Hi, one think/thing added:
Maik Justus schrieb: > > 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. In this calculation I forgot to split the centripetal force into a component along the blade (the only direction a force can attac) (which is F *(cos a) ) and at the flapping hinge this force has to be splitted into a component along the mast (the lift) and one perpendicular to it (which makes a rotational moment). the one perpendicular has a (cos a) factor, the parallel one a (sin a) factor. This results: r0 w0 w0 (cos a) for the component perpendicular to the mast and r0 w0 w0 (sin a) parallel to the mast. So teetering has no effect to the forces (in first order). Somehow I lost one (cos a) term in my older calculations which rides me into confusion, but now it seems to be clear to me. But the translational roll moment has still the wrong sign... Does someon know, if the Bell 206 (or the r 22) has this roll moment and how large is it? > 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 a little bit eased, Maik _______________________________________________ Flightgear-devel mailing list [EMAIL PROTECTED] http://mail.flightgear.org/mailman/listinfo/flightgear-devel
