2010 - How much does change in tire pressure affect tire circumference?

29984 Views 25 Replies 17 Participants Last post by TTiimm, 4 mo ago

SierraHotel

Discussion Starter · Dec 20, 2012

We know that it is important to maintain tires with close to the same circumference on AWD cars, and pay a lot of attention to keeping identical tires with similar tread depths on the car. I have often seen it mentioned on the forum that maintaining proper tire pressures is also important as tire pressures affect tire circumference. I have not noticed any numbers regarding how much a drop in tire pressure in a tire affects circumference... so I did a little experiment.

I attached a bar clamp to a tire with a long nail taped to the end to act as a pointer. I taped a ruler to a board and positioned the board vertically next to the point of the nail. I inflated the tire (a cold 225/55 17 Geolander G95) to 36 psi and noted the pointer position on the ruler. Then I dropped the pressure in 2 psi increments down to 20 psi...noting the change in position every 2 psi. Then I reinflated back up to 36 psi in the same 2 psi increments.

The pointer dropped 1/32" every 2 psi from 36 psi to 26 psi, then dropped about 2/32" every 2 psi from 26 psi to 20 psi. The measurements were the same while reinflating.

So.... a 2 psi change in tire pressure (above 26 psi) results in a 1/32" change in radius... and therefore a 2/32" change in diameter. 3.14 X 2/32" (pi x d) = a little over 3/16" change in circumference. Double that for pressures between 26 psi and 20 psi.

http://i1190.photobucket.com/albums/z450/SierraHotel058/IMG_0039_zpsb836a9bf.jpg

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KenBourque

· #21 · 4 mo ago

Under load, the tire flattens on the ground. For this to happen, the tire must abrade (slip) on the ground. Consequently, the loaded circumference is less than the unloaded one. The rolling radius is therefore less than the half the diameter of the unloaded tire. As the wheel rotates, the distance from the center of the wheel to the ground remains the same as new rubber is positioned under the center of the wheel. Hence, it's the loaded radius that must be used in calculating the circumference. This reduction in the rolling radius is what I've never been able to find published probably because for the same nominal radii of tires, the width of those various tires is variable, and this determines how much the tire will be deformed (non circular) under load.

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Kevin

· #22 · 4 mo ago

G'day & Welcome @KenBourque

Could you please update your Public Profile; vehicle details at least; thanks. For detail instructions pls refer: Your forum "Public Profile" - how to fill this...

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KenBourque

· #23 · 4 mo ago

I think very little if you're within say ±5 lbs. However, it's not the circumference that determines the rolling distance on the ground, it's the radius under load that must be used which is not half of the nominal diameter. As I said in my post, I don't know how to accurately calculate this.

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jadatis

· #24 · 4 mo ago

I think to know how it goes.
From the surface-length on the ground, and the loaded radius ( unloaded radius minus deflection) you can calculate the angle the tire makes and the surface length.
Then( 360 / calculated ange) x surfacelength is rolling circumference.

Then different tires have different overgoing curves from unloaded radius to flat on the ground ( see paint image I made), wich gives different surface length at same deflection. So best is to measure surface length bij sliding piece of paper at front and rear surface and measure it.

Now roughly you can calculate for other pressure the new surface length, and from that the new deflection. This than use to calculate the new rolling-circumference.

Now where does the missing treathlength goes to.
My idea is that before going to the ground, and after going from ground, the treathsegment moves a little backward to the underground. This has to be compensated by the same distance forward moving to get at the same point.

example measured treathlength off the ground 314, rolling circumference 306.
Then 314 minuus 306 is 8 / 4 = 2
Treathsegment goes frontward 2, then moves backward 2, stands still while on the ground, moves back when off the ground at the back 2, then moves forward 2 and still no forward moving to the ground.

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jadatis

· #25 · 4 mo ago

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TTiimm

· #26 · 4 mo ago (Edited)

How does a vehicle's front/rear weight difference affect dire diameter? Does the fact that the more weight is on the front of the vehicle impact diameter? If more weight changes the tire's diameter, will a higher pressure be required?

Keep in mind that driving isn't always done on straight roads at a constant speed. Braking (temporarily) shifts a majority of the vehicle's weight to the front. Similarly, cornering unweights the inside wheels and weights the outside wheels. I've lifted the outside, rear wheel in my Forester under hard braking and cornering, to the point where the outside rear wheel locked up.

(not my car, just an example)

How do we adjust pressure and keep Subaru differentials from burning up as weight distribution, and therefor tire diameter change as we drive - brake, turn etc?

My answer is to put keep the tires rotated, put 34 PSI in the front, 31 PSI in the rear and drive. Anything else is overthinking it. I'm more concerned with maximizing contact patch and therefor adhesion. Tire diameter is metadata, interesting but not a primary metric.

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