How Do mecanum wheels handle the bumps?

[JesseK]

Ether, the problem with your logic is that Mecanum wheel rollers are not effectively locked when pushing against a hard object. Properly designed rollers still 'roll' at the off angle due to a good bushing that allows the roller to free-roll anyway. The underlying issue isn't that Mecanums can't push; even 70% torque of 4 CIMs is enough to push an object to get out of most situations. It's even good enough to play defense in most situations since defense can usually be done just by turning another robot or getting in the way of another robot.

Conventional wisdom says Mecanums can't push; this is due to software control and a driver's lack of understanding of how a Mecanum design works holistically, OR that the drive train was designed with speed in mind and no high-speed drive train can push worth a durn. We've found that with 4 CIMs a good balance is 10-11 fps before losses. Under the covers, when on a diagonal, software tells 2 motors to drive the diagonal while the other two motors sit idle. This allows the diagonal strafe. Yet when pushing while on this diagonal, the bot effectively has 70% * 1/2 (due to 2 motors out of 4) = 35% of the original torque of the 4 CIM setup. Ergo, Mecanum drive trains with true holonomic control only ever apply 35% to 70% of 4CIMs to any direction. Thus 'Mecanums can't push' should be turned into 'Mecanum drivers don't know how to drive' or 'the whole holonomic idea of Mecanums is terrible for competition robots'.

Teams who try to do a full field-centric Mecanum holonomic drive can probably anecdotally attest to this. This year 1885 did Mecanum, yet we only did standard tank/skid steer with the ability to strafe 90 degree from forward -- nowhere in between. Rarely did we get into a situation where we needed to push, yet we also didn't have a problem since it wasn't setup for true holonomics.

[Alan Anderson]

Originally Posted by Ether

When you are pushing directly forward, the rollers are effectively locked.

All the forward torque applied to the wheel is transmitted to the carpet.

The torque is not reduced by cos(45).

None of these assertions is true.

The rollers do spin noticeably when you're driving straight forward. This is where the obvious loss of pushing power manifests itself.

If you go through the vector analysis of how much torque is pushing in which directions for an ideal rectangular-layout mecanum or omniwheel drivebase, you find that ~70% goes to movement in the forward direction, and ~70% goes to trying to squash or stretch the frame in the sideways direction.

All the torque is available in the diagonal direction. The torque in the forward direction is only cos(45) of the maximum.

[Ether]

Originally Posted by Chris is me

My post assumes the rollers of a mecanum wheel spin.

If they don't, the number is greater than 71%, but then you don't have a mecanum wheel anymore, at least one as efficient in any other direction.

Hi Chris,

My reply to your post also assumes we are talking about mecanum wheels with rollers which are free to spin. Whether or not they actually spin in a given scenario is a different question.

For example, consider the following thought experiment:

There are two identical robots, RobotA and RobotB, which are absolutely identical in every detail except that RobotA has mecanum wheels and RobotB has standard wheels. The tread on RobotB's standard wheels is the same material as the rollers on RobotA's mecanum wheels. The standard wheels have the same effective diameter as the mecanum wheels.

RobotA and RobotB are each facing a brick wall, and there is a load cell on the front of each robot to measure how hard it is pushing on the wall. Each robot is given a slowly increasing identical forward command.

I claim the following happens:

- Up to a certain point, both robots push with exactly the same force. There is no 71% factor involved.

- If the motors are sufficiently powerful, a point will be reached where the wheels start to slip. RobotA, with the mecanum wheels, will reach this point before RobotB, with the standard wheels.



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[Alan Anderson]

Originally Posted by JesseK

Under the covers, when on a diagonal, software tells 2 motors to drive the diagonal while the other two motors sit idle. This allows the diagonal strafe. Yet when pushing while on this diagonal, the bot effectively has 70% * 1/2 (due to 2 motors out of 4) = 35% of the original torque of the 4 CIM setup. Ergo, Mecanum drive trains with true holonomic control only ever apply 35% to 70% of 4CIMs to any direction.

On the diagonal, all the torque of those two motors is working in the direction of travel. You get 100% of two motors, or 50% of a 4 motor traditional drivebase.

(The original LabVIEW omni/mecanum drive vi took that into account and reduced the maximum power in the forward/sideways directions to 70% of what it could be. That gave 50% of full motor torque in every direction. It was a good idea for making consistently-controllable omnidirectional movement easy, but it was a terrible idea for being able to call on the most performance when necessary.)

[Chris is me]

Originally Posted by Ether

Hi Chris,

My reply to your post also assumes we are talking about mecanum wheels with rollers which are free to spin. Whether or not they actually spin in a given scenario is a different question.

Mecanum wheel rollers should spin in all circumstances when the robot is attempting to move "forward". I don't think your illustration of traction limited pushing is quite correct, but I'll let someone smarter than me clarify independent of mecanum wheel rollers always spinning.

[JesseK]

Originally Posted by Ether

I claim the following happens:

- Up to a certain point, both robots push with exactly the same force. There is no 71% factor involved.

- If the motors are sufficiently powerful, a point will be reached where the wheels start to slip. RobotA, with the mecanum wheels, will reach this point before RobotB, with the standard wheels.



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Again, the assertions assume the rollers pit all of the force forward. The reality is that the rollers, even when not spinning, will put part of the force inward toward the robot. The force vector calculations easily show this. The resulting forward force is based on the fact that one component of both force vectors faces the other side of the robot; these two components cancel each other out. The forward component of the force vector is what pushes on the hypothetical wall.

It is the same logic as the situation where a box sits on an inclined plane. Gravity pull the box straight down, yet only a component of the gravitational force creates the force normal to the ramp that keeps the box on the ramp. the other component is parallel to the ramp. Same logic, different situation.

Interestingly enough, the speed of a Mecanum drive train is the same as standard skid steer, and perhaps that's the confusion?

[Ether]

Originally Posted by Chris is me

Mecanum wheel rollers should spin in all circumstances when the robot is attempting to move "forward".

The rollers will not spin in all circumstances when the robot is attempting to move "forward".

If the robot is pushing against a brick wall, and the wheels are not turning, then the rollers will not be spinning. Everything looks static. Try it.

I don't think your illustration of traction limited pushing is quite correct, but I'll let someone smarter than me clarify independent of mecanum wheel rollers always spinning.

This question is easily resolved by running a simple test. Drive your mecanum-wheeled robot on a carpet against a wall and push forward. Up to a point, the wheels will be completely static. Only when the wheels break free and start to spin, will the rollers spin also.


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[Ether]

Originally Posted by JesseK

Again, the assertions assume the rollers pit all of the force forward. The reality is that the rollers, even when not spinning, will put part of the force inward toward the robot. The force vector calculations easily show this. The resulting forward force is based on the fact that one component of both force vectors faces the other side of the robot; these two components cancel each other out. The forward component of the force vector is what pushes on the hypothetical wall.

http://www.chiefdelphi.com/forums/sh...6&postcount=33

Refer to RobotA mentioned in the thought experiment in the post at the link shown above.

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. The forward component is 10 pounds (the exact same forward component you would get with a standard wheel). It cannot be otherwise.

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.


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[Ether]

Originally Posted by Alan Anderson

None of these assertions is true.

If you go through the vector analysis of how much torque is pushing in which directions for an ideal rectangular-layout mecanum or omniwheel drivebase, you find that ~70% goes to movement in the forward direction, and ~70% goes to trying to squash or stretch the frame in the sideways direction.

...The torque in the forward direction is only cos(45) of the maximum.

http://www.chiefdelphi.com/forums/sh...0&postcount=38

Hi Alan,

Please take a look at my post at the above link and see if I can convince you otherwise.


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[JesseK]

Originally Posted by Ether

http://www.chiefdelphi.com/forums/sh...6&postcount=33

Refer to RobotA mentioned in the thought experiment in the post at the link shown above.

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. The forward component is 10 pounds (the exact same forward component you would get with a standard wheel). It cannot be otherwise.

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.


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I will concede in that the split second before the rollers start to slip there is an instant the wheels put all force directly forward. Yet this split instant of an isolated experiment doesn't translate to assisting a driver in pushing due to the dynamics of the holistic system. The driver doesn't go up to something, stop, and THEN push the object. The only scenario in which this may minutely apply is in a situation of pinning. Yet even then, the instant the robot moves it turns into a dynamic situation -- the rollers slip, the tractive forces only translate 70% torque, and thus the bot remains pinned.

[NyCityKId]

Originally Posted by alicen

mecanums make it very simple to avoid pinning by scooting off in whatever direction you choose, it is also surprising how well you can push/pin, but only in the forward direction. It's easy to be pushed when pushed from the side, but from front or back you can just strafe away. the biggest pro (to me at least) between mechs and crab is that mechs are much simple mechanically, than crab, and are usually lighter as well. over other drive trains, mechs have the maneuverability. They can go any direction they please, and turning is not difficult, like it can be for tank or skid steer.


active suspension -- basically it is some kind of force that keeps the wheels touching the ground. if there's a dip under only one wheel, the suspension pushes that wheel further down while the rest of the wheels a pulled a bit higher. Think of the shocks on a bicycle, it's a similar concept. For our team, the way we implement suspension is by using springs captured between a point on the wheel module and the frame. If you'd like, i can dig up a picture of the robot for you

i'd love to see a picture of your suspension. it would be really helpful when building next years robot

[Ether]

Originally Posted by JesseK

Heh, this doesn't look like it has an end in sight.

This is a conversation worth having. It's worth the trouble.

I will concede in that the split second before the rollers start to slip there is an instant the wheels put all force directly forward. Yet this split instant of an isolated experiment doesn't translate to assisting a driver in pushing due to the dynamics of the holistic system.

It's not about split-seconds. When a mecanum-wheeled robot is pushing against a brick wall (or any other immobile object) in the forward direction with insufficient torque on its wheels to cause the wheels themselves to turn, then neither are the rollers turning, and 100% (not 71%) of the torque being applied to the wheels must be reacted by the carpet in the plane of the wheels. If the motors are providing 40 in-lbf of torque to each wheel in the forward direction (against the wall), then the carpet reacts with 10 pounds of force on each wheel (on an 8" diameter wheel) in the forward direction in the plane of the wheel, just as it would with a standard wheel. That 10 pounds of force IS the forward component. The forward component is not created by dividing the 10 pounds by the square root of 2.

This is a real-world effect, and it does explain the pushing force of a stalled mecanum-wheeled robot in the forward direction.


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[NyCityKId]

thanks to everyone who is replying. i'm learning a lot about drivetrains. unfortunately im wishing i paid more attention in trigonometry -_-.... but please continue. this is good stuff

[Ether]

Originally Posted by JesseK

the instant the robot moves it turns into a dynamic situation -- the rollers slip, the tractive forces only translate 70% torque

We can discuss the physics which apply to a moving robot once we can agree on the (simpler) physics of a stationary robot whose wheels are not slipping.


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[efoote868]

Originally Posted by Ether

I claim the following happens:

- Up to a certain point, both robots push with exactly the same force. There is no 71% factor involved.

- If the motors are sufficiently powerful, a point will be reached where the wheels start to slip. RobotA, with the mecanum wheels, will reach this point before RobotB, with the standard wheels.

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.

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.

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


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.
No matter which way the wheels spin, they're fighting each other.