Obviously any nitwit knows that tire pressure has to do with supporting the vehicle. Why else do NASCAR racing teams engage in tire pressure adjustments at every pit stop? The reason your car’s tire pressure does not seem to change is simply because you don’t have an accurate pressure gauge. Indeed, the pressure in your tires changes radically due to the heat which builds up due to the flexing of the tire’s structure. This is called hysteresis. Pressure obviously increases with temperature since air expands when it gets hot. This is why you see racing drivers swerving their cars at the start of a race. They are trying to heat up the tires to get the maximum contact patch. I bet if I squeezed some sensitive body part you’d figure out how stupid this question is, and how ignorant most of the people answering are. Maybe instead of physicists you should try asking some engineers. Not that I have anything against physicists, I just wish I could find some today. I scared the daylights out of a friend a year ago by explaining to him exactly how to construct a thermonuclear bomb.
Bunco.
More bunco…
Calculate how many square inches of surface area there are in a given tire. Now place 35 pounds of air pressure on EACH square inch of surface. (Pounds per square inch). The forces involved are tremendous. The weight of the car being supported is insignificant compared to the load the air pressure itself imposes on the tire.
For those who haven’t read the previous replies, some of the replies are by people who have actually measured, with a $50 accurate-to-tenths (or, better) tire pressure gauge, a real pressure change with tire loading, and unloading. The internal loading of a representative tire, at 32 psi, is about 60,000 pounds, total. An external loading of 1,000 pounds, is 1.6% of the 60,000 pound internal loading. Thus, 1.6% of 32 psi = 0.53 psi. This is added to the 32 psi which yields: 32.53 psi loaded. Of course, the internal volume would be decreased by a volume which would yield a change of 1.6% surface pressure. // The formula for the surface area of a dough-nut (torus) is: 4 pi squared times R times r. Little r is the radius of the dough-nut. Big R is the distance from the middle of the hole to the middle of the dough-nut. www.webcalc.net/calc/0046.php will perform the calculation for you … just put in the values.
Dear Mr. Click and Mr. Clack:
I think the answer is a little different.
Of course you probably want to know what the question is. It?s the one about why the air pressure in a tire doesn?t change whether the car is on the ground or not.
My answer involves an experiment and two parts, but none of it is real complicated.
?.. and here?s the experiment:
What you will need to do this is a tire gauge and a jack, and something to pump the tire back up when you are finished.
With the car sitting on the ground, let the air out of one of the tires until it?s almost flat. The trick here is to leave just enough air in the tire to keep the car just a little bit off the ground. Now what you want to do is use the tire gauge and measure the air pressure in the tire.
Next, jack the car off the ground so that the weight of the car is no longer on the tire that you just deflated. And (you guessed it) measure the air pressure in the tire again.
What you?ll notice is that the tire had more air pressure in it when the car was on the ground, but not so much after it had been jacked up. In fact, the higher air pressure that you measured is all it takes to keep the car off the ground, and for a regular passenger car it isn?t very much.
What?s that? You want to know why you don?t notice a difference when the pressure in the tire is up to 35 psi?
It?s a two part answer. Here?s part one; it?s mathematical, with arithmetic and stuff like that, and it?s about why it takes so little air pressure to lift the car. I know what you?re probably thinking. ?I don?t like math and what on earth does this have to do with why the tire pressure doesn?t change when I lift the car off the ground?? ?. Right? Well, just be patient for a few moments longer. We really do need to figure this out before we can answer the question.
If we have a car that weighs, say 3,000 pounds, and the weight is distributed so that 60% is in the front where the engine is, and 40% is in the back where the empty trunk is, then we have weight on the front tires that is 60% of 3,000 pounds, which equals 1,800 pounds. Next we have to divide that by 2, because we have two tires, so each front tire has only 900 pounds of weight on it.
Here comes more math stuff.
You can actually figure out how much air pressure it?s going to take to lift 900 pounds, (and trust me, this is pretty important to the final answer). What we need to do is take the projected area of the wheel hub and divide it by the weight that?s on the tire. ?Projected area? is a term that engineers and design guys use a lot when we figure loads on bearings, and the force generated by things like an air cylinder. In this case all you have to do is take the diameter of the wheel hub, On my Honda CRV it about 14?, and multiply it by the width of the hub, which is right around 6?, and that gives us a projected area of 84 square inches. Then if we divide 900 pounds by 84 square inches we get an air pressure of a whapping 10.7 pounds per square inch?. and that?s all it?s going to take to lift the corner of the car.
What?s that? You?re ready for the final answer? Well O.K., and there is only one more math step.
While the car is still up on the jack, pump up the tire to the 35 psi we talked about earlier. Now we need to go to the formula we just went through and work it backwards. When you put 35 pounds of air pressure in the tire you can actually lift a weight that is equal to 35 psi multiplied by the projected area of the hub, which in our example is 84 square inches, so we have (35) X (84), or a total of 2,940 pounds. Now when you let the car down off the jack and put the 900 pounds of load onto the tire, it just yawns. In order to see an increase in tire pressure you would have to load it to something over 2,940 pounds. It?s only at that weight and higher that you would overcome the force holding the car up and cause an increase in air pressure. You can calculate it for yourself. Go ahead, try a load of, say, 3,500 pounds and see how much air pressure it would take.
Of course I want to be notified in advance if you try that as an experiment?. I want to be in the next town, or county, or state, or traveling overseas. Ya, that would be nice. Overseas.
Best Regards,
Jim Greathouse
I can’t understand this silly argument. Placing pressure on any object raises its temperature, because the energy has to go somewhere. Therefore, the pressure increases. Your argument is essentially the opposite of the perpetual motion machine, it’s the notion that a tire is a black hole. It is usually black, but if there’s a hole in it it’s likely to go flat and the last time I checked a flat tire won’t support a car. A tire is essentially a closed system, hopefully if it’s not leaking, and adding energy to any closed system increases the pressure. I have a digital tire gauge and I can detect the difference in pressure between the tires on the sunny side of the car. There’s that naughty “T” thing. Maybe you should talk to somebody on a racing team. You have added negative entropy to some environment, namely the tire, so it must respond by increasing both its pressure and its temperature. When Megan Mullally sits on a balloon it pops. Thus, it doesn’t matter if the volume changes, if you decrease the entropy of a closed system something has to give. Objects of any kind get hot when they’re crushed. It’s all a matter of understanding the laws of thermodynamics. Given this silly argument, I wonder if anyone can understand how thermonuclear devices work. Hidden clue, it isn’t the styrofoam. That just keeps the pollution out of the channel.
I am absolutely loving this thread.
Clearly there are some folks who don’t know the right answer (After all,there isn’t a universal agreement here!!) - so some of you guys are just flat wrong. I’m hoping all of those who are convinced they are right realize that there is a group who equally think they are right.
This is just fun to watch.
I did this years ago (It’s reported in an earlier post), but I don’t have access to the proper tire pressure gauge to duplicate it today. When my new gauge arrives from Tire Rack in the next day or so, I’ll post the results of my latest experiment.
The volume of the air in tires (and thus the pressure) would only change if the pressure (the force per square inch) was greater than the one inside tires. With 2000 pound car it’s 500 per tire and if the area of tire contact with the ground is e.g. 100 sq. inch it’s just 5 PSI per with is less then 30-40PSI in typical tire, so the tire won’t give in.
Even if my numbers are wrong and the car weight is much greater (4000), and the contact area smaller (50 sq. inch), the pressure from the weight would be 20 - still too small. Even when we get to 30-40 PSI from the car weight, probably the rubber in tire would give in first, before we saw the change in volume.
And finally if we got to this point the difference could be too small to tell using regular gauge.
If you really want to see the difference in the pressure just heat it up - remember the p*V/T is constant.
Good news: The pressure gauge arrived. It?s an Accutire Model MS-5510B. It measures to the nearest 0.1 psi.
I used a 1995 Ford Mustang 2.3L (4 cyl) equipped with P205/65R15 Continental CH95?s.
Results: Loaded / Unloaded
LF ? 29.7 / 29.5
RF ? 28.5 / 28.3
LR ? 27.0 / 26.8
RR ? 29.4 / 29.2
Notice the consistency!
What should happen now is that others should also report their tests to verify the principle.
You guys are hopeless. You gave the right answer several times, without actually identifying it as the right answer. I’m sure this was because you were just speculating about possibilities, without actually knowing anything. The tire pressure does drop when you raise the car on the lift. It just doesn’t drop by enough to register clearly on a cheap pressure gauge. Tires are designed that way so they won’t be too bouncy. As you increase the load on a tire, it mostly deforms, so that the contact patch increases in area. It can’t do this, however, without raising the pressure slightly. Otherwise, there would be no restoring force to bring the tire back to its original shape when the load is removed.
-Jim
CapriRacer’s Ford Mustang weights about 3400 pounds. Each tire carries 850 pounds load. The total internal pressure against the inner sidewalls is approximately 38,000 pounds (32 psi x 1200 square inches). The 850 pounds is 2% of 38,000 pounds. 2% of 29 psi is 0.6 psi. The air pressure change, measured by CapriRacer, is 0.2 psi. The difference of 0.4 psi represents the portion of the load (566 pounds) which is transmitted through the tread and sidewalls to the wheel. Now, why don’t any of the tire engineers respond to the question? Shy?
“…Now, why don’t any of the tire engineers respond to the question? Shy?..”
I’ve been responding!!! And if you look at the number of posts on this thread alone, you’ll convince yourself that I am not shy!
But really, I’ve had to restrain myself. Especially on the thread about what holds the tire up - the air or the tire.
Nevertheless, it is good exercise in trying to frame the discussion so that the other party is convinced. Clear, concise, and convincing - that’s the way it should be.
Oh, and I forgot that a lot of the force coming from the weight goes into the suspension first
Now try the reverse, it will be consistently lower, too.
Each time you use the gauge you let out some air.
Folks, I’ll give another example - if you sat on the gas tank you wouldn’t expect the pressure inside to increase, would you? All the force goes into your butt. Same here, the suspension and the tire rubber gets the absorbs the force, and after that you can only change the volume of the air in the tire if the pressure from the outside is greater that the one inside.
Hi again from the old tired, retired Chem. E. who knows a little about tires.
I’m enjoying the thread. There are a number of people who seem to understand the issues and many more convinced that they do. I’m curious as to how this will play out. Will you guys figure out the right answer from all of this? I really have no doubt (well maybe only a little). Will you get Wolfgang to give us a write-up detailing the answer? That should generate many more posts.
I do have some comments about some of the earlier comments and an extension of the challenge (which should prove helpful in understanding what’s going on).
One comment was that you should have asked engineers rather than physicists. Initially, I thought no, but then again maybe so. Physicists should be able to answer this question quite easily. It does require some thought but physics is a very practical science and mechanics is the heart of physics. However, most physicists are working on much more complex issues and really wouldn’t spend much time on this type of problem. Engineers, on the other hand do a lot of this type of calculation. Force, force vectors and how they are transferred and distributed across areas are very important to us. Otherwise, you get leaning towers (pressure on one side of the structure being greater than the ground can support causing the ground to slowly give way), collapsed structures and other bad things happening. This should be right down the alley for civil engineers (structural) and mechanical engineers.
I have a few comments about the energy contained in the walls of the tires. As you inflate the tires there is a relatively small amount of expansion. It’s small because the tire is reinforced with steel belts and polyester tire cord. It results in only a small amount of energy (ft-lbs) going to the tire. The energy contained in the expansion does not play a significant role.
The total surface area of the air chamber of the tire along with the pressure applied on it and the resultant total force (which is quite large) also does not play a role whether this includes metal rim or not. To balance a force, an equal, opposite force must be applied. For 3000 lbs down, 3000 lbs up must be applied. The tires apply 3000 lbs of force to the ground (over the area of contact). The tire treads apply 3000 lbs up to the air in the tire air chamber. The tires outer wall probably carries about 10 lbs of the weight (they’re designed to retain pressure with minimal expansion and to flex many times without building up too much heat or losing mechanical heat; being rigid is counter to this.)
Pressure and the ability of the tire to deform to distribute the applied force over a sufficient area (pressure), without deforming too much, are the critical issues.
As has been mentioned several times in the thread, the pressure in the tire changes only when a greater PRESSURE is applied to the tire (yes, increased temperature would increase pressure, but it isn’t a factor in this scenario). Just as pumping the tire up with a bicycle pump, the pressure in the tire doesn’t change until the pressure in the bicycle pump exceeds the pressure in the tire. (Air flows from higher pressure to lower pressure). The pressure need only act over a small area.
Now for the extension: What if you added the additional weight to the tires by putting the weight on top of the tires rather than transmitting it through the metal rim? Using the same 3000 lbs (minus the weight of the tires) on the tires with 40 psig pressure what would be the results?
Answer: (1) The bottom of the tire would deform exactly the same as it did before.
(2) The top of the tire would now deform to look like the bottom (it would have slightly less weight - the weight of the tire and rim, but this won’t be detectable by simple observation.
(3) The pressure in the tire will increase almost twice as much as before (but still small).
Why is this the case?
I’ll explain tomorrow.
Best regards,
Cal Hammond
When the tire is supporting 3,000 lbs of weight, the bottom of the tire flattens but other parts of the tire bulge outward thus maintaining a relatively equal volume of space within the tire thus not changing the air pressure reading. However, if enough weight is applied, the tires may expand to the point of bursting thus indicating that the answer discussed above is valid only over a certain relevant range of applied weight.
If CapriRacer, and anyone else with an accurate (to tenths psi, or better) tire pressure gauge, will do some empirical research, it might answer some questions. [ Remember Galileo, when the Church said, "We don’t care if you can prove there are other heavenly bodies. We’re not going to look through your telescope. We KNOW better!]. Raise the subject tire. Let all of the air out of one tire. Set the tire down (load). Take an air pressure measurement, and record it. Raise tire, again. Put 5 psi air pressure into tire. Lower tire and take tire air pressure reading. Continue the 5 psi increments, and readings, up to 35 psi. Post the results back, and we’ll look through your telescope. // Actually, a MATERIALS ENGINEER’S input could be helpful to the discussion.
When I was a lad, learned to ride a bicycle without also learning the graceful “stand on the left pedal, push off with the right foot and only straddle the bike when in motion”. Oh no; I learned to push the bike and leap onto it like the B movie cowboys. This worked until one day, just after filling the tires at the local S… gas station, my mounting leap resulted on one of the tires exploding. So, I don’t think Boyle’s law applies here: at some point the pressure does go up with the applied weight or force.
Actually, the Pope’s astronomers confirmed Galileo’s observations and his predictions. The arguement was more about why they moved that way…blindly following laws of nature or following God’s will. Not that much different with the tire/wheel. Try the tire with no air, then, if you must, try air with no tire. One might suspect that air and tire together hold up car. Peace. jkd
The original question (and conclusion) was, “Why doesn’t tire [air] pressure change when the tire is lifted off the ground?”
So far, theoretical calculations and empirical tests, show that there IS an internal air pressure change. The empirical values don’t match the theoretical values. Why? What is happening on the way between the tire tread and the wheel rim? Where is the force going? HOW is the force transmitted around a semi-circle (the sidewalls) from the tread to the wheel? Where is the flaw in the theoretical values? Simple questions; but, inquiring minds and all that.