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Re: [Phys-l] Another tire question

Bob,

Because you say essentially the same thing I said before, except that
what I call the (wheel+bead) you call the wheel, your comments indicate
that you did not get my first post on this topic. Did you get that
post? If not, I am definitely "talking" past you. The message read:

Here's my take on the situation. I think it's good to first order.

Before you put the tire on the wheel, the radius of each bead is smaller
that the radius of the part of the rim on which it is designed to
reside. When you put the tire on the wheel, assuming the bead is in its
ultimate resting place on the wheel, the bead will be stretched.
Assuming no bunching, there will be uniform tension in the bead. That
tension will remain uniform at one and the same value throughout the
rest of this discussion because the length of the bead will not change
and the tension in any segment of the bead is determined by its length
and its spring constant.

Considering forces only in the plane of the circle formed by the bead,
the wheel will be exerting a radially outward force-per-length loading
on the bead. That loading times the length of any infinitesimal segment
of the bead is the "in the plane of the circle formed by the bead"
component of the normal force on that segment of the bead. The Newton's
3rd law partner to this loading is a radially inward directed loading
exerted on the wheel by the bead. It pushes inward on the surface of
the wheel. The bead can only push on the wheel. The bead is not
trapped.

When you inflate the tire, the sidewalls of the tire exert a radially
outward directed loading on the bead. The net radially outward loading
on the bead never changes, rather, the radially outward loading exerted
on the bead by the wheel decreases as the radially outward loading
exerted on the bead by the sidewalls increases. This normal loading
exerted on the bead by wheel adjusts itself to whatever it has to be to
maintain the bead in the shape of a circle with the radius of the wheel.

When you lower the car down to the floor of the garage so that it is
resting on the floor, the radially outward loading exerted by the
sidewalls on the bead becomes lesser on the parts of the bead on the
lower half of the wheel for reasons already mentioned in this thread.
The net radially outward loading on the bead is the same on all parts of
the bead as it always was but on the lower half of the bead, more of
that loading is being provided by the wheel (than was being provided by
the wheel after inflation but prior to setting the car on the floor).
The wheel is pushing downward on the lower half of the bead harder than
the wheel is pushing upward on the upper half of the bead. By Newton's
third law, the bead is pushing upward on the lower half of the wheel
harder than the bead is pushing downward on the upper half of the wheel.
The net force exerted by the bead on the wheel is thus upward. This
counteracts the downward force exerted on the wheel at the center of the
wheel by the axel.

The wheel is supported because it is being pushed upward by the lower
half of the bead harder than it is being pushed downward by the upper
half of the bead.

The (wheel+bead) is being supported because it is being pulled upward by
the sidewalls in contact with the upper half of the (wheel+bead) harder
than it is being pulled downward by the sidewalls in contact with the
lower half of the (wheel+bead).


-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu [mailto:phys-l-
bounces@carnot.physics.buffalo.edu] On Behalf Of LaMontagne, Bob
Sent: Friday, November 09, 2007 12:44 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Another tire question

Is this another case of talking past each other? I assume that what we
are calling the "bead" is simply the collar that contacts the rim of
the
tire. If that assumption is correct, then the collar "hangs" from the
sidewall (which I assume is the visible side portion of the tire when
it
is on the wheel.) I think the hangup here is trying to envision how a
"bead" which cannot physically grab the wheel can exert an upward
force.
The sidewall certainly can pull the collar (bead?) up - and the collar
(bead) really becomes part of the rim after mounting simply because of
tension pulling it against the rim since it is acting like a taut
elastic band. The bead acts like a sling but with a radially inward
tension acting on its entire circumference. The upper sidewall pulls
up
on this sling.

Bob at PC

-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu [mailto:phys-l-
bounces@carnot.physics.buffalo.edu] On Behalf Of Jeffrey Schnick
Sent: Friday, November 09, 2007 11:49 AM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Another tire question

-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu [mailto:phys-l-
bounces@carnot.physics.buffalo.edu] On Behalf Of Rick Tarara
Sent: Friday, November 09, 2007 10:35 AM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Another tire question


----- Original Message -----
From: "Jeffrey Schnick" <JSchnick@Anselm.Edu>

The axle "hangs" from the top rim via the top spokes. The rim
"hangs"
from the top sidewall of the tire.

No. The bead hangs from the top sidewall. The rim is supported
from
below by the bead.


Do you mean the rim/wheel is pushed up from below by the bead,

Yes, as in my previous message in this thread:


<https://carnot.physics.buffalo.edu/archives/2007/11_2007/msg00071.html>


or that a
rim
on the top of the wheel that traps the bead is pushed up by the
bead.
The
latter is, IMO, the same as hanging from the bead which is hanging
from
the
sidewall.

The bead is not trapped in the sense that, if it weren't for the
tension
in the bead, you could pull any piece of it radially outward, away
from
the center of the circle formed by the bead.
The bead's position is constrained by the hub in the following
sense:
Viewing a rear wheel a position behind the car, directly behind the
rear
wheel you are looking at: The rim is shaped so that the right bead
can't
slide rightward off the rim, and the rim is shaped so that the left
bead
can't slide leftward off the rim.

Look at the bead seat in the diagram
<http://www.alloywheelsindia.com/images/crosssection.jpg>
referenced by John Denker in the message (in this thread) at


<https://carnot.physics.buffalo.edu/archives/2007/11_2007/msg00034.html>


Seems to me that an answer to this is to carefully look at a car
wheel--
does
anyone have one available or can you get down to a HS auto shop?
IS
the
bead of a mounted tire trapped or not (I'm thinking not). If not
trapped,
then I can't see how the wheel can hang from the bead. If it is
trapped,
that is certainly a possibility.


Rick

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