Longitudinal stability and control In this chapter, chapter, we will start to investi investigate gate the stability stability of the entire entire aircraft. aircraft. This can b e split split up into two parts: longitudinal and lateral stability. In this chapter, we will only look at longitudinal stability.
1 1.1 1.1
Stick Stick fixed fixed long longitu itudin dinal al stabil stabilit ity y Effec Effects ts of the wing wing and the the tail tail on stabil stabilit ity y
To start our investigation in the stability of an aircraft, we reexamine the moment equation. In an earlier chapter, we found that xcg − xw C m = C mac + C N − C N N wα (α − α0 ) Nh c¯
V h V
2
S h lh = 0, 0, S c¯
(1.1)
where C where C N Nh is given by C N Nh = C N N hα
dε (α − α0 ) 1 − dα
+ (α (α0 + ih ) + C N Nh δ δ e .
e
(1.2)
We can also rewrite the moment equation to C m = C = C mw + C mh . In this equation, C equation, C mw is the contribution due to the wings. Similarly, C mh is the contribution from the horizontal tailplane. They are both given by 2 x cg − xw V h S h lh C mw = C mac + C N (α ( α − α ) and C = − C . (1.3) N wα mh N N h 0 c¯ V S c¯
Taking a derivative of the moment equation will give us C mα = C mαw + C mαh , where C mαw
xcg − xw = C N N wα c¯
and
C mαh = −C N Nh α
dε 1− dα
V h V
2
S h lh . S c¯
(1.4)
To achieve stability for our aircraft, we should have C mα < 0. Usually Usually,, the wing is in front of the CG. We thus have xcg − xw > 0 and also C mαw > 0. The wing thus destabilize destabilizess the aircraft aircraft.. Luckily Luckily,, the horizontal tailplane has a stabilizing effect. This is because C mαh < 0. To achieve stability, stability, the stabilizing effect effect of the tailplan tailplanee should should be bigger bigger than the destabilizing destabilizing effect of the wings. We should should thus thus have have |C mαw | < |C mαh |.
1.2
(1.5)
Effects Effects of the the center center of grav gravit ity y on stabili stability ty
We will now examine examine the effects effects of the CG on the stabilit stability y. To do this, we suppose xcg increases (the CG moves to the rear). Howeve However, r, the other parameters parameters (including (including δ e ) stay constant. constant. The movemen movementt of the CG causes C causes C mα to increase. At a certain point, we will reach C mα = 0. When the CG moves beyond this position, the aircraft becomes unstable. Let’s examine the point at which C mα = 0. We remember, remember, from a previous chapter, chapter, that this point is called the neutral point. And, because the stick deflection is constant ( δ e is constant), we call this point the stick the stick fixed neutral point. point. Its x Its x coordinate is denoted by x nfix . To find it, we can use C mα
xnfix − xw = C N + C N Nw α Nh α c¯
1
dε 1− dα
V h V
2
S h xnfix − xh . S c¯
(1.6)
After some mathematical trickery, we can find the position of the stick fixed neutral point, with respect to the wing. It is given by xnfix − xw C N hα = c¯ C N α
dε 1− dα
V h V
2
S h lh . S c¯
(1.7)
From this, we can also derive the position of the stick fixed neutral point, with respect to the aircraft CG. This is given by xcg − xnfix C mα = C N α . (1.8) c¯ xcg −xn
fix The quantity is known as the (stick fixed) stability margin. It is an indication of how much c ¯ the CG can move, before the aircraft becomes unstable.
1.3
The elevator trim curve
Now let’s examine the effects of the elvator deflection δ e . We know from a previous chapter that the elevator deflection necessary to keep the aircraft in equilibrium is δ e = −
1 (C m + C mα (α − α0 )) . C mδe
0
(1.9)
δ e depends on α. To see how, we plot δ e versus α. We usually do this, such that the y axis is reversed. (Positive δ e appear below the horizontal axis.) Now we examine the slope of this graph. It is given by dδ e C mα =− . dα C mδe
(1.10)
We always have C mδe < 0. To make sure we have C mα < 0 as well, we should have dδ e /dα < 0. The line in the δ e , α graph should thus go upward as α increases. (Remember that we have reversed the y axis of the graph!) δ e also depends on the aircraft velocity V . To see how, we will rewrite equation (1.9). By using C N ≈ W C N α (α − α0 ) ≈ ρV , we find that S 1 2
2
1 δ e = − C mδe
C mα W C m + . C N α 12 ρV 2 S 0
(1.11)
We can now plot δ e against V . (Again, we reverse the δ e axis.) We have then created the so-called elevator trim curve. Its slope is given by dδ e 4W 1 C mα = . dV ρV 3 S C mδe C Nα
(1.12)
To have C mα < 0, we should have dδ e /dV > 0. The line in the graph should thus go downward. Also, if you want to fly faster in a stable aircraft, you should push your stick forward.
2 2.1
Stick free longitudinal stability The stick free elevator deflection
Previously, we have assumed that δ e is constant. The pilot has his stick fixed. But what will happen if the pilot releases his stick? It would be nice if the aircraft remains stable as well. 2
Let’s suppose the pilot releases the stick. In that case, aerodynamic force will give the elevator a certain stick free elevator deflection δ efree . To find δ efree , we examine the moments H e about the elevator hinge point. (Or, to be more precise, we look at the non-dimensional version C he .) Contributing to this hinge moment are the horizontal tailplane, the elevator and the trim tab. By using a linearization, we find that C hefree = C hα αh + C hδ δ efree + C hδt δ te = 0. (2.1) It follows that the stick free elevator deflection is δ efree = −
C hδt C hα αh − δ t . C hδ C hδ e
(2.2)
From this, we can also derive that
dδ e dα
free
C h =− α C hδ
dε 1− dα
.
(2.3)
The elevator deflection thus changes as the angle of attack is changed.
2.2
Differences in the moment due to the stick free evelator
The free elevator deflection effects the contribution C mh of the horizontal tailplane to the moment C m . Let’s investigate this. We can remember that
C mh = − C N hα αh + C N hδ δ e
V h V
2
S h lh . S c¯
(2.4)
We now substitute δ e by δ efree . If we also differentiate with respect to α, and work things out, we will get 2 C hα dε V h S h lh C mαh = − C N hα − C N hδ 1− . (2.5) free C hδ dα V S c¯
If we compare this equation to the right side of equation (1.4), we see that only C Nh α has changed. In fact, we can define C h C Nh α = C N hα − C Nh δ α . (2.6) free C hδ If we use C Nh α
free
, instead of C Nh α , then our stability analysis is still entirely valid.
Let’s take a closer look at the differences between C N hα
free
and C Nh α . This difference is the term
C
C N hδ C hhα . We know that C N hδ > 0. The term C hδ is interesting. If it would be positive, then it can be δ shown that the elevator position is unstable. So, we have to have C hδ < 0. Finally there is C hα . This term can be either positive or negative. If it is positive ( C hα > 0), then the stick free aircraft will be more stable than the stick fixed aircraft. If, however, it is negative ( C hα < 0), then it will be less stable, or possibly even unstable.
2.3
The stick free neutral point
Let’s find the stick free neutral point x nfree . Finding x nfree goes similar to finding x nfix . In fact, we can adjust equations (1.7) and (1.8) to C N hα xnfree − xw free = c¯ C Nα
3
dε 1− dα
V h V
2
S h lh , S c¯
(2.7)
xcg − xnfree . (2.8) c¯ In this equation, we have C Nα free ≈ C N α . This is because the elevator has a negligible influence on C N α , compared to the influence of the wing. C mαfree = C Nα free
We can also find the position of the stick free neutral point, with respect to the stick fixed neutral point. Subtracting equation (1.7) from equation (2.7) gives C N hδ C hα xnfree − xnfix =− c¯ C N α C hδ
2.4
dε 1− dα
V h V
2
S h lh C mδ C hα = S c¯ C N α C hδ
dε 1− dα
.
(2.9)
Elevator stick forces
Now we will examine the stick forces which the pilot should exert. We denote the stick deflection by se . By considering the work done by the pilot, we find that F e dse + H e dδ e = 0. From this follows that the stick force F e is given by F e = −
dδ e dδ e 1 H e = − C he ρV h2 S e c¯e . dse dse 2
(2.10)
By the way S e is the elevator surface and c¯e is the mean elevator chord. If we massively rewrite the above equation, we can eventually find that dδ e F e = − S e c¯e dse
2
V h V
1 W 1 C h ρV 2 + C hα 2 S C Nα
0
.
(2.11)
We see that F e consists of two parts. One part varies with the airspeed, while the other part does not. By the way, the coefficients C h and C hα are given by
0
C hδ C hδ C mac − C N hα (α0 + ih ) + C hδt δ te , free C mδ C N hδ xcg − xnfree C hδ C h =− C mαfree = − δ C N α . C mδ C mδ c¯
C h = − 0
C hα
(2.12)
(2.13)
We see that C h depends on δ te . To simplify our equation, we can apply a small trick. We define δ te to be the value of δ te for which C h = 0. It follows that
0
0
0
δ te
1 = C hδt
0
C hδ C hδ C mac + C N hα (α0 + ih ) . free C mδ C N hδ
(2.14)
We can now rewrite the stick deflection force as dδ e F e = S e c¯e dse
2
V h V
W C hδ xcg − xnfree 1 2 − ρV C hδt δ te − δ te S C mδe c¯ 2
0
.
(2.15)
The control forces, which the pilots need to exert, greatly determine how easy and comfortable it is to fly an airplane. The above equation is therefore rather important. We can also derive something else from the above equation. Let’s define the trim speed V tr to be the speed at which F e = 0. We now examine the derivative dF e /dV at this trim speed. (So at F e = 0.) If it is positive (dF e /dV > 0), then the aircraft is said to have elevator control force stability in the current flight condition. It can be shown that this derivative is given by
dF e dV
F e =0
dδ e = −2 S e c¯e dse
V h V tr
2
W C hδ xcg − xnfree 1 . S C mδe c¯ V tr
It’s the job of the designer to keep this derivative positive. 4
(2.16)
3 3.1
Longitudinal control Special manoeuvres
Previously, we have only considered steady flight. Now we suppose that we are performing some special manoeuvre. We will consider both a steady pull-up manoeuvre and a horizontal steady turn. During these manoeuvres, we will have a certain load factor n = N/W . There are two parameters that are important for the manoeuvres. They are the elevator deflection per g, denoted by dδ e /dn, and the stick force per g, denoted by dF e /dn. Both these parameters should be negative. And they may not be too high or too low either.
3.2
The elevator deflection per
g
We will now find an expression for dδ e /dn. Let’s suppose we’re initially in a horizontal steady flight. But after a brief moment, we’ll be in one of the special manoeuvres. In this brief moment, several aircraft parameters have changed. Let’s examine the change in normal force ∆C N and the change in moment ∆ C m . The change in normal force is effected by the angle of attack α and the pitch rate q . This gives us ∆C N =
1 2
∆N = ρV 2 S
1 2
W q c¯ ∆n = C ∆α − C ∆ . N Z α q V ρV 2 S
(3.1)
Similarly, the change in moment is effected by the angle of attack α, the pitch rate q and the elevator deflection δ e . This gives us q c¯ ∆C m = 0 = C mα ∆α + C mq ∆ + C mδe ∆δ e . (3.2) V You may wonder, why is ∆C m = 0? This is because both in the initial situation and the final situation, we have a steady manoeuvre. There is thus no angular acceleration present. The moment must thus stay constant. From the first of the above two equations, we can find the derivative of α with respect to n. It is given by C Zq d qV c¯ dα 1 W = + . (3.3) dn C Nα 12 ρV 2 S C Nα dn From the second of these equations, we can find that dδ e 1 =− dn C mδe
d qV c¯ dα C mα + C mq dn dn
.
(3.4)
Inserting the value of dα/dn will eventually give us dδ e 1 =− dn C mδe
C mα W + C Nα 12 ρV 2 S
C mα C Zq d qV c¯ + C mq C N α dn
.
(3.5)
We will determine the term d qV c¯ /dn later, since it depends on the type of manoeuvre that is being performed.
3.3
The stick force per
g
It’s time to find an expression for dF e /dn. From equation (2.10), we can derive that
dF e dδ e 1 2 dαh dδ e =− ρV h S e c¯e C hα + C hδ dn dse 2 dn dn 5
.
(3.6)
We already have an expression for dδ e /dn. The expression for α h is a bit tricky. This is because we also have a rotation q . If we take this into account, we will have
dε αh = (α − α0 ) 1 − dα
+ (α0 + ih ) +
l h q c¯ . c¯ V
(3.7)
The derivative of α h , with respect to n, will then be dαh = dn
qc ¯
dε dα l h d V 1− + . dα dn c¯ dn
(3.8)
Luckily, we still remember dα/dn from equation (3.3). From this, we can derive an equation that’s way too long to write down here. However, once we examine specific manoeuvres, we will mention the final equation.
3.4
The pull-up manoeuvre
Let’s consider an aircraft in a pull-up manoeuvre. When an aircraft pulls its nose up, the pilot will experience higher g-forces. This will thus cause the load factor n to change. To be able to study pull-up manoeuvres, we simplify them. We assume that both n and V are constant. If this is the case, the aircraft’s path will form a part of a circle. The centripetal accelaration thus is N − W = mV q . By using n = N/W and W = mg, we can rewrite this as q c¯ g¯ c = 2 (n − 1). V V
(3.9)
Differentiating with respect to n gives d qV c¯ g¯ c 1 W = 2 = , 1 dn V 2µc 2 ρV 2 S
where
µc =
m W = . ρS c¯ gρS ¯ c
(3.10)
By using this, we can find the elevator deflection per g for a pull-up manoeuvre. It is
dδ e 1 =− dn C mδe
W C mα 1 2 ρV S C N α 2
C Zq 1+ 2µc
C mq + 2µc
.
(3.11)
Often the term C Zq /2µc can be neglected. This simplifies matters a bit. We can also derive a new expression for the stick force per g. We will find that dF e dδ e W = dn dse S
V h V
2
C hδ S e c¯e C mδe
C mαfree C Nα
+
C mqfree 2µc
.
(3.12)
In this equation, we can see the parameters C mαfree and C mqfree . These are the values of C mα and C mq when the pilot releases his stick. They are given by C mαfree = C Nw α (The relation for C mαh
xcg − xw + C mαh free c¯
free
3.5
and
C mqfree = C mq − C mδe
C hα lh . C N α c¯
(3.13)
was already given in equation (2.5).)
The steady horizontal turn
Now let’s consider an aircraft in a steady horizontal turn. It is performing this turn with a constant roll angle ϕ. From this, we can derive that N cos ϕ = W
and
N − W cos ϕ = mV q. 6
(3.14)
If we combine the above relations, and rewrite them, we will get q c¯ g¯ c = 2 V V
n−
1 n
.
(3.15)
Differentiating with respect to n will then give us d qV c¯ 1 W 1 = 1+ 2 1 2 dn 2µc 2 ρV S n
.
(3.16)
By using this, we can find the elevator deflection per g for a horizontal steady turn. It is dδ e 1 =− dn C mδe
W C mα + 1 2 ρV S C Nα 2
C mα C Zq C mq + 2µc C Nα 2µc
1 1+ 2 n
.
(3.17)
Again, we may often assume that C Zq /2µc ≈ 0. This again simplifies the equation. We also have the stick force per g. In this case, it is given by dF e dδ e W = dn dse S
V h V
2
C hδ S e c¯e C mδe
C mαfree C N α
+
C mqfree 2µc
1 1+ 2 n
.
(3.18)
It is interesting to see the similarities between the pull-up manoeuvre and the steady horizontal turn. In fact, if the load factor n becomes big, the difference between the two manoeuvres disappears.
3.6
The manoeuvre point
An important point on the aircraft, when performing manoeuvres, is the manoeuvre point. It is defined as the position of the CG for which dδ e /dn = 0. First we will examine the stick fixed manoeuvre point x mfix . To have dδ e /dn = 0 for a pull-up manoeuvre (neglecting C Zq /2µc ), we should have xcg − xnfix C mq C mα C mq + = + = 0. C Nα 2µc c¯ 2µc
(3.19)
If the above equation holds, then the CG equals the manoeuvre point. We thus have xmfix − xnfix C mq =− c¯ 2µc
xcg − xmfix C mα C mq = + . c¯ C Nα 2µc
and also
(3.20)
(Remember that the above equations are for the pull-up manoeuvre. For the steady turn, we need to multiply the term with C mq by an additional factor 1 + 1/n2 .) By using the above results, we can eventually obtain that dδ e 1 W xcg − xmfix =− . (3.21) 1 dn C mδe 2 ρV 2 S c¯
By the way, this last equation is valid for both the pull-up manoeuvre and the steady horizontal turn. We can also find the stick free manoeuvre point xmfree . This goes, in fact, in a rather similar way. We will thus also find, for the pull-up manoeuvre, that C mqfree xmfree − xnfree =− c¯ 2µc
C mαfree C mqfree xcg − xmfree = + . c¯ C N α 2µc
and
(For the steady turn, we again need to multiply the term with C mqfree by 1 + 1/n2 .)
7
(3.22)
