过特征多边形始末端点的四次均匀B样条曲线及其应用

Quadratic Uniform B-Spline Curve of Passing Start Point and End Point of the Characteristic Polygon and Its Application

摘要: 本文构造了过控制多边形始点和终点的四次均匀B样条曲线，并以此四次均匀B样条曲线为轴线的管道光滑拼接了轴线异面管道。该方法相较于过始末端点的三次均匀B样条曲线为轴线的管道光滑拼接轴线异面管道，通过给定位于曲线上的一些点，反算出B样条曲线的特征多边形的顶点构造的B样条曲线为轴线的管道光滑拼接轴线异面的管道有其不同的特性。具有一定的应用价值。

Abstract: In this paper, we construct a quartic uniform B-spline curve of passing start point and end point of the control polygon. Tubes whose axes are in non-coplaner are smoothly blended by tube taking a quartic uniform B-spline curve as its axis. Compared with the tube with cubic uniform B-spline curve as the axis passing through the start and end points, the tubes whose axes are in non-cop- laner are smoothly blended by the tube. By giving some points located on the curve, this method can calculate the vertices of the characteristic polygon of the B-spline curve. The constructed B- spline curve is used as the tube with axis, and tubes are smoothly blended by the tube whose axis is in non-coplaner which has different characteristics. It has certain application value.

1. 引言

我们在文 [1] 中，根据实际问题给定B样条曲线的控制多边形的顶点，构造轴线异面管道的拼接管道。在文 [2] [3] [4] 中，假定B样条曲线过某些固定的点，反过来求B样条曲线的实际控制顶点，构造轴线异面管道的拼接曲面。两种方法具有明显的优缺点：前者，拼接管道由多段光滑拼接的管道构成，光顺性较好，应用广泛，不仅能应用于轴线异面圆管道的拼接，还能应用于椭圆管道的拼接，实际应用时加工复杂。后者，拼接管道的构造可根据需要尽量简单，由一段或两段光滑拼接的管道构成，加工容易。但是，应用范围受限，不能应用于轴线异面椭圆管道的拼接。

本文研究四次均匀B样条曲线过控制多边形始末端点的条件，并讨论光滑拼接轴线异面管道拼接问题。

设

${\Phi }_1:\{\begin{array}{l}x=x_1+a{N}_{11}\cos \phi +a{B}_{11}\sin \phi ,\\ y=y_1+b_{1}s+a{N}_{12}\cos \phi +a{B}_{12}\sin \phi ,\\ z=a{N}_{13}\cos \phi +a{B}_{13}\sin \phi .\end{array}$ 和 ${\Phi }_2:\{\begin{array}{l}x=a{N}_{21}\cos \phi +a{B}_{21}\sin \phi ,\\ y=y_2+a{N}_{22}\cos \phi +a{B}_{22}\sin \phi ,\\ z=z_2+c_{2}s+a{N}_{23}\cos \phi +a{B}_{23}\sin \phi .\end{array}s\in \left[0,1\right],\phi \in \left[0,2\pi \right]$

是两个待拼接的管道表达式，其中a是管道的半径， $N_i=\left(N_{i1},N_{i2},N_{i3}\right),B_i=\left(B_{i1},B_{i2},B_{i3}\right),i=1,2$ 分别是 $s=1$ 和 $s=0$ 时的法矢和副法矢。管道轴线的表达式为

$L_1:\{\begin{array}{l}x=x_1+0\cdot s,\\ y=y_1+b_{1}s,\\ z=0+0\cdot s,\end{array}$ 和 $L_2:\{\begin{array}{l}x=0+0\cdot s,\\ y=y_2+b_{\text{2}}\cdot s,\\ z=z_2+c_{2}s.\end{array}$

2. 过始末端点的四次均匀B样条曲线

定义1设

$r_i\left(s\right)={\displaystyle \underset{j=0}{\overset{4}{\sum }}{N}_{j,4}\left(s\right)V_{i+j}}$

是四次均匀B样条曲线段。其中 $N_{0,4}\left(s\right),N_{1,4}\left(s\right),N_{2,4}\left(s\right),N_{3,4}\left(s\right),N_{4,4}\left(s\right)$ 为B样条基， $V_i,V_{i+1},V_{i+2},V_{i+3},V_{i+4}$ 为特征多边形的顶点。

为了使所求B样条曲线过特征多边形的始点 $V_0$ 和终点 $V_n$，我们在 $V_0{V}_1$ 的反向延长线上取 $V_{-1}$，在 $V_{n-1}{V}_n$ 的延长线上取 $V_{n+1}$。在(1)中，可令 $i=0,s=0$，得

$r_0\left(0\right)=\frac{1}{24}{V}_{-1}+\frac{11}{24}{V}_0+\frac{11}{24}{V}_1+\frac{1}{24}{V}_2,$

并令 $\frac{1}{24}{V}_{-1}+\frac{11}{24}{V}_0+\frac{11}{24}{V}_1+\frac{1}{24}{V}_2=V_0$，解得

$V_{-1}=13V_0-11V_1-V_2.$ (2)

同理，令 $i=n-2,s=0$，得

$r_{n-2}\left(0\right)=\frac{1}{24}{V}_{n-2}+\frac{11}{24}{V}_{n-1}+\frac{11}{24}{V}_n+\frac{1}{24}{V}_{n+1},$

并令 $\frac{1}{24}{V}_{n-2}+\frac{11}{24}{V}_{n-1}+\frac{11}{24}{V}_n+\frac{1}{24}{V}_{n+1}=V_{n+1}$，解得

$V_{n+1}=13V_n-11V_{n-1}-V_n.$ (3)

式(2)和(3)即为四次均匀B样条曲线过控制多边形的始末端点的条件。

先构造过始末端点的光滑拼接轴线的四次均匀B样条曲线。

例1设 $V_0=\left(5,-5,0\right),V_1=\left(5,0,0\right),V_3=\left(0,5,0\right),V_4=\left(0,5,5\right)$，分别取 $V_2=\left(\frac{5}{2},\frac{5}{2},0\right),V_2=\left(\frac{5}{2},\frac{5}{2},1\right)$ 和 $V_2=\left(\frac{5}{2},\frac{5}{2},-1\right)$。带入公式(2)和(3)，分别得 $V_{-11}=\left(\frac{15}{2},-\frac{135}{2},1\right)$， $V_{51}=\left(-\frac{5}{2},\frac{15}{2},64\right)$， $V_{-12}=\left(\frac{15}{2},-\frac{135}{2},0\right)$， $V_{52}=\left(-\frac{5}{2},\frac{15}{2},65\right)$， $V_{-13}=\left(\frac{15}{2},-\frac{135}{2},1\right)$， $V_{53}=\left(-\frac{5}{2},\frac{15}{2},66\right)$。用以上三组点序列为顶点 $V_{-1i}{V}_0,V_1,V_{2i},V_3,V_4,$ $V_{5i}\left(i=1,2,3\right)$ 的四次均匀B样条曲线拼接轴线异面管道的表达式为

$r_1\left(s\right):\{\begin{array}{l}{r}_{11}\left(s\right)=\{\begin{array}{l}{x}_{11}=\frac{5}{16}{s}^4-\frac{5}{6}{s}^3-\frac{5}{6}s+5,\\ y_{11}=-\frac{35}{16}{s}^4+\frac{55}{6}{s}^3-15s^2+\frac{85}{6}s-5,s\in \left[0,1\right],\\ z_{11}=\frac{5}{24}{s}^4-\frac{1}{3}{s}^3-\frac{1}{3}s.\end{array}\\ r_{12}\left(s\right)=\{\begin{array}{l}{x}_{12}=\frac{5}{12}{s}^3-\frac{5}{8}{s}^2-\frac{25}{12}s+\frac{175}{48},\\ y_{12}=-\frac{5}{24}{s}^4+\frac{5}{12}{s}^3-\frac{5}{8}{s}^2+\frac{35}{12}s+\frac{55}{48},s\in \left[0,1\right],\\ z_{12}=-\frac{1}{24}{s}^4+\frac{1}{2}{s}^3+\frac{1}{4}{s}^2-\frac{1}{2}s.\end{array}\\ r_{13}\left(s\right)=\{\begin{array}{l}{x}_{13}=-\frac{5}{16}{s}^4+\frac{5}{12}{s}^3+\frac{5}{8}{s}^2-\frac{25}{12}s+\frac{65}{48},\\ y_{13}=\frac{5}{16}{s}^4-\frac{5}{12}{s}^3-\frac{5}{8}{s}^2+\frac{25}{12}s+\frac{25}{12},s\in \left[0,1\right]\\ z_{13}=\frac{25}{12}{s}^4+\frac{1}{3}{s}^3+\frac{3}{2}{s}^2+\frac{4}{3}s.\end{array}\end{array}$ (4)

$r_2\left(s\right):\{\begin{array}{l}{r}_{21}\left(s\right)=\{\begin{array}{l}{x}_{21}=\frac{5}{16}{s}^4-\frac{5}{6}{s}^3-\frac{5}{6}s+5,\\ y_{21}=-\frac{35}{16}{s}^4+\frac{55}{6}{s}^3-15s^2+\frac{85}{6}s-5,s\in \left[0,1\right]\\ z_{21}=0.\stackrel{}{\underset{}{}}\end{array}\\ r_{22}\left(s\right)=\{\begin{array}{l}{x}_{22}=\frac{5}{12}{s}^4-\frac{5}{8}{s}^3-\frac{25}{12}s+\frac{175}{48},\\ y_{22}=-\frac{5}{24}{s}^4+\frac{5}{12}{s}^3-\frac{5}{8}{s}^2+\frac{35}{12}s+\frac{55}{48},s\in \left[0,1\right]\\ z_{22}=-\frac{5}{24}{s}^4.\end{array}\\ r_{23}\left(s\right)=\{\begin{array}{l}{x}_{23}=-\frac{5}{16}{s}^4+\frac{5}{12}{s}^3+\frac{5}{8}{s}^2-\frac{25}{12}s+\frac{65}{48},\\ y_{23}=\frac{5}{16}{s}^4-\frac{5}{12}{s}^3-\frac{5}{8}{s}^2+\frac{25}{12}s+\frac{175}{48},s\in \left[0,1\right]\\ z_{23}=\frac{15}{8}{s}^4+\frac{5}{8}{s}^3+\frac{5}{4}{s}^2+\frac{5}{6}s+\frac{5}{24}.\end{array}\end{array}$ (5)

$r_3\left(s\right):\{\begin{array}{l}{r}_{31}\left(s\right)=\{\begin{array}{l}{x}_{31}=\frac{5}{16}{s}^4-\frac{5}{6}{s}^3-\frac{5}{6}s+5,\\ y_{31}=-\frac{35}{16}{s}^4+\frac{55}{6}{s}^3-15s^2-\frac{85}{6}s-5,s\in \left[0,1\right]\\ z_{31}=-\frac{5}{24}{s}^4+\frac{1}{3}{s}^3+\frac{1}{3}s.\end{array}\\ r_{32}\left(s\right)=\{\begin{array}{l}{x}_{32}=\frac{5}{12}{s}^4-\frac{5}{8}{s}^2-\frac{25}{12}s+\frac{175}{48},\\ y_{32}=-\frac{5}{24}{s}^4+\frac{5}{12}{s}^3-\frac{5}{8}{s}^2+\frac{35}{12}s+\frac{55}{48},s\in \left[0,1\right]\\ y_{32}=\frac{11}{24}{s}^4-\frac{1}{2}{s}^3-\frac{1}{4}{s}^2-\frac{1}{2}s.\end{array}\\ r_{33}\left(s\right)=\{\begin{array}{l}{x}_{33}=-\frac{5}{16}{s}^4+\frac{5}{12}{s}^3+\frac{5}{8}{s}^2-\frac{25}{12}s+\frac{65}{48},\\ y_{33}=\frac{5}{16}{s}^4-\frac{5}{12}{s}^3-\frac{5}{8}{s}^2+\frac{25}{12}s+\frac{175}{48},s\in \left[0,1\right]\\ z_{33}=\frac{5}{3}{s}^4+\frac{4}{3}{s}^3+s^2+\frac{1}{3}s+\frac{2}{3}.\end{array}\end{array}$ (6)

是三条三段连续的四次B样条曲线。 $r_1\left(s\right),r_2\left(s\right),r_3\left(s\right)$ 分别与异面轴线拼接效果如图1：

Figure 1. When V₂ takes $V_2=\left(\frac{5}{2},\frac{5}{2},1\right),V_2=\left(\frac{5}{2},\frac{5}{2},0\right)$ and $V_2=\left(\frac{5}{2},\frac{5}{2},-1\right)$ respectively, the effect diagram

图1. V₂分别取 $V_2=\left(\frac{5}{2},\frac{5}{2},1\right),V_2=\left(\frac{5}{2},\frac{5}{2},0\right)$ 和 $V_2=\left(\frac{5}{2},\frac{5}{2},-1\right)$ 时，与异面轴线拼接效果图

由例1可以看出：

1) 用过始末端点的四次均匀B样条曲线光滑拼接异面轴线，与 [1] 一样也需要三段曲线；

2) 曲线位置与 $V_2$ 的位置相关， $V_2$ 在上，拼接曲线位于上侧； $V_2$ 居中，拼接曲线在中间； $V_2$ 在下边，拼接曲线在下侧；

3) 从拼接效果看，轴线拼接效果没有太大差异，光滑度很好。

3. 过始末端点的四次均匀B样条曲线及其在轴线异面管道拼接中的应用

基于轴线光滑拼接的异面管道拼接方法，在 [5] [6] [7] [8] [9] 中进行了详细讨论。下面考察过控制多边形始末端点的四次均匀B样条曲线在管道拼接中的效果。

例2设

${\Phi }_1:\{\begin{array}{l}x=5+N_{11}\cos \phi +B_{11}\sin \phi ,\\ y=-5+s+N_{12}\cos \phi +B_{12}\sin \phi ,\\ z=N_{13}\cos \phi +B_{13}\sin \phi .\end{array}$ 和 ${\Phi }_2:\{\begin{array}{l}x=N_{21}\cos \phi +B_{21}\sin \phi ,\\ y=5+N_{22}\cos \phi +B_{22}\sin \phi ,\\ z=5+s+N_{23}\cos \phi +B_{23}\sin \phi .\end{array}s\in \left[0,1\right],\phi \in \left[0,2\pi \right]$

构造以轴线 $r_2\left(s\right)$ 为轴线的圆管道 ${\Phi }_2\left(s,\phi \right)$ ：

${\Phi }_{21}\left(s,\phi \right):\{\begin{array}{l}{x}_1=x_{21}+N_{11}\left(s\right)\cos \phi +B_{11}\left(s\right)\sin \phi ,\\ y_1=y_{21}+N_{12}\left(s\right)\cos \phi +B_{12}\left(s\right)\sin \phi ,s\in \left[0,1\right],\phi \in \left[0,2\pi \right]\\ z_1=z_{21}+N_{13}\left(s\right)\cos \phi +B_{13}\left(s\right)\sin \phi .\end{array}$

${\Phi }_{22}\left(s,\phi \right):\{\begin{array}{l}{x}_2=x_{22}+N_{21}\left(s\right)\cos \phi +B_{21}\left(s\right)\sin \phi ,\\ y_2=y_{22}+N_{22}\left(s\right)\cos \phi +B_{22}\left(s\right)\sin \phi ,s\in \left[0,1\right],\phi \in \left[0,2\pi \right]\\ z_2=z_{22}+N_{23}\left(s\right)\cos \phi +B_{23}\left(s\right)\sin \phi .\end{array}$

${\Phi }_{23}\left(s,\phi \right):\{\begin{array}{l}{x}_3=x_{23}+N_{31}\left(s\right)\cos \phi +B_{31}\left(s\right)\sin \phi ,\\ y_3=y_{23}+N_{32}\left(s\right)\cos \phi +B_{32}\left(s\right)\sin \phi ,s\in \left[0,1\right],\phi \in \left[0,2\pi \right]\\ z_3=z_{23}+N_{33}\left(s\right)\cos \phi +B_{33}\left(s\right)\sin \phi .\end{array}$

其中， ${\Phi }_{21}\left(s,\phi \right),{\Phi }_{23}\left(s,\phi \right),{\Phi }_{23}\left(s,\phi \right)$ 是三段光滑拼接的圆管道， $N_i=\left(N_{i1}\left(s\right),N_{i2}\left(s\right),N_{i3}\left(s\right)\right)$ 和 $B_i=\left(B_{i1}\left(s\right),B_{i2}\left(s\right),B_{i3}\left(s\right)\right),i=1,2,3$ 是构造的B样条曲线的主法矢和副法矢。其拼接效果如图2：

Figure 2. When $V_2=\left(\frac{5}{2},\frac{5}{2},0\right)$, circular pipe stitching effect drawing

图2. $V_2=\left(\frac{5}{2},\frac{5}{2},0\right)$ 时，圆管道拼接效果图

例3设

${\Phi }_1:\{\begin{array}{l}x=5+N_{11}\cos \phi +0.6B_{11}\sin \phi ,\\ y=-5+s+N_{12}\cos \phi +0.6B_{12}\sin \phi ,\\ z=N_{13}\cos \phi +0.6B_{13}\sin \phi .\end{array}$ 和 ${\Phi }_2:\{\begin{array}{l}x=N_{21}\cos \phi +0.6B_{21}\sin \phi ,\\ y=5+N_{22}\cos \phi +0.6B_{22}\sin \phi ,\\ z=5+s+N_{23}\cos \phi +0.6B_{23}\sin \phi .\end{array}s\in \left[0,1\right],\phi \in \left[0,2\pi \right]$

构造以轴线 $r_2\left(s\right)$ 为轴线的椭圆管道 ${\Phi }_2\left(s,\phi \right)$ ：

${\Phi }_{21}\left(s,\phi \right):\{\begin{array}{l}{x}_1=x_{21}+N_{11}\left(s\right)\cos \phi +0.6B_{11}\left(s\right)\sin \phi ,\\ y_1=y_{21}+N_{12}\left(s\right)\cos \phi +0.6B_{12}\left(s\right)\sin \phi ,s\in \left[0,1\right],\phi \in \left[0,2\pi \right]\\ z_1=z_{21}+N_{13}\left(s\right)\cos \phi +0.6B_{13}\left(s\right)\sin \phi .\end{array}$

${\Phi }_{22}\left(s,\phi \right):\{\begin{array}{l}{x}_2=x_{22}+N_{21}\left(s\right)\cos \phi +0.6B_{21}\left(s\right)\sin \phi ,\\ y_2=y_{22}+N_{22}\left(s\right)\cos \phi +0.6B_{22}\left(s\right)\sin \phi ,s\in \left[0,1\right],\phi \in \left[0,2\pi \right]\\ z_2=z_{22}+N_{23}\left(s\right)\cos \phi +0.6B_{23}\left(s\right)\sin \phi .\end{array}$

${\Phi }_{23}\left(s,\phi \right):\{\begin{array}{l}{x}_3=x_{23}+N_{31}\left(s\right)\cos \phi +0.6B_{31}\left(s\right)\sin \phi ,\\ y_3=y_{23}+N_{32}\left(s\right)\cos \phi +0.6B_{32}\left(s\right)\sin \phi ,s\in \left[0,1\right],\phi \in \left[0,2\pi \right]\\ z_3=z_{23}+N_{33}\left(s\right)\cos \phi +0.6B_{33}\left(s\right)\sin \phi .\end{array}$

其中， ${\Phi }_{21}\left(s,\phi \right),{\Phi }_{22}\left(s,\phi \right),{\Phi }_{23}\left(s,\phi \right)$ 是三段光滑拼接的管道， $N_i=\left(N_{i1}\left(s\right),N_{i2}\left(s\right),N_{i3}\left(s\right)\right)$ 和 $B_i=\left(B_{i1}\left(s\right),B_{i2}\left(s\right),B_{i3}\left(s\right)\right),i=1,2,3$ 是构造的B样条曲线的主法矢和副法矢， $x_{2i},y_{2i},z_{2i}\left(i=1,2,3\right)$ 由(5)式给出。

其拼接效果如图3：

Figure 3. When $V_2=\left(\frac{5}{2},\frac{5}{2},0\right)$, the effect picture of elliptical pipeline splicing

图3. $V_2=\left(\frac{5}{2},\frac{5}{2},0\right)$ 时，椭圆管道拼接效果图

例4设

${\Phi }_1:\{\begin{array}{l}x=5+N_{11}\cos \phi +0.6B_{11}\sin \phi ,\\ y=-5+s+N_{12}\cos \phi +0.6B_{12}\sin \phi ,\\ z=N_{13}\cos \phi +0.6B_{13}\sin \phi .\end{array}$ 和 ${\Phi }_2:\{\begin{array}{l}x=N_{21}\cos \phi +0.6B_{21}\sin \phi ,\\ y=5+N_{22}\cos \phi +0.6B_{22}\sin \phi ,\\ z=5+s+N_{23}\cos \phi +0.6B_{23}\sin \phi .\end{array}s\in \left[0,1\right],\phi \in \left[0,2\pi \right]$

构造以轴线 $r_3\left(s\right)$ 为轴线的椭圆管道 ${\Phi }_3\left(s,\phi \right)$ ：

${\Phi }_{31}\left(s,\phi \right):\{\begin{array}{l}{x}_1=x_{31}+N_{11}\left(s\right)\cos \phi +0.6B_{11}\left(s\right)\sin \phi ,\\ y_1=y_{31}+N_{12}\left(s\right)\cos \phi +0.6B_{12}\left(s\right)\sin \phi ,s\in \left[0,1\right],\phi \in \left[0,2\pi \right]\\ z_1=z_{31}+N_{13}\left(s\right)\cos \phi +0.6B_{13}\left(s\right)\sin \phi .\end{array}$

${\Phi }_{32}\left(s,\phi \right):\{\begin{array}{l}{x}_2=x_{32}+N_{21}\left(s\right)\cos \phi +0.6B_{21}\left(s\right)\sin \phi ,\\ y_2=y_{32}+N_{22}\left(s\right)\cos \phi +0.6B_{22}\left(s\right)\sin \phi ,s\in \left[0,1\right],\phi \in \left[0,2\pi \right]\\ z_2=z_{32}+N_{23}\left(s\right)\cos \phi +0.6B_{23}\left(s\right)\sin \phi .\end{array}$

${\Phi }_{33}\left(s,\phi \right):\{\begin{array}{l}{x}_3=x_{33}+N_{31}\left(s\right)\cos \phi +0.6B_{31}\left(s\right)\sin \phi ,\\ y_3=y_{33}+N_{32}\left(s\right)\cos \phi +0.6B_{32}\left(s\right)\sin \phi ,s\in \left[0,1\right],\phi \in \left[0,2\pi \right]\\ z_3=z_{33}+N_{33}\left(s\right)\cos \phi +0.6B_{33}\left(s\right)\sin \phi .\end{array}$

其中， ${\Phi }_{31}\left(s,\phi \right),{\Phi }_{32}\left(s,\phi \right),{\Phi }_{33}\left(s,\phi \right)$ 是三段光滑拼接的管道， $N_i=\left(N_{i1}\left(s\right),N_{i2}\left(s\right),N_{i3}\left(s\right)\right)$ 和 $B_i=\left(B_{i1}\left(s\right),B_{i2}\left(s\right),B_{i3}\left(s\right)\right),i=1,2,3$ 是构造的B样条曲线的主法矢和副法矢， $x_{3i},y_{3i},z_{3i}\left(i=1,2,3\right)$ 由(6)式给出。

其拼接效果如图4：

Figure 4. When $V_2=\left(\frac{5}{2},\frac{5}{2},-1\right)$, the effect picture of elliptical pipeline splicing

图4. $V_2=\left(\frac{5}{2},\frac{5}{2},-1\right)$ 时，椭圆管道拼接效果图

由例2、例3和例4可以看出，在 $V_0,V_1,V_3,V_4$ 取定的情况下，用我们所构造的管道拼接轴线异面管道，还需要优化以下几点：

1) 拼接效果与 $V_2$ 的选取有关，对于椭圆管道拼接需要优化 $V_2$ 的选取方法；

2) 用三段连续的四次均匀B样条曲线为轴线的管道拼接轴线异面椭圆管道，段与段之间能够光滑拼接。要想取得视觉上的光顺效果，还需要用 [8] 中给出的方法，逐步旋转椭圆管道的长半轴与短半轴来实现；

3) 用三段连续的四次均匀B样条曲线为轴线的管道拼接轴线异面椭圆管道，对于异面椭圆管道的长半轴和短半轴的方向有特殊要求，也需要逐步旋转每段拼接椭圆管道的长半轴与短半轴来实现。

4. 结束语

过始末端点的四次B样条曲线为轴线的管道拼接轴异面管道需要三段管道，光顺性取决于控制顶点 $V_2$ 的选取。对于椭圆管道拼接更为明显。需要优化 $V_2$ 点的选取。尽管相较于插值于两端点的三次有理B样条曲线、插值于制定点的三次B样条曲线和四次B样条曲线段数多，但是段与段之间的光顺程度较好，且能够应用于椭圆管道的拼接。所以，比以上三种情形具有更好的应用前景。

基金项目

国家自然科学基金项目资助(11561052)。

参考文献