> 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.

Note that this is based on exactly the same formula that I mentioned 
recently for drag of a body. The only difference is a different 
constant due to weird UK units (snails cousins) AND an extra velocity 
term to turn drag into power. ie Power = Drag x Velocity.

My formula was Drag = 0.5 x Cd x Rho x A x V^2

Rho = air density = 1.3 kg/M^3
A = frontal area in metres
V = velocity i metres/second
Drag = drag force in Newtons (eg Kg x g ~~= 10 x kg).
Cd = coefficient of drag = improvement relative to a flat plate.

So

STARTER FOR 10 POINTS

Give an flat plate object with Cd = 1.
Intuitively derive the above formula.
A word description would suffice, probably with a few simple formulae.
Once you realise that this is totally intuitive it does marvels for 
seeing places that it can be used.
No Reynolds numbers and the like here. Just intuitive common sense and 
a few basic formulae.

It works "well enough" for raindrops, skydivers tucked in or spread 
out, parachutes, falling field mice, cars and more.



        Russell McMahon




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