I learned through a comment on this question about multiple engines powering a single propeller that the V-22 Osprey can power both rotors from a single engine, or use both engines to power a single rotor. What are the mechanical systems that make this possible?


3 Answers 3


It is possible to drive the two rotors of the V-22 Osprey from a single engine using a segmented driveshaft connecting both engine shafts via gearboxes:

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The drive shaft is depicted in blue in the left wing. Source

Each engine drives its own rotor and the connecting shaft. If one fails the same rotation speed is maintained, but each rotor receives now half the power of the remaining engine.

A similar mechanism is also used on the CH-47 Chinook family:

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This is a CH-47 Chinook green drive shaft:

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On aircraft with two distant rotors, it is vital to have both rotating, as if one slows or stops the aircraft is unbalanced to the extend it crashes.

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    $\begingroup$ What happens if an engine fails/seizes in a way that the connecting shaft becomes immovable? Does that cause both rotors to fail, or is there a mechanical linkage that can disconnect from the failed engine? $\endgroup$
    – nanofarad
    Jan 29, 2017 at 2:24
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    $\begingroup$ @hexafraction It wouldn't matter, either one is an unflyable aircraft. The V-22 cannot fly on a single prop/rotor and neither can the CH-46/47. $\endgroup$
    – Ron Beyer
    Jan 29, 2017 at 3:22
  • $\begingroup$ @hexafraction , if the engine fails, normally the shaft is not affected. If, for some reason, the shaft rotation is in prejudice, both rotors will lose RPMs in synchronization (this will eventually cause lift loss and a crash because, even with autorotation, the breaking force can exceed the upwind traction of the maneuver). $\endgroup$ Jun 28, 2017 at 23:46
  • $\begingroup$ If, in other scenario, the shaft breaks apart at some point, rotors will be out of synchronization. In synchropters and tandem ones, they will probably collide with each other; in tilt rotors they don't touch their selves, BUT even if they are free from the shaft motion, THEORY says an autorotation can be attempted, but strong and progressive roll oscilations will happen as rotors differences of RPMs causes the aircraft to swing side do side (the upwind flow raising when one wing falls, causing the rotor to recover lift and switch the other side to do the same, like a seesaw). In this case... $\endgroup$ Jun 28, 2017 at 23:51
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    $\begingroup$ ...the aircraft is strongly in danger to the proportion of the height, because this side to side rolls will become more and more violent until the aircraft rolls over the point of autorotation efficiency (rotors being feed by sides rather from beneath) and a last roll will cause it to, eventually, invert and crash. Smaller heights can (again, in theory) increase the survivability because the aircraft will have no time to develop violent oscilations, reaching ground at a possible lower vertical speed. All of this is valid for speeds under ~60 knots, where wings are not providing enough lift. $\endgroup$ Jun 28, 2017 at 23:57

Tilt rotors and virtually every other VTOL aircraft with more than one lift rotor (Chinook, K-Max) or proprotor (AW609) must employ a shaft between these lift units, so when one engine is out the other keep the units turning will half power (which doesn't mean half rotations per minute, but half maximum torque). That's why available power reserve on the engines of these aircraft is important; choosing engines that are 'just enough' for the normal job may deny their certification for obvious safety reasons. A failure or a simply deceleration of one rotor/proprotor would cause disastrous asymmetry of lift along the aircraft center of mass.

Also, many of these aircraft require the rotors to be permanently synchronized (and, therefore, connected) to simply avoid collision between their counter rotating blades. In tilt rotors, rotor sync is also beneficial when talking about vibrations, acoustics and symmetry of aerodynamic flow through the airframe. Finally, connected proprotors are also a solution towards simplicity regarding yaw: the AW609, for example, doesn't have a rudder, since it can yaw by simply changing the blades pitch of one proprotor (or changing both in opposite angle values).

That's why every aircraft with connected rotors/props and two or more power plants slowly spins all the blades even when only one engine is starting. Since the first transmission of output inside the engine is done through air without a mechanic shaft connection (using the free turbine solution), the engine doesn't suffer that much from overload. Over pressure in the combustion chamber may be an issue, but is easily addressed with bleeding valves and/or FADEC protections.

The details of those connection shafts are not very new, resembling the shafts that connects main and tail rotors on heavy helicopters of traditional design, but far more robust and having no reduction gears. They are basically assembled with the less amount of segments possible. The connections between them must be as precise as possible, withstanding some difference of torque between the segments in order to avoid over twist efforts that could lead to failure. Also, they can't absorb differences of angular momentum (like the springs on a manual car transmission clutch does) since this can easily evolve to a resonance between the rotors.


The cross shaft of the V-22 is normally unloaded. It only takes up the load if an engine fails, at which point everything continues as normal albeit with reduced power. Alternatively, if the shaft is damaged, but you've still got two good engines, the aircraft flies as normal because the shaft isn't being used anyway. On the CH-47, though, the shaft is always under load because that's how the front rotor is powered by the engines in the rear. If that shaft is damaged, it will result in an immediate loss of the aircraft, even with two good engines.

  • $\begingroup$ For the CH-47, why couldn't a driveshaft breakage be dealt with by completely shutting down both engines and autorotating down? $\endgroup$
    – Vikki
    Jul 5, 2018 at 22:23

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