均匀B样条曲线的表达式

1 矩阵表达

对于均匀B样条曲线，其曲线方程的矩阵表达为：
c ( t ) = [ 1   t   ⋯   t k ] M k [ P i − k P i − k + 1 ⋮ P i ]   t ∈ [ 0 , 1 ] (1) c(t) = [1\ t\ \cdots\ t^k] M_k

\ t\in[0,1] \tag{1} c(t)=[1 t ⋯ tk]Mk​⎣ ⎡​Pi−k​Pi−k+1​⋮Pi​​⎦ ⎤​ t∈[0,1](1)
其中$t$为参数，$P_{i-k},P_{i-k+1},\cdots ,P_i$为$k+1$个控制点，而$M_k$由下式给出：
M k = [ m 0 , 0 m 0 , 1 ⋯ m 0 , k ⋮ ⋮ ⋯ ⋮ m k , 0 m k , 1 ⋯ m k , k ] (2) M_k =

[m0,0m0,1⋯m0,k⋮⋮⋯⋮mk,0mk,1⋯mk,k]" role="presentation" style="position: relative;">⎡⎣⎢⎢m0,0⋮mk,0m0,1⋮mk,1⋯⋯⋯m0,k⋮mk,k⎤⎦⎥⎥$$\left[\begin{array}{cccc}{m}_{0,0}& m_{0,1}& \cdots & m_{0,k}\\ ⋮& ⋮& \cdots & ⋮\\ m_{k,0}& m_{k,1}& \cdots & m_{k,k}\end{array}\right]$$

\tag{2} Mk​=⎣ ⎡​m0,0​⋮mk,0​​m0,1​⋮mk,1​​⋯⋯⋯​m0,k​⋮mk,k​​⎦ ⎤​(2)
而矩阵$M_k$中的每个元素$m_{i,j}$可以由下式直接给出，
$$m_{i,j}=\frac{1}{(k-1)!}{C}_k^{k-i}\sum _{s=j}^k(-1{)}^{s-j}{C}_{k+1}^{s-j}(k-s{)}^{k-i}\text{\hspace{1em}(3)}$$
由上式，很容易得出，
M 1 = [ 1 0 − 1 1 ] M_1 =

[10−11]" role="presentation" style="position: relative;">[1−101]$$\left[\begin{array}{cc}1& 0\\ -1& 1\end{array}\right]$$

M1​=[1−1​01​]
M 2 = [ 1 1 0 − 2 2 0 1 − 2 1 ] M_2=

[110−2201−21]" role="presentation" style="position: relative;">⎡⎣⎢1−2112−2001⎤⎦⎥$$\left[\begin{array}{ccc}1& 1& 0\\ -2& 2& 0\\ 1& -2& 1\end{array}\right]$$

M2​=⎣ ⎡​1−21​12−2​001​⎦ ⎤​
M 3 = 1 2 ⋅ [ 1 4 1 0 − 3 0 3 0 3 − 6 3 0 − 1 3 − 3 1 ] M_3 = \frac{1}{2} \cdot

[1410−30303−630−13−31]" role="presentation" style="position: relative;">⎡⎣⎢⎢⎢1−33−140−63133−30001⎤⎦⎥⎥⎥$$\left[\begin{array}{cccc}1& 4& 1& 0\\ -3& 0& 3& 0\\ 3& -6& 3& 0\\ -1& 3& -3& 1\end{array}\right]$$

M3​=21​⋅⎣ ⎡​1−33−1​40−63​133−3​0001​⎦ ⎤​
M 4 = 1 6 ⋅ [ 1 11 11 1 0 − 4 − 12 12 4 0 6 − 6 − 6 6 0 − 4 12 − 12 4 0 1 − 4 6 − 4 1 ] M_4 = \frac{1}{6} \cdot

[1111110−4−1212406−6−660−412−12401−46−41]" role="presentation" style="position: relative;">⎡⎣⎢⎢⎢⎢⎢⎢1−46−4111−12−612−41112−6−1261464−400001⎤⎦⎥⎥⎥⎥⎥⎥$$\left[\begin{array}{ccccc}1& 11& 11& 1& 0\\ -4& -12& 12& 4& 0\\ 6& -6& -6& 6& 0\\ -4& 12& -12& 4& 0\\ 1& -4& 6& -4& 1\end{array}\right]$$

M4​=61​⋅⎣ ⎡​1−46−41​11−12−612−4​1112−6−126​1464−4​00001​⎦ ⎤​
计算矩阵$M_k$的matlab程序为，

clear all
clc
close all


%%
k = input('请输入一个大于等于1的整数：');
M = zeros(k+1,k+1);
for i = 0 : k
    for j = 0 : k
        a = 1 / factorial(k - 1);
        a = a * nchoosek(k, k - i);
        b = 0;
        for s = j : k
            b = b + (-1)^(s - j) * nchoosek(k + 1, s - j) * (k - s)^(k - i);
        end
        a = a * b;
        M(i+1, j+1) = a;
    end
end
M %小数表示
format rat
M %分数表示
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23

2 多项式表达

对于均匀B样条曲线，其曲线多项式表示为：
$$c(t)=w_0{P}_{i-k}+w_1{P}_{i-k+1}+\cdots +w_k{P}_i\text{\hspace{1em}(4)}$$
其中$t$为参数，$P_{i-k},P_{i-k+1},\cdots ,P_i$为$k+1$个控制点，而$w_0,w_1,\cdots ,w_k$是$t$的多项式，可以理解为控制点的权重，由下式给出，
$$w_i=\frac{1}{(k-1)!}\sum _{j=0}^{k-i}(-1{)}^j{C}_{k+1}^j(t+k-i-j{)}^{k}i=0,1,\cdots ,k\text{\hspace{1em}(5)}$$
当$k$取4时，有5个控制点，每个控制点的权重分别为，
$$w_0=\frac{1}{6}\cdot \sum _{j=0}^4(-1{)}^j\cdot C_5^j\cdot (t+4-j{)}^4$$
$$w_1=\frac{1}{6}\cdot \sum _{j=0}^3(-1{)}^j\cdot C_5^j\cdot (t+3-j{)}^4$$
$$w_2=\frac{1}{6}\cdot \sum _{j=0}^2(-1{)}^j\cdot C_5^j\cdot (t+2-j{)}^4$$
$$w_3=\frac{1}{6}\cdot \sum _{j=0}^1(-1{)}^j\cdot C_5^j\cdot (t+1-j{)}^4$$
$$w_4=\frac{1}{6}\cdot \sum _{j=0}^0(-1{)}^j\cdot C_5^j\cdot (t-j{)}^4=t^4$$
将$w_0,w_1,w_2,w_3,w_4$展开，合并同类多项式，可得，
w 0 = 1 6 ⋅ ( t 4 − 4 t 3 + 6 t 2 − 4 t + 1 ) = 1 6 ⋅ [ 1   t   t 2   t 3   t 4 ] [ 1 − 4 6 − 4 1 ] w_0 = \frac{1}{6}\cdot (t^4 - 4t^3 + 6t^2 - 4t + 1)=\frac{1}{6} \cdot[1\ t \ t^2 \ t^3 \ t^4]

w0​=61​⋅(t4−4t3+6t2−4t+1)=61​⋅[1 t t2 t3 t4]⎣ ⎡​1−46−41​⎦ ⎤​
w 1 = 1 6 ⋅ ( − 4 t 4 + 12 t 3 − 6 t 2 − 12 t + 11 ) = 1 6 ⋅ [ 1   t   t 2   t 3   t 4 ] [ 11 − 12 − 6 12 − 4 ] w_1=\frac{1}{6} \cdot (- 4t^4 + 12t^3 - 6t^2 - 12t + 11) = \frac{1}{6} \cdot[1\ t \ t^2 \ t^3 \ t^4]

[11−12−612−4]" role="presentation" style="position: relative;">⎡⎣⎢⎢⎢⎢⎢⎢11−12−612−4⎤⎦⎥⎥⎥⎥⎥⎥$$\left[\begin{array}{c}11\\ -12\\ -6\\ 12\\ -4\end{array}\right]$$

w1​=61​⋅(−4t4+12t3−6t2−12t+11)=61​⋅[1 t t2 t3 t4]⎣ ⎡​11−12−612−4​⎦ ⎤​
w 2 = 1 6 ⋅ ( 6 t 4 − 12 t 3 − 6 t 2 + 12 t + 11 ) = 1 6 ⋅ [ 1   t   t 2   t 3   t 4 ] [ 11 12 − 6 − 12 6 ] w_2=\frac{1}{6} \cdot (6t^4 - 12t^3 - 6t^2 + 12t + 11) = \frac{1}{6} \cdot[1\ t \ t^2 \ t^3 \ t^4]

[1112−6−126]" role="presentation" style="position: relative;">⎡⎣⎢⎢⎢⎢⎢⎢1112−6−126⎤⎦⎥⎥⎥⎥⎥⎥$$\left[\begin{array}{c}11\\ 12\\ -6\\ -12\\ 6\end{array}\right]$$

w2​=61​⋅(6t4−12t3−6t2+12t+11)=61​⋅[1 t t2 t3 t4]⎣ ⎡​1112−6−126​⎦ ⎤​
w 3 = 1 6 ⋅ ( − 4 t 4 + 4 t 3 + 6 t 2 + 4 t + 1 ) = 1 6 ⋅ [ 1   t   t 2   t 3   t 4 ] [ 1 4 6 4 − 4 ] w_3=\frac{1}{6} \cdot (- 4t^4 + 4t^3 + 6t^2 + 4t + 1) = \frac{1}{6} \cdot[1\ t \ t^2 \ t^3 \ t^4]

[1464−4]" role="presentation" style="position: relative;">⎡⎣⎢⎢⎢⎢⎢⎢1464−4⎤⎦⎥⎥⎥⎥⎥⎥$$\left[\begin{array}{c}1\\ 4\\ 6\\ 4\\ -4\end{array}\right]$$

w3​=61​⋅(−4t4+4t3+6t2+4t+1)=61​⋅[1 t t2 t3 t4]⎣ ⎡​1464−4​⎦ ⎤​
w 4 = 1 6 ⋅ t 4 = 1 6 ⋅ [ 1   t   t 2   t 3   t 4 ] [ 0 0 0 0 1 ] w_4=\frac{1}{6} \cdot t^4=\frac{1}{6} \cdot[1\ t \ t^2 \ t^3 \ t^4]

[00001]" role="presentation" style="position: relative;">⎡⎣⎢⎢⎢⎢⎢⎢00001⎤⎦⎥⎥⎥⎥⎥⎥$$\left[\begin{array}{c}0\\ 0\\ 0\\ 0\\ 1\end{array}\right]$$

w4​=61​⋅t4=61​⋅[1 t t2 t3 t4]⎣ ⎡​00001​⎦ ⎤​
将上述$w_0,w_1,w_2,w_3,w_4$代入公式(4)，整理可得B样条曲线的表达式为，
c ( t ) = [ 1   t   t 2   t 3   t 4 ] ⋅ 1 6 ⋅ [ 1 11 11 1 0 − 4 − 12 12 4 0 6 − 6 − 6 6 0 − 4 12 − 12 4 0 1 − 4 6 − 4 1 ] ⋅ [ P i − k P i − k + 1 ⋮ P i ] c(t) = [1\ t\ t^2\ t^3\ t^4]\cdot \frac{1}{6}\cdot

[1111110−4−1212406−6−660−412−12401−46−41]" role="presentation" style="position: relative;">⎡⎣⎢⎢⎢⎢⎢⎢1−46−4111−12−612−41112−6−1261464−400001⎤⎦⎥⎥⎥⎥⎥⎥$$\left[\begin{array}{ccccc}1& 11& 11& 1& 0\\ -4& -12& 12& 4& 0\\ 6& -6& -6& 6& 0\\ -4& 12& -12& 4& 0\\ 1& -4& 6& -4& 1\end{array}\right]$$

\cdot

[Pi−kPi−k+1⋮Pi]" role="presentation" style="position: relative;">⎡⎣⎢⎢⎢⎢Pi−kPi−k+1⋮Pi⎤⎦⎥⎥⎥⎥$$\left[\begin{array}{c}{P}_{i-k}\\ P_{i-k+1}\\ ⋮\\ P_i\end{array}\right]$$

c(t)=[1 t t2 t3 t4]⋅61​⋅⎣ ⎡​1−46−41​11−12−612−4​1112−6−126​1464−4​00001​⎦ ⎤​⋅⎣ ⎡​Pi−k​Pi−k+1​⋮Pi​​⎦ ⎤​
该表达式与第1节矩阵表达中公式(1)取$k=4$时一致。

3 参考文献

[1] http://t.csdn.cn/WJQkU