B样条曲线曲面介绍

B样条基函数

B样条基函数的定义

由de Boor和Cox分别导出B样条基函数的递推定义，B样条基函数可以表示为
$\begin{aligned} N_{i,0}(u) &= \begin{cases} 1 ,\quad u_i \leqslant u < u_{i+1} \\ 0 ,\quad 其他 \end{cases} \\ N_{i,p}(u) &= \dfrac{u-u_i}{u_{i+p}-u_i}N_{i,p-1}(u)+\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u) ,\quad p > 0 \end{aligned}$
并约定$0/0=0$。式中$p$表示B样条的幂次，$u$为节点，下标$i$为B样条的序号。

上式表明，任意$p$次B样条基函数可由两个相邻的$p-1$次B样条基函数的线性组合构成。

B样条基函数的性质

• 如果$u\notin\left[u_i,u_{i+p+1}\right)$，则$N_{i,p}(u)=0$。
• 当$u\in\left[u_i,u_{i+1}\right)$时，$\sum\limits_{j=i-p}^{i}N_{j,p}(u)=1$。

B样条基函数的导数

B样条基函数的求导公式为
$N^{'}_{i,p}(u)=\dfrac{p}{u_{i+p}-u_i}N_{i,p-1}(u)-\dfrac{p}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u).$
下面通过对$p$利用数学归纳法来证明求导公式。当$p=1$时，$N_{i,p-1}(u)$ 和$N_{i+1,p-1}(u)$ 在每个节点区间内或者为$0$，或者为$1$，因此$N^{'}_{i,p}(u)$ 等于$\dfrac{1}{u_{i+1}-u_i}$或者$-\dfrac{1}{u_{i+2}-u_{i+1}}$。我们假设当$p-1(p>1)$时求导公式成立。根据求导法则$(fg)^{'}=f^{'}g+fg^{'}$，对基函数
$N_{i,p}(u)=\dfrac{u-u_i}{u_{i+p}-u_i}N_{i,p-1}(u)+\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u)$
求导，得到
$\begin{aligned} N^{'}_{i,p}(u)=& \dfrac{1}{u_{i+p}-u_i}N_{i,p-1}(u)+\dfrac{u-u_i}{u_{i+p}-u_i}N^{'}_{i,p-1}(u) \\ &-\dfrac{1}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u)+\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+1}}N^{'}_{i+1,p-1}(u) \\ =& \dfrac{1}{u_{i+p}-u_i}N_{i,p-1}(u)-\dfrac{1}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u) \\ &+\dfrac{u-u_i}{u_{i+p}-u_i}\left(\dfrac{p-1}{u_{i+p-1}-u_i}N_{i,p-2}(u)-\dfrac{p-1}{u_{i+p}-u_{i+1}}N_{i+1,p-2}(u)\right) \\ &+\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+1}}\left(\dfrac{p-1}{u_{i+p}-u_{i+1}}N_{i+1,p-2}(u)-\dfrac{p-1}{u_{i+p+1}-u_{i+2}}N_{i+2,p-2}(u)\right) \\ =& \dfrac{1}{u_{i+p}-u_i}N_{i,p-1}(u)-\dfrac{1}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u) \\ &+\dfrac{p-1}{u_{i+p}-u_i}\dfrac{u-u_i}{u_{i+p-1}-u_i}N_{i,p-2}(u) \\ &+\dfrac{p-1}{u_{i+p}-u_{i+1}}\left(\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+1}}-\dfrac{u-u_i}{u_{i+p}-u_i}\right)N_{i+1,p-2}(u) \\ &-\dfrac{p-1}{u_{i+p+1}-u_{i+1}}\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+2}}N_{i+2,p-2}(u). \end{aligned}$
由于
$\begin{aligned} \dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+1}}-\dfrac{u-u_i}{u_{i+p}-u_i}=& -1+\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+1}}+1-\dfrac{u-u_i}{u_{i+p}-u_i} \\ =& -\dfrac{u_{i+p+1}-u_{i+1}}{u_{i+p+1}-u_{i+1}}+\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+1}} \\ & +\dfrac{u_{i+p}-u_i}{u_{i+p}-u_i}-\dfrac{u-u_i}{u_{i+p}-u_i} \\ =&\dfrac{u_{i+p}-u}{u_{i+p}-u_i}-\dfrac{u-u_{i+1}}{u_{i+p+1}-u_{i+1}}. \end{aligned}$
于是得到
$\begin{aligned} N^{'}_{i,p}(u)=& \dfrac{1}{u_{i+p}-u_i}N_{i,p-1}(u)-\dfrac{1}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u) \\ &+\dfrac{p-1}{u_{i+p}-u_i}\left(\dfrac{u-u_i}{u_{i+p-1}-u_i}N_{i,p-2}(u)+\dfrac{u_{i+p}-u}{u_{i+p}-u_{i+1}}N_{i+1,p-2}(u)\right) \\ &-\dfrac{p-1}{u_{i+p+1}-u_{i+1}}\left(\dfrac{u-u_{i+1}}{u_{i+p}-u_{i+1}}N_{i+1,p-2}(u)+\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+2}}N_{i+2,p-2}(u)\right) \\ =& \dfrac{1}{u_{i+p}-u_i}N_{i,p-1}(u)-\dfrac{1}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u) \\ &+\dfrac{p-1}{u_{i+p}-u_i}N_{i,p-1}(u)-\dfrac{p-1}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u) \\ =& \dfrac{p}{u_{i+p}-u_i}N_{i,p-1}(u)-\dfrac{p}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u). \end{aligned}$
证毕。

B样条曲线曲面

B样条曲线的定义

$p$次B样条曲线的定义为
$P\left(u\right)=\sum\limits_{i=0}^{n}N_{i,p}(u)V_i,\quad a\leqslant u\leqslant b$
这里$\left\{V_i\right\}$是控制点，$\left\{N_{i,p}(u)\right\}$是定义在非周期(并且非均匀)节点矢量
$U = \left\{ \underbrace{a,\cdots,a}_{p+1}, u_{p+1},\cdots,u_{m-p-1}, \underbrace{b,\cdots,b}_{p+1} \right\}$
(包含$m+1$个节点)上的$p$次B样条基函数。由$\left\{V_i\right\}$构成的多边形称为控制多边形。

有理B样条曲线曲面

NURBS曲线的定义

一条$p$次NURBS曲线的定义为
$P\left(u\right)=\dfrac{\sum\limits_{i=0}^{n}N_{i,p}(u)\omega_iV_i}{\sum\limits_{i=0}^{n}N_{i,p}(u)\omega_i},\quad a\leqslant u\leqslant b$
这里$\left\{V_i\right\}$是控制点(它们形成控制多边形)，$\left\{\omega_i\right\}$是权因子，$\left\{N_{i,p}(u)\right\}$是定义在非周期(并且非均匀)节点矢量$U$上的$p$次B样条基函数。其中
$U = \left\{ \underbrace{a,\cdots,a}_{p+1}, u_{p+1},\cdots,u_{m-p-1}, \underbrace{b,\cdots,b}_{p+1} \right\}$