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Motors for a scale model plane (airfix not rc)

  1. May 21, 2012 #1
    I'm looking to put motors into a scale model plane and was just wondering whether the motors I've picked will be ok and what sort of supply voltage I would need.

    I'm going to use 4 of these.7


    Operating Voltage: 3.7V
    Diameter: 7.0mm
    Length: 25.5mm
    Output Shaft Diameter: 0.95mm

    I assume I would just wire them in parralel to get them all to run at a similar speed and match the polarities to get them running in the same directions.

    As they just for display could I just use 2 AA batteries which would give me a supply voltage of 3V or do I need to supply them with exactly 3.7V?

  2. jcsd
  3. May 21, 2012 #2
    i would assume that the operating voltage is the voltage which causes them to run at their maximum performance of not causing damage to themselves, reducing the voltage would reduce the RPM of the motors, since voltage is the same as rate of magnetic polarity change , this is Faraday's Law ... the torque on the motors are proportional to the current flowing through them, this is the Lorentz Force Law

    basically, the motors will start off with nothing but Lorentz force from a standstill, it will draw alot of current and speed up quickly... as the motor's rate of rotation increases, it will experience more and more of Faraday's Law , and less and less Lorentz force, because it will create its own voltage which cancels out your AA batteries' voltage, reducing the voltage differential across the motor, until it has reached its maximum speed where by the motor is generating 3 volts, hence no current will flow because there is no voltage differential, so there will be no Lorentz force and no torque...

    in reality, there will be friction in the motor (not to mention whatever you are using them for) , which tends to increase as the RPM increases so the Lorentz force will be required to overcome this and there will be a continuous draw of current
    secondly, there is resistance in the motor itself, which is Ohm's law, which means that there is a limit on how much current will flow through the motor, which means how much torque the motor will have ... because the resistance in the motor is constant (if you dont let it heat up) , the higher the voltage, the higher the current , and the more torque you will get, but if you have too high a voltage, you will damage the commutators by causing them to spark ,this is why the motor is specified at 3.7 volts... since you are running with less voltage than the specification of the motor, this will not be a worry for you...
    and lastly, the batteries themselves have resistance, which means that the higher the output current, the lower the voltage ... if you have the maximum amount of current coming out of the battery, i.e. a short circuit, the internal resistance of the batteries will receive all 3v of the battery, and technically the battery isnt giving out voltage

    but the other concern is having too much current flow through the motors... by Faraday's Law, the slower the motor turns, the less the voltage it generates to counter the voltage you supply, and by Ohm's law, this large voltage differential will create a large current flow... if for whatever reason one of the motors is not able to turn, it will 'suck' power from all the other motors (since they are wired in parallel) and all your other motors will turn at just enough RPM to generate a voltage that matches whatever voltage is left after the batteries' internal resistance, because your AA batteries will be practically in a short circuit condition...
    if your batteries were more powerful (techincally less internal resistance , more current output at the same voltage) , or if you have enough AA batteries wired in parallel (to divide the internal resistance), the high current would overheat and damage the stuck motor ... but with a single pair of AA batteries, having a motor with insufficient power to start turning would just drain your AA batteries quickly and nothing else
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