Spherical wheels

[joseff]

I'm with LifeSpitter here, to apply any sort of force, a friction device (ie tire) must slip. The greater the force, the greater the slip. Prime example: traction control system and ABS. Both regulate slip such that it is at its maximum value possible, without skidding. An example of a slip-minimization system would be a "granny" traction control, say, in a Toyota Camry. Not very high performance.

you can displace the wheel laterally or longitudinally: the "flat spot" or contact patch does not have to be located under the axis of the weight forces of the car.

Uh, no way. The car's weight vector has to pass within the boundaries defined by the tire contact patches, or the car tips over. Maybe you're referring to <i>pneumatic trail</i> instead?

...lateral forces at the tire do not act at the center of the contact patch, but at a distance behind the nominal contact patch. This distance is called the pneumatic trail and varies with speed, load, steer angle, surface, tire type, tire pressure and time. A good starting point for this is 30 mm behind the nominal contact patch.

Quoted from here

[Ciro Pabón]

We could " expand " the contact patch of a spherical tire by filling it with a light liquid polymer foam that would respond in density to an electrostatic charge... Sorry--- I did not intend to detract attention from the technical depth of the conversation--up too early and just too many cups of strong coffee--and too much time on my hands. Thank you all for your indulgence.

Actually, Carlos, an F1 tire works exactly that way: the lowest density material "melts" away and adhere to the surface. This happens in response to temperature, not to electrostatic charge, so this is what you try to control. If you don't believe this (I know I wouldn't! ), this is the theory and here is an explanation. Please, read the explanation: you'll be amazed.

I like to think that the work of Mr. Persson is behind the success of Bridgestone tyres and helped somehow Ferrari since 2001, when he published, but it could be wishful thinking. Anyway, I like to post his photo, because he solved the old mistery of why a tire grips the road and the mistery behind the friction "coefficient" so here it goes again (sorry...):



An F1 tire is not degraded as long-distance tires are: it is "sucked dry". Keep the colombian coffee up!

As for joseff, I am sorry, master, but I still don't see the light. I am still confused, looking at the diagram I made. As you displace the wheel to the left, as I tried to draw on the right of the image, you are exerting a greater force on the right side of the car.

This component of lateral and vertical force, that I also tried to draw with the three arrows in the image, is what "falls" on the center of the contact patch and precisely, what makes the car move laterally. The wheel can spin around an axis perpendicular to this force without problems, at least in the Jhon Hopkins design I shown.

In my confused state of mind, I try to make this comparison: imagine a car with four wheels you can change the camber only (without changing the direction of the wheel as they point straight ahead only). Would this car turn? And... what is the sound of one hand clapping?

[joseff]

One more thing just popped into my head: in a conventional tire, you're pretty much guaranteed a radially-even wear. Even if the camber setup is all wrong, you still end up with a uniformly round tire. You can't do this on a spherical tire.

Imagine going hundreds of km on a highway going straight south for example. Given that the tires' orientation stays constant, you'll end up with a shallow "ring". Yes, even given curves on the highway.

[Ciro Pabón]

One more thing just popped into my head: in a conventional tire, you're pretty much guaranteed a radially-even wear. Even if the camber setup is all wrong, you still end up with a uniformly round tire. You can't do this on a spherical tire.

Imagine going hundreds of km on a highway going straight south for example. Given that the tires' orientation stays constant, you'll end up with a shallow "ring". Yes, even given curves on the highway.

C'mon joseff... think again. What wheels could be easy to "rotate" to allow an uniform wear than the "proposed" "Jhon-Hopkins-type-maglev-wheels"? (we need a shorter name). Actually, you can use the entire surface around the wheel, wich means you use all the material in it, "sidewalls" (to say it somehow) included.

You do not need a wrench, all you have to do is to instruct the computer to rotate it evenly. You go the first 100 kilometers on one thread, rotate the wheel slightly and there you go: another new thread (almost new: they intersect, I know).

You would end your southbound journey with a tyre that has less wear than the "conventional" ones (hey, go South, by all means! It is "hotter" here! ).

In fact, you could extrapolate this even more: the computer could "balance" the wheel by making it rotate around the appropriate axis every such number of kilometers, for the rubber to wear until balance is restored.

You could even "flat-spot" the wheel and then use other "threads" around the periphery that are not "flat-spotted".

The haptic interface example I showed (one hundred posts ago) is a demonstration of the degree of accuracy with which you can "feel" and compensate or influence what the magnetic spherical wheel is doing while it rotates.

Next argument, please.

What I find hard to imagine is the "wet" design: how in heaven can you move the water sideways with a wheel that can rotate around any axis? Mmmmm.... perhaps you can reserve a "thread" with a wet design and rotate the wheel appropriately to use it when it rains.

[Guest]

In my confused state of mind, I try to make this comparison: imagine a car with four wheels you can change the camber only (without changing the direction of the wheel as they point straight ahead only). Would this car turn?

Well heres a good one....

O.K. as far as I understand tire dynamics, this car would NOT turn, but it would not travel exactly in the direction parallel to the tire heading. The vehicle would travel one or two degrees (maybe even less than that) to the left or right (depending on camber bias) of the tire heading. This is simply because when you add a camber angle to a tire, you create a camber thrust. If all four tires are cambered equally then the vehicle still won't turn. Maybe if you cambered the front tires only (ex: for a left hand turn you would need left-front tire camber+ and right-front tire camber-) then the vehicle could "turn" with all four wheels pointing forward. I put "turn" in quotations because the radius of this turn would be so freakin enourmous you might as well be going straight.

So, now that we know the effects of camber thrust, even if you could drastically change camber angle on a spherical wheel, the tire would still need to have a slip angle in order to generate significant lateral force. Camber helps, but not enough to turn the vehicle by itself.

As a side note, camber thrust itself is generated by a slipping of the tire near the rear of the contact patch, and by the sidewall stiffness of the carcass. Basically, if you tilt a tire and roll it, the path that a single particle on the tire surface is trying to travel through is not a straight line relative to the ground. It is actually a curve. The friction between tire and ground opposes this hypothetical particle's tendancy to travel in a curve and forces it to travel in a straight line through the majority of the contact patch. This results in a lateral force on the tire sidewall, ala camber thrust.

As a second side note, this is really difficult to explain without a drawing,... so I hope I haven't confused everyone.

[Ciro Pabón]

Hey, thanks. Very clear for me. Now I KNOW you are right: I've tried with an old gyroscope frame that allows me to simulate three-degrees of freedom. Sorry, not until I can "feel" it, I understand it. I am convinced now that there is no escape to slip by "twisting" some part of the tire. Is the only way to move the car laterally.

It remains the essential difference between holonomic and non-holonomic systems to differentiate spherical wheels from cylindrical ones.

What about the sound of the one hand?

[taleed]

Audi RSQ
[img]

Sweet car and interesting topic. I never knew...

[Ciro Pabón]

Sorry to bring this subject again, courtesy of ZenTM, at GP.com. The car at the top left of link is the one I couldn't find previously (the "huge wheel from bottom to top of cabin Peugeot design" that I stated somewhere I saw last year):

Peugeot Design Contest


Perhaps this thread wasn't as useless as I thought...

[DaveKillens]

Actually, I was wondering why the entire chassis isn't one huge vertical wing. With spherical wheeels, things like left, right, forward, and backward are only relative. The car could just as easily go sideways with the chassis as fore and aft. Not imagine that with a vertical wing, you can turn it relative to the direction of travel to achieve lift in the desired direction.

[jackhookss]

Rules are made to be broken

[Jersey Tom]

Not in physics they aren't.

[Belatti]

Not in physics they aren't.

=D>

[riff_raff]

... to apply any sort of force, a friction device (ie tire) must slip.

Technically, there's difference between friction and traction. I would argue that a tire contact is more akin to traction than friction. Friction is limited by force. Traction is limited by power transfer. The grip of a tire is due to traction and not friction.

riff_raff

[PlatinumZealot]

What is traction? You know I did a course on that and I didn't learn a thing.. (didn't like the lecturer so I never went to the classes).

I know there was some friction and something with ground penetration with some cone thing. There was also some slip involved. You are right Riff-raff, the driving wheels made a difference to the overall traction of the vehicle.. I don't remember any more though.. heheh

[Jersey Tom]

A large component (but not the only component) of rubber "grip" is from hysteresis and energy loss... as opposed to the classic 'friction' taught for most 'rigid' materials.