Torque needed to generate lateral force with eccentric

My engineering school days are way back, and I'm looking for some help
with what I think is a fairly basic problem, but I'm not sure where to
I want to move an object with high precision over let's say 3mm, and
space under the object is very limited.
I want to use a small motor with gearbox to drive an eccentric, which
would push on the object as the eccentric rotates towards higher
radius. Let's say the object itself is sitting on the eccentric on one
point of contact.
My question is: how do I calculate the torque I need on the motor for
the eccentric to generate enough lateral force (upwards) to fight the
load (weight of the object)?
Let's say, for example that I use an eccentric of 15mm in diamter,
with an eccentricity of 9-6 (that would give me the 3mm motion over a
full turn)
I believe it is calculated using moments, but not sure where to start
Any suggestion?
Thanks for your kind help!
Reply to
Loading thread data ...
Yes, you can use moments -- assuming no friction, the weight on the point of contact with the eccentric, times the horizontal displacement between the point of contact and the eccentric shaft axis, will be the torque due to lifting.
Or calculate the speed that the object will lift or fall with respect to the speed of the shaft, and divide -- that's the effective moment arm.
But the real kicker here is "assuming no friction". If that eccentric is rubbing, then most of the torque will be due to friction. You'll have to take that into account.
Reply to
Tim Wescott
Thanks Tim,
Great point about the friction... ...any hint on how to calculate/estimate that? I assume it depends on the materials used, and there must be standard friction coefficients or something, right?
Thanks for the help. Regards
Reply to
Sure, there are lots of ways to calculate friction. Each one gives a different answer, and most of them are wrong!!
Seriously, yes it can be done to a rough approximation. But friction coefficients aren't simple -- a single material doesn't have one, only pairs of materials, and the surface finish matters, and wear matters, and lubrication, and, and, and.
The answer to friction questions is generally "get an expert" and/or "testing, testing, testing". I'm not an expert at this -- I'm usually in the happy position of standing there with my hands in my pockets, smirking, while the experts beat their brains out on friction issues, although if they fail to overcome them then the smirk disappears when someone says "Tim, can you get this system working right anyway?".
Reply to
Tim Wescott
OK... then I'll just consider 80% of my torque is lost to friction and I should be safe, right? ;-)
Reply to
It should come clear if you draw a force diagram. Accounting for friction will be a major difficulty whatever you calculate. Have you considered actual tests on a full-scale model?
Reply to
Jerry Avins
Do I hear 95? Going once ...
Build the system with a pair of good bearings where the gearbox will be. Mount the cam* on a shaft run through the bearings. Attach a lever arm to the shaft and determine what weight is needed to move the object in a series of positions along its path. Remember that the cam will exert a lateral as well as a lifting thrust.
Jerry ________________________________________ *
"Cam" because an eccentric won't provide constant distance for constant angle. You may prefer a snail cam (Archimedean spiral).
Reply to
Jerry Avins
Jerry's an old retired fart so he may not realize that there are computer aided design tools that will do all of this in mathemagic land, without you ever needing to build anything.
Which doesn't explain why, when I see mechanical types congregating in a lab, they're generally doing measurements to determine friction, or to verify friction calculations.
Note that in addition to any difficulties that friction presents to you for determining the motor size, it also makes it hard to precisely position that cam, which in turn makes it hard to precisely locate your object. Nothing in life is free...
Reply to
Tim Wescott
With a distance of movement of 3mm or so, and if your required profile of motion is suitable for it, you could make the cam by putting a roller or ball bearing assembly onto an off-centre round shaft. The bearing outer can interface with the object and roll around the off-centre shaft to minimise the friction. I think others have given clues for the rest of the calculations you need.
Reply to
Paul E. Bennett
I know I thought that real hard, even if I didn't actually point it out.
Even roller bearings have friction effects, and depending on how tightly you need to locate within that 3mm a roller bearing (or, for that matter, any motor/cam arrangement) may not be suitable.
But a roller bearing is probably going to be better than a sliding contact for all but the least (or most) possible required precision.
Reply to
Tim Wescott
By the way: your total stroke is 3mm, but what precision are you looking for?
+/- 0.1mm is going to be easy, but 0.01mm is starting to put some pretty stringent demands on bearings, surface finishes, and whatnot.
Reply to
Tim Wescott
No! There are books and tables that have the coefficient of friction between different materials.
formatting link
is easy to find the static friction values but harder to find the dynamic friction vales. Then you must find a way to blend the together if you want do know a dynamic response other wise you can simply assume the friction is dynamic and the angular speed is constant. Do some searching for coefficient of friction. There is lots of information. Don't worry too much about being exact. ( Never let perfect be the the enemy of good ). Just know that there may be variances. The lubricated metal on metal coefficient of friction is 0.16. This is much lower than the 0.8 or 80% you mentioned.
There are many ways to generate friction coefficients. If you use modeling software they may include more than one model for friction. I don't care if my estimate for friction is 0.15 or 0.17 instead of 0.16 because I will try them all. The kinetic friction will be less and maybe as low as 0.1 but I can't find the right answer. If you use 0.16 you should be safe. Now how much force will be required to accelerate the mass? Obviously we need to know the mass but also the change in radius with respect to the rotation and then we need to know the rotational speed. The chain rule will provide the acceleration as a function of angle and frequency. In your case the eccentric will make contact at slightly different angles so it will vary even at a constant speed.
This sounds like a test system and I/we get involved with a lot of them. Usually a linear motor or hydraulic cylinder applies a sinusoidal force but at high frequencies moving something between two cams seems like the easiest but not the most flexible way to go.
When modeling remember that the sum of forces is equal to 0 unless something is accelerating and the energy in must be equal to the work done+heat generated. If everything balances then you probably have it right.
Search the internet. There is a lot of info
Peter Nachtwey
Reply to
Instead of a cam (with or without a rolling surface) and gearbox, I would first consider a fine-pitch screw. If there's no room under the object, it might be possible to put the screw(s) to one side. I once built a carriage for wafer examination that implemented fine focus with three short screws bearing sprockets and linked by roller chain. I still have the dies we made for the 1/8x100 thread (and the commercial taps that made them). The focus assembly was about 1/2" thick.
If friction is reduced to the point that the drive is more than 50% efficient, power will be needed to keep it from lowering under its own weight. A block and tackle is more efficient than 50%, a self-locking chain hoist is less.
Reply to
Jerry Avins
I agree with Jerry. I love a good statics problem, but if I had a similar problem I would no doubt use a stepper motor with a lead screw.
Reply to

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.