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Old 01-06-2008, 11:02 PM   #1
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Vehicle weight vs. rolling resistance

It's been suggested that a lighter car will have less rolling resistance because the weight doesn't squash the tires down as much.
For a long time I accepted that as being intuitively true. It seemed right.

Today I finally crunched some numbers and figured it out. The answer came to me when I realized that the size of a tires contact patch is a function of the weight it carries and the tire pressure, and that rolling resistance is determined by how much the contact patch deforms with load. The size of the contact patch indicates how much load and deformation it has. Right?

Here comes the math:
I'm using my car as an example. It weighs 2650 pounds and has a 61/39% weight distribution. Tire pressure is 50psi.
2650 pounds x 61% = 1616.5 pounds on the front wheels.
1616.5/2 = 808.25 pounds on each front tire.
808.25/50psi tire pressure = 16.165 square inches of contact patch.

Lets say that I remove 50 pounds from the car.
2600 pounds x 61% = 1586 pounds on the front wheels.
1586/2 = 793 pounds on each front tire.
793/50psi = 15.86 square inches of contact patch.

What tire pressure do I need to run in order to get the same size contact patch as a 50 pound lighter car?

808.25 (weight carried by one front tire of the stock weight car) divided by 15.86 square inch (contact patch size of a 50 pound lighter car) = 50.96psi tire pressure.

Or if my car was 50 pounds lighter, how much could I reduce pressure and have the same size contact patch as the heavier car?
793/16.165 = 49.06psi.

Not a big difference! Not yet anyway....

Dave W.
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Old 01-07-2008, 09:20 AM   #2
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um...why don't you run the numbers for, say, 3500 lb car, 80k lb tractor trailer (use rear 16 wheels assuming a balanced trailer...ignore the steer wheels for now) FYI they're often 90-100 psi, etc

Deformation of the tire isn't proportional to just the contact area, there's also a function of how tall and wide the tire is...a larger tire deflects less to achieve a given surface area than a smaller tire. a wider tire will have less deflection because more surface area is being spread laterally. Another question would be where is the point of diminishing returns between deformation (internal friction) losses of a narrow tire vs overall area of a wider tire? I'm just sayin...

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Old 01-07-2008, 10:50 AM   #3
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Originally Posted by kamesama980 View Post
I'm just sayin...

Take one car with a specific weight. Add a substantial amount of weight to it and then adjust the tire pressure to make the psi the same per the above idea. The bearings at each corner are still bearing the additional weight which generate frictional loads.

I think there are too many variables. Some tires have less rolling resistance than others, period. It doesn't come just from additional psi.

Still, good work on the thoughts though.
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Old 01-07-2008, 07:05 PM   #4
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The pressure factor is only one of the many elements in the total equation.

Compounds of the materials used to form the tread and sidewall , construction techniques and materials used all play a part.

Rolling resistance also includes brake and bearing drag as well as the drag from oils and lubricants in the rotating parts of the drive line.

Most of this information is not easily come by from tyre companies since it is what helps them stay ahead of the competition in the market place.

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Old 01-07-2008, 09:43 PM   #5
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Just put it up to 60psi and call it a day
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Old 01-07-2008, 09:45 PM   #6
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Thanks for the feedback.
A wise man once told me, 'If you can't describe it with numbers, then you don't know anything.' With this exercise I now know one new thing.

Let me backtrack a bit so you know where I'm coming from. I started thinking about the relationship between rolling resistance and weight because I found a few areas where I can remove some weight from my car. There's a slight sacrafice to the weight savings, so I wanted to quantify the benefits. I know there are many other things that add rolling resistance besides the tires, but tires are more elastic than wheelbearings, and tire pressure is adjustable and measureable, whereas the other components would remain unchanged.

Also unchanged are the tires on my car. If I remove weight, then the tires that carry my lightened car are the same tires that carried my car when it was heavier. In other words, in this exercise I wasn't trying to find optimal tire size based on different vehicle weight. With a given tire, it's contact patch is determined by load and inflation pressure. Yes, I'm ignoring the tires casing stiffness.

So essentially the info I posted has distilled the problem down to just two variables: weight and pressure. It's simplified. Gotta start somewhere.

Finding the optimum size tire for a given load and inflation pressure is a tough one. I know that my tires are listed as being 205mm wide, so if the contact patch is 15.86 square inches, then the contact patch would be 8" x 1.98" assuming it's a perfect rectangle, which it isn't. A real contact patch in this case would be a wide oval. If I were to mount a narrower 185mm tire on my car (same car) and use the same inflation pressure, the contact patch would be 7.28" x 2.18". A slightly rounder oval. Anyone know if that's a better shape or not? Should the contact patch look like a slice of Pepperoni, or a slice of Biscotti?
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Old 01-08-2008, 12:34 AM   #7
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I found a Michelin site once that basically said for optimal r.r. the contact patch should be "square". I know from bicycle sites and those super-high-mileage competitions that it IS possible to go too skinny- that is, if the contact patch is longer than it is wide, r.r. goes up. My own bicycling experience seems to confirm that. (seems, due to my lack of dedicated testing of it)

So... anything from "square" to longer-than-wide = bad for r.r. The data I have been unable to find is, what is going on when you go from "square" to wider-than-long? The Michelin comment would lead me to believe that r.r. would go up. I think that is a plausible assumption to make for many reasons, and I also think r.r. doesn't go up as fast as it does on the "narrow patch" side of the equation. Basically, a slightly wider tire shouldn't hurt much if at all while a too-narrow one will really underpeform.

I have done a lot of research into what would be optimal for my application if using the above assumptions.

Seems wierd, but the FIRST thing to do is pick the minimum tire psi you think you will run. Then plug in the maximum weight you expect to load the car to- on each end- and come up with the contact patch areas. Figure the length of the sides as if it were "square" and voila! there's your MINIMUM tread width.

Note I said tread width and NOT section width. DRW I believe in your last paragraph you assumed a 205xxxxx tire to have a 205mm tread width and that is not the case!!! You need to find tread width data or go measure it- it's a lot narrower than section width.

On my car I can go only about 20mm narrower in the front BUT if I accept lower weight limitations (like my years of history with this car suggests I can) I can go 60mm narrower on the rear!!!!

One thing I found interesting was that, when I calculated the tire sizes in this manner, their load ratings were quite closely matched to the loads I calculated!

Next thing to do is either find more data, or invest in the idea, get appropriate wheels and tires, and test it.
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Old 01-08-2008, 06:00 AM   #8
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Originally Posted by theclencher View Post
Note I said tread width and NOT section width. DRW I believe in your last paragraph you assumed a 205xxxxx tire to have a 205mm tread width and that is not the case!!! You need to find tread width data or go measure it- it's a lot narrower than section width.
Indeed, section width does not equal tread width. I went from 195 Michelins to 185 General LRR tires. 185/195 = 0.95, so you would think my new treads would be 5% narrower. But actual measurements showed the new treads to be 81% the width of the old tires (5.25"/6.5"), or 19% narrower.

My tire data in table form:
section - treadwidth
195 - 157
185 - 133
Roll on,

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Old 01-08-2008, 06:12 AM   #9
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Previous poster's comment about a square patch is sort of correct.

But I TOTALLY agree with tread width vs. section width. You could check this if you could see under the car through glass - shows contact patch perfectly, although most of us don't have a setup with several inch thick glass to look through.

Let's look at this from the bicycle point of view. If the patch is "square" then you have minimum deflection from the tires themselves. If it's longer than wide, you have underinflated tires. Basically the tire deflects more, giving a longer patch, and you have to "unstick" more tire, resulting in more friction - higher rolling resistance.

Let's look at the opposite view - wider than long. Vehicle view is best for this, but start thinking of a bicycle tire and you'll see why this makes sense. Imagine a car tire on a bike - super high pressure (say 100 psi) so that it's a wide but thin ribbon in contact with the road. See anything wrong with the wide tire on the bike, though? Scale it up to car sizes - the tires would be measured in feet or yards wide, and you could see that on turns the outside of the tire would travel farther than the inside, etc. So no good for vehicles.

Square patches result in the following:
1) difference between inside and outside travel of tire on turns
2) STOPPING POWER. Here's something that people don't take into account when trying to maximize for FE on tires. You have to stop SOMETIME. If it were a closed course, with no braking, you'd see bike-tire like geometry on vehicle tires.
3) EVASION CONTROL. Basically when you over-inflate tires, you'll decrease the size but also alter the shape of the contact patch. The higher the pressure, the more like a circle the contact patch turns into. For FE, the best shaped contact patch would be a circle, and the smaller the better. But because we need to evade vehicles, obstacles and turns (think about it for a minute) in the road, we need more friction. We need to overcome inertia in order to effect turns. Otherwise the vehicle stays going in the same direction - can't turn.

Basically safety wins out over FE, because we have to turn and we have to stop. The best FE would be ball bearings on a smooth surface. However, they would have the LEAST ability to turn, and stop, not to mention accelerate.

Upping PSI will give you better FE, worse turning and stopping power. Plain & simple. More load with more PSI will also alter the SHAPE of the contact patch, so a straight mathematical model won't work. It sort of works, but adding 500 pounds to a vehicle and then inflating higher will give you a more circle-like contact patch. Think rounding off the corners first, but it's more circle-like.

Basically the tire manufacturers work within limits. Taking reflexes into account - they have to underinflate over *optimum* conditions a few PSI. Taking turns into account - same thing. They are trying to balance people driving like idiots with fuel economy with stopping power with trying to get you to buy their product faster, and a few other factors.

My opinion - a few more PSI will help FE, but you have to balance that with how quickly that makes you buy new tires, as the tread in the middle wears faster, plus it decreases stopping & evasion power.

I hope this was somewhat coherent.
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Old 01-08-2008, 07:34 AM   #10
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Upping psi won't necessarily give you worse handling or stopping. I remember an article that was linked from here that spoke of a police department upping their trainer vehicles to 60psi, and that gained quite a bit of performance in all aspects.

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