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Old 11-26-2013, 09:00 AM   #73 (permalink)
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I just realized that one can easily calculate/estimate the power requirements to overcome the pure air shear forces underneath the car.

(It is also known as Coutte flow: Couette flow - Wikipedia, the free encyclopedia)
(I was just too lazy to actually run the numbers before).
Assuming you have a car with a width of 1.7 m and a length of 4 m and decide to lower it just down to 8 cm above ground (anything lower than this wouldn't make any practical sense).

At 120 km/h one ends up with:
P = F*v = 18 * 10^-6 Pa*s * 33 m/s * 4 m * 1.7 m / 0.08 m * 33 m/s = 1.67 W
to overcome the shear forces underneath the car.
(Viscosity of air is: 18 * 10^-6 Pa*s Viscosity - Wikipedia, the free encyclopedia )
(1.67 W is about 0.01% of what is needed to overcome the rest of the resistance at that speed).

Granted that this is assuming laminar flow and one would actually deal with more resistive turbulent flow, but one would simply not reach a practical point where lowering has a detrimental effect, since the beneficiary effects of having less frontal area and lesser exposed wheel-wells would be higher.
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