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I'm trying to calculate the distance moved by an object rotating on a spindle, if that spindle is also rotating around a larger spindle (similar to earth moving around the sun). Hopefully you can see the image attached (but info summarized below anyway) It's related to motion of a polishing machine - see youtube link below at time 1:20 so you get the idea. i.e. what is the total distance a part would move given the following info. Also would like to calculate the angular and linear velocity

  • time = 10 minutes
  • larger circle radius = 50 mm
  • larger circle RPM = 40 RPM
  • smaller circle radius = 20 mm
  • smaller circle RPM = 30 RPM

https://www.youtube.com/watch?v=HU7CqFej17I&t=1m24s (see 1:24)

schematic

student91
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Paddy Mc Veigh
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1 Answers1

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You can describe the position of the particle as follows:

point on big circle:

$p_b = \Big(50\sin(2\pi\cdot40t),\ 50\cos(2\pi\cdot40t)\Big)$

point on small circle if it were centered on the origin:

$p_s = \Big(20\sin(2\pi\cdot20(t+t_0)),\ 20\cos(2\pi\cdot20(t+t_0))\Big)$

point on small circle with center on big circle:

$p_p = p_b+p_s = \Big(50\sin(2\pi\cdot40t)+20\sin(2\pi\cdot20(t+t_0)),\ 50\cos(2\pi\cdot40t)+20\cos(2\pi\cdot20(t+t_0))\Big)$

We can write this as

$$p_x = 50\sin(2\pi\cdot40t)+20\sin(2\pi\cdot20(t+t_0))$$ $$p_y = 50\cos(2\pi\cdot40t)+20\cos(2\pi\cdot20(t+t_0))$$

(You can use your favorite plotting software to look at this path if you want)

Here:

  • $t$ is in minutes
  • $t_0$ is an offset from where you start on small circle ($t_0=0$ means starting all the way out)
  • The position of the point is described in mm.

You should then do this numerically.

Here's what WolframAlpha came up with: It gives an exact answer with value $96000 E(5/9)\approx126923.5167$ where $E$ is this funny function.

Trying different values for $t_0$ doesn't make a difference, the total distance is always 126923.5 mm.

student91
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