How to separate the stability derivatives: $C_{M\dot \alpha}$ and $C_{Mq}$.
I can imagine a $q=const, \dot \alpha=0$ movement like this
But what a $\dot\alpha=const, q=0$ movement would be look like?
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Sign up to join this communityHow to separate the stability derivatives: $C_{M\dot \alpha}$ and $C_{Mq}$.
I can imagine a $q=const, \dot \alpha=0$ movement like this
But what a $\dot\alpha=const, q=0$ movement would be look like?
$q = 0$ means no change of pitch angle. You also have a constant rate change in AoA ($\dot{\alpha} = const$).
The difference between the pitch angle ($\theta$) and AoA ($\alpha$) is the flight path angle ($\gamma$).
$\theta = \alpha + \gamma$
From this it follows that the flight path angle changes opposite to the change of change of AoA.
$\dot{\gamma}=-\dot{\alpha}$
This is not a manoeuvre that will happen in real life. An approximation of such a situation can happen when after a brief but strong pull-up the pitch angle is kept constant while the path angle of the aircraft is still increasing. The rate of change of the AoA will not be constant in such a manoeuvre.
If you add wind (a gust), the sought-after situation would be flying into a vertical gust. If the vertical wind speed changes linearly with the distance flown and the aircraft has sufficient inertia and/or low static stability, you get no pitch movement and a constant angle-of-attack rate, albeit only for a short time.