# I My tire physics recolletion is a bit off -- Question about the coefficient of friction μK

1. Mar 25, 2017

### quickquestion

Okay so I've been digesting a lot about tire physics, it is a lot to digest. I have read about 50 pages worth of long and complex equations and hours upon hours of videos. But before I get to the complex refinements, I want to make sure I have an understanding of the basic behavoir.

In all physics lessons, it is implied μK is a constant. That once a tire's lateral and longitudal force exceeds the weight on the tire, the friction coefficent changes from Static (mU) to kinetic. (μK). But it is implied that μK is always constant, does not change based on speed or slip angle of the tire. I have been fiddling around making my own car simulation, but the tires do not behave right. It behaves similar to a car but has some big problems. Brian Beckman, in his book ten years ago, says his physics are experimental and this is active research...I am not sure if someone has ever figured out the combined grip equation 10 years later...so I view this post as experimental and part of the active research.

So, in plain english, having a simply static μK doesn't seem to be working out. So I'm thinking theres something more to this. Could someone provide an equation that returns the simple μK (simple please as I have the brain drain right now.) It doesn't have to be 100% realistic or detailed, just something that appears to have the behavoir and handling of a car at least within 96%-97% accuracy. Simpler is better.

Also, this equation should take place on a 2d flat surface.

So basically, an equation that returns the μK, only using as input: World velocity of the tire, World acceleration of the tire, and World angle of tire.

2. Mar 25, 2017

### haruspex

Perhaps that is not what you meant. The issue is whether the force parallel to the road exceeds the normal force times μs.

Can you be more specific about the divergence of reality from simple theory?

From a bit of reading, received wisdom is that wider tires give better grip, even though it is not directly supported by the standard simple equation.
As far as I can determine, variation in road surface needs to be considered. Some parts might offer less grip, while small undulations may cause fluctuations in the normal force. Either of these may lead to insufficient static friction transiently, and thus to the transition to kinetic friction.
To minimise this risk, a large contact patch is needed. That can only be achieved by lowered tire pressure. A low pressure in a standard width tire might cause excessive flexing, heating the tire and making grip poorer. A wider tire might cope with the reduced pressure better, not least by being able to shed more heat.

Last edited: Mar 26, 2017
3. Mar 27, 2017

### quickquestion

Well, from what I gathered from the lectures, once the force exerted by the tire exceeds a certain value, the friction coefficient of changes from 1 to 0.8.

So what I have been doing, is along the lines of something like this:
If tireforce>frictionforce
then
fcoefficient=0.8 else fcoefficient=1

And then how I apply this to the tire, would be to reduce the amount a tire can grip by this amount, like this:
ApplyForce(LateralCancellationDirection,carmass/4*LateralCancellationDistance*fcoefficient).

But as you can see, that doesn't produce any kind of realistic drifting or slipping behavoir, it only provides a very subtle slip effect, the back end never slides or slips around like a real car. When the car has total grip, it corners 10% with no slip, So then I changed it to something like: ApplyForce(carmass/4*LateralCancellationDistance*fcoefficient/(1+centrifugalforce)*tire.slipanglePajeckPeak) And it behaved somewhat more like a real car, but still felt quite off.

Last edited: Mar 27, 2017
4. Mar 27, 2017

### haruspex

Yes, the maximum frictional force without slipping is normal force * the coefficient of static friction. Once the required force exceeds that, slipping starts and the frictional force becomes normal force * coefficient of kinetic friction. Whether those two numbers are 1 and 0.8 or some other pair of values depends on circumstances.
I assume by carmass you mean car weight, not mass, and that this is for horizontal ground.
What is LateralCancellationDistance? On horizontal ground the force is carweight/4*fcoefficient,
The direction of the force (LateralCancellationDirection?) is directly opposite to the direction the tire would slide if there were no friction.

I'm not at all sure how to bring slip angle into this. Although it is called slip angle, it is not actually to do with slipping. Creep angle might have been a better term. It certainly applies even before exceeding maximum static friction. An analogy would be walking in a straight line, except that as you put each foot down you put it a cm to the left of the obvious position. So although your feet are pointing one way you creep to the left.

The 1+centrifugalforce in your second equation makes no sense. It is dimensionally inconsistent (a force plus a dimensionless number) and anyway the "required force" you calculated should already take that into account.

5. Mar 27, 2017

### quickquestion

Well yes, it was a garb equation. Putting it to /(1+centrifigual) was an act of desperation just to see its effect. But I noticed studying car behavoir, that the wheel appears to lose grip as a function of the centrifugal force of the car on it. When you whip a car back and forth, each time you whip it the whip becomes more and more effective, and the behavoir is very similar to a merry go round or circus ride. So, when I added the /(1+centrifugal) to the equation, it actually improved the realism of the simulation (although the simulation was overall, still garb.) This is because, the centrifugal should (according to me) actually decrease the weight on the tire, something this thesis paper does not account for. https://nccastaff.bournemouth.ac.uk/jmacey/MastersProjects/MSc12/Srisuchat/Thesis.pdf

Also, I've been doing some thinking and I believe my original methology used the wrong approach and had an error.
My original methodology (that I described here) was to have a function that was based around grip...Each tire had 25% grip and so it caused the car to turn well at low speeds of about 10 miles an hour. But the problem was, I plugged the 0.8 (kinetic friction) directly into the grip...so instead of 25% grip it would be 25%*0.8. This is a mistake, because the actual kinetic friction force should be 0.8*weight and not simply setting the counterforce to -totalcaraccelerationforce*0.8.

I have a headache right now and I need to go eat, havent eaten a real meal in hours. But I shall run more tests and return to you tomorrow. I forsee running into more problems, even though I found one of my mistakes I think I will run into more issues. I may have climbed the hill, but I'm not sure I climbed the mountain.

And yes, I meant to put carweight instead of carmass. Also, this creep angle thing brings me more questions than answers...Slip angle, for me means the contact patch is actually twisted relative to the tire, so to determine from a game standpoint what the slip angle is, is harder than simply getting the angle of the visual tire. Also, I am not sure what effect the slip angle has on the kinetic friction coefficient...According to Pajeck formula the slip angle seems to affect the amount of force the car can handle before it slips. But physicists say the kinetic coefficient is always static, so its confusing. Maybe its because, the slip angle inherently means there is less "rolling give" in the tire, making the tire less availiable to de-stress, when you turn the tire at an angle and such thats more stress on the axle, since the tire cannot roll with it any longer.

Last edited: Mar 27, 2017
6. Mar 27, 2017

### haruspex

The centrifugal force creates a torque on the car, tipping it towards the outside of the curve. That decreases the normal force on the tires on the inside of the curve but increases it on the tires on the outside. The total normal force must still equal the weight of the car.
Remember that as you steer a car the axis of rotation lies on the line of the rear axle. In a sharp turn, there is a lot more demand for traction on the front wheels than on the rear.

7. Mar 28, 2017

### quickquestion

Yes that is quite true. And on a drift, the axis of rotation is inbetween the front tires.

I corrected the mistakes of my past equations, but still have problems. The car does behave more similar to a car now, but, it is way too slippy. It suffers from a problem in some old Gamecube games, it is when you are driving the car and you begin to spin out, and when you correct the spin out, the angular rotation speed (the car's omega) goes to zero, so you think you're safe to drive, but as soon as you let go of the tires, it begins to spin out again, like you never reset the angular rotation speed to zero.

So I went back to the basics, trying to understand the theory behind a tire that grips 100%. Here is what I'm not understanding...lets say you have a perfect tire, with a uS and uK of 1000, that grips 100%. And you want to put this in your engine, just to test that your simulation can grip 100% and behaves properly, (before you move on to adding slip and weight and other things). But here's the problem...
Lets say you have a max steering angle of 90 degrees (just to test the engine and see that the tires can make the car stop 100%.) And so, this is very easy to plot the equation, you just set the force to apply to the left front tire to be -carvel*phy_mass/2 and the right tire to also have the same force applied to it. This will stop the car dead in it's tracks, which is what you want before you move on to testing the advanced stuff and adding slip behavior.
Now...stage 2...The steering angle is 45 degrees...so now you use a dotproduct to figure out the lateral cancellation distance...So now its still simple...you just set the left front tire to apply a force of -dotproduct*phy_mass/2 and the same for the right front tire.

But now...adding the back tires...its not so simple...
Because now the back tires have also to cancel out the lateral velocity of the car...so now your equation for each tire looks like -dotproduct*phy_mass/4...But this is wrong because...now the front tires will not stop the car dead in it's tracks any longer...
To make matters even worse...imagine your car is skidding at a 40 degree angle...how would you compute the amount of counterforce to apply to each tire so that the car stops dead in it's tracks? I have no idea.

8. Mar 28, 2017

### jack action

With a rolling rubber tire, the friction coefficient is not constant. In the thesis you refer to in post #5, it says on p. 17:
This is wrong, because the lateral force is NOT linear with load. This is why load transfer is bad for vehicle: a car with the weight equally divided among identical tires (25%-25%-25%-25%) will always outperformed any other weight distribution.

The coefficient of friction decreases as load increases.

Once you reach slipping, i.e. kinetic friction coefficient, not only the friction coefficient drops noticeably, but we then go back to a more traditional definition of friction, i.e. fixed rubber sliding on a surface instead of rolling. So the kinetic coef. is not as affected by load as the static coef. is.

As for slip angle, I prefer to see it as a consequence of the lateral force more that the cause. A lateral force will cause a slip angle, for which the magnitude will depend on the tire construction. The slip angle will modifying the actual rolling direction of the tire (i.e. steering angle) and so will create an apparent lateral velocity to the tire rolling axis.

The normal weight is the important factor that affects lateral grip capabilities.

Also the friction coefficient of a tire is not necessarily 1.0, it can be higher or lower.

References:

Last edited by a moderator: May 8, 2017
9. Mar 28, 2017

### haruspex

The magnitude of the velocity has nothing to do with it, except in regard to centripetal force.
If the car is already skidding at this point then the direction of velocity matters, but not the magnitude.
Where "dotproduct" is what, precisely?
Is that 90 degrees to the chassis or 90 degrees to the velocity?

Your diagram appears to show a vehicle already skidding, with front wheels at right angles to the direction of skid.
I assume there is no braking. If so, you are right to draw the force arrows as you have, perpendicular to each tire.
Each friction force will equal normal force * μk, but the normal force might be different for each tire. In the diagram, the right hand tires will have greater normal force, particularly front right, because of the tendency of the car to roll on its suspension.

10. Mar 28, 2017

### quickquestion

Oh, dot product is the same as lateral cancellation distance, it just tells you the lateral cancellation vector (basically it is the vector from this article right here: http://www.iforce2d.net/b2dtut/top-down-car Note: this article produces unrealistic physics.)

Also, maxsteeringangle refers to the tire direction relative to the front axis of the chassis. In normal cars its about 30-40 degrees, but in my beta-testing mode I put it to 90 degrees to test whether or not the tires put the car to a complete stop (they don't, thus my simulation sucks.)

Oh ok, I get what you are saying. What you are saying remains true for real life, but for game solutions it becomes a problem. In a game, I need to simulate the friction force. Ok let me give an example of why this becomes difficult.

Let's say I computed the weight of the car, and I know the normal force (weight) on each tire. Say my car is going 10 miles per hour, and so the front tires should stop the car completely. If I just plug "normalforce*uk" into my game...the game won't understand what to do. It will just give the car a backwards force exceeding the actual velocity of the car. So then I move to stage 2...instead of that I simply clamp it as a min...forcetoadd=-min(FrictionForce,carforce/4),. But see there is a problem...I don't know in what relation to add carforce to each tire...Because if I divide it evenly, the front tires dont have enough grip. And carforce is not related to the normalforce, thus I cannot multiply it by the normalized weight either.

It does at low speeds, unless I am thinking about this incorrectly. Because if the acceleration force of the car does not exceed the frictionforce, if you apply a backwards friction force it will actually create a net gain of energy and make the car go backwards, instead of just stopping in its tracks. I could be making a mistake though.

Last edited: Mar 28, 2017
11. Mar 28, 2017

### quickquestion

This is actually a relief for me...I was having trouble accepting the idea that the uK of a tire remains constant. I got this idea after watching various youtube videos teaching physics, such as this video: www.youtube.com/watch?v=bkZ8bGYYAPQ

However, I want to state this, that the statement "lateral force changes on the load of the tire" in no-way implies that mU is dynamic. By the statement, it could been intended to mean that lateral force could change purely based on weight (N) alone.

Would you happen to know the gradient of the coefficient as it changes in relation to the weight, speed, acceleration, centrifugal, and other factors it has?
(From what I understand, centrifugal is different from a typical lateral force...lateral force is determined by wheel angle in relation to acceleration...centrifugal is related to velocity angle in relation to previous velocity.)

I share this sentiment! Physcists seem to based their game engines on a causal loop - they take the curves of empirical data points (such as Pajeck's formula and/or slipangle) and cause the game to simply duplicate the empirical data points...they are using the "caused" to cause and missing the middle man (or rather, the godman.)

True. Some lectures tell me it can be as high as 1.7. There is a website out there that lists the different coefficients for truck tires and bicycle tires and such. Currently my game just uses 1.7 uS and 0.8 for uK (Ferrari-ish tires, or sports car tires, if I remember correctly.)

References:

12. Mar 28, 2017

### haruspex

My comment was in relation to velocity. You are confusing velocity with acceleration.
To calculate the friction force on a tire:
- determine the acceleration being demanded of it; this includes any attempt at braking, steering, centripetal acceleration and pressing the accelerator; the acceleration so determined has a magnitude and a direction; note that different accelerations may be desired for different tires
- determine the normal force on the tire; take into account the slope if not horizontal; ideally, take into account the torque arising from the fact that friction is at ground level while the acceleration required acts on the mass centre;
- multiply the normal force, FN, by the coefficient of static friction, μs, and compare the result with the magnitude of the desired frictional force;
- if the desired frictional force is the lower, the actual frictional force will equal that;
- if the desired frictional force is greater than FNμs then the actual frictional force will be FNμk.
- in all cases, the actual friction is in the same direction as the desired acceleration.

13. Mar 28, 2017

### quickquestion

Hmm, but if I do it that way, there is a problem. Tire order.
If I am going 10 miles an hour sideways (when the grip is disabled for testing purposes), and I want to enable the grip...I switch the button to enable the grip physics and then I apply it to the front tire...The front tire makes the car come to a complete stop, and none of the other tires do anything. Or what happens is, the front left tire causes a torque, and then the other tires are now using up their frictionforce quota trying to stop a torque that shouldn't exist in the first place.

14. Mar 28, 2017

### haruspex

I do not understand what prevents the software from applying the friction to all tires simultaneously.

15. Mar 28, 2017

### quickquestion

It can, but I don't know how to calculate what ratios.
The problem here lay in this statement: - "determine the acceleration being demanded of it; this includes any attempt at braking, steering, centripetal acceleration and pressing the accelerator; the acceleration so determined has a magnitude and a direction; note that different accelerations may be desired for different tires"

Consider the case of 2 90 degree steering angle front tires and the car is going forwards (this is not a real life car, most cars do not have tires pointed 90 degrees...its just a test car for testing purposes.)
Left has an acceleration of say, 10 mph. So if it cancels that, that leaves no work for the right tire to do...causing a torque. But lets say, we split the work for both tires...Works fine in our situation...but lets say we have other situations...such as all four tires with non-90 degree angles and accelerations...How do we know "What" or "how" to split the acceleration as between each tire?

16. Mar 28, 2017

### jack action

Just concentrate on the normal load for now. Actual data is almost impossible to get as tire manufacturers rarely share them.
You can look at this web page; look also at the sources at the bottom of the page.

17. Mar 28, 2017

### quickquestion

So, you are saying the uK changes based on the normal load? Because the equation they teach in highschool is uN...but are you saying, for in-terms of realistic tire behavoir, the equation should be more like u/(1+N/2)*N?

Ha, that is exactly the website I was talking about.

18. Mar 28, 2017

### jack action

From the Wikipedia link I gave you earlier:
So, instead of the classical $F=\mu N$, you could use $F=\mu N^{0.8}$. Let's say you know - using the classical method - that for a certain value of $N_0$, the coefficient of friction is $\mu_0 \left(=\frac{F_0}{N_0}\right)$, then you can estimate that $\mu = \mu_0 N_0^{0.2}$, such that $F=\left(\mu_0 N_0^{0.2}\right) N^{0.8}$.

19. Mar 28, 2017

### quickquestion

Yeah...that 0.8 exponent thing seems pretty nifty, I think it may come in handy. Thanks.

One thing though, I'm looking over your equations and I'm not sure which to use.
Do I use F=uN.8 or F=(uN.2)*N.8
Guessing from the Wikipedia article, I would use F=uN.8?

Also, I'm guessing that when the slip angle is more, the friction coefficient is less also...But the tire manufactures wont share the data...so I'm guessing the final friction equation would be
fF=uN.8/(1+speed/100+abs(abs(slipangle)-3)/90) or something like that?

Last edited: Mar 28, 2017
20. Mar 28, 2017

### haruspex

You are probably right about this, but my reading of the thread is that there are rather more fundamental errors in the approach quickquestion is using now. I suggest getting the basics right first.