Is it possible for control surface flutter to occur at low Reynolds and low speed? If yes,why don't Ultralights have counter weights or tail sting mass to immune the control surfaces against flutter just like how most G.A planes and aerobatic planes do?
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asked Aug 21, 2018 at 5:16
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Flutter requires 3 factors to come together: flexibility of the structure that a control surface is attached to; the natural frequency of said structure; and an undamped moving mass attached to the structure with an internal CofG offset from the hinge point of the mass such that inertia helps magnify motion of the mass. If you have an ultralight with a really floppy, flexible wing that can twist easily, with a heavy, unbalanced aileron, and you excite motion in the aileron, which excites sympathetic motion in the wing itself, you can probably get it to flutter at 40 mph.
I say undamped moving mass because a lot of jets with hydraulically operated control surfaces don't use mass balancing but depend on the damping of the hydraulic actuators, and if that isn't sufficient, dedicated flutter dampers that work like motorcycle steering dampers connecting the surface to the structure.
Anyway, ultralights and other lighter airplanes are able to avoid the factors I mentioned because, in designing it to hold up the airplane in the first place, the wing/stabilizer is inherently stiff enough, and/or the oscillating mass attached is light enough, combined with the lack of potential energy in the system at the low speeds they operate at, so it's not a problem.
You see lots of light airplanes (the Aeronca Champ for example) that operate up to 100 mph + without balanced surfaces because the structural stiffness of the strut/wire braced wing/tail and the light control surfaces makes it impossible for the control surface to excite sympathetic motion in the wing/stabilizer, making balancing unnecessary.
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3$\begingroup$ Just a tip, @DavidTeahay - It's recommended to let your question sit for at least 24-hours before accepting an answer. Many people don't bother to read a question with an accepted answer, so you may miss out on additional information. NOTE: no implication whatsoever that this answer is wrong or even not very good (I don't have enough knowledge to say one way or the other), just that others may not come along now that there's a check mark. $\endgroup$– FreeManAug 21, 2018 at 13:42
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