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2021-02-20 23:09:27 +01:00
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2020-03-13 17:40:22 +01:00
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<title>Motion and Force Requirements for the Nano-Hexapod</title>
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2021-02-20 23:09:27 +01:00
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2020-03-13 17:40:22 +01:00
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2020-04-17 10:25:44 +02:00
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<a accesskey="h" href="./index.html"> UP </a>
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2020-03-13 17:40:22 +01:00
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2021-02-20 23:09:27 +01:00
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<a accesskey="H" href="../../index.html"> HOME </a>
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2020-03-13 17:40:22 +01:00
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</div><div id="content">
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<h1 class="title">Motion and Force Requirements for the Nano-Hexapod</h1>
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<div id="table-of-contents">
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<h2>Table of Contents</h2>
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<div id="text-table-of-contents">
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<ul>
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<li><a href="#orgecd6b81">1. Soft Hexapod</a>
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<ul>
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<li><a href="#org292e705">1.1. Example</a></li>
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</ul>
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</li>
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</ul>
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</div>
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<div id="outline-container-orgecd6b81" class="outline-2">
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<h2 id="orgecd6b81"><span class="section-number-2">1</span> Soft Hexapod</h2>
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<div class="outline-text-2" id="text-1">
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<p>
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As the nano-hexapod is in series with the other stages, it must apply all the force required to move the sample.
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</p>
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<p>
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If the nano-hexapod is soft (voice coil), its actuator must apply all the force such that the sample has the wanted motion.
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</p>
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<p>
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In some sense, it does not use the fact that the other stage are participating to the displacement of the sample.
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</p>
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<p>
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Let’s take two examples:
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</p>
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<ul class="org-ul">
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<li>Sinus Ty translation at 1Hz with an amplitude of 5mm</li>
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<li>Long stroke hexapod has an offset of 10mm in X and the spindle is rotating
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Thus the wanted motion is a circle with a radius of 10mm
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If the sample if light (30Kg) => 60rpm
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If the sample if heavy (100Kg) => 1rpm</li>
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</ul>
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<p>
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From the motion, we compute the required acceleration by derive the displacement two times.
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Then from the Newton’s second law: \(m \vec{a} = \sum \vec{F}\) we can compute the required force.
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</p>
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</div>
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<div id="outline-container-org292e705" class="outline-3">
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<h3 id="org292e705"><span class="section-number-3">1.1</span> Example</h3>
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<div class="outline-text-3" id="text-1-1">
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<p>
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The wanted motion is:
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</p>
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\begin{align*}
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x &= d \cos(\omega t) \\
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y &= d \sin(\omega t)
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\end{align*}
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<p>
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The corresponding acceleration is thus:
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</p>
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\begin{align*}
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\ddot{x} &= - d \omega^2 \cos(\omega t) \\
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\ddot{y} &= - d \omega^2 \sin(\omega t)
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\end{align*}
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<p>
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From the Newton’s second law:
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</p>
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\begin{align*}
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m \ddot{x} &= F_x \\
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m \ddot{y} &= F_y
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\end{align*}
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<p>
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Thus the applied forces should be:
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</p>
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\begin{align*}
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F_x &= - m d \omega^2 \cos(\omega t) \\
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F_y &= - m d \omega^2 \sin(\omega t)
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\end{align*}
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<p>
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And the norm of the force is:
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\[ |F| = \sqrt{F_x^2 + F_y^2} = m d \omega^2 \ [N] \]
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</p>
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<p>
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For a Light sample:
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</p>
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<div class="org-src-container">
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<pre class="src src-matlab"> m = 30;
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d = 10e<span class="org-type">-</span>3;
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w = 2<span class="org-type">*</span><span class="org-constant">pi</span>;
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F = m<span class="org-type">*</span>d<span class="org-type">*</span>w<span class="org-type">^</span>2;
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<span class="org-constant">ans</span> = F
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</pre>
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</div>
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<pre class="example">
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11.844
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</pre>
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<p>
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For the Heavy sample:
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</p>
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<div class="org-src-container">
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<pre class="src src-matlab"> m = 80;
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d = 10e<span class="org-type">-</span>3;
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w = 2<span class="org-type">*</span><span class="org-constant">pi</span><span class="org-type">/</span>60;
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F = m<span class="org-type">*</span>d<span class="org-type">*</span>w<span class="org-type">^</span>2
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<span class="org-constant">ans</span> = F
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</pre>
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</div>
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<pre class="example">
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0.008773
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</pre>
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</div>
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</div>
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</div>
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</div>
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<div id="postamble" class="status">
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<p class="author">Author: Dehaeze Thomas</p>
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2021-02-20 23:09:27 +01:00
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<p class="date">Created: 2021-02-20 sam. 23:08</p>
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2020-03-13 17:40:22 +01:00
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