Mismatched wheels

[Caleb Sykes]

How bad is it to have wheels of different sizes that are all powered together without any reduction compensation?

For example, say we have a 6WD drop-center drive with a 9" center wheel and 8" diameter wheels in the front and back.

My speculation would be that, as long as most of the weight is on the center wheels, you would lose the static friction on the front and back wheels, and have kinetic friction instead. The reverse would be true if most of the weight were on the front/back wheels. Since this arrangement would cause the wheels to slip more, wear on the wheels would also be a bigger concern than it would be for matched wheels.

Any theory or practical experience would be appreciated.

[piersklein]

This would be a very bad idea. The inner wheel would spin much much faster than the outsie wheels because Linear Speed=angular speed * radius. In this case, either the middle wheel would slip and cause a lot of friction or the outer wheels would cause friction from not spinning fast enough

[T3_1565]

You would only be driving with the 2 biggest wheels. The smaller wheels would not help the drive. If I'm reading what your asking correctly of course.

[Jared Russell]

This only works as long as the difference in diameters of the driven wheels is very small. I've seen plenty of cases where a team achieved a 1/8"-3/16" center drop by having all axles in line and using a slightly larger custom center wheel(s) than on the corners.

A 1" difference in diameter is much larger than I would recommend for this purpose.

[Kevin Sevcik]

You'd kill your battery pretty quickly. Driving straight forward, the motors would have to provide enough torque to overcome the static or kinetic friction of the wheels you're slipping. Even without much weight on them, the outer wheels will take a decent amount of torque to slip like that.

I think the reason it works in the case of slight differences in diameter is because there's enough give in the carpet-tread system, that the tread and carpet just elastically deform to accommodate the differential as they roll from starting contact to full contact to releasing contact. The incoming tread and carpet is always undeformed, so it's not like there's any build up of differential. Elastically deforming the tread and carpet for that little bit still requires some force, but it's almost certainly less force than sliding friction.

[Ken Streeter]

Originally Posted by Jared Russell

A 1" difference in diameter is much larger than I would recommend for this purpose.

Agreed.

However, we've done this before, with a not-too-difficult proviso, that can make it work perfectly:

The trick is to change the belting / chain / gears (whatever it is that connects your center axle to the outer axles) so that the linear speed of the wheels is the same. This is a matter of adjusting the pulley / sprocket / gear sizes for the different-sized wheels. Essentially, you'll have a different reduction ratio to the 9" wheels than you would to the 8" wheels. In theory this is easy.

In reality, it's a little trickier (but not too much), in that you need to source (or make) pulley / sprocket / gear sizes that give the desired ratios. However, that's not an unsolvable problem.

[Caleb Sykes]

Thanks for the replies, I have heard of teams with slightly mismatched wheels, but nothing as extreme as this. I thought about it a little more and realized that the slipping of the wheels will cause them to fight against each other badly. I made an excel spreadsheet that shows how traction varies as a function of the robot's CoM.

What confused me was that this arrangement really has no negative impact if the CoM stays directly over the center wheel at all times. However, as the robot tilts one way or the other, the wheels really start to fight each other, and if the CoM is in the perfectly wrong spot, the robot will only be able to provide (mu_s - mu_k)/2 times as much force as it's maximum.

Hopefully the excel spreadsheet makes sense, let me know if there are any questions.

[StephenNutt]

I would connect the 8" wheels with a smaller sprocket/pulley. Simply divide the number of teeth of the sprocket/pulley on the 9" wheel by 9 and then multiple by 8. So if the sprocket/pulley on the centre wheel is 36 teeth, use 32 teeth on the 8" wheels.

[MoistRobot]

Originally Posted by StephenNutt

I would connect the 8" wheels with a smaller sprocket/pulley. Simply divide the number of teeth of the sprocket/pulley on the 9" wheel by 9 and then multiple by 8. So if the sprocket/pulley on the centre wheel is 36 teeth, use 32 teeth on the 8" wheels.

I think you mean that the other way around- the 8" wheels need to spin faster than the 9" to maintain the same linear speed as the 9"