Відео

reduce collective pitch and slowly decrease pressure on your cyclic. Stay away from your VNe speed and don’t maneuver too aggressively to avoid this in total.

@wyattdean5192 Is Vne is hard wall ias limit or is there a 10 knot buffer ? Where helicopter starts shaking and bucking. Or does RBS occur catastrophically at Vne? Which is more likely to RBS . Larger diameter 2 bladed rotor with lower rotor RPM, or smaller 4 or 5 bladed articulated rotor with higher rotor RPM? Think Bell 202, 204, vs . MD-500/OH-6A.

@ Yes the VNe is basically a hard wall limit because as you go faster the positive stall region on the rotor disk grows which is at the tip of your retreating side. It’s always best to avoid flying at or above your VNe not because it’s 100% guaranteed, but it’s extremely likely when flying nearer above the speed. RBS can happen to any helicopter, 2 bladed or more, semi-rigid or fully articulated they are all prone to RBS when at high speeds, this is why most helicopters don’t fly super super fast.

@ You’ll notice RBS when the helicopter pitches up and or rolls to the retreating side un commanded.

Yes, you are right! The angle of incidence decreases as the blade flaps up because the distance between the swash plate and the pitch horn has to stay the same since they are linked. However, the change in induced flow has a much more powerful effect on the lift gradient across the disk for three reasons: 1) because it changes the resultant AOA more than the pitch links change it, 2) because of the relative powers of the different terms in the lift equation, and 3) because blade flapping also causes a spanwise differential change in the induced AOA. Whatever the flapping angle happens to be, there is progressively more up/down displacement of the blade the farther you get from the hub. Any change in angle of incidence from feathering would at best be the same along the span, and at worst most ineffective at the tip if the blade design featured any built-in twist. The explanation of all of this is as follows: Tilting the swash plate induces cyclic feathering that in turn induces flapping that tilts the rotor disk. Building up forward airspeed as a continued result of forward thrust induces even more flapping, and it's this "even more" part that is the topic of this video. Yes, the blades also feather due to increased flapping as airspeed builds and this feathering oscillation does tend to abate lift dissymmetry, but the relative amount of equalization you get out of it is kind of akin to pedaling your bike going down a steep hill. It helps a little, but only a little because the whole blade can feather only as much as the very inner part of the blade root is flapping, which isn't much in terms of vertical displacement right at the pitch horn. Whereas, the rest of the blade can flap A LOT, especially way out on the end and especially with high forward airspeed. Remember that, despite blade twist and tip vortices acting counter to dissymmetry, the outer portion of the blades still produce the lion's share of lift due to the "velocity squared" term in the lift equation combined with the fact that blade rotational speed increases geometrically from the center of rotation. This is how the lift equation shows us that flapping is more powerful than feathering when it comes to equalizing lift. The coefficient of lift term is not squared - a significant mathematical disadvantage with respect to induced AOA. To form a mental picture of what I'm saying, keep in mind that in addition to teetering about a hinge the blades are flexible and so you can visualize that most of the flexing and flapping is happening out in the region where most of the lift is being generated. It therefore follows that the highest-lift part of a blade flexes and diverges more sharply into the induced flow (which, by the way, magically* has similarly-increased opposite-direction velocity in the same region), so the induced AOA and consequently the resultant AOA change the most precisely in the location where the maximum lift is being produced. It's kind of beautiful in a way. * There's no such thing as magic. The Army manual generalizes it like this: "1-121. In forward flight, two factors in the lift equation, blade area and air density, are the same for the advancing and retreating blades. Airfoil shape is fixed for a given blade, and air density cannot be affected; the only remaining variables are blade speed and AOA. Rotor RPM controls blade speed. Because rotor RPM must remain relatively constant, blade speed also remains relatively constant. This leaves AOA as the one variable remaining that can compensate for dissymmetry of lift. This compensation is accomplished through a combination of or individually by blade flapping and cyclic feathering." The Navy helo aerodynamics book also talks about it some, but in a bit more detail. The FAA HFH mentions that feathering happens but does not explain it. For helicopter pilots, it's sufficient to comprehend that the blades flap and that this moving around tends to even all kinds of things out.

They’re probably connected through a planetary gear of a different ratio or something.

You are now discovering the difference between angular velocity and linear velocity. They have the same angular velocity. The entire blade rotates 360 degrees in the same amount of time, but since the tip of the blade is farther from the center, it must move faster to cover a greater distance in the same amount of time as the root. The root moves around the center in a small circle while the tip moves in a much bigger circle, but it takes the same amount of time. Angular velocity is degrees over time, or rotations over time (RPM). Linear velocity is distance over time. Time is constant but the distance covered is much higher for the tip than the root due to a higher radius from the center. This results in a much higher linear velocity for the tip.

It makes all the physical sense because it's rotation, not linear motion. So V=ω d where ω is angular velocity and d is distance to the blade axis

I’m not entirely sure. As it is a high speed blade I would expect to be symmetrical however I would also assume all blades are different depending on the intended aircraft weight and rotor speed

Haha, I also had some good time with Bill Barrow in Redhill.

I find the original explanation more easy to understand. I don't really understand the induced airflow, but I was teached that the mecanical link pulled back the AoA of the blade as it was naturally getting up.

Apologies, I’m about to start again in the next week or so as I just haven’t had the time to get these done.

Looking at the left side of the curve you can see that it rises slightly as you move from the hover. This rise in power requirement is due to the aircraft moving away from the ground cushion and losing the benefit of the big high pressure cushion beneath. This is between 0-12 Kts approx. At about 12-20 Kts the airflow across the disc is sufficient enough to start effecting the induced flow and reduce its effect. This is the point where the power required starts to drop off due to translational lift and the power curve changes direction and goes downwards.

Yes, but that is not that big mistake. Are you sure that phase lag is 72 degrees for R22. I taught that phase lag is dependant of flapping hinge offset, and for R22 (semi rigid), there is not offset.

can you pls explain the difference in phase lag angle for a heavier blade and lighter blade?

I have added a longer vector on the adv side but the increase change of AoA is not relevant for flapping to equality as the aim is to show that the disc does not roll to the retreating side but instead flaps back because the thrust reduces slowly as the blade lifts. Phase angle is something different I’m afraid and is more related to inputs from the cyclic and the reaction in relation to the control orbit. Flapping to equality does take place then but it is not due to phase lag. Phase lag is always 90 degrees but your advance angle might be different on different helicopters.

This explanation is not rubbish. It is very much correct. The lift force should not be drawn at right angles from the plane of the foil, but at right angles from the relative wind, like he has done. Together with the drag it will create the resultant force. He is always drawing the drag backwards in his diagrams, however, with the angle of the lift pointing forwards enough (90 degrees of the relative wind) the resultant force is pointing forward, dragging the rotor forward driving the rotordisk. This is a correct explanation and I would advise you to watch it again and maybe find some other explanation which will show you very much the same thing.

The rotation of the rotor is one part of the relative wind, the other is the air coming from below due to the descending profile during a autorotation. This relative wind is at an angle causing the lift vector to point forward. Due to it being a rotor the outward part of the rotor is experiencing a much higher horizontal velocity the the part closer to the rotorhub. Due to this difference in speed at the ends of the rotor the resultant aerodynamic force (combination of drag and lift) point to the rear, decelerating the blade. However, closer to the rotorhub the horizontal airflow velocity is less, however the vertical velocity remains the same. This increases the angle of the relative wind causing the lift vector to point more forward. The resultant force will therefor still be pointing forward, even though the drag is still pointing to the rear, accelerating the blade. Setting the correct pitch on the rotorblade will cause a equilibrium between the driving and driven part of the rotor. You are trading altitude for rotorspeed.

I finally got my head round this. What I know understand is that the aerodynamic neutral plane of an aerofoil is perhaps 3 deg down from the geometric plane. Therefore a positive incidence to the aerodynamic neutral plane is in fact a negative 1deg. I define the aerodynamic neutral plane of an aerofoil is when the foil goes neither up nor down in an airflow.

Apologies. Poorly annotated. It should just say TQ not TR. the TQ comes down twice. TQ is always coloured yellow in my diagrams.

No gyroscopic precession no. The high and low points are due to the 360 degree nature and that an input on one side has an equal and opposite input on the other side of the disc. The high and low will be opposite each other and the points at which they reach the top and bottom is simply due to the inability for the blade to instantaneously move to the furtherest extent.