Dynamic and static friction on tires 2

[Tunalover]

In physics and statics they always told us that the friction force F is given by F=µN where N=the force normal to the contact surface and µ=the coefficient of friction. The friction force is independent of the contact surface area. This seems over-simplistic because intuition and observation show that wider tires on cars provide better stopping and starting. What does the friction force REALLY depend on? I know that a bicycle tire width on a car will not stop the car faster than the a wider tire! Somebody correct me if I'm wrong about this.



Tunalover

 

[chicopee]

Larger wheels may mean larger rims, larger bearings leading to heavier vehicles all of which will contribute to greater rolling resistance eventho. the coefficient of friction under static and dynamic conditions remain constant.

 

[GregLocock]

To properly answer this would require a book. Your assertion " wider tires on cars provide better stopping and starting" is wrong in detail.

As an example, using a proprietary database of several thousand tires and a tire model of limited ability, a mean tire (Z=0) 26555R17 on an 8.5 rim has a peak longitudinal mu of 1.05, at an FZ of 6750 N, and an OD of 773 mm. A mean tire 24570r17 on 7.5 rims has an OD of 775, and a peak longitudinal mu of 1.062 at the same Fz.

The fundamental reason is that for longitudinal traction a longer narrower contact patch offers a more progressive slip velocity distribution at optimum slip.

We have discussed the mechanisms of tire grip before, I suggest you search this forum.








Cheers

Greg Locock


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[GregLocock]

Incidentally, although we use 'mu', we actually modify mu for different Fz, so the old equation you were taught at school would be better represented as Fx/Fz is approximately constant for small changes in Fz.

Cheers

Greg Locock


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[JStephen]

"What does the force REALLY depend on?"
My guess: Normal pressure, sliding/rolling speed if not static, composition, temperature and surface roughness of both surfaces, cleanliness of the surfaces, nature and extent of any contaminants on the surface.

The physics equation is usually presented as being for "dry" friction. For something like tires, you're intentionally trying to come up with increased friction, so the farther you get from that "ideal" case, the better.

 

[BrianPetersen]

Wider tires will not provide more grip under circumstances such as travelling at higher speeds through a puddle (the wides will want to hydroplane while the skinnies will want to cut through to what's underneath), nor when trying to go through snow (the wides will want to push a "wedge" of snow ahead of them and get you stuck, where the skinnies will cut through and be more able to push the snow to either side).

If your car has typical compliant street suspension which allows body roll, and independent suspension that keeps the wheels more-or-less upright to the body, those wheels will not be upright to the road under body-roll conditions. Skinnies and tires with a rounded tread profile might not be bothered too much by this. Wide, low-profile tires will try to ride up on one corner of the tread, so you are not getting the contact patch that you think you have. Many production cars do this intentionally with the front suspension as one of many factors to ensure that the car understeers.

This situation is far from a simple one.

 

[drawoh]

To properly answer this would require a book. Your assertion " wider tires on cars provide better stopping and starting" is wrong in detail.

...

I was told by my statics teacher in college that Newton's laws of friction do not apply to tires. He claimed that they vulcanize with the pavement, which is why tires can exceed one 1g acceleration. Does this sound intelligent?

I just Googled "aluminium friction". According to The Engineering Toolbox, aluminium on aluminium has a friction coefficient of 1.05 to 1.35.

--
JHG

 

[3DDave]

Vulcanization is cross-linking within the rubber, so probably not.

http://en.wikipedia.org/wiki/Vulcanization

Rubber does mechanically interlock with rough surfaces; the softer the rubber the more it can deform to interlock. Interlocking is what allows steel-on-steel parts to have very high levels of resistance to motion, as screw thread parts demonstrate.

The high friction of aluminum on aluminum is due to galling, where the oxides of aluminum are scrapped off and the crystals in one piece bond with the crystals in the other pieced. It doesn't apply to anodized aluminum, for example. It's a form of welding.

 

[Tunalover]

There MUST be at least some simple, generalities that can be made relating contact area and friction. What are they?

Tunalover

 

[GregLocock]

The more real estate you have to work with the easier it becomes to manage the compromise between (wet and dry)*( lateral and longitudinal) grip, weight, rolling resistance, cornering stiffness, breakaway characteristic, life, cold weather behavior, cost, ride comfort,noise, and probably half a dozen things I've forgotten and a dozen i don't know about.

As such in general a larger contact patch will tend to have a higher peak friction, because a softer compound can be used, because the shear stresses are lower, for a given life. But a wider tire doesn't guarantee a larger contact patch, it just gets shorter.











Cheers

Greg Locock


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[LittleInch]

You've already stated it in your OP - " The friction force is independent of the contact surface area."

The problem is that in going from a thinner tyre to a wider tyre, there are many things going on which affect friction - tyre compound, tyre deformation, tyre deflection (especially if going around corners) and all the things greg mentions.

If you try to build a tyre to handle the same load, but at 25% of the width / contact area, you will have a completely different type of tyre and characteristic, so the width issue is not simple to compare like with like.

Tyre grip or performance varies hugely with surface, dryness, temperature - so many variable and issues that there is no "simple" generalities.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[GregLocock]

I've remembered a way of exploring it yourself. You really need to buy the book

http://www.amazon.com/Tire-Vehicle-Dynamics-Third-Edition/dp/0080970168

, but here is a contact patch modelling program, treadsim, which produces the sort of plot that I spend all day working with

http://greglocock.webs.com/vehicledynamics.htm

you'll need matlab or octave



Cheers

Greg Locock


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[JStephen]

"This seems over-simplistic because..."
"There MUST be at least some simple, generalities that can be made relating contact area and friction. What are they?"

You can't have it both ways!

It might help a lot to tell us what it actually is that you're actually trying to do with this information. As it is, you basically have an "Explain friction" post, which can generate a lot of answers without giving you the information you're after.

 

[JackAction]

Tires do not obey «pure» friction law. I like this answer found on another forum, talking about a coefficient of adherence instead of a coefficient of friction. It does the simplest résumé I can think of to explain why contact area matters with tires:

Coefficient of friction is almost fixed.
Coefficient of adherence is not.

Coefficient of friction is a local property.
It has basically two values: a standing value, good when there is no sliding between surface, and a sliding value.
Transition between these two is not totally abrupt, but it is quite sharp anyway.

When a wheel is moving, it will touch the road through an area of a certain extension, which is called contact patch. For several reasons, too long to be discussed now, speeds inside the contact patch are different from point to point.
Some areas of the contact patch will not move relative to the road surface; they will then experience the standing friction coefficient.
Other areas will slide, and will esperience the sliding friction coefficient.

The force a wheel is able to exert depends on the distribution of these two different coefficients and on the distribution of load in the contact patch.

Dividing this force by the total load you have a coefficient of adherence, which varies a lot with speed because speed causes changes in load distribution and in the distribution of friction coefficients inside the patch.

These phenomena are not exclusive of rubber: railway wheels will behave in exactly the same way.

Other more detailed answers attempting to illustrate the phenomena can be found on the following pages:

[URL unfurl="true"]http://astro.physics.sc.edu/~rjones/phys101/tirefriction.html[/url]
[URL unfurl="true"]http://newton.dep.anl.gov/newton/askasci/1993/physics/PHY2.HTM[/url]
[URL unfurl="true"]http://insideracingtechnology.com/tirebkexerpt1.htm[/url]

 

[Nereth1]

Note two key issues:

1) There is differing normal force on differing subsections of the tyres contact patch. There are also differing shear forces throughout the contact patch to be transmitted through friction, because rubber is extremely flexible compared to the stiffness of the static friction bond and the ground. Therefore, parts of the contact patch will slip as they have less normal load and more shear forces. So the contact patch is partially static and partially dynamic - this, amongst other more difficult for me to explain reasons, really ruins any simple attempt at a single friction coefficient.

2) Note that there is no rule stating things cannot have a coefficient of friction greater than 1. Most interfaces don't but the 1:1 ration is just arbitrary. I recall a post on this very forum years ago describing an experiment to convince people of such involving a piece of some rubber or another on an inclined plane of glass, and the rubber didn't slip until well past 45 degrees.

 

[gruntguru]

I think this is the best answer to your question. I am sure you are not asking about all the dynamic characteristics of tyres - just the relationship between mu and tyre size.

"As such in general a larger contact patch will tend to have a higher peak friction, because a softer compound can be used, because the shear stresses are lower, for a given life." (GregLocock)

Engineering is the art of creating things you need, from things you can get.

 

[Tmoose]

D=1/2 at^2
T^2 = 2d\a
T = ( 1320 ft / 32.2 ft/sec^2)^.5
T = (41)^.5 seconds
T = 6.4 seconds = the quickest any vehicle will traverse the quarter mile driven thru its tires if max rubber u = 1.0. (and neglecting aero drag)

That was sometimes heard back in the 50s and 60s.

It took the Top Fuel dragsters until the 70s to defy high school physics and exceed 1 g average acceleration for 1320 feet

 

[GregLocock]

Shrugs, i have measured data for the ratio of wheel torque to vertical force on real production tires during ABS stops and the front tires regularly exceed a mu of 1.1, and I have seen 1.2

Perhaps the drag boys had suspension problems, or their tire pressures were too low.

Cheers

Greg Locock


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[Lou Scannon]

Keep in mind, for a dragster acclerating down the quarter mile, the available thrust to the tire may not always be limited by mu, all the way along the 1/4 mile, so an average acceleration of 1 g for 6.4 implies somewhat higher peak accelerations.

"Schiefgehen will, was schiefgehen kann" - das Murphygesetz

 

[JStephen]

On a dragster, they have a big wing over the rear tires specifically to generate downward force for additional traction. Who knows what the normal force actually is?