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Helicopter Rotation, Torque and Angular Momentum

  1. Apr 18, 2015 #1
    The direction of the torques in the following working will be found using: [itex]\vec{\tau} = \vec{r} \times \vec{F}[/itex].

    When viewed from above, the counterclockwise rotation of the blades produces a torque out of the page:
    2553tds.jpg

    As the angular momentum (right-hand corkscrew rule) is also out of the page, the body will begin to rotate clockwise to conserve the angular momentum.

    The direction of the torque associated with this rotation of the body will be into the page:
    15pk8l3.jpg

    Hence we have a torque into the page, and a torque out of the page. Q1) If these torques are in opposite directions, why are these not sufficient to cancel out the rotation and keep the body steady?

    The tail rotor is added to do this job (as I have been told infinitely-many times). By considering a side view of the helicopter, and a tail rotor with a clockwise rotation, it will produce a torque that is out of the page in this view.
    r00znt.jpg

    If the helicopter is re-considered from a birds-eye view, the tail rotor has produced a sideways torque. Hence the torque on the body due to the main rotor, and the torque of the tail rotor are perpendicular - how can they cancel one another out?

    I know that I have a misunderstanding somewhere, hence why I have written this thread. I find rotations by far the most difficult part of mechanics to grasp and am really struggling with it, plus I need to be able to extent this knowledge to the behaviour of gyroscopes too (not looking good).

    If you are able to clear the above problem up for me, and give me any advice on how to best solve rotations problems I would be very grateful.
     
  2. jcsd
  3. Apr 18, 2015 #2

    Orodruin

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    The torque around its own axis is not the only force acting on the helicopter from the rotors. There is also the upward force from the big rotor, you know, the one lifting the helicopter ... and the corresponding force from the small rotor, which is perpendicular to the rotor itself. This latter force is going to provide an additional torque about the axis of the big rotor.
     
  4. Apr 18, 2015 #3

    A.T.

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    These equal-opposite torques act on two different bodies: rotor and helicopter. If you treat rotor and helicopter as one body, this torques becomes internal, and the external torque is from the air (drag on the blades). This external torque by the air needs to be canceled by some other external forces, than those on the main rotor.


    The tail rotor also produces a horizontal thrust force, which creates a vertical torque that cancels the vertical torque from the air on the main rotor.

    Fig4-2.JPG

    The reaction torque from the trail rotor (that you show in the side view) is canceled by a combination of main rotor thrust force and gravity. The sideways thrust force of the tail rotor also has to be canceled by the main rotor thrust force to avoid sideways drift.

    f0628-03.gif

    Your idea of two rotors canceling their axial torques directly, applies to counter rotating twin main rotor configurations, but not to the tail rotor configuration:
    http://www.aerospaceweb.org/question/helicopters/q0034.shtml
     
    Last edited: Apr 18, 2015
  5. Apr 18, 2015 #4

    A.T.

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    For more explanations watch these videos:











     
  6. Apr 19, 2015 #5
    The part that solved my confusion, and that nowhere seems to emphasise adequately is that the torque that the tail rotor provides is about the axis of the big rotor. This is what I was missing. Thank you.
     
  7. Apr 19, 2015 #6
    Thank you, this was a really good explanation.
     
  8. Apr 19, 2015 #7

    rcgldr

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    This means a helicopter needs to lean a bit into the direction of thrust from the tail rotor in order to achieve a stationary hover. If there's a crosswind, that may be more significant than the tail rotor thrust.

    In case it wasn't clear in video #4 from the post above, the gyroscopic precession is a 90 degree delay in the reaction of the main rotor. For the model helicopter in the video, when viewed from above, the rotor rotates clockwise, so it may be best to use "left hand" rule to explain this, since the left thumb points in the direction of the main rotor's axis (upwards when the helicopter is right side up and level), and closing the left fingers corresponds to the main rotor's direction of rotation. So assume a pitch down aerodynamic torque is applied by the air to the rotor in response to cyclic control input. With "left hand rule", this corresponds to the thumb pointed right, and the equivalent of the cross product of the upwards angular momentum vector and the right pointed torque vector, the helicopter rolls to the right (clockwise as viewed from behind), 90 degrees "behind" the torque vector. (You can add a short torque vector to the end of a long angular momentum vector to get an idea of which direction the angular momentum vector is going to rotate towards). To get a pitch down response, a left (counter clockwise when viewed from behind) aerodynamic torque is applied by the air to the rotor in response to cyclic input, and the helicopter pitches down, again 90 degrees "behind" the torque vector.

    To compensate for the delay, the aerodynamic torque applied to the rotor needs to be advanced by 90 degrees, but the linkage required to do this is simple. Each rotor blade's pitch just needs to correspond to the tilt of the swash plate. Say the swash plate is pitched down (tilted forward): when a rotor blade is pointed right (relative to the helicopter), it's pitched more upwards, when the rotor blade is pointed left, it's pitched more downwards, and when the rotor blade is pointed forwards or backwards, it pitch is zero relative to the collective pitch. The upwards pitch on the right and downwards pitch on the left results in an aerodynamic torque to the left (counter clockwise when viewed from behind), resulting in a pitch down response.
     
    Last edited: Apr 19, 2015
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