Introduction: Measure the Drag Coefficient of Your Car

Picture of Measure the Drag Coefficient of Your Car

The purpose of this experiment is to determine your vehicle's drag coefficient Cd and coefficient of rolling resistance Crr. This is done by measuring your vehicle's speed as a function of time while coasting in neutral.

Why would you want to know Cd and Crr for your vehicle? Well, suppose you're interested in modifying your vehicle for improved fuel efficiency. You might consider modifications such as air dams, wheel skirts, removing mirrors, switching to low rolling resistance tires, etc. Cd and Crr offer a quantitative method of comparing vehicle performance before and after these types of modifications to see if you made any improvement.

For other experiments you can do on your car see my website

Step 1: Equipment

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You will need the following equipment:
  • a vehicle (and someone with a driver's license)
  • a clock or stopwatch
  • a pen and paper (and someone other than the driver to record data)
  • a flashlight (driving at night avoids traffic)
  • a long stretch of flat road with little traffic or wind
  • Excel or another spreadsheet application. I prefer OpenOffice Calc because I like to support open source software, but its Solver function does not handle non-linear systems (yet) so you'll have adjust input variables manually by an iterative process to fit your model to the data (it's not too hard).
  • The spreadsheet I created to analyze the results. Download here: Drag_Coefficient.xls

Step 2: Background Information

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First, let's define some quantities:

   Fd is the force on the vehicle due to air resistance (drag) in Newtons
   Frr is the force on the vehicle due to rolling resistance in Newtons
   F is the total force on the vehicle in Newtons
   V is the vehicle's velocity in m/s
   a is the vehicle's acceleration in m/s^2
   A is vehicle frontal area in m^2
   M is vehicle mass including occupants in kg
   rho is the density of air which is 1.22 kg/m^3 at sea level
   g is the gravitational acceleration constant which is 9.81 m/s^2
   Cd is the vehicle's drag coefficient we want to determine
   Crr is the vehicle's coefficient of rolling resistance we want to determine

Now for some formulas:

   Fd = -Cd*A*0.5*rho*V^2 (formula for force due to air resistance or drag)
   Frr = -Crr*M*g (formula for force due to rolling resistance)
   F = Fd + Frr (total force is the sum of Fd and Frr)
   F = M*a (Newton's second law)

Note that both Fd and Frr are negative indicating that these forces act opposite to the direction of the velocity. Note also that Fd is increases as the square of velocity. This is why driving at high speeds is much less efficient than driving at low speeds. Combining these formulas with a bit of algebra gives us the acceleration due to air and wind resistance as a function of velocity:

   a = -(Cd*A*0.5*rho*V^2)/M - Crr*g

Note that the acceleration is negative indicating that air and wind resistance will cause the velocity to decrease.

I created a spreadsheet based on these formulas to generate a model of velocity vs time that can be compared to actual data. The model values for Cd and Crr can thus be adjusted until the model matches the data. This adjustment can be done manually, by overwriting the values of Cd and Crr with new values till the model matches the data, or it can be done using a "Solver" function.

Step 3: Procedure

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You can determine Cd and Crr from the same set of test data by measuring velocity with respect to time as your vehicle coasts in neutral. Note that Crr will not be pure rolling resistance but will include some drive-train resistance as well.

1. Drive to a flat road with little traffic or wind.

2. Have the passenger ready with stopwatch and paper to record data.

3. Have the driver accelerate up to above 70 km/h or so, and shift into neutral.

4. Record data as follows. The driver should indicate when the speed drops to exactly 70 km/h. At this time (t=0) the passenger should start the clock. The passenger should indicate every 10 seconds after that and the driver should call out the current speed to the nearest whole km. The passenger should record this value next to each time.

Aside: If you have a digital camera capable of recording several minutes of low resolution video (as most people seem to have these days), the process is much easier and more accurate. You don't need any equipment except the digital camera. Simply have your passenger record a video of your speedometer during the coast down tests, or find some way of mounting the camera so you can do the recording without an assistant. Using a free program such as Avidemux ( you can play the video back on your computer frame by frame and view the timestamp at desired speeds.

5. Repeat the test in the opposite direction.

6. Repeat the test in both directions twice more (6 trials in all, 3 in each direction). All these values will be averaged for a more accurate analysis.

7. Download the spreadsheet I created Drag_Coefficient.xls and enter all your data following the instructions included. The spreadsheet averages data from all 6 trials to create a single data set representing velocity (V actual) as a function of time. It then generates it's own model for velocity (V model) based on entered constants and initial guesses for Cd and Crr. Excel's "Solver" function can be used to adjust Cd and Crr in order to minimize the error between the model and actual data. If you are using OpenOffice Calc (which I highly recommend and which you can download for free from, unfortunately, the solver function currently only handles linear systems, so you will have to adjust the input values manually to minimize the error between the model and the data. Once the error is minimized and the model data matches the actual data as best it can, then Cd and Crr are correct.

Step 4: Results

Picture of Results

Here are the quantities I measured for my car (a 1992 Geo Metro):

   M = 1000 kg (about 850kg curb weight plus 150 kg of occupants)
   A = 2.3 m^2 (a good approximation based on measurements of my car)

A plot of velocity vs time is shown below. It is based on the averages from my 6 trials. You can see that the model curve closely matches the data points. The values of Cd and Crr for the model are:

   Cd = 0.370
   Crr = 0.0106

Therefore, these are the drag coefficent and coefficient of rolling resistance for my car.

These values are nice to know. However, in practice, if you want to compare performance before and after making modifications to your car, you can get faster results just by measuring the time to decelerate from speed A to speed B. Pick high to medium speeds if your modifications are likely to affect drag. Pick medium to low speeds if your modifications are likely to affect rolling resistance. Don't forget to take multiple measurements in each direction and average the results.

For more experiments you can do on your car see my website

Update 2009-01-02:
I've learned a lot since originally posting this instructable 16 months ago. I've played with measuring Cd and Crr under different conditions on a number of vehicles and other experimenters have picked apart and tweaked my spreadsheet for their own uses.

My experience is that there IS a mistake in one of the underlying assumptions of the model: namely that the force of rolling resistance is constant independent of V. Vehicles are designed with negative lift (so they get pushed into the road more at higher speeds, improving handling) so the force of rolling resistance also has a component that varies with V2 like the drag force. The force of rolling resistance also includes a small component of viscous force (drivetrain) which varies with V.

The model assumes that the drag force is related only to V2 and that the force of rolling and drivetrain resistance is constant. In reality the force of rolling and drivetrain resistance is also related to V2 and V. So a better model of the force on a moving vehicle is:

    F = iV2 + jV + k where i, j, and k are constants.

A curve based on that model more closely matches actual coast down data indicating it is a more accurate model. But after solving for i, j and k, there is no way to extract meaningful values of Cd and Crr since by definition, they assume i is related only to drag, and j is 0, neither of which is entirely true.

As mentioned above, If you want to compare performance of a vehicle before and after making mods, the change in coast down time itself is MUCH more meaningful than any change in Cd or Crr extracted from the coast down data.


SO2 (author)2014-07-29

I thought of the same trick years ago and am glad I'm not the only one to ponder such things. Because high speed does reveal drag better, I also considered this test: find a straight smooth road that's rarely traveled. Drive straight at about 82 mph. Have a video camera, with on-screen hh:mm:ss showing, shooting a gps. Let off the gas. Go back and watch the video. How many seconds did it take to coast down from say 80 to 40.

Top mph and mpg are influenced by drivetrain friction and rolling (tire/wheel) resistance, so leaving them in the equation isn't that bad. Another thing this factors in, is front area + Cd, or total drag, contrary to the title of the article.

Myself, I'd like to eliminate human error, even without high speed. Go to a tall hill, release the brakes with aforementioned gps / camera, and note the stopping location/time achieved. I'd do this with a super sleek rental car, and a blunt one. Look up the Cd in R/T or carfolio and interpolate to estimate your car, based on the time/distance you observed. Try to establish the baseline with semi boring cars. The interesting cars are more likely to have bogus Cd numbers.

Sorry if it was mentioned, but there are two very critical ideas I didn't see, scanning the feedback. One is turns. If the best hill you have features a subtle curve, try your best to take it smoothly and consistently. Second one is road surface. If you establish a baseline 12 years ago, and take a project car to the same hill, be on the lookout for different quality of pavement. Whether you coast, tow, or down-hill roll, a substantial difference in pavement smoothness (because of repaving) means all bets are off.

drrbc (author)2014-06-12

SCT has a free data logger/analyser from which you should be able to calculate this. Also, though I've not tried it yet, (works with any ELM327 OBD2) has just added a logger to their app. Check the App Store to see if there are any more apps that have an OBD2 interface.

jumpjack2 (author)2014-02-20

How could I "emulate" Excel solver using VBA or Javascript? If I could figure out this, I think using AppInventor I could actually create the android app which calculates the Cx and Crr I was talking about!

jumpjack2 (author)2014-02-19

That's all we need to build a smartphone app which, using onboard GPS data, can calculate Cx and Crr.

Unfortuately I am not able to write it! :-(

irl13 (author)2013-05-22

Wow, Wow, just wow! What's going on here, the speculation on the effects of tire pressure and rolling resistance is out of control. I know this is an old topic but I can't just let this stuff lie.

Your post here and and aero civic are great by the way. Job well done and I admire your scientific approach to everything. But this tire business is irking me. I'll add as many references as I think are required but most of this I learned from print books.


Increased tire Pressure:
Reduces Rolling Resistance (Duh!)
Increases fuel economy (all the way to bursting point)
Decreases chances of hydro planing

This is just simple physics The rest of the factors that are being discussed here have a lot more factors than just tire pressure involved

Firstly, Manufacturer tire pressure ratings are based on a multitude of performance factors, many of which have not been discussed here. Safety, cabin noise, vibration, tire compound, tire tread, tire geometry, car load, car suspension, intended car speeds, longevity, consumer market, typically expected road surfaces of main market, the list goes on. no one tire pressure is optimal regardless of what's printed on your car. These pressures just optimise a number of these factors. They are not safety limits, by any margin and 10-15% difference wont start causing crashes. Most people can barely notice a 30% change. (You just need to walk around a local parking lot to see how many people have no idea what's going on with their tires)

Tire pressure is not the only factor affecting rolling resistance and hence fuel economy. Check this out:
Anyone tempted by slick tires? Hard compounds? Weird tread patterns? Narrow tires? Different sidewall thickness? Heavier tires? Lighter tires? Different gases for inflation?

Braking and cornering
Mainly a function of 'contact patch' of tire. There's no simple answer for 'square' car tires and the sheet of paper with paint tactics mentioned here have a massive margin of error. Get a temperature gauge and have a look at the first link here if you want to start being scientific. On ashphalt, there is an optimum pressure to maximise the contact area between tire and road. Too low is as bad as too high. on rough surfaces: the lower the better. make the rubber flex into all the contours of the surface. This applys to both acceleration and braking but negatively affects cornering.
ABS systems skew all this a bit (If anyone's interested ABS systems affect all this in a variety of ways, some very good others very bad, like ice!)
Ice is a category all on it's own, whole different ball game and different ways to deal with it.

For the high pressure guys and their sceptics: this guy runs max pressure on his tires and measures the wear. nothing out of the ordinary after 22k miles

Mythbusters tackled the fuel economy myth but didn't really deal with the safety issue other than to say it's not recommened (true scientists as always, sigh)

to the author
"A simple proof: Suppose the coefficient of friction is 0.8 (typical of rubber on concrete). Let's compare a 1x1 inch square of rubber with 100 lb of weight evenly distributed on it to a 10x10 inch square of rubber with 100 lb of weight evenly distributed on it. In the first case, each square inch of rubber supports 100lb so each square inch can tolerate a shear force of 0.8*100=80lb before slipping. Multiplying this by the number of square inches (1) gives a maximum total shear force of 80lb. In the second case each square inch only supports 1lb, so each square inch can tolerate a shear force of 0.8*1=0.8lb before slipping. Multiplying this by the number of square inches (100) gives a maximum total shear force of 80lb. I say again... the size of the footprint makes no difference to the force that can be transmitted without slipping."

Yes, this is Amontom's 2nd law of STATIC friction. But car tires roll and can't be analysed in this way. Dynamically, the coefficient of friction changes according to tire load (which is not static, unless the car is. The load shifts from tire to tire with every twitch of the car) and contact patch is a factor both parallel and tangential to the wheel rotation. Here's an intro:
Conventional ideal system kinematics don't apply here. It's best just to look at reliable models for reference, F1 or motorsport enthusiasts for instance. Car and tires manufacturers never release useful data unfortunately

to leroy:

"So I guess the conclusions to be drawn are: (1) the larger the footprint, the less likely you are to wear out the tires (2) The size of the footprint has no effect on the rolling friction (friction retarding the car's motion in the nonskidding condition) or on the locked-up-brakes friction (skidding condition) (3) the size of the footprint also has no bearing on how long it takes to brake to a stop--for either the skidding or nonskidding case I had a look at your website."

Are you seriously suggesting we should put the narrowest tires possible on our cars, as small as can bear the car weight and this will have no effect on rolling resistance or braking distance? That race car engineers are mad for putting 12 inch wide tires on their cars. Think about that for a minute

dariuswilkinson (author)2011-06-03

This is a great post mate. Thanks for sharing this to car enthusiasts.

NeilBlanchard (author)2011-04-24

If you video record your speedometer, then you don't have to do the timing and data recording while driving.

My biggest challenge is estimating the rolling resistance of my tires. I wish I had been able to do a test before making the mods, to calibrate things, because I know the "factory" Cd.

Oh, and if you take a frontal photograph with a reasonably long (telephoto) lens, with a measuring tape that is readable at the high point of the roof, you can insert it into a CAD program (I used DataCAD) and then trace the outline, for a pretty accurate frontal area. I included the side view mirrors, even though i believe these are excluded from the "factory" number. (One of my mods is to replace the mirrors with video cameras.)

Sincerely, Neil

kstruve (author)2010-04-16

I wonder if you may have better results measuring the drag due to air resistance if you started your coastdowns at a much higher speed, since at high speeds, most of the total drag is from the air.  I might start the coastdowns at something like 100 mph, then stop at around 50 mph.  This may help isolate the Cd from the Crr more effectively, since Crr is much closer to a constant force, where Cd is an exponential force.

iwilltry (author)kstruve2010-05-10

Definitely the results will be more accurate coasting down from higher speeds, for exactly the reasons you say. But there are few places (at least near me) with level ground where one can safely do coast down testing from such high speeds.

stevepuk (author)2009-07-17

Using a GPS device would greatly simplify the whole thing.

al75k (author)2009-01-01

Hi, I think there is a mistake in your car's frontal area. All sites that I found said it is about 18 ft2 or about 1.8 m2. I think you might have added the area between the ground and the car. Here are some of these sites: (This calculator is not accurate)
Wikipedia Geo Metro has dimensions

So I put the correct area in your Excel file and it gave me a Cd of .47 and Crr .010. So it seems to me like there is a mistake here. I also found a PDF that shows equations for what you are doing, maybe it will help:

iwilltry (author)al75k2009-01-02

Thanks for the correction. I didn't measure the frontal area with any great accuracy, just a tape measure and some quick estimates so I would not be surprised if it is off. I've played with measuring Cd and Crr a bit more since originally posting this instructable and other experimenters have picked apart and tweaked the spreadsheet for their own uses so I'm 99.9% confident there is not a mistake in the spreadsheet.

However, I believe there IS a mistake in one of the underlying assumptions: that the force of rolling resistance is constant independent of V. The rolling resistance measured in a coast down test includes a component from viscous forces (drivetrain) which vary with V. Vehicles are also designed with negative lift (so they get pressed into the road at higher speeds for improved handling) so there is also a component of force from rolling resistance that varies with V2.

The model assumes the drag force is related only to V2 and the force of rolling and drivetrain resistance is constant. The reality is that the force of rolling and drivetrain resistance is also related to V2 and V.

So a better model of the force on a moving vehicle would be:

F = iV2 + jV + k where i, j, and k are constants.

A curve based on that model will much more closely match actual coast down data. But after solving for i, j and k, there is no way to extract meaningful Cd and Crr values since they are part of a completely different model that assumes i is affected only by Cd, and j is always 0, neither of which appear to be true.

To illustrate the point, if you do a coast down test while holding your doors open (which should affect only Cd and not Crr) you'll find that the indicated Crr changes too. That is a clear indication that the model itself is lacking. For this reason I don't advise relying on Cd or Crr values calculated from coast down data. If you really want an accurate Cd value you need to eliminate rolling resistance and viscous forces from the test (think wind tunnel). But by the same token, knowing an accurate Cd value isn't particularly useful. It will allow you to calculate the force on your car in a wind tunnel but it won't allow you to accurately calculate the force on your vehicle on the road.

If you want to compare the performance of a vehicle before/after making mods, the change in coast down time itself is MUCH more meaningful than any change in Cd or Crr extracted from the coast down data.

glubok (author)iwilltry2009-01-02

My analysis of your data shows a very nice linear relationship between arc tan (v/K1) and time which supports your original model where force is proportional to v2 and to a constant. Maybe you should look at my modification of your spreadsheet. Another way to measure rolling resistance would be to measure the angle of repose. That would require putting the vehicle on a ramp that could be inclined by jacking it up until the vehicle begins to move. It's easy to determine the component of the vehicle weight projected onto the ramp which would propel the vehicle forward.

iwilltry (author)glubok2009-01-02

I would like to see your spreadsheet. I'll private message you.

It may be that the viscous force proportional to V is small, but there may still be a component of the force of rolling resistance that is proportional to V2 due to negative lift at high speeds. This should result in an overestimate of Cd and an underestimate of Crr which seems to be the case (the Cd spec for this vehicle is supposed to be around 0.36)

It would be useful to compare the rolling resistance calculated to that measured by inclined plane as you suggested, or by pulling the vehicle at constant speed (a walking pace) with a spring scale on level ground.

itjstagame (author)2008-10-01

More on topic, do you know how to solve for Cid and Crr as a curve? I made a program to count my VSS and injector pulses and extrapolate MPG, etc, etc, and I want a button I can hit to say start timer when I drop below 60MPH and end at 30MPH and do a coast. It will record exact speed and exact time. I can simply plug back and forth with trial and error but it's proving really annoying to basically rebuild the 'Solver' on my own. Is there some integration or some formula to solve the entire curve?

iwilltry (author)itjstagame2008-10-02

To match the entire curve you would have to write your own "solver" equivalent (ie a recursive function to keep tweaking Cd and Crr values until the error between the model and the actual data is as low as possible).

But you can get pretty good results without matching the whole curve. Rearranging the formula for drag and rolling resistance gives

M*a = -Crr*M*g -Cd*A*0.5*rho*V2

You can look at a 1 second interval near 60 MPH and another 1 second interval near 30 MPH. Using your VSS data you can calculate "V" and "a" in each case. Everything else is known except Cd and Crr. You will have two equations (one for 60MPH and one for 30MPH) and two unknowns (Cd and Crr). Use basic algebra to solve for Cd and Crr in terms of known values.

glubok (author)iwilltry2008-12-25

I have separated variables leading to an equation of the form:

dv/(v2 + K12) = K2dt

where K1 and K2 are constants:

K1 = sqrt(2mgCrr/(rhoACd)), and

K2 = -0.5rhoACd/m

This was integrated over v from v0 to v and over t from 0 to t. The result was:

arc tan(v/K1) = K1K2t + arc tan(v0/K1)

The assumption of values for Cd and Crr, establishes K1 and K2. Then, plotting arc tan(v/K1) versus time should produce a straight line with slope K1K2 and zero intercept arc tan(v0/K1). The least squares fit will produce estimates of slope and intercept which can be used to calculate K1 and K2. We have thus created an iterative process which should converge upon an estimate of K1 and K2. The definitions of K1 and K2 can be then used to establish the corresponding estimate of Cd and Crr. By rearranging those definitions,

Cd = -2mK2/(rhoA) , and

Crr = -K12K2/g.

I modified your Excel spreadsheet accordingly. Using your data, the process converged upon

Cd = 0.3902 and Crr = 0.01079 .

Then I used a guessing approach like yours to find the coefficients resulting in minimum error in the least squares fit. The results were:

Cd = 0.4261 and Crr = 0.01022 .

These differ slightly from your results:

Cd = 0.3697 and Crr = 0.01057 .

I don't know which is the best estimate of the coefficients. It's hard to argue with iterative convergence though. Maybe they are close enough to each other that it doesn't matter.

The modified Excel spreadsheet is available to anyone. I wrote it using manual iteration, since I don't know how to do recursive programming in the Excel language. (It's easy in other languages.) Also, I have a Word document that may make the above a little clearer. I'm a little restrained by the format capabilities here.

iwilltry (author)glubok2009-01-02

Thanks for your post. I suspect your results are more accurate than mine since I believe you've avoided the "quantization" error inherent in my simulation (my model looks at 5 second time slices, and assumes constant acceleration = F/m within each time slice). A quick modification to expand my table to use 1 second time slices yields values of Cd = 0.393 and Crr = 0.01058. My model also assumed an initial velocity equal to that of the first data point. Removing that constraint and using Excel's solver function to find the initial velocity for best fit, along with the values of Cd and Crr yields Cd = 0.4047 and Crr = 0.01039, values much closer to those you determined.

In any case, since posting this instructable, I've come to the realization that the assumed model, despite its wide acceptance is a poor one. See update I just added to the end of the instructable. Therefore I don't have much faith in the values generated by this spreadsheet except as extreme approximations.

Leroy (author)2008-10-01

Thank you for the nice instructional for measuring drag and rolling resistance coefficients. I haven't tried them yet, so cannot comment on their accuracy.

Overinflating vs. safety:

The main consideration is the footprint of the tire. The footprint is the amount of rubber in contact with the asphalt. It is the flattened bottom of the tire as it sits on the asphalt.

Why is the footprint so important? When force is exerted on the pavement to change the car's motion--when you hit the brakes, for example -- the force must be transmitted through the footprint to the tires and to the mass of the car. Obviously, if the footprint is partly ice, the car will go sliding when you hit the brakes or otherwise try to accelerate (in the physics sense of "change its inertia") it.

Consider the car sitting on the pavement. Its footprint is a certain size, in square meters. Consider the car driving on the pavement at 70 km/hr. The axle is moving forward at 70 km/hr.
The bottom of the tire is rolling on the pavement. Therefore its speed relative to the pavement is zero km/hr. (If it had speed relavative to the pavement, it would be skidding and would leave black rubber on the pavement. ) The tire is spinning so very, very fast to make this speed of zero on its bottom possible. So the speed of the tires' footprints is still zero km/hr.

Hence, so long as the tire air pressure hasn't changed, and the car's weight hasn't changed, then the tires' footprints are EXACTLY the same size as
they were with the car sitting at rest.

I think you can see that the larger the tires' footprints are:

(1) the easier it is to accelerate the car (e.g. braking to a stop) without something breaking

(2) the greater the car's coefficient of rolling resistance

(3) the less shear force each square cm of tire will have to bear when subjecting your car to a given acceleration (again, in the physics sense of changing your car's velocity)

The larger the area in contact with the pavement, the safer the car will be, at the expense of fuel efficiency.

Now the relevant question is, as you change the air pressure in the tires, does the tire's footprint change significantly, or does it remain about the same?

I don't know the answer. I think it would probably depend on the particular tire. However, using the principle I stated above, it would be EASY to test
whether or not the tire footprint changes with tire air pressure.

First equalize the air pressure in all four tires and measure it with a tire pressure gauge. Record the air pressure.

Get a big piece of cardboard, like a cardboard box from a refrigerator, and a can of spray paint and a Sharpie marker. (You can get cardboard for free from any retail distributor or from neighbors.) Park the car so its tires rest on the cardboard. Lie down on the cardboard with your spray paint. Spray the cardboard completely around each tire. Then move the car forward and trace the outline with your Sharpie. Mark the four outlines with #1 to signify your first air-pressure test.

When the paint dries, you're ready for your second test. Change the air pressure in your tires--either inflate it to higher pressure or let some air out. Record the air pressure. Then flip the cardboard over and repeat the test.

You can check as many different pressures as you want to see if the tire footprint changes size with air pressure--until one of your tires bursts, hahaha!

iwilltry (author)Leroy2008-10-02

I disagree. A larger footprint won't transfer greater force to the road before slipping. The force of friction is unrelated to the size of the footprint. It is only related to the normal force (the weight on the wheel).

A simple proof: Suppose the coefficient of friction is 0.8 (typical of rubber on concrete). Let's compare a 1x1 inch square of rubber with 100 lb of weight evenly distributed on it to a 10x10 inch square of rubber with 100 lb of weight evenly distributed on it. In the first case, each square inch of rubber supports 100lb so each square inch can tolerate a shear force of 0.8*100=80lb before slipping. Multiplying this by the number of square inches (1) gives a maximum total shear force of 80lb. In the second case each square inch only supports 1lb, so each square inch can tolerate a shear force of 0.8*1=0.8lb before slipping. Multiplying this by the number of square inches (100) gives a maximum total shear force of 80lb. I say again... the size of the footprint makes no difference to the force that can be transmitted without slipping.

This assumes a constant coefficient of friction. However, surface conditions aren't always constant and there can be localized areas (ex patches of ice) where the coefficient of friction is lower. What matters is the average coefficient of friction over the entire footrpint. I agree a larger footprint is safer for inconsistent surface conditions. Clearly a 1x1" block of rubber will have more trouble dealing with a 2x2" patch of ice than will a 10x10" block of rubber assuming they are supporting the same load.

But under non-freezing conditions on asphalt roads the surface is relatively consistent and the size of your tire's footprint will have little effect on your stopping or accelerating ability. Here's a better test: Inflate your tires to maximum pressure, go to an empty parking lot and measure the distance it takes you to stop with wheels locked up. Deflate your tires to half the maximum pressure and repeat, noting whether your stopping distance changes.

Leroy (author)iwilltry2008-10-02

You're right, Rob, that a larger footprint won't transmit more braking force to a car. Your simple proof illustrates this lucidlly. The force should be the same no matter how large the footprint. I guess I just meant to say that the larger the footprint, the larger the area over which the shear force acts. Hence, the lower the shear force per unit area. To use your example of the 100 # weight distributed on a one square inch base and on a larger 100 square inch base: For the smaller footprint, each square inch supports 100 psi of weight pressure and 80 psi of shear (braking) pressure. For the larger footprint, each square inch supports 1 psi of weight and 0.8 psi of shear. Hence, the larger footprint applies the same braking force to the car with lower material stress to the tire. So I guess the conclusions to be drawn are: (1) the larger the footprint, the less likely you are to wear out the tires (2) The size of the footprint has no effect on the rolling friction (friction retarding the car's motion in the nonskidding condition) or on the locked-up-brakes friction (skidding condition) (3) the size of the footprint also has no bearing on how long it takes to brake to a stop--for either the skidding or nonskidding case I had a look at your website. Nice projects, especially the solar hot water heater. It reminds me of an instructable I saw here on building a small solar hot water heater using the black condensor coil off the back of a junked refrigerator. Take care

itjstagame (author)2008-10-01

First off: "4) "Bicycling" a car requires specific tires and rims, and suspension reinforcement, just ask any vehicle-stunt coordinator. Inflation at 100 PSI is simply a farce, as even with racing tires, this would cause an immediate blowout, potentially killing the person inflating the tire to this pressure. " What? It's like you don't understand loading and PSI. You mentioned before also that more people in the car with tires at 55psi would be more dangerous, but this is less so. If my car is 2500lbs and has 40psi tires there's a certain footprint on the road, if I load with 6 'large boned' people I'll have 3500lbs and with my same 40psi tires I'll have a MUCH larger footprint. The same footprint as if I had my original 2500lb vehicle using tires at 28.5psi, that's a big difference! To maintain the same footprint at 3500lbs I would need 56psi in my tires. I'd say almost all SUV and truck manuals tell you to increase pressure on the tires (sometimes higher than max sidewall) depending on the weight of load you're loading. I say most because I haven't owned every truck and SUV in the world, but everyone I have owned has stated this. So if you want to take a heavy vehicle (like a 4500lb Crown Vic) that normally has the load distributed across 4 tires at 50psi each, and ride it on only 2 wheels, in order to NOT have those tires get squashed and look like 25psi tires and potentially fail the seal on the rim, etc you HAVE to put 100psi into them. Rims seals with pressure, they will never lose a seal due to increased pressure unless you actually except the stress of the tire material itself and the force of air itself punctures it. Rubber innertubes would pop almost instantly at that pressure, but the heavy, multi ply tires of today will have no issue.

mecos (author)2008-09-02

Prometheus: I thought we were talking tire inflation pressure, not soap-box politics? To all others: Truth be known you will cause differential wear to your tires depending on how you inflate them (toward the shoulder on lower pressure at lower speed and depending on average speed and sidewall stiffness; in the center of the tread at higher pressure or even lower pressure at higher speed depending, again, on the stiffness of the side wall). Tire manufacturers specify a maximum pressure based on their testing. Will you kill yourself over-inflating your tires? Probably not. But you will void your warrantee. Will it affect performance? Insignificantly, unless your driving a race car. Auto manufacturers specify tire pressure based on their criteria(reliability vs comfort and performance). It's more of a balance, but will never exceed the tire manufacturer's limits. Statistically, speaking (and if you want I can site the references later) most tire "blow-outs" occur with "UNDER-inflated" tires. This is due to the build-up of heat from friction in the sidewall. Most often the sidewall is the failure point of tires, not the tread (Firestone not withstanding, which is completely different). Yes, there are time when you should over/under inflate your tires? Sure, but they are rare enough that most of us need not worry about it. Those of you who need to do this know who you are and what to do. If you want more mileage from your gas follow these steps. 1) Control your acceleration. You're not in the NHRA. 2) Control your speed. Driving 5 MPH slower can save you up to 5%. 3) Maintain your motor. Since that is where the gas is going in the first place. 4) Since this article is about drag coefficient and not tire pressure: GENERALLY, the smaller and more shapely your vehicle, the better the coefficient of drag. 5) GOD BLESS AMERICA AND DEMOCRACY! GO VOTE! Whatever!!!

Prometheus (author)mecos2008-09-03

See above, the comment must have been shuffled with another topic.

PA32R (author)2008-08-30


DJglass (author)2008-04-24

The simplest change you can make to your car to improve rolling resistance and fuel economy is to get your tyre pressures correct. By Correct I mean start with the manufacturer recommended pressures (on a plate or sticker inside the drivers door) then add between 2 and 5 psi on top of that. More if it is cold weather and the tyres won't warm up. Remember to check them often.

iwilltry (author)DJglass2008-04-24

If you are only interested in decreasing rolling resistance, then the higher the pressure, the better. You can go well beyond 5 PSI over the listed pressure. The listed pressure for my tires is 32 PSI but I run them at 45. Above that I have not seen any increase in coast down times for my car. The main sacrifice at higher pressures is comfort (shock absorption). One would also expect uneven wear and less traction, but I have not noticed these to any significant degree, and the increase in fuel economy would likely make up for the cost of early tire replacement.

Prometheus (author)iwilltry2008-04-25

Running your tires at 45PSI is extremely dangerous. You may notice a little writing on the sidewall, "Do not exceed 40PSI when mounting". The minuscule improvement in rolling resistance you get is greatly offset by the danger of explosion, which will do a lot more than make your rolling resistance dramatically worse. Also, as the tire heats up, pressure will increase, further increasing the chance of a blowout. Get your tire pressure CORRECT. The manufacturer has those recommendations because they have calculated the loads between front and rear. Too low a pressure overstresses the sidewalls, seriously reduces tire performance with side loads, increases the temperature of the tire, dramatically shortens tire life, and largely increases rolling resistance. Too high a pressure reduces tire life even further, can lead to tire failure, and can actually increase wear on suspension components. What might have remained a nail hole can result in a catastrophic failure. The roads are not racetracks, and you do not need to meddle with your tire pressure other than the norm except in extremes. Heavy snow, then sure, decrease your tire pressure, as heat buildup and top speed are not factors. Just be sure to bring them back up to pressure when on pavement again.

iwilltry (author)Prometheus2008-04-27

The change in rolling resistance going from 32 PSI to 45 PSI is hardly minuscule. I know because I've measured it. I also know that Hypermilers across the country routinely run their tires at well over the manufacturer's maximum ratings (search the forums at and none seem to have experienced any problems with blow outs. It's been argued that higher pressure may actually reduce the risk of a blow out because the tire holds it's shape better and doesn't experience as much fatigue or temperature increase from constant flexing. I do not know what dangers may or may not lurk beyond the manufacturer's rated pressure, but I do know there isn't a magic pressure X above which it is dangerous and below which it is safe. The maximum rating is simply the point at which the manufacturer chooses to limit their own liability. You may exceed that point, but you do so at your own risk. My personal experience is that the risk is small. Your experience may lead you to a different conclusion.

bikerbob2005 (author)iwilltry2008-08-14

narrow tires have less resistance than wide ones not inflate more than 10% of the max. printed on the sidewall.long time ago we would mount 17 ply aircraft tires to the vw and they would last forever think about 90 psi on them. drag coefficient and frontal area are not anything like the same thing. a vw beatle (old ones ) has more frontal area than a B-52 does but the BUFF has a much better Cf scarier part? the bug has a better glide ratio

richms (author)iwilltry2008-04-24

You are compromising your ability to stop however. I run my tyres below what they specify slightly because it makes a huge difference in braking before losing traction. Recommended pressure is 220kPa, I would be at about 190. No idea what those are in pounds but I think its about 28 on the gauge when I am checking them.

iwilltry (author)richms2008-04-24

As I said, one would expect less traction (which includes braking ability). However, I have not noticed a significant change with my car. Others' results may vary.

DJglass (author)iwilltry2008-04-25

Sure you can run more than that. Tyres can take a surprising amount of pressure. Those pressures should result in even and longer wear unless your alignment is out or you have excessive camber. The range I suggested is taking into account the comfort sacrifice. As the opposite of the above, less traction but better cornering ability.

Prometheus (author)DJglass2008-04-25

What results in even and longest wear is the correct pressure. Too low a pressure and the treadwall will wear at the edges because the sidewalls are effectively trying to travel faster than the center of the tread pattern. Too high, and it wears down the centerline because the treadwall bulge has to handle all the traction while little to no load is spread to the rest of the treadwall. If you want to decrease resistance, rolling up your windows while driving will prove to have over 400% more effect than your tire pressure

iwilltry (author)Prometheus2008-04-27

What data are you basing these statements on?

I've measured coast down time's for my car ('92 Geo Metro) with windows open vs windows closed and at 32 PSI vs 45 PSI tire pressure and found there was hardly any difference with windows open vs closed, but there was a noticeable increase in coast down time at 42 PSI vs 32 PSI. This is for coast down from 70 km/h to 20km/h. At higher speeds I could imagine the windows having a more significant effect since drag force is proportional to velocity squared, while force due to rolling resistance is relatively constant, independent of speed. At city speeds, though, increasing tire pressure has a greater effect (at least for my car).

In any case, it doesn't matter much to me which has more effect. They both have an effect so I do both.

I expected my tires to wear more in the center but it simply hasn't happened. On further research, other hypermilers have reported similar experiences. See There is a photo on that page showing tire wear at 45 PSI after 2000km on brand new tires. The "molding bumps" show even wear.

Prometheus (author)iwilltry2008-04-30

I'll just add that you are comparing two different tire constructions, and opposite sides of the tire spectrum, if you will. Commercial freight tires are simply built for two purposes:

1) To protect the rim from the road
2) To handle the immense weight on them with the durability to go long distances in less-than-ideal conditions.

Passenger car tires are built far differently...for one they are not regroovable and generally not designed to be re-capped. They are also not overbuilt for durability like commercial tires are. That durability includes overpressure, overweight, rough roads, and heavy side loads. Consumer-level car tires have to be built in a purpose-built lightweight design, and still offer some shock-absorption. If a consumer-grade tire blows-out, the vehicle still retains a great deal of controllability. If a long-hauler pulling doubles blows a front tire, the disaster makes news and nearly a day's worth of traffic hardships.

The sidewalls and treadwalls are not just thicker, but proportionally tougher for commercial vehicles because weight is less important than sheer armor to a plethora of potential abuse the tire may suffer. The rubber compound for commercial tires is much harder and has a higher threadcount for the structural ply, and because of this, they can handle more abuse.

You can take your chances by grossly over inflating your tires based on a trucker's trick, but this does not apply to passenger cars. There are far better ways to improve fuel efficiency, such as a K&M filter, increased exhaust flow with an oversize catylitic converter and a high-flow muffler, proper tuning and maintenance, buying a *quality* motor oil like Castrol, instead of the convenience-store brand, as well as intelligently matching OEM specs for oil viscosity with the actual wear on your engine.

First, I drove a 1981 Datsun 510 with a history of oil leakage. after solving the oil leak, I automatically regraded to 40W oil, contrary to the OEM spec of 30W, and gas mileage improved over 12% when used as a newspaper delivery vehicle on a fixed route (which made quantifying actual gas mileage rather easy), and a 7-10hp increase.

Then I drove a 1991 Nissan Sentra 2.0L SR20DE with the 4-spd gearbox. Base HP was 133bHp off the factory floor, with an estimated 28mpg by the EPA IIRC. After the catalytic converter suffered temperature shock from a "freeway lake", I simply patched it with a form that allowed moderate bypass, only when the muffler was being obstructive. With barely any tuning at all, it became a 155bHp 40MPG monster, and torque came out of it that would dust a V8 Cadillac STS 10 years and over 200,000 miles it's junior.

I might add that I would haul over 700lbs of newspapers (+250 GAVWR) and a gas-mileage loss was barely mentionable. Of course, I inflated the rear tires to maximum inflation to handle the load, but that had little to do with gas mileage. Properly tuning the engine, and letting it breathe properly, made more difference than any tire-inflation method could hold a candle to.

Finally, I used quality sport-grade tires. Try this with some cheap-o tires and you could find yourself in a world of hurt. I have raced enough cars for long enough to know what tire pressure means. Higher pressures can reduce your overall traction, besides putting you at risk for a blowout. There are far better ways to improve gas mileage and/or reduce drag than to risk a tire blowout. Try an airdam on the front and a corrective-spoiler on the rear of hatchbacks. Better still, buy a fastback or a sedan. Don't buy some american-made oil-burner like a Suburban and then complain about gas mileage, buy into efficient design and using your brain before you purchase a vehicle to see if it is really what you need.

Larger vehicles are not safer by any means, because the best way to keep from getting hurt in an accident is to avoid a collision in the first place. It starts with a lighter vehicle with better control, and continues with not talking on the cell-phone while you drive...some call it "paying attention and being responsible", and ends with getting a basic education into reality and realizing that automobiles are not magic, but actual machines, and Jesus is never your co-pilot with the way that you drive. Appreciate the machine more and you learn how to operate it better. This is why it is so much safer to fly than to drive....That, and it requires a proven IQ to get a pilot's license.

You go ahead and inflate your tires to dangerous levels that ompromise your actual control over the vehicle for the sake of a 0.1MPG increase, which will certainly be offset by the cost of replacing tires and possibly repair-work for accident damage. I'd rather inflate tires properly and look at the biggest loss of efficiency, the internal-combustion engine itself. I gained 12MPG simultaneously with an increase of 12 bHp with a simple exhaust-system mod. Perhaps tires aren't the best place to start to increase efficiency...

Just my thoughts on an observed science through my previous career. My previous experience and observational science should carry some validity, just as well as I respect your own. Finally, keep in mind that automatic transmissions suffer more "neutral" drag than a "stickshift" does, due to reliance on hydraulic resistance instead of mechanical opposition.

iwilltry (author)Prometheus2008-04-30
I'm not sure what made you think I was referring to commercial freight tires. I think perhaps you misinterpreted what I meant by "hypermiler". A hypermiler is simply a person who attempts (often obsessively) to extract every possible mile from each tank of gas. They drive ordinary passenger vehicles and use many strategies for increasing mileage, some of which you've mentioned. There is an instructable on the topic: How to Become a Hypermiler

I'm not suggesting that exceeding recommended tire pressure is the only way to increase mileage, but it's certainly one of the easiest ways which does make it a good starting point for anyone willing to accept the risk (whatever they believe it may be). My experience is that the gains are much more than the 0.1 MPG you suggest and I question whether it compromises control or causes premature tire wear. The main compromise I've experienced is comfort.

In this article which I posted in another comment, Sergent Dave Storton, Director of the San Jose Police Academy, claims (among other things):
  • Tires do not balloon at higher pressures. They keep their shape and wear evenly.
  • Responsiveness improves at higher pressures.
  • Risk of hydroplaning decreases at higher pressures.
  • Significant cost savings result at higher pressures.
For these reasons in 1999 the San Jose Police Dept inflated all tires on their training fleet to 50 PSI (recommended maximum was 44). There is an image in that article of a stunt vehicle riding on two wheels with those same tires inflated to 100 PSI.

I can see that you have a lot of experience, and I respect that. I simply don't agree with your conclusions. Nor do I expect you to agree with mine. So let's disagree and leave it at that.


Prometheus (author)iwilltry2008-05-02

I will agree to disagree, but I have one last closing argument, and with this I'll leave readers to their own decisions, as some recent instructables here have begun to promote dangerous behavior:

1) If tires do not change shape at higher pressures, then how would hydroplaning resistance be changed by above-normal inflation pressures? Those two facts simply contradict one another. Tires do "balloon" slightly even if the effect is not immediately apparent by visual inspection. Few police officers have any background in racing, where one of the first things a pit-crew needs to understand is the exact effects of a tire-pressure difference of as little as 0.5PSI.

2) By "responsiveness", it is referring to cornering response and the "quickness" of the tire with lateral loads. Most drivers do not put their tires under such duress where this is helpful. Higher pressures inherently make the lateral stiffness of any given tire increase. This is part of the reason why it is easier to slide a car on underinflated tires. Inflation pressures are one of many ways stunt-drivers in cinema gain a desired performance from a particular vehicle, such as an over-steering Camaro as opposed to it's normal understeering handling characteristic.

3) Police vehicles often use special safety-rims specifically designed for their purpose. You cannot buy these rims with your stock Crown-Vic. They also use "sport-grade" tires meant to handle speeds in excess of 150MPH, and they are expected to perform reliably to a Crown-Vic's top speed of 135MPH.

4) "Bicycling" a car requires specific tires and rims, and suspension reinforcement, just ask any vehicle-stunt coordinator. Inflation at 100 PSI is simply a farce, as even with racing tires, this would cause an immediate blowout, potentially killing the person inflating the tire to this pressure. Typical sidewall-ply construction has two plies of polyester, even on sport tires, which will shear at that pressure. It is more the fact that the sidewall plies will shear from the bead itself, rather than the sidewall bursting.

5) The site you mentioned (that I have read through) is not officially endorsed by law-enforcement or the DOT. It is more a fan-site if anything. It also mentions that they do not exceed the maximum inflation pressure for anything other than training vehicles (which are operated under closely-controlled conditions on a closed "skid-pad" course). As it clearly states, 44 PSI is the maximum inflation pressure of the tire and not exceeded by anything other than training vehicles. The "Interceptor" version of a Crown-Vic gains over 400 lbs with the engine/transmission package, rollbar, communications, and body reinforcement, necessitating pressures above stock factory standard. Training vehicles are intended to have expendable parts such as tires, and the lack of traction from such overinflated tires helps to get the feel of an apparent loss-of-control. This is why I endorse ice-rinks as a driving test so that people can really understand how a car reacts with severely limited traction. This ingrains the motions with the practice and makes a better driver overall. I agree with that as I had the rear tires of my front-wheel-drive delivery vehicle inflated to maximum to handle the 600-800lb load I was placing on the rear axle daily, and only because I was meeting the maximum load of the tires. If it were simply my personal car alone, standard pressure would suffice based on the typical load the tires would withstand, even if I had a little illegal fun now and then when noone was looking (a 4-wheel drift around an onramp hairpin is pretty impressive-looking, especially with a '91 Sentra). I can take anyone to school when I perform a drift spanning 1/4 mile on a straight street of icy roads, and demonstrate how I can drive without chains in icy conditions insurmountable by an AWD vehicle with a typical driver. It's not the vehicle, but the driver, and in the United States of American Emirates, driving skill is not endorsed or even recognised. This is why I got ticketed for defensive driving, which was later thrown out in court, despite how this is encouraged by the local PD's PR department.

6) If there was no effect of wear on overinflated tires, then where does this apparent myth (supplied with actual pictures) come from that overinflated tires wear down the middle? I suppose that the wear line found in any basic instruction of examining tire wear is based on a manufactured picture of the effects of overinflation? I gauge you smarter than that, I just think you might have read too much into the wrong information.

If you question tire-wear diagnosis, I invite you to view illustrations accepted by the entire industry as it relates to tires. If tires do not ballon at higher pressures, then this tire diagnosis is completely incorrect along with my actual experience and research done by the industry since day one.

I might also add that since the quoted tests were done with high-grade tires, inflating your $20 tires to 50PSI is only asking for trouble. If you don't know what an "A-car" is, or in cop-speak "a 4", you may soon find out. Maximum rated pressure is not simply a limit of manufacturer liability, it is often best heeded because that limit of liability has a good reason behind it.

Now, if we're talking about tires for motorcycles, mopeds, or even bicycles, where the tires are specifically designed to "balloon", then yes, maximum inflation pressure is usually my standard, as the last thing you want on a two-wheeler is sidewall flex. But radial tires meant for 4-wheel operation run on a completely different set of rules and operating conditions, being required to retain their shape without the benefit of loading based on centrifugal correction of the vehicle, or also the lack of a substantial lateral load (as with "balloon tires" on two-wheelers). Radial tires are expected to handle lateral loads by resisting "bead-roll", or the tire deflecting laterally under a lateral load. This can be clearly observed by taking a generic american car and forcing hard cornering on generic passenger tires. The astute observer will clearly see the tires attempting to roll of the rim on the front (with your typical american car), and the outside circle of the tire riding on the sidewall because the treadwall has failed to retain it's place.

Typical tires come with 3 steel on the centerline and one for each side of the tire. Depending on actual scale of the tire, some come with as many as 7 steel belts, which is why changing a rear tire for a Ferrari F-50 can cost upwards of $900, due to it's low-profile and relatively high pressure, but replacing a cop tire costs as little as $180.

If you don't believe that radials can change shape, observe a top-fuel drag race. Those tires are steel-belted, but go from 20" wide and 5" high to 8" wide and 10" high during a burnout. Steel belts are made to have stretch in any tire, to retain durability. In fact, steel belts in tires have inspired the construction of CVT belts in automotive CVT's due to the harsh conditions they must survive.

I will concede that I mixed up "hypermiler" with "high-miler" though. But I will end this by saying, that in my many years and over 3 million miles of driving experience, and enough racing-training and upbringing to know the effect of any tuning anywhere on a car, that ignoring maximum inflation pressure of a tire is inviting fate to hand you a Darwin award. It's fine if you decide to inflate to a tires maximum inflation, this is not unsafe....but exceeding that is risking the merciless hand of fate. Various tires have their various rating for one reason or another, and I strongly suggest that under no circumstances that anyone attempt to inflate their tires to beyond the nameplate limits. To do so is juggling fate with contempt, and highly ill-advised, as the odds always favor "the house"...and fate owns "the house" in life. Not odds you want to play with, as those who win the Darwin award often don't live to brag about it, as if anyone should be proud of getting it anyway.

I say that this project is subject to your own discretion, and performed at your own risk, holding noone accountable but yourself if anything goes wrong. As a professional, I'd advise against taking this verbatim, but it is your own risk you take in your hands. While well-founded, I dispute the safety of some of the suggested steps based on my experience. Exceeding the safe limits prescribed falls only on you, and at your own risk.

iwilltry (author)Prometheus2008-05-02

3 points: 1. This "project" is about measuring the drag coefficient and rolling resistance of a vehicle. There is little risk involved in the project. The concept of increasing tire pressure to reduce rolling resistance has been proposed and discussed only in the comments. 2. I have not personally claimed that tires don't change shape at higher pressures. I have only related content from an article which I referenced. I believe the point the author was trying to make is that tires do not balloon as much as most people might think. I have even said that I fully expected my tires to wear unevenly as a result of over-inflation, but I found that was not the case. Other hypermilers who over-inflate their tires have been similarly surprised by the lack of uneven wear. I agree that tires MUST balloon to some degree as a result of over-inflation. If not, then there would also be no decrease in rolling resistance which is the goal of over-inflating the tires in the first place. 3. We agree about what are the effects of increasing tire pressure. We just disagree about the magnitude of each effect resulting from a pressure increase. Going from 32 PSI to 45 PSI I have experienced that the effect on wear and traction is minimal for my car and the effect on mileage is significant (more than the 0.1MPG you keep suggesting). The decrease in comfort, while significant, is tolerable. The effect on responsiveness (lateral traction) is positive. These four effects I can see and measure to some degree. However, I cannot measure the increase in likelihood of a blowout. To say that exceeding the maximum pressure is risking the merciless hand of fate is hardly empirical or quantitative. Perhaps you can provide some data (I did a quick web search but could not find anything). The risk of a blowout MUST be a continuous function of tire pressure. The risk I'm willing to accept is almost certainly more than the liability the manufacturer is willing to accept. For example, I might happily risk 2 in 10000 chances of a blowout during the lifetime of my tires while the manufacturer might choose to limit their liability at 1 in 10000 chances (I have no idea of what the actual numbers are... just illustrating a point). Based on your experience, at what pressure above the maximum rating would you expect double the chances of a blowout during the life of the tire? I've enjoyed reading your comments as you've provided a lot of information I have not seen elsewhere.

Prometheus (author)iwilltry2008-05-03

I am happy that we can have friendly debate about this, and that you clearly see nothing personal about it as I have seen in a previous social project on this site. To summarize all above points, obviously higher tire pressures will reduce rolling resistance because the energy required to flex the tire through it's "contact-patch" has to be assumed somewhere, and by default it is taken by the energy consumed in rolling resistance. Your validity on that point is assumed and agreed to be correct to my knowledge. Again, I don't disagree with inflating to a maximum inflation pressure, but going over that is still dangerous. I, nor any scientific board, can accurately quantify the odds of a blowout based upon overpressure of any given tire. While tires are made to a standard, each type/model/style of tire varies, but most importantly that each individual tire varies due to subtle and relatively undetectable circumstances by the process of mass-production. Just as 2-liter soda bottles meet a minimum standard for pressure tolerance, so do tires. But while one 2-liter soda bottle may burst at 170PSI, another may burst at 220PSI. The fact being that they all meet a minimum bursting pressure of 80PSI. Yes, of course manufacturers are going to limit their liability by understating the maximum pressures proven in testing, that's simply a corporate common sense. Firestone forgot that for one moment and a media uproar spawned because of it. The "heat" from that has not settled down yet, but the minimum standards, while since increased, have not gone so far as to even test such levels as 100PSI for a car in a "bicycling" stunt. I largely question the site for it's content more than I question you on quoting it. The internet is the "new TV", but it still applies to not believe everything you see on TV (and updated , 'the internet too'") It is entirely hard to quantify where, when, and how a tire may fail in any circumstance, but after much observation of a hapless criminal fleeing the police and losing a tire, it is most often that the beads stay on the rim while the rest of the tire takes an early vacation. Also knowing tire construction, the bond of the tire carcass to the bead itself is often the weakest bond to the wheel assembly. If you observe top-fuel drag cars when a tire fails, it is always the sidewall that fails to keep the treadwall connected. A careful look at your tire sidewall shows required information, such as sidewall and treadwall plies, and their basic composition. I have witnessed a blowout during tire mounting cue to a simple switch failure, and the tire failed right at the bead-line.This is the last place you want any tire to fail as this is the least-controllable blowout. As far as actually doubling the chances of a blowout on a street tire, I still cannot say, but cheaper tires will almost always fail more easily. On approximation, every 8PSI over the safe maximum doubles the risk of blowout up to about 50PSI, which then triples for every 5PSI over that, and increases 10-fold for every PSI over 60. Tire size itself makes a large difference, as the larger the internal cubic space of the tire, the lesser the pressure required to retain shape. Monster trucks with those huge 800lb car-crushing tires are inflated to a whopping 6-8PSI, which is about as hard as they can get without being solid. So by that means, it is hard to quantify a blowout threshold without knowing tire-size, load and inflation parameters, rim size, construction, brand name and style, temperature rating, tread rubber grade and composition, sidewall composition, thread-count and consistency for side and treadwalls, etc. Even a statistical test could not produce reliable odds.

Prometheus (author)richms2008-04-25

220kPa = 31.91 PSI 190kPa = 27.56 PSI

kPa - PSI Converter

It may seem that it makes a difference and it does, but it can actually reduce traction since not all of the tire is meeting the road equally. What you are experiencing is a greater feel of when the tire is about to break traction, and the sidewall twisting between the tread and rim more. Your tire size has a lot to do with it.

It is better to learn how to brake properly (you'd be surprised that almost all drivers on the road don't know how to), than to try altering tire pressures to modify that. Resisting the impulse to slam on the pedal goes a long way to shortening stopping distance, as well as knowing when to let off.

DJglass (author)richms2008-04-25

Sure it increases traction... in a straight line. Grip under acceleration and braking increase, but car control suffers, you are more likely to spin or understeer. If you add load (passengers, luggage) it will get worse. Without knowing the road conditions you drive in mostly that could be perfectly fine though.

vmohanraja (author)2008-08-08

Hi really good effort.. I have some doubts in that!! 1.How u calculate the frontal area of the car( Near to windshield area or Front bumper area) 2. how u calculate Cd & Crr .please give explanation.

iwilltry (author)vmohanraja2008-08-09

1. The frontal area is the cross-sectional swept area of the car. Imagine you drove your car through a snow bank. The frontal area is the area of the hole that it would leave behind. There are lots of ideas for how to measure it accurately. Try doing a search on "measure frontal area" in the forums at or For my car I just made a rough estimate based on simple measurements. 2. Technically I did not "calculate" Cd and Crr. I determined them by trial and error. I guessed some initial values. Using Excel I plotted a curve of what the velocity of the car would do based on those values. Then I adjusted the values (using Excel's solver function) until the predicted velocity curve matched the actual data. All the formulas are given, and you can download and check the Excel spreadsheet to see how I used them.

zpersichetti (author)2008-06-07

"Piglit says, Science is Fun" roflao!

wi-fi astronomer (author)2008-05-22

You could do this solo with a camcorder aimed at the speedometer and the clock and a switch attached to the clock's battery. You go to just above desired speed, shift to neutral, then at desired start-of-run speed turn on the clock and glide. When finding out the mass of the car, ensure the gas tank is filled exactly as it was when weighed as you do the glide run. Other than ensuring proper tyre inflation, there is little you can do about rolling drag. At full speed, there is mostly aerodynamic drag anyways. Reducing aerodynamic drag will improve highway mileage more than improving rolling friction. Have fun!

iwilltry (author)2008-05-02

This is not an appropriate forum for discussions of politics or religion. To address the one part of your post that is relevant to my instructable, I will say that my interest in improving fuel economy has nothing to do with economic crunch and everything to do with the damage that fossil fuels do to the world both directly through their consumption and indirectly through competition for resources. I'd be quite happy to see the price of fuel double, if it meant people would use less and if the proceeds were put towards undoing some of the damage done.

iwilltry (author)2008-04-27
The benefits and possible dangers of increasing tire pressure have recently become a hot topic in the comments here. Here is a link to an article by Sergent Dave Storton, Director of the San Jose Police Academy, that may lend credibility to some of my arguments.

Some of the points he makes:
  • Tires do not balloon at high pressures. They will still wear evenly.
  • Responsiveness is MUCH improved at higher pressures.
  • Risk of hydroplaning decreases at higher pressures.

He also says: "In 1999 the San Jose Police Department realized a significant cost savings by increasing the pressure in the training fleet to 50 psi. They soon followed up by increasing the pressure in the patrol fleet to 44 psi. For liability reasons, most agencies are reluctant to exceed the maximum pressure listed on the tire for actual patrol vehicles, but they reap the cost saving when going to 50 psi on training vehicles."
nature223 (author)2008-04-26

if you have low tire pressure,a dragging brake,altered body work,you will record differently then the models basic baseline Cd. numbers which ARE availiable if you look hard enough for your particular car...

Prometheus (author)2008-04-25

While I cannot say this method is entirely accurate, it would measure a quantifiable *change*. "The wind resistance of any object varies as the square of the speed". Because of that, it might make sense to gauge your car coasting from 60mph to 30mph (or a metric equivalent), and then use the same formula to determine an actual drag coefficient. Nice project overall though....

About This Instructable




Bio: I have a B.A.Sc and M.Eng. from the University of British Columbia, specializing in electromechanical design, but mostly I like to tinker ... More »
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