Omni vs Mecanum CoF ?

[gpetilli]

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.

[Chris is me]

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.

[Oblarg]

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.

[Ether]

Originally Posted by Oblarg

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.

Assuming "perfect" omni & mec and floor surface, that is true.

However, due to roller axial free play and carpet compliance there will be some motion of the mec rollers, even in the forward direction.

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.

In the strafe direction, those frictional forces in the mec rollers require more torque to be applied to the wheel in order to get the same motive torque. This causes increased losses due to carpet stretching, compared to the forward direction. See Page4 of this document.

[gpetilli]

Originally Posted by Oblarg

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.

Agreed, that does contribute to the forward vs. strafe for mecanum. I don't see the connection to why the mecanum CoF is apparently lower than equivalent Omni in forward.

[Ether]

Originally Posted by gpetilli

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

Originally Posted by gpetilli

I don't see the connection to why the mecanum CoF is apparently lower than equivalent Omni in forward.

The numbers don't mean anything if you don't know how the tests were conducted.

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).

[gpetilli]

Originally Posted by Ether

The numbers don't mean anything if you don't know how the tests were conducted.

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).

AndyMark is good about posting forward vs. side for their wheels, so I believe that the CoF comparison is valid (never expected it tested at 45deg). What I was less sure of is that the CoF measurement would have exhibited the sqrt(2)/2, but I believe your diagram predicts that it would. IFI does not report forward vs. side for their mecanum and given the similarity in CoF to the Omni, I am highly suspicious that they are reporting the CoF for the roller material, not the wheel assembly.

[JVN]

Originally Posted by gpetilli

AndyMark is good about posting forward vs. side for their wheels, so I believe that the CoF comparison is valid (never expected it tested at 45deg). What I was less sure of is that the CoF measurement would have exhibited the sqrt(2)/2, but I believe your diagram predicts that it would. IFI does not report forward vs. side for their mecanum and given the similarity in CoF to the Omni, I am highly suspicious that they are reporting the CoF for the roller material, not the wheel assembly.

We report CoF for an overall drive assembly, weighted to 150 lbs (120 lb robot + 14 lb battery + 20 lb bumpers = 154 lbs). In this way, we are reporting the average CoF of a drive train "in application."

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).

[gpetilli]

Originally Posted by JVN

We report CoF for an overall drive assembly, weighted to 150 lbs (120 lb robot + 14 lb battery + 20 lb bumpers = 154 lbs). In this way, we are reporting the average CoF of a drive train "in application."

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).

Thanks John for the quick reply. That helps clarify things, and I agree that incline method is best.

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.

[JVN]

Originally Posted by gpetilli

Thanks John for the quick reply. That helps clarify things, and I agree that incline method is best.

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.

I will look into posting the sideways CoF. I don't believe we tested that when we did our experimentation last year.

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?

[gpetilli]

Originally Posted by JVN

I will look into posting the sideways CoF. I don't believe we tested that when we did our experimentation last year.

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?

Conceptually I agree that mecanum should have equal forward and side CoF. I am trying to reconcile why AndyMark specifies 0.7 forward and 0.6 side for their mecanum and 1.0 forward for their Omni. I thought I had it, but more data seems to just make things more confusing.

The dynamic CoF is interesting for two reasons. First, I am trying to predict where the transition to/from traction limited will occur (possibly not of practical importance). Second, many teams seem to depend on the wheels slipping during a pushing match (or drive into a wall) to limit the motor current. Four CIMs each drawing a stall current of 133A will obviously pop the breakers. The question is, what gear ratio gets the current to about 40A per CIM when driving into a wall. I believe we need the dynamic CoF for that calculation. This is true for traction wheels as well as holonomic wheels.

[Oblarg]

Originally Posted by Ether

Would you mind posting your calculation please

Well, this is embarrassing: I can't find the scratch paper I did this on earlier, and I'm differing by a factor of two upon repeating the calculation. At any rate:

We'll omit the efficiency figure for the calculation; you can multiply by an estimate at the end if you wish.

Let's round up to 160lbs, so we have 40lbs per wheel. Assume all weight is distributed equally among the wheels. CoF is 1, so our maximum friction force from a single wheel is 40lb.

Let Wf be the free speed of a CIM, and Ts be the stall-torque of a CIM.

Say our drive is geared to a top linear speed S, with effective stalled-torque-at-wheel T. Let our wheel radius be denoted r (we need not specify a value, as it is divided out later). Then (Wf * 2 * pi * r)/S = T/Ts. Let T/r = 40lb, i.e. the force to stall our wheel is precisely equal to the available friction force. Then we have (Wf * 2 * pi * r)/S = (40lb * r)/Ts, which yields S = (Wf * 2 * pi * r)*Ts/(40lb * r). Our wheel radius term cancels, leaving S = (Wf*2*pi*Ts)/(40lb).

If you plug in and calculate, you end up with ~25 feet per second, which is a factor of two off from what I got last time. For the life of me, I can't find where this calculation is wrong, though the result is much more surprising than what I had previously believed, and does not really mesh with my experience of wheel slippage while driving...

[yash101]

I do not know if this really helps, but my Vex Omnibot is able to climb walls (and end up flipping itself over) because it has a very high traction. To me, I think that omni wheels give a better traction than mecanum because it is pulling apart the carpet in the front and pushing it together in the back so it seems like the wheels would be a lot more grippy. Also, as I have noticed, omni is great for turning in place. This may be the size causing this, but I find that Omni is more responsive than mecanum! I can change the direction instantly and I won't have to wait for the rollers to stop coasting. However, this could be because of the robot size differences, and thus the inertia!

[AlecS]

Originally Posted by Oblarg

Say our drive is geared to a top linear speed S, with effective stalled-torque-at-wheel T. Let our wheel radius be denoted r (we need not specify a value, as it is divided out later). Then (Wf * 2 * pi * r)/S = T/Ts. Let T/r = 40lb, i.e. the force to stall our wheel is precisely equal to the available friction force. Then we have (Wf * 2 * pi * r)/S = (40lb * r)/Ts, which yields S = (Wf * 2 * pi * r)*Ts/(40lb * r). Our wheel radius term cancels, leaving S = (Wf*2*pi*Ts)/(40lb).

I think you may need to take a look at your derivation there, it's difficult to see where you're going.

It would probably be simpler to create a formula for the required gear ratio to slip the wheels at stall torque. (Gear ratio is lacking in your above formula). The formula would be as follows,

Assuming 4 wheels and 4 CIMs;

Desired drive force = (Gear Ratio * Stall torque)/Radius

Desired drive force/(stall torque * radius) = gear ratio

Once you have the required gear ratio you can then calculate the speed you would achieve.

Using the numbers you provided, this would give a gear ratio of about 4:1, with a top speed of ~23FPS, assuming 0 inefficiencies or losses. However, in the real world, with losses and FRC batteries, the maximum speed to slip the wheels is going to be substantially lower.

[Ether]

For one-CIM-per wheel:

amps=(5760*Istall*mu*V*W)/(pi*eff*Sfree*Tstall)

or

V=(pi*amps*eff*Sfree*Tstall)/(5760*Istall*mu*W)


where:

amps = current required to slip the wheel

Istall = CIM spec stall amps

Tstall = CIM spec stall oz_in

Sfree = CIM spec free RPM

mu = static coefficient

V = robot ft/sec speed at CIM free RPM

W = weight in lbs on the wheel

eff = drivetrain torque efficiency fraction

So, plugging in Oblarg's numbers:

amps=(5760*133*1.0*25*40)/(pi*1.0*5310*343.4) = 133.7

You're not going to get 133.7 amps in each of 4 motors, so if you gear for 25 ft/sec you definitely will not slip.


Let's try some more reasonable numbers and try again:

mu=1.0 V=12 W=37.5 eff=0.8

amps=(5760*133*1.0*12*37.5)/(pi*0.8*5310*343.4) = 75.2

Will the wheels slip? Well, 75 amps per motor times 4 motors is 300 amps. If your battery is strong your wiring is in good shape and the CIMs aren't too hot you might be able to push 300 amps thru the 4 CIMs.


Let's try V=10:

amps=(5760*133*1.0*10*37.5)/(pi*0.8*5310*343.4) = 62.7

OK, maybe that will slip.