Some notes on aircraft and spacecraft stability and control

Some notes on aircraft and spacecraftstability and control

Michael Carley, [email protected]

A lonely impulse of delightDrove 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 inviscidsubsonic 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’, NASACR-2144, 1972.

• Thompson, Ambler and Taylor, Barry N., ‘Guide for the use of the internationalsystem 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 Ruledthe World, Faber & Faber, 2010.

You are responsible for finding and obtaining these documents. They will not bedistributed to the class. They can all be downloaded via the university’s systems. Youshould print out the first two. The third is long and we will not be using all of it sowait until we come to use it before printing out the parts you need. You will onlyneed Appendix B of NIST SP-811. The microlight paper has a lot of information oninstability of both flexible and rigid wing aircraft. Empire of the Clouds is an accountof British aeronautical engineering in the decade or so after 1945, and is well worthreading 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 31.1 Equilibrium and stability . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Functions of aircraft controls . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Forces and moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4 Trim and stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Aerodynamic centre and neutral point . . . . . . . . . . . . . . . . . . . 7Definitions of static and c.g. margins . . . . . . . . . . . . . . . . . . . . 9

1.5 Basic aerofoil and control characteristics . . . . . . . . . . . . . . . . . . 10Aerofoils and wings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10Control forces and moments . . . . . . . . . . . . . . . . . . . . . . . . . 11Control hinge moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Longitudinal static stability 152.1 Some basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.2 Downwash . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.3 Stick fixed stability and c.g. margins . . . . . . . . . . . . . . . . . . . . 182.4 Stick free stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.5 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.6 Stick free stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3 Flight testing 233.1 Kn—elevator angle to trim . . . . . . . . . . . . . . . . . . . . . . . . . . 233.2 What does the pilot feel? . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.3 Kn’—tab angle to trim . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4 Tailless aircraft 294.1 Stick fixed stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294.2 Static margin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

i

ii CONTENTS

5 Stick forces 335.1 Analysis to calculate stick forces . . . . . . . . . . . . . . . . . . . . . . . 335.2 More flight testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355.3 Modification of stick forces . . . . . . . . . . . . . . . . . . . . . . . . . . 36

6 Manoeuvre stability 396.1 Analysis of a steady pullout . . . . . . . . . . . . . . . . . . . . . . . . . 396.2 Stick fixed manoeuvre point . . . . . . . . . . . . . . . . . . . . . . . . . 426.3 Stick fixed manoeuvre stability . . . . . . . . . . . . . . . . . . . . . . . 436.4 Stick free manoeuvre stability . . . . . . . . . . . . . . . . . . . . . . . . 436.5 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436.6 Tailless aircraft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446.7 Tailless aircraft manoeuvre point . . . . . . . . . . . . . . . . . . . . . . 466.8 Tailless aircraft manoeuvre margins . . . . . . . . . . . . . . . . . . . . . 476.9 Relationships between static and manoeuvre margins . . . . . . . . . . 47

Conventional aircraft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47Tailless aircraft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

6.10 Modification of stick free neutral and manoeuvre points . . . . . . . . . 48

7 Compressibility effects 517.1 High speed effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

II Dynamic stability 55

8 Dynamic behaviour of aircraft 578.1 Axes and notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 578.2 Aerodynamic derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . 578.3 Longitudinal symmetric motion . . . . . . . . . . . . . . . . . . . . . . . 61

9 Normal modes of aircraft 63Phugoid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63Short period oscillation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

9.1 Lateral motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65Dutch roll . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66Spiral mode and roll subsidence . . . . . . . . . . . . . . . . . . . . . . . 66

9.2 Dihedral effect and weathercock stability . . . . . . . . . . . . . . . . . 67Dihedral effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67Weathercock stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

III Spacecraft dynamics and control 71

10 Getting around: Orbits 7310.1 The two-body problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

CONTENTS iii

Elliptical orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7610.2 Orbital maneouvres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

Hohmann transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78Orbital capture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

11 Getting things done: Spacecraft control 8111.1 Attitude control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

Gravity gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82Sun-synchronous orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

11.2 Manouevring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

IVProblems 85

12 Problems 87

Basic equations 101

List of Figures

1.1 Phases of flight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Terminology for study of stability . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Axes and sign conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4 Sign conventions for longitudinal stability . . . . . . . . . . . . . . . . . . . 71.5 Trim and stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.6 Centre of pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.7 Incremental loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.8 Centre of gravity and aerodynamic centre relationships . . . . . . . . . . . 91.9 Lift curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.10 Loading due to control deflection . . . . . . . . . . . . . . . . . . . . . . . . 111.11 Measurement of control/tab deflections . . . . . . . . . . . . . . . . . . . . 121.12 Pressure distributions due to deflections, a1 > a2 > a3. . . . . . . . . . . . . 121.13 Measurement of control surface area . . . . . . . . . . . . . . . . . . . . . . 131.14 Control hinge moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.1 Stick fixed stability (conventional aircraft) . . . . . . . . . . . . . . . . . . . 162.2 Trailing vortices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.3 Effect of downwash on tailplane . . . . . . . . . . . . . . . . . . . . . . . . 172.4 Stick free elevator conditions . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.1 A typical weight and balance envelope . . . . . . . . . . . . . . . . . . . . . 233.2 Elevator angle to trim at various lift coefficients . . . . . . . . . . . . . . . . 243.3 Measurement of neutral point location . . . . . . . . . . . . . . . . . . . . . 253.4 What the pilot experiences . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.5 Tab angle to trim at varying lift coefficients . . . . . . . . . . . . . . . . . . 273.6 Measurement of stick free neutral point location . . . . . . . . . . . . . . . 27

4.1 Canard configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294.2 Control surfaces for tailless aircraft . . . . . . . . . . . . . . . . . . . . . . . 304.3 Representation of tailless aircraft . . . . . . . . . . . . . . . . . . . . . . . . 30

5.1 Aerodynamic assistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

6.1 Manoeuvre conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406.2 Manoeuvre conditions for a tailless aircraft . . . . . . . . . . . . . . . . . . 45

iv

LIST OF FIGURES v

6.3 Aerodynamic forces during pitching motion . . . . . . . . . . . . . . . . . 466.4 Modification of neutral and manoeuvre points . . . . . . . . . . . . . . . . 486.5 Effects of positive springs and bob-weights . . . . . . . . . . . . . . . . . . 496.6 Modification of stick force per g . . . . . . . . . . . . . . . . . . . . . . . . . 49

7.1 Compressibility effects on lift curve slope and aerodynamic centre . . . . . 527.2 Compressibility effects on zero lift pitching moment and zero lift angle . . 527.3 Aeroelastic effects on lift curve slope . . . . . . . . . . . . . . . . . . . . . . 537.4 Aeroelastic effects on tailplane and elevator . . . . . . . . . . . . . . . . . . 537.5 Variation of downwash with Mach number . . . . . . . . . . . . . . . . . . 547.6 Variation of pitching moment with Mach number . . . . . . . . . . . . . . 547.7 Variation of stick forces with Mach number . . . . . . . . . . . . . . . . . . 54

8.1 Notation for analysis of dynamic stability . . . . . . . . . . . . . . . . . . . 58

9.1 Phugoid oscillation trajectory . . . . . . . . . . . . . . . . . . . . . . . . . . 649.2 Short period oscillation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 659.3 Rolling subsidence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 679.4 Stability of the lateral modes . . . . . . . . . . . . . . . . . . . . . . . . . . . 689.5 Dihedral effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 689.6 Wing sweep effects on Lv . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 699.7 Wing-fuselage interference effects on Lv . . . . . . . . . . . . . . . . . . . . 699.8 Use of twin fins at high speed . . . . . . . . . . . . . . . . . . . . . . . . . . 709.9 Intake effects on Nv . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

10.1 The two-body problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7310.2 Polar coordinate system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7410.3 The geometry of an ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7610.4 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 . . . . . . . . . . . . . . . . . . . . 8712.2 Full-scale and model aircraft. CMp is measured by the balance, which re-

strains the model in pitch. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

Part I

Static stability

1

Chapter 1

How aircraft fly

Aircraft fly by generating a lift greater than or equal to their weight. They do this byholding a wing at a certain angle of attack, or incidence. Longitudinal control is thestudy of how to set, maintain or change that angle of attack; stability is the study ofwhether and how that angle of attack will remain fixed when the aircraft is subjectedto 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 thanits weight in order to leave the ground. In cruise, the aircraft operates at a constantspeed and constant lift. Finally, when the aircraft lands, it needs to reduce its speedwithout 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—raisingor 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 therelationship 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 anaircraft responds to small disturbances in flight and how it can be designed so that itremains 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 thatit rotates and changes the wing incidence or so that the net moment is zero and theaircraft flies at constant incidence. This is done via the aircraft controls. Before we goany 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 aircraftinto 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, itinitially 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 whatthe 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 stableand/or unstable, in various combinations, including the case of a system which isstatically 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 controlsThe 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 5Response

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 changein pitching moment on the aircraft. This allows the pilot to adjust the aircraftincidence;

ailerons these change the lift on each wing when they are deflected. They move inopposite directions—one goes up when the other goes down so that the lift onone wing increases and the other decreases. This generates a change in rollingmoment and allows the aircraft to rotate about its axis to initiate turns, or allowsit to oppose disturbances due to crosswind or gusts;

rudder changes the side force on the vertical tailplane (or fin), generating a change inyawing moment, rotating the aircraft about a vertical axis. This can be used toresist yawing moments due to engine failure and crosswind and to aid in spinrecovery and turn co-ordination.

In steady, level flight in still air, the rudder and ailerons will be undeflected whilethe 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 usedsimultaneously. The controls can be operated directly by the pilot, through a systemof mechanical actuators, possibly with aerodynamic or power assistance, or controlsmay be fully powered using a hydraulic or electrical system. These systems can bemechanically 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

xRoll

y

Pitch

Figure 1.3: Axes and sign conventions

1.3 Forces and momentsForces and moments on an aircraft are due to the mass of the aircraft, a function ofhow it is built and loaded, and the aerodynamic forces generated in flight. The massdistribution of the aircraft gives rise to inertial forces and moments as described byIsaac Newton and to gravitational forces.

You should already know that the basic aerodynamic forces include the lift anddrag on the aircraft. More generally, we consider the static forces and moments dueto linear velocities; the damping forces and moments caused by angular velocities(such as pitching and rolling); the control forces generated when the pilot operates acontrol surface.

By considering the lateral symmetry of most aircraft, it is clear that in forwardlevel flight, the forces on the aircraft act in the plane of symmetry. This means thatany symmetric disturbance, such as operation of the elevator, will generate horizontaland vertical motion of the aircraft and rotation in the vertical plane only. At thispoint, then, we study longitudinal symmetric motion of the aircraft and considerlongitudinal stability.

1.4 Trim and stabilityFigure 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 anglebetween the direction of flight and the Zero Lift Line (ZLL). When α is zero, the ZeroLift 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 paralleland 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 aircraftare in equilibrium and the aircraft is said to be trimmed or in trim. This is the basicproblem of control: is it possible, using the tailplane, to generate a pitching momentso 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 subjectedto a disturbance which changes its incidence, does it tend to return to the equilibriumposition?

This can be restated: if the aircraft pitches nose up, the change in aerodynamicmoment about the centre of gravity ∆Mcg should be negative in order to push thenose back down or, ∂Mcg/∂α < 0.

Figure 1.5 shows various ways Mcg can vary with α, including how it is possibleto trim an unstable aircraft and how is possible for an aircraft to be stable withoutbeing 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 andnot on the position of the centre of gravity so we can consider the aerodynamic loads


CM

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 anaerofoil 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, movesas the incidence varies, so it is not very useful as a reference point in calculationsinvolving changing incidence.

L

Figure 1.6: Centre of pressure

To make life easier, we can give up the requirement that the moment about ourreference point be zero and, instead, allow it to have some finite value as long as thereference point is fixed and the moment is constant. We can do this by looking at theincremental 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 changein moment about that position. This point is the aerodynamic centre and is the pointabout 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 thecentre 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 ispositive and the aircraft is statically unstable.

3. If the centre of gravity is at the aerodynamic centre, dMcg/dα is zero and theaircraft is neutrally stable.

When we talk about the properties of a whole aircraft, rather than just a wing oraerofoil, we use the term ‘neutral point’. This is the position of the centre of gravityfor which the aircraft is neutrally stable. It is a purely aerodynamic property, whichdepends 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. Thisis equivalent to the gradient ∂CM/∂α. When this is large, the aircraft experiences alarge 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 = (C2L + 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, scaledon 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 hncis 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 byMach number or aeroelastic effects so that Hn ≈ Kn.

1.5 Basic aerofoil and control characteristicsSo far, we have considered aircraft stability without considering the behaviour ofaircraft. In order to deal with realistic problems, we need to know something abouthow aerodynamic surfaces and bodies behave in flight.

Aerofoils and wings

As always, we work in terms of non-dimensional quantities using standard referencevalues 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. Thesubscript in each coefficient is upper case because the coefficients are those for threedimensional 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 isthat we choose a reference incidence such that the lift is zero when α = 0. This willnot always be true in other calculations and you should check the conventions usedwhen you take data from published sources. In this course, we will only considerlinear aerodynamics, i.e. the part of the lift curve where dCL/dα is constant. This is areasonable assumption for most aircraft most of the time, but will not be correct nearstall or in high speed manoeuvres.

Control forces and moments

We control an aircraft by moving a control surface such as an elevator. In order tohave some clue what the result of a control input will be, we need to know how farthe surface must be moved in order to generate, indirectly, the pitching moment weneed and we need to know what moment will be needed to rotate the control surfaceabout its hinge point.

If the incidence of the wing or tailplane as a whole is held constant, deflecting thecontrol surface in the positive sense is equivalent to introducing extra camber. Thisgenerates extra lift and a pitching moment which is usually nose-down, because ofthe 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 andnegative pitching moment (if the tail pushes up, it forces the nose down). If there is atab, an extra small surface on the end of the elevator, this too will generate positive liftand 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 forconvenience:

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

ηβ

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

∂ηη +

∂CM0

∂ββ.

Control hinge moments

When an aircraft control system is designed, we will need to know what force isrequired to move a control, whether it is being moved directly by the pilot or througha powered actuation system. The force needed depends on the moment required torotate 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 ofthese 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 deflectionto 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 momentwhen all other deflections are zero, given the symbol b0. The pressure distributionsassociated 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

Hingeline

∆Cp

x/c

Hingeline

∆Cp

x/c

Hingeline

∆Cp

x/c

ηβ

Figure 1.14: Control hinge moments

Chapter 2

Longitudinal static stability

Given the definitions and background information of Chapter 1, we are in a positionto start doing some calculations for the stability of aircraft. We consider two basiccases, the ‘stick fixed’ and ‘stick free’. In the first case, conceptually, we move theelevator to the position required for trim and then fix the stick so that the elevatorremains at the set deflection, ignoring the question of what force is required to keepit in place. The stability problem can then be phrased: ‘with the stick fixed, how doesthe aircraft respond to a perturbation?’ In the ‘stick free’ case, we use the tab to adjustthe moment on the elevator so that it comes to an equilibrium position which trimsthe aircraft. In this case, zero force is required to keep the control fixed. Using thiszero-force case as a reference, we can work out the stick force required for any othertab setting.

2.1 Some basicsFigure 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 whichgenerates most of the lift, and the tailplane which generates a small amount of lift butwhich can be adjusted to change the pitching moment on the aircraft as a whole. Thetailplane 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 wakeeffect 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 canneglect TzT and DzD, and

Mcg = M0 − (h0 − h)cL− LT l.

15

16 CHAPTER 2. LONGITUDINAL STATIC STABILITY

Datum

W

hc

LWBNh0c

M0

LT

l

TzT 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 operatingconditions and CM0 is known from the aircraft geometry. Then, if V is known, CLT

can be calculated and from that the elevator deflection; if CLTis known (because

the tailplane shape has already been decided), V can be calculated, and the size ofthe tailplane fixed. The tail volume coefficient represents the ‘effectiveness’ of thetailplane at generating a moment. It contains the size of the tailplane ST and the leverarm l which, combined, tell us the moment which the tailplane can generate.

2.2 DownwashAny lifting wing generates a downwash, due to the trailing vortex system, shown inFigure 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 ofattack, Figure 2.3.

ZLL tailplane

ZLL WBN

ηT

Free stream

αε

Resultant flow

Figure 2.3: Effect of downwash on tailplane

From Figure 2.3:

αT = α+ ηT − ε.

For an untwisted wing, the downwash angle ε is proportional to the lift on the wing,meaning that in the linear regime, it is also proportional to α:

ε =dε

dαα+ ε0,


with ε0 only present for a wing where the zero lift angle of attack varies along itsspan (i.e. a wing with a varying cross-section or camber along its length or withtwist). Combining the previous equations:

αT = α+ ηT −(ε0 +

dε

dαα

)= α

(1− dε

dα

)+ (ηT − ε0).

From §1.5:

CLT= a1αT + a2η + a3β,

so that:

CLT= a1(α+ ηT − ε) + a2η + a3β,

and:

CLT= a1α

(1− dε

dα

)+ a1(ηT − ε0) + 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− dε

dα

)CL + a1(ηT − ε0) + a2η + a3β. (2.4)

In deriving (2.3), we made no assumptions about how CLTwas generated, so (2.4)

can be substituted to give:

CM = CM0 − (h0 − h)CL − V

[a1

a

(1− dε

dα

)CL + a1(ηT − ε0) + 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 symbolwith an overbar). In designing aircraft, there will be a limit on the maximum elevatordeflection. Given this maximum η, we can use the flight conditions to estimate V andso size the tailplane.

2.3 Stick fixed stability and c.g. marginsThe 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 toCL:

−dCM

dCL

= (h0 − h) + Va1

a

(1− dε

dα

).

The neutral point is the centre of gravity position where dCM/dCL ≡ 0:

hn = h0 + Va1

a

(1− dε

dα

),

where h0 is the neutral point of the aircraft less tail. Adding a tail has moved theneutral point back by an amount V (a1/a)(1− dε/dα), increasing the stability (as youmight 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 wholeaircraft stable and controllable. The stability requirement is specified as a minimumvalue of h − hn; the control requirement is stated, in effect, as a maximum pitchingmoment to be generated.

2.4 Stick free stabilityThe ‘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 neededto get there. As the first step towards calculating the forces needed to move a controlsurface, we consider the ‘stick-free’ problem, where the elevator is allowed to floparound until it reaches an equilibrium position where the moment on it, and so thestick 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 andno tab is required. In both cases, however, we need to know the forces on the elevatoreither to keep them within the limits of a pilot’s strength or to size the actuators.

2.5 AnalysisThe 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− dε

dα

)CL + a1(ηT − ε0) + a2η + a3β,


and

CLT=a1

a

(1− dε

dα

)CL + a1(ηT − ε0)−

a2

b2(b0 + b1αT + b3β) + a3β.

Incorporating the expression for α from §2.2, page 18)

αT =CL

a

(1− dε

dα

)+ (ηT − ε0),

and

CLT=

(a1 −

a2b1b2

) (1− dε

dα

)CL

a+

(a1 −

a2b1b2

)(ηT − ε0)

+

(a3 −

a2b3b2

)β − a2b0

b2.

To simplify our notation, we define two new variables (both given on the datasheet in the appendix):

a1 = a1

(1− a2b1

a1b2

),

a3 = a3

(1− a2b3

a3b2

),

so that:

CLT= a1

(1− dε

dα

)CL

a+ a1(ηT − ε0) + a3β −

a2b0b2

.

We already know the tailplane lift coefficient, (2.3), so that we can find the lift ifthe stick is released and the elevator comes to equilibrium. The resulting pitchingmoment is:

CM = CM0 − (h0 − h)CL − V CLT,

and:

CM = CM0 − (h0 − h)CL − V

[a1

a

(1− dε

dα

)CL + a1(ηT − ε0) + a3β −

a2b0b2

],

which allows us to find the tab angle to trim with zero stick force, β.

2.6 Stick free stabilityTo find the stick-free stability properties, we differentiate the pitching moment equa-tion:

dCM

dCL

= −(h0 − h)− Va1

a

(1− dε

dα

),

2.6. STICK FREE STABILITY 21

and the static margin stick free:

K ′n = −dCM

dCL

= (h0 − h) + Va1

a

(1− dε

dα

).

This is related to the static margin stick fixed:

Kn = (h0 − h) + Va1

a

(1− dε

dα

),

with the two static margins being equal if:

a1 = a1 = a1

(1− a2b1

a1b2

),

or

a2b1a1b2

= 0.

Since a1 and a2 are both positive, and b2 is negative for correct feel of the elevator, thiscan only happen if b1 = 0, which can be achieved using aerodynamic balancing or bymoving 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 h′n and the static and c.g. margins stick-freeH ′

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 theproduction model to find what the real behaviour is. In the early stages of design, weuse approximate analyses and semi-empirical methods (for example, ESDU sheets) toestimate the aerodynamic parameters such as lift curve slopes, largely because earlyin 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 ourestimates. When the first few aircraft are produced, we must test them to see whatthe real behaviour of the real aircraft is.

This information is used in setting the limits to be observed in service—the ‘flightenvelope’ of Figure 3.1. Before flight, the aircraft weight and centre of gravity areplotted on the diagram and must lie within the limits indicated. If they do not, thenthe 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 trimGiven that, for a trimmed aircraft:


hc

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− dε

dα

)CL + a1(ηT − ε0) + a2η + a3β,

CM = 0 = CM0 − (h0 − h)CL − V

[a1

a

(1− dε

dα

)CL + a1(ηT − ε0) + a2η + a3β

].

This can be differentiated (§2.3) to examine the stick-free stability:

Kn ≈ Hn = −∂CM

∂CL

= (h0 − h) + Va1

a

(1− dε

dα

).

We also know that:

η =1

V a2

CM0 − (h0 − h)CL − V

[a1

a

(1− dε

dα

)CL + a1(ηT − ε0) + a3β

],

and that there is a relationship between η and Kn because:

dη

dCL

= − 1

V a2

[(h0 − h) + V

a1

a

(1− dε

dα

)]= − Kn

V a2

.

Figure 3.2 shows η plotted against CL, while Figure 3.3 shows the relationshipbetween dη/dCL and h.

CL

η

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-fixedis: fly the aircraft straight and level at various speeds, recording the elevator angle totrim. This is repeated for various different centre of gravity positions, yielding a plotlike 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 whereKn = 0, the neutral point.

3.2. WHAT DOES THE PILOT FEEL? 25

h

dη/dCL

h1h2h3

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 forceand for elevator deflection (because they know how far the stick has moved) and forspeed (because they can see out the window or look at the instruments). We can seehow the elevator angle to trim varies with speed, to look at what the pilot feels inflying 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 happensbecause at low speed, the control surfaces cannot generate enough force to balance themoment about the centre of gravity. Likewise, above a certain speed, small changesin 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 thelimits in question will be a function of the aircraft type and of the skill assumed of thepilot.

26 CHAPTER 3. FLIGHT TESTING

V

η

c.g. moving forward

Figure 3.4: What the pilot experiences

3.3 Kn’—tab angle to trimTo find the neutral point stick-free, we can use the same approach as in the stick-fixedcase, but using the tab to trim, rather than the elevator. We know that:

CM = CM0 − (h0 − h)CL − V CLT= 0,

and

CLT=a1

a

(1− dε

dα

)CL + a1(ηT − ε0) + a2η + a3β.

From §2.5:

η = −b0 + b1αT + b3β

b2,

and CLT=a1

a

(1− dε

dα

)CL + a1(ηT − ε0) + a3β −

a2b0b2

,

so that:

CM = CM0 − (h0 − h)CL − V

[a1

a

(1− dε

dα

)CL + a1(ηT − ε0)−

a2b0b2

].

Rearranging to find the tab angle to trim:

β =1

V a3

CM0 − (h0 − h)CL − V

[a1

a

(1− dε

dα

)CL + a1(ηT − ε0) + a3β −

a2b0b2

],

and differentiating with respect to CL:

dβ

dCL

= − 1

V a3

[h0 − h+ V

a1

a

(1− dε

dα

)].

3.3. KN ’—TAB ANGLE TO TRIM 27

We know, however, that:

K ′n = (h0 − h) + V

a1

a

(1− dε

dα

),

and that

dβ

dCL

= − K ′n

V a3

.

So to find the neutral point stick free, we vary the aircraft speed at fixed centre ofgravity, trimming with the tab, giving us Figure 3.5. We then plot the gradients fromthat figure against CL, Figure 3.6 to find h′n

CL

β

h1

h2

h3

Figure 3.5: Tab angle to trim at varying lift coefficients

h

dβ/dCL

h1h2h3

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 usea 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, thebasic equations are the same as in the conventional case, but the tail arm l about theaerodynamic centre is negative, Figure 4.1.

LF LWBN

l

M0

W

hc

h0c

Figure 4.1: Canard configuration

On a tailless aircraft, there is no separate control surface for pitch control, with theelevators and ailerons being combined into ‘elevons’. These are moved in oppositedirections for roll control and in the same direction for pitch control, Figure 4.2.

4.1 Stick fixed stabilityFigure 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 tailcontribution to include. On tailless aircraft, the pitching moment required to trim isgenerated by the control surfaces which also have a large effect on lift. This makesthe 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 pitchcontrol and differentially for roll

η

LWBN

M0

W

hc

h0c

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

∂ηη − (h0 − h)c0L,

and non-dimensionalizing:

CM = CM0 +∂CM0

∂ηη − (h0 − h)CL,

which can be rearranged to yield the elevon angle to trim.

4.2 Static marginThe 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, theircontrol systems are powered, so that there is no ‘stick free’ condition: we do not needto consider this case.

Chapter 5

Stick forces

So far, we have not considered how what force or, equivalently, torque, will be neededto 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 requiringexcessive 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 essentialto 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 thedifferent controls under various circumstances.

Aileron Elevator RudderStick Wheel Stick Wheel (Push)

Maximum all-out effort 2 hands 400 530 800 980 1780 NMaximum permissible effort 2 hands — 360 440 440 890 N

1 hand 220 220 310 310 NMaximum comfortable effort 2 hands — 130 — 180 270 N

1 hand 90 90 130 130 NLargest 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 ishigher than the maximum aileron force and that the maximum rudder force is higherthan both. The controls are said to be ‘harmonized’ if the aileron, elevator and rudderforces have the ratio 1:2:4 for a given control response. For example, the rudder forcefor 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 thestick 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 = meρV 2

2SηcηCH .

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 thefull range of deflections needed; too low and a small stick deflection will generatea large acceleration on the aircraft with a risk of overloading the structure. To startwith, we need the hinge moment to trim, which we can derive from our previousanalysis 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− dε

dα

)CL

a+ a1(ηT − ε0) + a2η + a3β,

and, on substituting η:

CLT= a1

(1− dε

dα

)CL

a+ a1(ηT − ε0) + a3β +

a2

b2(CH − b0).

This equation is the general form of one we have already derived for the zero stickforce case. Given that:


and substituting for CLT:

CM = CM0 − (h0 − h)CL − V

[a1

a

(1− dε

dα

)CL + a1(ηT − ε0) + a3β +

a2

b2(CH − b0)

],

which can be re-arranged to find β:

V a3β = CM0 − (h0 − h)CL − V

[a1

a

(1− dε

dα

)CL + a1(ηT − ε0)−

a2b0b2

], (5.1)

or hinge moment to trim:

Va2CH

b2= CM0 − (h0 − h)CL − V

[a1

a

(1− dε

dα

)CL + a1(ηT − ε0) + a3β −

a2b0b2

].

(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

b2CH

)= V a3β,

yielding:

CH =b2a2

a3(β − β)

so that:

Pe = meρV 2

2Sηcη

b2a2

a3(β − β),

so that the stick force to trim is proportional to the difference between the actual tabangle and the tab angle to trim.

5.2 More flight testingIn theory we might use the stick forces to calculate the stick free neutral point, begin-ning from:

CH =b2a2

a3(β − β),

which, under differentiation with respect to lift coefficient at constant tab angle, yields:

∂CH

∂CL

=b2a2

a3∂β

∂CL

.

In §3.3, we found that:

dβ

dCL

= −K′n

V

1

a3

,

so that:

dCH

dCL

= −b2K′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 cannotmeasure the force accurately enough for a reliable estimate, due to errors introducedby such things as friction in the system.

36 CHAPTER 5. STICK FORCES

5.3 Modification of stick forcesThere are three main methods which can be used to modify the stick forces to bringthem into the correct range for control of the aircraft:

Gearing between stick and control surface, but this is limited because of the range ofelevator movement required.

Power assistance, which can ‘share’ the load or supply all of the force required tomove the control, with a feedback system to give the pilot ‘feel’ for the controlinput.

Aerodynamic methods of modifying the loads include adding surface ahead of thehinge line (a ‘horn balance’), moving the hinge line or adding tabs which aregeared 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 agiven elevator deflection, dPe/dη:

Pe = meρV 2

2SηcηCH ,

but,

CH = b0 + b1αT + b2η + b3β,

5.3. MODIFICATION OF STICK FORCES 37

so that

dPe

dη= me

ρV 2

2Sηcηb2.

To reduce the stick force, we want to reduce b2, but dPe/dη b2, must be negative forcorrect feel of the controls. Reducing b2 is useful at high speed (because of the effectof V 2) but at low speed, the pilot may not have enough feel for the controls and othermethods 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 nextstep is to examine longitudinally symmetric manoeuvres (i.e. manoeuvres that affectthe left and right hand side of the aircraft equally). The most straightforward exampleof this is a steady ‘pullout’ at constant velocity. Somewhat surprisingly, the elevatorangle required for pitch trim in a steady banking turn can also be calculated in thesame 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 pulloutConsider 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 wouldtake to complete a full circle at constant speed. If the aircraft completed a full circle itwould pitch though 2π radians and would therefore cover a distance of 2πr, where ris 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 thepitch rate, q, is:

q =2π

2πV/ng=ng

V.

This pitching will cause the tail of the aircraft to move down relative to the incomingair. This causes the incidence at the tailplane to increase by an amount:

∆αT =qlTV,

where lT is the tail arm measured from the centre of gravity. We already have an expres-sion for q, so we get:

∆αT =nglTV 2

.

Unfortunately, this expression has a V 2 term, and hence will vary with the flightconditions. 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 =ρgSlTW

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 theaircraft in the steady pullout is (1 + n)CL:

CLT=a1

a

(1− dε

dα

)(1 + n)CL + a1(ηT − ε0) + a2η + a3β + a1∆αT.

Hence,

CLT=a1

a

(1− dε

dα

)(1 + n)CL + a1(ηT − ε0) + a2η + a3β + a1

nCL

2µ1

.

The basic pitching moment equation is still valid, since it makes no assumptionsabout the source of the lift and moments—it is simply the result of non-dimensionalizinga 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− dε

dα

)(1 + n)CL + a1(ηT − ε0) + a2η + a3β + a1

nCL

2µ1

].

For the straight and level flight of the aircraft, in trim, we have previously derivedthe equation:

CM = 0 = CM0 − (h0 − h)CL − V

[a1

a

(1− dε

dα

)CL + a1(ηT − ε0) + a2η + a3β

].

(6.1)


In trim, the pitching moments acting on the manoeuvring aircraft will be zero ifthe aircraft is undertaking a steady manoeuvre. If we now look at the expression fortrim 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− dε

dα

)(1 + n)CL + a1(ηT − ε0) + a2(η + η) + a3β + a1

nCL

2µ1

]. (6.2)

These equations (6.1) and (6.2) are very similar. We can therefore subtract onefrom the other and get:

0 = −(h0 − h)nCL − V

[a1

a

(1− dε

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− dε

dα

)+

a1

2µ1

].

This must always be negative, otherwise the aircraft would pitch nose-down whenthe 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− dε

dα

)+

a1

2µ1

],

hm = h0 + V

[a1

a

(1− dε

dα

)+

a1

2µ1

].

This should be compared with the neutral point location stick fixed, hn, which wehave previously shown to be:

hn = h0 + Va1

a

(1− dε

dα

).

Therefore, the stick fixed manoeuvre point is a distance a1c/2µ1 aft of the stickfixed 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 ofgeometry.

6.3. STICK FIXED MANOEUVRE STABILITY 43

6.3 Stick fixed manoeuvre stabilityThe 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− dε

dα

)+

a1

2µ1

].

Hence,

h = h0 +V a2

CL

∆η

n+ V

[a1

a

(1− dε

dα

)+

a1

2µ1

].

Therefore,

Hm = −V a2

CL

∆η

n.

This result is important because it demonstrates that there is a relationship betweenthe 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 aboveexpression. 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 trimthe aircraft in a steady pullout or bank, but tells us nothing about the stick forcesrequired for the manoeuvres (just as stick fixed analysis told us nothing of the stickforces for straight and level flight). ‘Stick free’ manoeuvre stability analysis will allowus to calculate these stick forces.

6.5 AnalysisIn §5.1, we derived an expression that allowed us to calculate the hinge moments fortrim in straight and level flight:

CM = 0 = CM0 − (h0 − h)CL − V

[a1

a

(1− dε

dα

)CL + a1(ηT − ε0) + a3β +

a2

b2(CH − b0)

].

The same process can be undertaken to find the hinge moments for trim in a steadypullout, CH + ∆CH . For the pullout, assuming that the tab is not used:

CM = 0 = CM0 − (h0 − h)(1 + n)CL−

V

[a1

a

(1− dε

dα

)(1 + n)CL + a1(ηT − ε0) + 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− dε

dα

)nCL + a1

nCL

2µ1

+a2∆CH

b2

].

Hence,

V a2

b2CL

∆CH

n= −(h0 − h)− V

[a1

a

(1− dε

dα

)+

a1

2µ1

].

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

h′m = h0 + V

[a1

a

(1− dε

dα

)+

a1

2µ1

].

The stick free manoeuvre margin, H ′m, is defined as:

H ′m = h′m − h,

which can easily be shown to be:

H ′m = − V a2

b2CL

∆CH

n.

The stick force per g is calculated from the hinge moment per g, in exactly thesame way as for straight and level flight:

∆Pe

n= me

ρV 2

2Sηcη

∆CH

n.

For handling safety the stick force required to pull high g should be appreciable toavoid accidentally exceeding the structural limitations of the aircraft. A typical valuefor a non-aerobatic aircraft is usually of the order of 20 N/g.

6.6 Tailless aircraftThe analysis for tailless aircraft is very similar to that for conventional aircraft. Theflight 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

∂ηη − (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 rateq we can write:

CM = 0 = CM0 +∂CM0

∂η(η + ∆η)− (h0 − h)(1 + n)CL +

∂CM

∂qq.

Again, we can subtract one equation from the other to get:

∆η =1

∂CM0/∂η

[(h0 − h)nCL −

∂CM

∂qq

],

where ∂CM/∂q is an aerodynamic derivative. There are a large number of aerodynamicderivatives that can be defined for any aircraft, and they enable us to calculate theaerodynamic 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 Sc20

∂M

∂q.

Hence,

∂CM

∂q=

1

ρV 2Sc0/2

∂M

∂q=

2c0Vmq

and we know that

q =ng

V.

Therefore,

∆η =1

∂CM0/∂η

[(h0 − h)nCL −

2c0Vmq

ng

V

].


As for the conventional aircraft, we have an expression that includes the flightvelocity. Again, we can remove this by using the definition of the lift coefficient andrearranging 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 undertakemanoeuvres.

6.7 Tailless aircraft manoeuvre pointAs for the conventional aircraft, the manoeuvre point is defined by the c.g. locationthat 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 neutralpoint for the aircraft (which is at h = h0). The damping in pitch has increased thestability.

6.8. TAILLESS AIRCRAFT MANOEUVRE MARGINS 47

6.8 Tailless aircraft manoeuvre marginsThe manoeuvre margin for a tailless aircraft, Hm, is defined identically to that for aconventional 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 manoeuvremargins

Conventional aircraft

We have shown that the static margins, stick fixed and stick free, for conventionalaircraft are:

Kn = (h0 − h) + Va1

a

(1− dε

dα

),

K ′n = (h0 − h) + V

a1

a

(1− dε

dα

).

Also, the manoeuvre margins for conventional aircraft are:

Hm = (h0 − h) + V

[a1

a

(1− dε

dα

)+

a1

2µ1

],

H ′m = (h0 − h) + V

[a1

a

(1− dε

dα

)+

a1

2µ1

].

Therefore,

Hm = Kn +V a1

2µ1

,

H ′m = K ′

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 thepitch rate. Note that since µ1 is a function of the air density the manoeuvre margindecreases 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 manoeuvringdue 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 adensity term in µ1.

6.10 Modification of stick free neutral and manoeuvrepoints

Two common ways of modifying the stick free neutral and manoeuvre points areshown 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 wouldcause a positive deflection of the elevator.

These two additions have no effect on the stick-fixed neutral or manoeuvre pointssince 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 neutralpoint aft but has no effect on the stick free manoeuvre point. In contrast, a positivebob-weight moves both the stick free neutral point and the stick free manoeuvre pointof an aircraft aft. These effects are shown in Figure 6.5.

By combining positive and negative springs and bob-weights it is possible tomove 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

N ′ M ′

Figure 6.5: Effects of positive springs and bob-weights

are directly proportional to the stick free static margin and the stick free manoeuvremargin this enables the stick forces to be modified by a simple mechanical additionto 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 thepilot and suitable stick loads for trim, but the low stick force per g resulting from thelow manoeuvre margin might cause a risk of inadvertently overstressing the aircraft.The addition of a negative spring together with a positive bob-weight would solvethis problem since the stick free static margin would be unchanged but the stick freemanoeuvre margin would increase. This is shown in Figure 6.6.

Positive bob-weight

Negative spring

N ′ M ′

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 willbe valid at all flight conditions. At higher angles of attack the aerodynamics becomenon-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 occurat high speeds, and the effect that this has on the control of the aircraft.

7.1 High speed effectsChanges 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 lowspeed to mid-chord at supersonic Mach numbers.

These effects are shown in Figure 7.1. Combined with these are the effects of Machnumber 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 dynamicpressure, 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 differentaltitudes, 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, asshown 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 thevolume 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 supersonicspeeds. 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 wingaircraft (subsonic jet transports—Boeing 707, 747, etc.) an up-elevator or stabilisationinput is provided by a Mach number sensing system. This is known as ‘Mach trim’. Ifthe nose down moment were allowed to take effect the stick force gradient would bereversed, and there is also a danger that the maximum allowable speed of the aircraftdue to structural limits would be exceeded. The stick forces for such an aircraft areshown 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

V a1

1.0

Rigid aircraft

Sea levelDecreasing altitude

M

V a2

1.0

Rigid aircraft

Sea level

Decreasing altitude

Figure 7.4: Aeroelastic effects on tailplane and elevator

54 CHAPTER 7. COMPRESSIBILITY EFFECTS

M

∂ε/∂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

Part II

Dynamic stability

55

Chapter 8

Dynamic behaviour of aircraft

In the first part of the course, we examined the static stability of aircraft, which meansthat we considered whether an aircraft tends to return to its equilibrium position aftera perturbation, §1.1. We are now going to analyze the dynamic stability of aircraft andsee 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 givenin Figure 8.1 and follow a logical order. Once the x, y and z-axes are defined we thenhave, 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 inspace, but moves with the aircraft. The origin of the axis system is at the centre ofgravity 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 whenperforming analysis of the dynamic modes of an aircraft, these degrees of freedom areexpressed as perturbation quantities in relation to steady straight flight (i.e. velocityperturbations u, v and w and rotational velocities p = φ, q = θ and r = ψ).

8.2 Aerodynamic derivatives

We now have a co-ordinate system that allows us to define any perturbation of theaircraft from straight and level flight. To continue, we need to find out what forcesare acting on the aircraft for a given perturbation1.

1The 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 Momentforce velocity velocity angle velocity of inertia

x X U u φ p A Ly Y V v θ q B Mz 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

∂uu+

∂M

∂vv +

∂M

∂ww +

∂M

∂pp+

∂M

∂qq +

∂M

∂rr.

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 allof the forces and moments acting on the aircraft (six equations). If, in addition, weknow the mass of the aircraft and its inertia in roll, pitch and yaw (A, B, C) we cancalculate 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 theaircraft and is not necessarily horizontal. The other quantities we need for the aircraftare:

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) = mv + Ω× (mv) = F +mg, (8.3a)

dh

dt= h + Ω× h = L, (8.3b)

where the boxed terms are required because the frame of reference is rotating. Theapplied moment L is:

L = Li +M j +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 dynamicallystable, so we need to make some approximations to see how the aircraft behaveswhen 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 = U i,

u = ui + vj + wk,

ω = pi + qj + rk.

For a small rotation χ,

χ = φi + θj + ψk,

ω = φi + θj + ψk.

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:

mu +m(χ×V1 + χ× g) = δF, (8.4a)

h = 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 w, the down-wash acceleration. Then:

δF = δXi + δY j + δZk,

δL = δLi + δM j + δNk,

and, for example,

δX =∂X

∂uu+

∂X

∂vv +

∂X

∂ww +

∂X

∂pp+

∂X

∂qq +

∂X

∂rr,

δM =∂M

∂uu+

∂M

∂vv +

∂M

∂ww +

∂M

∂ww +

∂M

∂pp+

∂M

∂qq +

∂M

∂rr.

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, forexample, cannot cause a response in yaw or roll. Also, the symmetric response to anasymmetric input has to be symmetric: if the aircraft rolls at a given rate, the pitchresponse must be the same whether it rolls in a positive or negative sense. These tworequirements 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

∂uu+

∂X

∂ww +

∂X

∂qθ)i + (

∂Y

∂vv +

∂Y

∂pφ+

∂Y

∂rψ)j

+ (∂Z

∂uu+

∂Z

∂ww +

∂Z

∂qθ)k,

δL = (∂L

∂pφ+

∂L

∂rψ +

∂L

∂vv)i + (

∂M

∂qθ +

∂M

∂uu+

∂M

∂ww +

∂M

∂ww)j

+ (∂N

∂pφ+

∂N

∂rψ +

∂N

∂vv)k.

8.3. LONGITUDINAL SYMMETRIC MOTION 61

We need one more assumption about the aircraft, which is that there is no inertialcoupling between yaw and roll. This means that the only moments of inertia weneed 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:

mu =∂X

∂uu+

∂X

∂ww +

∂X

∂qq −mgθ, (8.5a)

m(w − Uq) =∂Z

∂uu+

∂Z

∂ww +

∂Z

∂q, (8.5b)

Bq =∂M

∂qq +

∂M

∂uu+

∂M

∂ww +

∂M

∂ww. (8.5c)

and

m(v + Ur) =∂Y

∂vv +

∂Y

∂pp+

∂Y

∂rr +mgφ, (8.6a)

Ap =∂L

∂pp+

∂L

∂rr +

∂L

∂vv, (8.6b)

Cr =∂N

∂pp+

∂N

∂rr +

∂N

∂vv. (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 tonote is that these equations are uncoupled so that longitudinal motion does not affectlateral and vice versa.

8.3 Longitudinal symmetric motion

The important information about the dynamic response of a system is the set ofmodes 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 ofparameters which satisfy the system. The most convenient form of solution is:

u = u0eλt, v = v0e

λt, θ = θ0eλt.

Inserting these assumptions into (8.5a), for example, yields:

mu0λeλt =∂X

∂uu0e

λt +∂X

∂ww0e

λt +∂X

∂qθ0λeλt −mgθ0e

λt.

2The following analysis, with different notation, is based on GRAHAM, W., ‘Asymptotic analysisof the classical aircraft stability equations’, Aeronautical Journal, February 1999, pp 95–103.


Now, we can divide through by expλt and, as always, non-dimensionalize the pa-rameters, to give the non-dimensional equations of motion:

(Λ− xu)u′ − xww

′ −(xqΛ

µc

− CL

2

)θ0 = 0, (8.7a)

−zuu′ + (Λ− zw)w′ − Λ

(1 +

zq

µc

)θ0 = 0, (8.7b)

−(mw

Λµc +mw

)w′ +

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,

mw =∂M∂w

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 statewithout proof that there is a solution with Λ and u′/θ of order one and w′/θ0 of order1/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 −Λ0 −mw Λ(bΛ−mq)/µc

u′w′

θ0

=

000

.This equation can only have a non-trivial solution if the determinant of the matrix iszero:

Λ2 − xuΛ−−zu

2CL = 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 constantenergy, 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 rateof change of vertical force with small changes in horizontal speed. Rememberingthat the z axis points vertically down, we can see that zu < 0 and the damping ispositive. Although it is not proven on the basis of these results, a statically stableaircraft always has a stable phugoid.

Short period oscillation

The second solution for longitudinal oscillation is for the case where Λ is of orderµ

1/2c , u0/θ0 is of order µ−1/2

c and w0/θ0 is of order one. In this case, the approximationto (8.7) is:

Λ− xu −xw −(xqΛµc

− CL2

)0 Λ− zw −Λ

0 −(mwΛµc

+mw

)Λ(bΛ−mq)

µc

u′w′

θ0

=

000

.Again, we find the natural frequency by requiring that the determinant of the matrixbe zero:

Λ(Λ− xu)

[Λ2 −

(zw +

mq +mw

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 +mw

b

), damping. (9.2b)

This is the short period oscillation and is a heavily damped mode with period typicallyof 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 typicallya few seconds, but must not be less than about 1.25s, otherwise there is a risk of PilotInduced Oscillation (PIO).

9.1. LATERAL MOTION 65

V

Figure 9.2: Short period oscillation

The frequency is proportional toK1/2n , and increases with dynamic pressure, ρV 2/2.

Therefore the aircraft will have the highest frequency SPO, and hence the shortesttime 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 = φ0e

λt, r = r0eλt,

and non-dimensionalize quantities in (8.6), which we do in the same way as beforeexcept that our reference length is now s, the wingspan:

(Λ− yv)v′ −

(ypΛ

µs

+CL

2

)φ0 +

(1− yr

µs

)r′ = 0, (9.3a)

−lvv′ + (aΛ− lp)Λ

µs

φ0 −lrµs

r′ = 0, (9.3b)

−nvv′ − npΛ

µs

φ0 +cΛ− nr

µs

r′ = 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,

v′ =v0

U, r′ =

mr0ρUS

, µs =m

ρSs.


Dutch roll

The first lateral mode we consider is Dutch roll which has oscillations of roughly equalmagnitude in pitch, yaw and roll. In this case, (9.3) reduce to: Λ 0 1

−lv aΛ2/µs 0−nv 0 cΛ/µs

v′φ0

r′

=

000

.As before the determinant of the matrix must be zero for a non-trivial solution toexist:

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 istypically a few seconds, but as for the SPO it should not have a period of less than1.25s due to PIO.

Dutch roll is not permitted to be divergent. Divergent Dutch roll can be ‘fixed’ bya yaw damper on the rudder which damps the yawing oscillation, and hence the rollresponse as well.

Spiral mode and roll subsidence

There are two further solutions to the dynamic equations which have small values ofΛ. These are dominated by yaw and roll with weak sideslip and the correspondingapproximations to (9.3) are: 0 CL/2 1

−lv (aΛ− lp)Λ/µs −lr/µs

−nv −npΛ/µs (cΛ− nr)/µs

v′φ0

r′

=

000

.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. Thefirst, Λrs, describes rolling subsidence which is a pure rolling motion that is generallyheavily damped, and is therefore usually stable. The damping is primarily from thewings, where the incidence along the wing is changed due to the roll-rate, as shownin 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 therolling subsidence mode is stable. This is generally the case. However, if Lp becomesnegative, 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 spiralmode of the aircraft. This is a combined yaw and roll motion which is allowed to beunstable (i.e. negatively damped) as long as it does not double amplitude in less thantwenty seconds, so that it can be controlled out.

The dynamics of the spiral mode are that if the aircraft rolls slightly, it will startto sideslip, and the fin then tries to turn the aircraft into the relative wind due to ayawing moment Nvv. However, the rolling moment due to sideslip Lvv tries to rollthe wings back level. Depending on which of the effects ‘wins’, the aircraft will bespirally unstable or stable, as can be seen from the numerator of (9.6).

9.2 Dihedral effect and weathercock stabilityThe 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 ‘dihedraleffect’ and ‘weathercock stability’ respectively. The effect of the two aerodynamicderivatives 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 tosideslip comes from dihedral (on an aircraft with unswept wings), as shown in Fig-

1Something 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 microlightaircraft’, 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 rollingmoment (and hence negative Lv).

ΓRelative 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 wingeffective incidence near the root. These contributions are negative for high mountedwings 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 highmounted wings (e.g. Harrier). Ground clearance issues may limit anhedral on lowwing 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 < 0Lv > 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 thesideways lift-force of the fin in sideslip, and should always be negative. However, asshown 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, sincethe fin’s lift curve slope is reducing. Therefore an aircraft with a large fin may bespirally 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 theireffectiveness, while at supersonic speeds this interference is progressively removed,increasing their effectiveness to combat the decreasing lift curve slope. This is shownin Figure 9.8.

For VSTOL aircraft (e.g. Harrier) engine air intake mass flow may give a negativecontribution to Nv making the aircraft directionally unstable in the hover and at lowforward speeds. This is because the air undergoes a change in direction to go downthe 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 theyoperate without friction and, most of the time, have no applied thrust. The control ofspacecraft is a problem in fixing their orbital parameters, so we start by analyzing theclassic 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 agravitational force of magnitude Gm1m2/|r1 − r2|2. We note, in passing, that onemass will usually be very much greater than the other (when the Earth orbits the sunor a satellite orbits the Earth, for example), though the analysis does not depend onthis 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:

m1r1 = − Gm1m2

|r1 − r2|3(r1 − r2), (10.1a)

m2r2 = − Gm1m2

|r1 − r2|3(r2 − r1). (10.1b)

We can extract one useful piece of information immediately by considering thecentre of mass of the system rc:

rc =m1r1 +m2r2

m1 +m2

.

Adding equations 10.1 shows that rc ≡ 0. In other words, the centre of mass of thesystem moves at constant velocity.

To find the relative motion of the masses, subtractm2 times( 10.1a) fromm1 times (10.1b)and define r = r2 − r1:

m1m2r = −Gm1m2(m1 +m2)

r3r,

r = −µrr3, (10.2)

where the gravitational parameter µ = G(m1 +m2). For most purposes, m1 m2 andµ ≈ Gm1. For Earth, µ = 3.98601× 105km3/s2.

x

y

θr

θ r

Figure 10.2: Polar coordinate system

To solve for r, we rewrite the system using polar coordinates, Figure 10.2. Thereare two unit vectors r and θ, the radial and azimuthal vectors respectively:

r = [cos θ sin θ], (10.3a)

θ = [− sin θ cos θ], (10.3b)

and differentiation will easily show that:

˙r = θθ, ¨r = θθ − θ2r, (10.4a)˙θ = −θr, ¨

θ = −θr− θ2θ. (10.4b)

10.1. THE TWO-BODY PROBLEM 75

Since r = rr:r = (r − rθ2)r + (2rθ + rθ)θ = − µ

r2r,

and, extracting components,

r − rθ2 = − µ

r2, (10.5a)

2rθ + rθ = 0. (10.5b)

Equation 10.5b can be rearranged to show that:

d

dt

(r2θ

)≡ 0.

This tells us that h = r2θ, the angular momentum, is constant. We can also derive anenergy conservation equation by taking the dot product of (10.2) with r:

dE

dt≡ 0, (10.6)

E =r.r

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 toθ. We also make the transformation r → 1/u. Then:

r =1

u,

r = − 1

u2

du

dθθ,

r = − 1

u2

d2u

dθ2θ2.

Inserting these terms into (10.5a) and noting that h = r2θ, the differential eqution ofmotion is:

d2u

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 ofθ, 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 10.11has the form of a conic section. In particular, for most spacecraft it is elliptical (orcircular).

Elliptical orbits

Equation 10.11 describes the trajectory of an orbiting body as a conic section. Thegeometry of an elliptical orbit is shown in Figure 10.3 which shows the notation andgeometrical 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 shapeof 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, parabolicorbits have e = 1 and hyperbolic orbits have e > 1.

a

b

a

FocusFocus

rθ

p

PerigeeApogee

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 toapogee 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

µ1/2a3/2 (Kepler’s third law), (10.13)

which tells us that once an orbital radius (or semi-major axis) is selected the period ofthe 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 thedebris are in a circular orbit.

Orbital radius a = (6400 + 200)km,

Orbital period T =2π

µ1/2a3/2,

= 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 maneouvresThe first thing we have to note that makes spacecraft different from aircraft comesfrom (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) constantbut we can vary E, r and v, though not independently—(10.14) is always true. If wechange the spacecraft velocity at some radius r, thenE will change and the spacecraftwill 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 Earthorbit.

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 alow earth orbit of radius a1, we can shift it to a higher orbit of radius a2 using thescheme shown in Figure 10.4. The transfer orbit is elliptical with semi-major axisa = (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 enterthe transfer orbit, we have to change from the circular orbit velocity v = (µ/a1)

1/2 tothe 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’), itscapacity to change velocity, as this limits its range of maneouvre. Absolutely every-thing that a spacecraft uses, including propellant for maneouvring, has to be liftedfrom the Earth’s surface, so it is vital to use economical transfer orbits. The onlyproblem 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 bea problem for other missions (Mars, for example). An especially important ∆v is thevelocity change required to escape the gravity of a planet. This is called the escapevelocity 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 tohave a spacecraft leave the orbit of one planet and approach the orbit of another. If allwe want is to fly past the planet, no more need be said. Indeed, a common methodof 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 toenter the orbit of the planet, however, we need to slow it down. Remember that thespacecraft is moving very quickly because it has reached the escape velocity for theplanet 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 inorbit about the sun. It is only as it comes close to the planet that it feels the effect ofthe planet’s gravity. At some point, as for a Hohmann transfer, we need to change thespacecraft velocity to match the velocity of a planetary orbit. These velocity changescan be quite large (for Mars Global Surveyor, for example, ∆v = 973m/s) so it isobviously 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 orbitalradius. If the spacecraft approaches the planet with speed v∞, its speed in the planetframe 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 costsmoney so we would like to find the radius which minimizes ∆v. Making trajectoryadjustments during the trip is relatively cheap, because very small adjustments cangive 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 inputis applied, subject only to gravitational effects. This makes life simple in one waysince basic Newtonian mechanics applies, without dissipation, but it does make thedesigner’s life quite difficult, since there is no damping. The two main functions of acontrol 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 examplesinclude:

earth observation can only be performed if a satellite can point a camera at a knownposition 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 theright direction to generate power. Depending on the reason for the alignment, and theaccuracy 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 thedesired attitude is wasteful of fuel and makes the system bounce back and forth. Inpractice, spacecraft are designed to be inherently stable, which is achieved by makingthem spin.

A mechanics textbook will give the Euler equations for a spinning body. If thebody 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 = Adω1

dt+ (C − A)ω2ω3, (11.1a)

M2 = Adω2

dt+ (A− C)ω3ω1, (11.1b)

M3 = Cdω3

dt+ (A− A)ω1ω2, (11.1c)

where Mi and ωi are the moment and angular velocity about the ith axis and A andC are the moments of inertia. For a body in space, there are no applied moments soMi ≡ 0 and this means that dω3/dt = 0 or ω3 = Ω and the body rotates at constantfrequency about its axis. Now, we can solve for the other two angular velocities:

dω1

dt= +

A− C

AΩω2,

dω2

dt= −A− C

AΩω1.

Differentiating the first and substituting the second equation:

dω1

dt2+ α2ω1 = 0,

where α2 = Ω(C − A)/A. This is a simple harmonic oscillator and the solution is:

ω1 = ω0 sinαt, (11.2a)ω2 = −ω0 cosαt. (11.2b)

This solution says that the spinning body rotates about one axis at a frequency Ωand oscillates about the other two. The axis of rotation actually swings around anddescribes a cone in space. If the spacecraft has been designed and set up properly, thespin 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 spacecraftis asymmetric, the variation in gravity over its extent is enough to generate a torquewhich draws it back into the required alignment. This is like having a pendulumwhich 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 placedinto an orbit which is synchronized with the sun. This means that the satellite willspend 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 theground will always be illuminated from the same direction. Finally, this means thatthe satellite can guarantee that its solar panels will be generating power: an importantconsideration 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, becausethings don’t look right.

An example is the docking of one spacecraft with another (supply flights to aspace station; Shuttle to Hubble; Apollo lander with orbiter). If one spacecraft isalready on orbit, the other must approach that orbit to rendezvous with it. There aretwo standard approaches.

The first is to arrive along the orbit in front of the other spacecraft. Because thismeans flying along the velocity vector of the main craft, it is called a ‘v-bar’ approach(as in v 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 theradius vector makes the docking craft move ahead of the main craft, if it approachesfrom 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 rendezvousingwith.

While a v-bar approach is favoured for simplicity, an r-bar manouevre is oftenfavoured because it means that the thrusters used for braking never point at the maincraft. 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. onHubble). The manouevre is carried out by adjusting the docking vehicle’s speed tochange its orbit slightly and bring it onto the main orbit from ‘above’ or ‘below’.

Part IV

Problems

85

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 netmoment 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 insteady straight and level flight.

Show that the pitching moment about the centre of gravity is given by

CM = CM0 − (h0 − h)CL − V CLT.

For the sailplane whose details are given in Table 12.1, calculate the value ofCLT

required for trim at 50kt EAS with a pilot weighing 0.75kN. The empty weightequipped is 2.5kN, with c.g. on the mean chord 0.45c aft of the leading edge ofc. 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 bothstable and able to trim at positive lift coefficient the overall pitching momentabout the centre of gravity must be positive at CL = 0 in that configuration.

87

88 CHAPTER 12. PROBLEMS

S = 28m2 ST = 1.4m2 c = 1.15ml = 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 isgiven by:

CLWBN= CL

[1 + (h0 − h)

c

l

]− CM0

c

l.

If the aircraft stalls when CLWBNreaches its maximum value, (CLWBN

)max say,then obtain an expression relating the stalling speed to the c.g. position at anyone given weight.

Hence calculate the c.g. shift that would increase the stalling speed by 1% ifc = 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.

hcW

hnc

CL

(CM0)hpc

hnc

CL

(CM0)(CMp)

Figure 12.2: Full-scale and model aircraft. CMp is measured by the balance, whichrestrains the model in pitch.

In (a) the full scale aircraft is in steady free flight with values CL, h, η for the liftcoefficient, c.g. position and elevator angle respectively.

In (b), the model of the same aircraft is suspended from a wind tunnel balanceat the sameCL and elevator setting as in (a). The balance measurement gives thepitching moment coefficient CMP about the balance pivot axis which is locatedat hp with respect to the same datum line as h.

a) Write down the moment equations for situations (a) and (b), and hencederive the relationship between the balance reading CMP , equivalent tothe steady free flight conditions, and interrelating hp, h and CL.

b) An aircraft model is found to have a zero-lift pitching moment coefficientof 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 astick-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 thestick-fixed neutral point and static margin.

W = 30kN S = 23m2 ST = 3.5m2

c = 1.96m l = 5.5mh0 = 0.25 c.g. is 0.61m aft of datumCM0 = -0.036 ηT = -1.5 ε = 0.48αa = 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 totrim (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 a2dη

dCL

= (h0 − h) + Va1

a

(1− ∂ε

∂α

).

An aircraft has the following values of the aerodynamic coefficients:

h0=0.25, a=3.5, a1=3.0, a2=1.5, ∂ε/∂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 ofCLof 1.0 (hfwd).

Hence find the minimum tail volume ratio such that with a c.g. shift of 0.15c thechange in elevator angle to trim is not more than 10 for a change of CL of 1.0and 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 wingzero lift line.

Minimum Kn= 0.05 c.g. range ∆h=0.50 h0=0.075 S = 232m2

c = 4.72m l=19.5ma=5.7 a1 = 2.7 a2 = 2.1CM0= -0.14 ε=0.16α.

The change in CM0 at landing flap setting ∆CM0 = −0.10. Note that the wingzero 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 inwhich the elevator angles to trim are found at certain speeds. In practice, theaeroplane is trimmed at a series of speeds by adjusting the tab setting, and boththe elevator angle and tab angle are observed. Since the theory which relatesthe stick-fixed static margin to the elevator angles to trim implicitly assumes aconstant tab angle, show that a correction must be applied to elevator anglesobtained 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 byan independent power control unit. These units are so designed that if a failureoccurs 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 givelongitudinal trim of the aircraft whose details are given below:

Weight = 850kN Speed = 70m/s EASWing area S = 358m2 (h0 − h) = 0.15CM0= +0.02 ∂CM0/∂η = -0.45a1 = 4.0 a2 = 0.95b1 = -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 thatthe tailplane setting relative to the wing zero lift line is directly proportional tothe tail load. Prove that the reduction in stick fixed static margin compared withthat of the rigid aircraft is given by:

∆Kn = Krigidn −Kflexible

n ,

= Va1

a

(1− ∂ε

∂α

) [1− 1

1 + 12ρV 2STa1f

]For the human-powered aircraft having the characteristics given below, findthe 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 of9.2m/s at sea level.

S = 28m2 ST = 1.4m2 l = 5.34m c = 1.14ma = 6.0 a1 = 4.5 ε = 0.20α.

[f = 0.1/N]

93

4. The control column of a low-speed aeroplane is connected to the elevator byan arrangement of cables which stretches when a stick force is applied. Thestiffness of the circuit is given by dHE/dη = ENm/rad where HE is the hingemoment 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) + Va1

a

(1− ∂ε

∂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 arigid aeroplane. What practical use might you make of this information?


1. What conditions define the stick-fixed and stick-free manoeuvre points of anaircraft?

From first principles, stating your assumptions, derive an expression for thestick fixed maneouvre point of a low speed aircraft of canard layout. Showwhether this is forward or aft of the corresponding neutral point and compareyour expression with that for a conventional aircraft.

2. Define the maneouvre point stick-free for a conventional aircraft. How does itdiffer from the corresponding neutral point?

Find the minimum stick force per g at sea level for the light aircraft whosedetails are given below. Comment on your result and find the c.g. positionrequired to give 22N/g. Suggest alternative means for increasing the existingvalue.

W = 2.7kN S = 7.6m2 l = 2.9mh0 = 0.238 c = 1.2m V = 0.34ε = 0.385α a = 3.865 a1 = 2.73a2 = 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

a2VH ′

m,

= 83.2H ′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 steadypullout at AN = 2.5 (n = 1.5) at 250kt EAS at a height where the air densityρ = 1.150kg/m3, the change of elevator setting compared with steady level flightunder 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 perg 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 isflying at 440kt EAS in an 8g pullout at a height where the relative density of theair 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=2.7CM0 = +0.03 ∂ε/∂α=0.38 h0 = 0.17 h = 0.50σ = 0.74

LWBN 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, thechange of elevator hinge moment coefficient-to-trim ∆CH between two lift co-efficients is given by

CH = − b2

a2V∆CLK

′n.

The aircraft described in the table below is making a zero stick force trimmedlanding approach at 155kt EAS. Calculate the value to which the speed may bereduced while keeping the stick force within 150N without altering the trim tabsetting, indicating clearly whether this is push or a pull force.

Weight W =785kN c.g. at h = 0.26Wing area S = 223m2 smc c = 5.68mTail area ST = 46.5m2 tail arm l = 15.66mElevator area Sη = 11.2m2 elevator mean chord cη = 0.908m

h0 = 0.16 CM0 = -0.06 ε = 0.38αa = 4.5 a1 = 2.75 a2 = 1.16b0 = 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 thehandling 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 anglerequired to pull 0.5g flying at 350kt EAS at an altitude where the relative densityσ = 0.374.

Explain, in simple physical terms, why this change of elevator angle would begreater 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 retractableforeplane. 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 higherweight 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 groundclearance problems, using the data in the table.

Calculate the elevon angles to trim of both versions of the aircraft. Comment onyour 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 = -0.25a1 = 3.0 a2 = 0.80 h0 = 0.61 c0 =27.4mlF = 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 centresection (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 withrespect to the centre section unit and that, for the lengths considered, ∂ε/∂α isconstant.

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 tobe affected only by the change of nose length ∆lN and is given by

∆h0 = −0.037∆lNc.

If a variant of the aircraft retains the original fuselage, but has its wing tipsextended, how could the longitudinal static stability be affected?


S = 223m2 c = 5.6mST = 46m2 l=15.5mh0 = 0.25 h = 0.20ε = 0.4α CM0=-0.06a = 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 curveslope, and comment on your answer.

b) The 1903 Flyer was stabilized in pitch by the addition of ballast to shiftthe centre of gravity forward. If 30% of the aircraft gross weight can becarried 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 aircraftperformance?

h0 ≈ 0 V = 0.134 CM0 = −0.141h = 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 thestatic margin stick-fixed and estimate the zero-lift pitching moment. Estimatethe dimensional and non-dimensional phugoid mode and SPO frequencies.

S = 18.58m2 s = 6.82m c = 3.13mh = 0.22m = 7056kg B = 10700kgm2 V = 331kt EASa = 3.5/rad ∂CM/∂α = −0.8/radZu = −332Ns/m Mw = −40.7NsZw = −14300Ns/mMq = −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 velocitychange 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 tolunar orbit and, for comparison, the escape velocity of Earth.

c) Comparing the ∆V required for transfer to lunar orbit and that to escapeEarth’s gravity, comment on the feasibility of a manned mission beyondthe 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. Calculatethe velocity change required to change its orbit to one with a perigee of174km and an apogee of 39000km.

b) Sketch the shape of the new orbit, indicating the point at which the burntakes place. Why is such an orbit used in practice?

Basic equations

101

103

Conventional aircraft Tailless aircraftSteady flight:

CM = CM0 − (h0 − h)CL − V CLTCM = CM0 +

∂CM

∂ηη − (h0 − h)CL

CLT=a1

a

(1− ∂ε

∂α

)CL + a1(ηT − ε0) + a2η + a3β

Steady pullout/bank:

CLT=a1

a

(1− ∂ε

∂α

)CL + a1(ηT − ε0) + a2η + a3β + a1∆αT

CM = CM0 +∂CM

∂ηη − (h0 − h)CL +

∂CM

∂qq

Static and manoeuvre margins:

Kn = (h0 − h) + Va1

a

(1− ∂ε

∂α

)Kn = h0 − h

Hm = Kn +V a1

2µ1

Hm = Kn −mq

µ1

K ′n = (h0 − h) + V

a1

a

(1− ∂ε

∂α

)H ′

m = K ′n +

V a1

2µ1

General relationships

µ1 =W

ρgSlTµ1 =

W

ρgSc0

a1 = a1

(1− a2b1

a1b2

)mq =

1

ρV Sc20

∂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

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Signature:____________________________ Date:_______________

Student Registration V 4November 2012 NFLC, Cranfield University

Documents

Some notes on aircraft and spacecraft stability and control