rsl12

kutulu, it CAN be broken down into those two equations, with very basic assumptions about the way the plane is flying. Going both ways, the plane will be travelling exactly the same speed relative to the air around it. Drag and all other relevant forces will be functions of quantities that will be exactly the same in both directions. The drag force will NOT be stronger on the return trip--it will be exactly the same. there is no reason for the drag to be greater, since the air molecules will be passing by the airplane at the same velocity as when the plane was going to wherever it was going.

Again, consider this scenario--a boat is travelling downstream a very very deep and wide river (deep and wide because I don't want viscocity and distance to ground to be an issue, as it isn't with an airplane other than to determine turbulance and drag issues. the river can be a bit turbulent--that's fine). If the boat has no engine on it, then the steady state solution is that it goes exactly the same speed as the river water, if the water is not turbulent, and some slightly slower speed if the water is turbulent (and i remind you that turbulence is a scalar quantity). if it were going any faster, drag would slow it down. if it were going any slower, drag would speed it up.

If the boat had an engine that could read a speed of 10 mph, then the boat could go downstream at a speed of 10 mph + the river current speed. If it were going upstream, it would be 10 - the current speed. same reasoning.

EDIT: actualy, with turbulence, the boat I was wrong. the boat will go either slightly slower *or* slightly faster, as it gets joslted a bit in all directions, so you have a random walk effect.

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