OK, I promised you a diatribe on my second point so here goes - should we expect the reaction to take place at exactly 90 degrees from the cyclic input?
We all know that there is a world of difference between text-book physics and practical engineering. We are invited to consider frictionless systems, to ignore air resistance and so on.
A good example of this is the projectile: basic mechanics predicts that a projectile will follow a parabolic path - but these are ideal objects: rigid, pointlike, not subject to aerodynamic forces etc. A real projectile such as a golf ball or a frisbee or an arrow shot from a bow will not follow an exact parabola because it is subject to a collection of forces and movements in addition to the constant acceleration of gravity.
Rotation and precession are affected in the same way. In an ideal system consisting of a completely rigid rotating object subjected to a single torque which is applied precisely perpendicular to the axis of rotation, then the reaction will indeed take place exactly 90 degrees after the point at which the torque is applied provided that there are no other forces or torques acting.
The real system consisting of rotorhead, hinges and rotor is not rigid: it can cone and teeter; the blades can bend and flex. It is not subjected to a pure torque caused by a single force applied at the rim; there is drag, there is lift, there is friction; the aerodynamic forces are distributed non-uniformly along the length of the blade and vary throughout the cycle, rather than being applied at a single point.
The collection of forces acting on the rotor is far more complicated than a pure torque acting perpendicular to the axis of rotation. Just as you wouldn't expect a ball thrown into the air to follow a precise parabola, we shouldn't expect a rotor system to have an exact textbook 90 degree phase shift.
A good example is the boomerang. To throw it correctly, you hold it with the arms vertical - not flat. As you throw it away from you it moves forward, spinning about a horizontal axis, so that the upper blade is moving forward through the air faster than the lower blade. The upper blade therefore generates more lift than the lower one.
For a right-handed thrower the lift is directed to the left and since the lift is greater for the upper limb this generates a torque which, looking from behind as the boomerang flies away from you, acts in an anticlockwise direction. Because the boomerang is rotating, the reaction to this torque occurs 90 degrees after the blade reaches the top of its travel - that is to say the front (far side from the thrower) moves to the left. Thus the boomerang follows a circular path, curving to the left and returning to the thrower.
So far so good, but as those of you who have boomerangs will know, by the time the boomerang returns to you, it is no longer vertical - it is flat, spinning about a vertical axis like a frisbee (or indeed a helicopter rotor !) and you can catch it between your hands using a 'crocodile jaw' motion. As it goes around its orbit, it gradually 'lies down', going from vertical to horizontal.
The point being that the reaction to the torque caused by asymmetry of lift does not take place at exactly 90 degrees - because this is a 'real world' system the non-ideal effects (drag etc.) cause you to have an approximation only to the textbook behaviour.
Now I'm not saying that 72 degrees is or isn't correct for the R22 - all I'm saying is that there are more factors to take into consideration.
[This message has been edited by Grainger (edited 29 June 2001).]