Thank you for the explanation. From what I have read about this, what I thought was that the flapping compensated the dissimetry if lift, but not totally, and the explanation to the nose up tendency when entering ETL or lifting the collective in flight was caused by some dissimetry of lift effect and the gyroscopic precession of the rotor, which would make the lift rotate 90º from the side of the rotor disk to the front. Is this wrong? How does the gyroscopic precession fit in this explanation?
Thanks, I think explaning it as a continuous action and a rate of flapping makes it easier to grasp, compared to many other explanations just giving states/snapshots.
Eyeopener. I understood that on side would Flap up and the other would flap down. But I always thought that (flapping up, for example) would explicitly decrease the angle of attack due to the linkage staying "stationary" and forcing the angle of attacht to decrease. But now I understand it actually initiates a vertical airflow.
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.
I will do a short video but for now, as soon as the advancing blade moves past the 6 o’clock position it is subjected to a higher velocity of the relative airflow than the retreating side. It’s lift increases and therefore starts to flap up from the 6 o’clock point. As an analogy, imagine you need to push a car. You start at the 6 o’clock point, it takes a bit to get the car going but it does move forward and accelerate. As momentum builds the car accelerates more and more, you get faster and faster reaching a max speed (rate of flapping up) at the 3 o’clock position. You then run out of energy. The acceleration starts to reduce and the car starts to slow, its doesn’t move backwards but it starts to decelerate. The same thing happens here, the blade accelerates upwards quickly until the 3 o’clock point then keeps going up but its rate of flapping up gradually slows as less and less of the blade is exposed to that additional airflow. At the 12 o’clock point, the relative airflow changes markedly and therefore there is no longer the force/lift required to hold the blade up at its high point. Down it comes. The changing relative airflow is therefore effecting how fast the blade is being pushed up, not how far it is being pushed up.
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.
This is an AMAZING explanation!!! Thank you so much!!!! Just to clarify though, when we are talking about velocity we are referring to the speed of the airflow and not the speed of the blade itself correct?
I always thought the angle of attack on the advancing blade was pulled down by the mecanical link on the main rotor mast. As the blade is allowed to move up thanks to an hinge. This movement also forces its angle to reduce, thus achieving equality ? Is it a wrong explanation ? I must admit I didn't really get the induced airflow shown in the video. I must watch again
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.
Sorry, your vector diagrams are wrong. You correctly showed the changes in Induced Flow due to flapping up and flapping down, but you left off the more important vector - the change in airflow. The RHS diagram needs an extra length added to the horizontal line (the rotational flow) for the extra airflow on the advancing side, and a negative length on the retreating side. On the RHS, if the IF stayed the same, the extra horizontal line will give a resultant vector coming in at a shallower angle, adding to the angle of attack, and on the retreating side, the shorter airflow line brings the IF arrow closer, elevating the resultant vector and reducing the AoA. But the flapping changes the IF line, adding to the RHS and subtracting from the LHS, so the AoA remains essentially the same. BUT the shorter horizontal vector means that there is more lift on the advancing side. Along comes Phase Lag - the rate of upward acceleration is at the max at 3 o'clock, but the mass of the blade means that the blade takes time to move, and the resultant high point is correctly shown at 12 o'clock. Remember that this is not gyroscopic precession, because the rotor system is not a gyroscope. The phase lag is usually around 90 degrees, but for light blades, like the R22, it is only 72 degrees.
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.
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.
Yes agree, that’s the best explanation I’ve seen covering all aspects.
Amazing analysis
that is the best explanation I have ever had. thank you so much for this tutorial.
The best explanation I have seen on this topic. Thank you very much and keep going
ealaa majhudatik
The best people are those who learn knowledge and teach it. Thank you for your efforts
Finally! Someone proficiently explains why flapping up reduces the aoa and vice versa. Thanks.
Great video lesson. Thank you sir.
best explanation I've seen, covers all the things that contribute and a great visual with the ruler
Great explanation, thanks.
Beautiful explanation 😊
A marvelous job!
Thank you for the explanation. From what I have read about this, what I thought was that the flapping compensated the dissimetry if lift, but not totally, and the explanation to the nose up tendency when entering ETL or lifting the collective in flight was caused by some dissimetry of lift effect and the gyroscopic precession of the rotor, which would make the lift rotate 90º from the side of the rotor disk to the front. Is this wrong? How does the gyroscopic precession fit in this explanation?
Brilliant 👍
Thanks, I think explaning it as a continuous action and a rate of flapping makes it easier to grasp, compared to many other explanations just giving states/snapshots.
Great job, thanks
Eyeopener.
I understood that on side would Flap up and the other would flap down. But I always thought that (flapping up, for example) would explicitly decrease the angle of attack due to the linkage staying "stationary" and forcing the angle of attacht to decrease. But now I understand it actually initiates a vertical airflow.
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.
Thank you for the lovely lesson, but can you please elaborate more on how the blade's highest and lowest points are at 12 o'clock and 6 o'clock...
I will do a short video but for now, as soon as the advancing blade moves past the 6 o’clock position it is subjected to a higher velocity of the relative airflow than the retreating side. It’s lift increases and therefore starts to flap up from the 6 o’clock point. As an analogy, imagine you need to push a car. You start at the 6 o’clock point, it takes a bit to get the car going but it does move forward and accelerate. As momentum builds the car accelerates more and more, you get faster and faster reaching a max speed (rate of flapping up) at the 3 o’clock position. You then run out of energy. The acceleration starts to reduce and the car starts to slow, its doesn’t move backwards but it starts to decelerate. The same thing happens here, the blade accelerates upwards quickly until the 3 o’clock point then keeps going up but its rate of flapping up gradually slows as less and less of the blade is exposed to that additional airflow. At the 12 o’clock point, the relative airflow changes markedly and therefore there is no longer the force/lift required to hold the blade up at its high point. Down it comes.
The changing relative airflow is therefore effecting how fast the blade is being pushed up, not how far it is being pushed up.
Great explanation,does the highest and lowest points have something to do with the gyroscopic precession?
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.
This is an AMAZING explanation!!! Thank you so much!!!! Just to clarify though, when we are talking about velocity we are referring to the speed of the airflow and not the speed of the blade itself correct?
Thnku
I always thought the angle of attack on the advancing blade was pulled down by the mecanical link on the main rotor mast.
As the blade is allowed to move up thanks to an hinge. This movement also forces its angle to reduce, thus achieving equality ?
Is it a wrong explanation ? I must admit I didn't really get the induced airflow shown in the video. I must watch again
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.
Sorry, your vector diagrams are wrong. You correctly showed the changes in Induced Flow due to flapping up and flapping down, but you left off the more important vector - the change in airflow. The RHS diagram needs an extra length added to the horizontal line (the rotational flow) for the extra airflow on the advancing side, and a negative length on the retreating side.
On the RHS, if the IF stayed the same, the extra horizontal line will give a resultant vector coming in at a shallower angle, adding to the angle of attack, and on the retreating side, the shorter airflow line brings the IF arrow closer, elevating the resultant vector and reducing the AoA. But the flapping changes the IF line, adding to the RHS and subtracting from the LHS, so the AoA remains essentially the same. BUT the shorter horizontal vector means that there is more lift on the advancing side.
Along comes Phase Lag - the rate of upward acceleration is at the max at 3 o'clock, but the mass of the blade means that the blade takes time to move, and the resultant high point is correctly shown at 12 o'clock. Remember that this is not gyroscopic precession, because the rotor system is not a gyroscope. The phase lag is usually around 90 degrees, but for light blades, like the R22, it is only 72 degrees.
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.
add the transverse flow roll or inflow roll to this shit and make things more clearcut lol..
then they say why are helicopter pilots so complicated!!