By Mike Lehmann, Mike Switzerland 22:11, 20 July 2006 (UTC) - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=964282

This question asks how coaxial helicopters achieve yaw. The accepted answer correctly states that this is done by increasing pitch on one rotor while decreasing pitch on the other rotor, which leaves lift the same and creates a yawing moment.

The post became active again on 20 July, when @Andy Ewen asked how the mechanism works exactly:

..a lot of the answers are stating the obvious but fail to explain HOW. Yes a torque differential between the rotors will effect a yaw but the real question is how is this achieved?

The post was deleted and not entered by OP as a separate question, hence this question. How is this achieved, exactly?

  • 1
    $\begingroup$ Don't know why the DV, this is a good question; it's even trickier when I consider a co-axial powered by a single powerplant (e.g., a Manzolini Libellula). $\endgroup$
    – ymb1
    Aug 3 '19 at 5:33
  • $\begingroup$ @ymb1 It's a boggle-copter, boggles the mind. $\endgroup$
    – Koyovis
    Aug 3 '19 at 6:17

Mechanically it's simple, and ingenious.

The following is from a 1966 patent titled: Coaxial rigid rotor helicopter and method of flying same (US3409249A) by United Technologies Corp. (Of course this is not the only way, but it's one way as invented by a major aerospace company over 50 years ago.)

I've done the laborious reading referencing numbers and text, and I'll summarize it here.

Swash control

enter image description here
(Don't mind the steering wheel, it's the inventor's idea of pushing the cyclic left/right.) I've added an arrow pointing to 274 in the illustration, it's a wheel, and it's the key.

Using the collective (254) translates that wheel forward and aft, which moves the rigging of both swash plates up/down collectively.

But using the anti-torque (yaw) pedals, rotates that same wheel by way of cable 328. The wheel rotation would move the rigging for each rotor opposite each other.

Below is a simplified elevation view for the effects of translation and rotation of the wheel on both swash plates:

enter image description here

Swash mounting

For yawing, the summary is that the up/down translating parts do not rotate. It is the attachments from plates to rotors that do rotate and are on bearings. Basically two of the same but in a concentric manner.

enter image description here

The two main element per rotor are bearings that permit the vertical translation of the fixed portion, and the rotation of the attachments to the rotor. Focus on labels 136 and 140 (added arrows for easier finding) for the upper rotor, and here's the quotation:

(...) Bearing 136 is positioned between ring member 132 and fixed housing 52 so as to prevent rotation of ring member 132 about the axis of rotation 28 but to permit translation of ring member 132 along axis 28. Swash plate mechanism 130 also includes inner gimbal ring 133 which is pivotable about axis 134, outer gimbal ring 135 which is pivotable about axis 137, and ring member 138, which is supported from ring member 135 by bearing 140 for rotation with blades 96 about axis 28.

(Emphasis mine.)

  • $\begingroup$ Mixing the pedals into the swashplate deflections is indeed complicated and ingenious. Does the patent also mention how the two swashplates are connected to the two rotors rotating in opposite directions? $\endgroup$
    – Koyovis
    Aug 3 '19 at 23:27
  • $\begingroup$ @Koyovis: It does. For yawing, the up/down translating parts do not rotate. It is the attachments from plates to rotors that do rotate and are on bearings. Basically two of the same but in a concentric manner (here's the drawing). $\endgroup$
    – ymb1
    Aug 4 '19 at 2:49
  • $\begingroup$ Might be a good addition to the answer. $\endgroup$
    – Koyovis
    Aug 4 '19 at 2:52
  • $\begingroup$ @Koyovis: Agreed. Planning on it later today. $\endgroup$
    – ymb1
    Aug 4 '19 at 2:55
  • $\begingroup$ @Koyovis: Done. See now. $\endgroup$
    – ymb1
    Aug 4 '19 at 6:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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