机械手末端速度计算（实例）

机械手末端速度计算(实例)

上一篇博文已经推导了相邻连杆i和连杆i+1间速度的传递

• 连杆i+1为旋转关节时有

$${}^{i+1}{w}_{i+1}{=}_i^{i+1}R{}^i{w}_i+\dot{\theta }i+1{}^{i+1}\hat{Z}i+1\text{\hspace{1em}(5-45)}$$

$${}^{i+1}{v}_{i+1}{=}_i^{i+1}R{(}^i{v}_i{+}^i{w}_i{\times }^i{P}_{i+1})\text{\hspace{1em}(5-47)}$$

• 连杆i+1为移动关节时有

$$\begin{gathered} {}^{i+1}{w}_{i+1}{=}_i^{i+1}R{}^i{w}_i{ \\ }^{i+1}{v}_{i+1}{=}_i^{i+1}R{(}^i{v}_i{+}^i{w}_i{\times }^i{P}_{i+1})+{\dot{d}}_{i+1}{}^{i+1}{\hat{Z}}_{i+1}\text{\hspace{1em}(5-48)} \end{gathered}$$

例子

如下图所示，计算出操作臂末端的速度，将它表达成关节速度的函数。给出两种形式的解答，一种是用坐标系{3}表示，另一种是用坐标系{0}表示。

连杆间的旋转变换矩阵为
$$\begin{gathered} {}_1^{0}R=\left[\begin{array}{ccc}\cos {\theta }_1& -\sin {\theta }_1& 0\\ \sin {\theta }_1& \cos \theta 1& 0\\ 0& 0& 1\end{array}\right]{ \\ }_2^{1}R=\left[\begin{array}{ccc}\cos {\theta }_2& -\sin {\theta }_2& 0\\ \sin {\theta }_2& \cos \theta 2& 0\\ 0& 0& 1\end{array}\right]{ \\ }_3^{2}R=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \end{gathered}$$

对连杆依次使用上一篇博文中的式（5-45）和（5-47）,就有
$${}^1{w}_1=\left[\begin{array}{c}0\\ 0\\ {\dot{\theta }}_1\end{array}\right]\text{\hspace{1em}(5-50)}$$

$${}^1{v}_1=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\text{\hspace{1em}(5-51)}$$

$${}^2{w}_2{=}_1^{2}R{}^1{w}_1+{\dot{\theta }}_2{}^2{\hat{Z}}_2=\left[\begin{array}{c}0\\ 0\\ {\dot{\theta }}_1+{\dot{\theta }}_2\end{array}\right]\text{\hspace{1em}(5-52)}$$

$$\begin{gathered} {}^2{v}_2{=}_1^{2}R{(}^1{v}_1{+}^1{w}_1{\times }^1{P}_2)=\left[\begin{array}{ccc}{c}_2& s_2& 0\\ -s_2& c_2& 0\\ 0& 0& 1\end{array}\right]\left(\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ {\dot{\theta }}_1\end{array}\right]\times \left[\begin{array}{c}0\\ l_1\\ 0\end{array}\right]\right) \\ =\left[\begin{array}{c}{l}_1{s}_2{\dot{\theta }}_1\\ l_1{c}_2{\theta }_1\\ 0\end{array}\right]\text{\hspace{1em}(5-53)} \end{gathered}$$

$${}^3{w}_3{=}_2^3{R}^2{w}_2+{\dot{\theta }}_2{}^2{\hat{Z}}_2{=}^2{w}_2\text{\hspace{1em}(5-54)}$$

$${}^3{v}_3{=}_2^{3}R{(}^2{v}_2{+}^2{w}_2{\times }^2{P}_3)=\left[\begin{array}{c}{l}_1{s}_2{\dot{\theta }}_1\\ l_1{c}_2{\dot{\theta }}_1+l_2({\dot{\theta }}_1+{\dot{\theta }}_2)\\ 0\end{array}\right]\text{\hspace{1em}(5-55)}$$

求速度相对于固定极坐标系的变换
$${}_3^{0}R{=}_1^{0}R{}_2^{1}R{}_3^{2}R=\left[\begin{array}{ccc}{c}_{12}& -s_{12}& 0\\ s_{12}& c_{12}& 0\\ 0& 0& 1\end{array}\right]\text{\hspace{1em}(5-56)}$$
通过这个变换得到
$${}^0{v}_3=\left[\begin{array}{c}-l_1{s}_1{\dot{\theta }}_1-l_2{s}_{12}({\dot{\theta }}_1+{\dot{\theta }}_2)\\ l_1{c}_1{\dot{\theta }}_1+l_2{c}_{12}({\dot{\theta }}_1+{\dot{\theta }}_2)\\ 0\end{array}\right]\text{\hspace{1em}(5-57)}$$

参考文献

[1] JOHN J.CRAIG. 机器人学导论: 第3版[M]. 机械工业出版社, 2006.