My 97 SR had 265/70's from the factory.

Don't need to search for the tire sizes...I know the SR was 265/70 OEM.

But are 265/70 THE SAME as 31x10.50? If so, then I guess my 31 Mich's were larger than the Yok's or whatever was on it because it was a little higher and the wheel wells were more filled...

265/70/15 is not the same as 31x10.5x15.

265mm/25.4 = 10.4331" wide

10.4331 x .7 (aspect ratio) = 7.3031" sidewall height (approx.)

7.3031 x 2 (sidewalls) = 14.6063"

14.6063" (2 sidewalls) + 15" (rim diameter) = 29.6063 tire diameter.

The formula doesn't allow for minor differences in tire specs, but it'll get you close enough for most calculations. If you need exact, then mark the tire on the floor and mark the floor. Drive straight for 10 revolutions and divide by 10 for distance covered per revolution. Then work back to find the diameter. Sorry, but I don't have scientific symbols set up on this keyboard, but for a hint it includes a slice of "pi". If you take the time to learn the formulae, it'll save you a lot of confusion.

Frank

Thanks FrankR! I with you on the math, just didn't think to run the calcs. (And that 29+" number comes back to me now...)

Would a physical roll-off give a number to compare to manufaturer's specs? At normal pressures, an automobile/truck tire will deform and alter the effective diameter to varying degrees depending on the KPa and GVW. Wouldn't the reduced diameter from hub to ground distort the calculation, or would the surface on the circumfrence end up tracking the correct distance anyway? <img src="/forums/images/graemlins/confused.gif" alt="" />

Tire dimensions assume proper inflation and no load, so yes - the vehicle's weight slightly distorts the dimensional characteristics of the tire, depending on air pressure.... easy to see with a flat tire. <img src="/forums/images/graemlins/wink.gif" alt="" /> The distortion is (by design) primarily in the cross sectional dimension (sidewall) rather than the circumference.

At low speeds, the tread circumference represents the total distance covered in one tire revolution. At high speeds, centrifugal force and developed heat causes the tire to "grow" slightly... the faster it spins, the more it grows, causing the distance covered by each tire revolution to slightly increase.

The GM manual calls for a change in ABS chip with tire size changes of more than 10% from stock. Their recommended method for measuring tire diameter is to mark the tire and pavement on a level spot, move the truck forward through several complete revolutions and divide the total distance travelled over the ground by the number of revolutions.

The slight distortion caused by vehicle weight on a properly inflated tire won't change the distance covered with one revolution. Think of a v-belt in either a circle or around a set of pulleys - the distance travelled by the outside edge of the belt is the same with one revolution of either example.

Frank

For a speedometer, the roll-off will give the practical numbers needed for accurate guage measurement, but I think these numbers will still differ from a theoretical calculation using factory specs. In the contact patch of a tire, the amount of tread that normally fills an arc is compressed into a line from the front to the rear of the tire's footprint. Wouldn't that compression lead to a smaller distance travelled (my tires look like they lose almost an inch in diameter) when rolled as opposed to, say an unmounted tire of the same size?

I don't think so, but if you want to prove it to yourself one way or the other, try it with an unmounted tire, then put it on your truck and repeat the drill. Make certain several complete revolutions are made.

Frank

Here's a link to dementions of tire sizes and their conversions. Tire conversions