on the patch

[Dan Masi]

> >Tihol Tiholov wrote:
> >>
> >> I'm no accomplished engineer but it seems that Dan is right about
> >> that formula being overly simple and "tireless".
> >
> >except that he isn't.
> >
> >--
> >Huw Powell

Oh, Jeez.  Now how did I manage to miss this little gem of wisdom?

I'm sorry about being wrong on this, and I understand that it's
important to be right at all costs, so I won't belabor this thread
with a lot more wrongness.  But I *will* add just a couple of
bits of trivial misinformation which will, of course, be declared
to be blatantly incorrect.  That's ok.

For tires with high inflation pressures that are relatively light
by design relative to the loads they carry (e.g. tractor-trailer tires,
racing bicycle tires), the Pt = L / Ac equation will hold *almost* true
throughout the designed tire inflation range [where Ac is the contact
patch area, Pt is the tire's inflation pressure and L is the load on
that tire].  Almost. the calculated pressure based on the load and
contact area will always be slightly higher than the inflation pressure.

Now, for passenger car tires the story is different.  They run at
lower pressures and have a much more significant structural component.
The relevant part of the tire's construction is the radial body cords or
plies, which run (you guessed it) radially, i.e. from sidewall bead to
sidewall bead.  These cords retain the tire's shape under load and are
put in tension by the air pressure within the tire.  As I mentioned
in an earlier mail, your car wheel is actually supported from both
above and below; above the wheel, these radial cords are actually
pulling up on the sidewall bead and thus pulling up on your wheel.
For those of you familiar with bicycle wheel construction, these
cords are no different than spokes with the concomitant tension/compression
cycles that they undergo as the upper spokes pull up on the hub.

Anyway, if you measure the average contact patch pressure for
a passenger tire (weight divided by contact area), you'll come
up with a pressure that is typically in the range of 1-1/2 to 2-1/2
*times* the tire's inflation pressure.  I.e. *higher* than the tire's
inflation pressure, significantly. The rest of the force holding
up the car is contributed by the tire's structural components, most
notably the force it takes to bend the shoulder material built
into the tire as the radial cords at the bottom of the tire move
from tension to compression. Also, because of these solid shoulder
foundations, contact pressures for a normally-loaded passenger tire
will increase somewhat near the shoulders and may be as much as twice
the contact pressure observed in the center of the patch.

And here's a real simple question for those that claim steadfastly that
the car is supported by "air, not the tire": what happens when you
draw a static free-body diagram of the wheel with air pressure (and the
force of gravity due to the mass of the car) acting on it?  Car fall down
and go booboo?

Sorry for the long post, and again, sorry to be so darned wrong.
And for that special someone who finds it somehow beneath him to
fill in details as to why I'm so darned wrong, let me save you the
risk of repeatetive stress injury on your "Reply" finger and write
yet another reply for you:

"No, Dan is incorrect".



Dan "that's my final answer" Masi