Paging All Physicists!

Maintenance/Repairs
21

Dear Bros:

You recently had a question about tire pressure and why it
doesn’t seem to change with the tire on or off the car. I too have
pondered this mystery and concluded the explanation lies in the
wheel and tire’s internal surface area. But how could I calculate
that area?

Since a tire is in the shape of a torus (that’s the doughnut, not
the Ford) I sought the opinion of a local expert but abandoned my
research when I noted his hand moving slowly from his handcuffs
to his Beretta (that’s the pistol, not the Chevy).

Retreating to my garage, I spied the wife’s 2007 Toyota
Camry fitted with 215x60x16 tires and realized I could approximate
the needed data. Knowing there are 25.4 millimeters per inch, and
rounding off any deciamls, I quickly calculated a tire diameter of
26 inches and a width of 8 inches.

Computing the area of the tire’s 26 inch diameter and
deducting the area of the 16 inch wheel, I arrived at 330 square
inches contained within each of the tire’s sidewalls. Then,
multiplying the 26 inch diameter by the 8 inch width and by pi
yielded an additional 653 square inches of area under the tire’s
tread. Finally, deducting an inch of width on each side for the
wheel rim and tire bead’s thicknesses, I multiplied 6 inches times
the wheel’s 16 inch diameter, times pi for an additional 301 square
inches of area on the inside surface of the wheel.

All of this adds up to a total internal surface area of
1,614 square inches. Multiply this by the manufacturer’s
recommended 30 pounds per square inch tire pressure and we see the
wheel/tire combination contains a total pressure of 48,420 pounds.
Wow!! No wonder blowouts can be so loud!

Given a weight of 4000 lbs for the wife’s Camry, we see
that jacking up one corner would affect that 48,420 lbs pressure by
about 1,000 lbs, or around 2% of the total pressure. This would
change the measured pressure by about 0.7 psi which is barely
noticeable on the typical mechanical tire gage. Mystery solved.

I hope this explanation satisfies Amy, her dad and
Wolfgang.

Sincerely,
Todd Smith
Suwanee Ga

22

I tried to post here once but it disappeared, so I’ll just give you the short form.

Forget about belts, tension, bearings and all the rest. It’s all about the internal surface area that contains the pressure. Using the 215x60x16 tire and wheel on my wife’s '07 Camry as an example, I computed a total surface area of 1,614 square inches inside the tire. Multiplied by the 30 pounds per square inch tire pressure we see a total pressure of 48,420 pounds inside the tire. Given a weight of 4000 lbs for the wife’s Camry, jacking up one corner would affect that 48,420 lbs pressure by about 1,000 lbs, or around 2% of the total pressure. This would change the measured pressure by about 0.7 psi which is barely noticeable.

23

It’s simple: P1V1/T1=P2V2/T2. Weight is irrelevant except insofar as it changes the tire’s internal volume. Scary as it is, Click, Clack, and Wolfgang are right!

24

I think the explanation is ridiculously simple. The air pressure doesn’t change because the tire expands imperceptibly from the weight of the car. The sides of the tires bulge out a little bearing the weight. This can be demonstrated with a balloon. Blow it up, tie the end in a knot and put a little weight on it. The internal pressure doesn’t change. The shape of the balloon changes.

25

Finally someone who remembered the old highschool formala PV=RT, where R is a constant. If the temperature does not change, and the volume remains the same through the tire expanding and contracting, the pressure will remain the same. If the tire volume could not change, and a very heavy weight was put on the car, the pressure would go up!.

26

A tire is NOT a baloon. The belting ensures that it can’t expand an appreciable amount after a certain threshold.

27

Here is what i came up with, and sent in an email to whoever it is that got it. They pointed me here…

Tire pressure: the math and the answer…

The question posed on 11/3/07 got me
thinking, so i figured i would do the math
and get a real answer.

Why doesn’t the tire pressure change when the
car is put on a lift?

It does, but not much.

I drive a 2007 subaru impreza wagon. It
weighs just over 3000 lbs, and has four tires
which are about 7" wide and about 25" around.
I run them at 35 PSI.

The load on each tire is, of course, 1/4th of
the total weight of the car, or about 750
lbs. At 35 psi, to support the car, the car
needs 21 square inches of contact patch per
wheel (pressure times area equals force).
This is a patch roughly 3 inches long by 7
inches wide. I went outside and measured the
patch, and it is indeed about 3 inches long.

The length of the contact patch has a name,
in mathematics: It’s called the chord, which
is a line connecting any two points on the
circumference of a circle. The largest chord
of any circle, of course, is the diameter.

Anyway, if you know the diameter of a circle
and a chord length, you can figure out the
area of the section the chord sets aside. In
the case of a cylinder, like a tire, you can
find the volume of that section. In my
vehicles case, the volume of the section set
aside by a 3" chord is about 130 inches
cubed. The volume of the entire tire is about
2114 inches cubed (the volume of the tire
minus the volume of the wheel or rim). 130
inches is a mere 6 percent of the total tire
volume. So, when i put my car down, i would
expect a change in tire pressure of six
percent of 35 psi, or 2.1 PSI.

It doesn’t change nearly that much, of
course. My calculation assumes that the tire
simply collapses in on itself, which is
doesn’t. In fact, the sidewalls will expand
outwards (very obvious when the tire is
extremely low on air) and the top of the
tire will expand out in all directions
slightly. Also, the tire will become out of
round, with the extra volume moving to the
front and rear of the contact patch. If those
three things combined accounted for only 4%
of the volume, the pressure change would drop
to about 0.6 PSI . The friction from the seal
in a normal pen-type pressure gauge is
probably enough to throw your reading by that
much.

Cheers.

As you can see, my results matched your measured change fairly well, so i think i’m right. I’d also like to point out that the pressure MUST go up to some degree, because the belting keeps the tire from expanding any more than it already has. Therefore, the only direction volume can go is down.

Also, the difference will be most noticeable in cars with exceptionally small tires per unit weight. I would imagine that a large truck will have very little change, where a ford festiva might even make the 1 PSI mark.

Also, my math doesn’t take into account the thickness of the tire tread, which will again lower the change in pressure.

28

The pressure in the tire will be constant whether the car is on the road, or lifted off the tires and suspended. The reason is in the ideal gas law with the assumption that the volume of the tires does not change, that air is an ideal gas (which is a very good approximation and means that the molecules are point particles and do not interact with each other), and that the temperature is constant.

PV = nRT is the ideal gas law where n represents the amount of air in the tire (number of moles), R is a constant, T the temperature, V the volume and P the pressure. If n, R, T and V are all constant then P must be constant too and it does not matter if the car is resting on the ground or suspended in the air.

29

I’m not a physicist, but this old tired, retired chemical engineer knows the answer.

Your (and Wolfgangs) answer that the volume doesn’t change begs the question. Why doesn’t the pressure (and thus the volume) change when you add 3000 lbs of force on the tires? Good answer off the top of your heads, but it really requires more thought and detailed analysis.

First, I want you to think about a simple model of what’s going on. This will explain the bulk of the issue then I’ll follow up with the remainder.

Consider a cylinder with a piston with a surface area of 100 sq. in. Add air to the cylinder until the air pressure reaches 40 psig (This requires that there be a retaining ring or the piston flies out of the cylinder;) The force the cylinder exerts on the retaining ring is 4,000 lbs (100 times 40). Add 3,000 lbs of weight to the piston. The piston will support the weight without increasing the pressure inside the cylinder. It doesn’t move! 0 change in pressure inside the cylinder. What changes? The force on the retaining ring drops to 1,000 lbs.

Now for more detail:

The tire works pretty much the same way. You have “base loaded” the tire so that it can support a substantial amount of weight. Now, if the 3000 lbs were added equally over the surface of the tire (which is substantially larger than 100 sq in), you would see no change in shape or pressure. But, the force (weight of the car) ends up acting on one side of the tire.

So, what happens?

Start with the car on the lift; all tires have 40 psig of pressure in them. Since, they are not supporting the car they are round. As the lift is lowered and the tires start to support more and more of the weight of the car, the bottom of the tires start to flatten and the sides bulge out a little. What is happening is that initial contact surface is so small it can’t bear the weight and deforms until the weight can be distributed over a larger area that it can bear.

A rough measurement of the front tires on my Honda Accord (V6) indicated about 50 sq in per tire contacting the ground. Now the force is distributed across the contact area and is transmitted to the inter-wall of the tire where it is balanced against the air pressure. The inter-area would be less than the outer area. So, the 100 sq. in. (for 4 tires) is reasonable estimate of the area that supports the weight of the car (it gets you in the ball park).

Now, there is a slight pressure change (volume change) in the tire. This is NOT due DIRECTLY to weight of the car but is a result of the necessity for the tire to deform to support the weight on the bottom side. This deformation can be seen as the bulge in the side wall at the bottom of the tire. Note, if this bulge were due to pressure increasing in the tire it would be all around the tire not just at the bottom. This bulge is essentially the flatting of the circle of the interior to more of an oval shape. This reduces volume and thus increases pressure. Overall, the change is small and would not be detectable by a normal tire gage.

This is also the beginning of the explanation as to why you can properly adjust the air pressure in a modern tire visually and need to always use a tire pressure gage (requires a little more discussion about how the force is distributed and how the area changes with increased pressure) .

Cal Hammond
BSChE

PS, Most are having trouble with this question because they are think of it as if you loaded a 100 sq. in. piston with 4000 lbs to balance 40 psig in the cylinder. Then an additional 3000 lbs are added to make the total force on the piston 7000 lbs. In this case the pressure inside the cylinder would have to increase to 70 psig which would be very easily measured with a tire gage. Even if you think of the force acting on 1000 sq. in. there would be a 3 psig increase (40,000 + 3,000 = 43,000; divided by 1000 = 43 psig). It’s hard to envision how the force could be distributed over such a large area on an automobile tire.

30

A tire is INDEED a balloon, to the extent that its shape can be changed. If not, it would always retain the same shape regardless of internal pressure or outside forces. After a “certain threshold,” the tire would explode if it could no longer expand, just like a balloon. The reason the pressure does not appreciably change when the tire is loaded: It expands to accomodate the weight of the vehicle, until equilibrium is reached. Jeremy’s answer above is spot on.

31

THIS IS THE CORRECT ANSWER!!

32

For all those who think the pressure doesn’t go up:

http://www.tirerack.com/accessories/airgauges/air13.jsp

This is an air pressure gauge that measures to the nearest 0.1 psi - costs $50.00.

After you bought it, try measuring your tires both loaded and unloaded. That will answer Amy’s - and Tom and Ray’s - and Wolfgang’s - and your questions.

33

I think I know a safer way for anybody to make a measurement. Rather than adding 300,000 pounds of sand in the trunk, reduce tire pressure! While on the car, let the air out of a tire until it is almost but not totally flat with the rim just barely supported off the ground. Measure the air pressure. Then jack up the car. The tire will fully expand, reducing air pressure. Make a new measurement. If you are really ambitious and have that great pressure gauge mentioned above, make several measurements at differnt pressures, and plot on log-log paper. You can then predict the exact delta change in pressure versus total pressure.

34

Nope, it’s not.

The material a baloon is made of can stretch, the material a tire is made of cannot (well, you could nitpick me here - the steel belting will expand slightly, but i think you will agree it will be a negligible amount). Therefore, when the tire is filled, it maintains its maximum possible volume in the absence of a external forces. ANY deformation can ONLY reduce the volume of the tire. As such, the tire does not act like a baloon.

Of course, the volume can be shifted around to a degree, but the change in volume will always be negative, not 0 or positive.

35

sniper1rfa, all substances have some elasticity, even steel. Elasticity of metal is demonstrated in the most basic Physics lab in college. Notice that I used the word “imperceptibly” above. Whether it is a tire, a balloon, or a propane tank, the internal pressure will not noticeably increase until the container reaches its elasticity threshold and stops expanding. My friend, you may not be able to see it with the naked eye, but steel-belted rubber is elastic. Pressure from increased air pushes against this elastic pressure but adding weight to the vehicle doesn’t significantly add to this engagement between air pressure and elastic forces. Admittedly, most balloons burst when they reach their elastacity threshold, but they can be used to demonstrate the principle that I describe.

36

I’m surprised that TOm and Ray couldn’t get this, because one of them has a PhD from MIT.

The Answer is fairly simple, in that under ideal gas laws, pressure won’t change without a change in volume or temperature.

In Freshman chem terms

PV = nRT

Where n = number of moles or avogadro’s constant, and R is the ideal gas constant.

37

Oh my God! I was riding down the interstate and what do I hear but Tom and Ray discussing gas laws… and they got it right! You clearly explained the effect of Boyles’ Law. And your physicist explained the Ideal Gas Law (or the Perfect Gas Law for all your British listeners).

I’m a Nationally Certified High School Chemistry Teacher, I plan to start a unit on gas laws in my Honors Chemistry Classes this Friday, and I almost had a wreck! Please, please, please… may I have your permission to use the audio file of this 10 minute discussion that you had with Amy in my introduction to gas laws now and in the future. I will give a “shameless” plug for your show to all my chemistry students each time I use it!

I agree that the tire was probably not deformed enough to change the volume and therefore the pressure. Tires are constructed with cords inside. The purpose of these cords is to prevent that kind of deformation (many years ago I used to sell and install tires).

Now, I strive to teach my students how to approach problems like those posed on your show in a scientific manner. In the language of high school science education, I teach students how to do the scientific inquiry process. Now, if I were helping one of my students to develop a hypothesis and to test your theory that the weight of the car was not great enough to significantly change the volume of the car tire and, therefore, to change the pressure of the air in the tire, here is what I would suggest. Lower the pressure of the air in the tire in short steps… say 2 psi for each test and measure the pressure in the tire on the rack and on the ground. As the pressure gets lower, the effect of the car’s weight should have a more significant effect on the pressure in the tire and eventually you will be able to measure the change.

38

On the first page of this thread, you’ll find my mathematical model of the situation, underneath a post with some number gathered via real world testing. The pressure does go up, which makes sense.

I had a big ol’ thing type up, with comparisons and stuff, and then i realized something. You’re first statement is wrong. fill a balloon, stick the end over a hose connected to a pressure guage, and squeeze it. The indicated pressure goes up. I’ll bet a cookie on that (seriously, if i’m wrong, i’ll mail you a BOX of cookies). Simple explanation – Before, the only force acting on the balloon was atmospheric pressure, and now it’s atmospheric pressure AND the load. To counteract the new force, the pressure must go up, and the only way for the pressure to go up is for the balloon the get warmer (it will, imperceptibly) and for the volume to go down. According to PV=nRT you could also add more air, but we aren’t creating matter here, and if we are i want a generator hooked up to your tires immediately. :wink:

39

Yes, the pressure goes up as you press on the balloon, but if it is only a small amount of weight (it would have to be really small to be on the same scale with a balloon as a car is to tires), your sensitive $50 gauge won’t be sensitive enough to measure the pressure change. The Balloon would have to get really distorted and near its bursting point before you perceive any noticeable increase in pressure.

40

http://www.airmichelin.com/pdfs/05%20-%20Aircraft%20Tire%20Ratings.pdf
Michelin says that their aircraft tire’s pressure increases 4% when loaded to the max load rating. For a typical passenger car, the max load rating is around 2500 lbs per tire. For an average car w/ 4 tires, you would have to load 5 tons of weight (10,000 lbs) to deflect the pressure by just 1.2 psi. When the tires are just loaded with the car’s own weight (~4000 lbs), the pressure difference is probably too small (0.4 ~ 0.6 psi) to be measured by a cheap pressure gauge.