This week on Car Talk we talked with Amy in Louisville. She and her dad have been discussing a profound automotive physics conundrum. Specifically: Why doesn’t a car’s tire pressure change when it’s lifted off the ground? Amy’s dad thinks it should because you’re taking 3,000 pounds of weight off the tires, and “un-scrunching” them. Yet, his trusty tire gauge isn’t measuring any change. We weren’t afraid to wade in with an answer, and got a little help from our pal Wolfgang. His credentials? A lot more than ours. He’s a physicist at the World’s Greatest University, just down the street from Car Talk Plaza in Our Fair City.
But what do you think? Did Wolfgang (and we) have it right… or would you like to go mano-a-mano with someone in a white lab coat at a lectern?
Share your thoughts here. And, trust us, we’ll love watching this little discussion develop!
Yours in hot air, no matter what the pressure,
Tom and Ray
Click and Clack, the Tappet Brothers
See the Question: “How is a car with tubeless tires supported?”. To calculate the inner surfaces of a tire, use the formula for a torus (doughnut). In response 49, you’ll see that, in the example, tire air pressure changes 2/3 of a psi. Of course, to detect this small amount, you’d need a very accurate tire air pressure gauge.
Someone posted it right in the post called Amy’s Tire question much more aptly than I did in a longwinded post on Tire pressure. I did include (in the wrong forum under tire pressure) some reasonings with the pressure per square inch, but mine was more on why tires didn’t explode I suppose in simple terms.
Edit after rereading the Amy’s tire question…
I reread one item, 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.
I guess I posted this in the wrong area. I’ll re-post it here.
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 :).
Dennis Agosta
Assistant Professor of Physics
The College of Idaho
The volume inside the tire does not change and the amount of gas inside the tire does not change whether the car is on a lift or not. One or the other must change for the pressure to change. BTW, I can’t read Wolfie’s answer. My computer doesn’t know what a *.SMIL file is.
JTsanders is right. Lifting the car changes nothing. The air is not supporting anything, the structure of the tire is. The air maintains the shape of the tire so the TIRE can support the car, not the air…
I would just say to give a brain a break. Maybe the brains should give us a break? A tire gage that reads in thousandths of a PSI would be too expensive and useless for me to buy. I wouldn’t mind borrowing it if it can read the tire pressure from my recliner. Better yet; just leave a message on the answering machine. I stored up a bottle of Summer air to put in the tires in January. Why does the pressure in the tank go down whenever I put some in the tire? Stay busy.
My Kubuntu machine got the file right away. What it really is, is an RA file, and will play on the correct audio/media player, which in Linux is mplayer. I have no idea which player one would need in Windows. It is a playback of part of a radio show by the brothers. Isn’t RA a Real Player file?
If you put a tire gauge on a metal piston and cylinder that has the volume of a tire and has 3,000 pounds on it, well I will bet that the pressure would go up! But a tire can flex, thus the compressed air simply transfers the forces to the side wall which bends outward the amount that it takes with 750 pounds loaded on it. But it would still seem to me that the pressure would go up some too. Has anyone really checked this with an accurate presure gauge like a digital one?
There must be tire pressure gages which read accurately to tenths of psi (pounds/square inch). The tire companies surely have them, and use them, when they perform tire tests. My calculation is that the force of 1,000 pounds (1/4 of total vehicle weight) will be transmitted, by the tread and sidewalls, to the air contained within the tire. The tire, pressurized to 32 psi (pounds per square inch), having an inner surface of 1500 square inches will have it’s pressure raised approximately 0.7 psi, to 32.7 psi. The 1500 square inches of inner tire surface has 32 psi air pressure against it, all the way around (sidewalls, tread area, inner wheel area), for a total force of 48,000 pound (1500 x 32 = 48,000). This high resistance of 48,000 pounds is needed because, in use, the tire will have impacts with pot holes, etc., with forces of several dozens, or more, gees (units of gravity). Visualize: the 4,0000 pound car is moving at 60 mph. The tire slams into the bottom, and side, of the pot hole. The force of impact is absorbed by the tire and resisted by the 48,000 pounds of force within the tire. Some of the force, of impact, is transmitted to YOU as a sizable “jar”.
I’m the guy who posted “Amy’s Tire Question”.
And, yes, tire manufacturers have more precise gauges - some to the nearest 0.01 psi - yes, that’s right 1/100th of a psi.
If you will read “Amy’s Tire Question”, you find we already know the answer. The loaded tire has more pressure than the unloaded tire. By between 0.2 psi and 0.5 psi.
WHY?
Obviously the loaded tire occupies less volume. In it’s undistorted shape, the air pressure in the tire tries to get to a state where the forces are balanced and at a minimum. This means maximum volume. Then when you load the tire, the weight changes the shape, and therefore the volume is reduced.
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.
I am afraid you are missing the point. Just how does the weight of the car get through the tire to the ground?
To understand what is happening, first realize that a tire is a tension/compression system. Air is compressed within the tire ?stretching the steel belts and rubber with tremendous force. The force is essentially the pressure, multiplied by the diameter and width. For a typical tire at operation pressure, you have about five thousand pounds force trying to rip the tire in two when the car is on Amy?s lift.
This is analogous to pre-stressed concrete, where steel rebar supports tension because the concrete can only support compression. Except here, the steel belted rubber goes limp in compression and only works in tension. The compressed air keeps the tension in the tire just like stretched steel rebar keeps compression in concrete.
When the car weight is applied to the wheels, it wants to compress the rubber of the tires. But it can?t compress the tires until it overcomes the five thousand pounds of tension caused by the compressed air. A typical car puts about one thousand pounds of weight on the tire, so you have four thousand pounds of tension left to go before your steel belted rubber tire goes limp and really changes shape.
So that?s the answer. The weight of the car lowered from a lift is transferred not to increased air pressure but to the steel belts and rubber of the tire. If you wanted to measure the change, it would be in the reduced stretch of the side walls, not increased air pressure.
The mechanisms of load transfers and transmission may be explained in different ways…such are the vagaries of language and perception for each person. My quick, back of the envelope calculations were a bit off. An “average” size tire has about 3000 square inches inner surface area. So, if it’s loaded with 1,000 pounds (at 32 psi), its pressure would change by, about, 0.35 psi, instead of the 0.7 psi, I previously stated. So, theoretical values (mine) meets empirical (measured) values (yours). Your measured values of 0.2 to 0.5 agree with the theoretical values (of course, in the “back of the envelope” approaches we’ve been using). Theory and empirical evidence are co-dependent. Neither is a whole without the other.
How about the basics? Here are a couple of things we know for sure:
1.) The volume inside a pressurized, unloaded tire is the maximum possible. (If it weren’t we could get free work out of it, and we’re wasting our time here. We should be building carbon-free tire power plants.)
2.) Because the volume is at a maximum, it follows that the pressure is at a minimum.
3.) Deforming the pressurized tire casing into ANY other shape requires external work - force through a distance.
4.) The external work has to be matched by an increase in stored energy in the contained air and the tire structure. (Never mind the heat for now.}
Now, about lowering a car from a lift, we can say the following:
1.) With our unaided eyeballs, we can watch the tires deform, at least locally in the region of the contact patch with the shop floor.
2.) The deformation has to be accomplished through work - force through distance.
3.) The mechanical energy stored in the tire has to go up.
As above, the mechanical energy can be stored in the air volume or in elastic deformation of the casing.
For a relatively inextensible tire casing (probably a good approximation to reality), most of the energy will be stored as p.dv work in the contained air as it decreases in volume from its maximum. This will be accompanied by a rise in pressure. It may be difficult to measure the pressure difference with even an accurate shop-grade tire pressure gage.
Maybe, this afternoon, I’ll try it starting with 5 or 10 psi. But probably not.
Ray intuitively stated, ?There would be no significant difference in the tire pressure reading between a loaded and unloaded tire (using a standard tire gauge)?, and Wolfgang zeroed in on the principles involved, ?deformation of the tire?, but he didn?t stress the principles involved.
There will be little, if any, change in the volume of a tire when going from unloaded to loading it with a vehicle the tire was engineered for. As the bottom of the tire flattens and reduces local volume in the tire, the rest of the tire expands proportionally compensating for the reduction.
The Car Talk, ?In-house Expert, Wolfgang?, would certainly be a great source for mathematical analysis of the principles involved, however, a better reference would be engineers at any of the many tire manufacturers who have to deal with preventing tires being overstressed from such issues as centrifugal and loading forces.
Signed,
Smuck with Physics Common Sense 
Any pressure placed on the tire will affect the tire pressure relative to the amount of pressure. Even if you tapped on the tire with a hammer you should be able to detect fluctuations in the pressure, if you have a sensitive enough gage. After all, each size tire is designed to carry a specific load. If the load exceeds the design limit we all know what happens - the internal pressure exceeds the limit and ka boom. You can take an inflated tire off the car and bounce it. The bounce is caused by a compression of the tire and it then rebounding back into shape. A perfectly ridged tire (no deformation) will not bounce if dropped on an equally perfectly ridged surface.
A long time ago, though I’ve had sampled several versions of pre-production tires that’s been designed(supposed) to have better controlability of its surface pressure/traction, their treds are almost triangle in appearance when it’s lifted off but goes “flat” when on the ground. As I recall, the air pressure tends to change within the limits as the tread gets deformed by the traction. and I also recall it is quite hard to control due to the lack of transient responce.
I’m not a physicist in a lab coat wearing no underwear guy but guessing that it might be doable making a tire(air pressure) that is susceptible to its traction but for some reasons manufacturers don’t do that.
It’s been a long time since I’ve had physics, but…
3000 pound car on four tires = 750 pounds per tire
Also, put 100 pound weight on one inch cube = 100 pounds/sq.in;
Put 100 pounds on 10 inch cube = 1 pound/sq inch
Therefore, how many sq inches of tire rest on the ground?
Hey Gang,
Here are just a few musings: I guess I find it more interesting to consider what is happening to the pressure in a spinning tire. It’s also worth reconsidering what happens to tire pressure when a car hits a bump. Air in the tire must compress, if only momentarily, but as this force is transmitted throughout the tire, the tire deforms and hopefully is able to rebound without blowing out. Pressure in the tire is therefore not uniform throughout the tire at the moment that the tire hits a bump if there is there actually a “local” compression. Tire deformation also produces heat (but if it’s too soft, then it lowers mileage). If the tire is too hard, tire cannot properly contact the road (and the ride is also less comfortable and maybe even less safe if the tire loses contact). As the tire rolls, it inevitably deforms, which heats up the tire and the kinetic energy of the molecules increases, which is part of the reason why we measure cold tire pressure.
Numbers are good:
Figure a 4000# car, 1000# per wheel.
A tire on my Jeep looks like a torus, D = 23" big diameter, about d = 6" small diameter. It has a volume v = (Pi D)(Pi d^2/4) = 2040 in^3
Static tire pressure is say p = 30 lb/in^2
Now lower the car to the shop floor off the lift. The tire deforms say u = 1/2 in in a radial direction relative to the axle.
The external work done on the tire is maybe 1/2 (1000 lb) (1/2 in) = 250 in-lb.
If we assume inextensible tire casing material, the casing elastic energy can’t change.
The external work is transferred to the contained gas as pdv work: 250 in-lb = 30 lb/in^2 dv. The change in volume, then is something like 250 in-lb/30 lb/in^2 = 8 in^3.
This represents a volume change of 8/2040 * 100 = 0.4%.
Process quibbles aside, this corresponds to a pressure increase of somewhere around 1/8th psi, not enough to be detected with a shop tire pressure gage.
The analysis suggests that if the tire pressure is low enough, you might to be able to detect the pressure change. Say p = 10 lb/in^2 is enough to support the tire without collapsing tread to bead. Radial deflection on lowering off the lift is far higher, say two inches, so the work is 1/2 * 1000 lb * 2 in = 1000 in-lb. 1000 in-lb = 10 lb/in^2 dv. dv = 100 in^3. This would correspond to an isothermal pressure change of 100 in^3 / 2040 in^3 * 10 lb/in^2 = 0.5 psi.