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Measuring the lifting power of an out runner motor

  1. Sep 15, 2008 #1
    Hi guys, how would one measure the lifting force that an electric out runner motor could put out. assuming one knows the rated power of the motor, how does one convert this into lifting force?

    Thanks, Ben.
     
  2. jcsd
  3. Sep 15, 2008 #2
    Electric motors don't output lifting force. Only mechanical torque at a rotational speed.
     
  4. Sep 15, 2008 #3
    Thanks mate but that isn't much help. Im looking for a series of equations that will allow me to follow the energy transfer chain to its conclusion: the force that the motor exerts through its prop.
     
  5. Sep 15, 2008 #4

    russ_watters

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    Force produced via torque is the torque times the length of the lever arm. If you have gearing, you include that by multiplying by the gear ratio.
     
  6. Sep 16, 2008 #5

    rcgldr

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    Since you mention lifting, I assume you mean a radio control helicopter. Obviously someone does the math, because radio control helicopter kits will specify motor requirements if the motor isn't included in the kit.

    The input power is watts delivered to the motor. Outunner motors are 85% to 90% efficient, so power output from motor is 85% to 90% of the wattage input. There are losses in the gearing, and main rotor, plus the tail rotor consumes power (about 5%). The power output from the main rotor equals thrust times air speed, although the air speed part can be tricky, since there is a rotor wash speed plus the air speed perpendicular (vertical) to the rotor. If the helicopter is in a high enough hover, net work done on the air is equal to mass of the affected air times it's speed squared when the air's pressure transitions back to ambient (using the air itself as a frame of reference), or more accurately, the integral sum of tiny masses of air times their speed^2 (at the moment their pressures transitions back to ambient).

    In terms of forces, motor torque is multiplied by the gearing into rotor torque, and rotor torque is converted into thrust. The amount of thrust depends on gearing, rotor size, and rotor pitch, minus any losses that occur during the conversion of forces.

    Generic info on helicopters:

    http://www.cybercom.net/~copters/helicopter.html

    http://www.absoluteastronomy.com/topics/Helicopter


    In case anyone here is curious about what an outrunner motor is:

    http://en.wikipedia.org/wiki/Outrunner
     
    Last edited: Sep 16, 2008
  7. Sep 16, 2008 #6

    russ_watters

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    Heh - I assumed an "outrunner" was a winch or something. So I guess we need to know that...
     
  8. Sep 16, 2008 #7

    rcgldr

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    One interesting feature is that the magnetic impulse moves around the motor much faster than the motor rotates, like a built in gearing ratio, allowing higher torque at lower rpm, good for direct drive propellers, but these also turn out to be efficient motors for use in geared applications like a helicopter.

    Although relatively new to radio control models, these motors have been used in cd/dvd rom drives for quite some time.
     
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