帮我编写一个PUMA机器人运动学仿真的程序
时间: 2023-10-19 16:05:17 浏览: 33
好的,下面是一个简单的PUMA机器人运动学仿真的MATLAB程序,仅供参考:
```matlab
% DH参数
a = [0, 425, 392, 0, 0, 0]; % 关节长度
alpha = [-pi/2, 0, 0, -pi/2, pi/2, -pi/2]; % 关节旋转角度
d = [0, 0, 0, 1093, 94.75, 0]; % 关节偏移量
theta = [0, pi/2, 0, 0, 0, 0]; % 关节角度
% 正向运动学
T = eye(4);
for i = 1:6
T = T * DH(a(i), alpha(i), d(i), theta(i));
end
% 打印末端执行器的位置和姿态
disp('End effector position:');
disp(T(1:3, 4)');
disp('End effector orientation (Euler angles in degrees):');
disp(rad2deg(rotm2eul(T(1:3, 1:3))));
% 逆向运动学
P = [800, 0, 1000]'; % 末端执行器的位置
R = eul2rotm([0, 0, pi/2]); % 末端执行器的方向
T = [R, P; 0, 0, 0, 1];
theta = puma_ik(T); % 计算关节角度
% 打印关节角度
disp('Joint angles:');
disp(rad2deg(theta));
% DH参数转换矩阵
function T = DH(a, alpha, d, theta)
T = [cos(theta), -sin(theta)*cos(alpha), sin(theta)*sin(alpha), a*cos(theta);
sin(theta), cos(theta)*cos(alpha), -cos(theta)*sin(alpha), a*sin(theta);
0, sin(alpha), cos(alpha), d;
0, 0, 0, 1];
end
% 逆向运动学
function theta = puma_ik(T)
a = [0, 425, 392, 0, 0, 0];
d = [0, 0, 0, 1093, 94.75, 0];
theta = zeros(6, 1);
P = T(1:3, 4);
R = T(1:3, 1:3);
O = P - R * [0, 0, d(6)]';
theta(1) = atan2(O(2), O(1));
c3 = (norm(O)^2 - a(2)^2 - a(3)^2) / (2 * a(2) * a(3));
s3 = sqrt(1 - c3^2);
theta(3) = atan2(s3, c3);
alpha = atan2(-O(3) - d(1), sqrt(O(1)^2 + O(2)^2)) - theta(1);
c2 = (a(2) + a(3)*c3)*cos(alpha) + a(3)*s3*sin(alpha);
s2 = (a(2) + a(3)*c3)*sin(alpha) - a(3)*s3*cos(alpha);
theta(2) = atan2(s2, c2);
R03 = [cos(theta(1))*cos(theta(2)+theta(3)), -sin(theta(1)), cos(theta(1))*sin(theta(2)+theta(3));
sin(theta(1))*cos(theta(2)+theta(3)), cos(theta(1)), sin(theta(1))*sin(theta(2)+theta(3));
-sin(theta(2)+theta(3)), 0, cos(theta(2)+theta(3))];
R36 = R03' * R;
theta(4) = atan2(R36(2, 3), R36(1, 3));
theta(5) = atan2(sqrt(1 - R36(3, 3)^2), R36(3, 3));
theta(6) = atan2(R36(3, 2), -R36(3, 1));
end
```
这个程序包括正向运动学和逆向运动学两部分。正向运动学计算末端执行器的位置和姿态,逆向运动学根据末端执行器的位置和姿态计算关节角度。注意,这个程序只是一个简单的示例,实际应用中需要更加完善和精细的处理。