Hi, guys!
I've read this thread for a couple of days now with great interest, because
I also wanted to know, why a non-locked tire sqeals and what it means that a
part of the contact patch is sliding while another part still has grip. But
unfortunately your answers didn't really enlighten my mind, so I've put
together my own theory.
<guess mode on>
You say, when the tire is near its limit a part of the contact patch loses
its grip. This part will be the one, which has the biggest friction force
and the least normal force exerted on it (which part it will be depends, if
it's while braking, cornering or accelerating or any combination of the
three). OK, but what happens then? This part of the contact patch will begin
to slide, i.e. to move relative to the track (sliding with reduced coeff.of
friction), but it will also move relative to the rest of the tire (and the
rest of the contact patch, which still has grip). Obviously this moving
relative to the rest of the tire can't go on forever (not even the few
seconds you spend braking for Parabolica ;-) otherwise the tire should be
ripped apart. The rest of the tire will exert a force onto the slipping part
of the contact patch, which wants to move it back in place (and which will
get bigger, the further the part has moved). Now this force should be
directed more or less against the friction force exerted by the track onto
the sliding part of the contact patch, thus reducing it maybe enough that
the sliding part will regain grip again. Anyway, as long as the whole tire
isn't locked it will force the part of *** which 'locked' to turn with
it. Of course, now the tire has turned a little and the contact patch
(partly) is made up of another piece of ***. The force, which wanted to
push the sliding part of *** back in place a few fractions of a second
ago will be more or less gone, so a part of the current contact patch will
start to slide again. And the whole procedure will repeat itself at a quite
high frequency. The whole tire will begin to oscillate at that frequency and
this oscillation will produce the sqealing sound.
<guess mode off>
What do you think? Reasonable? Complete nonsense?
Please let me know!
Peter
Vienna, Austria
> I'm confused, let me know if I'm thinking about this right. When tires
> squeal, what's producing the sound? It has to be *** sliding across
> asphalt, no? As long as no sliding is occurring, 100% of the tire's normal
> contact patch is gripping the road. When you're cornering and the tires
> start to "squeal a little", some portion of the contact patch (say 10%)
must
> be sliding while the remainder of the contact patch (90%) is still
> maintaining grip. Then when the percentage gripping falls below some
> critical point, the tire loses grip completely and you slide.
> It would seem to be a little different with straight-line braking. Here
the
> tire is trying to slide straight ahead instead of sideways. There's
always
> another contact patch right next to the current one. In order to make the
> tire squeal, you would have to either lock the wheel, or reduce the size
of
> the contact patch. The only way to reduce the size of the contact patch
is
> to reduce the weight on the tire. Whenever you're breaking, you're
putting
> more weight on the front tires and less on the rears. This could cause
the
> rears to "squeal a little" without locking the fronts or having the car
> slide noticeably.
> Does any of this sound right?
> > x-no-archive: yes
> > I think most people will agree that when cornering hard, tyres will
> > start to squeal a little before any noticeable sliding ocurrs, so this
> > _must_ apply to braking, too.
> > R.
