I was browsing Stack (great way to kill time at work) and I came across this question. One of the comments confuses me somewhat.

@Zlelik No, helicopter blades cannot exceed the speed of sound - even ignoring the massive damage that would cause, it would cause them to lose all lift, losing control of the craft (don't forget that while one blade is supersonic, the opposing blade isn't). Not everything has a linear relationship, you know - you need to understand how things scale. Neither 14 m @ 392 RPM nor 32 m @ 132 RPM give you a good idea o how 25 m @ 500 RPM would behave. – Luaan

(Emphasis mine)

Now I'm merely a layman, but it seems to me that if you're dealing with opposing rotors, they should both travel at the same speed when spinning. (Visually testable with a pencil, both ends should travel at the same speed when then pencil is spun, assuming the rotational center is the center of the pencil)

The only exception I can think of, is that in this particular case the author is specifically describing supersonic speeds. I'm aware that the speed of sound is different depending on the density of the atmosphere. Is it possible that this statement is true because the 'first' blade is leaving a vacuum that the 'second' blade travels through? (Thus changing the speed of sound for the second blade)

My intuition tells me that this is incorrect, as it should make it impossible to fly, much less control a rotor driven aircraft (You'd only generate lift with half of your rotor), in addition to the fact there is no real 'first' or 'second' here as they are equidistantly opposed in both directions of the rotor assembly.

So, do rotor blades really move at different speeds? If so, how is this possible?

Being a layman I will likely not understand any of the math behind this, however if you could include it for the sake of those that may, and summarize for those of us that don't, it would be much appreciated.

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    $\begingroup$ When a helicopter is moving the leading rotor has a greater airspeed than the trailing rotor (the leading rotor being the one travelling in the same direction as the helicopter), so it is possible for the leading rotor to be travelling faster than the speed of sound while the trailing rotor is not. $\endgroup$
    – BillDOe
    Jun 6, 2018 at 18:50
  • $\begingroup$ I'm not that familiar wit the technology behind rotary wing aircraft, but if they use counter-rotating blades they don't necessarily need to rotate at the same speeds if the moments of inertia of the two are different and the object is to cancel or achieve some net angular momentum - perhaps for better vertical stability and/or maneuverability. You have my curiosity aroused now. $\endgroup$
    – docscience
    Jun 6, 2018 at 18:51
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    $\begingroup$ I think that the writer was referring to the case when the helicopter is moving forward. $\endgroup$
    – user93237
    Jun 6, 2018 at 18:52
  • $\begingroup$ @BillDOe That is an excellent point that I failed to take into account. I was thinking exclusively of the downward movement of air pulled through the rotor. Could you go ahead and make your comment an answer? If there is no better alternative by tomorrow, I'll go ahead and accept it. Thanks. :) $\endgroup$
    – Lord Drake
    Jun 6, 2018 at 18:59

1 Answer 1


First, let's get terminology out of the way: in a helicopter the rotor traveling in the direction of the aircraft is referred to as the advancing rotor, while the one traveling in the opposite direction is referred to as the retreating rotor. Of course, in a hover all blades have the same airspeed. However, in forward motion the advancing rotor has a greater airspeed than the retreating rotor, since its airspeed would be the sum of the rotors linear velocity plus the aircraft's airspeed plus ambient wind speed. If the aircraft flies fast enough, the advancing rotor can travel faster than the speed of sound while the retreating rotor does not. (Also, though you didn't specifically ask, the retreating rotor can stall.)

  • $\begingroup$ Nice note about the retreating rotor stalling. :) $\endgroup$
    – Lord Drake
    Jun 7, 2018 at 18:34

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