How Do mecanum wheels handle the bumps?

[Clinton Bolinger]

Originally Posted by joeweber

2008 Team 2337 made it to the finals in Atlanta and I thought they were using Mecanums.

Yes in 2008 and this year in 2010 we used Mecanums. We did make it into the eliminations of our division but unfortunately we did not make it to Einstein.

This year as a mid fielder, I think we play very aggressive defense and can push people around with our mecanums. Also defense isn't always about having the ability to just push people, there is more to it.

-Clinton-

[Alan Anderson]

Originally Posted by Ether

If the motor is providing a torque of 40 in-lbf to a wheel, and the wheel is not slipping, then the carpet MUST be reacting with a force of 10 pounds (for a 8" diameter wheel) in the plane of the wheel, acting against the wheel trying to push it forward.

Nope. The force is directed diagonally. An ideal mecanum roller can only apply force to the carpet in the direction parallel to the roller axis. Any component of force perpendicular to the roller axis acts to spin the roller instead.

And yes, with the mecanum wheel there are additional vector forces at play. But they do not subtract from the 10 pounds. I encourage those of you who are using the "vector" argument to think a bit more about how the vectors are created.

The vectors don't subtract from the 10 pounds, but they do move it away from straight ahead.

Try your thought experiment with a single mecanum wheel. Better yet, try a real experiment with a single mecanum wheel, applying a spinning torque to it with your hand, and trying to hold it in place at the same time. You'll definitely find as much force pointing sideways as pointing forward.

[NyCityKId]

Originally Posted by Ty Tremblay

If mecanum drive trains lack pushing power, why not have the opportunity to lock the rollers, effectively turning mecanum into 4 wheel tank drive?

While I was away in Thailand during build season, 190 did just that.

how would you be able to lock the rollers? please send me more information on this or point me towards any websites that might help me figure this out.

anyone can respond to this

[Joe Ross]

Originally Posted by NyCityKId

how would you be able to lock the rollers? please send me more information on this or point me towards any websites that might help me figure this out.

One way is presented in this paper: http://ftp.mi.fu-berlin.de/Rojas/omn...eter-Tlale.pdf

[Ether]

Originally Posted by efoote868

If you take a mecanum wheel and spin it, you'll find that it pushes you on a 45. Its hard to describe the feel until you actually do it.

I have done it. It reinforces everything I am saying in this thread.

Originally Posted by efoote868

Consider a wheel with rollers free to spin on the same axis as the main axle. It doesn't matter how grippy the material is for the rollers, these wheels can't apply a force no matter how hard they drive or how fast they spin.

Agreed.

Originally Posted by efoote868

Consider a wheel with rollers free to spin on an axis perpendicular to the main axle (traditional trick/omni wheels). It doesn't matter how hard they drive, they'll have the same coefficient of friction as wheels without the roller but made with the same material.

Agreed.

Originally Posted by efoote868

Mecanum wheels are in the middle - they spin on a 45.

Agreed. But your unstated but seemingly implied conclusion is not correct.

Refer to the scenario described in this post...

http://www.chiefdelphi.com/forums/sh...8&postcount=42

... and consider the following:

In the case of 45 degree rollers, the 40 in-lbf of "forward" torque applied to each wheel creates a forward force on each wheel of 10 pounds (due to the reaction force of the carpet) AND a sideways force of 10 pounds (also due to the reaction force of the carpet). These two vectors add (via vector addition) to a force along the axis of the roller of 10*sqrt(2). THAT is how the vector forces work.

Consider your second example above with wheels with rollers free to spin on an axis perpendicular to the main axle. The 40 in-lbf of torque applied to each wheel will create a forward force on each wheel of 10 pounds (due to the reaction force of the carpet) and NO sideways force.

Consider your first example above with wheels with rollers free to spin on the same axis as the main axle. If you try to apply torque to a wheel, all it will do is spin.

Consider a new example with rollers aligned not at a 45 angle, but a 60 degree angle (relative to axis of the wheel). The 40 in-lbf of "forward" torque applied to each wheel creates a forward force on each wheel of 10 pounds (due to the reaction force of the carpet) AND a sideways force of 10/sqrt(3) pounds (also due to the reaction force of the carpet). These two vectors add (via vector addition) to a force along the axis of the roller of 20/sqrt(3). THAT is how the vector forces work.

The forward force on each wheel in all cases (except the degenerate case where the roller axis is parallel to the wheel axis) is 10 pounds. 100% of the applied torque shows up as the corresponding forward force.


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[Ether]

Originally Posted by Alan Anderson

Nope. The force is directed diagonally.

Yes, the force is directed diagonally, along the axis of the roller. But not the force you imagine. See link at bottom of this post.

An ideal mecanum roller can only apply force to the carpet in the direction parallel to the roller axis. Any component of force perpendicular to the roller axis acts to spin the roller instead.

Of course. No disagreement here.

The vectors don't subtract from the 10 pounds, but they do move it away from straight ahead.

Nope. The 10 pounds is not "moved away" from straight ahead. The 10 pounds is the forward component of the (larger) force acting along the roller axis.

Try your thought experiment with a single mecanum wheel. Better yet, try a real experiment with a single mecanum wheel, applying a spinning torque to it with your hand, and trying to hold it in place at the same time. You'll definitely find as much force pointing sideways as pointing forward.

This does not contradict anything I have said in any of my previous posts. See this link for more detailed explanation:

http://www.chiefdelphi.com/forums/sh...5&postcount=50



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[Ty Tremblay]

Originally Posted by NyCityKId

how would you be able to lock the rollers? please send me more information on this or point me towards any websites that might help me figure this out.

anyone can respond to this

It's not simple by any means, but I will try my best to describe what 190 has done this year. I don't have any pictures or CAD models, so bear with me.

190 designed a mecanum wheel with double internal yokes (this means that the end pieces of the rollers are cantilevered, and the middle piece is supported on both ends) that is driven like any other mecanum wheel via sprocket on one side. On the other side of the wheel, however, is a friction plate with a face profile that matches the side of the rollers closest the the wheel's axis of rotation. When the friction disk is pressed into the rollers via pneumatic cylinder, the disk prevents the rollers from spinning while allowing the wheel to rotate. When they want to return to mecanum drive, they disengage the friction plate.

I will now attempt to illustrate this on my computer...

EDIT: Here's the pic.

[Alan Anderson]

Originally Posted by Ether

In the case of 45 degree rollers, the 40 in-lbf of "forward" torque applied to each wheel creates a forward force on each wheel of 10 pounds (due to the reaction force of the carpet) AND a sideways force of 10 pounds (also due to the reaction force of the carpet). These two vectors add (via vector addition) to a force along the axis of the roller of 10*sqrt(2). THAT is how the vector forces work.

Where does the extra 40% force come from?

It seems from your explanation that you might be able to put a carpet-treaded wheel underneath the mecanum wheel at a 45 degree angle and get greater torque at its axle than you put into the original mecanum wheel. Then use that torque to drive another mecanum wheel driving another carpet wheel and keep multiplying forces ad infinitum. It's apparent that something is wrong with your model.

[Ether]

Originally Posted by Alan Anderson

Where does the extra 40% force come from?

Same place the extra force comes from if I use a first-class lever to lift a heavy rock. There is no "conservation of force" law in physics.

It seems from your explanation that you might be able to put a carpet-treaded wheel underneath the mecanum wheel at a 45 degree angle and get greater torque at its axle than you put into the original mecanum wheel.

Of course I could. Or I could save myself all that trouble and use a gearbox instead :-)

Or I could just use the circumference of a small standard wheel to drive the circumference of a larger standard wheel. This would create greater torque on the larger wheel than was input to the smaller wheel.

[quote]
Then use that torque to drive another mecanum wheel driving another carpet wheel and keep multiplying forces ad infinitum.
/quote]

Yes, you could do that. Just like I could use the "load" end of one first-class lever to drive the "effort" end of a another second-class lever, ad infinitum.

http://www.sciencebyjones.com/first_...rs_drawing.gif

There is no "conservation of torque" law in physics either.

It's apparent that something is wrong with your model.

Nope.


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[JesseK]

There is no "conservation of torque" law in physics either.

Ether there is a Law of Conservation of Energy. While "bit for bit" in your arguments you may not break it, the reality is that this system is more dynamic than a single lever system. You've also failed to point out that your lever system has the end result of the final object moving less and less due to the torque conversion, whereas Alan is implying that both Mecanum wheels in the multiplied system of your model have the same speed.

Power = Torque * Rotational Speed, and is Energy per Unit Time -- in the example of pushing against a wall, the mechanical input power into the wheel is constant. The rollers translate that power 45 degrees from the angle at which it's applied to the wheel (due to the direction of the traction and tangental free spin). The power at 45 degrees cannot be greater than the input power, lest the Law CoE be broken. Thus, to figure out how much tractive force is applied axialy to the roller (given that the roller free spins tagental to its axle), vector components are necessary.

Assuming that when the wheel spins a minute amount to push forward on the force sensor, the roller spins an equal amount, and
Assuming that there are two wheels oriented such that their roller axles are tangental to each other (and 45 degrees from the axis wheel rotation)

The forces generated into the robot are absorbed by the frame.
The forces generated forward of the robot are absorbed by the sensor.

The two forces, when summed via vector analysis, much equal the sum forces at the 45 degree angle of the respective rollers.

Otherwise, the law of CoE is broken since this is in effect, a Power problem.

[fritzdejongh]

I believe Ether is right. I think you can see the "extra" force as coming from the sideways forces holding the wheels together (these forces must be there, or the wheels would fly apart). This force also gets reduced to the component along the roller, which then has a forward component. The bottom line is, if nothing is slipping sideways, the full torque is being converted to a forward force. It's kind of reminiscent of tacking in a sailboat.

If you're going at 45 degrees, two wheels slip completely so you only get power from half the wheels, so a swerve drive is probably better for a general angle.

[Ether]

Originally Posted by JesseK

Ether there is a Law of Conservation of Energy.

Of course there is.

Alan is implying that both Mecanum wheels in the multiplied system of your model have the same speed.

I think you are inferring something that Alan did not imply. I assumed that Alan knew better than that. Of course the hypothetical "carpet tread" wheel that Alan used in his argument would be turning more slowly since it has greater torque, just as the larger wheel in my example would be turning more slowly than the smaller wheel which is driving it.

I encourage you to mull over the vector analysis that I presented and see if it doesn't make sense to you.


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[Alan Anderson]

Originally Posted by Alan Anderson

Where does the extra 40% force come from?

Originally Posted by Ether

Same place the extra force comes from if I use a first-class lever to lift a heavy rock.

But you didn't add a lever to the system, or change the one that already existed. The wheels are the same diameter. The same torque on the axle should produce the same force at the edge, right?



I'm trying to understand the vectors as you are trying to describe them, and I'm failing utterly. Can you explain how they work when a mecanum drivebase is traveling at 45 degrees? I wouldn't expect more than 100% of the force to be available in that case either.

[Ether]

Originally Posted by fritzdejongh

I believe Ether is right. I think you can see the "extra" force as coming from the sideways forces holding the wheels together (these forces must be there, or the wheels would fly apart). This force also gets reduced to the component along the roller, which then has a forward component.

It might be a bit more accurate to state it thus: the torque applied to the wheel creates a force along the axis of the roller. This force can be split into forward and sideways components for sake of analysis.

In the case of the 45 degree rollers mentioned in my prior post...

http://www.chiefdelphi.com/forums/sh...5&postcount=50

... the force created by the 40 in-lbf drive torque on the wheel creates a force of 10*sqrt(2) along the roller axis.

The force is 10*sqrt(2) because the wheel is constrained from moving sideways by the counter-balancing sideways force from the wheel on the other side of the bot.

This 10*sqrt(2) force can be split into forward and sideways components of 10 pounds each.

The bottom line is, if nothing is slipping sideways, the full torque is being converted to a forward force.

If the command is forward on all 4 wheels and the robot is not moving, yes. We haven't yet discussed the case where the robot is moving forward.


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[Alan Anderson]

Originally Posted by Ether

Of course the hypothetical "carpet tread" wheel that Alan used in his argument would be turning more slowly since it has greater torque,...

Of course it would...except for one thing. The two wheels are in contact with one another, so they're turning at the same rate (assuming they're the same size). As long as you insist that the rollers don't spin while the robot is being driven forward, your description fails to obey the laws of physics with which I am familiar.