How Do mecanum wheels handle the bumps?

[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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