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If an aircraft like a Cessna 172 is dropped from a helicopter, what absolute minimum altitude above the ground will be required to safely land the aircraft?

helicopter crane carrying cessna 172

CONDITIONS:

  • Weather: Perfect conditions, 0 knot wind.
  • Ground: 0f ASL.
  • Gross weight: Max gross weight.
  • Engine: off.
  • Speed: 0 knot(standstill drop).
  • Pitch Attitude: 0 degrees(as in the picture).
  • The helicopter is just an "abstraction"(there is no downwash exerted on the airplane)

PS: if you can simulate this on XPlane your results will also be welcomed(I don't have it).

Also, if you want to answer from a slightly different scenario like a stall, though not the same as the scenario described above, your answer will also be welcomed.

NOTE: The question is purely about physics / aerodynamics. Not planning on doing this in real life.

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    $\begingroup$ Btw, you did not actually specify the pitch attitude of the airplane at the moment of release. (You drew a picture but did not say that it was supposed to exactly constrain the situation under consideration.) The pitch attitude of the airplane might matter a lot. Google "hang glider balloon drop", this is a zero-airspeed situation at least as far as horizontal airspeed goes, and it has been found to be the best practice to drop the glider in a strongly nose-down pitch attitude. $\endgroup$ Commented Feb 3 at 18:17
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    $\begingroup$ @quietflyer that video was very useful to see the physics in a real life example. It looks he is at 0 degree attitude though. The recovery was very quick btw, almost instant, really surprised. $\endgroup$
    – Gabe
    Commented Feb 3 at 18:48
  • $\begingroup$ Glad I could contribute something useful -- keep in mind though that the moment of rotational inertia of a hang glider might be much lower than that of a Ces. 172-- PS you may not have watched the same video that I did, I know some of the training manuals for hang gliding recommend a rather nose-low pitch attitude for balloon drops-- $\endgroup$ Commented Feb 3 at 18:49
  • $\begingroup$ watched this: youtube.com/watch?v=YB1IH2CgGB4 $\endgroup$
    – Gabe
    Commented Feb 3 at 19:39
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    $\begingroup$ Not to be pedantic, but groundspeed is irrelevant. Forward airspeed is what matters. $\endgroup$ Commented Feb 4 at 0:03

2 Answers 2

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Minimum altitude is 0 ft: simply release the plane and let it sit pretty on the ground. Call it cheating, but it fits the scenario perfectly.

If you want it high in the air, what probability of landing is required? Aircraft can sometimes land in the strangest of circumstances. So let's say there's at least a 1% chance at any altitude.

If it's being dropped from an actual helicopter, rotor downwash and attitude will affect the outcome. Safe practice for stall recovery normally requires 3,000 ft, recovery itself takes less than 1,000 ft - so that's a safe altitude, but it's done in a plane with some forward airspeed.

If the "helicopter" is an abstraction, and the conditions are better, this depends on just how good the conditions are. At the right speed and attitude on the glideslope, any altitude will work.

With the new input - 0 airspeed, no engine:

A non-aerobatic GA airplane like the 172 is at risk. At zero initial airspeed, the control surfaces have zero authority; the plane might spin or tumble and never resume controlled flight. Or it might fall nose-down, but that's a gamble.

4+ generation fighters and high-end aerobatic aircraft would most likely recover. They have all-moving tails, oversized control surfaces, can take the g-loads involved, and fighters add thrust vectoring and computers.

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    $\begingroup$ Stalls are often done at 3000 feet, but yikes, a power off stall recovery, even without reapplying power, doesn’t take more than a couple hundred feet. $\endgroup$ Commented Feb 3 at 7:33
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    $\begingroup$ Not disputing that. But your original statement of “Stall recovery normally requires at least 3,000 ft.” was wildly inaccurate. $\endgroup$ Commented Feb 3 at 7:47
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    $\begingroup$ Or, edit the drawing so it drops with the nose pointing down... $\endgroup$
    – Ralph J
    Commented Feb 3 at 8:29
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    $\begingroup$ A C172 dropped from standstill is more than likely recoverable. Worst case is it enters a spin, which is recoverable if C172 is within weight&balance. More likely it just goes 45+ degrees nose down almost immediately (due to tailplane generating a force), and picks up airspeed in a non-spin spiral until controls regain authority. $\endgroup$
    – Roman
    Commented Feb 3 at 18:05
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    $\begingroup$ "So let's say there's at least a 1% chance at any altitude." - not really. If the altitude is not enough for it to start flying, but enough to be utterly destroyed, there is zero chance. Drop it from 100 feet, and unless you prepared something below it, it will 100% be destroyed. $\endgroup$
    – vsz
    Commented Feb 5 at 7:54
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Not much in principle. In freefall, the 172 would reach its stall speed in a little over a hundred feet. You'd need to make sure to get the nose down as soon as possible. Then you need to gain a little more speed and pull out of the dive. Probably less than 1000 feet in total to recover to controlled flight as long as you can get the nose down relatively quickly. Then you still need however much altitude is required to find a safe landing spot. To do this remotely safely, you'd want a very healthy margin.

The plane would most likely enter a spiral at first. Both the wings and tailplane are behind the center of gravity, so unlike in normal flight they both produce a nose-down moment when they have air coming from below, and at 90 degrees angle of attack they are producing a lot of upward force along the vertical axis for a given vertical speed- significantly more than they would produce in horizontal flight at the same speed!

So the nose should drop pretty quickly even without pilot input. With the engine off, the airframe is completely symmetrical and there is no yawing or rolling moment. Eventually one wing will drop by chance and the plane will enter a spiral. But as the nose drops, the airspeed increases and the pilot regains control authority. Then they only need to recover from the ensuing dive.

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    $\begingroup$ "at 90 degrees angle of attack they are producing a lot of upward force along the vertical axis for a given vertical speed- significantly more than they would produce in horizontal flight at the same speed!". It's comparable. Airfoils right before stall have a lift coefficient of ~1.5 and a flat plate drag (a good approximation for the falling plane) is ~1.3. It's counter-intuitive that an airfoil with a modest downward deflection of air can produce as much lift force as a flat plate produces drag! $\endgroup$ Commented Feb 4 at 4:22

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