B樣條曲線曲面介紹

B樣條基函數

B樣條基函數的定義

由de Boor和Cox分別導出B樣條基函數的遞推定義，B樣條基函數可以表示爲
$\begin{aligned} N_{i,0}(u) &= \begin{cases} 1 ,\quad u_i \leqslant u < u_{i+1} \\ 0 ,\quad 其他 \end{cases} \\ N_{i,p}(u) &= \dfrac{u-u_i}{u_{i+p}-u_i}N_{i,p-1}(u)+\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u) ,\quad p > 0 \end{aligned}$
並約定$0/0=0$。式中$p$表示B樣條的冪次，$u$爲節點，下標$i$爲B樣條的序號。

上式表明，任意$p$次B樣條基函數可由兩個相鄰的$p-1$次B樣條基函數的線性組合構成。

B樣條基函數的性質

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

B樣條基函數的導數

B樣條基函數的求導公式爲
$N^{'}_{i,p}(u)=\dfrac{p}{u_{i+p}-u_i}N_{i,p-1}(u)-\dfrac{p}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u).$
下面通過對$p$利用數學歸納法來證明求導公式。當$p=1$時，$N_{i,p-1}(u)$ 和$N_{i+1,p-1}(u)$ 在每個節點區間內或者爲$0$，或者爲$1$，因此$N^{'}_{i,p}(u)$ 等於$\dfrac{1}{u_{i+1}-u_i}$或者$-\dfrac{1}{u_{i+2}-u_{i+1}}$。我們假設當$p-1(p>1)$時求導公式成立。根據求導法則$(fg)^{'}=f^{'}g+fg^{'}$，對基函數
$N_{i,p}(u)=\dfrac{u-u_i}{u_{i+p}-u_i}N_{i,p-1}(u)+\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u)$
求導，得到
$\begin{aligned} N^{'}_{i,p}(u)=& \dfrac{1}{u_{i+p}-u_i}N_{i,p-1}(u)+\dfrac{u-u_i}{u_{i+p}-u_i}N^{'}_{i,p-1}(u) \\ &-\dfrac{1}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u)+\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+1}}N^{'}_{i+1,p-1}(u) \\ =& \dfrac{1}{u_{i+p}-u_i}N_{i,p-1}(u)-\dfrac{1}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u) \\ &+\dfrac{u-u_i}{u_{i+p}-u_i}\left(\dfrac{p-1}{u_{i+p-1}-u_i}N_{i,p-2}(u)-\dfrac{p-1}{u_{i+p}-u_{i+1}}N_{i+1,p-2}(u)\right) \\ &+\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+1}}\left(\dfrac{p-1}{u_{i+p}-u_{i+1}}N_{i+1,p-2}(u)-\dfrac{p-1}{u_{i+p+1}-u_{i+2}}N_{i+2,p-2}(u)\right) \\ =& \dfrac{1}{u_{i+p}-u_i}N_{i,p-1}(u)-\dfrac{1}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u) \\ &+\dfrac{p-1}{u_{i+p}-u_i}\dfrac{u-u_i}{u_{i+p-1}-u_i}N_{i,p-2}(u) \\ &+\dfrac{p-1}{u_{i+p}-u_{i+1}}\left(\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+1}}-\dfrac{u-u_i}{u_{i+p}-u_i}\right)N_{i+1,p-2}(u) \\ &-\dfrac{p-1}{u_{i+p+1}-u_{i+1}}\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+2}}N_{i+2,p-2}(u). \end{aligned}$
由於
$\begin{aligned} \dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+1}}-\dfrac{u-u_i}{u_{i+p}-u_i}=& -1+\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+1}}+1-\dfrac{u-u_i}{u_{i+p}-u_i} \\ =& -\dfrac{u_{i+p+1}-u_{i+1}}{u_{i+p+1}-u_{i+1}}+\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+1}} \\ & +\dfrac{u_{i+p}-u_i}{u_{i+p}-u_i}-\dfrac{u-u_i}{u_{i+p}-u_i} \\ =&\dfrac{u_{i+p}-u}{u_{i+p}-u_i}-\dfrac{u-u_{i+1}}{u_{i+p+1}-u_{i+1}}. \end{aligned}$
於是得到
$\begin{aligned} N^{'}_{i,p}(u)=& \dfrac{1}{u_{i+p}-u_i}N_{i,p-1}(u)-\dfrac{1}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u) \\ &+\dfrac{p-1}{u_{i+p}-u_i}\left(\dfrac{u-u_i}{u_{i+p-1}-u_i}N_{i,p-2}(u)+\dfrac{u_{i+p}-u}{u_{i+p}-u_{i+1}}N_{i+1,p-2}(u)\right) \\ &-\dfrac{p-1}{u_{i+p+1}-u_{i+1}}\left(\dfrac{u-u_{i+1}}{u_{i+p}-u_{i+1}}N_{i+1,p-2}(u)+\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+2}}N_{i+2,p-2}(u)\right) \\ =& \dfrac{1}{u_{i+p}-u_i}N_{i,p-1}(u)-\dfrac{1}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u) \\ &+\dfrac{p-1}{u_{i+p}-u_i}N_{i,p-1}(u)-\dfrac{p-1}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u) \\ =& \dfrac{p}{u_{i+p}-u_i}N_{i,p-1}(u)-\dfrac{p}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u). \end{aligned}$
證畢。

B樣條曲線曲面

B樣條曲線的定義

$p$次B樣條曲線的定義爲
$P\left(u\right)=\sum\limits_{i=0}^{n}N_{i,p}(u)V_i,\quad a\leqslant u\leqslant b$
這裏$\left\{V_i\right\}$是控制點，$\left\{N_{i,p}(u)\right\}$是定義在非週期(並且非均勻)節點矢量
$U = \left\{ \underbrace{a,\cdots,a}_{p+1}, u_{p+1},\cdots,u_{m-p-1}, \underbrace{b,\cdots,b}_{p+1} \right\}$
(包含$m+1$個節點)上的$p$次B樣條基函數。由$\left\{V_i\right\}$構成的多邊形稱爲控制多邊形。

有理B樣條曲線曲面

NURBS曲線的定義

一條$p$次NURBS曲線的定義爲
$P\left(u\right)=\dfrac{\sum\limits_{i=0}^{n}N_{i,p}(u)\omega_iV_i}{\sum\limits_{i=0}^{n}N_{i,p}(u)\omega_i},\quad a\leqslant u\leqslant b$
這裏$\left\{V_i\right\}$是控制點(它們形成控制多邊形)，$\left\{\omega_i\right\}$是權因子，$\left\{N_{i,p}(u)\right\}$是定義在非週期(並且非均勻)節點矢量$U$上的$p$次B樣條基函數。其中
$U = \left\{ \underbrace{a,\cdots,a}_{p+1}, u_{p+1},\cdots,u_{m-p-1}, \underbrace{b,\cdots,b}_{p+1} \right\}$