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):
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:
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:
$M=L(x_{cg}-x_{ac})+M_{wing}+L_{tail}(x_{cg}-x_{ac_{tail}})$
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:
$\frac{∆M}{∆\alpha}<0$
where $∆$ means "variation". Applying this definition to the previous expression we get:
$\frac{∆C_M}{∆\alpha}=\frac{∆C_L}{∆\alpha}(x_{cg}-x_{ac})+\frac{∆C_{L_{tail}}}{∆\alpha}(x_{cg}-x_{ac_{tail}})\frac{V_{tail}²S_{tail}∆\alpha_{tail}}{V²S∆\alpha}$
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.
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