Pressure and Car Tires

Doing a lot of driving this week, and since the interstates I'm driving (I-49 and I-10) are deathly boring, I have a lot of time to think. Being a physics nerd, I do a lot of thinking about physics. This trip the physics thing I'm thinking about is tire pressure.

i-5b33d761ca922ec1bc516432280b3a90-car.pngFig 1. A car. From the Wikipedia article "car". Research!

Tire pressure made a brief blip during the news a while back when it as a contributor to engine efficiency made a side appearance in the presidential election in regard to global warming. Climate gets done to death around here though, so let's talk about pressure as what it really is fundamentally: force per area. Well, in thermodynamics it's more useful and more interesting to think of it as the derivative of energy with respect to volume, but we'll save that for later.

Stand bare-footed on a piece of styrofoam and it may well hold you up if it's reasonably strong or if you're thin. Put on some high heels (guy readers may wish to get some privacy first) and do the same thing, and you'll punch a hole right through it. In both cases you've applied the same quantity of force. The difference is that instead of being spread over the whole area of the sole of your foot, it's concentrated into the much smaller area of the spike. That what pressure is - force divided by area.

Your car tires have to hold up your car, and they're typically inflated to something like 30 pounds per square inch? A typical car might weigh, I dunno, a thousand pounds. Probably considerably less if it's a small car, but that's way out of the area of my expertise. Now we'd like to recover the area of tire required to hold up the car. Pressure is force per area, so to get area we need force divided by pressure. Pretending the car is a thousand pounds, that's a total of around 33 square inches worth of rubber in contact with the ground holding up your car, providing every but of the acceleration and steering you rely on. 33 square inches is a square around 6 inches, so it's not much.

And that's why tire care is important. Because there's not a whole lot of it between you and sliding into a tree.

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While it doesn't affect your point much, even small cars like a Honda Civic are closer to 2500 lb.

No, pounds are a unit of force, not mass. Those conversion tables of pounds to kilograms make the parochial assumption of a 1g gravitational field ... On the moon, you would only weigh <30 pounds rather than your Earth weight of 175, but you still mass 75 kg in either location.

By Scott Simmons (not verified) on 15 Jan 2009 #permalink

Andy: Pounds are a unit of force. The unit of mass in that system is the slug.

Thinking of how much tire area is in contact with the road makes it obvious why fuel economy improves with a higher tire pressure. Less tire in contact with the road means less frictional resistance. Of course, you don't want to overdo it, because without frictional resistance you would go flying off the road at the next curve, or sliding through the intersection at the next red light--especially in conditions where friction is low to begin with, such as light rain on top of built-up oil (Southern California in November/December) or compact snow/ice on the roadway (most of the northern half of the country at various times during the winter).

By Eric Lund (not verified) on 15 Jan 2009 #permalink

Apologies. I did not know that.

If I might be permitted to season my humble pie, to make it taste better. There is a unit of mass called the pound (sometimes pound-mass) and a unit of force called the pound (sometimes called pound-force). This does rather beg for an error of a factor of g.

Of course mass is only proportional to gravitational force in a uniform field...

Less tire in contact with the road means less frictional resistance.

Not to a first approximation. (Coefficient of friction, anyone?)

However, higher pressures do mean that the tire deforms less as it rolls and the less it deforms the less energy is absorbed by the solid parts of the tire, which among other things reduces the heating of the tire. Cooler tires are less "sticky" as well, so there is less energy absorbed by the tread as it makes and breaks contact with the road (also related to deformation, see above.)

Curiously, I'm having tire-pressure issues myself lately when driving from Phoenix (temps in the mid-20C range) to the White Mountains, where the temps are often below 0C. That's as much as a 10% change in temperature, and it results in a 10% change in tire pressure -- which is enough to set off the low-pressure warning.

By D. C. Sessions (not verified) on 15 Jan 2009 #permalink

Mass is not proportional to gravitational force. The mass of an object is defined and doesn't change unless it loses something.

Tire "care" isn't the real problem apart from tire pressure. If you have crappy tires they will not work well even if you take care of them.

Chris P

Pound-mass is sort of a made up unit of convenience used mostly because nobody has a frame of reference for how much mass a slug represents. (~.031 slugs on the Earth's surface weighs 1 pound, or .031 slugs = 1 lbm)

In any of my engineering classes in college (interesting that none of my physics instructors would ever use the unit pound-mass) I would always convert anything figures supplied in pound-masses into slugs right away, it was too easy to screw up the math along the way if you leave them in pound-masses.

By Karl Withaky (not verified) on 15 Jan 2009 #permalink

I once saw an educational video where the four tires of a car were inflated to equal pressure, then the car was rolled onto sheets of typing paper. The kids then traced the footprint of each tire. After retrieving the 'prints', they estimated the areas, then combined them to calculate the weight of the car from the pressure value.

You could also work the other way and estimate how much a car weighs based on the tire area in contact with the road and the pressure in the tires: 4 tires x 6in x 6in x 30psi ~ 4300 lbs

A few people have mentioned this, but I believe the way it works is that "pounds" can be taken as a unit of force or mass depending on the convention. If it's force, I think it's the "slug" that's the unit of mass. If it's mass, I think it's "pound-force" that's obviously force.

Which is why science pretty much sticks to metric these days, even though I still think in imperial in my daily life.

In this entry just assume that "pounds" is a unit of force, corresponding to the reading on a usual scale.

Okay, let's get really picky and pedantic here. Because we're physicists. Pressure can apply a force in all directions, like a compressed gas in a scuba tank. It's a scalar, as you would expect from dE/dV. But when you're standing on styrofoam in high heels, the force is downwards only - You've got a vector acting in one direction. Thus, the force/area in this case is really a _stress_, not a pressure. It's about the only case I can think of where even the best Ph.D.s routinely get sloppy with the anal-compulsively proper terminology.

Please forgive a question (naive) , from someone who likes science but is not a physicist. I don't understand why tire pressure does not change whether it is off the car, or on the car? If the tire is supporting the car, why doesn't the tire pressure go a lot!

Yo! Physicists, wake up!

In the customary US system (inherited from the Brits), the pound is both a unit of mass (lbm) and a unit of force (lbf): F = ma/gc, where gc is 32.174 lbm ft/sec2 lbf. This is called the lbm-lbf-ft-sec system of units and is the most common system of units in use in the US. This is why the dumbshits as NASA crashed two (TWO) half billion dollar satellites (plus launch costs) onto Mars. The dumbshits must have studied physics at an Ivy League school.

Lest you think only American engineers do this, in Europe a very common system of units is the kgm-kgf-m-sec system: gc = 9.8066 kgm m/sec2 m.

By the way, chemists, biologists, geologists and physicians use the cgs system because it is more convenient than the mks or SI systems. Only the dumbshits working in electromagnetism need the SI system.

There are many other systems out there, too: slug-lbf-ft-sec, lbm-poundal-ft-sec, etc. Because our civilization has history and our predecessors actually built buildings and made things that are still around, a practicing engineer confronts all of them and needs to suck it up and deal with them. Only Ivory Tower dumbshits can live in la la land.

By Bob Sykes (not verified) on 16 Jan 2009 #permalink

I don't understand why tire pressure does not change whether it is off the car, or on the car?

Because the amount of air in the tire doesn't change (assuming the tire doesn't leak and that there has been no attempt to inflate/deflate the tire), and the volume that the air occupies doesn't change much. At typical temperatures and pressures we can treat air as an ideal gas, which means that the quantity PV/NT is a constant (here P is absolute pressure, V is volume, N is the amount of air, and T is absolute temperature). Since V, N, and T don't change (at least not very much), P does not change (at least not very much).

If you're a physicist, N refers to the number of molecules, and the constant of proportionality is Boltzmann's constant, which is about 1.38*10-23 J/K (kelvin is the unit of absolute temperature, the Kelvin scale is the Celsius scale with the zero point shifted to absolute zero). If you're a chemist, N is the number of moles of gas (for a pure substance, a mole is the amount of atoms/molecules needed to make the mass of the sample in grams equal to the atomic weight, in units where 12C is defined to be exactly 12, of the atom or molecule in question), and the constant is the ideal gas constant R = 8.314 J/(mol K).

By Eric Lund (not verified) on 16 Jan 2009 #permalink

@14 Bob Sykes: Wow, Bob, bitter much? We get it, you didn't get into a good university -- get over it and move on with your life.

Thanks Eric, for writing a very helpful theoretical explanation. I understand the math, but the problem still does not make intuitive sense to me. Is the problem similar to a balloon? If i rest a book on the balloon, will not the air pressure inside the balloon go up? What if i pile up another and another book on the balloon? Won't the balloon eventually burst? If I pile heavy weights on a car, is it not possible that the tires could burst?


Frank: The difference between piling books on top of a balloon and putting the tire on the car is the effect on the surface in which the volume of air is contained. Most balloons form an approximately spherical shape because the sphere maximizes the volume for a given surface area. To start off with another shape you need to design the balloon appropriately; inner tubes are designed to be toroidal because their function--to be placed inside a tire that must fit around a wheel/axle assembly--requires that shape. When you put enough weight on the balloon, it will change its shape significantly. Most likely, the surface of the balloon stretches to try to maintain a constant volume, until it is stretched so thin that the material cannot maintain its integrity and is torn apart. (It's also possible that the volume is reduced if the material cannot stretch quickly enough, but my intuition is that this is a secondary effect.) The tire deforms, too, but the walls of your typical inner tube are thicker than the balloon's and are therefore better able to handle the effects of deformation, which is probably less (for a given weight) for a toroid than for a sphere. Putting bicycle tires on an SUV could well lead to bursting inner tubes for the same reason the balloon bursts under the weight of the books; this is why getting the proper size tire for your vehicle is important.

By Eric Lund (not verified) on 16 Jan 2009 #permalink

Excellent explanation, Eric. Thanks.

First, as others noted, you really need to know some order of magnitude values better than a factor of 2. There are few cars with a mass less than a metric ton.

Second, Andy was right about lb being a unit of mass in the avoirdupois system of weights and measures that is in common use in the US, but wrong about its relevance to psi. Eric and others (including Matt) are wrong about the pound, but this is understandable given that the error has been passed on as gospel in physics books since the middle of the last century.

Pressure in psi is defined as lbf/in^2, which has a conversion to Pa via the definition of the lbf (using the lb mass and the internationally defined value of g) and the inch.

When used on commercial products, the pound is a unit of mass in the avoirdupois system. It would a different value if it was used to refer to precious metals, which use the troy system, except they normally use troy ounces. For grins, notice that the troy ounce is bigger than the avoirdupois ounce, but the troy pound is smaller than the avoirdupois pound.

I'd recommend looking in an engineering book if you want to know what is actually used, since physicists never actually use pounds for anything. What you tend to find in engineering is lb for the mass, and either lbf, kip, or poundal for force. The force of gravity on 1 lb is 1 lbf (as someone showed above, shenanigans are required in the calculation of F=mg if anyone really bothered to do it) or 32 poundals. The force of gravity on 1000 lb of mass is 1 kip, which seems to be the most widely used unit in civil engineering based on what former students are being taught to do and the NTSB analysis of the Minnesota bridge collapse.

Memo to Bob Sykes: I have a torque wrench that is calibrated in ft-lbf and m-kgf. Shudder.

By CCPhysicist (not verified) on 17 Jan 2009 #permalink