In Chapter 02: Aerodynamics of Flight of Helicopter Flying Handbook, on page 10, it says:

Induced flow velocity is increased, resulting in a decrease in AOA.

I don't understand how airspeed can change AOA. Looking at other posts on the topic, it seems that everybody assumes that to be true without an explanation. Can anyone tell me why?

PS: the statement above explains why there is a positive stall region in a helicopter's retreating blades when the rotor speed is too fast.

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    $\begingroup$ Depends on what's being held constant. To maintain a constant amount of lift, less AOA is required at higher airspeed. Or (in the case of a fixed wing), shove in the power, hold AOA constant, and start climbing. Go back to your source and see what assumptions (or controls) are being applied -- is blade pitch being controlled so as to keep the lift vector constant? $\endgroup$
    – Ralph J
    May 16 '19 at 23:53
  • $\begingroup$ @Holding Arthur my answer addresses the main question you seem to be asking-- do you also want to ask why changes in forward airspeed affect the angle-of-attack of the forward and retreating blades and may cause part of the rotor disk to be stalled? That would best be a separate question I think. $\endgroup$ May 17 '19 at 4:16
  • $\begingroup$ @quietflyer You answer is good. Let's save that for another time. $\endgroup$ May 17 '19 at 9:16

On page 2-10 under the heading "Induced Flow (Downwash)", the book is making the point that if you have a wing moving forward horizontally at some fixed speed relative to the ground and some fixed positive pitch attitude relative to the horizon, and you surround the wing with a descending airmass, the wing's angle-of-attack will be less than if the airmass were not descending. A rotor blade of a helicopter can be viewed as such a wing, at least at a given instant in time.

The book is using the term "induced flow velocity" specifically to refer to the downward velocity of the airmass caused (induced) by the rotor blades. I.e., downwash.

The exact sentence you quoted was under the subheading "Out of Ground Effect (OGE)". Since the ground is no longer acting like a "floor" to impede the downflow of air through the rotor system, you have more downflow than you would in ground effect.

If you were hovering in ground effect and the ground suddenly disappeared, the downflow velocity would increase and the angle-of-attack of rotor blades would decrease.


The induced flow is the flow pulled through the helicopter blades, which is largest in the hover out of ground effect.

from Leishman Principles of Helicopter Aerodynamics

Adding a vertical speed pointing straight down to the relative blade speed, decreases the angle with the resulting velocity vector.

enter image description here


To express it at the most basic level:

  • Lift is created by deflecting a flow of air downwards. Either a large mass of air is deflected a little or a small mass of air is deflected a lot.
  • The angle of attack controls (within limits) how large the deflection angle is.
  • Speed determines how much air can be deflected per unit of time.

The higher the speed, the higher the mass of air available for deflection. For the same lift, a higher mass of air needs less deflection, therefore, less angle of attack.


There's a small but significant mistake in the FAA handbook. As your airspeed increases there is a reduction in induced flow, resulting in a decrease in blade AOA.

Another way of thinking about it is to consider a fixed wing airplane as it speeds up. To maintain the same amount of lift as it accelerates, the coefficient of lift and angle of attack go down. The same is true for the advancing blade of helicopter. Until you reach the bottom of the power required bucket the reverse flow region on the retreating blade is quite small and not too significant.


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