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<title>Motion and Force Requirements for the Nano-Hexapod</title>
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<h1 class="title">Motion and Force Requirements for the Nano-Hexapod</h1>
<div id="table-of-contents">
<h2>Table of Contents</h2>
<div id="text-table-of-contents">
<ul>
<li><a href="#org70e526c">1. Soft Hexapod</a>
<ul>
<li><a href="#org3326d69">1.1. Example</a></li>
</ul>
</li>
</ul>
</div>
</div>
<div id="outline-container-org70e526c" class="outline-2">
<h2 id="org70e526c"><span class="section-number-2">1</span> Soft Hexapod</h2>
<div class="outline-text-2" id="text-1">
<p>
As the nano-hexapod is in series with the other stages, it must apply all the force required to move the sample.
</p>
<p>
If the nano-hexapod is soft (voice coil), its actuator must apply all the force such that the sample has the wanted motion.
</p>
<p>
In some sense, it does not use the fact that the other stage are participating to the displacement of the sample.
</p>
<p>
Let&rsquo;s take two examples:
</p>
<ul class="org-ul">
<li>Sinus Ty translation at 1Hz with an amplitude of 5mm</li>
<li>Long stroke hexapod has an offset of 10mm in X and the spindle is rotating
Thus the wanted motion is a circle with a radius of 10mm
If the sample if light (30Kg) =&gt; 60rpm
If the sample if heavy (100Kg) =&gt; 1rpm</li>
</ul>
<p>
From the motion, we compute the required acceleration by derive the displacement two times.
Then from the Newton&rsquo;s second law: \(m \vec{a} = \sum \vec{F}\) we can compute the required force.
</p>
</div>
<div id="outline-container-org3326d69" class="outline-3">
<h3 id="org3326d69"><span class="section-number-3">1.1</span> Example</h3>
<div class="outline-text-3" id="text-1-1">
<p>
The wanted motion is:
</p>
\begin{align*}
x &= d \cos(\omega t) \\
y &= d \sin(\omega t)
\end{align*}
<p>
The corresponding acceleration is thus:
</p>
\begin{align*}
\ddot{x} &= - d \omega^2 \cos(\omega t) \\
\ddot{y} &= - d \omega^2 \sin(\omega t)
\end{align*}
<p>
From the Newton&rsquo;s second law:
</p>
\begin{align*}
m \ddot{x} &= F_x \\
m \ddot{y} &= F_y
\end{align*}
<p>
Thus the applied forces should be:
</p>
\begin{align*}
F_x &= - m d \omega^2 \cos(\omega t) \\
F_y &= - m d \omega^2 \sin(\omega t)
\end{align*}
<p>
And the norm of the force is:
\[ |F| = \sqrt{F_x^2 + F_y^2} = m d \omega^2 \ [N] \]
</p>
<p>
For a Light sample:
</p>
<div class="org-src-container">
<pre class="src src-matlab">m = 30;
d = 10e<span class="org-type">-</span>3;
w = 2<span class="org-type">*</span><span class="org-constant">pi</span>;
F = m<span class="org-type">*</span>d<span class="org-type">*</span>w<span class="org-type">^</span>2;
<span class="org-constant">ans</span> = F
</pre>
</div>
<pre class="example">
11.844
</pre>
<p>
For the Heavy sample:
</p>
<div class="org-src-container">
<pre class="src src-matlab">m = 80;
d = 10e<span class="org-type">-</span>3;
w = 2<span class="org-type">*</span><span class="org-constant">pi</span><span class="org-type">/</span>60;
F = m<span class="org-type">*</span>d<span class="org-type">*</span>w<span class="org-type">^</span>2
<span class="org-constant">ans</span> = F
</pre>
</div>
<pre class="example">
0.008773
</pre>
</div>
</div>
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<div id="postamble" class="status">
<p class="author">Author: Dehaeze Thomas</p>
<p class="date">Created: 2020-04-17 ven. 09:35</p>
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