> Typical car has a CD of about 0.4, something slick will be down about 0.2
> You can figure out how much hp it takes to go down the road at a certain
> speed
> with the formula HP=(CD*Q*A*(V^3)*6.8)/10^6
>
> CD = drag coe say 0.4 to 0.2 range
> Q = density of air in slugs or 0.0024
> A = frontal area of the car basically the cross section area. For example
> a
> small SUV might be 4.5 feet wide and 6 high with 1 foot of ground
> clearance
> so the area would be 22.5 The VW Vortex was aprox 13.45.
> V = speed to test at in mph. Ie if you want to see what hp it requires at
> highway speeds then use 70 or whatever. The amount of power climbs quickly
> from 45 on up as the air pressure builds as the cube of the speed.


Nice formula! Doesn't the force increase proportional to the square of
speed? I think if you double speed, the force goes up four times, AND you
cover twice as much distance in the same period of time, so the energy per
unit time is 8 times (force times distance).

In any case, the formula and typical numbers are interesting! It really
looks like the vast majority of the losses are aerodynamic. How much
improvement can we get by using "perfect tires?"

Harold

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