Quote:
Originally Posted by spyderbite9001
Sorry to burst ur bubble man, but your looking at the wrong equation. For air resistance yes, it is a function of the square of your velocity (v^2) however, when we talk about gas mileage we should be more concerned with the power req'd to gain/maintain the objective velocity. To calculate the req'd force or power we look at the equation
Power = Force Drag * V supposing that
Force Drag = (1/2)*density fluid*(v^ 3)*area*drag coefficient
and therefore the growth in power needed, which directly translates to gas mileage, is in fact, exponential.
Reference: Drag (physics) - Wikipedia, the free encyclopedia
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No, sorry. Fuel mileage is "distance/fuel." This is distance/energy as far as we are concerned, since fuel is energy for us. Invert this for energy/distance. Then notice that energy is equivalent to work. This leads to work/distance. Work is force*distance, so now we have (force*distance)/distance or force. Therefore, mileage is proportional to the inverse of force. Air drag force is proportional to the square of speed, so mileage is a function of the inverse of the square of speed.
Besides, what you have for power (which is correct by the way) is still not exponential, it's just a power function with a different power, 3 instead of 2. If it were exponential, it would increase as a fixed number raised to the power of speed.
You're missing the fact that while power required to overcome aerodynamic drag does increase with the cube of speed, distance covered increases with the first power of speed, so energy required to overcome aerodynamic drag to cover distance increases with the square of speed.