clear all; clc; du = pi/180; a = [0+0.001, 185+0.0079, 0+0.005, 120+0.12]; alpha = [pi/2+0.003, 0+0.001, pi/2+0.005, pi/2]; d = [0+0.001, 0+0.0079, 90+0.005, 0+0.12]; theta = [90du+0.02, 0, 0.023, 0.08]; beta = zeros(1, 4)+0; L1(1) = Link('d', d(1), 'a', a(1), 'alpha', alpha(1), 'qlim', [180du, 365du], 'modified'); L1(2) = Link('d', d(2), 'a', a(2), 'alpha', alpha(2), 'qlim', [3du, 63du], 'modified'); L1(3) = Link('d', d(3), 'a', a(3), 'alpha', alpha(3), 'qlim', [60du, 120du], 'modified'); L1(4) = Link('d', d(4), 'a', a(4), 'alpha', alpha(4), 'qlim', [230du, 326du], 'modified'); Needle = SerialLink(L1, 'name', 'Needle'); T1 = DH(1, a(1), alpha(1), d(1), theta(1)+beta(1)); T2 = DH(2, a(2), alpha(2), d(2), theta(2)+beta(2)); T3 = DH(3, a(3), alpha(3), d(3), theta(3)+beta(3)); T4 = DH(4, a(4), alpha(4), d(4), theta(4)+beta(4)); T = T1 * T2 * T3 * T4; % Step 2:利用微分变换原理计算机器人各个连杆机构之间的微小原始偏差 delta_a = 0.001; % a参数的微小偏差 delta_alpha = 0.001; % alpha参数的微小偏差 delta_d = 0.001; % d参数的微小偏差 delta_theta = 0.001; % theta参数的微小偏差 delta_beta = 0.001; % beta参数的微小偏差 delta_T1 = DH(1, a(1)+delta_a, alpha(1), d(1), theta(1)+beta(1)) - T1; delta_T2 = DH(2, a(2)+delta_a, alpha(2), d(2), theta(2)+beta(2)) - T2; delta_T3 = DH(3, a(3)+delta_a, alpha(3), d(3), theta(3)+beta(3)) - T3; delta_T4 = DH(4, a(4)+delta_a, alpha(4), d(4), theta(4)+beta(4)) - T4; % Step 3:计算误差矩阵 delta_T = delta_T1 * delta_T2 * delta_T3 * delta_T4; % Step 4:将误差矩阵转化为误差值 delta_x = delta_T(1,4); delta_y = delta_T(2,4); delta_z = delta_T(3,4); % 输出末端位姿误差 fprintf('末端位姿误差:\n'); fprintf('Delta x: %.6f mm\n', delta_x1000); fprintf('Delta y: %.6f mm\n', delta_y1000); fprintf('Delta z: %.6f mm\n', delta_z1000);想要输入一组角度值然后得到末端位姿误差,帮我改写一下程序。
时间: 2023-09-10 19:12:53 浏览: 42
好的,可以尝试如下改写程序:
```matlab
clear all; clc;
% Step 1:机器人模型建立
du = pi/180;
a = [0+0.001, 185+0.0079, 0+0.005, 120+0.12];
alpha = [pi/2+0.003, 0+0.001, pi/2+0.005, pi/2];
d = [0+0.001, 0+0.0079, 90+0.005, 0+0.12];
% 定义机器人的末端执行器姿态
theta = zeros(1, 4);
beta = zeros(1, 4);
% 根据 DH 参数创建机器人模型
L1(1) = Link('d', d(1), 'a', a(1), 'alpha', alpha(1), 'qlim', [180*du, 365*du], 'modified');
L1(2) = Link('d', d(2), 'a', a(2), 'alpha', alpha(2), 'qlim', [3*du, 63*du], 'modified');
L1(3) = Link('d', d(3), 'a', a(3), 'alpha', alpha(3), 'qlim', [60*du, 120*du], 'modified');
L1(4) = Link('d', d(4), 'a', a(4), 'alpha', alpha(4), 'qlim', [230*du, 326*du], 'modified');
Needle = SerialLink(L1, 'name', 'Needle');
% Step 2:输入角度值
theta = [10*du, 20*du, 30*du, 40*du];
% Step 3:计算机器人运动学正解
T = Needle.fkine(theta);
% Step 4:利用微分变换原理计算机器人各个连杆机构之间的微小原始偏差
delta_a = 0.001; % a参数的微小偏差
delta_alpha = 0.001; % alpha参数的微小偏差
delta_d = 0.001; % d参数的微小偏差
delta_theta = 0.001; % theta参数的微小偏差
delta_beta = 0.001; % beta参数的微小偏差
delta_T1 = DH(1, a(1)+delta_a, alpha(1), d(1), theta(1)+delta_theta+beta(1)) - DH(1, a(1), alpha(1), d(1), theta(1)+beta(1));
delta_T2 = DH(2, a(2)+delta_a, alpha(2), d(2), theta(2)+delta_theta+beta(2)) - DH(2, a(2), alpha(2), d(2), theta(2)+beta(2));
delta_T3 = DH(3, a(3)+delta_a, alpha(3), d(3), theta(3)+delta_theta+beta(3)) - DH(3, a(3), alpha(3), d(3), theta(3)+beta(3));
delta_T4 = DH(4, a(4)+delta_a, alpha(4), d(4), theta(4)+delta_theta+beta(4)) - DH(4, a(4), alpha(4), d(4), theta(4)+beta(4));
% Step 5:计算误差矩阵
delta_T = delta_T1 * delta_T2 * delta_T3 * delta_T4;
% Step 6:将误差矩阵转化为误差值
delta_x = delta_T(1,4);
delta_y = delta_T(2,4);
delta_z = delta_T(3,4);
% Step 7:输出末端位姿误差
fprintf('末端位姿误差:\n');
fprintf('Delta x: %.6f mm\n', delta_x*1000);
fprintf('Delta y: %.6f mm\n', delta_y*1000);
fprintf('Delta z: %.6f mm\n', delta_z*1000);
```
其中,Step 2 中输入的角度值可以根据实际情况进行修改。同时,Step 4 中的计算偏差的代码需要进行微小偏差的计算,同时需要针对每个连杆机构单独计算。最后,Step 6 中将误差矩阵转化为误差值时需要乘以1000,转换为毫米单位。
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