hydroplaning

[Orin Eman]

> > wider tires = wider but shorter contact patch.
> > 
> > until you are riding on the rims, your "weight per square 
> > inch" is what
> > you measure with your tire pressure gauge...
> 
> A simple thought experiment might convince you that it's not 
> quite that simple.
> 
> Drive your car over a lift, put gauges on the tires, and
> observe the pressure.  Now operate the lift enough that it
> takes just some of the car's weight.  What did the pressure
> do?  Ok... now take even more of the car's weight on the
> lift.  What's the pressure doing?  Ok... now take enough
> weight on the lift so that the tires are just brushing the
> pavement.  Zero psi???  'course not.  In fact, probably little
> difference from the first measurement.

The original statement is correct.  For most tires, weight
supported by the sidewalls is negligable.  The original
statement might be better put as 'load per square inch'
to get the weight of the car out of it - the lift experiment
changes the load on the tires.

Now the no load on the tires case?  You have a divide by zero
problem.  0 load, 0 contact patch.  You would have to take
the limit as load goes to zero of load/size of contact patch
to make your point.

> Here's another thought.  My car has, say, 800 lbs. supported
> by each tire.  Using the simple load x area = pressure
> assumption, that'd be a contact patch of 20 sq. in. at 40 psi.
> For 8" wide tires, the rectangular length of the patch would
> be 2.5".  Ok, sounds fine.  Now reduce the pressure to 10 psi.
> The expected length would now be 10".  Not likely.  But, ok,
> reduce the pressure to 5psi. I guarantee you won't see a 20"
> long contact patch!

No, as someone else pointed out, it would be on the rims.
 
> The problem with the simple load x area assumption is that
> it implies that the tire doesn't exist.  The air is *not*
> supporting the car.  The *tire* is supporting the car, and
> the air is giving the tire it's shape.  The sidewall at
> 6 o'clock is supporting the wheel from underneath, the sidewall
> at 12 o'clock is pulling the wheel upwards.  The air inside the
> tire is supporting the shape of the tire.

Let's take a look how this really works...

With no load on the wheel, the air is pushing the tread outwards,
putting the sidewalls in tension.  This tension is even all the
way around.  It's like a spoked wheel where the sidewalls are
like the spokes.  Now, as you put load on the wheel, the
tension in the bottom sidewalls is reduced - the rim pushes
down and the road (Newtons laws) pushes up the same amount.
When you reach a certain load, the tension in the sidewall
becomes zero and the sidewall starts to bulge.  Now what was
generating that tension?  It was the air pressure against the
inside of the tread.  What has balanced the air pressure against
the inside of the tread?  The pressure of the road against
the outside of the tread.  To hold the car up, the tension
in the lower sidewall has to reduce by the load.  For the
forces to balance (wheel isn't going up or down) we must
have a contact patch of size which will reduce the tension
by the load, ie load/air pressure.  This is the original
result: air pressure = load/size of contact patch.

There are some that would say the rim stands on the sidewall...
since compression is the same as a reduction in tension.
(That should be good for some arguments.)  Whatever,
the upper half of the tire has negligible role as the tire
is loaded.  The pressure stays the same, so the tension in
the sidewall stays the same.  It is the reduction in the
tension in the lower sidewalls that supports the car.
Counter-intuitive?  Sure.

Orin.