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Efficiency of a DC Motor Gearbox

  1. Aug 19, 2011 #1
    Does the efficiency of a gearbox just reduce the torque output of the motor? Or does it also reduce the speed?

    For example I have a motor with 100 N*m torque at 100 RPM, it then goes into a gearbox with a 80% efficiency and a 2:1 ratio. So the torque would be 2*100*0.8 Nm and the RPM 100/2, correct?
  2. jcsd
  3. Aug 19, 2011 #2


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    Yes, the ratio of the gearbox stays the same, but the losses in the gearbox due to friction etc form part of the load on the motor and reduce the useful output available to actual loads.
  4. Aug 19, 2011 #3
    Motor aren't giving out Torque and RPM, like say, a light bulb is giving out light.
    In fact, Torque is produced only when output is connected.
    So, basically it goes like this.
    In a motor without gear box, if you connect an external load of 100 Nm then motor runs at 100 rpm say.
    In the same motor with gear box, if you connect an external load of 100 Nm, then it behaves as (100 Nm + 20 Nm (of gear-box) = 120 Nm), so, either the motor don't run at 100 rpm now (if its DC or induction motor) or It will still run at 100 rpm but consume 20% more current and hence 20% more power. (if it is synchronous motor)
  5. Aug 19, 2011 #4
    I believe this is a post on mechanics, rather than a post on electrical engineering. Anyways, here's the answer (no need to bother about if its a synchronous or induction motor here, from the definition of the problem):

    You have some output power P_1=T_1*omega_1 for some motor (shaft power, not electrical) which is the product of torque and angular velocity. Then you know that the formula for efficiency (of anything) is Efficiency=P_out/P_in. On the output of the gearbox, you also get a mechanical power, which we will define as P_2=T_2*omega_2 (which is P_out here). Knowing that Efficiency=P_2/P_1, you know that T_1*omega_1*Efficiency=T_2*omega_2. With the ratio of gears being omega_1/omega_2, you can find the ratio of torques, with the part of torque lost in the gears.

    So, from the point of view of the motor, we still have P_1=T_1*omega_1 on the shaft. Of course, if (as previous poster pointed out) you which to keep the same output torque on the load (100 Nm with gearbox), this will mean you are changing the speed (induction motor). But I think your question was more about power transmission (right?).

  6. Aug 19, 2011 #5
    One thing, I would like to get clarified. I am in the thought that loss in gear box is like loss in water head in pipe, it occurs only when things are moving.
    Suppose, there is a gear-box. It has input-shaft and output-shaft. If I apply 100 N-m to its input shaft, I still need 100 N-m torque to be applied on the output shaft in opposite direction to keep the shaft stand-still. Anything less and the shaft starts rotating.
    However, during running condition, it can occur that, input torque is 100 N-m and output is 80 N-m and the shaft is rotating at constant speed.
  7. Aug 19, 2011 #6
    Thank you all for the helpful replies. I think I have a good conceptual picture of how it works now.
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