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Say I have a motor with a certain amount of torque, $T.$ It is turning a ball screw mechanism like this .

Say, I mount something on the nut, I want to calculate the mass of an object I can lift upwards, i.e. the force that the actuator can generate.

I think this equation is what I want: $$T=\frac{Fl}{2\pi\nu}$$ source

But I don't understand:

$\bullet$ What is the "lead" $l\,?$ Is it the millimeter pitch distance between grooves?

If so, using that tiny motor above (I have a similar one) it seems that $F = 2\cdot \pi \cdot 0.26~\mathrm{Nm / 8mm}$ which is $204$ Newtons, so equating to $F = mg$ it seems I could lift $204~\mathrm N/9.8~\mathrm{m/s^2} = 20.8~\mathrm{kg}\,?$ This seems really heavy for that small motor, so I figure something is wrong with my interpretations and/or calculations.

$\bullet$ How is the wikipedia formula derived?

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3 Answers 3

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lead is the pitch of the screw,it probably needs to be in meters if the rest of the equation is SI units.

Yes you can lift a very big load with a screw and a small force - that's why your car has a screw thread jack to change wheels.

Really a screw is just a slope (inclined plane in physics speak) so the equation should be pretty easy to derive.

edit: An easy way to check is to consider the energy. In fact whenever you aren't sure in a physics calculation ALWAYS consider the energy, it's often the simplest way.

One turn of the thread moves the object 8mm vertically.

If you have a Torque of 0.26Nm then one turn of the motor is like providing a force of 0.26N at a radius of 1m, and energy is force * distance.

So a force of 0.26N around a circle of 1m radius is 2*pi*1.0m*0.26N = 1.6J

The energy to lift 20kg vertically 8mm = 20kg * g * 0.008m = 1.6J

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  • $\begingroup$ So you think my calculations, when we add in some losses due to efficiency, are probably in the right ball park? Even being able to lift 10 kg I find surprising, but I ordered the part to test it out, so I'll do some "applied physics" and confirm.. $\endgroup$
    – JDS
    Commented Oct 4, 2016 at 16:58
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I think this is far more a question for EngineeringSE, but just to hopefully clear up two points

enter image description here

The lead on a thread.

The derivation follows, but it's not exactly the same as Wikipedia.

enter image description here

$\eta_{thread} \eta_{thrust}$ might be combined as efficiency in the Wikipedia equation.

If so, using that tiny motor above (I have a similar one) it seems that F = 2*pi * 0.26Nm / 8mm which is 204 Newtons, so equating to F = m*g it seems I could lift 204N/9.8m/s2 = 20.8 KG?? This seems really heavy for that small motor, so I figure something is wrong with my interpretations and/or calculations.

My apologies, I am an idiot, Martin's answer spells it out.

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  • $\begingroup$ Up-vote for providing a picture and basic terminology, and for a try to explain some subtle details. Down-vote for the ugly Word-document (learn TeX) which one hardly can call an explanation. So no vote :) $\endgroup$
    – LRDPRDX
    Commented Jan 21, 2022 at 10:00
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It is simple:

F = T*2*Pi*eff/l

Where: F: Force, newtons; T: Torque, newton-meters; l: lead, meters; eff: ballscrew efficiency

In most cases, efficiency can be set safely to 90% with enough margin.

So, for example, a ball screw with 5mm lead, driven by 2Nm motor will give:

2*2*Pi*0.9/0.005 = 2262N = 231kg

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