This was discussed in the Car Questions forum awhile back, no definitive answer. I have a suggestion for you or anyone here that really wants to know the answer.
First you need access to a Manometer like this one
Then, with a car off the ground, let all the air out. Remove the valve core to insure that the tire is at 1 bar or at local barometric pressure at the time. Hook up the manometer without the core, use a screw on type adapter.
Now set the car down and measure the pressure. Consult with your local physicist or mathematician to get the maximum change in tire volume. This is better (safer) than putting 300,000 lbs. of sand in the vehicle.
You could repeat the same experiment but start with 5 psi and measure the change, repeat again with a starting pressure of 10 psi. With three points, the mathematician could plot the change in volume vs change in pressure so you could figure out what the change in volume would be at 30 psi or any other starting pressure. From that, you could plot the change in pressure due to setting the car down. I’m guessing it would be in the range of .01 to .001 psi at 30 psi starting pressure.
You could even repeat the experiment with various size vehicles and various size tires. Imagine the amount of trivial information you would now have if you did all this.
I’m mildly curious about the answer, but not enough to do this myself, especially if I have to buy an expensive manometer. Maybe the caller’s father might want to do this since it seems to be really bugging him. Maybe it would be a good experiment for a budding engineering student at Northern Michigan University, might make an interesting thesis paper.
This was discussed in the Car Questions forum awhile back, no definitive answer. I have a suggestion for you or anyone here that really wants to know the answer.
I already sent an email, but couldn’t find the right category. I’ve wrote a lenthy letter explaining it, but registered just to answer this here. This is as simple as I can put it I believe, without getting overly complex. No physics paper or anything required. This is a really simple principle, but I stink at simple explanations.
Today you answered a question in regards to the Lady who wanted to know why a car’s tire pressure doesn’t change when it is on the ground vs. when it is in the air without any weight on the tires. You guys answered her question correctly, but didn’t address the reason WHY.
To put it simply, the reason why is due to the basic Physics principle of equal but opposite. Hence a force will be reacted to by an equal but opposite force. For example, if you hit the ground with a hammer, the hammer will bounce off the ground (probably, as long as you hit it hard enough) because the ground hits the hammer back with a force equal to what the hammer hit the ground with.
This applies to tires as well. The air inside the tires are pressing against the walls of the tires at the set PSI. Or if you had 40 PSI in the tires, then while the air is pushing against the walls of the tires at 40 PSI, the walls of the tires are pushing back at the air at 40 PSI. Meanwhile, outside the tires are pushing out with the same air pressure as the outside world.
Now some would look at a balloon, and try to make sense of this. Looking at a balloon, the first thing to note is that the walls of the balloon are not as strong as the walls of a tire. You try putting 40 PSI in a balloon that you can blow up with your mouth, and more than likely it’s going to blow up. This is because the walls of the balloon do not have the strenth to push back at the air with the equal but opposite strength. So it goes to the rest of the atmosphere where it DOES have the strength, as will be explained afterwards (sort of). However the walls of the balloon do have the strength to push back with a small amount of force, and the balloon will not explode until someone or something exerts pressure enough to exceed the PSI the walls of the balloon can take. So if I put pressure OUTSIDE that was greater than the pressure the walls of the balloon could push back with it would explode. Since the air on the inside is equal strength to the air on the outside, once that happens, the air is of equal strength with itself.
How does this apply to tires? Well, we know from the above that when the air in the tires is at 40 PSI, that the tires are pressing back at 40 PSI. Meanwhile, outside, the atmosphere is nowhere near the concentrated 40 PSI (taken for the atmospheric pressure that is applicable to both the tire and outside pressure of course) of the tires. Hence, as you aptly put it, the pressure outside (even with weight on the tires) is nowhere close to causing a difference. It’s the pressure, NOT the weight that causes changes.
For example, let’s take a thousand pound car (just for an example). That would mean that you would have 250 pounds on each tire. Each tire has a total tire area of, let’s say it’s a foot wide, and circumference is 7 feet (okay, a very SMALL tire, but hey, there are some rear emergency tires that are pretty small). That means that the tire has 84 square inches of area. 250 / 84 = ~8.928 inches of pressure on the tire. As you can see, no where near the 40 PSI inside the tire.
Or let’s take a 8000 pound car. That’s 2000 pounds per tire on that vehicle, that’s still only ~23.8 pounds. Still not close to that 40 PSI.
This is also the same reason why pressurized tanks don’t suddenly have more pressure on the inside air than on the outside when they are pressured.
This leads to what would burst a tire. First of course would be anything that would wear down the tire wall strength to the point where the inside pressure exceeds the wall strength of the tire. They can no longer contain the air and you have an emergency on the highway.
The second would be a situation where the pressure on the outside would exceed the pressure on the inside. An example of this would be where you drive over a nail point. Take the 1000 pound car, suddenly that entire 250 pounds per tire is on that one point of the nail. That nail point is obviously less than an inch wide, hence you suddenly have 250 lbs on the outside, which exceeds the 40 PSI (or even more appropriately, whatever the PSI the tire walls can take) of the tire and you have an emergency on the highway.
Other things, which get even more complex…
Another note, is the ambient air pressure, or the air pressure outside, which is reflected in the tire. This also applies to the air in and out of the tire. The tire has this pressure as the force pushing in and out of the tire. Becuase this pressure is equal to the pressure inside (because you filled it at that air pressure), they are both pushing against the tire walls on opposite sides. However, if you go up, such as in an aircraft, without pressurization, the pressure outside the tire decreases while the pressure inside the tire decreases. Now you might not have as equal a force on the outside tire walls as you once did. Keep this up, and get high enough, and you might actually blow the tire from the inside out, since you don’t have that ambient air pressure on the outside of the tire.
Vice versa for going deeper into the earth, where the pressure on the outside increase, only this time it’s more like caving in the tire since the ambient air pressure on the inside isn’t as strong as the force on the outside.
If the last two paragraphs were two confusing disregard as they were only to show other methods of where the tire could explode, and won’t be encountered under normal road conditions.
Hopefully that is a lengthy enough but simple enough explanation (without too much math) to explain the girls question.
Looking at this purely theoretically, tire pressure is simply the difference between the pressure inside the tire and outside the tire, lets call it the pressure differential. Virtually nothing else matters. When weight is applied, such as letting the car down off the jack, the added weight causes the tire to displace (or bulge) the exact amount to account for the added weight, but the pressure inside the tire doesn’t change because the pressure differential has not changed.
If you blow up a balloon and then sit on it, assuming it doesn’t break, it will displace or stretch to account for the added weight, but the amount it stretches will be purely determined by the added weight, the pressure inside and outside the balloon will not change. The difference between the tire and the balloon is simply that the balloon is much more stretchable and therefore the psi inside the balloon isn’t mucn different than outside, and the added weight of a person sitting on it is much greater in proportion than the weight of a car on a tire.
The answer to this question can be found in any freshman physics text - look up gauge pressure in the index. The gauge is only going to measure the difference between the pressure inside the tire and outside the tire. If you take the tire off the car and let out the air, is the pressure inside zero? The gauge will read zero, but there is still atmospheric pressure inside and outside the tire.
Increase the pressure outside by having it bear the weight of the truck and the pressure inside increases by the same amount. The difference in pressure between inside and outside hasn’t changed, so neither will the gauge reading.
I think a form of this question will appear on the next physics exam I give my students :).
Assistant Professor of Physics
The College of Idaho
I reread one item (in relation to original post_, there is one thing I should point out. The pressure isn’t related to the weight, hence the weight or the bearing isn’t really the thing determining the pressure. The thing is the amount of pressure, which if all atmospheric pressure staying the same, should be relavant to the amount of space left in the tire. (area vs volume) If the area amount decreases, THEN you would see the PSI increase, but if it decreases, it would increase. In that way, and the pressure on the outside, is relavant.
Or as put above, the pressure outside and inside, is relavant, and not the particular weight, though in a related way.
Here’s the answer:
Let’s say your tires are pressurized at 40 pounds per square inch. That’s 40 pounds on all 500 square inches of the tire. So you have a total of 20,000lbs on each tire. 1/4 of the car’s weight, or about 800 lbs. is on each tire. That’s enough to change the pressure by about 1.5psi. But if the tire expands under the extra pressure, the answer will change. If the tire were very flexible and expanded 2.7% in the above example, the pressure would not increase at all. But if it were that flexible, it wouldn’t be able to hold 40psi without ballooning to monster truck size.
Here’s another way to solve it without figuring out the surface area of the tire:
Look at the bottom of the tire. Thre are ~20 square inches touching the ground. Why? Because you need 40psi times 20 square inches to support the 800lbs. on that tire. Compare that area to the entire surface area of the tire. Pretty small, huh? So the weight of the car isn’t significant compared to the pre-existing force against the tire.
Oh yea, I got a third answer, also mentioned on the show. Pressure is proportional to temperature and volume. You jack up the car. Did the volume change significantly? Did the tire get freezing cold? No? Okay, the presure hasn’t changed significantly.
So far there are many good answers, but no actual experiments. Everyone who has had a small trailer with two 20" wheels, and tires rated for about 500 lbs each knows they can take a one ton (2000 lb) load down the highway easily. (We have never tried two tons down the road, but it will hold it in place.) The issue then becomes whether they encounter an impact that either punctures the tire, or breaks the tire sidewall away from the rim. At slow speeds it rarely occurs, but at higher speeds and hitting a large concrete break in the road, the tire can break away from the rim.
So the answer is that the excessive load can be supported almost infinitely by the air, but the seal on the rim is the weakest point. The internal pressure of the tire remains the same right up until impact.
So, the idea of placing tubes into otherwise tubeless tires is not that extreme to overload tires, since the air will be held in as the seal on the rim is breaking down.
IMHO: consider the surface area of the contact patch [between the tire and the ground] times the tire pressure … for simplicity, presume all four tires have the same contact patch area and the same weight on each wheel. In that case, the product (tire pressure x contact area) x 4 (four wheels) should equal the weight of the vehicle. This is the reason airplane tires can have over 100 psi of nitrogen - the surface patch area x number of tires x pressure = weight. When you have 300,000 lbs of airplane on 8 main and 2 nosewheels (consider a WC-135 at max T/O weight)… Back to our issue: The contact patch with the ground requires a flattening of the tire. This logic establishes the tires do deform to some extent. To what extent does this change the total internal volume of the tire? Consider the sidewall flex (readily observed in radial tires) v. the flattening at the ground. Do these compensate for each other, leaving the internal volume unchanged? Quite possibly. Now, when the vehicle is jacked up, the tire might re-assume an ideal torus shape. If we accept the internal volume did not change, then the air pressue inside the tire will still be whatever it was before the lift was operated.
The issue really boils down to: how much does the internal volume of a tire change? Perhaps someone who designs tires can provide an authoritative statement ???
PS - aircraft tires are serviced with nitrogen for several reasons, including fire protection: if a retractable gear a/c has a fire in the wheel well and the tire is compromised, then the tire won’t be adding any oxygen to the fire - only inert nitrogen. Similar idea for a brake fire on the ground (fixed gear or retractable) … no spray of oxygen-containing air onto the fire. Servicing vehicle tires with nitrogen seems a bit overkill IMHO … but I could learn something new every day.
Even if the volume does change the pressure should not increase by more than 1.5 psi. 1.5 psi was assuming the tires don’t bulge out at while on the ground any more than they do when the car is raised. i.e., it assumes the volume is slightly less when the car is on the ground. If the volume stays exactly the same, then the change in pressure would be 0psi. But those would also be some awfully flexible tires that would inflate like a balloon if you tried to put 40psi in them.
Nitrogen has some minor advantages compared to air that make it better for any application. The main one is no water vapor. Water vapor increases the tire pressure by a random amount when your wheels are hot, depending on how much water is in the air. Even though you measure the pressure when your tires are cold, it’s the hot pressure that determines your handling, fuel economy, etc. So nitrogen is better than air insofar as it is important to have an accurate tire pressure. i.e., it’s not worth the trouble for your average car owner since nitrogen is harder to find than air. But for industrial use, race track use, etc. picking nitrogen over air is a no-brainer.
Exactly. Without the load, the air inflates the tire to the optimum shape that has the greatest volume, reducing the pressure to the lowest possible point. Any load deforms the tire. ANY deformation at all will not be as perfect, reducing volume and therefore increasing pressure. Furthermore, the pressure with be further increased by heating caused by the work performed on compressing the air into a smaller volume. However, in practice, the total amount of change will be so little it would take an accurate computer model to figure out. It is like calculating the number of angels that can dance on the head of a pin.
I agree with the poster who said that tire pressure will increase slightly when a tire is made to support a load. With a high enough load, the tire will explode as did my overloaded trailer tire. With that in mind, one or both of you brothers are engineering graduates from MIT, are you not? What do you say?
There’s another thread posted by Tom and Ray in the "Car Questions area called “Paging all Physicists”
But here’s the result of actual measurements:
Tire manufacturers have more precise gauges - some to the nearest 0.01 psi - yes, that’s right 1/100th of a psi.
The loaded tire has more pressure than the unloaded tire. By between 0.2 psi and 0.5 psi.
I hope everyone realizes this is not a theoretical study. It is based on observations of loaded and unloaded vehicles using a pressure gauge capable of meauring to the nearest 0.1 psi. I suggest that folks who don’t believe this should perform the experiment themselves. However, you need to get a pressure guage capable of at least 0.1 psi increments.
Hope this helps.
BTW, you don’t need to be a Physicist to do this. In fact it helps if you are not. Physicists tend to do a theoretical analaysis and a simple experiment would be better.
Can you cite the study to support those figures. If it is the one I’m thinking of, it was not done by the manufacturer and it was not done using standard scientific protocols. I have done a similar experiment and did not get the same results.
The tire is flexible enough that the volume changes when the weight of the car is added. One can easily demonstrate that flexibility by observing the volume change when adding or removing significant amounts of air from the tire. Pressure trumps tire rigidity.
What do you mean “scientific protocols”.
I did this myself with a simple lift and a pressure gauge. You can do the same. Before you try to refute the results, please try it!!
I think a lot of these answers are getting toward the correct view of this situation, but I don’t believe any of them, on the radio or on this site,have hit on the most significant factor. Barring other evidence, people tend to assume that systems are linear. This one is not.
If you were to take a tire (an imaginary one with a perfect bead seal and really elastic rubber compound) and totally evacuate the air, the volume would be close to zero. As one added air, the volume would increase in a reasonably linear fashion, as would the pressure. With that tire, I believe one would see a significant pressure increase when the weight of the car was added by removing the jack, since the volume would then be reduced.
With an actual tire, the relationship between pressure, volume, and number of air molecules is extremely nonlinear, partly due to the rubber compound itself, but more due to the cords – especially the radial cords, whose whole goal in life is to keep the diameter (and hence also the volume) stable no matter what.
Think of a balloon inside of a steel sphere. As you blow it up, the pressure will increase, but so will the volume, as a function of the balloon’s elasticity. And both will be more or less linear. But when the balloon’s outside diameter reaches parity with the inside diameter of the steel sphere, you can blow your brains out and not increase the volume. Only the pressure will increase, based on the compressibility of the air only, and no longer dependent on the elasticity of the balloon.
Going back to the tire, if you could move to a more linear portion of the volume v pressure curve, say by making the pressure really low (just enough to keep the bead seal), I suspect you would see a measurable increase in pressure when the car is lowered from the jack, since the tire’s volume would be decreased. I hope someone will try this. It would not be difficult, if you had a good floor jack and some time.
The truth in any situation is always easier to spot if you can eliminate the extraneous factors, the converse of which the puzzler manages to prove almost weekly!
Short analysis: Inflate the tires with the car on a lift… they are round. Lowered, the car settles… flattening the tires until the contact patch area times the tire pressure equals the car’s weight. That flattening (a chord) reduces the tire’s volume by the tire’s width times the segment area (a segment being the part of the circle cut off by the chord).
Calculation result: Using a 3200# car with tires: p = 33 psi, OD = 24", ID = 16", and width = 6", I got a volume decrease of about 0.4% for a corresponding 0.13 psi pressure increase.
In real life, the tires probably deform everywhere as they take the car’s weight… stretching out as the pressure increases, so the net change is less than the 0.13 psi I calculated. Probably not much less, though. Tires don’t seem very stretchy.
Doesn’t the outside pressure remain constant at 1 atmosphere regardless of whether or not the tire is bearing weight? I think the question pertains more to the elasticity of the tire and whether or not the volume remains constant as the weight of the vehicle is applied. Does the tire deform such that it’s volume remains constant as weight is applied? If so, the delta between exterior and interior pressure (i.e., the measure of tire pressure as you stated) also remains constant. But if the deformity in the shape of the tire due to added weight causes the volume to decrease, then the delta will correspondingly increase. It appears that Johnly and CapriRacer are on the right track. Those who probably really know the answer are the design engineers at Goodyear, Goodrich, Firestone, et. al. Maybe someone from a manufacturer could weigh in on this question.
I forgot to mention that my simplification for my calculation was that each tire’s cross section is rectangular… that is, the sidewalls are flat and the inner surface (against the rim) and outer (the tread) are simple, flat bands.
The bulging sidewalls of a real tire would increase the basic volume and lower the percent change I got. The bulging tread would, I think, increase the percentage change. The volume of the rim would also increase my basic volume and decrease the percentage change. Sidewalls seem to bulge more than tread, so they probably have a greater effect.