Never mind, I think I just answered my own question. I failed to notice that
both the differential ratio and the final drive ratio were in the
equation.
Cheers,
Joe (C)
Joe Curry wrote:
>
> While both these equations may come up with the same figure, I suspect that
>they are based on the gearbox having a top gear of 1:1 ratio. What about
> my gearbox that has 2 more gears that are both overdrive?
>
> Inquiring minds want to know!
> Joe (C)
>
> BillDentin@aol.com wrote:
> >
> > In a message dated 08/08/2002 3:45:54 PM Central Daylight Time,
> > mjb@autox.team.net writes:
> >
> > > Tire Diameter (in) x RPM x 188.5
> > > Speed (mph) = -------------------------------------------
> > > final drive ratio x gearbox ratio x 63360
> > >
> > > I just have this spreadsheet I wrote up years ago where you plug in a few
> > > bits of info and you get the chart of speeds in the various gears. I
> > > haven't
> > > actually sat down with a calculator and the equation for quite some time.
> > >
> >
> > Mark:
> >
> > We have similar to the above on a spread sheet as well, for the several
> > different tires sizes and drive trains we run. I am not very secure about
> > matters mathematical (although I like to multiply), but I am interested. We
> > normally get our tires from SASCO SPORTS, and the formula they gave us was:
> >
> > MPH=(((ENGINE RPMs*TIRE DIA.)*0.002975)/(DRIVE RATIO*RING&PINION RATIO))
> >
> > Seems to be a variation. I've not tried the two equations yet to see if
>they
> > come up with a similar answer, but 188.5 divided by 63360 = .002975, so I
> > guess both would get similar answers.
> > .
> > At one time BFG had given me an equation that involved tire revolutions per
> > mile. I seem to follow the logic on that one a little better, but I've
> > misplaced the equation.
> >
> > Bill Dentinger
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