This image below about the torque reaction of a helicopter has confused me a bit. I would like to ask whether i understand the principles here correctly. In the picture we see a helicopter whose main propeller is spinning Counter Clock Wise (CCW), as is indicated by the up most arrow. We know that the torque is $ \vec{T} = \vec{r}\times\vec{F} $, so according to the right hand rule i want to calculate the direction of the torque of the main propeller here, or better yet to understand why it is in the direction illustrated.

We know that the main propeller is producing a force downwards to the air. And then the air is countering this force, by producing a lift force that lifts the helicopter upwards. Now, my problem is which one of the two forces should i consider when i calculate the Torque with the above equation? I believe i should take into account the force-action (from the helicopter to the air) and not the force-reaction (from the air to the helicopter). If i consider the force-action-dowwards using the right hand rule (i know how to do that correctly, i'm sure) then the direction of the torque should be that of the image below. However the direction that the main rotor is spinning confuses me whether i'm doing the right thing.

So finally in order to compensate for this torque, the tail rotor must produce the torque reaction in the opposite direction of the main torque.

My main question is whether my reasoning about the direction of the main torque is correct. Thanks in advance.

enter image description here


1 Answer 1


If, as shown in the diagram, the main rotor is moving in the counter-clockwise direction, then the body of the helicopter will try to twist in the opposite direction (i.e., it will want to turn clockwise). The tail rotor thus needs to provide a thrust which gives a counter-clockwise torque in order to counter and cancel the clockwise torque on the helicopter body from the main rotor. In order to do this, the tail rotor needs to blow air towards the left in the diagram shown, which then results in a thrust on the helicopter tail to the right, as shown in the diagram.

  • $\begingroup$ Thank you for answering. Yes, you are right. This is a more theoretical approach which is absolutely right and i completely understand it. However what do you think about my approach? $\endgroup$
    – KeyC0de
    Commented Sep 20, 2015 at 23:24
  • $\begingroup$ The only suggestion I have is that your "Tail rotor thrust" arrow should be straight, rather than curved. Thrust is a force, not a torque, although it does (in this case) produce a torque. If you're going to stay consistent, label the arrow "Torque produced by tail rotor" or something similar. $\endgroup$ Commented Sep 21, 2015 at 0:33
  • 1
    $\begingroup$ @RestlessC0bra: The 'lift' forces do not enter into the consideration of the torque here since those forces are parallel to the torque axis being considered. The forces involved here are the 'drag' forces on the main rotor as it turns around. Due to those drag forces the helicopter exerts a (counterclockwise) torque on the main rotor and, in turn, the main rotor exerts an equal and opposite torque (i.e., a clockwise torque) on the helicopter body. $\endgroup$
    – user93237
    Commented Sep 21, 2015 at 1:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.