Date:         Fri, 11 Sep 2009 17:58:07 -0400
Reply-To:     Edward Maglott <emaglott@BUNCOMBE.MAIN.NC.US>
Sender:       Vanagon Mailing List <vanagon@gerry.vanagon.com>
From:         Edward Maglott <emaglott@BUNCOMBE.MAIN.NC.US>
Subject:      Re: Rolling resistance
Comments: To: Kim Brennan <kimbrennan@MAC.COM>
In-Reply-To:  <33392408-3B6D-485F-A469-47AD415B9364@mac.com>
Content-Type: text/plain; charset="us-ascii"; format=flowed

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.