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  • Andreas

Secrets of stability: Cm-alpha

Updated: Oct 6, 2022

If you ask an aerospace engineer, what the fundamental requirement for static longitudinal stability of an airplane is, he or she will tell you that “Cm-alpha needs to be negative”. If you face a pilot with the same question, most likely you will be shown the typical “marble in the bowl” sketch and some “wing/tail - up/down force” picture. The engineer’s answer is physically correct but sounds somewhat diffuse, the pilot’s answer sounds more straight-forward, but might be quite detached from real airplane physics.

This article aims to provide an easy-to-comprehend, but still accurate explanation of what really matters for static longitudinal stability.

Figure 1: VariEze aircraft with canard-configuration

First of all, let us fix the term “static stability”. Many pilot training books refer to the “initial” behaviour of an airplane after a disturbance, when considering static stability [1] [2]. This is undoubtedly a consequence of static properties, however the very word “static” implies that there is no time-axis here, so strictly speaking there is no “initial phase”. A more accurate description would be that a statically stable airplane develops a moment opposite to the disturbance direction when displaced from equilibrium [3] [4].

In the following figures, the CG will be between the wing center of pressure (CP) and the tail for illustrative purposes (i.e. the tail will also provide upward lift). For simplicity, the following assumptions are made:

  • Wing downwash is ignored

  • Angles are exaggerated for display

  • Drawings are not to scale

  • Units are symbolic

Figure 2 depicts an airplane in equilibrium. Note the "unusual configuration", as the CG is behind the wing CP, this is done to emphasize that there is no need for a downward force at the tail to achieve static longitudinal stability.

The wing is exposed to 4° AOA and produces 3 units of lift. With the wing arm of 1 unit, that produces a wing pitching moment of 3 units. The tail is flying at 2° AOA and produces 1 unit of lift. With an arm of 3 units, that produces a tail pitching moment of 3 units, in the opposite direction.

Figure 2: Airplane in equilibrium (CG behind wing CP)

Now let us imagine, the airplane gets disturbed nose-up and therefore changes the AOA of BOTH surfaces by 2°. The wing just experienced a change in AOA from 4° to 6°, which changes the lift from 3 units to 4.5 units, as the lift coefficient is more or less proportional to AOA below the critical AOA. That produces a new wing pitching moment of 4.5 units. And the tail? Here, the 2° change caused the AOA to increase from 2° to 4°. That doubles the lift produced by the tail, resulting in 2 units. With the tail arm, that now produces a tail pitching moment of 6 units….

Figure 3: Airplane developing a restoring pitching moment (CG behind wing CP)

…..magic, right?

All of a sudden, the situation is no longer balanced, and the airplane is subject to a restoring pitching moment. If you run the example with different numbers, you will find that there is a secret behind this:

In this example the airplane is statically stable, as long as the surface behind the CG is exposed to a lower AOA than the surface in front of the CG.

If you move the CG back to the point, where the tail and the wing are exposed to the same AOA, you will reach the airplane’s “neutral point”, and there will be neutral stability.

In most real situations, the tail will produce some amount of downward lift, therefore the situation is even more easily understood. You can perform the same calculation with the tail producing a downward force and the CG in front of the wing CP, yielding similar results.

The aforementioned explanation makes it clear that there is no need for a downward lift component at the tail for static stability.

Although we made some simplifying assumptions, I find this concept quite useful as a mental model, as it universally applicable and also works for canard-type airplanes.

So what is Cm-alpha after all?

In aerospace engineering, there are many coefficients, and unsurprisingly there is also one for the pitching moment. It is called Cm. By convention, a positive Cm refers to nose-up pitching moment [3]. Now, if we think about what could change Cm, we will find parameters, such as airplane configuration, CG-position, Mach-Number etc. In particular, Cm also changes as a function of AOA. If we now express the change of Cm, based on a change in AOA, we are talking about a partial derivative, which is called Cm-alpha in aerospace engineering. Figure 4 shows a typical Cm vs AOA graph.

Figure 4: Airplane pitching moment coefficient vs. AOA

The point, where Cm is zero, is the “Trim point” [4] [5]. If the airplane is disturbed towards the nose-up direction, Cm has to decrease, in order to create a negative (nose-down) pitching moment. In other words:

The “slope” of this curve, has to be negative for static stability [5].

That is, what the engineer was referring to in our question at the beginning. If the CG is at the neutral point of the airplane, the slope is zero and the airplane will therefore never develop a restoring moment. With that in mind, we can also get rid of the idea, that “there needs to be a tailplane for static stability”. The criteria really is the Cm vs. alpha (AOA) diagram, and as long as the slope is negative, the airplane will be statically stable [3] [4] [5].

This post is the first of a number of “secrets of stability” posts which I will add to the site over time, aiming to improve the understanding of airplane stability…



[1] FAA, Airplane Flying Handbook, US Department of Transportation, 2016

[2] P.J. Swatton, The Principles of Flight for Pilots, Wiley, 2011

[3] C. Perkins & R. Hage, Airplane Performance Stability and Control, Wiley, 1949

[4] D. Hull, Fundamentals of Airplane Flight Mechanics, Springer, 2007

[5] M. V. Cook, Flight Dynamics Principles, Elsevier, 2007

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