What Equation Defines μK in Tire Physics?

[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).

 

[quickquestion]

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 (Csf in your figure).

Ok, got that part, makes sense.

Ok, so I guess to reformulate my question, would be to temporarily abandon the bicycle model and go back to the original car model.

 

[jack action]

You use the bicycle model and then split the result equally between the 2 wheels of the axle. If one wheel exceeds its limit (because of weight transfer), then only the other one can increase until its own limit (which have more weight on it).

 

[quickquestion]

Ok, so back to the bicycle model, assuming in an ingame engine we see a moving body (the bike) but do not have access to the information of where the last force was applied, the only information we have access to is the velocity of the car and the angular velocity of the car. We only see a moving car, but no knowledge of the origins of forces that were applied to the car.
So basically, I want to make sure I understand the bicycle model 100%, then move onto integrating 2 bicycles and putting them together as one car.
Basically, imagine you have a custom bicycle where the back tire can rotate independently of the chassis.
Just to make sure I got the bicycle physics down pat, here is a picture to illustrate:

 

[jack action]

The only way I can think of, that it can be done:

1. With the forces and moments acting on the vehicle, find the accelerations of the vehicle. Beginning here, it doesn't matter anymore how your wheels are positioned;
2. With the found accelerations and the initial velocities, find the new velocities;
3. With the velocities and the initial position, you find the new position;
4. Repeat.

You cannot find forces by knowing velocities. One is a statics analysis, the other is a kinematics analysis.

The stuff you say makes no sense at all: «Would tire A's lateral force be 75% of [velocity] 30». A force cannot be based on a velocity. They are different things, independent of each other.

In the case you presented, If you know the linear velocity of one wheel, it is necessarily the same for the other (they are link by the frame). If there is an angular velocity about a known point, then there will be another velocity component added to each wheel, which will vary in direction and magnitude.

Here is a classical representation of an object that has a linear ($v_B$) and angular ($\omega_{@B}$) velocities. You can imagine points A & B to be the tires of a vehicle. Once you know the velocities of the vehicle, it doesn't matter what are the angle of the tires, they have to follow.

 

[A.T.]

Just to make sure I got the bicycle physics down pat, here is a picture to illustrate:

I don't see any physics in your diagrams. You just seem to make up some percentages, without any justification. Try a free body diagram.

 

[quickquestion]

I don't see any physics in your diagrams. You just seem to make up some percentages, without any justification. Try a free body diagram.

The pic in #38 is a free body diagram, it displays the 2 ground forces (friction forces) resisting forward motion of the body. The one crucial thing it is missing, is it doesn't say the actual force which caused the velocity to be 30, it just says the current velocity is 30.

The reason I mention this, is because many physics engines don't allow you to access what forces are acting on an object, they only allow you to apply a force. But if a car hits a wall, there is no function to get the force applied on a car.

But there is an error in my diagram which I will point out. The error is that the rear friction force is tilted when in the description it says braking is applied. I will correct the diagram and x-out the error and show what it should be:

The only way I can think of, that it can be done:

1. With the forces and moments acting on the vehicle, find the accelerations of the vehicle. Beginning here, it doesn't matter anymore how your wheels are positioned;
2. With the found accelerations and the initial velocities, find the new velocities;
3. With the velocities and the initial position, you find the new position;
4. Repeat.

Ok, so I'm thinking if I track the accelerations of the tire (I should be able to in the game engine) I can find the acceleration of the tires. The problem here is, Force=mass*acceleration applies to a general body. If the front tire is accelerating 3 units, and the back tire is accelerating 2 units, obviously the force for the front tire is not carmass*3, nor is it tiremass*3. And that, I think, is the problem I'm having trouble wrapping my head around...what is the "practical mass" of the tire.

Also, the diagram and text I put in post 38# is just bad...I was getting kind of tired and I should have used accelerations instead of velocities.
Wait no I remember why I put velocity instead of acceleration now. It was for the ice scenario...Imagine you are running a simulation, and the car is on ice, just sliding along without any acceleration. And you flip the switch and change the surface material to concrete...The simulation will not know how to handle this situation, because the friction force should be applied regardless if no other force was applied to the car.
(The diagram was still bad though, because it had an error.)

 

[jack action]

The problem here is, Force=mass*acceleration applies to a general body. If the front tire is accelerating 3 units, and the back tire is accelerating 2 units, obviously the force for the front tire is not carmass*3, nor is it tiremass*3. And that, I think, is the problem I'm having trouble wrapping my head around...what is the "practical mass" of the tire.

You just find the force ($ma$) for the entire car. Then it becomes something like the "wind force" we discussed earlier and it acts at the CoG. Now you just refer to the car geometry to split the force between left & right (longitudinal force) or front & rear (lateral force) set of wheels. Then split each found force again 50/50 between the two wheels of each set. If the maximum friction value of one of the tire of the set is exceeded, use the kinetic friction value instead (sliding) and put the difference on the other tire. That should be pretty close to what happens in reality.

 

[A.T.]

The pic in #38 is a free body diagram, it displays the 2 ground forces (friction forces) resisting forward motion of the body.

The force distribution between front and rear wheel will mainly depend on how high the center of mass is. Where is that in your diagram or your derivation?

 

[quickquestion]

You just find the force ($ma$) for the entire car. Then it becomes something like the "wind force" we discussed earlier and it acts at the CoG. Now you just refer to the car geometry to split the force between left & right (longitudinal force) or front & rear (lateral force) set of wheels. Then split each found force again 50/50 between the two wheels of each set. If the maximum friction value of one of the tire of the set is exceeded, use the kinetic friction value instead (sliding) and put the difference on the other tire. That should be pretty close to what happens in reality.

Ok, I am going to break it down into steps and to show you some problems I'm running into.
1. You just find the force ($ma$) for the entire car.
In the ice scenario, a car is on a testbed simulated ice surface, sliding sideways 90 degrees. According to Isaac Newton, it has no force, since it is not accelerating, just sliding at a constant velocity. So when I hit a switch and change the surface material from ice to concrete, the force that we are talking about now is the frictionforce on the ground. Which, can only be this: frictionforce to apply= min(maxfrictionforce,carvelocity*carmass).
Once again, velocity is needed here (Because according to Newton, no actual force is yet associated with the car, since it was sliding at a steady velocity. There is no actual force on the car whatsoever, except the hypothetical friction force which we are about to apply, but haven't applied yet. So according to Isaac, the car force is zero. Thus, in order to determine what the frictionforce should be, we can't actually reference a force, the only way we have to determine it is by thinking of the hypothetical future force that should be applied on it by the end of the frame.)
2. Then it becomes something like the "wind force" we discussed earlier and it acts at the CoG.
Assuming that our force is: hypothetical frictionforce to apply=min(maxfrictionforce,carvelocity*carmass), then if the COG is in the middle, then if both tires are angled symmetrically, then we divide by 2. So it would be tireA applied friction force=min(maxfrictionforce/2,carvelocity*carmass/2)
So it would be tireB applied friction force=min(maxfrictionforce/2,carvelocity*carmass/2)
But...if one tire has no grip, or is angled, or off the ground, etc...Then all or most of the burden goes on the gripping tire. So there has to be some kind of equation to determine the amount of burden. So if tire A loses grip for some reason, or is angled 90 degrees and doesn't resist much, the grip of tireB would look something like: tireA applied friction force=min(maxfrictionforce*0,carvelocity*carmass*0) (zero burden), tireB applied friction force=min(maxfrictionforce,carvelocity*carmass)
So there is something I am missing here besides just the COG, I am missing the equation to determine the ratios of burden.
3. If the maximum friction value of one of the tire of the set is exceeded, use the kinetic friction value instead (sliding) and put the difference on the other tire.
That would cause an imbalance. If I apply the maximum friction value to tire A, then there is no friction work left for tire B to do, creating a torque.

The force distribution between front and rear wheel will mainly depend on how high the center of mass is. Where is that in your diagram or your derivation?

Oh...I didn't know. That could be important. What is the equation for that?

 

[A.T.]

What is the equation for that?

Draw a free body diagram from the side. The moments on the bike must balance in that plane, unless it's flipping over.

 

[quickquestion]

Here is my diagram so far:

Now, the principle of moments is:

The principle of moments

When an object is balanced (in equilibrium) the sum of the clockwise moments is equal to the sum of the anticlockwise moments.

Force 1 x distance 1 from pivot = Force 2 x distance 2 from pivot

F1xd1 = F2xd2

But this implies that an object can become imbalanced. I am wondering, what are these imbalance conditions? When you say "flipping over", I assume that this is an imbalance condition?

Also, these equations would simplify in my case to, this, I think? (Could be wrong but I think I'm right.)
FrictionForce1=TotalFrictionForce*(totaldistance/distance1)
FrictionForce2=TotalFrictionForce*(totaldistance/distance2)

I'm trying to figure out what role the height of COG has yet to play in these equations (Assuming that this is a top down 2d bike simulation.)In any case, I kind of want to basically put aside all of the above texts and diagrams I wrote and think about this problem in a new way.
The discussion I was having with Jack action was along the lines of a BMX finger bike...you push the bicycle with your finger, and so it's easy to see the friction burden is mostly on the tire that is being pushed by your finger.

But I want to think about this in a new way. I want to introduce a new component to the discussion, which is the center of rotation. Imagine a lego beam, where you can stick an axle anywhere on the roof of the car...so on a top-down view you can rotate it any manner you wish. I will illustrate in the following diagram:
So here it is, the problem simplified.

Once I figure that part out...
The second component to the problem is, this:

 

[jack action]

In the ice scenario, a car is on a testbed simulated ice surface, sliding sideways 90 degrees. According to Isaac Newton, it has no force, since it is not accelerating, just sliding at a constant velocity. So when I hit a switch and change the surface material from ice to concrete, the force that we are talking about now is the frictionforce on the ground. Which, can only be this: frictionforce to apply= min(maxfrictionforce,carvelocity*carmass).
Once again, velocity is needed here (Because according to Newton, no actual force is yet associated with the car, since it was sliding at a steady velocity. There is no actual force on the car whatsoever, except the hypothetical friction force which we are about to apply, but haven't applied yet. So according to Isaac, the car force is zero. Thus, in order to determine what the frictionforce should be, we can't actually reference a force, the only way we have to determine it is by thinking of the hypothetical future force that should be applied on it by the end of the frame.)

If there is a relative velocity between the tire and the ground, then you are sliding and you have to use the full kinetic friction force. It then becomes an input force to your system, regardless you have other forces or not. This force will decelerate your vehicle.

Assuming that our force is: hypothetical frictionforce to apply=min(maxfrictionforce,carvelocity*carmass), then if the COG is in the middle, then if both tires are angled symmetrically, then we divide by 2. So it would be tireA applied friction force=min(maxfrictionforce/2,carvelocity*carmass/2)
So it would be tireB applied friction force=min(maxfrictionforce/2,carvelocity*carmass/2)
But...if one tire has no grip, or is angled, or off the ground, etc...Then all or most of the burden goes on the gripping tire. So there has to be some kind of equation to determine the amount of burden. So if tire A loses grip for some reason, or is angled 90 degrees and doesn't resist much, the grip of tireB would look something like: tireA applied friction force=min(maxfrictionforce*0,carvelocity*carmass*0) (zero burden), tireB applied friction force=min(maxfrictionforce,carvelocity*carmass)
So there is something I am missing here besides just the COG, I am missing the equation to determine the ratios of burden.

It is imperative that you learn how to do a free body diagram to answer that question. This is how you built the equations to determine those unknowns. In the first part of your statement, you are actually making a free body diagram, only you are doing it intuitively. (It is true that you have to split the force in two, but why?)

That would cause an imbalance. If I apply the maximum friction value to tire A, then there is no friction work left for tire B to do, creating a torque.

You only apply the maximum friction value to tire A if the total friction force needed is greater than that value. There will not be an imbalance or a torque created, because both forces are colinear.

 

[quickquestion]

You only apply the maximum friction value to tire A if the total friction force needed is greater than that value. There will not be an imbalance or a torque created, because both forces are colinear.

Well, if the car is going at a non-perpendicular angle, the forces wouldn't be colinear. So if I applied the full friction force to the front tire, instead of the back, it would have a different behavoir than if I applied the full friction force to the back and none to the front.

I have been doing some deep thinking about this and I have decided that I need to do a free body diagram, but at a molecular level. Here is my theory.

Spoiler: Here is my theory...it is only a theory, not sure if fact.

The friction force is a force, but not a single thrust force in the sense that the high school diagrams describe...it is actually a set of 2 forces, frontwards and backwards, like a gear network. Essentially, it is a damping force/field, not a thrust force.

 

[jack action]

Well, if the car is going at a non-perpendicular angle, the forces wouldn't be colinear.

Yes it always will, if you treat the lateral and longitudinal components of the force (with respect to the vehicle) separately.

 

[quickquestion]

Well the 2 forces would be parallel, but they wouldn't be on the same line. If the car is turned 45 degrees (in world space) but the tires are 90 degrees in world space, the 2 lines of friction force on the tires would be parallel but spaced apart.