6轴机器人DH建模、仿真、正逆解代码

一.模型

二.MDH模型

使用修改DH模型、改进DH模型与标准DH模型主要区别在于末端，改进DH并没有建到工具坐标系，这里将4、5、6轴的坐标系都建在同一点，因此机器人连杆的参数只需要4个值a1、a2、a3、d4。a2、d4分别代表大臂与小臂的长度。要完全得到末端或者工具尖点的位姿还需要再右成一个工具坐标系6Te

  • DH图


    1604040111741.jpg

  • DH参数表

θ d a α
0 0 0 0
0 0 109 -90
0 0 250.5 0
0 239.5 115.5 -90
0 0 0 90
0 0 0 -90

  • MATLAB仿真效果


    image.png

  • MATLAB代码

clear;
clc;
%建立机器人模型
%       theta    d           a        alpha     offset
ML1=Link([0       0           0           0        0    ],'modified'); 
ML2=Link([0       0           109       -pi/2      0    ],'modified');
ML3=Link([0       0           250.5       0        0    ],'modified');
ML4=Link([0       239.5       115.5     -pi/2      0    ],'modified');
ML5=Link([0       0           0          pi/2      0    ],'modified');
ML6=Link([0       0           0         -pi/2      0    ],'modified');
modrobot=SerialLink([ML1 ML2 ML3 ML4 ML5 ML6],'name','6axis');
teach(modrobot)
  • 一般6轴机器人初始状态为大臂垂直与地面，主要是通过给2轴一个初始角度-90°来实现，下图第一个模型为各轴0°时的姿态，第二图模型为一般回零后机器人保持的姿态。
    1

    2

三.6轴机器人运动学正解

  • 代码
#include <Eigen/Dense>
/* 正解函数
* @joint 6轴各关节值结构体，单位弧度
* @return 返回一个4*4的位姿矩阵
*/
Eigen::Matrix4d IFKinematics::fKine(const Joint & joint) 
{
    double j1 = joint.j1,
           j2 = joint.j2,
           j3 = joint.j3,
           j4 = joint.j4,
           j5 = joint.j5,
           j6 = joint.j6;

    double s1 = sin(j1), c1 = cos(j1),
           s2 = sin(j2), c2 = cos(j2),
           s3 = sin(j3), c3 = cos(j3),
           s4 = sin(j4), c4 = cos(j4),
           s5 = sin(j5), c5 = cos(j5),
           s6 = sin(j6), c6 = cos(j6);

    Eigen::Matrix4d T01, T12, T23, T34, T45, T56, T0_6;
    T01 << c1,    0, -s1, a1 * c1,
           s1,    0,  c1, a1 * s1,
            0,   -1,   0,       0,
            0,    0,   0,       1;
    T12 << c2,  -s2,   0, a2 * c2,
           s2,   c2,   0, a2 * s2,
            0,    0,   1,       0,
            0,    0,   0,       1;
    T23 << c3,    0, -s3, a3 * c3,
           s3,    0,  c3, a3 * s3,
            0,   -1,   0,       0,
            0,    0,   0,       1;
    T34 << c4,    0,  s4,       0,
           s4,    0, -c4,       0,
            0,    1,   0,      d4,
            0,    0,   0,       1;
    T45 << c5,    0, -s5,       0,
           s5,    0,  c5,       0,
            0,   -1,   0,       0,
            0,    0,   0,       1;
    T56 << c6,  -s6,   0,       0,
           s6,   c6,   0,       0,
            0,    0,   1,       0,
            0,    0,   0,       1;
    T0_6 = T01 * T12 * T23 * T34 * T45 * T56;
    return T0_6;
}

四.6轴机器人运动学逆解

  • 代码
#include <Eigen/Dense>
#include <math.h>

#define EPSINON 1e-4  // 精度，小于它就认为是0
#define ALLOW_MAX_DIFF (1 / 180 * pi)  /* 距离上一轴角度的最大允许偏差 */

/* 各轴运动角度范围 */
const double MAX_JOINTS[6] = { 90, 30, 90, 135, 135, 180}; 
const double MIN_JOINTS[6] = { -90, -135, -135, -135, -135, -180};
const double pi = 3.1415926

/* 角度值转弧度值 */
inline double toRadian(double degree)
{
    return degree * pi / 180;
}

/* 逆解函数
* @Joint 6轴各关节值结构体，单位弧度
* @lastJoints 上一次计算的关节值结构体
* @pose 要解逆解的位姿矩阵
*/
Joint IFKinematics::sixJointsIK(Eigen::Matrix4d pose, Joint lastJoints) 
{
    double nx = pose(0, 0), ox = pose(0, 1), ax = pose(0, 2),
           ny = pose(1, 0), oy = pose(1, 1), ay = pose(1, 2),
           nz = pose(2, 0), oz = pose(2, 1), az = pose(2, 2),
           px = pose(0 ,3), py = pose(1, 3), pz = pose(2, 3);

    /** joint 1 */
    double j1[2] = {0, 0};
    j1[0] = -atan2(-py, px);
    j1[1] = -atan2(-py, px) + pi;

    /** joint 3 */
    double j3[4] = {0, 0, 0, 0};
    /* temp, x, y, z, A are temp valiables */
    double temp[2], A[2];
    double x = a2 * d4,
           y = a2 * a3,
           z = a2 * a2 + a3 * a3 + d4 * d4;
    for (int i = 0; i < 4; i+=2) {
        temp[i/2] = a1 - px * cos(j1[i/2]) - py * sin(j1[i/2]);
        A[i/2] = z - pz * pz - temp[i/2] * temp[i/2];
        j3[i]   = -atan2(-y, x) + atan2(A[i/2],sqrt( (4 * x * x + 4 * y * y - A[i/2] * A[i/2])));
        j3[i+1] = -atan2(-y, x) + atan2(A[i/2],-sqrt((4 * x * x + 4 * y * y - A[i/2] * A[i/2])));
    }


    /** joint 2 */
    double j2[4] = {0, 0, 0, 0};
    double m[4]  = {0, 0, 0, 0};
    double n[4]  = {0, 0, 0, 0};
    for (int i = 0; i < 4; ++i) {
        m[i] = a2 + a3 * cos(j3[i]) - d4 * sin(j3[i]);
        n[i] = a3 * sin(j3[i]) + d4 * cos(j3[i]);
        j2[i] = atan2(-pz * m[i] + n[i] * temp[i/2],-pz * n[i] - m[i] * temp[i/2]);
    }


    /** joint 5 */
    double j5[8] = {0, 0, 0, 0, 0, 0, 0, 0};
    for (int i = 0; i < 4; ++i) {
        m[i] = (ax * cos(j1[i / 2]) * cos(j2[i]) + ay * sin(j1[i / 2]) * cos(j2[i]) + az * -1 * sin(j2[i]));
        n[i] = -(ax * cos(j1[i / 2]) * sin(j2[i]) + ay * sin(j1[i / 2]) * sin(j2[i]) - az * -1 * cos(j2[i]));
    }
    for (int i = 0; i < 8; ) {
        x =  n[i/2] * cos(j3[i/2]) - m[i/2] * sin(j3[i/2]);
        y = sqrt(pow((ay * cos(j1[i / 4]) - ax * sin(j1[i / 4])),2) + pow((m[i/2] * cos(j3[i/2]) + n[i/2] * sin(j3[i/2])),2));
        j5[i++] = atan2( y, x);
        j5[i++] = atan2(-y, x);
    }


    /** joint 4 */
    double j4[8] = {0, 0, 0, 0, 0, 0, 0, 0};
    for (int i = 0; i < 8; ++i) {
        if (fabs(sin(j5[i])) >= EPSINON) {
            y = ((ay*cos(j1[i/4])-ax*sin(j1[i/4]))) / sin(j5[i]);
            x= (-m[i/2]*cos(j3[i/2])-n[i/2]*sin(j3[i/2]))/(sin(j5[i]));
            j4[i] = atan2(y , x);
        } else {
            j4[i] = 0;
        }
    }


    /** joint 6 */
    double j6[8] = {0, 0, 0, 0, 0, 0, 0 ,0};
    for (int i = 0; i < 8; ++i) {
        double u = nx * sin(j1[i/4]) - ny * cos(j1[i/4]);
        double v = ox * sin(j1[i/4]) - oy * cos(j1[i/4]);
        j6[i] = atan2(cos(j4[i]) * u - cos(j5[i]) * sin(j4[i]) * v, cos(j4[i]) * v + cos(j5[i]) * sin(j4[i]) * u);
    }

    /** 8组逆解结果 */
    double ans[8][6];
    for (int i = 0; i < 8; ++i) {
        ans[i][0] = j1[i / 4];
        ans[i][1] = j2[i / 2];
        ans[i][2] = j3[i / 2];
        ans[i][3] = j4[i];
        ans[i][4] = j5[i];
        ans[i][5] = j6[i];
    }

    /** 在选择8组解中的最优解，主要根据三个原则：连续原则、限位原则、最短路径原则 ，连续原则暂时不使用 */
    double last[6] = {lastJoints.j1, lastJoints.j2, lastJoints.j3, lastJoints.j4, lastJoints.j5, lastJoints.j5};
    bool isOutLimit[8] = {false, false, false, false, false, false, false, false};
    //bool isContinuity[8] = {true, true, true, true, true, true, true, true};

    /** 排除掉超过运动范围的解 */
    for (int i = 0; i < 8; ++i) {
        for (int j = 0; j < 6; ++j) {
            if (ans[i][j] < toRadian(MIN_JOINTS[j]) || ans[i][j] > toRadian(MAX_JOINTS[j])) {
                isOutLimit[i] = true;
                break;
            }
        }
    }

    /** 运动连续：排除掉离上一轴角度过大的解，需要设定最大允许偏差值 ALLOW_MAX_DIFF */
//    for (int i = 0; i < 8; ++i) {
//        for (int j = 0; j < 6; ++j) {
//            if (abs(ans[i][j] - last[j]) > ALLOW_MAX_DIFF) {
//                isContinuity[i] = false;
//                break;
//            }
//        }
//    }

    /** 最短路径原则：选出距离上一次姿态各轴所需最小的运动量，给各轴分配一个权重，使尽量动小轴或姿态轴 */
    double weight[6] = {0.2, 0.25, 0.2, 0.15, 0.12, 0.08};
    //double weight[6] = {1, 1, 1, 1, 1, 1};  // 各轴相同权重
    double sum[8] = {0, 0, 0, 0, 0, 0, 0, 0};
    double minSum = INT_MAX;
    int opt = -1;
    for (int i = 0; i < 8; ++i) {
        if (isOutLimit[i]) {
            //printf("第%d解超限\n", i + 1);
            continue;
        }
        for (int j = 0; j < 6; ++j) {
            sum[i] += fabs(ans[i][j] - last[j]) * weight[j];
        }
        if (sum[i] < minSum) {
            minSum = sum[i];
            opt = i;
        }
    }

    /** 最优解 */
    Joint res;
    if (opt >= 0) {
        /// 因为4轴和6轴平行,因此4轴转-x°和6轴转x°效果相同,因此如果遇到这种情况就将他们矫正
        if (-ans[opt][3] - ans[opt][5] <= 1e-2) {
            ans[opt][3] = 0;
            ans[opt][5] = 0;
        }
        res = Joint(ans[opt][0], ans[opt][1], ans[opt][2],
                    ans[opt][3], ans[opt][4], ans[opt][5]);
    } else {
        printf("无合适逆解\n");
        res =  lastJoints;
    }
    return res;
}
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