For one of our math whizzes out there.

Say you have a 275-60-15 inch tire mounted and aired up.

Laying flat on the ground, let's say measures 28 inch diameter.

Now mount the wheel on the car, lower to where is is sitting on the pavement.

As tire will flatten out a bit from the weight of the car, perhaps it is now 27 inches from the pavement to the top of the tire.

How does this effect the rolling circumference of the tire?

If no flattening, 1 full revolution would move the auto 28 inches times pi of 3.1416 equal 88.9 inches.

How far will the car move forward in one revolution allowing for 1 inch less diameter due to this flattening?

Often have wondered about this and how it effects odometer calculations.