Bezier曲线构造

曲线公式

$$\begin{gathered} 曲线：C(u)=\sum _{i=0}^n{B}_{n,i}(u)P_i \\ 基函数：B_{n,i}=\frac{n!}{i!(n-i)!}{u}^i(1-u{)}^{n-i} \end{gathered}$$

曲线性质

1.Bezir曲线的阶数由控制点个数决定，n+1个控制点，则对应阶数为n. 例如3次Bezir曲线需要4个点，2次Bezir曲线需要3个点.

2.Bezir曲线通过第一个点和最后一个点.

3.所有的样条基函数均为非负数.

4.Bezir曲线完全位于给定的n+1个控制点的凸包内

5.仿射不变性

三次Bezier曲线

三次Bezier曲线是最常用的曲线，这里注重讲一下。

由曲线公式可得：
$$C(u)=\sum _{i=0}^3{B}_{3,i}(u)P_i=(1-u{)}^3{P}_0+3(1-u{)}^{2}u{P}_1+3(1-u)u^2{P}_2+u^3{P}_3,0\le u\le 1$$
设：
$$\begin{array}{rlr}{J}_{3,0}(u)& =(1-u{)}^3& =1(1-u{)}^3{u}^0\\ J_{3,1}(u)& =3(1-u{)}^{2}u& =3(1-u{)}^2{u}^1\\ J_{3,2}(u)& =3(1-u)u^3& =3(1-u{)}^1{u}^2\\ J_{3,3}(u)& =u^3& =1(1-u{)}^0{u}^3\\ C(u)& =J_{3,0}{P}_0+J_{3,1}{P}_1+J_{3,2}{P}_2+J_{3,3}{P}_3& \end{array}$$

部分性质

u=0 u=1处的值：
$$\begin{array}{rlr}{J}_{3,0}(0)& =1J_{3,0}(1)& =0\\ J_{3,1}(0)& =0J_{3,1}(1)& =0\\ J_{3,2}(0)& =0J_{3,2}(1)& =0\\ J_{3,3}(0)& =0J_{3,3}(1)& =1\end{array}$$
基函数和：
$$J_{3,0}(u)+J_{3,1}(u)+J_{3,2}(u)+J_{3,3}(u)=1,0\le u\le 1$$
基函数一次导数：
$$\begin{array}{rlr}{J}_{3,0}^{{}^{\prime }}(0)& =-3& J_{3,0}^{{}^{\prime }}(1)=0\\ J_{3,1}^{{}^{\prime }}(0)& =3& J_{3,1}^{{}^{\prime }}(1)=0\\ J_{3,2}^{{}^{\prime }}(0)& =0& J_{3,2}^{{}^{\prime }}(1)=-3\\ J_{3,3}^{{}^{\prime }}(0)& =0& J_{3,3}^{{}^{\prime }}(1)=3\end{array}$$
基函数一次导数：
$$\begin{array}{rlr}{J}_{3,0}^{{}^{\prime \prime }}(0)& =6& J_{3,0}^{{}^{\prime \prime }}(1)=0\\ J_{3,1}^{{}^{\prime \prime }}(0)& =-12& J_{3,1}^{{}^{\prime \prime }}(1)=0\\ J_{3,2}^{{}^{\prime \prime }}(0)& =0& J_{3,2}^{{}^{\prime \prime }}(1)=-12\\ J_{3,3}^{{}^{\prime \prime }}(0)& =0& J_{3,3}^{{}^{\prime \prime }}(1)=6\end{array}$$
基函数最大值：
$$基函数最大值出现在：J_{3,k}(u),u=\frac{k}{3},k=0,1,2,3$$

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