I've looked at the other questions concerning this topic, such as:

But I'm still confused about why transport aircraft generally have the center of gravity before the aerodynamic center.

In this resource from Airbus, they write and show:

enter image description here


Another example Airbus sketches the following picture


And writes:

As explained earlier, the more distant the CG and the CP, the bigger the pitch-down moment. Since for aircraft stability reasons the CP is always located behind the CG, a forward CG increases the distance between the CP and the CG. A CG position further forward than the most forward position of the operational envelope can affect the safety of the flight in many ways.

Why is the bold statement true? Couldn't you put the c.o.g. behind the aerodynamic center, and trim the aircraft with an upward lift force on the tail?

I understand that positioning the lift ahead of the c.o.g. is unstable (since the destabilizing moment increases), but can't this be fixed with an appropriate tail design?

  • $\begingroup$ Remember that you want it to be stable at multiple airspeeds, not just one. So the balances between pitching moment between wing and tail, and and the lift forces between wing and tail would need to track each other. Also, apparently, lifting tails are generally less efficient since the main wing is more efficient at producing lift forces and the balance of the numbers normally makes it more efficient to have the main wing produce a little more lift to counteract the downward tail. $\endgroup$
    – DKNguyen
    Commented Apr 6, 2023 at 0:25

7 Answers 7


It is this way so the aircraft will be stable. Think about a dart -- the CG (the heavy brass weight) is far forward, while the aerodynamic center (the fins) are very far aft.

The dart can not trim -- so it is an incomplete example. However, the fins provide great stability.

Get yourself a dart, a foam target board, and a safe place to throw. Then do some experiments... Start by throwing the dart normally. Then throw it oriented 90-degrees to the direction of your toss. Then 180 (backwards). What happens?

Get a friend to do the tossing, but move so you can see the flight. If the friend back up a bit - and throws somewhat slowly, you will be able to see the dynamic response of the dart through the air.

The dart is very stable -- because the aerodynamic center is behind the CG.

  • $\begingroup$ I wish I had more upvotes. So often big words and math formulas get in the way of basic understanding very simple, observable, and demonstratable concepts. This is something you can explain to a 5 year old. (no insult to the OP intended...) $\endgroup$ Commented Aug 10, 2023 at 16:31
  • $\begingroup$ @MichaelHall: albeit being formally correct for a dart, this answer cannot be extend to an airplane since an airplane possesses not only a tail but also a wing and a fuselage that, by an aerodynamic point of view, complicate things quite a bit. For example the dart analogy cannot explain why a canard is stable too, even if "the tail" is "in front": according to this answer the Wright brothers couldn't have flown... I agree with you that "often big words and math formulas get in the way" but sometimes that's the only way to understand/explain things in a correct and general way. $\endgroup$
    – sophit
    Commented Aug 12, 2023 at 19:04
  • $\begingroup$ @sophit, agree completely. In this case the question was well thought out and deserved the details provided in the other excellent answers such as yours. Other times i wonder sometimes if people over answer based on my perception of the asker's base knowledge... $\endgroup$ Commented Aug 12, 2023 at 19:29
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    $\begingroup$ @sophit The question has been modified since I answered it. The question was not about all aircraft or canard aircraft. The question is clearly about transport category aircraft -- all of which have been a wings forward, tail aft design (except the Concord & TU-144). When engineers first look at static stability, we ignore the fuselage aerodynamically -- considering only the lifting surfaces. I am usually quick to go to equations and rigorous theory -- I am certainly not shy from it. However, this question 'Why is the CG in front of the AC' can be concisely answered 'For stability'. $\endgroup$ Commented Aug 12, 2023 at 20:25
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    $\begingroup$ I also ignored the fuselage in my answer 😉 I definitely didn't want to criticise your answer, I just wanted to broaden the comment of @MichaelHall 🖖 $\endgroup$
    – sophit
    Commented Aug 12, 2023 at 21:11

I'm shocked - shocked! - to see this inaccuracy from Airbus. Or am I?

Of course the bold statement is utter nonsense, but is repeated over and over. Which doesn't make it true, but very hard to dispel.

Improvements in efficiency over the years have demonstrated that at least the engineers understand the matter correctly. But I'm afraid they have no time to constantly check on the nonsense their marketing apprentices publish.

Why do most commercial aircraft have the Center of Gravity before the Aerodynamic Center?

Not most, but all of them have their Center of Gravity ahead of the Aerodynamic Center. This is a certification requirement. However, with relaxed static stability, it is quite common for the center of pressure of the wing to be forward of the center of gravity, so that the tail is slightly positively loaded.

Couldn't you put the c.o.g. behind the aerodynamic center, and trim the aircraft with an upward lift force on the tail?

You can put the center of gravity (CG) behind the center of pressure (CP) of the wing. Only in landing configuration, when the CP of the wing is shifted backwards due to flap extension, will downforce on the tail be required. However, for natural static stability you need to have the CG ahead of the aerodynamic center at all times.

I understand that positioning the lift ahead of the c.o.g. is unstable

Actually, both lift and center of gravity should be at the same lengthwise station in order to enable straight flight. Stability is a consequence of the lift vector derivative over angle of attack.

  • $\begingroup$ Aerodynamic Center and Neutral Point are different things? There is a a lot of confusion over that. $\endgroup$
    – John K
    Commented Mar 31, 2023 at 20:42
  • $\begingroup$ @JohnK Where did I say that? No, they should be the same. I thought I only mentioned the center of pressure (which indeed is a different thing). $\endgroup$ Commented Apr 1, 2023 at 6:17
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    $\begingroup$ @JohnK Oops, I mixed up center of pressure (CP) and neutral point (NP). What is ahead of the center of gravity (CG) is the center of pressure of the wing (at least in Airbus airplanes with relaxed static margin). Of course the neutral point of the whole airplane is still aft of the CG. With the large shift in the CP with extended flaps, this changes and now the CP of the wing is also aft of the CG (which changes only a little when flaps are deflected). In steady flight in all phases the CG will coincide with the CP of the whole airplane. $\endgroup$ Commented Apr 1, 2023 at 18:56
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    $\begingroup$ @JohnK I should read the questions more carefully. I only focused on the bold line from the Airbus quote. Now I understand your doubts, and thank you for pointing out my mistake. $\endgroup$ Commented Apr 1, 2023 at 18:59
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    $\begingroup$ @ROIMaison I was referring to the AC of the whole airplane. Of course the tail makes the pitch moment gradient over Alfa negative but also shifts the AC of the whole airplane back behind the center of gravity (which is essentially the same). $\endgroup$ Commented May 1, 2023 at 13:01

Actually, the diagrams are sort of a "tongue in cheek" way of saying we don't fly staticly stable planes because we want to save fuel costs

"your winnings, sir"

Over the life of the aircraft, flying staticly neutral but directionally stable can indeed result in considerable savings.

can it be done safely, even with swept wings?

Here's where it's important to realize that these huge airliners don't scale very well with smaller aircraft when it comes to stall recovery.

That's why it is so important to understand all sources of pitching moments, including large underslung fans that, by design, create a "pitchup" moment in cruise so not as much tail downforce needs to be applied to counteract wing moment.

Unfortunately, these gigantic fans, low and slow, can produce so much pitching moment at TOGA thrust that the tiny tail cannot keep up, especially if the CG is too far back.

So, it's a bit more complicated, but I might not mind as a passenger, as long as the builders of the aircraft aren't fooling themselves.


I have experience flying flying free-flight model airplane hand-launched gliders with the CG typically located at 60% of the chord or even further back. This allows the planes to shoot up 200 feet or more at tremendous speed, then come out into a slow circling glide, all with no moveable control surfaces. It also causes the planes to crash a lot, not recover well from disturbances and be extremely sensitive to small warps. So it is possible to have just barely enough longitudinal stability with an aft CG and lifting tail, but at a loss of the kind of safety that human-carrying craft require. The more aft the CG, the more problems. Free-flight contest models carry this to a ridiculous degree. Traditional contest free-flight models are aerodynamically optimized for the greatest possible speed difference between the climb and the glide. The fact that this leads to a lot of crashing has caused modern designs to incorporate control surfaces automatically actuated at the transition to glide. Contests are won by rapidly getting high enough to ride thermals at the same speed a hawk does.


The reason for placing the CoG in front of the neutral point is the behaviour when the Angle of Attack increases: for passive stability, we want gravity to provide a nose-down moment: $dC_L/d\alpha$ must be < 0, as explained in this answer

An intuitive case occurs when the wing enters a stall: the lift vector decreases, and if the CoG is behind the CoL, the reduction of the lift vector causes a nose-up moment. This further stalls the aeroplane, which is not what we want.

  • $\begingroup$ But, arguably, gravity does not create a moment at all, right? I mean, sure, in the other answer you linked to it does, but that's due to an arbitrary choice of station zero way out in front of everything else, right? To purely work a torque balance problem, we are free to pick any point we want to resolve the torques around, but if there is any acceleration going on, then we must pick the CG to resolve everything around, unless we want to complicate the picture by introducing fictitious inertial forces ("centrifugal force"). Doesn't that pretty much sum it up? $\endgroup$ Commented Apr 2, 2023 at 3:39
  • $\begingroup$ Are you saying that you want gravity to create a nose-down moment about the Aerodynamic Center, perhaps for reasons alluded to in this answer to a related question aviation.stackexchange.com/a/50380/34686? And if there is any acceleration tangent to the flight path, shouldn't you really say that you want "gravity + centrifugal force" to create a nose-down moment about the Aerodynamic Center?" (I think... actually still thinking that last one over...) $\endgroup$ Commented Apr 2, 2023 at 3:49
  • $\begingroup$ @quietflyer the aeroplane was trimmed before losing lift. After the stall there is a change in moment and in vertical force due to the loss of lift, therefore it descends and rotates, and we want the rotation to be nose down. Indeed pick any point for the moment equilibrium - the tailplane is a convenient one for this case. $\endgroup$
    – Koyovis
    Commented Apr 2, 2023 at 7:26

An airplane (or any other mechanical system) is stable when, after something has perturbed its attitude, it tends to go automatically back to its initial condition.

In particular, an airplane is longitudinally stable if an increase in the AoA (the perturbation) will create a nose-down pitching moment which makes the AoA decrease, thus restoring the initial condition (picture source):

 Longitudinal stability depiction

What are the sources of pitching moment for an airplane? To simplify a bit the discussion we limit the analysis to a conventional airplane (i.e. an airplane with a well defined wing, fuselage and tailplane) and we omit, as is customary, the contributions coming from drag, engine(s) and fuselage. We have then:

  • lift $L$ of the wing; lift acts in the aerodynamic center of the wing which lies at a distance $x_{ac}$ from the nose; the aerodynamic center is at 25% of the chord at subsonic speeds and 50% at supersonic speeds;
  • aerodynamic pitching moment $M_{wing}$ of the wing; it is normally negative and more or less constant with the AoA;
  • lift $L_{tail}$ of the horizontal tailplane, which acts at a distance $x_{ac_{tail}}$ from the nose.

The following figure (source, modified by me) shows schematically these loads:

A380 longitudinal stability

Now we only need a point in respect to which to calculate the total pitching moment $M$: we use the CG of the airplane since it gives some useful results. Then summing up all these terms in respect to the CG we get:


If we equal this equation to zero, we get the conditions needed to equilibrate i.e. trim our airplane. Anyway to get its stability we need to go a step further and apply the previous definition: the airplane is stable if any perturbation that changes the AoA creates a pitching moment $M$ that brings the AoA back to its initial trimmed value. This means that the aircraft is stable if a positive change of AoA creates a negative change of moment $M$, i.e. if their ratio is negative:


where $∆$ means "variation". Applying this definition to the previous expression we get:


which is the "stability equation" of the airplane. A couple of comments before going further:

  • as usual in the aerospace world, moments and forces have been adimensionalised via $½\rho V²S$ and written in coefficient form;
  • $M_{wing}$ disappeared; as said, wing's pitching moment is basically constant in respect to AoA and therefore its variation is simply zero;
  • the very last term just translate the fact that the aerodynamic characteristics of the horizontal tailplane must obviously be given in respect to its $(½V_{tail}²S_{tail}∆\alpha_{tail})$ while the adimensionalisation has been done in respect to the wing's $(½V²S∆\alpha)$; worthy of remark is the fact that both the speed $V_{tail}²$ and the AoA $\alpha_{tail}$ seen by the horizontal tailplane are different than the ones of the wing; this is due to the downwash of the wing impinging on the tailplane which reduces both.

Now we only have to set $\frac{∆C_M}{∆\alpha}<0$ to get the aeromechanical settings needed to have a stable airplane. Since we have the sum of two terms, in order for this equation to be negative at least one of these terms must be negative and more negative than the other.

We start from the tailplane:

  • $\frac{∆C_{L_{tail}}}{∆\alpha} \rightarrow$ this is the slope of the lift curve which is always positive (bigger $\alpha$, bigger $C_l$);
  • $\frac{V_{tail}²S_{tail}∆\alpha_{tail}}{V²S∆\alpha}\rightarrow$ this is the ratio of geometric entities and is also positive;
  • $(x_{cg}-x_{ac_{tail}})\rightarrow$ if the horizontal stabiliser is located on the tail, then this term is negative; bingo! The horizontal tailplane gives a negative contribution to stability and therefore it is always stabilising.

Now the wing:

  • $\frac{∆C_L}{∆\alpha}\rightarrow$ the slope of the lift curve is positive;
  • $(x_{cg}-x_{ac})\rightarrow$ this term is positive or negative according to the relative position of CG and AC; if the CG is in front of the AC then this term is negative (i.e. stabilising) and viceversa.

What do we choose now? CG first or AC first?

But I'm still confused about why transport aircraft generally have the center of gravity before the aerodynamic center.

If we want to be conservative then we put the CG in front of the AC: should the tailplane lose some of its effectiveness, the airplane will remain stable. Plus, if the CG helps in stabilising the airplane then a smaller tailplane can be used giving less drag and weight.

Couldn't you put the c.o.g. behind the aerodynamic center, and trim the aircraft with an upward lift force on the tail?

Yes definitely, we can we put the CG between the wing and the tailplane but then $(x_{cg}-x_{ac})$ becomes positive (i.e.destabilising) and the tailplane needs to be bigger to compensate also for this instability due to the CG. Anyway this configuration with the CG between wing and tailplane entails that the horizontal tailplane produces a positive lift relieving a bit the job of the wing.

  • $\begingroup$ Don't forget those little ventrals in the back. They help too. $\endgroup$ Commented Apr 4, 2023 at 19:12
  • $\begingroup$ @RobertDiGiovanni: only talking about longitudinal stability here. $\endgroup$
    – sophit
    Commented Apr 4, 2023 at 19:21
  • $\begingroup$ Though not quite as good as endplating a horizontal stabilizer, as AoA increases, you may find a pocket of higher pressure there. $\endgroup$ Commented Apr 4, 2023 at 19:49

It has nothing to do with stability in normal forward flight, and a slightly rear CG with up-lift on the tail will work just fine and be rather efficient.

The problem is recovery from upset like a stall or spin. With a rear CG the aircraft will tend to stay tail low and cannot reduce the AOA enough to recover the forward airpeed needed for control authority in any axis. And so you fall, fall down to the ground.

look at the top down planform of the aircraft and imagine the distribution of vertical drag on the surfaces of an airplane that you just drop at zero forward airspeed.(an absolute stall)

CG forward and the drag on the tail surface causes the nose to drop and the plane develops some relative forward airspeed. That is forward airspeed relative to the AOA, not necessarily horizontal over to the ground. With an aft CG the tail drops first or equally and the plane can never develop airflow in the direction needed to effect pitch or yaw.(related to a flat spin)

The way around this CG efficiency-safety tradeoff is with a canard planform. Though the canard form has some other engineering tradeoffs. (Canard is not a term for a forward horizontal stabilizer. Canard is French for duck, because a mallard duck in flight similarly has a planform with very little tail aft of the main wings and a long neck out front.)


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