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Originally Posted by Brian Carlton
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.
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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.