No, not the frictional losses, if I understand you correctly. I'm obviously having difficulty in being clear, which is my fault. Imagine that one has a steel wheel which has a flat spot similar to the radial where wheel meets flat road, level grade. But, the steel wheels geometry doesn't change from flat surface to circle instantaneously. So, we begin to roll this steel wheel forward. We first have some difficulty (power = X), then as we hit pure circle geometry we only have the rolling resistance, with all the variables that contribute to that rolling resistance (here Power to roll the wheel = X-a, where "a" is some constant). In the rubber radial, I think, we have the same basic phenomenon going. Though, with the caveat that the first non-flat spot is also circular (for some small distance) but that arc belongs to a circle of much greater diameter than our 20" diameter (for example) wheel. That arc always just ahead of the flat spot belonging to a circle diameter of >20", which in a continuous fashion moves towards the arc belonging to the 20" diameter circle uses some of the power put to the wheel by the axel. This is the basis by my saying the tire of relatively non-rigid geometry is essentially always going "uphill" in the same manner as if the relatively rigid steel wheel "experiences" going uphill, though the radial does it all of the time and the steel wheel only goe's uphill when the road grade is uphill. I do understand that even in my analogy that the flexing rubber has it's own intrinsic frictional loss as it rolls. apologize for moving towards a point, correct or not, that moves perhaps too far away from Post #1. I agree, Duke has provided us with the mathematical versus the intuitive physics, which is invaluable.

Absolutely, and I believe there are free body diagrams and equilibrium in dynamics too. I am interested in knowing where is that extra force applied, and how it is transmitted to the wheel. If you are right, you are right. I want to learn.

JL

There is no "uphill". The steel wheel has a single contact point and there is no deflection in the wheel. There are no losses associated with the deflection.

The rubber tire must be deflected as it moves forward. The losses are not great but they are more than the steel wheel.

If there was an "uphill", you would never be able to push an automobile yourself........the slightest "uphill" and you're toast.......as you probably are aware.

I frequently push the SD in and out of the garage because I don't want to bother to start it.........it usually takes most of my capabilities to do that...........and I'm not going uphill.

Yes, the "uphill" I was referring to in the rigid wheel is an actual uphill gradient of the road surface. The single point contact of the rigid wheel is what makes it easier to push your "car". So, I agree with you on that.
I do disagree if there was an "uphill" (the uphill instrinsic to the tire, but with 0% gradient) in the case of the radial, one would still be able to push the car on level gradient. It is more difficult, by far, than the rigid wheel, but that is the difference created by the non-rigid wheel, exclusive of frictional losses versus the rigid wheel. Yes, the difficulty in pushing your car out of the garage on flat gradient, versus what work you would experience with a rigid wheel is the "uphill" (poorly named, I know) created by the radial. It doesn't approach trying to push your car up a 6% gradient, but phenomenologically it is similar and a portion of that difficulty is the "uphill" intrinsic to the non-rigid wheel. The arc length of the non-rigid wheel just ahead of where it going to next meet the pavement is greater than the arc length of the non-rigid wheel at the top of the tire. It is the change in arc length which is the metric associated with what I've referred to as "uphill", in the mathematical sense.

I agree, but, I don't advocate the term "uphill" to describe the phenomenon.

The effort is expended to effect the angular change of the tire contact patch.

Yes, "uphill" was a bad term. At the end of the post I awoke enough to say "arc length". Arc length directly yields the tangent vectors, which is, I think, synonymous with "angular change".

Thanks for helping me get this. And thanks to all the contributors to this thread... I appreciate the chance to eavesdrop on your insights.

__________________
-Marty

1986 300E 220,000 miles+ transmission impossible
(Now waiting under a bridge in order to become one)

Reading your M103 duty cycle:
http://www.peachparts.com/shopforum/831799-post13.html
http://www.peachparts.com/shopforum/831807-post14.html

Anytime.

Help is what we do here.