The first prototypes of the Cierva autogyro used a disk tilt mechanism rahter than cyclic pitch by means of a swashplate to achieve roll and pitch (DOI: 10.13140/RG.2.2.19168.94729).

Let's first assume the rotor blades are rigidly connected to the rotor hub and rotation axis. Moreover, let's assume we have two blades. Then, in order to tilt the rotor disk, one needs to apply a moment which scales with the angular momentum of the blades. Hence, the bigger the rotor and/or rotational velocity, the larger the moment required to tilt the disk.

Now, suppose a hinge is added whilst the blades remain rigidly connected to eachother like a seesaw. For the time being, assume the rotor spins in a room filled with stagnant air. What would the required moment be to tilt this rotor by means of disk tilt?

I am stuck on this conceptually: Suppose we move the rotor axis exactly and instantaneously when the rotor is positioned as in A. Since there is a hinge, nothing happens except the rotor will preces to such that the blades are perpendicular to the rotor axis.

Now suppose the instantaneous moment is applied when the rotor is at position B? Then, the blades act as a rigidly connected blade without hinge, right? Does the gyroscopic moment need to be overcome in this situation (omit the infinite moment for infinite acceleration)? Or are both situations the same and am I missing something?

hinged rotor in two situations A&B


1 Answer 1


The effect is identical to a situation where you hang the machine with its rotor head bolted rigidly to the ceiling of your garage, with the tiller of the rotor in your hand. You push the tiller to the side, and you find you are able to move the fuselage off the rotor's axis, say, 6 inches to a foot, before the effort gets too big.

So you have the ability, just by directly tilting the rotor off perpendicular to the fuselage, to move the C of G of the machine off axis some amount and impart a rolling or pitching moment as long as you maintain the CofG offset from the rotor's axis some amount.

When you do it while flying, any displacement will get a roll or pitch response started, and just holding, say, 30 lbs of side force will keep the C of G offset from the rotor axis and maintain the roll at some value. It's basically weight-shift control, like a hang glider.

Just about all small sporting gyros work this way, by tilting the rotor head directly, because the forces required a relatively low.

Once you get to a certain size, the forces just get too large and ponderous, and the only options are to hydraulically boost the tilting system (I don't know if that's ever been done) or articulate the rotor with a swash plate, by effectively making the blades themselves to the work.

Even with a articulating system, once you get to a certain size, you need to hydraulically boost that system as well to have manageable control forces, and most helicopters use boosted controls once the all up weight is in the 2000-3000 lbs range.

  • $\begingroup$ I think the example is a strange and counter-intuitive one. If I am not mistaken, what you're saying is that the rotor disk remains in the same position (as if it is hanging on a ceiling) and you push yourself and the aircraft using the stick connected to the rotor? This implies that the force you need to apply is only related to the aircraft's weight. And what would happen when the rotor is spinning, but the fuselage is on the ground? Friction on the ground prevents the fuselage from moving, but I bet the rotor disk will. What moment is then required? $\endgroup$
    – lWindy
    Mar 2, 2023 at 7:09
  • $\begingroup$ Go back to hang glider, or trike ultralight sitting on the ground on its gear. Next to it is a similar size autogyro, like the very early Benson with a tiller bar coming down from the rotor head. Tilt the wing on the trike by moving the control bar left, tilt the rotor on the gyro to the right by moving the control tiller to the left. Wing, spinning disc, same thing. In the air, same relationship but now you have the C of G suspended from the wing or rotor disc.Move the contro, tilt the wing/disc off the vertical axis, machine rolls.To maintain the roll, hold the input to keep the CG off axis. $\endgroup$
    – John K
    Mar 2, 2023 at 14:11

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