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Say I'd like to calculate the motor specs required for lifting a 1kg weight at a rate of 1 m/s. How much torque and RPM should the motor have, assuming a wheel radius of 1cm?

My calculations are as follows:

Force = mass * acceleration = 1kg * 9.8 m/s^2 = ~10N

Torque = Force * radius = 10N * 1cm = 10N * 0.01m = 0.1 Nm = 1 kgF-cm

Rotational speed = 1 m/s * 60/(2 * pi * r) = ~320 RPM

What I don't understand is, should I be looking for rated values or no-load values of torque and speed? For example, should I be looking for a motor with rated torque of 0.1 Nm/1 kgF-cm, with rated speed of 320 RPM?

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  • \$\begingroup\$ The no-load value of torque is 0 by definition (neglecting friction) and the no-load speed is the speed with no load ... but you have a load, so... And 320rpm is pretty slow, so you are probably looking for a geared motor. Choose one with the rated speed you need at your supply V. Higher rated torque than you need won't hurt, except the budget. \$\endgroup\$
    – user16324
    Commented May 27, 2023 at 13:56

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Technically, the torque you require is determined by the weight of your load AND the time you are willing to wait for it to accelerate up to speed. If your motor torque is sized exactly to the lever arm and weight of your load all it will be able to do is hold the load in place against gravity. It requires additional torque to accelerate to some speed, after which it can reduce the torque to produce just enough torque to exactly counter gravity to maintain that speed.

Rated motor values are just the operating conditions that the motor designer has deemed an good balance between performance, mass, cost, heat, power density for the expected usage. If the expected usage changes (i.e. the SAME motor is intended to be operated in hot conditions), then the rated values would likely be different. Therefore rated values are helpful for sizing a motor but not helpful for determining specific performance.

What you really want for motor performance though is the motor speed-torque curve and these are fortunately quite easy to get for DC motors. They shape for DC motors (brushed and brushless) all generally look like this:

enter image description here https://islproducts.com/design-note/how-to-read-dc-motor-gear-motor-performance-curves/

All you need is the stall torque and no-load RPM for the same voltage. If you have the no-load current and stall-current at that same voltage you can also determine the current draw for a particular torque.

  1. Stall torque happens at zero RPM. So that's one data point.
  2. Zero torque happens at no-load RPM. So that's a second data point.
  3. Since the torque-speed curve for a motor is linear, you just connect points 1 and 2. That's your torque speed curve.
  4. The same thing for current. No-load current happens at no-load RPM and zero torque. So that's one data point.
  5. Stall current happens at zero RPM and stall torque. So that's a second data point.
  6. Connect points 4 and 5.

And since these were all the same voltage they can all track each other. Now if you want to operate at a different voltage, these roughly scale linearly. So half the voltage means half the RPM, half the stall torque, and half the current to construct a new series of curves at the new voltage, physically limited by what the motor can handle before it overheats and burn out or overspeeds and throws bearings and rotor out (you can't just keep scaling up the voltage forever to get more and more speed, torque, and power).

The power and efficiency curves also have the same shape. The power curve could be directly calculated from the torque and speed curves but you probably don't need to go that far. It's pretty obvious from the shape anyways. The efficiency curve is less obvious but still follows the same shape and there are some rather accurate rules of thumb:

  • Maximum efficiency occurs at 1/7ths stall torque or 6/7ths no-load RPM (this is the same point).
  • Maximum power occurs at 1/2 stall torque or 1/2 no-load RPM (again this is the same point).

So you want to operate somewhere between maximum power and maximum efficiency depending on your objectives. Note that efficiency rapidly falls off as you underload the motor below 1/7ths stall torque (the friction losses and such remain fairly constant but your output power is dropping and so the losses being to dominate). Therefore, it is in your best interest to operate with slightly more torque than 1/7th stall torque than slightly less since the efficiency drops off much more slowly in that direction.

NOTE: You generally do not know where the rated torque, speed, and current values are on these curves. As I mentioned they are chosen by the designers based on various criteria and expected usage so their positions can vary on these curves. You can't reconstruct these curves based off the rated values.

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  • \$\begingroup\$ Thanks for the elaborate answer. This stuff was hard to understand from books/articles. \$\endgroup\$ Commented May 27, 2023 at 4:35
  • \$\begingroup\$ @VenkataNarayana I had my fractions reversed for torque and speed at max efficiency. Corrected. 6/7th stall torque is close to burning out your motor. \$\endgroup\$
    – DKNguyen
    Commented May 27, 2023 at 4:36
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    \$\begingroup\$ @VenkataNarayana It's not that hard. If the motor provides stall torque and no-load RPM, just pick a motor that has double your required torque and double your required speed for max power, or alternatively 7 times your required torque and 7/6 times your required speed for max efficiency. Or anything in between the ratios for max power and max efficiency. You get to choose the trade-off. Just stay between those bounds unless your application is particularly short duty cycle and requires an unusually small motor. \$\endgroup\$
    – DKNguyen
    Commented May 27, 2023 at 4:38
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    \$\begingroup\$ @VenkataNarayana Some applications such as a rover require balancing various conditions. For example, cruising versus overcoming an obstacle. Cruising is best done at maximum efficiency while you might want accelerating to speed to be done at maximum power and for climbing an obstacle to overload the motor a bit beyond the torque for maximum power to keep the motor a reasonable size. You can do that because climbing an obstacle should be infrequent and short duration. That's when you have to find a motor that gets close to all three points as best you can. \$\endgroup\$
    – DKNguyen
    Commented May 27, 2023 at 4:42

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