# Can a free-falling aircraft with no horizontal (or minimal horizontal velocity) start up and fly to safety?

Inspired by this Grand Theft Auto gif, an aircraft with its engines initially idling is dropped from a high altitude (such as from the top of a building) with no or minimal horizontal velocity. Can it successfully throttle up and fly to safety? How about for

1. A fixed-wing aircraft, and
2. A helicopter
• @Federico 1. Just two. 2. I understand games and movies are not necessarily realistic, but they sometimes are plausible. – isanae Jul 12 '15 at 21:29
• @Pondlife Thanks. I'll remove the first question since it seems to be "yes". – isanae Jul 12 '15 at 21:31
• Impossible for a helicopter. Auto-rotation can only be entered with RPM already present. The blades would simply fold up as the helicopter fell and the engine will be unable to overcome the drag to get them back to flying speed (and on most helicopters, they would be bent beyond use anyway). – Simon Jul 12 '15 at 21:42
• This is not an exact duplicate of the question mentioned. I am rewording it to make it more legible, so that the differences from the question mentioned can be clearly seen!! – Victor Juliet Jul 13 '15 at 8:03
• @VictorJuliet Thank you. I didn't know how else to phrase the question and was surprised to see this marked as a dupe, since 1) there was no mention of helicopters and 2) this is about a stationary plane falling. But I don't know anything about aircraft, so I didn't argue. – isanae Jul 13 '15 at 8:27

Rather obviously, both of these situations are far from likely to occur. As a result, the following is just idle (but hopefully informed!) speculation — I don't recommend experiments!

1. I'm going to be a little bit cheeky in answering this question -- you've asked if it's possible for “a fixed wing aircraft to [...] get the engines up to speed and fly off.” I'll answer “Yes, but it depends on the aircraft.” For the fixed wing aircraft I usually fly, a K-21 glider, getting the engines up to speed is a doddle — it’s powered by gravity (and essentially stays aloft by the sun). What matters here is the stall speed of the aircraft in question, and how quickly the pilot could recover from the ‘upset’ of being kicked off the top of a tall building, and land somewhere sensible (which can only realistically happen once the aircraft is not stalled). If the relative airflow over the K21’s wings is about 35 knots (40 mph, 65 kph) or greater, they generate enough lift to keep it airborne (and going about 30 m forward for every meter down).

If we assume that you’re a free-falling body, Newton (in the form of $v=u+at$) says that it’ll take about 1.8 seconds to reach that speed, and you’ll fall ($v^2=u^2+2as$) about 16 meters during that time.

Of course, you’ll be pointing nose down at the ground, accelerating, and needing to do something about it — which will take time, and, more importantly distance. But, in the grand scheme of sky-scrapers, mountains and cliffs, 16.2 m isn’t that high. Incidentally, for this reason, bungee launching is a traditional method of getting gliders airborne in parts of the world with big hills — a team of runners using glorified elastic bands shoot brave pilots into the blue yonder, as illustrated by the picture below (at the Long Mynd).

So, given that a hill is enough height for a sailplane to take off, possibly for a few hours on a good day, I’m reasonably sure that something like the John Hancock Centre would provide more than enough time for a pilot to recover from an odd attitude, loop forward to a nose-down dive, recover smoothly from it, and fly away (from the big building behind him).

The gif you linked to featured what looked a lot like a small private jet. I defer to the others on this site with a lot more turbofan experience than I, but I’ll just say this: given an hour that I spent in a 777 simulator, twenty minutes were spent going from “cold and dark” to pushback, I suspect you’d have a much harder time doing it. Of course, rapid engine starts are possible in some aircraft. It depends a lot on the exact circumstances you're asking (and passenger jets tend not to be designed for inverted flight, for example).

1. By all accounts, this situation isn’t probably going to be very recoverable. If the blades are stationary when the aircraft starts falling, they’ll be fully stalled, resulting in a low-rotor RPM stall, which, depending on the brand of helicopter may not “be recoverable” (e.g., even with a working engine, Robinson helicopters are generally not recoverable from a very low rotor RPM stall). Autorotation is the driven motion of the rotor by the air falling through it, but, as pointed out, fully-stalled blades at any angle will not generate much torque (or lift), and therefore be unable to arrest the rate of descent of the falling aircraft. I presume that the only exception would be a fuselage that could provide lift through another means, such as the V-22 Osprey.
• The "brand" of helicopter does not matter. For any conventional helicopter that requires the rotor to be turning to generate lift, you will slam into the ground. Let's assume that you can stop the blades from folding upwards and let's assume that somehow you are lucky and the craft remains upright. The airflow is now entirely upwards through the disc and even with full collective applied, the AoA will be something greater than 90 degrees. Even with a slightly negative pitch, which some helicopters can adopt, the AoA is never going to be recoverable. – Simon Jul 13 '15 at 7:44
• ...since you have no way of adjusting the attitude. – Simon Jul 13 '15 at 7:50
• Autorotation actually can spin up a rotor as evidenced by autogyros. It just requires horizontal flow across the rotor, not vertical, so the helicopter would have to be pointing nose-down during the fall. However unlike fixed-wing aircraft it lacks control surfaces that could ensure that attitude. – Jan Hudec Jul 13 '15 at 18:55
• In the planes I've flown (a single-engine turboprop and a twin turbofan), both took 20 seconds or more from the beginning of the airstart sequence to the availability of power (of course, if the engines are idling as in the question, then you would just need the airspeed to get lift and then pull up and add power – SSumner Aug 2 '15 at 2:06
• @SSumner You've only flown turbines?? You didn't learn in a piston? At any rate, you can get power much faster than that in a piston. Like, less than 2 seconds from when you switch the ignition to 'start' (not that you should do that, but you can.) – reirab Aug 3 '15 at 3:11

Unlike in a RC or a maple tree seed, recovering from zero-RPM rotor is not possible in a helicopter because the "driving region" of the rotor system can't produce enough rotation to generate lift in the "driven region."

In fact, once a heli rotor RPM is reduced below about 80%, it stops flying

• I've upvoted your answer, but because it seems to confirm an older one with more details, I've accepted that one. Thank you for taking the time to write this. – isanae Nov 21 '15 at 22:37
• Landak's "full stall" answer is just wrong. – rbp Nov 21 '15 at 22:39
• Can you elaborate? I don't understand the difference between what both of you are saying. – isanae Nov 21 '15 at 22:50
• Non-pilots have a hard time understanding what "stall" means, and even fixed-wing pilots don't understand rotor my wing stalls. – rbp Nov 21 '15 at 22:55

Blade stop autos can be done with model helicopters, proving it is theoretically possible. There are a few differences in a model helicopters rotor design that make it more likely, but here is a video showing it done.

At 1:55 the blades are slowed to the point they are no longer spinning and the helicopter drops like a shot duck, until the operator reduces pitch enough to get them spinning again - without power.

You'd probably not be able to physically withstand the stresses, but I think it's technically possible.

• Models can do a lot of things full-scale aircraft can't, because the scaling does not work linearly. In this case I suspect the difference is that model rotor has relatively lower moment to inertia compared to the aerodynamic forces than a full-scale craft. – Jan Hudec Nov 4 '15 at 7:05
• Not really. The major differences between the model and full size rotor mechanics are the model has far more blade pitch movement (most models like that pictured have blade angle range close to 12 degrees plus and minus), and the rotor head on the model is far more stiff (which results in a less stable, less comfortable flight, but allows for more control). Both of these things make it easier to accomplish flight from a drop, but I still think it's theoretically possible. I doubt anybody will ever try it with a functional helicopter. – Greg Taylor Nov 5 '15 at 15:42

In a fixed wing aircraft this is a common aerobatic maneuver called a tailslide. Many airplanes can safely perform this maneuver, and if you google around, you can easily find video.