The amplitude of the flapping may not be large, but that's probably due to the presence of a feathering input. Most of the dissymmetry of lift is compensated by the feathering, so flapping is reduced to a minimum...
THREE things make-up the cyclic feathering.
The most apparent and intuitive is the pilot's control stick input toward the half of the rotor disk with the most lift. As your speed increases, most helicopters require more lateral cyclic.
The next most apparent and not quite as intuitive is the designed-in features and rigging of the rotor head that feather the blades without any pilot input as they climb and descend in relationship to the rotor mast and hub.
The hardest thing to see and understand is the varying vector sum, or "resultant" relative wind that is striking a rotor blade as the blade makes each 360 degree revolution. If a single blade is frozen at 360 points and analyzed at each, the resultant relative wind's vertical component at each point reaches max at 3:00 and 9:00 O'Clock, and tapers off to zero over the nose and tail as that vertical component changes from negative to positive. On the advancing half, blade climbing within the parcel of air results in an increasing, negative, vertical component of resultant relative wind and decreases lift. On the retreating half, blade descending within the parcel of air results in an increasing, positive, vertical component of resultant relative wind and increases lift. Even though this third part of cyclic feathering doesn't involve a blade actually twisting in its feather bearings like the first and second do, it is definitely a change in angle of attack so it is cyclic feathering by definition.
Now. To better understand how the climbing and descending of the blade compensates for dissymmetry of lift, and how all that acts on the helicopter, let's isolate the first 90 degrees of the advancing half of the rotor and look at it closer using some hypothetical, unitless lift numbers. Let's say 0 to 100. Over the tail and nose, not considering gyroscopic precession (for the moment), there is no lift force created by the rotor because the blade is at "zero-lift pitch." At 3:00 O'Clock, the blade has feathered to "minimum pitch" and the downward force is 100.
ALL OF THESE HYPOTHETICAL NUMBERS ARE
AFTER THE PILOT INPUT AND THE OTHER TWO THINGS THAT AUTOMATICALLY FEATHER THE BLADES HAVE STOPPED THE PITCH-UP AND LEFT ROLL THAT IS CAUSED BY DISSYMMETRY OF LIFT (American CCW Helicopter).
If
1) Lift over the tail is 0
2) 10 degrees later lift is 10 downward
3) 20 degrees later lift is 22 downward
4) 30 degrees later lift is 34 downward
5) 40 degrees later lift is 46 downward
6) 50 degrees later lift is 58 downward
7) 60 degrees later lift is 70 downward
8) 70 degrees later lift is 82 downward
9) 80 degrees later lift is 93 downward
10) 90 degrees later at 3:00 O'Clock, downward lift is max at 100
Now if we apply gyroscopic precession (reaction is 90 after the applied force) to see how the rotor actually RESPONDS to those forces.
1) No force up OR down
2) A miniscule left roll of 1, and small nose-down force of 9
3) A small left roll of 7, and a larger nose-down force of 15
4) A larger left roll of 14, and a larger yet nose down force of 20
5) About equal left roll and nose down forces of 23 and 23
6) About equal left roll and nose down forces of 23 and 23
7) A larger left roll of 45 and a smaller nose down force of 25
8) A larger still left roll of 70 and a nose down force of 12
9) A left roll force of 90 and a tiny nose down force of 3
10) A large left roll force of 100
Totals
Left Roll Force = 373
Nose Down Force = 260
As I said earlier, these forces are needed to equalize lift on the advancing vs. retreating halves of the rotor disk.
This is what keeps the helicopter level. Similar but opposite forces are present on the retreating side.
