the pont s defned as a weghted sm of the closest 4 srrondng ponts. Below are shown the weghts for each pont along =0 8 Blendfncton B 1 () for pont p 1 p 0 p 1 p 2 p p 4 p 5 p 6 p 7 p 8 The weght fncton (blend fncton) B () for a pont p can ths be wrtten as a translaton of a bass fncton B(t). B () = B t (-) B(t): 100% Or complete B-splne crve p() can ths be wrtten as: t p( ) = B ( ) p

53 Generalzng Splnes We can extend to splnes of any degree Data and condtons do not have to be gven at eqally spaced vales (the knots) Nonnform and nform splnes Can have repeated knots Easest mplemented by jst repeatng a ctrl pont Can force splne to nterpolate ponts (Cox-deBoor recrson gves method of evalaton (also known as decastelja-recrson, see page 579, RTR :rd Ed. for detals)) DEMO of B-Splne crve: (make dplcate knots)

54 NURBS Nonnform Ratonal B-Splne crves and srfaces add a forth varable w to x,y,z Can nterpret as weght to gve more mportance to some control data Can also nterpret as movng to homogeneos coordnate (Reqres a perspectve dvson NURBS act correctly for perspectve vewng Qadrcs are a specal case of NURBS)

55 NURBS NURBS s smlar to B-Splnes except that: 1. The control ponts can have dfferent weghts, w, (hegher weght makes the crve go closer to that control pont) 2. The control ponts do not have to be at nform dstances (=0,1,2,...) along the parametersaton. E.g.: =0, 0.5, 0.9, 4, 14, NURBS = Non-Unform Ratonal B-Splnes The NURBS-crve s ths defned as: p() = n =0 n =0 B ()w B ()w p Dvson wth the sm of the weghts, to make the combned weghts sm p to 1, at each poston along the crve. Otherwse, a translaton of the crve s ntrodced (whch s not desrable)

56 Concder a control pont n dmensons: The weghted homogeneos-coordnate s: The dea s to se the weghts w to ncrease or decrease the mportance of a partclar control pont NURBS [ ] z y x,, = p = 1 z y x w q

57 NURBS The w-component may not be eqal to 1. Ths we mst do a perspectve dvson to get the three-dmensonal ponts: p( ) n 1 B = dw 0, p( ) = q( ) = n w( ) = 0, d Each component of p() s now a ratonal fncton n, and becase we have not restrcted the knots (the knots does not have to be nformly dstrbted), we have derved a nonnform ratonal B-splne (NURBS) crve B w

58 NURBS Allowng control ponts at non-nform dstances means that the bass fnctons B p () are beng streched and non-nformly located. E.g.: Each crve B p () shold of corse look smooth and C 2 contnos. Bt t s not so easy to draw smoothly by hand... (The sm of the weghts are stll =1 de to the dvson n prevos slde )

59 NURBS Srfaces - examples

60 NURBS If we apply an affne transformaton to a B-splne crve or srface, we get the same fncton as the B-splne derved from the transformed control ponts. Becase perspectve transformatons are not affne, most splnes wll not be handled correctly n perspectve vewng. However, the perspectve dvson embedded n the NURBS ensres that NURBS crves are handled correctly n perspectve vews. Qadrcs can be shown to be a specal case of qadratc NURBS crve; ths, we can se a sngle modelng method, NURBS crves, for the most wdely sed crves and srfaces WHAT IS IMPORTANT?