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  • Sept. 18, 2008 5:03 p.m. SVreX UltraDork

    John Brown wrote:

    When you step on a balloon you do not change the internal dimension until one of inner walls of the chamber makes contact with another inner wall. Once your footprint causes a large segment of the balloon to contact itself then the reduced chamber capacity causes the air to stretch against the chamber and eventually tear the chamber.

    I'm telling you, that is not correct. (see above post).

    Basic geometry says you are wrong.

    Any shape that has a cross section that is not a circle can be enlarged without increasing the exterior area. A sphere, cone, or cylinder cannot, as their cross sections are circles, and a circle maximizes the area in relation to the perimeter.

    If you don't believe me, go try it with your wife's tupperware.

    However, it has very little relevance to the question at hand.

    The answer is that there is a ratio of carcass strength to payload that would correctly answer the question. Generally speaking, it has very little impact. But a tire most certainly can be exploded due to increases in pressure from overloading in excess of what the carcass can stand.

    The airplane examples are also not very relevant. Their carcass strength to payload ratio is vastly beyond that of a car.

  • Sept. 18, 2008 5:31 p.m. SVreX UltraDork

    Since no one else seemed to want to do this scientifically, I did.

    I had my car on the lift. Mounted 2 different types of tires on the rear. Couldn't use the front, suspension is off the car.

    Left side: Kumho Ectsa Supra 195/45/R15 78V (mild street performance tire, narrow stiff sidewalls)

    Right side: Barum Bravuris radial 195/60/R14 82H (basic crappy street tire)

    Off the ground, I pumped them both to exactly 50 psi.

    Set it down. Kumho remained @ 50 psi, Bravuris increased to 52 psi.

    Picked it up. Re-checked pressures. Both read 50 psi again.

    Decreased pressure to 20 psi and tried again.

    Set it down. Kumho remained @ 20 psi, Bravuris increased to 22 psi.

    Picked it up. Re-checked pressures. Both read 20 psi again.

    Conclusion: weak tire carcass increased in pressure when weight was added.

    NOTE: this was the rear of a front engined AWD car. Most of the weight was in the front. If the car was cornering hard, the weight differences could have been 3 times the static weight.

    THEREFORE: Don't drive on cheap tires.

  • Salanis

    Sept. 18, 2008 5:38 p.m. Salanis Dork

    If the bulged sidewall shape of a tire were the most efficient use of internal volume to surface area, that is the shape that the tire would take when inflated and not loaded. The fact that it changes from its ideal shape means that internal volume in that area is changing.

    Either the rest of the tire needs to expand to make up the lost volume, or the pressure will increase. Most likely a combination of the two.

    As I understand, tires don't stretch a whole lot, what with all the cords and such.

  • Jack

    Sept. 18, 2008 5:46 p.m. Jack SuperDork

    Thanks SVreX,

    I was hoping to have time to check it out over the weekend.

    It's not just cheap tires. Good snow tires have very soft sidewalls, so the tread can "grab" discontinuities in the snow cover. They would be affected too.

    Jack, feeling redeemed.

  • Varkwso

    Sept. 19, 2008 6:29 a.m. Varkwso Reader

    Most changes in tire pressure are due to heat changes and the partial pressures of a wet gas (unless filled with dry nitrogen - still changes based on heat but not exponentially). Squeezing or expanding the volume of the air can change pressure. Tires are designed to deflect and to still maintain volume. Loaded or unloaded pressure changes are negligible at best.

  • foxtrapper

    Sept. 19, 2008 7:38 a.m. foxtrapper SuperDork

    John Brown wrote:

    A 1L object remains a 1L object no matter what shape you make it unless you cause it to leak.

    Lets demonstrate this with some square blocks, 1 inch square to make it simple.

    We'll take a few blocks and arange them in three rows of three: 123 456 789 The total distance around this shape is 12 (3+3+3+3) The total volume (area actually) is 9 (3x3)

    Now lets rearange those blocks. 123456789 The total distance around this shape is 20 (9+1+9+1) The total volume (area actually) is 9 (9x1)

    It we wanted to hold the outside size the same as the inital shape, we'd have to lop some of the blocks off. 12345 The total distance around this shape is 12 (5+1+5+1) The total volume (area actually) is 5 (5x1)

    Do you now understand that changing a shape very definately changes it's volume?

  • Jensenman

    Sept. 19, 2008 8:11 a.m. Jensenman UltimaDork

    SVreX wrote:

    Since no one else seemed to want to do this scientifically, I did.

    I had my car on the lift. Mounted 2 different types of tires on the rear. Couldn't use the front, suspension is off the car.

    Left side: Kumho Ectsa Supra 195/45/R15 78V (mild street performance tire, narrow stiff sidewalls)

    Right side: Barum Bravuris radial 195/60/R14 82H (basic crappy street tire)

    Off the ground, I pumped them both to exactly 50 psi.

    Set it down. Kumho remained @ 50 psi, Bravuris increased to 52 psi.

    Picked it up. Re-checked pressures. Both read 50 psi again.

    Decreased pressure to 20 psi and tried again.

    Set it down. Kumho remained @ 20 psi, Bravuris increased to 22 psi.

    Picked it up. Re-checked pressures. Both read 20 psi again.

    Conclusion: weak tire carcass increased in pressure when weight was added.

    NOTE: this was the rear of a front engined AWD car. Most of the weight was in the front. If the car was cornering hard, the weight differences could have been 3 times the static weight.

    THEREFORE: Don't drive on cheap tires.

    Dood, you obviously got WAY too much time on your hands.

  • John Brown

    Sept. 19, 2008 9:03 a.m. John Brown UltimaDork

    foxtrapper wrote:

    John Brown wrote:

    A 1L object remains a 1L object no matter what shape you make it unless you cause it to leak.

    Lets demonstrate this with some square blocks, 1 inch square to make it simple.

    We'll take a few blocks and arange them in three rows of three: 123 456 789 The total distance around this shape is 12 (3+3+3+3) The total volume (area actually) is 9 (3x3)

    Now lets rearange those blocks. 123456789 The total distance around this shape is 20 (9+1+9+1) The total volume (area actually) is 9 (9x1)

    It we wanted to hold the outside size the same as the inital shape, we'd have to lop some of the blocks off. 12345 The total distance around this shape is 12 (5+1+5+1) The total volume (area actually) is 5 (5x1)

    Do you now understand that changing a shape very definately changes it's volume?

    Got it.

  • 16vCorey

    Sept. 19, 2008 9:43 a.m. 16vCorey Dork

    John Brown wrote:

    foxtrapper wrote:

    John Brown wrote:

    A 1L object remains a 1L object no matter what shape you make it unless you cause it to leak.

    Lets demonstrate this with some square blocks, 1 inch square to make it simple.

    We'll take a few blocks and arange them in three rows of three: 123 456 789 The total distance around this shape is 12 (3+3+3+3) The total volume (area actually) is 9 (3x3)

    Now lets rearange those blocks. 123456789 The total distance around this shape is 20 (9+1+9+1) The total volume (area actually) is 9 (9x1)

    It we wanted to hold the outside size the same as the inital shape, we'd have to lop some of the blocks off. 12345 The total distance around this shape is 12 (5+1+5+1) The total volume (area actually) is 5 (5x1)

    Do you now understand that changing a shape very definately changes it's volume?

    Got it.

    I was just going to say, "Buy a 2 liter of Coke, unscrew the cap, squeeze the bottle. You now have Coke all over you because of the change in volume."

  • John Brown

    Sept. 19, 2008 9:57 a.m. John Brown UltimaDork

    The Coke statement includes a loss of pressure. My statement involves retaining structural integrity and transferring the shape but retaining a 1L interior.

  • 16vCorey

    Sept. 19, 2008 10:04 a.m. 16vCorey Dork

    You're right, I forgot about that part.

  • wlkelley3

    Sept. 19, 2008 11:24 a.m. wlkelley3 Reader

    No I don't understand because you lost 5 blocks. Of when you take away thing volume will change. Compression don't take anything away, just rearranges it. your last example you would have 2 shapes of equal size with the same distance and each would have a reduced volume but together would equal the same.

  • Salanis

    Sept. 19, 2008 11:48 a.m. Salanis Dork

    Yar matey, the simplest explanations be the ones from when we were but we lads. Get ye a bottle of bubbles. Aye, bubbles. Blow some pretty bubbles and watch them float in the sea breeze.

    What shape do the bubbles try to form?

    Aye matey! They be formin' inta spheres, because spheres offer the most internal volume fer a given surface area. Any elastic surface ye inflate will try ta shape itself ta use the minimum surface area fer maximum volume.

    If yer bubbles changed ta not be spheres, they'd need ta keep the same internal volume. The only way fer them ta do that is ta stretch and increase the surface area.

    Ye have pressure, volume, and surface area. If'n ye change one, ye have ta change at least one other. Both volume and surface area are related ta shape.

    Ye're not changin' the volume of air in a tire. So if'n ye apply external pressure and deform it, either the density increases (higher pressure) or the surface area does (the tire stretches).

    Yar matey. Learn some basic physics ye scurvy scallawags!

  • wlkelley3

    Sept. 19, 2008 11:59 a.m. wlkelley3 Reader

    Arrr! Now that's something I can sink my teeth into. I understand that.

  • d_jabsd

    Sept. 19, 2008 3:25 p.m. d_jabsd Reader

    Salanis wrote:

    Yar matey, the simplest explanations be the ones from when we were but we lads. Get ye a bottle of bubbles. Aye, bubbles. Blow some pretty bubbles and watch them float in the sea breeze.

    What shape do the bubbles try to form?

    Aye matey! They be formin' inta spheres, because spheres offer the most internal volume fer a given surface area. Any elastic surface ye inflate will try ta shape itself ta use the minimum surface area fer maximum volume.

    If yer bubbles changed ta not be spheres, they'd need ta keep the same internal volume. The only way fer them ta do that is ta stretch and increase the surface area.

    Ye have pressure, volume, and surface area. If'n ye change one, ye have ta change at least one other. Both volume and surface area are related ta shape.

    Ye're not changin' the volume of air in a tire. So if'n ye apply external pressure and deform it, either the density increases (higher pressure) or the surface area does (the tire stretches).

    Yar matey. Learn some basic physics ye scurvy scallawags!

    Happy talk like a pirate day! yarggg!

  • bravenrace

    Sept. 21, 2008 9:39 a.m. bravenrace HalfDork

    Okay, sorry for beating a dead horse, but I've been following this thread with curiousity. I'm an engineer, and I should know the answer to this, but engineering school was 20 years ago, and I haven't given it a lick of thought since, because I've never needed it for what I do. Anyways, I was moving some dirt with my front end loader this morning and decided to do a somewhat uncontrolled test. I say that because what I did was measure the air pressure in one of my front tires with, without, and then with approximately 1 ton of clay in the bucket. I did this with an ordinary tire gauge (not real accurate), and I had no way of verifying that each bucket of clay was the same weight, but here's what I found. With the first bucket of clay, the pressure in the tire was 29 psi. after emptying it, it was 28 psi. After refilling it, it was 29.5 psi. Now, this in itself would suggest that the load can change the pressure, but the amount that it changed was probably within the tolerance of error on the gauge (alhtough the gauge is virtually new). What makes sense to me is that one, we are measuring pounds per square inch. If the inches of internal surface area don't change, then the psi shouldn't either. Now, I'm making the assumption that the tire doesn't change in volume from stretching, ant the temperature or amount of air doesn't change either. However, I don't agree with some of the remarks in this thread that suggest that the pressure in a generic container will not change pressure with change in shape. That would be correct if the volume doesn't change, but theoretically, if you had a container that was crushed without breaking, the shape and internal volume would change and thus the pressure would also. So what I've said is that I measured a change in pressure with load, but I'm not sure I believe it! Glad I could help add clarity to this subject!:

  • Sept. 21, 2008 12:12 p.m. SVreX UltraDork

    The pressure of a gas in a container is the quantity of gas fighting the strength of the container.

    A ball with a given amount of air measures, perhaps, 10 psi. Add more air, gain more pressure.

    Remove the air and put the same amount in a larger ball, you'll have less pressure.

    Therefore, if you can leave the air in the first ball but make the ball bigger (like the second one), you will have less pressure.

    Therefore, if the carcass of the ball stretches, the pressure reduces. If the carcass contracts, the pressure increases.

    The wrong assumption in this thread is that the dimensions of the tire carcass doesn't change. It does, dependent on the strength of the carcass and the load that is applied to it.

    Picture that ball. Squeeze it 'till it pops. Increased pressure.

    The amount that a tire compresses on the bottom when load is applied is not necessarily equal to the amount the sidewalls increase. It's rubber, folks. And it was designed by engineers to meet certain criteria. Stay within the design criteria, and it will perform as designed. Surpass those criteria and it will fail.

    Two of the design criteria are cost of manufacture, and profit margin. Cheap tires are weaker than expensive ones.

  • Sept. 21, 2008 12:35 p.m. SVreX UltraDork

    Examples of design criteria:

    1- Cheap car tires: approximately $30, need to support perhaps 1000 lbs. cornering force while maintaining pressures within 10% of target pressure.

    2- Performance race tires: Approximately $100, need to support perhaps 5000 lbs. cornering force while maintaining pressures within 2% of target pressure.

    3- Tractor tires: Approximately $300, need to support perhaps 2000 lbs. load lifting force while allowing pressures to vary a lot, as they have no springs and the tire is the suspension system.

    4- Cargo aircraft : Approximately $2000, need to support perhaps 50000 lbs. loaded force while maintaining pressures within 1% of target pressure.

  • foxtrapper

    Sept. 21, 2008 8:15 p.m. foxtrapper SuperDork

    wlkelley3 wrote:

    No I don't understand because you lost 5 blocks. Of when you take away thing volume will change. Compression don't take anything away, just rearranges it. your last example you would have 2 shapes of equal size with the same distance and each would have a reduced volume but together would equal the same.

    You actually got it, without realizing it. Changing shape changes volume. That's the point I'm making.

    A nice square 3x3 shape has a skin of 12, and a volume of 9.

    Step on that nice square and flatten it. No stretching, just flattening it down. So it keeps it skin size of 12. Lets squash it 1 block high. That's going to be 5 blocks wide. The skin stays the same, the volume is reduced now to just 5.

    So, a nice skin (tire profile), squashed down (unstretched), has less volume.

  • Salanis

    Sept. 21, 2008 9:40 p.m. Salanis Dork

    foxtrapper wrote:

    So, a nice skin (tire profile), squashed down (unstretched), has less volume.

    And to go back to my earlier point:

    When you inflate a tire, but don't load it, it naturally assumes the most efficient shape that provides maximum internal area for the lowest surface area (lowest pressure).

    If you load that tire, it deforms the sidewall and changes the shape of the tire. The surface area stays the same (or similar) but internal volume is decreased. Pressure therefore must increase.

    With a car at rest, it doesn't deform the tire very much though. Pressure may only increase a tiny amount. It might not even be enough to measure with the sort of tools we'd have available, but it will increase.

    As pointed out, the amount of load on a tire increases substantially when the tire is operating. This will greatly increase the pressure, but we don't have a means to measure it at these times.

    What would you have to do to mimic that loading? I'm not up for crunching numbers on the load increase of a tire, but I suspect you'd have to about triple or quadruple the weight of the car to be equivalent of the forces a tire sees while cornering at 1g.

  • Varkwso

    Sept. 23, 2008 10:10 a.m. Varkwso Reader

    Compressing air in a tire using the force of gravity and a constant 14.7 psi on the outside of the tire by changing the available volume in the tire?

    Physics does not back up that hypothesis...where is the pirate when you need one?

  • foxtrapper

    Sept. 23, 2008 11:01 a.m. foxtrapper SuperDork

    Salanis,

    Why are you saying the load increases greatly on a tire when the tire is operating? I'm presuming you mean the car is being driven by that description. The weight of the car doesn't change. Are you talking about slamming into potholes and the like?

    Potholes could/would/should deform the tire momentarily, producing a momentary spike in pressure. Easy enough to calculate the value simply by the change in volume.

    Tires don't inflate to a naturally most efficient shape. They are constrained by the carcass design. We've gotten away from balloon shapes a long time ago. Today, the tread stays flat.

    And if you want to really play with the "hmm" factor, consider sag. As in the weight of the car through the rim (not the tire, the metal rim). The rim is squishing down on the bottom of the tire, but is it pulling down on the top of the tire? Beware, the question isn't as simple as it sounds.

  • Salanis

    Sept. 23, 2008 11:23 a.m. Salanis SuperDork

    foxtrapper wrote:

    Why are you saying the load increases greatly on a tire when the tire is operating? I'm presuming you mean the car is being driven by that description. The weight of the car doesn't change. Are you talking about slamming into potholes and the like?

    When a tire is operating at it's peak it certainly does.

    Mass does not change. The "stuff" stays constant. Weight is a measurement of force. The force applied to a tire when it hits its limits of adhesion (say through accelerating in a turn) is pretty substantial.

    Let's say the traction limits on your car limit you to 1g of acceleration. Say you're using that to turn.

    The car actually has more than 1g of force on it. Because the force of gravity pulling your downwards remains constant. You have a 1g force pulling you down and a 1g force pulling you sideways. I'm not sure exactly how those add together, but I suspect it's tangentially. So roughly 1.7 total G's of force applied to your car.

    Now. Anyone on this board should realize that an accelerating vehicle will undergo weight transfer. So not only is there 1.7 times the normal amount of force on a tire, but more of it is shifted to the outside wheels. It's probably not even distributed evenly over two, either. Chances are one tire has more weight on it than the other three.

    I couldn't tell you what percentage of the total force on a car each tire receives. I think it's safe to say that the tires working hardest are receiving over twice as much force when working than when resting.

    Plus, if we're talking about cornering, tires will deform as the sidewall flexes sideways. They'll be more of a parallelogram, which has less internal area.

    The tire sag you mentioned, I think, is what I was originally talking about with the tire changing shape. If you look at the side of an unloaded tire, it is a circle. But when you load it, on an axle, the bottom smooshes out. Not a circle, not as efficient.

  • foxtrapper

    Sept. 24, 2008 5:30 a.m. foxtrapper SuperDork

    Salanas,

    You could be so much clearer if you'd have just said the word cornering to start with. Not obfuscating with terms like "when the tire is operating". Yes, of course a tire is distorted and the load changed when cornering. As well braking, accelerating, going up hills, etc. Even when cooled by splashing through a puddle.

    No, I'm not talking about the tire changing shape at the bottom when I referred to sag, that would be squish, a compressive force. I mean sag when I say sag. As in hanging from the top of the upper bead of the tire. Where the support of the air in a tire is acting is an interesting physics question that some of us have tossed around in the lounge, with no resolution.

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