88 lines
3.2 KiB
Org Mode
88 lines
3.2 KiB
Org Mode
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#+TITLE: Stewart Platform - Static Analysis
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:DRAWER:
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#+HTML_LINK_HOME: ./index.html
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#+HTML_LINK_UP: ./index.html
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#+HTML_HEAD: <link rel="stylesheet" type="text/css" href="./css/htmlize.css"/>
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#+HTML_HEAD: <link rel="stylesheet" type="text/css" href="./css/readtheorg.css"/>
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#+HTML_HEAD: <script src="./js/jquery.min.js"></script>
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#+HTML_HEAD: <script src="./js/bootstrap.min.js"></script>
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#+HTML_HEAD: <script src="./js/jquery.stickytableheaders.min.js"></script>
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#+HTML_HEAD: <script src="./js/readtheorg.js"></script>
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#+PROPERTY: header-args:matlab :session *MATLAB*
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#+PROPERTY: header-args:matlab+ :comments org
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#+PROPERTY: header-args:matlab+ :exports both
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#+PROPERTY: header-args:matlab+ :results none
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#+PROPERTY: header-args:matlab+ :eval no-export
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#+PROPERTY: header-args:matlab+ :noweb yes
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#+PROPERTY: header-args:matlab+ :mkdirp yes
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#+PROPERTY: header-args:matlab+ :output-dir figs
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:END:
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* Jacobian
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** Relation to platform parameters
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A Jacobian is defined by:
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- the orientations of the struts $\hat{s}_i$ expressed in a frame $\{A\}$ linked to the fixed platform.
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- the vectors from $O_B$ to $b_i$ expressed in the frame $\{A\}$
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Then, the choice of $O_B$ changes the Jacobian.
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** Jacobian for displacement
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\[ \dot{q} = J \dot{X} \]
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With:
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- $q = [q_1\ q_2\ q_3\ q_4\ q_5\ q_6]$ vector of linear displacement of actuated joints
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- $X = [x\ y\ z\ \theta_x\ \theta_y\ \theta_z]$ position and orientation of $O_B$ expressed in the frame $\{A\}$
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For very small displacements $\delta q$ and $\delta X$, we have $\delta q = J \delta X$.
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** Jacobian for forces
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\[ F = J^T \tau \]
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With:
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- $\tau = [\tau_1\ \tau_2\ \tau_3\ \tau_4\ \tau_5\ \tau_6]$ vector of actuator forces
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- $F = [f_x\ f_y\ f_z\ n_x\ n_y\ n_z]$ force and torque acting on point $O_B$
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* Stiffness matrix $K$
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\[ K = J^T \text{diag}(k_i) J \]
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If all the struts have the same stiffness $k$, then $K = k J^T J$
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$K$ only depends of the geometry of the stewart platform: it depends on the Jacobian, that is on the orientations of the struts, position of the joints and choice of frame $\{B\}$.
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\[ F = K X \]
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With $F$ forces and torques applied to the moving platform at the origin of $\{B\}$ and $X$ the translations and rotations of $\{B\}$ with respect to $\{A\}$.
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\[ C = K^{-1} \]
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The compliance element $C_{ij}$ is then the stiffness
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\[ X_i = C_{ij} F_j \]
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* Coupling
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What causes the coupling from $F_i$ to $X_i$ ?
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#+begin_src latex :file coupling.pdf :post pdf2svg(file=*this*, ext="png") :exports both
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\begin{tikzpicture}
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\node[block] (Jt) at (0, 0) {$J^{-T}$};
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\node[block, right= of Jt] (G) {$G$};
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\node[block, right= of G] (J) {$J^{-1}$};
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\draw[->] ($(Jt.west)+(-0.8, 0)$) -- (Jt.west) node[above left]{$F_i$};
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\draw[->] (Jt.east) -- (G.west) node[above left]{$\tau_i$};
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\draw[->] (G.east) -- (J.west) node[above left]{$q_i$};
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\draw[->] (J.east) -- ++(0.8, 0) node[above left]{$X_i$};
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\end{tikzpicture}
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#+end_src
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#+name: fig:block_diag_coupling
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#+caption: Block diagram to control an hexapod
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#+RESULTS:
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[[file:figs/coupling.png]]
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There is no coupling from $F_i$ to $X_j$ if $J^{-1} G J^{-T}$ is diagonal.
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If $G$ is diagonal (cubic configuration), then $J^{-1} G J^{-T} = G J^{-1} J^{-T} = G (J^{T} J)^{-1} = G K^{-1}$
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Thus, the system is uncoupled if $G$ and $K$ are diagonal.
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