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This is purely a thought experiment / theoretical question, not asking about practical engine out or best glide procedures.

Say you are cruising at 150 knots and suddenly lose all power. You best glide (V-glide) speed is 75 knots (the numbers don't matter). Would you glide further (straight ahead, no turns) if you:

a) Pitch up to quickly drop your airspeed to 75, then immediately pitch to maintain V-glide (gaining altitude in the process), or

b) Hold level flight until your airspeed bleeds off, then immediately pitch to maintain V-glide (gaining more forward distance in the process), or

c) Some other maneuver?

I believe I've seen an answer to this before but I cannot find it :/ Apologies if this is a re-post.

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    $\begingroup$ Not a pilot, but I've always read to pitch up up best glide speed. It makes sense, in that the longer you are not at best glide speed, the shorter your glide will be. $\endgroup$
    – RetiredATC
    Commented Nov 30 at 17:50
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    $\begingroup$ In small one engine planes I have flown (Piper Cherokee, Cessna 172, ... ) the standard procedure is to trade speed for height. Height gives you more options in where you want to aim for your crash landing, and only slightly decreases the glide distance. $\endgroup$
    – ghellquist
    Commented Nov 30 at 20:34

3 Answers 3

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Once you lose all power, you become, in the first approximation, a ballistic projectile.

More accurately, you're aeroballistic, but the rules remain similar. The longest-range trajectory will be one where you first trade KE for PE (potential energy), then PE for KE, at an equal rate, resulting in a roughly-symmetrical ballistic arc. However, due to drag, it's isochronously symmetrical, i.e. arc segments take equal time rather than distance.

Such an arc looks roughly as follows:

enter image description here

With a difference that alpha will be determined by your aerodynamics. Overall, your goal will be to lose as little total energy as possible at any point in time. For an artillery shell, the optimal AoA is zero, and alpha is 45 degrees.

For a real airplane, the calculation will be somewhat complex, but comes down to maintaining the best possible lift-to-drag ratio at all times.

The simplest rule is: while above your climb speed, climb, then glide, while minimizing drag. Stick to your best climb configuration until halfway between your best-glide and best-climb speed, and gradually transition to your best-glide configuration at that point.

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    $\begingroup$ The words "ballistic projectile" made the answer very clear! Never thought of it that way. $\endgroup$
    – codenoob
    Commented Dec 1 at 15:42
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    $\begingroup$ Ballistic projectile is patently false. $\endgroup$
    – Mark Jones Jr.
    Commented Dec 1 at 18:57
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    $\begingroup$ I think ”ballistic projectile” is wrong by definition. An aircraft that loses its power is very much steerable and generates plenty of lift - that is the whole point of this question. $\endgroup$
    – busdriver
    Commented Dec 1 at 20:34
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    $\begingroup$ @Therac a controllable projectile isn't in ballistic flight. Obviously every "ballistic" missile has some non-ballistic period of flight, otherwise they wouldn't go anywhere (e.g. an ICBM has an initial burn period), but "ballistic" refers to the non-controllable portion of flight. $\endgroup$
    – Acccumulation
    Commented Dec 1 at 20:45
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    $\begingroup$ "The longest-range trajectory will be one where you first trade KE for PE (potential energy)" Cite? The optimum trajectory for a ballistic trajectory is based on being, well, ballistic. In a ballistic trajectory, you don't have any choice as to whether KE is being traded for PE or vice versa. The "optimum" is assuming you have a choice in initial angle, and finding what the best choice there is. $\endgroup$
    – Acccumulation
    Commented Dec 1 at 20:48
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If the wind were calm, you would glide farther straight ahead if you were to pitch up to turn excess kinetic energy into potential energy. You can even fly (briefly) under stall when rounding off the top of your climb as your stall speed decreases as the wings are unloaded.

If you bleed off the airspeed, then you are losing that energy at a less efficient point of operation.

And if you pull up too hard, you might scrub lots more energy than you realize in the high-gee maneuver.

If the wind isn't calm, then there are certain choices to be taken into account to maximize your glide range, but that's something we glider pilots think about a lot more than normal folx.

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Somewhere between a) and b).

An aircraft best converts its kinetic energy into lift when the wing is at optimal angle of attack. Here the lift to drag ratio is highest.

If one is cruising at 150 knots, the angle of attack will be 1/4 that at 75 knots. So, bringing the wing to optimal angle of attack for optimal V best glide will cause the aircraft to climb.

However, in a real situation, if the engine went out suddenly, the aircraft would sink, and the pilot, in pitching to Vbg, would initiate a climb from that point, perhaps crossing the original altitude.

pitch of stick/trim tab should less than or equal to stick/trim tab position at Vbg

Where ever Vbg is reached would be the end of the climb. Attempting to gain more altitude (up to the point of stall) would be a waste of kinetic energy, trading less efficiently for more potential energy which will be needed to accelerate up to and to maintain Vbg.

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