Hi,
I try to explain my confusion:

Lets think the heli is in forward flying and the rotor is spinning
counter clock wise (seen from top, like the bo 105). Relative to air the
rotor blades at the left side are slower and on the right side are
faster. So they produce more force at the right side and less on the
left side. Because of the gyroscopic effect the result is 90° shifted,
so the blade will be up in front and down in back (flapping). The heli
will rise the nose if you don't correct the cyclic input. (Up to this
point I think I understood it.) 

m is the mass of one pointlike blade 
r is the radius to this mass (which vary with flapping and teetering)
r0 the radius without flapping and teetering
w (omega) the rotational speed of the blade
a flapping angle


The angular momentum ( m r w) is conservated. If the blade flapps up, r
will be changed (r=r0 cos a, if we assume the hinge in the rotor
center), so w has to be increased (that is what we know as teetering). 
If we assume a positive cone angle of the rotor (and this is the normal
case) than the rotorblade will be faster in front position than in back
position. And will produce more force in front position than in back
position. 
And with thy gyro we have a roll moment, but to the right and not to the
left as our helicopter has in real.

It is very easy to run into more confusion. Lets think of a rotor system
like the Jet Ranger has. The rotor can flap free around the center, so
the only force to the mast is the centripetal force. The centripetal
force is m r w w. Without teetering this force is m r0 w w cos a. If the
rotor is tilted to the front and has a positive cone angle the blade
will produce higher centripetal force in front position than in back
position. A effective force to the rotormast pointing forward results.
The CG is lower than the rotor head, so this result in a forward roll
moment. ok.
But what is the result of the teetering? If m r w is constant, than w
must be w0/(cos a) (remember: r=r0*cos a). And the centripetal force is
then m r0 w0 w0/(cos a), which means, that a forward tiltet rotor will
produce an effective force pointing backward, which is obviously
nonsense. My only idea to solve this: The teetering hinge at every real
heli is not in the rotor center. It is out of center. Every teetering
angle produces a force to the blade which "wants to reduce the teetering
angle", so the change of w will be smaller (the missing angular momentum
goes to the rotor mast). But if you look in detail into this, you find,
that this is a osscillator, which is nearly in resonance. One solution
is to think of a big damping constant, but the teeter hinge at model
helicopters needs no damping. Maybe the damping can be explained
aerodynamically? I don't know. 

Or is there a big error in this calculations?
Confused,
Maik



Andy Ross wrote:
> 
> David Megginson wrote:
> > Maik Justus wrote:
> > > > Also the rolling tendency in translational lift is missing.
> > >
> > > That is a very complicate thing. Allways if I think about I run into
> > > confusion.
> >
> > Is it just a gyroscopic effect?
> 
> If I'm not misunderstanding the terminology, this is the rolling
> moment due to airspeed along the plane of the rotor.  One side is
> moving faster than the other, and produces more force.
>  
> But like everything with the rotor, it does involve gyro effects.
> Outside of plain aerodynamic forces, none of the forces or moments on
> a helicopter act on the rigid body of the airframe.  They all cause
> the rotors to tilt or flap (or even bend, if you really are into
> modelling this stuff), which *then* causes (and feels) a force/moment
> on the body.
> 
> And since the rotor is spinning, it produces all sorts of
> non-intuitive behavior like the 90° precession phase shift (try to
> roll it left, it tilts forward, etc...).  It's ugly. :)
> 
> Andy
> 
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