Omni vs Mecanum CoF ?

[Ether]

Originally Posted by gpetilli

[does] mecanum drive [have] the same theoretical loss of traction as Omni-wheel drive

Yes.

Look at Figure1 on Page2 of this document.

You will see that the ratio of forward motive force to traction force is the same for both omni and mec:

mec: (tau/r)/(tau*sqrt(2)/r) = 1/sqrt(2)

omni: (tau/(r*sqrt(2)))/(tau/r) = 1/sqrt(2)

[gpetilli]

Originally Posted by Ether

Yes.

Look at Figure1 on Page2 of this document.

You will see that the ratio of forward motive force to traction force is the same for both omni and mec:

mec: (tau/r)/(tau*sqrt(2)/r) = 1/sqrt(2)

omni: (tau/(r*sqrt(2)))/(tau/r) = 1/sqrt(2)

Okay, so then if I increase the gear ratio by sqrt(2)/2 to increase the available torque for the Omni-wheel drive, it should match the pushing force and traction of the mecanum (minus larger friction loss of rollers for Omni). That leaves the major advantage of mecanum as the mechanical mounting being square. Omni has the advantage of lower weight, cost and complexity.

Do we assume that AndyMark included the sqrt(2)/2 in the CoF specification or is there a difference in materials? IFI reports CoF of 1.1 for Omni-wheels and 1.0 for mecanum - still an difference, but not the magic ratio.

That said, I have one extra design variable available with Omni-wheels because I am not locked into 45deg mounting by the wheel manufactures. If I build an asymmetric Killough with wheels at 30deg, I get 22% more forward torque and traction (0.866 vs. 0.707) than can be achieved with mecanum. Am I missing something here? Of course this comes at a price, I get 0.5 vs. 0.707 = 30% less side torque and traction. More forward traction is highly desirable for better forward acceleration. Having even poor staffing capability has benefits over KOP 6-wheel tank drive, if the other tradeoffs can be managed.

[Ether]

Originally Posted by gpetilli

Okay, so then if I increase the gear ratio by sqrt(2)/2 to increase the available torque for the Omni-wheel drive, it should match the pushing force and traction of the mecanum

To a first approximation. Keep in mind that friction in the rollers reduces the available forward pushing force of omni, but increases the available forward pushing force of mec: roller friction causes an omni to lose traction but causes a mec to gain traction (in the forward direction). See the second-to-last paragraph on Page3 of this document.

If I build an asymmetric Killough with wheels at 30deg, I get 22% more forward torque and traction (0.866 vs. 0.707)

Decreasing the toe-in angle of omni indeed increases the forward force (for a given wheel torque), and the available traction.

[Oblarg]

Originally Posted by gpetilli

More forward traction is highly desirable for better forward acceleration.

I believe free acceleration (i.e. not pushing against another robot) is almost never traction-limited in FRC; you'll pretty much only ever slip your wheels if you gun it from a dead-stop, and only for a few fractions of a second.

I did some crude calculations recently, and found that for a 150lb, 4-CIM robot with a wheel CoF of about 1, you are traction-limited at a dead-stop for any gearing below ~12 feet/second.

So, if we have a mecanum geared for 12 feet/second, then, which is a pretty standard gearing, we'll only be traction-limited until our motor torque drops to ~70% of stall torque, which corresponds to ~30% of top speed.

With this in mind, I don't think you're going to see all that much practical change in your acceleration with increased CoF.

[Ether]

Originally Posted by Oblarg

I did some crude calculations recently, and found that for a 150lb, 4-CIM robot with a wheel CoF of about 1, you are traction-limited at a dead-stop for any gearing below ~12 feet/second.

Would you mind posting your calculation please?

[gpetilli]

Originally Posted by Ether

Would you mind posting your calculation please?

I guess I agree with Oblarg that there is minimal difference in acceleration. I tried my drag race predictor (attached) for both drives and they are fairly close (would be even closer with non-standard gearboxes). However, I believe if both were in a pushing match, the Omni would still have a 20% traction advantage.

Omni 30deg, 12.76:1 Vfinal=11.58FPS, Accel= 28.6ft/s^2, T5ft=0.70s, T15ft=1.57s

Mecanum 0deg, 10.71:1 Vfinal=11.94FPS, Accel=27.4ft/s^2, T5ft=0.67, T15ft=1.52s

[Ether]

Originally Posted by Oblarg

I did some crude calculations recently, and found that for a 150lb, 4-CIM robot with a wheel CoF of about 1, you are traction-limited at a dead-stop for any gearing below ~12 feet/second.

Originally Posted by Ether

Would you mind posting your calculation please?

Originally Posted by gpetilli

I guess I agree with Oblarg that there is minimal difference in acceleration.

I'm not sure why you linked to my post instead of Oblarg's for that comment.

I'm still interested in how he did the calculation, regardless.