

| Subject: | Re: Rolling resistance |
| From: | Edward Maglott |
| Date: | Fri, 11 Sep 2009 17:58:07 -0400 |
I'm having trouble getting this. Are you saying is that it's the rotating mass and not the diameter that uses more energy? So if you had a larger diameter tire/wheel combination that had the same mass as a smaller one, there would be no change in the amount of energy required to spin it? Where is the friction that you refer to, and why is it increased with a larger diameter tire? It seems like the whole package from the tranny input shaft all the way to the tire determines how many engine revolutions equal 1 mile. If you have a smaller tire and higher top gear in the tranny to equal the same overall ratio, would you use less energy to go the same distance? Thanks, Edward At 01:59 AM 9/11/2009, Kim Brennan wrote: >I didn't phrase it particularly well. For best fuel economy you want >lightweight tires/wheel/rotating mass. The rest of the unsprung weight >is irrelevant towards fuel economy (though yes, you are correct it has >a lot to do with handling and ride.) > >Simply put, a low mass rotating body is much easier (i.e. uses a lot >less energy) to spin than a higher mass rotating body. Mass always >matters. If we lived in a world without friction, Newton's laws of >motion would rule supreme. But we live in a world with friction, so >what is in motion does not stay in motion. And to keep it in motion, >requires energy. This applies both to vehicles traveling at a steady >speed, as well as vehicles accelerating.
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