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Hi,

I am trying to figure out maximum linear force I can produce with an electric motor through use of lead screw and nut. I have come up with an equation which seems correct but resulting forces are way higher than my intuition believes is correct.

I tackled the problem from the perspective of energies:

F' * p = M/r * 2*π*r

F = 2*π*M/p * (1-k) //k stands for coefficient of friction between lead screw and nut)

Radius falls out of equation, which seems fine as force is provided by torque, which is radius agnostic. I used a 7€ stepper for source of torque which can provide 0.42 Nm of torque and a lead screw with 8 mm pitch and 4 mm radius, I assumed 15% losses. My equation tells me that such system can provide 280 N of force, which seems wayyy too much for a cheap motor.

I have attached a python code that calculates linear force in case it helps

Can someone tell me if my equation is either correct or flawed?

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I think you did the math right.

The amount of work the torque does in one turn is that torque multiplied by 2*pi radians.

W = T*(2*pi) = (0.42 N*m)*(6.283185)= 2.64 J

Then equating that amount of work to F*x, with x=8 mm = 8.0e-3 m

W = F*x <==> F=W/x

F= W/x = (2.6389378 J)/(0.008 m) = 330 N

and, you know, that is the maximum for F, in the case no work is lost to friction.

In the case where work is lost to friction, maybe the equation

W = (Fout + Ffriction)*x

can account for work done and work lost to friction. In other words, Fmax = Fout +Ffriction, and Fmax=W/x, and the actual force you get out, capable of doing external work, is some proper fraction of Fmax=W/x = 330 N

If that number seems high, maybe it is because the quote for torque is a lie, because motor manufacturers often exaggerate, or quote their numbers in funny units, like newton*centimeters, or other stupid tricks like that.

Or maybe the amount of work lost to friction will be much higher than you expect, like maybe instead of only losing 15%, it is more like 15% is the amount you get to keep, as useful work output. In that case, Fout would be 0.15*330 = 49.5 N

A scissor jack Effectively a linear screw actuator can lift much ore than 2 tonnes with a humans feeble input. The linear thrust from your screw will be very high.

It's true. In fact I have experienced this magic of lifting up a car with my bare hands, and a scissor jack. I mean, my hands supplied the work, and the jack supplied the mechanical advantage.

Actually, I guess a lot of people have experienced this magic as well, since changing a tire is kind of a common skill.

Speaking from my experiments I can only say calculating a possible force is a good start but nothing beats the real world check.
Theoretical numbers can give you a reslut you can expect under perfect conditions.
Usually we still encounter losses and things that can't be calculated with ease and accuracy.
One big problem with lead screws is the type.
A standard metric pitch gives tons of torque but also tons of problems to get a smooth movements with little friction.
The standard ACME thread used for example on most 3D printers or tiny milling machines is a compromise that reduces friction while also offering a higher transport speed if you use a 2mm or higher pitch.
Last but not least there are the ball screw systems, here the shaft itself forms a bearing with the actual nut.
Gives the most for spped and accuracy in many areas of operation.

Apart from the pure math that was already done by my friends here you also need to cater for the physical forces.
The more torque you apply or need the more sturdy the screw system needs to be.
For example driving the screw directly might cause the bearings in the motor to fail.
For higher torque the lead screw is often secured in its own bearing system, quite often even thrust bearing on either end.
The longer it is the more bending and wobbeling can become a problem too with high forces.
Simple leadscrews are designed to transport and if it is more than their lengths is restricted unless high quality steel was used.

On an 8mm pitch you can do fast movements but not deal with much forces.
I might use such a pitch for the positiong of something fast and light, which would require considerable force but I would not use it to move something by force ;)
A pitch that high is more like a ramp so you end of with lots of wear if too much force is used.
Please double check your pitch as it is not really easy to find 8mm pitch screws these days for cheap.
And I think your motor might be labeled in Ncm and not Nm.
A good sized Nema 17 will provide around 0.4Nm torque but this number is fictional if you ask me.
From my experience they only come close to this for holding torque but not for any real torque when moving.
Around 2amp is the upper max these motors can tolerate without driver or motor overheating badly.
With speed the torque reduces.
Loosing steps for example is quite common at higher speeds due to this problem.
Some of it can be compensated for by using dedicated drivers and code.
But from my experience I would say a 0.4Nm stepper is only good for around 0.25Nm at reasonable speeds and around 0.1 - 0.15Nm for higher speeds.
And this only works if the stepper is "locked" all the time, meaning the coils stay active at max power throughout operation.

At this point I just assume your lead screw is not as stated in your request as 8mm pitch seems to be a quite useless thing on a 4mm diameter screw unless high speed operation with no real accuracy was the goal.