I'm building a 3d printer of size 500 x 500 x 500 build area. For the Z axis, I'm planning to use this Linear actuator.

The maximum weight Z axis might encounter is 15 Kg due to it being a clay printer. A single linear actuator can, according to the specs, lift 10 Kg. So I'm planning to use two of this.

My question is a ball screw of pitch 4mm or 5mm, will it be able to Maintain it's position when the motor is de-energized under a load of 15 kg shared by two systems.

What effect the diameter of rod has on it??

Is there any way to find that??

  • $\begingroup$ The question about load is design dependant. What kind of design you want? A Prusa "printhead moves up" or something like a Hypercube with the print platform moving down? Can you give us at least a sketch? $\endgroup$ – Trish Sep 19 '18 at 7:50

My question is a ball screw of pitch 4mm or 5mm, will it be able to Maintain it's position when motor is deenergized under a load of 15 kg shared by two systems.

The detent torque of a typical NEMA 23 stepper varies between around 3 and 7 N·cm. This is the torque produced when the windings are not energized.

Using this leadscrew torque calculator, you can find that the torque required for a 12 mm diameter, 4 mm pitch leadscrew to hold up a 75 N load, is around 5 N·cm - assuming there is no friction. If there is friction, then the required torque will be lower.

So, the torque required is almost equal to or possibly even higher than the detent torque. Therefore, you should not count on a de-energized motor holding up the build platform. In practice, you might see that friction is enough to hold up the build platform, but that any disturbance (such as somebody bumping into the printer) is enough to get the leadscrews to start spinning and have the platform drop like a rock.

What effect the diameter of rod has on it?

Increasing the pitch also increases the torque required (so, go with a lower pitch leadscrew). The diameter does not affect the torque directly, but having a larger diameter increases the friction and so is beneficial.

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    $\begingroup$ @Trish That's not what the question is about though. Besides, I'm not sure that's a very practical design. $\endgroup$ – Tom van der Zanden Sep 19 '18 at 8:54
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    $\begingroup$ @Trish Note that comments are not for secondary discussion or providing alternative answers. The question only asks whether the given leadscrew will support the given weight, and my answer addresses that. It doesn't ask for alternative solutions. $\endgroup$ – Tom van der Zanden Sep 19 '18 at 9:38
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    $\begingroup$ Using zero coefficient of friction in your calculations would be inaccurate. No ball/lead screw mating will have zero friction. Depending on the material used in both will modify the friction coefficient. Using the lowest measurement given (0.04), you'd have 9.59 N-cm of torque, going up to 28.9 N-cm on the upper end. With two stepper motors, this would be doubled. Secondly, the linear actuators the OP showed also have guide rails. This, too, increases the amount of holding power of these when the motors are not energized. Realistically, I don't think it'll be an issue to hold things up. $\endgroup$ – Pᴀᴜʟsᴛᴇʀ2 Sep 19 '18 at 12:17
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    $\begingroup$ @Pᴀᴜʟsᴛᴇʀ2 I have a printer that uses two steppers (albeit nema 17) with 8mm pitch leadscrews to hold up a small bed (200mmx300mm) and has accompanying linear guides. I know from experience that it drops like a rock if you even look at it the wrong way when de-energized. Admittedly I have double the pitch and steppers with half the detent torque, but my bed only weighs around 1-2kg. I would not rely on the friction of components that are supposed to provide smooth motion to hold my 15kg print bed in place. $\endgroup$ – Tom van der Zanden Sep 19 '18 at 12:51
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    $\begingroup$ @Athul The larger pitch means increased torque, making it less stable. The bigger diameter means more friction, making it more stable. It is impossible to know which one will be more stable without knowing the exact friction coefficient. $\endgroup$ – Tom van der Zanden Sep 20 '18 at 7:41

Let's pull out physics...

We have a load of 15 kg, which is a weight force of 147.15 N, PLUS the bed. Let's take into account about two kilograms of bed and cables that need to be carried too, so 166.77 N. That is in the margin of error of 150 N, which simplifies calculations a lot, so let's just assume that. $$F_{g_\text{total}}=166\text N$$

Let's assume we use 2 lead screws, one left, one right, and both go exactly vertical. Then we get, that the load is half, so $$F_{g_\text{per lead}}=83\text N$$

Now, let's look at the specs of the lead screw: Pitch is 4 mm, Diameter is 12 mm. For the calculator we need Pitch Diameter (=Diameter) and the Thread Density (=Threads/cm), so I pull up the pitch conversion table and look under 4 mm Pitch. Also, we check the table and grab an intermediate "steel on steel" friction coefficient of 0.2 for dry, or 0.15 for oiled.

The results I get with those numbers are $\tau_\text{dry}=-0.0478\text{ Nm}$ and $\tau_\text{oiled}-0.0222\text{ Nm}$ respectively. That's the torque that needs to be applied to our motors to move it down. Wait, that number is negative?! Yes, it is, and that is actually not a bad sign:

Let's assume a higher high pitched thread. Like, REALLY high: 1 rotation per inch. One thread. It is pretty much "straight". The calculator drops us a result of $\tau_\text{dry,super-steep}=0.273\text{ Nm}$ for 0.2 friction coefficient that the motor would need to keep working against to not have it moving.

So, when do we stay, when to we move? Well, if $\tau_\text{calculated}>\tau_\text{detent}$, it moves down on its own. if $\tau_\text{calculated}<\tau_\text{detent}$ it stays put for friction. The $\tau_\text{detent}= 0.027\text{ Nm}$ for a single length NEMA 23. Our negative Torque just means we need to apply some extra torque to overcome friction before we get the bed going.

That means, with the 15kg load (+2kg bed), it is to be expected that the bed stays up. If you would use a more slick combination of materials, you might start to slide.

Always remember, there is the caveat of overcoming the detention strength with a short impact, or applying force directly to the rods when bumping into them. If the machisnism starts to spin, the friction from the nut on the bar and from the motor is all that can provide a breaking to the machine.


Adding to the other more scientific answers here. The backdriving force will move the lead screws depending on the stiction between the lead screw and the nut. If it was friction less then it will always slide. Basically you need to keep the lead screw from turning while the machine is off. You can make a basic electromagnetic clutch which will add some resistance to turning by using double pole double throw relays to short the motor leads to ground when the power is off. You can test this yourself by just connecting all the lead wires together and then try turning the motor spindle by hand.

Concept of an electromagnetic brake


  • $\begingroup$ Without a plan for how these devices work or are constructed, this answer is of quite low quality. Please consider expanding it by linking to relevant literature or plans, or including such into your answer. $\endgroup$ – Trish Sep 20 '18 at 16:36

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