已知六个自由度关节空间的角度和dh参数求到末端位置的转换代码
时间: 2023-03-26 21:01:00 浏览: 67
以下是一个求解六个自由度关节空间的角度和dh参数到末端位置的转换代码的示例:
#include <iostream>
#include <cmath>
using namespace std;
const double PI = 3.14159265358979323846;
// DH参数
double dh[6][4] = {
{, , , },
{, , , },
{, , , },
{, , , },
{, , , },
{, , , }
};
// 末端位置
double end_pos[3] = {, , };
// 计算转换矩阵
void calc_transform(double alpha, double a, double d, double theta, double T[4][4]) {
T[][] = cos(theta);
T[][1] = -sin(theta) * cos(alpha);
T[][2] = sin(theta) * sin(alpha);
T[][3] = a * cos(theta);
T[1][] = sin(theta);
T[1][1] = cos(theta) * cos(alpha);
T[1][2] = -cos(theta) * sin(alpha);
T[1][3] = a * sin(theta);
T[2][] = ;
T[2][1] = sin(alpha);
T[2][2] = cos(alpha);
T[2][3] = d;
T[3][] = ;
T[3][1] = ;
T[3][2] = ;
T[3][3] = 1;
}
// 计算末端位置
void calc_end_pos(double angles[6]) {
double T[4][4];
double T_total[4][4];
double alpha, a, d, theta;
for (int i = ; i < 6; i++) {
alpha = dh[i][];
a = dh[i][1];
d = dh[i][2];
theta = dh[i][3] + angles[i];
calc_transform(alpha, a, d, theta, T);
if (i == ) {
for (int j = ; j < 4; j++) {
for (int k = ; k < 4; k++) {
T_total[j][k] = T[j][k];
}
}
} else {
double T_temp[4][4];
for (int j = ; j < 4; j++) {
for (int k = ; k < 4; k++) {
T_temp[j][k] = ;
for (int l = ; l < 4; l++) {
T_temp[j][k] += T_total[j][l] * T[l][k];
}
}
}
for (int j = ; j < 4; j++) {
for (int k = ; k < 4; k++) {
T_total[j][k] = T_temp[j][k];
}
}
}
}
end_pos[] = T_total[][3];
end_pos[1] = T_total[1][3];
end_pos[2] = T_total[2][3];
}
int main() {
double angles[6] = {, , , , , };
calc_end_pos(angles);
cout << "末端位置: (" << end_pos[] << ", " << end_pos[1] << ", " << end_pos[2] << ")" << endl;
return ;
}
希望这个代码能够帮助到你!