I know for static stability reasons, the derivative of the moment coefficient of an aircraft has to be negative.

What I do not understand, why the moment of coefficient (cm) at all is allowed to be positive? Because when it is positive the aircraft pitches up.

Is it because when cm>0 I use my tails for trimming ? But is this "static stability" when I actively use my tail ?

I struggle with:
alpha = 0 ° --> e.g. cm=0.3 : the tail is used for trimming so that it counteracts the positive moment of the wing

alpha = 2° -- > e.g. cm=0.4 : now the pilot could also use the tail for trimming the higher moment

Thank you for your hints


1 Answer 1


An airplane achieves steady flight at a given airspeed only if the total pitching moment about its CG is zero. This is achieved with a particular fixed tail angle/elevator angle/canard angle. If the AOA corresponding to the total lift at this airspeed is 0 AOA, then so be it. It's just that for typical airplanes at operational speeds (away from $V_{FE}$ for example), the AOA is typically larger than zero; this means that, for a typical plane, the pitching moment at 0 AOA must be larger than zero at the tail angle, elevator angle, etc. corresponding to the trim speed for static stability.

Why larger than zero? Static stability means that when perturbed from the trim AOA (at which the pitching moment is zero), the resulting pitching moment will return the aircraft back to its original trim without pilot compensation. Take a look at the graph below; note that the moment is zero at AOA=1.5 (i.e. the trim condition).

enter image description here

If the pitching moment at zero AOA is negative (the orange line), then the airplane will diverge from its trim point when perturbed. That's the opposite of static stability.


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