Friction when driving gear systems from the "other" end. 2

[StephenMontgomerSmith]

I have a gear system, where one shaft S1 drives a second shaft S2. The gear system works equally well whichever direction I turn S1.

Now I want to turn S2 to drive S1. I find that the friction becomes quite a bit greater.

I did quite a bit of web searching, and I also asked Mechanical Engineering Professors. I didn't find anything, except some very technical papers on friction in gear systems. I have also done some thinking about it. The only difference I can see is the the gears slide over the same surfaces, but in the opposite direction.

 

[handleman]

What is the ratio between S1 and S2? What sort of system is it? A worm drive has a very, very high friction when you try to back-drive it...

 

[StephenMontgomerSmith]

The ratio is something like 10 to 1 (10 turns of S1 correspond to 1 turn of S2). It is a system of four gears, one pair is helical, and the other pair is spur. (So there is a third intermediate shaft.)

But I have another example whose ratio I am not totally sure of, and it is all spur gears.

The first system is this:

https://www.pololu.com/product/4758

The second system is this:

https://www.amazon.com/gp/product/B072R5G5GR/ref=ppx_yo_dt_b_asin_title_o02_s00?ie=UTF8&psc=1

I feel it is related to the following phenomenon. If you drag a pencil at an angle across a surface, and hold the pencil at an angle, then the friction will be quite a bit less if you drag the pencil behind the motion rather than pushing it ahead of the motion. But standard friction theory says that friction is coefficient of friction times normal force, and clearly this formula doesn't work in this instance. It seems to me that it has something to do with the pencil not being held rigidly, so that when you push the pencil ahead of the motion, the friction will tend to push the pencil into the surface, thus creating more normal force, and conversely if the pencil is dragged behind the motion, then the friction pushes the pencil away from the surface, thus reducing the normal force.

 

[handleman]

OK... Just for the sake of numbers, let's ignore all the internal friction and just consider the friction of the input and output bearing/seals.

Let's say each one has a friction torque by itself of 1Ncm. So, you grab S1 and spin it. You will feel 1Ncm resistance from S1's seal, and 0.1Ncm resistance from S2's seal due to the mechanical advantage of the 10:1 gear reduction. Now, try to grab S2 and spin it. You will feel 1Ncm resistance from S2's seal and 10Ncm from S1's seal.

Obviously, the actual sources of friction are spread through the various bearings, grease, seals, gear faces, etc. but the general principle remains. You should expect to feel at least something like 10x the resistance when back-driving a 10:1 reducer.

 

[drawoh]

StephenMontgomerSmith,

The reflected inertia of a gear drive varies as the square of the gear ratio. I have played with gear drives, driving them backwards, and I have been surprised at the torque I needed to exert. When I crunched the numbers, it all made sense.

Even friction is a problem. If you drive a 10:1 gear train from the fast end, the friction torque is reduced by a factor of ten. If you drive from the slow end, you multiply the friction torque by ten. I would expect that you would notice this.

--
JHG

 

[StephenMontgomerSmith]

What a simple answer! I would call myself stupid for missing this, but a whole bunch of very smart people also missed this.

Maybe it was the way I originally phrased the question. I wanted two motors working in opposition to each other. So there was a torque on both shafts S1 and S2. The friction was very different depending on whether the system was turning in one direction versus the other, and the difference was huge.

Since friction is non-linear (it just depends upon the direction of travel, not the speed of the travel), I can now see that the effect of changing from one direction to the other is going to be very large.

I was able to overcome this particular problem by removing the gears, and using direct drive. But these motors aren't good enough for this, and their encoders aren't accurate enough to get decent control.

 

[IRstuff]

Then, you probably have the wrong motors for your application.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

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