First I would like to thank everyone who has been bearing with me and so patiently contributed to this thread! Last night I came about a line in the book by Padfield "Helicopter Flight Dynamics" that finally gave me the impression that here was an explanation which I completely understood, see link below. Let me first start with a tiny bit of math. I still have difficulties to understand Jean Claude's idea of a false and true rotation center. For me the rotation center always is the shaft, which I find easy to understand because if you follow math rigorously it leads you to this point of view. I found it very helpful to rigorously stick to math in cases of highly complex, three dimensional motions because you only have to handle the math correctly to arrive at the correct result. The centrifugal acceleration is described by the term:

ac = oM x (oM x r)

where oM is the angular velocity of the shaft about the direction of the shaft (oM actually is a vector, as is r), r is the location vector of the point you consider and x is the vector cross product. Now oM x r is the velocity V of the particle. Let us say that this, in our case, is a tip weight, and by virtue of the cross product is perpendicular to both the axis of rotation and r. So one may write the formula as

ac = oM x V

where V is the tip velocity, as mentioned before. But again oM x V is perpendicular to both oM and V, so following the math the rotation is about the axis oM and ac is perpendicular to oM. Consequently Jean-Claude's "False Rotation Center" is my choice and correctly describes the physics, which is important in what is to follow.

On the page I found, Padfield offers a formula for the out of plane blade acceleration (sic!) azb. Omitting the fuselage pitch/roll contributions the formula is:

azb = rb*( -oM^2*beta - beta2dot)

where beta is the total flapping angle with respect to the shaft and beta2dot the flapping acceleration of the blade. Padfield writes:

Quote: The term azb is the normal acceleration out of plane of rotation, and there are three important components. The first is the acceleration due to blade flapping (note by JH: this is beta2dot), the final term in the expression. When the blade flaps up and down once every revolution

**this term is equal and opposite **to the oM^2*beta term in azb, the component normal to the blade of the centripetal acceleration. /Quote.

I am very glad Padfield rewrote this part of the book, because in the old version, I have in print, this is not stated as clearly. So finally now every piece of the puzzle is in place:

a) there

** is** an acceleration of the blade due to the aerodynamic forces which leads to an out of plane blade acceleration with a 90° phase lag and a change in blade flap angle with respect to the shaft. So rotor flapping

**does exist** in a physical sense!

b) by the odd physics of this very complex system, that is a rotor, this blade acceleration is exactly canceled at every point of the path by the component of centripetal acceleration which is perpendicular to the blade (see sketch below)

If we write the longitudinal flapping angle as:

beta = bN + b1c*cos(oM*t)

and differentiate twice we get

beta2dot = -b1c*oM^2*cos(oM*t)

If we plug this into Padfield's formula for blade acceleration we get:

azb = rb*{

**(**- oM^2*[bN + b1c*cos(oM*t)]

**)** -

**(**-b1c*oM^2*cos(oM*t)

**)** }

expanding the square brackets gives

azb = rb*{ - oM^2*bN + ( - oM^2* b1c*cos(oM*t)) - (-b1c*oM^2*cos(oM*t)

**)** }

or

azb = rb*{ - oM^2*bN - oM^2* b1c*cos(oM*t) + b1c*oM^2*cos(oM*t)

**)** }

and the two cyclic oM^2*b1c*cos(oM*t) terms cancel. The remaining constant coning angle (bN) term is balanced by the constant part of blade thrust (please note that in the sketch only the cyclic part of centripetal acceleration is shown, you would have to add the constant part that balances the averaged blade thrust, responsible for the blade coning angle). This is identical to the result that Jean-Claude presented, in that there is no resultant acceleration out of the plane of the blade, but for me it is much more rigorous and thus much better to understand. The result actually means that my initial idea was wrong, since the offset blade weight also moves with zero acceleration perpendicular to the blade and there will be no blade torsion. The remaining terms in the azb formula are, in my opinion, too small to give the effect I was considering in this thread.

Of course a direct observation, like the one related by Fergus (thanks goes out for that!), is a strong argument, although I seem to remember that the aircraft with which Ken made his hands off landings is not the one he flew for his hands off photo demonstrations, where his hands off flying was much longer than a few seconds. The one remaining question for me is, why Ken Wallis stuck so stubbornly to his offset blade balancing weights, (something that very obviously deteriorates blade performance) when it would have been much easier to do all the balancing inside the blade.

I had a lot of fun in this thread and have finally understood one part of rotary wing physics I had been struggling with, ever since I started out on my quest to get to grips with rotary wing stability and control.

https://books.google.de/books?id=PjhuDwAAQBAJ&pg=PA161&lpg=PA161&dq=blade+normal+acceleration&source=bl&ots=AhSDi8mqds&sig=ACfU3U3P4Ht3ISTiqeaCmWC70jwXdZa2-w&hl=en&sa=X&ved=2ahUKEwjwpoy-vJ7zAhVlhP0HHZ1qAQ4Q6AF6BAgXEAM#v=onepage&q=blade normal acceleration&f=false