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.) ?

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

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

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

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

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.

and we care because????

just kidding! at least your noggin is working unlike me

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