# Is tire pressure different if you measure with tire off the car (as compared to when it is on the car and holding the weight of the car)?

September 18, 2009 1:34 PM Subscribe

Is tire pressure different if you measure with wheel off the car (as compared to when it is on the car and holding the weight of the car)?

wouldn't the pressure from the ground that partially squashes the tire decrease the volume, thus increasing the pressure? (Assuming the tire is thick enough to prevent the tube from expanding in other directions)

good question, I'm sure someone will be able to site something accurate

posted by Think_Long at 1:46 PM on September 18, 2009

good question, I'm sure someone will be able to site something accurate

posted by Think_Long at 1:46 PM on September 18, 2009

Yes, the pressure will be higher when the tire is supporting the car. If you want to imagine a free body diagram, simplify things and consider a car resting on one big tire. If you take the car off, it's supporting no weight - it does not have to exert any force on the axel to keep the car in static equilibrium. Air is a compressible fluid - pressure will increase when the tire bears a load because it's the "pushing back" of the air that keeps the car body from falling.

posted by phrontist at 2:16 PM on September 18, 2009

posted by phrontist at 2:16 PM on September 18, 2009

The elasticity of the tire is also a factor. The more elastic, the less the pressure increase.

posted by phrontist at 2:18 PM on September 18, 2009

posted by phrontist at 2:18 PM on September 18, 2009

I would guess that it's going to be the same, within a very few fractions.

The pressure for a given quantity of air is entirely dependent on the volume of the container. We know that the tire deforms under the weight of the car; we can see this in the bulge along the bottom of the sidewall near the ground and in the increased contact patch where the rubber meets the road. We also know the tire can deform to a fairly large extent, witness what it looks like when it is flat.

When the tire is off the car, the tire will deform in other directions--perhaps the tread will get a crown in it that is not present when the tire is on the machine. If the tire is stiff enough that it doesn't deform much, then the volume is essentially constant.

Please clue me in if this is wrong.

posted by maxwelton at 2:40 PM on September 18, 2009

The pressure for a given quantity of air is entirely dependent on the volume of the container. We know that the tire deforms under the weight of the car; we can see this in the bulge along the bottom of the sidewall near the ground and in the increased contact patch where the rubber meets the road. We also know the tire can deform to a fairly large extent, witness what it looks like when it is flat.

When the tire is off the car, the tire will deform in other directions--perhaps the tread will get a crown in it that is not present when the tire is on the machine. If the tire is stiff enough that it doesn't deform much, then the volume is essentially constant.

Please clue me in if this is wrong.

posted by maxwelton at 2:40 PM on September 18, 2009

A more simple thought experiment: Imagine a balloon. Now put a car on top of it. Does the air pressure increase?

At a rough estimate I think the surface area of an average tire would be about 1200 square inches. Times 4 of them is 4800 square inches. So a 4800lb car would increase tire pressure by 1psi when it rests on the tires.

posted by sfenders at 3:18 PM on September 18, 2009

At a rough estimate I think the surface area of an average tire would be about 1200 square inches. Times 4 of them is 4800 square inches. So a 4800lb car would increase tire pressure by 1psi when it rests on the tires.

posted by sfenders at 3:18 PM on September 18, 2009

Modern car tyres have steel belts that are not going to stretch at all (at least for the typical pressures that tyres are inflated to).

Thought experiment: if you fill a tyre with water instead of air, will it work the same? (Except for being much heavier.)

So, how does a tyre transmit the force from the ground to the wheel/axle? Let's assume rubber is infinitely stretchy, and only the steel belts provide the strength. When a tyre carries a load, it deforms a bit where it touches the ground. The sidewalls therefore make slightly larger angles at the bead where they're attached to the wheel. This decreases the radial (downward) component of the belt's tension reducing the downward force in that area of the wheel.

So if a tyre is fully inflated, there should be no change in the internal pressure when it's loaded. (Of course there are second order effects like stretchiness and stiffness of the belts, strength of the rubber, etc. that may have a small effect on the pressure.)

posted by phliar at 3:26 PM on September 18, 2009

Thought experiment: if you fill a tyre with water instead of air, will it work the same? (Except for being much heavier.)

So, how does a tyre transmit the force from the ground to the wheel/axle? Let's assume rubber is infinitely stretchy, and only the steel belts provide the strength. When a tyre carries a load, it deforms a bit where it touches the ground. The sidewalls therefore make slightly larger angles at the bead where they're attached to the wheel. This decreases the radial (downward) component of the belt's tension reducing the downward force in that area of the wheel.

*Voila!*the radial force at the lower end of the wheel is lower than the radial force elsewhere on the wheel's perimeter ⇒ upward force on the axle.So if a tyre is fully inflated, there should be no change in the internal pressure when it's loaded. (Of course there are second order effects like stretchiness and stiffness of the belts, strength of the rubber, etc. that may have a small effect on the pressure.)

posted by phliar at 3:26 PM on September 18, 2009

Think_Long and maxwelton: yes, a tyre deforms under load; but that doesn't say necessarily say anything about its volume.

posted by phliar at 3:32 PM on September 18, 2009

posted by phliar at 3:32 PM on September 18, 2009

yeah, I just wasn't sure if the deformation compensated completely for the volume in other directions, or not. I accept your answer with enthusiasm

posted by Think_Long at 3:35 PM on September 18, 2009

posted by Think_Long at 3:35 PM on September 18, 2009

Crap, I just thought of the perfect example -- the Lunar Rover. Its tyres were filled with

Inflating a tyre makes sure it won't buckle under load. The Lunar Rover's "tyre belts" are actually its whole tyre, which shows that the weight is supported entirely by the sidewall flexing, and not by any "pushing back" that the air inside the tyre is doing. This also implies the Rover's tyres' must have been much stiffer than the belts in a car tyre.

posted by phliar at 3:38 PM on September 18, 2009

*vacuum!*Inflating a tyre makes sure it won't buckle under load. The Lunar Rover's "tyre belts" are actually its whole tyre, which shows that the weight is supported entirely by the sidewall flexing, and not by any "pushing back" that the air inside the tyre is doing. This also implies the Rover's tyres' must have been much stiffer than the belts in a car tyre.

posted by phliar at 3:38 PM on September 18, 2009

It should be the same.

p = absolute pressure

V = volume of gas

n = physical quantity of the gas

R = gas constant = 8.314472 J(oule) K(elvin)

T = temperature

R, T, and n stay the same. The volume changes

posted by Civil_Disobedient at 3:40 PM on September 18, 2009 [1 favorite]

*pV = nRT*(Ideal gas law)p = absolute pressure

V = volume of gas

n = physical quantity of the gas

R = gas constant = 8.314472 J(oule) K(elvin)

^{−1}mol^{−1}T = temperature

R, T, and n stay the same. The volume changes

*slightly*as pressure is applied, but if you've ever taken a look at the bottom of a tire, you'll notice it's really not that much.posted by Civil_Disobedient at 3:40 PM on September 18, 2009 [1 favorite]

*if you fill a tyre with water instead of air, will it work the same?*

Yes, it would. Pressure would increase by same amount, only deformation would be less as water doesn't compress so much.

True that this ignores the simplifying assumption that the tire sidewalls aren't rigid enough to support any of the weight of the car, but this is close enough to true as makes no difference.

posted by sfenders at 3:41 PM on September 18, 2009

To be clear, what I mean is that the variance in measurement will be greater than the variance in gas volume, so while it will be

posted by Civil_Disobedient at 3:42 PM on September 18, 2009

*slightly*not-the-same, you won't be able to tell using a run-of-the-mill tire pressure gauge, anyway.posted by Civil_Disobedient at 3:42 PM on September 18, 2009

*The volume changes slightly as pressure is applied, but if you've ever taken a look at the bottom of a tire, you'll notice it's really not that much.*

Well, I think it ought to increase pressure by only 0.5 psi for a typical car. And normally they're around 30psi to begin with, so 1.6% -- not much. And some amount of it not at the bottom.

posted by sfenders at 3:44 PM on September 18, 2009

Yes, I was using the standard high-school assumptions like the belts are weightless, infinitely thin, perfectly flexible, etc.

However, you write that "deformation would be less as water doesn't compress so much". This is directly opposite to my argument, which is that (to first-order, using a perfect tyre) the internal volume of the tyre

I can probably come up with some references, but off the top of my head there's a book by Jobst Brandt called The Bicycle Wheel that has detailed analyses of all kinds of wheels.

posted by phliar at 3:50 PM on September 18, 2009

However, you write that "deformation would be less as water doesn't compress so much". This is directly opposite to my argument, which is that (to first-order, using a perfect tyre) the internal volume of the tyre

*does not change under load*, which means it makes no difference what you fill it with.I can probably come up with some references, but off the top of my head there's a book by Jobst Brandt called The Bicycle Wheel that has detailed analyses of all kinds of wheels.

posted by phliar at 3:50 PM on September 18, 2009

Of course, if you take a non-pressurised tire, it can support some weight. Maybe a hundred pounds to deform it as much as a car would? A guess based on old experience leaning on empty tires. So I guess it is enough to make a small difference.

posted by sfenders at 3:50 PM on September 18, 2009

posted by sfenders at 3:50 PM on September 18, 2009

When I get home I'm going to perform a little experiment: fill two identical balloons, one with water and one with air, so they're the same size. Put the same book on top of each balloon. By looking at the deformation of the balloon, can you guess which one is filled with water? My bet is that they will both deform equally.

posted by phliar at 3:56 PM on September 18, 2009

posted by phliar at 3:56 PM on September 18, 2009

We measure tire pressure with the vehicles on the lifts (i.e. no weight from the vehicle on the tires) at our shop. The tire pressure is not going to vary to an appreciable degree as others with more scientific knowledge have pointed out above.

posted by redsnare at 4:25 PM on September 18, 2009 [1 favorite]

posted by redsnare at 4:25 PM on September 18, 2009 [1 favorite]

My knee-jerk physics say yes - it's going to be affected. It's not going to be zero.

Agree with others - it won't be measurable on your run of the mill tire guage... I'd imagine the variance in pressure due to temperature would be a far bigger factor.

posted by TravellingDen at 5:24 PM on September 18, 2009 [1 favorite]

Agree with others - it won't be measurable on your run of the mill tire guage... I'd imagine the variance in pressure due to temperature would be a far bigger factor.

posted by TravellingDen at 5:24 PM on September 18, 2009 [1 favorite]

Putting a car on a balloon does not increase the pressure in the balloon. The balloon can support a weight equal to the pressure in the balloon multiplied by the contact area. As the load on the balloon increases, the contact area increases. If the contact area increases beyond the elastic limit of the balloon, the balloon bursts with the same pressure that it started with. Similarly the pressure in a car tire does not increase with load; it is the contact area that increases. The tire deforms by getting shorter and wider but, for ordinary pressures and loads, the volume doesn't change. The experiment can be readily done on a bicycle where a load much greater than the vehicle can be readily applied.

posted by llc at 6:20 PM on September 18, 2009 [1 favorite]

posted by llc at 6:20 PM on September 18, 2009 [1 favorite]

I am not so sure, but thinking it through...

Balloons are pretty elastic, so they stretch a lot to get a small increase in pressure. 6-inch diameter balloon would be 450 square inches surface area. I've no idea what might be the average pressure of a balloon, but say 20psi. You'd need 45lb weight to get a 0.5% (0.1psi) increase in pressure and decrease in (gas) volume. The same pressure might give pretty close to zero volume change with water, but even with more weight than the average balloon could stand the two should look about the same I'd think. A volume difference not easily observable by eye, while the shape deformation due to the uneven distribution of weight is easily seen.

But where are you going with the "ideal" balloon? Perfectly elastic does not tell you elastic constant which could be anything. One that takes exactly zero effort to stretch would just flatten to a disc if you put a big flat weight on top. One that takes infinite effort to stretch beyond some non-zero unstressed surface area (but is still perfectly flexible), if you put a book on top I suppose there'd be a kind of very small flat spot there, the rest remaining perfectly round; and the pressure inside more obviously increasing. So depending on the elastic constant, the tire or balloon might take various shapes.

In any case though, the only thing causing all the stretching of the sides of the balloon is increased pressure of the fluid inside it. What else could it be, with the ideal perfectly-flexible balloon material? And that must depend on the amount of force you're pushing on the thing with. What else would cause it? You're increasing the average pressure on the outside of the balloon, by pushing on part of it. That reduces the difference in pressure between inside and outside, and therefore its volume ought to decrease.

If you could apply that increased pressure evenly around the whole thing, instead of just on some small part of the surface, it's easy enough to see why the volume change does have to depend on how much work it takes to compress the fluid inside.

posted by sfenders at 6:33 PM on September 18, 2009

Balloons are pretty elastic, so they stretch a lot to get a small increase in pressure. 6-inch diameter balloon would be 450 square inches surface area. I've no idea what might be the average pressure of a balloon, but say 20psi. You'd need 45lb weight to get a 0.5% (0.1psi) increase in pressure and decrease in (gas) volume. The same pressure might give pretty close to zero volume change with water, but even with more weight than the average balloon could stand the two should look about the same I'd think. A volume difference not easily observable by eye, while the shape deformation due to the uneven distribution of weight is easily seen.

But where are you going with the "ideal" balloon? Perfectly elastic does not tell you elastic constant which could be anything. One that takes exactly zero effort to stretch would just flatten to a disc if you put a big flat weight on top. One that takes infinite effort to stretch beyond some non-zero unstressed surface area (but is still perfectly flexible), if you put a book on top I suppose there'd be a kind of very small flat spot there, the rest remaining perfectly round; and the pressure inside more obviously increasing. So depending on the elastic constant, the tire or balloon might take various shapes.

In any case though, the only thing causing all the stretching of the sides of the balloon is increased pressure of the fluid inside it. What else could it be, with the ideal perfectly-flexible balloon material? And that must depend on the amount of force you're pushing on the thing with. What else would cause it? You're increasing the average pressure on the outside of the balloon, by pushing on part of it. That reduces the difference in pressure between inside and outside, and therefore its volume ought to decrease.

If you could apply that increased pressure evenly around the whole thing, instead of just on some small part of the surface, it's easy enough to see why the volume change does have to depend on how much work it takes to compress the fluid inside.

posted by sfenders at 6:33 PM on September 18, 2009

*6-inch diameter balloon would be 450 square inches surface area*

Err, that'd be a 6-inch radius actually. So you'd only need a 10lb weight to get 0.5%... still probably hard to see and measure for any weight the average balloon might hold.

posted by sfenders at 6:44 PM on September 18, 2009

*As the load on the balloon increases, the contact area increases.*

Ah, I see why it's counter-intuitive: You're increasing surface area of the balloon, while decreasing volume. If the friction on the contact patch were zero, I guess all parts of the balloon surface would be stretched evenly, but this would still be true: volume decreases while surface area increases, as it gets further from a sphere. Again easily to picture at the limit where it goes sorta disk-shaped. With normal friction at the contact patch, less of the material has to do more of the stretching, so it just breaks sooner.

posted by sfenders at 7:02 PM on September 18, 2009

The rear tires on my tractor are filled with a saline solution and except for being a much heavier, are indistinguishable from the same tires filled with air.*

* The filling isn't perfect, of course, being about 80-90% fluid and the rest air, inflated to the normal pressure.

posted by maxwelton at 7:54 PM on September 18, 2009

* The filling isn't perfect, of course, being about 80-90% fluid and the rest air, inflated to the normal pressure.

posted by maxwelton at 7:54 PM on September 18, 2009

Working in an automotive shop, to satisfy my own curiosity, I have measured this with a digital gauge within a tenth of a psi. For all practical purposes, there is no difference.

A tire is a very solid and sturdy structure and its interior dimensions aren't going to change just from the weight of the car. When the tire deforms at the contact patch, it deforms equally opposite to it. If the bottom of the tire is squashed, the top is a little bit bulged. So, if your mass of air is constant and the volume of space it occupies is constant, then the pressure will be constant, too.

Look at it this way:

The air isn't holding the car off the ground. The structure of the tire is holding the car up and the air is just helping to support that structure.

In all reality, exterior temperature and fluctuations in ambient barometric pressure will change your practical tire pressure much more than adding weight to the vehicle.

You could fill your car with Ikea furniture and you'd see less change than if the ambient temperature dropped twenty degrees.

Even the heat from the friction of the tire rotating across the road surface will increase the pressure more than adding weight to the vehicle.

posted by Jon-o at 8:06 PM on September 18, 2009 [5 favorites]

A tire is a very solid and sturdy structure and its interior dimensions aren't going to change just from the weight of the car. When the tire deforms at the contact patch, it deforms equally opposite to it. If the bottom of the tire is squashed, the top is a little bit bulged. So, if your mass of air is constant and the volume of space it occupies is constant, then the pressure will be constant, too.

Look at it this way:

The air isn't holding the car off the ground. The structure of the tire is holding the car up and the air is just helping to support that structure.

In all reality, exterior temperature and fluctuations in ambient barometric pressure will change your practical tire pressure much more than adding weight to the vehicle.

You could fill your car with Ikea furniture and you'd see less change than if the ambient temperature dropped twenty degrees.

Even the heat from the friction of the tire rotating across the road surface will increase the pressure more than adding weight to the vehicle.

posted by Jon-o at 8:06 PM on September 18, 2009 [5 favorites]

No difference. I've measured it several times over the last fifty years.

posted by torquemaniac at 8:35 PM on September 18, 2009 [4 favorites]

posted by torquemaniac at 8:35 PM on September 18, 2009 [4 favorites]

Well, we can apply some real numbers to determine the possible boundaries of the problem. From Civil_Disobedient we know that the pressure depends only on amount of air, temperature and volume. Since the amount of air and temperature are constant, the pressure is proportional to volume. If we know the change in volume, we know the change in pressure.

So let's approximate the change in volume when you load a tire. Let's first reduce it to a two dimensional problem, looking at the tire from the side. What you have is a circle with one flattened side cut off. From simple geometry you can calculate the difference in area between a true circle and a circle with the flattened segment removed. The ratio of the areas of the removed segment to the area of the true circle is proportional to the pressure change between a loaded and unloaded tire.

So I plugged in the numbers from my car. The wheel radius is 11.75 inches and the flattened length (apothem) is 11.25. That is, the tire is flattened by 1/2 inch. So we want to know how the area of the flattened piece we slice off the circle compares to the whole circle since this is proportional to the change in volume. You can do do the arithmetic or just plug the numbers into this handy calculator. It shows that the segment that is sliced off the circle is 2.27 square inches compared to the area of the circle which is 433.74 square inches. So the change in area is less than 0.5%. Now we really should compare not to the area of the entire circle but only the area of the air filled tire itself, so we need to subtract the area of the 8 inch radius steel wheel. When we do that we have a round tire area of 232 square inches compared to a flattened segment of 2.27 inches.

From this we can conclude the the very maximum change in volume and therefore pressure change is only 1%. For a tire pressure of 30 pounds this means that the pressure change when loaded could be no more than 0.3 pounds, not something that you can read on a standard tire gauge.

But we need to take one more step. We only looked at two dimensions, the round side of the tire. Now let's look in the other dimension, from the front of the car. An unloaded tire looks approximately like a thin vertical rectangle. When the tire is loaded, the rectangle is shorter by 1/2 inche, but the side walls bulge out on the bottom, increasing the area of the rectangle. I'm not going to try to quantify this change in area in this view, but just assume that the bulge and the shortening cancel out. You are still left with a maximum change in pressure of 1% or about 0.3 pounds. You aren't going to be able to measure that with a standard tire gauge.

So if the question is whether is makes a difference if you adjust the tire pressure before you mount the tire on the car, the answer is that it makes no practical difference. The change can only be tenths of a pound in the worst case.

And I see that Jon-o has actually made the measurement and it was undetectable. This means that decreasing volume in one dimension is counteracted by increasing volume elsewhere so that the volume of the tire does not change noticeably.

posted by JackFlash at 8:39 PM on September 18, 2009

So let's approximate the change in volume when you load a tire. Let's first reduce it to a two dimensional problem, looking at the tire from the side. What you have is a circle with one flattened side cut off. From simple geometry you can calculate the difference in area between a true circle and a circle with the flattened segment removed. The ratio of the areas of the removed segment to the area of the true circle is proportional to the pressure change between a loaded and unloaded tire.

So I plugged in the numbers from my car. The wheel radius is 11.75 inches and the flattened length (apothem) is 11.25. That is, the tire is flattened by 1/2 inch. So we want to know how the area of the flattened piece we slice off the circle compares to the whole circle since this is proportional to the change in volume. You can do do the arithmetic or just plug the numbers into this handy calculator. It shows that the segment that is sliced off the circle is 2.27 square inches compared to the area of the circle which is 433.74 square inches. So the change in area is less than 0.5%. Now we really should compare not to the area of the entire circle but only the area of the air filled tire itself, so we need to subtract the area of the 8 inch radius steel wheel. When we do that we have a round tire area of 232 square inches compared to a flattened segment of 2.27 inches.

From this we can conclude the the very maximum change in volume and therefore pressure change is only 1%. For a tire pressure of 30 pounds this means that the pressure change when loaded could be no more than 0.3 pounds, not something that you can read on a standard tire gauge.

But we need to take one more step. We only looked at two dimensions, the round side of the tire. Now let's look in the other dimension, from the front of the car. An unloaded tire looks approximately like a thin vertical rectangle. When the tire is loaded, the rectangle is shorter by 1/2 inche, but the side walls bulge out on the bottom, increasing the area of the rectangle. I'm not going to try to quantify this change in area in this view, but just assume that the bulge and the shortening cancel out. You are still left with a maximum change in pressure of 1% or about 0.3 pounds. You aren't going to be able to measure that with a standard tire gauge.

So if the question is whether is makes a difference if you adjust the tire pressure before you mount the tire on the car, the answer is that it makes no practical difference. The change can only be tenths of a pound in the worst case.

And I see that Jon-o has actually made the measurement and it was undetectable. This means that decreasing volume in one dimension is counteracted by increasing volume elsewhere so that the volume of the tire does not change noticeably.

posted by JackFlash at 8:39 PM on September 18, 2009

*You could fill your car with Ikea furniture and you'd see less change than if the ambient temperature dropped twenty degrees.*

Unless your furniture weighs somewhere well over 1000 pounds, sure. But the air is indeed doing a large part of the job of "holding the car off the ground". This is evident from the fact that if you let any of it out, the car gets that much closer to the ground.

Also, this.

posted by sfenders at 8:54 PM on September 18, 2009

*sfenders: Also, this.*

You know, you can patent just about anything, but that doesn't mean that it actually works. As we have seen from the evidence above, both by geometry and by Jon-o's actual empirical results, there is no readily measurable pressure difference between a load of 0 pounds and 4000 pounds. So the patent claim does not work unless they are capable of measuring pressure changes of much less than a tenth of a pound. Plus, if you read the claims, they introduce a hypothetical constant Cf for each tire with no clue as to how you determine that constant. Presumably it depends on the dimensions of the tire, the construction of the tire, tread and sidewall flexibility, age, wear, etc. You also need to measure the temperature of the tire at the same time as you measure the pressure to an accuracy of much better than 1%. So there is some doubt as to the practical application of the claims.

posted by JackFlash at 9:25 PM on September 18, 2009 [1 favorite]

Well, nothing is certain in this life, but I'll take the patent and my understanding of physics over Jon-o's test. It says 0.5-1psi is what they measured as typical for trucks, and confirms the basic correctness of my initial theory and calculations above minus any estimation errors. 0.3psi would be reasonable for a car. Tell me how wide your tires are, and I'll tell you how much your car weighs, if it's got even weight distribution and your measurements are accurate.

Although, those are some relatively low-profile tires I suppose, so maybe the sidewalls do support more weight than the negligable amount it should be for taller ones at good pressure. It will of course depend on the tire to some extent, as they do make run-flats and Lunar Rover tires.

posted by sfenders at 9:37 PM on September 18, 2009

Although, those are some relatively low-profile tires I suppose, so maybe the sidewalls do support more weight than the negligable amount it should be for taller ones at good pressure. It will of course depend on the tire to some extent, as they do make run-flats and Lunar Rover tires.

posted by sfenders at 9:37 PM on September 18, 2009

*This is evident from the fact that if you let any of it out, the car gets that much closer to the ground.*

The structure of a tire is basically like an arch and the force of the vehicle is transferred from the tread through the sidewalls and then into the rim and suspension. Just like an arch, the air underneath it isn't really subjected to the force of the mass supported by the arch. In a tire, the air pressure exists, more or less, to hold the tire firmly against the bead of the rim and to ensure that, since the tire is flexible, it maintains its correct shape for its application.

If you load up your car or truck, you need to

*increase*the tire psi before hand to ensure the tire maintains its correct shape. If your tire pressure drastically increased as a result of added weight, you wouldn't have to add psi before adding cargo or passengers.

posted by Jon-o at 6:35 AM on September 19, 2009

*In a tire, the air pressure exists ... [to] maintain its correct shape for its application.*

Yes. And the more weight, the more pressure it takes to do that.

*If your tire pressure drastically increased as a result of added weight, you wouldn't have to add psi before adding cargo or passengers.*

Exactly backwards. If, for a given degree of tire flexibility, in some alternate universe where our laws of physics don't apply, adding weight changed the interior volume and therefore also both the shape and pressure of the tire by more than it does in reality, you'd need to add that much more additional pressure to compensate and keep it in shape.

And if on the other hand the tire has more flexibility than that of the average car; if it's more like a balloon -- like many bicycle tires for example -- you again need more pressure to keep it in approximately the right shape for a given ratio of tire interior surface area to load, even though the pressure you put on it from the outside (and therefore the pressure increase on the inside) is the same. Which is why road bicycles using tires with thin flexible sidewalls for very light weight often have pressures somewhere above 90psi.

posted by sfenders at 9:25 AM on September 19, 2009

*even though the pressure you put on it from the outside (and therefore the pressure increase on the inside) is the same.*

So, are you maintaining that there's a significant and relevant increase in internal psi when the tire assembly is supporting the weight of a vehicle?

posted by Jon-o at 10:39 AM on September 19, 2009

Jon-o, I think it ought to be around 0.3 to 0.6 psi for the average car. Enough that I really can't understand why it wouldn't show on a guage accurate to 0.1 pounds. Not enough to be relevant to anything much, except possibly in racing, and in answering the question.

posted by sfenders at 11:04 AM on September 19, 2009

posted by sfenders at 11:04 AM on September 19, 2009

*sfenders: Jon-o, I think it ought to be around 0.3 to 0.6 psi for the average car.*

What is the basis for this "belief"? You need to demonstrate with some certainty that the volume of the tire changes and to what extent.

posted by JackFlash at 11:49 AM on September 19, 2009

*I really can't understand why it wouldn't show on a guage accurate to 0.1 pounds.*

Because there's no change in pressure. Since the interior volume of the tire doesn't change, there's no compressive force on the air mass inside.

When you fill a tire up with 30psi, the entire job of that air is to clamp the tire bead to the rim and prop up the sidewall. You would have to substantially reduce the volume of the tire just to raise it a single psi.

posted by Jon-o at 2:47 PM on September 19, 2009 [1 favorite]

Well there has to be

By how much exactly it increases I'm not entirely certain, hope someone comes along who knows for sure. But the idea is that you've increased average pressure around the outside of the tire by pressing on part of it. I think it should decrease in volume by about the same amount as if you'd increased outside pressure by pressing on all of it at once at that average amount, as in a pressure chamber.

posted by sfenders at 5:01 PM on September 19, 2009

*some*change in pressure. Tire material under compression isn't typically itself strong enough to hold much weight, as you can confirm by leaning on an un-mounted tire. Contact patch area times air pressure equals load on the tire, as seems obvious and is confirmed all over the web as a generally valid approximation. Contact patch size does not grow as quickly as load, as you can confirm by sitting on one corner of a car, adding maybe 30% to its load if you weight as much as I do, and seeing how much it changes. So pressure increases. Measuring the geometry like JackFlash did also seems to confirm it, though it's tricky to be accurate. A bunch of engineers have written various patents that depend on it. And increasing the pressure on the outside of any sealed flexible container is going to increase the pressure on the inside, that's just common sense.By how much exactly it increases I'm not entirely certain, hope someone comes along who knows for sure. But the idea is that you've increased average pressure around the outside of the tire by pressing on part of it. I think it should decrease in volume by about the same amount as if you'd increased outside pressure by pressing on all of it at once at that average amount, as in a pressure chamber.

posted by sfenders at 5:01 PM on September 19, 2009

As posters with real world experience have mentioned, the difference is going to be very small. I didn't mean to imply otherwise - without doing some calculations it would be hard to say. But I'm quite sure there would be some difference.

posted by phrontist at 5:11 PM on September 19, 2009

posted by phrontist at 5:11 PM on September 19, 2009

PV = NRT

If the volume of the tire doesn't change (steel-belted helps assure this change is minimal), the gas amount inside doesn't change (N1= N2), the temperature doesn't change, and the gas constant R doesn't change, the pressure bloody well isn't going to change significantly.

Intuitive answers are not helpful. Gas gauge experience is helpful. And they support the validity of the centuries-old Ideal Gas Law.

QED.

posted by IAmBroom at 6:53 PM on September 19, 2009 [1 favorite]

If the volume of the tire doesn't change (steel-belted helps assure this change is minimal), the gas amount inside doesn't change (N1= N2), the temperature doesn't change, and the gas constant R doesn't change, the pressure bloody well isn't going to change significantly.

Intuitive answers are not helpful. Gas gauge experience is helpful. And they support the validity of the centuries-old Ideal Gas Law.

QED.

posted by IAmBroom at 6:53 PM on September 19, 2009 [1 favorite]

To reitterate the one's that aren't saying "But there must be", there is no measurable difference on a gauge that reads to 0.1psi accuracy. I have checked this extensively.

The air pressure isn't necessarily supporting the weight of the car as much as supporting the sidewall so that IT can. With no air pressure in the tyre buckles, and THIS is why it can't hold the car up. The majority of the vertical loading is taken by the sidewall in static loading- ie until cornering or braking/power loadings are taken into account.

posted by Brockles at 7:16 PM on September 19, 2009 [2 favorites]

The air pressure isn't necessarily supporting the weight of the car as much as supporting the sidewall so that IT can. With no air pressure in the tyre buckles, and THIS is why it can't hold the car up. The majority of the vertical loading is taken by the sidewall in static loading- ie until cornering or braking/power loadings are taken into account.

posted by Brockles at 7:16 PM on September 19, 2009 [2 favorites]

Well, thanks Jon-o and Brockles. Sorry it took me so many posts, but now the experiment's been replicated, guess I can accept it.

So yeah, the pressure has to increase only enough to keep the sidewalls from buckling more than they do under the new load. No idea why it'd be so little, but obviously it ain't much for car tires.

Maybe the heavy trucks mentioned in the patent text had tires like this.

posted by sfenders at 4:08 AM on September 20, 2009

So yeah, the pressure has to increase only enough to keep the sidewalls from buckling more than they do under the new load. No idea why it'd be so little, but obviously it ain't much for car tires.

Maybe the heavy trucks mentioned in the patent text had tires like this.

posted by sfenders at 4:08 AM on September 20, 2009

*the pressure has to increase only enough to keep the sidewalls from buckling more than they do under the new load.*

Still no, though. The pressure will increase only if the load is greater than the pressure in the tyre is able to support. The tyre will change shape when you place a localised load on it (the ground) but won't necessarily change

*size*unless the load is higher than the pressure - this is why the tyre buckles and squashes when there is no air in it - the load is higher than the pressure in the tyre is able to support.

The standard pressure in a tyre is able to support a load higher than the static loading of a car sat on it. The load required to distort the tyre (and hence change the pressure) is much higher. Just because the tyre changes shape doesn't mean it changes in volume (which is the only way the pressure would change).

posted by Brockles at 7:33 AM on September 20, 2009

I don't think it's particularly likely the tire would change in shape without changing in volume at least a little bit. But certainly possible, sure. If the sidewalls are buckling outwards due to load, not due to air pressure, the pressure change could even be negative. As it probably would be for some range of load if you had square tires.

Without any way to calculate it it could be anything up to 0.09; so the question remains unanswered. Rubber-like materials have non-linear elasticity, so it's probably hard to calculate.

posted by sfenders at 5:43 AM on September 21, 2009

Without any way to calculate it it could be anything up to 0.09; so the question remains unanswered. Rubber-like materials have non-linear elasticity, so it's probably hard to calculate.

posted by sfenders at 5:43 AM on September 21, 2009

*I don't think it's particularly likely the tire would change in shape without changing in volume at least a little bit.*

Your gut feeling versus physics? With several real world examples thrown in because the physics didn't seem to be convincing you?

*so the question remains unanswered.*

Incorrect. The question is answered, you just refuse to believe it. As long as the pressure in the tyre is in the nominal 'correct range' the force that pressure is able to support is greater than the weight of the car - ie the volume will not change (so nor will the pressure). The localisation of the loading (the contact patch) means the tyre will deform, but not change in volume.

In a flat tyre, the pressure in the tyre may change, but then all bets are off.

This question is answered, you're just arguing because you aren't convinced. That's not all that relevant.

posted by Brockles at 7:26 AM on September 21, 2009 [1 favorite]

Sorry, but

The tire is held round above the contact patch by elastic material of tread and/or belt pulling outwards from around the circumference. The sidewalls are held up by this as well, plus whatever exact (non-zero) degree of flexibility they have left under the stresses applied to them.

posted by sfenders at 8:28 AM on September 21, 2009

*The standard pressure in a tyre is able to support a load higher than the static loading of a car sat on it ... so the volume will not change*is not an answer based on physics, it's just nonsense. If you have a piston pushing into a cylinder, where the only thing stopping it is gas pressure inside, the pressurization can be anything you like, and no matter how high it is, increasing the force pushing in the piston will increase that pressure, at equilibrium. That does not change at all if say you make the bottom of the cylinder flexible.The tire is held round above the contact patch by elastic material of tread and/or belt pulling outwards from around the circumference. The sidewalls are held up by this as well, plus whatever exact (non-zero) degree of flexibility they have left under the stresses applied to them.

posted by sfenders at 8:28 AM on September 21, 2009

*no matter how high it is, increasing the force pushing in the piston will increase that pressure, at equilibrium.*

No, it won't. The pressure will only increase if the force on the piston is greater than that exerted on it by the gas pressure inside.

You seem to be trying to view this system as air pressure inside the tyre being equal to outside it. The tyre is already pressurised and is already exerting a significant force on the inside of the tyre and pushing out. Until the force from the weight of the car and the ground is higher than this force, the pressure in the tyre will not rise at all, although the tyre may flex and move to accommodate that lesser force. The equilibrium change is purely in force distribution, rather than any flex == greater force on the tyre (and hence a higher internal pressure).

Imagine a tyre inflated to 150 psi. It'd remain a perfect circle if you put it on a car. If it doesn't deform at ALL, do you accept that the internal pressure will not change when the car is lowered onto the ground?

Tyre deformation does not necessarily mean a volume change.

posted by Brockles at 8:39 AM on September 21, 2009

*The pressure will only increase if the force on the piston is greater than that exerted on it by the gas pressure inside.*

...at

*equilibrium*, I said. That is, the force pushing down on the piston was already enough to stop it flying out the top, and not so much to let it push in further. The piston, pushing down into the cylinder, will rest comfortably there, supported by the pressure inside the cylinder. If the piston itself weighs 10lbs, and it's supported only by the pressure inside the cylinder, and it's cross-sectional surface area is one inch, then the pressure inside is 10psi. If you put a one-pound weight on top the piston, it moves down a bit until pressure inside increases to exactly 11psi, and there it is in equilibrium.

Of course this is not exactly like what happens with the tire. You've got two things going on:

Imagine a box shaped like a milk carton. With slightly flexible but fairly strong sides bulging out just a little. If you put more air in it, sides are going to bulge out more... and the top of the box is pulled down as the whole thing gets rounder and closer in shape to a sphere. Conversely, push down on the top, and the sides bulge out more, and the pressure inside

*decreases*.

On the other hand, imagine a sphere where the top half is rigid and the bottom half flexible. Push down on the top when it's resting on the ground, and no matter the air pressure, the volume and pressure inside

*increases*.

With a tire, it's a combination of both, with one cross-section being round like the sphere, and the other being square like the box. So to find out the pressure change you'd have to calculate the relative importance of each, along with the stresses in the tire material and how they end up distributed, and what they do to its compression strength and flexibility.

posted by sfenders at 9:08 AM on September 21, 2009

*With a tire, it's a combination of both, with one cross-section being round like the sphere, and the other being square like the box.*

Holy crap, no it isn't. You clearly have no idea how a tyre flexes.

*That is, the force pushing down on the piston was already enough to stop it flying out the top, and not so much to let it push in further. The piston, pushing down into the cylinder, will rest comfortably there, supported by the pressure inside the cylinder. If the piston itself weighs 10lbs, and it's supported only by the pressure inside the cylinder, and it's cross-sectional surface area is one inch, then the pressure inside is 10psi. If you put a one-pound weight on top the piston, it moves down a bit until pressure inside increases to exactly 11psi, and there it is in equilibrium.*

Your model is entirely flawed. Imagine a piston at full extension (ie a shock absorber fully extended) and THEN pressurise the cylinder. To over two atmospheres. That is the correct model for this example.

If the tyre is correctly pressured, it is exerting a force on the inside of the tyre that is

*above that which the weight of the car is able to compress*. The forces required to keep the tyre fully round do not affect this and are largely irrelevant. The tyre deforming does not in any way imply that it has reduced in volume by the amount of the flat section - that is just rubbish. You are arguing a physics model that doesn't exist in this example. The tyre is preloaded (essentially) by the air pressure in it (like the shock absorber example), and until the load on it from the car is above that load, there will be no further compression of the tyre. Deformation DOES NOT equal compression or change of volume (for small deformations of the tyre as in this example).

When the car is cornering or braking hard, the tyre will be further deformed and the pressure may rise. But it needs much higher loads than that of the normal static loading of the tyre. Yes, compressing a tyre such that the volume changes will raise the pressure - no-one has claimed otherwise - but what everyone is saying is that the tyre deforms but does NOT compress or change volume by the action of just being place on the ground. It needs more force on the tyre than the static loading of the car's weight. Incidentally, the point at which the tyre does compress to any sizeable extent (rather than just deform) is at an extremely high loading and is the point at which the tyre will no longer be able to grip effectively, so it is WELL outside the realm of the static loading example here. One of the major parameters of tyre dynamics is ensuring their ability to flex

*without*changing volume.

The balance of the two forces (the one from the internal air pressure of the tyre vs the weight of the car) is the element of physics that you are constantly missing. So to go back to the original question:

*Is tire pressure different if you measure with wheel off the car (as compared to when it is on the car and holding the weight of the car)?*

If the tyre is properly inflated, no it isn't.

posted by Brockles at 9:26 AM on September 21, 2009

*The tyre is preloaded (essentially) by the air pressure in it (like the shock absorber example)*

Yes. But it doesn't have as hard a maximum amount of extension as the shock absorber does. It depends on the non-linear stretchiness of rubber like I mentioned. If you keep on adding pressure to an unloaded tire it'll keep on getting bigger right up until it bursts, just at a smaller and smaller rate. Not by much at the limit, granted.

But we can observe that it does deform when you put normal load on it. So the stresses on different parts of the rubber do change enough to move them; which means not all parts of it are exactly at their elastic limit. And if it doesn't change volume despite that, it's because of the geometry I described. Which may well be taken into account in design, for the optimal sidewall height to get minimum pressure change under load.

posted by sfenders at 10:01 AM on September 21, 2009

*But it doesn't have as hard a maximum amount of extension as the shock absorber does.*

On what are you basing this assumption? What does 'as hard a maximum extension' even mean? The physical stop to the tyre's expansion?

*If you keep on adding pressure to an unloaded tire it'll keep on getting bigger right up until it bursts, just at a smaller and smaller rate.*

This is not true. There are great big steel bands in them that prevent this. You have serious gaps in your knowledge of the physical characteristics of a tyre, and this is why your reasoning is flawed.

*which means not all parts of it are exactly at their elastic limit.*

What makes you think that ANY of the tyre is at it's elastic limit? It'd be a pretty poor tyre that was anywhere near it's elastic limit during anything other than catastrophically high conditions.

posted by Brockles at 11:00 AM on September 21, 2009

*This is not true. There are great big steel bands in them that prevent this.*

Combination of steel and rubber and whatever else the thing is made of, it doesn't matter. The stronger the steel, the more of the increase is from the sidewalls bulging out. Even steel isn't in reality infinitely strong. But practically it's close, and that is what I meant by being "

*at the elastic limit*": when you pump more air in, it doesn't get any substantial amount bigger, because the stuff it's made of can't stretch more. It can be practically there and still be flexible. Not elastic limit as in the point it breaks, elastic limit as in the point where it takes a whole lot of force to stretch it any more, so maybe the wrong term.

I got one more: Take a sphere instead of a tire shape, inflated right to the point where it can't possibly expand in volume any more because of the steel belts in it; and yet the stuff it's made of is thin enough so it's still flexible even despite it being stretched to the absolute max. If you pressed equally on all parts of the surface it wouldn't compress at all, like with the shock absorber at max extension. But if you press on one side and it deforms in any way, its volume is less and the pressure goes up. If it deforms so much that you can see the change, it must be measurably less in the way JackFlash calculated. The sphere is *the* most efficient way to contain the most volume with a given amount of surface area. Any shape it takes that's not a sphere, unless the stuff its made of stretched even more, which it can't, is therefore less volume.

posted by sfenders at 11:14 AM on September 21, 2009

But your sphere example is utterly irrelevant - just as my example of a tyre inflated to (if it were possible) 150 psi is. It's not how a tyre works, nor in the range of how a tyre works. The tyre deforms where it touches the ground, but also to a corresponding amount (ie keeps the same volume) by bulging the sidewall all around the tyre (mostly by the contact patch).

At what point do you just accept that you were wrong?

posted by Brockles at 11:29 AM on September 21, 2009

At what point do you just accept that you were wrong?

posted by Brockles at 11:29 AM on September 21, 2009

*The tyre deforms where it touches the ground, but also to a corresponding amount (ie keeps the same volume) by bulging the sidewall all around the tyre*

All I'm saying is those two happen to come pretty close to equalling out only because of the shape of the tire. And they are close to equal to different degrees for different tire profiles and sizes, as it should be and as illustrated by the guy with the patent who measured big (0.5 to 1.0psi) pressure change for trucks. There's no mysterious physical principle that would make the one exactly equal the other. If the tire was some other shape, say a sphere, or square, they quite obviously wouldn't be anywhere close.

I totally accept that I was wrong before, earlier in the thread, but I'm right now. :)

posted by sfenders at 11:39 AM on September 21, 2009

*All I'm saying is those two happen to come pretty close to equalling out only because of the shape of the tire.*

*There's no mysterious physical principle that would make the one exactly equal the other.*

'By design', I believe it is called. By extensive and exhaustive testing and development for many decades, to achieve a characteristic that is required for consistent and beneficial tyre behaviour in terms of grip and handling. Tyres work that way for a reason - you haven't 'cracked it' through some description of spheres and geometry happy coincidence.

Right about what now? You now understand what it is about a tyre that makes you wrong before, but describing it makes you somehow right? If it makes you sleep better, I guess... ;)

posted by Brockles at 11:49 AM on September 21, 2009

Indeed. If you hadn't kept insisting that the change was *exactly* zero until some threshold, instead of merely designed with geometry so as to be as close as possible to zero in the normal range of load, I'd probably never have got there and been able to sleep at night. Feigned ignorance in order to teach me a lesson, I assume. So thanks for that.

posted by sfenders at 12:12 PM on September 21, 2009

posted by sfenders at 12:12 PM on September 21, 2009

*If you hadn't kept insisting that the change was *exactly* zero until some threshold*

Can't find that anywhere in my own or anyone else's posts, as it happens. But, you know, feel free to use an askme to crowd out the people that were actually answering the question withn facts and applicable knowledge with wild guesses and false calculations based on flawed knowledge to work through your own understanding. I'm sure it was helpful somehow to someone.

posted by Brockles at 12:23 PM on September 21, 2009

*I'm sure it was helpful somehow to someone.*

maybe, maybe not. but I was surprised to find that this thread was still going - and it is a pretty entertaining argument.

posted by Think_Long at 3:03 PM on September 21, 2009

Yeah, an entertaining argument to be sure, and I think it's over, but I can add one more thing.

There are other design considerations besides consistent handling and grip, and they must explain why tires for heavy trucks are shaped in a way that will show a larger increase in pressure between loaded and not loaded, as understanding the mechanism makes clear that they are, being typically much taller. If the idea was just to always keep pressure change under load to a minimum for good handling, all tires for a given rim size would have the same profile by now, and they don't. The tires on a truck are opting instead for a greater load with the least material, a greater range of load, more shock absorption, or something else I can't think of. Car tires make the same kind of trade-off on a smaller scale, and the more they deviate from the probably near-ideal (for consistent grip) shape of the tires Brockles is no doubt most familiar with, the more pressure change they'll show when you put a car on them.

Also of note that they're only going to be at that perfect shape where more deformation equals minimum pressure change somewhere (presumably) in the middle of their normal load. So if that's the case, the unloaded shape is relatively far from that ideal shape, and the first hundred pounds above zero load changes pressure by more than an extra hundred pounds when they're already in the range of normal load. For just the same reason that increasing beyond normal load again starts increasing pressure faster. Unless that ideal shape is for whatever mathematical reason necessarily close to the unloaded shape, as seems possible; in which case the ratio of increased load to increased pressure just constantly goes up from there, in non-linear fashion.

posted by sfenders at 4:31 PM on September 21, 2009

There are other design considerations besides consistent handling and grip, and they must explain why tires for heavy trucks are shaped in a way that will show a larger increase in pressure between loaded and not loaded, as understanding the mechanism makes clear that they are, being typically much taller. If the idea was just to always keep pressure change under load to a minimum for good handling, all tires for a given rim size would have the same profile by now, and they don't. The tires on a truck are opting instead for a greater load with the least material, a greater range of load, more shock absorption, or something else I can't think of. Car tires make the same kind of trade-off on a smaller scale, and the more they deviate from the probably near-ideal (for consistent grip) shape of the tires Brockles is no doubt most familiar with, the more pressure change they'll show when you put a car on them.

Also of note that they're only going to be at that perfect shape where more deformation equals minimum pressure change somewhere (presumably) in the middle of their normal load. So if that's the case, the unloaded shape is relatively far from that ideal shape, and the first hundred pounds above zero load changes pressure by more than an extra hundred pounds when they're already in the range of normal load. For just the same reason that increasing beyond normal load again starts increasing pressure faster. Unless that ideal shape is for whatever mathematical reason necessarily close to the unloaded shape, as seems possible; in which case the ratio of increased load to increased pressure just constantly goes up from there, in non-linear fashion.

posted by sfenders at 4:31 PM on September 21, 2009

Well, for anyone still interested in this thread, you might find this link from Michelin aircraft tires informative (PDF).

It says that a loaded aircraft tire will increase its tire pressure by about 4%. But there is a key difference. Automobile tires are designed for a deflection of about 17% and aircraft tires are designed for a deflection of 33% to 35%. In other words, aircraft tires are designed to run flatter. This means that you get a bigger volume change when you subtract the flattened segment from the full circle. Note that the relationship is non-linear. Doubling the percent deflection in this example quadruples the volume change.

So if I crank through the previous geometry exercise using a deflection of 35%, lo and behold, I get a volume decrease of 4.1% which is close to Michelin's change in tire pressure of 4%.

This supports my original calculation that automobile tires, which have a relatively modest deflection when loaded, will have somewhat less than a 1% change in tire pressure. This is less than is measurable with a standard tire gauge.

This also explains why a big truck tire might have up to 1 pound of pressure change. Truck tires run around 100 psi so a 1% change would be about a pound.

posted by JackFlash at 6:48 PM on September 21, 2009

It says that a loaded aircraft tire will increase its tire pressure by about 4%. But there is a key difference. Automobile tires are designed for a deflection of about 17% and aircraft tires are designed for a deflection of 33% to 35%. In other words, aircraft tires are designed to run flatter. This means that you get a bigger volume change when you subtract the flattened segment from the full circle. Note that the relationship is non-linear. Doubling the percent deflection in this example quadruples the volume change.

So if I crank through the previous geometry exercise using a deflection of 35%, lo and behold, I get a volume decrease of 4.1% which is close to Michelin's change in tire pressure of 4%.

This supports my original calculation that automobile tires, which have a relatively modest deflection when loaded, will have somewhat less than a 1% change in tire pressure. This is less than is measurable with a standard tire gauge.

This also explains why a big truck tire might have up to 1 pound of pressure change. Truck tires run around 100 psi so a 1% change would be about a pound.

posted by JackFlash at 6:48 PM on September 21, 2009

*There are other design considerations besides consistent handling and grip*

Yes there are. This is why your assertion that all tyres would have the same proportions to maintain the desired deflection vs pressure control is flawed and simplistic.

*If the idea was just to always keep pressure change under load to a minimum for good handling, all tires for a given rim size would have the same profile by now*

It isn't just about geometry, but also the construction of the tyre (made up of many layers of different materials and rubber compounds, reinforcement etc even the mould process) and the particular compromises needed for the application - the weight of the car it is to be fitted to, the ride vs handling compromise demanded (or tolerated) by that market, road noise and grip vs economy, likely weather or road conditions the tyre will be used in etc. There are many many ways the pressure control could be achieved within a range of these parameters, as witnessed by the large variety of tyre manufacturers and range of tyres they produce. Some are better, some are worse at optimising the various parameters, but to suggest it is simply about geometry is utterly inaccurate.

*Also of note that they're only going to be at that perfect shape where more deformation equals minimum pressure change somewhere*

**(presumably)**in the middle of their normal load.Again, it is not about the shape of the tyre but the tyres ability to change shape without changing volume. You are making an enormous amount of presumptions based on very little knowledge and making some wild claims and conclusions based purely on them. That's not very helpful.

posted by Brockles at 6:33 AM on September 22, 2009

It's still all about the shape of the tire, though. I think you mean to say the shape is more complicated than the simplified ideal. Which is true, but by how much...?

The volume change is exactly, no matter the tire construction, dependent on the difference in shape before the additional load, and after it. If the tires are of such a strength and construction that they flatten the outer part of the tire, which is under tension along its circumference, at the contact patch by a specific amount, you know how far the sidewalls have to have deflected regardless of their relative overall strength. Making the sidewalls relatively stronger similarly means less flattening of the tread for a given load. The shape for a given amount of deflection stays the same, though the amount of load required to achieve it varies. The non-linear elasticity of rubber-like materials plus whatever reinforcement means the outer circumference of the tire changes barely at all when you add more pressure: It's a nearly fixed length of material, long as it's under sufficient tension, which it always is with a static load. So the overall strength of the sidewalls does not make a difference to shape at any given deflection.

However, yeah, they can make that sidewall strength non-uniform across their height, and that does make a difference. Good point. My car tires have a fairly substantial amount of tread above the contact patch on the side, which isn't in contact with the ground under static load. But above that area, I suspect sidewalls are of nearly uniform strength across most of their height. So, effectively the radius you'd use to estimate the degree to which they agree with simple assumption based on ratio of sidewall height is smaller. Ever more slightly as the tires get taller and the sides more uniform. Having a less-flexible part of sidewall near the outer circumference (or inner, same principle) would just do about the same thing as increasing the thickness of the tread, which has to be part of the calculation.

You can observe how close a tire is to that simplified geometric ideal by observing how uniformly its sidewall is curved. Observing some convenient big trucks, they look pretty close. Despite perhaps being slightly more than average designed for cornering performance, my car tires aren't all that far from it either, with about 2/3rds of the length between rim and ground having a nice round curve. The degree to which my own tires conform to the simplified ideal would be increased a bit by factoring in the thickness of the bottom of the tire. If that's a quarter-inch, it's about 75% of the height that's close to the ideal, plus whatever degree of flexibility the non-uniform (by an amount significant enough to see, and thus have a substantial effect on this) parts of the sidewall have. But again, it's going to get quickly closer than that to the simplified ideal when you have taller tires. And further from it, with other influences taking over, when you have very low-profile tires with very non-uniform sides.

Now, when you get into actually using the tire, with it dynamically flexing and handling sideways loads, it's another story, of course. The flexibility of the tread area, and the exact shape of the side become a whole lot more important. So I expect they design non-uniform sidewall strength mostly for reasons entirely different from optimizing pressure change under static load, though it still puts limits on what they can do, in a way that depends on the shape and degree of uniformity of flexibility under load of whatever they come up with.

JackFlash:

They have *much* less than 1% change, as was measured! Aircraft tires get squashed a lot, so they're similar to a 3-dimensional version of your calculation. Truck tires (of the sort that have tall sides relative to other dimensions) are that much more like your method of calculation. As you can see by imagining infinitely tall sidewalls, where it'd ideally be exactly your calculation. Car tires are in-between, so your calculation isn't close to the right one, as confirmed by experiment.

posted by sfenders at 9:11 AM on September 22, 2009

The volume change is exactly, no matter the tire construction, dependent on the difference in shape before the additional load, and after it. If the tires are of such a strength and construction that they flatten the outer part of the tire, which is under tension along its circumference, at the contact patch by a specific amount, you know how far the sidewalls have to have deflected regardless of their relative overall strength. Making the sidewalls relatively stronger similarly means less flattening of the tread for a given load. The shape for a given amount of deflection stays the same, though the amount of load required to achieve it varies. The non-linear elasticity of rubber-like materials plus whatever reinforcement means the outer circumference of the tire changes barely at all when you add more pressure: It's a nearly fixed length of material, long as it's under sufficient tension, which it always is with a static load. So the overall strength of the sidewalls does not make a difference to shape at any given deflection.

However, yeah, they can make that sidewall strength non-uniform across their height, and that does make a difference. Good point. My car tires have a fairly substantial amount of tread above the contact patch on the side, which isn't in contact with the ground under static load. But above that area, I suspect sidewalls are of nearly uniform strength across most of their height. So, effectively the radius you'd use to estimate the degree to which they agree with simple assumption based on ratio of sidewall height is smaller. Ever more slightly as the tires get taller and the sides more uniform. Having a less-flexible part of sidewall near the outer circumference (or inner, same principle) would just do about the same thing as increasing the thickness of the tread, which has to be part of the calculation.

You can observe how close a tire is to that simplified geometric ideal by observing how uniformly its sidewall is curved. Observing some convenient big trucks, they look pretty close. Despite perhaps being slightly more than average designed for cornering performance, my car tires aren't all that far from it either, with about 2/3rds of the length between rim and ground having a nice round curve. The degree to which my own tires conform to the simplified ideal would be increased a bit by factoring in the thickness of the bottom of the tire. If that's a quarter-inch, it's about 75% of the height that's close to the ideal, plus whatever degree of flexibility the non-uniform (by an amount significant enough to see, and thus have a substantial effect on this) parts of the sidewall have. But again, it's going to get quickly closer than that to the simplified ideal when you have taller tires. And further from it, with other influences taking over, when you have very low-profile tires with very non-uniform sides.

Now, when you get into actually using the tire, with it dynamically flexing and handling sideways loads, it's another story, of course. The flexibility of the tread area, and the exact shape of the side become a whole lot more important. So I expect they design non-uniform sidewall strength mostly for reasons entirely different from optimizing pressure change under static load, though it still puts limits on what they can do, in a way that depends on the shape and degree of uniformity of flexibility under load of whatever they come up with.

JackFlash:

*This supports my original calculation that automobile tires, which have a relatively modest deflection when loaded, will have somewhat less than a 1% change in tire pressure.*They have *much* less than 1% change, as was measured! Aircraft tires get squashed a lot, so they're similar to a 3-dimensional version of your calculation. Truck tires (of the sort that have tall sides relative to other dimensions) are that much more like your method of calculation. As you can see by imagining infinitely tall sidewalls, where it'd ideally be exactly your calculation. Car tires are in-between, so your calculation isn't close to the right one, as confirmed by experiment.

posted by sfenders at 9:11 AM on September 22, 2009

*So the overall strength of the sidewalls does not make a difference to shape at any given deflection.*

Sorry, that's wrong. I should've said the difference it makes to shape is exactly determined with some calculus, when the sidewalls are of uniform strength. I mean, I never claimed the calculation was simple, just that it's there and depends on the tire geometry. It does also depend on sidewall strength under whatever pressure it's at. Still, tires are not normally designed to run "flat", with the substantially squishy shape you get when they aren't adequately strong. So some slightly-less-complicated calculus can probably be used if the tires aren't flat.

posted by sfenders at 10:00 AM on September 22, 2009

*It's still all about the shape of the tire, though.*

In determining the volume, then of course it is. But that's not at all what you have been saying. You seem to be constantly modifying your position, yet still presenting it as fact and essentially the same.

This is less a discussion and more of a series of refutations of your guessing and declarations of tyre design based on assumptions and 'well it can't be that hard' science.

*I suspect sidewalls are of nearly uniform strength across most of their height.*

You have no evidence to support this assumption. It is also a false assumption. Like most of your answers, your later 'science' entirely hinges on a wild assumption you make out of the blue that you then present as part of a 'factual' explanation. Your answers are dangerously close to noise in this thread.

You can observe how close a tire is to that simplified geometric ideal by observing how uniformly its sidewall is curved.

You can observe how close a tire is to that simplified geometric ideal by observing how uniformly its sidewall is curved.

Can you really? Even thought the sidewall doesn't just flex at the bottom? How exactly can you observe that and extrapolate data to support your theory?

Face it now, you're just making this shit up. Trying to keep this thread in the realms of real engineering and less about you trying to justify your earlier, hopelessly wrong, emphatic declarations is getting more and more tiring.

posted by Brockles at 1:10 PM on September 22, 2009

*How exactly can you observe that and extrapolate data to support your theory?*

A good question. I can look at those truck tires and observe that their sides are straight when unloaded, and uniformly curved across their height when loaded, to a very close approximation at least. I can compare the curve they make with the one my car tire makes, and see they aren't anywhere near different enough to account for a whole lot in terms of tire volume, compared to that 1%. And, I can examine an old tire and observe that this one here at least has sidewalls of uniform thickness and uniform unloaded flexibility, again not necessarily exactly but close as I can tell, after half an inch from the bottom and from the bead. Which happens to match exactly my earlier eyeball estimate on a mounted tire of the same size.

And no, I didn't "modify my position" except to accept the wrinkle of non-uniform sidewalls which is the only theoretical out the counter-argument has, even if it doesn't change much in reality by the time you get to the tallest tires on ordinary cars. I've just been trying to say the same damn thing in different ways the past two days. So, see ya next time.

posted by sfenders at 2:49 PM on September 22, 2009

By the way, I double checked again today.

A 235/45 R18 tire on a 3700lb car doesn't change pressure when measured within 0.1psi.

I measured 36.2psi raised and 36.2 loaded.

And even if I used a 205/70 14, with big dopey sidewalls and a little bitty tread, it'd still have the same result.

posted by Jon-o at 5:50 PM on September 22, 2009

A 235/45 R18 tire on a 3700lb car doesn't change pressure when measured within 0.1psi.

I measured 36.2psi raised and 36.2 loaded.

And even if I used a 205/70 14, with big dopey sidewalls and a little bitty tread, it'd still have the same result.

posted by Jon-o at 5:50 PM on September 22, 2009

If anyone needs confirmation, I'd be happy to lift my ye olde classic with 265/70 14s on it tomorrow to check.

posted by maxwelton at 9:18 PM on September 22, 2009

posted by maxwelton at 9:18 PM on September 22, 2009

This thread is closed to new comments.

posted by Palamedes at 1:44 PM on September 18, 2009