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基于最小二乘的椭圆拟合理论

芥子书屋 2022-03-12 阅读 57
算法

基于最小二乘的椭圆拟合理论

设平面任意位置椭圆方程为

1=b11*x-c12+2*b12*x-c1*y-c2+b22*y-c22

Pi(xi,yi)(i=1,2,…,N)为椭圆轮廓上的N(N≥5)个测量点,依据最小二乘原理,所拟合的目标函数为:

Fb11,b12,b22=i=1Nb11*xi-c12+2*b12*xi-c1*yi-c2+b22*yi-c22-12

展开上式得:

Fb11,b12,b22=i=1Nb11*xi-c12+b22*yi-c22-12+2*b12*xi-c1*yi-c22+2*b11*xi-c12+b22*yi-c22-1*2*b12*xi-c1*yi-c2

上式中的每项单独进行分解:

b11*xi-c12+b22*yi-c22-12

=b112*xi-c14+b22*yi-c22-12+2*b11*xi-c12*b22*yi-c22-1

=b112*xi-c14+b222*yi-c24-2*b22*yi-c22+1+2*b11*xi-c12*b22*yi-c22-2*b11*xi-c12

2*b12*xi-c1*yi-c22=4*b122*xi-c12*yi-c22

2*b11*xi-c12+b22*yi-c22-1*2*b12*xi-c1*yi-c2

=2*b11*xi-c12*2*b12*xi-c1*yi-c2+2*b22*yi-c22*2*b12*xi-c1*yi-c2-4*b12*xi-c1*yi-c2

=4*b11*b12*xi-c13*yi-c2+4*b22*b12*xi-c1*yi-c23-4*b12*xi-c1*yi-c2

方程中有三个未知数:b11,b12,b22,需要三个方程,则要求采样点至少为三组

∂F∂b11=∂F∂b12=∂F∂b22=0

0=∂F∂b11=i=1N2*b11*xi-c14+2*xi-c12*b22*yi-c22-2*xi-c12+4*b12*xi-c13*yi-c2

化简上式

i=1Nb11*xi-c14+ i=1Nb22*xi-c12*yi-c22+ i=1N2*b12*xi-c13*yi-c2= i=1Nxi-c12

                                                                                                           ---------公式(1)

0=∂F∂b22=i=1N2*b22*yi-c24-2*yi-c22+2*b11*xi-c12*yi-c22+4*b12*xi-c1*yi-c23

化简上式

i=1Nb11*xi-c12*yi-c22+i=1Nb22*yi-c24+i=1N2*b12*xi-c1*yi-c23= i=1Nyi-c22

                                                                                                                                                                                                                                             ---------公式(2)

0=∂F∂b12=i=1N8*b12*xi-c12*yi-c22+4*b11*xi-c13*yi-c2+4*b22*xi-c1*yi-c23-4*xi-c1*yi-c2

化简上式

i=1N2*b11*xi-c13*yi-c2 + i=1N2*b22*xi-c1*yi-c23 +i=1N4*b12*xi-c12*yi-c22= i=1N2*xi-c1*yi-c2

                                                                                                                                                                                                                                             ---------公式(3)

根据化简的三个公式左侧组成矩阵M,根据三个公式右侧组成矩阵L即可求出三个未知数:b11,b12,b22

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