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Using a potentiometer to detect position

  1. Apr 23, 2010 #1
    I am reading up on this since it was suggested that I do this for a project. I am supposed to track the position of a steel ball on a beam as the beam is tilted up and down. Now unless I am mistaken, the potentiometer has to physically touch the entity whose position is to sense right?

    Or am I mistaken there? Because if that is the case, I don't see the pot as a good option for this application.
     
  2. jcsd
  3. Apr 23, 2010 #2

    minger

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    Unless I'm mistaken, you're not necessarily using a potentiometer to measure the position, rather the system you have is being represented as a pot. A potentiometer in its simplest case is a variable resistor. You are trying to model your system as a variable resistor, and then calibrate the position of the ball using the measured voltage....I'm assuming.

    In this case, well yes, there appears that the ball needs to be connected to the circuit somehow. This setup reminds me a lot of EGR valves. The position of (some) EGR values is found by the piston being connected to a metal set of fingers which slides on a conductive plate. The voltage is sent to the ECU which calculates the position.

    Maybe a little more information on your setup would help?
     
  4. Apr 23, 2010 #3
    Ok minger, I think that makes a little more sense to me. Essentially the system layout is as follows:

    ballbeam.jpg

    There is a feedback control system that takes the ball's position and velocity as input via this "potentiometer" setup in order to command the motor. The beam has a metal track in which the ball sits which keeps it from rolling "sideways" off of the beam. That is the ball constrained to move on the xy-plane only.

    I am thinking that this metal track can serve as the base for our potentiometer setup. That is a current can be run through it, or a portion of it and the steel ball could be used to complete the circuit as it passes.
     
    Last edited: Apr 23, 2010
  5. Apr 23, 2010 #4

    berkeman

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    You will need the track to be something like two parallel rods of some resistive material. You would then measure the resistance seen between the rods at each end, and use those two resistances to calculate the position of the ball.

    Alternately, you could use two metal rods, and measure the inductance at each end with a moderate frequency AC inductance measurement. The inductances are small, though, so it would take a moderate frequency to get reasonable impedances. Fun project.
     
  6. Apr 23, 2010 #5
    This might be helpful.

    http://dspace.mit.edu/bitstream/handle/1721.1/32805/57587831.pdf?sequence=1 [Broken]
     
    Last edited by a moderator: May 4, 2017
  7. Apr 23, 2010 #6
    What kind of accuracy are you looking for? A potentiometer might be difficult to use since the contact resistance will most likely change over time and with humidity.
     
  8. Apr 23, 2010 #7

    Averagesupernova

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    I would not use anything that requires mechanical contact to the ball. I would use something like a nonconductive v-shaped trough. On each side of this trough I would have foil strips buried within the trough. The metal ball should affect the capacitance between the foil strips. If the strips are tapered from one end of the trough to the other I would think the varying capacitance could indicate the position of the ball. Just a thought, I've never tried anything like this but it's a place to start or at least provide seed for a better idea. Of course parasitic capacitance of the rest of the framework is a concern.
     
  9. Apr 23, 2010 #8

    Integral

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    If you take the time to read the link provided by skeptic2 you will find an excellent discussion of how to achieve the measurement using a "potentiometer" . The track consists of 2 conductive rails, one is a resistor like material (Graphite?) on which a consistent voltage is maintained so the voltage varies with position along the track. The other track is allowed to float and is attached to a Voltmeter. The ball makes contact with both tracks so becomes the wiper of a potentiometer. The wiper voltage is set by the position of the ball on the track.

    Sweet.
     
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