# Under (tire) Pressure

If you live on flat terrain like I do, you might not get a chance to experiment with your car coasting down hills in neutral. It's kind of dangerous even if you can. But let's say you're on the top of your driveway and beginning from a stop you coast down to the street below. If the total drop is 10 feet, that gravitational potential energy will be converted to kinetic energy and you'll end up going something like 17 miles per hour. To do the calculation yourself, just set the potential energy at the top equal to the kinetic energy at the bottom, and solve for the velocity:

The mass cancels; it doesn't matter how heavy the falling thing is, its speed is completely determined by how far it falls.

But your car doesn't go that fast when it gets to the bottom of the hill. It probably isn't even very close. I've done the experiment myself and my car isn't anywhere near the predicted speed. Either physics is wrong or we're doing something wrong. Well, I won't keep you in suspense. It's not the physics.

Not all the energy goes from potential to kinetic. Some of it will be lost to friction - both in the tires, axle, and air resistance. But especially at the low speeds of coasting from a standing start, friction should be very small. Where else might the energy have gone? Well, entirely apart from the forward motion of the car the wheels are also spinning. If a giant picked the coasting car straight up, the wheels would keep spinning. That spinning is mass in motion, so it has kinetic energy. So actually, our expression which turns potential energy into kinetic energy needs another term:

That last term is the rotational kinetic energy. The letter I denotes the moment of inertia, which is a way to quantify the way mass is distributed in different shapes - in this case a tire. The Greek letter omega is the angular velocity in radians per second. One radian per second is 1/(2π) times the number of revolutions per second. Now we want to figre out what the moment of inertia of our tires are. This isn't possible yet because we haven't related the linear velocity v to the angular velocity omega. Fortunately there's an easy way to relate the two, straight out of what you learned in trig.

Where r is in this case just the radius of the tire. Plus this into the equation and you'll get a complete formula that accurately relates the various energies involved in this situation:

Now solve this for I, and you can actually experimentally calculate the rotational inertia of your tires.

Assuming I haven't screwed up the algebra, plugging in your starting height and your final velocity will give you the moment of inertia of your times. Sadly the equation is not linear, so if you don't start from a standstill you'll have to redo this derivation in a somewhat more tedious manner.

Now if we're looking at a situation where we have very lightweight tires (or heck, we're a stone falling from the sky), I will equal zero. This is only possible if the two terms in parentheses are equal - but that's exactly the situation described in the very first equation. Things are perfectly consistent, as expected. Neat, huh?

If you like, try the experiment and calculate the moment of inertia of your tires. Given the informal nature of the setup I've used English units above, but you'll definitely want to convert everything to metric to make the math easier if you actually want to try it in real life. If you can do so safely, give it a shot!

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A (tire + rim)'s mass distribution is thus important. Large mass at large radius eats mileage every time you brake. SUV's and asynchronous traffic lights are the mother's milk of traffic politics. "We must regulate you (with fat user fees) because you are wastrel and polluting."

When a truck tire fails during use the shrapnel is amazingly energetic.

Friction, accuracy of measurement and a load of other things will mean that this is a waste of time. Given the distribution of mass in the tire just taking the outer diameter minus 1 or 2 inches using that to calculate the effective diameter will give you a real answer.

At low speeds rolling resistance is EVERYTHING. Friction is the main contributor to drag until you get well above 10 miles an hour.

This blogger goes astray when he starts trying to talk about the reality of physics. Too many numbers - too little reality.

Having designed and modelled bicycle and solar car dynamics this blog is quite uselessly wrong.

Truck tire shrapnel is due to the high air pressure - usually over 100 psi.

Chris P

@Chris: That's the difference between an engineer in and physicist in my mind. All of the engineers I've met can't solve a basic general relativity problem. While there is something to be said for increased accuracy by factoring in friction and what not, the public doesn't really care. They'll find relativity or quantum mechanics more interesting than friction. Shoot, one can even go about building stuff like a solar car without going through all the work of dealing with friction, I know because a couple of my friends have.

By Erik Remkus (not verified) on 11 Aug 2008 #permalink

Chris P, it seems to me that your pretension has gotten the better of you because you've quite missed the point. This is a blog. It's for popularization. Its aim is to get people thinking about physics, and it fulfills that purpose quite well. It may be wrong, but it is not "uselessly" so.

I suggest that you stop wearing your ass as a hat. Or maybe find another group of anonymous people on the internet to try to convince of your genius.

At low speeds rolling resistance is EVERYTHING. Friction is the main contributor to drag until you get well above 10 miles an hour.

This blogger goes astray when he starts trying to talk about the reality of physics. Too many numbers - too little reality.

Numbers are reality, sport.

As it happens, I have had the opportunity to push stalled cars on numerous occasions, and it's an experimental fact that the overwhelmingly dominant factor in how hard it is to push the car is the slope of the incline it's sitting on. While the rolling friction is the dominant frictional effect, friction is here a small force compared that of gravity.

But in fact nothing in the world stops the interested reader from modifying the experiment to measure the distance a car takes to roll to a stop on flat ground from some given speed. You can find the coefficient of rolling friction. That way you can quantify friction and add a μmg sin(θ) term to the third equation for greater acracy.

By Matt Springer (not verified) on 11 Aug 2008 #permalink

So this is a blog where the physics is made up - then why is it called "Built on facts". Fact is rolling friction is predominant at less than 10 mph.

As an engineer I have no need to solve a relativity problem - if I did I'd talk to my astrophysics major daughter.

You can "build" a solar car without worrying about friction but it won't win anything. A solar car is all about friction - rolling and aerodynamic.

A blog that popularizes BS is dishonest. Either get the physics right or don't pretend to know what your are talkiing about.

Assuming the wheels represent at most 10% of the mass of the car, you find less than a 5% change in the velocity of the car at the bottom of the hill when including the rotation of the wheels. In the example, that would mean the car would still be going 16+ mph. Friction is definitely going to be a much larger effect.

Of course, we could always assume a frictionless axle and engine (make sure to apply liberal amounts of WD-40 before doing this experiment).

So this is a blog where the physics is made up - then why is it called "Built on facts". Fact is rolling friction is predominant at less than 10 mph.

Ok, here's Physics 101, day 1. Acceleration is the net sum of forces, divided by the mass. On a car rolling down the hill, there's two forces: gravity and friction. Gravity pulling the car down the slope, friction pulling it the other way.

The car starts rolling down the slope. Guess which force must be stronger?

By Matt Springer (not verified) on 11 Aug 2008 #permalink

Assuming the wheels represent at most 10% of the mass of the car, you find less than a 5% change in the velocity of the car at the bottom of the hill when including the rotation of the wheels. In the example, that would mean the car would still be going 16+ mph. Friction is definitely going to be a much larger effect.

The mass of the tire is of less interest than the moment of inertia - low-mass objects can still have high moments of inertia depending on their configuration. I'm not at all convinced friction will dominate that effect, as the slowing effect seems to be undiminished even on very high slopes in my experience.

But hey, as my snarky friend Chris says above, the facts are the issue. I proposed a method of measuring friction a few comments up, and an interested person could certainly calculate both using that method.

By Matt Springer (not verified) on 11 Aug 2008 #permalink

So now even Matt Springer is into name calling to win a facts debate. Saying a person is "snarky" when you base it on one post isn't going to get you far. You win engineering debates by facts.

We seem to be continuing in the use of myths. WD40 is NOT a lubricant. WD40 is a water displacer. It's lubricating properties extend to the point of lubricating the cutting tool when machining aluminum because it contains Stoddard's solvent. You'll notice it doesn't say that on the can.

Matt if you are going to say "rolling down the hill" then why not be smart and say that the hill has to be such that its angle has to exceed the friction angle before it starts to roll. Any hill will not do.

And no I will not comment on relativity posts.

Chris, if you're calling Matt's basic physics post here BS, you very clearly have not experienced much BS on the Internet.

And the "snarky" comment was clearly deserved. At least to this casual observer.

Snarly is an adjective, one which applies to most of us at one time or another.

Anyway, I don't know exactly how friction and rotational inertia will compare in terms of their magnitudes. I suspect that at low speeds inertia will not be insignificant. But this is as you say largely an engineering problem and one which can be checked.

Set the car rolling slowly on a flat surface and calculate μ. Then repeat the experiment I suggest in the post proper and add in the frictional term. Then you'll have data for both the friction and the inertia, which will in either case be more accurate than just one of the effects by itself. (It's a μmg cos(θ) term, I made a typo a few comments above.)

By Matt Springer (not verified) on 11 Aug 2008 #permalink

The mass of the tire is important because the largest moment of inertia for a circular object is M r^2. Then, at worst, the velocity at the bottom will be reduced by a factor of SQRT[m/(m+M)], where M is the mass of the tires and m is the total mass of the car.

Minor correction, though: the formula you give for friction is for the wheel sliding against the ground, which is not the friction that comes into play. The important friction comes from the wheel assembly and engine (i.e. a resistance to turning). Model that as a constant torque and this becomes an interesting rotational dynamics problem. As in, what internal resistance/torque is necessary to keep the car from rolling down an incline of angle theta?

If you thought I was serious in applying WD-40 to a car axle so that you can do this experiment, then I have a bridge to sell you. Cheap. (and in a prime location!)

Ah but Matt, you weren't using forces, you were using ENERGY. You need to factor in the work done by friction, which depends on the length of the driveway (and thus indirectly on the slope if related to the work done by gravity). The issue was never whether it *would* roll down the hill, it was what the speed would be at the bottom of the driveway.

The problem with your proposed method of measuring I for the tire is that it will be extremely noisy. You are using the small difference between two large numbers (a huge problem unless you measure h and v to more sig figs than you likely planned to get) in addition to ignoring other rotating mass (the drivetrain, including parts of the transmission) as well as rolling friction. Remember, uncertainties add even when you subtract.

If you really want a reliable answer, take the tire off the car and do the much simpler measurement of its radius of gyration (the effective radius Chris P refers to). Just remember that a rolling car tire and wheel can easily break your arm or leg. Don't try to stop it.

Your proposed way of measuring the rolling frictional work was OK right up until you made a rather glaring error putting the sine of theta into it. That car would never stop on a level surface. Think before you post.

And your last comment is simply nonsense. The mass of the tire is very important to the size of the effect. The point CS made (and Uncle Al referred to) was the familiar one about minimizing what practitioners call the "rotating mass". Call the rotating mass M and the radius of gyration R, and you get that the rotational inertia looks like f*(1/2)mv^2 where f = (M/m)(R/r)^2 is [almost but not quite] the fraction of energy that ends up in rotation. In the worst case (R=r for a hoop) it is just M/m, decreasing to half of that if R=0.707*r (as it would be if it acted like a solid disk). The small size of the rotating mass sets a limit on the size of the effect.

Aside:
The tradeoff between speed and safety is the reason the NHRA will not let racers touch the wheels or tires on T/F and FC cars. They don't want the wheel shaved to reduce the rotating mass at the risk of it coming apart and killing spectators. The same risk from doing the same thing to what passes for a flywheel in the clutch system is guarded against by safety structures teams cannot tamper with.

Jon wrote:-

And the "snarky" comment was clearly deserved. At least to this casual observer.

Fine, but at least you didn't include your last name so I or anybody else can't hold it against you in a job interview.

I thought getting your facts wrong was much worse that pointing it out - particularly when the blog invokes the word "facts".

Most of the rolling friction for tires comes from deforming due to the weight of the car, so an mgcos(theta) effect would be expected. I would expect axle loads on the bearings are next in magnitude. Answer from any engineers?

That deformation is reduced with high tire pressure, and increased (dramatically) at low pressure. That is the reason all high mpg runs are done with highly inflated tires (and why my wife's Honda Insight has relatively narrow, high pressure tires). The deformation will be minimal if you take the tire off of the car, hence my proposed measurement of I.

PS - This is also the reason everyone but John McCain knows that low tire pressure is a major problem, and (problem)^2 if you are in high c.o.g vehicle like an SUV where it also increases the risk of death due to a rollover. Let's leave deliberate ignorance and ad hominem attacks to the politicians and stick to physics here. My advice to you, Matt, is that the future engineers in your classroom will know more than you about real things within 12 months of leaving your classroom. Listen and learn.

By CCPhysicist (not verified) on 11 Aug 2008 #permalink

No, no, no! I am using ideal, non-deformable tires! ;P

Stupid reality. Why does it have to go and make everything complicated?

CCPhysicist

Yes the next contributor (assuming the engine isn't in the loop) is from the wheel bearings.

One of the interesting things is that while rails are often touted as having a lower rolling resistance than roads, when it comes to lightly loaded vehicles the picture is less clear. It's hard to get real data on the rolling coefficients of wheels on steel rails.

So far the human powered vehicles have a higher top speed (81 mph) on road than on rail.

Rail was supposed to give the ultimate solution - although the same paper did do the calculations for riding on the Moon.

Your proposed way of measuring the rolling frictional work was OK right up until you made a rather glaring error putting the sine of theta into it. That car would never stop on a level surface. Think before you post.

And your last comment is simply nonsense. The mass of the tire is very important to the size of the effect. The point CS made (and Uncle Al referred to) was the familiar one about minimizing what practitioners call the "rotating mass". Call the rotating mass M and the radius of gyration R, and you get that the rotational inertia looks like f*(1/2)mv^2 where f = (M/m)(R/r)^2 is [almost but not quite] the fraction of energy that ends up in rotation. In the worst case (R=r for a hoop) it is just M/m, decreasing to half of that if R=0.707*r (as it would be if it acted like a solid disk). The small size of the rotating mass sets a limit on the size of the effect.

A few comments down, you overlooked that I mentioned that the sine term was a typo. It should be a cosine term, of course!

What I was going for with the mass comment was that the mass by itself doesn't make a difference - it's the mass along with the way it's distributed. As you rightly point out, the radius of the tire sets an upper limit on this effect.

Now if I can summarize, the controversy here in the comments seems to be on the relative sizes of the effects due to the friction (of the various varieties) and the moment of inertia of the tires. I thought the friction was smaller than it apparently it is according to several of you. While I'm perfectly willing to defer to the engineers on that practical problem, I wold still need a little more data to convince me that that the friction completely dominates even at low speeds and high road slopes. But the argument that CS brings up about the constant torque is pretty convincing.

On a number of occasions, I've entered uphill onramps at speed with my car in neutral. The car seems to slow significantly less quickly than it ought. Friction can't be responsible for this effect, since it's pointing in the wrong direction to keep the car going.

There's two immediate possibilities that spring to mind - first, the rotational inertia of the tires. Second, my judgment that the car is not slowing quickly enough may simply be an illusion of my mind.

In the absence of any empirical data however, I'll defer to the engineers and I thank you, Chris, etc for the advice and corrections despite any, um, snark. ;) I'm very appreciative of the level of knowledge that you all bring.

It should be easy enough to calculate. Let's see, my Miata tires weigh 20 pounds each, total of 80 pounds. The car weighs 2200 pounds so the wheels are around 3.7% of the car weight.

Assume all the wheel weight is on the rim (not so bad an approximation, probably; there are other things on my car that turn when I roll down a hill than the wheels, i.e. parts of the differential and the axles, etc.). My intuition says that the KE of the wheel will be twice what it should be (and I'm not willing to work it out so please correct), and the overall effect will be a 3.7% increase in KE at a given speed; a 1.8% decrease in speed.

On the other hand, how do I estimate rolling friction on a car? Is it going to be proportional to velocity? My guess is that one looks for a slope where one's car just barely rolls down at some given speed, and the speed and slope determine the rate of loss to friction.

For the slide down that slope, of course the KE stays constant and so does not contribute. At that velocity, one requires a steeper slope for the car to put energy into rotational KE.

By Carl Brannen (not verified) on 11 Aug 2008 #permalink

I don't have my car design books here at work to give you the numbers but the rolling resistance is not simple. There are, I believe, at least three coefficients going up to the square of velocity.

For the inertia of a car wheel things are more complicated than bicycles. In general for bicycle you just double the wheel weight because it is mostly concentrated at the rim.

Simple to visualize because the whole bike is moving forward at the required velocity but you also have to provide energy to rotate the rims to that same velocity.

Chris P

Apropos your comment at 6:33 PM, Matt, there were no replies from you when I started composing the comment that appeared at 3:37, only the ones through 3:09. I could probably type fast enough to write that entire screed between 3:23 and 3:37, but I read it several times and tweaked it at least twice before hitting "Post".

I think the biggest issue is the first one I raised. It will be very hard to see the effect of rotating mass in comparison to friction forces in the proposed experiment if you don't do better than a car's uncalibrated digital speedometer to measure speed. I'd start by measuring I (or equivalently R) and use that to design the experiment.

PS - How do you know how quickly your car should slow down? At what speed? Automatic or manual? What is the coefficient of drag? (To Carl: top up or down, headlights up or down. I've got one too.)

I would measure the drag horsepower (there is a coast-down way to measure that) as a function of speed. Drag is very complicated and the energy losses to "friction" vary by huge factors with speed. BTW, a measurement of drag horsepower also addresses Uncle Al's speculation a few weeks ago about speed and efficiency. I know our Insight turned 94 mpg at a steady 35 mph, but only 63 mpg at 55 mph (with AC running) under otherwise optimal conditions.

By CCPhysicist (not verified) on 11 Aug 2008 #permalink

OK - the approximate Crr for radial tires at low speeds is .015. The reason for doubts about your experiment are based on history. Coast down testing of bicycles has been going on for many years and has got to the point of being very well instrumented.

Results to this day, attempting to measure air drag and rolling resistance, have seen huge variances. Rolling resistance depends not only on the tires but also the road surface. Wind and road slope have variability that greatly affects the results. Given all these problems it would be quite inadvisable to suggest people could detect wheel inertia effects.

For those who might want to measure the rolling resistance and drag coefficient and have a flat level road with no wind at their disposal the formulas are given on page 325 of the Bosch Automotive Handbook Vol3. As the bicyclists have found - the road has to be really level.

Wow, this is funny. Physics trolls. I had no idea. I don't get those on my blog (but I do get Linux trolls, which is similar)