None of the posts here and in other threats about tyres really satisfy me. It seems like most of the people have a false understanding of what happens with a tyre. Especially the term “slip” leads to most of the misunderstandings. Obviously most of you think of a tyre as of a totally solid piece. In fact it is made of rubber so it can deform quite a lot. Rubber changes the things quite a lot that’s why a train (steel) has other characteristics than a car (rubber).
First thing we will notice is that rubber can generate much more friction than metal. The second thing is that the friction coefficient isn’t a fix number anymore. It is a variable figure. So applying the equations from Coulomb may lead to wrong results in some cases. But that’s not the topic now.
I'd think that ~15% slower figure is an average because for small amounts of time the wheels are locked. Could be mistaken.
No.
I've read a couple of times ( can't remember exactly where ) that when "treshold braking" the tires spin ~15% slower than the car's actual speed.
If so, what's the logical explanation? What I can imagine is that as the tire is rolling slower the contact patch is deforming, rubber is deformed absorbing energy braking the car. Coefficient of Friction increases when rubber is deformed?
You are quite close.
Imagine a piece of rubber like an eraser resting on a surface. When you try to push it in one direction you put a force on it. The rubber will get deformed until the forces in the rubber are in equilibrium with the applied force (F1). When you put more force on the rubber it will deform even more similar to a spring. The coefficient of frictions stays constant and it is still static friction.
When you try to push even more you will reach a point where the friction between the rubber and the surface is not big enough to hold the rubber on its place. The rubber will start to slide (V). Then static friction changes to dynamic friction which is much smaller. So the force (F2) who is deforming the rubber gets also smaller what will cause the rubber to bounce back a little bit.
There is one special case possible we know as “stick-slip”. After the rubber element has bounces back it will stop its movement along the surface for one instance. At this instance it will have static friction again and will start to deform again. The rubber element will always switch between static and dynamic friction. This causes oscillations in the rubber element and the squeezing noise. It’s something like the transition between static friction and dynamic one and warning the driver that he is close to the limit.
On a tyre these things get more complicated. I read 2 books about that stuff now and got it like this.
Basically we have lots of those rubber elements and arrange them now on a circle. What we want is static friction because this will give us the highest friction coefficient and that’s what we have most of the time. A tyre is somehow similar like the chain of a tank. While the elements touch the ground they don’t move or slide along it (static friction). Just that the rubber on a tyre is flexible. While entering the contact patch the rubber elements get deformed and moved slightly. Therefore the contact patch gets slightly bigger than the one of a metallic wheel. During the contact patch they stay on the ground. When they are leaving the contact patch they bounce (sliding on the surface) back to their original position.
The trick is that they are arranged on a circular pattern and every element moves slightly. When you add up this movement you will get an effect like the top layer of the tyre would spin a bit faster or slower (depends whether you are braking or accelerating) than the rest does. In fact the elements bounce back after they have left the contact patch, so the tyre sticks together.
Another thing to take into account is that the tyre radius is slightly smaller on the contact patch. Changes of the radius also changes the demanded revolutions of the tyre of course. I guess these has already being taken into account be definition of the slip ratio.
There are several different definition of the slip ration. SAE defines it like this:
SR=(Ω-Ω_0)/Ω_0
Ω: angular velocity of the driven or braked wheel
Ω0: angular velocity of the free rolling wheel (no force applied)
So when you get such a figure like slip ration = 0,15 it doesn’t mean that the wheel is locked 15% of the time or whatever you might think…..
It also doesn’t mean that the tyre is sliding over the surface. There are still areas where the tyre sticks to the surface. SAE defines spinning with a SR of +1 and locked with SR of -1.
Last, but not least. Does the same apply to accelerating?
The basic principle is the same but the force acts in a different direction.
During acceleration the force acts in the other direction as the movement of the elements and therefore slows them down while they enter the contact patch. So the tyre has to spin faster.
During braking the force acts in the same way as the elements move. So the tyre has to spin slower.
I really hope this helps because it took me several hours to find out and writte here.
