A L O N E LY I M P U L S E O F D E L I G H T

D R O V E T O T H I S T U M U LT I N T H E C L O U D S ;

W I L L I A M B U T L E R Y E AT S , A N I R I S H A I R M A N F O R E S E E S H I S D E AT H

L’ A V I O N A V A I T G A G N É D ’ U N S E U L C O U P, À L A S E C O N D E M Ê M E O Ù

I L É M E R G E A I T, U N C A L M E Q U I S E M B L A I T E X T R A O R D I N A I R E . PA S

U N E H O U L E N E L’ I N C L I N A I T. C O M M E U N E B A R Q U E Q U I PA S S E L A

D I G U E , I L E N T R A I T DA N S L E S E AU X R É S E RV É E S. L A T E M P Ê T E ,

AU - D E S S U S D E L U I , F O R M A I T U N AU T R E M O N D E D E T RO I S M I L L E

M È T R E S D ’ É PA I S S E U R , PA R C O U R U D E R A F A L E S , D E T R O M B E S D ’ E A U ,

D ’ É C L A I R S , M A I S E L L E T O U R N A I T V E R S L E S A S T R E S U N E FA C E D E

C R I S TA L E T D E N E I G E .

A N T O I N E D E S A I N T- E X U P É R Y, V O L D E N U I T

T H E R E W I L L B E T I M E TO AU D I T

T H E A C C O U N T S L AT E R , T H E R E W I L L B E S U N L I G H T L AT E R

A N D T H E E Q U AT I O N W I L L C O M E O U T AT L A S T.

L O U I S M AC N E I C E , AU T U M N J O U R N A L
MICHAEL CARLEY

A I R C R A F T S TA B I L I T Y
AND CONTROL
This is not a textbook

This is not a textbook and should not be read as one. It is a set of

notes for a third year unit at the University of Bath, introducing
aircraft stability and control to aerospace engineering students.
The aim is to develop an understanding of concepts, but only if the
notes are read in conjunction with other material, and combined
with attendance at lectures. These notes will not be much use on
their own. You will have to work hard on ideas which will not be
obvious, and were not obvious to the smart people who developed
them. You will often feel stupid and confused, and you will wonder
why you are doing this. You are doing this because it is worth
it: you are taking on a difficult topic which has confused bright
people for over a century, but in which it is possible to make a
contribution.
Feeling stupid means you are working on something worth the
trouble: if you want to feel clever, read the Daily Mail.

Produced using the tufte-book class for LATEX on 8th January 2020.
 Contents

1 How aeroplanes fly and how pilots fly them 1

2 Longitudinal control and static stability 15

3 How to design a tailplane 23

4 Flight testing and aircraft handling 27

5 Piloting: stick forces 31

6 Manoeuvre 37

7 Aircraft configurations and control 47

8 High-speed flight: compressibility effects 53

9 Dynamic behaviour of aircraft 57

10 How aircraft wobble: normal modes 63

11 Flying aeroplanes 71

12 Questions 77
 ii CONTENTS

13 Exam questions 83

A Finding out more 95

1
How aeroplanes fly and how pilots fly them

Recently several papers have been published dealing with airplane

stability and control problems from the pilot’s point of view. Al-
though the airplane [sic] designer should not be expected to have any
views on a subject so completely in the pilot’s domain, he [sic] does
have the responsibility of translating the pilot’s requirements into
concrete airplane proportions.1 1
Otto C. Koppen. Airplane stability
and control from a designer’s point of
view. Journal of the Aeronautical Sciences,
The design of aircraft is a problem of translating mechanical
7(4):135–140, 1940
and system requirements into a physical form which can be used
by a pilot. While the operational parameters can usually be easily
stated—speed, payload, range, say—there is still a need to design
in such a way that a suitably-trained human can control the air-
craft. This requires that the aircraft present itself to the pilot as
standard controls, to which the aircraft will respond in a stand-
ard and predictable manner. This is true whether we are talking
about high performance aircraft or the most basic of microlights:
the fundamental task of the designer is to make the aircraft behave
“properly” from the point of view of the pilot. In an extreme case,
an aircraft may be completely uncontrollable by a human pilot; in
the worst case, it may be controllable over almost, but not quite,
all of the design flight regime. In any case, our role as designers is
to analyze the effect of aircraft configuration on aircraft response
to control inputs and to perturbations in flight, such as gusts, and
then to understand how to translate the pilot’s needs into a flyable
configuration.
To start, we consider the problem of longitudinal control which
is the question of how to maintain, or change, an aeroplane’s in-
cidence, or angle of attack. You should know that lift on a wing or
other body is controlled by the incidence, the angle between a ref-
erence line on the body and the relative velocity of the flow.2 The 2
If you believe that wings generate
most basic tasks a pilot must perform in an aircraft translate into lift because of the “Bernoulli effect”
and air speeding up to keep step with
control about the pitching axis (Figure 1.5 shows the definition of itself, you should not be taking this
the aircraft axes): steady level flight, change of incidence to change unit.
speed and/or height, recovery from stall. Most of the time, most pi-
lots want to maintain steady level flight with a minimum of effort,
to leave mental capacity available for other tasks such as navigation,
observation, or a cup of tea. For a small part of their time, pilots
want to change the state of the aircraft. Then, the pilot wants an
2 How aeroplanes fly and how pilots fly them

aircraft to respond predictably to a given input. The first of these

requirements is for stability; the second is for control.

1.1 Equilibrium and stability

The study of stability and control can be viewed as the problem of

setting and maintaining equilibrium. In steady level flight or steady
climb, for example, the net force and moment on an aircraft are
zero and the aeroplane advances in unaccelerated motion.
First, we define equilibrium: a body is in equilibrium when the
net force and net moment acting on it are both identically zero. An
aircraft which is in equilibrium is said to be in trim, or trimmed.
Stability relates to the tendency of a system to return to equi-
librium if it is disturbed in some way. Static stability refers to the
instantaneous response of a system when perturbed: a statically
stable system will initially move back towards its equilibrium state.
A dynamically stable system will eventually recover its equilibrium, Statically unstable

Response
though not necessarily immediately. Figure 1.1 illustrates the two Neutral stability
cases and also that of neutral stability, where the system remains in Time
Statically stable,
the state to which it has been perturbed. dynamically stable
Statically stable,
dynamically unstable
L
Figure 1.1: Equilibrium and static and
α dynamic stability
Zer
o li
θ ft li
ne
T Horizontal
D

Figure 1.2: Notation for longitudinal

W stability: the dashed line indicates the
flight direction

Figure 1.2 shows the notation for the analysis of aircraft stability.
The two angles shown are the incidence α and the inclination θ.
The second of these is the angle between a reference line on the
aircraft and the horizontal and in practice is of little interest to us in
analyzing stability and control, though it is important to a pilot, to
whom it is known as “attitude”. The incidence, on the other hand,
is of great interest and is the angle between the reference line and
the direction of flight. As a reference, we take the zero lift line (ZLL)
which is the angle of attack at which the lift is zero. This choice
makes future analysis a little more compact, because then CL = aα,
but be careful in consulting other work since the reference system
might be different.
Resolving forces and moments from Figure 1.2,

T − D − W sin θ = 0; (1.1a)
L − W cos θ = 0; (1.1b)
Mcg = 0, (1.1c)
1.2. AERODYNAMICS OF WINGS AND CONTROLS 3

where c.g. refers to “centre of gravity” and coordinates are taken

in a frame of reference attached to the aircraft. By taking moments
about the centre of gravity, we remove the effects of the mass dis-
tribution of the aircraft and (1.1c) is a statement about the balance
of aerodynamic moments only. If we are to relate scale-test data to
full-size aircraft, this is a very useful thing. A pilot brings an air-
craft into, or out of, trim by modifying the aerodynamic moments
through use of the control surfaces; our analysis lets us deal with
the things a pilot changes without worrying about details of the
mass distribution.
Having found an equilibrium, we would like to know if it is
stable. In aeronautical terms, this can be stated as the requirement
that when an aircraft is pitched nose-up (nose-down) by a perturb-
ation, the change in moment must be such as to pitch it nose-down
(nose-up). In other words, ∂Mcg /∂α < 0: the change in moment is
in the opposite direction to the change in incidence.

∂C M /∂α > 0
C Mcg

Neutrally stable
cannot trim

CL < 0
α

∂C M /∂α < 0

∂C M /∂α < 0
Figure 1.3: Trim and stability beha-
viour

Figure 1.3 shows some possibilities for equilibrium and stability

in terms of moment and lift coefficients C Mcg and CL .

1.2 Aerodynamics of wings and controls

We cannot study equilibrium and stability in the abstract: at some

point we must soil our hands with reality and think about actually-
existing aircraft and how they behave. In designing any moderately
complex system, we usually reduce the elements of the system to a
small number of parameters in order to keep the problem tractable.
In this case, we do not look at the details of how a wing works, or
how the pressure distribution changes when a control is deflected,
but only at the overall effect on forces and moments.
As always in aerodynamics we deal in non-dimensional quant-
ities, normalized on air density ρ, velocity V, wing planform area
S and, where necessary, wing mean chord c, or root chord c0 for
tailless aircraft. The ‘mean chord’ is usually the ‘mean aerodynamic
chord’, or m.a.c. This is a way of representing the wing which gives
the same force and moment on the aircraft as the real wing. Fig-
ure 1.4 shows some typical planforms and their m.a.c.s. Note that

Figure 1.4: Wing planforms (rectangu-

lar, delta, bi-elliptical, and swept and
tapered) with their mean aerodynamic
chords and neutral points
4 How aeroplanes fly and how pilots fly them

the length and position of the mean chord are both important, since
they are used in computing moments as well as forces. Coefficients
of lift, drag, and moment are thus given by:

L D M
CL = , CD = , CM = . (1.2)
ρV 2 S/2 ρV 2 S/2 ρV 2 Sc/2

As noted above, CL = aα, where a is the lift curve slope for the
wing or other body. As well as stating where our reference angle
lies for incidence, this also says that we deal in linear aerodynamics.
We do this for two reasons. The first is that, up to stall, aerody-
namic behaviour is linear: lift is proportional to incidence. Should
an aircraft reach stall, the linearity of the lift curve is not our main
concern. Secondly, we design aircraft to be linear, to make them
flyable. A pilot wants a linear response to control inputs: a given
change in stick force should always give the same change in aircraft
response.

Figure 1.5: Axes and sign conventions

for control deflections

Once an aircraft has been built, the aerodynamic properties

are fixed, essentially by the choice of wing section and planform,
though other effects will need to be considered. This means that the
lift curve slope of the wing is constant 3 except in the case of control 3
If flaps or other devices are deployed,
surfaces—elevator, rudder, and aileron—which can be deflected to the lift curve slope changes, but from
one fixed value to another.
change the aerofoil section and thus its properties. You can think of
this deflection as a change in section camber, with a corresponding
change in lift curve slope.
Figure 1.5 shows these primary control surfaces on a conven-
tional aircraft, with the corresponding sign conventions. We take
a deflection as positive if it generates a positive increment in force.
Ailerons work differentially so the deflection is that of both sur-
η
faces, with a positive deflection being that which generates a posit-
ive rolling moment. β
Figure 1.6 shows the sign conventions for deflection of the elev-
Figure 1.6: Measurement of elevator
ator, the tailplane control surface, and the tab, whose purpose will and tab deflections, η and β
be explained later. The deflection η is measured from the zero lift
line of the tailplane and β from the elevator reference line. As noted
above, we deal in linear aerodynamics, so we can write the tailplane
1.2. AERODYNAMICS OF WINGS AND CONTROLS 5

lift coefficient

CLT = a1 αT + a2 η + a3 β, (1.3)
∂CLT ∂CLT ∂CLT
a1 = , a2 = , a3 = ,
∂α T ∂η ∂β

where αT is the tailplane incidence, which is not the same as the

aircraft incidence. The deflections in (1.3) are the point where a pi-
lot intervenes in the system. The tailplane deflection η is set by the
pilot’s moving the control, and likewise the tab angle β. On many
aircraft, the pilot may also have control of αT if the aircraft has an
all-moving tailplane, such as the X-1, or a trimming tailplane, which
is deflected to trim the aircraft, with an elevator for short-term con-
trol inputs.
6 How aeroplanes fly and how pilots fly them

We do not need to know the details of the aerodynamics of con-

trol surfaces in order to design a tailplane, but we should know
something of how they work. Figure 1.7 shows the change in pres- ∆C p
sure coefficient over a surface for changes in, respectively, incid-
ence, elevator deflection, and tab angle. You can see that changes
in αT or η give quite large changes in pressure distribution, corres-
ponding to quite large changes in CLT , making the tailplane useful
as a means of adjusting moments on the aircraft. The tab on the
other hand seems to have little effect on the pressure distribution or x/c
Tailplane, αT
tailplane lift. Why have it?

∆C p

x/c
Elevator, η

∆C p

x/c
Tab, β
Figure 1.7: Pressure distribution
changes with control deflection
1.2. AERODYNAMICS OF WINGS AND CONTROLS 7

Figure 1.8 shows the same data as Figure 1.7 but with the addi- Hinge line
tion of the hinge line, where the elevator joins the tailplane proper.
From the shaded regions on the plots, you can see how a deflection ∆C p
modifies the moment about the hinge line. Changing η in order
to change CLT requires quite a large moment; a change in β also
gives quite a large change in moment, but with a small change in
lift coefficient. This moment is called the hinge moment, and is
expressed in non-dimensional form as hinge moment coefficient
x/c
MH Tailplane, αT
CH = , (1.4)
ρV 2 Sη cη /2
Hinge line
given, on linear aerodynamics, by
∆C p
CH = b0 + b1 αT + b2 η + b3 β. (1.5)

The quantities Sη and cη , the elevator area and chord respect-

ively, are measured behind the hinge line, as shown in Figure 1.9,
which also shows an aerodynamic balance, surface ahead of the
hinge line which has the effect of reducing the hinge moment for a x/c
given deflection. This is one way to alter the stick force required of Elevator, η
the pilot, as we will see in Chapter 5. Hinge line
The hinge moment is fundamental to the control of aircraft, be-
cause it corresponds to the force felt by the pilot when they move a ∆C p
control. This means that the hinge moment must be small enough
to allow a pilot to move the control surface to any required deflec-
tion, but not so small that there is a risk of moving a surface too
far and over-accelerating the aeroplane. The control force must
also conform to the needs of human physiology and psychology:
if control forces are too small, the pilot cannot accurately perceive x/c
changes in force even for quite large changes in deflection.4 The Tab, β
Figure 1.8: Pressure distribution
design of a control system for an aircraft is more than simply the and moment changes with control
sizing of surfaces with respect to some performance criteria; the deflection
configuration must also present itself to the pilot in a usable form.
Even if the controls are powered, the hinge moment is used to size
the actuators which drive the surfaces, so it remains an important
consideration in design.
Aerodynamic
balance

Hinge line

Sη
Figure 1.9: Measurement of control
geometry
4
Think Goldilocks.
8 How aeroplanes fly and how pilots fly them

1.3 Aerodynamic centre

When we come to calculate the moments on an aircraft, it is not

enough to know what lift and moment are generated by a wing or
control surface; we must also make some choice about where they
act. When we compute the moment generated by a lifting body, we
do so by viewing it as a force (lift mostly) and a moment placed at
some reference point, so that the moment about some other point at
a distance x is

M ( x ) = M0 + Lx.

We have some freedom in where we take our reference and we

should make best use of it. The obvious reference point on a wing
is the centre of pressure, the point about which the aerodynamic
moment is zero. Then, M0 ≡ 0 and M = Lx, which makes life easy.
The problem with this reference point is that the centre of pressure
moves with changes in incidence. Given that stability and control
are largely concerned with controlling α, the centre of pressure is
not very useful as a reference, since it moves as our aircraft pitches.
Instead of the centre of pressure, then, we use an alternative ref-
erence point called the aerodynamic centre or, for a whole aircraft, the
neutral point. This is a point about which the moment is independ-
ent of incidence, dM/dα ≡ 0, and

M ( x ) = M0 + L( x − xn ),

where now M0 is the zero-lift pitching moment and subscript n de- L

notes “neutral point”. The neutral point of the whole aircraft is one
of the most fundamental properties from the point of view of flight. M0
If we think of the total lift on the aircraft acting at the neutral
point, we can sketch some possible relationships between centre-of- W
gravity position and static stability, Figure 1.10. Remember that if Stable

the aircraft pitches nose-up, lift increases, which generates a pitch- L

ing moment about the centre of gravity. The sign of that moment
depends on the relative positions of neutral point and centre of M0
gravity. In Figure 1.10, the first figure shows a stable aircraft be-
cause an increase in lift tends to push the nose down: ∂Mcg /∂α < 0; W
Neutrally stable
the second is neutrally stable because changes in lift generate no
change in moment; the third is unstable because an increase in L
incidence causes an increase in pitching moment which keeps push-
M0
ing the nose up, ∂Mcg /∂α > 0.

W
1.4 Measures of stability: static and c.g. margins Unstable
Figure 1.10: Centre of gravity and
neutral point positions
Being engineers, and comfortable with numbers, we would like to
have numerical measures of aircraft stability, if only to bring clarity
to the situation. Since, for a stable aircraft ∂Mcg /∂α < 0, we can
use ∂Mcg /∂α as a measure. In non-dimensional terms, we call this
measure the static margin,5 5
This is really important. Memorize it.
1.4. MEASURES OF STABILITY: STATIC AND C.G. MARGINS 9

dC Mcg
Kn = − , (1.6)
dCR

where the resultant force coefficient CR = (CL2 + CD2 )1/2 , and the

negative sign means that a stable aircraft has a positive margin,

which is easier to visualize.
In practice, because CD  CL and CR ≈ CL , we can use an
approximation to Kn , the c.g. margin

dC Mcg
Hn = − . (1.7)
dCL

In the rest of these notes, we will use the approximation Kn ≈ Hn

without further comment.
10 How aeroplanes fly and how pilots fly them

1.5 Secondary flight controls

Throughout these notes we consider the so-called ‘primary’ con-

trol surfaces: elevator, rudder, ailerons, and elevons. We should
not forget, however, that control is also affected and/or effected by
the ‘secondary’ control systems, such as high-lift devices including
flaps and leading edge devices, Figure 1.11. These are not normally
used as ‘controls’, as they are usually moved from one configura-
tion to another and then left in place, but they do have an effect on
the aircraft characteristics. In particular, flaps can have a large effect
on pitching moment and lift curve slope of the wing, resulting in
changes in the stability and handling properties.6 6
Take a look at the accident report for
While not strictly ‘controls’, we should also take into account G-CHNL to see how flap deployment
moved the neutral point, making
the effect of the engine(s) on handling characteristics. Changes in a marginally stable aircraft very
thrust can have a large effect on aircraft pitching moment and, on unstable.
multi-engined aircraft, on yaw. Differences in flow over the wing
in the wake of a propeller, and engine torque, can also cause quite
large rolling moments which must be balanced using the rudder
and ailerons or by some other means, such as by making the wing
shorter on one side than on the other.7 7
The Italian Macchi C.202 had exactly
this feature.
1.5. SECONDARY FLIGHT CONTROLS 11

Changes in power or thrust can alter the pitching

moment on the aircraft, depending on the distance
of the thrust line from the c.g. There are also rolling
moment effects caused by engine torque and prop-
wash over the inboard region of the wing.

Leading edge devices allow the wing to operate

at an angle of attack where it would normally
stall, giving increased lift at low speed. Unlike
flaps, they do not affect the the zero-lift incidence
because they do not change the camber.

Spoilers disrupt the flow over

part of the wing to reduce lift
and/or increase drag. They are
used to dump lift on landing, to
increase descent rate, and, ap-
plied differentially, for roll con-
trol.

Flaps, fitted to the trailing edge, are used to increase

drag and/or lift. In control terms, they change the
lift curve slope a, the zero-lift pitching moment C M0 ,
and the zero-lift incidence (or rotate the zero-lift
line).

Figure 1.11: There is more to control

than the control surfaces.
12 How aeroplanes fly and how pilots fly them

1.6 What pilots do

The point of aircraft stability and control is that we want to build

aeroplanes which are stable and can be controlled. This means that
whatever might be the aerodynamic design of the control system,
an aircraft must present itself to a pilot in a flyable form: it should
respond to a pilot input in the manner a competent pilot expects.
The task of a designer is to conceal the design of the control system
in such a way that the pilot need know nothing of the connection Stick

between the controls and the elevator, but is free to think only of
the effect of controls on the aircraft.
A fixed-wing aeroplane is controlled through one of a standard
set of controls which allow a pilot to move the elevator and ailerons
with hand movements, and the rudder through a pair of pedals.
Figure 1.12 shows sketches of some typical control arrangements.
For now, we only think about the elevator, though the principles are
the same for the other controls. Chapter 5 on stick forces discusses
the mechanical design of control systems, and talks about how
aerodynamic load, i.e. hinge moment, is fed back to a pilot as a
stick force which is part of the information available about the state Yoke

of the aircraft. The movement of the stick reflects in some sense the
movement of the control. For example a typical range of elevator
deflection is ±30◦ ; this range of deflection should be achievable
using the full range of stick movement. If the pilot can move the
elevator beyond its limit, they risk stalling the tailplane which is in-
variably disastrous; if the pilot cannot move the elevator to its limit,
they do not have full control of the aircraft. There is a limit, how-
ever, to how precisely a human can control an increment of stick
displacement or stick force. Remember that an increment of dis-
Sidestick
placement or force is a control input which accelerates the aircraft. Figure 1.12: Some standard pilot
If an aircraft is to be responsive and capable of large accelerations, controls: the rudder pedals are almost
universal for yaw control, but the pitch
it should require quite small stick forces to generate a response; if it and roll controls can vary markedly
is to be stable and docile, even large stick forces should not produce between aircraft
excessive accelerations. A numerical statement of the ease of hand-
ling of an aircraft is the quality of control a pilot is expected to be
able to exercise over stick displacement and force over the range of
aircraft accelerations.
Historically, the design of aircraft controls has been concerned
with making an aircraft stable, and controllable by a human pi-
lot. With the development of fly-by-wire the mechanical linkage
between the pilot control system and the control surfaces was re-
placed by a computational intermediary, but the design principle
remained the same: the aircraft must be flyable by a human being
without excessive mental or physical workload.8 8
The meaning of ‘excessive’ varies
The question of how to integrate ergonomics, or human perform- with type of aircraft and phase of
flight: what would be excessive for
ance, into aircraft design is far too big to be considered in detail in a transport aircraft in cruise might
this course, but we will take some account of it, and you should be perfectly normal for a fast jet on
landing.
bear it in mind as we proceed. We need to think about what it
means to be a competent pilot and how aircraft should be designed
1.6. WHAT PILOTS DO 13

to meet the needs and expectations of competent pilots. This means

that an aircraft should respond to control inputs in the manner ex-
pected by a pilot, and should not stress the pilot psychologically
or physiologically. We will look in passing at some crashes. Very
often these are attributed to “pilot error”; in those cases, engineers
should ask themselves “what led to this error and how could it
have been avoided?” In part, if a pilot error has led to a crash, some
of the responsibility lies with the designers who made it possible
for such an error to lead to catastrophic failure.
2
Longitudinal control and static stability

Longitudinal stability and pitch control of an aircraft are the most

basic properties which concern a pilot. Most of the time, a pilot
wants to hold an aircraft at a constant incidence, and does so by
moving a control surface to the “right” position for moment equi-
librium. In order to change the state of flight, the pilot moves the
control surface to some other position to impose a finite moment on
the aircraft, and force it to rotate. The basic instrument for analysis
of the aircraft is thus a moment equation, derived from a free body
diagram. LT

Datum
LWBN
h0 c̄

2.1 The moment equation for aircraft T

zT
zD
D

M0

Figure 2.1 shows the relevant forces and moments acting on an hc̄ l

W
aircraft, with the wing-body-nacelle (WBN) lift placed at the aero-
Figure 2.1: Free body diagram for an
dynamic centre ch0 and the tailplane (T) contribution placed at aeroplane
the aerodynamic centre of the tailplane. The resulting equation for
moment about the centre of gravity is

Mcg = M0 − LWBN (h0 − h)c − L T [(h0 − h)c + l ] − Tz T + Dz D

= M0 − (h0 − h)c( LWBN + L T ) − L T l − Tz T + Dz D
= M0 − (h0 − h)cL − L T l − Tz T + Dz D .

We can safely assume that lift is much greater than drag and that
the combination of drag and thrust is negligible so that the equa-
tion can be simplified,

Mcg = M0 − (h0 − h)cL − L T l.

Non-dimensionalizing on Sc(ρV 2 /2) and noting that the tailplane

lift coefficient is based on tailplane area ST ,
LT
CLT = , (2.1)
ρV 2 ST /2
ST l
and C Mcg = C M0 − (h0 − h)CL − CLT .
S c
Defining the tail volume coefficient,

ST l
V= (2.2)
S c
16 Longitudinal control and static stability

results in the fundamental equation of aircraft stability and control:

C Mcg = C M0 − (h0 − h)CL − VCLT . (2.3)

This is the most basic equation you need to know 1 since it con- 1
You really, really need to know this.
tains within it all the behaviour of the aircraft which need concern Engrave it on your heart with an
obsidian dagger. This shall be the
us. Once an aircraft is built and flying, the control problem is how whole of the law.
to adjust CLT for trim, C Mcg ≡ 0; to design a tailplane, we begin by
finding the value of V which allows us to achieve stable trim over
the required operating range.
To examine the effect of control deflection, we need to include
ZL
some detail about the behaviour of the tailplane, and how CLT is L tai
related to pilot input and aircraft operating condition. The big lpl
ηT a ne
effect we have to include is that of tailplane incidence being affected ZLL W Free stream
BN
by downwash, the deflection of the freestream flow caused by lift e
on the wing.
Res
From Figure 2.2, the tailplane incidence is made up of the aircraft αT ultan
t flo
incidence α and the angle at which the tailplane is attached to the w
aircraft ηT , modified by the effect of downwash angle e: Figure 2.2: Effect of downwash on
tailplane incidence
αT = α + ηT − e.

For an untwisted wing, e is proportional to the lift on the wing,

meaning that in the linear regime, it is also proportional to α,

de
e= α + e0 ,
dα
with e0 only present for a wing where the zero lift angle of attack
varies along the span. Combining these equations,
   
de de
α T = α + ηT − e0 + α = α 1− + ( ηT − e0 ) ,
dα dα

and from (1.3),

CLT = a1 αT + a2 η + a3 β,

so that

CLT = a1 (α + ηT − e) + a2 η + a3 β,

and
 
de
CLT = a1 α 1 − + a1 (ηT − e0 ) + a2 η + a3 β.
dα

We know that

CL = aα,

where a is the overall lift curve slope of the aircraft so that

 
a de
CLT = 1 1 − CL + a1 (ηT − e0 ) + a2 η + a3 β. (2.4)
a dα
2.2. AIRCRAFT STABILITY 17

In these notes, unless otherwise stated, we assume that e0 = 0.

In practice, we can always do this as long as we modify the value
of ηT to take account of zero-lift downwash.
The value of CLT in (2.3) includes no assumptions about how the
lift is generated, so it can be expanded,
   
a1 de
C M = C M0 − (h0 − h)CL − V 1− C L + a 1 ( ηT − e0 ) + a 2 η + a 3 β .
a dα
The most basic thing this equation lets us do is calculate the control
input required to trim the aircraft. For example, if we have all the
required information about the operating condition of the aircraft,
we can calculate the elevator deflection needed for moment equilib-
rium, η, where the overbar denotes a trim quantity. Also, by using
the moment equation as a relation between incidence and pitching
moment, we can calculate Kn , our measure of stability.

2.2 Aircraft stability

In §1.4, we stated our measure of stability for an aircraft,

dC Mcg
Kn ≈ Hn = − , (1.7)
dCL

which, from (2.3),

dCLT
= ( h0 − h ) + V , (2.5)
dCL
into which we can substitute the expression for CLT from the pre-
vious section. This gives us a means of calculating measures of
stability under different conditions.
The most basic case is that where the aircraft pitches with the
controls locked, known as the stick-fixed condition. Then, from (2.4),
 
dCLT a1 de
= 1− ,
dCL a dα
 
a1 de
K n = h0 − h + V 1− . (2.6)
a dα
This tells us how stable an aeroplane is for a given position of
centre of gravity. For stability, Kn > 0, and for any required min-
imum stability margin (2.6) tells us how far back (aft) we can place
the centre of gravity and still meet the requirement.
To summarize this with regard to the aircraft, we call the centre
of gravity position where Kn = 0 the neutral point hn , so that the
static margin is the non-dimensional distance between the centre of
gravity and the neutral point:
 
a de
h n = h0 + V 1 1 − , (2.7)
a dα
and Kn = hn − h.

We will see in later chapters how hn can be measured on an aircraft

using flight-test data, so that we do not need to rely on estimates of
18 Longitudinal control and static stability

aerodynamic parameters to find the safe loading conditions for an

aeroplane. Given that we can always measure or estimate the centre
of gravity position on an aircraft, this is sufficient information to
meet a minimum stability requirement.
As well as the stick-fixed case, we consider the stick-free, where
the control is free to move until it reaches moment equilibrium,

CH = b0 + b1 αT + b2 η + b3 β = 0,
b0 + b1 αT + b3 β
and η = − ,
b2
yielding
    
a2 b1 de CL a b
CLT = a1 − 1− + a 1 − 2 1 ( ηT − e0 )
b2 dα a b2
 
a2 b3 a2 b0
+ a3 − β− .
b2 b2
For concision, we introduce some auxiliary variables,
   
a2 b1 a2 b3
a1 = a1 1 − , a3 = a3 1 − , (2.8)
a1 b2 a3 b2
so that, in the stick-free case,
 
a de a2 b0
CLT = 1 1 − C L + a 1 ( ηT − e0 ) + a 3 β − .
a dα b2
Inserting CLT into the pitching moment equation gives
   
a1 de a2 b0
C M = C M0 − (h0 − h)CL − V 1− C L + a 1 ( ηT − e0 ) + a 3 β − ,
a dα b2

which allows us to find the tab angle to trim with zero stick force, β.
The reason we have a trim tab is that it gives the pilot a means of
zeroing the stick force. The tab has very little effect on the tailplane
lift, but has quite a large effect on the elevator hinge moment. By
moving the tab to deflection β, the pilot can remove their hands
from the controls in order to perform other tasks, reducing their
physical and mental workload. Also, by zeroing the stick force for
the required flight condition, the pilot can use small control inputs,
which gives them much finer control over the aircraft than if they
had to use some large force to hold the stick in its trim position.2 2
You can try this yourself: hold your
The measure of stability is the same as it ever was, so we can hand steady palm upwards. Now do
the same thing with a heavy book on
calculate a static margin using your palm. Can you keep your hand
  stationary? Now do the same, but try
dCLT a de
= 1 1− , moving your hand a small distance
dCL a dα under full control.

which gives the static margin stick-free,

 
a de
Kn0 = h0 − h + V 1 1− , (2.9)
a dα
with a corresponding neutral point stick free
 
0 a1 de
h n = h0 + V 1− . (2.10)
a dα
Note that stick-free values are denoted by a prime symbol.
2.3. AIRCRAFT CONTROL 19

2.3 Aircraft control

The two control conditions which we have seen, stick-fixed and

stick-free, give different trim behaviour to the aircraft. In the stick-
fixed case, the pilot is actively holding the stick to generate the
elevator deflection needed for trim; in the stick-free case, at least
notionally, they can release the stick and the aircraft will continue
to fly at the same incidence, because the elevator is already in aero-
dynamic equilibrium.

Variable Significance Table 2.1: Significance of solution for

trim equation variables for C M = 0 and
h Forward centre of gravity limit all other variables held fixed
CL In flight, trim speed
V Tailplane size
η Stick-fixed, elevator angle to trim
β Stick-free, tab angle to trim
ηT Tailplane setting (possibly to trim)

Using the trim equation for design, we can size a tailplane for a
given operating condition or, given the final aircraft geometry, we
can set operating limits or calculate behaviour in flight. There are
six variables which we can change in the trim equation, h, CL , V, η,
β, and tailplane setting ηT . On aircraft with an all-moving tailplane
we can also change ηT in flight. If the aircraft is in trim, C M ≡ 0,
so fixing five of the variables gives us a solution for the sixth. They
can be interpreted using Table 2.1. In each case, a solution for the
variable can be found and has a particular significance in flight. For
example, an aircraft flying at a given weight and centre of gravity
position will have a trim speed determined by the tab angle β.
Likewise, for a given speed and weight, take-off speed and MTOW
for example, the forward limit for centre of gravity can be found by
solving for h with maximum elevator deflection.
In terms of flying the aeroplane, a useful quantity to consider is
∂CL /∂η = −Va2 /Kn . It relates the change in lift coefficient to the
change in elevator deflection. Elevator deflection is perceived by
the pilot as stick deflection, so this corresponds to how far a pilot
needs to move the stick to change the lift coefficient, or speed, by
a given amount. Also, since pilots typically fly on attitude, i.e. by
using a visual reference to establish the aircraft incidence, and CL
corresponds directly to incidence, this is also a measure of how
changes in Kn alter the perceived handling qualities of the aircraft.3 3
Now relate this to changes in aircraft
speed.
20 Longitudinal control and static stability

2.4 Tailless aircraft

The analysis up to now has been developed for conventional air-

craft, which have a tail. Not all aeroplanes do, and in the interests
of inclusivity, we should welcome them into the group, Figure 2.3.
Figure 2.4 shows the control surfaces on a tailless aircraft. The
rudder operates as on a conventional layout, but elevators and
ailerons are combined into “elevons” which operate differentially
for roll control, and together in pitch.

Vulcan: delta wing

Saab Gripen: close-coupled canard

The variables relevant to analyzing a tailless aircraft are shown

in Figure 2.5. Clearly, there is no tailplane contribution to include
in calculating the pitching moment, but there is a complication
because elevon deflection generates a change in lift coefficient and
a change in pitching moment. The coupling of these two effects can
make tailless aeroplanes quite challenging to control, especially on
Pegasus Quantum 15-912: flex-wing
landing.
Figure 2.3: Some tailless aircraft
planforms
L

Figure 2.4: Control surfaces for tailless

aircraft: elevons operate together for
η
pitch control and differentially for roll
M0

hc0

h0 c0 W Figure 2.5: Representation of tailless

aircraft

The lift coefficient for a tailless aircraft looks a bit like the cor-
responding expression for a tailplane, because there is a control
deflection to include,

CL = a1 α + a2 η,

but there is no tab, because tailless aircraft usually have none.

As before, we can analyze a free body diagram to find a pitching
2.4. TAILLESS AIRCRAFT 21

moment equation,

∂M0
Mcg = M0 + η − (h0 − h)c0 L,
∂η

which we then non-dimensionalize to give

∂C M0
C M = C M0 + η − ( h0 − h ) C L .
∂η

Our definition of static margin is the same as before so

dC M
Kn = − = h0 − h,
dCL

with no need to consider the stick-free case because such aero-

planes usually have powered controls.
3
How to design a tailplane

All things considered, an aeroplane needs a wing: most aircraft

cannot fly without one. After that, it usually needs a tailplane: most
aircraft cannot fly for very long without one, though see Figure 3.1.
Engines come later: push come to shove, you can always glide. The
functions of a tailplane are to stabilize the aircraft and allow the
pilot to control it, with a minimum weight and drag penalty.
The requirements for stability and control can be stated in vari-
ous ways, but we assume that we are given some combination of
a minimum value of Kn , and a forward centre of gravity limit, or
Figure 3.1: Wings are obligatory; fins
centre of gravity range, the distance between the forward and aft are optional
limits. The desire for minimum weight is an obvious one for any
part of an aircraft, and can be restated as wanting the smallest tail-
plane area we can get away with. The minimum drag requirement
comes from the problem of trim drag. As you should know, gener-
ating lift inevitably means generating drag. Given that a conven-
tional tailplane mostly generates negative lift, in order to generate
a nose-up pitching moment, there is a double drag penalty: the
drag generated by the tailplane itself, and the extra drag on the
wing which has to produce a lift greater than the aircraft weight to
compensate for the down force on the tail. Minimizing the tailplane
area minimizes the drag and weight penalty.

3.1 Basic tailplane sizing

The most basic design requirement is to size a tailplane for a given

minimum static stability margin Kn , and centre of gravity range
∆h = haft − hfwd . We can write two equations
 
a de
Kn = h0 − haft + V 1 1 − , (3.1a)
a dα
C Mcg = C M0 − (h0 − hfwd )CL
   
a1 de
−V 1− CL + a1 (ηT − e0 ) + a2 η + a3 β = 0.
a dα
(3.1b)

The first of these equations should be obvious: the aft limit on

centre of gravity fixes the minimum stability margin. The second
is a trim requirement based on the pitching moment which can
24 How to design a tailplane

be generated by the tailplane. The maximum moment is required

when the centre of gravity is at its forward limit. There is also a
limit on the elevator deflection to trim. The first limit is that the
elevator may not be deflected beyond some point fixed by aerody-
namic considerations such as stall. This limit is usually about 30◦.
A second limit is set by the need for the pilot to have a reason-
able range of options at any point. For example, if the aircraft is
trimmed with a forward centre of gravity in the flight condition set
by (3.1b), the pilot might want to manouevre by changing η. If η is
already at its aerodynamic limit, the pilot has no options available.
Another way to state this constraint is to impose a limit on CLT to
keep it well within its linear operating range, avoiding the risk of
tailplane stall.
In any case, there will be a flight condition given in terms of CL
and η. The three equations can then be combined to find V, hfwd ,
and haft . Given basic information about the aircraft geometry, l,
c, and S, the tailplane area ST can be determined from V. Some
calculations of this type are given as tutorial questions at the end of
the notes.
Rotation
∆h2

V
Kn
3.2 Scissors plots
∆h1
When the tailplane and centre of gravity range must conform to Landing Nose wheel
multiple, possibly conflicting, requirements, the standard design
h
method is the scissors plot, which is a graphical method for determ- Figure 3.2: A simple scissors plot
ining a tailplane area and centre of gravity range for an aircraft.
The approach is to rearrange the various constraint equations
to give V as a function of h, plot them, and read from the plot
the value of V which gives the required range of h. Typical con-
straints are the stability limit on Kn , the limiting cases for trim at
low speed 1 such as landing approach and climb after take-off, and 1
Why are these important?
other important operating conditions such as take-off rotation and
the need to have sufficient load on the nose wheel to be able to
steer the aircraft on the ground.
Figure 3.2 shows a simple fictional scissors plot, with two centre-
of-gravity ranges indicated. The first, ∆h1 , is quite a narrow range
and the limiting cases are the take-off rotation and stability con-
straints. If the designer wants a larger range, ∆h2 , the aft constraint
is no longer aerodynamic but the requirement to keep sufficient
load on the nose wheel. This knowledge of which constraint is driv-
ing the tailplane size can be used in subsequent design iterations
for the whole aircraft.

3.3 Designing a useful tailplane

The scissors plot gives us an estimate of tailplane area for a given

set of aerodynamic parameters, a, a1 , etc., but does not tell us if the
tailplane is “good” in some sense. The aerodynamic parameters
themselves will usually be estimated using published methods,
3.3. DESIGNING A USEFUL TAILPLANE 25

such as ESDU, or by copying a tailplane which is known to work

well. In order to be useful, however, the tailplane must be usable
by a pilot, which imposes constraints on the hinge moment coef-
ficients, and it must continue to function when things are going
wrong.
In aeronautical terms, “going wrong” usually translates as
“stall”. Stall occurs when the wing reaches its maximum lift coef-
ficient because its incidence is too great. Stall is a simple prob-
lem to deal with, and is one of the first things student pilots learn
about: the pilot moves the stick forward to reduce incidence which
also has the effect of increasing the aircraft speed. Implicit in this
procedure, however, is the assumption that the tailplane has not
stalled. If the tailplane stalls, the pilot has no control over the angle
of attack of the aircraft and cannot recover. The tailplane must be
designed to stall later than the wing. In practice, this means that
it has slightly more sweep and consequently a lower value of lift
curve slope, and may have a constrained choice of ηT .
In discussing the scissors plot, we talked about how to select
a suitable value of V. The value we obtain is one suitable for the
stability and control requirements we have set for the design but
it is not enough information for us to size the tailplane proper. If
we assume the wing geometry is fixed, we have no choice of S and
c, so we have to achieve a particular value of ST l. Obviously, if we
make l large enough, we can make ST as small as we like, but we
cannot choose l arbitrarily. On any aircraft which must fit a partic-
ular footprint, there is a maximum length for the aeroplane, which
places one limit on l. There are further constraints which arise from
structural considerations and wanting to avoid “dead space” in a
fuselage, between the rear bulkhead and the empennage. The major
exception to these limits is glider or sailplane design, where it is
structurally feasible to have the tailplane on a long boom or long
slender fuselage.
4
Flight testing and aircraft handling

Having designed an aircraft to have given stability characteristics,

we must test the production model to find what the real beha-
viour is. In the early stages of design, we use approximate analyses
and correlations and semi-empirical methods (for example, ESDU
sheets) to estimate the aerodynamic parameters such as lift curve Pencil and paper

log e
slopes, largely because early in design we have not fixed the exact
shape and size of the aircraft or of its subsystems. When we have a Computers
detailed geometry, we can use computational methods to refine our
estimates. When the first few aircraft are produced, or after modi- Wind tunnel
fications to a design, we test them to see what the real behaviour of
the real aircraft is. Flight test
Figure 4.1 gives an indication of how e, the error or uncertainty
in estimated aircraft properties, varies with the cost of different
methods. The simplest methods using pencil and paper are cheap
but have a relatively large uncertainty, which is considered accept- log £, $, €
able because the methods introduce uncertainties no greater than Figure 4.1: Accuracy versus cost for
different methods of estimating aircraft
the uncertainty in the input data. In other words, the precision of properties
the method matches its accuracy. Computational methods give es-
timates with less uncertainty but take longer and cost more. Wind
tunnel testing gives data based on physical testing, but in idealized
conditions with uncertainties introduced by rig and interference
aircraft weight

effects and model scaling: it is also very costly. Finally, flight testing
gives the least uncertainty but is the most expensive way to gather
data.
This information is used in setting the limits to be observed in
service—the ‘flight envelope’ of Figure 4.2. Before flight, the aircraft
weight and centre of gravity are plotted on the diagram and must
lie within the limits indicated.1 If they do not, then the weight
must be reduced or the centre of gravity must be moved by adding c.g. position
Figure 4.2: A typical weight and
ballast. This guarantees that the aircraft will fly within the limits set
balance envelope for a small aircraft
at the design stage. The rear centre-of-gravity limit, the vertical line 1
Federal Aviation Administration
on Figure 4.2, is fixed by the minimum stability requirement; the Flight Standards Service. Aircraft
weight and balance handbook. US
forward limit is set by the maximum moment which the tailplane
Department of Transportation, 2007
can generate in order to maintain pitch equilibrium in all phases of
flight.
28 Flight testing and aircraft handling

4.1 Measuring stick-fixed stability

Flight testing depends on being able to use quantities we can meas-

ure to estimate the things we want to know. For a given aircraft,
we know the shape of the aeroplane, because we can measure it.
For a given flight, we know the centre of gravity position and the
aircraft weight, because we can calculate them or measure them
on the ground. For a given flight condition, we know the aircraft
speed and the control deflections. This gives us h, CL , η, and β,
and we know that C M = 0 in trim. How far can we get with this
information?
As always, we start from the fundamental moment equation,

C M = C M0 − (h0 − h)CL − VCLT ,

and the tailplane lift coefficient,

 
a de
CLT = 1 1 − CL + a1 (ηT − e0 ) + a2 η + a3 β,
a dα

giving

η̄
C M = 0 = C M0 − (h0 − h)CL −
   
a de
V 1 1− C L + a 1 ( ηT − e0 ) + a 2 η + a 3 β .
a dα CL
h1
As in §2.2, we differentiate to find the static margin,
  c.g. forward h2
∂C a de
Kn ≈ Hn = − M = (h0 − h) + V 1 1 − , h3
∂CL a dα Figure 4.3: Elevator angle to trim at
various lift coefficients
and since we can also calculate the elevator angle to trim,
 h3 h2 h1
1
η= C M0 − (h0 − h)CL h
Va2
dη̄/dCL

   
a1 de
−V 1− C L + a 1 ( ηT − e0 ) + a 3 β ,
a dα

we find that η and Kn are related,

   Figure 4.4: Measurement of neutral
dη 1 a1 de Kn point location
=− ( h0 − h ) + V 1− =− . (4.1)
dCL Va2 a dα Va2

Figure 4.3 shows η plotted against CL , while Figure 4.4 shows

the relationship between dη/dCL and h. It is worth noting that the
elevator angle to trim at zero lift is independent of centre-of-gravity
position, as the moment equation makes clear.2 2
Can you think of a physical reason
Given this information, one way of finding the aircraft neutral why this should be so?

point stick-fixed is: fly the aircraft straight and level at various
speeds, recording the elevator angle to trim. This is repeated for
various different centre of gravity positions, yielding a plot like
Figure 4.3. To find the neutral point, plot the gradients of the lines
of Figure 4.3, as in Figure 4.4. Extrapolating to dη/dCL gives the
centre of gravity position where Kn = 0, the neutral point hn .
4.2. WHAT DOES THIS MEAN FOR THE PILOT? 29

4.2 What does this mean for the pilot?

Equation 4.1 expresses a relationship between stability and control,

from the pilot’s point of view. A more stable aircraft (one with large
Kn ) requires larger changes in control deflection for given changes
in incidence (α or CL ) or speed. The limits on elevator deflection
and operating condition for the aircraft thus impose a limit on the
maximum static margin which will allow the aircraft to be flown
by a pilot. Likewise, a very small value of Kn makes it difficult for
the pilot to control the aircraft, because small changes in control
deflection can give quite large changes in incidence, or speed.
We can also directly examine the relationship between speed and
elevator deflection. The elevator angle to trim is a function of speed
via the lift coefficient,
dη dη dCL
= .
dV dCL dV

We know that
L
CL = ,
ρV 2 S/2
η̄
giving

dCL 2L 2C
=− 3 = − L,
dV ρV S/2 V V
c.g. forward
and
dη 2C dη 4 W 1 Kn
=− L = , Figure 4.5: What the pilot experiences
dV V dCL ρ S V 3 Va2

which is sketched in Figure 4.5.3 3

What happens to the curves in Fig-
From Figure 4.5, it is clear that the aircraft is uncontrollable be- ure 4.5 as the wing loading changes?

low some minimum flight speed—it is not possible to move the

elevator far enough to trim. This happens because at low speed, the
control surfaces cannot generate enough force to balance the mo-
ment about the centre of gravity. Likewise, above a certain speed,
small changes in η lead to large changes in trim speed and the air-
craft is also very hard to control. The useful range of speeds for an
aircraft lies between these two limits, although the limits in ques-
tion will be a function of the aircraft type and of the skill assumed
of the pilot.
These effects become apparent to the pilot as “sloppy” handling
at low speed, which is taken as a sign of incipient stall. One way
to interpret this poor handling is as the result of insufficient ρV 2
for the wing and control surfaces to generate enough moment or
force for the controls to be responsive. One of the basic properties
of any aircraft is thus the minimum control speed, which is largely
determined by the size of the control surfaces with respect to the
available dynamic pressure.
30 Flight testing and aircraft handling

− β̄
4.3 Measuring stick-free stability

To find the neutral point stick-free, we can use the same approach
as in the stick-fixed case, but using the tab to trim, rather than the CL
elevator. Once again, h1
h2
C M = C M0 − (h0 − h)CL − VCLT = 0,
h3
and Figure 4.6: Tab angle to trim at varying
  lift coefficients
a1 de
CLT = 1− CL + a1 (ηT − e0 ) + a2 η + a3 β, h3 h2 h1
a dα
h
so that

−d β̄/dCL
b0 + b1 αT + b3 β
η=− ,
b2
 
a de a2 b0
CLT = 1 1− C L + a 1 ( ηT − e0 ) + a 3 β − ,
a dα b2 Figure 4.7: Measurement of stick free
neutral point location
and
   
a1 de a2 b0
C M = C M0 − (h0 − h)CL − V 1− C L + a 1 ( ηT − e0 ) + a 3 β − .
a dα b2

This gives the tab angle to trim for the flight condition,

1
β= C M0 − (h0 − h)CL
Va3
   
a1 de a2 b0
−V 1− C L + a 1 ( ηT − e0 ) − ,
a dα b2

which can be differentiated,

  
dβ 1 a1 de
=− h0 − h + V 1− .
dCL Va3 a dα

and related to Kn0 ,

 
a de
Kn0 = ( h0 − h ) + V 1 1− ,
a dα

with

dβ K0
=− n .
dCL Va3

So to find the neutral point stick free, we vary the aircraft speed
at fixed centre of gravity, trimming with the tab, giving us Fig-
ure 4.6. We then plot the gradients from that figure against CL ,
Figure 4.7, to find h0n .
5
Piloting: stick forces

The physicist, philosopher and motorcycle mechanic Matthew

Crawford has written extensively on how we experience the world
and draw non-verbal information from the sensory data which
physical reality feeds back to us, for example through our sense
of touch. After talking about the way an ice-hockey player uses
the information transmitted by his stick, Crawford looks at the
more generic problem of using a probe, for example to feel inside a
component to check its surface finish.

Consider the experience of using a probe to explore an unseen space,

or the way a blind person feels his [sic] way by tapping with a stick.
At first you feel the varying pressure of the probe against your palm
and fingers, and you have to interpret this pressure, mapping it in
some as yet uncertain way onto a spatial representation that you
are developing of the object. But as you learn to use the probe, your
awareness of this pressure at the handle end is transformed into
something quite different. What you have eventually is a direct,
unmediated sense of the probe’s tip touching the objects you are
exploring [page 47].1 1
Matthew B. Crawford. The world
beyond your head: How to flourish in an
Part of the experience of controlling a machine, such as a car or an age of distraction. Viking, 2015

aeroplane, is using the information which the machine transmits to

the driver or pilot. This is fundamental to our experience of con-
trol, and is something which we internalize early in learning how
to drive or fly. Part of the problem of design in aircraft control is
balancing the sometimes conflicting requirements for aerodynamic
control of the aircraft, information transmission to the pilot, and al-
lowing the pilot to move control surfaces without excessive physical
effort.
Crawford’s thoughts on the equivalent problem in car design are
worth reading.

A car that interposes layers of electronic mediation between the

driver and the road demands an effort of interpretation by the driver,
because each of those layers is based on a representation that has
no inherent, necessary relationship to the states being represented.
Some committee of engineers had to make a whole series of decisions
about how the pedal pressure felt by a driver in a car with brake-by-
wire, for example, should map onto the braking force delivered and,
crucially, the readiness of the system to keep delivering it. Should
the pedal effort change with sustained or heavy braking, to convey
32 Piloting: stick forces

the fact that those little DC motors doing the work are getting hot?
Brake rotors get hot under heavy use and, in doing so, become less
effective. This fact gets conveyed to the driver in a necessary and
lawlike way with the familiar “brake fade” in conventional hydraulic
brakes [page 82].2 2
Matthew B. Crawford. The world
beyond your head: How to flourish in an
The majority of aircraft, even large ones, have a direct mechanical age of distraction. Viking, 2015

linkage between the pilot controls and the corresponding control

surface. As well as being mechanically simpler than “electronic
mediation”, with the resulting advantages of reliability and ease of
maintenance, direct linkages give the pilot physical feedback from
the control surface, which is incorporated into the pilot’s picture of
how the aircraft is behaving.3 3
To get an idea of how important the
pilot’s construction of the state of the
aircraft is, read some accounts of the
5.1 Aerodynamics, stick force, and piloting AF447 crash.

The question which then arises is how to design a control system

which gives the aerodynamic forces and moments required to allow
a pilot to predictably and reliably control an aircraft, without the
risk of accidental overloading. Allowable control forces are laid out
in the regulations governing aircraft certification and operation.
Table 5.1 gives the important numbers.

Pitch Roll Rudder

Stick Wheel Stick Wheel (Push)
Temporary 267 111 222 334 133 222 667 N
Prolonged 44.5 44.5 — 22 22 — 89 N
application One Two One Two
hand hands hand hands
Table 5.1: Allowable control forces,
from EASA CS 23.143, CS 25.143.
Good design practice is to make sure that the maximum rud-
der force is greater than the maximum elevator force which is in
turn greater than the maximum aileron force. A further criterion
is to aim for the controls to be ‘harmonized’, meaning that aileron,
elevator, and rudder forces required for a given control response
have the ratio 1:2:4. For example, the rudder force for a 10◦ /s yaw
should be twice the elevator force for a 10◦ /s pitch.
The pilot input to the system, from a designer’s point of view,
is the stick force to trim Pe , which is the force required on the pilot
control to balance the hinge moment at the control surface,

ρV 2
Pe = me Sη c η C H ,
2
where me is the gearing ratio between the stick and control deflec-
tions.
The stick force to trim must lie within reasonable limits over the
operating range of the aircraft: too high and the pilot will not be
able to move the elevator over the full range of deflections needed;
too low and a small stick deflection will generate a large accelera-
tion on the aircraft with a risk of overloading the structure. The first
5.1. AERODYNAMICS, STICK FORCE, AND PILOTING 33

piece of information we need is the hinge moment to trim, which

depends on the flight condition and on the tab setting.
We already know that

CH = b0 + b1 αT + b2 η + b3 β,

which gives η as a function of CH ,

CH − b0 − b1 αT − b3 β
η= .
b2
Tailplane lift coefficient is
 
de CL
C L T = a1 1 − + a1 (ηT − e0 ) + a2 η + a3 β,
dα a

and so
 
de CL a
C L T = a1 1− + a1 (ηT − e0 ) + a3 β + 2 (CH − b0 ),
dα a b2

a general form of the stick-free expression with CH 6= 0.

The pitching moment equation is then

C M = C M0 − (h0 − h)CL
   
a1 de a2
−V 1− CL + a1 (ηT − e0 ) + a3 β + (CH − b0 ) ,
a dα b2

which can be re-arranged to find β,

Va3 β = C M0 − (h0 − h)CL

   
a de a2 b0
−V 1 1− C L + a 1 ( ηT − e0 ) − , (5.1)
a dα b2

or hinge moment to trim,

a2 C H
V = CM0 − (h0 − h)CL
b2
   
a de a2 b0
−V 1 1− C L + a 1 ( ηT − e0 ) + a 3 β − . (5.2)
a dα b2

Now, if we subtract (5.2) from (5.1),

 
a2
V a3 β − CH = Va3 β,
b2

yielding

b2
CH = a3 ( β − β )
a2
and
ρV 2 b2
Pe = me Sη c η a3 ( β − β ),
2 a2

so that CH and the stick force to trim depend linearly on the differ-
ence between the current tab setting and the tab angle to trim for
the flight condition.
34 Piloting: stick forces

In theory we could find the stick-free neutral point from meas-

urements of stick force, via
b2
CH = a3 ( β − β ),
a2

which gives

∂CH b ∂β
= 2 a3 .
∂CL a2 ∂CL

In §4.3, we found that

dβ K0 1
=− n ,
dCL V a3
so that

dCH b K0
= − 2 n.
dCL Va2

In principle, by measuring the stick force or hinge moment at

different flight conditions, we can work out the stick free neut-
ral point. In practice, however, we cannot measure the stick force
accurately enough for a reliable estimate, because of errors intro-
duced by such things as friction in the system. We can, however,
use a measurement of hinge moment, taken at the control proper,
to find dCH /dCL and perform the required analysis. Note once
again the effect of static margin—a stability measure—on the con-
trol characteristics of the aircraft where the relationship between
hinge moment and incidence is a function of Kn0 .
5.2. MODIFICATION OF STICK FORCES 35

5.2 Modification of stick forces

Having designed an aircraft with suitably-sized control surfaces,

it can happen that the stick forces do not lie in a range usable by
a pilot. In this case, there are a number of means of modifying
the stick forces to make the aircraft controllable. A couple of these
require no aerodynamic redesign. The first approach is to change
the gearing ratio between the stick and the control deflections, but
this is limited because it can affect the range of control movement
available. A more flexible approach is to add power assistance to
reduce or increase the pilot input, or to eliminate it, though this
then requires a feedback system to give the pilot force information.
Figure 5.1 shows some aerodynamic methods for modifying stick Horn balance
force. The first two modify the moment required for a given control
deflection by adding surface ahead of the hinge line (aerodynamic
balancing) or by moving the hinge line. Hinge line
The aim is to change the hinge moment required for a given
increment in elevator deflection, dPe /dη. In this case,
horn balance
ρV 2
Pe = me Sη c η C H ,
2
but since hinge location

CH = b0 + b1 αT + b2 η + b3 β,
dPe ρV 2
= me Sη cη b2 .
dη 2 geared tab

To reduce the stick force, we want to reduce b2 , but b2 dPe /dη,

must be negative for correct feel of the controls. Reducing b2 is
useful at high speed (because of the effect of V 2 ) but at low speed,
the pilot might not have enough feel for the controls and other anti-balance tab
Figure 5.1: Aerodynamic assistance
methods of reducing the stick force may be needed.
The third and fourth methods for modifying stick force in Fig-
ure 5.1 are to gear the tab to the movement of the elevator. In one
case, as the elevator moves, the tab deflection changes in such a
way that it reduces the hinge moment on the elevator; in the other,
the tab movement increases the hinge moment. By a suitable choice
of gearing, it is possible to modify the stick forces to make the air-
craft fully controllable by the pilot within a normal range of force.
6
Manoeuvre

The history of all hitherto existing stability has been the history of
equilibrium. Most of the time, this is what a pilot wants, but it is
clearly important to be able to change state of flight, or manouevre.
The most basic manouevre is a steady ‘pullout’ where a descend-
ing aircraft makes a transition to horizontal flight, at a constant
speed. We can approximate the dynamics of the process in terms of
flight on a circular path in the vertical plane, which will allow us to
analyze the behaviour of the aircraft and the relationship between
control input and aircraft acceleration. This is also a good approx-
imation to the state of an aircraft in a banked turn.
Aircraft acceleration is important for two reasons. First, the max-
imum acceleration which can be imposed on an aircraft is a state-
ment of agility: how rapidly can the aeroplane change from one
flight condition to another? Second, there is a maximum acceptable
acceleration which arises from structural considerations or, for aero-
batic aircraft which can sustain high loads, the peak acceleration
which a pilot can tolerate.
38 Manoeuvre

6.1 Steady pullout

Figure 6.1: Manoeuvre conditions

/ 2π
Vt g
r= 2 /n
V
=

)W
+n
(1

g
n)
L=

1+
Pu

m(
llou
t

=
W
L=W t = 2πV/ng,
∆θ = 2π

t = 0, ∆θ = 0
W = mg

Our basic model for a manouevre is shown in Figure 6.1, which

illustrates an aircraft flying around a circle in the vertical plane.
Also shown in grey is the more common case of a descending air-
craft making a transition to level flight, which can be modelled
as motion along part of the vertical circle. We take the aircraft as
moving at constant speed V on a circle of radius r. Clearly, this is
accelerated motion and there is an additional force on the aircraft
to keep it moving on a circular path. We call the increment in force
nmg, where g is acceleration due to gravity, and n is then the num-
ber of “gees pulled”. In steady level flight, n ≡ 0, and L = W = mg.
In the manouevre, there is a change in lift, L = W = m(1 + n) g
6.1. STEADY PULLOUT 39

and the new lift coefficient is CL + ∆CL ,

nW
∆CL = nCL =
ρV 2 S/2

where CL is the lift coefficient in the straight and level case.

We want to relate the dynamics of the aircraft to the acceleration
n, which we can do by looking at the details of the manouevre, Fig-
ure 6.1. In travelling round the circle, the aircraft pitches through
an angle of 2π. We know the aircraft speed, so we can find the time
required to traverse the loop,

2πr
t= .
V
The acceleration on the aircraft is related to its speed and the radius
of the circle,

V2
r= ,
ng
so
2πV
t= ,
ng

and the aircraft pitch rate is

2π ng
q= = .
t V
The aircraft is rotating about its centre of gravity at angular
velocity q, which means that there is a change of incidence at the
tailplane,

qlT
∆αT = ,
V
where lT is the tail arm measured from the centre of gravity.1 Inserting 1
This is important: moving the centre
the pitch rate, of gravity affects the stick force re-
quired for a given acceleration, which
is why aerobatic piston engine aircraft
nglT
∆αT = , are taildraggers.
V2
and non-dimensionalizing to remove the explicit dependence on V,

nCL
∆αT = ,
2µ1
µ1 = W/ρgSlT .

The term µ1 is called the longitudinal relative density. Note that it

uses the true density at altitude, and not the sea-level value.2 2
Why is this so?
The pitch-induced change in tailplane incidence generates a
corresponding change in tailplane lift coefficient, a1 ∆αT , and the
total lift coefficient is then
 
a de nC
CLT = 1 1 − ( 1 + n ) C L + a 1 ( ηT − e0 ) + a 2 η + a 3 β + a 1 L .
a dα 2µ1
40 Manoeuvre

We now have the elements we need to complete the pitching

moment equation,

C M = C M0 − (h0 − h)(1 + n)CL

   
a1 de nCL
−V 1− ( 1 + n ) C L + a 1 ( ηT − e0 ) + a 2 η + a 3 β + a 1 .
a dα 2µ1

In steady level flight,

C M = 0 = C M0 − (h0 − h)CL
   
a de
−V 1 1− C L + a 1 ( ηT − e0 ) + a 2 η + a 3 β . (6.1)
a dα

For a trimmed steady manoeuvre, the pitching moment is zero,3 3

Do you believe this?
and we can write the elevator angle to trim as η + ∆η, so that

0 = C M0 − (h0 − h)(1 + n)CL

  
a de
−V 1 1− (1 + n ) C L
a dα

nCL
+ a1 (ηT − e0 ) + a2 (η + ∆η ) + a3 β + a1 . (6.2)
2µ1

In practice, what we want to know is the elevator deflection per

g, or the control input required for a given acceleration, which we
can find by subtracting (6.1) from (6.2),
   
a1 de nCL
0 = −(h0 − h)nCL − V 1− nCL + a1 + a2 ∆η ,
a dα 2µ1

which we rearrange to find

∆η
    
CL a1 de a1
=− ( h0 − h ) + V 1− + . (6.3)
n Va2 a dα 2µ1

This must always be negative, otherwise the aircraft would pitch

nose-down when the pilot pulls back.

6.2 Stick fixed manoeuvre stability

Much of the information contained in the expression for ∆η/n is

based on calculations which may have no independent verification
until flight testing has taken place. In the same way as we stated
static margins in terms of distance between centre of gravity and a
neutral point, we can express manoeuvre characteristics in terms
of a manoeuvre point, whose position can be estimated on a real
aircraft using measured data.
When ∆η/n = 0 the centre of gravity is at the stick fixed man-
oeuvre point, so
   
a de a
h m = h0 + V 1 1 − + 1 .
a dα 2µ1

Returning to the definition of stick-fixed neutral point, we can see

that hm is a1 V/2µ1 aft of hn .
6.3. STICK FREE MANOEUVRE STABILITY 41

The stick fixed manoeuvre margin, Hm , is then defined in the obvi-

ous way,

Hm = hm − h.

From §6.1,
∆η
    
C a de a
=− L ( h0 − h ) + V 1 1− + 1 ,
n Va2 a dα 2µ1
and,
Va2 ∆η
   
a de a
h = h0 + +V 1 1− + 1 ,
CL n a dα 2µ1
so that there is a relationship between the stick-fixed manoeuvre
margin and the elevator angle to trim for a given acceleration,

Va2 ∆η
Hm = − ,
CL n
in the same way that the static margin stick-fixed is related to the
elevator angle to trim in steady level flight, (4.1).

6.3 Stick free manoeuvre stability

Given the stick-fixed manoeuvre characteristics of an aircraft, we

can determine how it will perform aerodynamically in a man-
oeuvre, but we must also consider the pilot input required for a
given acceleration.
From §5.1, we know the hinge moment required for trim in
steady level flight,

C M = 0 = C M0 − (h0 − h)CL
   
a de a2
−V 1 1− CL + a1 (ηT − e0 ) + a3 β + (CH − b0 ) .
a dα b2
To assess the pilot input needed for manoeuvre, we require the
change in hinge moment ∆CH for a given acceleration, which we
can find from the usual moment equation, with the manoeuvre lift
coefficient (1 + n)CL ,

C M = 0 = C M0 − (h0 − h)(1 + n)CL −

  
a1 de
V 1− (1 + n ) C L
a dα

nCL a
+ a 1 ( ηT − e0 ) + a 3 β + a 1 + 2 (CH + ∆CH − b0 ) .
2µ1 b2
Subtracting these expressions

Va2 ∆CH
   
a de a
= −(h0 − h) − V 1 1 − + 1 .
b2 CL n a dα 2µ1
Adopting the usual notation, we define the stick free manoeuvre point,
h0m , the centre-of-gravity position where ∆CH /n = 0,
   
a de a
h0m = h0 + V 1 1 − + 1 .
a dα 2µ1
42 Manoeuvre

0 , is
The corresponding stick free manoeuvre margin, Hm
0
Hm = h0m − h,
0 Va2 ∆CH
Hm =− .
b2 CL n
The stick force per g is calculated from the hinge moment,

∆Pe ρV 2 ∆CH
= me Sη c η .
n 2 n
The stick force per g is a fundamental piloting property of the air-
craft and must lie within reasonable limits to avoid the risk of a
pilot accidentally overloading the aeroplane. For aerobatic aircraft,
on the other hand, it can be relatively low, but with a requirement
for greater skill on the part of the pilot.

6.4 Tailless aircraft

Tailless aircraft have similar dynamics to conventional aeroplanes,

but we need to introduce some extra notation to allow for the dif-
ferent configuration. We already know that for a tailless aircraft,

∂C M0
C M = 0 = C M0 + η − ( h0 − h ) C L ,
∂η
and when the aircraft is in a steady pullout with acceleration g and
pitch rate q,
∂C M0 ∂C
C M = 0 = C M0 + (η + ∆η ) − (h0 − h)(1 + n)CL + M q.
∂η ∂q
Subtracting one equation from the other gives the change in
elevator angle for the manoeuvre,
 
1 ∂C
∆η = (h0 − h)nCL − M q ,
∂C M0 /∂η ∂q

where ∂C M /∂q is an aerodynamic derivative.4 When we look at the 4

Does a conventional aircraft have a
dynamic behaviour of aircraft, we will come across more aero- corresponding aerodynamic derivat-
ive? What is it?
dynamic derivatives which are used to express the relationship
between the motion of the aircraft and the resulting forces and
moments. For consistency, these derivatives are expressed in
a standard non-dimensional form. In this case ∂C M /∂q is non-
dimensionalized as
1 ∂M
mq = .
ρVSc20 ∂q

Then,
∂C M 1 ∂M 2c
= = 0 mq
∂q ρV 2 Sc0 /2 ∂q V

and we know that

ng
q= ,
V
6.5. STATIC AND MANOEUVRE MARGINS 43

so that
 
1 ρgc0 S
∆η = (h0 − h)nCL − mq nCL .
∂C M0 /∂η W

The longitudinal relative density, µ1 , for a tailless aircraft is

W
µ1 = ,
ρgSc0

so that
∆η mq
 
1
= ( h0 − h ) − CL .
n ∂C M0 /∂η µ1

This gives us a means of defining a manoeuvre point, the centre-

of-gravity position where ∆η/n = 0, Forces oppose motion
mq
h m = h0 − .
µ1

Consulting Figure 6.2 should convince you that mq < 0, because

the aerodynamic forces introduced by a positive pitch rate generate Figure 6.2: Aerodynamic forces during
a negative moment: the aerodynamic loads oppose the rotation, and pitching motion
the pitch damping increases the stability of the aircraft, which is
also clear from the definition of hm .
The manoeuvre margin for a tailless aircraft, Hm , is defined in
the same way as before

Hm = hm − h,

and
mq mq
Hm = (h0 − h) − = Kn − .
µ1 µ1

Then, as for the elevator deflection on a conventional aircraft, the

elevon angle per g is proportional to Hm ,

∆η Hm CL
= .
n ∂C M0 /∂η

6.5 Static and manoeuvre margins

We have shown that the static margins, stick-fixed and stick-free, for
conventional aircraft are
 
a1 de
K n = ( h0 − h ) + V 1− ,
a dα
 
a de
Kn0 = (h0 − h) + V 1 1 − .
a dα

Also, the manoeuvre margins for conventional aircraft are

   
a de a
Hm = (h0 − h) + V 1 1 − + 1 ,
a dα 2µ1
   
a de a
Hm0
= ( h0 − h ) + V 1 1 − + 1 ,
a dα 2µ1
44 Manoeuvre

and
Va1
Hm = Kn + ,
2µ1
0 Va1
Hm = Kn0 + .
2µ1

Manoeuvre points are aft of the corresponding neutral points, be-

cause of the stabilizing effect of pitch damping.
We have shown that the static margin for tailless aircraft is

K n = h0 − h

and that the manoeuvre margin is

mq
Hm = (h0 − h) − .
µ1

So,
mq
Hm = Kn − .
µ1
6.6. PILOTING QUALITIES: CHANGING THE STICK FORCE 45

6.6 Piloting qualities: changing the stick force

It can happen that an aircraft is aerodynamically acceptable in

manoeuvre, but does not have good handling properties from the
point of view of the pilot. This leads us to modify the primary
control input, the stick force, by changing the relationship between
the input force and the aircraft response.5 5
USAF Test Pilot School. Flying
Adding a spring into the circuit, Figure 6.3, generates a moment qualities textbook, USAF-TPS-CUR-86-
02, volume II. United States Air Force,
on the stick which corresponds to a control force. The tension T on Edwards Air Force Base, 1986
the spring is approximately constant over the range of stick travel,
so that the force P is also approximately constant. This means that
a spring in the system imposes a constant increment, or decrement,
on the control force perceived by the pilot.
A bob-weight on the other hand introduces a moment which
varies with aircraft acceleration. From Figure 6.3, the stick force P = T `1 / `2
`2
generated by the weight varies as W (1 + n) and the force required
on the stick changes as the aircraft accelerates, a modification of the `1 T
stick force per g. Increasing the stick force per g can be interpreted Spring
as shifting h0m aft.
P = ( 1 + n )W ` 1 / ` 2
In practice, springs and weights are used in combination to give
`1 `2
the required handling properties. For example, if a weight is intro-
duced in order to change the stick force per g, a spring may be used
to balance the moment from the weight at zero acceleration.
W (1 + n )
Bob weight
Figure 6.3: Modification of stick force
and stick force per g using spring and
bob-weight respectively
7
Aircraft configurations and control

There are many different tail configurations in use, in response

to the many requirements which must be satisfied by any aircraft
design. It is not sufficient for a tail to generate the moments re-
quired to allow a pilot to control the aircraft, and to maintain some
measure of stability: it must be integrated onto a particular aircraft
which has a particular role. This leads to various configurations.
The most common tail unit, or empennage, is a combination of a
vertical fin, with a rudder, and a horizontal tailplane, with an el-
evator and tab. On some aircraft, there is no tab and the whole
tailplane is moved for trim. Even this basic layout has variations,
however, depending on the vertical positioning of the horizontal
tailplane. Further variations include combining the horizontal and
vertical surfaces into a ‘V’ or ‘Y’ shape. These choices are often
made for reasons which are not directly related to stability and
control, but to other requirements of the overall design.
48 Aircraft configurations and control

Figure 7.1 shows a few otherwise conventional aircraft with

tails of varying outlandishness. Engineering being engineering, the
reasons for choosing such configurations do not always arise from
stability considerations alone, but are often attempts to provide
the required handling and stability properties while meeting other
requirements. For example, the tee tail, shown here on a Grob 109,
may be chosen on transport aircraft such as the A400M or C17 in
order to leave room for cargo handling equipment and to avoid
aerodynamic interference during airdrops, or on rear-engined air-
craft such as the Boeing 717 to keep the tailplane clear of the engine
exhausts.1 It does, however, also act as an “end plate” on the fin
1
We draw a discreet veil over three-
which can help improve its aerodynamic efficiency, and a tee tail engined aircraft.
can allow a greater lever arm for the tailplane while keeping the
fuselage proper the same length. Tee tail aircraft, however, are par-
ticularly prone to a phenomenon called “deep stall” which led to
a number of fatal crashes in aircraft under test before the prob-
lem was recognized. This occurs when the outboard region of the
wing stalls, shifting the tip vortices inboard. This leads to increased
downwash on the tailplane, which lies roughly in the plane of the
tip vortices. If the aircraft is stable in pitch beyond the stall, there
may be insufficient control authority for the pilot to recover by
pushing the nose down.2 2
Malcolm J. Abzug and E. Eugene
Also shown in Figure 7.1 are a vee and an inverted-vee tail. Larrabee. Airplane stability and control:
A history of the technologies that made
These have the advantage of having fewer surfaces to build, and aviation possible. Cambridge University
reduced interference drag, but with the disadvantage of greater Press, Cambridge, 2002

control complexity since the elevators are used for both pitch and
yaw control. Vee tails are also prone to stall in sideslip which has
been proposed as a possible cause of the high accident rate of the
Beech 35. It appears that the inverted-vee tail of the Predator was
not chosen for pure stability reasons but to protect the propeller
during landings, by acting as a bumper.2
Finally, at the bottom of Figure 7.1 are a pair of quite unusual
designs, the triple vertical tail of the OV-1 light attack aircraft and
the vertical tail of the C-2 carrier supply aircraft, which has four
vertical surfaces, and three rudders.
 49

Grob 109 Handley–Page Jetstream

Sukhoi 27 De Havilland Vampire

Fouga Magister General Dynamics Predator

Grumman OV-1 Mohawk C2 Greyhound

Figure 7.1: Tailplane configurations

on aircraft of otherwise conventional
layout
50 Aircraft configurations and control

7.1 Canard aircraft

The first real aeroplane, the Wright Flyer, was a canard but the now
conventional arrangement with a tailplane was soon found to be
better for most purposes. There are a number of canard aircraft in
operation, however, so they clearly have their uses. Probably the
leading designer of canard aircraft is the legendary Burt Rutan, Rutan VariEze
founder of Scaled Composites. Two of his designs, the home-built
VariEze and the Beechcraft Starship, are shown in Figure 7.2, with
the Piaggio Avanti, a three-surface aircraft.
The principal advantages of a canard configuration lie in the
design of highly manoeuvrable aircraft, where their disadvantages
are outweighed by the possibilities of post-stall control and su-
Beechcraft Starship
permanoeuvrability, which is why a canard layout is often seen on
modern fast jets.
In more conventional flight regimes, the main reason for choos-
ing a canard is that the aircraft becomes very difficult to stall. If
the angle of attack increases sufficiently, the canard stalls first, and
the lift on the wing pushes the nose back down, giving an inherent
stability which recovers from the incipient stall. If the pitch rate is
too high, however, the aircraft can rotate past the canard stall angle Piaggio Avanti
to the point where the wing enters dynamic stall and the canard Figure 7.2: Some canard aircraft

does not have sufficient control authority to recover. This can be

mitigated using a three-surface layout with both a tailplane and a
canard as on the Piaggio Avanti of Figure 7.2.3 3
Malcolm J. Abzug and E. Eugene
Larrabee. Airplane stability and control:
A history of the technologies that made
aviation possible. Cambridge University
7.2 Weight-shift and microlight aircraft Press, Cambridge, 2002

A large number of aircraft operated by recreational pilots fall into

the microlight category, which broadly speaking means a maximum
takeoff weight of 300kg for a single-seat and 450kg for a two-seat
aircraft.4 These aircraft include powered parachutes, as well as 4
There is more to the definition than
flex-wing and conventional aeroplanes. weight.

The dynamics of such small aircraft can be quite different from

the behaviour expected of larger designs, because of the importance
of added mass effects: this is especially true of Human Powered
Aircraft (HPA). When a body moves in air, its apparent mass and
moment of inertia include a contribution from the loads imposed
by aerodynamic effects. For a relatively dense, or heavy, aircraft
these effects represent only a small percentage of the overall mass
and can be neglected. For aircraft which are already quite small,
these effects may have a large influence on the dynamics of the
aeroplane. Given that such aircraft have lightweight, i.e. flexible,
structures, deformation effects must also be accounted for in the
dynamics, resulting in behaviour which is not what might be intuit-
ively expected.
Deformation effects are especially important on flex-wing aircraft
where the elastic properties of the wing (“sail”) must be managed
in order to maintain safe handling and acceptable performance, in
7.2. WEIGHT-SHIFT AND MICROLIGHT AIRCRAFT 51

particular to ensure that the wing holds a reasonable profile over

the whole speed and incidence range.5 The delta wing configura- 5
G. B. Gratton. The weightshift-
tion used for flex-wings is inherently stable in all three axes, but controlled microlight aeroplane.
Proceedings of the IMechE, 215 Part G:
there is one particular instability called the “tumble mode” which 147–154, 2001
invariably leads to loss of the aircraft and is almost always fatal.6 6
G. Gratton and S. Newman.
In this case, the aircraft rotates rapidly, at up to 400◦ /s, about its The ‘tumble’ departure mode in
weightshift-controlled microlight air-
pitch axis, usually because of aircraft modification or an attempt to craft. Proceedings of the IMechE, 217 Part
fly beyond the aircraft or pilot’s capability. The acceleration leads to G:149–166, 2003
structural failure and destruction of the airframe.
8
High-speed flight: compressibility effects

So far in these notes, the assumption, tacit or otherwise, has been

that the aerodynamics are linear and the constants are constant, in
effect the assumption of low speed flight. In the transonic flow re-
gime, compressibility effects can lead to large changes in the stabil-
ity and control characteristics of aircraft with possibly catastrophic
results. In the 1940s, when aircraft began to enter the transonic
regime, the study of these effects began, largely in order to invest-
igate anomalous handling, as seen, in almost correct form, in The
Sound Barrier.1 1
Malcolm J. Abzug and E. Eugene
Larrabee. Airplane stability and control:
A history of the technologies that made
aviation possible. Cambridge University
Press, Cambridge, 2002
54 High-speed flight: compressibility effects

8.1 High speed effects

Figure 8.1 summarizes the principal effects which modify the con-
trol of aircraft at high speed. First, there is the motion of the neutral
point h0 from wing quarter chord to half chord as Mach number
M increases. This raises two problems. The first is the change in
h0 with speed as the aircraft passes through M = 1. This leads
to control problems for a pilot as the pitching moment changes
with no corresponding control input. The second problem is that
the change in h0 leads to a large increase in Kn . Remembering the
relation between elevator angle to trim and static margin (page 28),
  
dη 1 a1 de Kn
=− ( h0 − h ) + V 1− =− , (4.1)
dCL Va2 a dα Va2
we can see that an increase in Kn increases the change in elevator
angle needed to trim for a given change in CL or, equivalently,
speed. A large increase in Kn can thus make the aircraft uncontrol-
lable because there is insufficient elevator travel to change speed or
to manoeuvre, as can be seen by considering changes in ∆η/n, (6.3).
The next two plots in Figure 8.1 show more bad things: the zero-
lift pitching moment coefficient changes with Mach number, as
does the wing zero-lift incidence. The combination of these ef-
fects means that purely by virtue of its changing speed, the aircraft
wants to pitch as it approaches and exceeds a Mach number of
unity.2 2
If the aircraft geometric incidence is
The second row of Figure 8.1 shows how changes in lift curve held constant and α0 changes, what are
the implications for stall?
slopes, a and a1 , and elevator effectiveness a2 manifest themselves.
Considering only the upper curve in each case, that for a rigid air-
craft, we can see large increases in lift curve slope around M = 1,
followed by a drop-off with increasing Mach number. This can
cause various problems for stability and control, as you can see by
looking at the expressions for stability and manoeuvre margins,
and also causes difficulties for aircraft design: an aeroplane de-
signed for good qualities with the high speed values of a and a1
may well not have good handling qualities at low speed, unless
special measures are taken. Then, it is clear from the plot of a2 that
above a certain speed, the elevator simply stops working, and can-
not be used for pitch control. It was some time before this effect
was recognized, and led to the use of all-moving tailplanes for lon-
gitudinal control. Note also the effect of aircraft deformation, which
is a function of altitude and the corresponding change in density,
and leads to further changes in the variation of the aerodynamic
coefficients.
The final row of Figure 8.1 shows, first, a large change in down-
wash which will have an effect on the tailplane behaviour, and then
the so-called “Mach tuck” phenomenon. As an aircraft reaches high
speed, the net pitching moment can decrease before increasing and
then returning to its equilibrium value. This leads to a control prob-
lem for the pilot, who may well input a wrong stick force.3 The 3
This is the grain of truth in a scene
solution to this problem is shown in the final plot of the figure: a from the David Lean film The Sound
Barrier, which is well worth watching
as an account of test flying.
8.1. HIGH SPEED EFFECTS 55

Mach sensor is used to generate an additional stick force which

depends on speed so that the stick force gradient remains “correct”
through the speed change.

h0 Figure 8.1: Compressibility effects on

C M0 α0 stability-related parameters

c/2
c/4 1.0 M 1.0 M

1.0 M
Rigid aircraft Rigid aircraft
a V̄a1 V̄a2 Rigid aircraft
Decreasing altitude

M M M
∂e/∂α

CM Push Mach trim input

M M
M Mach tuck Pull Uncorrected stick force
9
Dynamic behaviour of aircraft

So far, we have only looked at the static stability of aircraft: how

they initially respond to a perturbation. To understand the flying
qualities of an aeroplane, we need to consider dynamic stability,
how the aircraft responds over time. Start with a simple example in
pitch. The motion of a rotating body is governed by

M = I θ̈,
or, for an aeroplane, Bα̈ − Mcg = 0,

where B is the moment of inertia about the pitch axis. Assuming

the controls are locked, Mcg is related to α via the static margin
stick-fixed,
ρV 2 Sc
Mcg (α) = C Mcg ,
2
ρV 2 Sc
=− Kn aα + Mcg (0),
2
so that
ρV 2 Sc
Bα̈ + Kn aα = 0, (9.1)
2
ignoring the zero-incidence pitching moment. This is the equation
of motion of a simple harmonic oscillator with natural frequency ω,

α = ejωt ,
ρV 2 Sc
ω2 = Kn a.
2B
If the static margin is negative, (9.1) becomes

ρV 2 Sc
α̈ − |Kn | aα = 0, (9.2)
2B
and the response to a perturbation is no longer oscillatory, but
grows exponentially,

α = eλt ,
ρV 2 Sc
λ2 = |Kn | a.
2B
This oscillation is a simple model for what happens when an air-
craft encounters a gust, or the pilot changes elevator deflection.1 1
How would this change if you as-
sumed a stick-free condition?
58 Dynamic behaviour of aircraft

We will look at this in more detail in the next chapter when we

consider the dynamics of the whole aircraft.
We can also look at another form of motion, where the aircraft
flies at constant incidence at varying speed. In this case, the aircraft
can be approximated as a mass acted on by a force perpendicular to
the flight path.

L Figure 9.1: Motion of a body under lift

V and gravity

Figure 9.1 shows the notation.2 We assume that the aircraft flies 2
The analysis presented here is based
at constant incidence with thrust balancing drag, so that the lift on Milne-Thomson, L. M., Theor-
etical aerodynamics, MacMillan, fourth
L = ρV 2 SCL /2 and varies only with speed V. The inclination of the edition, 1966, pp 376–378.
flight path is θ so the net force normal to the aircraft is given by

W V2
L − W cos θ = , (9.3)
g R
where R is the radius of curvature of the path. We also know that
energy is conserved so that V 2 /2 − gz is constant, with z taken
positive downwards. We can choose an origin for z such that the
total energy is zero and then V 2 = 2gz: the aircraft trades kinetic
and potential energy (height) so we can write speed in terms of
height and v.v.
If we take V1 as the speed the aircraft would have in steady level
flight at the prescribed CL , we can rewrite (9.3),
z 2z
− cos θ = , (9.4)
z1 R
and since
1 dθ dz
= and sin θ = − ,
R ds ds
where s is the arc length along the flight path and R is the radius of
curvature of the trajectory, (9.4) can be rewritten

d  1/2  z1/2
z cos θ = , (9.5)
dz 2z1
and integrating gives a solution for cos θ and R,

1 z  z 1/2
cos θ = +C 1 , (9.6a)
3 z1 z
z1 1 C  z1 3/2
= − , (9.6b)
R 3 2 z
where Cz1/21 is the constant of integration. We cannot solve directly
for the flight path, called the phugoid, but we can say something
about its behaviour as a function of C.
9.1. ANALYSIS OF AIRCRAFT DYNAMICS 59

First, if the aircraft flies at constant height z = z1 , it is obvious

that C = 2/3. If the aircraft is to fly in a loop, at some point cos θ =
−1, which can only happen if C < 0, because z and z1 are never
negative. When C = 0, R = 3z1 and the flight path is a sequence of
semi-circles. When 0 < C < 2/3, the flight path is an oscillation in
z. Some possibilities are shown in Figure 9.2, found by numerically C = 2/3
solving the equations of motion,
C = 1/3
V 2 − cos θ
V̇ = − sin θ, θ̇ = . (9.7)
V
These two examples illustrate the essential properties of lon- C=0
gitudinal dynamics of an aeroplane. To properly understand the
handling qualities of an aircraft, we need to develop a systematic
analysis of its dynamics, incorporating rotation and translation in C = −1/3
three axes and the coupled effects of aerodynamic interactions. Figure 9.2: Phugoid flight paths

9.1 Analysis of aircraft dynamics

Figure 9.3: Notation for analysis of

dynamic stability

Axis Perturbation Mean Perturbation Rotation Angular Moment Moment

force velocity velocity angle velocity of inertia
x X U u φ p A L
y Y V v θ q B M
z Z W w ψ r C N

The first step is, as always, to define our notation. Figure 9.3
shows the system of axes. The axes are attached to the aircraft,
rather than to an inertial frame, and have their origin at the centre
of gravity. The table of quantities gives the notation for the dis-
placements, rotations, forces, moments and moments of inertia. It is
literally as easy as A-B-C. In practice, to examine problems of sta-
bility we will linearize the system and write quantities as the sum
of a mean value for steady level flight, and a small perturbation.
The logic of our analysis is the same as for any dynamic prob-
lem: identify the forces and moments which act on a free body,
insert these in the appropriate dynamic equations, and calculate the
motion of the body. The difficulties arise from the aircraft’s having
six degrees of freedom and from coupling between motion in those
60 Dynamic behaviour of aircraft

degrees of freedom. We need to develop a systematic way of mod-

elling this coupling, ideally using quantities which can be measured
in flight as well as calculated on the ground.3 3
The analysis which follows is taken
As always in this course, we assume linearity, so we can work to from Milne-Thomson, L. M., The-
oretical aerodynamics, MacMillan and
first order in perturbation quantities. For example, Company, 1966.
∂M ∂M ∂M ∂M ∂M ∂M
M = M0 + u+ v+ w+ p+ q+ r.
∂u ∂v ∂w ∂p ∂q ∂r
The moment in this case is written as a sum of inputs from the
velocities and angular velocities on all three axes. The constants,
which are effectively the first terms in a Taylor series, are called
“aerodynamic derivatives” or “stability derivatives”. One of them,
∂M/∂q has already arisen in §6.4, with regard to pitch damping
during manoeuvres. If we know these quantities, and the state of
the aircraft, we can compute its motion using Newtonian dynamics.
This is feasible using computational methods, but does not give us
insight into the qualitative behaviour of an aeroplane, so we will
have to do some maths.
We know that there are two sets of forces on the aircraft, aerody-
namic F, and gravitational g,
F = Xı̂ + Y̂ + Zk̂, (9.8)
mg = mg1 ı̂ + mg2 ̂ + mg3 k̂, (9.9)
where the components of g are needed because the reference frame
is fixed to the aircraft and rotates about three axes. We also need to
know the motion of the aircraft,
v = uı̂ + v̂ + wk̂, velocity,
Ω = pı̂ + q̂ + rk̂, angular velocity,
h = h1 ı̂ + h2 ̂ + h3 k̂, angular momentum.
The equations of motion are then
d
(mv) = mv̇ + Ω × (mv) = F + mg, (9.10a)
dt
dh
= ḣ + Ω × h = L, (9.10b)
dt
where the boxed terms are required because the frame of reference
is rotating. The applied moment L is
L = Lı̂ + M̂ + N k̂.
We now approximate these equations to examine how the aircraft
responds when it is perturbed from steady level flight.
In steady flight, we write
v = V, Ω = 0, F + mg = 0,
and add small perturbations so that
V = V1 + u,
V1 = Uı̂,
u = uı̂ + v̂ + wk̂,
ω = pı̂ + q̂ + rk̂.
9.1. ANALYSIS OF AIRCRAFT DYNAMICS 61

For a small rotation χ,

χ = φı̂ + θ̂ + ψk̂,

ω = φ̇ı̂ + θ̇̂ + ψ̇k̂,

and likewise, the perturbation forces are

F + δF, m(g + δg),

so it can be shown that

δg + χ × g = 0.

Inserting these assumptions into (9.10) gives the equations of mo-

tion for small perturbations,

mu̇ + m(χ̇ × V1 + χ × g) = δF, (9.11a)

ḣ = δL. (9.11b)

Some reasonable assumptions will now help us to make the

system tractable. First, we can assume that forces and moments
depend on velocities but not on accelerations, except for M, which
has a dependence on ẇ, so

δF = δXı̂ + δY̂ + δZk̂,

δL = δLı̂ + δM̂ + δN k̂,

and, for example,

∂X ∂X ∂X ∂X ∂X ∂X
δX = u+ v+ w+ p+ q+ r,
∂u ∂v ∂w ∂p ∂q ∂r
∂M ∂M ∂M ∂M ∂M ∂M ∂M
δM = u+ v+ w+ ẇ + p+ q+ r.
∂u ∂v ∂w ∂ẇ ∂p ∂q ∂r
We also assume that the aircraft is laterally symmetric so that
symmetric perturbations cause symmetric responses: a pitch dis-
turbance cannot cause yaw or roll. Furthermore, the symmetric
response to an asymmetric input is symmetric: a given roll rate
has the same response in pitch whether the roll rate is negative or
positive. Taking these assumptions together,
∂Y ∂Y ∂Y ∂L ∂L ∂L ∂N ∂N ∂N
= = = = = = = = ≡ 0,
∂u ∂w ∂q ∂u ∂w ∂q ∂u ∂w ∂q
∂X ∂X ∂X ∂Z ∂Z ∂Z ∂M ∂M ∂M
= = = = = = = = ≡ 0.
∂p ∂r ∂v ∂p ∂r ∂v ∂p ∂r ∂v
Eliminating zero terms,
   
∂X ∂X ∂X ∂Y ∂Y ∂Y
δF = u+ w+ θ̇ ı̂ + v+ φ̇ + ψ̇ ̂
∂u ∂w ∂q ∂v ∂p ∂r
 
∂Z ∂Z ∂Z
+ u+ w+ θ̇ k̂,
∂u ∂w ∂q
   
∂L ∂L ∂L ∂M ∂M ∂M ∂M
δL = φ̇ + ψ̇ + v ı̂ + θ̇ + u+ w+ ẇ ̂
∂p ∂r ∂v ∂q ∂u ∂w ∂ẇ
 
∂N ∂N ∂N
+ φ̇ + ψ̇ + v k̂.
∂p ∂r ∂v
62 Dynamic behaviour of aircraft

With the further assumption that there is no inertial coupling

between yaw and roll, we find that the only moments of inertia
we need consider are A, B and C.
Perturbing from horizontal flight and expanding the products
in (9.11) gives

∂X ∂X ∂X
mu̇ = u+ w+ q − mgθ, (9.12a)
∂u ∂w ∂q
∂Z ∂Z ∂Z
m(ẇ − Uq) = u+ w+ , (9.12b)
∂u ∂w ∂q
∂M ∂M ∂M ∂M
Bq̇ = q+ u+ w+ ẇ. (9.12c)
∂q ∂u ∂w ∂ẇ

and
∂Y ∂Y ∂Y
m(v̇ + Ur ) = v+ p+ r + mgφ, (9.13a)
∂v ∂p ∂r
∂L ∂L ∂L
A ṗ = p+ r+ v, (9.13b)
∂p ∂r ∂v
∂N ∂N ∂N
Cṙ = p+ r+ v. (9.13c)
∂p ∂r ∂v

The first of these systems of equations covers symmetric motion,

e.g. pitch oscillations, while the second covers lateral motion, such
as yaw and roll. An important point to note is that these equations
are uncoupled so that longitudinal motion does not affect lateral and
vice versa.
10
How aircraft wobble: normal modes

Given the equations of motion for an aircraft, we would like to

extract some solutions which characterize the dynamic behaviour.
In any dynamic system, these solutions are the normal modes and
arise from an eigenvalue analysis of the equations of motion. Since
longitudinal and lateral motion are uncoupled, we can treat them
separately as two three-degree-of-freedom systems, which is rather
simpler than dealing with the full six-degree-of-freedom problem.1 1
The following analysis, with different
notation, is given in Graham, W.,
‘Asymptotic analysis of the classical
aircraft stability equations’, Aeronaut-
10.1 Longitudinal symmetric motion ical Journal, February 1999, pp 95–103.

Normal modes for the multi-degree of freedom system are found

as a natural frequency and a set of amplitudes for the motion. We
begin in the usual manner by inserting assumed forms for the solu-
tion of (9.12),

u = u0 eλt , v = v0 eλt , θ = θ0 eλt ,

so that, for example,

∂X ∂X ∂X
mu0 λeλt = u0 eλt + w0 eλt + θ0 λeλt − mgθ0 eλt .
∂u ∂w ∂q

Non-dimensionalizing,

xq Λ CL
 
(Λ − xu )u0 − xw w0 − − θ0 = 0, (10.1a)
µc 2
zq
 
−zu u0 + (Λ − zw )w0 − Λ 1 + θ0 = 0, (10.1b)
µc
m
ẇ
 Λ(bΛ − mq )
− µc + mw w0 + θ0 = 0. (10.1c)
Λ µc

where the non-dimensional parameters are given on the data sheet

and primes denote velocities scaled on U, u0 = u0 /U, w0 = w0 /U.
The first solution we consider is a low frequency oscillation. We
state without proof that there is a solution with Λ and u0 /θ of order
one and w0 /θ0 of order 1/µc . This means that the vertical motion
is negligible or, equivalently, the incidence is almost constant.2 We 2
What does this imply about the
can rewrite (10.1) in matrix form, with the negligible terms in each aircraft attitude or inclination?
64 How aircraft wobble: normal modes

equation removed:
    
Λ − xu 0 CL /2 u0 0
  0  
 −zu 0 −Λ   w  = 0 .

0 −mw Λ(bΛ − mq )/µc θ0 0

This equation can only have a non-trivial solution if the determin-

ant of the matrix is zero,
−zu
Λ2 − x u Λ + CL = 0,
2
which gives
 !2 1/2
(− xu ) − xu
Λ=− ± jΩph 1 −  .
2 2Ωph

This is a solution for oscillatory motion (note the imaginary part in

Λ) with

−zu CL 1/2
 
Ωph = , natural frequency, (10.2a)
2
− xu
cph = , damping. (10.2b)
2Ωph

This is the phugoid mode, a lightly damped long period oscillation,

which we examined in simplified form in Chapter 9. The incidence
is almost constant and the aircraft varies altitude at constant energy,
trading potential for kinetic energy and back again, Figure 10.1.

zmax , Vmin Figure 10.1: Phugoid oscillation

trajectory

zmin , Vmax

An important point to note is that the damping is proportional to

(− xu ), the rate of change of horizontal force with horizontal speed,
which depends largely on drag.3 Since drag acts in the opposite 3
What else might it depend on?
direction to velocity, xu < 0 and the damping is positive, stabilizing
the motion.
10.2. LATERAL MOTION 65

The second solution for longitudinal oscillation is for the case

−1/2
where Λ is of order µ1/2
c , u0 /θ0 is of order µc and w0 /θ0 is of
order one. In this case, the approximation to (10.1) is

xq Λ CL
  
Λ − xu − xw − µc − 2  u0
   
0
0 = 0 .
    
 0
 Λ − zw −Λ 
  w   
Λ(bΛ − mq )
− mµẇcΛ + mw
    θ0 0
0 µc

Again, we find the natural frequency by requiring that the determ-

inant of the matrix be zero,
mq + mẇ zw mq − mw µc
   
Λ ( Λ − x u ) Λ2 − z w + Λ+ = 0,
b b
which, on solving the quadratic, gives a result for the non-dimensional
natural frequency and damping:

µc (−mw ) + mq zw 1/2
 
Ωspo = , natural frequency, (10.3a)
b
mq + mẇ
 
1
cspo =− zw + , damping. (10.3b)
2Ωspo b

This is the short period oscillation, which you saw at the start of
Chapter 9. It is a heavily damped mode with period typically of
a few seconds. The aircraft pitches rapidly about its centre of grav-
ity which continues to fly at almost constant speed in a straight
line. The periodic time is typically a few seconds, but must not
be less than about 1.25s, otherwise there is a risk of Pilot Induced Figure 10.2: Short period oscillation

Oscillation (PIO).4 4
If you are in the humour, you might
The frequency is proportional to Kn1/2 , and increases with dy- like to try modelling Pilot Induced
Oscillation.
namic pressure, ρV 2 /2. Therefore the aircraft will have the highest
frequency SPO, and hence the shortest time period, at high speed
with the centre of gravity in the furthest forward position.5 The 5
Using this information, could you
SPO is always stable for a statically stable aircraft. relate the static margin stick-fixed to
the aerodynamic derivatives?

10.2 Lateral motion

In the case of lateral motion, we insert the assumed form for the
solution

v = v0 eλt , φ = φ0 eλt , r = r0 eλt ,

into (9.13), and non-dimensionalize, using wingspan s as our refer-

ence length,
y p Λ CL
   
yr
(Λ − yv )v0 − + φ0 + 1 − r 0 = 0, (10.4a)
µs 2 µs
Λ lr
−lv v0 + ( aΛ − l p ) φ0 − r 0 = 0, (10.4b)
µs µs
npΛ cΛ − nr 0
−nv v0 − φ0 + r = 0. (10.4c)
µs µs
66 How aircraft wobble: normal modes

Again, the non-dimensional quantities are given on the data sheet.

The first lateral mode we consider is Dutch roll which has oscil-
lations of roughly equal magnitude in pitch, yaw and roll. In this
case, (10.4) reduces to
    
Λ 0 1 v0 0
2
 −lv aΛ /µs 0  φ0  = 0 .
    
−nv 0 cΛ/µs r0 0

As before the determinant of the matrix must be zero for a non-

trivial solution,

Λ2 (cΛ2 + µs nv ) = 0,

and the frequency of the oscillation is, on the approximations we

are using,
 µ n 1/2
s v
Ωdr = . (10.5)
c
In Dutch roll, yawing oscillation (analogous to the longitudinal
SPO) causes alternating sideslip. This in turn causes a rolling oscil-
lation via Lv v. The periodic time is typically a few seconds, but as
for the SPO it should not have a period of less than 1.25s to avoid
PIO.
Dutch roll is not permitted to be divergent. Divergent Dutch roll
can be ‘fixed’ by a yaw damper on the rudder which damps the
yawing oscillation, and hence the roll response as well.
There are two further solutions to the dynamic equations which
have small values of Λ. These are dominated by yaw and roll with
weak sideslip and the corresponding approximation to (10.4) is
    
0 CL /2 1 v0 0
 −lv ( aΛ − l p )Λ/µs −lr /µs  φ0  = 0 .
    
−nv −n p Λ/µs (cΛ − nr )/µs r0 0

The requirement for a non-trivial solution is then that

anv Λ2 + [lv (n p − cCL /2) − l p nv ]Λ + (lv nr − lr nv )CL /2 = 0.

The two roots of this equation can be approximated as:

Figure 10.3: Motion of an aircraft
(−l p )nv + (−lv )[cCL /2 + (−n p ) undergoing Dutch roll
Λrs = − , (10.6)
anv
and
CL l v n r − lr n v
Λsm = − . (10.7)
2 (−l p )nv + (−lv )[cCL /2 + (−n p )]

Note that both of these roots are real and so they do not describe
oscillations. The first, Λrs , describes rolling subsidence, a pure rolling
motion which is generally heavily damped, and is usually stable.
The damping is primarily from the wings, where the incidence
along the wing is changed by the roll-rate. This is experienced by
the pilot as a lag in roll response. Roll control is not like pitch and
10.2. LATERAL MOTION 67

yaw control because the control input sets roll rate rather than roll
angle. A lag in response means that the required roll rate is not
reached immediately, and the pilot must change the control input
slightly early to stop the aircraft rolling at the required roll angle.
This roll-rate results in a rolling moment L p p. Therefore, if L p
is negative the rolling subsidence mode is stable. This is usually
the case. However, if L p becomes positive, usually due to non-
linearities in the lift curve slopes at high roll rates, auto-rotational
rolling can occur. This is what happens when an aircraft spins.
68 How aircraft wobble: normal modes

The second root Λsm , which is much smaller than Λrs , corres-
ponds to the spiral mode of the aircraft. This is a combined yaw
and roll motion which is allowed to be unstable (i.e. negatively
damped) as long as it does not double amplitude in less than
twenty seconds, so that it can be controlled out. The spiral mode
normally happens so slowly that it can only be perceived visually
or using instruments, but not by the pilot’s inner ear, so that it can
be fatal in reduced visibility when no visual reference is available
for aircraft attitude.
The dynamics of the spiral mode are that if the aircraft rolls
slightly, it will start to sideslip, and the fin then tries to turn the air-
craft into the relative wind due to a yawing moment Nv v. However,
the rolling moment due to sideslip Lv v tries to roll the wings back
level. Depending on which of the effects prevails, the aircraft will
be spirally unstable or stable, as can be seen from the numerator
of (10.7).

Figure 10.4: Spiral mode

10.3. DIHEDRAL EFFECT AND WEATHERCOCK STABILITY 69

10.3 Dihedral effect and weathercock stability

The aerodynamic derivatives Lv and Nv establish whether an air-

craft is stable or unstable in rolling subsidence and Dutch roll. Lv Unstable Dutch roll
and Nv are known as the ‘dihedral effect’ and ‘weathercock stabil- − Lv
Increasing
ity’ respectively. The effect of the two aerodynamic derivatives on altitude
the lateral stability of the aircraft is shown in Figure 10.5.
Lv is known as the dihedral effect since the majority of the
rolling moment caused by sideslip comes from dihedral (on an
aircraft with unswept wings), as shown in Figure 10.6. Positive di-
All lateral modes stable
hedral combined with positive sideslip results in a negative rolling
moment (and hence negative Lv ).
Unstable spiral mode
Nv
Figure 10.5: Stability of lateral modes

Figure 10.6: Dihedral effect

Wing sweep has a large, negative, effect on Lv because of re-

duced or increased effective sweep for positive sideslip. This is
shown in Figure 10.7.

Figure 10.7: Wing sweep effects on

Lv : reduced effective sweep in the
direction of sideslip generates a higher
lift

Wing–fuselage interference effects give contributions to Lv be-

cause sideslip changes the effective incidence near the wing root.
These contributions are negative for high mounted wings and posit-
ive for low mounted wings, as shown in Figure 10.8.
A reasonable value of Lv may be achieved by using anhedral
with swept and high mounted wings (e.g. Harrier). Ground clear-
70 How aircraft wobble: normal modes

Figure 10.8: Wing-fuselage interference

effects on Lv

High wing

Low wing

ance issues may limit anhedral on low wing aircraft, resulting in an

unstable Dutch roll mode.
The aerodynamic derivative Nv is known as weathercock stabil-
ity since it is, effectively, the tendency of an aircraft to turn into the
wind. It is produced mainly by the side force of the fin in sideslip,
and should always be negative. However, as shown in Figure 10.5,
if Nv is too large the aircraft may be spirally unstable.
11
Flying aeroplanes

These notes are mainly intended to introduce the ideas you need
in order to design the mechanical elements of aeroplanes. As in
many engineering problems, the requirements are a translation into
numerical form of a set of human needs, in this case the require-
ment that a complex machine be controllable by a human being in
order to carry out some set of functions. This human element of the
design question is what makes the difference between adequate air-
craft and great ones. The field is normally called “handling proper-
ties” or “flying qualities” and is the area where mechanical design,
aerodynamics, physiology, psychology, and ergonomics intersect.
The idea that there is such a thing as flying qualities and that
these qualities can be specified numerically is not an obvious one,
and it is worth reading a history of how these qualities were first
recognized and defined and then stated as ranges of numerical
values.1 In short, over a period of about twenty five years after 1
Walter G. Vincenti. Establishment of
the First World War, test pilots and research engineers working design requirements: Flying-quality
specifications for American aircraft,
together developed an understanding of what it means to fly an 1918–1943. In What engineers know and
aeroplane in terms which allow for the discussion of the qualities of how they know it: Analytical studies from
aeronautical history. Johns Hopkins,
the aircraft, so that it becomes possible to properly design to make Baltimore, 1990
the aircraft useable by a human being. By 1949, one textbook was
dealing with “the comparatively new art of designing the airplane
[sic] for adequate flying qualities”: the existence of flying qualities
had been recognized and engineers were being taught to design
for them, rather than hoping the aircraft’s first pilot survived long
enough to report on the aircraft’s handling.
In Vincenti’s words:
Flying qualities comprise those qualities or characteristics of an
aircraft that govern the ease and precision with which a pilot is able
to perform the task of controlling the vehicle. Flying qualities are
thus a property of the aircraft, though their identification depends on
the perceptions of the pilot.

Vincenti gives examples of aircraft from the mid-thirties which

were considered quite adequate at the time, but would not now
be thought flyable such as the Martin M-130 flying boat which
was commercially successful but whose range was limited by the
length of time the pilots could withstand the control forces, and the
P-35, in service with the US Army Air Corps, but which Charles
72 Flying aeroplanes

Lindbergh found too sensitive to be easily flown.

The problem was stated by an officer at the US Navy’s Bureau of
Aeronautics quoted by Vincenti:
At present we simply specify that the airplane [sic] shall be perfect in
all respects and leave it up to the contractor to guess what we really
want in terms of degree of stability, controllability, maneuverability
[sic], control forces, etc. He [sic] does the best he [sic] can and then
starts building new tails, ailerons, etc. until we say we are satisfied.

The development of flying handling qualities definitions and

specifications resulted in the numerical requirements which are
stated in regulations and specifications, for example, the stick force
requirements of Table 5.1. For an aircraft to be flyable, there are
also requirements on stick force gradient (modified by weights,
page 45), linearity of response, and dynamic response, so that an
aircraft shifts predictably from one state of flight to another and
behaves predictably in any phase of operation.
11.1. THE COOPER–HARPER SCALE 73

11.1 The Cooper–Harper scale

The piloting qualities of an aircraft relate to the human experience

of an aircraft’s response to inputs. In order to design an aircraft,
we need some way of specifying which properties of the aircraft
form part of the flying qualities and of specifying an acceptable
range for those properties so that the handling of the aircraft can be
assessed at the design stage. In these notes, we have looked at such
numerical parameters as stick deflection and stick force per g, and
at how we might design for particular values. These are examples
of the general approach developed by Robert Gilruth of NACA. In
the words of two historians of aircraft stability,2 2
Malcolm J. Abzug and E. Eugene
Larrabee. Airplane stability and control:
Gilruth’s seminal achievement was to rationalize flying qualities by A history of the technologies that made
separating airplanes into satisfactory and unsatisfactory categories aviation possible. Cambridge University
for some characteristic, such as lateral control power, by pilot opin- Press, Cambridge, 2002
ion. He then identified some numerical parameter that could make
the separation. That is, for parameter values above some number,
all aircraft were satisfactory, and vice versa. The final step was to
develop simplified methods to evaluate this criterion parameter,
methods that could be applied in preliminary design.

Thus by 1943 there were numerical criteria relating the design of

an aeroplane to the qualities required by pilots, and methods for
applying these criteria at the design stage.
The other side of flying qualities specification is the require-
ment to measure, in some sense, the handling of the aircraft from
the point of view of a pilot. The standard method for this is the
Cooper–Harper scale, developed at NASA in the 1960s.3 3
George E. Cooper and Robert P.
Harper, Jr. The use of pilot rating in
the evaluation of airplane handling
qualities. Technical Report TN D-5153,
NASA, 1969
74 Flying aeroplanes

Aircraft Pilot Rating

characteristic demands

Excellent Pilot compensation not a factor for desired 1

Highly desirable performance

Good Pilot compensation not a factor for desired 2

Negligible deficiencies performance

Fair—some mildly Minimal pilot compensation required for 3

unpleasant deficiencies desired performance

Minor but Desired performance requires moderate 4

annoying deficiencies pilot compensation

Moderately objectionable Adequate performance requires considerable 5

deficiencies pilot compensation

Very objectionable but Adequate performance requires extensive 6

tolerable deficiencies pilot compensation

Major deficiencies Adequate performance not attainable with 7

maximum tolerable pilot compensation;
controllability not in question

Major deficiencies Considerable pilot compensation is required 8

for control

Major deficiencies Intense pilot compensation is required 9

to retain control

Major deficiencies Control will be lost during some portion 10

of required operation

Figure 11.1: The Cooper–Harper scale

The scale, Figure 11.1, assigns numerical scores to the aircraft for handling qualities

in particular phases of flight so that a test pilot can report the ad-
equacy of the aircraft for a given task in terms of the workload
which it imposes on a pilot. The scale can then be used in spe-
cifications and regulations, and can also be mapped to contours
of frequency and damping of the aircraft modes to link numerical
data, which can be estimated at the design stage, to pilot percep-
tion. An early example of this is the “bullseye” or “thumbprint”
plot of iso-opinion contours, shown in Figure 11.2, which has been
replotted from the published original.4 4
Malcolm J. Abzug and E. Eugene
Larrabee. Airplane stability and control:
A history of the technologies that made
aviation possible. Cambridge University
Press, Cambridge, 2002
11.1. THE COOPER–HARPER SCALE 75

Very highly responsive to

point of being too
1.4 sensitive; pilot-induced
Very responsive; oscillations; light forces;
sensitive; too quick; pilot likes stick motion
pilot-induced
1.2 oscillations; difficult to
trim; tracking very
difficult; force light;
stick motion small
1.0
Natural frequency f /Hz

Slow response;
Proposed have to reverse
boundary control to stop;
difficult to
0.8 manoeuvre;
sluggish; dead;
Little delayed or force a little high;
slow initially, more motion
then overshoots; than pilot likes
0.6 trim and track
little difficult;
force little high

0.4

Close to unflyable; safe in straight

and level or very small
0.2 manoeuvres; aeroplane tends to go
unstable hands off; feels near
neutral point

0
0.1 0.2 0.3 0.4 0.6 0.8 1.0 1.5 2.0
Damping ratio ζ
Figure 11.2: Iso-opinion contours for
the short-period oscillation from tests
The curves on the plot are contours of “constant opinion” (of on a variable stability F-94F, replotted
flying qualities) as rated by pilots on a variable stability F-94F. By from data in Abzug and Larabbee
varying the stability of the aircraft, the contours could be plotted
against the properties (natural frequency and damping ratio) of
the short period oscillation. This gives designers an indication of a
range of properties, within the heavy boundary, which give good
handling for a particular type of aircraft, allowing them to account
for flying properties at the design stage, and to see how changing
those properties will affect the pilot’s perception of the aircraft’s
qualities.
Modern approaches are more sophisticated and incorporate
models of human psychology and physiology but still work on the
principle of linking predicted dynamic properties of an aircraft to
a human assessment of the more intangible qualities of the design.
In design, much of the assessment of flying qualities is now carried
out by having pilots “fly” simulations of the aircraft so that the
handling properties can be tuned before moving into production.
12
Questions

These questions are intended to give you practice in applying the

methods and theory of the text, and to encourage you to think
about how what you are learning fits into the general context of
aircraft design, and engineering more generally. You should work
through the questions in this section in order, but you can take the
questions in §12.1 in any order you please, depending on when you
get around to watching the films recommended.

1. For the two situations shown in Figure Q1 cal- low, calculate the value of CLT required for trim
culate the values of LW and L T required to give at 50kt EAS with a pilot weighing 0.75kN. The
both a total lift equal to the aircraft weight and empty weight equipped is 2.5kN, with c.g. on
give zero net moment about the aircraft c.g. the mean chord 0.45c aft of the leading edge of
[LW = 99.3kN, L T = 0.7kN, LW = 95.3kN, c. The pilot c.g. is assumed to be 0.8m ahead of
L T = 4.7kN] the leading edge of c.
LT [CLT = −0.552]
LW l =15m
x= S = 28m2 ST = 1.4m2 c = 1.15m
0.3m l = 5.35m h0 = 0.25 C M0 = −0.11
3. Distinguish between stability and trim. Show
that for an aircraft to be both stable and able to
M0 =40kNm trim at positive lift coefficient the overall pitch-
W=100kN
LT ing moment about the centre of gravity must be
LW l =15m positive at CL = 0 in that configuration.
x= 4. From first principles, show that the portion of
0.3m
the total lift coefficient (CL ) provided by the
wing-body-nacelle (WBN) group of a conven-
tional aircraft is given by:
M0 =40kNm W=100kN
 
c c
Figure Q1: Aircraft with different centres of CLWBN = CL 1 + (h0 − h) − C M0 .
l l
gravity
2. Draw the system of forces and moments acting If the aircraft stalls when CLWBN reaches its max-
on a conventional aeroplane in steady straight imum value, (CLWBN )max say, then obtain an
and level flight. expression relating the stalling speed to the c.g.
position at any one given weight.
Show that the pitching moment about the centre
of gravity is given by Hence calculate the c.g. shift that would in-
crease the stalling speed by 1% if c = 5.6m,
C M = C M0 − (h0 − h)CL − VCLT . l = 15.5m and (h0 − h) = 0.05.
For the sailplane whose details are given be- [∆h = −0.0566, ∆hc = −0.317m]
78 Questions

5. Consider the two situations shown in Fig- 6. The data shown below apply to an aircraft in
ure Q5. steady level flight at 200kt EAS. Calculate the
CL elevator angle required for longitudinal trim.
Also obtain the stick-fixed neutral point and
hn c̄
static margin.
W = 30kN S = 23m2 ST = 3.5m2
c = 1.96m l = 5.5m
(CM0 )
hc̄ h0 = 0.25 c.g. is 0.61m aft of datum
W
CL C M0 = -0.036 ηT = -1.5◦ e = 0.48α
a = 4.58 a1 = 3.15 a2 = 1.55
hn c̄
[η = -1.658◦ , hn =0.4027, Kn = 0.0915]

7. The centre of gravity range for an aircraft is

found by considering that the
(C )
h p c̄ (CM p ) M0 (a) aft c.g. limit (haft ) is determined by the min-
Figure Q5: Full-scale and model aircraft. C M p is imum stability condition (minimum Kn );
measured by the balance, which restrains the (b) forward c.g. limit (hfwd ) is determined by
model in pitch. the maximum elevator angle to trim (while
retaining enough elevator movement for
In (a) the full scale aircraft is in steady free
manoeuvre).
flight with values CL , h, η for the lift coefficient,
By considering the static forces and moments on
c.g. position and elevator angle respectively.
an aircraft in symmetric flight, find an expres-
In (b), a model of the same aircraft is suspended
sion for the static margin stick-fixed, Kn , and
from a wind tunnel balance at the same CL and
show that:
elevator setting as in (a). The balance measure-  
ment gives the pitching moment coefficient C MP dη a1 ∂e
Kn = −Va2 = ( h0 − h ) + V 1− .
dCL a ∂α
about the balance pivot axis which is located at
h p with respect to the same datum line as h. An aircraft has the following values of the aero-
(a) Write down the moment equations for situ- dynamic coefficients:
ations (a) and (b), and hence derive the rela- h0 =0.25, a=3.5, a1 =3.0, a2 =1.5, ∂e/∂α=0.4.
tionship between the balance reading C MP ,
Find the relationship between the c.g. position
equivalent to the steady free flight conditions,
and the tail volume ratio:
and interrelating h p , h and CL .
(b) An aircraft model is found to have a zero- (a) for a static margin of 0.05 (haft );
lift pitching moment coefficient of 0.027 for (b) for the change in elevator angle to trim to
a particular elevator angle. The pitching mo- be 10◦ for a change in CL of 1.0 (hfwd ).
ment is measured about the wind-tunnel axis Hence find the minimum tail volume ratio such
of rotation P and has a slope: that with a c.g. shift of 0.15c the change in el-
evator angle to trim is not more than 10◦ for
dC M p
= 0.15; lift curve slope a = 5.851. a change of CL of 1.0 and the static margin is
dα
never less than 0.05.
Determine the position of the c.g. of the
full-scale aircraft relative to P if a stick-fixed [V = 0.764]
margin of 0.11 is required (c = 3.96m). 8. A transport aircraft with conventional tail is
If the wing loading is 2.25kN/m2 in steady to have zero elevator angle in cruising flight
level flight with the above elevator angle, at 560km/h EAS (mass 100,000kg), with the
what is the airspeed if the air density c.g. in the mid position. The landing approach,
is 1.030kg/m3 . out of ground effect, is made with flaps down
[0.537m forward of P, 133.3m/s TAS.] at 210km/h (mass 90,000kg), and the maximum
 79

elevator movement permitted for trimming is Weight = 850kN Speed = 70m/s EAS
η = ±10◦ . Using the data below, calculate the Wing area S = 358m2 ( h0 − h ) = 0.15
minimum tailplane area suitable for this air- C M0 = +0.02 ∂C M0 /∂η = -0.45
craft, and the tailplane setting ηT relative to the a1 = 4.0 a2 = 0.95
flaps-up wing zero lift line. b1 = -0.7 b2 = -1.05

Assume that each elevon contributes equally to

Minimum Kn = 0.05 c.g. range ∆h = 0.50
a2 and to ∂C M0 /∂η.
h0 = 0.075 S = 232m2
c = 4.72m l = 19.5m [ηfailed = −9.61◦ , ηoperating = −13.15◦ ]
a = 5.7 a1 = 2.7
11. A conventional aircraft flying at low speed has
a2 = 2.1
a flexible rear fuselage such that the tailplane
C M0 = −0.14 e = 0.16α.
setting relative to the wing zero lift line is dir-
ectly proportional to the tail load. Prove that the
The change in C M0 at landing flap setting
reduction in stick fixed static margin compared
∆C M0 = −0.10. Note that the wing zero lift
with that of the rigid aircraft is given by:
incidence angle changes by 10◦ when the flaps
are lowered to the landing setting. rigid
∆Kn = Kn − Knflexible ,
" #
[ST = 68.5m2 , ηT = −3.92◦ ]

a1 ∂e 1
=V 1− 1−
a ∂α 1 + 12 ρV 2 ST a1 f
9. The static margin, stick-fixed may be obtained
in practice from flight tests in which elevator For the human-powered aircraft having the
angles to trim are found at certain speeds. In characteristics given below, find the fuselage
practice, the aeroplane is trimmed at a series of flexibility f (degrees deflection per Newton)
speeds by adjusting the tab setting, and both that reduces the stick-fixed static margin by 0.05
the elevator angle and tab angle are observed. compared to the rigid case when flying at a
Since the theory which relates the stick-fixed speed of 9.2m/s at sea level.
static margin to the elevator angles to trim im- S = 28m2 ST = 1.4m2 l = 5.34m c = 1.14m
plicitly assumes a constant tab angle, show that a = 6.0 a1 = 4.5 e = 0.20α.
a correction must be applied to elevator angles [ f = 0.1◦ /N]
obtained in this way such that
12. The control column of a low-speed aeroplane
a3 is connected to the elevator by an arrangement
η corrected = η + β
a2 of cables which stretches when a stick force is
applied. The stiffness of the circuit is given by
where η and β are the observed elevator and tab dHE /dη = ENm/rad where HE is the hinge
angles to trim at a given speed. Suggest how moment and η is the elevator deflection, the
you would determine a3 /a2 in flight. stick being held fixed.
10. A tailless aircraft is controlled in pitch by six Show that the stick-fixed c.g. margin (as op-
elevons. Each elevon is actuated by an inde- posed to the “elevator fixed” c.g. margin) is
pendent power control unit. These units are so given by:
designed that if a failure occurs the affected el-   
a ∂e a2 b1 1
evon is able to move until its hinge moment is K n = ( h0 − h ) + V 1 1− 1−
a ∂α a1 b2 1 − λ
zero.
Assuming the failure of one such unit, calculate where
CL SE
the elevon angles that will give longitudinal λ= .
b2 Sη cη W
trim of the aircraft whose details are given be-
low: It should be assumed that the aircraft is initially
in trim with the tab angle adjusted to give zero
stick force.
80 Questions

Show how this margin is related to the stick 16. An aircraft of conventional layout is controlled
fixed and stick free c.g. margins of a rigid aero- in pitch by an all-moving tailplane, having no
plane. What practical use might you make of separate elevators (see Figure and table). Show
this information? that the tail angle per g is given by
∆ηT C Hm
13. Which conditions define the stick-fixed and =− L
stick-free manoeuvre points of an aircraft? n Va1
From first principles, stating your assump- where the symbols have their usual meanings.
tions, derive an expression for the stick fixed Hence calculate the tail angle, tail load and
maneouvre point of a low speed aircraft of ca- pivot moment when the aircraft is flying
nard layout. Show whether this is forward or aft at 440kt EAS in an 8g (n = 7) pullout at a height
of the corresponding neutral point and compare where the relative density of the air σ = 0.74.
your expression with that for a conventional Comment on your results.
aircraft.
W=175kN V=440kt EAS S=33.2m2 ST =19.1m2
14. Define the maneouvre point stick-free for a
l=5.25m c = 2.39m a=3.8 a1 =2.7
conventional aircraft. How does it differ from
C M0 = +0.03 ∂e/∂α=0.38 h0 = 0.17 h = 0.50
the corresponding neutral point?
σ = 0.74
Find the minimum stick force per g at sea level
for the light aircraft whose details are given LWBN LT
below. Comment on your result and find the l
c.g. position required to give 22N/g. Suggest 0.144m
alternative means for increasing the existing
value. Pivot point

W = 2.7kN S = 7.6m2 l = 2.9m [∆ηT = −4.72◦ , ηT = −4.85◦ , L T = 224.6kN,

h0 = 0.238 c = 1.2m V = 0.34 MP = −32.34kNm, CLT = 0.3739]
e = 0.385α a = 3.865 a1 = 2.73 17. For a conventional aircraft show that if the
a2 = 2.16 b1 = −0.282 b2 = −0.536 tab setting remains unaltered, the change of
elevator hinge moment coefficient-to-trim ∆CH
The permitted c.g. range is from 0.22c to 0.28c.
between two lift coefficients is given by
The stick force per g is given by
b2
∆CH = − ∆CL Kn0 .
Pe W b2 0 a2 V
Q= = − m e Sη c η H ,
n S a2 V m The aircraft described in the table below is mak-
0
= 83.2Hm N/g for this aircraft. ing a zero stick force trimmed landing approach
at 155kt EAS. Calculate the value to which the
[Q = 5.8N/g, for Q = 22N/g, h = 0.0853] speed may be reduced while keeping the stick
15. The table below shows data for a tailless air- force within 150N without altering the trim tab
craft. When it performs a steady pullout at setting, indicating clearly whether this is push
A N = 2.5 (n = 1.5) at 250kt EAS at a height or a pull force.
where the air density ρ = 1.150kg/m3 , the W =785kN h = 0.26
change of elevator setting compared with steady S = 223m2 c = 5.68m
level flight under the same conditions is 3.20◦ . ST = 46.5m2 l = 15.66m
Calculate mq if the static margin is known to Sη = 11.2m2 cη = 0.908m
be 0.05. h0 = 0.16 C M0 = -0.06 e = 0.38α
a = 4.5 a1 = 2.75 a2 = 1.16
W = 160kN S = 50m2 c0 = 10m b0 =0 b1 = -0.133 b2 = -0.16
∂C M0 /∂η = −0.5 Kn = 0.05
The stick–elevator gearing ratio me = 1.0m/rad.
[mq = −0.264] [118 kt, pull force]
 81

18. Using the approach of §4.2, and the results 21. The aircraft described in the Figure and
of §5.1, derive a formula for dPe /dV, the gradi- table below is to have its capacity increased
ent of stick force with flight speed. What does by lengthening the cylindrical portion of the
this tell you about the handling qualities of an fuselage by 6m. The centre section (including
aircraft? the wings), the nose and the tail portions are to
19. What are stick-fixed and stick-free manoeuvre remain unaltered.
points and what is the significance of stick force It is assumed that the c.g. position will be ad-
per g. justed to remain unchanged with respect to
Using the data of question 17, calculate the the centre section unit and that, for the lengths
change of elevator angle required to pull 0.5g considered, ∂e/∂α is constant.
flying at 350kt EAS at an altitude where the Calculate how the additional fuselage length is
relative density σ = 0.374. to be inserted ahead of and behind the centre
Explain in physical terms why this change of section, if the low speed stick-fixed static mar-
elevator angle would be greater at a lower alti- gin is to be unaltered. The movement of the
tude when flying at the same lift coefficient. aerodynamic centre of the aircraft less tail is as-
sumed to be affected only by the change of nose
[∆η = −1.005◦ ] length ∆l N and is given by
20. The tailless aircraft shown in the figure has ∆l N
been fitted with a small retractable foreplane. ∆h0 = −0.037 .
c
At low speeds this foreplane is extended and,
operating in a semi-stalled condition at constant If a variant of the aircraft retains the original
setting, it generates a constant lift coefficient fuselage, but has its wing tips extended, how
CLF = 1.2 (based on S F ). Use of the foreplane could the longitudinal static stability be af-
enables the aircraft to take off at a higher weight fected?
than the original aircraft without the foreplane.
Calculate the increment in take-off weight that
may be achieved when using the foreplane,
by considering the trimmed lift at 200kt EAS,
if the incidence is restricted to 12◦ by ground
S = 223m2 c = 5.6m
clearance problems, using the data in the table.
ST = 46m2 l=15.5m
Calculate the elevon angles to trim of both ver- h0 = 0.25 h = 0.20
sions of the aircraft. Comment on your results. e = 0.4α C M0 =-0.06
[With foreplane: η = −0.5◦ ; L = 1842kN; a = 4.5 a1 = 2.75
without foreplane: η = −5.8◦ ; L = 1557kN]
[4.1m ahead of wing, 1.9m aft]
c0
22. (a) The 1903 Wright Flyer was a canard con-
lF figuration of conventional layout, summar-
ized in the table below. Calculate the stick-
fixed neutral point, assuming that the wing
and canard have approximately equal lift
curve slope, and comment on your answer.
hc0 (b) The 1903 Flyer was stabilized in pitch by
h0 c0 the addition of ballast to shift the centre
of gravity forward. If 30% of the aircraft
S = 438m2 S F = 9.4m2 C M0 = +0.002 gross weight can be carried as ballast, where
∂C M0 /∂η = -0.25 a1 = 3.0 a2 = 0.80 should it be placed to move the centre of
h0 = 0.61 c0 =27.4m l F = 13.26m gravity to the wing leading edge. What effect
hc0 = 15.34m would this have on the aircraft performance?
82 Questions

h0 ≈ 0 V = 0.134 C M0 = −0.141 12.1 Aircraft in the movies

h = 0.3c W ≈ 340kg
The 1903 Wright Flyer. The datum is the wing 1. Watch The First Of The Few.
leading edge.
(a) What is the most unrealistic scene in the
23. The table below contains flight test data for film?
the X-15 spaceplane. Calculate the static margin (b) Who was R. J. Mitchell?
stick-fixed and estimate the zero-lift pitching
(c) What kind of men are the pilots in the film?
moment. Estimate the dimensional and non-
dimensional phugoid mode and SPO frequen- (d) What is expected of the female characters in
cies. the film?
S = 18.58m2 s = 6.82m (e) What impression did you get of Britain, and
c = 3.13m h = 0.22 the Royal Air Force, from the film?
m = 7056kg B = 10700kgm2
V = 331kt EAS a = 3.5/rad 2. Watch The Sound Barrier.
∂C M /∂α = −0.8/rad Zu = −332Ns/m (a) What is the most unrealistic scene in the
Mw = −40.7Ns Zw = −14300Ns/m film?
Mq = −158600Nms
(b) Who was Geoffrey de Havilland Jr?
[Ωph = 0.0946 (0.052rad/s); Ωspo = 10.074
(c) What kind of men are the pilots in the film?
(5.5395rad/s)]
24. NASA CR-2144, Aircraft Handling Qualit- (d) What is expected of the female characters in
ies Data, contains stability information for ten the film?
aircraft. For the Boeing 747: (e) What impression did you get of Britain, and
(a) calculate the static margin stick-fixed; its aircraft industry, from the film?
(b) estimate the zero-lift pitching moment;
3. Watch The Right Stuff. Even better, then read
(c) estimate the phugoid, SPO and Dutch roll
the book.
periods;
(d) estimate the time constants for rolling sub- (a) What is the most unrealistic scene in the
sidence and the spiral mode. film?
25. How would you modify the pitching moment (b) Who was Pancho Barnes?
equation (2.3) to give a first approximation to
(c) What kind of men are the pilots in the film?
near- and post-stall behaviour?
(d) What is expected of the female characters in
26. How would you modify the pitching moment
the film?
equation (2.3) to model the behaviour of an air-
craft on the ground, in particular at the point of (e) Are modern astronauts chosen to have the
take-off rotation? same characteristics as the pilots in the film?
27. Analyze the stability and control characterist- (f) What impression did you get of the US, and
ics of a Spitfire. its space programme, from the film?
13
Exam questions

These are sample questions taken from previous years’ exam pa-
pers, intended to show the style of question and give an idea of
how to approach the exam. You should also look at the papers
themselves to see how they are structured, rather than rely only on
the questions here.

1. (a) Show that the static margin stick-fixed of an aircraft is re-

lated to its control characteristics via
 
dη a1 ∂e
Kn = −Va2 = ( h0 − h ) + V 1− .
dCL a ∂α

[13 marks]
(b) Data are given in Table Q1 for a transport aircraft with
conventional tail. The nominal cruise speed of the aircraft
is 300kt. Estimate the minimum tail volume coefficient re-
quired if the pilot is to be able to change the aircraft speed by
up to 50kt without using the trim tab and with a change in
elevator deflection of no more than 5◦ .
[12 marks]
(c) Discuss qualitatively the aircraft response if the pilot should
make an abrupt 5◦ change in elevator deflection. How will
this response vary with centre of gravity position?
[8 marks]

Minimum Kn = 0.05 c.g. range ∆h = 0.40

h0 = 0.075 e = 0.16α
a = 5.7 a1 = 2.7 a2 = 2.1
m = 100 × 103 kg S = 232m2
Table Q1
84 Exam questions

2. (a) For a conventional aircraft show that if the tab setting

remains unaltered, the change of elevator hinge moment
coefficient-to-trim ∆CH between two lift coefficients is given
by
b2
∆CH = − ∆CL Kn0 .
a2 V
[12 marks]
(b) The aircraft described in the Table Q2 is making a zero stick
force trimmed landing approach at 155kt EAS with flaps up.
Calculate the value to which the speed may be reduced while
keeping the stick force within 150N without altering the trim
tab setting, indicating clearly whether this is push or a pull
force. The stick–elevator gearing ratio me = 1.0m/rad.
[8 marks]
(c) To a good approximation, flap deployment changes the lift
curve slope a by a factor 1.2 × (1 − d sin2 δ f ) where δ f is the
flap deflection and d = 0.25 for this aircraft. Estimate the stick
force required to make the same change in aircraft speed as in
part b with 40◦ flap deflection. Comment on your answer.
[6 marks]
(d) In view of your answer to part c, would you recommend any
changes to the operating procedure of the aircraft with regard
to control?
[7 marks]

W = 785kN h = 0.26
S = 223m2 c = 5.68m
ST = 46.5m2 l = 15.66m
Sη = 11.2m2 cη = 0.908m
h0 = 0.16 C M0 = -0.06 e = 0.38α
a = 4.5 a1 = 2.75 a2 = 1.16
b0 =0 b1 = -0.133 b2 = -0.16
Table Q2
 85

3. (a) Show from first principles that the elevator deflection per
gee for a conventional aircraft is given by

Va2 ∆η
Hm = − ,
CL n

where symbols have their usual meanings.

[12 marks]
(b) The Junkers 87 “Stuka”, Figure Q3, was a German dive
bomber of the Second World War. Its mode of attack was
to enter a vertical dive at an airspeed of 500km/h, drop its
bomb, and then enter a 6g pull out (n = 5) from low altitude.
Using the estimated data in Table Q2, estimate the centre of
gravity position which would be required to perform the pull
out if the aircraft mass after dropping its bomb is W = 3800kg
and the change in elevator deflection for the manoeuvre is 15◦ .
Comment on your answer.
[14 marks]
(c) Estimate the minimum height where the pull out could be
initiated.
[7 marks]

Figure Q3: Ju-87 Stuka (Kaboldy, CC BY-SA 3.0, Wikimedia)

e = 0.2α S = 32m2 c = 2.1m
a = 4.5 a1 = 4 a2 = 1.1
l = 5.5m ST = 4m2 h0 = 0.25
Table Q3
86 Exam questions

4. (a) Show that the pitch angle oscillation of an aircraft is gov-

erned by

ρV 2 Sca
Bα̈ + Kn α = 0,
2
where B is the pitching moment of inertia about the centre of
gravity and other symbols have their usual meanings. From
the equation determine the frequency of short period oscilla-
tion (SPO).
[12 marks]
(b) The Hawker Typhoon, Figure Q4, was a successful Second
World War ground attack aircraft. Using the approximate data
given in Table Q4, estimate the SPO frequency and period for
a Typhoon flying at 430km/h with static margin Kn = 0.05.
Comment on your answer.
[6 marks]
(c) The Hispano–Suiza 20mm cannon fitted in the Typhoon had
a muzzle exit velocity of 850m/s. The muzzles lay 1.8m ahead
of the centre of gravity. If the short period oscillation had an
amplitude of 5◦ , estimate the angular deflection of the traject-
ory of the round caused by the SPO and the corresponding
error in the trajectory for a target at a distance of 500m. Com-
ment on your answer.
[10 marks]
(d) What implications does SPO have for “pointing accuracy” in
aircraft?
[5 marks]

Figure Q4: Hawker Typhoon (Wikipedia Commons)

S = 29.6m2 c = 2.3m B = 6000kgm2 a = 4.5

Table Q4
 87

5. (a) Show from first principles that the stick-fixed static margin
of a canard aircraft is:
l F S F a1
K n = h0 − h −
c S a
where symbols have their usual meanings.
[15 marks]
(b) One way to model the post-stall behaviour of a lifting sur-
face is to treat it as having a negative lift curve slope. Table Q5
gives approximate data for a hypothetical small canard air-
craft. Calculate the static margin stick-fixed before and after
stall of the foreplane, assuming that post-stall behaviour can
be modelled by changing the sign of a1 . Comment on your
answer, with particular reference to the handling qualities of
the aircraft.
[8 marks]
(c) Canard aircraft can be prone to dynamic stall, where the
foreplane incidence is increased by the pitch rate of the air-
craft. How can this situation be avoided, through aircraft
design or through centre-of-gravity restrictions? What effects
will possible solutions have on the usefulness of the aircraft?
[10 marks]

S=5m2 S F = 0.65m2 c = 0.6m l F = 4m

a = 4.7 a1 = 2.3 h0 c = 4.5m hc = 4.2m
Table Q5
88 Exam questions

6. (a) The control column of an aeroplane is connected to the elev-

ator by a flexible cable whose stiffness is given by dHE /dη =
E Nm/rad where HE is the hinge moment and η is the elev-
ator deflection, the stick being held fixed.
Show that the stick-fixed c.g. margin (as opposed to the “elev-
ator fixed” c.g. margin) is given by:
  
a ∂e a2 b1 1
K n = ( h0 − h ) + V 1 1 − 1−
a ∂α a1 b2 1 − λ

where
CL SE
λ= .
b2 Sη cη W
It should be assumed that the aircraft is initially in trim with
the tab angle adjusted to give zero stick force.
Show how this margin is related to the stick fixed and stick
free c.g. margins of a rigid aeroplane.
[15 marks]
(b) By considering the variation of Kn with flight speed V, show
that for a divergent elevator (b1 < 0), the c.g. margin reduces
with increased speed.
[12 marks]
(c) How could the aircraft design be modified to mitigate the
change in margin with speed?
[6 marks]
 89

7. (a) Show that the pitch angle oscillation of an aircraft is gov-

erned by the equation

ρV 2 Sca
I α̈ + Kn α = 0,
2
where I is the pitching moment of inertia about the centre of
gravity and other symbols have their usual meanings. From
the equation find the frequency of short period oscillation.
[12 marks]
(b) Table Q7 contains basic data for a small electrically-powered
UAV used for mapping and agricultural observation. For ad-
equate image quality, it has been found that the aircraft oscil-
lation frequency must be limited so that the distance scanned
by the camera as it shoots an image is not more than 0.5m. If
the camera exposure time is 1/1000s, estimate the speed at
which the image frame sweeps the ground, and if the UAV
is to operate at an altitude of 80m, the maximum acceptable
pitch oscillation frequency. From this frequency, estimate the
required static margin stick-fixed for the UAV. You may neg-
lect the effect of flight speed on the swept area of the image.

V = 9m/s S = 0.16m2 c = 0.2m

a = 4.2 m = 0.5kg I = 0.005kgm2
Table Q7

[8 marks]
(c) If the static margin stick-fixed Kn < 0, the aircraft is statically
unstable. Qualitatively, what would determine the degree to
which it could be made statically unstable and still be control-
lable?
[8 marks]
(d) What are the implications for aircraft handling of the result
that SPO frequency is proportional to the square root of static
margin stick-fixed?
[5 marks]
90 Exam questions

8. The C130 Hercules, Figure Q8, is a military aircraft used for

transport and parachute dropping of vehicles and other supplies.
Estimated aircraft parameters are given in Table Q8. The cargo
hold has a capacity of 20,000kg, and extends from 5.3m to 17.6m
from the datum, which is the aircraft nose.

(a) Show that the static margin stick fixed is given by:
 
dη a1 ∂e
Kn = −Va2 = ( h0 − h ) + V 1− .
dCL a ∂α

[10 marks]
(b) A Hercules normally carries out a drop at a flight speed
of 125kt EAS. Size the tailplane such that the minimum static
margin, stick fixed, is never less than 0.05, and so that the
change in elevator angle to trim during drop of a full payload
is no more than 15◦ . Assume ∆h = 0.5.
[10 marks]
(c) If the centre of gravity, with payload, is at the aft limit, can
the drop be performed safely?
[7 marks]
(d) What dynamic behaviour would you expect of the aircraft
immediately after the drop has been carried out?
[6 marks]

Figure Q8: C130 Hercules (Wikimedia Commons)

W = 65 × 103 kg (including payload of 20 × 103 kg) S = 162.1m2 e = 0.2α

c = 4m h0 c = 13.6m l = 12.8m a = 5.8 a1 = 4.2 a2 = 2.0
Table Q8: C130 Hercules data
 91

9. The first variable geometry aircraft, capable of changing its

wing sweep in flight, was the Bell X-5, shown in Figure Q9 with
estimated parameters given in Table Q9. Lengths are measured
from the aircraft nose.

Figure Q9: Bell X-5 variable geometry aircraft (Wikimedia

Commons).
Aircraft length l = 10.1m Chord c0 = 2.5m
Wing area S = 16.26m2 Mass m = 4500kg
C M0 = +0.05 ∂e/∂α = 0.4
Unswept Swept
h0 c0 = 3.9m h0 c0 = 5.2m
U = 150kt EAS U = 630kt EAS
a = 4.8 a=4
a1 = 3.0 a1 = 2.5
Table Q9: Bell X-5 parameters.

(a) Given that the centre of gravity range is from 3m to 4m aft

of the aircraft nose, find the tail area and setting which give
a static margin stick fixed Kn = 0.05 in the unswept case and
trim with zero elevator deflection in the swept case. Assume
that the tailplane lift acts 10.0m from the aircraft nose.
[25 marks]
(b) What are the control issues affecting aircraft which operate
at high speed and how can the problems be alleviated?
[8 marks]
92 Exam questions

10. (a) In the 1952 David Lean film The Sound Barrier, a test pilot
experiences “control reversal” at a flight speed near the speed
of sound. Describe the control effects of high speed flight,
with reference to the pilot’s perception of handling, and state
what phenomenon actually occurs as an aircraft approaches
and exceeds the speed of sound.
[10 marks]
(b) In the 1983 film The Right Stuff, based on Tom Wolfe’s 1979
book, Chuck Yeager is shown making the first supersonic
flight in the X-1. Which principal design feature made the
aircraft controllable at high speed and why was it necessary?
[10 marks]
(c) The 1942 film The First Of The Few is a fictionalized account
of the development of the Spitfire by R. J. Mitchell. It includes
a scene of the first flight of the aircraft in which it is shown
flying high-g manoeuvres. What are the design considerations
relevant to the control of high-performance aerobatic aircraft
and how are the flying qualities of such aircraft assessed?
[14 marks]
 93

11. The Supermarine Spitfire is a very well-known aeroplane which

saw extensive service in the 1940s. Table Q11 gives summary
data for the Mark 1 variant. The value for ` T is the tailplane
distance from the rear centre of gravity limit, haft . The datum is
the leading edge of the mean aerodynamic chord.

(a) Mark 1 Spitfire models were subjected to a series of wind

tunnel tests in 1945. Measurements were taken of the pitching
moment about haft on a model with no tail and it was found
that C M0 ≈ −0.05 and ∂C M /∂CL ≈ 0.149. Estimate the pos-
ition of the wing-body-nacelle neutral point h0 , the tailplane
volume coefficient, and the resulting static margin stick-fixed
at the aft centre of gravity limit. Comment on your answer.
[15 marks]
(b) With the centre of gravity at the aft limit, estimate the man-
oeuvre margin stick-fixed and the elevator deflection required
to pull 4g (n = 4) at a speed of 200kt. You may use the result

∆η C
= − L Hm .
n Va2
Comment on your answer.
[10 marks]
(c) What considerations do you think should be taken into ac-
count in designing the control system of the Spitfire, with
regard to handling qualities and response to pilot input?
[8 marks]

S = 22.482m2 ST = 3.135m2
c = 1.99m `T = 5.462m haft = 0.340c
a = 4.6 a1 = 2.83 a2 = 2.15
e = 0.037 + 0.3α
W = 3000kg
Table Q11
94 Exam questions

12. Figure Q12 shows the notation for motion of an aircraft at con-
stant incidence and lift coefficient, acted upon by lift L perpen-
dicular to the trajectory, and by gravity acting vertically down-
wards.

(a) Show that the trajectory is governed by the equations

1 z  z 1/2
cos θ = +C 1 ,
3 z1 z
z1 1 C  z1 3/2
= − ,
R 3 2 z
where R is the radius of curvature of the trajectory, and z is
shown on Figure Q12. [12 marks]
(b) Depending on the value of the constant of integration C, the
trajectory can have four qualitatively different behaviours.
Describe these cases, and sketch the trajectories to which they
correspond.
[12 marks]
(c) Describe qualitatively the phugoid response of real aircraft,
with reference to dynamic stability, damping, and stick free
effects.
[9 marks]

L
V
θ

W
Figure Q12: Motion of a body under lift and gravity
A
Finding out more

A.1 Further reading

• Abzug, Malcolm J. & Larrabee, E. Eugene, Airplane stability and

control: A history of the technologies that made aviation possible, Cam-
bridge University Press, 2002.

• Brown, Eric, Wings on My Sleeve: The World’s Greatest Test Pilot

Tells his Story, W & N, 2007. Brown moved from naval combat
flying to test flying during the Second World War and still holds
the record for the greatest number of aircraft types flown by
one pilot. You can hear him interviewed on Desert Island Discs,
http://www.bbc.co.uk/programmes/b04nvgq1.

• Culick, F. E. C., ‘The Wright brothers: First aeronautical engin-

eers and test pilots’, AIAA Journal, 41(6):985–1006, 2003.

• Hamilton-Paterson, James, Empire of the Clouds: When Britain’s

Aircraft Ruled the World, Faber & Faber, 2010.

• Hamilton-Paterson, James, Marked for Death: The First War in

the Air, Pegasus Books, 2016, the human story of how aviation
developed during the First World War.

• Langewiesche, William, ‘The human factor’, Vanity Fair, Octo-

ber 2014, https://bit.ly/2FyE5eB, an account of the crash of
AF447 in 2009, with a lot of detail about the human factors en-
gineering of cockpits.

• Markham, Beryl, West With The Night, the memoirs of one of

the first female pilots, who made the first non-stop flight from
England to North America.

• Mercurio, Jed, Ascent, Vintage, 2008, a novel based on a fictitious

Soviet lunar mission, with excellent accounts of flying and a
physically correct space emergency. The author is an ex-RAF
medical doctor who trained as a pilot, which gives him great
insight into flying and people who fly.

• de Saint-Exupéry, Antoine, Courrier sud/Southern Mail, Vol de

nuit/Night Flight, Pilote de guerre/Flight to Arras, classics of avi-
ation by one of the great pilot-authors.
96 Finding out more

• Salter, James, The Hunters, Penguin Classics, 2007, a novel by one

of America’s finest writers based on his time as a fast jet pilot in
the Korean War.

• United States Air Force Test Pilot School, Flying Qualities Text-
book, Volume II, Part 1, AF-TPS-CUR-86-02, April 1986, a big
book (more than 700 pages) but the chapters on longitudinal sta-
bility and flight testing are manageable and well worth reading.

• Vanhoenacker, Mark, Skyfaring: A Journey With a Pilot, Vin-

tage, 2015. A commercial pilot on the experience and mechanics
of flying. Already a classic, which stands comparison with the
finest writing on aviation.

• Vincenti, Walter G., What Engineers Know and How They Know
It, Johns Hopkins University Press, 1990. This is a collection
of studies looking at how aeronautical engineers acquired and
acquire knowledge of the systems they work on. You should
read all of it, but chapter 3 on how flying quality specifications
for aircraft evolved up to 1945 is especially relevant.

• Wolfe, Tom, The Right Stuff, regularly republished, an account of

the early days of the American space programme: one of the best
books written on test flying and the people who do it.

• ‘What really happened aboard Air France 447’, Popular Mechan-

ics, December 2011, http://bit.ly/1PtPlde, including a tran-
script from the cockpit voice recorder.

A.2 Data sources

• Heffley, R. K. & Jewell, W. F., Aircraft handling qualities data,

NASA CR-2144, 1972, is a collection of flight test data, includ-
ing aerodynamic derivatives, for a ten aircraft: NT-33A, F-104A,
F-4C, X-15, HL-10, Jetstar, CV-880M, B-747, C-5A, and XB-70A.

• Mair, W. A., High-speed wind-tunnel tests on models of four single-

engined fighters (Spitfire, Spiteful, Attacker and Mustang), ARC R&M
2535, 1951, is a collection of reports giving detailed aerodynamic
data, including tailplane information, for four single-engined
aircraft of the 1940s.

A.3 Accident reports

The reports of accident investigations are often used in aeronautical

engineering to help understand how things can go wrong and how
we can design and operate aircraft safely and reliably.

• The Aviation Safety Network has a database of accidents since 1919,

which you can search using various criteria, for example ‘Centre
of Gravity outside limits’. The network’s website is https:
//aviation-safety.net/
A.4. WEB SITES 97

• Air Accidents Investigation Branch, Report No: 2/2000. Report on

the incident to Fokker F27-600 Friendship, G-CHNL, near Guernsey
Airport, Channel Islands on 12 January 1999, The Stationery Of-
fice, 2000. This is the report of an accident caused directly by
incorrect loading leading to a centre-of-gravity position which
was too far aft. The analysis section of the report explains very
clearly the sequence of events which led to the crash, and how
different factors interacted to cause it. The report is available
from http://bit.ly/2bAwr4D.

A.4 Web sites

• Hush Kit, http://hushkit.net/, is an excellent, and well-

written, aviation site covering various aspects of historical and
modern aircraft, including the top ten best-looking British,
French, Swedish, Australian, Soviet, German, Japanese, and
Latin-American aircraft.

• Dr Brett Holman of the University of New England in Australia

maintains an excellent research blog, http://airminded.org/,
covering his work on the history of aviation and attitudes to it.

• Mark Vanhoenacker (see above) also has a website associated

with his book at http://www.skyfaring.com/.

A.5 Further watching

As well as the movies listed on page 82, there are some other films
and television programmes worth seeing to develop your know-
ledge of aviation and its culture.

• Cold War, Hot Jets, part 1 (fighters) http://bit.ly/1Ha0PuX;

part 2 (bombers) http://bit.ly/1QBYnSp.
98 Finding out more

A.6 Picture credits

All images are the work of the author, except those listed below.
The URLs link to the original image with full information on au-
thorship and usage rights.

Figure 2.3

1. Avro Vulcan, James Humphreys, http://bit.ly/1PxASt2

2. Saab Gripen, Matthias Kabel, http://bit.ly/1TRuPjn
3. Pegasus Quantum, Adrian Pingstone, http://bit.ly/1E0TkZP

Figure 7.1

1. Grob 109, public domain, http://bit.ly/1LhGaal

2. Jetstream, Dyvroeth, http://bit.ly/1Jowr4E
3. Sukhoi 27, Dmitry A. Mottl, http://bit.ly/1E3jVoO
4. Vampire, Timothy Swinson, http://bit.ly/1NhhMsI
5. Fouga Magister, Tim Felce, http://bit.ly/1J0pjZ5
6. Predator UAV, US Air Force/Lt Col. Leslie Pratt, http://bit.
ly/1MwlJbk

7. Grumman OV-1D, Valder137, https://bit.ly/2LdlpWc

8. C2A, US Navy, https://bit.ly/2uElnw1

Figure 7.2

1. Rutan VariEze, Stephen Kearney, http://bit.ly/1Jox3r7

2. Beechcraft Starship, Ken Mist, http://bit.ly/1UR0cgA
3. Piaggio Avanti, Tibboh, http://bit.ly/1MANkcZ
TO ME SHE IS ALIVE AND TO ME SHE SPEAKS. I FEEL THROUGH THE

SOLES OF MY FEET ON THE RUDDER-BAR THE WILLING STRAIN

A N D F L E X O F H E R M U S C L E S . T H E R E S O N A N T, G U T T U R A L V O I C E O F

H E R E X H A U S T S H A S A T I M B R E M O R E A R T I C U L AT E T H A N W O O D A N D

S T E E L , M O R E V I B R A N T T H A N W I R E S A N D S PA R K S A N D P O U N D I N G

PISTONS.

S H E S P E A K S T O M E N O W , S A Y I N G T H E W I N D I S R I G H T, T H E N I G H T

I S FA I R , T H E E F F O R T A S K E D O F H E R W E L L W I T H I N H E R P O W E R S .

I F LY S W I F T LY. I F LY H I G H — S O U T H - S O U T H W E S T, O V E R T H E N G O N G

H I L L S . I A M R E L A X E D. M Y R I G H T H A N D R E S T S U P O N T H E S T I C K

I N E A S Y C O M M U N I C AT I O N W I T H T H E W I L L A N D T H E W A Y O F T H E

PLANE. I SIT IN THE REAR, THE FRONT COCKPIT FILLED WITH THE

H E A V Y TA N K O F O X Y G E N S T R A P P E D U P R I G H T I N T H E S E AT, I T S

R O U N D S T I F F D O M E F O O L I S H LY R E M I N D I N G M E O F T H E P O I S E D

R I G I D I T Y O F A PA S S E N G E R O N F I R S T F L I G H T.

B E RY L M A R K H A M , W E S T W I T H T H E N I G H T