mecanum forces
 

I started this thread to respond to a discussion in another thread which does not allow attachments.

This is a somewhat simplified version of the analysis presented in this paper.


Given:

- robot with 6" mecanum wheels

- each of the 4 wheels is being driven forward with a torque of 30 inch pounds

- robot is stalled against a brick wall

- assume rollers have zero spinning friction

- assume there is enough traction so that the wheels are not spinning


Analysis of front right wheel:


Figure 1

The net force, Fn, of the floor acting on the roller in contact with the floor, in the plane of the floor, must be aligned as shown along the roller axis (45 degrees).

Reason: if Fn were in any other direction, the roller could not be static.

At this point, we don't know the magnitude of Fn, just its direction.


Figure 2

Fn can be split into the two red orthognal components Ff (forward) and Fs (sideways).

The vector sum of Ff and Fs is Fn


Figure 3

The magnitude of Ff must be 10 lbs ( = 30 inch_lbs divided by 3 inches radius ).

Reason: The wheel is not accelerating. The net torque on the wheel in the plane of the wheel must be zero.


Figure 4

The magnitude of Fs is also 10 lbs.

Reason: Fs = Ff since mec roller is 45 degree angle


Figure 5

We now know that the magnitude of Fn must be 14.14 lbs

Reason: Fn is the vector sum of its components Ff and Fs.

Re: mecanum forces
 

Bookmarked. Very useful for future stuff.