Tyre normalised braking grip vs slip ratio

[Christopher Catto]

Ok. This should appeal to those of you who know their tyres or got hold of some of the Calspan data.

I'm trying to get a small model going that calculates braking performance of a car and the tyre model is obviously the toughest part. The model is not aimed at being very accurate, but more for working out some ideas and putting them into practice.

I am basing much of the tyre model on a Mu vs slip ratio graph in RCVD (page 38, fig 2.17, "Normalized braking force vs Braking slip ratio"�) and plotting a similar curve with discrete data. I adjust some coefficients to make the curve be proportionally similar but with different maximum (to take into account stickier rubber compounds for example).

QUESTION 1:
Does anybody know how slip ratio increases with applied braking force from the brakes? Is it possible to work out a relationship or is it incorrect to assume there is a relationship? I am not interested in the tyre grip, I already know that from the Mu vs slip ratio graph!

I know that by transposing the Mu vs SR curve (so normalised braking force is on the x-axis and slip ratio on the y-axis) I can work out the instantaneous tyre coefficient of friction for a given slip ratio. But there are two problems:

1) This only defines the braking force from the tyre, and it is quite obvious that the available braking torque from brakes or motor could be higher

2) Mu vs SR is a curve with two y values for each x value. In other words, the curve bends back on itself. (see these diagrams: http://img510.imageshack.us/img510/3307/brakingslipratiodiagramnh3.th.png (http://img510.imageshack.us/my.php?image=brakingslipratiodiagramnh3.png) ).



So, I would like to know views from those who looked closely at some tyre rig and data. My thinking is that:

-I am considering for now the case with no weight transfer (wheel on rig, rather than wheels in a car). I know how to cope with the weight transfer, that's no problem.

-THEORETICALLY as the braking torque from the brakes (or motor) increases, the tyre goes up to its best slip ratio (usually 0.15 to 0.25). This is clear: braking force vs tyre grip are equal.

-Once the braking torque is higher, the tyre cannot respond any more and it starts slipping, to the point where eventually it will start totally skidding when slip ratio is -1. Not clear!

-The problem is that at SR = -1 the tyre has a grip of let's assume 75% (ref. Milliken) of what it is at its optimum. So this creates two distinct paths! (see this diagram: http://img108.imageshack.us/img108/4652/forcefrombrakesdiagrampl9.th.png (http://img108.imageshack.us/my.php?image=forcefrombrakesdiagrampl9.png) )

-Does this imply that the available force from the brakes has to drop to the available grip from the tyre in order for it to take its slip ratio back towards 0?

-The above statement implies that when the brake is applied progressively harder, the tyre goes beyond its peak and starts to slip considerably. It also implies that when braking is backed off the tyre will not gain traction until a certain threshold. Surely some important phenomenon is missing.

-The slip ratio only defines the relationship between the angular velocity of the braked wheel and that of the free-rolling wheel. No mention of braking torque/force from the wheels.

QUESTION 2:
Does anybody know if slip ratio can in any way be related to tyre grip and the braking force from the brakes?

My brain tells me that this is not possible and that it definitely is not the ratio of the two. I am wondering what information might be required to make this step. In other words, is there any way of creating a certain slip ratio purely by calculation or does the process require feedback (wheel speed sensors)?

Why am I asking these stupid questions when most ABS systems simply measure the tyre rotating speed (and therefore explicitly measure slip ratio) http://fsae.com/groupee_common/emoticons/icon_rolleyes.gif? Because I wanted to try if there is any way of calculating optimum braking forces for a system by having some tyre data and without a wheel speed sensor (since this is a theoretical model rather than a system installed in a car).

Hope this isn't too boring! http://fsae.com/groupee_common/emoticons/icon_wink.gif and that the diagrams can be viewed.
Cheers for any help
Chris

[Jevon]

Here is a good paper that may help.
Stability and Bifurcation of Longitudinal Vehicle Braking (http://www.egr.msu.edu/dvrl/preprints/Olson-etal_nd05PRE.pdf)

Jevon

[flavorPacket]

Chris,

Correct me if I'm wrong, but in steady state doesn't the tractive force equal the braking force (within a few percent, due to friction in the rotating assembly)? So, if a wheel locks, then the braking force will equal mu(static) x F(normal).

I don't think that's what you're asking though. You want to know how transient braking forces generate slip ratio, right?

[murpia]

I picked out a few pieces from the post, because I can see some confusion...

Originally posted by Christopher Catto:
-THEORETICALLY as the braking torque from the brakes (or motor) increases, the tyre goes up to its best slip ratio (usually 0.15 to 0.25). This is clear: braking force vs tyre grip are equal.

-Once the braking torque is higher, the tyre cannot respond any more and it starts slipping, to the point where eventually it will start totally skidding when slip ratio is -1. Not clear!

-Does this imply that the available force from the brakes has to drop to the available grip from the tyre in order for it to take its slip ratio back towards 0?

-The slip ratio only defines the relationship between the angular velocity of the braked wheel and that of the free-rolling wheel. No mention of braking torque/force from the wheels.
Tyres generate a slip ratio because of an applied torque, it's not independent, so the tyres do not go to their 'best' slip ratio unless you are able to control the torque absolutely precisely. Obviously other factors, primarily vertical load, influence the torque / slip ratio relationship.

If you apply enough torque to go over the top of the 'peak' in order to avoid the wheel locking / spinning you have to reduce torque to a level below the 'peak' and reapproach the limit from the 'stable' side. This is what ABS & traction control attempt to do, and what skilled drivers do. Possibly 'drift' drivers operate on the 'unstable' side of the curve, as would a drag driver controlling a burnout or someone doing 'doughnuts'...

Regards, Ian

[murpia]

Originally posted by Jevon:
Here is a good paper that may help.
Stability and Bifurcation of Longitudinal Vehicle Braking (http://www.egr.msu.edu/dvrl/preprints/Olson-etal_nd05PRE.pdf)

Jevon
That's a very in-depth paper... Anyway, one statement from the conclusion is relevent here:

'...as one continues to increase the front brake torque, the front wheels eventually lock up, but will exhibit hysteresis when backing off the brake...'

Regards, Ian

[Christopher Catto]

thanks for the good answers!

about 1 hr after my post i had a long discussion with another engineer about it and i came to my own conclusion (which is still under some aspects not 100% correct) that for my model I had assumed a CONTINUOUS line but in fact the plot is discontinuous if considering classical braking (mashing the pedal with your foot and just waiting what happens http://fsae.com/groupee_common/emoticons/icon_rolleyes.gif )

in order to be come up with something usable in a model i realised that after max grip (at a given optimum slip ratio) the wheel will probably lock up, if we neglect some of the other tyre complexities like heat, hysteresis and wearing/deforming. so the graph would have a discontinuity (a bit like some root locus plots where the solution jumps across to different areas of the plot) after the max grip location. This is not very clever because I bet the tyre machines are sofisticated enough to avoid this but I do not know if they apply continuous braking or if in fact they do apply pulsing at some frequency to cause very fast lock-grip phenomena.

i implemented my idea (I am not making the model so complex until I actually decide to shift it to the next level) so that when the max available braking force is higher than the max tyre available grip then the tyre locks up. So that in effect means that my model does not consider the zone with slip ratios from cca. -0.2 to -1. This does mean that if the driver applies high braking the tyres slip, then re-grip due to weight transfer and then slip again and this repeats itself to cause oscillation. This is something I will refine by looking at the iteration method and maybe applying some damping term in the equations. there is always an issue if you assume that the instant you apply the brake the car is level and only then starts to have weight transfer and pitch (chicken or the egg problem!)

The zone between max grip and terminal grip (skid) I labelled the "ABS zone", in other words the zone where slip ratio is probably controlled by brake pulsing at various frequency and pulse duration.

feel free to throw some other ideas. it's probably enough for me to implement just some damping. I'll work on that now.

[ben]

Maybe I'm just skim reading, but I can't really see what the big issue is? You mash the pedal and supply a large brake torque to the wheel, the tyre will generate a slip ratio to allow the tyre to generate a braking force at the level you've requested. If this force exceeds the peak friction coefficient the wheel will lock.

If the wheel locks the driver will reduce (hopefully) brake pressure and the wheel will regain traction. Given that the friction coefficient will change dramatically in such extreme thermal conditions I'm confused as to what sort of model you could usefully make of this situation.

Ben