Must a tire slip to generate force?

[Shredcheddar]

Hi, I'm a new poster but have been lurking for a while. I value the wealth of knowledge on this forum, so I feel like it's the appropriate place for such a specific and academic question as the following.

I've been studying tire behavior, and every time I've come across a longitudinal slip percentage versus driving force curve, the intercept is always the origin (0,0) and the initial slope is always linear.

Here is an example curve, from and excerpt of "The Racing and High-Performance Tire":


In other words, the implication is that a tire MUST have some finite velocity difference between the tread and the ground to generate any force. I know racing is really only concerned with high forces, but when thinking about very small driving or braking forces, it seems reasonable to imagine a condition at the tire/road interface where no velocity difference exists.

Think of a street car slowly pulling away from a stop sign. It's not scientific, but the way my mind's eye explains it, the elasticity of the tire would seem to prevent any velocity difference at the interface for very small driving torques.

So are there any studies or conclusions about this? Is there any sort of argument for the idea that a tire must have a slip velocity to generate even minute longitudinal forces? Certainly, it could be that since the generated forces are minute, the slip velocity could be equally minute (yet still exist). I'm just looking for some sort of physical argument to support such an implication.

[Jersey Tom]

Couple things...

The intercept won't exactly be at (0,0). Not for longitudinal nor lateral force. It will always be a little off... sometimes significantly.

Don't get the slip term confused with sliding. In the linear range the footprint is more or less stuck to the ground. When drive or brake torque is applied, the leading edge of the footprint decelerates, travels at some velocity through the footprint, and then accelerates back out.

At zero "slip" the tire is coasting. Free rolling. With your example of a car slowly pulling away, that little bit of drive torque still causes the tire to "wind up" just a little, not even enough to visually perceive, and transmit torque to the ground.

Once the footprint enters the "sliding" range where you've exceeded the compound's grip capacity, THEN the rubber is really being scrubbed against the pavement.

And don't be fooled into thinking racing is all about high force / limit behavior!

[Ian P.]

To sumarize T J (or is it J T??)
I think the answer to your question is ... yes.

I P

[Roland Ehnström]

Definately yes, a tire must slip to produce force, that's simply the way it works. It's true for any tire, even the ones on your bicycle.

By the way, the graph for side-force is very similar to the graph for driving force above, and studying it explains why some drivers can go faster through a bend than others. The best drivers can ride the very top of the slip-curve through the whole bend, with all four tires producing maximum side-force for the maximum amount of time in the bend. Slower drivers will slide too much or too little, meaning their tires don't produce the maximum amount of side-force, meaning they have to slow down more to make the bend.

[NDR008]

Yup, tyre slip is also what lets a motogp bike lean in so hard, if you had to calculate the forces in primitive methods of frictional forces, there is no way a motogp bike could lean the angles they do. It is the forces generated from slip ratios that produces that grip - which could be interpreted (wrongly) as a coefficient of friction greater than 1.

You'll have different slip-ratio graphs for loading values too.

[Shredcheddar]

Good responses, guys, thank you. Jersey Tom, when you say that "there is always and offset," I assume you are referring to the tire construction effects such as ply-steer and conicity (for the lateral force curve, but I assume it's similar in the longitudinal direction)?

My question is partially based on my poor ability to visualize the necessity for slip in the longitudinal direction. Slip angles are more intuitively obvious - they are something you can both see as well as feel through the steering wheel. With longitudinal slip versus "zero slip" (free rolling as JT mentioned), the only difference is an applied torque, which is usually not as obvious visually unless the tire is operating at high levels of slip.

So I can accept that a tire must always be in slip to generate force (even if it does not appear so), but JT could you further explain the mechanism you mentioned by which a tire "stuck to the ground" generates a slip velocity? Are there localized areas of slip velocity that are associated with (or produced as a consequence of) the tire's elastic deformation (you called this "wind up")?

So if we view the carcass winding up while the footprint lags behind due to the resistive force generated at the ground, is this an acceptable way to mentally picture the slip velocity at lower applied torques?

NDR008: I was under the impression that bikes corner by a considerable combination of camber thrust and slip-generated force. Looking at the steer angle of the front tire (very little!), there is no way the required cornering force could be generated by slip angle alone. So certainly the angles that they lean at are both indicative of the cornering forces involved, as well as a part of their generation! At least that is how I understand it.

And there has been a lot said about tire friction coefficients reaching values in excess of 1. I'm not sure if you were talking about how the % slip graph could be misleading or what you meant by that? Anyways, I have keen interest in the Hallum tread momentum theory that Ciro has referred to on this forum... I simply need to get my hands on some substantial material.

[Jersey Tom]

You can get tire friction coefficients above 3 (under the right conditions) without even getting into tread momentum.

Regarding longitudinal slip.. when I refer to the tire being "stuck to the ground" that's probably a bit misleading. What I mean to say is the friction level of the compound hasn't been exceeded, or anywhere near, and its not being spun against the pavement at high speed. There is going to be local "slip."

Long story short, yes, you do need slip to generate force (with the exception of rolling resistance, plysteer, camber thrust, etc). It's just so small it's not perceivable. Even with steering and slip angle.. keep in mind a few tenths of a degree out of alignment and your car will pull significantly to one direction.

Or if you have high performance tires on and you're cruising on an on-ramp, and put in say 15 degrees of steer to get you through. Steering ratio is going to be probably 15:1 anyway, at the very least. Only 1 degree of slip angle.

[xpensive]

When being far from an expert in this area, I think the graph on the left above is in a way an example of what people in my line of work sometimes refers to as static- and dynamic friction, or "stick-slip" if you wish.

And way beside the point, I cannot help noting how amazingly similar the graph is to the typical characteristics of steel, stress versus strain. First there's a linear part of pure elastic elongation, then yield point, whereafter the material "floats" at a lower level of stress.
Except for the cold-hardening part up to rupture of course.

[Conceptual]

So, a tyre standing on top of a hill has slip forces when the wind blows it enough to make it roll the rest of the way down?

I guess this brings me to a question about the nature of friction. Is it opposing forces, or is it electromagnetic on a molecular level? Like the 2006 F1 tyres, molecular velcro...

[xpensive]

Be careful not to confuse friction with inertia now, but there is always an element of rolling-friction between the tyre and the surface.

[Shredcheddar]

So, a tyre standing on top of a hill has slip forces when the wind blows it enough to make it roll the rest of the way down?

Hmm. I'd have to classify the forces on the tires as purely rolling resistance and rotating friction forces. The defining thing about slip is that it is a response to an applied torque on the wheel/tire.

So when wind blows the car down the hill, there is an applied force on the car to make it accelerate from its static position, but the force is not applied at the road surface. The force is applied at the aerodynamic center of pressure of the plane normal to the wind direction, and the component of this force in the direction of the free-rolling tire accelerates the vehicle in a straight line.

So the force reaction on the tire represents the sum of the rolling resistance and the force required to overcome static friction. But it is important to note that there is no applied torque at the wheel.

You can think of applied torque as a torque that must be reacted through the chassis. This is because it is applied by an object that is mounted to the chassis (brake caliper, drive axle).

Although technically the friction force in the wind example involves a reaction at the wheel center, and there is still a moment present. This moment will be reacted by friction about the wheel hub, I would think (the actual distribution of the forces involved is indeterminate). So technically those would be reacted through the chassis as well. But those are obviously not what is being referred to as an applied torque.

I guess this brings me to a question about the nature of friction. Is it opposing forces, or is it electromagnetic on a molecular level? Like the 2006 F1 tyres, molecular velcro...

Again, this only comes from the (sparse) reading I've done, but the nature of "friction" will be different than the nature of a tire. When we speak of Coulomb friction, that just refers to what is generally regarded as metallic friction theory, which I believe limits the coefficient of friction to a maximum of 1. Tires break the 1-g barrier because they have other methods by which they produce force, such as adhesion like you mentioned ("molecular velcro"). There's also hysteresis, but I don't really understand how hysteresis grip works. I'm getting there, hopefully. More books, more reading, more time.

[Conceptual]

So, a tyre standing on top of a hill has slip forces when the wind blows it enough to make it roll the rest of the way down?

Hmm. I'd have to classify the forces on the tires as purely rolling resistance and rotating friction forces. The defining thing about slip is that it is a response to an applied torque on the wheel/tire.

So when wind blows the car down the hill, there is an applied force on the car to make it accelerate from its static position, but the force is not applied at the road surface. The force is applied at the aerodynamic center of pressure of the plane normal to the wind direction, and the component of this force in the direction of the free-rolling tire accelerates the vehicle in a straight line.

So the force reaction on the tire represents the sum of the rolling resistance and the force required to overcome static friction. But it is important to note that there is no applied torque at the wheel.

You can think of applied torque as a torque that must be reacted through the chassis. This is because it is applied by an object that is mounted to the chassis (brake caliper, drive axle).

Although technically the friction force in the wind example involves a reaction at the wheel center, and there is still a moment present. This moment will be reacted by friction about the wheel hub, I would think (the actual distribution of the forces involved is indeterminate). So technically those would be reacted through the chassis as well. But those are obviously not what is being referred to as an applied torque.

I guess this brings me to a question about the nature of friction. Is it opposing forces, or is it electromagnetic on a molecular level? Like the 2006 F1 tyres, molecular velcro...

Again, this only comes from the (sparse) reading I've done, but the nature of "friction" will be different than the nature of a tire. When we speak of Coulomb friction, that just refers to what is generally regarded as metallic friction theory, which I believe limits the coefficient of friction to a maximum of 1. Tires break the 1-g barrier because they have other methods by which they produce force, such as adhesion like you mentioned ("molecular velcro"). There's also hysteresis, but I don't really understand how hysteresis grip works. I'm getting there, hopefully. More books, more reading, more time.

THANKS!

I remember watching an 11 part series on NOVA about String Theory, and they mentioned friction for about 3 seconds in an 11 hour series. I thought that the friction influence would have answered a few of their puzzles, but they didn't mention it more than once.

That, and the idea struck me that Gravitons and Neutrons were actually the same thing, since it would completely remove the need for the theoretical "Strong" Nuclear force... But then again, I am less educated in this field than most others, so it is all speculation....

Anyways, thanks again!

[Jersey Tom]

So, a tyre standing on top of a hill has slip forces when the wind blows it enough to make it roll the rest of the way down?

I guess this brings me to a question about the nature of friction. Is it opposing forces, or is it electromagnetic on a molecular level? Like the 2006 F1 tyres, molecular velcro...

No.

[Miguel]

First of all, I'm a bit of a nitpicker, so for me 99% of what I see is electromagnetic, 0.99% is gravitational and the rest is nuclear strong and weak.

I remember watching an 11 part series on NOVA about String Theory, and they mentioned friction for about 3 seconds in an 11 hour series. I thought that the friction influence would have answered a few of their puzzles, but they didn't mention it more than once.

That, and the idea struck me that Gravitons and Neutrons were actually the same thing, since it would completely remove the need for the theoretical "Strong" Nuclear force... But then again, I am less educated in this field than most others, so it is all speculation....

Anyways, thanks again!

OK, in the Standard Model elementary particles are "points". This seems to work pretty well, because otherwise the gyromagnetic factor of the electron wouldn't be calculated to such precision (11 significant figures). Now, don't quote me on this but from what I've heard, in String Theory elementary particles are strings, not points, in not 4 dimensions but 11, and there are some symmetry transformations (imagine turning a virgin CD, it just doesn't matter) that can carry us from one particle to another.

In any case, gravitons and neutrons would never be the same. First of all, gravitons are elementary, massless and Spin 2 point particles, while neutrons are compound massive spin 1/2 particles. So a neutron is made of two down quarks and one up quark. Even if I know nothing about string theories, while I could believe that a transformation would bring you from a graviton to a quark (and a large part of me complains anyway*), I can't see how you would get three quarks in a bound state.

BTW: Last year, one of the great advancement of physics was being able to calculate the mass of a proton from it's components (up up down quarks). While this may seem trivial to you, it's an extremely complicated calculation that has required lots of time, a big bad computer and brilliant scientists. The numbers they get seem to indicate that our model for the nuclear strong force (Quantum Chromodynamics) is pretty good.

* Graviton to quark means boson to fermion (supersymmetry, OK), massless to massive (smells quite a lot, one moves at the speed of light while the other just can't) and finally propagator to particle (these take part in totally different places in the "equation of the system", the Hamiltonian). The later one is the one my ignorance has a hard time swallowing.

PS: I did my last QFT calculation like 4 years ago, so this is all very rusty in my mind.

EDIT: The pdf needs subscription, but I hope the comments by Adrian Cho don't. The reference of the proton paper (in case anybody is interested) is
Science 322, 1224 (2008)

[Jersey Tom]

The way I think of this, by the way, is as such...

Engine (or brake) generates torque -> Torque applied to wheel and tire -> Torque reacted by footprint by means of longitudinal force -> Strain and localized slip result -> At some point the torque input is greater than the force that can be reacted by the footprint, and slip rapidly increases.