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rsl12 · 07-26-2004, 02:12 PM

kutulu, this is incorrect. a plane will be push by two horizontal forces (yes you philosophy majors, there are more than 2 forces, i don't need to hear about it): the force exerted by the engines, and the force of drag. If the plane is going at a constant speed, that means that the two forces cancel each other out. assuming that the weight of the plane and the force of engines remain the same in both directions, we only need to look at drag. drag is a function of the speed that the plane is travelling relative to the air about it. whether or not the wind is a tailwind or headwind is irrelevant, because the plane will reach the same velocity relative to the wind, no matter whether it's coming from the front or behind. (winds coming from the side are a different matter.) in other words:

Fe+Fd = 0

where Fe = vector force of the engine
Fd = vector force of drag, which is a function of Vp, vector velocity of the plane relative to the air around it (the turbulence of which we assume is constant). If the engines are turned off and the plane miraculously still floats, and the (laminar) air is going 500 mph, the plane will (eventually) also go 500 mph, so that the equation balances. Basically, the plane will go as fast as necessary to balance this equation (ie, as fast as necessary to achieve the Vp (velocity vector) value that satisfies the equation. The equation is unsolvable if the wind is not exactly facing towards or away from the plane, or if the plane is rising or sinking. (Then you need to take into account the rudders or whatever else planes use to steer themselves). But other than those issues, the equation is independent of the wind speed relative to the ground.

Basically, what I'm saying is that the question asked is perfectly fine as it is, even though it's worded poorly.

07-26-2004, 02:12 PM	#16 (permalink)
rsl12 On the lam Location: northern va	kutulu, this is incorrect. a plane will be push by two horizontal forces (yes you philosophy majors, there are more than 2 forces, i don't need to hear about it): the force exerted by the engines, and the force of drag. If the plane is going at a constant speed, that means that the two forces cancel each other out. assuming that the weight of the plane and the force of engines remain the same in both directions, we only need to look at drag. drag is a function of the speed that the plane is travelling relative to the air about it. whether or not the wind is a tailwind or headwind is irrelevant, because the plane will reach the same velocity relative to the wind, no matter whether it's coming from the front or behind. (winds coming from the side are a different matter.) in other words: Fe+Fd = 0 where Fe = vector force of the engine Fd = vector force of drag, which is a function of Vp, vector velocity of the plane relative to the air around it (the turbulence of which we assume is constant). If the engines are turned off and the plane miraculously still floats, and the (laminar) air is going 500 mph, the plane will (eventually) also go 500 mph, so that the equation balances. Basically, the plane will go as fast as necessary to balance this equation (ie, as fast as necessary to achieve the Vp (velocity vector) value that satisfies the equation. The equation is unsolvable if the wind is not exactly facing towards or away from the plane, or if the plane is rising or sinking. (Then you need to take into account the rudders or whatever else planes use to steer themselves). But other than those issues, the equation is independent of the wind speed relative to the ground. Basically, what I'm saying is that the question asked is perfectly fine as it is, even though it's worded poorly. __________________ oh baby oh baby, i like gravy.