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I've been reading several sources on the location of the aerodynamic center

Many (such as for example Wikipedia) determine the aerodynamic center by writing out the equation for the moment for an arbitrary point, and then apply some definitions to get to the end result: the aerodynamic center lies at the quarter chord, see below:

enter image description here

However, I'm looking for a more physically intuitive explanation, probably starting from thin airfoil theory. The way I understand it, ultimately the aerodynamic center is determined by how the distribution of lift changes with angle of attack. Thus, the distance the center of lift moves w.r.t. the aerodynamic center cancels out the resulting moment due to the change in lift with angle in attack.

Can we explain why this point ends up to be at the quarter chord location?

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It has to do with the Birnbaum distribution. The center of its area is at one quarter.

In potential flow theory, lift can be calculated as the linear superposition of a contribution from camber and one from angle of attack. While the camber-related part of lift is constant, the angle-of-attack related part varies linearly with this parameter. This means that changing the angle of attack adds or subtracts circulation, the chordwise distribution of which is described by the Birnbaum distribution. The important part is the self-similarity of the angle of attack dependent lift resulting from this addition or subtraction of the same chordwise distribution.

The center of pressure of the resulting angle-of-attack dependent part is constant and at 25% of chord for 2D flow and wings of large aspect ratio, at least as long as flow remains attached and viscous effects are negligible.

The pictures below (own work) show the circulation over chord produced by camber alone (so AoA is 0°) on the left and the Birnbaum distribution alone on the right (so using a symmetric airfoil). Total circulation is simply the sum of both. Copied over from this answer thanks to the helpful suggestion of ROIMaison.

picture one

picture two

picture three

ultimately the aerodynamic center is determined by how the distribution of lift changes with angle of attack

That is precisely right. Do not look at the lift distribution, but at its change over angle of attack, because that is what determines the center of pressure. The constant part caused by camber does not change with angle of attack and the part caused by angle of attack has its center always at 25% (under the conditions mentioned above).

BTW, I wish we had more people like Aaron Swartz. That article is now 97 years old and these monopolists still want 42 € from me for a copy. Sorry, no details from it – I could not find any online.

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  • $\begingroup$ I'm not sure which images you mean exactly (I do have access to the paper). However, in one of your previous answers you show the Birnbaum distribution for different angles of attack, I think it illustrates your point clearly, and would be useful to include here as well. $\endgroup$ – ROIMaison Sep 30 at 7:44
  • $\begingroup$ @ROIMaison: Thank you, I forgot about this answer completely. I did those graphs ages ago using OmniGraffle. I am a happy Inkscape user since maybe 15 years now and didn't touch OmniGraffle and those pictures for maybe 20 years. $\endgroup$ – Peter Kämpf Sep 30 at 8:32
  • $\begingroup$ I'm using the formulae from the Birnbaum paper to create the three parts of the circulation. If I add them, I get something that looks like the graphs you show, but if I then determine the center of pressure, it does not become the 1/4 chord point (it's just shy of 40% chord). I noticed there are coefficients in front of the three formulae, should I adjust the weights of the three functions, if yes, how? $\endgroup$ – ROIMaison Sep 30 at 10:23
  • $\begingroup$ @ROIMaison now we see how important the down-nuetral-up force of tail is to regulate CP for cambered wings as AOA increases, and also how the "plank" wing is possible on the Helios. $\endgroup$ – Robert DiGiovanni Sep 30 at 12:45
  • $\begingroup$ @ROIMaison I do not have access to the paper and only know of Birnbaum distributions from secondary literature. Sorry, but I cannot help you with those equations. And you certainly had to assure that you would not give copies of the paper away before you received it, right? $\endgroup$ – Peter Kämpf Sep 30 at 15:25
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Intuitively, the aerodynamic center of a wing is at 1/4 MAC because the lift created by the front top curved surface of the wing and the front bottom of the wing roughly equals the lift created by the back bottom of the wing throughout the range of non-stalled AOA.

From this we can deduce that the top of the wing behind MAC is not a major contributor to lift.

A tail can be used as needed to help keep the net center of lift from moving too much with changes in AOA. Tail requirement is greatly dependent on wing design.

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