CX-RIDE FLAPPING TO EQUALITY Helicopter principles of flight

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  • เผยแพร่เมื่อ 6 พ.ย. 2024

ความคิดเห็น • 28

  • @nicholasbarbier3310
    @nicholasbarbier3310 2 ปีที่แล้ว +2

    that is the best explanation I have ever had. thank you so much for this tutorial.

  • @husseinabdelhamid5124
    @husseinabdelhamid5124 7 หลายเดือนก่อน +1

    The best explanation I have seen on this topic. Thank you very much and keep going

  • @tactcom7
    @tactcom7 ปีที่แล้ว +1

    Finally! Someone proficiently explains why flapping up reduces the aoa and vice versa. Thanks.

  • @giorgosgalanikas6490
    @giorgosgalanikas6490 2 ปีที่แล้ว +1

    Amazing analysis

  • @djamelmeddahi4227
    @djamelmeddahi4227 10 หลายเดือนก่อน

    ealaa majhudatik
    The best people are those who learn knowledge and teach it. Thank you for your efforts

  • @picopilot
    @picopilot 3 ปีที่แล้ว

    best explanation I've seen, covers all the things that contribute and a great visual with the ruler

  • @johnbelvedere4521
    @johnbelvedere4521 2 ปีที่แล้ว +1

    Great explanation, thanks.

  • @Kkross90
    @Kkross90 5 หลายเดือนก่อน

    Beautiful explanation 😊

  • @davidwallace5738
    @davidwallace5738 3 ปีที่แล้ว +1

    Great video lesson. Thank you sir.

  • @pylaterreur
    @pylaterreur 4 หลายเดือนก่อน

    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.

  • @heli35
    @heli35 6 หลายเดือนก่อน

    A marvelous job!

  • @orchidahussuhadihcro9862
    @orchidahussuhadihcro9862 2 หลายเดือนก่อน

    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

  • @msvijayendrarao4150
    @msvijayendrarao4150 5 หลายเดือนก่อน

    Great job, thanks

  • @jacklathey5223
    @jacklathey5223 ปีที่แล้ว

    Brilliant 👍

  • @TiffanynNathanJosiah
    @TiffanynNathanJosiah 3 ปีที่แล้ว

    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?

  • @ernestocheguevara4032
    @ernestocheguevara4032 2 ปีที่แล้ว

    Great explanation,does the highest and lowest points have something to do with the gyroscopic precession?

    • @cx-ride9136
      @cx-ride9136  8 หลายเดือนก่อน

      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.

  • @emeshrangala4419
    @emeshrangala4419 3 ปีที่แล้ว +1

    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...

    • @cx-ride9136
      @cx-ride9136  3 ปีที่แล้ว

      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.

  • @merijnvanschaik4989
    @merijnvanschaik4989 11 หลายเดือนก่อน

    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.

    • @orchidahussuhadihcro9862
      @orchidahussuhadihcro9862 2 หลายเดือนก่อน

      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.

  • @Grumpy412
    @Grumpy412 7 หลายเดือนก่อน

    add the transverse flow roll or inflow roll to this shit and make things more clearcut lol..

  • @vinaypathak9982
    @vinaypathak9982 3 ปีที่แล้ว

    Thnku

  • @robhaylock7742
    @robhaylock7742 ปีที่แล้ว +1

    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.

    • @cir9968
      @cir9968 ปีที่แล้ว

      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.

    • @badri149
      @badri149 ปีที่แล้ว

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

    • @cx-ride9136
      @cx-ride9136  8 หลายเดือนก่อน

      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.

  • @Grumpy412
    @Grumpy412 7 หลายเดือนก่อน

    then they say why are helicopter pilots so complicated!!