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Motion and Force Requirements for the Nano-Hexapod

Table of Contents

1 Soft Hexapod

As the nano-hexapod is in series with the other stages, it must apply all the force required to move the sample.

If the nano-hexapod is soft (voice coil), its actuator must apply all the force such that the sample has the wanted motion.

In some sense, it does not use the fact that the other stage are participating to the displacement of the sample.

Let’s take two examples:

  • Sinus Ty translation at 1Hz with an amplitude of 5mm
  • 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) => 60rpm If the sample if heavy (100Kg) => 1rpm

From the motion, we compute the required acceleration by derive the displacement two times. Then from the Newton’s second law: \(m \vec{a} = \sum \vec{F}\) we can compute the required force.

1.1 Example

The wanted motion is:

\begin{align*} x &= d \cos(\omega t) \\ y &= d \sin(\omega t) \end{align*}

The corresponding acceleration is thus:

\begin{align*} \ddot{x} &= - d \omega^2 \cos(\omega t) \\ \ddot{y} &= - d \omega^2 \sin(\omega t) \end{align*}

From the Newton’s second law:

\begin{align*} m \ddot{x} &= F_x \\ m \ddot{y} &= F_y \end{align*}

Thus the applied forces should be:

\begin{align*} F_x &= - m d \omega^2 \cos(\omega t) \\ F_y &= - m d \omega^2 \sin(\omega t) \end{align*}

And the norm of the force is: \[ |F| = \sqrt{F_x^2 + F_y^2} = m d \omega^2 \ [N] \]

For a Light sample:

m = 30;
d = 10e-3;
w = 2*pi;
F = m*d*w^2;
ans = F
11.844

For the Heavy sample:

m = 80;
d = 10e-3;
w = 2*pi/60;
F = m*d*w^2
ans = F
0.008773

Author: Dehaeze Thomas

Created: 2020-03-13 ven. 17:39