Exploring Tire Physics: Seeking an Equation for μK

[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 there's 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.

 

[haruspex]

a tire's lateral and longitudal force exceeds the weight on the tire,

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

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.

 

[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.

 

[haruspex]

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

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.

ApplyForce(LateralCancellationDirection,carmass/4*LateralCancellationDistance*fcoefficient).

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.

 

[quickquestion]

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.

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, haven't 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 that's more stress on the axle, since the tire cannot roll with it any longer.

 

[haruspex]

the centrifugal should (according to me) actually decrease the weight on the tire

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.

 

[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.

 

[jack action]

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.

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:

The lateral force depends on the load of the tire. In order to find lateral forces at different loads, we could assume that the lateral force is linear with load and create a normalized lateral force diagram by dividing the lateral force by 5KN.

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:
Tire load sensitivity on Wikipedia;

http://farnorthracing.com/autocross_secrets4.html][/PLAIN]
The key to this process though is that while more force pressing the tire into the road provides more grip, the amount of grip generated per added unit of normal force declines on a curve. For example, 100 lbs of normal force (which I'll call load from now on) might provide 100 lbs of grip, 200 lbs of load might provide 150 lbs of grip, and 300 lbs of load might provide 175 lbs of grip. Always more grip with more load, but also always less grip per unit load as grip increases.

This property means that for any pair of tires, they will provide maximum grip when equally loaded.

The following graph explains the most important thing you need to know about all tires on all production cars. Since it is so important a concept, I'll state it a few different ways, hopefully one of them clicks and makes sense with the reader. Nothing in any of the chapters to follow, will make any sense until you get this, a slightly exaggerated tire load response curve:

Most succinctly, as we increase the load on a tire, the total grip the tire provides goes up, but the amount of grip we get per unit of load (this is also known as coefficient of friction, or "mu"), goes down. The more we increase load, the greater the rate of grip drop-off; it just gets worse and worse. On the other side, as we decrease load, we lose grip, but we don't gain as much mu from the load decrease, as we lose from an equal load increase.

 

[haruspex]

the force to apply to the left front tire to be -carvel*phy_mass/2

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.

lateral cancellation distance

Please define this term.

dotproduct*phy_mass/2

Where "dotproduct" is what, precisely?

max steering angle of 90 degrees

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 * μ_k, 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.

 

[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.)

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.

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 don't have enough grip. And carforce is not related to the normalforce, thus I cannot multiply it by the normalized weight either.

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.

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.

 

[quickquestion]

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.

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.

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.

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.)

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.

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.)

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.

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:
Tire load sensitivity on Wikipedia;[/QUOTE]

 

[haruspex]

if the acceleration force of the car does not exceed the frictionforce

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, F_N, by the coefficient of static friction, μ_s, 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 F_Nμ_s then the actual frictional force will be F_Nμ_k.
- in all cases, the actual friction is in the same direction as the desired acceleration.

 

[quickquestion]

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, F_N, by the coefficient of static friction, μ_s, 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 F_Nμ_s then the actual frictional force will be F_Nμ_k.
- in all cases, the actual friction is in the same direction as the desired acceleration.

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.

 

[haruspex]

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.

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

 

[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 let's say, we split the work for both tires...Works fine in our situation...but let's 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?

 

[jack action]

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?

Just concentrate on the normal load for now. Actual data is almost impossible to get as tire manufacturers rarely share them.

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.

You can look at this web page; look also at the sources at the bottom of the page.

 

[quickquestion]

Just concentrate on the normal load for now. Actual data is almost impossible to get as tire manufacturers rarely share them.

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

You can look at this web page; look also at the sources at the bottom of the page.

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

 

[jack action]

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

From the Wikipedia link I gave you earlier:

In practice, the maximum horizontal force Fy that can be generated is proportional, roughly, to the vertical load Fz raised to the power of somewhere between 0.7 and 0.9, typically.

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 $F=\left(\mu_0 N_0^{0.2}\right) N^{0.8}$.

 

[quickquestion]

Ohhh...I thought that link was an advertisement that appeared at the bottom of certain posts.
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 won't 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?

 

[haruspex]

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 $F=\left(\mu_0 N_0^{0.2}\right) N^{0.8}$.

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.

 

[jack action]

Ohhh...I thought that link was an advertisement that appeared at the bottom of certain posts.
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 won't 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?

First, note the $0$ subscript, it is a fixed value based on "real data" that you could find (like "$\mu_0$ = 1.7 @ $N_0$ = 500 lb"; units are important) to give you a realistic increment in friction coefficient.

As @haruspex mentioned, I'm sure you have a lot of misunderstanding, but I must admit I have trouble following your posts and that is why I tried to go another way to see if it triggered better questions from you. But I'm baffled at this last equation you posted:

fF=uN^.8/(1+speed/100+abs(abs(slipangle)-3)/90)

Where did you get this? I don't think speed can have that much influence on the friction force. If you look at the Pacejka formula, speed is not even a factor.

 

[quickquestion]

Some website (or a book, can't remember) listed that the amount of slip increases as the speed on the road increases. It was only by a small value, about .1 after a 10 mph increase.

And yes, like haruspex said, I am trying to master the basics before I move on to the more advanced stuff. First I am trying to just figure out the theory of how to get a car to grip 100% before I experiment with adding slip. What I am currently hung up on, is how to determine the amount of counterforce to apply to each tire. I know that Fmax=uN, and so for the purposes of beta-testing I set N to a ridiculously high value, like 1000, so the car will have 100% grip. But see the problem I have is, let's say the car is skidding at a 40 degree angle, and the front tires are turned at an angle of say, 30 degrees. The car, as a uniform object, has a velocity direction at a 40 degree angle...But I don't know in what proportions the velocity should be handled for each tire... What I mean by this problem can be demonstrated by the following problem...
Say they are 4 tires, and the car is going forwards, and then I suddenly tilt the front tires to 90 degrees...Apparently...the engine should "know" to consider the velocity of the front tire as Half of the totalcar velocity, and the left tires input velocity to be handled as Half of the total car velocity. And that the back tires should be ignored...this is easy. But when the car is going at angles and the front tires are angled different from the rear...how does it know "which proportion" of input velocity is to be decided for each tire...I can't just say "Use half of the car velocity for the front 2 tires" or "Divide the total velocity by 4 for each tire". This is what I'm having trouble understanding.

 

[jack action]

The car, as a uniform object, has a velocity direction at a 40 degree angle...But I don't know in what proportions the velocity should be handled for each tire...

The velocity for each wheel can be found knowing the vehicle velocity $v$ and its yaw angular velocity $\Omega$ ($v+\Omega d$; $d$ being the distance from the cg to the wheel).

I think the problem you referring to is for a force. When there is a purely lateral or longitudinal force act on a vehicle, the wheels are aligned, so the reaction forces can easily be split between both sides, just like a simple beam with 2 supports. When the force is at another angle, it can be more completed because you are now looking at a beam with 4 supports. So you have to separate the force into lateral and longitudinal components and analyze them separately.

 

[A.T.]

Say they are 4 tires, and the car is going forwards, and then I suddenly tilt the front tires to 90 degrees...Apparently...the engine should "know" to consider the velocity of the front tire as Half of the totalcar velocity.

What do you mean by "tilt the front tires to 90 degrees"? The steering cannot steer the wheels that far, but if it instantaneously did, they would stop turning and slide.

 

[quickquestion]

What do you mean by "tilt the front tires to 90 degrees"? The steering cannot steer the wheels that far, but if it instantaneously did, they would stop turning and slide.

Well I meant for a simulated car that has custom steering that you can set to 90 degrees (instead of the usual 30 degrees). At low speeds, if the simulation is behaving correctly, on tires with a high uS value and are at 90 degrees, the car should stop all forward movement.

The velocity for each wheel can be found knowing the vehicle velocity $v$ and its yaw angular velocity $\Omega$ ($v+\Omega d$; $d$ being the distance from the cg to the wheel).

I think the problem you referring to is for a force. When there is a purely lateral or longitudinal force act on a vehicle, the wheels are aligned, so the reaction forces can easily be split between both sides, just like a simple beam with 2 supports. When the force is at another angle, it can be more completed because you are now looking at a beam with 4 supports. So you have to separate the force into lateral and longitudinal components and analyze them separately.

Yes this is exactly it. The physics papers tell me that the friction force (not the rolling force, but the sideways cornering force) should be Fmax=uN. But they never tell me what ratios to apply it to each tire. The papers would work for say...a unicycle, but the ratios to apply to each tire is a mystery.

If there is no ackerman steering and only the front tires are turned, but the backwheels are going the same direction as the car, the ratio is easy - 0 lateral force applied to the back wheels, and 1/2 the Fmax applied to each front wheel. But when the car is sliding at non-uniform angles and the car tires are not the same angle as each other...the ratio is a mystery.

To find the ratio it is not simply by using a normalized weight distrubtion. For instance, if the weight is balance 25% weight on each tire, and you simply say Ff=uN*weightpercentage, it won't help, because imagine that the backwheels are going the same direction as the car, but the front wheels are angled...If the cars weight was still 25/25/25/25 the front tires would only apply 25/25 and not stop the car properly. So the problem is much more complex, and I'm not sure what the solution is. How would I go about analyzing it as a beam with four supports?

 

[jack action]

The magnitude of the (static) friction force doesn't depend on the weight or the friction coefficient, only the maximum static friction force does. So the value can be anything between zero and the maximum ($\mu_s N$). So you can analyze the vehicle as a static structure which is solidly pinned down. If any of the friction forces goes beyond the maximum value, that is when you have to change the conditions and set those forces to their kinetic friction values and redo your force analysis. You also know at this point that these tires can slide and the tire is no more "pinned down" and will move sideways if the other tires cannot hold the car either.

 

[quickquestion]

The magnitude of the (static) friction force doesn't depend on the weight or the friction coefficient, only the maximum static friction force does. So the value can be anything between zero and the maximum ($\mu_s N$).

Yes, I knew that already.

So you can analyze the vehicle as a static structure which is solidly pinned down. If any of the friction forces goes beyond the maximum value, that is when you have to change the conditions and set those forces to their kinetic friction values and redo your force analysis. You also know at this point that these tires can slide and the tire is no more "pinned down" and will move sideways if the other tires cannot hold the car either.

Yes I know, but how do I find out how to "ration" out the force.
Consider this. If a car is going 3 mph sideways, here's the problem...We only have a net max friction force of say, 3 units. So we must choose which tires to ration out the force to. We can only ration out so much force, if we do full force for each tire, the tire will actually have too much friction force and the car will go backwards and possibly violate the conservation of energy (possibly having more output than input - flubber tires.)

So... if the front tires are the only ones doing the lateral cancelling, then we apply /2 the front tires, none to the back. But if its a mix of it, then we may need to give 25% importance the the front tire only. But I don't know how to get the computer to automatically calculate these ratios. If I give 25% to all tires at all times, it is incorrect. But if I give 50% to the front left and 50% to the front right it is also incorrect. I have a net force (which I call the car force) that I have to divie up for each tire.

For example, if the car is going 3mph, then the friction force should full cancel out any lateral motion. If the front tires are doing all the cancelling, then its simple, we simply give 50% importance to each front tire. But if the back tires are doing some cancelling on their own...we might say give 25% importance to each front and back tire. But I don't know how to determine these dynamic ratios.

 

[jack action]

It is a statics problem: $\sum F_x = ma_x$, $\sum F_y = ma_y$, $\sum M = I\alpha$. Whatever net force is acting on your vehicle (please, stop talking about speed), you can decompose this force into 2 components: lateral and longitudinal. You analyze each force independently. This way, you can combine the wheels from the same axle (with the lateral force) or the same side (with the longitudinal force) and split the reaction 50/50 between the 2 wheels. Then you add the reaction forces for both analysis on each wheel.

If your wheels are angled with respect to the vehicle, you have to consider properly the reaction which is both lateral and longitudinal with respect to the wheel. So a pure lateral force on the vehicle will be partly lateral on the steered wheels and also partly rolling (considering rolling resistance, braking, acceleration).

Search for the bicycle model, where the left and right wheels of each axle are combined to a single one (the grey wheels in following figure):

 

[A.T.]

At low speeds, if the simulation is behaving correctly, on tires with a high uS value and are at 90 degrees, the car should stop all forward movement.

Then what did you mean by the below part?

...the engine should "know" to consider the velocity of the front tire as Half of the totalcar velocity, ...

 

[quickquestion]

Then what did you mean by the below part?

Okay, what I was saying was, if a car has a low force applied to it, that means you don't apply the full Fmax in the opposite direction to stop it's movement.

Spoiler: here's why.

Because if you do, instead of 100% cornerning grip, you may have 150% or 200% (as if you are hitting and bouncing off an invisible wall - the car will actually go backwards like it hit something.) And if you have 201% cornering grip, you actually have flubber physics where the car bounces off with more energy than had in the first place.

So basically, if a car has low force on it, you cannot apply the full Fmax to each tire, rather you can only apply the exact amount of force it takes to make the car stop moving (Force=mass*acceleration, so the amount of force you have to work with is carmass*car_velocity/60) Because you want to apply a force (an impulse at 1 frame) which makes the car stop moving. If the framerate is 60 fps, then it's /60. If the framerate is 30 its /30. If the car velocity is per frame and not per second then the equation changes. Anyway, the main point of it is, you have a set amount of force needed to stop the car. And you have to figure out how much of it to apply to each tire in order to get the right behavior. So lectures (like from iforce2d) don't give the right behavoir because they don't accurately divie up the force for each tire. It is not as simple as saying "Oh apply a counter force to each tire to stop all lateral movement." Because what most of these lectures don't tell is the ratio of the carforce (counter movement force) that must be divied unto each tire.

Search for the bicycle model, where the left and right wheels of each axle are combined to a single one (the grey wheels in following figure):Search for the bicycle model, where the left and right wheels of each axle are combined to a single one (the grey wheels in following figure):

Using a bicycle model seems to be the real deal, though that picture is a lot for me to digest at the moment (Just last night I forgot the lock combo I have been using everyday.) But from what I can sense, Jack and I seem to getting towards the same page. I am actually going to start working on modelling bicycle physics and then work on the problems from there. My guess is, you get the net lateral velocity on the chassis, and then get the longitudinal velocity of the chassis, and go from there? Still not understanding fully, but hopefully it will "click" soon.

 

[quickquestion]

Ok, so this a picture to basically demonstrate the problem I'm having:

Basically, to Jack, this picture explains why the 50/50 thing won't work...in diagram 2, the front tire should be given 100% lateral stopping responsibility...in diagram 1 it should be 70/30 or something along those lines.

For now on, I am using "lateral stopping responsibility" as a terminology.

 

[jack action]

For now on, I am using "lateral stopping responsibility" as a terminology.

Please stop using this made up terminology, it is very difficult to follow. You should be talking about force, moment, mass, inertia and acceleration at this point.

On the top figure the central red arrow if a side force that pushes on the car. It can be the wind (it would act at the center of pressure) or the reaction force due to an acceleration (it would act at the center of gravity). The reaction force from the tires are in blue and are purely lateral forces for each tire. The rear/front ratio of the tire force magnitudes depends only if the wind force is closer to one axle or the other.

I put 2 dotted red arrows to represent the portion of the wind force acting on each axle (which are equal and opposite to the tire forces).

The bottom figure is the same situation, but with a steer angle on the front axle. I did not put the wind force, just its 2 components which are the same as the in top figure.

The rear axle is exactly the same. For the front, the wind force component is further split in two: a lateral (green) and longitudinal (brown) forces with respect to the tire. The lateral component is opposed by the blue tire force (equal and opposite). The longitudinal component is in the rolling direction of the tire. It will tend to move back the front tire and pivot the entire car around the rear axle. If the car is rolling, then rolling resistance have to taken into consideration. If there is a front wheel torque applied (braking or acceleration), it must also be taken into consideration. If there is an acceleration torque on the front wheel, it can cancel out the wind force component that wants to roll the front tire backward.

If there was an acceleration torque on the rear wheel (horizontal), it would be reacted by the front tire (horizontal) which, once split into two tire components, would increase the lateral tire force but would also have a longitudinal component to prevent the front wheel from backing as well.

 

[quickquestion]

Ok, most of this I am getting, and also I want to say thanks for the help. I will highlight stuff in red I am having trouble understanding and put a descriptive explanation under it explaining what I don't understand.

On the top figure the central red arrow if a side force that pushes on the car. It can be the wind (it would act at the center of pressure) or the reaction force due to an acceleration (it would act at the center of gravity). The reaction force from the tires are in blue and are purely lateral forces for each tire. The rear/front ratio of the tire force magnitudes depends only if the wind force is closer to one axle or the other.

Since wind force is only for discussion purposes, I'm wondering how we would apply this knowledge in terms of frame-by-frame simulated dynamics. In diagram 2 of my post (post #31) it shows a front steering angle of 90. Should we use wind-force as a metaphor, would it be safe to say that the location of the "wind-force" is solely on the front tires? Is this a safe-metaphor to use, and if so, how then do we get the game engine to "know" the central location of the wind-force in both diagrams 1 and 2 of post #31? (if, instead, we replace "wind-force" with instead, sliding behavior where the "locus" of the force is unknown.") Not sure what the properly terminology is and what not, perhaps chassis pressure or something, but something to better illustrate the ratio of lateral force which should be "divied" upon each tire. Which I would also assume has some influence from angular velocity and weight distribution of the car. The difficulty I'm having is getting the game engine to "know" and handle these different situations, such as giving 100% lateral force ability to the front tires in diagram 2 of post #31. (Even if the weight is mostly on the back tire, the fronts should be the only ones doing the work of lateral cancellation, in this case of diagram 2 of post #31.)

I put 2 dotted red arrows to represent the portion of the wind force acting on each axle (which are equal and opposite to the tire forces).

The bottom figure is the same situation, but with a steer angle on the front axle. I did not put the wind force, just its 2 components which are the same as the in top figure.

The rear axle is exactly the same. For the front, the wind force component is further split in two: a lateral (green) and longitudinal (brown) forces with respect to the tire. The lateral component is opposed by the blue tire force (equal and opposite). The longitudinal component is in the rolling direction of the tire. It will tend to move back the front tire and pivot the entire car around the rear axle. If the car is rolling, then rolling resistance have to taken into consideration. If there is a front wheel torque applied (braking or acceleration), it must also be taken into consideration. If there is an acceleration torque on the front wheel, it can cancel out the wind force component that wants to roll the front tire backward.

If there was an acceleration torque on the rear wheel (horizontal), it would be reacted by the front tire (horizontal) which, once split into two tire components, would increase the lateral tire force but would also have a longitudinal component to prevent the front wheel from backing as well.

Assuming you are referring to the bottom diagram of post #32?

If there was an acceleration torque on the rear wheel (horizontal), it would be reacted by the front tire (horizontal) which, once split into two tire components, would increase the lateral tire force but would also have a longitudinal component

I assume what you are referring to would look something like this?

If there was an acceleration torque on the rear wheel (horizontal), it would be reacted by the front tire (horizontal) which, once split into two tire components, would increase the lateral tire force but would also have a longitudinal component

to prevent the front wheel from backing as well.

I am not sure what this means.

 

[jack action]

Should we use wind-force as a metaphor

Use the terms lateral and longitudinal, both either with respect to the car or the tire.

would it be safe to say that the location of the "wind-force" is solely on the front tires?

In this case, since there are on 2 forces involved, they must be opposite and equal.

how then do we get the game engine to "know" the central location of the wind-force in both diagrams 1 and 2 of post #31?

It should be a characteristic of the vehicle. If it is an aerodynamic force, it acts at the center of pressure of the car. If it is inertia (an acceleration), it acts at the center of gravity. Both of these locations depend on the vehicle design.

Assuming you are referring to the bottom diagram of post #32?

Yes. It is similar to your diagram #2.

I assume what you are referring to would look something like this?

Yes.

I am not sure what this means.

 

[jack action]

Should we use wind-force as a metaphor

Use the terms lateral and longitudinal, both either with respect to the car or the tire.

would it be safe to say that the location of the "wind-force" is solely on the front tires?

In this case, since there are only 2 forces involved, they must be opposite and equal. And the car can move if the front wheels are sliding.

how then do we get the game engine to "know" the central location of the wind-force in both diagrams 1 and 2 of post #31?

It should be a characteristic of the vehicle. If it is an aerodynamic force, it acts at the center of pressure of the car. If it is inertia (an acceleration), it acts at the center of gravity. Both of these locations depend on the vehicle design.

Splitting the force between the two axles or between the left and right wheels is just a matter of basic geometry. Right in the middle is 50/50, centered on one axle or the other is 100/0 or 0/100 and anything else in between varies in proportion between those extremes.

Assuming you are referring to the bottom diagram of post #32?

Yes. It is similar to your diagram #2.

I assume what you are referring to would look something like this?

Yes.

I am not sure what this means.

If you push sideways on the vehicle with the front wheels steered, the front wheels will back up. If you apply torque to the rear wheels, it should counteract the phenomenon (C_sf in your figure).