53 replies to this topic

[MclarenF1]

After speaking with one of my professors at school about a tires' coefficient of friction and the causes of load sensitivity I began wondering why and how this happens. The equation F=muN holds true only in a physics book. And from that point of view as the normal force increases the coefficient of friction should also increase (to a point). Think of the road as a rough surface and the tire being pressed into the valleys of this surface. Effectively the surface area of the contact patch should increase to the point where all valleys are filled, then the coefficient of friction would remain constant. We all know that this is not the case. So my question is, what other factors are contributing to the coefficient of friction of the tire/road interface? Temperature, carcass construction, pressure, etc. From what I have read about this it seems that no one really knows why tires act the way they do. Now I don't think that we will be able to answer this complex problem on this discussion board, I am just curious as to what everyone else has heard/read about this subject. Thanks.

Erik

[MRC]

One theory is that as the tire's material gets pushes harder into the road asperities, that the effective hardness of the rubber goes up, within the asperity pockets. Don't know if this worth a damn, just a theory from an old prof. There is some information on metal's friction interactions through differing normal loads.
There are multiple tribological journals available. You might do a search at your university library. These journals might have some information for you.

[Pioneer]

Temperature. Its all about temperature.

Had a recent thread on this... probably still live if you look back a few weeks.

[Ali_G]

McLaren: Grip is caused by the interaction of electro magnetic fields of different atoms.

Different types of atoms have different fields and hence different abouts of grip levels.

In fact the tyres never acctually touches the ground. It hovers slightly above it at all times.

Niall

[MRC]

Ali_G, could you send a me a pound of whatever you are smoking?

[Ali_G]

Sorry, that should be Electric Fields.

Electrons whciha are negatice repel electrons of other atoms.

Niall

[BRIAN GLOVER]

Then, dont bogart that joint my friend, roll me another one, just like the other one.

I cried with laughter. Man.

I dont care who you call yourself, dont walk on the water while I'm fishing.

Originally posted by MRC
Ali_G, could you send a me a pound of whatever you are smoking?

[lateralforce]

Originally posted by MclarenF1
After speaking with one of my professors at school about a tires' coefficient of friction and the causes of load sensitivity I began wondering why and how this happens. The equation F=muN holds true only in a physics book. And from that point of view as the normal force increases the coefficient of friction should also increase (to a point). Think of the road as a rough surface and the tire being pressed into the valleys of this surface. Effectively the surface area of the contact patch should increase to the point where all valleys are filled, then the coefficient of friction would remain constant. We all know that this is not the case. So my question is, what other factors are contributing to the coefficient of friction of the tire/road interface? Temperature, carcass construction, pressure, etc. From what I have read about this it seems that no one really knows why tires act the way they do. Now I don't think that we will be able to answer this complex problem on this discussion board, I am just curious as to what everyone else has heard/read about this subject. Thanks.

Erik

I guess in static condition what you said should hold true. However, tires generally operate under rolling condition where the differential between tire rolling speed and its speed of travel, and also the direction of heading and direction of travel produces tractive and lateral forces respectively.

When a tire is rolling on the road, there are two kinds of mu's, one is peak mu, and another one is sliding mu.

These mu's are affected by vertical load, inflation pressure, carcass stiffness, surface friction and rolling speed.

Why the mu (traction force/load) doesn't increase with vertical load is because the peak and slide friction forces don't increase proportionally with load. Why? This has a lot to do with the complex dynamic interactions between rubber molecules and the surface aggregate. IMHO, due to the high non-linearity of rubber compounds even in the molecular level, the shear force generation is not in proportion to the increase of normal load.

This complexity is why I think there exist no mathematical model that can describe tire behavior adequately in order to completely replace the traditional process of building and testing tires for its force and moments properties.

[Aubwi]

Notice that the conventional model for frictional force on solid objects doesn't take into consideration what happens when one or more of the objects is actually breaking apart like a tire does, as it wears out and forms "marbles" on the track. That must put some kind of limit on the frictional force, which in turn would of course affect mu.

Actually what Ali said is technically correct, although its not of much use in practice in our massively macro-atomic world.

[DataFly]

Originally posted by MRC
Ali_G, could you send a me a pound of whatever you are smoking?

If you slam your hand as hard as you can against the desk, the atoms/molecules in your hand never actually touch those in the desk; the electron clouds around the atoms/molecules in each object repell each other with a force that strengthens as they approach each other. As your hand approaches a certain distance from the desk, it's downward force will approach the upward force of the electron cloud repulsion and will place your hand in equilibrium. In other words, your hand is floating above the desk, like the tire above the track, by a few millionths (or billionths - not exactly sure) of a meter.

And just to make sure that this post isn't entirely off topic...I think both Aubwi and lateralforce are correct.

[Toby Padfield]

DataFly,

The atomic level is closer to 10^-9 m than 10^-6 m. Most people that routinely deal with materials/chemicals/etc. on the atomic scale actually use the Angstrom unit, which is 10^-10 m.

I won't make any attempt to get back on topic...

Toby

[olschak]

Here's a very general approach to this matter. ( I studied Physics once so I know there's so much more I do not know about this ) The friction equation here is a definition rather than a law. Every definition comes along with a set of circumstances in which it is applied. The question everybody's asking is how does the friction change with a changing environment given that definition. The answer here is of course complex. But going back to the definition the underlying statement here is, that the coefficient does not change significantly with the contact patch or anything but is characteristic for the two materials. Put in easy words: say you're walking bare feet on the street the level of friction does not depend so much on temperature, the way you're walking or anything like that. It's really depending on the skin and the type of tarmac.
F1 of course goes somewhat beyond that. Still it is fairly ok to say given the tarmac, the tire and the force down ( that is the weight plus whatever you get from aerodynamics ) you have a certain traction. The idea behind the definition is that you cannot do any real tricks unless you change the tire compound or let it rain.

[Aubwi]

I don't think there's actually anything wrong with F = mu N, under any circumstances. But I think of mu as sort of a meaningless variable. If mu N gives you the wrong F, then you must simply assume that mu has changed. It sounds like some kind of cheap cop-out, but I think that's the way mu is meant to be used.

[david_martin]

Originally posted by Ali_G
McLaren: Grip is caused by the interaction of electro magnetic fields of different atoms.

Different types of atoms have different fields and hence different abouts of grip levels.

In fact the tyres never acctually touches the ground. It hovers slightly above it at all times.

Niall

I am going to work on the assumption that this was a rather lame attempt at an April fools joke, and then go and thank the folk at Jelsoft for the ignore list.

[DataFly]

Read what Aubwi, Toby Padfield, and I said again.

[Christiaan]

I'm going to think out aloud here,
About four years ago I read about Von Misses (excuse the spelling) stresses and the stress distribution over the contact area on two meshing cog wheels. The book was Mechanical Engineering Design by Shigley. The contact patch has an interesting stress distribution pattern which directcly affects the amount of power transmitted from one surface to the next without slippng. Add to that the effects of hysterisis and you can see that it gets even more complex, esp given that the hystresis charactoristics of rubber are dependant on temperature and frequency of cyclic loading. You can appreciatte that at high speeds spot on a tyre would be subjected to a cyclic load and maybe it never fully regains its shape after deflecting. This further complicates the tyre dynamics. I would actually be interested in building a model to determine the stress condition of the contact patch.

[DOHC]

I found part of this discussion exceedingly academic. First, we have to distinguish different scales, the microscopic, the mesoscopic and the macroscopic scales. If we are discussing tires and tarmac, you can forget about the microscopic/atomic discussions above. That belongs to a different branch of materials science, and contact mechanics is best discussed on the mesoscopic and macroscopic scales.

No, tires and hands don't "float above" a surface, basically for two reasons: first, the surfaces we discuss are nonsmooth already at the mesoscopic scale (so "above" has no clear meaning) and second, two objects are by definition in contact when they're within a range where contact forces are not negligible (so "float" has no clear meaning either; to say that a surface "floats above" another is just a sort of incantation, not very scientific as it has no explanatory value unless we are talking about hovercraft). There are several different kinds of contact forces, friction being one, adhesion (at the mesoscopic level) being another. Both depend on surface smoothness: smoother surface generally means less friction and more adhesion.

MclarenF1's view of modelling the rough surface with some "valleys" and Christiaan's meshing cogs are good models, and by considering the von Mises stresses it can get pretty sophisticated. But the friction also depends on "dirt" in the contact patch. Particles in the contact patch can increase or decrease friction. Then water. If it rains, friction is greatly reduced, because water forms a thin film that fills out the valleys in the contact patch, acting effectively as a lubricant which prevents too close contact between rubber and asphalt and therefore reduces friction. Note that this is a matter of what happens on the mesoscopic scale---I'm not talking about aquaplaning, which is a macroscopic, hydrodynamic phenomenon. The very idea of lubrication is thin film hydrodynamic friction reduction, and it's treated in a separate scientific discipline, tribology.

In all, contact mechanics is a rather active field of research, in particular when it comes to rubber and other polymers, which tend to have a very complex behaviour. Note that polymers are indeed "macromolecules", so it is only natural to choose models of rubber/asphalt contact that emphasize mesoscopic phenomena.

Hope this can be of some help.

[Aubwi]

Well, I thought the dicussion of electrostatic forces was meant to be a bit humorous. But thanks for the info.

What is the difference between friction and adhesion? Friction is caused by the bumps in the road digging into the tire, something like the meshing cogs? And what causes adhesion?

One other thing that's been bugging me is why does an overinflated tire produce less grip? Smaller contact patch, of course. But why less grip???

[david_martin]

Adhesion is really one of the meso-scale mechanisms which gives rise to the macroscopic phenomena we call friction. The adhesion theory of friction was first proposed by Bowden and Tabor at the Cavendish Laboratory in about 1940. It was initially proposed to explain metallic friction, but later extended to other types of friction couples. At the heart of the theory is that concept that the friction couple has a characteristic "adhesive" junction strength between contacting asperities in shear and an equivalent normal hardness. In metallic friction this shear strength was hypothesised to be due to pressure welding, in non-metallic cases to be caused by chemical bonding, dipole attractions and other molecular phenomena. The frictional traction is presumed to be proportional of the real area of contact, the adhesive junction shear strength, and the effective normal hardness of the contacting asperities. Mathematically, this can be expressed as



Equation 1 shows the relationship between the tangential force (S), the real contact area (alpha), and the adhesive junction shear strength (tau). Similarly Equation 2 shows the relationship between the normal force (P), the real contact area (alpha), and the normal hardness (p). These can be combined as shown in Equation 3 to yield the Amontons-Coulomb law most people are familiar with.

The adhesion theory suggests that the friction coefficient is observed to remain approximately constant in many cases because, in macroscopically elastic contact, the real area of contact should increase linearly with applied normal load. Hence the ratio S/P should remain constant, and therefore mu, by Equation 3. From Equation 3 it should also become apparent how deviations from Coulomb friction can occur - either the real contact area is non-linear with respect to the normal load or the ratio of the two properties tau and p is not contant. There are all sorts of ways that both scenarios occur in practice...

[Aubwi]

OK, so by equation 1, an overinflated tire produces less grip because of a smaller real area of contact. I suppose you could also say the same for an underinflated tire, since the sidewalls are bearing more than their fair share of the load.

What about a smaller/narrower tire? To get the contact patch flat on the ground under the same weight, a smaller tire would have to have a higher ideal pressure, and thus a smaller real area of contact. So I suppose that debunks the idea that tire size does not matter as far as grip is concerned.

For wheels made of wood, the real area of contact would not change significantly depending on wheel size. Only weight would increase the real area of contact, as the tiny bits of wood actually in contact with the road surface get smooshed.

This is all starting to make sense to me, but correct me if I'm wrong.

[lateralforce]

Let me give another shot No micro-macro-meso jumbo mumbo, I promise

Tire friction is dependent on contact area, contrary to the law of friction for other stuffs. So the more contact area you got, the more friction you have. However...... (there's always a 'however')

Tire does not make the most efficient use of available contact area. Theoretically, a smooth baloon will have more mu as more load is pressed on it, but for tires the contact area does not increase proportionally with the amount of normal load applied. Even though the force does gets build up, it doesn't build up in the same proportion as normal force. Therefore you don't get increasing mu as the load increases.

Also during cornering, as slip angle increases, the area in the footprint that is not contributing to work due to zero slippage (due to high normal force concentration) increases, therefore cornering power reaches a plateu after a certain point and saturates.

I read this in the book called "Tire Mechanics" published by the USDOT think. I'll check again..I have it at work.

[DOHC]

Originally posted by Aubwi
the sidewalls are bearing more than their fair share of the load

The sidewalls always carry the full load (the weight of the vehicle).

[DOHC]

david_martin,

You refer to the friction law as "Amontons-Coulomb law" and in another thread I saw it referred to as "Armonton's law". When I studied engineering physics it was only called "Coulomb's law". Ok, that was almost 30 years ago, but anyway who is this Armonton or Amonton?

Thanks, btw, for a good post above!

[david_martin]

Originally posted by DOHC
david_martin,

You refer to the friction law as "Amontons-Coulomb law" and in another thread I saw it referred to as "Armonton's law". When I studied engineering physics it was only called "Coulomb's law". Ok, that was almost 30 years ago, but anyway who is this Armonton or Amonton?

Guillame Amontons (1663-1705) was the first modern author to identify the classical friction relationship we know today, when he published "De la resistance causee dans les machines" in 1699. Coulomb expanded and refined the idea and posulated that friction was due both to the roughness of surfaces and attractive forces between them in 1785. In fact da Vinci had recognised the basic premise of friction as we know it in one of his notebooks from 1508, but it was not rediscovered from that source until well after Admontons and Coulomb had published their defining works.

[DOHC]

Thanks a lot David!

[Aubwi]

Originally posted by DOHC


The sidewalls always carry the full load (the weight of the vehicle).

Structurally, yes, but surely the normal force is spread pretty much evenly accross the full width of the contact patch.

[HSJ]

Originally posted by MRC
Ali_G, could you send a me a pound of whatever you are smoking?

Stuff that makes him right? He is right you know... Touch your hand. Go ahead! You can't do it, you only feel like you can.

[Ben]

Yeah - maybe he is. But we generally look a little more macroscopically when considering tyres in any practical sense.

Ben

[Orac]

Originally posted by DOHC


The sidewalls always carry the full load (the weight of the vehicle).

I've always thought that the air in the tyre carried the full load. Take the air away and the sidewalls don't carry much load at all.

[MclarenF1]

Originally posted by Orac


I've always thought that the air in the tyre carried the full load. Take the air away and the sidewalls don't carry much load at all.

What happens if you get rid of the sidewalls?;)

[Orac]

Originally posted by MclarenF1


What happens if you get rid of the sidewalls?

The air escapes, and is no longer able to sustain the weight of the car. Same as a puncture.

The sidewalls are obviously a structural component as well, in that they provide rigidity to the tyre carcass as a whole and give you somewhere to hang the tread face. Having said that, I know absolutely nothing about the technology surrounding tyre carcass construction.

I'll still back my understanding of the fact that it's the air that holds the car up. I've always been told that you can imagine the air in the tyre as a series of columns, each of them one inch square. If the pressure of the air in the tyre is 30psi (30 pounds per square inch) each column of air will support 30 pounds of weight.

[DOHC]

Come on guys, air never carries any load at all unless we're talking hovercraft.

The way a tire works is that the sidewalls carry the entire load, unless you have a puncture.

The role of the air is to add a positive (stretching) tension to the sidewall---as the sidewalls are "soft" they can't take much of a negative (compressing) tension without deforming. (This is why a puncture is called a "flat".) So you pump up the tire with air (or nitrogen) to a pressure high enough that the sidewalls are able to have a postive tension strong enough to still be positive when the mechanical load (weight of the car) is on. As a side benefit, you get a spring/damper action from the tire. The purpose of filling the tire with gas of high pressure is to preload the sidewalls so that they will be able to carry the mechanical load.

There are other materials that work in the opposite way, concrete for example. It can take huge negative (compressing) tension but very little positive (stretching) tension. So when you build bridges, you put in reinforcement bars, and preload those by stretching them. Once the concrete has reacted chemically and stabilized, you release the stretching load on the reinforcement bars. Then they contract and put a preload (compressing) tension on the concrete. In that way the concrete structure can take far greater loads without experiencing stretching tension that might cause fractures and jeopardize the entire structure.

Then, back to the wheels/tires. Look at a bicycle spoke wheel. The spokes are thin steel wire. They can take huge stretching loads but easily collapse under compression. So you fix the spokes geometrically in such a way that they are a bit "off" from being completely radial and preload them with a stretching tension. Just like the sidewalls in the tire, the spokes of course carry the entire load of bike and rider. In turn, that load is transmitted to the ground via the bike's tire sidewalls. The reason for using spokes in bike wheels is that you get an incredibly light yet very strong construction.

If you look at old wheels with perfectly radially aligned wooden spokes, those spokes are not preloaded. They are under compressing tension. Of course, those wheels are not very strong.

I hope this is understandable, I have tried to avoid too technical terms.

[kos]

DOHC, I'm sorry if I misunderstood you, but I really think that it's the air in the tyre that holds the weight of the car.

Air is acting as a spring that bears the load applied to wheel axle, and you need sidewalls because air is transferring applied pressure into all directions.

For your example with concrete - concrete is preloaded to avoid it's fracturing when construction is placed under stretching stress, but it's not the concrete that counteracts stretching forces. Stretching forces are counteracted by reinforcements bars, and preloading insures that under all operating circumstances, concrete will not be placed under positive stress and reinforcement bars will not be placed under negative stress.

Preloaded component of construction will always counteract only the same type (positive/negative) of stress with wich it was preloaded. Precompressed spring will not counteract stretching, in fact it's "trying" to stretch itself back to the normal length, so it's up to the component that precompressed the spring (and thus "pre-stretched" itself) to counteract stretching forces.

[DOHC]

Originally posted by kos
Stretching forces are counteracted by reinforcements bars, and preloading insures that under all operating circumstances, concrete will not be placed under positive stress

Exactly, that's what I said. And likewise, replace reinforcement bar -> air; concrete -> tire sidewalls; concrete never placed under positive stress -> tire sidewalls never placed under negative stress. Then you have the complete analogy.

Similarly, you have an analogy that the spokes on a bike's wheel behave like the tire sidewalls too. The spokes take the full load of bike + rider.

I don't know what you mean by saying that "the air in the tyre holds the weight of the car". What I'm saying about the sidewalls taking this load is that there is a force in the sidewalls, corresponding to exactly the mechanical load sustained by the tire + load from pressurized gas. Without the preloading, the tire can't carry its mechanical load and you have a "flat", puncture or not. Note that the mechanical load in the sidewalls is nonuniform around the wheel---it's mainly present at the bottom, at the point of contact, where you also see the rubber a bit deformed.

Air is acting as a spring that bears the load applied to wheel axle, and you need sidewalls because air is transferring applied pressure into all directions.

This quote is what is wrong in the argument that air takes the load. Think of a section straight through the wheel's diameter. Make the tire very wide (infinetly wide) to "get rid of the sidewalls". The gas pressure upwards on the rim is the same as the pressure exerted downwards onto the tread part of the tire. But there's the force of gravity (car's weight) on the rim, where does it go? It has to be transmitted through the side walls, because there is no other connection between rim and road.

[kos]

Ok, I now see your point sorry.

But I still think that analogy with reinforced concrete is not accurate.

As I understand, in reinforced concrete it's characteristics for negative stress are ~ the same as for pure concrete, and they are not dependant of the preloading. With the tyre, the maximum weight load the tyre can hold is proportional to the pressure of the air inside it.

And I think we can not speak about sidewall acting like a compressed rod between rim and the tread and air pressure "prestretching" it. Sidewall is more like a bended rod, so by applying pressure force to the inner side of this "bended rod" air is increasing bending stiffeness of the sidewall, or, more accurately, pressure of the air is counteracting force of the weight that tries to bend sidewall.

Excuse me for the lack of proper english terms...

[DOHC]

Yes, I agree that concrete is a bit different from rubber, I only meant that the idea of preloading structures is similar---it changes the material's behaviour entirely and it can operate in a regime that would otherwise be impossible.

[imaginesix]

Originally posted by DOHC
Come on guys, air never carries any load at all unless we're talking hovercraft.

I believe it is correct to state that air supports the load of the car.

By following the load path from the pavement to the wheel, we can state that obviously the ground supports the tire, since the tire is the only part of the car that comes in contact with it. Also, the tire supports the wheel, since the air pressure is evenly distributted 360deg around the rim of the wheel, thus it can not be transmitting load in any one direction over another.

The problem is that the tire sidewall is in tension all around the tire as previously stated, and since it is not 'hooked' to the wheel as in bicycle spokes, how can the sidewall prevent the wheel from simply falling to the ground? The answer of course is the pressurised air. The load path then becomes:

Ground -> Tire (primarily the tread) -> Pressurised air - > Tire (primarily at the sidewalls) -> Wheel.

Thus the the air supports 100% of the load, as does the tire.

It is unfair to say that just because the air pre-loads the tire sidewalls, it is not in itself a load-bearing element of the structure, just as it would be unfair to say that re-bar is not a load-bearing part of pre-stressed concrete. If such were the case, then the air/re-bar could be removed without collapsing the structure, which is clearly untrue.

[Aubwi]

It seems a bit academic to me, but maybe a better way to phrase the question would be "which part of the wheel transmits the normal force"? I think that when the tire is properly inflated, it's about equal between the sidewall and the air. Otherwise the contact patch would not be flat.

Overinflated: the center bulges out, so the sidewalls don't transmit as much force.

Underinflated: and the sidewalls are no longer in tension, and the center bulges inward, so the air is not transmitting as much force.

You can imagine a tire with extremely thick sidewalls. Take all the air out of the tire and the sidwalls are transmitting ALL the force, assuming those thick sidewalls are strong enough to keep the rim off the ground.

[Marco94]

You could ofcourse chose to ignore DOHC's explanation, but it's the only one that is correct!

[RDV]

DOHC has expressed it well ( re pressures & sidewall w/ stressed concrete analogies) , but I though this thread was about friction coeffs?

[imaginesix]

Originally posted by Marco94
You could ofcourse chose to ignore DOHC's explanation, but it's the only one that is correct!

I'm certainly not on very solid ground in my understanding of the mechanics of tire loading, but I have a big problem seeing how the air can simply be dismissed as a load-bearing element in such a structure.

I have posted my reasoning for this, and anyone is free to criticise and correct it. Of all the explanations given throughout the thread, I don't see any explaining away the importance of the air pressure.

[RDV]

... then consider tyre as an air-spring , with the sidewalls acting as the container... in that sense air does hold the whole shebang up , and structurally it is the air that maintains the shape that holds the the car off the ground. .......now can we go back to friction coefficients?.

[imaginesix]

Originally posted by RDV
.......now can we go back to friction coefficients?.

What's stopping you?

[RDV]

What's stopping you?

Touchee!!

[RDV]

Sorry about misspeling , actually french for a hit as in fencing , havent got right keyboard for accents...

but all kidding asside , if you need more discussion on tyre structures feel free to enquire , I work with them ( racing ones)

[Aubwi]

OK. Do racing tires typically have significantly more grip in either the longitudinal or lateral direction?

Another thing I've been wondering is how much the grip difference between the front and rear tires typically is.

[Crazy Canuck]

Originally posted by Aubwi
OK. Do racing tires typically have significantly more grip in either the longitudinal or lateral direction?

No, in pure lateral acceleration or pure logitudinal acceleration the grip level available from the tire is the same. The grip available from a tire can be depicted as a circle on a "g-g" diagram [x-axis=lat acc, y-axis=long acc]. This graph is sometimes called a Friction Circle.


As for friction co-efficients:

F1 (front) = 1.875
CART (rear) = 1.8
Eagle GT-S (shaved for racing) = 1.15
Eagle ZR (street) = 0.987

I got this information from Goodyear via SAE. It is old but should give a decentidea of what the friction coeffs are.

CC

[Ben]

My understanding is that race tyres (most tyres as well?) have a slightly higher peak friction coefficient in the longitudinal direction. Therefore the simplified representation of tyre force generation which is the 'friction circle' should be more acurately referred to as a 'friction ellipse'.

A paper from Leeds Uni on lap time simulation uses a model where the longitudinal force is 10% greater than the lateral force. I don't know how accurate this is though.

Ben

[RDV]

Another thing I've been wondering is how much the grip difference between the front and rear tires typically is

Question needs to be quantifyed, as the whole dynamic balance equation depends on wether the vehicle is front/rear/4wheel drive, tyre sizes and construction are same/different, weight distribuition , and geometrical effects as different cambers and toes.

Assuming a equaly cambered , identical tyres ,50/50% weigth distribuition , the mere fact that one is accelerating or braking will alter the vertical force on the tyre hence changing the radius of the friction "elipse" , (for indeed the friction circle is really an elipse) . Combined lateral and long induce quite different shapes, use of Gough diagrams give some description of each tyre performance .

Difficult question to answer, but in theory equal tyres with all other factors same would have same elipses for same Fz ..... but in real life all other factors tend to be diferent.

A solid rubber ( or other elastomer) tyre would have a friction "circle ", but a pneumatic tyre as we know it will be elliptical, 10% +/-3% delta lat/long is about ballpark on data measured on race cars ( am a bit more vague on production tyres as not much experience)

For comprehension of the elipse in Fy:z think of a tyre carcass as (in the case of a bias ply ) as a tread held in place by a transverse catenary (the cords) ; as these are loaded , because of the cylindrical shape of tyre, they have different geometrical deformation depending if lat acc or longi, if contact patch was on a sphere (structuraly ) we would have the circle.

Radials end up being even more complex as they are realy bias-belted , and not true radials at all.

[Crazy Canuck]

Originally posted by Ben
My understanding is that race tyres (most tyres as well?) have a slightly higher peak friction coefficient in the longitudinal direction. Therefore the simplified representation of tyre force generation which is the 'friction circle' should be more acurately referred to as a 'friction ellipse'.

A paper from Leeds Uni on lap time simulation uses a model where the longitudinal force is 10% greater than the lateral force. I don't know how accurate this is though.

Ben

Yes, I agree...upon further investigation I found that there is a difference in the longitudinal and lateral grip levels, but not too much. I'm speculating here but one would think that a tyre manufacturer could build a tyre with significatly higher grip in a particular direction. Although the only place i could see this being useful is on a drag strip!


CC