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Rotary - turbulence

The issue with some helicopters is not flatland gusts (provided we maintain translational lift), but rather a large vertical gust component. With some (mostly two-blade) rotor designs, substantially unloading the rotor disc by such turbulence, or by doing wingovers or loops, can cause the whole contraption to cease to function as a flying machine.

Glenswinton, SW Scotland, United Kingdom

Most of the lift on a helicopter comes from the rotor tips which are typically moving at 3-400mph so intuitively a gust of 10-20 knots shouldn’t make a big difference

Intuitively it’s the opposite for me, since lift depends on square of speed, and so if we consider V the speed without gust and Vg the relative speed of the gust, lift depends on (V + Vg)^2 = V^2 + 2V*Vg + Vg^2, the impact of a gust would be stronger the larger is V.

I don’t have much experience on helis though (just a trial lesson)!

If indeed they are more stable, Is it possible that the giant gyroscope at the top plays its part on stabilising?

Try putting some numbers in. E.g v=50, vg=20. The ratio between lift with and without a gust is 1.96.

Next try 400 and 420. The ratio is 1.1.

In absolute terms, if all else were equal, the increase in lift would be greater for the faster rotor. In practice it will be at a lower angle of attack so it’s the ratio of the change in lift rather than the absolute value that is important.

Jacko, I think you’re referring to the nasty habit of teetering rotors of ‘mast bumping?’

If eel it’s the absolute increase in lift that matters, since it’s going to dictate the magnitude of the deviation from equilibrium.

If you are in equilibrium (zero acceleration), and you get gust which produces lift. Now you are no longer in equilibrium and the acceleration should depend only on the force produced by the gust (since the other forces are in equilibrium). The magnitude of the equilibrium forces shouldn’t matter at all, should it?

Last Edited by Noe at 05 Aug 09:40

You are mostly right, but the absolute increase in lift will be lower for the faster aerofoil because in practice it will be 1) of smaller surface area and 2) at a lower angle of attack than a lower speed aerofoil providing the same amount of lift.

The equilibrium forces depend on the mass of the helicopter so not entirely irrelevant i.e. a small force on a large helicopter will be less destabilising than the same force applied to a small helicopter.

Last Edited by kwlf at 05 Aug 10:48
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