1
$\begingroup$

I am trying to determine how helicopters pitch and roll (i.e. change attitude). I have heard two explanations.

1) Pitch/roll change is achieved by tilting the tip path plane and therefor tilting the thrust vector. By tilting the thrust vector, a moment is generated about the center of gravity which pitches/rolls the helicopter.

2) Pitch/roll is achieved by imparting a moment onto the rotor, which can be viewed as a giant gyroscope, causing the helicopter to precess with a 90 degree phase lag (see page 2-16 here).

In Helicopter Theory by Wayne Johnson (© 1980), section 5-5 pg 191, he writes:

Thus, there are two ways to view the rotor flap dynamics. First, the rotor blade can be considered a system with natural frequency of 1/rev, so that aerodynamic moments due to cyclic pitch excite the system at resonance. The response has a phase lag of exactly 90° (i.e. one-quarter of a cycle, which at 1/rev means an azimuth angle of 90° also). Alternatively, the rotor can be considered a gyro, with the flap hinges at the center of rotation forming the gyro gimbal. A control moment on the disk due to cyclic pitch then precesses the rotor with a 90° phase lag characteristic of a gyro until the flap damping produces a moment to stop the precession.

So, are the two explanations (1 and 2) equivalent, or are they both characteristics present? Wayne Johnson seems to suggest the former but gave no proof. Thanks in advance!

$\endgroup$

3 Answers 3

1
$\begingroup$

This animation shows it better than most. The swash plate changes the plane of the rotor disc to change the plane of the lift vector. It's the same thing as an airplane turning by banking. When an airplane banks it imparts a lateral thrust component to the vertical lift component and it moves sideways, and because it's moving forward as it does so, it moves in a horizontal arc, or a turn. A helicopter's rotor is just able to "bank" in all directions at any speed.

The swash plate directly flies the blades to the path that will put the overall disc in the desired plane of rotation. The blades have to fly to their position using the air; if you had the machine running in a vacuum, the cyclic would have no effect. However, gyro precession is a factor insofar as the input/result of the swashplate's effect on the blade's physical path is offset 90 degrees.

That is, to get the blade tip to its highest point in the circle, the highest blade angle, or highest lift force in the rotational cycle, is imparted 90 degrees before that point. To tilt the rotor disc forward say, the blade that is on its way to its high spot over the tail boom will have reached its highest blade angle while passing at 90 deg on the left.

On the Bell rotor in the animation you can clearly see this. The rotor disc angle follows the swash plate angle directly - swash plate tilts forward, rotor tilts forward. However, the linkage being driven by the swash plate acts on blade angle with a 90 degree offset (the control rods coming up on each side of the hub) to account for precession.

The Bell teetering system is interesting because it imparts no bending forces whatsoever to the rotor mast. The machine's body is suspended below the rotor hub as if it was a tennis ball on a string. When you move the stick you are making the rotor disc fly around on its own account by tilting this way and that and the machine suspended under it follows along. In fact if a bending moment WAS applied to the mast, by having the teetering stops contact the mast while it's spinning, it comes apart and you become a falling object (this is "mast bumping" which cause the deaths of a number of Huey pilots in the 60s and Robinson pilots in the 90s).

Rigid and articulated rotor systems do impart some rolling and pitch moments into the air frame and the rotor head, mast and suspension system for the transmission are designed to transfer these loads.

A good fixed wing analogy between an articulated and teetering rotor is to compare regular airplanes vs hang gliders and trikes. An articulated/semi rigid rotor is a bit like a regular aircraft where the body is directly acted on by the wings, although with a flexible connection between the wings and fuselage. The Bell system is more like a hang glider or trike where the wing and body are on a fully flexible connection and you basically fly the wing around and the body underneath follows along without ever reaching the physical limit of the connection. It's not a perfect analogy but sufficient for visualization.

$\endgroup$
3
  • $\begingroup$ your statement "The Bell teetering system is interesting because it imparts no bending forces whatsoever to the rotor mast" makes me think that both explanations account for how helicopters pitch/roll such that explanation 2 describes why the tip path plane tilts and explanation 1 explains how a roll/pitch moment is generated. $\endgroup$
    – eball
    Commented Jul 11, 2019 at 16:19
  • $\begingroup$ Also, I'm not sure if this is what you were suggesting, but a rotor in a vacuum would not precess as there would be no way to apply a torque to the rotor (via aerodynamic forces) to induce the precession. $\endgroup$
    – eball
    Commented Jul 11, 2019 at 16:33
  • $\begingroup$ Yes in a vacuum you'd have to apply the tilting force at the hub itself. $\endgroup$
    – John K
    Commented Jul 11, 2019 at 17:01
0
$\begingroup$

The 90º phase lag of a rotor has nothing to do with gyroscopes or precession, and is only due to resonance. A good explanation, by James Bennett, chief engineer of Cierva Autogiro Co. and a first-class expert in the field can be found here: https://archive.org/details/journal00britgoog/page/n136

$\endgroup$
9
  • $\begingroup$ please consider including the most relevant details of the paper here $\endgroup$
    – Federico
    Commented Jul 11, 2019 at 13:06
  • $\begingroup$ @xxavier, thank you for the resource! I skimmed the first several pages and it seems to be similar to the explanation Wayne Johnson gives starting in section 5-5 pg 184, which concludes "By this means the pilot can control the attitude of hte helicopter, using cyclic pitch (swashplate tilt) to produce moments about the helicopter center of gravity by tilting the thrust vector". This is explanation 1 I offered above. $\endgroup$
    – eball
    Commented Jul 11, 2019 at 13:10
  • $\begingroup$ @Federico Please give the link... $\endgroup$
    – xxavier
    Commented Jul 11, 2019 at 22:26
  • $\begingroup$ @eball Direction control of the lift vector is always by tilting the disk, and that tilt is achieved (not always, but usually) via a swashplate or a spider. In the general case, for zero hinge offset, the control action on the rotor and the response lag by 90º. That lag is often explained as gyroscopic precession, also as 'aerodynamic' precession and as a resonance effect. The latter is (in my opinion) the best explanation... $\endgroup$
    – xxavier
    Commented Jul 12, 2019 at 6:55
  • $\begingroup$ @xxavier, so are explanations 1) and 2) equivalent (i.e. are they explaining the same physics)? I would question this conclusion because when you do not have zero hinge offset, the response lags by something less than 90º, which can only be explained by explanation 1) because gyroscopic precession always occurs 90º out of phase. $\endgroup$
    – eball
    Commented Jul 12, 2019 at 12:52
0
$\begingroup$

Explanation 2) is not an option to explanation 1). The page in the FAA handbook only points out that the rotor has 90° phase lag when applying forces for deflection. Has no consequences for the rotor cyclic pitch, tipping the rotor forward still speeds the helicopter up. It only has consequences for where to place the cyclic pitch force actuators.

$\endgroup$
1
  • $\begingroup$ the rotor is a large spinning mass, when you apply cyclic input, the lift simultaneously increases and decreases on opposite sides of the rotor due to change in blade pitch. When a moment is applied to a spinning mass, the mass will precess at an angle perpenducular (90°) to the applied torque and angular momentum vector. Precession is a behavior observed in gyroscopes. I would think they are connected in some way. $\endgroup$
    – eball
    Commented Jul 11, 2019 at 16:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .