Ball screw: Maintaining Z axis position when step motor is OFF

Ball screw: Maintaining Z axis position when step motor is OFF

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.

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

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 τdry=−0.0478 Nm${\tau }_{\text{dry}}=-0.0478\text{ Nm}$ and τoiled−0.0222 Nm${\tau }_{\text{oiled}}-0.0222\text{ Nm}$ 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 τdry,super-steep=0.273 Nm${\tau }_{\text{dry,super-steep}}=0.273\text{ Nm}$ 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 τcalculated>τdetent${\tau }_{\text{calculated}}>{\tau }_{\text{detent}}$, it moves down on its own. if τcalculated<τdetent${\tau }_{\text{calculated}}<{\tau }_{\text{detent}}$ it stays put for friction. The τdetent=0.027 Nm${\tau }_{\text{detent}}=0.027\text{ Nm}$ 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.

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