The only thing that one needs to know is that P1V1/T1=P2V2/T2. Since change in tire volume is negligibele at best, one is left with P1/T1=P2/T2. Therefore, temperature is the only thing that can chage pressure within a tire.
Mostly it’s going into overly simple theoretical calculations. Mine was, anyhow.
A tire is a very, very complicated system. That’s why companies like goodyear can get away with such a limited product field. The calculations required to pin the pressure change down would be pretty big and ugly and would take way too long.
Basically, the calculation is “how much volume do you lose with a loaded tire?”. To find it, You would have to calculate the true change in the shape of the tire, which is something i personally haven’t got a clue how to do. All that rubber (multiple kinds) and belting (multiple kinds and directions) as well as the interaction of the inflexible rim and the flexible tire, and the fact that everything is round and funny-shaped…
It is WAY faster, and maybe even cheaper, to simply build the tire and measure it.
As “sniper” pointed out, the previous calculations obviously are too simplistic. He says he doesn’t have a clue how to calculate the change in the shape of the tire.
Well, here’s how: Finite Element Modeling
All the major tire manufacturers have FEM’s of some of their tires, and the change in volume coould be calculated using that.
So do we have any tire FE modelers in the audience?
I’ll bet not. They are on the computer all day, and surfing the net is probably a low interest activity for them.
Perhaps, modelling of the shape isn’t necessary. Remember Archie Meedees, the Greek? You know, water displacement, weight, and volume relationships? Well, for change in tire volume, one (someone) could put the tire in a tank of water. Measure the water level (volume), load the tire, and measure the water level (volume). This would give the gross volume of the tire loaded and unloaded. This would probably be the way that Archie would have done it. /// We KNOW the volume change with pressure is not linear. A 32 psi tire has not twice the volume it has at 16 psi./ The tire is a semi-solid. As a semi-solid, it transmits some of the force (the captured air, also, transmits some of the force) from the tread, through the semi-round sidewall (under compression on one half of the tire, and tension on the other half), to the wheel. It’s easier to understand the tension forces, on the top half, going around the bulge of the sidewall, to the wheel, than it is to understand the compressive forces, on the bottom half, going around the bulge of the sidewall to the wheel. / The sidewalls aren’t free-standing walls. They are supported by the tread on the bottom, the wheel on the top, and the air (under pressure) in the middle. How much support each provides, why, and how, is for a “court” to find out.
Well Jim, I see that you are likening a tire’s lifting ability to a closed cylinder, with a rigid bottom, rigid walls, and the wheel hub as a rigid top completely covering the top of the wheel, and acting as the lifting element of the car corner. The wheel hub does NOT enclose the top of the tire. So, the other calculations can’t be correct. I DO like the idea of actually measuring the tire volume and its air pressure under different loads and air pressures. An atomic fission bomb was theorized. When one was built and exploded, the theory was proven.
It doesn’t matter–compare this to how many angels fit on the head of a pin! Will that pin move further into the cushion?
In theory, there’s no difference between theory and practice. In practice, there is.
If you compare my deltaPs estimated using energy methods, assuming and inextensible tire membrane (~0.5psi), to the values CapriRacer measured (~0.2psi), the correspondence seems astonishingly good to anyone who has spent any time trying to reconcile theory with practice. What the results say is that the differential elastic energy stored in the tire structure has the same order of magnitude as the differential energy stored in the compressed air.
The appropriate conclusion to draw, I think, is that the external work done on the tire in setting the car down on the floor is very small compared to the energy already stored in the stretched membrane and in the contained gas of the inflated tire, and as a result, the changes in properties, either an increase in the air pressure or the visible expansion of the tire, are so small they’re hard to measure. Notice, though, that the energy balance guarantees that both DO occur.
As a physicist by degree, I wanted to point out that the load on each tire is not the same. Most vehicles carry more weight over the front tires, and that is why the front tires always appear lower in pressure even at the same P as the back tires. So… front and back tires could be compared for volume change to double check the theories/empirical evidence. Also, has anybody tried detecting the temp fluctuation rather than theorizing it? Not sure how that could be accomplished.
I posted briefly on the other thread concerning this issue (“Tire pressure”), having not seen any of the other posts there or on this thread. I have since read some of the posts on both threads (c’mon, you CAN’T expect me to read every single one!) and I get a sense that what people are arguing about are issues that are really tangential. As a near-physicist (it was my college major; I ended up going to medical school) I feel like I have some qualifications to discuss this.
First–the business about whether the structure of the tire holds the car up, rather than the air itself. Let’s also conceptualize the “ideal” tire as a hollow sphere. There is a spectrum of structural possibilities.
At one end you have a perfectly hard sphere–steel-reinforced concrete, say–where the structure holds up the load, and the air pressure inside is superfluous. Hell, you could have a perfect vacuum inside (no, not really! I’m not going to defend that assertion so don’t get off on a tangent arguing with me about it.) In this case, the volume does not change at all when the load is applied.
At the other end of the spectrum would be an infinitely expandable and deformable sphere with air inside. In this case, the sphere will expand until the outside pressure and inside pressure are at equilibrium. By definition, such a sphere will expand until the inner pressure is atmospheric pressure.
Right away, you can see that the real tire is somewhere in between. I’m so used to metric that I forget what 1 atm. is in English --isn’t it 14.7 psi? It’s definitely not the 25-37 psi of most tires. All this means is that the tire wall has some structural property that allows it to hold in pressures greater than one atm. without expanding.
So what doesn this mean in terms of supporting a load? Let’s idealize the car as a perfect 3000 lb sphere, and put it on our perfectly undeformable concrete sphere. They touch at a single point, so the force of the car’s mass in the earth’s gravitational field is imparted onto the “tire” through that point. It is distributed in the wall of the sphere itself around to where the "tire’ touches the ground (again, at a single point). At this point of contact, the weight of the car is transmitted to the ground. Obviously it can’t really be a point–you’d have the 3000 lb force applied to an area of zero, which would give you an infinite pressure at that point, and so the area of contact must be big enough so that the force per area is within the structural limits of the material. As a wild guess, let’s suppose that reinforced concrete, or some other material, could support a 3000 lb car with a contact are of only one mm square (or 1/16 in square, to avoid mixing units). If this were the situation, I would acknowledge that this is truly the structure of the “tire” holding up the load.
Take the other extreme: the unpoppable soap-bubble. The air pressure within is all that holds the car up, and the skin itself does not generate any “reactive” forces. How does such a structure hold up a car? Let’s idealize the car now as a rectangular brick. Take the 3000 lb brick and put it on the perfectly spherical soap bubble which is pressured to 14.7 psi. What happens?
At initial contact, the brick contacts the bubble at a single point, with zero area, so there is nothing holding up the car. As the car flattens the bubble, and the area of contact becomes nonzero, the bubble imparts a force upward on the car that is calculated by the area of contact. If the volume didn’t change as the bubble flattened, then the area would end up being 3000 lb/15 psi = 200 sq in. If this is a circular area, the radius would be just below 8 in.
Of course, the volume of the bubble changes. How much? It depends on the size of the bubble. If it’s the size of the earth, then however far the brick has to sink into the surface to make a 14X14 sqin contact, the volume change is negligible compared with the shpere’s volume, so effectively, the pressure change is negligible too.
With a realistically small bubble–beach ball size or so, the volume change is significant, and is going to be rather complex. You are left with a cylinder that is x inches in diameter, y inches high, plus a donut-shaped part around it that is also y inches high . . . how do you figure out the equation for that object’s volume? It would have to include how much the material stretches as it deforms . . . it get’s WAY too complicated at that point! That’s when it’s time for a reality check, an “empiric” analysis to see whether it is necessary to go through such mental calesthenics to determine what really happens.
So suppose, as I did in my original post, that with the real tire, the deformation that does take place does not change the volume significantly. Does that assumption lead to any predictions that don’t fit with reality? I figured 3000 lb supported by four tires, or 750 lb supported by one tire. If a pressure of 30 psi is supporting 750 lb, that pressure must be applied to a 25 sq in area. That’s a rectangle with dimensions of 5X5 in, or 6X4 in, which isn’t too out of line with what a tire actually looks like when the car is sitting on the ground.
I was satisfied enough at this point that I stopped thinking about it. But I realize now that it doesn’t at all settle the issue about volume change and pressure change. You can take a theoretical approach to prove or disprove the hypothesis that the volume, and therefore pressure, change significantly. Or you can take an experimental approach, and simply measure the pressure in the two different conditions. I gather from a number of the posts that this has been done and shows that the pressure change is on the order of 0.1 psi, or < 1% of the pressure.
So the argument isn’t over what really happens–it’s over why the change is so small even though it is “obvious” that the volume changes a lot. I think that this is where people feel the need to bring in all of these arguments about the structure of the tire, etc.
However, if you could analyze how a tire shape changes as it flattens, and come up with an equation for the volume as a function of the flattened area of contact, my bet is that it would tell you that a donut shaped tube with a diameter of 27 in and a width of 7 or 8 in or whatever, flattened to make a contact surface area of 25 sq in, has a volume that is within 1% of the original volume. If you think I am wrong, I would submit that your opinion is based entirely on your “feeling” that the squashed tire has a significantly lower volume, just because it “looks like” it should be. Only when you actually figure it out, can you say for sure. Since the empiric data supports the opposite, I don’t feel the need to do the work.
Again, I composed my (extremely overlong!) post (immediately above) after only skimming the previous posts. Now that I’ve taken the time to read them im more detail, I see that some have made similar arguments to my own, while others point out aspects of tire properties that render my analysis incomplete, or at least simplistic.
I think the question on the minds of laypeople or would-be physicists as myself is not whether the pressure changes, but WHY it DOESN’T change as much as you’d expect it to. The simple answer is that an auto tire is not a balloon, or a soap bubble, and that the things about it that make it different from a soap bubble all tend to make it less likely to change volume than a balloon would. I would extend that a bit to say that even if it WERE like a balloon, if you try to roughly estimate HOW MUCH the volume change would be, it is probably less than you’d predict–i.e., the pressure wouldn’t go from 30 psi to 50 psi.
Enough!
this isn’t rocket science people.
now i’m sure you physicists have better ways to calculate the volume of a torus than my high school education but don’t over complicate things. we all know a tire can hold between 500 and 1000+ lbs right? but the car companies know that too. heavy cars gasp! have bigger tires, so any example gives the same result.
they also know distorton=heat=failure so they design tires that need little distortion to hold the car.
we all know the tire must exert that weight/force to the ground and vise versa.
we know that in practice the pressure won’t change by a measureable amount.
so if you figure the area needed to exert that weight based on your tire pressure and dived by tire width you get the length of your contact patch assuming the contact patch is square(it’s not but it’s pretty close). you can then compare to a real tire with a car on it(were can i find one of those)
to verify your results.
then find your tire diameter and cordial height. you’ll find that although the tire has a decent sized contact patch the car sank maybe 1/4 inch probably less.
again check the nearest tire with a car on it.
what do ya know, it checks out.
you can calculate the resulting volume change or close to it as i’m sure side wall flex is hard to figure, even for a physics master but i think anyone can see it will be small.
then convert pressure to absolute and multiply resulting “compression ratio” to find approx pressure change.
but the question was why not as much change as you’d expect.
the answer, 4000lbs or 1000 lbs per tire at 8 inches wide and 26-28" tall isn’t much. the relatively large radius can increases in area very rapidly with little compression. once you understand what is required to support the weight and how the tire shape changes under load it becomes obvious that little pressure change is required.
i ran the #'s and came up with a small change in pressure(very small) maybe not as accurate as someone with better knowledge in mathematics. i left out some details i know but the #'s aren’t important. what’s important in understanding this is pressure as it relates to surface area and how much area a tire produces with a small volume change.
Isn’t it fairly easy to understand why the pressure in the tires does not increase when the car is on the ground if we imagine that the car is being supported by steel tanks instead of tires? The pressure in the tanks wouldn’t change whether the car was on the ground or not. Would it?
See attached pdf.
Thank you Bubulus for your brilliant, brief, slightly obnoxious, and unnerving replies. I used to work with people like you and miss that a lot. Meanwhile, others may sit on a balloon until it pops due to a pressure increase and then can think about what happened.
The gauge reads the difference between the pressure in the tire and the pressure of the atmosphere. The latter is not changed by the pavement pushing on the bottom of the tire.
And this is just an offset that you can subtract by appropriate labelling of the marks on the gauge, right? Zero can be relabelled 14.7 psi.
So the gauge is still measuring the air pressure inside the tire.
The air cannot maintain the shape of the tire without exerting pressure on it. If you let the air out, the tire goes flat.
So if the air holds up the tire, and the tire holds up the car, then how can the air not be bearing the weight of the car?
If the material of the sidewalls was stiff enough to do the job, you would not need to inflate the tire at all.
In my post somewhere below (pdf attachment), I show how the air pressure in a tube that is a completely flexible balloon can hold up a weight
placed on top of it.
bfaul,
First let me thank you for the pdf file. You obviously spent quite a bit of time on it - and I think it illustartes the point you are trying to make - the change in volume will be small.
But allow me to point out that there are some initial assumptions you make that render the analysis - oh…let’s say… not valid for analyzing what is going on with the tire itself.
For example: Tires are not circular toroids. They more closely resemble the second diagram, where “l” is the rim width and it is fixed. For the sake of simplicity you could also choose to have the tread be width “l” as well - and it is also unchanging. That reduces the problem to 2 arcs and their cross sectional area. (For those who want to follow along, the pdf file is near the end of this thread.)
Nevertheless, we all seem to agree about the answer to Amy’s question - and it is that the pressure change - and therefore the volume change - is small.
But what we seemed to disagree on is whether or not the air is holding up the vehicle. So I submit that if you do a free body diagram of the rim, you will find that the air pressure is the same 360 degrees around the rim, so the air pressure is not exerting a net upwards force on the rim.
Ergo, the air is not holding up the vehicle - the tire is.
http://www.bridgestonetrucktires.com/us_eng/real/magazines/98V3Issue4/v3i4Tech.asp
This website has a helpful cross sectional image. (See attached jpeg) Yes the pressure is
the same all the way around, but the tire warps so the area of horizontally-oriented surface
(tire surface + rim) over which the air pressure acts is not the same. The air is pushing up on the tire sidewall
on the bottom, leading to an upward force. This has to be the case, because as you decrease the air pressure
the bottom part of the tire collapses; it is not stiff enough to bear the weight on its own.
You can’t say on the one hand that the tire bears the weight entirely, and on the other hand, say
that it needs the air pressure to “maintain its shape.”
Thank you for posting that! I feel soooooo vindicated.
My pleasure, but I hope we are here to learn and I am open minded to persuasion. Now, of course, I’m not saying that the air pressure bears all of the weight. There is some stiffness or resistance to bending in the
sidewall material - and I did not include this in my original post. This is, of course, a property of the sidewall material. But the air pressure does not
magically endow the sidewall material with enhanced mechanical stiffness; it just supplies an additional force. The fact that the tire goes flat when deflated is the key piece of evidence here.