I was reading a paper that looks at deep stall for T-tail configurations, and figure 2 suggests that the first trim condition for the aircraft in question occurs at approximately 21 degrees. Surely this would be a bad condition to fly at as it would result in stall due to flow separation over the wing, and if this doesn't occur the nose would be pointed so far up the aircraft would constantly be in climb? Excuse me if I'm missing something fundamental here. I've linked the paper here: https://research-information.bris.ac.uk/ws/portalfiles/portal/312655220/2022_SciTech_deep_stall.pdf
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1$\begingroup$ I would not take that graph too seriously - it is only meant to show the general principle. $\endgroup$– Peter KämpfApr 2 at 16:24
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1$\begingroup$ The curve in figure 2 typically shifts up and down more or less proportionally to the elevator deflection. Most probably that particular curve has been drawn for an elevator with a quite positive (donwward) deflection. Pity that the source of the picture is not given. $\endgroup$– sophitApr 2 at 22:30
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$\begingroup$ @sophit As Peter pointed out, the graph is for illustration only, therefore it is described as "Typical pitching moment". It is an example the author drew himself (therefore no source) and most likely quickly drew himself. $\endgroup$– U_flowApr 3 at 9:00
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$\begingroup$ @U_flow: I perfectly understand that. Anyway that is a scientific paper and if you make something up you should clearly write it otherwise people starts to wonder about the content, as this question demonstrates. $\endgroup$– sophitApr 3 at 9:54
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$\begingroup$ Thanks all, that clarifies things for me a lot more! :) $\endgroup$– cagedApr 3 at 16:19
1 Answer
the first "trim" condition...occurs at 21 degrees
It's a shame they did not show the whole Cm vs AoA curve. The aircraft is most likely already stalled here.
What they are showing is the overall aircraft Cm for various AoA if no control inputs are added, relative to a given AoA and trim setting where pitch rate = 0
One would expect positive Cm at stall for certain asymmetric airfoils$^1$, while many others remain negative$^4$.
The graph shows increasingly negative Cm (greater static stability) as the stall deepens and lift producing airflow over the wing collapses. At this point the still unshadowed tail (and rear fuselage) is providing enough pitch down force (by way of basic longitudinal stability) to over come upward pitching forces of the forward fuselage and any portion of remaining wing lift ahead of the CG.
Deep stall occurs (Cm > 0) when turbulence from the wing greatly reduces the stabilator coefficient of lift. The plane will no longer pitch down even though it's angle of attack is very high.
Cm remains positive until a second "stable" zone appears, but at a hopelessly high AoA.
It would behoof the engineers to provide enough stabiliator control authority to keep Cm negative post stall. The model in this study may have been purposely created to be stall prone$^3$ in order to generate the graph showing the dangers of deep stall.
$^1$ even if the Cm of the wing remains negative, an increasing Cm, combined with the trim setting of the tail, can cause an aircraft to become staticly unstable as pitch increases. Wing sweep with wing tip washout$^5$ can greatly improve pitch stability$^2$.
$^2$ Slats are also a great way to enhance the safety of swept wings in high AoA situations by increasing wingtip "downwash" (lowering local wingtip AoA). Stalls on swept wings usually start at the wing tips, causing Cm to go positive, leaving the stabilator as the last line of defense.
$^3$ exceeding aft CG limits is an excellent way to do this
$^4$ a benefit of "olde tyme" undercamber, which can be created with a setting of slats and a little flap
$^5$ early Dunne D.8 aircraft criticized as being "too stable"
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$\begingroup$ The value of the pitching moment coefficient only shows how far the airplane is away from a trimmed condition. To learn about stability you need to look at the derivative over lift coefficient or angle of attack. A negative derivative shows positive static stability. $\endgroup$ Apr 3 at 16:30
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$\begingroup$ @PeterKämpf I agree with that, and it highlights the need for (available) control authority beyond trim. Thankfully, the graphs help me while I continue to learn the calculus. $\endgroup$ Apr 3 at 16:40