It seems like at this point it would be game over, since the spinning rotor is the only thing keeping it in the air. Is there any way to land a helicopter in this condition without crashing?

  • $\begingroup$ In some helicopters with high inertia rotor systems (basically heavy and store a lot of energy simply through rotation), auto rotations can be non-events. The B47 could be landed, picked up, turned through 180 degrees and put down again all without the engine. The other great thing about engine failures in many helicopters is that they need a lot less space than a fixed wing. I know some really good pilots who need something about the size of a cricket pitch. $\endgroup$
    – Simon
    Commented Mar 15, 2014 at 14:41

2 Answers 2


Helicopters are able to do something called autorotation if all thrust is lost.

In a helicopter, an autorotative descent is a power-off maneuver in which the engine is disengaged from the main rotor system and the rotor blades are driven solely by the upward flow of air through the rotor. In other words, the engine is no longer supplying power to the main rotor.


(Look in particular at the direction of flight arrows. Remember there's a lift vector coming of the blades as well in both cases.)

Basically, the collective is decreased immediately when thrust is lost for whatever reason. This allows the airflow coming through the blades to push them around, like a reverse windmill in a way, keeping the blades spinning. This also generates lift, keeping the helicopter generating some lift (while descending of course).

At the instant of engine failure, the main rotor blades are producing lift and thrust from their angle of attack (AOA) and velocity. By lowering the collective pitch, which must be done immediately in case of an engine failure, lift and drag are reduced, and the helicopter begins an immediate descent, thus producing an upward flow of air through the rotor system. This upward flow of air through the rotor provides sufficient thrust to maintain rotor rpm throughout the descent. Since the tail rotor is driven by the main rotor transmission during autorotation, heading control is maintained with the antitorque pedals as in normal flight.

Once the helicopter reaches ground, the pilot will pull up. Since the blades were spinning a bit faster than needed for a hover since he was flying forward the whole procedure, he simply pulls up and levels out carefully close to the ground until the helicopter touches down.

This being said, you have to be careful, since if you're flying to slow you can't pull it off.

When landing from an autorotation, the only energy available to arrest the descent rate and ensure a soft landing is the kinetic energy stored in the rotor blades. Tip weights can greatly increase this stored energy. A greater amount of rotor energy is required to stop a helicopter with a high rate of descent than is required to stop a helicopter that is descending more slowly. Therefore, autorotative descents at very low or very high airspeeds are more critical than those performed at the minimum rate of descent airspeed.

This dangerous conditions are shown in a height-velocity diagram as is seen below, in this case apparently from an R44.


Here's a little video to get an idea of how it looks for the pilot:

Autorotation is often time critical and has to be done right as there's only one shot and as such is practised by pilots on a regular basis.

All extracts in this section are from "Helicopter Emergencies and Hazards" by the FAA, an excellent and very extensive publication on the topic.


Many helicopters have more than one engine driving the rotor system. On those, losing an engine is probably less serious than it would be in a single-engine craft.

All helicopters that I'm aware of are capable of autorotation. In an autorotation, the pilot uses the air rushing up through the rotor disk to keep the blades spinning. This involves changing the collective angle of attack of the blades to near zero to minimize drag, plus some other control inputs.

While doing this, the pilot is trading potential energy (altitude) for kinetic energy (keep the rotors spinning).

Immediately prior to landing, the pilot changes the collective angle of attack of the rotors back to the normal "push air down" setting. This rapidly slows the rotors, but if done correctly there's enough cushion to land the vehicle.

For a much more detailed description of how to execute an auto-rotation landing, see How does a helicopter pilot execute an auto-rotation landing?

Per FAR 61.87(f)(15), A student pilot who is receiving training for a helicopter rating must receive and log flight training for the following maneuvers and procedures:

... (15) Simulated emergency procedures, including autorotational descents with a power recovery and power recovery to a hover;

  • 1
    $\begingroup$ I'm pretty sure helicopter pilots are taught autorotation prior to solo, but can't confirm right now. $\endgroup$ Commented Mar 13, 2014 at 22:28
  • $\begingroup$ This article has a pretty good illustration of what you're talking about. And the article also seems to confirm your comment. todayifoundout.com/index.php/2012/05/… $\endgroup$
    – Jae Carr
    Commented Mar 13, 2014 at 22:29
  • 5
    $\begingroup$ @DanPichelman they do, 61.87(f)(15) requires it. I have 0.3 in an R22 and even I have had it demonstrated and practiced. $\endgroup$
    – casey
    Commented Mar 13, 2014 at 22:45
  • $\begingroup$ How did this get 17 upvotes? Sorry, but it's just plain misleading. 2nd paragraph, wrong. With the exception of a wider "allowed RPM" band, the rotor spins at the same speed as in normal flight. 3rd paragraph, misleading. the pilot does not "tilt the rotors" back to a "normal" setting. The pitch of each blade is increased collectively to increase the angle of attack and therefore the lift. But this is only possible if you have already reduced the rate of descent significantly and flared to convert the remaining speed into rotor RPM. $\endgroup$
    – Simon
    Commented Oct 30, 2015 at 18:18
  • $\begingroup$ @Simon - thanks for the corrections. I've updated the answer so it's hopefully more accurate. I will be happy to make further changes if I've still missed something. $\endgroup$ Commented Oct 30, 2015 at 18:37

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