Plot B-Spline Curve by MATLAB
1. Plot B-Spline Basis Function
When draw Bezier curve, you only need to know Bernstein basis function, life was easier.Bernstein basis function solely as a function of the number of control points. Now you have a lot more flexibility, but you also have a lot more to worry about. In addition to control points, the B-Spline basis function must account for the degree of the cruve, as well as the ranges defined by the knot vector. The resulting basis functions are defined not by Bernstein polynomials, but by the Cox-de Boor recursion formulas. [Ref. : Focus on Cruves and Surfaces]
當畫Bezier曲線時，事情要簡單些，因Bezier曲線由Bernstein基函式確定，而Bernstein基函式只與控制頂點數有關。當畫B-Spline曲線時，有了更多的靈活性，即有了區域性修改能力，但是你需要考慮的事就更多了。除了控制頂點外，B-Spline曲線的基函式必須解釋曲線的次數和節點向量定義的範圍。即B-Spline曲線的基函式由Cox-de Boor遞迴方法定義：
Imagine I want to draw a fourh-order(k=4) cubic curve with 4 control points and I choose a knot vector of [x]=[0,0,0,0,1,1,1,1]. The knot vector forces each control point to affect the entire curve.
假如我想畫一個四個控制頂點形成的四階三次曲線，選擇節點向量為[x]=[0,0,0,0,1,1,1,1]。節點向量迫使每個控制頂點的改變影響到整個曲線。
使用B樣條基函式的遞迴公式畫出各階基函式的圖形。MATLAB程式碼如下：
1: %------------------------------------------------------------------------- 2: % Imagine I want to draw a fourth-order cubic curve with 4 control points 3: % and I choose a knot vector of [x]=[0,0,0,0,1,1,1,1]. 4: %------------------------------------------------------------------------- 5: 6: t=0:0.01:1; % knot vector range 7: 8: %-------------------------------------------------------------------------9: % 1. First-order basis functions for k=4 [x]=[0,0,0,0,1,1,1,1]
10: % N11=0; 11: % N21=0; 12: % N31=0; 13: % N41=1; 14: %------------------------------------------------------------------------- 15: 16: N11=0; 17: N21=0; 18: N31=0; 19: N41=1; 20: 21: subplot(2,2,1); 22: plot(t,N11,t,N21,t,N31,t,N41); 23: 24: %-------------------------------------------------------------------------25: % 2. Second-order basis functions for k=4 [x]=[0,0,0,0,1,1,1,1]
26: % N12=0; 27: % N22=0; 28: % N32=1-t; 29: % N42=t; 30: %------------------------------------------------------------------------- 31: 32: N12=0; 33: N22=0; 34: N32=1-t; 35: N42=t; 36: 37: subplot(2,2,2); 38: plot(t,N12,t,N22,t,N32,t,N42) 39: 40: %-------------------------------------------------------------------------41: % 3. Third-order basis functions for k=4 [x]=[0,0,0,0,1,1,1,1]
42: % N13=0; 43: % N23=(1-t)^2; 44: % N33=2t(1-t); 45: % N43=t^2; 46: %------------------------------------------------------------------------- 47: 48: N13=0; 49: N23=(1-t).^2; 50: N33=2*t.*(1-t); 51: N43=t.^2; 52: 53: subplot(2,2,3); 54: plot(t,N13,t,N23,t,N33,t,N43); 55: 56: %-------------------------------------------------------------------------57: % 4. Fourth-order basis functions for k=4 [x]=[0,0,0,0,1,1,1,1]
58: % N14=(1-t)^3; 59: % N24=3t(1-t)^2; 60: % N34=3(1-t)t^2; 61: % N44=t^3; 62: %------------------------------------------------------------------------- 63: 64: N14=(1-t).^3; 65: N24=3*t.*(1-t).^2; 66: N34=3*(1-t).*t.^2; 67: N44=t.^3; 68: 69: subplot(2,2,4); 70: plot(t,N14,t,N24,t,N34,t,N44);
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用MATLAB畫出各階基函式如下圖所示：
三次均勻B樣條基函式Ni,3(u)的圖形由MATLAB生成如下所示:
生成此圖形的MATLAB程式碼如下:
1: %------------------------------------------------------------------------------ 2: % 均勻B樣條基最簡單的形式是取節點為整數：Ti=i(i=0,1,2,...,n), 3: % 令：t-ti=u，則引數u的取值範圍為[0,1]。則得Ni,3(u)如下式： 4: % | u^3 / 6; u=[0,1] 5: % | (-3u^3 + 3u^2 + 3u + 1) / 6; u=[0,1] 6: % Ni,3(u) = | (3u^3 - 6u^2 + 4) / 6; u=[0,1] 7: % | (-u^3 + 3u^2 - 3u + 1) / 6; u=[0,1] 8: % 9: %------------------------------------------------------------------------------ 10: 11: u=0:0.01:1; 12: N03=u.^3/6; 13: N13=(-3*u.^3 + 3*u.^2 + 3*u + 1)/6; 14: N23=(3*u.^3 - 6*u.^2 + 4) / 6; 15: N33=(-u.^3 + 3*u.^2 - 3*u + 1) / 6; 16: 17: line(u, N03, 'Color', 'r');
18: line(u+1, N13, 'Color', 'g');
19: line(u+2, N23, 'Color', 'b');
20: line(u+3, N33, 'Color', 'y');.csharpcode, .csharpcode pre { font-size: small; color: black; font-family: consolas, "Courier New", courier, monospace; background-color: #ffffff; /*white-space: pre;*/ } .csharpcode pre { margin: 0em; } .csharpcode .rem { color: #008000; } .csharpcode .kwrd { color: #0000ff; } .csharpcode .str { color: #006080; } .csharpcode .op { color: #0000c0; } .csharpcode .preproc { color: #cc6633; } .csharpcode .asp { background-color: #ffff00; } .csharpcode .html { color: #800000; } .csharpcode .attr { color: #ff0000; } .csharpcode .alt { background-color: #f4f4f4; width: 100%; margin: 0em; } .csharpcode .lnum { color: #606060; }
2. Plot B-Spline Curve
已知平面上五個頂點向量V0(0,1), V1(1,1), V2(1,0), V3(1,-1), V4(2,-1), 要求構造一條三次均勻B樣條曲線,並做出圖形.
使用MATLAB程式碼如下:
1: %----------------------------------------------------------- 2: % Plot Cubic Uniform B-Spline Curve.3: % Just for Testing, Welcome your advice: [email protected]
4: % 5: % Date : 2011-12-28 21:31 6: % 7: %----------------------------------------------------------- 8: 9: % Knot Vector range. 10: u=0:0.01:1; 11: 12: % Control Points.13: % You can change the control point's number and value
14: % to test the effect. 15: V0=[0 1]; 16: V1=[1 1]; 17: V2=[1 0]; 18: V3=[1 -1]; 19: V4=[2 -1]; 20: 21: % Basis Functions. 22: N03=(-u.^3 + 3*u.^2 - 3*u + 1) / 6; 23: N13=(3*u.^3 - 6*u.^2 + 4) / 6; 24: N23=(-3*u.^3 + 3*u.^2 + 3*u + 1)/6; 25: N33=u.^3/6; 26: 27: % Calculate every segment. 28: r0x=N03 * V0(1) + N13 * V1(1) + N23 * V2(1) + N33 * V3(1); 29: r0y=N03 * V0(2) + N13 * V1(2) + N23 * V2(2) + N33 * V3(2); 30: r1x=N03 * V1(1) + N13 * V2(1) + N23 * V3(1) + N33 * V4(1); 31: r1y=N03 * V1(2) + N13 * V2(2) + N23 * V3(2) + N33 * V4(2); 32: 33: % Plot the Control Polygon.34: plot(V0(1), V0(2), 'Marker', 'o'); hold on;
35: plot(V0(1), V0(2), 'Marker', 'o'); hold on;
36: plot(V1(1), V1(2), 'Marker', 'o'); hold on;
37: plot(V2(1), V2(2), 'Marker', 'o'); hold on;
38: plot(V3(1), V3(2), 'Marker', 'o'); hold on;
39: 40: % Plot the Uniform B-Spline Curve.41: line('XData', r0x, 'YData', r0y, 'Color', 'r');
42: line('XData', r1x, 'YData', r1y, 'Color', 'g');
.csharpcode, .csharpcode pre { font-size: small; color: black; font-family: consolas, "Courier New", courier, monospace; background-color: #ffffff; /*white-space: pre;*/ } .csharpcode pre { margin: 0em; } .csharpcode .rem { color: #008000; } .csharpcode .kwrd { color: #0000ff; } .csharpcode .str { color: #006080; } .csharpcode .op { color: #0000c0; } .csharpcode .preproc { color: #cc6633; } .csharpcode .asp { background-color: #ffff00; } .csharpcode .html { color: #800000; } .csharpcode .attr { color: #ff0000; } .csharpcode .alt { background-color: #f4f4f4; width: 100%; margin: 0em; } .csharpcode .lnum { color: #606060; } 生成圖形如下所示:
如圖所示，B樣條曲線由兩段組成，控制頂點由“0”標出。
試求由特徵頂點V0=[-1,1], V1=[1,1], V2=[1,-1], V3=[-1,-1]決定的閉合的三次均勻B樣條曲線，並做出圖形。適當修改上述MATLAB程式碼，即可得到所求B樣條曲線，如下圖所示：
3.結論:
對於均勻B樣條基函式，由於節點向量均勻遞增，所以在每兩個節點組成的區間的距離相等。利用基函式的的遞推公式可以計算出每個區間上的函式表示式。對於上例中的均勻B樣條基每個區間取值範圍都是從0到1，通過偏移X軸，可畫出B樣條的基函式。
通過MATLAB畫出B樣條曲線的基函式，操作簡單，便於對B樣條基函式的理解。在理解B樣條基函式後，會對B樣條曲線的理解更加深刻。繼續加油！！