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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?

menjaraz
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Gaurang Gatlewar
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  • 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. – Jae Carr Jun 13 '14 at 14:08
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    @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? – Terry Jun 13 '14 at 15:25
  • @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 ;). – Jae Carr Jun 13 '14 at 17:30
  • @Gaurang are you talking about how you could find it for an arbitrary aircraft or a specific aircraft like the Boeing 737? – HardcoreBro Jun 14 '14 at 01:15
  • Yes.A method for any arbitrary aircraft irrespective of its type or shape. – Gaurang Gatlewar Jun 14 '14 at 04:52
  • Aerodynamic is everywhere: Find out here how BMW aerodynamic test center carry out optimisation of the car’s aerodynamic qualities. Wind tunnel is the mantra... – menjaraz Jun 14 '14 at 13:36
  • @menjaraz if you have the money for it :P Otherwise simulation is. (and is anyway before the wind tunnel) – Federico Jun 16 '14 at 08:31
  • @Frederico: Indeed, simulation is the way to go. Nevertheless simulation or not, one has to be good at CFD prior to playing/harnessing them. I know of an amateur built wood and fabrics ultralight designed using homemade wind tunnel. – menjaraz Jun 16 '14 at 10:27
  • @menjaraz ["Federico", no 'r' after the 'F', or I will not get notifications ;) ] Yes, you can also use a homemade windtunnel, but your observation about CFD stands also for wind tunnels. – Federico Jun 16 '14 at 10:31
  • @Federico: Sorry for the misspelling. Is there an opensource library you recommend (modelisation/simulation)? – menjaraz Jun 16 '14 at 10:38
  • @menjaraz for CFD? I would not know. In an answer XFoil was suggested for wings, but I do not know for full bodies. I think you should better ask Peter Kämpf. – Federico Jun 16 '14 at 10:48
  • @Federico: That would be for sure computationally intensive - Another layer of complexity... – menjaraz Jun 16 '14 at 11:04
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    You can find out where the aerodynamic centre is during a flight test. Are you looking for the procedure to do so? – DeltaLima Jun 16 '14 at 11:51

2 Answers2

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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.

Peter Kämpf
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Can't you use the stability derivation?

derivation1

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

DeltaLima
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    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. – Peter Kämpf Oct 20 '23 at 14:21
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    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) – DeltaLima Oct 20 '23 at 15:09
  • @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. – sophit Oct 20 '23 at 19:46
  • @sophit I never tried to decipher the writings. – Peter Kämpf Oct 21 '23 at 17:23