Add schematic to explain the dynamics
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@@ -159,6 +159,29 @@ The comparison of the two transfer functions is shown in Figure [[fig:comparison
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#+caption: Comparison of the transfer functions from $\bm{\mathcal{F}}$ to $\mathcal{\bm{X}}$ and from $\bm{\mathcal{F}}_{\text{ext}}$ to $\mathcal{\bm{X}}$ ([[./figs/comparison_Fext_F_fixed_base.png][png]], [[./figs/comparison_Fext_F_fixed_base.pdf][pdf]])
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[[file:figs/comparison_Fext_F_fixed_base.png]]
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This can be understood from figure [[fig:1dof_actuator_external_forces]] where $\mathcal{F}_{x}$ and $\mathcal{F}_{x,\text{ext}}$ have clearly the same effect on $\mathcal{X}_{x}$.
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#+begin_src latex :file 1dof_actuator_external_forces.pdf
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\begin{tikzpicture}
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\draw[ground] (-1, 0) -- (1, 0);
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\draw[spring] (-0.6, 0) -- (-0.6, 1.5) node[midway, left=0.1]{$k$};
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\draw[actuator] ( 0.6, 0) -- ( 0.6, 1.5) node[midway, left=0.1](F){$\mathcal{F}_{x}$};
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\draw[fill=white] (-1, 1.5) rectangle (1, 2) node[pos=0.5]{$m$};
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\draw[dashed] (1, 2) -- ++(0.5, 0);
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\draw[->] (1.5, 2) -- ++(0, 0.5) node[right]{$\mathcal{X}_{x}$};
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\draw[->] (0, 2) node[]{$\bullet$} -- (0, 2.8) node[below right]{$\mathcal{F}_{x,\text{ext}}$};
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\end{tikzpicture}
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#+end_src
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#+name: fig:1dof_actuator_external_forces
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#+caption: Schematic representation of the stewart platform on a rigid support
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#+RESULTS:
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[[file:figs/1dof_actuator_external_forces.png]]
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** Comparison with a flexible support
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We now add a flexible support under the Stewart platform.
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#+begin_src matlab
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@@ -229,6 +252,35 @@ The comparison between the obtained transfer functions is shown in Figure [[fig:
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#+caption: Comparison of the transfer functions from $\bm{\mathcal{F}}$ to $\mathcal{\bm{X}}$ and from $\bm{\mathcal{F}}_{\text{ext}}$ to $\mathcal{\bm{X}}$ ([[./figs/comparison_Fext_F_flexible_base.png][png]], [[./figs/comparison_Fext_F_flexible_base.pdf][pdf]])
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[[file:figs/comparison_Fext_F_flexible_base.png]]
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The addition of a flexible support can be schematically represented in Figure [[fig:2dof_actuator_external_forces]].
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We see that $\mathcal{F}_{x}$ applies a force both on $m$ and $m^{\prime}$ whereas $\mathcal{F}_{x,\text{ext}}$ only applies a force on $m$.
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And thus $\mathcal{F}_{x}$ and $\mathcal{F}_{x,\text{ext}}$ have clearly *not* the same effect on $\mathcal{X}_{x}$.
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#+begin_src latex :file 2dof_actuator_external_forces.pdf
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\begin{tikzpicture}
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\draw[ground] (-1, 0) -- (1, 0);
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\draw[spring] (0, 0) -- (0, 1.5) node[midway, left=0.1]{$k^{\prime}$};
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\draw[fill=white] (-1, 1.5) rectangle (1, 2) node[pos=0.5]{$m^{\prime}$};
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\draw[spring] (-0.6, 2) -- (-0.6, 3.5) node[midway, left=0.1]{$k$};
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\draw[actuator] ( 0.6, 2) -- ( 0.6, 3.5) node[midway, left=0.1](F){$\mathcal{F}_{x}$};
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\draw[fill=white] (-1, 3.5) rectangle (1, 4) node[pos=0.5]{$m$};
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\draw[dashed] (1, 4) -- ++(0.5, 0);
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\draw[->] (1.5, 4) -- ++(0, 0.5) node[right]{$\mathcal{X}_{x}$};
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\draw[->] (0, 4) node[]{$\bullet$} -- (0, 4.8) node[below right]{$\mathcal{F}_{x,\text{ext}}$};
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\end{tikzpicture}
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#+end_src
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#+name: fig:2dof_actuator_external_forces
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#+caption: Schematic representation of the stewart platform on top of a flexible support
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#+RESULTS:
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[[file:figs/2dof_actuator_external_forces.png]]
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** Conclusion
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#+begin_important
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The transfer function from forces/torques applied by the actuators on the payload $\bm{\mathcal{F}} = \bm{J}^T \bm{\tau}$ to the pose of the mobile platform $\bm{\mathcal{X}}$ is the same as the transfer function from external forces/torques to $\bm{\mathcal{X}}$ as long as the Stewart platform's base is fixed.
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