Numerical Analysis Timothy Sauer Second Edition

Transcription

1 Numerical Analysis Timothy Sauer Second Edition

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3 3.5 Bézier Curves Compile a list of 121 hourly temperatures over five consecutive days from a weather data website. Let x0=0:6:120 denote hours, and y0 denote the temperatures at hours 0,6,12,...,120. Carry out steps (a) (c) of Computer Problem 14, suitably adapted. 3.5 BÉZIER CURVES Bézier curves are splines that allow the user to control the slopes at the knots. In return for the extra freedom, the smoothness of the first and second derivatives across the knot, which are automatic features of the cubic splines of the previous section, are no longer guaranteed. Bézier splines are appropriate for cases where corners (discontinuous first derivatives) and abrupt changes in curvature (discontinuous second derivatives) are occasionally needed. Pierre Bézier developed the idea during his work for the Renault automobile company. The same idea was discovered independently by Paul de Casteljau, working for Citroen, a rival automobile company. It was considered an industrial secret by both companies, and the fact that both had developed the idea came to light only after Bézier published his research. Today the Bézier curve is a cornerstone of computer-aided design and manufacturing. Each piece of a planar Bézier spline is determined by four points (x 1,y 1 ), (x 2,y 2 ),(x 3,y 3 ),(x 4,y 4 ). The first and last of the points are endpoints of the spline curve, and the middle two are control points, as shown in Figure The curve leaves (x 1,y 1 ) along the tangent direction (x 2 x 1,y 2 y 1 ) and ends at (x 4,y 4 ) along the tangent direction (x 4 x 3,y 4 y 3 ). The equations that accomplish this are expressed as a parametric curve (x(t),y(t)) for 0 t 1. y 3 (x 2, y 2 ) (x 3, y 3 ) 2 (x 4, y 4 ) 1 (x 1, y 1 ) x Figure 3.14 Bézier curve of Example The points (x 1,y 1 ) and (x 4,y 4 ) are spline points, while (x 2,y 2 ) and (x 3,y 3 ) are control points. Bézier curve Given endpoints (x 1,y 1 ),(x 4,y 4 ) control points (x 2,y 2 ),(x 3,y 3 ) Set b x = 3(x 2 x 1 ) c x = 3(x 3 x 2 ) b x d x = x 4 x 1 b x c x b y = 3(y 2 y 1 ) c y = 3(y 3 y 2 ) b y d y = y 4 y 1 b y c y. 179

4 180 CHAPTER 3 Interpolation The Bézier curve is defined for 0 t 1by x(t) = x 1 + b x t + c x t 2 + d x t 3 y(t) = y 1 + b y t + c y t 2 + d y t 3. It is easy to check the claims of the previous paragraph from the equations. In fact, according to Exercise 11, and the analogous facts hold for y(t). x(0) = x 1 x (0) = 3(x 2 x 1 ) x(1) = x 4 x (1) = 3(x 4 x 3 ), (3.25) EXAMPLE 3.15 Find the Bézier curve (x(t), y(t)) through the points (x, y) = (1, 1) and (2, 2) with control points (1,3) and (3,3). The four points are (x 1,y 1 ) = (1,1),(x 2,y 2 ) = (1,3),(x 3,y 3 ) = (3,3), and (x 4,y 4 ) = (2,2). The Bézier formulas yield b x = 0,c x = 6,d x = 5 and b y = 6,c y = 6,d y = 1. The Bézier spline x(t) = 1 + 6t 2 5t 3 y(t) = 1 + 6t 6t 2 + t 3 is shown in Figure 3.14 along with the control points. Bézier curves are building blocks that can be stacked to fit arbitrary function values and slopes. They are an improvement over cubic splines, in the sense that the slopes at the nodes can be specified as the user wants them. However, this freedom comes at the expense of smoothness: The second derivatives from the two different directions generally disagree at the nodes. In some applications, this disagreement is an advantage. As a special case, when the control points equal the endpoints, the spline is a simple line segment, as shown next. EXAMPLE 3.16 Prove that the Bézier spline with (x 1,y 1 ) = (x 2,y 2 ) and (x 3,y 3 ) = (x 4,y 4 ) is a line segment. The Bézier formulas show that the equations are x(t) = x 1 + 3(x 4 x 1 )t 2 2(x 4 x 1 )t 3 = x 1 + (x 4 x 1 )t 2 (3 2t) y(t) = y 1 + 3(y 4 y 1 )t 2 2(y 4 y 1 )t 3 = y 1 + (y 4 y 1 )t 2 (3 2t) for 0 t 1. Every point in the spline has the form (x(t),y(t)) = (x 1 + r(x 4 x 1 ),y 1 + r(y 4 y 1 )) = ((1 r)x 1 + rx 4,(1 r)y 1 + ry 4 ), where r = t 2 (3 2t). Since 0 r 1, each point lies on the line segment connecting (x 1,y 1 ) and (x 4,y 4 ). Bézier curves are simple to program and are often used in drawing software. A freehand curve in the plane can be viewed as a parametric curve (x(t),y(t)) and represented by a Bézier spline. The equations are implemented in the following Matlab freehand drawing program. The user clicks the mouse once to fix a starting point (x 0,y 0 ) in the plane, and 180

5 3.5 Bézier Curves 181 three more clicks to mark the first control point, second control point, and endpoint. A Bézier spline is drawn between the start and end points. Each subsequent triple of mouse clicks extends the curve further, using the previous endpoint as the starting point for the next piece. The Matlab command ginput is used to read the mouse location. Figure 3.15 shows a screenshot of bezierdraw.m. Figure 3.15 Program 3.7 built from Bézier curves. Screenshot of MATLAB code bezierdraw.m, including direction vectors drawn at each control point. %Program 3.7 Freehand Draw Program Using Bezier Splines %Click in Matlab figure window to locate first point, and click % three more times to specify 2 control points and the next % spline point. Continue with groups of 3 points to add more % to the curve. Press return to terminate program. function bezierdraw plot([-1 1],[0,0], k,[0 0],[-1 1], k );hold on t=0:.02:1; [x,y]=ginput(1); % get one mouse click while(0 == 0) [xnew,ynew] = ginput(3); % get three mouse clicks if length(xnew) < 3 break % if return pressed, terminate end x=[x;xnew];y=[y;ynew]; % plot spline points and control pts plot([x(1) x(2)],[y(1) y(2)], r:,x(2),y(2), rs ); plot([x(3) x(4)],[y(3) y(4)], r:,x(3),y(3), rs ); plot(x(1),y(1), bo,x(4),y(4), bo ); bx=3*(x(2)-x(1)); by=3*(y(2)-y(1)); % spline equations... cx=3*(x(3)-x(2))-bx;cy=3*(y(3)-y(2))-by; dx=x(4)-x(1)-bx-cx;dy=y(4)-y(1)-by-cy; xp=x(1)+t.*(bx+t.*(cx+t*dx)); % Horner s method yp=y(1)+t.*(by+t.*(cy+t*dy)); plot(xp,yp) % plot spline curve x=x(4);y=y(4); end hold off % promote last to first and repeat Although our discussion has been restricted to two-dimensional Bézier curves, the defining equations are easily extended to three dimensions, in which they are called Bézier space curves. Each piece of the spline requires four (x,y,z)points two endpoints and two control points just as in the two-dimensional case. Examples of Bézier space curves are explored in the exercises. 181

6 182 CHAPTER 3 Interpolation 3.5 Exercises 1. Find the one-piece Bézier curve (x(t),y(t)) defined by the given four points. (a) (0,0), (0,2), (2,0), (1,0) (b) (1,1), (0,0), ( 2,0), ( 2,1) (c) (1,2), (1,3), (2,3), (2,2) 2. Find the first endpoint, two control points, and last endpoint for the following one-piece Bézier curves. { { x(t) = 1 + 6t 2 + 2t 3 x(t) = 3 + 4t t 2 + 2t 3 (a) y(t) = 1 t + t 3 (b) y(t) = 2 t + t 2 + 3t 3 (c) { x(t) = 2 + t 2 t 3 y(t) = 1 t + 2t 3 3. Find the three-piece Bézier curve forming the triangle with vertices (1,2),(3,4), and (5,1). 4. Build a four-piece Bézier spline that forms a square with sides of length Describe the character drawn by the following two-piece Bezier curve: (0,2) (1,2) (1,1) (0,1) (0,1) (1,1) (1,0) (0,0) 6. Describe the character drawn by the following three-piece Bezier curve: (0,1) (0,1) (0,0) (0,0) (0,0) (0,1) (1,1) (1,0) (1,0) (1,1) (2,1) (2,0) 7. Find a one-piece Bézier spline that has vertical tangents at its endpoints ( 1,0) and (1,0) and that passes through (0,1). 8. Find a one-piece Bézier spline that has a horizontal tangent at endpoint (0,1) and a vertical tangent at endpoint (1,0) and that passes through (1/3,2/3) at t = 1/3. 9. Find the one-piece Bézier space curve (x(t),y(t),z(t)) defined by the four points. (a) (1,0,0),(2,0,0),(0,2,1),(0,1,0) (b) (1,1,2),(1,2,3),( 1,0,0),(1,1,1) (c) (2,1,1),(3,1,1),(0,1,3),(3,1,3) 10. Find the knots and control points for the following Bézier space curves. (a) x(t) =1 + 6t 2 + 2t 3 y(t) =1 t + t 3 z(t) = 1 + t + 6t 2 (b) x(t) =3 + 4t t 2 + 2t 3 y(t) =2 t + t 2 + 3t 3 z(t) = 3 + t + t 2 t 3 (c) x(t) =2 + t 2 t 3 y(t) =1 t + 2t 3 z(t) = 2t Prove the facts in (3.25), and explain how they justify the Bézier formulas. 12. Given (x 1,y 1 ),(x 2,y 2 ),(x 3,y 3 ), and (x 4,y 4 ), show that the equations x(t) = x 1 (1 t) 3 + 3x 2 (1 t) 2 t + 3x 3 (1 t)t 2 + x 4 t 3 y(t) = y 1 (1 t) 3 + 3y 2 (1 t) 2 t + 3y 3 (1 t)t 2 + y 4 t 3 give the Bézier curve with endpoints (x 1,y 1 ),(x 4,y 4 ) and control points (x 2,y 2 ),(x 3,y 3 ). 182

7 3.5 Bézier Curves Computer Problems 1. Plot the curve in Exercise Plot the curve in Exercise Plot the letter from Bézier curves. (a) W (b) B (c) C (d) D. 3 Fonts from Bézier curves In this project, we explain how to draw letters and numerals by using two-dimensional Bézier curves. They can be implemented by modifying the Matlab code in Program 3.7 or by writing a PDF file. Modern fonts are built directly from Bézier curves, in order to be independent of the printer or imaging device. Bézier curves were a fundamental part of the PostScript language from its start in the 1980s, and the PostScript commands for drawing curves have migrated in slightly altered form to the PDF format. Here is a complete PDF file that illustrates the curve we discussed in Example %PDF obj /Length 2 0 R stream m c S endstream 2 0 obj obj /Type /Page /Parent 5 0 R /Contents 1 0 R 5 0 obj /Kids [4 0 R] /Count 1 /Type /Pages /MediaBox [ ] 3 0 obj /Pages 5 0 R /Type /Catalog xref f n n n n n trailer /Size 6 /Root 3 0 R startxref 1000 %%EOF 183