Tire Pressure and Sidewalls

The previous model I used assumed that the pressure of the tire
against the surface was uniform and equal to the pressure in the
tire. That isn't the case. The pressure between tire and surface
varies due to the local deformation of the tire. Estimating that
is difficult, but a simple thought experiment shows it surely
exists.

Put the loaded wheel resting on a smooth flat surface in a pressure
chamber. Assume that the tire tread is smooth so that the contact
with the surface is effectively gas tight. Consider the approximate
model, with F = P*A. Because of the gas tight contact, P is the
*absolute* pressure.

Now begin increasing the pressure in the pressure chamber.
As it increases, the relative pressure in the tire decreases.
The cords in the tire, though flexible, are stiff in tension
so the tire volume only changes (decreases) slightly. That
leads to a small and presumably negligible increase of the
absolute pressure in the tire. The sag of the tire should
not change appreciably.

However, when the absolute pressure in the chamber reaches
the absolute pressure in the tire, one can simply open the
valve and nothing will change. At that point it is clear
that the tire is effectively flat and is no longer supporting
the wheel. What happened?

The answer is that as the relative pressure in the tire decreases,
the tensile stress in the sidewall also decrease. This decreases
the support that sidewalls contribute. I don't expect that the
tire will suddenly collapse when the pressures equalized, but
rather will do so over some range. How wide that range is
is not clear.

--
Joe Riel

Am 10.03.2016 um 16:47 schrieb Joe Riel:
Put the loaded wheel resting on a smooth flat surface in a pressure
chamber. Assume that the tire tread is smooth so that the contact
with the surface is effectively gas tight. Consider the approximate
model, with F = P*A. Because of the gas tight contact, P is the
*absolute* pressure.

Now begin increasing the pressure in the pressure chamber.
As it increases, the relative pressure in the tire decreases.
The cords in the tire, though flexible, are stiff in tension
so the tire volume only changes (decreases) slightly. That
leads to a small and presumably negligible increase of the
absolute pressure in the tire. The sag of the tire should
not change appreciably.

So far, all is clear.

However, when the absolute pressure in the chamber reaches
the absolute pressure in the tire, one can simply open the
valve and nothing will change. At that point it is clear
that the tire is effectively flat and is no longer supporting
the wheel.

As long as the valve is closed, the tire is supported by the ir volume
inside.
Only after opening the valve, the air will be able to move freely
between inside and outside and the tire will stop being supported.

Rolf Mantel writes:

Am 10.03.2016 um 16:47 schrieb Joe Riel:
Put the loaded wheel resting on a smooth flat surface in a pressure
chamber. Assume that the tire tread is smooth so that the contact
with the surface is effectively gas tight. Consider the approximate
model, with F = P*A. Because of the gas tight contact, P is the
*absolute* pressure.

Now begin increasing the pressure in the pressure chamber.
As it increases, the relative pressure in the tire decreases.
The cords in the tire, though flexible, are stiff in tension
so the tire volume only changes (decreases) slightly. That
leads to a small and presumably negligible increase of the
absolute pressure in the tire. The sag of the tire should
not change appreciably.

So far, all is clear.

However, when the absolute pressure in the chamber reaches
the absolute pressure in the tire, one can simply open the
valve and nothing will change. At that point it is clear
that the tire is effectively flat and is no longer supporting
the wheel.

As long as the valve is closed, the tire is supported by the ir volume
inside.
Only after opening the valve, the air will be able to move freely
between inside and outside and the tire will stop being supported.

When the pressure equalizes, opening the valve makes no difference.
How can it? The pressure on both sides of the valve are equal.

--
Joe Riel

Am 10.03.2016 um 17:05 schrieb Joe Riel:
As long as the valve is closed, the tire is supported by the air volume
inside.
Only after opening the valve, the air will be able to move freely
between inside and outside and the tire will stop being supported.

When the pressure equalizes, opening the valve makes no difference.

Not statically.

How can it? The pressure on both sides of the valve are equal.

Assume you keep the valve shut. You press the the tire, the tire volume
changes and the pressure inside the tire changes. You stop the tire
pressure, the presure inside the tire causes the tire to return to its
old shape, you return to the old state.

Assume you open the valve. You press the tire, the tire volume changes,
the presure inside the tire changes, air leaves the tire. You release
the tire pressure, air does not return to the inside of the tire but the
tire volume decreases instead.

On 3/10/2016 9:47 AM, Joe Riel wrote:
The previous model I used assumed that the pressure of the tire
against the surface was uniform and equal to the pressure in the
tire. That isn't the case. The pressure between tire and surface
varies due to the local deformation of the tire. Estimating that
is difficult, but a simple thought experiment shows it surely
exists.

Put the loaded wheel resting on a smooth flat surface in a pressure
chamber. Assume that the tire tread is smooth so that the contact
with the surface is effectively gas tight. Consider the approximate
model, with F = P*A. Because of the gas tight contact, P is the
*absolute* pressure.

Now begin increasing the pressure in the pressure chamber.
As it increases, the relative pressure in the tire decreases.
The cords in the tire, though flexible, are stiff in tension
so the tire volume only changes (decreases) slightly. That
leads to a small and presumably negligible increase of the
absolute pressure in the tire. The sag of the tire should
not change appreciably.

However, when the absolute pressure in the chamber reaches
the absolute pressure in the tire, one can simply open the
valve and nothing will change. At that point it is clear
that the tire is effectively flat and is no longer supporting
the wheel. What happened?

The answer is that as the relative pressure in the tire decreases,
the tensile stress in the sidewall also decrease. This decreases
the support that sidewalls contribute. I don't expect that the
tire will suddenly collapse when the pressures equalized, but
rather will do so over some range. How wide that range is
is not clear.


Very good, THX.

--
Andrew Muzi
www.yellowjersey.org/
Open every day since 1 April, 1971

Rolf Mantel writes:

Am 10.03.2016 um 17:05 schrieb Joe Riel:
As long as the valve is closed, the tire is supported by the air volume
inside.
Only after opening the valve, the air will be able to move freely
between inside and outside and the tire will stop being supported.

When the pressure equalizes, opening the valve makes no difference.

Not statically.

Statically there will be no change.

How can it? The pressure on both sides of the valve are equal.

Assume you keep the valve shut. You press the the tire, the tire
volume changes and the pressure inside the tire changes. You stop the
tire pressure, the presure inside the tire causes the tire to return
to its old shape, you return to the old state.

If you the change the load on the bike, so that the tire deformation
changes, then the pressure could change so they are no longer at
equilibrium. In that sense, the situation is not the same as
with the valve closed. But that doesn't change the static situation
which I was describing (i.e. constant load).

Assume you open the valve. You press the tire, the tire volume
changes, the presure inside the tire changes, air leaves the tire. You
release the tire pressure, air does not return to the inside of the
tire but the tire volume decreases instead.

--
Joe Riel

Joe Riel writes:

The previous model I used assumed that the pressure of the tire
against the surface was uniform and equal to the pressure in the
tire. That isn't the case. The pressure between tire and surface
varies due to the local deformation of the tire. Estimating that
is difficult, but a simple thought experiment shows it surely
exists.

Put the loaded wheel resting on a smooth flat surface in a pressure
chamber. Assume that the tire tread is smooth so that the contact
with the surface is effectively gas tight. Consider the approximate
model, with F = P*A. Because of the gas tight contact, P is the
*absolute* pressure.

Now begin increasing the pressure in the pressure chamber.
As it increases, the relative pressure in the tire decreases.
The cords in the tire, though flexible, are stiff in tension
so the tire volume only changes (decreases) slightly. That
leads to a small and presumably negligible increase of the
absolute pressure in the tire. The sag of the tire should
not change appreciably.

However, when the absolute pressure in the chamber reaches
the absolute pressure in the tire, one can simply open the
valve and nothing will change. At that point it is clear
that the tire is effectively flat and is no longer supporting
the wheel. What happened?

The answer is that as the relative pressure in the tire decreases,
the tensile stress in the sidewall also decrease. This decreases
the support that sidewalls contribute. I don't expect that the
tire will suddenly collapse when the pressures equalized, but
rather will do so over some range. How wide that range is
is not clear.

In reflecting on this I wonder if the tire will sag with the decreasing
relative pressure in the tire pretty much as it does in a normal tire as
the pressure is reduced. That seems likely in that it is the tension in
the sidewalls that is the key, and that certainly is decreasing with the
relative pressure.

If that is the case, then P*A doesn't really hold, because P has to be
the absolute pressure in the tire (doesn't it?), and if anything it will
increase. Presumably the decreasing surface tension in the tire has to
be added to the pressure in the tire to get the pressure at the
tire/surface interface. Something to think about.

--
Joe Riel

On Thu, 10 Mar 2016 07:47:12 -0800, Joe Riel wrote:
The previous model I used assumed that the pressure of the tire
against the surface was uniform and equal to the pressure in the tire.
That isn't the case. The pressure between tire and surface varies due
to the local deformation of the tire. Estimating that is difficult,
but a simple thought experiment shows it surely exists.

Put the loaded wheel resting on a smooth flat surface in a pressure
chamber. Assume that the tire tread is smooth so that the contact
with the surface is effectively gas tight. Consider the approximate
model, with F = P*A. Because of the gas tight contact, P is the
*absolute* pressure.

Now begin increasing the pressure in the pressure chamber. As it
increases, the relative pressure in the tire decreases. The cords in
the tire, though flexible, are stiff in tension so the tire volume
only changes (decreases) slightly. That leads to a small and
presumably negligible increase of the absolute pressure in the tire.
The sag of the tire should not change appreciably.

However, when the absolute pressure in the chamber reaches the
absolute pressure in the tire, one can simply open the valve and
nothing will change. At that point it is clear that the tire is
effectively flat and is no longer supporting the wheel. What
happened?

The answer is that as the relative pressure in the tire decreases, the
tensile stress in the sidewall also decrease. This decreases the
support that sidewalls contribute. I don't expect that the tire will
suddenly collapse when the pressures equalized, but rather will do so
over some range. How wide that range is is not clear.

The tire has stiffness contributed by the bead hoops, the stiffness of
the fabric of the casing and the stiffness of the tread material and of
the rubber impregnated into the sidewall. Note how an unmounted tire
retains a shape of its own because it has residual stresses from the
manufacturing process, shaping a flat stip of fabric into an incomplete
toroid and molding/bonding a strip of rubber to the casing- as the
external and internal pressures exqualize, the tire will deform to the
shape compatible with those internal stresses.

But that raises an interesting question. If I am riding my bke with 100
psi in the tires in ambient sea-level air pressure, the rim stands above
the road on the tire. If the ambient air pressure could be increased to
100 PSI- without killing me in the process- would the tire lose its
ability to support the load?

On 3/10/2016 2:41 PM, Tim McNamara wrote:
On Thu, 10 Mar 2016 07:47:12 -0800, Joe Riel wrote:
The previous model I used assumed that the pressure of the tire
against the surface was uniform and equal to the pressure in the tire.
That isn't the case. The pressure between tire and surface varies due
to the local deformation of the tire. Estimating that is difficult,
but a simple thought experiment shows it surely exists.

Put the loaded wheel resting on a smooth flat surface in a pressure
chamber. Assume that the tire tread is smooth so that the contact
with the surface is effectively gas tight. Consider the approximate
model, with F = P*A. Because of the gas tight contact, P is the
*absolute* pressure.

Now begin increasing the pressure in the pressure chamber. As it
increases, the relative pressure in the tire decreases. The cords in
the tire, though flexible, are stiff in tension so the tire volume
only changes (decreases) slightly. That leads to a small and
presumably negligible increase of the absolute pressure in the tire.
The sag of the tire should not change appreciably.

However, when the absolute pressure in the chamber reaches the
absolute pressure in the tire, one can simply open the valve and
nothing will change. At that point it is clear that the tire is
effectively flat and is no longer supporting the wheel. What
happened?

The answer is that as the relative pressure in the tire decreases, the
tensile stress in the sidewall also decrease. This decreases the
support that sidewalls contribute. I don't expect that the tire will
suddenly collapse when the pressures equalized, but rather will do so
over some range. How wide that range is is not clear.

The tire has stiffness contributed by the bead hoops, the stiffness of
the fabric of the casing and the stiffness of the tread material and of
the rubber impregnated into the sidewall. Note how an unmounted tire
retains a shape of its own because it has residual stresses from the
manufacturing process, shaping a flat stip of fabric into an incomplete
toroid and molding/bonding a strip of rubber to the casing- as the
external and internal pressures exqualize, the tire will deform to the
shape compatible with those internal stresses.

But that raises an interesting question. If I am riding my bke with 100
psi in the tires in ambient sea-level air pressure, the rim stands above
the road on the tire. If the ambient air pressure could be increased to
100 PSI- without killing me in the process- would the tire lose its
ability to support the load?


Right, as Mr Riel noted, it's relative pressure and changes
in that relative pressure, such as in a vacuum/pressure
chamber, could be used to isolate that effect from the
casing stiffness and other factors.

--
Andrew Muzi
www.yellowjersey.org/
Open every day since 1 April, 1971

Tim McNamara writes:

On Thu, 10 Mar 2016 07:47:12 -0800, Joe Riel wrote:
The previous model I used assumed that the pressure of the tire
against the surface was uniform and equal to the pressure in the tire.
That isn't the case. The pressure between tire and surface varies due
to the local deformation of the tire. Estimating that is difficult,
but a simple thought experiment shows it surely exists.

Put the loaded wheel resting on a smooth flat surface in a pressure
chamber. Assume that the tire tread is smooth so that the contact
with the surface is effectively gas tight. Consider the approximate
model, with F = P*A. Because of the gas tight contact, P is the
*absolute* pressure.

Now begin increasing the pressure in the pressure chamber. As it
increases, the relative pressure in the tire decreases. The cords in
the tire, though flexible, are stiff in tension so the tire volume
only changes (decreases) slightly. That leads to a small and
presumably negligible increase of the absolute pressure in the tire.
The sag of the tire should not change appreciably.

However, when the absolute pressure in the chamber reaches the
absolute pressure in the tire, one can simply open the valve and
nothing will change. At that point it is clear that the tire is
effectively flat and is no longer supporting the wheel. What
happened?

The answer is that as the relative pressure in the tire decreases, the
tensile stress in the sidewall also decrease. This decreases the
support that sidewalls contribute. I don't expect that the tire will
suddenly collapse when the pressures equalized, but rather will do so
over some range. How wide that range is is not clear.

The tire has stiffness contributed by the bead hoops, the stiffness of
the fabric of the casing and the stiffness of the tread material and of
the rubber impregnated into the sidewall. Note how an unmounted tire
retains a shape of its own because it has residual stresses from the
manufacturing process, shaping a flat stip of fabric into an incomplete
toroid and molding/bonding a strip of rubber to the casing- as the
external and internal pressures exqualize, the tire will deform to the
shape compatible with those internal stresses.

But that raises an interesting question. If I am riding my bke with 100
psi in the tires in ambient sea-level air pressure, the rim stands above
the road on the tire. If the ambient air pressure could be increased to
100 PSI- without killing me in the process- would the tire lose its
ability to support the load?

The answer is yes. Basically the surface tension in the sidewall (for a
cut along the big circumference) is p*r, with p being the relative
pressure and r the small radius. Just as a wheel stands on its bottom
spokes, it also stands on the bottom sidewalls.

Consider the fully inflated, unloaded tire. The pressure is p.
The average surface tension in the sidewall is T0 = p*r0.
Now load the wheel. The p remains essentially unchanged.
The vertical force throuch the two lower sidewalls is essentially
the applied load. This can be computed via

f = 2*int((T0-T)*dl)

where

f = applied load
T0 = unloaded sidewall surface tension
T = loaded sidewall surface tension

But

T0 - T = p*(r0 - r)

where
T = local surface tension of sidewall at bottom of wheel
r = local radius of sidewall at bottom of wheel

The point of this is that support from the sidewall comes
from a reduction in its tension in the loaded region.

Now in the pressure chamber, as the pressure increases, the relative
pressure, p, decreases. But f and r0 remain constant. To maintain a
constant force, the local radius has to decrease. That is

p*(r0 - r) is constant

A decrease in the local radius means that the lower sidwalls are flexing
out, i.e. the tire is sagging down.

--
Joe Riel