Originally Posted by
FL_Gator
I think the proposed explanation is that the deformation of the tire under weight is consistent for the entire revolution, thus the reduced radius is consistent for the entire revolution and therefore the effective circumference loaded is less than an unloaded circumference.
Sort of.
Edit: note that "patch" below refers to the
contact patch (the part of the tire that is touching the road).
Note that the result you are looking for is the
horizontal distance the center of the axle moves with a full rotation of the wheel.
This is related to the patch radius (the distance of the axle to the ground directly under it). With a load, the patch radius is smaller than the tire radius.
The axle is rotating around an "imaginary" circle that has the patch radius. That is, this circle would be the size of the wheel if the wheel was perfectly rigid (no patch deformity).
The
(C) distance the axle moves horizontally in a full rotation is
equal to the
(C) circumference of the patch circle.
(Imagine the patch radius below rotating full circle.)
The only number that matters is the vertical patch radius. You really only need to measure this number (accurately). That's kind of hard to do. So, people measure the roll-out distance instead (which is the circumference of the patch-radius circle).
What is going on elsewhere with the tire (the tire radius) doesn't matter at all. (The tire radius isn't strictly constant anyway.)
======================================================================
Note that the actual radius to use is the "rolling" (or effective) radius but it's hard to explain what is going on with that. (This was somewhat clarified below.)
The rolling radius would be forward of the contact patch and (generally?) would be larger than the patch radius but smaller than the tire radius.