How much could this motor lift and how fast? If I connect a belt to its axis about 1cm away or even 10cm away from its tip.

Torque: 1.5Nm Peak Torque: 10Nm Power: 9-30A, to 1000W.


What kind of motor would I need to be able to lift 50kg about 1m high, in say 4 seconds? And to control finely the exact position within the 1m, sometimes it should lift to 20cm sometimes to 30cm. Is a stepper or servo motor capable of this task? Would a linear actuator be better, but I read those have low duty cycle, and they are slow so they would lift it slowly and wait for next time?

  • $\begingroup$ Really depends on your setup. Might be helpful if you would add a sketch of it. Just a couple of rough estimations: For the power output: P=mgh/t = 50*9.81*1/4~=122 W For the torque: T = mgx -> max(x) = T/(m*g)=10/(50*9.81)=2 cm So the power output might suffice, just notice that 10Nm is the peak torque and you already have a maximum leverage of around 2cm. But the power output is roughly in linear correlation with revs. $\endgroup$
    – Andrew
    Jun 4, 2018 at 9:47
  • $\begingroup$ How do other people do this, when they build a SCARA robot, which should just have a 1M high rail, and be able to lift 50kg fast to any point along the arm? So I need 122W for 50kg to be raised 1m in 4 seconds? Is that what your first calc showed? $\endgroup$
    – rapadura
    Jun 4, 2018 at 10:07
  • $\begingroup$ I'm probably not the right person to ask on the topic of robotics. These 122 W are only an estimation for the actual mechanical work performed over the time interval of 4 s. I would say as the required 122 W are much lower than the max power output of the engine (1kW), power is probably not going to be the limiting factor. $\endgroup$
    – Andrew
    Jun 4, 2018 at 10:42
  • $\begingroup$ You need to account for the weight and reduced inertia of each link too! $\endgroup$
    – joojaa
    Sep 3, 2018 at 13:31

2 Answers 2


Torque is measured in N M (Newton Meters)

A Newton is kg m / s^2

So torque kg m^2 / s^2

Torque = r x F = r x m x a

s is distance
t is time
a is acceleration
r is radius = 1/10 meter

s = 1/2 a t^2 = 1 m

2 / 4^2 = a = 1/8

plug it in

1.5 = 1/10 m 1/8
m = 1.5 x 10 x 8 = 1200 kg

that seems too high to me but that is what I get

  • $\begingroup$ @GürkanÇetin 2/16 = 1/8 $\endgroup$
    – paparazzo
    Aug 8, 2018 at 18:45
  • $\begingroup$ Well the robot itself needs to have some weight too, abd generally the weight of the robot far exceeds the thing its lifting. Because of rigidity needs. $\endgroup$
    – joojaa
    Sep 3, 2018 at 13:33

The height has nothing to do with the selection of the motor. Whichever motor type you choose (servo or step motors will allow you control the exact position while simple induction motor just cannot) - you should verify it has sufficient torque capabilities for the desired lifting velocity.

The needed torque in your case is calculated as follows: 1. for a constant speed: the load mass * g * the distance of the weight from the motor axis center (in meters) 2. for accelerating lifting: the load mass * the linear acceleration (in m/t^2) * the distance of the weight from the motor axis center (in meters)

the results for both cases are in Nm units.

Next, you need to convert the desired lift speed to the motor angular velocity. Basically, all you need to do is to divide to linear speed (m/sec) by pulley radius (m) to have it in rad/sec.

Each motor has a specific torque-speed curve. You should make sure your motor can supply the needed torque for the desired speed. Furthermore, you should relate to the motor peak torque only when this torque is applied for a short period of time. Otherwise, the motor will heat-up and wouldn't be able to supply enough torque.

  • $\begingroup$ For 1., it's load mass, not weight, and it's the effective horizontal distance that matters, not just "distance." For 2, it's the moment of inertia, not weight times distance. You can approximate moment of inertia with the parallel axis theorem, but then it's mass time distance squared. Also, I think that's a typo on the last line - "Otherwise, the motor will heat-up and would be able to supply enough torque." $\endgroup$
    – Chuck
    Aug 3, 2018 at 18:15
  • $\begingroup$ thanks. for 1: I corrected It. for 2: the two approaches lead to the same value. If you take the moment of inertia, you indeed take it as the mass times the distance squared. However, then you multiply it by the angular acceleration - which equals to the linear acceleration divided by the same distance... $\endgroup$ Aug 3, 2018 at 18:27
  • $\begingroup$ In part 1, the force exerted upon the mass by the motor (F=m·a) must be more than F=m·g, where g is acceleration due to the Earth's gravity. This is necessary so that work can be done—i.e., so that the motor can lift the mass against the counteracting force of the Earth's gravity pulling the mass down, F=-m·g. If the motor exerts force F=m·g upon the mass, and the Earth exerts force F=-m·g on the mass, the net force ΣF upon the mass is zero, which means no work can be done (W=F·d=0) and therefore the mass is held stationary (its velocity is zero): motor <-m·g-[mass]-m·g-> Earth $\endgroup$ Dec 4, 2022 at 9:30

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