Mecanum. What's Best

[Jared Russell]

Originally Posted by Chris is me

You can also turn a crab drive well by connecting diagonal pairs of wheels together (like 1717 did last year). This lets you both strafe and orient your wheels in a "diamond" pattern for turning on a dime.

This configuration lets you crab and turn in place, but turning while translating is difficult.

I actually took some time last year to analyze 25 different possible steering/driving configurations for crab drives. Basically, it combined all of the typical driving and steering configurations in a giant matrix, and figured out what drive modes (crab, tank steer in both long and wide, snake/car steering) would be possible (both with and without skidding the wheels) in each. I should probably wrap it all up into a white paper when I get a chance. I looked at:

4 drive transmissions (one per wheel)
-Steered all together
-Steered in pairs (F and B, L and R, diagonal)
-Steered individually
2 drive transmissions (F and B)
-Steered all together
-Steered in pairs (F and B, L and R, diagonal)
-Steered individually
2 drive transmissions (L and R)
-Steered all together
-Steered in pairs (F and B, L and R, diagonal)
-Steered individually
2 drive transmissions (diagonal)
-Steered all together
-Steered in pairs (F and B, L and R, diagonal)
-Steered individually
1 drive transmission driving all wheels
-Steered all together
-Steered in pairs (F and B, L and R, diagonal)
-Steered individually

The only configuration that gives you perfect (no skidding necessary) crab and turning under all possible circumstances is the 4 drive transmissions/4 steering transmissions case.

[Ether]

Originally Posted by Jared341

The only configuration that gives you perfect (no skidding necessary) crab and turning under all possible circumstances is the 4 drive transmissions/4 steering transmissions case.

For the mathematically inclined, the above result is equivalent to Equation 3 which is derived in this white paper. It shows the necessary X and Y components of velocity at each wheel to achieve any desired vehicle simultaneous translational and rotational motion.

Using the results from Equation 3, the wheel angle for the nth wheel is -ATAN2(Vyn,Vxn), and the wheel rotational speed for the nth wheel is sqrt(Vxn^2+Vyn^2)/r.

[joek]

just remember, you want an X, not an O