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#1
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Omni vs Mecanum CoF ?
I am looking at the differences between Omni-wheel drive (Killough) and Mecanum drive. I think I understand the cos(45) difference in forward torque and corresponding difference in velocity. I propose that rotating the Omni-wheel could be modeled as a wheel with radius of 1/cos(45) and if this larger wheel Omni-wheel was geared for the same maximum velocity, the available torque would be the same as the original mecanum wheel.
The discussions to date have suggested that there is a relative loss of available traction for the omni-wheel system since the torque of the wheel to the carpet is sqrt(2)/2 larger and will brake traction sooner than the mecanum wheel. Looking at the AndyMark specs for their wheels the Coefficient of Friction for the 6" Omni-wheel is 1.0 where as the CoF for the 6" mecanum is 0.7 F&R - the same sqrt(2)/2 ratio!!! For strafe, the mecanum CoF is only 0.6 which is additional sqrt(2)/2 loss (which I believe is expected). From a practical standpoint, it would appear that there is no traction advantage for macanum over Omni-wheel drive with the wheels available to FRC. Assuming AndyMark uses the same rubber for both wheels, does this suggest that mecanum drive has the same theoretical loss of traction as Omni-wheel drive? Can anyone explain or dispute these observations? I know that swerve does not suffer from these losses, so lets keep the dialog limited to Omni-drive vs. mecanum drive. |
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#2
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Re: Omni vs Mecanum CoF ?
I think you're right. In a 45 degree omni drive you can model the wheel as if it were a bigger diameter which explains a lot of the speed and torque phenomena.
Just subjectively it seems like mecanum platforms tend to do a better job of resisting motion, but not by much. I think it's just easier to spin omni drives into a position favorable for pushing. |
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#3
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Re: Omni vs Mecanum CoF ?
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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) Last edited by Ether : 09-12-2013 at 12:53. |
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#4
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Re: Omni vs Mecanum CoF ?
One thing to keep in mind is that in the omni-drive, rollers will be turning when you are moving forward or backward. In the mecanum drive, they will not. Whenever the rollers turn, you have non-negligible frictional losses - in fact, if you work out the geometry, you'll see that frictional losses in the spinning of the rollers is the only reason that mecanums strafe slower than their forward/backward movement.
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#5
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Re: Omni vs Mecanum CoF ?
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However, due to roller axial free play and carpet compliance there will be some motion of the mec rollers, even in the forward direction. Quote:
Last edited by Ether : 09-12-2013 at 13:15. |
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#6
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Re: Omni vs Mecanum CoF ?
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#7
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Re: Omni vs Mecanum CoF ?
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I suspect the 1.0 number for the omni was tested in the plane of the wheel, not at a 45 degree angle. Someone from AM please correct me if this is not true (I'm sure AM posted their test procedure somewhere but I can't find at right now). |
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#8
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Re: Omni vs Mecanum CoF ?
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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. |
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#9
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Re: Omni vs Mecanum CoF ?
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#10
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#11
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Re: Omni vs Mecanum CoF ?
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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. Last edited by Oblarg : 09-12-2013 at 15:37. |
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#12
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Re: Omni vs Mecanum CoF ?
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These tests were done with locked wheels (not locked rollers) using the tilted incline method (which we believe allows for greater accuracy than the pull test method). |
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#13
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Re: Omni vs Mecanum CoF ?
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Would I be possible to update the specifications with the side CoF for your wheels? It would help greatly with part selection verses other vendors. Is there a good way for measuring dynamic CoF? I am simply assuming 85% of static. BTW: we just received shipment of 4 VEXpro 6" Omnis and we were very impressed with the design and build quality. Assuming the 2014 game is appropriate, we intend to use them for an asymmetric Killough drive. |
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#14
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#15
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Re: Omni vs Mecanum CoF ?
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Our omni directional wheels should have very very low CoF Side-Side, since we're pretty happy with how freely the rollers spin. ![]() Unless I'm missing something, using the locked-wheel test, shouldn't the Mecanum Side-Side be identical to front-back? (Isn't that the simplifying virtue of a 45-degree angle?) Regarding a method to measure dynamic CoF -- one method I've used in the past is: Get the robot sliding, using the incline test. Slowly reduce the amount of tilt until it stops moving. Measure the angle and calculate like normal. Out of curiosity -- what sort of design are you doing for FRC which requires dynamic CoF? |
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