Thank you very much for the help, but
Motorherp:
Quote:Original post by Motorherp
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.
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 Motorherp
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.
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 SpaceDude
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.
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".