Mechanics of a car turning left

[RobertArvanitis]

Imagine a carom shot in a perfect world.
A ball travels diagonally from the lower left-hand corner towards the right cushion and bounces off to head towards the upper left-hand corner.
The ball has X and Y momentum. On the collision, the cushion absorbs and then return all the -X momentum to the ball, while the Y momentum remains unchanged.
Now imagine a car turning left in a perfect world.
One might suppose the same thing happens as in the first case, except it’s between the tire and the road.
It also seems to happen between different parts of tire and road, as the wheel turns along the track of the turn.
Anyone have thoughts on details of tire-road interaction, or ideas on where else to look for such details?

 

[FactChecker]

There is practically no similarity between a single impact of a billiard ball on the cushon and the turning force of a tire. The force on a tire is proportional to the angle between the tire pointing direction and the rolling direction. It is a continuous force due to the tread distortion.

 

[RobertArvanitis]

True but not quite the point of the question.
Start with a rocket. side thrusters can change the orientation of the rocket but not actually turn it left. It would keep going straight, turned sideways.
So the momentum in the forward direction must be canceled, and new momentum to the right, created.
Thinking about the basics, the pool ball "loses" its Y energy to the cushion and gets it returned as -Y.
A more complicated version of that happens between tire and road. That's what I'm interested in .

 

[Herman Trivilino]

Start with a rocket. side thrusters can change the orientation of the rocket but not actually turn it left. It would keep going straight, turned sideways.

What you describe requires a force couple, that is, a pair of equal but opposite forces, with lines of action separated by a distance known as the lever arm. This would require two side thrusters. Use instead just one and the rocket's direction of motion will change.

So the momentum in the forward direction must be canceled, and new momentum to the right, created.

You mean, like, for a 90-degree turn? Isn't that what you imagine is happening when a billiard ball hits a rail?

 

[FactChecker]

You could say that the linear momentum is continuously converted by contact tread distortion from one direction to another. Energy is transferred into contact tread distortion and then transferred back into momentum in another direction as the contact tread goes back to normal position.

 

[RobertArvanitis]

Factchecker - Yes, got to be something like that.
Have you ever seen the details of tire alignment - caster, camber and toe in? Race cars get "tuned" for different events. Hoping to come across the similar force diagrams for a simple turn.
(There's already material for over and under steer).

 

[jim mcnamara]

Hmm. A good starting question. But I think the Physics of tire-road interactions is complex in racing. @Ranger Mike can probably help.
As can @jack action ...

They would have a line on force diagrams during turns

 

[Roberto Teso]

Anyone have thoughts on details of tire-road interaction, or ideas on where else to look for such details?

Here you can download a booklet written by Massimo Guiggiani, professor of applied mechanics at Pisa University, for Michelin.

The Tyre

 

[A.T.]

Imagine a carom shot in a perfect world.
A ball travels diagonally from the lower left-hand corner towards the right cushion and bounces off...
Now imagine a car turning left in a perfect world.
One might suppose the same thing happens as in the first case, except it’s between the tire and the road.

Bouncing requires elastic deformation, while turning doesn't. The tire deformation mostly dissipates energy, and doesn't give it back like the cushion. A better analogy for the car turning would be a train on tracks, or a steel ball in a curved groove.

And it doesn't really matter which way the turn goes.

 

[Ranger Mike]

The real short answer is that we are dealing with (centrifugal force) a vehicles momentum.
Newton's first law of motion is often stated as - An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

In this example assume no vehicle suspension, assume a flat pavement corner. The vehicle will continue traveling in a straight line. Gravity is acting on the vehicle as well pulling the vehicle straight toward the center of the earth.
Weight always acts along the global vertical gravitational vector.
Once the steering wheel is turned, the wheel direction is changed. The tires in contact with the pavement resist the straight line force. When the tire contact patch maintains enough contact the vehicle will change direction. If the tires do not have enough traction or grip the tire will slide due to loss of traction.

There are a lot of other variables involved but this is the go kart scenario and is easiest to discuss.
On a flat track, while cornering, weight and centrifugal force act on the vehicle. Technically cornering force is a component of the lateral force taken in the direction perpendicular to the wheel moving direction, i.e. considering the side-slip angle. For easier understanding we consider lateral force acting perpendicular to the tire rotation plane. Centrifugal (lateral) forces are perpendicular to the ground and have zero contribution to the vertical force ( down force) felt by the tires (other than left/right distribution).
see Race Car Suspension Class in the mechanical engineering forum of this fine website for detailed explanation.

 

[BvU]

Hi, no physics from me here. Instead, my two cents goes to @vr-marco . He created some time-usurping videos that gave me the impression he has this figured out !

Disclaimer: Not intended as supporting the purported commercial; I just liked it .

Hi there!
Is anyone familiar with the Tametire tire model by Michelin? I understand it is based on the brush model and can take into account thermal and pressure properties of a tire. I was able to find only a very basic article online, but not a full description of how this model is able to generate Fx, Fy and Mz. Anyone happens to have more details on it? Many thanks in advance.
Marco.

Edit: for those wondering, I am working on realistic vehicle simulations. I have few videos on my YouTube channel (Marco Ghislanzoni) which apparently I cannot link here because they are flagged as spam.

 

[vr-marco]

Hi, no physics from me here. Instead, my two cents goes to @vr-marco . He created some time-usurping videos that gave me the impression he has this figured out !

Disclainer: Not intended as supporting the purported commercial; I just liked it .

Thanks for the mention @BvU . Not sure whether I have figured it all out, I feel like there is always something new to learn on the subject every day. I am in good company though. Professor Pacejka, who devoted his whole life to the study of vehicle tires, also thought there was a lot still to do in the field.

Cornering is one of the most complex aspects of vehicle dynamics. At macroscopic level, the vertical tire rotation caused by the steering system creates a "side-walking effect" of the contact patch. This side-walk generates a lateral reaction force on the tire due to the interaction with the ground. The force is transmitted to the body of the car through the steering system and generates a yaw moment, so the car goes around the curve. This is an extreme simplification though.

At microscopic level, during cornering the contact patch undergoes to a distortion. The bristles of the thread are bent by the side-walk and, as they try to get back to their original orientation due to the visco-elastic property of the material, they transmit a force to the ground. The ground reacts with an equal but opposite force (minus what is turned into heat) which produces a lateral cornering force and an aligning moment. The aligning moment is what make the steering wheel go back to center when you let it go after a cornering.

This said, there are a lot of other factors and conditions that affect the cornering forces and therefore the lateral vehicle dynamics. Camber is only one of them. One of the major effects is due to the vertical load on the tire. During cornering part of the vehicle weight is shifted to the outer wheels. This puts more load on the respective tires. One would expect that more load means more friction, so a better cornering, but actually it is the opposite. Past a certain load, the tire undergoes to more distortion and its capacity to generate lateral force diminishes. Another effect is the combination of longitudinal and lateral dynamics (so called friction circle). As the tire can only generate so much force in total (μ * F_z), if the driver brakes or accelerates during cornering, this ends up subtracting useful force from the amount available for cornering, and the car will very likely end up against the guardrail (understeer).

Here my two cents: tire dynamics (and vehicle dynamics in general) is quite complicated. You can surely come up with a very simple model to describe one specific behavior or aspect, but it will very likely fail as soon as you move away from the ideal conditions. By the way, this is true also for the more complicated models, so you are in good company! My suggestion is to pick up some good literature on the subject and familiarize with it. Then you will both start to grab the complexity of the subject and get new ideas.

 

[Ranger Mike]

VR, good points all.

Cornering is one of the most complex aspects of vehicle dynamics. At macroscopic level, the vertical tire rotation caused by the steering system creates a "side-walking effect" of the contact patch. This side-walk generates a lateral reaction force on the tire due to the interaction with the ground. The force is transmitted to the body of the car through the steering system and generates a yaw moment, so the car goes around the curve. This is an extreme simplification though.



A lot of things happen in a turn ( when cornering at speed) and totally agree! It is complex 4-D action. Resistance force is a vector from the tire cotnact patch to the Roll Center of the suspension. This Roll Center is connected to the Center of Gravity thru a moment arm (think lever). This counters the momentum force. The reaction is the rolling of the car body toward the outside. This roll over motion puts down force on the tire and increases traction. The tire thinks there is additionl weight on it( the weight transfer alaogy) but it is really force in the from of down force. And as you so graphially stated, the down force can be too great and overcome the tire traction and you have understeer or a push condion and this is not good!

You counter this problem with proper location of the roll center, proper selection of th esprings and dampers ( shock absorbers) better gripping and larger tires, proper camber settings and a lot of other settings.

 

[FactChecker]

Here my two cents: tire dynamics (and vehicle dynamics in general) is quite complicated. You can surely come up with a very simple model to describe one specific behavior or aspect, but it will very likely fail as soon as you move away from the ideal conditions. By the way, this is true also for the more complicated models, so you are in good company! My suggestion is to pick up some good literature on the subject and familiarize with it. Then you will both start to grab the complexity of the subject and get new ideas.

In real-time simulations of airplane ground-handling, we used very simplified models just to keep things reasonably believable without overwhelming the computers. I don't think that any real-time simulation tries to accurately model the detailed physics that would be needed to evaluate tire charactoristics. The Michelin document referenced by @Roberto Teso is quite interesting.

 

[vr-marco]

In real-time simulations of airplane ground-handling, we used very simplified models just to keep things reasonably believable without overwhelming the computers. I don't think that any real-time simulation tries to accurately model the detailed physics that would be needed to evaluate tire charactoristics. The Michelin document referenced by @Roberto Teso is quite interesting.

Yes, I am familiar with the booklet, it is actually very well written. There is a lot to learn from it. BTW, Michelin currently has one of the best tire models out there, the TaMeTire model. I am currently trying to know more about it, but the literature on it is really scarce.

You are right regarding the trade-off between model complexity and simulation time. If simulation time wouldn't be a problem (together with computational power) one could simply use a FEM approach and obtain a real-like behavior in an emergent manner, including thermal effects and advanced material properties. Having to simulate tires at > 400Hz though imposes some limitations.

At the end of the day each and every model is just a representation of reality, only as good as the boundary conditions set for it hold true. No need to go for a complicated model if a simple one explains what you need given your operating conditions.

 

[vr-marco]

Anyone have thoughts on details of tire-road interaction, or ideas on where else to look for such details?

Once you are done with booklet by Michelin pointed out by Roberto (@Roberto Teso ) above, this is a classical in the field:

Tire and Vehicle Dynamics - 3rd Edition
Authors: Hans Pacejka
eBook ISBN: 9780080970172
Hardcover ISBN: 9780080970165
Imprint: Butterworth-Heinemann
Published Date: 9th April 2012
Page Count: 672

 

[jack action]

I'm going to side with @A.T. here and say that the scenario of car turning compares to a train on tracks or a ball in a groove. The only difference is that instead of having an object pushing laterally, you have a lateral friction force. Because there can be lateral motion, it means that the 'tracks' can move sideways. Worst, the front & rear 'tracks' are independent of each other and can move in different directions.

I don't think that you need the complexity of tire modelling to explain the phenomena, as you can also turn with solid wheels. Even ideal wheels who don't deform and who conform perfectly to linear friction theory. It is just a matter of doing a free body diagram, combined with evaluating constraints, i.e. free rolling of the wheels.

Imagine the top view of a bicycle with its 2 wheels angled at 45° between each other. If there is an exterior force acting on it (not necessarily from a wheel torque), from any direction, the bicycle will turn. The only way it wouldn't turn is if the force is acting at a wheel axis, parallel to it.

 

[vr-marco]

I'm going to side with @A.T. here and say that the scenario of car turning compares to a train on tracks or a ball in a groove.

Sorry to be picky, but a train negotiates a curve using a completely different principle, namely the conic profile of its wheels and the variation of the working radius. It has little to do with the friction between the wheel and the track or the track pushing the wheels laterally.

 

[Robert Garner]

What about those off roaders that use the cupped tires to throw dirt rearwards to gain forward thrust like a rocket, then drift around the curves using both that thrust and tire cornering characteristics as well to get around the corner?

 

[A.T.]

What about those off roaders that use the cupped tires to throw dirt rearwards to gain forward thrust like a rocket, then drift around the curves using both that thrust and tire cornering characteristics as well to get around the corner?

That's even more messy than turning with full traction.

 

[Ranger Mike]

You covered a lot of ground there...(bad pun)..off roaders set up their chassis with zero offset as they must turn right and left. races running on left turn only tracks have a lot of offset on the chassis to assist the turning process. In both cases there is some degree of " drift" but the results sought is to be able to corner better than the other driver. it is not so much the tires cornering characteristics as th chassis set up to deal with momentum and track resistance.

 

[A.T.]

You covered a lot of ground there...

The next logical step is turning dynamics while driving on water...

 

[Chestermiller]

I have some practical experience modeling the structural mechanics of automobile tires under various modes of loading. Our interest was in the inter-laminar shear stresses that develop between the various anisotropic tire cord layers comprising the structure under load, and the tension variations along the tire cords in the layers. It was a very difficult mechanical system to model for several reasons. First, the layers are anisotropic (and begin and end at various locations along the contour), second, contact problems with surfaces are always difficult, and third, there are large changes in tire contour curvature which result in numerical problems involving prediction of unrealistic wiggles in the surface. Even the problem of tire contact with the ground (without rotational inertia, or cornering loads) was very difficult. We finally got it done, but it was a bear of a problem.

 

[sophiecentaur]

Bouncing requires elastic deformation, while turning doesn't.

I can't quite get a grasp of that idea - if you really mean it in an extreme way. Bouncing is 'easy' and can often be described in terms of coefficient of restitution (the schoolboy's friend) which can be anything from 0 to 1. In the case of a cornering tyre, there must be some restoring force from the tyre foot in order for the tyre to return to its original shape. You seem to be suggesting that the coefficient of restitution (or some sort of equivalent) has to be near zero. Why would that be necessary?
I can see your remark could apply to racing slick tyres which are taken almost to melting to achieve good stiction.

a train negotiates a curve using a completely different principle, namely the conic profile of its wheels

Isn't the conic profile there to give a suitable differential action on corners when the axel is necessarily rigid?

 

[A.T.]

Bouncing requires elastic deformation, while turning doesn't.

You seem to be suggesting that the coefficient of restitution (or some sort of equivalent) has to be near zero.

Saying that something is not required doesn't imply that it must be zero.

 

[vr-marco]

Isn't the conic profile there to give a suitable differential action on corners when the axel is necessarily rigid?

The conic profile of the wheels is what makes the train go around a curve without the need for lateral friction (the inner flange is just a safety mechanism). The fact that the axle is rigid is just a mechanical simplification made possible by the mentioned cornering dynamics. A train would work as well with independent wheels (in fact there are some designs without rigid axle) or with a differential (which is in fact not needed and would therefore be a waste of weight and money).

 

[A.T.]

The fact that the axle is rigid is just a mechanical simplification made possible by the mentioned cornering dynamics.

Explained nicely here:

 

[sophiecentaur]

Saying that something is not required doesn't imply that it must be zero.

I was interested if you had any idea about the order of magnitude involved. Plus, bouncing doesn't actually require elastic collision. Perhaps you could have expanded on your original statement.

The conic profile of the wheels is what makes the train go around a curve

Is that enough of a description, I wonder? The cones are in opposite senses for the inner and outer wheels. I can't picture what's actually going on to provide a centripetal force.

 

[FactChecker]

The conic profile of the wheels is what makes the train go around a curve without the need for lateral friction (the inner flange is just a safety mechanism). The fact that the axle is rigid is just a mechanical simplification made possible by the mentioned cornering dynamics. A train would work as well with independent wheels (in fact there are some designs without rigid axle) or with a differential (which is in fact not needed and would therefore be a waste of weight and money).

I don't understand how it would work without a rigid axle. Wouldn't the outer, larger-diameter wheel just rotate slower if the axle is not rigid?

 

[Ranger Mike]

Please see APTA PR-M-S-015-06 Standard for Wheel Flange Angle for Passenger Equipment. This is a the standard for railroad wheel contact angle to prevent low speed climb. Similar specs on the actual rail profile. Summary - there is a detailed profile of the wheel and the rail to provide maximum safety and most durability considering the mass of the train and horizontal land vertical forces applied to both as the train moves at speed. So the wheel is not a cone shape as such under high magnification as it may appear. The flange shape of the wheel has certain profiles at certain points to provide maximum contact minimum wear.

 

[sophiecentaur]

A train would work as well with independent wheels

That sounds right but the above video implies that the axle needs to be rigid for directional stability. To work with a rigid axle, the rotation rate must be proportional to 1/radius of curvature for both wheels, if slipping is to be avoided so the slopes on the cones need to be appropriate. The video doesn't actually say how the optimum cone angles are calculated; they will depend on the gauge of the rails, I guess.