<blockquote>
  <p>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.</p>
</blockquote>

<p>The detent torque of a <a href="https://motion.schneider-electric.com/downloads/quickreference/NEMA23.pdf" rel="noreferrer">typical NEMA 23 stepper</a> varies between around 3 and 7 N·cm. This is the torque produced when the windings are not energized.</p>

<p>Using this <a href="https://www.daycounter.com/Calculators/Lead-Screw-Force-Torque-Calculator.phtml" rel="noreferrer">leadscrew torque calculator</a>, 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.</p>

<p>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.</p>

<blockquote>
  <p>What effect the diameter of rod has on it?</p>
</blockquote>

<p>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.</p>

CC BY-SA 4.0
2018-09-19T08:19:18.707Z
This is all good and well, but one could also increase the load bearing ability with some design tricks, for example not mounting the Z straight but slanted. This does change the steps/mm for the benefit of turning some of the vertical weight force into horizontal force trying to shift the plate, which will be taken care of by other parts of the construction (the rails)
Trish
CC BY-SA 4.0
2018-09-19T08:54:07.593Z
@Trish That's not what the question is about though. Besides, I'm not sure that's a very practical design.
Tom van der Zanden
CC BY-SA 4.0
2018-09-19T09:05:02.667Z
it is if you need to reduce the load desperately and can use a pin-in-slot or a slider-on-rail as the connection between the Z-Motor and the bed. It is not the most simple solution, but uses the Aristotele law on levers: we get a longer way for each increment of height, but we need to apply less force. With a 45° slanted Z-rail one could achieve a theoretical bearing of double the calculated load, *but* one has to account for the same force trying to move the bed horizontally, demanding a very stiff back rail/side rails to counteract
Trish
CC BY-SA 4.0
2018-09-19T09:35:58.223Z
a side effect of tilting the Z-actuators would be to allow smaller Z-stepping as the number of steps/mm increases.
Trish
CC BY-SA 4.0
2018-09-19T09:38:54.087Z
@Trish Note that [comments are not for secondary discussion](https://3dprinting.stackexchange.com/help/privileges/comment) 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.
Tom van der Zanden
CC BY-SA 4.0
2018-09-19T12:17:56.157Z
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.
Pᴀᴜʟsᴛᴇʀ2
CC BY-SA 4.0
2018-09-19T12:51:30.607Z
@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.
Tom van der Zanden
CC BY-SA 4.0
2018-09-20T07:36:32.713Z
@TomvanderZanden I couldn't find threaded road with large dia and 1.25 mm pitch. Currently I'm using 8 mm dia 1.25 mm pitch. It holds position but I know in time it's gonna bend. Also problem with threaded rods are they wobble a lot. I think I'm doing something wrong when calculating torque. With pitch dia = 12 mm, Force = 75 N, thread density 6.35 in coefficient of friction = 0.2, The value I got was 14.1 and -4.32 Ncm
Athul
CC BY-SA 4.0
2018-09-20T07:37:01.983Z
Will it be more stable for 16 dia and 5 mm pich
Athul
CC BY-SA 4.0
2018-09-20T07:41:22.563Z
@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.
Tom van der Zanden
CC BY-SA 4.0
2018-09-20T07:51:18.240Z
@TomvanderZanden Can you explain how you got 5 Ncm as torque. With pitch dia = 12 mm, Force = 75 N, thread density 6.35 in coefficient of friction = 0.2, The value I got was 14.1 and -4.32 Ncm
Athul
CC BY-SA 4.0
2018-09-20T08:11:18.663Z
@Athul Because I used a 0 coefficient of friction. 0.2 is on the high end of possible friction coefficients. We don't know the exact friction coefficient, and I don't want to make any guesses. Therefore I was very conservative in the calculation. If I use the calculator with a 0.1 coefficient of friction with the parameters from my own printer it tells me it should easily hold up the bed, yet it doesn't.
Tom van der Zanden
CC BY-SA 4.0
2018-09-20T08:54:29.513Z
@TomvanderZanden Any idea what negative value specifies. Is it for just representing down ward torque?? If Calculated value is negative and detent torque is positive, what does that mean?
Athul
CC BY-SA 4.0
2018-09-20T08:57:14.133Z
If the calculated value is lower than the detent torque then in theory the axis will stay in position.
Tom van der Zanden
CC BY-SA 4.0
2018-09-19T08:06:01.357Z
2018-09-19T08:06:01.357Z
Tom van der Zanden
<p>Let's pull out physics...</p>

<p>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. $$F_{g_\text{total}}=166\text N$$</p>

<p>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 $$F_{g_\text{per lead}}=83\text N$$</p>

<p>Now, let's look at the specs of the lead screw: Pitch is 4 mm, Diameter is 12 mm. For the <a href="https://www.daycounter.com/Calculators/Lead-Screw-Force-Torque-Calculator.phtml" rel="nofollow noreferrer">calculator</a> we need Pitch Diameter (=Diameter) and the Thread Density (=Threads/cm), so I pull up the pitch <a href="https://www.newmantools.com/tech/pitchconversions.htm" rel="nofollow noreferrer">conversion table</a> 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.</p>

<p>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:</p>

<p>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.</p>

<p>So, when do we stay, when to we move? Well, if $\tau_\text{calculated}&gt;\tau_\text{detent}$, it moves down on its own. if $\tau_\text{calculated}&lt;\tau_\text{detent}$ it stays put for friction. The $\tau_\text{detent}= 0.027\text{ Nm}$ for a <a href="https://motion.schneider-electric.com/downloads/quickreference/NEMA23.pdf" rel="nofollow noreferrer">single length NEMA 23</a>. Our negative Torque just means we need to apply some extra torque to overcome friction before we get the bed going.</p>

<p>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. </p>

<p>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.</p>

CC BY-SA 4.0
2018-09-19T13:12:51.193Z
2018-09-20T16:40:06.893Z
2018-09-20T16:40:06.893Z
Trish
<p>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.</p>

<p><a href="https://i.sstatic.net/Rycew.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Rycew.gif" alt="Concept of an electromagnetic brake"></a></p>

<p><a href="https://en.wikipedia.org/wiki/Electromagnetic_brake" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Electromagnetic_brake</a></p>

CC BY-SA 4.0
2018-09-20T16:36:22.757Z
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.
Trish
CC BY-SA 4.0
2018-09-20T15:57:47.370Z
2018-09-21T19:19:48.463Z
2018-09-21T19:19:48.463Z
user77232
<p>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 <a href="https://www.3dprintronics.com/Linear-Actuator-Ball-Screw-1204-p108536003" rel="nofollow noreferrer">Linear actuator</a>. </p>

<p>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.</p>

<p>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.</p>

<p>What effect the diameter of rod has on it??</p>

<p>Is there any way to find that??</p>

CC BY-SA 4.0
2018-09-19T07:50:54.323Z
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?
Trish
<diy-3d-printer><printer-building><z-axis>
Ball screw: Maintaining Z axis position when motor is OFF
Athul
# Ball screw: Maintaining Z axis position when motor is OFF

(Asked by: username_0 on 2018-09-19)

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](https://www.3dprintronics.com/Linear-Actuator-Ball-Screw-1204-p108536003).

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??

**Tags:** diy-3d-printer, printer-building, z-axis

> **Comments:**
> **username_1**: 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?  

---

## 3 answers

### Answer by username_2 on 2018-09-19. Score: 7

> 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](https://motion.schneider-electric.com/downloads/quickreference/NEMA23.pdf) varies between around 3 and 7 N·cm. This is the torque produced when the windings are not energized.

Using this [leadscrew torque calculator](https://www.daycounter.com/Calculators/Lead-Screw-Force-Torque-Calculator.phtml), 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.

> **Comments:**
> **username_1**: This is all good and well, but one could also increase the load bearing ability with some design tricks, for example not mounting the Z straight but slanted. This does change the steps/mm for the benefit of turning some of the vertical weight force into horizontal force trying to shift the plate, which will be taken care of by other parts of the construction (the rails)  
> **username_2**: @username_1 That's not what the question is about though. Besides, I'm not sure that's a very practical design.  
> **username_1**: it is if you need to reduce the load desperately and can use a pin-in-slot or a slider-on-rail as the connection between the Z-Motor and the bed. It is not the most simple solution, but uses the Aristotele law on levers: we get a longer way for each increment of height, but we need to apply less force. With a 45° slanted Z-rail one could achieve a theoretical bearing of double the calculated load, *but* one has to account for the same force trying to move the bed horizontally, demanding a very stiff back rail/side rails to counteract  
> **username_1**: a side effect of tilting the Z-actuators would be to allow smaller Z-stepping as the number of steps/mm increases.  
> **username_2**: @username_1 Note that [comments are not for secondary discussion](https://3dprinting.stackexchange.com/help/privileges/comment) 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.  
> **username_3**: 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.  
> **username_2**: @username_3 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.  
> **username_0**: @username_2 I couldn't find threaded road with large dia and 1.25 mm pitch. Currently I'm using 8 mm dia 1.25 mm pitch. It holds position but I know in time it's gonna bend. Also problem with threaded rods are they wobble a lot. I think I'm doing something wrong when calculating torque. With pitch dia = 12 mm, Force = 75 N, thread density 6.35 in coefficient of friction = 0.2, The value I got was 14.1 and -4.32 Ncm  
> **username_0**: Will it be more stable for 16 dia and 5 mm pich  
> **username_2**: @username_0 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.  
> **username_0**: @username_2 Can you explain how you got 5 Ncm as torque. With pitch dia = 12 mm, Force = 75 N, thread density 6.35 in coefficient of friction = 0.2, The value I got was 14.1 and -4.32 Ncm  
> **username_2**: @username_0 Because I used a 0 coefficient of friction. 0.2 is on the high end of possible friction coefficients. We don't know the exact friction coefficient, and I don't want to make any guesses. Therefore I was very conservative in the calculation. If I use the calculator with a 0.1 coefficient of friction with the parameters from my own printer it tells me it should easily hold up the bed, yet it doesn't.  
> **username_0**: @username_2 Any idea what negative value specifies. Is it for just representing down ward torque?? If Calculated value is negative and detent torque is positive, what does that mean?  
> **username_2**: If the calculated value is lower than the detent torque then in theory the axis will stay in position.  

---

### Answer by username_1 on 2018-09-19. Score: 4

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](https://www.daycounter.com/Calculators/Lead-Screw-Force-Torque-Calculator.phtml) we need Pitch Diameter (=Diameter) and the Thread Density (=Threads/cm), so I pull up the pitch [conversion table](https://www.newmantools.com/tech/pitchconversions.htm) 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](https://motion.schneider-electric.com/downloads/quickreference/NEMA23.pdf). 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.


---

### Answer by username_4 on 2018-09-20. Score: 2

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](https://i.sstatic.net/Rycew.gif)](https://i.sstatic.net/Rycew.gif)

<https://en.wikipedia.org/wiki/Electromagnetic_brake>

> **Comments:**
> **username_1**: 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.  

---

