matlab练习程序(旋转矩阵、欧拉角、四元数互转)

匿名 (未验证) 提交于 2019-12-02 23:26:52

欧拉角转旋转矩阵公式:

旋转矩阵转欧拉角公式:

旋转矩阵转四元数公式,其中1+r11+r22+r33>0:

四元数转旋转矩阵公式,q0^2+q1^2+q2^2+q3^2=1:

欧拉角转四元数公式:

四元数转欧拉角公式:

matlab代码如下:

clear all; close all; clc;  %欧拉角 x = 0.5; y = 0.6; z = 0.7; Ang1 = [x y z];  %欧拉角转旋转矩阵 Rx = [1      0      0;     0 cos(x) -sin(x);     0 sin(x) cos(x)]; Ry = [cos(y)  0 sin(y);     0       1      0;     -sin(y) 0 cos(y)]; Rz = [cos(z) -sin(z) 0;     sin(z) cos(z)  0;     0      0       1]; R = Rz*Ry*Rx; R1 = R;  %旋转矩阵转欧拉角 x = atan2(R(3,2),R(3,3)); y = atan2(-R(3,1),sqrt(R(3,2)^2+R(3,3)^2)); z = atan2(R(2,1),R(1,1)); Ang2 = [x y z];  %旋转矩阵转四元数 t=sqrt(1+R(1,1)+R(2,2)+R(3,3))/2; q=[t (R(3,2)-R(2,3))/(4*t) (R(1,3)-R(3,1))/(4*t) (R(2,1)-R(1,2))/(4*t)]; Q1 = q;  %四元数转旋转矩阵 R=[ 2*q(1).^2-1+2*q(2)^2    2*(q(2)*q(3)-q(1)*q(4)) 2*(q(2)*q(4)+q(1)*q(3));     2*(q(2)*q(3)+q(1)*q(4)) 2*q(1)^2-1+2*q(3)^2     2*(q(3)*q(4)-q(1)*q(2));     2*(q(2)*q(4)-q(1)*q(3)) 2*(q(3)*q(4)+q(1)*q(2)) 2*q(1)^2-1+2*q(4)^2]; R2 = R;  %欧拉角转四元数 q = [cos(x/2)*cos(y/2)*cos(z/2) + sin(x/2)*sin(y/2)*sin(z/2) ...     sin(x/2)*cos(y/2)*cos(z/2) - cos(x/2)*sin(y/2)*sin(z/2) ...     cos(x/2)*sin(y/2)*cos(z/2) + sin(x/2)*cos(y/2)*sin(z/2) ...     cos(x/2)*cos(y/2)*sin(z/2) - sin(x/2)*sin(y/2)*cos(z/2)]; Q2 = q;  %四元数转欧拉角 x = atan2(2*(q(1)*q(2)+q(3)*q(4)),1 - 2*(q(2)^2+q(3)^2)); y = asin(2*(q(1)*q(3) - q(2)*q(4))); z = atan2(2*(q(1)*q(4)+q(2)*q(3)),1 - 2*(q(3)^2+q(4)^2)); Ang3 = [x y z];  Ang1 Ang2 Ang3  R1 R2  Q1 Q2

参考:

https://blog.csdn.net/lql0716/article/details/72597719

https://www.cnblogs.com/21207-iHome/p/6894128.html

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