Quote:
Originally Posted by Arminius
Blow on your hand as hard as you can. It's just a little weaker than that, but with more volume because of the size of the holes - each about the size of a half dollar coin.
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Assume the area of the back of your car dragging the low pressure region is A. If there is a negative pressure back there of -1 inches of water, how high, in inches of water, will the pressure of the air coming out of the two hole have to be to for the force of the air coming out of the holes to equal the force on the back of the car due to negative pressure? This isn't exactly the right calculation, but it is one we can do quickly.
Let A equal 1, which is at least order of magnitude for a passenger car. The diameter of a half dollar coin is, um, 5 cm(?), which is .05m. Area of each coin is pi * .05 * .05 / 4 ~= (3 * 2.5 / 4) *10^4 ~= .0002 m^2. You have two such holes, so .0004 m^2. See where this is going? To completely negate the drag pressure the outgoing pressure would have to 10000/4 = 2500 inches of H20. Assume your car fan is really powerful and you have 10 inches of H20 pressure difference (this is pretty doubtful), at most you negate 1/250th of the drag.