John, thank you for the brilliant idea!
I was wondering how to compensate for
the airdrag and had looked at some methods.
Your "neutral gear + wait and see" approach
appeals to me on so many levels, as I had
looked at previous results of wind drag
from manufacturers figures and wind tunnel
tests of the common race bikes. Your way
also takes into account tyre drag, brake
drag, rider wind drag, chain losses, etc etc.
Very nice. I do so love a real world test
over a bunch of formulae. :o)

Thanks again!
-Roman



John Gardner wrote:
>
> Roman Black wrote:
>
> > If speed normally goes from 50mph to 60mph in X
> > seconds, does this mean that 50mph to 60mph in
> > 0.90909 X seconds means a 10% greater hp in that
> > range?
>
> Probably not. Aerodynamic drag varies as the square
> of velocity, and accounts for most of the power at
> speeds over 30-45 mph, depending on the slipperiness
> of the vehicle.
>
> If you log rpm, velocity, and acceleration, and you know
> vehicle weight accurately then HP and torque can be cal-
> culated, once you have a set of data representing the
> aerodynamic deceleration of the vehicle - To generate this
> simply accelerate to top speed, shift into neutral, and
> log velocity and deceleration as you slow.
>
> (Straight, level, no wind, of course...), then
>
> If acceleration is in "g" units,
>
>        Force = Weight (lbs) X (acceleration + deceleration)
>
> If speed is in feet/second, since 1 HP = 550 ft/lb/sec,
>
>        HP = (Force X Velocity) / 550
>
>        Torque = 5250 X HP / RPM
>
> Doing it real-time with a PIC sounds like fun. Hope this is
> of some use.
>
> regards, Jack
>
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