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In the game DCS World flying the Ka-50, when flying forward at high speeds, the bottom rotor will flap up on the right side of the helicopter (also note that the top rotor is flapping up on the left, though not as much):

enter image description here

But according to this diagram:

enter image description here

while the maximum up-flap velocity is on the right side, the actual max up-flap displacement is in front of the aircraft. Thus, shouldn't the bottom rotor be flapping up in front of the helicopter, instead of on the right? This effect only occurs at high speeds, at lower speeds it's either not there or barely perceptible:

enter image description here

Is this up-flap behavior correct, and if so, what is causing it?

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  • $\begingroup$ I was literally posting the same question just now - this somehow didn't come up in the search. But yeah, same logic in the question, which unfortunately I lack the answer for. $\endgroup$ – Volk Dec 19 '20 at 20:45
  • $\begingroup$ As far as my understanding goes, the rotors should never flap differently from each other, and should not flap at all at steady flight. $\endgroup$ – Jan Hudec Dec 20 '20 at 14:42
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This is does appears to be partially correct behavior for a Ka-50 in forward flight (in the low speed example)--but not because of gyroscopic procession. It also may be plagued with transient artifacts from the DCS flight mechanics model.

According to DCS, the Ka-50 rotor is modeling using blade-element momentum theory with some inflow model, so the basics of its aerodynamics should be correct. That said, this shouldn't account for blade/vortex/body interactions or other higher order effects. Based on the same source, DCS implies that they use a rigid-blade model (that is, a rigid blade rotating on flap/lag hinges) for their rotor blades, which allows them to model blade motion and couple their aerodynamic model with some rotor dynamics. This is impressive for a sim to have, but remember that they are missing significant interactions between rotors, let alone the body. They can reproduce the performance from certain phenomena in flight, but they don't go into detail of what else goes into their model, so I cannot comment any further. However, I can say the following.

Gyroscopic precession does exist, this is true, but it is a much simplified explanation of why helicopter rotor blades exhibit a phase lag that delays a blade's physical response to an applied load. Rotor blades are not gyroscopes, and, in fact, the phase lag for rotor blades is not always 90$^\circ$ but varies depending on the design of the rotor system. As for why the FAA and other pilot-oriented groups continue to teach that gyroscopic precession is the reason why it occurs: I have no authoritative answer, but, as with how the FAA incorrectly teaches about lift, teaching the "actual" reason why an aircraft does what it does will not necessarily teach you how to fly the aircraft better (unless you have an extremely well-developed understanding of fluid mechanics). Getting the explanation "good enough" (so that you have a useful hermeneutic for how the vehicle interacts with the air is sufficient to enable you to use the aircraft safely) seems to be the mantra--which makes some sense. Else, no one* would be allowed to fly!

Regardless, to the point. Let's talk about phase lag, how to control a helicopter using a rotor, and how those two come together.

Phase Lag and Helicopter Controls

Phase lag refers to the difference in when a force is applied to a dynamic system in resonance and when its physical effect manifests itself. This gets a little complicated in a helicopter's rotor system since, in order to control the vehicle, the rotor itself has to be tilted in the desired direction of motion. Keep in mind that the rotor (I speak in the singular, but I'll apply this to the Ka-50 in a minute) is the primary source of propulsion for the helicopter, so, to move anywhere, the rotor has to be pointed in the direction of motion, as illustrated below. The purple line is the plane connecting the tips of the rotor blades (or the tip path plane) during hovering flight, and the orange line is the tip path plane in forward flight.

enter image description here

You can also see this in this video of a hovering UH-60 Blackhawk (note the level tip path plane) as compared to this video of a Blackhawk in forward flight (note the tip path plane tilted in the direction of motion--more evident during the first half of the video). A caveat, though: the rotor on the Blackhawk is actually set at a small angle forward to decrease parasitic drag and decrease the body pitch angle in forward flight. This is particularly evident in the first video, wherein the helicopter is actually pitched upward when the rotor tip path plane is horizontal. I'm using the UH-60 as an example, but this is a fairly common practice.

In short, you have a single force vector, and you have to point that force vector in the direction you want to push the helicopter.

DISCLAIMER: if you don't want to learn about phase lag and just want the answer, skip to the last section now. A digression on phase lag is ahead.

Hence, whether the rotor is articulated or some form of rigid, it is designed such that the blades can flap as needed. The Ka-50 rotor does have flapping hinges, as you can see in the image below from how the blade is slightly angled relative to the blade grip extending from the rotor hub.

enter image description here

So where does this leave us? I'll explain phase lag in a minute, but the confluence of phase lag and blade flapping is important to grasp first. For instance, if I put in a control input to a rotor to make the vehicle move to the left, I would expect the rotor disc to tilt to the left by way of the blades flapping up on the right side of the helicopter and down on the left side. However, since the physical response and the aerodynamic loads are out of phase, I have to put in the control input prior to when I want the effect to manifest. If this effect was gyroscopic, I would have to pitch the blades as they passed over my tail or nose (90$^\circ$ prior to the flapping effects I want to see). However, if we look at a Robinson R-22, it ends up being more like 72$^\circ$ prior instead of 90$^\circ$. And it's not the only one (though typical values for articulated rotors do end up being closer to 90$^\circ$, if memory serves)! However, this is because the effect is not due to gyroscopic precession, but is a result of rotor design.

A flapping rotor system is the product of the a variety of forces, including mechanical damping (think of a dashpot, although that particular item is not necessarily the case), stiffness (think of a spring), aerodynamic damping (resistance to motion proportional due to changing aerodynamic forces), aerodynamic forcing (think lift, or what's causing the flapping in the first place), centrifugal or inertial forces (what's trying to keep the blade level), and "coriolis" forces, to name a few. These interact with a rotor blade that is hinged with some amount of offset from the center of rotor as well, which we can see on the picture of the Ka-50 rotor hub, which also changes how the blade responds to aerodynamic forces. As a result, we can tune the rotor system we design to respond in a particular way to aerodynamic forcing. However, it changes the phase lag of the system.

Visualizing the Effect

One way to visualize this (if you like math -- if not, I'll provide a physical example too) is using an Argand diagram, which can be used to show components of a spring/mass/damper system (a rotor system is an example). This would come from the Fourier transform of an equation having the following form, where $I$ is inertia, $b$ is the damping coefficient (higher the coefficient's magnitude, the more resistance to motion there is), $k$ is the stiffness of the flapping hinge (the "restorative" force in the diagram), $\beta$ is the flapping angle, $\dot{\beta}$ is the flapping velocity, and $\ddot{\beta}$ is the flapping acceleration. $F(\Psi)$ is the aerodynamic forces that create the blade flapping motion.

$I\ddot{\beta}+b\dot{\beta}+k\beta=F(\Psi)$

For example, the diagram below shows a system where the spring (rotor system: the restoring force which is dependent on actual flap angle) is dominant, making the system's physical response pace closely with the input force. Conceptually, that makes sense -- if I have a stiff spring, the amount of force I use to compress the spring will match very closely with the spring's compression (in the absence of a damper or much inertia). I can get that from the Argand diagram from doing vector addition with the 3 components I have plotted. If I did so, the resultant vector would lie very close to the "restorative" axis, or to the right side of the horizontal axis.

enter image description here

However, what if the damping is much larger and closer in magnitude to the restorative vector? This more usual for a helicopter with articulated rotor blades. If so, we get an Argand diagram that looks more like this:

enter image description here

If I do my vector addition here, I see that the resultant vector, and thus the physical manifestation of my input force, is $90^\circ$ out of phase with my input. A physical analogy would be if I connected a spring to a pneumatic cylinder and then tried to, alternately, compress and extend it. The amount of force I apply is no longer directly proportional to the compression or extension of the spring, but is out of phase with the spring's physical motion due to the presence of the damper. If I had a spring with a lot of mass, you could see how that would further complicate the problem.

Putting it all together: what does this mean for the Ka-50?

Right: so, in the pictures provided (copied below), you show the Ka-50 rotor flapping by small amounts during lower speed forward flight (quoted IAS was 60 kph).

enter image description here

Note that both rotor discs are flapping up as they pass over the tail. As said, in order to maintain speed, the Ka-50's rotors need to be tilted in the direction of motion, though this should only be a transient feature as the vehicle pitches into the direction of motion. If they were not, parasitic drag (and other sources) would slow the vehicle down. To maintain speed, a force has to be continually applied in the direction of motion. Hence, the aerodynamic forces are applied earlier in the rotation to induce both rotors to flap up as the pass over the tail of the aircraft. The 60 kph case is also useful because flapping due to advancing/retreating blade effects are relatively small as compared to the second case, shown below.

enter image description here

In this case, the rotors are obviously tilted left and right, respectively: which is strange. Judging from the horizon lines in both pictures, it would appear that the helicopter is pitched down more in this higher speed case as well, which makes sense, since the blade flapping angle over the tail seems to be relatively small. The vehicle also seems to be in a shallow right bank.

As a result, and I'll be perfectly honest, I'm not entirely sure why the rotors are tilted in opposite directions, and this is where I doubt the veracity of their model. Due to the high speed, which should be leading to some interesting aerodynamics, I'm placing a bet on their modeling starting to break down. If they are using an inflow model that approximates wake motion, some of the assumptions of that model may be breaking down at this high of an advance ratio, leading to unexpected blade motion. An Jan Hudec says in the comments, both rotor planes should be parallel to each other.

*If I may perpetuate the legend: no one, that is, except Peter Kämpf.

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  • $\begingroup$ Rather than saying rotors are not gyroscopes I would say they are more complex case than for what the simple 90° gyroscopic precession is derived. When you start deriving the gyroscopic effect by observing a point on the gyroscope, you do the same thing as for the rotor blade, but then you make two assumptions: that the force is small compared to the rate of rotation, and that the axis does not resist the precession, neither being true of helicopters. If you consider a gyroscope that is only starting to spin, the phase lag will change from 0° to 90° continuously. $\endgroup$ – Jan Hudec Dec 20 '20 at 21:32
  • $\begingroup$ The flapping in any steady state, whether hover, forward flight, sideways flight etc. only depends on the centre of gravity. For forward flight, the rotor and body pitch forward together, the rotor just leads a bit with flapping as the active element, and the body lags due to inertia. But it must end in the same relative position as for hover, because the lift is still perpendicular to the rotor plane of rotation and acting through the hub, and the centre of gravity must be in its action line, otherwise there would be pitching moment (ok, centre of drag shifts it a bit). $\endgroup$ – Jan Hudec Dec 20 '20 at 21:41
  • $\begingroup$ Most helicopters just mount the rotor tilted forward relative to the body to make the forward pitch less pronounced in forward flight, at the cost of noticeable backward pitch in hover. $\endgroup$ – Jan Hudec Dec 20 '20 at 21:42
  • $\begingroup$ That's good point about gyroscopic procession being a simplified version of the explanation: I've updated to mention that. However, I disagree about the flapping angles in steady state always being the same, regardless of airspeed. For a given shaft tilt angle, the cyclic flapping angles will change with airspeed, as will the cyclic control angles, in order to keep the aircraft in trim at that speed. This is in response to asymmetric changes in lift production around the rotor in forward flight, an effect that DCS's aero model (which includes an inflow model) should be able to capture. $\endgroup$ – Marius Dec 21 '20 at 4:12
  • $\begingroup$ However, I would definitely agree that I have some doubts about the images being of a vehicle in steady state flight. I think there's a good chance that the latter image captures some transients, which is throwing things off for me. $\endgroup$ – Marius Dec 21 '20 at 4:13
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The rotor system in flight is very complex all parts affecting everything else, the basic answer would be the blades must move up an down to balance the forces on the rotor system to prevent to much lift on the advancing side, which would roll the helicopter to the retreating side [they also move forwards and back to bance the flapping motion for physics reasons, you can look up Lead lag hinges]

The reason they flap more at higher speeds, is due to "dissymmetry of lift" as the aircraft speeds up, the lift difference between the advancing blade, and retreating blade is more pronounced, requiring more flapping to equalize it.

I do see your confusion, but the behavior is correct even if it seems strange.

Good explanation: https://www.thehelicopterstudyguide.com/what-is-flapping/

A video of flapping:

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  • $\begingroup$ I understand what flapping is, but I don't understand why it occurs on the right side instead of in front. Based on the 2nd image I linked, the right side is just the max up-flap velocity; the front is the actual max up-flap displacement. This makes sense because the blade is advancing into the wind on the entire right side, so the upward movement slows down, but it's still moving upwards. Every resource I've found online simply says "because dissymmetry of lift" and leaves it at that, but I want to understand why it causes this and why it conflicts with the info in the 2nd image. $\endgroup$ – Daniel T. Dec 19 '20 at 11:25
  • $\begingroup$ The physics is beyond me, but some possible factors include the force pulling the blades strait out [and therefore down] is vary powerful, also remember that the rotor "plane" of most helicopters is offset both forwards and sideways for a better cruse attitude [mast is tilted], which may affect how it looks. hope someone else can expand. $\endgroup$ – wanna-beCanadianPilot Dec 19 '20 at 11:40
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    $\begingroup$ The flapping diagram shows the theoretical case where the flapping would place the blade tip track if the swash plate was set at "neutral" so to speak. Since that tip track is not desirable for forward flight, the swash plate is angled "automatically" by cyclic input to put the resulting disc plane where it needs to be for forward level flight. $\endgroup$ – John K Dec 19 '20 at 18:23
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TL;DR, it's just a game and is not really realistic.

Flapping isn't there to allow the rotor to compensate for increased lift on the advancing blade, it is there to make the helicopter controllable.

The helicopter cyclic control adjusts the blade pitch depending on the position of the blade along the cycle using a swashplate. When the pilot pushes the cyclic forward to transition from hover to forward flight, the swashplate tilts so that the pitch on the retreating side increases, making the blade flap up and reach highest position aft, and decreases on the advancing side, making the blade flap down and reach lowest position forward. The swash plate itself is tilted forward and the scissor links that connect the blade to it take care of the necessary lead; the end effect is that the rotor tries to keep itself aligned with the swash plate.

The lift always acts perpendicular to the plane of rotation and through the hub, so as the rotor tilts (by flapping), the action line of lift tilts with it, gets behind the centre of gravity, and the whole craft pitches forward.

Now only as the helicopter picks up speed, the increased relative speed will make the advancing blade produce more lift (though the flow will be slightly from above due to the tilt, reducing the effect) and the retreating one will produce less lift. And you are correct that this will lead to the rotor flapping up forward and down aft.

But this flapping will just add to the flapping caused by the swash plate. The rotor will now rotate tilted aft compared to the swash plate. If the pilot centred the cyclic, the rotor would pitch up, the craft would follow and slow down (and climb unless the pilot also lowers collective).

But as long as the pilot keeps the cyclic pushed forward to maintain the forward speed, the rotor will rotate roughly perpendicular to the shaft, the slight offset depending on position of centre of gravity and centre of drag. And the advancing and retreating blades will be producing the same lift, the higher air speed on advancing side compensated by lower angle of attack commanded by the forward tilted swashplate.

Last, for a coaxial twin-rotor helicopter like the Ka-50, the rotors will always be co-planar in any manoeuvre. I've said that the lift always acts perpendicular to the disk, so whenever the rotors were not co-planar, they would be trying to pull the shaft in different directions, creating no other effect than bending force on the shaft. So the control system is designed to induce the same tilt in both rotors.

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