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avoiding vibration of car

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So far I haven't able to find a solution to avoid vibration of my car. The tyre's lateral force is opposite to the lateral velocity of the tyre. So the lateral velocity is oscillating. Increasing the timestep the situation gets even worse. I know that I should damp it somehow, but I cannot distinguish the oscillation from the normal movement. Have somebody an idea, how to avoid this effect?

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I assume the force you mension is a friction force. If so then I think you may be making a mistake when calculating the magnitude of the friction force. This is because the friction force should be proportional to the velocity and should never be able to reduce it by more than 100%. As a result the velocity should never oscillate due to friction alone. Please could you list all the forces you are applying to the tyre and I might be able to be of more help.

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Idd, you must check when applying friction that the friction force does not cause the wheel to actually start moving in the opposite direction.

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Well, ignoring one dimension, the vertical axis, and assuming I don't accelerate or brake, but I can steer, the forces, torques that acts:

On the body (and thus the wheels):
- total lateral force of the tyres
- total longitudinal force of the tyres
- air resistance (no effect on this phenomenon)
On each wheel:
- longitudinal force as reaction torque
- rolling resistance torque (no effect too)

And these forces on the body create torque of course, which beside the forces moves, rotates the car. so it oscillates linearly and rotationally too, I think.

There is also a longitudinal vibration when the car is resting, but I have idea for that.
I only worry about lateral, and thus rotational vibration.

Theoritically friction cannot reduce the velocity more than 100 %, but with stiffer tyres, larger timesteps it may occur (the velocity simply changes too much within a timestep, and change sign).
I am able to detect that the sign if the lateral velocity of a tyre changes. This can be caused by oscillating, but also by normal motion. I cannot find out which is just the case.
Ohhh, I should filter out the high frequencies, with an FIR, or IIR filter. With fixed timestep it works easily. The body cannot do so much angular rotation change. Exception is the collision.

But I am waiting forward to your opinion, suggestion.

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First ignore what I said about friction being proportional to velocity, I was thinking air friction. Secondly, them forces seem sensible so I'm geussing the problem is maybe a misunderstanding in the application of friction. Are you distinguishing between static and kinetic friction? Also do you always apply the friction to directly oppose the motion of the tyre or are you applying it in the lateral and longitudinal directions?

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The tyre model calculates the lateral and the longitudinal forces. Lateral is perpendicular to the wheel, the other is parallel. The lateral force is always opposite the lateral motion.
The input of it are the velocity vector of the wheel, the angular velocity of it, the normal force of it, and the heading (forward) vector of it. From these I can find out the slip angle too.
You are right that with tyres there is static and kinetic friction. Kintic friction is for the sliding part of the contact patch, static for the non-sliding one. I calculate them in the tyre model. But in case of static friction, the force should be also applied, because in this case the deformation of the tyre is what transmits the force. Believe me.
I think, the force I get from the model should be applied, because it works OK, when I go with 60 km/h or any. At extremely low speed have I problems. Should I use a different, simpler tyre model with such small speeds?
I think this oscillation also is present when going faster, but it gets negligable in this case. As the speed decreases, it has more and more effect on the slip angle. When the car is resting, the slip angle is very-very oscillating. With a small timestep (1 ms) it's not visible, but with 0.01 s ...

As for other frictional forces, I have solved it well.

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Don't you already know what the timestep is when you calculate the friction force? If so, you can find out if the magnitude of your friction force will cause the wheel to slide in the opposite direction (before you actually come to apply it).

You should also be able to calculate the force necessary to stop the wheel movement completely within 1 time step, so you can just set the friction force equal to this magnitude.

In any case you will want to deal with it while you are calculating the magnitude of the friction force, rather than after you have resolved all your forces. Because as you said yourself, you won't be able to tell the difference anymore between friction force and input from player for example.

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Quote:
 Original post by SpaceDudeDon't you already know what the timestep is when you calculate the friction force? If so, you can find out if the magnitude of your friction force will cause the wheel to slide in the opposite direction (before you actually come to apply it).You should also be able to calculate the force necessary to stop the wheel movement completely within 1 time step, so you can just set the friction force equal to this magnitude.

I am afraid, it is very complex. The car can be on slope, or on any surface with any friction coefficient.

Quote:
 In any case you will want to deal with it while you are calculating the magnitude of the friction force, rather than after you have resolved all your forces. Because as you said yourself, you won't be able to tell the difference anymore between friction force and input from player for example.

No, I cannot tell if a force is due to player input or this nasty oscillating. This because player input creates torque on wheels (gas, brake, handbrake), or creates lateral force implicitely (steering). So I (and I think everybody) am unable to tell the reason of a lateral force. "Normal" lateral force can generated without user input too: imagine a slope where the car is rolling on it, perpendicular to the slope direction. In this case the car will slowly slip down the slope (even if there is adhension/tracktion), slowly because of the lateral force.

Thanks for the suggestion anyway.

I personally would apply the calculated forces, and after that from the velocity's sign change I would detect this oscillation somehow (?). Like with air resistance, or Columb friction.
The main difference is that I have 4 tyres not 1, and their motion are not independent.
This is not a simple simulator. It is based on real forces and torques. Please try it, then you will see the situation better. The URL is in my profile.

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It's tire, not tyre. o.O

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Ok first off friction always acts to directly oppose the motion, hence although you calculate the friction in the seperate directions to account for different friction coefficients, the resulting forces should be applied opposite to the tyres velocity direction. Secondly if the motion force is small enough that it lies below the maximum static friction then the static friction force should be scaled to cancel out the motion force. Hence static friction should never result in a reversal of direction. Last but not least, when kinetic friction comes into play it means that there is already a large motion force. For the kinetic friction force to reverse the velocity it will have to be extremely large which suggests you are miscalculating the grounds normal force or that you have an unrealistic friction coefficient (I suggest about 0.8). You might want to check your units just to make sure you're not inputting values which are magnitudes of order out. Let me know if any of this helps.
On second thoughts concerning the direction of application of the friction force, you might be right in applying it in the two seperate directions. Try both and see which one looks best.

[Edited by - Motorherp on July 1, 2004 3:10:39 AM]

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Quote:
 Original post by szinkopaI personally would apply the calculated forces, and after that from the velocity's sign change I would detect this oscillation somehow (?). Like with air resistance, or Columb friction.The main difference is that I have 4 tyres not 1, and their motion are not independent.

Well that sounds like a hack to me... and IMO, if you want your game to act realistic, you should do things properly.

I hate to repeat what has already been said several times but just make sure that the friction force does not cause a change in velocity in the first place, then you don't need to worry about it later on.

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Thank you very much for the help, but

Motorherp:

Quote:
 Original post by MotorherpOk first off friction always acts to directly oppose the motion, hence although you calculate the friction in the seperate directions to account for different friction coefficients, the resulting forces should be applied opposite to the tyres velocity direction.

This is true for the kinetic friction. The dirction of static friction depends on other forces acting on the object. But tyres are a particular case, they are not rigid.
In my tyre model the contact patch has a static and a sliding region. For the sliding region the friction force I apply is opposite to the direction the tyre is moving (and I decompose it to longitudinal and lateral components). This is OK, isn't it?
At the static region the tyre is deforming in both directions due to the tracktion. From the degree of deformtion (slip angle, slip ratio), I get the static longitudinal and lateral forces.

Quote:
 Original post by Motorherp Secondly if the motion force is small enough that it lies below the maximum static friction then the static friction force should be scaled to cancel out the motion force. Hence static friction should never result in a reversal of direction. Last but not least, when kinetic friction comes into play it means that there is already a large motion force.

Sorry, from this I cannot see, what you mean on motion force exactly.

Quote:
 Original post by Motorherp For the kinetic friction force to reverse the velocity it will have to be extremely large which suggests you are miscalculating the grounds normal force or that you have an unrealistic friction coefficient (I suggest about 0.8).

In my game if I skid in a corner, and say lock the wheels, then I begin to slide in a direction on flat ground (and possibly rotate). The sliding force is, what slows my car until it stops. And when it stops, then it begin to oscillate, but this is not visible, I know it from displaying slip angle. The friction coefficient is speed difference dependent (this is the realistic approach). On asphalt it is 1.1 at speed diff. of 0 (static friction), and decreases linearly, and reaches 0.6 at 40 km/h speed difference. After that it doesn't decrease more. I saw this graph in a paper somewhere. It is measured data.

Quote:
 Original post by Motorherp You might want to check your units just to make sure you're not inputting values which are magnitudes of order out. Let me know if any of this helps.

I think my units are OK, because at higher speeds it behaves normal, and can be controlled quite well.

Quote:
 Original post by MotorherpOn second thoughts concerning the direction of application of the friction force, you might be right in applying it in the two seperate directions. Try both and see which one looks best.

Sorry, I don't understand this cornering. Do you mean that I calculate longitudinal and lateral forces separatly? It doesn't matter for the motion of the body, because all of them will be applied on it. However, only the longitudinal component acts on the tyre as a reaction torque. That's why it have to be separated.

SpaceDude:

Quote:
 Original post by SpaceDudeWell that sounds like a hack to me... and IMO, if you want your game to act realistic, you should do things properly.I hate to repeat what has already been said several times but just make sure that the friction force does not cause a change in velocity in the first place, then you don't need to worry about it later on.

You may have misunderstoond something. I check the sign change of the object's velocity for these frictional forces, like air resistance, or brake torque, or rolling resistance. And this method is what you mention too. And I think it MUST be done such way. I check the sign change, because these forces cannot change the sign of any coordinate of the velocity, and cannot add energy to the system (hence cannot increase the abs() of any coordinate of the velocity). What's the hack in this thought? This results realistic behaviour for rigid bodies.

But tyres are different again. The deformability. The deformation (for the static region of the contact patch) works like a damped spring, so it is not simply a frictional force.
For example when a car begins turning, the angular velocity of it is increasing. A frictional force cannot do that (increasing energy). The turning is due to the adhension of the static region of the contact patch. Here the tyre is deformed, and that's what turns the car. Roughly like a spring force. If the steered tyres don't stick, then you cannot turn.

Both:

Thanks again for both your help. Maybe the situation and the work of the tyre is clearer now. Any more help is appreciated.
I think, I have to do something with the force of the static region. Motorherp's 2nd sentence might be the most valuable thought, if I could understand "motion force".

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Unfortunately I've never done any work on friction with deformable bodies so I may be off the mark but here goes anyway:
Quote:
 Original post by szinkopaThe dirction of static friction depends on other forces acting on the object. But tyres are a particular case, they are not rigid.

The applied friction at any one point is either static or kinetic and can't be both at once. Static friciton acts to completely cancel out the components perpendicular to the ground normal of all forces acting on the body. In a soft body, ie tyre, this would explain why you get areas of deformation due to some parts of the tyre sticking and others slipping. Hence what you need to do is sum up all the perpendicular components of all acting forces and if the resulting force's magnitude is less than the max static friction then simply zero this force (equivalent to applying an opposite friction force of equal magnitude).
Quote:
 Original post by szinkopaSorry, from this I cannot see, what you mean on motion force exactly.

By motion force I simply mean the force which is driving your object. Don't worry about my last statement from my previous post, just mind dribble.
It seems to me it might be a misunderstanding of static friction which is causing your problem. There is obviously a lot more going on in your simulation than this though, a lot of which I have no prior knowledge off, so I'm probably wrong and the problem might lie else where. It sounds like an interesting subject though and I might tackle it myself some time in the future but at the moment I've got my hands tied with ragdoll and a V-clip collision engine. Just let me know if there is anything else I can help with.

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I think this is just an inheret limitation in the Pacejka type tire modeling. I had a similar problem where the brakes would cause my wheel to overshoot zero angular velocity a bit, and it would oscillate a while before finally coming to zero. Even using the "known dt to solve for the exact torque required to reach zero velocity" method didnt work 100% of the time to dampen it, as there are other torques effecting the wheel that may not be known at any given time.
If I read your problem correctly, its that the car 'jitters' at low velocities? This could also be due to the slip angle calculations. What I've always done is use the wheels velocity vector and "this end foreward" vector to derive its slipAngle. However at very low speeds, the velocity vector fails to be valid, as attempts to normalize it could result in a divide by zero type thing.
I think the solution would involve some kind of 'sticktion'.
Realize that real tires actual apply a force even when they arnt sliding, however in most computer models, they only work WHEN sliding (ie: its velocity vector differes from that of its direction, creating a slipAngle). Note that this also effects your ability to park on hills without sliding down them.

I'm not 100% sure how you would determine the force of this 'sticktion'. You may need to find the lateral component of all the outside FORCES acting on each wheel. Then have the tire counteract those up to a set amount (based upon the load on that tire). If it isnt enough, the tire will slide and you can revert to the slipAngle calculations.

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CombatWombat:

What do you mean on Pacejka-type tyre model? Empirical models?
My one is based on physics, with some simplifications, and is a bit CPU-hungry :(.
As for sticking, I have noticed in V-Rally 3, that the car simply stops below a certain speed. It's a bit strange.
Your idea good, I will think about it.

Everybody:

I think the root of everything is the angular "jittering" of the wheels at low speed. With 0.01 s timestep this is quite "spectecular". This jittering creates longitudinal force at each wheel, and thus torque on the body, which rotates it. And that's why lateral force is created (I don't know better word for this, please help). And then everything is oscillating.

So first I have to eliminate this angular jittering, and after that I will see what happens.

Thanks for your help. I needed this discussion to realize the root of the problem.

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Quote:
 Original post by DeyjaIt's tire, not tyre. o.O
Not in my country it's not.

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Quote:
 Original post by szinkopaYou may have misunderstoond something. I check the sign change of the object's velocity for these frictional forces, like air resistance, or brake torque, or rolling resistance. And this method is what you mention too. And I think it MUST be done such way. I check the sign change, because these forces cannot change the sign of any coordinate of the velocity, and cannot add energy to the system (hence cannot increase the abs() of any coordinate of the velocity). What's the hack in this thought? This results realistic behaviour for rigid bodies.But tyres are different again. The deformability. The deformation (for the static region of the contact patch) works like a damped spring, so it is not simply a frictional force.For example when a car begins turning, the angular velocity of it is increasing. A frictional force cannot do that (increasing energy). The turning is due to the adhension of the static region of the contact patch. Here the tyre is deformed, and that's what turns the car. Roughly like a spring force. If the steered tyres don't stick, then you cannot turn.

I agree that the friction cannot cause a change in sign of the velocity, and it cannot add energy to the system. But what you where talking about (if i understood your correctly) is that you will add all your forces together (friction, wheel torque, air resistance, rolling resistance, gravity, ground reaction force, etc...) then calculate a new velocity for your car. And check if the new velocity of your car has changed sign or not. At this point it is too late to check for a change in sign since any of the other forces could have caused it rather than friction.

What i am suggesting is that you see what effect the friction force ALONE has on the car, or the wheel more accurately. You see what i mean?

>>For example when a car begins turning, the angular velocity of it is increasing. A frictional force cannot do that (increasing energy).

I dissagree, it is precisely the friction between the wheel and the ground which causes the car to turn. The energy does not increase, it is simply transfered from linear velocity to angular velocity.

>>Here the tyre is deformed, and that's what turns the car. Roughly like a spring force.

I downloaded your demo, good work btw it looks great. But i don't see any deformation of the wheel? do you mean to wheel moving relative to the car due to the springs on the axle?

>>In my tyre model the contact patch has a static and a sliding region.

It is concevable that part of the tire will be static while another part is sliping for a VERY brief moment at the point where the wheel makes the transition from static to sliding if your wheel is able to deform. But i don't think this is what you are talking about, since i can't see any deformation on your wheels. And quite frankly that would just be overkill trying to model that effect.

So please explain what you mean by, static and sliding region.

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Quote:
Original post by SpaceDude
Quote:
 Original post by szinkopaYou may have misunderstoond something. I check the sign change of the object's velocity for these frictional forces, like air resistance, or brake torque, or rolling resistance. And this method is what you mention too. And I think it MUST be done such way. I check the sign change, because these forces cannot change the sign of any coordinate of the velocity, and cannot add energy to the system (hence cannot increase the abs() of any coordinate of the velocity). What's the hack in this thought? This results realistic behaviour for rigid bodies.But tyres are different again. The deformability. The deformation (for the static region of the contact patch) works like a damped spring, so it is not simply a frictional force.For example when a car begins turning, the angular velocity of it is increasing. A frictional force cannot do that (increasing energy). The turning is due to the adhension of the static region of the contact patch. Here the tyre is deformed, and that's what turns the car. Roughly like a spring force. If the steered tyres don't stick, then you cannot turn.

I agree that the friction cannot cause a change in sign of the velocity, and it cannot add energy to the system. But what you where talking about (if i understood your correctly) is that you will add all your forces together (friction, wheel torque, air resistance, rolling resistance, gravity, ground reaction force, etc...) then calculate a new velocity for your car. And check if the new velocity of your car has changed sign or not. At this point it is too late to check for a change in sign since any of the other forces could have caused it rather than friction.

What i am suggesting is that you see what effect the friction force ALONE has on the car, or the wheel more accurately. You see what i mean?

OK. I explain it more exactly. Taking a dirction/axis 3 conditions must be fulfilled to stop the object along this direction:
- The sign of the velocity has changed
- The magnitude of the total frictional forces in this direction is greater than the magnitude of the total other forces in this direction
- The total frictional forces in this direction is opposite to the total other forces in this direction

This is, what I check. I think it's enough. But tell me, if I have missed anything.

Quote:
 >>For example when a car begins turning, the angular velocity of it is increasing. A frictional force cannot do that (increasing energy).I dissagree, it is precisely the friction between the wheel and the ground which causes the car to turn. The energy does not increase, it is simply transfered from linear velocity to angular velocity.

Well, it may be right. But only static friction can do that. The static region of the contact patch get deformed, and this is, what transmits the force to the car from the road. Sliding friction cannot do this. It is opposite to the velocity.

Quote:
 >>Here the tyre is deformed, and that's what turns the car. Roughly like a spring force.I downloaded your demo, good work btw it looks great. But i don't see any deformation of the wheel? do you mean to wheel moving relative to the car due to the springs on the axle?

I don't display the deformation on the wheel. It's just a static object. The deformation is described by numbers in the model. And the vertical red bars show the slip angle, the horizontal red bars show the size of the static region. I think the deformation wouldn't be that spectacular. Its not that much

Quote:
 >>In my tyre model the contact patch has a static and a sliding region.It is concevable that part of the tire will be static while another part is sliping for a VERY brief moment at the point where the wheel makes the transition from static to sliding if your wheel is able to deform. But i don't think this is what you are talking about, since i can't see any deformation on your wheels. And quite frankly that would just be overkill trying to model that effect.So please explain what you mean by, static and sliding region.

Deformation and the size of the static and sliding region is modelled in the background. It could be displayed on the wheel too. But I haven't seen any game where the wheel deformed.
But this is a static model yet, it doesn't simulate the transitions unfortunately. BTW, the basic Pacejka model is also static.
So the explanation: the main point is that static and sliding regions are present at the same time, and this is not just for a moment. When you take a turn with your car, the friction between the road and the tyre deforms the tyre laterally. Like on this picture:

As you can see, the deformation is growing along the contact patch. And when the stress caused by the deformation reaches the max. static friction, it ends. Here begins the sliding region. The tyre "deforms back" here and is sliding on the road.
Hope it is clear now.

And there is still the longitudinal deformation, when accelerating or braking.

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Quote:
 Original post by szinkopaOK. I explain it more exactly. Taking a dirction/axis 3 conditions must be fulfilled to stop the object along this direction: - The sign of the velocity has changed - The magnitude of the total frictional forces in this direction is greater than the magnitude of the total other forces in this direction - The total frictional forces in this direction is opposite to the total other forces in this directionThis is, what I check. I think it's enough. But tell me, if I have missed anything.

Ok i think i missunderstood what you said earlier... that looks good, so the problem is solved?

>>Hope it is clear now.

ah ok i think i understand what you mean about static and sliding region now... wow i didn't think you would need to model it in such detail to get realistic results.

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It's not solved yet for the tyres. As I earlier wrote, I will eliminate the angular jittering first, because I suspect that for the oscillation.
These 3 criterias I wrote have been used for the "traditional" frictional forces, torques with success for a long time.

As for the tyre modelling, deformation should be included in a good model somehow, this is the main point of it. One can approach tyre modelling in real time in 2 ways:
- empirical: formulas based on measurements, magical parameters, no physics, faster computation
- analytical: based on physical formulas with less or more simplifications, more sensible parameters, slower computation

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