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Torque required to lift a load independent of screw engagement?

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Dikuza

Aerospace
Jan 18, 2017
35
Hey all,

I'm trying to do a simple calc to determine the torque required to lift a load (essentially a power screw).

I have found several different equations from the following sources:

1. 2. 3. below image (from undergrad textbook)

My question is two part:

1. Why are there so many equations that seemingly claim to calculate the same thing? Some with or without lead and lead angle involved, others with additional terms.

2. Why does the torque not depend on screw engagement? On one hand, I feel like more threads engaged would result in a higher required torque due to more surface area in contact, thus more friction to overcome. On the other hand, I feel like with more threads, the load is distributed over a greater surface area and so should be less. None of the above equations, however, have any input of thread engagement, rather just lead and lead angle.

Sorry for the lengthy question. Any help is much appreciated, thanks in advance!

P.S.
Does anyone have a good explanation/video on why a screw with multiple thread starts (higher lead) travels more linearly per revolution? I'm having a hard time visualizing this.

 
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Multistart screw nut doesn't travel 'more linearly', but it does travel 'more', per rev.

I am interpreting the phrase 'more linearly' here to mean lead error, not lead, which is of course some integral multiple of the single-start thread's lead for the same pitch.




Mike Halloran
Pembroke Pines, FL, USA
 
Sorry, I lost you Mike. A screw with a higher lead travels farther per revolution into its mating component, correct? This is what I'm having a hard time visualizing.
 
It is assumed that friction is not area dependent. So more thread engagement does not mean more friction.

Friction force = coefficient of friction x force normal to the surface. Thread friction is calculated based on that portion of the load normal to the thread surface. Lead angle would be in the calculation of force normal to the thread. None of these calculations involve the amount of area of thread contact.

Ted
 
First problem, I think: to a first approximation, friction is not dependent on surface area.
... at least not until the area gets small enough to induce Brinelling, or large enough to allow hydrodynamic pressurization, or if the lubricant includes odd things like powders or clays.

Second problem, I think: All the different equations exist because there are many different 'models' of how friction works outside the laboratory, and none of them work outside of the small world in which and for which they were developed.

My EE friend loved to derive everything from 'first principles', which is possible for a lot of electronic stuff, less so for mechanical stuff.

If you need good numbers for a screw jack, take advantage of all the sales engineers you can find to preselect a few models, which may turn out to be similar. Social engineer your way into their engineering departments to get help with anything the s.e. can't help you understand.



Mike Halloran
Pembroke Pines, FL, USA
 
You can confirm the higher feed yourself by wrapping one paper strip around a rod and then wrapping two of them side by side. With the same pitch, more starts = more advance.

In a regular screw jack there is no change in the number of threads that are engaged. Once there are enough in the design for the maximum load + safety factor, there is no advantage to using more threads. More engagement would cost more and increase the chance for interference, because pitch is never perfect. I've read of cases where threads lock with as few as 10 threads.

Compare the equations and see if there isn't a section that derives them to show what contributions were considered. Sometimes in these sort of equations there are terms that are assumed to be equal and those are cancelled or are assumed to be close to 0 or 1, whereas in other equations the terms are kept for the general case.
 
You can confirm the higher feed yourself by wrapping one paper strip around a rod and then wrapping two of them side by side. With the same pitch, more starts = more advance.
I think that would depend on your definition of "pitch". If you measure from one thread peak to the next on a double start thread, you are probably only measuring half the pitch since you are measuring from one helix to the other instead of measuring between two points on the same helix.

je suis charlie
 
GG - it's a way to get the experimenter to reason this out.
 
I would start at a more fundamental level of conservation of energy. It takes x amount of energy to raise a load from point A to point B. Divide by the amount of time to raise the load and you have the minimum power required (before friction, etc.). From here, you know you have a rock-bottom minimum power from which to derive ideal torque.

Not your final answer but a good stake in the ground.
 
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