I have a rather simple question just to "double check" if I understood blade flapping right. So, I understand that a helicopter, in forward flight faces an asymmetric lift. Blade flapping is there to compensate this.

As far as I understand this, the higher lift on one side leads the rotor blade to tilt upwards (aka flap) thereby also tilting the lift vector and decreasing its purely vertical component. It flaps just enough so that in total it creates a symmetric lift situation.

Is that correct? All of this happens just by the nature of it? The pilot or flight controller don't have to do anything here?

My doubt here is that, if I apply an axial force on a object that rotates around a centre, it will reach its maximum "tilt" or in this case flapping angle, only 90° after the force was applied. This seems to contradict my understanding of blade flapping. So that’s why I am confused.

Context: I am an aerospace engineer (with emphasis on space) who tries to learn more about helicopters. My curriculum didn’t contain anything about helicopters.

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    $\begingroup$ Just to answer the part about the pilot's role in blade flapping, as a former helicopter pilot, the pilot doesn't have to do anything. The flapping occurs naturally, due to the nature of the assymetry of lift. The vertical movement of the blades "cancels" the assymetric lift that would otherwise be generated. $\endgroup$
    – Gaston
    Nov 9 at 13:03
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    $\begingroup$ Not just flapping, there is lead and lag too. Look for some videos shot from the root of a blade along its length as it spins around. You'll find that the blade moves around a lot. $\endgroup$ Nov 9 at 16:18

2 Answers 2


Blade flapping is both due to an "automatic" effect and a "controllable" effect.

  • The "automatic" effect is exactly as you've understood it. The advancing blade sees a total speed higher than the retracting one and therefore it creates a higher lift (and drag and pitching moment) than the retracting one. The opposite is true for the retracting balde. In order to compensate for this difference between the advancing and the retracting side of the rotor, the blades are left free to flap: flapping upward reduces the AoA seen by the blade and therefore the higher lift gets reduced as well. So the flapping is not only "automatic" but self stabilising too. This loop looks something like this: higher speed on advancing blade $\Rightarrow$ higher lift $\Rightarrow$ flapping upward $\Rightarrow$ reduction of AoA $\Rightarrow$ decrease of lift. The maximum flapping angle happens with a <90° delay: if the maximum lift happens when the blade is, say, on the right of the helicopter then the maximum flapping happens when the blade passes over the front of the fuselage (for a rotor rotating anticlockwise as seen from above).

  • The "controllable" effect is based on the same idea but controlled by the pilot via the "longitudinal" stick. If the pilot pushes the stick forward, the blade's AoA (aka feathering) is incremented when the blade passes over the left side so that its maximum flapping angle is reached when it passes over the tailboom. The total effect is that the rotor and its thrust tilt forward, just like commanded by the pilot.

The <90° delay between disk tilting and the feathering that causes it depends mainly on the distance of the flapping hinge from the rotor hub: for a teethering rotor the hinge is exactly by the rotor hub and the delay is 90°; for other rotor constructions the delay is smaller.

Note also that, albeit similar, this delay is caused by the fact that the feathering of the blade is in resonance with the flapping movement: so you don't have to think about someone pushing the border of a spinning top rather to someone pushing a swing 😉


Blade flapping is indeed automatic but note that the cyclic control of the pilot will fine-tune it such that lift is equal on both sides of the rotor disk.

It is not the maximum upward travel of the blade or the tilting of the lift vector which reduces its lift, but the motion itself. If you look at the flow components, you get one from the circular motion of the blade ($\omega$ x r), a second one from the forward motion of the whole helicopter (v) and a third one from the upward motion of the blade (v$_z$). This upward motion reduces the angle of attack of the blade, and this reduction in angle of attack is what reduces blade lift.

Note that the high position of the blade at the end of the forward half of the rotor disk gives it an ideal position to increase angle of attack on the backward-moving side of the rotor disk by moving down again. This increases lift and is the other but equally important side of flapping.

Angle of attack on both sides of a helicopter rotor

v$\;:\;\;\;\;$Flow speed from the movement of the helicopter through air (light blue).
$\omega$ x r: Flow speed from the rotation of the rotor (dark blue).
v$_z:\;\;$ Vertical speed component from the up/down movement of the rotor blade (red)
The green arrow shows the resulting flow speed which is the vector sum of the speed components.

  • $\begingroup$ So if I understand correctly the upward and downward movement vz (due to flapping) is the key because it changes alpha just right to change lift. Thats.... insane! and very well explained! Thanks Peter :) Its so crazy that this "just happens" on its own. $\endgroup$
    – Clex
    Nov 10 at 7:50
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    $\begingroup$ @Clex I just added a first sentence to answer the question directly. Yes, flapping does most of the work but it is up to cyclic control to fine-tune the rotor so it produces equal lift on both sides. $\endgroup$ Nov 10 at 7:53
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    $\begingroup$ @Clex: I've had this up for three days and nobody noted that the speeds on the backward-moving side were wrong. Fixed. $\endgroup$ Nov 12 at 22:38
  • $\begingroup$ I saw it but didn't think much of it/was to afraid to ask. But I thought the advancing blade is moving upward and the "backward" moving blade is moving downward... now I am confused again. $\endgroup$
    – Clex
    Nov 13 at 10:05
  • $\begingroup$ Ah yes since the blade moves upwards the air, relative to it moves downards, and that is vz. Now I got it :) $\endgroup$
    – Clex
    Nov 13 at 10:24

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