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Every pilot is familiar with the basic model of aircraft forces:

lift, weight; thrust, drag

If lift is greater than weight, the plane climbs, if less, descends. If thrust is greater than drag, the plane accelerates, if less, decelerates. In unaccelerated flight, all must be equal.

In an engine-out situation, there is no thrust, so if a constant glide ratio is to be maintained, I thought the plane must be pitched forward so the forward component of lift does the job thrust would normally do in cancelling drag.

However, in the Airplane Flying Handbook, 4-3, figure 4-2, it shows an example lift-drag diagram that shows $L/D_{max}$ ($= V_{BG}$) to be about $6°$ AoA. Wouldn't that result in lift having a backwards component and slowing down the aircraft to the point of stall?

What am I missing here?

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    $\begingroup$ While I was posting this question, I worked out the answer: The aircraft is pitched nose-down, but AoA is still positive due to the relative wind. Weight is what counteracts drag. I'll leave this question here though incase anyone has a good explanation or diagram to share. $\endgroup$ – Zaz Aug 9 '17 at 0:19
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    $\begingroup$ AoA and pitch angle are not the same thing. You can have +6 degrees AoA and still be descending. $\endgroup$ – Riccati Aug 9 '17 at 0:20
  • $\begingroup$ It is the same as this question -- see this answer-- aviation.stackexchange.com/questions/56352/… $\endgroup$ – quiet flyer Nov 13 '18 at 10:56
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Angle of attack is not equal to pitch angle. Yes you can decompose weight into a forward and a perpendicular component - to airstream! Or you can draw lift relative to airstream.

enter image description here

In this drawing, pitch angle is zero and AoA = 6 deg. Lift now has a forward component relative to gravity.

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If at zero airspeed and zero absolute altitude this were to happen, then yes the aircraft would stall and be incapable of flying. Then again, at this point, it would not matter, AS THE AIRPLANE IS ALREADY ON THE GROUND IN THIS STATE.

In all seriousness, though, the airplane would not slow and stall unless you attempted to remain at a constant altitude. In a constant airspeed glide, however, where the airplane is descending at a shallow angle relative to the ground, it can be shown that the force of gravity is no longer perpendicular to the longitudinal axis of the aircraft and will have a component vector which is parallel to the longitudinal axis and opposed to the direction of both the induced and parasite drag vectors.

Another way to think of this is to consider the energy state of the airplane at the time the engine quits. Its total energy is the sum of its potential energy, proportional to its absolute altitude, and its kinetic energy, proportional to the square of the groundspeed. Once engine thrust is lost, that energy state begins to be drained away by the force of drag applied over a distance. In order to maintain airspeed, the aircraft must begin to convert its potential energy into kinetic energy, causing the airplane to descend slowly until it reaches the ground.

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