This topic has been discussed already several times, but I do not really get it. I think this one is a quite nice explenation: enter link description here

So I understand that the Aerodynamic Centre does not change with angle of attack. But what I do not get: how this helps in a pre-design of my aircraft?

Because: for stability calculation, I have to consider the forces acting on the whole aircraft and the corresponding lenghts (moments).

So regard a stationary horizontal flying aircraft: when I make the equilibrium of moments around the centre of the gravity, I have to regard the force F_cp,w at the centre of pressure of the wing (resulting lifting force) with distance l_cp (from centre of gravity) AND the force F_cp,t (resulting lifting force at the tail) with it's lever arm l_cp,t .

The balance becomes around the centre of gravitiy: F_cp,w * l_cp,w = F_cp,t * l_cp,t

When the angle of attacks change, F_cp,w*l_cp,w will change since the centre of pressure will move and the resulting lift will change.

So what I assume: there is a trick with which I can make the moment of equilibrium around the centre of gravity and where l_cp,w remains constant = Aerodynamic Centre

But I can not imagine how this is working with the balance of moments.

Can someone help? I am really confused.

Thank you very much

  • 1
    $\begingroup$ Maybe you want to surround your formulas with $ signs as in $F_{cp,w}$ etc. to make it appear as $F_{cp,w}$. $\endgroup$
    – PerlDuck
    Commented Nov 17, 2019 at 14:01

1 Answer 1


Aerodynamic Centre (AC) is particularly handy when we analyse the dynamics, that is, changes. For instance, changes due to disturbances: stability.

First thing, for analysis, we split all forces into forces acting where we want plus moments caused by the fact that these forces actually act somewhere else. And then when we write our balance equation, we need to balance both the forces and the moments.

We want forces to be at CG? Fine, let's declare that lift happens at CG. But then we acknowledge that apart from just lift, there will be an additional moment with respect to CG.

Now, due to linearity of aerodynamics at small angles of attack (AoA), this moment will change linearly with AoA. This effectively means that there will be a certain fixed point at which the moment will not change; or in other words, a point at which all the additional lift (due to pure AoA) can be thought to be happening. This point is the AC.

This is great, because:

  • We can express this extra AoA-related lift as $L_{w\alpha} \cdot \alpha$, where $\alpha$ is AoA and $L_{w\alpha}$ is the wing lift derivative by AoA, which is an intrinsic measurable property of a wing (or airfoil/aircraft).
  • The moment produced by this lift (with respect to CG) is that lift times its arm $x_{wac} - x_{cg}$ ($x$ is a relative distance (coordinate) of a point with respect to some datum, typically the front of the mean aerodynamic chord (MAC), which is another fixed property). So, our moment from lift will be $L_{w\alpha} \cdot \alpha \cdot (x_{wac} - x_{cg})$.
    • Remember though that for all but symmetric airfoils there will also be a static moment $M_0$ due to the camber-related lift also not acting at our CG.
  • In the above moment equation everything except for $\alpha$ is constant for a given wing/airplane. When we analyse longitudinal dynamics, which is all about AoA changes, we need the derivative by AoA. This leaves us with a constant, which unequivocally characterises our static stability.

Let's add the tail. It's lift will be trivial, $L_{t\alpha} \cdot \alpha_t$ plus some static lift at the chosen zero AoA. The arm of the AoA-related lift with respect to CG will also be constant: the distance between CG and AC of the tail. (Given that the tail airfoil is often symmetric, the whole lift arm is simply the distance to 1/4 tail MAC).

So, when we equate the moments for a straight-and-level flight (and ignore the downwash effect for a moment, which changes the tail AoA $\alpha_t$ from the base AoA), we get $$(L_{w\alpha} (x_{wac} - x_{cg}) + L_{t\alpha} (x_{tac} - x_{cg})) \cdot \alpha = const \Leftrightarrow M_\alpha \cdot \alpha = const$$

The $const$ represents the static moments to be trimmed out; but what's important for the dynamics analysis is that we get a constant moment derivative $M_\alpha$: everything in the brackets is constant thanks to the 'constancy' of AC. We can break it up as $M_\alpha = L_\alpha \cdot (x_{ac} - x_{cg})$, which defines the overall AC of the wing + tail system, often called neutral point (NP).

Now if we want a statically stable aircraft, i.e. the one which negates the changes to AoA, we need, by definition, a negative $M_\alpha$ (and that's regardless of the trim $const$). From there, you can easily derive the requirements for the CG position with respect to AC. This very convenient, and this is why we need AC more than anything else. (In aircraft design, unlike pure airfoil/wing aerodynamics, CP practically never appears anywhere). Finding the AC is not always trivial though.

(Of course, in practice engineers use dimensionless coefficients rather than actual forces/moments, but I didn't want to introduce more entities than necessary here).

  • $\begingroup$ I will go through in detail your answer tomorrow Zeus - so far: thank you.What is more: when I understand it correctly: for longitudinal stability the CP of the WING is allowed to be in front of the CoG of the AIRCRAFT as long as my tail can balance the destabilizing moment of the wing, right? $\endgroup$
    – Helmut K.
    Commented Nov 18, 2019 at 20:10
  • 1
    $\begingroup$ Yes, this is a possible situation. For stability analysis, AC is everything, CP is nothing. CP in front of CG s not destabilising by itself; to see what is stabilising and what is not, you need to consider derivatives. A typical wing alone will be unstable, but a long tail will easily 'outmoment' it. $\endgroup$
    – Zeus
    Commented Nov 19, 2019 at 4:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .