Tire Pressure and Sidewalls

Frank Krygowski writes:

On Sunday, March 27, 2016 at 10:07:01 PM UTC-4, Radey Shouman wrote:
Frank Krygowski writes:

On Friday, March 25, 2016 at 3:37:56 PM UTC-4, Radey Shouman wrote:
Frank Krygowski writes:

Or consider a simpler system: A soft clay (or plasticine) ball thrown
at a wall, or dropped on a rigid floor. The result is a perfectly
inelastic collision. The clay ball had kinetic energy before the
collision; it has none afterward. What did the (negative) work on the
ball to remove the kinetic energy? You can't say "the wall was
stationary, so no work occurred."

This is basic high school physics, Frank. Work was done in deforming
the ball. No work was done on the wall, unless it was somehow damaged
or moved.

Good! We have progress! Work was indeed done on the ball, even though
the wall didn't move. Now reconsider your position regarding the
tire, please - which was "The ground is rigid and does not move. It
does no work."

No. Work was not done on the ball. Some bits of the ball did work on
other bits. Put another way, energy was transferred, and transformed,
within the ball -- but was not transferred across the plane of the wall.

The same is true of the wheel: energy is transferred and transformed
within the wheel and tire, but no work is done on the ground, nor does
the ground do any work. This is true, of course, only so long as the
ground does not move. If the wheel were dropped on, or ridden through,
gravel or mud, then work would be done on that surface.

I'm not sure what the difficulty is with this point.

Well, one difficulty is that when you were bouncing a tire on the
stationary ground, you insisted that work was done on the air in the
tire by that ground. When I propose dropping a clay ball on the
ground or throwing it against an immobile wall, you suddenly say that
the ground or wall can't do work on the moving object.

Frank, here you are making **** up. I never said that work was done on
the dropped bicycle wheel by the ground. Please try not to be an ass.

I propose drawing the system boundary around the moving ball and evaluating
its energy. (It's easier for the case of horizontal travel, where we deal
only with kinetic energy and work.) Where does the kinetic energy go? What
did the negative work to remove it?

Don't bother to answer. It's time to end this. But understand that this
stuff was what I did for decades.

That's why I am mystified by your confusion. Perhaps a medical issue?

Frank, I conceded, several times, that a change in *pressure* was not
necessary.

Ah. OK, that's good too. I don't see it upthread, but it's still good.
Or hell, just go back to the tire filled with pressurized water. It
works fine, as Joe and Jobst have reported. It absorbs shock, it
rolls OK; it's just heavy. Yet the water has no compressibility.

One of these days I may be bored enough to try it.

But answer this: why do you say "pressurized" water? What maintains
the pressure? The water, after all, incompressible.

The pressure is applied by the elasticity of the containing structure,
i.e. the tire. If the elastic container had a very low stiffness
(e.g. a latex balloon) then quite a bit of water would have to be
added to the container to generate much rise in pressure. Since this
elastic container (the tire) has high stiffness in tension, when Joe
had it filled, attempting to pump more water in produced a sudden very
stiff resistance to the pump handle. Pressure rose almost immediately
to the amount required to resist the pump's motion.

I don't believe that for a minute. Try inflating a bicycle tire from 30
psi to 100 psi, how much does the diameter change? The tire casing is
*very* stiff with respect to stretching. A much simpler explanation is
that some air remained in the tube. It's not easy to completely
evacuate air from a system, and there was no real reason for either Joe
Riel or Jobst Brandt to try. I don't recall that they said they
completely submerged the tubes and squeezed the air out, which would
have been only a start.

Only a small quantity of air is needed to greatly change the behavior of
the system, think of the difference between a hydraulic brake system
before and after bleeding.

I think you could get a good idea by moistening the outside of the
tire and letting it bounce on dry concrete. Examine the footprint.
Or for a permanent record, ink the tire surface and drop it onto a
sheet of paper. You should get a print like those in _Bicycling
Science_ 3rd edition, p. 218.

(Oh, BTW, in that chapter of that book, dealing with tires, there
seems to be no mention at all of the air compression that you're
focusing on. Lots of very technical discussion, free body diagrams,
derivation of equations, fitting equations to experimental curves,
etc. etc. But no mention of the compressibility of the air, AFAICT.
That might indicate how non-critical that effect really is!)

Does it say that the tire volume might increase or decrease, or
whatever, when the tire is loaded?

Nothing, IIRC. It's just not important.

Perhaps it's just too obvious to belabor.

I think we've belabored it enough. Some day you might try what Joe and Jobst did.

What they did is interesting, but it does not reflect the behavior of
pneumatic tires.

One more time, my contention is that an unloaded, inflated pneumatic
bicycle tire assumes a shape very close to an ideal torus that uniquely
maximizes it's volume subject to the constraints of the fixed area of
the tire casing and the location (for a clincher tire) of the tire
beads.

I continue to be surprised that you disagree with this.

--

On 3/28/2016 9:48 AM, Radey Shouman wrote:
Frank Krygowski writes:


I think we've belabored it enough. Some day you might try what Joe and Jobst did.

What they did is interesting, but it does not reflect the behavior of
pneumatic tires.

One more time...

One _more_ time?

my contention is that an unloaded, inflated pneumatic
bicycle tire assumes a shape very close to an ideal torus that uniquely
maximizes it's volume subject to the constraints of the fixed area of
the tire casing and the location (for a clincher tire) of the tire
beads.

I continue to be surprised that you disagree with this.

I don't believe I've ever seen that before, let alone disagreed with it!

I remember instead things like "the pressure must increase," or "work
must be done on the air in the tire" but later "The ground is rigid and
does not move. It does no work" etc. etc.

The above looks merely like an attempt to state something that has not
been proven wrong.

That's OK. As a teacher, I was praised for my patience. But even my
patience has run out.

--
- Frank Krygowski

On Monday, March 28, 2016 at 9:48:08 AM UTC-4, Radey Shouman wrote:
Snipped
Frank, here you are making **** up. I never said that work was done on
the dropped bicycle wheel by the ground. Please try not to be an ass.


That's typicla Frank. Now you know why so many here have killfiled Frank and consider him a TROLL of the 1st order.

The only time I see Frank's crap now is if I see it in someone else's post.

Cheers

Frank Krygowski writes:

On 3/28/2016 9:48 AM, Radey Shouman wrote:
Frank Krygowski writes:


I think we've belabored it enough. Some day you might try what Joe and Jobst did.

What they did is interesting, but it does not reflect the behavior of
pneumatic tires.

One more time...

One _more_ time?

my contention is that an unloaded, inflated pneumatic
bicycle tire assumes a shape very close to an ideal torus that uniquely
maximizes it's volume subject to the constraints of the fixed area of
the tire casing and the location (for a clincher tire) of the tire
beads.

I continue to be surprised that you disagree with this.

I don't believe I've ever seen that before, let alone disagreed with it!

But of course you have. Step by step:

1) The unloaded tire assumes the unique volume-maximizing shape,
subject to some constraints.

2) When loaded, none of the constriants are relaxed. Put another way,
all of the constraints continue to hold.

3) But, the shape is different from the volume-maximizing shape.

4) Therefore, the volume *must* decrease.


You might enjoy reading this paper:

F. Koutny (1976)
"A Method for Computing the Radial Deformation Characteristics of Belted Tires".
_Tire Science and Technology: August 1976, Vol. 4, No. 3, pp. 190-212.

doi: http://dx.doi.org/10.2346/1.2167222

Abstract:

This work proceeds from the idea that the air pressure in the tire is
the preponderant component of elasticity. It is possible to construct a
simple relationship between the stress and the reduction of the volume
after deformation from the equivalence of the work of the deforming load
to the work done on the air content of the tire. The description of the
procedures for calculating the volume in the case of belted tires forms
the essential portion of the present report. A comparison is given
between calculated and measured deformation characteristics and results
of measurements of pressure in deformed, belted tires.

It seems to be highly regarded, I found it referenced in a NHTSA
monograph on tires. Full disclosu I have not read it. I have not
been able to procure it through the local libraries, and haven't stumped
up $25 for it.

I remember instead things like "the pressure must increase," or "work
must be done on the air in the tire" but later "The ground is rigid
and does not move. It does no work" etc. etc.

I think I claimed that loading a tire would result in a decrease in
volume. I'm sorry that your self-vaunted patience does not extend to
try to actually *comprehend* the things to which you respond.

The above looks merely like an attempt to state something that has not
been proven wrong.

So which part of it is wrong now?

That's OK. As a teacher, I was praised for my patience. But even my
patience has run out.

--

Sir Ridesalot writes:

On Monday, March 28, 2016 at 9:48:08 AM UTC-4, Radey Shouman wrote:
Snipped
Frank, here you are making **** up. I never said that work was done on
the dropped bicycle wheel by the ground. Please try not to be an ass.


That's typicla Frank. Now you know why so many here have killfiled
Frank and consider him a TROLL of the 1st order.

The only time I see Frank's crap now is if I see it in someone else's post.

I believe Frank is often, though hardly always, right. Frequently not a
pleasant fellow, however.


--

On 3/28/2016 6:01 PM, Radey Shouman wrote:

But of course you have. Step by step:

1) The unloaded tire assumes the unique volume-maximizing shape,
subject to some constraints.

2) When loaded, none of the constriants are relaxed. Put another way,
all of the constraints continue to hold.

3) But, the shape is different from the volume-maximizing shape.

4) Therefore, the volume *must* decrease.

That is correct. When the tire is loaded, the volume decreases and the
pressure increases. It would be different if there were no constraints
on the tube from the tire.

But the sidewalls don't deform much. There's also the issue of increased
heat when a tire is under-inflated, which will cause the air to expand
and since the sidewalls are constraining the volume, the pressure will
increase.

I think I claimed that loading a tire would result in a decrease in
volume.

That is true, but it's a small decrease in volume.

I'm sorry that your self-vaunted patience does not extend to
try to actually *comprehend* the things to which you respond.

Be gentle. Not everyone understands PV=nRT.

Am 26.03.2016 um 22:44 schrieb Phil W Lee:
Radey considered Fri, 25 Mar 2016
20:02:24 -0400 the perfect time to write:

Phil W writes:
Sorry? The top surface of a water bed is at ambient pressure. The
container does not bulge like a tire casing, there is no outward
pressure to speak of. Not relevant at all.

So according to your logic, a water bed is incapable of supporting
your weight.

Maybe we are accidentally mxing up two different conversations here?

1) How does a wheel (or water bed) carry weight?
2) What is the meaningful spring constant?

I might like to point out that an iron railroad wheel is perfectly able
to carry a large amount of weight but that the iron railroad wheel does
not offer any suspension.

It is clear that a tire filled with water is able to carry a bicycle's
weight but not having ridden one, I am unable to determine whether this
wheel offers any suspension theoretically or practically.

In order to discuss pneumatic suspension, the changes in pressure
resulting by a change in force *are* essential.

Rolf Mantel

Am 23.03.2016 um 16:56 schrieb Frank Krygowski:
On 3/22/2016 11:34 PM, James wrote:

However a small increase in tension over a few spokes does occur, which
is what I wrote above and provided a reference to support my claim.
You claimed "No increase in tension anywhere is required or present.",
yet an increase in tension is present, as I described.

I agree it's present, though very small. It's not really required for
supporting the load, though. It's an inconsequential secondary effect.

I think this is really the essence of our misunderstandings:

in order to calculate the state or shape of the equilibrium ("support
the load"), you find the place where all higher-order effects add up to
zero, they can be disregarded.

in order to understand the dynamics that hold the equilibrium
("springiness") you need to show how or why these higehr-order effects
balance; you disregard the static part any only analyse the higher-order
parts. Natrually, these higher-order parts are small; the don't
influence the final result of the equilibrium but they give the
explanation *how* the equilibrium is achieved.

Rolf Mantel writes:

Am 26.03.2016 um 22:44 schrieb Phil W Lee:
Radey considered Fri, 25 Mar 2016
20:02:24 -0400 the perfect time to write:

Phil W writes:
Sorry? The top surface of a water bed is at ambient pressure. The
container does not bulge like a tire casing, there is no outward
pressure to speak of. Not relevant at all.

So according to your logic, a water bed is incapable of supporting
your weight.

Maybe we are accidentally mxing up two different conversations here?

I agree that several things are being mixed up. I would like to think
it's accidental. We have had to move verrryyy slowly, because agreement
on some points I would have thought commonplace has been difficult.

1) How does a wheel (or water bed) carry weight?
2) What is the meaningful spring constant?

I might like to point out that an iron railroad wheel is perfectly
able to carry a large amount of weight but that the iron railroad
wheel does not offer any suspension.

I agree. "Run-flat" rubber tires have also become practical, you might
see them rolling under those big black suburbans in your choice of
world leader's motorcade.

It is clear that a tire filled with water is able to carry a bicycle's
weight but not having ridden one, I am unable to determine whether
this wheel offers any suspension theoretically or practically.

Nor can I. A tire filled with water and carefully evacuated of air
would behave differently from a normal pneumatic tire. For example, I'm
fairly sure that the area of the contact patch would remain more or less
constant, and that the internal pressure of the tire would vary directly
as the load.

In order to discuss pneumatic suspension, the changes in pressure
resulting by a change in force *are* essential.

I agree.

This particular contentious eddy of a discussion on tire spring
constants began with a comment by Joe Riel:

-----------------------%----------------------------
From: Joe Riel
Subject: Tire Pressure and Sidewalls
Newsgroups: rec.bicycles.tech
Date: Thu, 17 Mar 2016 18:44:27 -0700

James writes:


Yep. As in my reply to Frank, this is a good reference.

http://scholarworks.rit.edu/cgi/view...ontext=article

That is badly done. The only behavior that is at all analgous to a
bike tire supportsis the contact with the road, and that isn't a
great model. The conclusion (that the load is supported by a volume
reduction) doesn't follow.
-----------------------%----------------------------

It seemed clear to me that, in fact, the load *is* supported by a volume
reduction, but several lines of argument have failed to convince Frank
or Phil of that.

sms writes:

On 3/28/2016 6:01 PM, Radey Shouman wrote:

But of course you have. Step by step:

1) The unloaded tire assumes the unique volume-maximizing shape,
subject to some constraints.

2) When loaded, none of the constriants are relaxed. Put another way,
all of the constraints continue to hold.

3) But, the shape is different from the volume-maximizing shape.

4) Therefore, the volume *must* decrease.

That is correct. When the tire is loaded, the volume decreases and the
pressure increases. It would be different if there were no constraints
on the tube from the tire.

But the sidewalls don't deform much. There's also the issue of
increased heat when a tire is under-inflated, which will cause the air
to expand and since the sidewalls are constraining the volume, the
pressure will increase.

I think I claimed that loading a tire would result in a decrease in
volume.

That is true, but it's a small decrease in volume.

I'm sorry that your self-vaunted patience does not extend to
try to actually *comprehend* the things to which you respond.

Be gentle. Not everyone understands PV=nRT.

It's hard to see how Frank could have maintained his position as an
engineering professor given the lack of understanding he has shown here.
I prefer to believe he is just not taking the time to actually read the
posts to which he responds, but fear that he may have a real
degenerative condition.

Were he a friend of mine I would try to convince him to have a checkup.

--