How does a helicopter rotor keep spinning in the same direction when autorotating without reversing the pitch

Question

I have read that if helicopter's engine fails, you set the pitch to minimum but not reverse pitch. The rotor would have to be tilted back to provide an angle of attack on the blades so it glides like a fixed wing aircraft but with the rotor spinning to keep the blades from buckling under the weight of the craft. I think it would have to go into a slight reverse pitch to spin the same way. That would also provide more angle of attack on the receding blades where it is needed because of the lower air speed over those blades. How does it keep rotating in the same direction with the air going up through it if it can't go into a slight reverse pitch?

Answer 1

Most rotors can go near zero pitch or slightly negative with the collective all the way down. But the blades don't have to go to negative pitch to get driven forward by the airflow coming up through the disc. They just have to be at a low enough pitch for the resultant angle of attack of the blade to produce a lift force with a strong forward (thrust) component somewhere along the blade.

This is happening in the mid span area of the blade, where the lift vector has a significant forward tilt, like an airplane wing gliding. Toward the tip, the lift vector is up and slightly aft, like an airplane wing under power, and all the lift force is vertical, while near the root, the blade is stalled and the velocity is too low to do much of anything.

The key bit is lift vectors produce forward thrust in the "driving" region of the blade because the blade is advancing forward at an appropriate velocity and it's the resultant airflow angle of speed + pitch, not the the pitch angle itself. This is achieved at lower collective pitch angle than while under powered flight, but it doesn't have to be negative. .

Answer 2

John K's explanation is correct. However, It may be interesting to add some text and an illustration taken from a classic book, 'The Autogyro and how to fly it' By Reginald Brie (1935):

Answer 3

For ease of exposition, let's consider a helicopter simply hovering.

The typical airfoil along the blade sees a velocity $V$ which is the sum of two components:

• an horizontal component $U$ due to the rotation of the blade
• a vertical component $V_i$ called induced velocity which is (more or less) due to the downwash of the blade(s) preceding the current one in its rotation; this term is typical for a rotary wing but is not normally encountered on a fixed wing.

As usual, the airfoil generates a lift $L$ perpendicular to the airflow $V$ and a drag $D$ parallel to it. Their sum is the total aerodynamic force $R$. In the rotary-wing world, this aerodynamic force $R$ is more usefully decomposed into a thrust $T$ parallel to the shaft of the rotor and a force $P$ perpendicular to it. The thrust $T$ is the force keeping the helicopter in hover while the force $P$ is the one that the engine has to win to keep the rotor spinning:

_{Aerodynamic force $R$ generated by a rotor in hover invested by airflow at speed $V$}

So far so good. What happens now when the engine quits? Nothing special. Due to its inertia, the rotor continues to rotate and nothing changes in the previous picture. Anyway, the force $P$ is now no more counterbalanced by the torque of the engine and therefore $P$ makes the rotor slow down. If the rotor slows down more than some 80% of its nominal rpm, the airfoil begins to stall increasing even further the magnitude of $P$ and leading to a no more recoverable catastrophic situation. The time from when the engine quits to when the rotor has lost most of its rotational speed is no more than a couple of seconds.

So the pilot has only a couple of seconds to reduce as much as possible $P$. How can this be achieved? Simply decreasing $\theta$ (by lowering the collective):

_{When the engine quits, $P$ must be reduced as quickly as possible to avoid the rotor to slow down too much and stall; this is achieved by reducing the pitch $\theta$ of the blade}

Now $\theta$ and $P$ have become lower but $T$ as well. The weight is no more compensated for by $T$ and therefore the helicopter starts sinking with a vertical speed $V_d$ (investing the rotor from below upwards):

_{The rotor sink with a speed $V_d$ which flattens $V$ increasing $\theta$ and tilting $R$ forward}

Note that the sinking speed $V_d$ makes $V$ "more horizontal" both increasing the AoA of the airfoil $\theta$ and tilting the total aerodynamic force $R$ a bit forward; this helps in reducing the component$P$. When $V_d$ is circa $>1.8V_i$, $R$ has tilted forward enough to have:

• a component $P$ null; and
• a component $T$ big enough to act as a parachute.

The helicopter has just entered the autorotation phase:

_{$V_d$ has increased so much that $R$ has become perfectly vertical: the rotor is in autorotation}

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Reality is, as usual, a bit more complicated but this simplified approach should help anyway to understand the autorotation manoeuvre.

Answer 4

From your answers, it seems to be the upward air hitting the upward sloping mid to rear underside of the blades as in the diagram below that keeps it spinning the same way with forward pitch and upward air flow, with lift provided by the air striking the bottom, and flowing over the top. Thanks fellows.