Everywhere I look I'm only able to find information about airfoils.

How do I find out the location of the Aerodynamic centre of an arbitrary aircraft irrespective of its type or shape?

  • $\begingroup$ Do you mean center of lift perhaps? Or center of gravity? There can't really be an aerodynamic center because aerodynamics is a reference to flow not force. $\endgroup$
    – Jae Carr
    Commented Jun 13, 2014 at 14:08
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    $\begingroup$ @JayCarr From Wikipedia: "The aerodynamic center is the point at which the pitching moment coefficient for the airfoil does not vary with lift coefficient (i.e. angle of attack), so this choice makes analysis simpler." I personally had never heard of the term and was surprised to find so much on it when I Googled "aerodynamic center." Live and learn, or maybe in my case live, learn, forget, learn again? $\endgroup$
    – Terry
    Commented Jun 13, 2014 at 15:25
  • $\begingroup$ @Terry or! Or I could be wrong! lol, not the first time, probably not the last. I should have just google it like you, this is what I get for being lazy ;). $\endgroup$
    – Jae Carr
    Commented Jun 13, 2014 at 17:30
  • $\begingroup$ @Gaurang are you talking about how you could find it for an arbitrary aircraft or a specific aircraft like the Boeing 737? $\endgroup$ Commented Jun 14, 2014 at 1:15
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    $\begingroup$ You can find out where the aerodynamic centre is during a flight test. Are you looking for the procedure to do so? $\endgroup$
    – DeltaLima
    Commented Jun 16, 2014 at 11:51

2 Answers 2


What you refer to as the aerodynamic centre is also called the neutral point, the point where pitch moments do not change at all angles of attack with attached airflow.

If you ask for an unswept body of high aspect ratio, the answer would be easy: At the quarter chord point in subsonic flow and at the half chord point in supersonic flow. Unfortunately, the neutral point shifts forward with decreasing aspect ratio until it sits right at the leading point (not edge) of a slender body, a body with infinitesimally small aspect ratio. It shifts slightly backwards with positive sweep, so you normally have to calculate correction factors which depend on

  • Sweep angle
  • Taper ratio
  • Ratio of fuselage width and span
  • Lengthwise position of the wing-fuselage intersection
  • High wing or low wing configuration (negligible influence)

If you don't have good wind tunnel data or a validated CFD model, you would use a collection of formulas and diagrams like DATCOM (see this link for a computerized version) to approximate a solution. Sorry, but I cannot give you a simple formula which would work out of the box.


Can't you use the stability derivation?


Differentiate to $\alpha$

$$ \begin{eqnarray} \frac{\text{d}C_m}{\text{d}\alpha} &=& 0 + \frac{\text{d}C_L}{\text{d}\alpha} \frac{l_{cg}}{c} - V_H\frac{\text{d}C_{L_H} } {\text{d}\alpha} \\ & &\\ &=& \frac{\text{d}C_L}{\text{d}\alpha} \frac{l_{cg}}{c} - V_H \frac{\text{d}C_{L_H} } {\text{d}\alpha_H} \frac{\text{d}\alpha_H } {\text{d}\alpha} \end{eqnarray}$$

Resulting in: $$ C_{m_\alpha} = C_{L_\alpha} \frac{l_{cg}}{c} - V_H C_{L_{H_\alpha}} \left( 1-\frac{\text{d}\epsilon}{\text{d}\alpha} \right) $$ derivation2

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    $\begingroup$ Hi Luca, and welcome here! May I suggest that you use MathJax to improve the readability of your answer? Also see here for general info on how to edit an answer properly. BTW, no downvote from me. I think the two who did were rude. $\endgroup$ Commented Oct 20, 2023 at 14:21
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    $\begingroup$ Hello Luca, also welcome from me. Like Peter suggested, I would encourage you to use MathJax to improve the readability of formulas. It may seem difficult to begin with, but once you get the hang of it, it's easy to use. To help you get started, I've converted the first couple of formula's from the second half of your answer to MathJax for you. You can see the code when you edit your answer. Good luck in converting the rest. If you need help, just ping (by adding an @ in front of my name in a comment to this answer) $\endgroup$
    – DeltaLima
    Commented Oct 20, 2023 at 15:09
  • $\begingroup$ @PeterKämpf: I didn't downvote either but this answer (longitudinal stability) has actually nothing to do (at least not directly) with the question (aerodynamic centre). I think is the task of who wrote this answer to highlight the link between the two concepts. $\endgroup$
    – sophit
    Commented Oct 20, 2023 at 19:46
  • $\begingroup$ @sophit I never tried to decipher the writings. $\endgroup$ Commented Oct 21, 2023 at 17:23

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