Question: Say I have a motor with a certain amount of torque, $T.$ It is turning a ball screw mechanism like [this](http://www.omc-stepperonline.com/threaded-rod-nema-17-external-linear-stepper-motor-34mm-body-100mm-t8-x-4-p-212.html) . 

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](https://en.wikipedia.org/wiki/Ball_screw#Equations)

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?

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

[![enter image description here](https://i.stack.imgur.com/ZulZH.png)](https://i.stack.imgur.com/ZulZH.png) 

The lead on a thread.

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

[![enter image description here](https://i.stack.imgur.com/5a1iW.jpg)](https://i.stack.imgur.com/5a1iW.jpg)

$\eta\_{thread} \eta\_{thrust}$ 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.