Omni vs Mecanum 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.

[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.

[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.

[Oblarg]

Originally Posted by AlecS

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

Equate force required to stall your wheel to maximum friction force on wheel.

Equate ratio of free speed of CIM to free speed of wheel to the ratio between the stall torque of the motor and the stall-torque of the wheel.

Relate stall-torque of wheel to force needed to stall the wheel.

Plug in and solve for free speed of the wheel.

The factor of 2*pi comes from the fact that we are specifying a linear velocity, not a rotational one.

[JVN]

Originally Posted by 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!

Hi Devyash,
During design it is important to know not only "what" occurs, but "why." When you mention anecdotal things that you've seen, you're mentioning "what" happens, but not "why."

Often, people don't even see the truth of what is going on. People have an idea in their head of what is supposed to happen, and as such their observations fill in the blanks to make it true. (This is called confirmation bias.)

We can't always trust ourselves to actually make good observations. This is another reason for truly trying to understand "why" things behave like they do.

I've noticed you've made a few posts where you make blanket statements based on anecdotal observations. You need to be careful when you do this. It might be better to really "dig in" to the physics, and try to get to the bottom of how things really work. It is REALLY cool, I promise you. It will change your view of the world.

If you're interested in adding some depth to your expertise...
This is a cool thread where some really sharp people are debating the vector physics behind Mecanum Wheels vs Omni-Directional Wheels. You should "search all posts" by Ether, and read some of the things he's linked and discussed.

In addition, feel free to ask lots of questions of the veteran designers on this forum; they'll be happy to help you out (since they probably started in the same place you did).

Regards,
John

[AlecS]

Ah, I think I see why I was confused, The initial equation was flipped.
Intial Eq:

(Wf * 2 * pi * r)/S = (40lb * r)/Ts

Eq:

S/(Wf) = (40lb * r)/Ts

This puts actual speed above free speed, and actual torque above stall torque, like you described. The previous version was not giving useful numbers, this one does.

You also need to do some unit conversion there,

If use Stall Torque in in/lbs (21.5);

S = (Wf/(40lb/Ts))*2pi)/60/12

This equation gives ~23FPS, same as what I got above. This number would appear correct, assuming 0 losses and perfect current supply.

[Oblarg]

Originally Posted by AlecS

Ah, I think I see why I was confused, The initial equation was flipped.
Intial Eq:

(Wf * 2 * pi * r)/S = (40lb * r)/Ts

Eq:

S/(Wf * 2 * pi * r) = (40lb * r)/Ts

This puts actual speed above free speed, and actual torque above stall torque, like you described. The previous version was not giving useful numbers, this one does.

You also need to do some unit conversion there,

If use Stall Torque in in/lbs (21.5);

((S/60)/12)/Wf*2pi = 40lbs/Ts

This equation gives ~23FPS, same as what I got above. This number would appear correct, assuming 0 losses and perfect current supply.

I always let WolframAlpha handle the units for me, because I'm a lazy bum

Well, looks like I was wrong the first time through about a week ago. I figure I probably had a radius in one part and a diameter somewhere else, or similar.

[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.

[Oblarg]

Originally Posted by 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. Unless your battery and wiring are in top shape, you're not going to get 75 amps per motor.


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.

Thanks for this. The calculation I did (obviously) placed no constraints on current draw on the motor.

[Ether]

Originally Posted by Oblarg

Thanks for this.

You're welcome.

Attached is the derivation of the formulas.