机械臂运动学逆解(牛顿法)

发布时间:2024年01月06日

机械臂运动学逆解(牛顿法)

??常用的工业6轴机械臂采用6轴串联结构,虽然其运动学正解比较容易,但是其运动学逆解非常复杂,其逆解的方程组高度非线性,且难以化简。
??由于计算机技术的发展,依靠其强大的算力,可以通过数值解的方式对机械臂的运动学逆解方程组进行求解。以下将使用牛顿法详解整个求解过程。

算法的过程

机械臂运动学正解方程组如式1所示。

[ f 11 f 12 f 13 f 14 f 21 f 22 f 23 f 24 f 31 f 32 f 33 f 34 0 0 0 1 ] = f ( θ 1 , θ 2 , θ 3 , θ 4 , θ 5 , θ 6 ) (1) \begin{bmatrix} f_{11}&f_{12}&f_{13} & f_{14}\\ f_{21}&f_{22}&f_{23} & f_{24}\\ f_{31}&f_{32}&f_{33} & f_{34}\\ 0&0&0 & 1\\ \end{bmatrix} =f(\theta_1, \theta_2, \theta_3, \theta_4, \theta_5, \theta_6) \tag1 ?f11?f21?f31?0?f12?f22?f32?0?f13?f23?f33?0?f14?f24?f34?1? ?=f(θ1?,θ2?,θ3?,θ4?,θ5?,θ6?)(1)
对于运动学正解,式1右边是已知量,对于运动学逆解,式1左边式已知量。采用牛顿法求解运动学逆解,已知机械臂末端姿态为 [ r 11 r 12 r 13 r 14 r 21 r 22 r 23 r 24 r 31 r 32 r 33 r 34 0 0 0 1 ] \begin{bmatrix} r_{11}&r_{12}&r_{13} & r_{14}\\ r_{21}&r_{22}&r_{23} & r_{24}\\ r_{31}&r_{32}&r_{33} & r_{34}\\ 0&0&0 & 1\\ \end{bmatrix} ?r11?r21?r31?0?r12?r22?r32?0?r13?r23?r33?0?r14?r24?r34?1? ?,构造目标函数,如式2所示。
F ( θ 1 , θ 2 , θ 3 , θ 4 , θ 5 , θ 6 ) = ( f 14 ? r 14 ) 2 + ( f 24 ? r 24 ) 2 + ( f 34 ? r 34 ) 2 + 0.08 ? ( f 11 ? r 11 ) 2 + 0.08 ? ( f 12 ? r 12 ) 2 + 0.08 ? ( f 13 ? r 13 ) 2 + 0.08 ? ( f 31 ? r 31 ) 2 + 0.08 ? ( f 32 ? r 32 ) 2 + 0.08 ? ( f 34 ? r 34 ) 2 (2) F(\theta_1, \theta_2, \theta_3, \theta_4, \theta_5, \theta_6) =(f_{14}-r_{14})^2+(f_{24}-r_{24})^2+(f_{34}-r_{34})^2+0.08*(f_{11}-r_{11})^2+0.08*(f_{12}-r_{12})^2+0.08*(f_{13}-r_{13})^2+0.08*(f_{31}-r_{31})^2+0.08*(f_{32}-r_{32})^2+0.08*(f_{34}-r_{34})^2 \tag2 F(θ1?,θ2?,θ3?,θ4?,θ5?,θ6?)=(f14??r14?)2+(f24??r24?)2+(f34??r34?)2+0.08?(f11??r11?)2+0.08?(f12??r12?)2+0.08?(f13??r13?)2+0.08?(f31??r31?)2+0.08?(f32??r32?)2+0.08?(f34??r34?)2(2)

目标函数的雅可比矩阵为 J = [ ? F ? θ 1 , ? F ? θ 2 , ? F ? θ 3 , ? F ? θ 4 , ? F ? θ 5 , ? F ? θ 6 ] J= [\frac{\partial F}{\partial\theta_1}, \frac{\partial F}{\partial\theta_2},\frac{\partial F}{\partial\theta_3},\frac{\partial F}{\partial\theta_4},\frac{\partial F}{\partial\theta_5},\frac{\partial F}{\partial\theta_6}] J=[?θ1??F?,?θ2??F?,?θ3??F?,?θ4??F?,?θ5??F?,?θ6??F?]
目标函数的雅克比矩阵为 H = [ ? 2 F ? θ 1 2 ? 2 F ? θ 1 ? θ 2 ? 2 F ? θ 1 ? θ 3 ? 2 F ? θ 1 ? θ 4 ? 2 F ? θ 1 ? θ 5 ? 2 F ? θ 1 ? θ 6 ? 2 F ? θ 2 ? θ 1 ? 2 F ? θ 2 2 ? 2 F ? θ 2 ? θ 3 ? 2 F ? θ 2 ? θ 4 ? 2 F ? θ 2 ? θ 5 ? 2 F ? θ 2 ? θ 6 ? 2 F ? θ 3 ? θ 1 ? 2 F ? θ 3 ? θ 2 ? 2 F ? θ 3 2 ? 2 F ? θ 3 ? θ 4 ? 2 F ? θ 3 ? θ 5 ? 2 F ? θ 3 ? θ 6 ? 2 F ? θ 4 ? θ 1 ? 2 F ? θ 4 ? θ 2 ? 2 F ? θ 4 ? θ 3 ? 2 F ? θ 4 2 ? 2 F ? θ 4 ? θ 5 ? 2 F ? θ 4 ? θ 6 ? 2 F ? θ 5 ? θ 1 ? 2 F ? θ 5 ? θ 2 ? 2 F ? θ 5 ? θ 3 ? 2 F ? θ 5 ? θ 4 ? 2 F ? θ 5 2 ? 2 F ? θ 5 ? θ 6 ? 2 F ? θ 6 ? θ 1 ? 2 F ? θ 6 ? θ 2 ? 2 F ? θ 6 ? θ 3 ? 2 F ? θ 6 ? θ 4 ? 2 F ? θ 6 ? θ 5 ? 2 F ? θ 6 2 ] H=\begin{bmatrix} \frac{\partial^2 F}{\partial\theta_1^2} & \frac{\partial^2 F}{\partial\theta_1 \partial\theta_2} & \frac{\partial^2 F}{\partial\theta_1 \partial\theta_3} & \frac{\partial^2 F}{\partial\theta_1 \partial\theta_4} & \frac{\partial^2 F}{\partial\theta_1 \partial\theta_5} & \frac{\partial^2 F}{\partial\theta_1 \partial\theta_6} \\ \frac{\partial^2 F}{\partial\theta_2 \partial\theta_1} & \frac{\partial^2 F}{\partial\theta_2^2} & \frac{\partial^2 F}{\partial\theta_2 \partial\theta_3} & \frac{\partial^2 F}{\partial\theta_2 \partial\theta_4} & \frac{\partial^2 F}{\partial\theta_2 \partial\theta_5} & \frac{\partial^2 F}{\partial\theta_2 \partial\theta_6} \\ \frac{\partial^2 F}{\partial\theta_3 \partial\theta_1} & \frac{\partial^2 F}{\partial\theta_3 \partial\theta_2} & \frac{\partial^2 F}{\partial\theta_3^2} & \frac{\partial^2 F}{\partial\theta_3 \partial\theta_4} & \frac{\partial^2 F}{\partial\theta_3 \partial\theta_5} & \frac{\partial^2 F}{\partial\theta_3 \partial\theta_6} \\ \frac{\partial^2 F}{\partial\theta_4 \partial\theta_1} & \frac{\partial^2 F}{\partial\theta_4 \partial\theta_2} & \frac{\partial^2 F}{\partial\theta_4 \partial\theta_3} &\frac{\partial^2 F}{\partial\theta_4^2} & \frac{\partial^2 F}{\partial\theta_4 \partial\theta_5} & \frac{\partial^2 F}{\partial\theta_4 \partial\theta_6} \\ \frac{\partial^2 F}{\partial\theta_5 \partial\theta_1} & \frac{\partial^2 F}{\partial\theta_5 \partial\theta_2} & \frac{\partial^2 F}{\partial\theta_5 \partial\theta_3} & \frac{\partial^2 F}{\partial\theta_5 \partial\theta_4} & \frac{\partial^2 F}{\partial\theta_5^2} & \frac{\partial^2 F}{\partial\theta_5 \partial\theta_6} \\ \frac{\partial^2 F}{\partial\theta_6 \partial\theta_1} & \frac{\partial^2 F}{\partial\theta_6 \partial\theta_2} & \frac{\partial^2 F}{\partial\theta_6 \partial\theta_3} & \frac{\partial^2 F}{\partial\theta_6 \partial\theta_4} & \frac{\partial^2 F}{\partial\theta_6 \partial\theta_5} & \frac{\partial^2 F}{\partial\theta_6^2} \\ \end{bmatrix} H= ??θ12??2F??θ2??θ1??2F??θ3??θ1??2F??θ4??θ1??2F??θ5??θ1??2F??θ6??θ1??2F???θ1??θ2??2F??θ22??2F??θ3??θ2??2F??θ4??θ2??2F??θ5??θ2??2F??θ6??θ2??2F???θ1??θ3??2F??θ2??θ3??2F??θ32??2F??θ4??θ3??2F??θ5??θ3??2F??θ6??θ3??2F???θ1??θ4??2F??θ2??θ4??2F??θ3??θ4??2F??θ42??2F??θ5??θ4??2F??θ6??θ4??2F???θ1??θ5??2F??θ2??θ5??2F??θ3??θ5??2F??θ4??θ5??2F??θ52??2F??θ6??θ5??2F???θ1??θ6??2F??θ2??θ6??2F??θ3??θ6??2F??θ4??θ6??2F??θ5??θ6??2F??θ62??2F?? ?

迭代步长 Δ Θ = ? H ? J T \Delta \Theta = -H*J^T ΔΘ=?H?JT

程序验证

clear;
clc;

rng(1);    %固定随机数种子

%构造运动学模型
syms a0 a1 a2 a3 a4 a5;
FK = FKinematics(a0, a1, a2, a3, a4, a5);

%构造目标函数
syms T14 T24 T34 T11 T12 T13 T31 T32 T33;
opt_F(a0, a1, a2, a3, a4, a5, T14, T24, T34, T11, T12, T13, T31, T32, T33) = (FK(1, 4) - T14)^2 + ...
       (FK(2, 4) - T24)^2 + ...
       (FK(3, 4) - T34)^2 + ...
       0.08 * (FK(1, 1) - T11)^2 + ...
       0.08 * (FK(1, 2) - T12)^2 + ...
       0.08 * (FK(1, 3) - T13)^2 + ...
       0.08 * (FK(3, 1) - T31)^2 + ...
       0.08 * (FK(3, 2) - T32)^2 + ...
       0.08 * (FK(3, 3) - T33)^2
opt_F = matlabFunction(opt_F);

%构造目标函数的雅可比函数矩阵
J(a0, a1, a2, a3, a4, a5, T14, T24, T34, T11, T12, T13, T31, T32, T33) = jacobian(opt_F, [a0 a1 a2 a3 a4 a5])
J = matlabFunction(J);

%构造目标函数的海塞矩阵
H(a0, a1, a2, a3, a4, a5, T14, T24, T34, T11, T12, T13, T31, T32, T33) = jacobian(J, [a0 a1 a2 a3 a4 a5])
H = matlabFunction(H);

T = FKinematics(2, 0.5, -1.6, 0.6, 1.5, -0.9)
X = IKinematics(opt_F, J, H, T);
X
T
FKinematics(X(1), X(2), X(3), X(4), X(5), X(6))


function T = FKinematics(x1, x2, x3, x4, x5, x6)
    
    T1 = urdfJoint(0, 0, 0.3015, 0, 0, 0, x1);
    T2 = urdfJoint(0.077746, -0.0869967, 0.1465, 1.5708, 1.5708, 0, x2);
    T3 = urdfJoint(-0.64, 0, -0.015, 0, 0, 0, x3);
    T4 = urdfJoint(-0.195, 0.9055, -0.072, -1.5708, 0, 0, x4);
    T5 = urdfJoint(0, 0, 0, -1.6876, -1.5708, -3.0248, x5);
    T6 = urdfJoint(0, 0, 0, -1.5708, 0, -1.5708, x6);
    T7 = urdfJoint(0, 0, 0.08, 0, 0, 0, 0);    %法兰盘的位姿
      
    T = T1 * T2 * T3 * T4 * T5 * T6 * T7;
end

function X = IKinematics(opt_F, J, H, T)
    X = [0; 0; 0; 0; 0; 0];
    X0 = [0; 0; 0; 0; 0; 0];
    min_opt_value = opt_F(X0(1), X0(2), X0(3), X0(4), X0(5), X0(6), T(1, 4), T(2, 4), T(3, 4), T(1, 1), T(1, 2), T(1, 3), T(3, 1), T(3, 2), T(3, 3));
    X_opt = X0;
    last_opt_value = min_opt_value;
    t0 = clock;
    for i = 1 : 1000
        i
        Jn = J(X0(1), X0(2), X0(3), X0(4), X0(5), X0(6), T(1, 4), T(2, 4), T(3, 4), T(1, 1), T(1, 2), T(1, 3), T(3, 1), T(3, 2), T(3, 3));
        Hn = H(X0(1), X0(2), X0(3), X0(4), X0(5), X0(6), T(1, 4), T(2, 4), T(3, 4), T(1, 1), T(1, 2), T(1, 3), T(3, 1), T(3, 2), T(3, 3));
        %[U, S, V] = svd(Hn);
        %det_X = V * inv(S) * U' * Jn';
        det_X = inv(Hn) * Jn';
        X0 = X0 - det_X;
        X0';
        opt_value = opt_F(X0(1), X0(2), X0(3), X0(4), X0(5), X0(6), T(1, 4), T(2, 4), T(3, 4), T(1, 1), T(1, 2), T(1, 3), T(3, 1), T(3, 2), T(3, 3));
        if(min_opt_value > opt_value) 
            min_opt_value = opt_value;
            X_opt = X0;
        end
        if(min_opt_value < 0.0001)
            break;
        end
        if(abs(last_opt_value - opt_value) < 0.00001)
            fprintf('陷入局部最小解,将重新生成迭代初始值');
            X0 = randn(6, 1);
        end
        last_opt_value = opt_value;
    end
    t = etime(clock,t0);
    fprintf('solve time: %f', t);
    T;
    X = X_opt';
    T1 = FKinematics(X_opt(1), X_opt(2), X_opt(3), X_opt(4), X_opt(5), X_opt(6));
end

function T = urdfJoint(x0, y0, z0, R0, P0, Y0, theta)
    r1 = [1       0        0;
          0 cos(R0) -sin(R0);
          0 sin(R0)  cos(R0)];
    r2 = [ cos(P0) 0 sin(P0);
                 0 1       0;
          -sin(P0) 0 cos(P0)];
    r3 = [cos(Y0) -sin(Y0) 0;
          sin(Y0)  cos(Y0) 0;
                0        0 1];
    r = r3 * r2 * r1;
    T0 = [r(1, 1) r(1, 2) r(1, 3) x0;
          r(2, 1) r(2, 2) r(2, 3) y0;
          r(3, 1) r(3, 2) r(3, 3) z0;
                0       0       0  1];
    T = T0 * [cos(theta) -sin(theta) 0 0;
              sin(theta)  cos(theta) 0 0;
                       0           0 1 0;
                       0           0 0 1];
end

注意事项

  1. matlab在构造雅可比函数、函数矩阵的时候比较慢;
  2. 使用四元数建立运动学模型,效率更低(暂时未发现什么原因);
  3. 可通过设置迭代的初始值,获得其它的逆解;
文章来源:https://blog.csdn.net/weixin_43879302/article/details/135395970
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