hydroplaning

[ben]

I apologise in advance to those of you who want to stop talking about
contact patches and hydroplaning, but I find this quite interesting,
and enough other people seem to as well that I thought I should weigh
in.

I'd really like to use metric units, but no-one else has, so at great
personal expense I've used dumb American units.  Not that the units
are at all important, but sometimes a "concrete" example is useful...?

With all due respect for Huw, George is right -- there are two
different definitions of pressure going on here, that the FAA result
as presented on this list didn't address.  Indeed, "tire pressure"
tends to refer to the air pressure inside the tire, but it could also
possibly be used to refer to the pressure of the rubber on the
pavement.  They are almost the same thing, but not quite.
Kindergarten physics demonstrates why they must be about the same, so
I won't bother with that.

Why are they different?  Because the tire is not infinitely flexible.
If you push at one point on a tire's tread, the force of your push is
spread out over a substantial volume of the rubber, in the form of
shear forces.

Here's a thought experiment.  Let's take a random tire, call it a
racing slick, under its own weight.  Put it on a sheet of glass or
some other very smooth surface.  The contact patch size will be almost
exactly the size that Huw and others have explained that it should be
(if it's a 25 pound tire/wheel at 25 psi, the contact patch will be
about 1 inch^2.  If we load it with 1000 pounds, the contact patch
will be about 40 inch^2, etc...).

Now, gently balance the tire on the head of a nail with a surface area
of, say, 0.1 inch^2.  What will happen?

1) The contact patch expands.  No: the contact patch can't expand,
   since only the nail is supporting the tire.  There's nowhere for it
   to expand to!

2) The tire will deform until the air pressure on the inside allows
   the small contact patch to support it.  Um, that would mean the
   tire goes up to 250 psi.  Assuming the tire doesn't burst, remember
   that the volume surrounded by the tire has now compressed by a
   factor of 10.  Even pressing against a flat surface can't compress
   it to 1/2 its volume, let alone 1/10.

3) The nail penetrates the tire, ruining the thought experiment.  No,
   I think 1/10 inch^2 can support a 25 pound tire.  I'm happy to try
   the experiment for real if anyone actually doesn't believe me, but
   it's moot anyway, since this is meant to be an extreme argument.
   Using real numbers (30 square inch contact patch supporting 1000
   pounds and reduced to 15 square inches on a gravel road, say) will
   definitely not be a problem, and the same argument applies.

4) The pressure of the nail is taken up by the (rather stiff and
   supportive) rubber as it flexes, supporting the tire.

Obviously, case (4) is what actually happens.  We have an internal
tire pressure of 25 psi, and a contact pressure of 250 psi, and the
difference is taken up by random forces (tension, compression,
shear...) inside the rubber.

Now, forget racing slicks.  Add some tread.  Make it aggressive, like
our nail from the previous example, but stuck to the tire rather than
to the road (there's no difference in the static case).  Guess what?
The same thing happens --- the external pressure between rubber and
pavement is greater than the internal air pressure inside the tire.
The parts of the tire between knobby bits are under certain stresses
(if this were not so, tires would not have rolling resistance, and
would not heat up with use, etc...).  Note further that the bits
between the knobs don't expand until they touch the road (ie. tires
have tread!), which also tells us that the point loads at the
tire/pavement interface are being spread around inside the tire.

> >I think it's the 'contact patch' thingie where the theory falls
> >apart. With an aggressive tread, less of the rubber makes contact at
> >the same pressures. Take the snow tire as an example. The contact
> >patch is relatively small compared to a touring tire at the same
> >pressure.

Yup.

> The "contact patch" is the rubber meeting the road, not the overall area
> over which that takes place.  If a tire has 50% "grooves" and 50% 
> "tread" it will use up twice the area on the road as a slick, in order 
> to have the same "contact patch."

This is a definition of "contact patch".  Careful!  Using this
definition is fine, but it means that internal pressure is not the
same as external pressure.  It's off by a factor of about 2 in this
case.

> >With an aggressive tread, less rubber actually meets the road at the
> >same tire pressures

If "tire pressure" refers to the air pressure inside the tire, then
this is correct.

> Incorrect.

Incorrect, Huw.  Sorry to pick on you, but you're usually right, so I
don't want people believing you here just on account of your
credibility :)


How does all this relate to hydroplaning?  If the water is around as
deep as the tread (probably a factor of roughly 3 or so?????) then
tread can help channel water out of the way, letting the rubber
contact the pavement.  Here a more aggressive tread will help, since
the water is under higher pressure between the knobs and the pavement
and will thus "get out of the way" faster, and has somewhere to get
out of the way _to_.  I don't know how deep the FAA's puddle is, but
if they hydroplane at a speed that depends only on the tire pressure,
then it is probably quite a bit deeper than the tread depth.

Cheers! :)

-Ben

-- 
Ben Pearre          1990 200TQA          http://hebb.mit.edu/~ben