Question

如何用matlab解以下矩阵方程

OTH =[-0.2924 -0.9563 0 0.6978;
0.9563 -0.2924 0 0.8172 ;
0 0 1 0
0 0 0 1]
A1=[ cos((pi*sita1)/180), -sin((pi*sita1)/180), 0, cos((pi*sita1)/180)
sin((pi*sita1)/180), cos((pi*sita1)/180), 0, sin((pi*sita1)/180)
0, 0, 1, 0
0, 0, 0, 1]
A2=[ cos((pi*sita2)/180), -sin((pi*sita2)/180), 0, cos((pi*sita2)/180)
sin((pi*sita2)/180), cos((pi*sita2)/180), 0, sin((pi*sita2)/180)
0, 0, 1, 0
0, 0, 0, 1]
A1*A2=OTH;
求sita1 sita2

Answer 1

matlab也不能解这么复杂的方程
手动加上matlab辅助应该应该比较好解决问题

首先，简单起见，先不用角度制，直接中弧度制
设c1=(pi*sita1)/180，c2=(pi*sita2)/180;

A1=[ cos(c1), -sin(c1), 0, cos(c1);
sin(c1), cos(c1), 0, sin(c1);
0, 0, 1, 0;
0, 0, 0, 1];
A2=[ cos(c2), -sin(c2), 0, cos(c2);
sin(c2), cos(c2), 0, sin(c2);
0, 0, 1, 0;
0, 0, 0, 1];
simplify(A1*A2)

[ cos(c1 + c2), -sin(c1 + c2), 0, cos(c1 + c2) + cos(c1)]
[ sin(c1 + c2), cos(c1 + c2), 0, sin(c1 + c2) + sin(c1)]
[ 0, 0, 1, 0]
[ 0, 0, 0, 1]

和OHT对比
OTH =[-0.2924 -0.9563 0 0.6978;
0.9563 -0.2924 0 0.8172 ;
0 0 1 0
0 0 0 1]
得到
cos(c1+c2)=-0.2924
sin(c1+c2)=0.9563
cos(c1 + c2) + cos(c1)= 0.6978
sin(c1 + c2) + sin(c1)= 0.8172

cos(c1)=0.9902
sin(c1)=0.1391

解得
c1~=8度
c1+c2~=107度

所以 c2=99度