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Old May 09, 2012, 07:55 PM
aeronaut999 is offline
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Originally Posted by BMatthews View Post
You're still tied to the idea that the plane has to push against the earth. This is simply not so.
Define the meaning of "has to". I don't mean that lift is tied to the creation of a push against the earth-- for example, a rudder creates sideways force without any push against the earth. But, the rudder can't make a net sideforce indefinitely without making the flight path curve. Curving flight is not a valid inertial reference frame.

What I mean by "has to", is that if the plane did not push against the earth, the plane's gravitational pull would accelerate the earth upwards. This obviously violates the laws of conservation of momentum. For example, in a few thousand years, the plane would accelerate the earth to a delta v comparable with the earth's orbital velocity around the sun. This is nontrivial. It doesn't make sense that this should happen. It doesn't happen. Because the earth does "feel" the aircraft's weight.

To be perfectly consistent, assume an earth far from any sun. Not describing a curved path through space. Only travelling in a straight line. This earth is a valid inertial reference frame. A plane creating lift over this earth would accelerate this earth upwards, by gravitational attraction, if the earth did not "feel" a downward push under the plane equal in magnitude to the plane's weight. This would violate the law of conservation of momentum.

This cannot be. So, we are forced to conclude that the earth does feel a downward push equal in magnitude to the upward gravitational attraction exerted by the plane, i.e. equal in magnitude to the plane's weight.

The reason the earth "feels" the wing's downforce but does not "feel" the rudder's sideforce, is that the rudder's sideforce is not aimed at the earth, while the wing's downforce is aimed squarely at the earth, except during transitory manuevers such as turns and loops. During a turn, a 1-G downforce component is still aimed at the earth. During a loop the average downforce component is still 1-G. So over the long run, the earth "feels" a 1-G downforce (equal to the weight of the plane) even during these maneuvers.

Surely this is obvious. If we have a lightweight framework enclosing a cube of air one mile per side, open on top, and a plane is flying in the cube of air, a scale under the cube of air measures more weight when the plane is engaged in a 2-G pullout, then when the plane is engaged in a zero-G pushover. Does anyone doubt that? It seems obvious to me.

Over the long run, on average, the scale "feels" a force equal to the weight of the plane.

Just as a scale under an athlete registers more weight when the athlete's legs are accelerating him/ her upward at 2G's, than when his/her leg muscles go slack and he/she accelerates earthwards with no resistance to gravity (0-G).

But over the long run, the scale registers a force equal to the athlete's weight.

I realize that this rocks some people's paradigms and they are forced as a result to use the "waste of time" argument. But, that is not a valid argument against the truth....

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Last edited by aeronaut999; May 09, 2012 at 08:46 PM.
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