智能机器人技术——正向运动学

本文最后更新于:2022年5月16日 上午

智能机器人技术作业记录

智能机器人技术——正向运动学

  1. 如下图所示有一处于初始位形的 PRRRRR 空间开链机器人, 试确定末端初始位形 \(M\) 、在 \(\{0\}\) 系描述的螺旋轴 Si 、$ 在 ${b} 系描述的螺旋轴 Bi (如讲义那样列表即可)。

解得:

相对基坐标系的\(\mathrm{PoE}\) :

\[ T(\theta)=e^{\left[\mathcal{S}_{1}\right] \theta_{1}} \cdots e^{\left[\mathcal{S}_{n-1}\right] \theta_{n-1}} e^{\left[\mathcal{S}_{n}\right] \theta_{n}} M \]

相对末端坐标系的\(\mathrm{PoE}\) :

\[ T(\theta)=M e^{\left[\mathcal{B}_{1}\right] \theta_{1}} \cdots e^{\left[\mathcal{B}_{n-1}\right] \theta_{n-1}} e^{\left[\mathcal{B}_{n}\right] \theta_{n}} \]

末端初始位形 \(M\)

\[ \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & L_1+L_2+L_3+L_4 \\ 0 & 0 & 1 & h \\ 0 & 0 & 0 & 1 \end{array}\right] \]

\(\{0\}\) 系描述的螺旋轴 $_{i} $:

\[ [\mathcal{S}]=\left[\begin{array}{cc} {[\omega]} & v \\ 0 & 0 \end{array}\right] \in \operatorname{se}(3) \longrightarrow \mathcal{S}=\left[\begin{array}{c} \omega \\ v \end{array}\right] \in \mathbb{R}^{6} \]

\[ \because v_i=-\omega_{i} \times q_i \\ \begin{array}{|c||c|c|c|} \hline i & \omega_{i} & q_i & v_{i} \\ \hline \hline 1 & (0,0,0) & NULL & (0, 1,0) \\ \hline 2 & (0,0,1) & (0, L_1,0)& (L_1, 0,0) \\ \hline 3 & (-1,0,0) & (0, L_1,h)& (0, -h,L_1) \\ \hline 4 & (-1,0,0) & (0,L_1+L_2,h)& (0, -h,L_1+L_2) \\ \hline 5 & (-1,0,0) & (0,L_1+L_2+L_3,h)& (0, -h,L_1+L_2+L_3) \\ \hline 6 & (0,1,0) & (0,0,h) & (-h, 0,1)\\ \hline \end{array} \]

\(\{b\}\) 系描述的螺旋轴 \(\mathcal{B}_{i}\)

\[ [\mathcal{B}]=\left[\begin{array}{cc} {[\omega]} & v \\ 0 & 0 \end{array}\right] \in \operatorname{se}(3) \longrightarrow \mathcal{B}=\left[\begin{array}{c} \omega \\ v \end{array}\right] \in \mathbb{R}^{6} \]

\[ \because v_i=-\omega_{i} \times q_i \\ \begin{array}{|c||c|c|c|} \hline i & \omega_{i} & q_i & v_{i} \\ \hline \hline 1 & (0,0,0) & NULL & (0, 1,0) \\ \hline 2 & (0,0,1) & (0, -L_2-L_3-L_4,0)& (-L_2-L_3-L_4, 0,0) \\ \hline 3 & (-1,0,0) & (0, -L_2-L_3-L_4,0)& (0, 0,-L_2-L_3-L_4) \\ \hline 4 & (-1,0,0) & (0,-L_3-L_4,0)& (0, 0,-L_3-L_4) \\ \hline 5 & (-1,0,0) & (0,-L_4,0)& (0, 0,-L_4) \\ \hline 6 & (0,1,0) & (0,0,0) & (0, 0,0)\\ \hline \end{array} \]

  1. 如下图所示有一处于初始位形的 RRRRPR 空间开链机器人, 试确定末端初始位形 \(M\) 、在 \(\{0\}\) 系描述的螺旋轴 Si 、$ 在 ${b} 系描述的螺旋轴 Bi (如讲义那样列表即可)。

解得:

相对基坐标系的\(\mathrm{PoE}\) :

\[ T(\theta)=e^{\left[\mathcal{S}_{1}\right] \theta_{1}} \cdots e^{\left[\mathcal{S}_{n-1}\right] \theta_{n-1}} e^{\left[\mathcal{S}_{n}\right] \theta_{n}} M \]

相对末端坐标系的\(\mathrm{PoE}\) :

\[ T(\theta)=M e^{\left[\mathcal{B}_{1}\right] \theta_{1}} \cdots e^{\left[\mathcal{B}_{n-1}\right] \theta_{n-1}} e^{\left[\mathcal{B}_{n}\right] \theta_{n}} \]

末端初始位形 \(M\)

\[ \left[\begin{array}{cccc} 1 & 0 & 0 & L_1 \\ 0 & 1 & 0 & L_3+L_4 \\ 0 & 0 & 1 & -L_5-L_6 \\ 0 & 0 & 0 & 1 \end{array}\right] \]

\(\{0\}\) 系描述的螺旋轴 $_{i} $:

\[ [\mathcal{S}]=\left[\begin{array}{cc} {[\omega]} & v \\ 0 & 0 \end{array}\right] \in \operatorname{se}(3) \longrightarrow \mathcal{S}=\left[\begin{array}{c} \omega \\ v \end{array}\right] \in \mathbb{R}^{6} \]

\[ \because v_i=-\omega_{i} \times q_i \\ \begin{array}{|c||c|c|c|} \hline i & \omega_{i} & q_i & v_{i} \\ \hline \hline 1 & (1,0,0) & (0, 0,0)& (0, 0,0) \\ \hline 2 & (0,0,-1) & (L_1,0,0)& (0, L_1,0) \\ \hline 3 & (0,1,0) & (L_1,0,L_2)& (-L_2, 0,L_1) \\ \hline 4 & (1,0,0) & (0,L_3,0)& (0, 0,-L_3) \\ \hline 5 & (0,0,0) & NULL & (0, 1,0)\\ \hline 6 & (0,1,0) & (L_1,0,-L_5) & (L_5, 0,-L_1) \\ \hline \end{array} \]

\(\{b\}\) 系描述的螺旋轴 \(\mathcal{B}_{i}\)

\[ [\mathcal{B}]=\left[\begin{array}{cc} {[\omega]} & v \\ 0 & 0 \end{array}\right] \in \operatorname{se}(3) \longrightarrow \mathcal{B}=\left[\begin{array}{c} \omega \\ v \end{array}\right] \in \mathbb{R}^{6} \]

\[ \because v_i=-\omega_{i} \times q_i \\ \begin{array}{|c||c|c|c|} \hline i & \omega_{i} & q_i & v_{i} \\ \hline \hline 1 & (1,0,0) & (0, -L_3-L_4,L_5+L_6)& (0, L_5+L_6,L_3+L_4) \\ \hline 2 & (0,0,-1) & (0, -L_3-L_4,0)& (L_3+L_4, 0,0) \\ \hline 3 & (0,1,0) & (0,0,L_2+L_5+L_6)& (-L_2-L_5-L_6, 0,0) \\ \hline 4 & (1,0,0) & (0,-L_4,L_5+L_6)& (0, L_5+L_6,L_4) \\ \hline 5 & (0,0,0) & NULL & (0, 1,0)\\ \hline 6 & (0,1,0) & (0,0,L_6) & (-L_6, 0,0) \\ \hline \end{array} \]

  1. 如下图所示有一处于初始位形的 RRRPRR 空间开链机器人, 试确定末端初始位形 \(M\) 、在 \(\{0\}\) 系描述的螺旋轴 Si 、$ 在 ${b} 系描述的螺旋轴 Bi (如讲义那样列表即可)。

解得:

相对基坐标系的\(\mathrm{PoE}\) :

\[ T(\theta)=e^{\left[\mathcal{S}_{1}\right] \theta_{1}} \cdots e^{\left[\mathcal{S}_{n-1}\right] \theta_{n-1}} e^{\left[\mathcal{S}_{n}\right] \theta_{n}} M \]

相对末端坐标系的\(\mathrm{PoE}\) :

\[ T(\theta)=M e^{\left[\mathcal{B}_{1}\right] \theta_{1}} \cdots e^{\left[\mathcal{B}_{n-1}\right] \theta_{n-1}} e^{\left[\mathcal{B}_{n}\right] \theta_{n}} \]

末端初始位形 \(M\)

\[ \left[\begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & 1 \end{array}\right] \]

\(\{0\}\) 系描述的螺旋轴 $_{i} $:

\[ [\mathcal{S}]=\left[\begin{array}{cc} {[\omega]} & v \\ 0 & 0 \end{array}\right] \in \operatorname{se}(3) \longrightarrow \mathcal{S}=\left[\begin{array}{c} \omega \\ v \end{array}\right] \in \mathbb{R}^{6} \]

\[ \because v_i=-\omega_{i} \times q_i \\ \begin{array}{|c||c|c|c|} \hline i & \omega_{i} & q_i & v_{i} \\ \hline \hline 1 & (0,0,1) & (0, 0,0)& (0, 0,0) \\ \hline 2 & (1,0,0) & (0,0,2)& (0, 2,0) \\ \hline 3 & (1,0,0) & (0,1,2)& (0, 2,-1) \\ \hline 4 & (0,0,0) & NULL & (0, 1,0) \\ \hline 5 & (0, \frac{\sqrt{2}}{2} , \frac{\sqrt{2}}{2} ) & (0,3,2) & (\frac{\sqrt{2}}{2}, 0,0)\\ \hline 6 & (0,0,-1) & (0,4,0)& (-4, 0,0) \\ \hline \end{array} \]

\(\{b\}\) 系描述的螺旋轴 \(\mathcal{B}_{i}\)

\[ [\mathcal{B}]=\left[\begin{array}{cc} {[\omega]} & v \\ 0 & 0 \end{array}\right] \in \operatorname{se}(3) \longrightarrow \mathcal{B}=\left[\begin{array}{c} \omega \\ v \end{array}\right] \in \mathbb{R}^{6} \]

\[ \because v_i=-\omega_{i} \times q_i \\ \begin{array}{|c||c|c|c|} \hline i & \omega_{i} & q_i & v_{i} \\ \hline \hline 1 & (0,0,-1) & (0, -4,0)& (4, 0,0) \\ \hline 2 & (-1,0,0) & (0,-4,-1)& (0, 1,-4) \\ \hline 3 & (-1,0,0) & (0,-3,-1)& (0, 1,-3) \\ \hline 4 & (0,0,0) & NULL & (0, 1,0) \\ \hline 5 & (0, \frac{\sqrt{2}}{2} , -\frac{\sqrt{2}}{2} ) & (0,-1,-1) & (\sqrt{2}, 0,0)\\ \hline 6 & (0,0,1) & (0,0,0)& (0, 0,0) \\ \hline \end{array} \]


本博客所有文章除特别声明外,均采用 CC BY-SA 4.0 协议 ,转载请注明出处!