Helicopter Rotation, Torque and Angular Momentum

In summary, the direction of the torques in the conversation was discussed in relation to a helicopter's rotors. The counterclockwise rotation of the blades produces a torque out of the page, while the angular momentum is also out of the page, causing the body to rotate clockwise. To cancel out this rotation, a tail rotor is added with a clockwise rotation, producing a torque that is out of the page in a side view. However, this torque is not enough to cancel out the rotation, so the main rotor's upward and sideways thrust forces play a role in keeping the body steady. The tail rotor's torque is also perpendicular to the main rotor's torque, which helps to balance the forces. Finally, the concept of gyroscopic
  • #1
epsilon
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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.
 
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  • #2
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.
 
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  • #3
epsilon said:
Q1) If these torques are in opposite directions, why are these not sufficient to cancel out the rotation and keep the body steady?
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.
epsilon said:
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

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
 
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  • #4
For more explanations watch these videos:











 
  • #5
Orodruin said:
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.

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.
 
  • #6
A.T. said:
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

Thank you, this was a really good explanation.
 
  • #7
A.T. said:
The sideways thrust force of the tail rotor also has to be canceled by the main rotor thrust force to avoid sideways drift.
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.
 
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FAQ: Helicopter Rotation, Torque and Angular Momentum

1. What is helicopter rotation?

Helicopter rotation refers to the movement or spinning motion of a helicopter around its vertical axis. This motion is achieved by the rotation of the helicopter's rotor blades.

2. What is the role of torque in helicopter rotation?

Torque is the force that causes an object to rotate around an axis. In the case of a helicopter, torque is generated by the engine and transmitted to the rotor blades, causing them to rotate and create lift, which results in helicopter rotation.

3. How does angular momentum affect helicopter rotation?

Angular momentum is the measure of an object's tendency to resist changes in its rotational motion. In the case of a helicopter, the spinning rotor blades have a high angular momentum, which helps to maintain the helicopter's stability and balance during rotation.

4. What factors can affect helicopter rotation?

There are several factors that can affect helicopter rotation, including the weight and distribution of the helicopter, the power of the engine, wind conditions, and the angle of the rotor blades. These factors can impact the amount of torque and angular momentum generated, thus affecting the helicopter's rotation.

5. How does a helicopter maintain its balance during rotation?

A helicopter maintains its balance during rotation through the use of various control systems, such as the tail rotor and cyclic and collective pitch controls. These systems allow the pilot to adjust the angle and speed of the rotor blades, thus controlling the amount of lift and torque generated, and maintaining the helicopter's balance during rotation.

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