Stabilitynotes1

Some notes on aircraft and spacecraft
stability and control
Michael Carley, m.j.carley@bath.ac.uk

A lonely impulse of delight
Drove to this tumult in the clouds; . ..

Before we begin ...
These notes contain most, but not all, of the content of the course. You will also need:
• ESDU, ‘Lift-curve slope and aerodynamic centre position of wings in inviscid
subsonic flow’, ESDU 70011.
• Culick, F. E. C., ‘The Wright brothers: First aeronautical engineers and test pi-
lots’, AIAA Journal, 41(6):985–1006, 2003.
• Heffley, Robert K. and Jewell, Wayne F., ‘Aircraft handling qualities data’, NASA
CR-2144, 1972.
• Thompson, Ambler and Taylor, Barry N., ‘Guide for the use of the international
system of units (SI)’, NIST Special Publication 811, 2008.
• Gratton, Guy and Newman, Simon, ‘Towards the tumble resistant microlight’,
In European Symposium of Society of Experimental Test Pilots, Dresden, Germany,
21–25 June 2006. Available from http://eprints.soton.ac.uk/43858/
• Hamilton-Paterson, James, Empire of the Clouds: When Britain’s Aircraft Ruled
the World, Faber & Faber, 2010.
You are responsible for finding and obtaining these documents. They will not be
distributed to the class. They can all be downloaded via the university’s systems. You
should print out the first two. The third is long and we will not be using all of it so
wait until we come to use it before printing out the parts you need. You will only
need Appendix B of NIST SP-811. The microlight paper has a lot of information on
instability of both flexible and rigid wing aircraft. Empire of the Clouds is an account
of British aeronautical engineering in the decade or so after 1945, and is well worth
reading as background to the present state of the industry.
You must also fill in and return the registration form for the flight test course.
Flight tests this year will be on the 30th and 31st of October. The flight test is a com-
pulsory element of the degree and essential for accreditation.

Contents
Contents i
List of Figures iv
I Static stability 1
1 How aircraft fly 3
1.1 Equilibrium and stability . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Functions of aircraft controls . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.3 Forces and moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.4 Trim and stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
Aerodynamic centre and neutral point . . . . . . . . . . . . . . . . . . . 7
Definitions of static and c.g. margins . . . . . . . . . . . . . . . . . . . . 9
1.5 Basic aerofoil and control characteristics . . . . . . . . . . . . . . . . . . 10
Aerofoils and wings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
Control forces and moments . . . . . . . . . . . . . . . . . . . . . . . . . 11
Control hinge moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2 Longitudinal static stability 15
2.1 Some basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.2 Downwash . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.3 Stick fixed stability and c.g. margins . . . . . . . . . . . . . . . . . . . . 18
2.4 Stick free stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.5 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.6 Stick free stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3 Flight testing 23
3.1 Kn—elevator angle to trim . . . . . . . . . . . . . . . . . . . . . . . . . . 23
3.2 What does the pilot feel? . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3.3 Kn’—tab angle to trim . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
4 Tailless aircraft 29
4.1 Stick fixed stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4.2 Static margin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
i

ii CONTENTS
5 Stick forces 33
5.1 Analysis to calculate stick forces . . . . . . . . . . . . . . . . . . . . . . . 33
5.2 More flight testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
5.3 Modification of stick forces . . . . . . . . . . . . . . . . . . . . . . . . . . 36
6 Manoeuvre stability 39
6.1 Analysis of a steady pullout . . . . . . . . . . . . . . . . . . . . . . . . . 39
6.2 Stick fixed manoeuvre point . . . . . . . . . . . . . . . . . . . . . . . . . 42
6.3 Stick fixed manoeuvre stability . . . . . . . . . . . . . . . . . . . . . . . 43
6.4 Stick free manoeuvre stability . . . . . . . . . . . . . . . . . . . . . . . . 43
6.5 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
6.6 Tailless aircraft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
6.7 Tailless aircraft manoeuvre point . . . . . . . . . . . . . . . . . . . . . . 46
6.8 Tailless aircraft manoeuvre margins . . . . . . . . . . . . . . . . . . . . . 47
6.9 Relationships between static and manoeuvre margins . . . . . . . . . . 47
Conventional aircraft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
Tailless aircraft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
6.10 Modification of stick free neutral and manoeuvre points . . . . . . . . . 48
7 Compressibility effects 51
7.1 High speed effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
II Dynamic stability 55
8 Dynamic behaviour of aircraft 57
8.1 Axes and notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
8.2 Aerodynamic derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
8.3 Longitudinal symmetric motion . . . . . . . . . . . . . . . . . . . . . . . 61
9 Normal modes of aircraft 63
Phugoid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
Short period oscillation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
9.1 Lateral motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
Dutch roll . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
Spiral mode and roll subsidence . . . . . . . . . . . . . . . . . . . . . . . 66
9.2 Dihedral effect and weathercock stability . . . . . . . . . . . . . . . . . 67
Dihedral effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
Weathercock stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
III Spacecraft dynamics and control 71
10 Getting around: Orbits 73
10.1 The two-body problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

CONTENTS iii
Elliptical orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
10.2 Orbital maneouvres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
Hohmann transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
Orbital capture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
11 Getting things done: Spacecraft control 81
11.1 Attitude control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
Gravity gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
Sun-synchronous orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
11.2 Manouevring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
IVProblems 85
12 Problems 87
Basic equations 101

List of Figures
1.1 Phases of flight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2 Terminology for study of stability . . . . . . . . . . . . . . . . . . . . . . . . 5
1.3 Axes and sign conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.4 Sign conventions for longitudinal stability . . . . . . . . . . . . . . . . . . . 7
1.5 Trim and stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.6 Centre of pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.7 Incremental loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.8 Centre of gravity and aerodynamic centre relationships . . . . . . . . . . . 9
1.9 Lift curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.10 Loading due to control deflection . . . . . . . . . . . . . . . . . . . . . . . . 11
1.11 Measurement of control/tab deflections . . . . . . . . . . . . . . . . . . . . 12
1.12 Pressure distributions due to deflections, a1 > a2 > a3. . . . . . . . . . . . . 12
1.13 Measurement of control surface area . . . . . . . . . . . . . . . . . . . . . . 13
1.14 Control hinge moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.1 Stick fixed stability (conventional aircraft) . . . . . . . . . . . . . . . . . . . 16
2.2 Trailing vortices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.3 Effect of downwash on tailplane . . . . . . . . . . . . . . . . . . . . . . . . 17
2.4 Stick free elevator conditions . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.1 A typical weight and balance envelope . . . . . . . . . . . . . . . . . . . . . 23
3.2 Elevator angle to trim at various lift coefficients . . . . . . . . . . . . . . . . 24
3.3 Measurement of neutral point location . . . . . . . . . . . . . . . . . . . . . 25
3.4 What the pilot experiences . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3.5 Tab angle to trim at varying lift coefficients . . . . . . . . . . . . . . . . . . 27
3.6 Measurement of stick free neutral point location . . . . . . . . . . . . . . . 27
4.1 Canard configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4.2 Control surfaces for tailless aircraft . . . . . . . . . . . . . . . . . . . . . . . 30
4.3 Representation of tailless aircraft . . . . . . . . . . . . . . . . . . . . . . . . 30
5.1 Aerodynamic assistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
6.1 Manoeuvre conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
6.2 Manoeuvre conditions for a tailless aircraft . . . . . . . . . . . . . . . . . . 45
iv

LIST OF FIGURES v
6.3 Aerodynamic forces during pitching motion . . . . . . . . . . . . . . . . . 46
6.4 Modification of neutral and manoeuvre points . . . . . . . . . . . . . . . . 48
6.5 Effects of positive springs and bob-weights . . . . . . . . . . . . . . . . . . 49
6.6 Modification of stick force per g . . . . . . . . . . . . . . . . . . . . . . . . . 49
7.1 Compressibility effects on lift curve slope and aerodynamic centre . . . . . 52
7.2 Compressibility effects on zero lift pitching moment and zero lift angle . . 52
7.3 Aeroelastic effects on lift curve slope . . . . . . . . . . . . . . . . . . . . . . 53
7.4 Aeroelastic effects on tailplane and elevator . . . . . . . . . . . . . . . . . . 53
7.5 Variation of downwash with Mach number . . . . . . . . . . . . . . . . . . 54
7.6 Variation of pitching moment with Mach number . . . . . . . . . . . . . . 54
7.7 Variation of stick forces with Mach number . . . . . . . . . . . . . . . . . . 54
8.1 Notation for analysis of dynamic stability . . . . . . . . . . . . . . . . . . . 58
9.1 Phugoid oscillation trajectory . . . . . . . . . . . . . . . . . . . . . . . . . . 64
9.2 Short period oscillation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
9.3 Rolling subsidence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
9.4 Stability of the lateral modes . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
9.5 Dihedral effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
9.6 Wing sweep effects on Lv . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
9.7 Wing-fuselage interference effects on Lv . . . . . . . . . . . . . . . . . . . . 69
9.8 Use of twin fins at high speed . . . . . . . . . . . . . . . . . . . . . . . . . . 70
9.9 Intake effects on Nv . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
10.1 The two-body problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
10.2 Polar coordinate system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
10.3 The geometry of an ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
10.4 Walter Hohmann and his transfer from low earth to geostationary Earth
orbit. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
11.1 Two approaches for a docking spacecraft . . . . . . . . . . . . . . . . . . . 83
12.1 Aircraft with different centres of gravity . . . . . . . . . . . . . . . . . . . . 87
12.2 Full-scale and model aircraft. CMp is measured by the balance, which re-
strains the model in pitch. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

Chapter 1
How aircraft fly
Aircraft fly by generating a lift greater than or equal to their weight. They do this by
holding a wing at a certain angle of attack, or incidence. Longitudinal control is the
study of how to set, maintain or change that angle of attack; stability is the study of
whether and how that angle of attack will remain fixed when the aircraft is subjected
to small perturbations, due to atmospheric turbulence, for example.
One way to think about this is to look at the phases of aircraft flight, Figure 1.1.
When an aircraft takes off, its speed is quite low and it must generate lift greater than
its weight in order to leave the ground. In cruise, the aircraft operates at a constant
speed and constant lift. Finally, when the aircraft lands, it needs to reduce its speed
without losing too much lift.
In each case, the issue for control of the aircraft is how it can maintain its inci-
dence at a given speed. When it takes off or lands, it does so by rotating—raising
or lowering its nose—in order to change the incidence of the wing, altering the rela-
tionship between speed and lift. Aircraft control is the study of how a pilot can fix the
relationship between speed and incidence.
When an aircraft cruises, it is desirable that it do so at constant speed and inci-
dence, so the controls are at a fixed setting. Aircraft stability is the study of how an
aircraft responds to small disturbances in flight and how it can be designed so that it
remains at a fixed incidence and speed without overworking the pilot.
In each case, the basic question is how to generate a moment on the aircraft so that
it rotates and changes the wing incidence or so that the net moment is zero and the
aircraft flies at constant incidence. This is done via the aircraft controls. Before we go
any further, we need to clarify what we mean by some basic terms.
1.1 Equilibrium and stability
The requirements of an aircraft control system are that it be able to bring the aircraft
into some required equilibrium and that it be able to maintain that equilibrium stably.
This statement of requirements contains some terms which have precise mean-
ings:
3

4 CHAPTER 1. HOW AIRCRAFT FLY
Take-off: the incidence increases to generate more lift at low speed.
Cruise: the aircraft flies at constant incidence and speed
Landing: the aircraft increases incidence to allow it to slow down.
Figure 1.1: Phases of flight
equilibrium a system is in equilibrium when the sums of all of the forces and mo-
ments acting on it are identically zero;
static stability a system is statically stable if, when disturbed from equilibrium, it
initially tends to return to the equilibrium configuration;
dynamic stability a system is dynamically stable if, when disturbed from equilib-
rium, it does eventually return to the equilibrium configuration.
The distinction between the static and dynamic stability of a system is simple,
though subtle. If we disturb a system, static stability deals with the question of what
the system does in the very short time just after the disturbance has been applied;
dynamic stability is the study of what happens after that, over long periods of time.
Figure 1.2 shows the response of systems which are statically and dynamically stable
and/or unstable, in various combinations, including the case of a system which is
statically stable but dynamically unstable. It also includes the case of neutral stability,
where the system remains in whatever configuration it has been shifted to.
1.2 Functions of aircraft controls
The function of an aircraft control system is to provide a means of changing the mo-
ments on an aircraft, to control, in this case, its incidence. There are many ways of

1.2. FUNCTIONS OF AIRCRAFT CONTROLS 5
Response
Time
Neutral stability
Statically and dynamically stable
Statically stable and dynamically unstable
Statically unstable
Figure 1.2: Terminology for study of stability
doing this, but for the conventional aircraft we consider for now, this is done by mov-
ing surfaces in order to change the aerodynamic forces on some part of the aircraft,
thereby changing the overall moment. These control surfaces are the:
elevator this changes the total lift on the tail when it is deflected, causing a change
in pitching moment on the aircraft. This allows the pilot to adjust the aircraft
incidence;
ailerons these change the lift on each wing when they are deflected. They move in
opposite directions—one goes up when the other goes down so that the lift on
one wing increases and the other decreases. This generates a change in rolling
moment and allows the aircraft to rotate about its axis to initiate turns, or allows
it to oppose disturbances due to crosswind or gusts;
rudder changes the side force on the vertical tailplane (or fin), generating a change in
yawing moment, rotating the aircraft about a vertical axis. This can be used to
resist yawing moments due to engine failure and crosswind and to aid in spin
recovery and turn co-ordination.
In steady, level flight in still air, the rudder and ailerons will be undeflected while
the elevator will probably be at some deflection which depends on the aircraft load-
ing. Under other conditions, or during a manouevre, all three controls may be used
simultaneously. The controls can be operated directly by the pilot, through a system
of mechanical actuators, possibly with aerodynamic or power assistance, or controls
may be fully powered using a hydraulic or electrical system. These systems can be
mechanically or electronically controlled (fly by wire or fly by light).
The sign conventions for the controls and motions are shown in Figure 1.3.

Positive up
Positive down
Positive left
Positive down
z
Yaw
x
Roll
y
Pitch
Figure 1.3: Axes and sign conventions
1.3 Forces and moments
Forces and moments on an aircraft are due to the mass of the aircraft, a function of
how it is built and loaded, and the aerodynamic forces generated in flight. The mass
distribution of the aircraft gives rise to inertial forces and moments as described by
Isaac Newton and to gravitational forces.
You should already know that the basic aerodynamic forces include the lift and
drag on the aircraft. More generally, we consider the static forces and moments due
to linear velocities; the damping forces and moments caused by angular velocities
(such as pitching and rolling); the control forces generated when the pilot operates a
control surface.
By considering the lateral symmetry of most aircraft, it is clear that in forward
level flight, the forces on the aircraft act in the plane of symmetry. This means that
any symmetric disturbance, such as operation of the elevator, will generate horizontal
and vertical motion of the aircraft and rotation in the vertical plane only. At this
point, then, we study longitudinal symmetric motion of the aircraft and consider
longitudinal stability.
1.4 Trim and stability
Figure 1.4 shows the basic configuration for the study of stability of an aircraft, la-
belled with the forces and moments and showing the corresponding sign conven-
tions. The orientation of the aircraft is labelled with two angles, θ and α. You should

1.4. TRIM AND STABILITY 7
not confuse these. The angle θ is the inclination of the aircraft which is the angle be-
tween the direction of flight and the horizontal; the angle α is the incidence, or angle
between the direction of flight and the Zero Lift Line (ZLL). When α is zero, the Zero
Lift Line is aligned with the flight direction and there is no lift acting on the aircraft,
whatever might be its inclination θ.
Horizontal
Flight direction
θ
Zero lift line
α
W
L
D
T
Figure 1.4: Sign conventions for longitudinal stability
To examine the equilibrium and stability of the aircraft, we resolve forces parallel
and perpendicular to the aircraft axis:
Parallel: T − D − W sin θ = 0, (1.1)
Perpendicular: L − W cos θ = 0, (1.2)
Moments about the c.g.: Mcg = 0. (1.3)
Moments about the centre of gravity (c.g.) cannot be due to the mass of the aircraft
(by definition). This means that if Mcg = 0, the aerodynamic moments on the aircraft
are in equilibrium and the aircraft is said to be trimmed or in trim. This is the basic
problem of control: is it possible, using the tailplane, to generate a pitching moment
so that the overall moment about the centre of gravity is zero?
The basic problem of static stability is then: when an aircraft in trim is subjected
to a disturbance which changes its incidence, does it tend to return to the equilibrium
position?
This can be restated: if the aircraft pitches nose up, the change in aerodynamic
moment about the centre of gravity ∆Mcg should be negative in order to push the
nose back down or, ∂Mcg/∂α < 0.
Figure 1.5 shows various ways Mcg can vary with α, including how it is possible
to trim an unstable aircraft and how is possible for an aircraft to be stable without
being able to trim at a useful incidence.
Aerodynamic centre and neutral point
The forces and moments acting on an aircraft depend on the shape of the aircraft and
not on the position of the centre of gravity so we can consider the aerodynamic loads

C
M
cg
α
Neutrally stable
cannot trim
∂CM /∂α > 0
∂CM /∂α < 0
∂CM /∂α < 0
CL < 0
Figure 1.5: Trim and stability
separately from the gravitational. Figure 1.6 shows a pressure distribution on an
aerofoil section. The loads can be considered to act at a point, the centre of pressure,
where the total aerodynamic moment is zero. The centre of pressure, however, moves
as the incidence varies, so it is not very useful as a reference point in calculations
involving changing incidence.
L
Figure 1.6: Centre of pressure
To make life easier, we can give up the requirement that the moment about our
reference point be zero and, instead, allow it to have some finite value as long as the
reference point is fixed and the moment is constant. We can do this by looking at the
incremental pressure distribution, sketched in Figure 1.7.
When the incidence is increased by some small amount, the incremental aerody-
namic load can be considered to act through a certain point, generating no change
in moment about that position. This point is the aerodynamic centre and is the point
about which dM/dα ≡ 0.
If we now think about the basic question of stability, we can consider what hap-
pens to an aircraft which pitches slightly nose up. Depending on the position of the
centre of gravity, relative to the aerodynamic centre, the aircraft will be stable, unsta-
ble or neutrally stable, Figure 1.8.

1.4. TRIM AND STABILITY 9
∆L
Figure 1.7: Incremental loads
W
L
M0
W
L
M0
W
L
M0
Figure 1.8: Centre of gravity and aerodynamic centre relationships
1. If the centre of gravity is forward of the aerodynamic centre, dMcg/dα is nega-
tive and the aircraft is statically stable.
2. If the centre of gravity is aft of (behind) the aerodynamic centre, dMcg/dα is
positive and the aircraft is statically unstable.
3. If the centre of gravity is at the aerodynamic centre, dMcg/dα is zero and the
aircraft is neutrally stable.
When we talk about the properties of a whole aircraft, rather than just a wing or
aerofoil, we use the term ‘neutral point’. This is the position of the centre of gravity
for which the aircraft is neutrally stable. It is a purely aerodynamic property, which
depends on the shape of the aircraft. For a wing, the neutral point and the aerody-
namic centre are identical.
Definitions of static and c.g. margins
A basic measure of stability is how quickly an aircraft responds to a perturbation. This
is equivalent to the gradient ∂CM /∂α. When this is large, the aircraft experiences a
large pitching moment when it is perturbed and quickly returns to its equilibrium.
Our measure of the innate stability of an aircraft is then the static margin:
Kn = −
dCMcg
dCR
(1.4)

where CR = (C2
L + C2
D)1/2
, the resultant force.
We can express this in a more easily visualized form by looking at the c.g. margin,
Hn, which is the distance between the neutral point and the centre of gravity, scaled
on the mean chord c:
Hn = −
dCM
dCL
= hn − h (1.5)
where hc is the displacement of the centre of gravity aft of the reference point and hnc
is the displacement of the neutral point aft of the reference point.
At low speed and low inclination, CR ≈ CL and CL, CM , CR are not influenced by
Mach number or aeroelastic effects so that Hn ≈ Kn.
1.5 Basic aerofoil and control characteristics
So far, we have considered aircraft stability without considering the behaviour of
aircraft. In order to deal with realistic problems, we need to know something about
how aerodynamic surfaces and bodies behave in flight.
Aerofoils and wings
As always, we work in terms of non-dimensional quantities using standard reference
values velocity V , density ρ, wing planform area S and wing mean chord c:
CL =
L
ρV 2S/2
, CD =
D
ρV 2S/2
, CM =
M
ρV 2Sc/2
. (1.6)
For tailless aircraft, the wing root chord c0 is often used as a reference length. The
subscript in each coefficient is upper case because the coefficients are those for three
dimensional bodies.
α
CL
Figure 1.9: Lift curve

1.5. BASIC AEROFOIL AND CONTROL CHARACTERISTICS 11
Figure 1.9 shows the lift curve slope of a wing. The important point to note is
that we choose a reference incidence such that the lift is zero when α = 0. This will
not always be true in other calculations and you should check the conventions used
when you take data from published sources. In this course, we will only consider
linear aerodynamics, i.e. the part of the lift curve where dCL/dα is constant. This is a
reasonable assumption for most aircraft most of the time, but will not be correct near
stall or in high speed manoeuvres.
Control forces and moments
We control an aircraft by moving a control surface such as an elevator. In order to
have some clue what the result of a control input will be, we need to know how far
the surface must be moved in order to generate, indirectly, the pitching moment we
need and we need to know what moment will be needed to rotate the control surface
about its hinge point.
If the incidence of the wing or tailplane as a whole is held constant, deflecting the
control surface in the positive sense is equivalent to introducing extra camber. This
generates extra lift and a pitching moment which is usually nose-down, because of
the form of the incremental pressure distribution, Figure 1.10.
x/c
∆Cp
Figure 1.10: Loading due to control deflection
This means that a positive control deflection generates a positive change in lift and
negative pitching moment (if the tail pushes up, it forces the nose down). If there is a
tab, an extra small surface on the end of the elevator, this too will generate positive lift
and negative moment. Figure 1.11 shows the notation for control surface deflection.
Since we are working on the basis of linear aerodynamics, each deflection con-
tributes linearly to the forces and moments, with the following symbols defined for
convenience:
a1 =
∂CLT
∂αT
, a2 =
∂CLT
∂η
, a3 =
∂CLT
∂β
, (1.7)
with the corresponding pressure distributions shown in Figure 1.12.

η
β
Figure 1.11: Measurement of control/tab deflections
a1 a2 a3
∆Cp
x/c
∆Cp
x/c
∆Cp
x/c
η
β
Figure 1.12: Pressure distributions due to deflections, a1 > a2 > a3.
The total tailplane lift coefficient is then:
CLT
= a1αT + a2η + a3β
and
CMac = CM0 +
∂CM0
∂η
η +
∂CM0
∂β
β.
Control hinge moments
When an aircraft control system is designed, we will need to know what force is
required to move a control, whether it is being moved directly by the pilot or through
a powered actuation system. The force needed depends on the moment required to
rotate the surface through a given deflection angle. The hinge moment coefficient is:
CH =
MH
ρV 2Sηcη/2
, (1.8)

1.5. BASIC AEROFOIL AND CONTROL CHARACTERISTICS 13
Hinge line
Aerodynamic balance
Sη
Figure 1.13: Measurement of control surface area
where Sη is the control surface area and cη is the control surface mean chord. Both of
these values are measured aft of the hinge line, as shown in Figure 1.13.
As before, we adopt a shorthand notation for the contribution of each deflection
to the total hinge moment:
b1 =
∂CH
∂αT
, b2 =
∂CH
∂η
, b3 =
∂CH
∂β
, (1.9)
and note that for non-symmetric tailplane sections there is usually a hinge moment
when all other deflections are zero, given the symbol b0. The pressure distributions
associated with each of these terms are shown in Figure 1.14.
The total hinge moment coefficient is then:
CH = b0 + b1αT + b2η + b3β. (1.10)

b1 b2 b3
Hinge
line
∆Cp
x/c
Hinge
line
∆Cp
x/c
Hinge
line
∆Cp
x/c
η
β
Figure 1.14: Control hinge moments

Chapter 2
Longitudinal static stability
Given the definitions and background information of Chapter 1, we are in a position
to start doing some calculations for the stability of aircraft. We consider two basic
cases, the ‘stick fixed’ and ‘stick free’. In the first case, conceptually, we move the
elevator to the position required for trim and then fix the stick so that the elevator
remains at the set deflection, ignoring the question of what force is required to keep
it in place. The stability problem can then be phrased: ‘with the stick fixed, how does
the aircraft respond to a perturbation?’ In the ‘stick free’ case, we use the tab to adjust
the moment on the elevator so that it comes to an equilibrium position which trims
the aircraft. In this case, zero force is required to keep the control fixed. Using this
zero-force case as a reference, we can work out the stick force required for any other
tab setting.
2.1 Some basics
Figure 2.1 shows the problem of stick fixed stability for a conventional aircraft, i.e.
one with a tail at the back. The aircraft has two lifting surfaces, the wing which
generates most of the lift, and the tailplane which generates a small amount of lift but
which can be adjusted to change the pitching moment on the aircraft as a whole. The
tailplane is set at angle ηT relative to the aircraft zero lift line.
On the usual assumption of linear aerodynamics, with small α and ηT and no wake
effect on the tailplane (but see §2.2):
Mcg = M0 − LWBN (h0 − h)c − LT ((h0 − h)c + l) − TzT + DzD
= M0 − (h0 − h)c(LWBN + LT ) − LT l − TzT + DzD
= M0 − (h0 − h)cL − LT l − TzT + DzD,
where the lift has been broken up into a wing-body-nacelle (WBN) and a tailplane (T)
contribution. Given that lift is much larger than drag (and likewise thrust), we can
neglect TzT and DzD, and
Mcg = M0 − (h0 − h)cL − LT l.
15

16 CHAPTER 2. LONGITUDINAL STATIC STABILITY
Datum
W
hc̄
LW BN
h0c̄
M0
LT
l
T
zT D
zD
Figure 2.1: Stick fixed stability (conventional aircraft)
We now non-dimensionalize the equation to write it in terms of the coefficients de-
fined in §1.5, with the addition of the tailplane lift coefficient:
CLT
=
LT
ρV 2ST /2
(2.1)
where ST is the area of the tailplane.
Thus, dividing by Sc(ρV 2
/2):
CMcg = CM0 − (h0 − h)CL − CLT
ST
S
l
c
and defining the tail volume coefficient:
V =
ST
S
l
c
(2.2)
we find:
CMcg = CM0 − (h0 − h)CL − V CLT
. (2.3)
This is the fundamental equation of aircraft stability and control. In control problems,
the aircraft is trimmed with CMcg ≡ 0, the lift coefficient is known from the operating
conditions and CM0 is known from the aircraft geometry. Then, if V is known, CLT
can be calculated and from that the elevator deflection; if CLT
is known (because
the tailplane shape has already been decided), V can be calculated, and the size of
the tailplane fixed. The tail volume coefficient represents the ‘effectiveness’ of the
tailplane at generating a moment. It contains the size of the tailplane ST and the lever
arm l which, combined, tell us the moment which the tailplane can generate.
2.2 Downwash
Any lifting wing generates a downwash, due to the trailing vortex system, shown in
Figure 2.2. This needs to be included in the stability calculation because it alters the

2.2. DOWNWASH 17
Figure 2.2: Trailing vortices
incidence at the tailplane and that change in incidence changes with aircraft angle of
attack, Figure 2.3.
ZLL tailplane
ZLL WBN
ηT
Free stream
α

Resultant flow
Figure 2.3: Effect of downwash on tailplane
From Figure 2.3:
αT = α + ηT − .
For an untwisted wing, the downwash angle is proportional to the lift on the wing,
meaning that in the linear regime, it is also proportional to α:
=
d
dα
α + 0,

with 0 only present for a wing where the zero lift angle of attack varies along its
span (i.e. a wing with a varying cross-section or camber along its length or with
twist). Combining the previous equations:
αT = α + ηT −

0 +
d
dα
α

= α

1 −
d
dα

+ (ηT − 0).
From §1.5:
CLT
= a1αT + a2η + a3β,
so that:
CLT
= a1(α + ηT − ) + a2η + a3β,
and:
CLT
= a1α

1 −
d
dα

+ a1(ηT − 0) + a2η + a3β.
This can be related to known quantities by including the relationship between inci-
dence and lift coefficient (Figure 1.9):
CL = aα,
where a is the overall lift curve slope of the aircraft. Upon substitution:
CLT
=
a1
a

1 −
d
dα

CL + a1(ηT − 0) + a2η + a3β. (2.4)
In deriving (2.3), we made no assumptions about how CLT
was generated, so (2.4)
can be substituted to give:
CM = CM0 − (h0 − h)CL − V

a1
a

1 −
d
dα

CL + a1(ηT − 0) + a2η + a3β

.
Given a flight condition (speed, aircraft weight, etc.), this equation allows us to cal-
culate the elevator angle to trim, η (trim quantities are written as the usual symbol
with an overbar). In designing aircraft, there will be a limit on the maximum elevator
deflection. Given this maximum η, we can use the flight conditions to estimate V and
so size the tailplane.
2.3 Stick fixed stability and c.g. margins
The static margin is defined in §1.4:
Kn ≈ Hn = −
dCM
dCL
,

2.4. STICK FREE STABILITY 19
so that to determine the stability of the aircraft, we differentiate (2.3) with respect to
CL:
−
dCM
dCL
= (h0 − h) + V
a1
a

1 −
d
dα

.
The neutral point is the centre of gravity position where dCM /dCL ≡ 0:
hn = h0 + V
a1
a

1 −
d
dα

,
where h0 is the neutral point of the aircraft less tail. Adding a tail has moved the
neutral point back by an amount V (a1/a)(1 − d/dα), increasing the stability (as you
might expect). The fundamental problem of designing the control system of an air-
craft is that of determining, via V , the size of the tailplane such that it makes the whole
aircraft stable and controllable. The stability requirement is specified as a minimum
value of h − hn; the control requirement is stated, in effect, as a maximum pitching
moment to be generated.
2.4 Stick free stability
The ‘stick-fixed’ analysis of the first part of this chapter only considers where the el-
evator needs to be in order to trim the aircraft, without looking at the forces needed
to get there. As the first step towards calculating the forces needed to move a control
surface, we consider the ‘stick-free’ problem, where the elevator is allowed to flop
around until it reaches an equilibrium position where the moment on it, and so the
stick force, is zero. On small aircraft, the moment on the elevator is modified aerody-
namically by moving the tab, on large ones, the elevator is moved by an actuator and
no tab is required. In both cases, however, we need to know the forces on the elevator
either to keep them within the limits of a pilot’s strength or to size the actuators.
2.5 Analysis
The hinge moment coefficient is:
CH = b0 + b1αT + b2η + b3β,
which is zero if the elevator (stick) is free. In this case, the elevator angle is:
η = −
b0 + b1αT + b3β
b2
.
From §2.2:
CLT
=
a1
a

1 −
d
dα

CL + a1(ηT − 0) + a2η + a3β,

and
CLT
=
a1
a

1 −
d
dα

CL + a1(ηT − 0) −
a2
b2
(b0 + b1αT + b3β) + a3β.
Incorporating the expression for α from §2.2, page 18)
αT =
CL
a

1 −
d
dα

+ (ηT − 0),
and
CLT
=

a1 −
a2b1
b2

1 −
d
dα

CL
a
+

a1 −
a2b1
b2

(ηT − 0)
+

a3 −
a2b3
b2

β −
a2b0
b2
.
To simplify our notation, we define two new variables (both given on the data
sheet in the appendix):
a1 = a1

1 −
a2b1
a1b2

,
a3 = a3

1 −
a2b3
a3b2

,
so that:
CLT
= a1

1 −
d
dα

CL
a
+ a1(ηT − 0) + a3β −
a2b0
b2
.
We already know the tailplane lift coefficient, (2.3), so that we can find the lift if
the stick is released and the elevator comes to equilibrium. The resulting pitching
moment is:
CM = CM0 − (h0 − h)CL − V CLT
,
and:
CM = CM0 − (h0 − h)CL − V

a1
a

1 −
d
dα

CL + a1(ηT − 0) + a3β −
a2b0
b2

,
which allows us to find the tab angle to trim with zero stick force, β.
2.6 Stick free stability
To find the stick-free stability properties, we differentiate the pitching moment equa-
tion:
dCM
dCL
= −(h0 − h) − V
a1
a

1 −
d
dα

,

2.6. STICK FREE STABILITY 21
and the static margin stick free:
K0
n = −
dCM
dCL
= (h0 − h) + V
a1
a

1 −
d
dα

.
This is related to the static margin stick fixed:
Kn = (h0 − h) + V
a1
a

1 −
d
dα

,
with the two static margins being equal if:
a1 = a1 = a1

1 −
a2b1
a1b2

,
or
a2b1
a1b2
= 0.
Since a1 and a2 are both positive, and b2 is negative for correct feel of the elevator, this
can only happen if b1 = 0, which can be achieved using aerodynamic balancing or by
moving the elevator hinge line, which will also change b2.
a2b1/a1b2 0 a2b1/a1b2 0 a2b1/a1b2 = 0
Figure 2.4: Stick free elevator conditions
Figure 2.4 shows the three possible cases for b1:
• when a2b1/a1b2 0, the aircraft is less stable stick free—the elevator is ‘conver-
gent’;
• when a2b1/a1b2 0, the aircraft is more stable stick free—the elevator is ‘diver-
gent’;
• when a2b1/a1b2 = 0, the aircraft is equally stable stick free—the elevator is ‘null’.
We define the neutral point stick free h0
n and the static and c.g. margins stick-free
H0
n in the same way as in the stick-fixed case.

Chapter 3
Flight testing
Having designed an aircraft to have given stability characteristics, we must test the
production model to find what the real behaviour is. In the early stages of design, we
use approximate analyses and semi-empirical methods (for example, ESDU sheets) to
estimate the aerodynamic parameters such as lift curve slopes, largely because early
in design, we have not fixed the exact shape and size of the aircraft or its subsystems.
When we have a detailed geometry, we can use computational methods to refine our
estimates. When the first few aircraft are produced, we must test them to see what
the real behaviour of the real aircraft is.
This information is used in setting the limits to be observed in service—the ‘flight
envelope’ of Figure 3.1. Before flight, the aircraft weight and centre of gravity are
plotted on the diagram and must lie within the limits indicated. If they do not, then
the weight must be reduced or the centre of gravity must be moved by adding ballast.
This guarantees that the aircraft will fly within the limits set at the design stage.
3.1 Kn—elevator angle to trim
Given that, for a trimmed aircraft:
,
hc̄
m
Aircraft safe to fly
Figure 3.1: A typical weight and balance envelope
23

24 CHAPTER 3. FLIGHT TESTING
and,
CLT
=
a1
a

1 −
d
dα

CL + a1(ηT − 0) + a2η + a3β,
CM = 0 = CM0 − (h0 − h)CL − V

a1
a

1 −
d
dα

CL + a1(ηT − 0) + a2η + a3β

.
This can be differentiated (§2.3) to examine the stick-free stability:
Kn ≈ Hn = −
∂CM
∂CL
= (h0 − h) + V
a1
a

1 −
d
dα

.
We also know that:
η =
1
V a2

CM0 − (h0 − h)CL − V

a1
a

1 −
d
dα

CL + a1(ηT − 0) + a3β

,
and that there is a relationship between η and Kn because:
dη
dCL
= −
1
V a2

(h0 − h) + V
a1
a

1 −
d
dα

= −
Kn
V a2
.
Figure 3.2 shows η plotted against CL, while Figure 3.3 shows the relationship
between dη/dCL and h.
CL
η̄
h1
h2
h3
c.g. forward
Figure 3.2: Elevator angle to trim at various lift coefficients
Given this information, one way of finding the aircraft neutral 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 3.2. To find the neutral point, plot the gradients of the lines of Figure 3.2,
as in Figure 3.3. Extrapolating to dη/dCL gives the centre of gravity position where
Kn = 0, the neutral point.

3.2. WHAT DOES THE PILOT FEEL? 25
h
dη̄/dCL
h1
h2
h3
Figure 3.3: Measurement of neutral point location
3.2 What does the pilot feel?
Pilots rarely know the lift coefficient of the aircraft: they will have a feel for stick force
and for elevator deflection (because they know how far the stick has moved) and for
speed (because they can see out the window or look at the instruments). We can see
how the elevator angle to trim varies with speed, to look at what the pilot feels in
flying the aircraft:
dη
dV
=
dη
dCL
dCL
dV
.
By definition,
CL =
L
ρV 2S/2
,
so that:
dCL
dV
= −
2L
ρV 3S/2
= −
2CL
V
,
and
dη
dV
= −
2CL
V
dη
dCL
=
2CL
V
Kn
V a2
,
which is sketched in Figure 3.4.
From Figure 3.4, it is clear that the aircraft is uncontrollable below 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
moment about the centre of gravity. Likewise, above a certain speed, small changes
in η lead to large changes in trim speed and the aircraft is also very hard to control.
The useful range of speeds for an aircraft lies between these two limits, although the
limits in question will be a function of the aircraft type and of the skill assumed of the
pilot.

26 CHAPTER 3. FLIGHT TESTING
V
η̄
c.g. moving forward
Figure 3.4: What the pilot experiences
3.3 Kn’—tab angle to trim
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 elevator. We know that:
= 0,
and
CLT
=
a1
a

1 −
d
dα

CL + a1(ηT − 0) + a2η + a3β.
From §2.5:
η = −
b0 + b1αT + b3β
b2
,
and CLT
=
a1
a

1 −
d
dα

CL + a1(ηT − 0) + a3β −
a2b0
b2
,
so that:
CM = CM0 − (h0 − h)CL − V

a1
a

1 −
d
dα

CL + a1(ηT − 0) −
a2b0
b2

.
Rearranging to find the tab angle to trim:
β =
1
V a3

CM0 − (h0 − h)CL − V

a1
a

1 −
d
dα

CL + a1(ηT − 0) + a3β −
a2b0
b2

,
and differentiating with respect to CL:
dβ
dCL
= −
1
V a3

h0 − h + V
a1
a

1 −
d
dα

.

3.3. KN ’—TAB ANGLE TO TRIM 27
We know, however, that:
K0
n = (h0 − h) + V
a1
a

1 −
d
dα

,
and that
dβ
dCL
= −
K0
n
V a3
.
So to find the neutral point stick free, we vary the aircraft speed at fixed centre of
gravity, trimming with the tab, giving us Figure 3.5. We then plot the gradients from
that figure against CL, Figure 3.6 to find h0
n
CL
β̄
h1
h2
h3
Figure 3.5: Tab angle to trim at varying lift coefficients
h
dβ̄/dCL
h1
h2
h3
Figure 3.6: Measurement of stick free neutral point location

Chapter 4
Tailless aircraft
In the previous chapters, we have considered conventional aircraft, those which use
a tail to provide pitching moment control. We can use the same methods to analyse
‘canard’ aircraft which have the control surface ahead of the wing. In this case, the
basic equations are the same as in the conventional case, but the tail arm l about the
aerodynamic centre is negative, Figure 4.1.
LF LW BN
l
M0
W
hc̄
h0c̄
Figure 4.1: Canard configuration
On a tailless aircraft, there is no separate control surface for pitch control, with the
elevators and ailerons being combined into ‘elevons’. These are moved in opposite
directions for roll control and in the same direction for pitch control, Figure 4.2.
4.1 Stick fixed stability
Figure 4.3 shows how a tailless aircraft is represented for the study of static stabil-
ity. The most obvious difference from the conventional case is that we have no tail
contribution to include. On tailless aircraft, the pitching moment required to trim is
generated by the control surfaces which also have a large effect on lift. This makes
the control of such aircraft quite complicated, especially on landing.
29

30 CHAPTER 4. TAILLESS AIRCRAFT
Elevon
Rudder
Figure 4.2: Control surfaces for tailless aircraft: the elevons operate together for pitch
control and differentially for roll
η
LW BN
M0
W
hc̄
h0c̄
Figure 4.3: Representation of tailless aircraft

4.2. STATIC MARGIN 31
The lift coefficient for a tailless aircraft is written:
CL = a1α + a2η,
with no tab included, since tailless aircraft rarely have them.
Taking moments about the centre of gravity:
Mcg = M0 +
∂M0
∂η
η − (h0 − h)c0L,
and non-dimensionalizing:
CM = CM0 +
∂CM0
∂η
η − (h0 − h)CL,
which can be rearranged to yield the elevon angle to trim.
4.2 Static margin
The definition of static margin is the same as in the conventional aircraft case:
Kn = −
dCM
dCL
= h0 − h.
Because tailless aircraft are usually large and have large control surfaces, their
control systems are powered, so that there is no ‘stick free’ condition: we do not need
to consider this case.

Chapter 5
Stick forces
So far, we have not considered how what force or, equivalently, torque, will be needed
to move a control surface into position or to hold it in place. This is an important ques-
tion because it must be possible for the pilot to control the aircraft without requiring
excessive stick force. On the other hand, the aircraft must not be too twitchy, respond-
ing excessively to small control inputs. If the controls are powered, it is also essential
to know what forces the actuators will need to generate, so that the hydraulic sys-
tem can be sized. Table 5.1 shows the maximum forces which can be applied to the
different controls under various circumstances.
Aileron Elevator Rudder
Stick Wheel Stick Wheel (Push)
Maximum all-out effort 2 hands 400 530 800 980 1780 N
Maximum permissible effort 2 hands — 360 440 440 890 N
1 hand 220 220 310 310 N
Maximum comfortable effort 2 hands — 130 — 180 270 N
1 hand 90 90 130 130 N
Largest full travel ±254 ±508 ±230 ±230 ±126 mm
Table 5.1: Maximum control forces
It is considered good practice to make sure that the maximum elevator force is
higher than the maximum aileron force and that the maximum rudder force is higher
than both. The controls are said to be ‘harmonized’ if the aileron, elevator and rudder
forces have the ratio 1:2:4 for a given control response. For example, the rudder force
for a 10◦
/s yaw is twice the elevator force for a 10◦
/s pitch.
5.1 Analysis to calculate stick forces
The input from the pilot for a given elevator deflection is the stick force, so that the
stick force to trim Pe, which is found from the moment on the control multiplied by
33

34 CHAPTER 5. STICK FORCES
the gearing ratio between the stick and the control deflection me, is:
Pe = me
ρV 2
2
SηcηCH.
To make the aircraft controllable, then, the stick force to trim must lie within rea-
sonable limits: 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 acceleration on the aircraft with a risk of overloading the structure. To start
with, we need the hinge moment to trim, which we can derive from our previous
analysis of the tab angle to trim, §2.5.
From the definition of hinge moment coefficient:
CH = b0 + b1αT + b2η + b3β,
we can rearrange to find η as a function of CH 6= 0:
η =
CH − b0 − b1αT − b3β
b2
.
From the data sheet:
CLT
= a1

1 −
d
dα

CL
a
+ a1(ηT − 0) + a2η + a3β,
and, on substituting η:
CLT
= a1

1 −
d
dα

CL
a
+ a1(ηT − 0) + a3β +
a2
b2
(CH − b0).
This equation is the general form of one we have already derived for the zero stick
force case. Given that:
,
and substituting for CLT
:
CM = CM0 − (h0 − h)CL − V

a1
a

1 −
d
dα

CL + a1(ηT − 0) + a3β +
a2
b2
(CH − b0)

,
which can be re-arranged to find β:
V a3β = CM0 − (h0 − h)CL − V

a1
a

1 −
d
dα

CL + a1(ηT − 0) −
a2b0
b2

, (5.1)
or hinge moment to trim:
V
a2CH
b2
= CM0 − (h0 − h)CL − V

a1
a

1 −
d
dα

CL + a1(ηT − 0) + a3β −
a2b0
b2

.
(5.2)

5.2. MORE FLIGHT TESTING 35
We could use (5.2) to work out the hinge moments directly, but it is more conve-
nient to use the stick-free case as a reference. Subtracting (5.2) from (5.1):
V

a3β −
a2
b2
CH

= V a3β,
yielding:
CH =
b2
a2
a3(β − β)
so that:
Pe = me
ρV 2
2
Sηcη
b2
a2
a3(β − β),
so that the stick force to trim is proportional to the difference between the actual tab
angle and the tab angle to trim.
5.2 More flight testing
In theory we might use the stick forces to calculate the stick free neutral point, begin-
ning from:
CH =
b2
a2
a3(β − β),
which, under differentiation with respect to lift coefficient at constant tab angle, yields:
∂CH
∂CL
=
b2
a2
a3
∂β
∂CL
.
In §3.3, we found that:
dβ
dCL
= −
K0
n
V
1
a3
,
so that:
dCH
dCL
= −
b2K0
n
V a2
.
In principle, by measuring the stick force or hinge moment at different flight con-
ditions, we can work out the stick free neutral point. In practice, however, we cannot
measure the force accurately enough for a reliable estimate, due to errors introduced
by such things as friction in the system.

36 CHAPTER 5. STICK FORCES
5.3 Modification of stick forces
There are three main methods which can be used to modify the stick forces to bring
them into the correct range for control of the aircraft:
Gearing between stick and control surface, but this is limited because of the range of
elevator movement required.
Power assistance, which can ‘share’ the load or supply all of the force required to
move the control, with a feedback system to give the pilot ‘feel’ for the control
input.
Aerodynamic methods of modifying the loads include adding surface ahead of the
hinge line (a ‘horn balance’), moving the hinge line or adding tabs which are
geared to the elevator, Figure 5.1.
Hinge line
Horn balance
Sη
a: horn balance b: hinge location
c: geared tab d: anti-balance tab
Figure 5.1: Aerodynamic assistance
Aerodynamic balancing is a means of changing the hinge moment required for a
given elevator deflection, dPe/dη:
Pe = me
ρV 2
2
SηcηCH,
but,
CH = b0 + b1αT + b2η + b3β,

5.3. MODIFICATION OF STICK FORCES 37
so that
dPe
dη
= me
ρV 2
2
Sηcηb2.
To reduce the stick force, we want to reduce b2, but dPe/dη b2, 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 may not have enough feel for the controls and other
methods of reducing the stick force may be needed.

Chapter 6
Manoeuvre stability
We have now completed our analysis of ‘straight and level’ static stability. The next
step is to examine longitudinally symmetric manoeuvres (i.e. manoeuvres that affect
the left and right hand side of the aircraft equally). The most straightforward example
of this is a steady ‘pullout’ at constant velocity. Somewhat surprisingly, the elevator
angle required for pitch trim in a steady banking turn can also be calculated in the
same way. This is because the radius of a typical banked turn is very large. Hence,
the asymmetry in the flow is small once the turn has been initiated.
6.1 Analysis of a steady pullout
Consider two conditions, shown in Figure 6.1:
1. an aircraft in steady, level flight at speed V ;
2. the same aircraft in a steady pullout at speed V .
In the steady pullout the aircraft has a radial (centripetal) acceleration V 2
/r = ng.
The difference in lift between (1) and (2) is nmg = nW. Hence:
∆CL = nCL =
nW
ρV 2S/2
where CL is the lift coefficient in the straight and level case.
The other key difference between the two cases is that in the steady pullout the
aircraft has an angular (pitch) velocity as well as a linear velocity. This angular ve-
locity can easily be found by considering the amount of time that the aircraft would
take to complete a full circle at constant speed. If the aircraft completed a full circle it
would pitch though 2π radians and would therefore cover a distance of 2πr, where r
is the radius of the circle. The time taken, t, at speed V would be:
t =
2πr
V
,
39

40 CHAPTER 6. MANOEUVRE STABILITY
V
CL1, L = W
W = mg
CL2 = (1 + n)CL1, L = (1 + n)W
W = m(1 + n)g
r
1: Steady, level flight 2: Steady pullout
Figure 6.1: Manoeuvre conditions
but by considering the centripetal acceleration we also know that:
r =
V 2
ng
.
Hence,
t =
2πV
ng
.
The aircraft has pitched through a total angle of 2π radians in this time. Therefore the
pitch rate, q, is:
q =
2π
2πV/ng
=
ng
V
.
This pitching will cause the tail of the aircraft to move down relative to the incoming
air. This causes the incidence at the tailplane to increase by an amount:
∆αT =
qlT
V
,
where lT is the tail arm measured from the centre of gravity. We already have an expres-
sion for q, so we get:
∆αT =
nglT
V 2
.
Unfortunately, this expression has a V 2
term, and hence will vary with the flight
conditions. We can get rid of this term by applying the definition of the lift coefficient,

6.1. ANALYSIS OF A STEADY PULLOUT 41
CL, for the straight and level case:
CL =
W
ρV 2S/2
.
Hence
V 2
=
W
ρSCL/2
.
Substituting this back into the expression for ∆αT results in:
∆αT =
ρgSlT
W
nCL
2
,
or
∆αT =
nCL
2µ1
,
where µ1 = W/ρgSlT , the longitudinal relative density.
The change in incidence of the tailplane causes the lift coefficient of the tailplane
to alter by an amount a1∆αT.Therefore, remembering that the lift coefficient of the
aircraft in the steady pullout is (1 + n)CL:
CLT
=
a1
a

1 −
d
dα

(1 + n)CL + a1(ηT − 0) + a2η + a3β + a1∆αT.
Hence,
CLT
=
a1
a

1 −
d
dα

(1 + n)CL + a1(ηT − 0) + a2η + a3β + a1
nCL
2µ1
.
The basic pitching moment equation is still valid, since it makes no assumptions
about the source of the lift and moments—it is simply the result of non-dimensionalizing
a free body diagram. Therefore, this revised expression for CLT
can be substituted.
Again, remembering that the lift coefficient in the steady pullout is (1 + n)CL:
CM = CM0 − (h0 − h)(1 + n)CL
− V

a1
a

1 −
d
dα

(1 + n)CL + a1(ηT − 0) + a2η + a3β + a1
nCL
2µ1

.
For the straight and level flight of the aircraft, in trim, we have previously derived
the equation:
CM = 0 = CM0 − (h0 − h)CL − V

a1
a

1 −
d
dα

CL + a1(ηT − 0) + a2η + a3β

.
(6.1)

In trim, the pitching moments acting on the manoeuvring aircraft will be zero if
the aircraft is undertaking a steady manoeuvre. If we now look at the expression for
trim in a steady pullout, and look at the change in elevator angle required for trim,
such that the elevator angle is now η + ∆η we get:
0 = CM0 − (h0 − h)(1 + n)CL−
V

a1
a

1 −
d
dα

(1 + n)CL + a1(ηT − 0) + a2(η + η) + a3β + a1
nCL
2µ1

. (6.2)
These equations (6.1) and (6.2) are very similar. We can therefore subtract one
from the other and get:
0 = −(h0 − h)nCL − V

a1
a

1 −
d
dα

nCL + a1
nCL
2µ1
+ a2∆η

. (6.3)
This can be rearranged to get the elevator deflection/g required for a steady pull-
out:
∆η
n
= −
CL
V a2

(h0 − h) + V

a1
a

1 −
d
dα

+
a1
2µ1

.
This must always be negative, otherwise the aircraft would pitch nose-down when
the pilot pulls back.
6.2 Stick fixed manoeuvre point
When ∆η/n = 0 the c.g. is at the stick fixed manoeuvre point. Hence, at h = hm:
0 = −
CL
V a2

(h0 − hm) + V

a1
a

1 −
d
dα

+
a1
2µ1

,
hm = h0 + V

a1
a

1 −
d
dα

+
a1
2µ1

.
This should be compared with the neutral point location stick fixed, hn, which we
have previously shown to be:
hn = h0 + V
a1
a

1 −
d
dα

.
Therefore, the stick fixed manoeuvre point is a distance a1c/2µ1 aft of the stick
fixed neutral point. It is worth noting that the location of the stick fixed manoeu-
vre point varies with altitude, since µ1 is a function of the air density as well as of
geometry.

6.3. STICK FIXED MANOEUVRE STABILITY 43
6.3 Stick fixed manoeuvre stability
The stick fixed manoeuvre margin, Hm, is defined as:
Hm = hm − h.
We showed in §6.1 that:
∆η
n
= −
CL
V a2

(h0 − h) + V

a1
a

1 −
d
dα

+
a1
2µ1

.
Hence,
h = h0 +
V a2
CL
∆η
n
+ V

a1
a

1 −
d
dα

+
a1
2µ1

.
Therefore,
Hm = −
V a2
CL
∆η
n
.
This result is important because it demonstrates that there is a relationship between
the stick fixed manoeuvre margin and the elevator angle to trim. There is, seemingly,
a discrepancy between the fact that hm moves with changing altitude and the above
expression. How can this discrepancy be explained?
6.4 Stick free manoeuvre stability
‘Stick fixed’ analysis has enabled us to calculate the elevator angles required to trim
the aircraft in a steady pullout or bank, but tells us nothing about the stick forces
required for the manoeuvres (just as stick fixed analysis told us nothing of the stick
forces for straight and level flight). ‘Stick free’ manoeuvre stability analysis will allow
us to calculate these stick forces.
6.5 Analysis
In §5.1, we derived an expression that allowed us to calculate the hinge moments for
trim in straight and level flight:
CM = 0 = CM0 − (h0 − h)CL − V

a1
a

1 −
d
dα

CL + a1(ηT − 0) + a3β +
a2
b2
(CH − b0)

.
The same process can be undertaken to find the hinge moments for trim in a steady
pullout, CH + ∆CH. For the pullout, assuming that the tab is not used:
CM = 0 = CM0 − (h0 − h)(1 + n)CL−
V

a1
a

1 −
d
dα

(1 + n)CL + a1(ηT − 0) + a3β + a1
nCL
2µ1
+
a2
b2
(CH + CH − b0)

,

where CL is, again, the lift coefficient in straight and level flight.
Equating these two expressions and cancelling identical terms results in:
0 = −(h0 − h)nCL − V

a1
a

1 −
d
dα

nCL + a1
nCL
2µ1
+
a2∆CH
b2

.
Hence,
V a2
b2CL
∆CH
n
= −(h0 − h) − V

a1
a

1 −
d
dα

+
a1
2µ1

.
The stick free manoeuvre point, h0
m, is defined as the c.g. position that gives ∆CH/n = 0.
Therefore,
h0
m = h0 + V

a1
a

1 −
d
dα

+
a1
2µ1

.
The stick free manoeuvre margin, H0
m, is defined as:
H0
m = h0
m − h,
which can easily be shown to be:
H0
m = −
V a2
b2CL
∆CH
n
.
The stick force per g is calculated from the hinge moment per g, in exactly the
same way as for straight and level flight:
∆Pe
n
= me
ρV 2
2
Sηcη
∆CH
n
.
For handling safety the stick force required to pull high g should be appreciable to
avoid accidentally exceeding the structural limitations of the aircraft. A typical value
for a non-aerobatic aircraft is usually of the order of 20 N/g.
6.6 Tailless aircraft
The analysis for tailless aircraft is very similar to that for conventional aircraft. The
flight conditions, as for conventional aircraft, are shown in Figure 6.2.
For a tailless aircraft in steady trimmed flight we have already derived the equa-
tion (§4.1):
CM = 0 = CM0 +
∂CM0
∂η
η − (h0 − h)CL.

6.6. TAILLESS AIRCRAFT 45
V
CL1, L = W
W = mg
V
CL2 = (1 + n)CL1, L = (1 + n)W
W = m(1 + n)g
r
q = ng/V
1: Steady, level flight 2: Steady Pullout
Figure 6.2: Manoeuvre conditions for a tailless aircraft
When the aircraft is in a steady pullout with radial acceleration ng and with pitch rate
q we can write:
CM = 0 = CM0 +
∂CM0
∂η
(η + ∆η) − (h0 − h)(1 + n)CL +
∂CM
∂q
q.
Again, we can subtract one equation from the other to get:
∆η =
1
∂CM0 /∂η

(h0 − h)nCL −
∂CM
∂q
q

,
where ∂CM /∂q is an aerodynamic derivative. There are a large number of aerodynamic
derivatives that can be defined for any aircraft, and they enable us to calculate the
aerodynamic behaviour of the aircraft. We will encounter more aerodynamic deriva-
tives when we examine the dynamic stability of aircraft. There is a standard non-
dimensionalised form for each of these parameters. The non-dimensional form of
∂CM /∂q is given the symbol mq, and is defined as:
mq =
1
ρV Sc2
0
∂M
∂q
.
Hence,
∂CM
∂q
=
1
ρV 2Sc0/2
∂M
∂q
=
2c0
V
mq
and we know that
q =
ng
V
.
Therefore,
∆η =
1
∂CM0 /∂η

(h0 − h)nCL −
2c0
V
mq
ng
V

.

As for the conventional aircraft, we have an expression that includes the flight
velocity. Again, we can remove this by using the definition of the lift coefficient and
rearranging such that:
V 2
=
W
ρSCL/2
.
Making this substitution results in:
∆η =
1
∂CM0 /∂η

(h0 − h)nCL − mqnCL
ρgc0S
W

.
The longitudinal relative density, µ1, for a tailless aircraft is defined as:
µ1 =
W
ρgSc0
.
Using this definition and rearranging results in:
∆η
n
=
1
∂CM0 /∂η

(h0 − h) −
mq
µ1

CL.
This expression can be used to calculate the elevon deflections required to undertake
manoeuvres.
6.7 Tailless aircraft manoeuvre point
As for the conventional aircraft, the manoeuvre point is defined by the c.g. location
that results in ∆η/n = 0. Therefore,
hm = h0 −
mq
µ1
.
The resulting aerodynamic forces due to a positive pitch rate are shown in Figure 6.3.
Forces oppose motion
Figure 6.3: Aerodynamic forces during pitching motion
These forces all oppose the motion of the aircraft. Hence, mq is always negative.
This means that the manoeuvre point for a tailless aircraft is always aft of the neutral
point for the aircraft (which is at h = h0). The damping in pitch has increased the
stability.

6.8. TAILLESS AIRCRAFT MANOEUVRE MARGINS 47
6.8 Tailless aircraft manoeuvre margins
The manoeuvre margin for a tailless aircraft, Hm, is defined identically to that for a
conventional aircraft:
Hm = hm − h.
Hence
Hm = (h0 − h) −
mq
µ1
= Kn −
mq
µ1
.
The elevon angle per g can therefore be written as:
∆η
n
=
HmCL
∂CM0 /∂η
.
The elevon angle per g is therefore directly proportional to the manoeuvre margin.
6.9 Relationships between static and manoeuvre
margins
Conventional aircraft
We have shown that the static margins, stick fixed and stick free, for conventional
aircraft are:
Kn = (h0 − h) + V
a1
a

1 −
d
dα

,
K0
n = (h0 − h) + V
a1
a

1 −
d
dα

.
Also, the manoeuvre margins for conventional aircraft are:
Hm = (h0 − h) + V

a1
a

1 −
d
dα

+
a1
2µ1

,
H0
m = (h0 − h) + V

a1
a

1 −
d
dα

+
a1
2µ1

.
Therefore,
Hm = Kn +
V a1
2µ1
,
H0
m = K0
n +
V a1
2µ1
.
The manoeuvre points of conventional aircraft are aft of the respective neutral points.
This is due to the stabilising influence of additional lift at the tailplane due to the
pitch rate. Note that since µ1 is a function of the air density the manoeuvre margin
decreases with increasing altitude.

Tailless aircraft
We have shown that the static margin for tailless aircraft is:
Kn = h0 − h
and that the manoeuvre margin is:
Hm = (h0 − h) −
mq
µ1
.
Therefore,
Hm = Kn −
mq
µ1
.
As for conventional aircraft, a tailless aircraft is more stable when manoeuvring
due to the stabilising effect of the pitch damping term mq (remember, mq is negative).
Again, the manoeuvre margin is reduced at high altitudes due to the presence of a
density term in µ1.
6.10 Modification of stick free neutral and manoeuvre
points
Two common ways of modifying the stick free neutral and manoeuvre points are
shown in Figure 6.4.
Bob weight
a: Spring b: Bob weight
Figure 6.4: Modification of neutral and manoeuvre points
A spring or bob-weight is defined as positive if it exerts a moment that would
cause a positive deflection of the elevator.
These two additions have no effect on the stick-fixed neutral or manoeuvre points
since the calculation of these locations does not require the consideration of hinge mo-
ments. However, it can be shown that a positive spring moves the stick free neutral
point aft but has no effect on the stick free manoeuvre point. In contrast, a positive
bob-weight moves both the stick free neutral point and the stick free manoeuvre point
of an aircraft aft. These effects are shown in Figure 6.5.
By combining positive and negative springs and bob-weights it is possible to
move the two stick free points independently of each other. Since the stick forces

6.10. MODIFICATION OF STICK FREE NEUTRAL AND MANOEUVRE POINTS49
N M
Spring
Bob-weight
N0
M0
Figure 6.5: Effects of positive springs and bob-weights
are directly proportional to the stick free static margin and the stick free manoeuvre
margin this enables the stick forces to be modified by a simple mechanical addition
to the system.
For example, an aircraft might have suitable stick free static stability but insuf-
ficient manoeuvre margins. This results in a stable aircraft with good feel for the
pilot and suitable stick loads for trim, but the low stick force per g resulting from the
low manoeuvre margin might cause a risk of inadvertently overstressing the aircraft.
The addition of a negative spring together with a positive bob-weight would solve
this problem since the stick free static margin would be unchanged but the stick free
manoeuvre margin would increase. This is shown in Figure 6.6.
Positive bob-weight
Negative spring
N0
M0
Figure 6.6: Modification of stick force per g

Chapter 7
Compressibility effects
Everything that we have examined so far has assumed linear aerodynamics, incom-
pressible flow, and rigid aircraft. In reality, of course, none of these assumptions will
be valid at all flight conditions. At higher angles of attack the aerodynamics become
non-linear (i.e. ∂CL/∂α not constant) and as the aircraft flies faster other effects be-
come important. In this section we will briefly outline the major changes that occur
at high speeds, and the effect that this has on the control of the aircraft.
7.1 High speed effects
Changes from the low speed case arise primarily from Mach number (compressibil-
ity) and distortion (aeroelastic) effects. The dominant effects of increasing Mach num-
ber come from:
• change of lift curve slope with Mach number;
• movement of the aerodynamic centre rearwards, from quarter-chord at low
speed to mid-chord at supersonic Mach numbers.
These effects are shown in Figure 7.1. Combined with these are the effects of Mach
number on zero-lift pitching moment and zero-lift angle.
The effect of aeroelasticity is to reduce lift curve slopes with increasing ρV 2
/2,
dynamic pressure. The loads acting on an aircraft are proportional to the dynamic
pressure, if the lift and drag coefficients are constant. The deflections are, similarly,
proportional to the forces. Hence, all aeroelastic effects depend on the dynamic pres-
sure. This results in changes in the aeroelastic response of the aircraft at different
altitudes, since the ambient air density varies with altitude. If we superimpose alti-
tude effects onto the variation of lift curve slope with Mach number for a typical (aft)
swept wing aircraft, we get a result as in Figure 7.3.
There are similar effects on the effectiveness of the tailplane and the elevator, as
shown in Figure 7.4. The downwash at the tail typically varies as shown in Figure 7.5,
decreasing to zero at high supersonic speeds, since any ‘downwash’ is confined to the
volume of air influenced by the wing.
51

52 CHAPTER 7. COMPRESSIBILITY EFFECTS
M
a
1.0 M
h
1.0
c/4
c/2
Figure 7.1: Compressibility effects on lift curve slope and aerodynamic centre
M
CM0
1.0 M
α0
1.0
Figure 7.2: Compressibility effects on zero lift pitching moment and zero lift angle
The usual result of the combined effects is that stability reduces as the Mach num-
ber nears unity and then increases, sometimes rapidly, to a higher value at supersonic
speeds. The variation of CM for a typical aircraft is shown in Figure 7.6.
To counteract the nose down pitching moment that often occurs on swept wing
aircraft (subsonic jet transports—Boeing 707, 747, etc.) an up-elevator or stabilisation
input is provided by a Mach number sensing system. This is known as ‘Mach trim’. If
the nose down moment were allowed to take effect the stick force gradient would be
reversed, and there is also a danger that the maximum allowable speed of the aircraft
due to structural limits would be exceeded. The stick forces for such an aircraft are
shown in Figure 7.7.

7.1. HIGH SPEED EFFECTS 53
M
a
1.0
Rigid aircraft
Sea level
Decreasing altitude
Figure 7.3: Aeroelastic effects on lift curve slope
M
V̄ a1
1.0
Rigid aircraft
Sea level
Decreasing altitude
M
V̄ a2
1.0
Rigid aircraft
Sea level
Decreasing altitude
Figure 7.4: Aeroelastic effects on tailplane and elevator

54 CHAPTER 7. COMPRESSIBILITY EFFECTS
M
∂/∂α
1.0
Figure 7.5: Variation of downwash with Mach number
M
CM
1.0
Mach tuck
Figure 7.6: Variation of pitching moment with Mach number
M
Push
Pull
1.0
Uncorrected stick force
Mach trim input
Figure 7.7: Variation of stick forces with Mach number

Chapter 8
Dynamic behaviour of aircraft
In the first part of the course, we examined the static stability of aircraft, which means
that we considered whether an aircraft tends to return to its equilibrium position after
a perturbation, §1.1. We are now going to analyze the dynamic stability of aircraft and
see how they respond over time to perturbations in flight.
8.1 Axes and notation
The axes and notation for the analysis of dynamic stability of an aircraft are given
in Figure 8.1 and follow a logical order. Once the x, y and z-axes are defined we then
have, for example, L, M and N—the moments about the x-, y- and z-axes, respec-
tively.
The axis system uses ‘body axes’ where the system is not locked in position in
space, but moves with the aircraft. The origin of the axis system is at the centre of
gravity of the aircraft, since all rotations take place about the c.g.
A rigid aircraft has six degrees of freedom. To simplify the equations used when
performing analysis of the dynamic modes of an aircraft, these degrees of freedom are
expressed as perturbation quantities in relation to steady straight flight (i.e. velocity
perturbations u, v and w and rotational velocities p = φ̇, q = θ̇ and r = ψ̇).
8.2 Aerodynamic derivatives
We now have a co-ordinate system that allows us to define any perturbation of the
aircraft from straight and level flight. To continue, we need to find out what forces
are acting on the aircraft for a given perturbation1
.
1
The analysis which follows is taken from MILNE-THOMSON, L. M., Theoretical aerodynamics,
MacMillan and Company, 1966.
57

58 CHAPTER 8. DYNAMIC BEHAVIOUR OF AIRCRAFT
z
x
y
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
Figure 8.1: Notation for analysis of dynamic stability
If we assume that the effects of each perturbation are linear (true for small pertur-
bations), then:
M =
∂M
∂u
u +
∂M
∂v
v +
∂M
∂w
w +
∂M
∂p
p +
∂M
∂q
q +
∂M
∂r
r.
The partial derivatives in this expression are known as ‘aerodynamic derivatives’ or
‘stability derivatives’. We have already met the derivative ∂M/∂q, often known as
‘pitch damping’, in our analysis of control deflections for tailless aircraft. Therefore,
if we know the aerodynamic derivatives and the perturbations, we can calculate all
of the forces and moments acting on the aircraft (six equations). If, in addition, we
know the mass of the aircraft and its inertia in roll, pitch and yaw (A, B, C) we can
calculate the acceleration of the aircraft, and hence its dynamic response.
The forces on the aircraft are the aerodynamic force F and mg:
F = Xi + Y j + Zk, (8.1)
mg = mg1i + mg2j + mg3k, (8.2)

8.2. AERODYNAMIC DERIVATIVES 59
where the components of g are needed because the reference frame is fixed to the
aircraft and is not necessarily horizontal. The other quantities we need for the aircraft
are:
v = ui + vj + wk, velocity,
Ω = pi + qj + rk, angular velocity,
h = h1i + h2j + h3k, angular momentum.
The equations of motion in translation and rotation are then:
d
dt
(mv) = mv̇ + Ω × (mv) = F + mg, (8.3a)
dh
dt
= ḣ + Ω × h = L, (8.3b)
where the boxed terms are required because the frame of reference is rotating. The
applied moment L is:
L = Li + Mj + Nk.
Equations 8.3 are the general equations of motion for an aircraft and could, in princi-
ple, be used to calculate the motion given enough information about the aerodynam-
ics and mass distribution. We, however, want to know if the aircraft is dynamically
stable, so we need to make some approximations to see how the aircraft behaves
when perturbed from steady flight.
In steady flight, we write:
v = V, Ω = 0, F + mg = 0,
and add the small perturbations so that:
V = V1 + u,
V1 = Ui,
u = ui + vj + wk,
ω = pi + qj + rk.
For a small rotation χ,
χ = φi + θj + ψk,
ω = φ̇i + θ̇j + ψ̇k.
Similarly, the perturbation forces are:
F + δF, m(g + δg),

and it can be shown that
δg + χ × g = 0.
Inserting these assumptions into (8.3) yields the equations of motion for small pertur-
bations:
mu̇ + m(χ̇ × V1 + χ × g) = δF, (8.4a)
ḣ = δL. (8.4b)
We can now simplify the system by making certain (reasonable) assumptions. First,
we assume that the forces and moments depend only on velocities and not on accel-
erations, with the exception of the dependence of pitching moment on ẇ, the down-
wash acceleration. Then:
δF = δXi + δY j + δZk,
δL = δLi + δMj + δNk,
and, for example,
δX =
∂X
∂u
u +
∂X
∂v
v +
∂X
∂w
w +
∂X
∂p
p +
∂X
∂q
q +
∂X
∂r
r,
δM =
∂M
∂u
u +
∂M
∂v
v +
∂M
∂w
w +
∂M
∂ẇ
ẇ +
∂M
∂p
p +
∂M
∂q
q +
∂M
∂r
r.
Secondly, we are assuming that the aircraft is symmetric so that a symmetric pertur-
bation can only cause a symmetric response. This means that a pitch disturbance, for
example, cannot cause a response in yaw or roll. Also, the symmetric response to an
asymmetric input has to be symmetric: if the aircraft rolls at a given rate, the pitch
response must be the same whether it rolls in a positive or negative sense. These two
requirements imply that:
∂Y
∂u
=
∂Y
∂w
=
∂Y
∂q
=
∂L
∂u
=
∂L
∂w
=
∂L
∂q
=
∂N
∂u
=
∂N
∂w
=
∂N
∂q
≡ 0,
∂X
∂p
=
∂X
∂q
=
∂X
∂r
=
∂Z
∂p
=
∂Z
∂q
=
∂Z
∂r
=
∂M
∂p
=
∂M
∂q
=
∂M
∂r
≡ 0.
Eliminating zero terms, we can write:
δF = (
∂X
∂u
u +
∂X
∂w
w +
∂X
∂q
θ̇)i + (
∂Y
∂v
v +
∂Y
∂p
φ̇ +
∂Y
∂r
ψ̇)j
+ (
∂Z
∂u
u +
∂Z
∂w
w +
∂Z
∂q
θ̇)k,
δL = (
∂L
∂p
φ̇ +
∂L
∂r
ψ̇ +
∂L
∂v
v)i + (
∂M
∂q
θ̇ +
∂M
∂u
u +
∂M
∂w
w +
∂M
∂ẇ
ẇ)j
+ (
∂N
∂p
φ̇ +
∂N
∂r
ψ̇ +
∂N
∂v
v)k.

8.3. LONGITUDINAL SYMMETRIC MOTION 61
We need one more assumption about the aircraft, which is that there is no inertial
coupling between yaw and roll. This means that the only moments of inertia we
need consider are A, B and C, the moments of inertia about the coordinate axes.
Now, assuming disturbed horizontal flight and expanding the cross products in (8.4)
yields the equations of motion for each translational and rotational component:
mu̇ =
∂X
∂u
u +
∂X
∂w
w +
∂X
∂q
q − mgθ, (8.5a)
m(ẇ − Uq) =
∂Z
∂u
u +
∂Z
∂w
w +
∂Z
∂q
, (8.5b)
Bq̇ =
∂M
∂q
q +
∂M
∂u
u +
∂M
∂w
w +
∂M
∂ẇ
ẇ. (8.5c)
and
m(v̇ + Ur) =
∂Y
∂v
v +
∂Y
∂p
p +
∂Y
∂r
r + mgφ, (8.6a)
Aṗ =
∂L
∂p
p +
∂L
∂r
r +
∂L
∂v
v, (8.6b)
Cṙ =
∂N
∂p
p +
∂N
∂r
r +
∂N
∂v
v. (8.6c)
The first of these sets 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.
8.3 Longitudinal symmetric motion
The important information about the dynamic response of a system is the set of
modes in which it oscillates2
. These can be found by the usual method of insert-
ing an assumed solution into the differential equations and finding combinations of
parameters which satisfy the system. The most convenient form of solution is:
u = u0eλt
, v = v0eλt
, θ = θ0eλt
.
Inserting these assumptions into (8.5a), for example, yields:
mu0λeλt
=
∂X
∂u
u0eλt
+
∂X
∂w
w0eλt
+
∂X
∂q
θ0λeλt
− mgθ0eλt
.
2
The following analysis, with different notation, is based on GRAHAM, W., ‘Asymptotic analysis
of the classical aircraft stability equations’, Aeronautical Journal, February 1999, pp 95–103.

Now, we can divide through by exp λt and, as always, non-dimensionalize the pa-
rameters, to give the non-dimensional equations of motion:
(Λ − xu)u0
− xww0
−

xqΛ
µc
−
CL
2

θ0 = 0, (8.7a)
−zuu0
+ (Λ − zw)w0
− Λ

1 +
zq
µc

θ0 = 0, (8.7b)
−
mẇ
Λ
µc + mw

w0
+
Λ(bΛ − mq)
µc
= 0. (8.7c)
The non-dimensional parameters are:
xu =
∂X
∂u
ρUS
, xw =
∂X
∂w
ρUS
, zu =
∂Z
∂u
ρUS
, zw =
∂Z
∂w
ρUS
,
xq =
∂X
∂q
ρUSc
, zq =
∂Z
∂q
ρUSc
, mu =
∂M
∂u
ρUSc
, mw =
∂M
∂w
ρUSc
,
mq =
∂M
∂q
ρUSc2 ,
mẇ =
∂M
∂ẇ
ρSc
,
b =
B
mc2
and
Λ =
mλ
ρUS
, µc =
m
ρSc
,
also given on the data sheet.

Chapter 9
Normal modes of aircraft
Phugoid
The first approximate 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, in this case, the vertical motion is negligible or, equivalently,
the incidence is almost constant. We can rewrite (8.7) in matrix form, with the negli-
gible terms in each equation removed:


Λ − xu 0 CL/2
−zu 0 −Λ
0 −mw Λ(bΛ − mq)/µc




u0
w0
θ0

 =


0
0
0

 .
This equation can only have a non-trivial solution if the determinant of the matrix is
zero:
Λ2
− xuΛ −
−zu
2
CL = 0.
Solving for Λ gives:
Λ = −
(−xu)
2
± jΩph

1 −

−xu
2Ωph
2
#1/2
,
which specifies oscillatory motion with:
Ωph =

−zuCL
2
1/2
, natural frequency, (9.1a)
cph =
−xu
2Ωph
, damping. (9.1b)
This solution defines the phugoid mode, which is a lightly damped long period os-
cillation. The incidence is almost constant and the aircraft varies altitude at constant
energy, trading potential for kinetic and back again, Figure 9.1.
63

64 CHAPTER 9. NORMAL MODES OF AIRCRAFT
hmax, Vmin
hmin, Vmax
Figure 9.1: Phugoid oscillation trajectory
An important point to note is that the damping is proportional to (−zu), the rate
of change of vertical force with small changes in horizontal speed. Remembering
that the z axis points vertically down, we can see that zu 0 and the damping is
positive. Although it is not proven on the basis of these results, a statically stable
aircraft always has a stable phugoid.
Short period oscillation
The second solution for longitudinal oscillation is for the case where Λ is of order
µ
1/2
c , u0/θ0 is of order µ
−1/2
c and w0/θ0 is of order one. In this case, the approximation
to (8.7) is:





Λ − xu −xw −

xqΛ
µc
− CL
2

0 Λ − zw −Λ
0 −

mẇΛ
µc
+ mw

Λ(bΛ − mq)
µc







u0
w0
θ0

 =


0
0
0

 .
Again, we find the natural frequency by requiring that the determinant of the matrix
be zero:
Λ(Λ − xu)

Λ2
−

zw +
mq + mẇ
b

Λ +
zwmq − mwµc
b

= 0,
which, on solving the quadratic, gives a result for the non-dimensional natural fre-
quency and damping:
Ωspo =

µc(−mw) + mqzw
b
1/2
, natural frequency, (9.2a)
cspo = −
1
2Ωspo

zw +
mq + mẇ
b

, damping. (9.2b)
This is the short period oscillation and is a heavily damped mode with period typically
of a few seconds. The aircraft pitches rapidly about its centre of gravity which con-
tinues 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 Oscillation (PIO).

9.1. LATERAL MOTION 65
V
Figure 9.2: Short period oscillation
The frequency is proportional to K
1/2
n , and increases with dynamic 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.
The SPO is always stable for a statically stable aircraft.
9.1 Lateral motion
In the case of lateral motion, we again need to insert the assumed form for the solu-
tion:
v = v0eλt
, φ = φ0eλt
, r = r0eλt
,
and non-dimensionalize quantities in (8.6), which we do in the same way as before
except that our reference length is now s, the wingspan:
(Λ − yv)v0
−

ypΛ
µs
+
CL
2

φ0 +

1 −
yr
µs

r0
= 0, (9.3a)
−lvv0
+ (aΛ − lp)
Λ
µs
φ0 −
lr
µs
r0
= 0, (9.3b)
−nvv0
−
npΛ
µs
φ0 +
cΛ − nr
µs
r0
= 0. (9.3c)
The non-dimensional quantities are:
yv =
∂Y
∂v
ρUS
,
yp =
∂Y
∂p
ρUSs
, yr =
∂Y
∂r
ρUSs
, lv =
∂L
∂v
ρUSs
, nv =
∂N
∂v
ρUSs
,
lp =
∂L
∂p
ρUSs2
, lr =
∂L
∂r
ρUSs2
, np =
∂N
∂p
ρUSs2
, nr =
∂N
∂r
ρUSs2
,
a =
A
ms2
, c =
C
ms2
,
v0
=
v0
U
, r0
=
mr0
ρUS
, µs =
m
ρSs
.

Dutch roll
The first lateral mode we consider is Dutch roll which has oscillations of roughly equal
magnitude in pitch, yaw and roll. In this case, (9.3) reduce to:


Λ 0 1
−lv aΛ2
/µs 0
−nv 0 cΛ/µs




v0
φ0
r0

 =


0
0
0

 .
As before the determinant of the matrix must be zero for a non-trivial solution to
exist:
Λ2
(cΛ2
+ µsnv) = 0,
and the frequency of the oscillation is, on the approximations we are using:
Ωdr =
µsnv
c
1/2
. (9.4)
In Dutch roll, yawing oscillation (analogous to the longitudinal SPO) causes alternat-
ing sideslip. This in turn causes a rolling oscillation via Lvv. The periodic time is
typically a few seconds, but as for the SPO it should not have a period of less than
1.25s due to 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.
Spiral mode and roll subsidence
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
approximations to (9.3) are:


0 CL/2 1
−lv (aΛ − lp)Λ/µs −lr/µs
−nv −npΛ/µs (cΛ − nr)/µs




v0
φ0
r0

 =


0
0
0

 .
The requirement for a non-trivial solution is then that:
anvΛ2
+ [lv(np − cCL/2) − lpnv]Λ + (lvnr − lrnv)CL/2 = 0.
The two roots of this equation can be approximated as:
Λrs = −
(−lp)nv + (−lv)[cCL/2 + (−np)
anv
, (9.5)
and
Λsm = −
CL
2
lvnr − lrnv
(−lp)nv + (−lv)[cCL/2 + (−np)]
. (9.6)

9.2. DIHEDRAL EFFECT AND WEATHERCOCK STABILITY 67
Note that both of these roots are real and so they do not describe oscillations. The
first, Λrs, describes rolling subsidence which is a pure rolling motion that is generally
heavily damped, and is therefore usually stable. The damping is primarily from the
wings, where the incidence along the wing is changed due to the roll-rate, as shown
in Figure 9.3.
∆α
Roll rate p
Loading
Rolling moment Lpp 0
Figure 9.3: Rolling subsidence
This roll-rate results in a rolling moment Lpp. Therefore, if Lp is negative the
rolling subsidence mode is stable. This is generally the case. However, if Lp becomes
negative, 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 spins1
The second root Λsm, which is much smaller than Λrs, corresponds 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 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 aircraft into the relative wind due to a
yawing moment Nvv. However, the rolling moment due to sideslip Lvv tries to roll
the wings back level. Depending on which of the effects ‘wins’, the aircraft will be
spirally unstable or stable, as can be seen from the numerator of (9.6).
9.2 Dihedral effect and weathercock stability
The aerodynamic derivatives Lv and Nv define whether an aircraft is stable or un-
stable in rolling subsidence and Dutch roll. Lv and Nv are known as the ‘dihedral
effect’ and ‘weathercock stability’ respectively. The effect of the two aerodynamic
derivatives on the lateral stability of the aircraft is shown in Figure 9.4.
Dihedral effect
Lv is known as the dihedral effect since the majority of the rolling moment due to
sideslip comes from dihedral (on an aircraft with unswept wings), as shown in Fig-
1
Something similar can happen to microlights, in the so-called ‘tumble’, which is almost always fa-
tal: GRATTON, G, NEWMAN, S., ‘The “tumble” departure mode in weightshift-controlled microlight
aircraft’, Proceedings of the Institution of Mechanical Engineers, Part G: Journal of Aerospace Engineering,
March 2003, 217(3), pp. 149–166.

−Lv
Nv
Increasing altitude
Unstable spiral mode
Unstable Dutch roll
All lateral modes stable
Figure 9.4: Stability of the lateral modes
ure 9.5. Positive dihedral combined with positive sideslip results in a negative rolling
moment (and hence negative Lv).
Γ
Relative wind
Positive sideslip
Lv 0
Figure 9.5: Dihedral effect
Wing sweep has a large, negative, effect on Lv due to reduced or increased effec-
tive sweep for positive sideslip. This is shown in Figure 9.6.
Wing–fuselage interference effects give contributions to Lv due to changes in wing
effective incidence near the root. These contributions are negative for high mounted
wings and positive for low mounted wings, as shown in Figure 9.7.
A reasonable level of Lv may be achieved by using anhedral with swept and high
mounted wings (e.g. Harrier). Ground clearance issues may limit anhedral on low
wing aircraft, resulting in an unstable Dutch roll mode.

9.2. DIHEDRAL EFFECT AND WEATHERCOCK STABILITY 69
Reduced effective sweep:
increased lift
Increased effective sweep:
reduced lift
Positive sideslip
Lv 0
Figure 9.6: Wing sweep effects on Lv
Lv 0
Lv 0
High wing Low wing
Figure 9.7: Wing-fuselage interference effects on Lv
Weathercock stability
The aerodynamic derivative Nv is known as weathercock stability since it is, effec-
tively, the ability of an aircraft to turn into the wind. It is produced mainly by the
sideways lift-force of the fin in sideslip, and should always be negative. However, as
shown in §9.2, if Nv is too large the aircraft may be spirally unstable.
The fin contribution to Nv generally reduces with increasing Mach number, since
the fin’s lift curve slope is reducing. Therefore an aircraft with a large fin may be
spirally stable at high speeds but unstable at low speeds. This can be solved by using
‘paired’ fins close together. At low speeds their mutual interference reduces their
effectiveness, while at supersonic speeds this interference is progressively removed,
increasing their effectiveness to combat the decreasing lift curve slope. This is shown
in Figure 9.8.
For VSTOL aircraft (e.g. Harrier) engine air intake mass flow may give a negative
contribution to Nv making the aircraft directionally unstable in the hover and at low
forward speeds. This is because the air undergoes a change in direction to go down
the intake, and hence a momentum change, giving a sideforce acting ahead of the

M
fin a1
1.0
Mach cone
Figure 9.8: Use of twin fins at high speed
aircraft c.g., as shown in Figure 9.9.
Intake flow
Sideforce
Nv 0
Flight speed
Nv Fin
Intake flow
Total
Directionally
unstable
Figure 9.9: Intake effects on Nv

Part III
Spacecraft dynamics and control
71

Chapter 10
Getting around: Orbits
Spacecraft are governed by different, simpler, equations than aircraft, because they
operate without friction and, most of the time, have no applied thrust. The control of
spacecraft is a problem in fixing their orbital parameters, so we start by analyzing the
classic problem of two bodies orbiting each other.
10.1 The two-body problem
Figure 10.1 shows the two body problem: two masses m1 and m2 interact with a
gravitational force of magnitude Gm1m2/|r1 − r2|2
. We note, in passing, that one
mass will usually be very much greater than the other (when the Earth orbits the sun
or a satellite orbits the Earth, for example), though the analysis does not depend on
this assumption.
r1
r2
r2 − r1
m1
m2
Figure 10.1: The two-body problem
73

74 CHAPTER 10. GETTING AROUND: ORBITS
The equations of motion for the two masses, in an inertial frame, are:
m1r̈1 = −
Gm1m2
|r1 − r2|3
(r1 − r2), (10.1a)
m2r̈2 = −
Gm1m2
|r1 − r2|3
(r2 − r1). (10.1b)
We can extract one useful piece of information immediately by considering the
centre of mass of the system rc:
rc =
m1r1 + m2r2
m1 + m2
.
Adding equations 10.1 shows that r̈c ≡ 0. In other words, the centre of mass of the
system moves at constant velocity.
To find the relative motion of the masses, subtract m2 times( 10.1a) from m1 times (10.1b)
and define r = r2 − r1:
m1m2r̈ = −
Gm1m2(m1 + m2)
r3
r,
r̈ = −
µr
r3
, (10.2)
where the gravitational parameter µ = G(m1 + m2). For most purposes, m1 m2 and
µ ≈ Gm1. For Earth, µ = 3.98601 × 105
km3
/s2
.
x
y
θ
r
θ̂ r̂
Figure 10.2: Polar coordinate system
To solve for r, we rewrite the system using polar coordinates, Figure 10.2. There
are two unit vectors r̂ and θ̂, the radial and azimuthal vectors respectively:
r̂ = [cos θ sin θ], (10.3a)
θ̂ = [− sin θ cos θ], (10.3b)
and differentiation will easily show that:
˙
r̂ = θ̇θ̂, ¨
r̂ = θ̈θ̂ − θ̇2
r̂, (10.4a)
˙
θ̂ = −θ̇r̂,
¨
θ̂ = −θ̈r̂ − θ̇2
θ̂. (10.4b)

10.1. THE TWO-BODY PROBLEM 75
Since r = rr̂:
r̈ = (r̈ − rθ̇2
)r̂ + (2ṙθ̇ + rθ̈)θ̂ = −
µ
r2
r̂,
and, extracting components,
r̈ − rθ̇2
= −
µ
r2
, (10.5a)
2ṙθ̇ + rθ̈ = 0. (10.5b)
Equation 10.5b can be rearranged to show that:
d
dt

r2
θ̇

≡ 0.
This tells us that h = r2
θ̇, the angular momentum, is constant. We can also derive an
energy conservation equation by taking the dot product of (10.2) with ṙ:
dE
dt
≡ 0, (10.6)
E =
ṙ.ṙ
2
−
µ
r
, (10.7)
the energy per unit mass.
What we really want to know is the shape of the orbit, i.e. r(θ) and the easiest way
to do this is to convert the derivatives with respect to t to derivatives with respect to
θ. We also make the transformation r → 1/u. Then:
r =
1
u
,
ṙ = −
1
u2
du
dθ
θ̇,
r̈ = −
1
u2
d2
u
dθ2
θ̇2
.
Inserting these terms into (10.5a) and noting that h = r2
θ̇, the differential eqution of
motion is:
d2
u
dθ2
+ u =
µ
h2
, (10.8)
with solution:
u = A cos(θ − φ) +
µ
h2
. (10.9)
It can be shown, using (10.7), that
A =
µ
h2

1 + 2
Eh2
µ2
1/2
,

so that the orbit is given by:
r =
h2
/µ
1 + e cos(θ − φ)
. (10.10)
The minimum value of r occurs when θ = φ. If we use this point as the origin of
θ, the equation for the orbit is:
r =
p
1 + e cos θ
, (10.11)
where p = h2
/µ and the origin is called the perigee (for Earth orbits). Equation 10.11
has the form of a conic section. In particular, for most spacecraft it is elliptical (or
circular).
Elliptical orbits
Equation 10.11 describes the trajectory of an orbiting body as a conic section. The
geometry of an elliptical orbit is shown in Figure 10.3 which shows the notation and
geometrical parameters for an elliptical orbit. The ellipse has two focii and the Earth
(say) is at one focus, which is the centre of the coordinate system. The size and shape
of the ellipse are defined by the semi-major and semi-minor axes a and b, respectively.
The eccentricity of the orbit is e and tells us how distorted the ellipse is:
b = a(1 − e2
)1/2
,
so that when e = 0, the orbit is circular. Elliptical orbits have 0 e 1, parabolic
orbits have e = 1 and hyperbolic orbits have e 1.
a
b
a
Focus
Focus
r
θ
p
Perigee
Apogee
Figure 10.3: The geometry of an ellipse

10.2. ORBITAL MANEOUVRES 77
From Figure 10.3, we can also find the radius to perigee a(1 − e) and the radius to
apogee a(1+e). It can be shown, using the geometric properties of the elliptical orbit,
that the energy E is:
E = −
µ
2a
, (10.12)
and the orbital period T is:
T =
2π
µ1/2
a3/2
(Kepler’s third law), (10.13)
which tells us that once an orbital radius (or semi-major axis) is selected the period of
the orbit is fixed.
Example: space debris
For a spacecraft orbiting the earth at an altitude of 200km, estimate the highest veloc-
ity at which it might be hit by space debris. Assume that both the satellite and the
debris are in a circular orbit.
Orbital radius a = (6400 + 200)km,
Orbital period T =
2π
µ1/2
a3/2
,
= 5336s.
Orbital velocity v = 2πa/T,
= 7.7km/s.
In the worst case, a spacecraft might, in principle, be hit by debris travelling at 2v =
15.4km/s.
10.2 Orbital maneouvres
The first thing we have to note that makes spacecraft different from aircraft comes
from (10.7). Rearranging that equation, the spacecraft velocity is:
v =

2E +
2µ
r
1/2
, (10.14)
so that the spacecraft velocity is fixed by E, µ and r. Now µ is (more or less) constant
but we can vary E, r and v, though not independently—(10.14) is always true. If we
change the spacecraft velocity at some radius r, then E will change and the spacecraft
will enter a different orbit. This fact can be exploited in spacecraft maneouvres.

Transfer
LEO
GEO
a1
a2
Figure 10.4: Walter Hohmann and his transfer from low earth to geostationary Earth
orbit.
Hohmann transfer
The Hohmann transfer is the minimum energy transfer between two circular orbits.
If we have a satellite—of any type, this also works for lunar or Mars missions—in a
low earth orbit of radius a1, we can shift it to a higher orbit of radius a2 using the
scheme shown in Figure 10.4. The transfer orbit is elliptical with semi-major axis
a = (a1 + a2)/2 and is tangential to the two circular orbits.
In order to change a spacecraft’s orbit, we have to change its velocity v. To enter
the transfer orbit, we have to change from the circular orbit velocity v = (µ/a1)1/2
to
the elliptical orbit velocity at that position:
v2
= µ

2
a1
−
1
a

.
The change in velocity ∆v1 is then:
∆v1 =

2µ
a1
−
2µ
a1 + a2
1/2
−

µ
a1
1/2
. (10.15)
To switch to the circular orbit at a2, the velocity change is
∆v2 =

µ
a2
1/2
−

2µ
a2
−
2µ
a1 + a2
1/2
. (10.16)
The fundamental measure of performance for a spacecraft is its ∆v (‘Delta vee’), its
capacity to change velocity, as this limits its range of maneouvre. Absolutely every-
thing that a spacecraft uses, including propellant for maneouvring, has to be lifted
from the Earth’s surface, so it is vital to use economical transfer orbits. The only
problem with a Hohmann transfer is that it is slow—it minimizes the total ∆v but

10.2. ORBITAL MANEOUVRES 79
maximizes the time. For transfer to GEO, this does not matter (5.3 hours) but can be
a problem for other missions (Mars, for example). An especially important ∆v is the
velocity change required to escape the gravity of a planet. This is called the escape
velocity and is found by setting the total energy E to zero. From (10.7):
vesc =

2µ
r
1/2
, (10.17)
which for a body on the surface of the Earth is about 11km/s.
Orbital capture
Without going into the details of how it might be done, it is obviously possible to
have a spacecraft leave the orbit of one planet and approach the orbit of another. If all
we want is to fly past the planet, no more need be said. Indeed, a common method
of speeding up spacecraft is to have them approach another planet and pick up grav-
itational energy to accelerate them in another direction. If we want the spacecraft to
enter the orbit of the planet, however, we need to slow it down. Remember that the
spacecraft is moving very quickly because it has reached the escape velocity for the
planet it has left so orbital capture is not an easy job.
The first point to note, as a spacecraft approaches another planet, is that it is in
orbit about the sun. It is only as it comes close to the planet that it feels the effect of
the planet’s gravity. At some point, as for a Hohmann transfer, we need to change the
spacecraft velocity to match the velocity of a planetary orbit. These velocity changes
can be quite large (for Mars Global Surveyor, for example, ∆v = 973m/s) so it is
obviously important to find optimal methods which save propellant.
If we want to enter a circular orbit about the planet, the orbital velocity is v1 =
(µp/rp)1/2
where µp is the gravitational parameter for the planet and rp is the orbital
radius. If the spacecraft approaches the planet with speed v∞, its speed in the planet
frame is:
v1 =

v2
∞ +
2µp
rp
1/2
, (10.18)
by conservation of energy. The velocity change required is then:
∆v =

v2
∞ +
2µp
rp
1/2
−

µp
rp
1/2
. (10.19)
Now we have to think about which orbit to choose. Propellant costs mass costs
money so we would like to find the radius which minimizes ∆v. Making trajectory
adjustments during the trip is relatively cheap, because very small adjustments can
give large changes in the entry point, so we are free to pick the best rp. Differentiat-

ing (10.19) with respect to rp, we find
rp =
2µp
v2
∞
, (10.20)
∆vmin =
v∞
21/2
. (10.21)
This is not the absolute optimum because a highly elliptical orbit will have a smaller
∆v again. One method for entering a circular orbit, as used by Mars Global Surveyor,
is to pick a highly elliptical orbit (low ∆v) for the planetary capture and use aerobrak-
ing (no propellant used, though a bit slow) to lower the orbit.

Chapter 11
Getting things done: Spacecraft control
Spacecraft are hard things to control because, unlike aircraft, they have no drag act-
ing on them: they will go on doing whatever they were doing until a control input
is applied, subject only to gravitational effects. This makes life simple in one way
since basic Newtonian mechanics applies, without dissipation, but it does make the
designer’s life quite difficult, since there is no damping. The two main functions of a
control system on a spacecraft are attitude control and manouevre or orbit change.
11.1 Attitude control
A satellite is usually required to point stably in a certain direction. Typical examples
include:
earth observation can only be performed if a satellite can point a camera at a known
position on the earth’s surface;
communications satellites need to beam information at a particular region (Corona-
tion Street is not very interesting to penguins);
scientific satellites such as Hubble need to point in a direction of interest to as-
tronomers.
Additionally, many satellites of various classes need to align their solar panels in the
right direction to generate power. Depending on the reason for the alignment, and the
accuracy required, there are a number of methods available for aligning spacecraft.
The brute force technique of firing a thruster to shift the spacecraft back on to the
desired attitude is wasteful of fuel and makes the system bounce back and forth. In
practice, spacecraft are designed to be inherently stable, which is achieved by making
them spin.
A mechanics textbook will give the Euler equations for a spinning body. If the
body is axisymmetric (i.e. the moment of inertia is the same about two of its principal
81

82 CHAPTER 11. GETTING THINGS DONE: SPACECRAFT CONTROL
axes):
M1 = A
dω1
dt
+ (C − A)ω2ω3, (11.1a)
M2 = A
dω2
dt
+ (A − C)ω3ω1, (11.1b)
M3 = C
dω3
dt
+ (A − A)ω1ω2, (11.1c)
where Mi and ωi are the moment and angular velocity about the ith axis and A and
C are the moments of inertia. For a body in space, there are no applied moments so
Mi ≡ 0 and this means that dω3/dt = 0 or ω3 = Ω and the body rotates at constant
frequency about its axis. Now, we can solve for the other two angular velocities:
dω1
dt
= +
A − C
A
Ωω2,
dω2
dt
= −
A − C
A
Ωω1.
Differentiating the first and substituting the second equation:
dω1
dt2
+ α2
ω1 = 0,
where α2
= Ω(C − A)/A. This is a simple harmonic oscillator and the solution is:
ω1 = ω0 sin αt, (11.2a)
ω2 = −ω0 cos αt. (11.2b)
This solution says that the spinning body rotates about one axis at a frequency Ω
and oscillates about the other two. The axis of rotation actually swings around and
describes a cone in space. If the spacecraft has been designed and set up properly, the
spin stabilizes its attitude and the coning motion is small.
Gravity gradient
A simple way to align a satellite is to use a gravity gradient device: if a spacecraft
is asymmetric, the variation in gravity over its extent is enough to generate a torque
which draws it back into the required alignment. This is like having a pendulum
which wants to swing into a ‘vertical’ orientation.
Sun-synchronous orbits
By exploiting the fact that the earth is not perfectly spherical, a satellite can be placed
into an orbit which is synchronized with the sun. This means that the satellite will
spend half its orbit in sunlight and will never have to work in a ‘twilight region’.
For an observation satellite, this also means it will pass a given point on the earth at

11.2. MANOUEVRING 83
V-bar
V-bar approach
R-bar
R-bar approach
Figure 11.1: Two approaches for a docking spacecraft
the same local time every day, which makes it easier to interpret images because the
ground will always be illuminated from the same direction. Finally, this means that
the satellite can guarantee that its solar panels will be generating power: an important
consideration for some systems.
11.2 Manouevring
Manouevring is a special case of changing orbits, but generally on a smaller scale.
The same rules apply, but our intuition is even more confused than usual, because
things don’t look right.
An example is the docking of one spacecraft with another (supply flights to a
space station; Shuttle to Hubble; Apollo lander with orbiter). If one spacecraft is
already on orbit, the other must approach that orbit to rendezvous with it. There are
two standard approaches.
The first is to arrive along the orbit in front of the other spacecraft. Because this
means flying along the velocity vector of the main craft, it is called a ‘v-bar’ approach
(as in v̄ for velocity). The second approach is along the radius vector to the main craft,
called an ‘r-bar’ approach. The ‘v-bar’ approach is easier because flying along the
radius vector makes the docking craft move ahead of the main craft, if it approaches
from below, as the orbital velocity at the lower altitude is a bit higher. Likewise,
approaching from above, you tend to fall behind the vehicle you are rendezvousing
with.
While a v-bar approach is favoured for simplicity, an r-bar manouevre is often
favoured because it means that the thrusters used for braking never point at the main
craft. In order to dock on a v-bar trajectory, you need to fire thrusters at the main craft,

84 CHAPTER 11. GETTING THINGS DONE: SPACECRAFT CONTROL
which can damage delicate components such as solar panels or instruments (e.g. on
Hubble). The manouevre is carried out by adjusting the docking vehicle’s speed to
change its orbit slightly and bring it onto the main orbit from ‘above’ or ‘below’.

Chapter 12
Problems
W=100kN
LW
x =
0.3m
LT
l =15m
M0=40kNm
W=100kN
LW
x =
0.3m
LT
l =15m
M0=40kNm
Figure 12.1: Aircraft with different centres of gravity
1. For the two situations shown in Figure 12.1, calculate the values of LW and LT
required to give both a total lift equal to the aircraft weight and give zero net
moment about the aircraft c.g.
[LW = 99.3kN, LT = 0.7kN, LW = 95.3kN, LT = 4.7kN]
2. Draw the system of forces and moments acting on a conventional aeroplane in
steady straight and level flight.
Show that the pitching moment about the centre of gravity is given by
.
For the sailplane whose details are given in Table 12.1, calculate the value of CLT
required for trim at 50kt EAS with a pilot weighing 0.75kN. The empty weight
equipped is 2.5kN, with c.g. on the mean chord 0.45c aft of the leading edge of
c. The pilot c.g. is assumed to be 0.8m ahead of the leading edge of c.
[CLT
= −0.552]
3. Distinguish between stability and trim. Show that for an aircraft to be both
stable and able to trim at positive lift coefficient the overall pitching moment
about the centre of gravity must be positive at CL = 0 in that configuration.
87

88 CHAPTER 12. PROBLEMS
S = 28m2
ST = 1.4m2
c = 1.15m
l = 5.35m h0 = 0.25 CM0 = −0.11
Table 12.1: Sailplane data
4. From first principles, show that the portion of the total lift coefficient (CL) pro-
vided by the wing, body and nacelles (WBN) group of a conventional aircraft is
given by:
CLW BN
= CL

1 + (h0 − h)
c
l

− CM0
c
l
.
If the aircraft stalls when CLW BN
reaches its maximum value, (CLW BN
)max say,
then obtain an expression relating the stalling speed to the c.g. position at any
one given weight.
Hence calculate the c.g. shift that would increase the stalling speed by 1% if
c = 5.6m, l = 15.5m and (h0 − h) = 0.05.
[∆h = −0.0566, ∆hc = −0.317m]
5. Consider the two situations shown in the diagrams below.
hc̄
W
hnc̄
CL
(CM0 )
hpc̄
hnc̄
CL
(CM0 )
(CMp )
Figure 12.2: Full-scale and model aircraft. CMp is measured by the balance, which
restrains the model in pitch.
In (a) the full scale aircraft is in steady free flight with values CL, h, η for the lift
coefficient, c.g. position and elevator angle respectively.
In (b), the model of the same aircraft is suspended from a wind tunnel balance
at the same CL and elevator setting as in (a). The balance measurement gives the
pitching moment coefficient CMP about the balance pivot axis which is located
at hp with respect to the same datum line as h.
a) Write down the moment equations for situations (a) and (b), and hence
derive the relationship between the balance reading CMP , equivalent to
the steady free flight conditions, and interrelating hp, h and CL.
b) An aircraft model is found to have a zero-lift pitching moment coefficient
of 0.027 for a particular elevator angle. The pitching moment is measured

89
about the wind-tunnel axis of rotation P and has a slope:
dCM p
dα
= 0.15; lift curve slope a = 5.851.
Determine the position of the c.g. of the full-scale aircraft relative to P if a
stick-fixed margin of 0.11 is required (c = 3.96m).
If the wing loading is 2.25kN/m2
in steady level flight with the above ele-
vator angle, what is the airspeed if the air density is 1.030kg/m3
.
[0.537m forward of P, 133.3m/s TAS.]

1. The data shown below apply to an aircraft in steady level flight at 200kt EAS.
Calculate the elevator angle required for longitudinal trim. Also obtain the
stick-fixed neutral point and static margin.
W = 30kN S = 23m2
ST = 3.5m2
c = 1.96m l = 5.5m
h0 = 0.25 c.g. is 0.61m aft of datum
CM0 = -0.036 ηT = -1.5◦
= 0.48α
a = 4.58 a1 = 3.15 a2 = 1.55
[η = -1.658◦
, hn=0.4027, Kn = 0.0915]
2. The centre of gravity range for an aircraft is found by considering that the
a) aft c.g. limit (haft) is determined by the minimum stability condition (min-
imum Kn);
b) forward c.g. limit (hfwd) is determined by the maximum elevator angle to
trim (while retaining enough elevator movement for manoeuvre).
By considering the static forces and moments on an aircraft in symmetric flight,
find an expression for the static margin stick-fixed, Kn, and show that:
Kn = −V a2
dη
dCL
= (h0 − h) + V
a1
a

1 −
∂
∂α

.
An aircraft has the following values of the aerodynamic coefficients:
h0=0.25, a=3.5, a1=3.0, a2=1.5, ∂/∂α=0.4.
Find the relationship between the c.g. position and the tail volume ratio:
a) for a static margin of 0.05 (haft);
b) for the change in elevator angle to trim to 10◦
for a change of CLof 1.0 (hfwd).
Hence find the minimum tail volume ratio such that with a c.g. shift of 0.15c the
change in elevator angle to trim is not more than 10◦
for a change of CL of 1.0
and the static margin is never less than 0.05.
[V = 0.764]
3. A transport aircraft with conventional tail is to have zero elevator angle in cruis-
ing flight at 560km/h EAS (mass 100,000kg), with the c.g. in the mid position.
The landing approach, out of ground effect, is made with flaps down at 210km/h
(mass 90,000kg), and the maximum elevator movement permitted for trimming

91
is η = ±10◦
. Using the data below, calculate the minimum tailplane area suit-
able for this aircraft, and the tailplane setting ηT relative to the flaps-up wing
zero lift line.
Minimum Kn= 0.05 c.g. range ∆h=0.50 h0=0.075 S = 232m2
c = 4.72m l=19.5m
a=5.7 a1 = 2.7 a2 = 2.1
CM0 = -0.14 =0.16α.
The change in CM0 at landing flap setting ∆CM0 = −0.10. Note that the wing
zero lift incidence angle changes by 10◦
when the flaps are lowered to the land-
ing setting.
[ST = 68.5m2
, ηT = −3.92◦
]

1. The static margin, stick-fixed may be obtained in practice from flight tests in
which the elevator angles to trim are found at certain speeds. In practice, the
aeroplane is trimmed at a series of speeds by adjusting the tab setting, and both
the elevator angle and tab angle are observed. Since the theory which relates
the stick-fixed static margin to the elevator angles to trim implicitly assumes a
constant tab angle, show that a correction must be applied to elevator angles
obtained in this way such that
ηcorrected = η +
a3
a2
β
where η and β are the observed elevator and tab angles to trim at a given speed.
Suggest how you would determine a3/a2 in flight.
2. A tailless aircraft is controlled in pitch by six elevons. Each elevon is actuated by
an independent power control unit. These units are so designed that if a failure
occurs the affected elevon is able to move until its hinge moment is zero.
Assuming the failure of one such unit, calculate the elevon angles that will give
longitudinal trim of the aircraft whose details are given below:
Weight = 850kN Speed = 70m/s EAS
Wing area S = 358m2
(h0 − h) = 0.15
CM0 = +0.02 ∂CM0 /∂η = -0.45
a1 = 4.0 a2 = 0.95
b1 = -0.7 b2 = -1.05
Assume that each elevon contributes equally to a2 and to ∂CM0 /∂η.
[ηfailed = −9.61◦
, ηoperating = −13.15◦
]
3. A conventional aircraft flying at low speed has a flexible rear fuselage such that
the tailplane setting relative to the wing zero lift line is directly proportional to
the tail load. Prove that the reduction in stick fixed static margin compared with
that of the rigid aircraft is given by:
∆Kn = Krigid
n − Kflexible
n ,
= V
a1
a

1 −
∂
∂α

1 −
1
1 + 1
2
ρV 2ST a1f

For the human-powered aircraft having the characteristics given below, find
the fuselage flexibility f (degrees deflection per Newton) that reduces the stick-
fixed static margin by 0.05 compared to the rigid case when flying at a speed of
9.2m/s at sea level.
S = 28m2
ST = 1.4m2
l = 5.34m c = 1.14m
a = 6.0 a1 = 4.5 = 0.20α.
[f = 0.1◦
/N]

93
4. The control column of a low-speed aeroplane is connected to the elevator by
an arrangement of cables which stretches when a stick force is applied. The
stiffness of the circuit is given by dHE/dη = ENm/rad where HE is the hinge
moment and η is the elevator deflection, the stick being held fixed.
Show that the stick-fixed c.g. margin (as opposed to the “elevator fixed” c.g.
margin) is given by:
Kn = (h0 − h) + V
a1
a

1 −
∂
∂α

1 −
a2b1
a1b2
1
1 − λ

where
λ =
CLSE
b2SηcηW
.
It should be assumed that the aircraft is initially in trim with the tab angle ad-
justed to give zero stick force.
Show how this angle is related to the stick fixed and stick free c.g. margins of a
rigid aeroplane. What practical use might you make of this information?

1. What conditions define the stick-fixed and stick-free manoeuvre points of an
aircraft?
From first principles, stating your assumptions, derive an expression for the
stick fixed maneouvre point of a low speed aircraft of canard layout. Show
whether this is forward or aft of the corresponding neutral point and compare
your expression with that for a conventional aircraft.
2. Define the maneouvre point stick-free for a conventional aircraft. How does it
differ from the corresponding neutral point?
Find the minimum stick force per g at sea level for the light aircraft whose
details are given below. Comment on your result and find the c.g. position
required to give 22N/g. Suggest alternative means for increasing the existing
value.
W = 2.7kN S = 7.6m2
l = 2.9m
h0 = 0.238 c = 1.2m V = 0.34
= 0.385α a = 3.865 a1 = 2.73
a2 = 2.16 b1 = −0.282 b2 = −0.536
The permitted c.g. range is from 0.22c to 0.28c. The stick force per g is given by
Q =
Pe
n
= −meSηcη
W
S
b2
a2V
H0
m,
= 83.2H0
mN/g for this aircraft.
[Q = 5.8N/g, for Q = 22N/g, h = 0.0853]
3. The table below shows data for a tailless aircraft. When it performs a steady
pullout at AN = 2.5 (n = 1.5) at 250kt EAS at a height where the air density
ρ = 1.150kg/m3
, the change of elevator setting compared with steady level flight
under the same conditions is 3.20◦
.
Calculate mq if the static margin is known to be 0.05.
W = 160kN S = 50m2
c0 = 10m
∂CM0 /∂η = −0.5 Kn = 0.05
[mq = −0.264]
4. An aircraft of conventional layout is controlled in pitch by an all-moving tailplane,
having no separate elevators (see Figure and table). Show that the tail angle per
g is given by
∆ηT
n
= −
CLHm
V a1

95
where the symbols have their usual meanings.
Hence calculate the tail angle, tail load and pivot moment when the aircraft is
flying at 440kt EAS in an 8g pullout at a height where the relative density of the
air σ = 0.74. Comment on your results.
W=175kN V =440kt EAS S=33.2m2
ST =19.1m2
l=5.25m c = 2.39m a=3.8 a1=2.7
CM0 = +0.03 ∂/∂α=0.38 h0 = 0.17 h = 0.50
σ = 0.74
LW BN LT
l
Pivot point
0.144m
[∆ηT = −4.72◦
, ηT = −4.85◦
, LT = 224.6kN, MP = −32.34kNm, CLT
= 0.3739]

1. 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 co-
efficients is given by
¯
CH = −
b2
a2V
∆CLK0
n.
The aircraft described in the table below is making a zero stick force trimmed
landing approach at 155kt EAS. 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.
Weight W =785kN c.g. at h = 0.26
Wing area S = 223m2
smc c = 5.68m
Tail area ST = 46.5m2
tail arm l = 15.66m
Elevator area Sη = 11.2m2
elevator mean chord cη = 0.908m
h0 = 0.16 CM0 = -0.06 = 0.38α
a = 4.5 a1 = 2.75 a2 = 1.16
b0 = 0 b1 = -0.133 b2 = -0.16
The stick–elevator gearing ratio me = 1.0m/rad.
[118 kt, pull force]
2. Using the approach of §3.2, and the results of §5.2, derive a formula for dPe/dV ,
the gradient of stick force with flight speed. What does this tell you about the
handling qualities of an aircraft?
3. What are stick-fixed and stick-free manoeuvre points and what is the signifi-
cance of stick force per g.
Using the data of the previous question, calculate the change of elevator angle
required to pull 0.5g flying at 350kt EAS at an altitude where the relative density
σ = 0.374.
Explain, in simple physical terms, why this change of elevator angle would be
greater at a lower altitude when flying at the same lift coefficient.
[∆η = −1.005◦
]
4. The tailless aircraft shown in the Figure has been fitted with a small retractable
foreplane. At low speeds this foreplane is extended and, operating in a semi-
stalled condition at constant setting, it generates a constant lift coefficient CLF =
1.2 (based on SF ). Use of the foreplane enables the aircraft to take off at a higher
weight than the original aircraft without the foreplane. Calculate the increment

97
in take-off weight that may be achieved when using the foreplane, by consider-
ing the trimmed lift at 200kt EAS, if the incidence is restricted to 12◦
by ground
clearance problems, using the data in the table.
Calculate the elevon angles to trim of both versions of the aircraft. Comment on
your results.
[With foreplane: η = −0.5◦
; L = 1842kN; without foreplane: η = −5.8◦
; L =
1557kN]
lF
c0
hc0
h0
S = 438m2
SF = 9.4m2
CM0 = +0.002 ∂CM0 /∂η = -0.25
a1 = 3.0 a2 = 0.80 h0 = 0.61 c0 =27.4m
lF = 13.26m hc0 = 15.34m
5. The aircraft described in the Figure and table below is to have its capacity in-
creased by lengthening the cylindrical portion of the fuselage by 6m. The centre
section (including the wings), the nose and the tail portions are to remain unal-
tered.
It is assumed that the c.g. position will be adjusted to remain unchanged with
respect to the centre section unit and that, for the lengths considered, ∂/∂α is
constant.
Calculate how the additional fuselage length is to be inserted ahead of and be-
hind the centre section, if the low speed stick-fixed static margin is to unaltered.
The movement of the aerodynamic centre of the aircraft less tail is assumed to
be affected only by the change of nose length ∆lN and is given by
∆h0 = −0.037
∆lN
c
.
If a variant of the aircraft retains the original fuselage, but has its wing tips
extended, how could the longitudinal static stability be affected?

S = 223m2
c = 5.6m
ST = 46m2
l=15.5m
h0 = 0.25 h = 0.20
= 0.4α CM0 =-0.06
a = 4.5 a1 = 2.75
[4.1m ahead of wing, 1.9m aft]
6. a) The 1903 Wright Flyer was a canard configuration of conventional layout,
summarized 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.
b) The 1903 Flyer was stabilized in pitch by the addition of ballast to shift
the centre of gravity forward. If 30% of the aircraft gross weight can be
carried as ballast, where should it be placed to move the centre of grav-
ity to the wing leading edge. What effect would this have on the aircraft
performance?
h0 ≈ 0 V = 0.134 CM0 = −0.141
h = 0.3c W ≈ 340kg
The 1903 Wright Flyer. The datum is the wing leading edge.

99
1. The table below contains flight test data for the X-15 spaceplane. Calculate the
static margin stick-fixed and estimate the zero-lift pitching moment. Estimate
the dimensional and non-dimensional phugoid mode and SPO frequencies.
S = 18.58m2
s = 6.82m c = 3.13m
h = 0.22
m = 7056kg B = 10700kgm2
V = 331kt EAS
a = 3.5/rad ∂CM /∂α = −0.8/rad
Zu = −332Ns/m Mw = −40.7Ns
Zw = −14300Ns/m
Mq = −158600Nms
[Ωph = 0.0946 (0.052rad/s); Ωspo = 17.683 (9.723rad/s)]
2. NASA CR-2144, Aircraft Handling Qualities Data, contains stability informa-
tion for ten aircraft. For the Boeing 747:
a) calculate the static margin stick-fixed;
b) estimate the zero-lift pitching moment;
c) estimate the phugoid, SPO and Dutch roll periods;
d) estimate the time constants for rolling subsidence and the spiral mode.

1. a) Given that the velocity of a body in planetary orbit is:
v =

2E +
2µ
r
1/2
,
describe the following spacecraft maneouvres and calculate the velocity
change required for each:
i. Hohmann transfer;
ii. orbital capture;
iii. planetary escape.
The parameters E and µ are the orbital energy per unit mass and the grav-
itational parameter, respectively.
b) A lunar module in low earth orbit travels to the moon on a Hohmann trans-
fer trajectory. Calculate the ∆V required to enter the Hohmann transfer to
lunar orbit and, for comparison, the escape velocity of Earth.
c) Comparing the ∆V required for transfer to lunar orbit and that to escape
Earth’s gravity, comment on the feasibility of a manned mission beyond
the moon.
2. a) A communications satellite for the high latitudes of the Northern hemi-
sphere is parked in a circular low earth orbit of altitude 174km. Calculate
the velocity change required to change its orbit to one with a perigee of
174km and an apogee of 39000km.
b) Sketch the shape of the new orbit, indicating the point at which the burn
takes place. Why is such an orbit used in practice?

103
Conventional aircraft Tailless aircraft
Steady flight:
CM = CM0 +
∂CM
∂η
η − (h0 − h)CL
CLT
=
a1
a

1 −
∂
∂α

CL + a1(ηT − 0) + a2η + a3β
Steady pullout/bank:
CLT
=
a1
a

1 −
∂
∂α

CL + a1(ηT − 0) + a2η + a3β + a1∆αT
CM = CM0 +
∂CM
∂η
η − (h0 − h)CL +
∂CM
∂q
q
Static and manoeuvre margins:
Kn = (h0 − h) + V
a1
a

1 −
∂
∂α

Kn = h0 − h
Hm = Kn +
V a1
2µ1
Hm = Kn −
mq
µ1
K0
n = (h0 − h) + V
a1
a

1 −
∂
∂α

H0
m = K0
n +
V a1
2µ1
General relationships
µ1 =
W
ρgSlT
µ1 =
W
ρgSc0
a1 = a1

1 −
a2b1
a1b2

mq =
1
ρV Sc2
0
∂M
∂q
a3 = a3

1 −
a2b3
a3b2

CH = b0 + b1αT + b2η + b3β
V =
ST
S
l
c
xu =
∂X
∂u
ρUS
, xq =
∂X
∂q
ρUSc
, mq =
∂M
∂q
ρUSc2
µc =
m
ρSc
, Λ =
mλ
ρUS
.

Cranfield University Flying Laboratory – Student Registration Form
All students participating in flights aboard the laboratory aircraft, operated by the National Flying
Laboratory Centre, must read and complete this form. Completion and submission of the form is
required to officially register students as members of a Cranfield University short course.
Please note:
1. Personal life insurance cover is usually restricted to scheduled or charter flights, and
does not extend to any other flying unless declared by the insured. Anyone concerned about
insurance are advised, in the first instance, to contact their own insurance company. The NFLC
will supply an interested insurance company with details of the CAA approved Air Operator’s
Certificate and exemption on request. In addition, special insurance cover is maintained for
students registered on a Cranfield course requiring flights in the laboratory aircraft, and details of
this cover may be obtained from Cranfield University’s Commercial Accountant.
2. The NFLC reserves the right to refuse permission for any person to fly if, in the opinion
of the crew, the person is not adequately briefed for the flight, or if the carriage of the person would
constitute a hazard to the aircraft or its occupants.
3. Students must be medically fit to fly at the time of the course, if there is doubt advice
from a doctor should be sought.
4. Student’s weight and sex must be entered (for load planning purposes).
5. The course will be planned to run to a strict timetable but operational considerations
may necessitate last-minute changes; in order to retain the necessary flexibility students should not
make any other commitments (such as part-time working) during the course so that they are
available should rescheduling be required.
6. The NFLC reserves the right to refuse permission for any person to fly if, in the opinion
of the crew, the person is wearing inappropriate footwear or clothing that is prejudicial to their
safety in the event of an aircraft emergency.
Please print your details clearly
University:___________________________ Course:______________________
Name:_______________________________ Weight:___________kg
Address:_______________________________ Sex, M or F: ____________
__________________________
__________________________
__________________________
Tel: __________________________ E-mail:_______________________
Nationality:__________________________
I certify that I have read and understood the notes above, and the details I have supplied are
correct.
Signature:____________________________ Date:_______________
Student Registration V 4
November 2012 NFLC, Cranfield University

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