[ Pobierz całość w formacie PDF ]
.The curve consists of n - 1 segments and the twoauxiliary control points of each segment are calculated automatically.Such a curve is usually pleasing to the eye and rarely needs to be edited.However,if it is not satisfactory, it can be modified by moving the auxiliary control points.Thereare 2(n - 1) of them, which allows for flexible control.A good program should displaythe auxiliary points and should make it easy for the user to grab and move any of them.The well-known drawing program Adobe Illustrator [Adobe 04] uses a similar ap-proach.The user specifies points with the mouse.At each point Pi, the user presses themouse button to fix Pi, then drags the mouse before releasing the button, which definestwo symmetrical points, X (following Pi) and Y (preceding it).Releasing the button isa signal to the program to draw the segment from Pi-1 to Pi (Figure 6.17).releaseXipressdragPiYi-1Xi-1Bezier segmentPi-1Figure 6.17: Construction of Xi and Yi by Click and Drag.Example: We apply this method to the six points P0 =(1/2, 0), P1 =(1/2, 1/2),P2 =(0, 1), P3 =(1, 3/2), P4 =(3/2, 1), and P5 =(1, 1/2).The six points yield three 214 6.B�zier Approximationcurve segments and the main step is to calculate the two intermediate points for eachof the three segments.This is trivial and it results in:X1 = P1 +(P2 - P0)/6 =(5/12, 2/3), Y1 =P2 - (P3 - P1)/6 =(-1/12, 5/6),X2 = P2 +(P3 - P1)/6 =(1/12, 7/6), Y2 =P3 - (P4 - P2)/6 =(3/4, 3/2),X3 = P3 +(P4 - P2)/6 =(5/4, 3/2), Y3 =P4 - (P5 - P3)/6 =(3/2, 7/6).Once the points are available, the three segments can easily be calculated.Each is acubic B�zier segment based on a group of four points.The groups are[P1, X1, Y1, P2], [P2, X2, Y2, P3], [P3, X3, Y3, P4],and the three curve segments areP1(t) =(1 - t)3P1 +3t(1 - t)2X1 +3t2(1 - t)Y1 + t3P2= (2 - t - 5t2 +4t3)/4, (1 + t)/2 ,P2(t) =(1 - t)3P2 +3t(1 - t)2X2 +3t2(1 - t)Y2 + t3P3= (t +7t2 - 4t3)/4, (2 + t + t2 - t3)/2 ,P3(t) =(1 - t)3P3 +3t(1 - t)2X3 +3t2(1 - t)Y3 + t3P4= (4 + 3t - t3)/4, (3 - 2t2 + t3)/2.The 12 points and the three segments are shown in Figure 6.18 (where the segmentshave been separated intentionally), as well as the code for the entire example.6.13 An Interpolating B�zier Curve: IIThe approach outlined in this section calculates an interpolating B�zier curve by solvingequations.Given a set of n+1 data points Q0, Q1,., Qn, we select n+1 values ti suchthat P(ti) =Qi.We require that whenever t reaches one of the values ti, the curve willpass through a point Qi.The values ti don t have to be equally spaced, which providescontrol over the speed of the curve.All that s needed to calculate the curve is tocompute the right set of n+1 control points Pi.This is done by setting and solving theset of n + 1 linear equations P(t0) =Q0, P(t1) =Q1,., P(tn) =Qn that s expressedin matrix notation as follows:�# �# �# �# �# �#Bn,0(t0) Bn,1(t0).Bn,n(t0)P0 Q0Bn,0(t1) Bn,1(t1).Bn,n(t1)P1 Q1�# �# �# �# �# �#�# �# �# �# �# �#=.(6.28).�# �# �# �# �# �#.Pn QnBn,0(tn) Bn,1(tn).Bn,n(tn)This set of equations can be expressed as MP = Q and it is easily solved by invertingM numerically.The solution is P = M-1Q.If we select t0 = 0, the top row ofEquation (6.28) yields P0 = Q0.Similarly, if we select tn = 1, the bottom row of 6.13 An Interpolating B�zier Curve: II 215Y2XP 331.4P (t)3P (t)2X21.2Y3P1 4P2Y10.8P (t)1X10.6P1P50.40.2P0.25 0.5 0.75 1 1.25 1.5(* Interpolating Bezier Curve: I *)Clear[p0,p1,p2,p3,p4,p5,x1,x2,x3,y1,y2,y3,c1,c2,c3,g1,g2,g3,g4];p0={1/2,0}; p1={1/2,1/2}; p2={0,1};p3={1,3/2}; p4={3/2,1}; p5={1,1/2};x1=p1+(p2-p0)/6;x2=p2+(p3-p1)/6;x3=p3+(p4-p2)/6;y1=p2-(p3-p1)/6;y2=p3-(p4-p2)/6;y3=p4-(p5-p3)/6;c1[t_]:=Simplify[(1-t)^3 p1+3t(1-t)^2 x1+3t^2(1-t) y1+t^3 p2]c2[t_]:=Simplify[(1-t)^3 p2+3t(1-t)^2 x2+3t^2(1-t) y2+t^3 p3]c3[t_]:=Simplify[(1-t)^3 p3+3t(1-t)^2 x3+3t^2(1-t) y3+t^3 p4]g1=ListPlot[{p0,p1,p2,p3,p4,p5,x1,x2,x3,y1,y2,y3},Prolog->AbsolutePointSize[4], PlotRange->All,AspectRatio->Automatic, DisplayFunction->Identity]g2=ParametricPlot[c1[t], {t,0,.9}, DisplayFunction->Identity]g3=ParametricPlot[c2[t], {t,0.1,.9}, DisplayFunction->Identity]g4=ParametricPlot[c3[t], {t,0.1,1}, DisplayFunction->Identity]Show[g1,g2,g3,g4, DisplayFunction->$DisplayFunction]Figure 6.18: An Interpolating B�zier Curve. 216 6.B�zier ApproximationEquation (6.28) yields Pn = Qn.This decreases the number of equations from n +1 ton - 1.The disadvantage of this approach is that any changes in the ti s require a recalcu-lation of M and, consequently, of M-1.If controlling the speed of the curve is not important, we can select the n +1equally-spaced values ti = i/n.Equation (6.28) can now be written�# �# �# �# �# �#Bn,0(0/n) Bn,1(0/n).Bn,n(0/n)P0 Q0Bn,0(1/n) Bn,1(1/n).Bn,n(1/n)P1 Q1�# �# �# �# �# �#�# �# �# �# �# �#=.(6.29).�# �# �# �# �# �#.Pn QnBn,0(n/n) Bn,1(n/n).Bn,n(n/n)Now, if the data points Qi are moved, matrix M (or, rather, M-1) doesn t have to berecalculated.If we number the rows and columns of M 0 through n, then a generalelement of M is given byn n!(n - i)n-jijMij = Bn,j(i/n) = (i/n)j(1 - i/n)n-j =.j j!(n - j)!nnSuch elements can be calculated, if desired, as exact rational integers, instead of (ap-proximate) floating-point numbers.Example: We use Equation (6.29) to compute the interpolating B�zier curve thatpasses through the four points Q0 =(0, 0), Q1 =(1, 1), Q2 =(2, 1), and Q3 =(3, 0).Since the curve has to pass through the first and last point, we get P0 = Q0 =(0, 0)and P3 = Q3 =(3, 0).Since the four given points are equally spaced, it makes senseto assume that P(1/3) = Q1 and P(2/3) = Q2.We therefore end up with the twoequations3(1/3)(1 - 1/3)2P1 + 3(1/3)2(1 - 1/3)P2 +(1/3)3(3, 0) =(1, 1),3(2/3)(1 - 2/3)2P1 + 3(2/3)2(1 - 2/3)P2 +(2/3)3(3, 0) =(2, 1),that are solved to yield P1 =(1, 3/2) and P2 =(2, 3/2).The curve isP(t) =(1 - t)3(0, 0) + 3t(1 - t)2(1, 3/2) + 3t2(1 - t)(2, 3/2) + t3(3, 0).Exercise 6.20: Plot the curve and the eight points. 6.14 Nonparametric B�zier Curves 2176.14 Nonparametric B�zier CurvesThe explicit representation of a curve (Section 1.3) has the familiar form y = f(x) [ Pobierz całość w formacie PDF ]
zanotowane.pl doc.pisz.pl pdf.pisz.pl centka.pev.pl
.The curve consists of n - 1 segments and the twoauxiliary control points of each segment are calculated automatically.Such a curve is usually pleasing to the eye and rarely needs to be edited.However,if it is not satisfactory, it can be modified by moving the auxiliary control points.Thereare 2(n - 1) of them, which allows for flexible control.A good program should displaythe auxiliary points and should make it easy for the user to grab and move any of them.The well-known drawing program Adobe Illustrator [Adobe 04] uses a similar ap-proach.The user specifies points with the mouse.At each point Pi, the user presses themouse button to fix Pi, then drags the mouse before releasing the button, which definestwo symmetrical points, X (following Pi) and Y (preceding it).Releasing the button isa signal to the program to draw the segment from Pi-1 to Pi (Figure 6.17).releaseXipressdragPiYi-1Xi-1Bezier segmentPi-1Figure 6.17: Construction of Xi and Yi by Click and Drag.Example: We apply this method to the six points P0 =(1/2, 0), P1 =(1/2, 1/2),P2 =(0, 1), P3 =(1, 3/2), P4 =(3/2, 1), and P5 =(1, 1/2).The six points yield three 214 6.B�zier Approximationcurve segments and the main step is to calculate the two intermediate points for eachof the three segments.This is trivial and it results in:X1 = P1 +(P2 - P0)/6 =(5/12, 2/3), Y1 =P2 - (P3 - P1)/6 =(-1/12, 5/6),X2 = P2 +(P3 - P1)/6 =(1/12, 7/6), Y2 =P3 - (P4 - P2)/6 =(3/4, 3/2),X3 = P3 +(P4 - P2)/6 =(5/4, 3/2), Y3 =P4 - (P5 - P3)/6 =(3/2, 7/6).Once the points are available, the three segments can easily be calculated.Each is acubic B�zier segment based on a group of four points.The groups are[P1, X1, Y1, P2], [P2, X2, Y2, P3], [P3, X3, Y3, P4],and the three curve segments areP1(t) =(1 - t)3P1 +3t(1 - t)2X1 +3t2(1 - t)Y1 + t3P2= (2 - t - 5t2 +4t3)/4, (1 + t)/2 ,P2(t) =(1 - t)3P2 +3t(1 - t)2X2 +3t2(1 - t)Y2 + t3P3= (t +7t2 - 4t3)/4, (2 + t + t2 - t3)/2 ,P3(t) =(1 - t)3P3 +3t(1 - t)2X3 +3t2(1 - t)Y3 + t3P4= (4 + 3t - t3)/4, (3 - 2t2 + t3)/2.The 12 points and the three segments are shown in Figure 6.18 (where the segmentshave been separated intentionally), as well as the code for the entire example.6.13 An Interpolating B�zier Curve: IIThe approach outlined in this section calculates an interpolating B�zier curve by solvingequations.Given a set of n+1 data points Q0, Q1,., Qn, we select n+1 values ti suchthat P(ti) =Qi.We require that whenever t reaches one of the values ti, the curve willpass through a point Qi.The values ti don t have to be equally spaced, which providescontrol over the speed of the curve.All that s needed to calculate the curve is tocompute the right set of n+1 control points Pi.This is done by setting and solving theset of n + 1 linear equations P(t0) =Q0, P(t1) =Q1,., P(tn) =Qn that s expressedin matrix notation as follows:�# �# �# �# �# �#Bn,0(t0) Bn,1(t0).Bn,n(t0)P0 Q0Bn,0(t1) Bn,1(t1).Bn,n(t1)P1 Q1�# �# �# �# �# �#�# �# �# �# �# �#=.(6.28).�# �# �# �# �# �#.Pn QnBn,0(tn) Bn,1(tn).Bn,n(tn)This set of equations can be expressed as MP = Q and it is easily solved by invertingM numerically.The solution is P = M-1Q.If we select t0 = 0, the top row ofEquation (6.28) yields P0 = Q0.Similarly, if we select tn = 1, the bottom row of 6.13 An Interpolating B�zier Curve: II 215Y2XP 331.4P (t)3P (t)2X21.2Y3P1 4P2Y10.8P (t)1X10.6P1P50.40.2P0.25 0.5 0.75 1 1.25 1.5(* Interpolating Bezier Curve: I *)Clear[p0,p1,p2,p3,p4,p5,x1,x2,x3,y1,y2,y3,c1,c2,c3,g1,g2,g3,g4];p0={1/2,0}; p1={1/2,1/2}; p2={0,1};p3={1,3/2}; p4={3/2,1}; p5={1,1/2};x1=p1+(p2-p0)/6;x2=p2+(p3-p1)/6;x3=p3+(p4-p2)/6;y1=p2-(p3-p1)/6;y2=p3-(p4-p2)/6;y3=p4-(p5-p3)/6;c1[t_]:=Simplify[(1-t)^3 p1+3t(1-t)^2 x1+3t^2(1-t) y1+t^3 p2]c2[t_]:=Simplify[(1-t)^3 p2+3t(1-t)^2 x2+3t^2(1-t) y2+t^3 p3]c3[t_]:=Simplify[(1-t)^3 p3+3t(1-t)^2 x3+3t^2(1-t) y3+t^3 p4]g1=ListPlot[{p0,p1,p2,p3,p4,p5,x1,x2,x3,y1,y2,y3},Prolog->AbsolutePointSize[4], PlotRange->All,AspectRatio->Automatic, DisplayFunction->Identity]g2=ParametricPlot[c1[t], {t,0,.9}, DisplayFunction->Identity]g3=ParametricPlot[c2[t], {t,0.1,.9}, DisplayFunction->Identity]g4=ParametricPlot[c3[t], {t,0.1,1}, DisplayFunction->Identity]Show[g1,g2,g3,g4, DisplayFunction->$DisplayFunction]Figure 6.18: An Interpolating B�zier Curve. 216 6.B�zier ApproximationEquation (6.28) yields Pn = Qn.This decreases the number of equations from n +1 ton - 1.The disadvantage of this approach is that any changes in the ti s require a recalcu-lation of M and, consequently, of M-1.If controlling the speed of the curve is not important, we can select the n +1equally-spaced values ti = i/n.Equation (6.28) can now be written�# �# �# �# �# �#Bn,0(0/n) Bn,1(0/n).Bn,n(0/n)P0 Q0Bn,0(1/n) Bn,1(1/n).Bn,n(1/n)P1 Q1�# �# �# �# �# �#�# �# �# �# �# �#=.(6.29).�# �# �# �# �# �#.Pn QnBn,0(n/n) Bn,1(n/n).Bn,n(n/n)Now, if the data points Qi are moved, matrix M (or, rather, M-1) doesn t have to berecalculated.If we number the rows and columns of M 0 through n, then a generalelement of M is given byn n!(n - i)n-jijMij = Bn,j(i/n) = (i/n)j(1 - i/n)n-j =.j j!(n - j)!nnSuch elements can be calculated, if desired, as exact rational integers, instead of (ap-proximate) floating-point numbers.Example: We use Equation (6.29) to compute the interpolating B�zier curve thatpasses through the four points Q0 =(0, 0), Q1 =(1, 1), Q2 =(2, 1), and Q3 =(3, 0).Since the curve has to pass through the first and last point, we get P0 = Q0 =(0, 0)and P3 = Q3 =(3, 0).Since the four given points are equally spaced, it makes senseto assume that P(1/3) = Q1 and P(2/3) = Q2.We therefore end up with the twoequations3(1/3)(1 - 1/3)2P1 + 3(1/3)2(1 - 1/3)P2 +(1/3)3(3, 0) =(1, 1),3(2/3)(1 - 2/3)2P1 + 3(2/3)2(1 - 2/3)P2 +(2/3)3(3, 0) =(2, 1),that are solved to yield P1 =(1, 3/2) and P2 =(2, 3/2).The curve isP(t) =(1 - t)3(0, 0) + 3t(1 - t)2(1, 3/2) + 3t2(1 - t)(2, 3/2) + t3(3, 0).Exercise 6.20: Plot the curve and the eight points. 6.14 Nonparametric B�zier Curves 2176.14 Nonparametric B�zier CurvesThe explicit representation of a curve (Section 1.3) has the familiar form y = f(x) [ Pobierz całość w formacie PDF ]
