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I'm trying to understand stability & control. Please correct me if I'm mistaken.
An aircraft will have longitudinal stability if the aerodynamic center (AC) is behind the center of gravity (CG). The AC is a point, where all the changes in the magnitude of the lift effectively take place.
The aircraft will be trimmed if the sum of the moments equals zero.

Question: is the center of pressure (CP) of the total aircraft (wing's lift and tailplane's aerodynamic force) or the AC, that must be positioned to coincide with the CG in order to trim the aircraft?
Do we change the location of the AC when we deflect the elevator?

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It depends on the aircraft in question, the arrangement of the lifting surfaces, and how they are designed to ensure both static and dynamic stability.

For a simple case, well consider a traditionally designed airplane with a wing set placed approx midway along a fuselage with a horizontal tailplane at the empennage.

The design provides both good static stability and good dynamic stability.

It does so by placing the center of gravity (CG) ahead of the center of lift (CP) for the main wing and causing the tailplane to produce a downwards lifting force to counterbalance the moment caused by the distance between the CG and the CP. While this is not as efficient a lifting arrangement as other designs like a canard foreplane, it forms a naturally stable platform in terms of longitudinal stability. The airplane will remain stable only if these two moments cancel each other out causing the Net moment about the CG to be zero AND if the design naturally returns to a zero moment about the CG when dynamically changing speeds and attitudes.

For trimming, here we will only consider pitch trim for longitudinal stability. An airplane trimmed for cruising at a specific airspeed and a particular altitude will have a net longitudinal moment about the CG of zero. If the airplane increases its airspeed, the tailplane arrangement is designed to increase its downward force, creating a nose up pitch moment. This in turn decreases the airspeed causing the the downward tailplane lift to decrease again, resulting in the nose pitching forward again until an equilibrium of forces and moments is again attained, which will be at the original airspeed for which the plane was trimmed for. To maintain straight and level flight at a higher airspeed the trimming action will decrease the amount of lift that tailplane produces at the new selected airspeed, again resulting in a net longitudinal moment about the CG to zero.

Longitudinal trimming can be accomplished several different ways the first of which is a design feature which can alter the angle of attack of the tailplane in flight to change the lifting force it creates. This is common on jetliners and other large aircraft but not common on light aircraft due to the excess weight of the structure. Another option for smaller aircraft is the installation of a movable servo or antiservo tab on the elevator surfaces. These tabs force the elevators to a new neutral position, again altering the lifting force from the tailplane and creating a trimmed condition.

In regards to the original question the net CP, that is the combined location of both the center of pressure of the wings and the center of pressure of the tailplane will coincide with the CG on the longitudinal axis of a trimmed airplane. If the airplane accelerates, the NET CP will move fwd of the CG, causing the airplane to nose up, without additional trimming to bring it back. Similarly, when the airplane decelerates, the NET CP will move aft of CG, causing the aircraft to nose down without additional trimming.

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  • $\begingroup$ Net CP seems easier to understand when elevator is deflected down, increasing lift in back and dropping the nose. A trim with CG forward of wing CP is elevator up. Would not accelerating from trim bring nose up? Just checking, answer made sense to me except last part. $\endgroup$ – Robert DiGiovanni Sep 28 '18 at 2:23
  • $\begingroup$ the NET CP will move forward of the CG, causing the airplane to nose down Is this sentence correct? Sounds backwards. $\endgroup$ – TomMcW Sep 28 '18 at 2:38
  • $\begingroup$ The problem is fixed $\endgroup$ – Carlo Felicione Sep 28 '18 at 2:48
  • $\begingroup$ No, looks like not fixed, the last paragraph is still contradictory. And anyway, this is not quite a correct explanation (why do you think the net CP would move with speed?) See footnote to my answer. $\endgroup$ – Zeus Oct 1 '18 at 7:03
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I'll offer a simpler and more direct answer: in a trimmed condition, it is a total CP that must coincide with the CG, and this is essentially by definition, and it doesn't "depend on the aircraft in question".

(Here for simplicity we restrict ourselves to the pitch motion and ignore possible effect of thrust and drag, which line may not exactly pass through CG and which moment will then need to be compensated).

In the same condition, AC will be behind CG (and CP) for a statically stable (in pitch) airplane. Imagine the airplane is disturbed and pitches up (or experiences an updraft; the fact is, its AoA temporarily increases). The extra lift due to increased AoA is applied at AC (now by the definition of AC), and since AC is behind CG, it creates a pitch-down moment which returns the aircraft to the original AoA until this extra lift is eliminated and everything returns to the former balance. This is the definition of (static) stability.(*)

From this it follows - and it is important to realise it clearly - that an airplane is trimmed for a certain AoA. Not the airspeed, not the pitch. At a given trim setting (for a steady level flight), you can fly at a higher speed and higher load, for example, in a turn or spiral.

Another thing that may help to avoid confusion is to understand that AC is a very theoretical, abstract point. It is defined purely for the convenience of stability analysis, and defined such that it doesn't move (within reasonable AoA). So in flight you can't "position" it at will, just as well as in most cases you can't move CG much. In a sense, all control is done by shifting the CP (of the whole aircraft).

At the same time, CP and CG can be though of as "real" points where a known real force is applied (although both are also abstractions in reality). When you need a balance, i.e. lack of total moment, you want the lift and gravity to act at the same point. (Remember we neglected moments from other forces, which are often small).


(*) Longitudinal stability is often incorrectly explained through speed: the aircraft pitches up, loses speed, then "wants" to return to the trimmed speed by pitching down to accelerate. This is wrong; the pitch-down moment arises immediately as the AoA grows, much sooner than any appreciable change of speed (if any) happens. When flight dynamicists speak of longitudinal static stability - and that's exactly where the concept of AC appears - they really speak about AoA stability. Airspeed is not even a factor there (or rather, change of airspeed is not). It is this AoA stability that makes the aircraft flyable by humans.

When we disturb only airspeed, as in Carlo's answer, e.g. by increasing thrust, a different process is involved. First (ignoring some fine effects), lift starts to increase quickly (as square of speed). But this increase doesn't come at AC; remember AC is only about AoA! Because we maintain the same AoA (at least initially), you get proportional increase of lift on the wing and tail so that the total balance remains, and the lift increases at CP=CG. As a result, the airplane starts to accelerate upwards (but not 'climb' in a normal sense). Now this means decrease of AoA, and, apart from damping of lift itself, this triggers the normal AoA response, that is, the attempt to increase it back to the trimmed AoA, i.e. pitch up.

Note I didn't mention the tail downforce. It's not really a requirement. It's just an exaggerated way to ensure a rearward AC position. But you don't need to involve it to explain stability if you already defined AC. The downforce is just an implementation detail, as programmers say.

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  • $\begingroup$ By definition net Cp must take into account the downforce of the tailplane, making it a required component of the net Cp. Also, not only does the lift from the main wing increase with the square of the speed for a given AoA, but so does the tailplane force, altering the net moment about the CG, causing the nose to pitch up. $\endgroup$ – Carlo Felicione Oct 1 '18 at 8:02
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    $\begingroup$ @Carlo, no. First, not the downforce, but just lift of the tailplane in general. It can be up or down, even for a statically stable airplane. Second, exactly because both the wing and the tail lift increases proportionally (say, doubles), the net moment doesn't change with airspeed alone. (Consider the classic example with uneven weight balances: if you double the weight on each arm, nothing will change). This is different to changes due to AoA: there lifts change disproportionally due to typically lower incidence of the tail; say, doubles on the wing and triples on the tail, and tail wins. $\endgroup$ – Zeus Oct 1 '18 at 9:25
  • $\begingroup$ @Zeus Do you mean the other way (last sentence)? For example when the AoA increases a bit by disturbance, both lifts increase, but lift on wing must be more dramatic for the whole plane to pitch down a bit to restore the AoA. $\endgroup$ – Hot.PxL Oct 14 '18 at 20:22
  • $\begingroup$ @Hot.PxL Yes in absolute terms the wing lift will likely increase more. But it is moments that matter here: we are talking about angular motion. The wing lift has a much shorter arm to CG (with respect to which we calculate everything). If we had a balance initially, then we can simply look at the proportions by which forces change. And to have a stabilising moment, the tail lift must increase by a greater percentage than the wing's. $\endgroup$ – Zeus Oct 14 '18 at 23:53
  • $\begingroup$ @Zeus Yes, I'm talking about moment (with respect to CG). When AoA increases a bit by disturbance (probably wind coming from below the airplane), the stability requires the plane to pitch down a bit to lower the AoA. For it to lower pitch, it must be the case that moment generated by the wing is greater (causing pitch down) than that of the tail (causing pitch up). Is my understanding correct? $\endgroup$ – Hot.PxL Oct 15 '18 at 2:58

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