Concerning your bulldozer comment, to an extent I agree that the run of its track (i.e., the distance traveled by a full revolution) will have negligible variances if slip and other factors are ignored. However, other than the fact that a tire revolves and is used on a vehicle, its design is nothing like a continuous track. Foremost, the form of a tire is largely supported by gas pressure not solely by the rigidity of its material. For radial tires, the layout of the belts causes deflection (bulging) of the sidewall under load, which is a reduction in the free / hanging radius. Your question appears to be "where is the circumference going if the radius is being reduced?". What I'm trying to explain is that as each section of the tire bears the load (i.e., rotates to a position--the bottom--where it has to support the weight of the vehicle),
the tread compresses AND the sidewall deflects, increasing the contact patch. There's a good description available from Google books (
link;
Tire Tread and Tire Track Evidence by William J. Bodziak, Pages 4 & 5, "How a Tire Carries the Load"). With that said, there is no fixed revs/mile value that will be completely accurate for all speeds as centrifugal force will lead to some tire growth. This effect is negligible for most street tires under normal conditions, so static loaded radius is used for calculating revolutions per mile.
Also, a correction to my above posts; I stated ~3% loss in radius. It should have read a ~3% loss in diameter / circumference / revs per mile (which is equivalent to a ~1.5% loss in radius). I've edited the posts accordingly.
Load and inflation both definitely effect the revolutions per mile, so I suggest you reevaluate your experimentation or measurement equipment. I've done a good amount of experimenting myself with this on the Marauder (four sets of wheels--3 I own--and many types of tires) and also on other vehicles.