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When the rotor disk is in the position shown by the dashed lines, the net Lift is at an angle. It should produce a forward pitching moment about CG (Center of gravity). That can cause the helicopter nose to go down with a further increase in the moment angle and further pitching. What stops/controls this from happening in the actual scenario?

  • $\begingroup$ Welcome to Aviation Stack Exchange. I made a small edit to clarify; you are welcome to roll it back if you don't feel it represents your intention for the question and clarifies it. Just click on the "edited 1 minute ago)" (or whatever) tab to do that. $\endgroup$ Commented May 2, 2020 at 13:07
  • $\begingroup$ @quietflyer - thanks for the edit. $\endgroup$
    – Raj Arjit
    Commented May 2, 2020 at 14:27

4 Answers 4


It's a complicated answer, with many contributing factors in a multi-dimensional matrix. In a helicopter flying at forward speed, there are several stabilising stability factors and several de-stabilising ones.

  1. Before the helicopter can fly at speed, it needs to take off in a hover.

    • The rotor provides the lift force to climb.
    • Gravity applies down force in the Centre of Gravity.
    • The rotor hub is hinged - one could picture the fuselage as hanging off of the rotor hub, like a pendulum.

    enter image description hereFrom this answer, own work

    In this situation, hinging freely, the CoG would move directly underneath the rotor hub without re-aligning the rotor - the helicopter would stay in place, and only the fuselage would tilt until the CoG is underneath the lift vector. There is a range of allowable horizontal CoG positions, which would result in a range of fuselage angles when climbing up in a straight hover. Of course, if the CoG is chosen excessively forward or rearward, this scenario would result in an impossible fuselage angle.

  2. But the rotor hub hinge does transfer torque, from rotor to the mast and vice versa. The fuselage is not free to dangle underneath the hub. From Prouty, Helicopter Performance, Stability and Control:, describing behaviour in a vacuum:

    enter image description here

    The teetering rotor in a vacuum has no tendency to align the rotor and mast perpendicular to each other, while the very common hinge offset rotor does. And in the atmosphere, the teetering rotor also wants to align the mast perpendicular tp itself via the torque of the flapping effect described here.

    So a helicopter taking off in a hover with the rotor horizontal, aligns both the fuselage and the rotor until the CoG is underneath the lift vector. If the CoG was not exactly underneath the hub centre, the fuselage rotation tilts the lift vector and the helicopter starts to drift, which needs to be compensated by cyclic stick input from the pilot. From FAA Rotorcraft Flying Handbook, Chapter 7 Weight and Balance:

    CG FORWARD OF FORWARD LIMIT. You can recognize this condition when coming to a hover following a vertical take-off. The helicopter will have a nose-low attitude, and you will need excessive rearward displacement of the cyclic control to maintain a hover in a no-wind condition.

    Note that the above means that the rotor wants to align to the forward lean of the fuselage, which tendency must be compensated for by rearward cyclic.

  3. In order to fly forward the pilot moves cyclic longitudinal forward, effectively tilting the rotor disk forward. As seen above, this also tilts the fuselage forward, moving the CoG to a new position further aft - this creates a nose-up moment, compensating for the nose-down moment of the thrust/drag torque.

    When the helicopter picks up forward speed, air pushes at the fuselage: there is parasitic drag. This wants to tilt both the fuselage and the rotor further forward.

enter image description here

  1. At increased airspeed, the rotor flapping provides an increased nose-up torque: the forward moving blade experiences higher lift at higher speed, and reaches a higher tip position when pointing forward. As explained in 2. above, this results in a fuselage nose-up moment as well due to the hinge coupling between rotor disk and fuselage.

  2. The helicopter trim angle at speed can be further influenced by incorporating a horizontal tail, which applies a nose-up torque at speed. The horizontal tail incidence is often variable, in order to provide a wide range of stabilising moments at cruise speed, high angle descent etc.

    The horizontal tail in cruise usually has a negative angle of attack, and is designed to make optimal use of the rotor downwash upon it. Horizontal tails are not required but highly desirable, for positive cyclic speed stability.

  3. But a horizontal tail is ineffective in the hover, and a rotor on top has negative speed stability, as explained in this answer - the main reason for the instability in the hover of conventional helicopters, requiring constant stick inputs to maintain position.

Summarising, the nose-down pitching moment at forward speed is compensated by:

  • The flapping angle of the rotor disk.
  • The hinge moment coupling, either from hinge offset or from aerodynamic coupling in case of a teetering rotor. This effect wants to keep the rotor mast aligned perpendicular to the rotor.
  • The horizontal tail.
  • The useable horizontal range of CoG location.
  • And not mentioned above: in some cases canting of the tail rotor.
  • $\begingroup$ Thanks for the detailed answer. One query - in point 2, if I balance moment about the CG, then we get a counter-clockwise moment due to thrust vector. The moment due to drag is zero about CG as the drag vector is passing through CG point, so how does this compensate? or is it being compensated by the moment of Lift vector about CG, which turns out to be clockwise? (For this to happen, the resultant of Lift and thrust vectror will have to pass through CG) $\endgroup$
    – Raj Arjit
    Commented May 4, 2020 at 12:48
  • $\begingroup$ The flapping is more "nose down" with speed, not nose up. There is indeed a natural "flap back" as described in #4, but that is more than countered by forward longitudinal cyclic. The end result is more nose down flapping and hub moment at high speed. BTW the fuselage typically adds more nose down moment at high speed as well. $\endgroup$
    – Mat
    Commented May 4, 2020 at 14:48
  • $\begingroup$ @Mat Yes the swash plate angle is more nose down with speed, selected by increased fwd cyclic longitudinal stick. $\endgroup$
    – Koyovis
    Commented May 5, 2020 at 7:38
  • $\begingroup$ @RajArjit The rotor hinge moment mentioned in point 3. balances out a CoG not in line with the lift vector. $\endgroup$
    – Koyovis
    Commented May 5, 2020 at 7:52

Let's assume that the when the helicopter fuselage is horizontal (skids level), the CG is directly under the rotor mast.

For a torque problem, we are free to choose any arbitrary pivot point, at least when rotational acceleration is zero and G-load is one, so no additional apparent inertial force due to rotational acceleration or G-loading acts upon the CG.

If we want to figure out how things are balanced in a steady-state hover or steady-state cruising flight, it may simplify things to take the top of the rotor mast, rather than the CG of the helicopter, as the pivot point for the torque calculation.

Note that in this case, the tilt of the rotor disk and resulting offset thrustline exerts no torque about our chosen pivot point, but the drag vector of the fuselage--presumably acting near the CG of the fuselage-- does exert a torque about our chosen pivot point. The two effects are interconnected-- in steady-state cruise, offsetting the drag vector is the only reason that the rotor disk must be tilted.

Assume for the moment that there is a completely rigid connection between the fuselage, rotor mast, and rotor disk. In that case, if the rotor disk tips, the fuselage tips. The rotor disk cannot tip relative to the fuselage.

Assume that the rotor itself has no mass. Draw the vector triangle of R (rotor force), L (lift or upward component of rotor force), and T (thrust or forward component of rotor force), with right angles between T and L, and R as the vector sum of T and L. Assume that the resulting net force R acts at the top of the motor mast and is pointing in exactly the same direction as the top of the rotor mast.

Assume all the mass and all the drag of the helicopter are concentrated at the CG of the fuselage. Draw the vector triangle representing W (weight), D (drag), and F (net force generated by fuselage) with right angles between W and D, and with F as the vector sum of W and D. This triangle must be a geometrically similar triangle to the previous one-- the one associated with the rotor.

This means that given all these assumptions, the net force F generated by the combination of Weight and Drag always is exactly in line with the rotor mast. Thus no torque about the top of the rotor mast is generated by the combined effects of Weight and Drag.

Or to put it another way, the net force R generated by the rotor system is directly in line with the CG of the fuselage after all.

This also means that, given all these assumptions, even though in reality the helicopter disk system is free to pivot to some degree with respect to the helicopter fuselage, or vice versa, (see Why does the helicopter's nose go down when the cyclic is pressed forward (in a hover)? for more ), in steady-state cruising flight there is no inherent tendency for the fuselage to hang at a different angle from the rotor disk than it would it there were no flexibility in the connection between the fuselage, rotor mast, and rotor disk system.

Note that if there were some tendency for the fuselage to hang at some angle other than "square" to the rotor disk system in cruising flight, this would affect the position of the cyclic stick required to obtain any given net pitch torque (or lack thereof) from the rotor disk, just as tipping the rotor blade system forwards or aft when the helicopter is at rest on the ground changes the cyclic angle of one blade relative to another. For example, if the fuselage were for some reason hanging straight earthwards from the top of the rotor mast even in high-speed cruise, the (tilted) rotor blade system would be generating a nose-up pitch torque even with the cyclic stick centered.

One effect that could cause the fuselage to hang not "square" to the rotor disk, but rather somewhat closer to straight "down" toward the earth, would be if the drag vector acts above the CG of the fuselage. This is likely often the case in reality.

This whole question is really all about whether the rotor blade system needs to generate a nose-up pitch torque in cruising flight, to prevent the helicopter from nosing down. The answer appears to be generally "no". Of course, if such a pitch torque were necessary, it would be simple enough to position the cyclic as needed to create such a torque.

Now, if you were to ask what happens in the first instant after we use the cyclic to tilt the rotor disk forward in a hover, when the forward airspeed vector and therefore the rearward Drag vector acting on the fuselage are not established yet-- as the top figure with the tilted rotor disk appears to illustrate-- that would be an entirely different question.

Note that none of this really addresses whether the system is stable or not.

Many assumptions have been made here, but maybe this answer will give you a starting point for looking at your problem.

  • 1
    $\begingroup$ Thanks for answering. May be this image can help. imgur.com/5f2EAsr I am assuming the CG is ahead (front) of the Lift vector. (Is this assumption correct?) Based on this assumption, the lift force generates a downward pitching moment Lx, which can increase further if the cyclic is increased, leading to higher value of x', and hence higher moment Lx'. How will the free body diagram for balancing the moments about CG, be - during forward flight. $\endgroup$
    – Raj Arjit
    Commented May 2, 2020 at 14:22
  • $\begingroup$ "Not ... true while system is accelerating", but true when it is not! Good step forward here, old school "Schriftsteller der Wissenschaft" take note! $\endgroup$ Commented May 2, 2020 at 16:05
  • $\begingroup$ @RajArjit -- answer completely revised. There's a lot more to the problem than I realized. Good question. And again, I'm just thinking it through, don't know much more about helicopters than I've included here. $\endgroup$ Commented May 2, 2020 at 17:22
  • $\begingroup$ See aviation.stackexchange.com/a/77811/34686 for a better answer $\endgroup$ Commented May 6, 2020 at 12:35

The main rotor often provides a pitch up moment about the CG, opposite to the diagram in the OP. This is due to it being located at a smaller station line, forward of the CG (even after accounting for tilt).

The main rotor is flapped forward at speed, with its thrust tilted forward. However, the vector is still often positioned and oriented to provide a nose up pitch moment about the CG.

The main rotor "hub moment" will indeed provide a nose down pitch moment.

The horizontal stabilizer on the tail will typically provide a large nose up pitch moment at speed.

See this link for such information.

  • $\begingroup$ If a helicopter is moving left, and I see a helicopter from the front, then the lift vector will provide a moment to rotate the helicopter counter-clockwise. $\endgroup$
    – Raj Arjit
    Commented May 3, 2020 at 8:41
  • $\begingroup$ Main rotor thrust (including the aero forces over all blades) typically acts on a line in front of the CG, providing a nose up pitch moment. The "hub moment" however is the moment the rotor creates about its own hub. With the blades flapped down over the nose this hub moment provides a nose down pitch moment (assuming there's a flap spring or hub restraint). $\endgroup$
    – Mat
    Commented May 6, 2020 at 14:56

In many ways helicopters are air planes in disguise.

First, you draw the rotor with a slight dihedral. Now draw the lift vector for each rotor and you find: it will actually try to pitch the aircraft back upright (after forward acceleration goes to 0 (constant speed))! Even if the rotor were perfectly straight, its lifting force will not create a pitching torque unless the center of gravity is off-set. (draw dotted lines for the rest of the helicopter too).

Center of drag (as viewed from the front) will also be a factor once the helicopter is moving. The downwash of the rotor on the tail also plays a role.

However, pitching forward, without increased power, will cause the heli to sink because the vertical lift is less. The upward drag aft of center of gravity will cause it to pitch down just like an airplane.

If you pitch forward and add enough power to climb, you may find the nose pitching up! So you add just enough for level flight.

The concept of "pitching forward" and adding "power" from hover is identical to V min sink rate compared to V "best glide" in airplanes.

Perhaps a bit like this: enter image description here

  • $\begingroup$ Thanks for answering. Maybe this image can help. imgur.com/5f2EAsr I am assuming the CG is ahead (front) of the Lift vector. (Is this assumption correct?) Based on this assumption, the lift force generates a downward pitching moment Lx, which can increase further if the cyclic is increased, leading to higher value of x', and hence higher moment Lx'. How will the free body diagram for balancing the moments about CG, be - during forward flight. $\endgroup$
    – Raj Arjit
    Commented May 2, 2020 at 14:23
  • $\begingroup$ A smart design would be counter balance with rotor wash downforce on tail. So make little fins there until it is right. Note "throttle" will be more for moving flight than hover in all cases. $\endgroup$ Commented May 2, 2020 at 14:31
  • $\begingroup$ May you please tell how is the tail's rotor wash controlled? As far as I know, only collective input can be given to tail rotor and that too in lateral direction. $\endgroup$
    – Raj Arjit
    Commented May 2, 2020 at 14:36
  • $\begingroup$ The down wash of the main rotor on the tail is what I was thinking. It will vary with throttle input. Other possibility would be a 3rd "pitch rotor". $\endgroup$ Commented May 2, 2020 at 14:41
  • $\begingroup$ @Raj Arjit after giving more thought, there are probably more things trying to pitch it backwards when power is added, so some forward pitch torque may be a solution, not a problem. $\endgroup$ Commented May 2, 2020 at 15:00

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