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Posted by: Jeff Krieger May 23 2003, 11:33 AM

If you apply a force of F lbs tangentially to, say, a flywheel's radius of R feet to produce a torque of F*R ft*lbs and causing the flywheel to rotate, would the power you apply be given by (F lbs)*(2*pi*R ft.)*(R.P.M.) ?

Posted by: madd_dogg_914 May 23 2003, 01:16 PM

Depends on hom much boost you are running . . . oh wait, you can't turbo a 914! MWUHAHAHAHAHAHAHA

Actually I think that you need to factor in the flux capacitor somewhere in that equation. Then divide by 2A

Actually I have no idea what I am talking about, but I am sure someone will chime in who does.

-Chris

Posted by: Aaron Cox May 23 2003, 06:18 PM

sorry....just got out of physics class.....hmmm hold on..ill dig thru this one!


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Posted by: Jeff Krieger May 23 2003, 06:20 PM

Hey, I've got the Halliday and Resnick physics text too! Unfortunately, I've forgotten most of it.

Posted by: Aaron Cox May 23 2003, 06:22 PM

it is my dads! his old college physics book! HP is on page 145 fyi ok?

Posted by: Jeff Krieger May 25 2003, 11:41 AM

I think I've figured it out. My equation above can be rewritten as (F*R)(lbs*ft)(2*pi)(RPM) or (torque)*(2pi)*(RPM) and since 1 hp =33000 lbs*ft/min we have hp = [(F*R)(lbs*ft)(2*pi)(RPM)] * [1hp/33000lbs*ft/min] = (F*R)*(RPM)/5252.1 = (torque)*(RPM)/5252.1 which is the well known equation for finding hp when the torque and RPM are known.

Posted by: Aaron Cox May 25 2003, 11:43 AM

and we care because????

just kidding! at least your noggin is working unlike me

Posted by: Jeff Krieger May 25 2003, 12:04 PM

Just talking to myself. Actually, I'm just glad that I didn't have to review line integrals to figure it out.