如图1所示的球形手腕(三个关节的轴线相交于一点)是常用的机械臂结构,我们希望在已知坐标系3至坐标系6的旋转矩阵的条件下求解3个关节值q3q3<script type="math/tex" id="MathJax-Element-658">q_{3}</script>,q4q4<script type="math/tex" id="MathJax-Element-659">q_{4}</script>,q6q6<script type="math/tex" id="MathJax-Element-660">q_{6}</script>。旋转矩阵可通过欧拉角集合、四元数或直接用3×3的矩阵给出。


这里写图片描述
图1 典型腕关节结构

        假设现在已知的旋转矩阵为:

R63=nxnynzsxsyszaxayazR36=[nxsxaxnysyaynzszaz]
<script type="math/tex; mode=display" id="MathJax-Element-661">R_{3}^{6}=\left[ \begin{matrix} {{n}_{x}} & {{s}_{x}} & {{a}_{x}} \\ {{n}_{y}} & {{s}_{y}} & {{a}_{y}} \\ {{n}_{z}} & {{s}_{z}} & {{a}_{z}} \\ \end{matrix} \right]</script>

        q3q3<script type="math/tex" id="MathJax-Element-662">q_{3}</script>,q4q4<script type="math/tex" id="MathJax-Element-663">q_{4}</script>,q6q6<script type="math/tex" id="MathJax-Element-664">q_{6}</script>的值为 :
        当q5(0π)q5∈(0,π)<script type="math/tex" id="MathJax-Element-665">{{q}_{5}}\in \left( 0,\pi \right)</script>

q4=atan2(ay,ax)q5=atan2(ax2+ay2,az)q6=atan2(sz,nz)(205)(206)(207)(205)q4=atan⁡2(ay,ax)(206)q5=atan⁡2(ax2+ay2,az)(207)q6=atan⁡2(sz,−nz)
<script type="math/tex; mode=display" id="MathJax-Element-666">\begin{align} & {{q}_{4}}=\operatorname{atan}2\left( {{a}_{y}},{{a}_{x}} \right) \\ & {{q}_{5}}=a\tan 2\left( \sqrt{{{a}_{x}}^{2}+{{a}_{y}}^{2}},{{a}_{z}} \right) \\ & {{q}_{6}}=a\tan 2\left( {{s}_{z}},-{{n}_{z}} \right) \\ \end{align}</script>
        当q5(π0)q5∈(−π,0)<script type="math/tex" id="MathJax-Element-667">{{q}_{5}}\in \left( -\pi ,0 \right)</script>
q4=atan2(ay,ax)q5=atan2(ax2+ay2,az)q6=atan2(sz,nz)(208)(209)(210)(208)q4=atan⁡2(−ay,−ax)(209)q5=atan⁡2(−ax2+ay2,az)(210)q6=atan⁡2(−sz,nz)
<script type="math/tex; mode=display" id="MathJax-Element-668">\begin{align} & {{q}_{4}}=\operatorname{atan}2\left( -{{a}_{y}},-{{a}_{x}} \right) \\ & {{q}_{5}}=a\tan 2\left( -\sqrt{{{a}_{x}}^{2}+{{a}_{y}}^{2}},{{a}_{z}} \right) \\ & {{q}_{6}}=a\tan 2\left( -{{s}_{z}},{{n}_{z}} \right) \\ \end{align}</script>

参考文献:
布鲁诺・西西里安诺.《机器人学:建模,规划和控制》 西安交通大学出版社 2015

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