You guys seem to enjoy physics questions.  How about this?

You inflate your vehicles tires to 30psi just as the manual and door frame label recommend, then you add 1000 lbs of weight in cargo (gravel, firewood, whatever).  You can SEE your tires look different.  When measured, do you think the pressure has changed?

No the shape deforms but the volume doesn't change any or maybe a very small amount.

Not significantly. Temperature is a larger contributor usually.

But more weight = more deformity on road contact portion of tire = less volume = greater pressure.

I looked it up:  pressure is equal to force divided by area.  So if the area changed then yes the pressure would change.  But simply making a tire lay flatter is not changing the volume just the shape.  So the answer for a vehicle tire is no.

This.

What'd your husband say ???

So no matter how much the truck weighs the tire pressure will never increase?

Please show me a free body diagram of this phenomenon.

If I step on a balloon it pops ...

Edit: genuinely curious.

The same amount of air is in the tire BEFORE the Load up as AFTER the load up.

Pressure will remain same all other things being constant.

If the truck drives at high speed, the tire pressure MAY increase slightly from heat generated by the greater flex of the deformed tire (due to heavier load)

I would guess yes.

Similar to squeezing a balloon.

I would imagine there is a point where pushing down on the tire would decrease the volume enough to increase the pressure significantly.

It's probably going to be a lot more than the vehicle itself could physically handle though.

Not what he said...or what i agreed to.
When the tire volume changes, pressure will change.
We're saying tire volume does not change or changes very little.
Your balloon changes volume considerably when you step on it.

It's an optical illusion that's effect can be calculated with the equation:

If eggplant = purple then ÷ by shoe

I would argue any increase in weight will increase the reaction force at the ground - increase the air pressure in the tire - decrease the volume as its compressible ...

There's no "slightly" about it. I inflate the rear tires of my truck to 71 PSI, per the door sticker. When I hitch up my trailer with 1100 lbs of tongue weight, and then I get on the highway, the tires will go up to 82 psi after 30 minutes or so. This is according to the TPMS display on my dash.  Total weight on the rear axle is about 5500 lbs.

And to the OP's question, no, the pressure does not change by simply adding the weight while parked. Again, the TPMS shows 71 psi before hitching IP, and 71 psi after.

Attached File

OH OH OH! I KNOW THIS ONE!

The answer is, of course:
Purple, cuz aliens don’t wear hats.

Heat will cause pressure to increase, but heavier loads do not, it simply displaces the air into a small volume vessel.  If you park your truck in the driveway and do not move it, but add a thousand pounds of ballast you have not increased the overall pressure of the tire, you have simply displaced into a smaller area.

Air expands with heat and contracts with cold.

30 PSI is 30 PSI no matter what happens, 30 PSI can be squeezed into a small vessel, which is exactly what happens when you put a load on a truck and that depends on how much side deflection your particular tire has, but is still has 30 PSI in that tire.

#treadmillthread

But pressure of a gas is related to volume that gas is in (area). If you constrict the volume that a gas is in the pressure will increase. If you increase the volume pressure will decrease.

The only other thing I can think of then is the top of the tire is also deforming so the overall volume is remaining the same.

Gas can compress. Solids and liquids do not. Tires are elastic to some degree. I would wager there is an increase in tire air pressure under load. Whether significant or not I do not know.

But even when the tire squishes, it doesn't change volume. Fill a tire with 100 percent water to standard pressure. It will still squish the same.

Ehhh...and I would argue back, you're confusing increase/decreasing volume with displacement of volume.
Doesn't matter...I've observed no change doing similar to what the OP suggests.

Why do you think the volume doesn't change?

Did you use Armor All on them?

The volume of air, does not increase, what happens is you have decreased the space that volume of air occupies, 30 PSI is 30 PSI.  In a sealed environment, it can exert more or less pressure due to the volume of that sealed environment as it decreases or increases, but the volume of the air will remain the same unless it is heated or cooled.

If you can show how you arrived there I'll allow it

I think it does, but it's a very small amount within the operating range of the tire.

If you put enough weight on it eventually the pressure will go up significantly as the volume would decrease, but you'd need to be so heavy the tire was deformed to reduce the volume significantly.

I'll admit it's been a long time since I've done any of this but that's not like any explanation I've ever heard. The amount of gas is in number of moles and the volume expressed is the volume of the container holding the gas, but the amount of gas isn't important for us in this question.

Boyle's law states pressure is inversely proportional to the volume.

What's important here is P1V1=P2V2 as the mass of the gas (what I think you're calling volume) is constant.

Here's an easy explanation. If pressure didn't increase under load in tires then you could put any object the frame of the vehicle would support on the tires and they wouldn't pop. Figure 87000 bigfeet in your brodozer. Yes the tires pop.

Edit. The tires pop because the pressure of the gas exceeds their structural integrity

PV=NRT

It has been over 40 years since I was in a physics class, but using different terms, it sounds like we are saying just about the same thing..

If enough weight is added to actually compress (I.e reduced the internal volume of) the tire, pressure will increase. Under normal conditions, the tire "squishing" is not changing the volume (significantly at least), it is simply changing the shape. The tire is slightly "shorter" but also slightly "wider".

That's my non-scientist take on it, anyway.

That's the ideal gas law which has a bit of Boyle's law in there.

But yea it also says as volume decreases pressure increases.

ETA: To expand on my previous post and the part of the formula that's relevant.
P1V1=P2V2
P1 is the initial tire pressure
V1 is the initial shape of the tire
P2 is the OP's question
V2 is the shape of the tire after load.

If the tire is squished and V2 decreases P2 must increase so the right side of the equation = the left.

Seems to me, if you sit on a balloon, it's going to pop.  

Look at the surface area of the contact patch of the tire touching the pavement.

Under a hugher load is the contact patch increasing?

By how much?   Can it be calculated what the PSI for the contact patch is in contact with the road?

Does it correlate to unloaded versus loaded?

The amount of pressure in PSI across that surface area likely relates to load carried.

Increase the contact patch under load and you have more surface area supported by the PSI of the tire.

I agree with everyone that says PSI only increasing if volume changes or temperature changes from increased friction.

This reminds me of the famous Abbot and Costello routine.

Who's on first, What's on Second and I don't know is on third!



Of course we could go with the famous Danny Kaye Routine...

What you're saying doesn't make sense to me.  "Volume" IS "the space... air occupies," they are the same thing.  If you decrease "the space something occupies" you are by definition decreasing its volume.

The answer to the question, is that NO, adding weight to the stationary vehicle does NOT increase the tire pressure.  The tire deforms - the shape changes - but the volume does not.  The tire will deform until the material is stressed beyond its elastic limit and fail - the pressure will not change until the tire bursts.  This is true as long as the temperature remains constant.

When you step on a balloon, you are similarly not changing its volume, only its shape.  It too will deform until some part of it is stressed beyond its elastic limit and bursts.  

Here's a neat correlary fact: the total area of your tire contact patches multiplied by the pressure in your tires, equals the weight of your vehicle.  When you increase the weight of your vehicle, you increase the tire contact patch area, not the tire pressure.  For example, a 4000lb vehicle, with all tires at 25psi, will have total contact patch area of 160 square inches, or an average of 40 sq inches per tire... such as a 5"x8" or 4"x10" contact patch.  This, however, assumes an equal weight distribution, which is unlikely - the contact patch will be bigger under the heavier corners and less under the lighter corners.

Below response by perfectsilence has changed my mind a bit - I agree a material can reach its elastic limit before it bursts, and pressure would increase during that time.  Good example with the water bottle.  I still think the answer to the OP's original question, adding 1000lbs, is NO, the pressure doesn't change.  Tires are pretty elastic, they will change shape but not volume until ridiculously overloaded.

Drink water from disposable water bottle. Replace cap. Squeeze in hand. Does the internal pressure increase? Yes, in direct proportion to the decrease in volume of the bottle.

The load itself doesn’t directly change the pressure in the tire. The load will deform the tire slightly, changing the tire’s volume, which in turn affects the air pressure inside the tire. If the tire was completely rigid, the internal pressure would not change.

ETA: Changes in shape will almost certainly change the internal volume of whichever shape is being modified. Else, when squeezing a balloon, or water bottle, or what have you, they would never burst, because no pressure increase would occur (and the pressure could only increase due to a change in container volume, assuming the container remains sealed and the air temperature does not change).

In the case of a tire, the deformation is likely too slight to cause a significant change in pressure.

It is late, I am tired, but we are basically saying the same thing, you just explained it much better than my old brain did!

Excellent point. Expanding on it, take the cap off a full bottle of water and squeeze it. The water sloshing all over your hand is proof the volume of the bottle changed.

And I still think this is true, unless you load the truck well beyond it's rated capacity.

Depends if you have summer air or winter air in them.

Yes

Put 1,000,000 lbs in the back of your truck- what happens?

The tires pop.

Why do they pop?

If there is no change in pressure, the tires will not pop.

Is this truck on a treadmill?

Txl

Lots of people smoking something this evening.

Yes.  Pressure increases.  It is not a linear increase to match the load, as the tire itself will deform as pressure increases.

When you step on a balloon you force all or some of the air into a space that is smaller proportionally to the amount of air you force into it than the space of the entire area of the balloon is to the entire amount of air, thus altering the ratio of force to area enough to produce enough pressure to pop the balloon.

<---------------barely graduated high school, so don't take my word on it.

Volume definitely changes.  Mass doesn’t change.  Density changes.

So many in this thread don’t understand the difference between the 3 and seem to think that volume is actually mass.

This.

NO, I'm specifically talking about volume of the cavity inside the tire.

The tire squishes a bit from weight. So long as you don't go stupid and actually massively squat the tires (which is what you are doing with the balloon analogy), the volume will not change (in any meaningful amount).

Lets say that you fill a tire 3/4 with fluid. Since fluid cannot be compressed (shut up physics nerds, I know), any compression from weight will be exaggerated, evident by a wildly swinging tire pressure from the remaining air compressing.

To make sure that I didn't exceed PSI specs on my tractor tires after I filled them 3/4 with fluid and max air pressure, I checked air pressure with the 3 point fully loaded.

Guess what: PSI never changed.

I've also taken semi-truck tires that are just mounted, filled to 100psi off-truck, installed on loaded trucks. Guess what: Still 100PSI.

As you can see from this graph, the HP required to compress air is not flat, or even linear. The air pushes right back. In the PSI ranges of most tires, it takes a lot more pressure to further compress the air than you get from the weight, hence the air holds the weight, not the tire. The air is exerting CONSTANT FORCE on the tire, and you will only go flat (collapsing the tire, thus reducing volume available for the air, compressing the air a bit), when you exceed the lifting capability of the air in the tire. You have 2 ways to lift a lot of weight with compressed air: high PSI or low PSI and high volume. Its easy to think about, because PSI is literally a measurement of force Pounds per Square Inch. So its measuring force over a given area. 100 PSI in a 1floz container doesn't contain much umph, while in a tire has a lot of potential energy (force).


Also, you can over-load the shit out of a tire, and have it sitting on the rims. The failure from over-squatting is how much the tire heats up from flexing that much over and over, or from an old/weak tire that just can't flex that much to begin with.

ETA: To think of things another way: as you push on the air inside the tire, the PSI of the air pushes back. Did you know that a tire mounted in a single wheel (1 wheel per side of axle) has a GREATER weight rating than when its mounted in on a dually? That is because as you push on the tire (air) with weight, the air changes the shape of the tire (because it doesn't want to compress, like in your air compressor tank where it has nowhere to go), and bulges at the sidewalls (weakest point of the tire, rims aren't going to allow compression that way). On a dually, if the bulges meet, boom. So the volume isn't changing (in any meaningful amount), its just changing the SHAPE of the tire, and it does that at the side-walls because that is where the tire is weakest.

Back when I was working for a different company, were were installing a ballasted roof on a school.  The roof was almost complete - we only had about 17-ish more tons of gravel on order at the pit.  They sent this total numbskull of a truck driver to go pick up some gravel and take it to the jobsite.  His single-axle dump truck could only hold about five tons of weight, so when he got to the gravel pit and they asked him how much of the remaining rock he wanted, he just said, "Hell, give me all of it."

He made it about a half mile before all six tires exploded and he was left sitting by the roadside on his rims.

So yes, if you put a shit-ton of weight on a truck, it DOES change the air pressure in the tires.  The pressure decreases to zero.

The answer is Yes.

How?

PV=nRT