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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 John Denker
Sent: Friday, November 09, 2007 1:06 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Another tire question

Let me summarize the physics as I understand it, starting with
some basic "review" items and segueing into some points that
haven't yet been completely agreed-upon.

1) The idea that the rim could support the car (or bike)
because of _pressure_ on the rim is dead on arrival. This
is basic physics. Pascal's principle says the pressure is
the same everywhere. And the rim doesn't change shape.

The applicability of Pascal's principle is pretty obvious in
the case of tubeless tires. I claim it also applies to tires
with inner tubes, on the grounds that the inner tube is very
soft and contributes nothing to the mechanics; it is just
there to plug small gaps so that the air doesn't leak out.

1a) In consequence, by process of elimination, we conclude
there must be a load-dependent force communicated from the
tire to the rim.

2) One of the most basic requirements is that when we put air
pressure in the tire, it doesn't immediately pop off the rim.

This has some immediate consequences. For one thing, it tells
us that the tire cannot be modeled as a uniform tube of squishy
rubber. The physics here starts with a scaling law. Suppose
(by way of contradiction) that we have a tube of squishy rubber,
and we put air pressure in it. To first order (and maybe more
orders than that) the tube will accommodate the pressure by
getting bigger. It gets scaled up by a length-factor of alpha.
That means the minor radius (r) of the torus gets bigger by a
factor of alpha ... and the major radius (R) also gets bigger
by a factor of alpha. Boom, the tire hops off the rim. Proof
by contradiction.

In contrast, the inner tube by itself *can* be modeled as a
uniform tube of squishy rubber. This is why you can't use
the inner tube by itself.

This tells us that a sew-up tire, even though from a distance
it "looks like" a plain old tube, most definitely is not.

3) There is a class of non-pneumatic tires where the stiffness
of the rubber in the sidewalls is essential to the functioning
of the tire.

Such tires have terrible performance. In particular, their
compliance is nonlinear in the wrong way; they are very
stiff for small deflections and then get less stiff. This
is analogous to the "beam collapse" problem: a structural
beam is very strong when compressed along its length, but
then, almost as soon as it starts to bend, it collapses.

I exclude non-pneumatic tires from further consideration. In
pneumatic tires, I assume that stiffness in the sidewalls has
been minimized.

4) In consequence of items (2) and (3) my model for ordinary
tire material is that it is mostly like a non-stretchy fabric.
It can flex in the out-of-plane direction, but in the plane
of the fabric it resists shear and greatly resists any
change in area.

5) Therefore I am mystified by one feature of
http://www.bluffton.edu/~edmistonm/Car.Tires.pdf

I call attention to the red arrows at the 12:00 position. As
we move from the upper row to the bottom row, the red 12:00
arrow gets longer. I don't see how this can be. Among other
things, there is no change in the 12:00 blue arrows, so this
change in the 12:00 red arrows means there is an unbalanced
force, i.e. an out-of-equilibrium situation that cannot
persist.

The blue arrows are the air pressure so I wouldn't expect any change in
them. The red arrows represent loading on the bead. An increase in
this loading would be accompanied by a decrease in the radially outward
component of the normal force exerted on the bead by the rim, hence the
increased length of the 12:00 red arrow does not mean there is an
out-of-equilibrium situation--the bead is still in equilibrium. It is
conceivable to me that the in-plane (in the plane of the circle defined
by the bead) component of the tension in the sidewall at the top of the
tire could increase. Consider the belt making up the circumference of
the tire, just inside the tread. When the car is on a lift, this belt,
viewed edge on, is in the shape of a circle. (More specifically, the
intersection of this belt with a plane parallel to the plane of the
circle formed by the bead, is a circle.) When you lower the car down so
that it is standing on the pavement, this belt, viewed edge on, is
roughly in the shape of a circle except for a portion of the bottom of
the belt which cuts across the circle along a secant instead of
following along the arc of the circle. If the air pressure in the tire
is to a good approximation the same before and after lowering the car to
the pavement, then it seems conceivable that the tension in this belt is
the same before and after lowering the car. This suggests that this
belt would have the same length before and after lowering the car.
Since the secant represents a shorter distance than the arc, this
suggests that the part of the belt that roughly conforms to a circle
after the car is lowered so that it is standing on the pavement is in
the shape of a circle of greater radius than that of the circle it
conformed to prior to lowering. This would mean that the sidewalls of
the tire above the rim are more vertical after lowering than they were
before lowering. This would mean that the radial loading exerted on the
bead by the sidewall at the top of the tire depicted by the red arrow at
the 12:00 position in
<http://www.bluffton.edu/~edmistonm/Car.Tires.pdf>
should indeed be longer after the car is lowered to the pavement.

As a related point, I *can* imagine ways that the 12:00
arrows could change, but it's tricky, and nobody has yet
hinted as to how this could happen. We can discuss it if
anybody is interested.

6) One way to accommodate the requirements of items (2), (3),
and (4) is to have a tire with a bead. Typically there is a
super-strong steel or kevlar cable in the bead, and there is
some sort of bead-retaining feature in the rim. Either the
bead sits on a shoulder, or the bead tucks inside some kind
of lip. I have revised
http://www.av8n.com/physics/img48/tire.png
to emphasize the bead. In this case it sits on the outside
of a shoulder, as is typical of car tires.

7) In the case of a shoulder-type design, it is imperative
that there be a huge zeroth-order tension in the bead wire,
even when the tire is uninflated, to keep the bead tight
against the shoulder when it is inflated.

As always, tension runs along the cable, and is the same
everywhere in the cable. A cable that has tension *and*
curvature exerts a force transverse to the cable, in the
direction of the curvature.

The total force exerted by the bead on the shoulder is the
sum of
-- the just-mentioned force related to the tension in the
bead cable. This is zeroth order, i.e. independent of
load, plus
-- forces produced by the side wall, related to the shape
of the tire and the air pressure enclosed therein. The
shape contains terms that depend on load.

8) In accordance with the "non-stretchy fabric" model, when
the tire is under load, the main thing that changes is
the cross-sectional shape. In particular, as shown at
http://www.av8n.com/physics/img48/tire.png
the shape of the 12:00 sector does not change much, but
the 6:00 sector is significantly flattened. This changes
the angle at which the sidewall meets the rim. The
cosine of this angle tells us the vertical component of
the force. The nonlinearity of the cosine gives us a
nonlinearity in the good direction, i.e. the effective
spring constant is softer for small deflections and
bigger for large deflections.

In this model, the arc length of the rubber (from bead
to bead, going around the minor radius of the torus)
does not change significantly as we go from unloaded to
loaded; it is the change in angle that does the job.

9) There are some nice pictures and some discussion of bike
tire structure at:
http://openairbicycles.com/page.cfm?pageID=88

On 11/09/2007 11:00 AM, Edmiston, Mike wrote:
I am not familiar with sew-up tires so I don't know many details.
However, this has reminded me of an earlier picture that John Denker
posted in which he showed a bulged tire on a rim. This is at

http://www.av8n.com/physics/img48/tire.png

In this drawing he shows the bottom cross-section as somewhat of a
horizontal ellipse whereas the top cross-section is somewhat of a
vertical ellipse.

The top is not meant to be elliptical. The rubber part is a
part of a circle, as accurately as I could draw it. I just
now re-checked this.

. In this case the air pressure (not the sidewall
tension) in the lower tire does contribute some upward force that is
not
countered by as much downward air pressure in the top cross-section.

Not a chance. Pascal's principle. The axle cares only about
the forces on the rim, and the rim doesn't change shape, and
the pressure is the same everywhere by Pascal's principle. The
only thing that changes the force on the rim as we go from the
unloaded (and symmetric) case to the loaded case is the angle
of the sidewall. Sidewall tension, with due regard to angle,
is the *only* physics in sight with the required dependence on
the amount of distortion.

I do not think this is the correct profile for an automobile tire
because when I look at mounted auto tired there is practically no
difference in the shape of the sidewall as comes away from the rim.
That is, the bulge is much lower (near the bottom tread). So I
think
the effect John is describing in this drawing is a very minor
contributor to auto tires.

I stand by my analysis. No physical mechanism has been adduced
whereby a bulge "lower down" can have any physical effect on the
rim ... except via the aforementioned change of angle.

It might make more contribution with the low profile tires becoming
more
popular, and with bicycle tires in which the sidewall is not very
long.

Becoming popular? It depends on the type of car. Fancy
sports cars have for eons had wide, low tires. It's hard
to afford such a car on a teacher's salary. But they sure
are fun to drive.

But when I look at the clincher tires on my bicycle, I still see the
bulge pretty low, and the sidewall profile near the rim is almost
the
same all around the rim. Perhaps for sew-up tires this is not the
case.
Maybe John's drawing is very much appropriate for sew-up tires.

Unless/until somebody contributes some new physics to the analysis,
I'm sticking with the model of angle-dependence of a non-stretchy
fabric ... as applied to all pneumatic tires, not just sew-ups.


On 11/09/2007 12:43 PM, LaMontagne, Bob wrote:
Is this another case of talking past each other?

Quite possibly.

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.

For car tires, yes. The bead is
a) strongly pretensioned, and
b) sitting on the outside of a shoulder.

Without the shoulder *and* the pretensioning, this design wouldn't
work at all.

Clincher-type bike tires use a different design, with the bead
clinched on the inside of a lip. (This doesn't require the huge
pretensioning that the car-tire design requires.)

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