6自由度机械臂python求正解代码

由于机械臂的结构和运动学方程的复杂性，求解机械臂的正解需要使用运动学模型和逆运动学算法。以下是一个简单的6自由度机械臂正解的python代码。

import numpy as np

def forward_kinematics(theta1, theta2, theta3, theta4, theta5, theta6):
    # 机械臂DH参数
    d1 = 0.089159
    a2 = -0.425
    a3 = -0.39225
    d4 = 0.10915
    d5 = 0.09465
    d6 = 0.0823

    # 转换矩阵
    T1 = np.array([
        [np.cos(theta1), -np.sin(theta1), 0, 0],
        [np.sin(theta1), np.cos(theta1), 0, 0],
        [0, 0, 1, d1],
        [0, 0, 0, 1]
    ])
    T2 = np.array([
        [np.cos(theta2), -np.sin(theta2), 0, a2*np.cos(theta2)],
        [np.sin(theta2)*np.cos(-np.pi/2), np.cos(theta2)*np.cos(-np.pi/2), -np.sin(-np.pi/2), -a2*np.sin(theta2)*np.cos(-np.pi/2)],
        [np.sin(theta2)*np.sin(-np.pi/2), np.cos(theta2)*np.sin(-np.pi/2), np.cos(-np.pi/2), -a2*np.sin(theta2)*np.sin(-np.pi/2)],
        [0, 0, 0, 1]
    ])
    T3 = np.array([
        [np.cos(theta3), -np.sin(theta3), 0, a3*np.cos(theta3)],
        [np.sin(theta3)*np.cos(-np.pi/2), np.cos(theta3)*np.cos(-np.pi/2), -np.sin(-np.pi/2), -a3*np.sin(theta3)*np.cos(-np.pi/2)],
        [np.sin(theta3)*np.sin(-np.pi/2), np.cos(theta3)*np.sin(-np.pi/2), np.cos(-np.pi/2), -a3*np.sin(theta3)*np.sin(-np.pi/2)],
        [0, 0, 0, 1]
    ])
    T4 = np.array([
        [np.cos(theta4), -np.sin(theta4), 0, 0],
        [0, 0, -1, -d4],
        [np.sin(theta4), np.cos(theta4), 0, 0],
        [0, 0, 0, 1]
    ])
    T5 = np.array([
        [np.cos(theta5), -np.sin(theta5), 0, 0],
        [0, 0, 1, d5],
        [-np.sin(theta5), -np.cos(theta5), 0, 0],
        [0, 0, 0, 1]
    ])
    T6 = np.array([
        [np.cos(theta6), -np.sin(theta6), 0, 0],
        [0, 0, -1, -d6],
        [np.sin(theta6), np.cos(theta6), 0, 0],
        [0, 0, 0, 1]
    ])

    # 求解正解
    T = T1 @ T2 @ T3 @ T4 @ T5 @ T6
    return T[:3, 3]  # 返回机械臂末端坐标

if __name__ == '__main__':
    theta1 = np.deg2rad(30)
    theta2 = np.deg2rad(-45)
    theta3 = np.deg2rad(60)
    theta4 = np.deg2rad(0)
    theta5 = np.deg2rad(90)
    theta6 = np.deg2rad(0)

    pos = forward_kinematics(theta1, theta2, theta3, theta4, theta5, theta6)
    print(pos)

上述代码中，我们首先定义了机械臂的DH参数和转换矩阵。然后使用numpy库中的矩阵乘法运算符@，将各个转换矩阵按顺序相乘，得到机械臂的正解矩阵T。最后返回矩阵T的前三行前三列，即机械臂末端的坐标。

在main函数中，我们给出了一组关节角度，调用forward_kinematics函数求解机械臂末端的坐标，并输出结果。