What are the critical steps a pilot follows to perform an auto-rotation landing?

  • 1
    $\begingroup$ @CGCampbell I agree, I edited the question to focus on the process of executing the landing, which I think (I may be wrong) is what the OP intended $\endgroup$
    – Pondlife
    May 7 '15 at 15:48
  • $\begingroup$ Gentlemen, I actually I am looking for specific procedures and why these procedures are done in order to accomplish a successful engine out autorotation landing. .... i.e. immediately lower the collective, establish a rate of descent, etc. thank you $\endgroup$
    – garyv440
    May 7 '15 at 20:32
  • $\begingroup$ What do you mean by critical? Do you mean survive, or 100% successful outcome? $\endgroup$
    – Simon
    May 8 '15 at 7:21
  • $\begingroup$ Gentleman , by using the word " critical" I am trying say,....having any reasonable chance at significantly lessening the impact and all aboard surviving,... this is also assuming there is enough altitude when the engine failure occurs, and a suitable place for landing are not concerns.....thank you $\endgroup$
    – garyv440
    May 8 '15 at 16:24

I am going to assume that by "critical", you mean survivable. I will also assume that the engine failure occurs at a height which will kill you.

It's not so much critical steps, it's critical outcomes. The single most critical outcome is to preserve, or regain, rotor RPM. What steps are needed to do this will depend on what the aircraft is doing when the engine quits and, to a lesser extent, the type of the helicopter. I will generalise for an "average" helicopter in straight and level flight.

The immediate action that is drilled into everyone from the very earliest hours as a student is to lower the collective. Doing this has three effects. It removes the positive pitch from the blades which removes most of the drag, it aligns the rotor thrust vector with the axis of rotor rotation therefore not using rotor energy for anything except lift and it causes a clutch to disengage the rotor from the engine allowing it to freewheel. From this point onwards, you are heading back towards earth.

The second immediate action at the same time as lowering the collective (assuming forward flight) is to pull back on the cyclic to flare. This loads the disc which causes it to "cone" upwards which reduces the disc diameter. Therefore, the centre of gravity of the disc moves inwards and due to conservation of angular momentum, the rotor RPM increases. For quite complex reasons, the nose of the helicopter will also pitch down when the collective is lowered so pulling back counters that tendency.

The third immediate action is to push in the pedal on the opposite side to the rotation of the blades. If the blades rotate counter clockwise (to the left as the pilot sees it) you push in right pedal to reduce the thrust which is produced by the tail rotor which is no longer countering the yaw caused by the drag from the powered rotor. This is less critical than the first two and although it can be dangerous and get you into an uncomfortable attitude, it is usually possible to recover from not pushing the pedal in immediately. If you've got rotor RPM, then you can sort it out.

Now you have successfully entered autorotation. From here, fly more or less normally to the next critical point which is about 50' off the ground.

What you've done is to ensure that the rotor has flying RPM and that you are managing energy by trading potential energy (height) for kinetic energy (rotor RPM). The conversion is done by the airflow which now comes from under the disc and "drives" the rotor to maintain RPM. The pitch is neutral, or maybe even negative but, the relative airflow is now upwards through the disc and therefore, the blades have a positive angle of attack and generate some lift. This stops the helicopter from falling. There is some drag generated as a consequence of generating that lift, but it is easily overcome by the power which is now driving the rotor from that upward airflow.

As long as you are descending, that conversion will happen and your RPM will be maintained. The controls are rigged such that with the collective fully down, RPM will remain in the normal range. Sometimes you have to tweak it a little with small amounts of collective, flares and turns but in general, you just fly towards your landing spot. The allowed RPM range is greater in autorotation. For example (and from memory), the R22 has a range of 97-103% in normal flight and 90-110% in autorotation.

You are now descending with a high rate of descent and usually, significant forward speed. You need to reduce both of those to get to a safe arrival. To do this, there are three more critical steps.

Starting at about 50 feet (depending on lots of factors but let's stick with the average helicopter which entered autorotation from straight and level with significant altitude), you flare the aircraft by pulling back on the cyclic. This will immediately begin to slow the aircraft. It will also begin to increase RPM (you are now converting speed into rotor kinetic energy).

At the same time, you increase collective to reduce the rate of descent by increasing the lift generated. This will rapidly increase the drag, but now the energy needed to maintain the RPM is coming from the flare which is converting speed into RPM. You also need to put in pedal to stop the aircraft from yawing as the drag increases on the rotor.

Providing you get the entry right and your flare reduces the speed and rate of descent to something survivable, then you will walk away. You might wreck the helicopter and break some bones, but arrive at 10 feet with only 20 knots and 150 feet per minute rate and you'll get away with it.

If you are well trained and in practice, then you will land safely and smoothly with no damage to machine or people.

In summary, critical steps:

Entry. Lever down, cyclic back, pedal in.

Arrival. Cyclic back, lever up, pedal in.

  • $\begingroup$ If we're trying to explain things to the layperson, phrases like "dump the collective" don't mean too much. $\endgroup$
    – Jamiec
    May 8 '15 at 10:45
  • $\begingroup$ @Jamiec Thanks. I've made some edits. I'd welcome any more input on "jargon". $\endgroup$
    – Simon
    May 8 '15 at 12:38
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    $\begingroup$ On the contrary, I thought the rest of this answer was concise and understandable. $\endgroup$
    – Jamiec
    May 8 '15 at 12:51
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    $\begingroup$ @ Simon....this is exactly the kind of knowledgeable answer I was hoping for...thank you sir....I love this site..... $\endgroup$
    – garyv440
    May 8 '15 at 16:26
  • $\begingroup$ @garyv440 You're most welcome. $\endgroup$
    – Simon
    May 9 '15 at 11:00

This is not intended to be anything like a full answer, but rather a lay-engineers comment to tack onto Simon’s very nice explanation.

When in descent mode with cyclic down the blades are in negative pitch to usual and airflow through them adds energy to them (up to some controlled limit) rather than transferring energy from them. The rotor becomes an energy store — a “winged flywheel.” Having the rotor at maximum allowed RPM in this mode maximises stored energy. Once you reach maximum allowed RPM the blades can be operated to maximise falling drag, subject to maintaining rotational speed — the rotor is akin to a large flat-plate. the machine will now fall at terminal velocity* for the maximum drag combination.

When you “flare” and raise the collective, blade pitch again becomes positive and you again have a powered helicopter. BUT it is powered by the inertial energy stored in the mass of the rotating blade assembly and you are going to use that up extremely fast — you have a few seconds of flying time, with rotor speed dropping as energy is taken from it. The flare procedure is designed to use the rotational stored energy in a manner which optimises the transition from pre flare velocities to post flare velocities.

* Auto-rotation terminal velocity:

I have not researched this so there may be reasons that it is wrong, but based on many other scenarios for bluff falling objects it seems likely that fall rate will be close to what is predicted by the classic drag equation so Rotor falling drag or machine mass is defined by

$$\frac{1}{2} \rho C_d A V^2$$


  • $\rho =$ air density ($1.2\ \mathrm{kg/m^3}$ near sea level)
  • $A =$ area ($\mathrm{m}^2$)
  • $V =$ velocity ($\mathrm{m/s}$)
  • $C_d =$ drag coefficient relative to flat plate, say 0.8 in this case

Therefore $\mathrm{mass} \cdot g = 0.6 \cdot 0.8 \cdot A \cdot V^2$

and terminal velocity

$$V_{\mathrm{terminal}} = \sqrt{\frac{\mathrm{mass} \cdot 9.8}{A/2}}$$

Terminal auto-rotate velocity in $\mathrm{m/s}$ is then

$$\sqrt{\frac{20 \cdot \mathrm{mass_{gross}}}{A_{\mathrm{rotor\ disk}}}}$$

While this is a guesstimate based on a range of assumptions, the general principle gives satisfyingly good albeit approximate results for objects as diverse as field, mice, bowling balls, skydivers and cargo parachutes. (It only works for raindrops when you realise that they generally fall as a flattened disk like shape when at terminal velocity.)


Robinson R22 Beta II, 620 kg gross weight, 151 inch rotor radius. Use 600 kg and 46 square metre rotor-disk area:

$$V_t = \sqrt{\frac{20 \cdot 600}{46}} = 16\ \mathrm{m/s} = 58\ \mathrm{km/h}$$

Looking further …

I can see I shouldn’t have started this. Fascinating. Time eating.
The above formula turns out to give a somewhat high auto-rotate velocity, which is good. Probably 50%+ higher than actual. Possibly due to lift from disk in forward gliding flight.

1,000 feet per minutes $\approx 5\ \mathrm{m/s}$. Various pages mention auto-rotate descent figures of 1,300 to 1,800 fpm.

Auto-rotation related calculator and MUCH MUCH more - superb. Includes comments -

  • A rotor in vertical autorotation has the same resistance as a parachute of the same diameter. This rate of descent is also approximately twice the hover-induced velocity.

  • 2500 ft/min is a reasonable upper limit for larger helicopters, i.e., 13 m/s

  • The $t/k$ ratio, which is the time in seconds that a rotor can lift the chopper when the engine stops. It is the ratio of $J\cdot\Omega^2$ divided by 4 times the power required in hover. (The 4 comes from the fact that one can only use half of the kinetic energy stored in the rotor system). Prouty uses a more complex formula that takes in account $C_l$ and $C_d$ of rotor system, but if you use the equation [$\mathrm{Power\ OGE} = (61\cdot10^{-3}/Dia_{rot}) \cdot \sqrt{m^3/ro})$ all in metric (with $ro = 1.225\ \mathrm{kg/m^3}$ at sea level], and divide the value obtained par 0.84 (for TR power, and Transmission losses), and plug this value into the t/k calculation, it works...

    So $t/k = \frac{J\cdot\Omega^2}{4\cdot\mathrm{Power\ OGE}}$ in seconds.

    The $t/k$ of the Robinson R22 is 0.8 (far too low I agree), and practically, you want $t/k$ around 1.2 to 1.7 sec, so roughly twice the Robinson.

  • The UltraSport-254 helicopter has an extremely low disk loading and an autorotation descent rate of 900 ft/min. It is said that in autorotation it can land then liftoff and re-land using only the inertia in the rotors. The Osprey V-22 has an extremely high disk loading. The test data indicate that the aircraft would impacted the ground at a rate of descent of about 3700 ft/min.

Robinson R22 discussion - informative. Comments on auto rotation and much more.

  • Due to its light weight and low inertia rotor system, the R22 is not forgiving of pilot error or sluggishness. After an engine failure, real or simulated, you and the instructor will have 1.6 seconds to lower the collective and enter an autorotation. Any delay beyond 1.6 seconds will be fatal as the rotor speed, once decayed below 80 percent, cannot be recovered. :-(

The Art of Autorotation
Extremely good auto-rotation tutorial and discussion with a number of pointers to arcane knowledge.

Video - R22 {almost} zero airspeed autorotation landing User comments useful.

R22 pricelist - just for interest

  • 1
    $\begingroup$ Thank you for your response and especially for the informational links. I also find it interesting and curious that the R22 has a reputation as an unforgiving helicopter, yet many of flight instruction schools I have researched, seem to offer the R22 as their most affordable model to train in. – $\endgroup$
    – garyv440
    May 9 '15 at 14:34
  • $\begingroup$ @garyv440 They are low cost (relatively). I've had a total of one flight as an intro to the craft with a few seconds at a time hands n with a hovering instructor watching the dual controls. He did not point out (not too surprisingly) that he had 1.6 seconds to enter autorotate after engine failure or you never can. [!!!] Decades ago we had a deer catching and shooting period of national madness. All who could chase deer with copters did. In one year we lost 30% of our rotary wing base in accidents ! [!!!]. The biggest single type lost was, no surprise , R22s. With people shooting and live ... $\endgroup$ May 9 '15 at 15:40
  • $\begingroup$ ... netting deer from copters and even (really!) jumping onto deer from copters all over our high country losses were inevitable. Deer prices were so good that the odd bullet hole also occurred. Latterly one of the Robinson brothers called to see what the heck we were doing . Conclusion - consistently being flown outside the envelope to maximise results. Losses are a lot lower these days :-). $\endgroup$ May 9 '15 at 15:43
  • $\begingroup$ Some of the user comments are useful. Many are not and a lot are wrong. For example, it is perfectly possible, and safe, to descend vertically. You only need airspeed at the bottom for the flare to store energy in the rotor for the collective pull to reduce ROD. it's even possible to descend backwards which feels extremely uncomfortable. As long as you have flying RPM and forward speed > about 35kts for the flare at the bottom, you're OK. youtube.com/… $\endgroup$
    – Simon
    May 13 '15 at 20:29

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