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Very much related to the many questions and great answers to lift and weight in a climb - if an F-16 climbs vertically at a constant speed, is thrust greater than weight?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Farhan
    Jun 6 at 15:13
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Yes, thrust force vector must have a force equal to (but opposite) the combined gravitational force and aerodynamic drag force (from velocity) to climb at constant speed.

In horizontal flight, thrust equals drag to fly at constant speed. But some of the drag comes from the wing creating lift. Since the larger wing moves "a lot of air a little", it is more efficient at creating vertical lift than pointing an engine skyward.

That same F-16 therefor needs less thrust to fly horizontally at the same speed. Climbing vertically, it is essentially an air breathing rocket.

So, since the wing is useless in a vertical climb, the pilot would reduce AOA to zero lift, meaning at the same velocity the vertical plane has less aerodynamic drag than the horizontal one!.

But, thrust required for lift and drag is much less in the horizontal because the wing creates vertical force far more efficiently than the engine.

One can see how this apportionment of thrust to flight requirements leads to a consideration of Drag as Cd0 (drag from velocity) and Cdi (induced drag from lift), perhaps as T0 and Ti.

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  • $\begingroup$ The first paragraph is very clear. $\endgroup$
    – Koyovis
    Jun 4 at 2:48
  • $\begingroup$ Another way to look at it is going down the runway before rotation. Less Cdi, better acceleration. I like trikes. $\endgroup$ Jun 4 at 2:52
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    $\begingroup$ The thrust vector is in the opposite direction from the sum of the gravitational and drag vectors, so they are not equal. At constant velocity, they have equal magnitudes. $\endgroup$ Jun 4 at 3:23
  • $\begingroup$ Noted and edited, thanks. $\endgroup$ Jun 4 at 3:28

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