Re: [Phys-l] Another tire question

["Edmiston, Mike" <edmiston@bluffton.edu>]

Yes, Rick has given a good response to Brian, so I will not add more
except to slghtly reword one phrase.

Where Rick says "(trapped bead pulling outwards)" I would say...
(trapped bead being pulled outward by the sidewall tension).


Michael D. Edmiston, Ph.D.
Professor of Chemistry and Physics
Bluffton University
Bluffton, OH 45817
(419)-358-3270
edmiston@bluffton.edu



-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu
[mailto:phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of Rick
Tarara
Sent: Thursday, November 08, 2007 9:38 AM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Another tire question

Brian (et. al.)

Let me rephrase Michael's argument so that maybe you can better
appreciate it (and I can see if I really understand it).

As Michael says, your half-tire argument IS a different problem in that
the air pressure pushes ONLY upwards on the bottom of the
wheel--Pressure x cross-sectional area, giving a Net Force upwards on
the wheel.  In the full wheel, there is an equal force downwards due to
the air pressure acting on the top half of the wheel--Net Force = zero
(discarding the TINY buoyancy force).  With the full, inflated tire, the
air pressure does not directly provide any of the lifting force on the
car although it is essential to the workings of the tire--this is, I
believe, then one of Denker's zeroth order effects.

In your example below, I think you are reverting to my VERY STIFF model
where we can talk about compressive forces in the sidewalls and at least
some of the upwards force coming from such.  BUT...in the tire models
being presented by Michael and others, the sidewalls cannot really
sustain compressive forces only tension forces and therefore become like
the spokes in the bike wheels.  Then ultimately, with no net force due
to the air pressure, the tire itself must produce a net upwards force on
the wheel and that can only be transmitted through tension which is
going to be radially outwards from the wheel (trapped bead pulling
outwards).  To have a net upwards force, that tension force must be
larger on the top of the wheel than the bottom--hence the image of the
wheel hanging from the top half of the tire.

Does that help any?

Rick

----- Original Message -----
From: "Brian Whatcott" <betwys1@sbcglobal.net>

> At 09:50 PM 11/7/2007,  Mike E., you wrote:
> > ..
> > > But the strange thing is, both upper and lower spokes take a share

> > > in supporting my weight.  How can that possibly be?
> >
> > It can't be.  This is a false statement.  The lower spokes are pulling

> > down, not pushing up.
> >
> > > Opposing spokes are in tension.     When an upper spoke sees
> > > greater tension, a lower spoke is also seeing reduced tension.
> >
> > Yes.  You said it.  The lower spokes are still in tension, although 
> > reduced.
> ...
> > Michael D. Edmiston, Ph.D.
>
>
> Hmmm... I see I am not dislodging you from the upper half support
> position, Mike.   I'll offer one more extreme case to illustrate,
> then leave the topic on its merits.
>
> Let me select a tire of extraordinary wall strength, but great 
> flexibility.
> Let me inflate this tire so hard, that regular tests of deflection and

> load bearing cannot distinguish it from a wheel made entirely of 
> steel.
>
> You might think of this as the limit of  high inflation.
>  An engineer or physicist could hardly be faulted for measuring the 
> load bearing contribution of the tire walls in the lower half, is that

> not the case?
>
> And at the other inflation limit, a flat tire demonstrably supports 
> the entire load on the lower portion of the tire.
>
> You already concur that a (lower)  half tire can support weight, but 
> you curiously spring to a "this is a balloon" explanation. Yes, it is!
>
> I rest.....
>
>
> Brian Whatcott    Altus OK    Eureka!

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