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

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?

• Nov 16, 2023 at 7:41
• For a stable autorotation, the angle of attack of the driving part of the blades has to stay around 12º. But that's the AoA, and not the pitch... The AoA is the sum of the inflow and pitch angles, and the inflow angle is a function of the horizontal and vertical relative winds as 'seen' by the relevant blade section. If the inflow angle is high enough, you can have a convenient AoA for autorotation, even if the pitch is negative... [img]i.imgur.com/DLy3Ffh.png[/img] Nov 17, 2023 at 13:24
• @Bianfable I didn't have a problem understanding windmilling. I just couldn't understand how it could rotate in the same direction when the wind was reversed, but not the pitch. From the answers I got, it seems to be due to the upward slope of most of the underside of the blades when the pitch is low. Nov 18, 2023 at 22:56

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. .

• Wow, is the rotor blade really that "twisted"? I'm comparing the chord line of the blade sections in the "driven region" and "driving" region diagrams, noting that the axis of rotation is the same same in each diagram. Wouldn't essentially the same diagrams apply, still showing a greater aoa on the "driving region" and also a more forward tilt of the "L" and "T" vectors for the "driving region", all due to the difference in "inflow" direction (which is due to the difference in rotational speed), even if the blade had no twist? Nov 21, 2023 at 13:22
• The "effective twist" is a product of the velocity variation. Imagine two sailplanes gliding past each other in opposite directions, and they hook wing tips and are forced to circle each other while gliding down. The hookup point has nearly zero airspeed and nearly 90 deg AOA. AOA will decrease out toward the tip as velocity increases until at the tip it's down to a few degrees. About where the fuselages are, the AOA will be a range and orientation that is able to keep propelling the glider forward in a glide, but forced to circle because you both foolishly added Velcro to your wing tips. Nov 21, 2023 at 13:57
• Re above-- ok, I see how the "inflow" (maybe not the best name as what is illustrated appears to be the resultant of the relative wind w/r/t helicopter body plus the forward velocity of the blade) comes from a very different angle at different points on the blade. Due to difference in forward velocity of blade at different points. Still doesn't explain why angle between chord line of blade and axis of rotation of blade which is represented as dashed line in diagram is so different between top and bottom of the right-hand diagrams. Nov 21, 2023 at 15:53
• Ah I see what you mean. Not sure. Nov 22, 2023 at 4:09
• @ Quiet flyer It would be easier to understand if the people producing these diagrams varied the angle of the relative wind flow, as that is what is different, and not the chord line which is the same in all sections. Also, I wish they wouldn't exaggerate the pitch. It doesn't show how the blade is propelled forward, as, at that angle, there would be no part of the underside sloping up toward the back, and the rotor would be driven backwards. Nov 29, 2023 at 4:42

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):

• +1 for going to the trouble to scan the pages lol. Nov 16, 2023 at 20:50
• @ xxavier From your diagram & a picture of a real chopper, the blade cross section looks symetrical like an aerobatic plane wing, sloping upward on most of the underside, which would explain the forward slope on the lift vector. I had always assumed the blades were shaped like the bidirectioal wing with camber I described hear:"aviation.stackexchange.com/questions/19122/…", and pitch set slightly negative when gliding, which would also give less AoA on the advancing blades & more on the receding ones. Thanks Nov 16, 2023 at 23:53
• @John K When you said symetrical, did you mean front to back like the bidirectional wing, or top to bottom like anaerobatic plane wing. Also, in the previous PS above, I meant to say driving (inner) region. Does the air flow backward over it. Nov 19, 2023 at 1:25
• Symmetrical as in same curvature top and bottom, like an aerobatic wing designed for inverted flight. Look at an old Bell-47 blade cross section, and the blade on a late model design and it'll be obvious. Nov 19, 2023 at 5:33

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

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

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.

• Coanda effect happens when a 1) turbulent 2) jet 3) blows upon 4) "half" of the surface of a 5) flat or slightly curved body. Except for very particular cases, coanda effect has normally nothing to do with aircraft. What happens in autorotation is simply that the composition of the sinking speed with the rotational speed gives a lift which is mainly vertical: the blade produces thrust with no torque (see the last picture of my answer). Nov 29, 2023 at 7:47
• Is that more like what happens? That should produce an upward & slightly forward L vector with the air deflected back at the mid to rear underside, and flowing around the blade to provide lift. If you rotate blade in your picture to the actual pitch, that's about the angle the air (V) would be striking the bottom. Nov 29, 2023 at 15:27
• @sophit why does it have to be turbulent? (one known example of coanda effect is water flowing on a spoon, where flow is laminar) Nov 30, 2023 at 14:28
• @jkztd: water flowing on a spoon is not due to the coanda effect but simply due to "surface tension". Coanda effect relies indeed on the creation of vortices where it gets in touch with the surface and a consequent low pressure region. Nov 30, 2023 at 18:36
• It's a bit hard to distinguish the difference between Coanda effect and Bernoulli effect. My understanding is that the Bernoulli effect attracts the fluid to the surface by dragging away any fluid in between, and holds it there if the surface is curved, creating the Coanda effect. The little low pressure zone that creates the vortex after the step in diag. 5 Here "en.wikipedia.org/wiki/Coand%C4%83_effect" just helps to pull it down to the surface, and wouldn't be necessary if the opening was on the surface. I don't think NOTAR helicopters use that step. Dec 2, 2023 at 5:44