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DC efficiency % and load

  1. Mar 4, 2009 #1
    Does anybody know of an equation or rule of thumb for power decrease as load increases?
    As an example an electric golf cart will go say 15mph with one 100lb driver. Put 4, 200lb people in it and it will go much slower, say 9mph. Is there a way to determine what that loss will be other than simple observation?
    Does an electric motor's efficiency rating play into it? One motor I am lookng at is 86% efficient. Does that mean 86% of the incoming electrical power is converted into mechanical power? If so, is that under no load conditons or some sort of industry standard load?
  2. jcsd
  3. Mar 4, 2009 #2


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    Staff: Mentor

    I don't think it's the power decreasing, it's that given a fixed amount of power, with a higher load, you go slower.

    Power is Work/Time, or Force*Distance/Time. So if the required force is going up, the time takes longer.

    Don't know what load the 86% is at -- it doesn't give some curve or more test details in the datasheet? You could call the manufacturer to find out, perhaps.
  4. Mar 4, 2009 #3
    I’m going to make some assumptions because this problem can’t be solved without them. To the extent my assumptions are wrong, the answer will be wrong.

    The first assumption is that the power out (angular velocity x torque) of the motor is constant. This won’t be strictly true because as the output shifts from higher angular velocity and lower torque to lower angular velocity and higher torque the windings and battery will carry more current, heat up and lose efficiency due to heat.

    The second assumption is that the torque required is directly proportional to the weight of the golf cart and cargo.

    From the problem description I don’t know if the four 200 lb people replace the driver or if they are in addition to the driver. I will assume the latter.

    So, if the cart plus 100 lbs. goes 15 mph and the cart plus 900 lbs. goes 9 mph, I think we can make a proportion where c is the weight of the empty cart.

    15/9 = (c + 900)/(c + 100)
    c = 1100 lbs.
    top speed = (12 * 15)/11 = 16.36 mph
    speed = (16.36 * 1100) / (load + 1100)

    Is this what you were looking for?
    Last edited: Mar 4, 2009
  5. Mar 4, 2009 #4


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    Gold Member

    Yes, its the conversion from electric to mechanical power. The efficiency is of course zero at no load (no output / a non zero current). The rated efficiency will peak at a given load. See the attached typical DC more operating curves.

    Good mfn data sheet:
    http://hades.mech.northwestern.edu/wiki/images/e/ee/Maxon_Motor_Guide.pdf [Broken]

    Attached Files:

    Last edited by a moderator: May 4, 2017
  6. Mar 4, 2009 #5


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    Science Advisor

    NEMA motors typically give full-load efficiency, not no-load. Most motors are designed to operate more efficiently at their designed load, so it makes sense to rate them based on full-load instead of no-load.

    Although they may give nominal efficiency as well for worst case scenarios.

    Last edited by a moderator: May 4, 2017
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