Tyre Slip Ratio

Can anyone explain to me what tyre slip ratio is?

I read that it is defined as the difference in angular velocity of freely rotating tyre versus that of the same tyre whilst braking. Although I "loosely" understand the principle of this being how a tyre generates fore and aft grip for accelerating / braking due to the tyre rubber flexibilty (similar to that of slip angle during cornering) I don't understand the definition. How can you compare the angular velocity of a freely rotating wheel with that of one that is continually slowing down???

Also I read this on Wikipedia which confuses me too:

"When a vehicle is being driven along a road in a straight line its wheels rotate at virtually identical speeds. The vehicle’s body also travels along the road at this same speed. When the driver applies the brakes in order to slow the vehicle, the speed of the wheels becomes slightly slower than the speed of the body, which is travelling along under its own inertia. This difference in speed is expressed as a percentage, and is called ‘slip ratio’. The ideal slip ratio for maximum deceleration is 10 to 30%."

"Slip ratio is calculated as follows: - Slip Ratio % = [(Vehicle Speed – Wheel Speed)/Vehicle Speed] x 100"

How can the rest of the car be travelling faster than the wheels other than during the transient period when the brakes are first applied and the sprung mass moves towards it's dived position? The wheels slow down so therefore the rest of the car does too?

Pretty sure I am just missing a simple point here so can anyone please point out the obvious for me!

Cheers

Fraz

It an interesting concept once you get your head round it & easy to understand how it effects performance on the track or road.

+15% slip (115% wheel speed to road speed) is peak for acceleration
-15% slip (85% wheel speed to road speed) is also peak for braking

Go much beyond the slip ratio & you reduce traction

There would be a similar peak for cornering too though its often a combination of at least two of the above, sometimes 3 if using lfb'ing

Also tyre design has a huge effect on the slip ratio, for example winter/snow tyres, have tall wobbly thin knobbles with lots of sipes in (small cuts) so the rubber will flex lots lots more, they have a much greater slip ratio up to 30% I believe. This provides a nice comfort zone in low grip enviroments so no sudden loss of traction just a nice transition. However their peak loading is much less than a summer tyre.

The opposite to that would be a slick with a very narrow window of slip, but equally a very high loading ability on the right surface.


The handling attributes of a car are often defined within these slip limits

Here's two examples in slow motion
http://www.youtube.com/watch?v=_c8KMDaSpEQ

http://www.youtube.com/watch?v=FCXVd5RGsKI

Thanks for the replies! All good stuff!

So if we measured the ammount of tyre revolutions our car makes per metre or mile or whatever and then use basic school physics to predict how far the car will travel at constant retardation from say 50mph to 0mph we will find that the car travels farther until it stops than we calculate all due to the strechiness of the rubber? The difference in this, as a percentage, is the slip ratio??

The Miliken definition which I think is taken directly from the SAE definition is that the angular velocity of the braked wheel differs from the free rolling wheel. I don't understand how you can compare these since when you start braking the angular velocity is constanly decreasing. All I can think of is that they are saying in a test environment with a tyre running on a conveyer belt at a constant linear road speed then when you then brake the wheel whilst maintianing the road speed you will change the angular velocity of the wheel relative to the free rolling one because it is working harder and therefore flexing more. Does this make any sense to anyone? Any other suggestions of how to compare free rolling angular velocity with braked angular velocity?

Is slip ratio variable with speed? Ie at high speeds the optimum slip ratio will differ as the car slows?

CPTsideways when you say "115% wheel speed to road speed" do you mean tangential linear velocity of the wheel?

Avocet I am not saying you are wrong so please do not swipe my head off with the nearest available angle grinder or anything but I don't think it is correct to say the the tyre is always sliding on the road surface to some degree until we are in a skidding situation. I once wrote something similar in a university report and was marked heavily badly for saying that. For sure the trailing elements of the contact patch once stretched by the braking force need to return to their equilibruim position and at as the tyre rotates and the weight comes off these elements at the rear of the contact patch they will partially slide to achieve this. But the main bulk of the contact patch that is providing our grip shouldn't be sliding at all for max grip and is one of these car dynamic deliberate mistakes such as slip angle, there isn't any actual sliding. That's been my take anyway. Please correct me if you think this wrong.

Where's me angle grinder?!

OK, I could very easily be wrong here and am keen to watch this thread develop as I may well learn something but I THINK that in any situation where the tyre is generating significant grip (fore-aft or laterally) the whole contact patch IS sliding across the road surface to a small extent. I don't know that this is true for a fact but the way I see it, if (let's consider braking), the tyre contact patch didn't slip at all relative to the road surface, the car wouldn't ever slow down. Let's say we're plodding alog at 50MPH and the tyre's peripheral speed (i.e. the linear velocity of the bit of tread opposite the contact patch at that instant in time) is also 50MPH. We apply the brakes, gently, and the car begins to slow down. The question is, how can it slow down unless the contact patch starts moving at something very slightly less than 50MPH? If we brea down the time taken to come to a complete stop into very small instants, I think the contact patch will always be moving a bit slower than the car until it comes to a stop.

Another thought, as I sit here looking at the rubber on my desk, is that it is stationary on there and isn't generating any force in the plane perpendicular to the direction in which it's weight acts. If, however, I start to push it with my finger in any direction, it will generate a frictional force between itself and the surface of the desk that tries to oppose the force I am applying to it to make it move.

Does that make any sense?

I am, as ever, more than willing to be shot down by reasoned argument here - and the good thing, is that I might even learn something in the process!

The way I see it, you need to think about this in terms of which regions of the contact patch are sliding as well as which parts are deflecting. In the straight forward braking case the speed differential between the rigid belt and the road surface is relatively constant over the whole contact patch, so as the belt moves back through the contact patch the road surface moves faster resulting in a deflection (between the belt and the adjacent bit of the road surface) which increases as you get further back through the contact patch. This introduces a sheer deflection into the tread which is relatively small at the front of the contact patch and increases as you get further back.

At the same time, the contact pressure varies over the contact patch with the main part held down by air pressure but less contact pressure around the edges.

Looking at the interface between the rubber and the road, as you get closer to the back of the contact patch with the sheer stress increasing and the contact pressure reducing, you eventually reach a point where the tyre does not have enough grip to maintain the deflection and starts to creep relative to the road. Ultimately as the contact patch lifts off the road and the contact pressure drops to zero, the tyre has crept far enough to completely relieve the sheer loads that were applied to it.

How close to the back edge this creeping starts depends on how much of the available grip is being used and hence what the speed differential is. If the speed differential is high, this creeping (sliding) could start well towards the front of the contact patch with the rest creeping (sliding) sufficiently to maintain the sheer deflection at the level that the tyre/road grip is able to maintain.

The lateral grip case is a bit more interesting because here the lateral deflection changes dramatically along the length of the tyre which is why we get the self-aligning torque. The fact that the lateral deflection (and hence lateral load on the wheel) is limited by the available grip is why the distribution of sheer stresses within the contact patch varies as the tyre approaches peak grip (steering gets lighter as the limit is approached) and also why wide short contact patches produce more lateral grip and less self-aligning torque that long narrow ones.

Holy thread resurrection!

Welcome aboard, Walter!

This might be a very boring thread, since regrettably we don't see much of Avocet or GreenV8s around these parts any more (though GreenV8s has at least started posting again on Pistonheads after a long absence).

Agreed; as I said above, there is undoubtedly an element of true sliding involved, though this ultimately serves to degrade and limit the grip being generated by 'creep'. Remember the 15% figure is peak slip angle, just before the car falls off the edge of its traction circle altogether.

You've lost me... you're saying that slip reduces acceleration therefore it increases traction?

Does not compute...

The typical graph for tyre slip angle vs. grip explains this, I think:



The first quarter of the graph (up to about the third mark on the horizontal scale in this instance), where there is a virtually linear relationship between grip and slip angle, suggests that the tyre is functioning pretty much entirely by 'creep'.

The second quarter of the graph (from 3 to 6.5 on the horizontal scale), where grip starts tailing off in proportion to slip angle, shows where some 'sliding' is starting to build up.

The third quarter of the graph from (6.5 to about 9) shows where the sliding has become substantial and, although the tyre hasn't let go altogether and is still generating some grip by 'creep', there is enough sliding going on to degrade the grip beyond its peak.

The final quarter of the graph (from 9 onwards) is all 'slide'; the tyres have ceased to generate any significant grip by 'creep' and what little grip is being generated is via simple sliding friction... there is no longer any relationship between grip and slip angle so the graph is literally flatlining.

...so, the direct answer to your question is that the % share of stretching (creep) vs. slipping/sliding depends on where on the tyre's slip angle curve you are. It varies between negligible (in the first part of the graph) to 100% (flatlining, after the tyre has let go altogether).

At peak grip, I guess you could estimate how much sliding is occurring by projecting the early, linear grip vs. slip angle relationship on upward, and the difference between this and the actual grip being generated at the peak shows how much 'sliding' is taking place at that point?