I am trying to select what stepper motor size (nema17,23,34...) to use for my project and quantity of motors needed. But am not sure how to apply the motors "torque" to real scenarios. The scenario I am interested in is moving a table, with some weight, a precise distance in the z direction using an 8mm lead screw (2mm pitch and 8 starts).

This may be a tall order but I have 3 scenarios below that I could really benefit in understanding how to apply stepper torque rating to calculate the max vertical lifting abilities for a typical Nema17. Basically what I am looking for is what is the maximum mass I can vertically lift at a reasonable velocity/acceleration (whichever is limiting) without taking the motor beyond its torque rating. I am not sure what assumptions are needed here.

Also I'd like to ask if I ensure that the stepper's acceleration and velocity is infinitely slow will my vertical lifting abilities be much higher?



A single stepper with a circular pedestal. Assuming the weights added to the pedestal were doughnut shaped and the pedestal itself is weightless, what is the max mass one could lift at reasonable speeds without experiencing missed steps or taking the motor beyond its torque rating? Scenario2


Same question, assuming a weightless platform 600mmx50mm with 2 steppers on each end what is the maximum mass we can lift? Assuming the weight is precisely in the center may be necessary.



And finally with a weightless platform approx. 600mmx800mm what is the maximum weight we can lift with 4 steppers.

  • 3
    $\begingroup$ see my comment on your previous question, concerning using a constant accel algorithm. To calculate the real possible load you would also need to know the pitch of the thread, and also have some idea of the losses involved. If your system is such that even one step is an overload, you are already dead, but you will find this out quickly on your first prototype. Once you can make one step reliably, all you need to consider is your min (starting) and max (final) pulse frequency, and the acceleration, which you can tune by trial and error to get smooth operation and acceptable speed. $\endgroup$
    – danmcb
    Commented Jun 8, 2021 at 14:13
  • 2
    $\begingroup$ I'd agree with @danmcb. But I'd add another consideration....because steppers don't know at what angle they're at when you power up. For those cases where two or more steppers must cooperate, some method of ensuring a level starting point should be worked out. $\endgroup$
    – glen_geek
    Commented Jun 8, 2021 at 14:53
  • $\begingroup$ @glen_geek yes. In an XYZ system this is typically done with limit switches. Normally you back up until the switch closes, then move slowly in the other direction until it opens. However in the mechanical design proposed here, something other arrangement needs to be made. $\endgroup$
    – danmcb
    Commented Jun 9, 2021 at 7:09
  • $\begingroup$ Parallel setups shown are tricky because of the over-constraint -- you have to be careful how the drivescrew nuts are mounted, to prevent uncontrolled side forces (and thus drastic increase in friction). In addition you may have binding (look up previous questions on that topic). The motor's torque curve will tell ideal low-speed (quasi-static) torque. Multiply by ratio derived from screw pitch, then derate by a healthy factor. Look up previous questions on power screws... $\endgroup$
    – Pete W
    Commented Jun 18, 2021 at 0:10

1 Answer 1


There are many parameters that you need to consider here, so I will just focus on the first scenario (i.e. A single stepper with a circular pedestal.

Assuming the weights added to the pedestal were doughnut shaped and the pedestal itself is weightless, what is the max mass one could lift at reasonable speeds without experiencing missed steps or taking the motor beyond its torque rating?

Torque calculation as a function of rpm and supply voltage

First of all, the rpm and the supply voltage are two critical factors. In the following image you can see that depending on the rpm and the Voltage you get significantly different torque.

enter image description here

Figure 1: Nema 17 torque vs rpm curves (source: moons motoros)

What is important to note here is the significant drop in Torque after 100rpm (for this particular model).

Calculating the maximum weight

In order to calculate the maximum weight the next thing that's required is the thread pitch (e.g. p = 1 mm/revolution). Then the most convenient thing is to use the following equations:

$$P = M\cdot \omega = F\cdot v \tag{eq.1}$$


  • $P$: power
  • $M$: torque in [Nm]
  • $\omega$: angular velocity [rad/s] ($=\frac{2\pi n}{60}$)
  • $F$: weight on [N]
  • $v$: velocity [m/s] (the velocity $v=\frac{n}{60}\cdot p$)

In the above equations I am assumming.

  • $n$: revolutions per minute of the shaft units:[rpm]
  • $p$: travel per revolution of the shaft (preferred unit [m/rev])

As a result you can rewrite Eq.1 as $$ M\cdot \frac{2\pi n}{60}= F\cdot \left(\frac{n}{60}\cdot p\right)$$

$$ F = \frac{2\pi }{p} \cdot M$$

This is the theoretical maximum weight if there is no acceleration. However, in reality it is lower for a number of reason, eg:

  • during the acceleration stage some of the torque is expended in the rotational energy of the rotational masses. usually this can be neglected in most application. (if you want to include that its easy assuming you know the acceleration profile, but IMHO its not worth the effort, because low accel will have almost no effect, high accels will only last a few seconds).
  • friction on the setup
  • cogging torque
  • ...

In general, the implementation parameters (e.g. alignment) are the most difficult to estimate.

  • $\begingroup$ So if I am following correctly with the following: M = 0.15Nm, p = 2mm, taking 9.8m/s2 for gravity. We have maximum mass of 0.8kg for scenario 1. This seems very low! Also wanted to mention the lead screw in the example is 2mm pitch and has 4 starts this means that for every 360 degrees we will move 8mm in the z direction, so should I use 8mm or 2mm in the equation..? $\endgroup$
    – Feynman137
    Commented Jun 18, 2021 at 3:43
  • 1
    $\begingroup$ you are right. As I was deriving (for nth-time ) the equation, I forgot to convert n [rpm] to rps (I will update) . Now with respect to the pitch, you need to use the travel per revolution in the formula, so 8mm/rev. $\endgroup$
    – NMech
    Commented Jun 18, 2021 at 5:40
  • 1
    $\begingroup$ So with the updated formulas for your example it should be about 12 kg (theoretical maximum). Given the [motor] (e-motionsupply.com/…) I guess this is a lot more that I'd expect. $\endgroup$
    – NMech
    Commented Jun 18, 2021 at 6:53
  • 1
    $\begingroup$ you don't need to. Its probably cheaper to buy a small gear set 4:1 (or a T2 pulley with the ame ratio) and you will be back to the original velocity. The lead screws will probably be a lot more expensive. $\endgroup$
    – NMech
    Commented Jun 18, 2021 at 11:30
  • 1
    $\begingroup$ Scenario 2 is trickier. If you get everything aligned right and if you put the weight in the middle its exactly as you said it (24kg). The thing is that those if's (especially the first one) are quite hard to achieve. If you there is a misalignment aligned, it is like trying to push a jammed door. $\endgroup$
    – NMech
    Commented Jun 18, 2021 at 17:54

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