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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 |
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Tire Pressure and Sidewalls
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. |
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Tire Pressure and Sidewalls
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 |
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Tire Pressure and Sidewalls
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. |
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Tire Pressure and Sidewalls
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 |
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Tire Pressure and Sidewalls
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 |
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Tire Pressure and Sidewalls
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 |
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Tire Pressure and Sidewalls
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? |
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Tire Pressure and Sidewalls
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 |
#10
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Tire Pressure and Sidewalls
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 |
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