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Can tilt-rotors like the V-22 execute an autorotation landing?

If so, are the conditions under which that would be possible (and survivable) different from a helicopter?

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The V-22 rotor disks are much lighter than a helicopter's. There's little energy available to use to cushion a landing, so although autorotation is theoretically possible, there's little point.

The Boeing V-22 handbook has this to say (see page 26):

The V-22 is a tiltrotor and does not rely on autorotation for a survivable power-out landing. The wide separation of the engines and the ability to drive both rotors with one engine make a power-out landing extremely unlikely. However, if required, the V-22 can glide for a predictable run-on landing in airplane mode, much like a turboprop

I've seen glide ratios of 2:1 and 4.5:1 quoted on the Internet, so it'll come down quite hard and destroy the rotors.

However, the design mitigates against this in various ways, or so the manufacturer claims. In the survivability section of the same handbook they say

V-22 crashworthiness is a function of design. Heavy components, such as the engines and transmissions, are located away from the cabin and cockpit area. The proprotors are designed to fray or “broomstraw” rather than splinter on impact with the ground. The energy-absorbing landing gear system is designed to attenuate most of the energy for hard landings up to 24 fps. The wing is constructed to fail outboard of the wing/fuselage attachment in a manner that absorbs kinetic energy and ensures the cabin area will not be crushed, thereby protecting the occupants. An anti-plow bulkhead prevents the nose from digging in on impact, and the fuselage provides a reinforced shell that is designed to maintain 85% of its volume during a crash. Aircrew and embarked troops receive additional protection from crashworthy seats that stroke vertically to absorb energy.

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    $\begingroup$ So it's literally a problem of lacking inertia for the landing flare? Can the disks at least slow the sink rate to give it a gentler power-off glide slope? Or is the optimal power-out landing done with props forward and feathered for minimal drag? (Or is the prop configuration irrelevant?) $\endgroup$ – feetwet Oct 29 '15 at 19:10
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    $\begingroup$ I imagine that attempting to autorotate will wreck the glide angle, and there's nothing I've seen that suggests autorotation is a good option. The manufacturer's handbook explicitly recommends a run-on landing, for which, as you say, the best option will be props forward and feathered. $\endgroup$ – user11933 Oct 29 '15 at 19:17
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This seems to be a hot topic - in principle, the V-22 should be able to fly and land in autorotation, but tests so far did not demonstrate this. The manufacturer's position is that autorotation was never part of the specification.

What it comes down to is the inertia of the rotating parts relative to the aircraft's mass. The requirement to fold the rotors put a limit on their diameter and consequently their inertia while increasing their disc loading. The inertia is too low to reduce the rate of descent enough to enable a safe and soft landing. Quote from Wikipedia:

While technically capable of autorotation if both engines fail in helicopter mode, a safe landing is difficult;[73] in 2005, a director of the Pentagon's testing office stated that in a loss of power while hovering below 1,600 feet (490 m), emergency landings "...are not likely to be survivable."

This is specific to the V-22 - other tiltrotors might well be able to land in autorotation if their rotor inertia and speed is high enough.

For all practical purposes, the V-22 can glide down in autorotation, but will not be able to perform a soft landing at the end of this glide.

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    $\begingroup$ So it is able to generate some lift by autorotating its propellers in the vertical position? I imagine there would be one envelope for gliding in forward-flight configuration (i.e., fixed-wing-style power-off landing), and a different envelope for autorotation. The "inertia" you refer to is literally the mass of the propeller blades (and hub)? $\endgroup$ – feetwet Oct 29 '15 at 19:07
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    $\begingroup$ @feetwet: The recommended power-out mode is to glide like an aircraft with vertical proprotor position, so they work as impellers and drive the hydraulic pumps and electrical generators. Classical autorotation is possible, but the sink rate is too high for a soft landing. Inertia is the integral of the mass over the distance from the center, so both mass and proprotor diameter are involved. $\endgroup$ – Peter Kämpf Oct 29 '15 at 20:41
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    $\begingroup$ I guess that's the "real" answer for the V-22. But it leaves me wondering whether, given that the props are linked to the hydraulics and generators, one couldn't backfeed enough power from batteries and pressurized reservoirs at the last moment to power a landing flare? $\endgroup$ – feetwet Oct 29 '15 at 21:54
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    $\begingroup$ @feetwet: Let's consider the energy change needed for that. The sink rate of the V-22 in autorotation is 18m/s, the mass is 20 tons. For landing you need to get rid of a kinetic energy of 0.9 kWh. Say you decelerate the V-22 over 2 sec by one additional g. This would require an energy of 1,600 kW on top - much more than any generator or battery can deliver. You need an operational aircraft engine, and that is without any losses or reserves, and without adding what you need for lift. Either at least one of the engines is running, or the V-22 hits the ground hard. $\endgroup$ – Peter Kämpf Oct 29 '15 at 22:33
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There is nothing that prevents tilt rotors from performing an autorotational landing, in theory. AugustaWestland AW 609 has already demonstrated this. The Bell XV-3 also did this.

However, V22 has not demonstrated autorotation in any practical sense. The descent rate is too high for safe landing. The failure of V 22 to autorotate is due to the high wing loading (which is ~50% high compared to the AW609) and the low inertia of the blades.

The manufacturer/operator has instead claimed two things- the requirement (of both engines failing together) is remote and that it can glide in any case. At the end of the glide (which is pretty steep compared with 'normal' aircraft), the structure and seats are expected to absorb the impact.

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    $\begingroup$ If hovering, it will need quite some height to get into a gliding state. Since there seems to be no autorotate procedure, and this capacity seems to be doubtful, then we should best ignore that option. This means that it will basically fall like a brick and crash if it does not have enough height to transition into "aircraft gliding state" the this only supports the claim mentioned earlier (made by "a director of the Pentagon's testing office") that a failure under a specific altitude in hover mode would not be surviveable. $\endgroup$ – Rolf Oct 30 '15 at 10:57
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I read there was nothing in the specs to require autorotation like a traditional helicopter. I also read that the glide path to accomplish an auto would be too steep to go to a run on landing, which is recommended for single engine out procedures. The only time there is not sufficient altitude for an auto is when you are at a hover below 1800 feet, which is rediculously high to hover and I am not sure what mission would require that task. Even in a normal helicopter flying low over the trees does not allow much time to decelerate, speed up the rotor in time, and cushion landing with rotor inertia momentum. Cushion is what the traditionalists are looking for. Only the most superior pilot will stop all forward movement and not hit any obstacles when terminating at the ground without power. The Osprey meets its specs. And, all the other points are mute without the best pilot that exists at the controls, which is only the Senior Instructor Test Pilots or Acrobatic Pilots, which are not a part of the training. Just stick with run on landings and avoid hard wood trees at the bottom. There is no need to fuss over the Osprey ability to fully autorotate when landing.

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  • $\begingroup$ You may add references and links to support your answer and to provide further reading. $\endgroup$ – Manu H Feb 16 at 11:33

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