Force generated by ball screw linear motor?

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?

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

• 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..
– JDS
Commented Oct 4, 2016 at 16:58

I think this is far more a question for EngineeringSE, but just to hopefully clear up two points

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

$\eta_{thread} \eta_{thrust}$ 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.

• 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 :) Commented Jan 21, 2022 at 10:00

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