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© Aditya A. Paranjape
Chapter 1
Introduction to Flight Mechanics
Review of Aerodynamics; Thrust
Aditya Paranjape
1 Perspective on Flight Mechanics
Figure 1: The big picture of aeronautics. Flight mechanics sits squarely at the intersection of aerodynamicsand dynamical systems theory.
Figure 1 gives a broad division of aeronautics. Notice the areas that fall in the intersection zones: flightmechanics, aeroelasticity and vibrations. As Sinha and Ananthkrishnan note in their textbook 1, flightmechanics “is the point of confluence of other disciplines with aerospace engineering and the gateway toaircraft design.” This becomes apparent when we consider that this is the only sub-discipline which dealswith the aircraft as a whole! 2
2 Review of Aerodynamics
We assume the reader has had a basic course on subsonic aerodynamics at the very least. We quickly reviewthe major concepts on aerodynamics required in the coming chapters.
2.1 Anatomy of a Fixed Wing Aircraft
Figure 2 shows the standard components of an aircraft. The functions of each component are as follows:
• Fuselage: carry passengers, payload and fuel
1Sinha and Ananthkrishnan, Elementary Flight Dynamics with an Introduction to Bifurcation and Continuation Methods,CRC Press, 2013.
2A useful allegory is that of an elephant and blind men. Flight mechanics is like a blind man sitting atop the elephant - hecan feel how the elephant moves but can’t tell just what makes it move.
1
Figure 2: The major components that make up a generic aircraft. Image source:http://dduino.blogspot.ca/2012/08/blu-baby-42-rc-plane-build-log.html
• Wings: produce lift, carry fuel and provide a mounting for engines
• Empennage: provide stability (more on this in the latter chapters), house the primary power unit
The aircraft is a highly integrated and connected system, and it is getting more so. For example, althoughit may appear that the engines can be designed independently of the airframe as long as they provide therequired value of thrust, this is not so any more. The aerodynamic properties of the wing around the engineare affected by the location and the geometry of the engine inlet to the point where they have to be designedtogether. The reader is encouraged to read the Wikipedia entry on Boeing 737-MAX and follow it up withtechnical papers for an illustration. The wings, despite simplistic exteriors, are incredibly complex: theycontain fuel tanks, actuators for the flaps and spoilers, and a whole slew of sensors for detecting the loaddistribution and structural deformation. All of these sensors provide real-time feedback to the flight controlsystem which enables it to maintain the aircraft to the highest possible degree of efficiency.
Figure 3 shows some unconventional configurations. The X-31, shown on the left, never made it pastthe experimental stage. Notice two major differences: that it has canards instead of a horizontal tail, and ithas vanes at the engine nozzle for thrust-vectoring. These design changes were made specifically to achieveimprovements in the agility of the airframe. The B-2 bomber, on the other hand, is actively deployed by theUS Air Force. It is a flying wing configuration with no empennage and integrated engines. The configurationwas designed specifically to ensure that it is stealthy, but also achieves a high degree of aerodynamic efficiency.Incidentally, both these configurations present significant challenges for control design.
Figure 4 shows the standard notation used for writing the equations of motion and for defining theattitude of the aircraft. Of particular importance here is the orientation of the aircraft with respect tothe wind velocity vector. The orientation is defined via two angles: the sideslip β which measures how farthe velocity vector is oriented from the plane of symmetry of the aircraft, and the angle of attack α whichmeasures how far the nose has pitched up from the projection of the velocity vector on the plane of symmetry.The flight speed is denoted conventionally by V∞, but we drop the subscript ‘∞’ unless it is required. Othersymbols will be introduced in later chapters when they are required.
2.2 Forces on an Aircraft
Figure 5 shows the forces acting on aircraft. The aerodynamic forces are resolved along and perpendicularto the wind velocity vector. The lift acts in the plane of symmetry perpendicular to the wind velocity vector.It is produced primarily by the wings, while the horizontal tail typically reduces the net lift for reasons
2
(a) X-31 (b) B-2
Figure 3: Unconventional aircraft configurations: an aircraft where the horizontal stabilizer is moved aheadof the wing and referred to as a canard, and an aircraft without a tail. Source: Wikipedia
xB
zB
yB
Roll
Yaw
Pitch
V∞
α
β
90 - β
Longitudinal plane
q
pr
δ
Figure 4: The standard flight mechanics notation.
which will become clear later in this book. The fuselage also produces a small amount of lift which can be,however, safely ignored in a basic setting.
The drag acts along the velocity vector and is contributed mainly by the fuselage (skin friction) and thewings (skin friction and induced drag). In addition, aircraft flying at transonic speeds experience wave dragdue to shockwaves produced locally over the aircraft.
The angle γ in Fig. 5 is called the flight path angle. Thrust typically acts at a fixed angle to the bodyx axis. For our analysis, we will assume that α is small enough so that thrust can be assumed to act alongthe velocity vector. This assumption, however, needs to be used with caution.
The lift and drag are written as
L =1
2ρ∞V
2∞SCL
D =1
2ρ∞V
2∞SCD
where S is the wing reference area, and CL and CD are functions of
• Angle of attack α and sideslip β, with α being the most important of all factors. We will ignore thesideslip.
• Reynolds number Re =ρ∞V∞c
µ, where c is the mean wing chord length and µ is the coefficient of
viscosity of air. The Reynolds number affects the individual terms CL and CD, as shown in Fig. 6.
3
Figure 5: Forces on an aircraft.
Clearly, low Reynolds number regime is highly nonlinear and in general not suitable for fixed-wingflight. This regime is highly conducive, however, for efficient flapping flight.
• Mach number Ma =V∞a
. We will not delve into the effects of Mach number in this course.
We will assume that the flow is at a sufficiently large Reynolds number and subsonic so that across theentire range of flight speeds, CL and CD are functions of just α. Furthermore, we will assume that α issufficiently small so that the lift-α relationship is linear:
CL = CL0+ CLαα
where CL0 depends on the wing camber and aspect ratio AR, while CLα depends on the aspect ratio:CLα = 2π
1+2/AR. Note that α0, the angle of attack at which lift is zero, does not change and depends only on
the aerofoil cross-section. The coefficient of drag in the low-α regime can be expressed in terms of CL:
CD = CD0+
1
πeARC2L
where CD0depends largely on the airfoil geometry (aside from Re), while e is the Oswald efficiency factor
(0 < e < 1).The lift-to-drag ratio L/D = CL/CD is an important design assessment metric for an aircraft. Since both
CL and CD are functions of α, we will try to locate an α such that CL/CD is maximized. Let F = CL/CD.Since CL is a linear function of α, maximizing F with respect to α is equivalent to maximizing F withrespect to CL
F =CLCD
=CL
CD0+ kC2
L
∂F
∂CL= 0 =⇒ 1
CD− 2kC2
L
C2D
= 0
=⇒ CD = 2kC2L or CD0
= kC2L
The L/D ratio is maximized when the angle of attack equals
α(L/D)max=
√CD0/k − CL0
CLα
Finally, we note that at high angles of attack, the CL − α relation is nonlinear so that we can no longerwrite CL = CL0
+ CLαα. A more accurate representation is of the form CL = k1 sin(kαα + k0), which isconsistent with Fig. 7.
4
(a) Lift-to-drag ratio (b) Lift
(c) Drag
Figure 6: Effect of Reynolds number on lift and drag. Source: Mueller, 1999 (first plot) and Sandia NationalLabs Report 802114 (last two plots)
Figure 7: Lift curve across a ±180 degree angles of attack range. Source: Sandia National Labs Report802114
3 Standard Atmosphere
The standard atmosphere is a set of relations between the pressure p, density ρ, temperature T and altitudehG. In particular, it gives mean values of p, ρ and T as functions of hG (‘G’ denotes geometric; above the
5
mean sea level). The standard atmosphere is used for
• Air speed calculation: determining true air speed from measured air speed (recall how a Pitot tubeworks).
• Altitude determination using static pressure
• Design: provide baseline values of temperature and density for designing air frames and propulsion
The atmosphere is split into layers of constant temperature lapse rate, as shown in Fig. 8. Pressure anddensity are calculated from the governing equations.
Recall the universal gas lawp = ρRT
and the ydrostatic equationdp = −ρg dhG
where hG = 0 denotes the mean sea level. These equations yield
dp
p= −g dhG
RT
These equations help derive p and ρ as functions of temperature and altitude. There is one caveat though: gdepends on altitude hG. To get around the difficulties posed by this dependence, we define the geopotentialaltitude h via the equation
g0 dh = g dhG =⇒ dp
p= −g0 dh
RT
where g0 is the gravitational constant at mean sea level (hG = 0). Sinceg
g0=
R2e
(Re + hG)2, we get
dh = R2e
dhG(Re + hG)2
=⇒ h = Re −R2e
Re + hG=
RehGRe + hG
The relationship between h and hG is nonlinear, and furthermore hG = 0 =⇒ h = 0 and h → Re ashG →∞.
Figure 8: Atmospheric properties as functions of altitude. Image source: noaa.gov
The temperature lapse rates are defined with respect to h, so that dT = aT dh. If aT 6= 0 and if thetemperature at the base of the layer, altitude h1, is known as T1, then
T (h) = T1 + aT (h− h1)
6
This gives the following equation for the layer with base at h = h1:
dp
p= − g0 dh
R(T1 − aTh1) +RaTh
whose solution is given by
p(h) = p(h1)
(T1 + aT (h− h1)
T1
)−g0/RaT= p(h1)
(T
T1
)−g0/RaTThese formulae are useful in the lowest layer of the atmosphere, namely the toposphere, where the temper-ature lapse rate is −6.5 deg/km (degrees Celcius or Kelvin). If aT = 0, then dp/p = −g0 dh/RT1 and weget
p = p(h1)exp
(−g0(h− h1)/RT1
)The universal gas law can now be used to calculate the density ρ(h) from p(h) and T (h). These are tabulatedin the International Standard Atmosphere (ISA) tables which are available quite readily.
The standard atmosphere finds application in two important areas: determining the altitude of an aircraftfrom pressure measurements, and determining the air speed accurately from Pitot tube measurements.
3.1 Determination of altitude
The standard atmosphere gives a staight-forward correspondence between p and h, while hG can be foundusing the following expression:
h =RehGRe + hG
=⇒ hG =Reh
Re − hTo be precise, the altitude calculated this way is called the pressure altitude which equals the altitude onlyunder ISA conditions: 1 atm pressure and 298 K temperature at sea level. Under non-standard conditions,the ground values of pressure and temperature are adjusted to compute the altitude.
3.2 Air Speed Measurement
The pitot tube measures the air speed of an aircraft directly. This is called the Indicated Air Speed (IAS).The IAS is calculated using Bernoulli’s equation:
V∞ =
√2(p0 − p)
ρ
where p0 is the total pressure and p is the static air pressure. In traditional aircraft instruments, p andρ, which are altitude-dependent, are set to their static (Ma = 0) sea-level values. Therefore, the ISAis generally far from the actual speed of the the aircraft which can be calculated by making systematic,step-wise corrections. Each step yields a refined value of the air-speed (see Fig. 9:
• The calibrated air speed (CAS) is obtained from the IAS by correcting for position errors, instrumenterrors, etc., using charts supplied by the manufacturer.
• The density ρ is corrected by accounting for compressibility effects at Mach numbers outside thelow-subsonic range. This gives the equivalent air speed (EAS):
VEAS =
√(2(p0 − p)ρsea
)= VCAS
√ρsea,Ma=0
ρsea
7
• The altitude is corrected to obtain the true air speed
VTAS =
√(2(p0 − p)
ρ
)= VEAS
√ρseaρ
• The ground speed is found by subtracting the speed of the head-wind Vground = VTAS − Vheadwind
Figure 9: The four speeds.
4 Propulsion
There are four typical propulsion mechanisms used in aircraft, depending on the flight regime:
• Jet engines: most commonly used device, up to mid-supersonic speeds (Ma < 3)
• Propeller and engine: moderate subsonic speeds and low altitudes
• Ramjets and Scramjets: high supersonic and hypersonic speeds; used in missiles (BrahMos)
• Rockets: used in a few cases for take off (Me262c), and in some others for braking assistance whilelanding (YMC-130 Hercules)
Figure 10 is a schematic diagram of a turbojet. The thrust produced by a turbojet is given by
Figure 10: Schematic diagram of a turbojet. Source:Wikipedia
T = (mair + mf )Ve − mairV∞ + (pe − p∞)Ae
where e denotes the exit of the nozzle; Ve is the exhaust velocity of the air, pe is the exit pressure, and Ae isthe nozzle exit area. The mass flow rate is given by mair = ρ∞AinV∞. The thrust is generally approximatedby
T ≈ mair(Ve − V∞) ≈ ρ∞AinV∞Ve
8
Clearly, thrust reduces with increasing altitude since ρ∞ decreases with increasing altitude. Likewise, thrustdecreases when the ambient temperature increases. However, jet engine thrust is unaffected by flight speedin the subsonic regime, i.e., until Ma = 1. The latter property will be used in the next chapter when weanalyse level flight.
The derivation of an expression for the thrust produced by a propeller is beyond the scope of this course.Figure 11 shows the forces on a propeller cross-section: the net force is found by integrating these across thelength of all blades.
Figure 11: Forces on a spinning propeller, at an arbitrarily chosen cross-section. Source:McCormick
The thrust produced by a propeller is strongly dependent on the advance ratio J = V∞nD , where n is
the propeller rotational rate, and D is the diameter. This has been illustrated in Fig. 12. Clearly, propellerthrust drops as a function of the flight speed due to the increasing advance ratio, unlike a turbojet. Propellerthrust also drops with decreasing air density (increasing altitude).
Figure 12: Nondimensional thrust as a function of the advance ratio for a typical propeller. Source: Mc-Cormick
9
Chapter 2
Straight and Level Flight
Aditya Paranjape
Topics covered: lift and drag in level flight; flight speed and altitude envelope;range and endurance.
1 Basics
Figure 1: Forces on an aircraft in wings-level flight.
Figure 1 shows the forces on an aircraft: the thrust acts along the longitudinal axis of the body;the lift acts perpendicular to the velocity vector in the plane of symmetry, while the drag acts alongthe velocity vector. The equations of motion can be written in terms of the speed V and the flightpath angle γ as:
mV = T cosα−D −mg sin γ (Linear)
mV γ = T sinα+ L−mg cos γ (Centripetal)
We are interested in equilibrium conditions (also called trims or trimmed flight conditions) foundby setting V = γ = 0:
T cosα = D +mg sin γ (Linear)
T sinα+ L = mg cos γ (Centripetal) (1)
We assume that α is small enough so that T cosα ≈ T and T sinα << L. This gives us theequations of motion of straight and wings-level flight:
T = D +mg sin γ = D +W sin γ
L = mg cos γ = W cos γ
1
In this chapter, we focus on the equilibrium equations for straight and level flight which are obtainedby setting γ = 0 and correspond to flight at constant speed and altitude:
T = D =1
2ρV 2SCD(α)
L = W =1
2ρV 2SCL(α)
Straight and level flight is the most common mode of flight and is employed during
• Regular cruising flight
• Holding pattern and loiter
• Step-descent while landing in bad weather
2 Analysis of the Lift Equation
Consider the lift equation
L = W =1
2ρV 2SCL(α)
For a given aircraft, W and S are constant, while ρ is constant at a given altitude. Isolating theconstant terms on one side, we get
V 2CL =2W
ρS=
2
ρ
W
S= constant (2)
The ratio W/S is called the wing loading, and it is one of the most important design parametersfor an aircraft. This relation helps us determine
• The value of α required to fly straight and level at a given speed.
• Alternately, the flight speed which will be achieved for a chosen value of α.
Figure 2 plots the flight speed as a function of α for several values of wing-loading. For a givenaircraft, there is a maximum attainable value of CL which is represented by CLmax. Recall thatthe corresponding angle of attack is referred to as the stalling angle of attack. The stall speed ofan aircraft is given by
Vstall =
√2
ρCLmax
W
S
This is an important value of the flight speed since, for all practical reasons, controlled flight atsub-stall speeds is impossible. A slew of dangerous flight conditions arise out of stall, such as spin,auto-rotation and deep-stall. Consequently, aircraft generally fly at speeds greater than or equalto Vmin ≈ 1.2 Vstall.
From Figure 2, notice that the stall speed increases as the wing-loading W/S increases. Aircraftuse flaps to increase the wing area during take-off and landing; this allows them to achieve a higher
2
Figure 2: The relationship between flight speed and CL during straight and level flight for an AirbusA330 at sea level.
wing loading for cruise flight at higher speeds. The cruise speed increases further at high altitudesdue to a reduction in the air density ρ (see Eq. (2)). As altitude increases, the stall speed increasesas well. Notice that Vstall ∝ 1/
√ρ and has been illustrated in Fig. 3. Interestingly, the ratio between
the stall speed at a given altitude and the stall speed at sea-level is independent of the aircraftgeometry and weight, since
Vstall(ρ)
Vstall(sea)=
√ρseaρ
The stall-speed doubles around an altitude of 12000 m, which is close to the cruising altitude ofmost commerical jet aircraft.
Figure 3: Increase in the stall speed with altitude. These numbers are obtained for the AirbusA330.
3
3 Analysis of the Drag Equation
Consider the equation for thrust-drag balance:
T = D =1
2ρV 2SCD, CD = CD0 + kC2
L
First, we observe that since CL =2W
ρV 2S, the coefficient of drag during level flight is given by
CD = CD0 + kC2L = CD0 + k
(2W
ρS
)2 1
V 4
The thrust required for level flight at speed V is equal to the drag and is given by
TR = D =
(1
2ρSCD0
)V 2︸ ︷︷ ︸
skin friction
+ k
(2W 2
ρS
)1
V 2︸ ︷︷ ︸induced drag
(3)
The above equation tells us that while skin friction drag increases with flight speed as expected,the induced drag reduces with increasing flight speed due to the reduction in CL. The drag CD
Figure 4: The drag as a function of the flight speed for the Airbus A330.
has been plotted in Fig. 4 (again, for the Airbus A330) as a function of the flight speed. The dragis minimum at a speed where the skin friction drag is exactly equal to the induced drag. Thisobservation can be confirmed theoretically from Eq. (3) by using ∂D/∂V = 0.
Since
CD = CD0 + kC2L = CD0 + k
(2W
ρS
)2 1
V 4
The condition for minimum drag yields
V 4Dmin =
k
CD0
(2W
ρS
)2
=⇒ VDmin =
(4k
ρ2CD0
)1/4√W
S, k =
1
πeAR(4)
4
The minimum-drag CL is given by
CDminL =
√CD0
k=√πeARCD0 =⇒ αDmin = C−1L (
√πeARCD0) (5)
where αDmin is the angle of attack at which the aircraft should fly in order to ensure that thelevel-flight drag is minimized.
We make the following observations for the minimum drag speed.
1. The angle of attack for minimum drag depends only on the shape of the wing planform, theaspect ratio of the wing and on CD0 . It is notably independent of the weight of the aircraftand the wing area.
2. The minimum drag speed increases with increasing wing loading. This is a consequence ofCDminL being independent of the wing loading.
3. The minimum-drag speed increases with altitude. This is, yet again, a consequence of Eq. (5).
4. The speed for minimum drag reduces with increasing CD0 . Therefore, the “smoother” theaircraft configuration, the faster it can fly with given engines (and fuel, but that will be takenup later in this chapter).
The minimum drag is given by
Dmin = W × CDmin
CDminL
= 2W√kCD0 = 2W
√CD0
πeAR(6)
We make the following observations
1. The value of minimum drag is independent of altitude
2. The minimum drag reduces as the aspect ratio AR increases and as the wings become moreelliptical (e→ 1).
4 Thrust and Power
The thrust required for sustained level flight equals drag, so that
TR =1
2ρV 2SCD = W × CD
CL= W ×
CD0 + kC2L
CL
Let TA denote the maximum available thrust. There are two level flight conditions at which allavailable thrust is required. The value of CL at these two conditions is found by solving thequadratic equation
kC2L +
(TAW
)CL + CD0 = 0
The above equation has two solutions; let us denote them as CSL ≤ CBL . We make the followingobservations.
1. The fastest speed at which an aircraft can fly is given by√
2W/ρSCSL .
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2. The slowest speed at which the aircraft can fly is given by max(Vstall,√
2W/ρSCBL ).
3. When CSL = CBL , we infer that the aircraft is flying at the maximum flyable altitude.
For a jet engine, the maximum thrust TA depends on the altitude and temperature, but is inde-pendent of the flight speed. However, the minimum drag is independent of altitude. Consequently,as altitude increases, the maximum available thrust reduces, but the drag does not decrease. Thesituation is illustrated in Fig. 5. The maximum flyable altitude can thus also be found by solvingTA(ρ) = Dmin. In Chapter 1, we pointed out that the maximum available thrust of a jet engine at
a given altitude obeys the law TA(ρ) = TA(sea)√
ρρsea
. Thus, the maximum flyable altitude of a jet
aircraft is found from the ISA tables for the density
ρ = ρsea
(Dmin
TA(sea)
)2
Everything else being equal, notice that ρ(max altitude) ∝ W 2; this forces a heavy aircraft to flylow and slow. We will show in the last part of this chapter how the range and endurance of anaircraft are affected because of this.
Figure 5: Computing the maximum flight speed for jet and propeller-powered aircraft.
For a propeller-powered aircraft, thrust reduces with altitude as well as speed. However, theshaft power PA does not change with flight speed, but reduces monotonically with altitude. Ananalytical calculation of the maximum flight speed and altitude of a propeller-powered aircraftrequire that we work with power rather than thrust.
The power required for straight and level flight at speed V is defined as PR = TRV , where TRis the thrust required to fly straight and level at V . To calculate the minimum power required for
level flight, we start by writing V =√
2WρSCL
and note that TR = W (CD/CL). This gives
PR =
√2W 3
ρS× CD
C3/2L
Clearly, the power required is minimized at the speed corresponding to
(C
3/2LCD
)max
.
6
A direct way to calculate the flight speed for minimum power, VPmin, is as follows. Recall that
TR =
(1
2ρSCD0
)V 2 + k
(2W 2
ρS
)1
V 2
Hence
PR =
(1
2ρSCD0
)V 3 + k
(2W 2
ρS
)1
V
Condition for minimum power:
∂PR∂V
= 0 =⇒ 3
(1
2ρSCD0
)V 2Pmin = k
(2W 2
ρS
)1
V 2Pmin
(7)
Thus,
VPmin =
(1
3
)1/4( 4kW 2
ρ2S2CD0
)1/4
=VDmin
31/4=VDmin1.3161
= 0.76VDmin
Recall that the power available in a propeller aircraft (PA(ρ)) is constant at a given altitude. Themaximum flight speed is found by solving
PA(ρ) =
(1
2ρSCD0
)V 3 + k
(2W 2
ρS
)1
V
This equation has two real positive roots; the larger real root yields the maximum flight speed atthat given altitude.
A commonly employed quantity in flight mechanic studies is the so-called specific excess power(SEP) which is defined as
SEP =PA −DV
m
The SEP depends on the flight speed and altitude, and measures the ability of the aircraft tochange its energy (more on this in Lecture 3). The maximum flight speed at a given altitude andthe maximum flyable altitude are both found by solving for SEP = 0. This is true for any aircraft,but is particularly convenient for propeller-powered aircraft.
5 Interregnum
Before proceeding, we need to find the value of CL and the corresponding level flight at which the
ratio C1/2L /CD is maximized
Let F =C
1/2LCD
=C
1/2L
CD0+kC2
L
Condition for maximum F :
∂F
∂CL= 0 =⇒
C−1/2L
2CD−
2kC3/2L
C2D
= 0
=⇒ CD = 4kC2L
=⇒ CD0 = 3kC2L
7
Figure 6: Plot showing the three speeds: minimum power, drag, and (T/V ).
Thus, skin friction drag is thrice as large as the induced drag. Corresponding speed:
V 4R =
12kW 2
ρ2S2CD0
= 3V 4Dmin
=⇒ VR = 31/4VDmin ≈ 1.316VDmin
Consider the ratio F = TRV . Then,
F =TRV
=1
2ρSV CD0 +
(2kW 2
ρS
)1
V 3
Minimize F :∂F
∂V= 0 =⇒ 1
2ρSCD0 =
6kW 2
ρSV 4R
which yields the expression we obtained on the previous slide:
V 4R =
12kW 2
ρ2S2CD0
Thus, the speed VR yields
(T
V
)min
5.1 The Three Speeds
Figure 6 shows the three speeds graphically in a drag versus speed plot. The speed for minimumpower does not have any associated straight line; rather, it occurs where a hyperbola of the formDV = constant is tangential to the drag curve. The three speeds have been tabulated in Table 5.1.
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Table 1: The three speeds
Condition on CL, CD Flight Speed Physical significance
maxCL
C3/2LCD
VPmin = 0.76VDmin Minimum power
maxCL
CLCD
VDmin Minimum drag (thrust required)
maxCL
C1/2LCD
VR = 1.316 VDmin Minimum (T/V ) ratio
6 Range and Endurance
• Definition: the range of an aircraft is the maximum distance that it can fly with the givenamount of fuel.
• Definition: the endurance of an aircraft is the maximum time for which the aircraft can flywith the given amount of fuel. The endurance is optimized during surveillance missions andwhile holding in traffic prior to landing.
In order to calculate the range and endurance, we must quantify the fuel consumption of theengines. This is done using two quantities:
• Thrust-specific fuel consumption: TSFC =|mf |gT
• Power-specific fuel consumption: PSFC =|mf |gP
TSFC and PSFC may additionally depend on the thrust/power settings, but we will assume theseto be constant. TSFC/PSFC are fundamentally engine parameters: among other things, theyinclude the propulsive efficiency of the engine. The reason why jet engines are characterised byTSFC, and turboprops by PSFC is apparent enough.
6.1 Endurance of a Jet Aircraft
The endurance is given by
tF =
∫ tf
0dt
The rate of fuel consumption can be linked to the rate of change of weight of the aircraft:
mf = −1
g
dW
dt
9
so that
dt =−dWmfg
Therefore,
tF =
∫ tF
0dt =
∫ W0
Wf
dW
mfg
where W0 is the initial weight of the aircraft, and Wf is the final weight, after fuel has been burnt.This formula for endurance is identical for all aircraft; the difference between jet and propeller
aircraft arises in how we model mf .Let cT denote the TSFC of the jet engine, so that
mfg = cTT
Also, since T = WCL/CD
,
tF =
∫ W0
Wf
1
cT
CLCD
dW
W
If we choose α so that CL/CD ratio is maximum, then
tF =1
cT
(CLCD
)max
ln
(W0
Wf
)
Note that the endurance is independent of the altitude and the geometric parameters of the aircraft!
6.2 Endurance of Propeller-Powered Aircraft
Recall that the endurance is given by
tF =
∫ tF
0dt =
∫ W0
Wf
dW
mfg
Let cP denote the PSFC of the engine, so that
mfg = cPP
Also, note that P = DV =W 3/2
C3/2L /CD
√2√ρS
.
We will choose an angle of attack to maximize the ratio C3/2L /CD, so that
tF =1
cP
C3/2L
CD
√2ρS
(1√Wf
− 1√W0
)
Clearly, the endurance is higher (everything else being equal) when the aircraft flies at a loweraltitude and has a large wing area (or a low wing loading).
10
6.3 Range of a Propeller-Powered Aircraft
The range is given by
R =
∫ τ
0V dt =
∫ Wf
W0
VdW
mfg
Yet again, this formula is the same for all aircraft; we will specialize it for jet aircraft and propelleraircraft.
For a propeller-powered aircraft, since mfg = cPP = cPDV , we get
R =
∫ Wf
W0
dW
cPD=
∫ Wf
W0
1
cP
CLCD
dW
W
This gives
R =1
cP
CLCD
ln
(W0
Wf
)
This formula is very much like that obtained for endurance of a jet aircraft: it is independent of thealtitude and the geometric parameters of the aircraft. Note, however, that the maximum altitudeis limited by thrust availability.
6.4 Range of a Jet Aircraft
Recall that
R =
∫ Wf
W0
VdW
mfg
Since mfg = cTT , we get
R =
∫ Wf
W0
dW
cT (T/V )
Again, T =CDCL
W and V =
√2W
ρSCL, so that
R =
∫ Wf
W0
√2
cT√ρS
C1/2L
CD
dW
W 1/2
=⇒ R =2√
2
cT√ρS
C1/2L
CD
(√W0 −
√Wf
)We observe that range improves at higher altitudes (lower ρ) and when S is small (i.e., the wingloading is high).
Table 6.4 summarizes the observations made for the range and endurance of aircraft.
11
Table 2: Summary of Range and Endurance
Parameter Jet Aircraft Propeller Aircraft
Optimum α (Endurance) Max CL/CD Max C3/2L /CD
Optimum α (Range) Max C1/2L /CD Max CL/CD
Optimum V (Endurance) VDmin 0.76VDminOptimum V (Range) 1.316VDmin VDmin
Altitude dependence (Endurance) Insignificant DecreasesAltitude dependence (Range) Increases Insignificant
Wing loading dependence (Endurance) Insignificant Low wing loadingWing loading dependence (Range) High wing loading Insignificant
6.5 Example
Consider a modern jet aircraft such as the Boeing 777 or the Airbus A330. The fuel to wright ratiois typically 0.5, so that W0/Wf ≈ 2. The maximum value of CL/CD is around 20. The aircraftseek to fly at speeds 30% higher than those for minimum drag in order to maximise their range.However, the stall speed of the aircraft and the available thrust require that they start their cruiseat relatively low altitudes under 30, 000 ft when fully loaded. As the aircraft burns fuel, it climbsup in steps of 2000 ft to comply with the flight rules. Towards the end of their cruise, they aretypically around 38, 000 − 40, 000 ft.
12
Chapter 3
Climbing Flight, Level Turns, Take off and Landing
Aditya Paranjape
1 Gliding Flight
In the absence of thrust, the equilibrium equations of motion are given by
L = W cos γ, D +W sin γ = 0
This gives the following equation for the flight path angle:
tan γ = −CDCL
or tan(−γ) =1
CL/CD=⇒ tan(|γ|) =
1
CL/CD
Note that γ < 0, since the aircraft descends; furthermore,
tan(|γ|) =loss in altitude
forward distance covered
What does this mean: for the shallowest glide, choose an angle of attack to maximize the lift-to-drag ratioRecall that CL/CD is maximum when
CD0 = kC2L, k =
1
πeAR
Thus, the flight speed is given by
V 2 =2W
ρSCL=⇒ V =
2
ρ
W
S
√k
CD0
The lowest flight path angle is given by
tan(|γ|min) =1
(CL/CD)max= 2√kCD0
We observe that for a long and shallow glide, the aircraft should
• Use a large aspect ratio, so that k is small
• Keep CD0small by minimizing skin friction and frontal area
Example: Let CD0 = 0.01, AR = 16, and e ≈ 1. This gives
k =1
πeAR=
1
16π
tan(|γ|min) = 2√kCD0 =
1
20√π
=1
35.45
1
Thus, if this glider is left from an altitude of 3000 m (about 10000 ft), then it will cover nearly 106 km onthe ground.
To calculate the optimum flight speed, assume W/S ≈ 330:
V 2 =2
ρ
W
S
√k
CD0
=10
2.4√π
W
S≈ 775 =⇒ V = 28 m/s = 101 km/h
• Gliding flight can directly yield the lift and drag coefficients of an airframe
• The flight path angle γ yields the CL/CD ratio
• CL can be obtained from V and γ:
V 2 =2W cos γ
ρS
• Therefore, data from several gliding flights will be of the form CD(CL): obtain CD0and k by fitting a
curve of the form CD0+ kC2
L
2 Climbing Flight
The objectives of this section are to:
• Determine the maximum rate of climb
• Determine the corresponding flight speed
Yet again, we start with the equilibrium equations of motion which are given by
L = W cos γ =1
2ρV 2SCL
T = D +W sin γ =1
2ρV 2SCD +W sin γ
The student should verify these by drawing a free-body diagram.The motion in the vertical plane is described by
x = V cos γ, h = V sin γ
so that the rate of climb is given by RC = h = V sin γ.We start our analysis with the drag equation,
RC = V sin γ =(T −D)V
W
At a given flight speed, the rate of climb is maximized by setting the thrust or the power available (for jetand propeller-powered aircraft, respectively) to the maximum value:
RCmax(V ) =(TAV −DV )
W=PA −DV
W=:
SEP
g
where TA and PA are the maximum available values of thrust and power, respectively.
2
2.1 Maximum RC in a Propeller Aircraft
In a propeller powered aircraft, PA is independent of the flight speed, and depends only on the altitude.Thus, the maximum climb rate is given by
RCmax(V ) =PA −DV
W
which means that the global maximum climb rate is
RCmax = maxV
RCmax(V ) =PAW−minV
DV
W
Notice, however, that
D =1
2ρV 2S(CD0
+ kC2L), CL =
2W cos γ
ρV 2S
The presence of cos γ is a complicating factor. Fortunately, most aircraft climb at angles where it is reasonableto approximate cos γ = 1. Once this approximation is made, notice the drag term is identical to that forstraight and level flight. Therefore, the maximum rate of climb is achieved by selecting
α :
(C
3/2L
CD
)max
, i.e., CL =
√3CD0
k
VPmin =
(4kW 2
3ρ2S2CD0
)1/4
= 0.76VDmin
Note that the speed at which the maximum RC is achieved depends strongly on the altitude. The drag atminimum power is given by
Pmin = 2ρV 3PminSCD0
= 2ρSCD0
(4kW 2
3ρ2S2CD0
)3/4
=4√
2
31/4(C
1/4D0
)k3/4
√W 3
ρS
Hence, the maximum rate of climb is given by
RCmax =PAW− 4√
2
31/4(C
1/4D0
)k3/4
√W
ρS
Note that the maximum cruising altitude can be found by obtaining the ρ at which RCmax(ρ) = 0. Themaximum rate of climb reduces with increasing altitude and increasing weight. Furthermore, the maximumflyable altitude reduces with increasing wing loading.
2.2 Maximum RC of a Jet Aircraft
In a jet aircraft, PA is a linear function of the flight speed:
RCmax(V ) =(TA −D)V
W
We find the speed V for RCmax by maximizing TAV −DV :
∂(TAV −DV )
∂V= 0
=⇒ TA =3
2ρV 2SCD0 −
2kW 2
ρSV 2
=⇒ V 2RCmax =
TA +√T 2A + 12kCD0
W 2
3ρSCD0
3
V
RC
Thrust constraint
Stallconstraint
FlightEnvelope
SteepestClimb
FastestClimb
FastestClimb
SteepestClimb
Figure 1: A sketch of the maximum rate of climb as a function of the flight speed. Notice that climbs arepossible at sub-stall speeds.
The speed for the maximum rate of climb increases with increasing altitude. The rate of climb, though,reduces with altitude. The maximum rate of climb can be found by substituting for VRCmax in the equationfor RCmax above. The resulting expression is cumbersome to write (and look at), but there is much to begained by plotting the resulting RCmax(V ) as shown in Fig. 1.
We make some quick observations about the speed for maximum RC:
• Since TA ≥ Dmin, we have that VRCmax > VDmin.
• CL in climbing flight:
CL(climb) =2W cos γ
ρV 2S= CL(level) cos(γ)
Thus, it is possible for the aircraft to climb at stall speed and at lower speeds provided sufficient thrustis available:
TA > Tstall =W
CLmax
(CD0 + kC2
Lmax
)• This gives a flight path angle envelope:
– V > Vstall: [0, γmax]; constraint on TA dictates γmax
– V < Vstall: [γmin, γmax], where cos γmin = ρV 2SCLmax
2W
• Minimum flyable speed: γmax = γmin > 0
2.3 Steepest Climb versus Fastest Climb
The flight path angle can be found from
sin γ =T −DW
Yet again, the maximum climb angle at a given V is achieved by setting T = TA, i.e., by ensuring that theexcess thrust (unlike excess power for rate of climb) is maximized.
Assuming that cos γ ≈ 1, so that D(V ) is the same as that under level flight conditions, it is evident thata jet aircraft climbs steepest at V = VDmin. That the steepest climb speed should be slower than that formaximum rate of climb is apparent from Fig. 1. For a jet aircraft, this is validated by the observations inthe previous section.
The speed at which a propeller-powered aircraft achieves the steepest climb is found by formally differ-entiating the expression for sin γ:
γmax =⇒ ∂(TA(V )−D)
∂V= 0
4
This exercise requires a model for TA(V ). Since TA(V ) is usually quite nonlinear, the exercise is cumbersomeand is left to the reader’s enthusiasm.
3 Level Turn Performance
R O
Source: Russell
Figure 2: Free-body diagram of a turning aircraft
So far, we have looked at flight in a straight line. Next, we move to turning manoeuvres which areperformed to bring about a prescribed change in aircraft heading. We would like to determine the maximumturn rate that can be achieved at a given flight speed and, by extension, the maximum achievable turn rateand the minimum attainable turn radius.
We make some assumptions regarding the turn:
• The aircraft is banked through a constant angle φ (see Fig. 2
• The sideslip angle is zero; i.e., β = 0
This is shown in Fig. 2. The free-body diagram can be used to deduce two equations of motion which arisefrom lift:
Weight balance: L cosφ = W
Centripetal force: L sinφ =WωV
g(1)
Depending on whether or not a third condition γ = 0 is met, we get two types of turns (both of which satisfyβ = 0 and φ = constant):
• Sustained turn: γ = 0 is satisfied. This entails that T = D.
• Instantaneous turn: γ need not be zero. The condition T = D is not imposed on the problem.
3.1 Instantaneous Turn
We start by analyzing Eq. (1). By dividing the weight balance equation by that for centripetal force, we get
tanφ =ωV
g
We define the load factor n = L/W . With this new terminology, we see that
ω =g
Vtanφ =
g
V
√n2 − 1 (2)
This is the fundamental equation which describes the turn rate in terms of the flight speed and the loadfactor. We make some important observations:
5
• As V increases, ω decreases for fixed load factor
• For a given V , the turn rate increases as n increases
Evidently, in order to increase the turn rate at a given V , it is necessary to maximize the load factor n.There are two fundamental constraints on the load factor:
• CLmax: the maximum achievable value nlift(V ) =
ρV 2SCLmax
2W. Load factors higher than nlift(V ) are
simply not attainable.
• Structural and human limits: regardless of the flight speed, the aircraft structure and the human pilotimpose a load factor bound denoted by nsafe. At high flight speeds, this bound is attainable, but unsafefor the aircraft.
Thus, the maximum turn rate at a given flight speed is found using
ωmax(V ) =g
V
√n2max(V )− 1, nmax = min(nlift(V ), nsafe) (3)
With this, we claim (and it is quite easy to prove) that the turn rate is maximized globally at a speed Vcwhich satisfies
nlift(Vc) = nsafe =⇒ Vc =
√2W nsafeρSCLmax
= Vstall√nsafe
This speed is called the corner speed. When V < Vc, the constraint due to CLmaxdetermines the maximum
achievable turn rate, while for V > Vc, the maximum achievable turn rate is determined by the safe loadfactor. Note also that Vc is directly proportional to Vstall. Thus, for a given value of nsafe, a low corner speedcan be achieved (for a high turn rate, from Eq. (2)) by designing for as small a Vstall as possible.
When the load factor n2 >> 1, we have that ω ≈ ng
V. We can thus write the global maximum instanta-
neous turn rate as
ωmax ≈gnsafeVc
=g
Vstall
√nsafe = g
√ρnsafeCLmax
2
√S
W
Thus, to maximize the instantaneous turn rate, we need the following design pointers:
• Keep the wing loading as low as possible. This is in sharp contrast to the wing loading required tomaximize range and flight speed. Thus, the most manoeuvrable aircraft seldom have a large range.Their flight speed may be quite high, but that requires powerful engines.
• The values of CLmax and nsafe should be as high as possible.
Notice also that the turn rate reduces as altitude increases.
3.2 Sustained Turn
A sustained turn differs from an instantaneous turn only in that the constraint T = D is obeyed, whichallows the turn to be performed at a constant altitude and maintained that way for a long time. Note thatEq. (1) does not change and consequently, the analysis described above holds for sustained turns as well,except for an additional constraint on CL (aside from CLmax and the one due to nsafe.
Consider an aircraft turning at a speed V and coefficient of lift CL. Then, the thrust required for levelflight, given by
T = D =1
2ρV 2S(CD0
+ kC2L),
6
increases monotonically with CL. Thus, the maximum sustainable value of CL at a given V , denoted byCL,T (V ), is given by
CL,T (V )2 =1
k
(2TAρV 2S
− CD0
)=⇒ CL,T (V ) =
√1
k
(2TAρV 2S
− CD0
)Hence, the maximum sustainable load factor is given by
nT (V ) =ρV 2S
2W
√1
k
(2TAρV 2S
− CD0
)Note that an additional constraint nT ≥ 1 is required for level flight. It is evident that nT first increaseswith V up to a critical speed Vc,T , and then decreases as speed is increased. The speed Vc,T plays the roleof the corner speed for the thrust constraint alone. It can be checked that Vc,T is given by
ρV 2c,TSCD0
= TA
The corresponding interpretation for Vc,T is that it is the speed at which the skin friction drag equals theinduced drag, and the net drag equals the maximum thrust available at that altitude. The straight-and-levelflight analogue is that Vc,T is VDmin and Vmax rolled into one.
Maximum load factor is now given by
nc,T =ρV 2
c,TS
2W
√TA
kρV 2c,TS
=TA2W
√1
kCD0
=TADmin
If n2T >> 1, so that the maximum turn rate is approximately
ωc,T ≈gnTVT
= g
(√ρS
k
)(√TA
2W
)=g
2
(√ρS
kW
)(√TAW
)Therefore, the sustained turn rate can be maximized by designing the aircraft with low wing loading, highthrust-to-weight ratio and large aspect ratio.
It is important to note that the sustained turn constraint operates alongside the bounds imposed byCLmax and nsafe. Therefore, the maximum permissible load factor for sustained turns at any given speed is
nmax(V ) = min(nsafe, nT (V ), nlift)
so that the maximum sustainable turn rate is
ω(V ) =g
V
√nmax(V )2 − 1
Figure 3 shows the maximum turn rate of the F/A-18 HARV as a function of the flight speed for eachindividual constraint. We have considered two cases of nsafe: 5 and 9. This plot is quite typical for mostaircraft. The following observations are made from the plot, which has been split conveniently into severalregions:
• The two curves DKL CP-G correspond to nsafe = 5 and 9, respectively. The CLmax constraint iscaptured by the curve A-C-D-CP, while the thrust is maximum on M-B-K-Q.
• Region 1: all constraints are satisfied, and represents sustained turns. The maximum sustained turnrate occurs at point T.
• Regions 2 and 5: in these regions, only the thrust constraint is violated. These regions correspond toinstantaneous turns. The maximum instantaneous turn rate occurs at points D or CP, depending onwhether nsafe = 5 or 9.
7
Figure 3: Maximum turn rate as a function of flight speed for the F/A-18 HARV. Source: Paranjape andAnanthkrishnan, AFM 2005.
• Region 3: the constraint on CL is violated. Therefore, turn rates in region 3 are simply not attainable.
• Region 4: If nsafe = 5, then only the safe load factor is exceeded. Sustained turns in this region arepossible, but not safe if nsafe = 5.
Note that such plots are altitude-specific. The overlap between the various regions may change as the altitudechanges.
3.3 Turn Radius
An important metric for studying the turning performance is the turn radius defined by
R =V
ω
Just as for steepest climbs, the minimum turn radius can be found by drawing a straight line from theorigin to an ω − V plot such as Fig. 3 and looking for the line that is tangential to the plot. Evidently, theminimum instantaneous turn radius is obtained along the CLmax branch. For sustained turns, it may occuron the branch for the nT constraint.
There is simple way to calculate the smallest instantaneous turn radius. It is to be expected that thesharpest turns are obtained for a large value of n >> 1. Thus, we can write
Rmin =V 2
gnlift=
2
ρgCLmax
W
S
4 Take-Off and Landing
Having seen all the major phases of flight, we turn our attention to the ground roll that accompanies take-offand landing. The objective here is to minimize the ground roll and enable aircraft to use shorter runways.
4.1 Take-Off
During take-off, the aircraft accelerates to a critical speed called VR, at which it rotates its nose upwardsand climbs. The speed VR is called the rotation speed and is typically 1.25Vstall. The ground roll distance,denoted by sTO, is found using
sTO =
∫ τTO
0
V dt =
∫ VR
0
V dV
V
The acceleration V is decided by a combination of thrust, drag and friction:
V =TA −D − µN
m= (TA − µW )− (D − µL)
8
where µ is the coefficient of friction and N = W −L is the normal force. Substituting for V in the equationfor sTO gives
sTO =
∫ τTO
0
V dt =
∫ VR
0
V dV
V=
∫ VR
0
mV dV
(TA − µW )− (D − µL)
Since the aircraft is at a near-zero angle of attack, CL = CL0 and CD = CD0 + kC2L0. Therefore,
D − µL =1
2ρV 2S
(CD0
+ kC2L0 − µCL0
)︸ ︷︷ ︸CDe
Thus,
sTO =
∫ VR
0
mV dV
(TA − µW )− 12ρV
2SCDe=
W
ρgSCDeln
(T − µW
T − µW − 12ρV
2RSCDe
)To minimize the ground roll distance, the aircraft needs:
• Large T/W ratio: ensures that ln(·) is small
• Smallest possible CDe, ensured by designing CL0 = µ/k ≈ 1.
• Small wing loading W/S. Flaps can be used so that CL,max is increased over the normal wings, W/Sis reduced, and CL0 is close to the optimum value.
Notice that the take-off length increases when ρ reduces; i.e., at high altitude and under hot and dryconditions.
4.2 Landing
The ground roll required to come to a complete halt during landing is given by
sL =
∫ 0
VTD
V dV
V
where VTD is the touch-down speed. The net acceleration:
|V | = D + TR + µN
m=D − µL+ TR + µW
m
where TR is the thrust from thrust-reversers.Following an approach similar to that used for calculating the ground roll for take-off, we get
sL =W
gρSCDeln
(TR + µW + 1
2ρV2TDSCDe
TR + µW
)≈ W
gρSCDeln
((TR/W ) + µ+ 1.5CDe/CL,max
(TR/W ) + µ
)CDe = CD0 + kC2
L0 − µCL0
To minimize the landing distance, the aircraft needs:
• Large CDe: flaps and wing spoilers are deployed for this purpose.
• As large a value of µ as possible. This is done using brakes, but only at low speeds to prevent excessivemechanical wear.
• As large TR as possible: this is accomplished using thrust reversers.
9
Chapter 4
Longitudinal Static Stability
Aditya A. Paranjape
1 Introduction
Thus far, we have looked at how aircraft can be designed to meet the desired performance specifications. Wealso know the angle of attack and thrust that have to be set in order to achieve the desired flight speed andclimb rate. The performance analysis, however, dealt with a point-mass model of the aircraft. Aside frommacro-scale metrics such as W/S and the aspect ratio, the geometry of the aircraft was all but ignored.
The fact is that the geometry is an essential element of what makes an aircraft fly. Consider the followingquestions, neither of which our prior analysis would enable us to answer but both of which are essential forsafe flight:
1. The performance analysis essential prescribes the angle of attack α at which the aircraft must be flown.Can this value even be attained by the aircraft? What does it take to attain a specific value of α?
2. If the aircraft is disturbed by an external gust and its angle of attack changes in the process, can theaircraft restore its angle of attack by itself? If not, what can be done to ensure that it does?
These questions fall within the purview of stability and control - two topics which will occupy us for theremainder of this course.
We start by reviewing a narrow but useful notion of stability, called static stability. Next, by derivingexpressions for rotational equilibrium and by analysing small perturbations about an equilibrium, we answerthe questions about stability and control.
2 Static Stability
2.1 Broad Definition
Broadly speaking, an aircraft is said to be stable if it is able to restore its flight speed, angle of attack, etc., tothe set values after encountering a disturbance. It is evident that stability refers to the long-term behaviourof aircraft.
Here is a different perspective on stability: suppose an aircraft is disturbed from its operating condition.Then, it may be said to be stable if the forces and moments instantaneously try to restore the operatingcondition. They may or may not succeed in doing so in the long-run. If the instantaneous response is in thecorrect direction, we say that the aircraft is statically stable.
An analogy is shown in Fig. 1. One may conjecture that the ball is stable inside the valley and unstableat the peak. Strictly speaking, the ball is only statically stable inside the valley. In the absence of friction,the ball would simply travel back and forth like the bob of a pendulum and is clearly not stable in thelong-run.
Formally, we say that a system is statically stable if, upon a disturbance ∆x, the force F produced bythe system is such that F ·∆x < 0. This is considered to be a necessary condition for long-term stability, butis not sufficient (as in the case of a ball in a valley). In fact, it is not difficult to produce an example wherea statically stable system is, in fact, unstable in the long run. This simple exercise is left to the reader.
1
Figure 1: Classic example of a ball in a valley or on a ridge.
2.2 Longitudinal Static Stability
In this chapter, we are interested in the pitching motion of an aircraft. As a matter of convention, thepitching moment on aircraft is denoted by M . We may write M(α) since M depends on α.
Definition: We define an equilibrium angle of attack α0 as one at which M(α0) = 0. This is theconventional definition of equilibrium: a state in which the net forces and moments on a body add up tozero. Suppose that ∆α denotes the perturbation about α0. This perturbation produces a pitching moment∆M(∆α). The aircraft is said to possess longitudinal static stability if ∆α ·∆M = 0.
Since M(α0) = 0, we can use Taylor series expansion to write
∆M = M(α0 + ∆α) =
[∂M
∂α
]α0
∆α ,Mα(α0)∆α
It follows that the aircraft is statically stable at α0 only if Mα(α0) = 0. Notice that we have defined stabilityas being a property not just of the aircraft as a whole, but also a pointwise property at each α0.
If we assume that the pitching moment M is a linear function of α, then we may write
M(α) = M0 +Mαα
In this case, the partial derivative Mα does not depend on the trim value α0.As a matter of convention, we define the non-dimensional coefficient of pitching moment Cm such that
M =1
2ρV 2ScCm
and we can further write
M0 =1
2ρV 2ScCm0, Mα =
1
2ρV 2ScCmα
Notice that
Cmα =∂Cm∂α
We now ask the question: under what conditions is it possible to trim and statically stabilize the aircraftat some α0 > 0. From Cm = Cm0 + Cmαα, we readily conclude that
1. In order to trim at some positive angle of attack, we require that Cm0 > 0. The corresponding trim
angle of attack is given by α0 = −Cm0
Cmα
2. In order for the trimmed condition to be stable, we require that Cmα < 0.
These are fundamental conditions for static trim and stability, and the rest of this chapter builds upon theseequations systematically.
2
Figure 2: Wing-only model for trim and stability analysis.
3 Longitudinal Static Stability and Aircraft Geometry
3.1 Wing-Body Only
Consider an aircraft with a wing and a fuselage, but no horizontal tail. Suppose the wing AC is at a distancexAC from the CG of the aircraft, as shown in Fig. 2. The pitching moment is derived only from lift, so that
M =1
2ρV 2S (cCmac + xACCL(α)) , CL = CL0
+ CLαα
We first determine the trim angle of attack, α0, by setting M(α0) = 0:
1
2ρV 2S
(cCmAC + xAC(CL0 + CLαα
0))
= 0
=⇒ α0 = −cCmAC + xACCL0
xACCLα
The trimmed value of lift (with CmAC < 0) is given by
L0 =1
2ρV 2S
(−cCmAC
xAC
)> 0, xAC > 0
< 0, xAC < 0
We conclude that the trimmed value of lift is positive only if xAC > 0, i.e., if the CG is behind the AC, asshown in Fig. 2. However, is this configuration stable? To determine the stability of this trim, suppose theAoA is perturbed by ∆α. The instantaneous pitching moment is given by
∆M = xACLα∆α =1
2ρV 2SCLαxAC∆α
To ensure static stability, we need ∆α ·∆M < 0, i.e., xAC < 0, which runs contrary to the requirement forpositive lift. Therefore, a wing-body combination that is flight-worthy (capable of producing positive lift)will be statically unstable. Since static stability is a necessary condition for stability, the aircraft will beunstable in the long-term as well. There are two ways to tackle this problem:
• Use a flap and active control: a flap changes lift as well as Cmac and offers a direct way to mitigate adisturbance.
• Use an airfoil cross-section with a negative camber at the trailing edge. This configuration ensuresthat Cmac > 0. Such airfoils, however, are not particularly suitable at high angles of attack.
• Use a combination of wing sweep and washout (reducing wing twist at outboard sections). This shiftsthe CG of the aircraft behind the aerodynamic centre of the inboard wing sections, so that stabilitycharacteristics improve substantially.
3.2 Longitudinal Control: Trimming with a Movable Horizontal Tail
Let α0 be the trim angle of attack of the aircraft. We will assume that the wing inclination angle iw = 0and that the tail is symmetric. The lift on the tail is given by
Lt =1
2ρV 2StCLα (α+ it)
3
Figure 3: Wing and tail schematic for trim and stability analysis.
Note: if the tail has an elevator instead of variable it as the control input, then
Lt =1
2ρV 2St (CLα (α+ it) + CLδeδe)
At trim, M = Mwing +M tail = 0 about the CG; i.e.,
Mwing + LwingxAC = (lt − xAC)Ltail
Thus, xAC (CL0 + CLαα) + cCmac =
(ltStS− xACSt
S
)CLα (α+ it)
i.e.,xACc
(CL0 + CLαα) + Cmac =
(VH −
StS
xACc
)CLα (α+ it)
The trim AoA is given by
α0 =1(
VH − xACc (1 + St/S)
)CLα
((xac/c)CL0 + Cmac)−(
VH − (St/S)xACcVH − xAC
c (1 + St/S)
)it (1)
As we will see shortly, VH >xACc
(1 + St/S). Therefore, as it increases (i.e., deflects downwards), the trim
α0 reduces, and vice-versa. Therefore, the tail-based control surface allows the aircraft to trim across a widerange of values of lift (and flight speeds). The same principle applies to elevator-based control where it isheld constant, but the elevator is deflected to achieve the desired moment.
Suppose that the angle of attack is perturbed by a small ∆α. We note the following change in theaerodynamic forces and pitching moment:
• Change in lift on the wing: ∆Lw = 12ρV
2SCLα∆α
• Change in lift on HT: ∆Lt = 12ρV
2StCLα∆α
• Net change in pitching moment
∆M = ∆LwxAC −∆Lt (lt − xAC) =1
2ρV 2ScCLα∆α
(xACc
(1 +
StS
)− VH
)We deduce that
Mα =1
2ρV 2ScCLα
(xACc
(1 +
StS
)− VH
)We infer that the aircraft is statically stable only if xAC < VH/(1 + St/S) = V ′
H . For a given aircraft, V ′H is
thus a constant, and so is c. Thus, the static stability condition imposes the requirement on how the aircraftis loaded: the CG location should satisfy xAC < cV ′
H . Since xAC > 0 when the CG is behind the AC, theexpression xAC > cV ′
H tells us how far behind the wing AC the CG is allowed to lie.Definition: We define the neutral point as the CG location at which xAC = cV ′
H , i.e., Mα = 0. Whenthe CG is located ahead of the neutral point, the aircraft is statically stable, and vice-versa. It is importantto note that the location of the neutral point depends only on the geometry and aerodynamic coefficients.
4
It is instructive to consider the non-dimensional CG location xAC/c. For static stability,
xACc
>StS
ltc
(1
1 + StS
)
Since St/S is usually between 0.25 and 0.3, we infer that the CG can lie at most a quarter of the way fromthe wing to the tail.
Definition: We define the static margin:
SM ,xNP − xAC
c
An aircraft is loaded on the ground with payload and fuel so that the static margin never decreases below athreshold. The static margin is used largely for operational purposes.
3.3 Effect of Downwash
Figure 4: Trailing edge vortices. Source: aerospaceweb.org.
Recall that trailing edge vortices are generated by finite wings (Fig. 4 as a consequence of lift production.The resulting downwash reduces the angle of attack of the horizontal tail:
αt = α+ it︸ ︷︷ ︸geometric
− ε︸︷︷︸downwash
The downwash can be written as ε := ε(α) = ε0 + εαα, where ε0 > 0 and εα > 0. Thus,
αt = α(1− εα) + it − ε0
The presence of εα changes Mα:
CMα =xACcCLα −
(VH −
StS
xACc
)CLα(1− εα)
Clearly, downwash reduces the longitudinal-stability of the aircraft, and the neutral point shifts forward to
satisfyxACc
= V ′H
((1 + (St/S))(1− εα)
1 + (St/S)(1− εα)
). The exact expression for ε is difficult to derive. The reader is
referred to a textbook for standard approximations, but is cautioned against using it for analysis without acareful review of the assumptions.
5
3.4 Statically Unstable Aircraft and Rearmost CG Location
It is interesting to note that several aircraft are designed to be statically unstable, and have to be stabilizedusing automatic control systems. This is especially true of fighter aircraft. In general, statically unstableaircraft are more manoeuvrable than their stable counterparts. There is another benefit to having a staticallyunstable airframe, namely that the net lift is higher. We perform a simple analysis to ascertain this. At agiven flight speed, the trim condition is
Lwxac +MAC = Ltlt
=⇒ Lnet = Lw + Lt = Lw(
1 +xAClt
)+MAC
lt
Clearly, the higher the value of xAC , the greater the net lift on the aircraft. This begs the question: how farback can the CG be allowed to shift? The answer to this question lies at the heart of control: the CG canbe permitted to move to the rear as long as the pitching moment from the horizontal tail is adequate.
It is evident that as the CG moves backwards, the pitch-up moment from the wing increases rapidly.In order to trim the aircraft, the horizontal tail must be able to provide a sufficient pitch-down moment.In particular, if the maximum desired angle of attack is αmax, then the horizontal tail must be able tobalance the pitching moment generated by the wing at αmax while ensuring that its own deflection is withinpermissible limits.
Let δ denote the control surface deflection (which could be either the elevator or the complete tail, orboth) and suppose that we need δ ∈ [δmin, δmax]. The CL of the tail can be written as
CtL = CLαα+ CLδδ
The corresponding control surface deflection is found using Eq. (1):
it =
(xACc
(1 +
StS
)− VH
)α0 +
(xAC/c)CL0 + CmacCLα (VH − (St/S)xAC/c)
(2)
If we wish to fly between angles of attack of αl and αu, then we need to ensure that δ(αl) and δ(αu) arewithin the deflection bounds. Since xAC > cV ′
H , we note that δ(α) increases with α. Thus, the rearmost CGlocation is then found by solving for xAC when
• δ(αl) = δmin and δ(αu) < δmax, or
• δ(αl) < δmin and δ(αu) = δmax
The first constraint is usually milder, and the second expression yields the rearmost CG location.
3.5 Positioning the Horizontal Stabilizer: T-Tails and Canards
Figure 5: Examples of T-tails and canards in aircraft. Source: Wikipedia.
Figure 5 shows two alternate longitudinal stabilizer configurations:
6
• T-tail: the horizontal stabilizer mounted on top of the tail. The arrangement keeps the stabilizer out ofthe downwash and makes room for rear engines and cargo doors. On the down-side, the static stabilityis reduced due to drag and the tail becomes highly susceptible to a dangerous phenomenon called adeep-stall by being in the wing wake at high α.
• Canard: the horizontal stabilizer is mounted ahead of the wing. Canards improve the maneuverabilityof the aircraft significantly. Since they are never in the wing wake, they are quite effective even whenthe aircraft pitches to high α. On the flip-side, the use of canards inherently reduces the static stabilityof the aircraft. They may also introduce a performance-degrading downwash on the wings.
It is interesting to note that a canard-wing combination is exactly like a wing-HT combination. The staticstability condition can be found using that for the wing - horizontal tail equation by replacing the wing withthe canard, and the horizontal tail by the wing. The analysis is left to the reader who should verify that
• The static stability condition translates into a lower bound on how far ahead of the wing the CG shouldbe placed.
• Imposing static stability degrades the control authority of the canard.
3.6 Effect of Nonlinearities
In the analysis presented so far, we ignored contributions from the fuselage and other external payload.Their contributions are usually nonlinear in α. Yet, the control input δ enters as a linear term, so that thetotal pitching moment on the aircraft at equilibrium can be written as
M = M(α) +Mδδ = 0 at trim
The incremental moment after a perturbation ∆α is given by
∆M =
[∂M
∂α
]α0
∆α = Mα(α0)∆α
Therefore, the condition for static stability can be written as Mα(α0) < 0, which is the same as the one wehad before with the difference being that the derivative needs to be evaluated at each trim condition.
3.7 Neutrally Stable Configurations and Trim
Let us return to the case where M is linear in α. The equilibrium pitching moment is given by
M = M0 +Mαα+Mδδ = 0 =⇒ α = −Mδ
Mαδ
If |Mα| is very small (close to zero), the angle of attack changes by large amounts even for small elevatordeflection. This is highly undesirable. Furthermore, if Mα = 0, then at equilibrium, M = M0 + Mδδ = 0.The angle of attack is nowhere in the picture! Therefore, without active control, the aircraft can trim at anyangle of attack, i.e., there is no control whatsoever on the trim value of α and CL.
4 Stick-Free Versus Stick-Fixed Stability
Longitudinal control surfaces are typically actuated by a combination of actuators:
• Electro-mechanical actuators connected to the flight computer
• Hydraulic actuators connected to the flight computer as well as the control column in the cockpit
7
• In small aircraft, mechanical wires and pulleys connected directly to the control column
In all cases except the electro-mechanical actuators, the force exerted by the pilot on the control column istransmitted to the actuator to move the control surface. The control surface exerts an opposing force onthe actuator which is duly transmitted back to the control column. Under equilibrium conditions, the twoforces are equal and cancel each other.
So far, we assumed that the elevator deflection is constant, for which the pilot would have to hold thecontrol column in one place manually. The stability that we have looked at so far is therefore called “stick-fixed” stability. In nominal trim flight, the pilot takes his hands off the control column after “arranging” fora certain amount of force to be applied to the elevator at all times. This is achieved by using a small flapon the elevator, called the trim tab. Thus, the elevator is no longer statically deflected and the dynamics ofthe elevator deflection further affect the stability of the aircraft. The stability (or lack thereof) is referred to“stick-free” stability (or instability). Although stick-free flight is rare in the present times, it is still widelyprevalent in general aviation and sports aircraft.
We start our analysis of stick-free stability with the pitching moment equilibrium equation:
M = M0 +Mαα+Mδδ
As we will show presently, the elevator/horizontal tail deflection can be written δ = H0 + Hαα, for someconstants H0 and Hα. Substituting into the pitching moment expression yields
M = (M0 +MδH0) + (Mα +MδHα)α
The static stability condition changes to ensuring that Mα+MδHα < 0. If MδHα > 0, then we need a muchmore negative Mα to ensure stick-free stability. Thus, stick-free stability is a much more stringent attributeto achieve than stick-fixed stability. Usually, Mδ < 0 for a horizontal stabilizer (can it ever be otherwise?).Thus, Hα < 0 is destabilizing and vice-versa. We need to derive an approximate expression for H.
(a) An actual trim tab (b) Schematic
Figure 6: Elevator trim tabs and a schematic representation. Source for the image on the left: Wikipedia.
A trim-tab is shown in Fig. 6 together with a schematic representation. In principle, the trim tab behaveslike an independent symmetric airfoil. The force on the tab produces a moment which adds on to the momentdue to the stick force. Our objective is to find the tab deflection angle δtab to achieve zero hinge moment.The hinge moment consists of two contributions:
• Moment from the elevator: Me = 12ρV
2SeCeLα(α+ δe)xe
• Moment from the trim tab: Mtab = 12ρV
2StabCtabLα (α+ δtab)xtab
Since equilibrium is achieved when Me +Mtab = 0, the trim elevator angle is given by
δe =
(−StabxtabC
tabLα
SexeCeLα− 1
)α−
(StabxtabC
tabLα
SexeCeLα
)δtab
A comparison with δ = H0 +Hαα clearly shows that Hα < 0. This proves that stick-free configurations areless stable than stick-fixed configurations.
8
Chapter 5
Lateral-Directional Static Stability
Aditya A. Paranjape
1 Elements of Lateral-Directional (L-D) Static Stability
Figure 1: The three primary angles of interest, all defined with respect to the velocity vector.
Figure 1 shows the three primary rotational degrees of freedom of an aircraft. Notice that these rotationsare defined with respect to the velocity vector. The resulting angles are also thus referred to as “wind axisangles.” The perturbation ∆α is the “defining perturbation” for longitudinal static stability. On similar lines,since there are two rotational lateral-directional degrees of freedom, we have two candidate lateral-directionalangles:
• Yaw: perturbations in the the sideslip angle ∆β are the direct analogue of ∆α for lateral-directionalstability.
• Roll: a static perturbation ∆µ (roll) does not alter the orientation of the velocity vector with respectto the aircraft, and hence does not produce incremental aerodynamic forces and moments by itself.Therefore, the perturbation ∆µ is not relevant to our discussion on static stability.
Since sideslip arises primarily as a result of the yawing motion, we expect that the yawing momentderivative Nβ would be of interest. It turns out that the rolling moment Lβ is of interest as well. Thisis because a perturbation in the body-axis roll angle (rotation about the body x axis) at non-zero anglesof attack produces a sideslip along with change in µ. This has been shown in Fig. 2, where the sideslipperturbation is given by ∆β = sinα0∆φ. Notice that the perturbation depends on the trim angle of attackand is zero when α0 = 0.
Important: we will use the symbol L for lift as well as rolling moment, following the flight mechanicconvention. The reader should pay attention to the context to avoid any confusion. Interestingly, thenotation differs for the non-dimensional coefficients of lift and rolling moment, which are denoted using asCL and Cl, respectively.
1
Figure 2: The change in sideslip due to a perturbation in the body-axis roll angle: ∆β = sinα∆φ. Notethat φ and µ are very different, if related, quantities.
1.1 General Theory
Figure 1 shows the positive sense of the angles. The sideslip is positive when the velocity vector is to theright of the aircraft. From Fig. 2, it is also clear that a roll to the right creates positive sideslip when α > 0.A positive sideslip therefore needs to be countered by (a) a positive yawing moment, and (b) a negativerolling moment.
Conditions for stability: Following the notation established for longitudinal stability, we infer that forstatic stability, we need Nβ > 0 and Lβ < 0.
The above condition is also usually written in terms of the corresponding non-dimensional quantities
Cnβ > 0 and Clβ < 0, where Nβ =1
2ρV 2SbCnβ and likewise for Lβ .
In what follows, we will derive the contributions from the wing and the tail to the two derivatives Nβand Lβ . The derivations require application of the analytical version of strip theory - the method can beeasily converted into a numerical form as well.
2 Contribution from the Vertical Tail
Figure 3: Vertical tail contribution to the rolling and yawing moments.
2
2.1 Yawing moment
Let bv, cv and Sv denote the span, chord and area of the vertical tail. From Fig. 3, it is clear that β playsthe role of angle of attack as far as the vertical tail is concerned. The yawing moment generated by the liftLv produced by the vertical tail tail is
N = Lv(lt − xAC)
where lt is the distance between the tail and the wing AC. Since the vertical tail is symmetric, the yawingmoment would be zero β = 0. The incremental yawing moment due to a small perturbation ∆β is thus givenby
∆N =1
2ρV 2Sv(lt − xAC)CLα∆β
=⇒ Nβ =1
2ρV 2Sv(lt − xAC)CLα > 0 (1)
Therefore, the vertical tail stabilizes the derivative Nβ .
2.2 Rolling Moment
The rolling moment needs to be calculated using strip theory because the spanwise aerodynamic centres areat varying distances from the x axis. Consider a strip of width dz whose aerodynamic centre is at a distancez from the rolling axis. Then, the incremental rolling moment from this strip is given by
d(∆L) = −1
2ρV 2cvCLαz dz∆β
Notice the negative sign: this comes from the fact that the lift on the vertical tail, which points to the leftwhen ∆β > 0, causes the aircraft to roll to the left.
By integrating from z = 0 to z = bv, we get
Lβ = −1
4ρV 2SvbvCLα < 0 (2)
Therefore, a vertical tail makes stabilizing contributions to the Lβ derivative as well provided it is locatedabove the fuselage.
3 Contributions from the Wing: Important Parameters
The effect of the wing on L-D stability arises from three geometric characteristics of the wing:
• Wing dihedral angle Γ (positive upwards for both wings)
• Wing sweep angle Λ (positive backwards for both wings)
• High wing versus low wing
These have been depicted in Fig. 4. The net influence of a wing is the sum total of the individual contributionsof these three effects, and we will analyse each in isolation.
3.1 Dihedral Effect
The basis of the dihedral effect is the following chain of events which has been visualised in Fig. 5.
1. A perturbation ∆β causes a perturbation in the local angle of attack ∆αl at each spanwise section ofthe wing.
3
Figure 4: Wing configuration angles. The dihedral and sweep angles shown here are in the positive sense.
Figure 5: The physical basis for the dihedral effect.
2. The resultant ∆αl is anti-symmetric; i.e., it has opposite signs on the two wings, and it is constant oneach wing.
3. The resultant asymmetry in the lift produces a rolling moment, and a yawing moment is produced dueto the AC of the wing being longitudinally separated from the CG (i.e., xAC 6= 0) (see Fig. 6).
As a matter of convention, the body axis components of V are denoted by u, v, w along the x, y and zaxes, respectively. The angle of attack is defined as tanα = w/u. The dihedral effect essentially allows v tocontribute to the local angle of attack. On the right wing, the local angle of attack is thus given by
αl,R ≈w + v sin Γ
u
The change due to the sideslip perturbation is, therefore, ∆αl,R = ∆β sin Γ, where we have additionallyassumed that α and β are sufficiently small and u ≈ V . By the same token, the change of the local angle ofattack of the left wing is ∆αl,L = −∆β sin Γ.
The difference in the angles of attack between the left and right wings gives rise to a rolling momentwhich can be found using strip theory to be
∆L =
(−1
4ρV 2SbCLα sin Γ
)∆β
Thus, Lβ = −1
4ρV 2SbCLα sin γ < 0 when Γ > 0. In other words, a positive dihedral angle improves static
stability in roll (Lβ), while a negative dihedral (called anhedral; Γ < 0) reduces static stability in roll.
3.2 Dihedral and Yawing Moment
The analysis in the previous section leads us to believe that the dihedral angle allows the wing to behavesomewhat like the vertical tail. Thus, we expect that the wing would contribute to improving Nβ as well.
4
Figure 6: The moment of the asymmetric lift about the CG produces a rolling moment and a yawing moment.
As we show, this does happen, but the CG’s proximity to the wing makes the wing a secondary contributorto Nβ .
Suppose the wing AC is ahead of the CG, and let xAC denote the distance between the AC and CG.Then, the yawing moment on the aircraft due to the dihedral is given by
N = (Liftleft − Liftright) sin ΓxAC
Since the difference in angles of attack is given by ∆β sin Γ, we get
N = −1
4ρV 2SxACCLα sin2 Γ∆β
=⇒ Nβ = −1
4ρV 2SxACCLα sin2 Γ
Thus, sign(Nβ) depends only on sign(xAC): if xAC > 0 then increasing the absolute value of the dihedralangle leads to a reduction in the static stability, and vice-versa. From the point of view of stability, yetagain, it helps to place the CG ahead of the wing AC.
4 Additional Notes on Lateral-Directional Stability
4.1 Wing-VT Combination
Just as in the case of longitudinal static stability, the wing dihedral setting destabilizes yaw (Nβ) if the CGis behind the wing, while the vertical tail has a stabilizing effect. The net Nβ is given by
Nβ = Nwingβ +N tail
β (3)
=1
2ρV 2SbCLα
(Sv(lt − xAC)
Sb− xAC
2bsin2 Γ
)(4)
Define the vertical tail volume ratio Vv =SvltSb
. This gives the following condition for static stability:
Static stability : Vc >xACb
(sin2 Γ
2+SvS
)Interestingly, as the dihedral angle increases, the rearmost CG location for static stability (which we couldcall the directional neutral point) moves closer to the wing.
5
4.2 L-D Departure Criterion
The criteria Nβ and Lβ are static stability criteria, but they are useful pointers for dynamic stability as wellas other complex phenomena. The term criterion Cnβ ,dyn was defined by Moul and Paulson (1958) as
Cnβdyn = Cnβ cosα−(IzIx
)Clβ sinα
It can been argued that Cnβ ,dyn < 0 can indicate the likelihood of two dangerous phenomena:
• Wing-rock: rapid, large-amplitude rolling oscillations at high angles of attack.
• Spin: combination of high α, almost vertical flight path and large angular rates.
The condition for the onset of these phenomena is treated very often as Cnβ ,dyn = 0. Notice that Cnβ andClβ both depend on α. If the wings have an anhedral (i.e., if Clβ > 0), a directionally-stable aircraft (i.e.,with Cnβ > 0 can experience departures at relatively low α.
5 Wing Sweep
Figure 7: A swept wing and the coordinate system required for analysis.
Consider a wing with sweep angle Λ, as shown in Fig. 7. Then, the lift on a small strip of the wingperpendicular to the leading edge
dLift =1
2ρ(V cos Λ)2c cos ΛCnLds =
1
2ρ(V cos Λ)2cCnL dy
where
• CnL is the coefficient of lift of the normal section
• dy = cos Λ ds, and y ∈ [0, b/2]
There are two contributors to the Lβ derivative: V cos(Λ) and CnL. We will deal with each of these in turn.
5.1 Velocity
Suppose that the aircraft is perturbed by a small sideslip angle ∆β. Then, the effective sweep angle of theright wing becomes (Λ−∆β), while that of the left wing becomes (Λ + ∆β). Thus, the rolling moment onthe aircraft is given by
L1 =
(1
2ρV 2cCnL
∫ b/2
0
ydy
)(cos2(Λ + ∆β)− cos2(Λ−∆β)
)=
1
4ρV 2SbCnL (− cos Λ sin Λ) ∆β
Therefore, we get
L1β = −1
4ρV 2SbCnL cos Λ sin Λ
{< 0, Λ > 0 (sweep-back)
> 0, Λ > 0 (sweep-forward)
6
5.2 Coefficient of Lift
We define the normal angle of attack
tanαn =w
u cos Λ=
tanα
cos Λ=⇒ αn ≈
α
cos Λ
Since we perturb only the sideslip and not α (the angle of attack of the FRL is held constant while takingthe partial derivative), we get the following expressions for the angle of attack on the two wings:
Right wing : αnR =α
cos(Λ−∆β),
Left wing : αnL =α
cos(Λ + ∆β)
The resulting rolling moment is given by
L2 =1
16ρV 2 cos2(Λ)SbCnLαα
(1
cos(Λ + ∆β)− 1
cos(Λ−∆β)
)=⇒ L2β =
1
8ρV 2SbCnLαα sin Λ
We now add the two contributions to obtain the derivative Lβ :
Lβ = L1β + L2β = −1
2ρV 2Sb
(CnL sin Λ cos Λ
2− CnLαα
sin Λ
4
)Note that CL = CnL cos2 Λ and CLα = cos ΛCnLα . This gives
Lβ = −1
2ρV 2Sb
(2CL − CLαα
4
)tan Λ
=⇒ Clβ = −2CL − CLαα4
tan Λ = −(CL + CL0
4
)tan Λ
For stability, we need Λ > 0 (sweep-back); a forward-swept wing is statically unstable.Note that, in most textbooks, the second term in Clβ is ignored. While the primary contribution does
come from the first term, the second term is not negligible and illustrates the importance of mathematicalrigour in lateral-directional dynamics in particular where it is difficult to isolate perturbations in any givendegree of freedom.
6 High Wing Versus Low Wing
Figure 8: Flow around the fuselage in the event of a lateral-directional perturbation affects the angle ofattack near the root of the wing.
7
The position of the wing relative to the fuselage reference line determines the direction of the flowexperienced by the wing in the event of a lateral-directional perturbation. Consider a high wing, such asthat in Fig. 8, and suppose that the aircraft gets a positive ∆β (without loss of generality). Then, the rightwing experiences an increase in angle of attack close to the root, while the left wing experiences a reductionin the angle of attack close to the root. This effect is almost negligible near wing tips.
The change in the angle of attack induces a negative rolling moment, so that Lβ < 0. The exact oppositehappens for low wings, and thus low wings are destabilizing.
The change in Clβ due to the location of the wings is difficult to derive analytically. Rather, we use thefollowing estimates:
∆Clβ = σ 0.00917 rad−1
{σ = −1 high wing
σ = 1 low wing
It should be noted that several books and research papers use the phrase “dihedral effect” to refer to allthree terms looked at above. They are quite similar in nature and from the perspective of design, it helpsto take a combined view of the three terms. While static stability is desirable, too much static stability canimpede maneuverability. Thus, aircraft with high wings use wings in an anhedral configuration, while lowwings are usually accompanied by a positive dihedral setting. The choice of sweep angle is dictated usuallyby Mach number considerations, and the other two parameters (dihedral and wing position) are chosen toprovide the remainder of the desired stability.
7 Lateral-Directional Control
7.1 Crosswind Trims
Lateral-direction trim is achieved using the rudder and ailerons together. The rudder and aileron deflectionsare conventionally denoted by δr and δa, respectively.
The rudder develops a yawing moment as well as a rolling moment. These exact contributions can bederived using strip theory and thin airfoil theory. We merely note that the corresponding derivatives Lδr < 0and Nδr > 0.
The ailerons, for their part, primarily generated rolling moment. As a matter of convention, positiveaileron deflection corresponds to a downward deflection on the left wing and an equal and opposite deflectionon the right wing. Hence, Lδa > 0. The yawing moment from the ailerons can be neglected at low angles ofattack, and we choose to do as much.
In order to trim the aircraft at a sideslip angle β, we need to ensure that
L = Lββ + Lδaδa + Lδrδr = 0
N = Nββ +Nδrδr = 0 (5)
which gives the following expressions for δa and δr:
δr = − NβNδr
β
δa =
(LδrNβLδaNδr
− LβLδa
)β (6)
Although aircraft usually trim at β = 0, a non-zero sideslip is used in two distinct scenarios: while tryingto reduce speed rapidly using the increased drag due to sideslip, and while trying landing with crosswindswhile keeping the nose pointed along the runway.
7.2 Control Problems Involving the Ailerons
The control authority of the ailerons, measured by Clδa , can be compromised when the portion of the wingscontaining the ailerons is stalled, or due to finite wing effects leading to reduced lift in the aileron portionof the wings.
8
In the case of stall, aileron deflection leads to Clδa < 0; i.e., a positive aileron deflection may produce anegative rolling moment. This situation, called aileron reversal, can arise either due to the aircraft pitchingto overly large angles of attack, or due to excessive wing loading at high speeds leading to sectional stall. Inthe former case, pilots are trained to use the rudder to recover the aircraft to a controlled condition, whilein the latter case, it usually suffices to reduce the speed by reducing thrust.
The control authority of ailerons can also be reduced also due to finite wing effects, i.e, when the liftdistribution is such that the lift is smaller near the wing tips. This may pose a problem at low flight speeds.One way to mitigate the problem is to use vortex generators to energize the flow in outboard areas of thewing and make the lift distribution more uniform. The same effect can be achieved by adding winglets orwing-tip fences. Alternately, spoilers may have to be employed to make up for the deficiency in the rollcontrol authority.
7.3 Lateral Control Departure Parameter (LCDP)
Earlier in this chapter, we noted that the criterion Cnβ ,dyn is a bellwether for lateral-directional instabili-ties. Spin prediction, in particular, is done using Cnβ ,dyn and another parameter called the lateral controldeparture parameter (LCDP) defined in its simplest form as
LCDP = Cnβ −CnδaClδa
Clβ > 0 for stability
Notice that Cnδa cannot be ignored at high angles of attack and plays an important role in determining thevalue of LCDP.
The safe flight envelope is thus usually approximated by Cnβ ,dyn > 0 and LCDP > 0. Different typesof departure phenomena arise when either one of the conditions fails to apply, spin being the result ofsimultaneous failure. Interestingly enough, despite the complex nature of the departure phenomena, thesusceptibility of the aircraft can be predicted reasonably well using purely static criteria.
7.4 Adverse Yaw
For the purpose of ensuring smooth turns, it is essential that the rolling motion induced by the ailerons beaccompanied by a yaw in the same direction; i.e., a positive roll should be accompanied by a positive yawand vice-versa. When the yaw is in the opposite sense of roll, it is referred to as adverse yaw.
Adverse yaw is a serious problem in tailless aircraft (e.g., flying wings such as the B-2 bomber). A direct
way to prevent adverse yaw is to design forCnδaClδa
> 0. However, since Cnδa is very sensitive to the wing
geometry, CG location and α, its sign cannot be taken for granted. In fact, a vertical tail goes a long way inpreventing adverse yaw (for reasons which will become apparent later in the book). Since tailless aircraft lacka vertical tail, it is helpful to place the CG behind the wing AC - the dihedral effect enusres that a positiverolling moment is accompanied by a positive yawing moment and vice-versa. However, an aft-located CGcompromises directional stability by making Cnβ < 0.
While this poses a stability-versus-control trade-off scenario, a bit of adverse yaw is actually helpful forpiloting. When the airframe has a tendency for adverse yaw, a pilot is required to use the rudder in thesame direction as the ailerons, which is a more natural pair of actions for the human body (a right on theyoke and right leg down for right rudder, and vice-versa). This illustrates how complex aircraft design cansometimes get, and the extent to which seemingly minor bells and whistles can have an incommensurateinfluence on the design decisions.
9
Chapter 6
Longitudinal Flight Dynamics and Control
Aditya Paranjape
The objective of this chapter is to study the stability and control of an aircraft constrained to move inits plane of symmetry. Although this is a restricted case of the complete aircraft motion, the results are infact valid for the complete aircraft due to the dynamic decoupling that we will see in the next chapter.
1 Review of Systems Theory
We start by reviewing the concepts of equilibrium solutions and stability of linear systems.
1.1 Equilibrium Solutions
Definition: Given a (linear or nonlinear) system x = f(x, u), we define an equilibrium solution as a pair(x0, u0) of constants that satisfy f(x0, u0) = 0.
For a linear system x = Ax+Bu, we have the following cases for an equilibrium solution:
1. If A is invertible, then for every u0, there is a unique equilibrium solution x0 = −A−1Bu0.
2. If A is non-invertible, then we have exactly one of the following two cases:
• There exists no equilibrium solution if B /∈ Range(A)
• There exist an infinite number of equilibria given by x0 = v + n, where v is any vector satisfyingAv = −Bu0 and n satifies An = 0.
1.2 Equilibrium Solutions and Stability
Definition: An equilibrium solution x0(u0) is said to be stable if a perturbation ∆x (e.g., caused by anexternal disturbance) is rejected by the system and x(t)→ x0 as t→∞.
Let x(t) = x0 + δx(t), where δx is assumed to be “small.” The small-perturbation dynamics about theequilibrium x0(u0) are given by
x(t) = δx =
(∂f
∂x
)x0
δx
The equilibrium x0 is stable if the Jacobian(∂f∂x
)x0
to be Hurwitz
The “simplest” type of differential equations are linear differential equations
x = Ax+Bu, A ∈ Rn×n
For linear systems, the matrix A is itself the Jacobian. An equilibrium solution is stable if and only if A isHurwitz; i.e., it is invertible with all of its eigenvalues in the open left half of the complex plane. Furthermore,if A is Hurwitz, there is a unique equilibrium point for a given u0 and it is given by x0 = −A−1Bu0.
Definition: An equilibrium solution is said to be unstable if it is not stable. In other words, stability of asystem is a dichotomous property.
1
1.3 Modal Decomposition of Linear Systems
Consider the uncontrolled linear system x = Ax. If the geometric multiplicity of A is equal to its algebraicmultiplicity, then the solution x(t) of the linear differential equation can be written in the form
x(t) =
N∑i=1
aieλitvi (1)
where the coefficients ai depend on the initial conditions (x(t = 0)). Notice that if λi is complex, then so isthe corresponding vi. The solution x(t) itself is real. If the algebraic multiplicity is more than the geometricmultiplicity, there is a similar expansion for x(t), except that it also involves terms of the form tkeλit whosecorresponding vector coefficients are not eigenvectors.
In Eq. (1), if the initial condition x0 is such that ai = 0 for all i 6= k for some k and ak 6= 0, thenthe solution x(t) = ake
λktvk; i.e., the solution x(t) continues to lie along vk for all t if it starts on vk. Asimilar situation occurs if the solution starts in a plane spanned by the eigenvectors of complex conjugateeigenvalues: it continues to remain in the eigenplane forever.
It is thus possible to view the solution of a linear system as a linear combination of its modal solutions.In certain circumstances, it is possible to obtain the modal solutions in a rather straight-forward way. Whatis important here is the stability of the individual modes: if we design the aircraft and its controllers tostabilise the individual modes, then we would have stabilised the complete dynamics. This simplifies theproblems of stability analysis of complex systems as well as control design.
1.4 Complex Conjugate Modes
A typical second-order system is of the form x+ cx+ kx = 0. Its eigenvalues are given by
λ =−c±
√c2 − 4k
2
Thus, the necessary and sufficient conditions for stability are c > 0 and k > 0. If c2 > 4k, both eigenvaluesare real; else, the eigenvalues are complex conjugate and given by
λ =−c± i
√4k − c2
2
We define the natural frequency of a second order system as ωn =√k, and its coefficient of damping as
ζ =c
2ωn. We get the following cases for different values of ζ:
• ζ < 0: the system is unstable (c > 0)
• ζ = 1: critical damping; the two roots are equal and given by λ = −ζωn
• 0 < ζ < 1: the system is stable and exhibits decaying oscillations. The damped natural frequency ofthe oscillations is defined as ωd = ωn
√1− ζ2.
• ζ > 1: the system is over-damped with real, stable roots. The system does not exhibit any oscillatorybehaviour.
On the modal plane corresponding to complex conjugate eigenvalues, the dynamics are identical to spring-mass-damper systems and we therefore extend the terminology and the associated concepts to the flightdynamic modes as well.
2
Figure 1: Free body diagram of an aircraft.
2 Equations of Motion in the Longitudinal Plane
The aircraft has three degrees of freedom in the longitudinal plane:
• Two translational degrees of freedom described by V∞ and γ
• One rotational degree of freedom with α and q as the state variables.
From the free body diagram in Fig. 1, it follows quite readily that
V =1
m(T cosα−D)− g sin γ
γ =1
mV(L+ T sinα−mg cos γ)
Recall the equations for level flight and climbing performance: these were found by setting V = γ = 0 (i.e.,constant V and γ).
The rotational dynamics are given by
q =M
Iyy(2)
where M is the pitching moment which we encountered earlier in Chapter 4.Finally, note that q = α+ γ. Thus, we get
α = q − 1
mV(L+ T sinα−mg cos γ)
In order to simplify the analysis, we assume that T sinα is negligible and T cosα ≈ T . This gives us thefollowing EoM:
V =1
m(T −D)− g sin γ
α = q − 1
mV(L−mg cos γ)
q =M
Iyy(3)
γ =1
mV(L−mg cos γ)
3 Linearization
Equilibrium solutions of Eq. (4) are found by setting V = γ = α = q = 0. Thus, flight speed, flight pathangle, and the angle of attack have to be constant. These are the familiar flight conditions encounteredearlier in Chapters 2 and 3. If γ = 0, we get level flight, while a non-zero γ corresponds to trimming climbor descent.
3
3.1 Preview of the Modal Decomposition
Before formally examining the modal structure of the longitudinal dynamics, it is instructive to look atthe time history of the longitudinal variables for a generic aircraft, found by simulating Eq. (4). The timehistories of V , α and γ are shown in Fig. 2.
(a) α (b) γ
(c) V
Figure 2: Time history of the longitudinal variables for a generic aircraft. Two complex conjugate modescan be readily discerned from the time history.
Two modes are clearly discernible in Fig. 2:
1. A fast mode in α with a time period of approximately 5 s. This mode is well-damped and is almostnot visible in V and γ.
2. A slow mode in V and γ which is under-damped and has a time period of approximately 55 s. It isalmost missing in the α plot.
It turns out that these modes are quite typical of longitudinal dynamics. The fast mode is also called theshort period (SP) mode, while the slow mode is called the phugoid mode. We will now formally deriveanalytical approximations to these modes.
3.2 Small Perturbation Model
The stability of an aircraft is as much a property of the dynamics as it is of specific trim solutions. Thesimplest longitudinal trims satisfy q = 0 and are parametrized by constant values of V 0, α0 and γ0. Weknow from our analysis in Chapters 2 and 3 that equilbrium solutions satisfy
L =1
2ρ(V 0)2SCL(α0) = W cos γ0
T = D(V 0, α0) +W sin γ0
Let x0 = [V 0, α0, γ0, 0] denote the equilibrium condition. In order to linearize about a trim, we perturb thestate variables by sufficiently small values ∆V , ∆α, ∆γ and ∆q.
4
The dynamics of the perturbed variables are linear, since we assume these perturbations to be smallenough for (∆V )2 and similar higher order terms to be zero. The dynamics of the perturbed variables, alsocalled the linearized dynamics, are given by
∆V
V 0= − g
V 0
(∆γ + 2
(CD(α0)
CL(α0)cos γ0
)∆V
V 0+ 2k cos γ0CLα∆α
)∆γ =
g
V 0
(sin γ0∆γ + 2 cos γ0 ∆V
V 0+CLαCL
∆α
)∆α = ∆q −∆γ
∆q =Mα
Iy∆α+
Mα
Iy∆α+
Mq
Iy∆q
where Mα = ∂M/∂α and so on. These are the same derivatives that we encountered in Chapter 4.Based on numerical results, and arguments based on physics, it is appropriate to hypothesise that there
exist two modes:
• A fast mode which can be approximated from the dynamics of ∆α and ∆q alone, and
• A slow mode which can be approximated from ∆V and ∆γ alone.
3.3 Short-Period (SP) Dynamics
Short period dynamics represent fast, but well-damped oscillations in pitch. The time scale of the oscillationsis fast enough so that the translational variables are more or less unchanged while the SP oscillations settledown. Thus, we can study the SP dynamics by simply ignoring the dynamics of V and γ. We can thus write
∆α = ∆q − g
V0
CLαCL
∆α
∆q =Mα
Iy∆α+
Mα
Iy∆α+
Mq
Iy∆q
These equations can be consolidated into a single second order ODE of the form
∆α+ 2ζSPωSP∆α+ ω2SP∆α
where
ω2SP = −
(Mα
Iy+Mq
Iy
g
V0
CLαCL
)2ζSPωSP = −
(Mα +Mq
Iy+
g
V0
CLαCL
)(4)
It turns out that the stability of the short period depends largely on ensuring that Mα < 0 and Mq < 0. Theother terms in Eq. (4) make secondary contributions to the stability, although their effect can be significantat high speeds (since V0CL ∝ 1/V0).
It is worth noting that the modal frequency increases quite rapidly with V0, as does the damping ratio.At low speeds, and high angles of attack, the SP mode is affected significantly by aerodynamic nonlinearities,especially stall.
3.4 Phugoid (P) Dynamics
The phugoid mode consists of poorly damped, long-period oscillations of the point-mass aircraft. We canderive an approximation for the phugoid mode along the lines of the derivation for the SP mode. We
5
start by assuming that ∆α = ∆q = 0. Furthermore, we normalize the perturbation ∆V by V 0 to keep itnon-dimensional. The phugoid dynamics are then given by[
∆VV 0
∆γ
]=
g
V 0
[−2CD(α0)
CL(α0) cos γ0 −1
2 cos γ0 sin γ0
] [∆VV 0
∆γ
](5)
It follows that the natural frequency of the phugoid mode is given by
ω2P = 2
( g
V 0
)2
cos γ0
(1− CD(α0)
CL(α0)sin γ0
)≈ 2
(g
V0
)2
Unlike ωSP , the natural frequency of phugoid ωP reduces with increasing flight speed and with increasingγ0. The time constant of the phugoid mode is typically several tens of seconds.
The damping ratio (with γ0 ≈ 0) is given by
ζP =1√2
(CDCL
)α0
The damping ratio decreases with increasing aerodynamic efficiency, and it typically around 0.04 − 0.07.Interestingly enough, the damping ratio is minimum at the speed for minimum drag, and increases on eitherside of it.
The formulae given here suggest that the phugoid mode should be generically stable, but it is not. Theformulae that we have derived are, thus, strictly approximate in nature and require considerable refinement.Accurate analytical prediction of phugoid is beyond the scope of this course. However, since the phugoidhas a large time constant and fairly small amplitudes, it is not uncommon for the phugoid mode to be leftuncompensated by active control, especially in combat aircraft.
6
Chapter 7
Equations of Motion
Aditya Paranjape
1 Translational Equations of Motion
Figure 1: The total momentum is found by writing the momentum of the element dm and then summingover the complete aircraft.
Consider the small element with mass dm with vector r from the origin of the body-fixed B-frame, asshown in Fig. 1. Then, the total velocity of the element is given by
VT = V + ω × r
The linear momentum of the element is given by
dp = dm VT = dmV + ω × (r dm)
Hence, the total momentum of the body is given by
p =
∮dp = mV + ω ×
∮r dm
We assume that the origin of B-frame is located at the centre of mass; i.e.,∮r dm = 0. Then, using Newton’s
second law, we get
F =
(dp
dt
)I
=⇒ F =
(d(mV )
dt
)I
Assuming that the mass m is constant, i.e., the rate of fuel burn is negligibly slow, we get
m
(dV
dt
)I
= m
(dV
dt
)B
+m(ω × V
)= F (1)
We express the vectors are in the body (B) frame:
• V = (u, v, w)
• ω = (p, q, r)
• Forces: (X, Y, Z)
1
Substituting into Eq. (1) This gives us the following equations:
u+ (qw − rv) =X
m
v + (ru− pw) =Y
m(2)
w + (pv − qu) =Z
m
The forces X, Y Z derive from three sources: propulsive, aerodynamic and gravity.
1.1 Propulsive
Throughout this course, we have assumed that the line of thrust coincides with the xB axis, so that Xprop =T , and Y prop = Zprop = 0.
1.2 Aerodynamic
Lift and drag are the two primary aerodynamic forces. However, their components L and D are written withrespect to the velocity vector. In particlar, they are written in what are called as the wind axes whose axesare defined as follows:
• xw axis: along V∞
• zw axis: along the cross product V∞ × yB
• yw axis: along the cross product zw × xwThe rotation from the wind frame to the body frame is achieved by two rotations:
• Rotation about zw by −β to bring the xw axis in the plane of symmetry of the aircraft. The wind axisframe is then tranformed into a new frame which we denote by W ′. Note that zw = zw′ .
• Rotation about the y axis of W’ frame by an angle α.
In the wind frame frame, the aerodynamic forces are given by
Faero,W =
−D0−L
Let R3(−β) denote the rotation from W to W ′ (the subscript denotes the axis of rotation). In particulargiven a vector r, its components in W ′ and W are related by rw′ = R3(−β)rw. With this notation, we get:
Faero,B =
cosα 0 − sinα0 1 0
sinα 0 cosα
cosβ − sinβ 0sinβ cosβ 0
0 0 1
Faero,W=⇒
Xaero
Y aero
Zaero
=
cosα cosβ − cosα sinβ − sinαsinβ cosβ 0
sinα cosβ − sinα sinβ cosα
−D0−L
=
L sinα−D cosα cosβ−D sinβ
−L cosα−D sinα cosβ
Notice that α and β are related to u, v w by the following relations:
tanα =w
u, sinβ =
v
V∞(3)
2
This can be verified by using the above rotation matrix from W to B: notice that the velocity vector is[V∞, 0, 0] in the wind frame. Thus,
u = V∞ cosα cosβ, v = V∞ sinβ, w = V∞ cosβ sinα
All of the forces that we have seen so far depend only on u, v, w. The next force, gravity, will require thatwe introduce additional states beyond the velocity and angular velocity components.
1.3 Gravity
The gravitational acceleration is defined by its components [0, 0 g] in the inertial frame (I). The transfor-mation from I to the body frame B involves a sequence of three rotations:
1. Rotation from I to I1: about zI by an angle ψ
2. Rotation from I1 to I2: about yI1 by an angle θ
3. Rotation from I2 to B: about xI2 by an angle φ
It follows that
gB = R1(φ)R2(θ)R3(ψ) gI
=
1 0 00 cosφ sinφ0 − sinφ cosφ
cos θ 0 − sin θ0 1 0
sin θ 0 cos θ
cosψ sinψ 0− sinψ cosψ 0
0 0 1
00g
=
−g sin θg sinφ cos θg cosφ cos θ
Thus, the components of the weight in the body frame are given by Xgrav
Y grav
Zgrav
=
−mg sin θmg sinφ cos θmg cosφ cos θ
(4)
2 Euler Angle Rates
We have introduced two new states: φ and θ. Notice that ψ does not affect the translational equations ofmotion in any way, and therefore ψ is decoupled from the rest of the dynamical equations.
First, we note that ψ, θ and φ are defined in three different frames: I, I1 and I2, respectively. Theangular velocities defined in the three frames are thus [0, 0, ψ]I , [0, θ, 0]I1 and [φ, 0, 0]I2 . Thus, the bodyaxis components of the sum of these vectors are given by p
qr
= R1(φ)
φ00
+R1(φ)R2(θ)
0
θ0
+R1(φ)R2(θ)R3(ψ)
00
ψ
=
1 0 − sin θ0 cosφ cos θ sinφ0 − sinφ cos θ cosφ
φ
θ
ψ
Before inverting the above matrix, note that its determinant is cos θ. Thus, the determinant is invertibleonly when |θ| < π/2. In that case, inversion yields
φ = p+ q tan θ sinφ+ r tan θ cosφ
θ = q cosφ− r sinφ
ψ = sec θ(q sinφ+ r cosφ)
3
3 Equations of Motion: Rotational
Consider Fig. 1 again. The angular momentum of the element dm is given by
dh = r × dp = dm(r ×
(V + ω × r
))The total angular momentum is then given by
h =
∮dh =
(∮dm r
)× V −
(∮dm (r×)2
)ω
The cross product r × ω can be written as
r × ω =
0 −r3 r2r3 0 −r1−r2 r1 0
ω = S(r)ω
Similarly, we can write:
(r×)2ω = S(r)S(r)ω =
−(r22 + r23) r2r1 r3r1r1r2 −(r23 + r21) r2r3r1r3 r2r3 −(r21 + r22)
ωThe moments of inertia are defined by
Principal moments : I11 =
∮ (r22 + r23
)dm
Cross moments : I12 = −∮
(r1r2) dm
Define the matrix J = {Iij} is given by
J = −∮
(S(r))2dm
The matrix J is also called the moment of inertia matrix (or moment of inertia tensor). With this notation,the total angular momentum can be written as
h = Jω (5)
Using Newton’s laws of motion, and with M denoting the vector angular moments, we get the generalequations of motion:
˙ω = J−1(−S(ω)Jω + M
)where ω = (p, q, r) and M = (L, M, N). Assume that the aircraft is symmetric about the xB − zB plane.This implies that I12 = I23 = 0. Furthermore, suppose we choose the principal axes so that I13 = 0 as well.With these simplifications, we get
p =
(I2 − I3I1
)qr +
L
I1(6)
q =
(I3 − I1I2
)rp+
M
I2(7)
r =
(I1 − I2I3
)pq +
N
I3(8)
where I1 = I11 and so on.
4
4 Consolidated Equations of Motion
The equations of motion of a rigid aircraft can be put together from the derivations in this chapter:
u = rv − qw +X
m
v = pw − ru+Y
m
w = qu− pv +Z
m
p =
(I2 − I3I1
)qr +
L
I1
q =
(I3 − I1I2
)rp+
M
I2
r =
(I1 − I2I3
)pq +
N
I3
φ = p+ q tan θ sinφ+ r tan θ cosφ
θ = q cosφ− r sinφ
ψ = sec θ(q sinφ+ r cosφ)
Notice that the none of the equations depend on ψ; thus, the equation for ψ is dropped from this list whilestudying the stability of an aircraft. The equation should be retained, however, when performance is to bestudied.
4.1 Translational Equations Rewritten in the Wind Frame
It is generally useful to replace the equations for u, v and w with the corresponding equations in V∞, α andβ since it is the latter set that is of relevance from the point of view of physics.
We start by noting that
u = V∞ cosα cosβ
=⇒ u = V∞ cosα cosβ − V∞ sinα cosβα− V∞ cosα sinββ
Likewise,
v = V∞ sinβ − V∞ cosββ
w = V∞ sinα cosβ + V∞ cosα cosβα− V∞ sinα sinββ
Substituting into the above equations of motion and replacing u, v and w in terms of V∞, α and β yields afamiliar-looking set of equations:
V∞ =1
m(T cosα cosβ −D −W sin γ)
α = q − tanβ(p cosα+ r sinα)− 1
mV∞ cosβ(T sinα+ L−W cos γ cosµ)
β = p sinα− r cosα+1
mV∞(−T cosα sinβ + Y +W cos γ sinµ)
where
• Y is the aerodynamic side-force acting on the aircraft.
• µ is the wind axis roll angle which we saw earlier in Chapter 5.
5
It remains to describe γ and µ in terms of α, β, θ and φ. Notice that the γ and µ terms arise out of anattempt to resolve the gravity vector in the wind frame. The Euler angles for the wind frame are χ (velocityheading), γ and µ. Therefore, we simply transform the gravity vector from the body frame, which we havealready done earlier, to the wind frame: − sin γ
cos γ sinµcos γ cosµ
=
cosβ sinβ 0− sinβ cosβ 0
0 0 1
cosα 0 sinα0 1 0
− sinα 0 cosα
− sin θcos θ sinφcos θ cosφ
=
cosβ cosα sinβ cosβ sinα− sinβ cosα cosβ − sinβ sinα− sinα 0 cosα
− sin θcos θ sinφcos θ cosφ
This gives
sin γ = sin θ cosβ cosα− sinβ cos θ sinφ− sinα cosβ cos θ cosφ
cos γ sinµ = cosα sinβ sin θ + cosβ cos θ sinφ− sinα sinβ cos θ cosφ
cos γ cosµ = cosα cos θ cosφ+ sinα sin θ
5 Special Solutions
It is a fact that most standard aircraft missions can be described as a concatenation of equilibrium solutionsof the 8 dynamical equations described above. It is for this reason that equilibrium solutions are important.
In the preceding chapters, we saw several flight conditions such as level flight, level turn and the steadysideslip maneuver. Each of these can be found as an equilibrium solution of the above equations of motion.The sideslip maneuver may also have a non-equlibrium element.
There are three types of variables that make up a maneuver:
1. The prescribed variables: the values of these variables are given. For equilibrium fligth, these areconstant.
2. To-be-determined (TBD) variables: TBD variables are state variables whose values are not prescribeda-priori. Rather, their values are found by solving the equations of motion.
3. Control variables: these are the control inputs, i.e., thrust, elevator, aileron and the rudder deflectionsin a conventional aircraft.
In general, note that at most 4 variables can be prescribed, corresponding to the four control inputs, exceptin the wings-level condition where lateral-directional variables vanish automatically and may be treated asbeing prescribed zero values.
5.1 Wings-Level Flight
In wings level flight, the distribution of the variables among the three sets is as follows:
• Prescribed zero variables: β = p = r = φ = 0
• Prescribed non-zero quantities: V∞, γ (i.e., θ − α)
• TBD variables: θ, α
• Control variables: δa = δr = 0; thrust and elevator need to be determined.
6
5.2 Level Turn
The equation for ψ enters the picture for turns: the equilibrium turn rate may be directly equated to ψ.
• Prescribed zero variables: β = 0, γ = 0
• Prescribed non-zero quantities: V∞, turn rate ψ
• TBD variables: θ, α, p, q, r, φ
Under equilibrium conditions, we note the following connection between the various quantities which helpswith the calculation of the TBD variables:
p = −ψ sin θ, q = ψ cos θ sinφ, r = ψ cos θ cosφ
This equation ensures that the Euler angle rates φ = θ = 0.An important point to note here comes from the α equation. When we studied level turns in Chapter 3,
we set L cosµ = W . Can you reconcile this with α = 0, i.e., L−W cosµ = mV∞q?
5.3 Steady Sideslip Maneuver
A cross-slipping maneuver is performed typically when an aircraft lands. It may be used to slow the aircraftduring descent as well; in fact, it is used routinely by gliders. The distribution of the variables is as follows:
• Prescribed (non-zero): β, γ
• Prescribed (possibly non-equilibrium): V∞(t); either a constant speed would be prescribed or a rateV∞. The thrust T can be adjusted to achieve this profile.
• TBD: φ, θ, α
• TBD (but zero): p = q = r = 0
The control variables and the TBD variables are determined as follows:
1. Aileron and rudder inputs by setting Cl = Cn = 0.
2. The aileron and rudder inputs, and the sideslip, generate a non-zero sideforce Y . Thus, the windaxis roll angle µ must take on a value which ensures equilibrium in β, i.e., β = 0. In fact, ignoringT cosα sinβ allows us to determine µ quite readily.
3. The angle of attack takes on a value found from L = W cos γ cosµ.
4. The thrust can be found from V (t): T = D +W sin γ +mV (t).
5. The pitch angle achieved in the process can be found from the γ and µ equations above.
5.4 Level Acceleration
A wings-level acceleration at constant altitude is an example of a non-equilibrium maneuver. Consider anaircraft accelerating at a fixed rate Vd at a fixed altitude (γ = 0). The maneuver can be profiled as follows:
• Prescribed variables: Vd, γ = 0
• Prescribed/TBD variables: β = p = r = φ = 0
• TBD: α, q, θ
• Control variables: δa = δr = 0; T, δe to be determined.
7
The thrust depends only on the instantaneous drag and acceleration:
T = mVd +1
2ρV 2SCD(α)
Notice that α changes at each instant to ensure that γ = 0:
CL(α) =2W
ρV 2S=⇒ α(t) = C−1L
(2W
ρV 2(t)S
)− CL0
Notice that α is time-varying since V is time-varying. Once α(t) is known, we can find α(t). The pitch rateq(t) is then given by
q(t) = α(t) =⇒ δe(t) =−1
Mδe
(M0 +Mαα(t) +Mqq(t) +Mαα(t))
Note that the variables calculated here represent the ideal trajectory; the actual trajectory would featureperturbations about the ideal values which subsequently need to be compensated for by automatic controlsystems.
6 Linearization and Decoupling
The flight dynamic equations derived above form a set of nonlinear differential equations. As we saw inChapter 6, the stability of the dynamics is not really a global property - rather it is a local property ofequilibrium solutions. Level flight equilibria are of primary interest to us. The results established for levelflight are generally valid for turning flight as well, although nonlinearities in the parent dynamics do bringabout some qualitative and quantitative changes in stability.
We will show here that the dynamics obtained by linearizing the equations of motion decouple of thelongitudinal states [u, w, q, θ] and the lateral-directional states [v, p, r, φ] from each other. Once we havelinearized the dynamics, we will replace [u, v, w] with [V∞, α, β].
A level flight equilibrium is specified by prescribing the values of u0, w0 and θ0. All other variablesare zero at equilibrium. Moreover, if the aircraft is symmetric, the only force and moment derivatives thatmatter are listed in Table 6. Note that CXθ , CYφ and CZθ arise exclusively due to the weight of the aircraft.The derivatives Xq and Zq are almost always non-zero, but can be ignored safely in low angle of attack flight.
Table 1: Aerodynamic derivatives; note that q∞ = 0.5 ρV 2∞.
Force/Moment Important Stability Derivatives
X = q∞SCX CXu , CXw , CXq , CXθY = q∞SCY CYv , CYp , CYr , CYφZ = q∞SCZ CZu , CZw , CZq , CZθL = q∞SbCl Clv , Clp , ClrM = q∞ScCm Cmu , Cmw , Cmq , CmwN = q∞SbCn Cnv , Cnp , Cnr
Let ∆u denote the perturbation in u, and likewise for other quantities. Substituting into the equationsof motion, we get two decoupled sets of linearized dynamics:
8
1. Longitudinal dynamics
∆u = −w0∆q +Xu
m∆u+
Xw
m∆w +
Xθ
m∆θ
∆w = u0∆q +Zum
∆u+Zwm
∆w +Zθm
∆θ
∆q =1
I2(Mu∆u+Mw∆w +Mw∆w +Mq∆q) (9)
∆θ = ∆q
The reader should verify, as an exercise, that these equations can be transformed into the wind-axisequations which were derived in Chapter 6. Note that Xθ = −W cos θ and Zθ = −W sin θ. Finally,the term Mw is the equivalent of Mα.
2. Lateral-directional dynamics
∆v = w0∆p− u0∆r +Yvm
∆v +Ypp
∆p+Yrm
∆r +Yφm
∆φ
∆p =1
I1(Lv∆v + Lp∆p+ Lr∆r)
∆r =1
I3(Nv∆v +Np∆p+Nr∆r) (10)
∆φ = ∆p+ tan θ0∆r
These equations can be written in the wind-axis frame quite readily. In fact, the only equation thatchanges is for ∆v, with ∆v = V∞∆β. This gives us
∆β = α0∆p−∆r +1
mV∞(Yβ∆β + Yp∆p+ Yr∆r) +
g
V∞cosα0∆φ
∆p =1
I1(Lβ∆β + Lp∆p+ Lr∆r)
∆r =1
I3(Nβ∆β +Np∆p+Nr∆r) (11)
∆φ = ∆p+ tanα0∆r
Since the linearization is about level flight, we replaced Yφ/m = g cos θ0 and we additionally assumedthat u0 ≈ V∞.
The stability of the complete dynamics about level flight can thus be determined by analysing thelongitudinal and the lateral-directional dynamics independently. In the next chapter, we will analyse thelateral-directional dynamics along the lines of our analysis for the longitudinal dynamics in Chapter 5.
9
Appendix: Tranformations between Coordinate Frames
Figure 2: Illustration: Rotation about x axis
Transformations between coordinate frames are achieved through a sequential set of rotations. Considera single rotation through an angle δ from frame ’A’ to frame ‘B’ about axis 1 as shown in Fig. 2. Let xa andxb denote the coordinates of a vector x in frames A and B, respectively.
xb =
1 0 00 cos δ sin δ0 − sin δ cos δ
xaDenote this transformation by
xb = R1BA(δ)xa
For the ithaxis:RiBA(δ) = RiAB(−δ)
i.e., the transpose of a rotation matrix is also its inverse.We can now give formulae for coordinate transformation under single rotations:
• Rotation about axis 1:
R1BA(δ) =
1 0 00 cos δ sin δ0 − sin δ cos δ
• Rotation about axis 2:
R2BA(δ) =
cos δ 0 − sin δ0 1 0
sin δ 0 cos δ
• Rotation about axis 3:
R3BA(δ) =
cos δ sin δ 0− sin δ cos δ 0
0 0 1
10
Chapter 8
Lateral-Directional Flight Dynamics
Aditya Paranjape
The linearized equations of motion of the lateral-directional dynamics were derived in the previous chap-ter:
∆β = α0∆p−∆r +1
mV∞(Yβ∆β + Yp∆p+ Yr∆r) +
g
V∞cosα0∆φ
∆p =1
Ix(Lβ∆β + Lp∆p+ Lr∆r)
∆r =1
Iz(Nβ∆β +Np∆p+Nr∆r) (1)
∆φ = ∆p+ tanα0∆r
We would like to understand the modal decomposition of the lateral-directional (L-D) dynamics along thelines of the longitudinal dynamics. To simplify our analysis, we will make the following assumptions:
1. α0 ≈ 0. While this assumption limits the validity of our results to low angles of attack, the analysisnevertheless captures the essential properties of the L-D modes.
2. Yp = Yr = Np = 0. These assumptions are based on physics and will not be discussed any further.
We will thus work with the following linearized equations:
∆β =YβmV∞
∆β −∆r +g
V∞∆φ
∆p =1
Ix(Lβ∆β + Lp∆p+ Lr∆r)
∆r =1
Iz(Nβ∆β +Nr∆r) (2)
∆φ = ∆p
Our first observation is that, unlike the longitudinal modes, there is no obvious way to identify the fast andthe slow modes by isolating the translational and the rotational degrees of freedom. Instead, the way toarrive at a modal decomposition is to analyse numerical results, postulate upon the nature of the modes andverify that the postulates are indeed correct.
It turns out that the L-D dynamics decompose into three modes which are visible in the root locus plotfor the F/A-18 HARV, shown in Fig. 1:
• Roll: fast and non-oscillatory; almost entirely p; period T ∼ O(0.1 s)
• Dutch roll: moderately fast; β + r; period T ∼ O(1 s)
• Spiral: slow and non-oscillatory; mostly r; period T ∼ O(10 s)
1
Figure 1: Low α root locus of the L-D dynamics of the F/A-18 HARV
1 Roll Mode
The roll dynamics are given by
∆p =LpIx
∆p+1
Ix(Lr∆r + Lβ∆β)
The roll eigenvalue is given by
λR =LpIx
with the accompanying stability condition Lp < 0. Clearly, the roll mode is stable as long as the aircraft isin the pre-stall regime.
Due to the fast roll dynamics, the roll rate converges rapidly to a value found by setting p ≈ 0. Thisvalue is called the static residual of the roll rate and is given by
∆ps = − 1
Lp(Lr∆r + Lβ∆β) (3)
The rest of the states “see” this value of roll rate in the course of their evolution.
2 Dutch Roll
The Dutch roll mode predominantly involves β and r. Its dynamics are given by
∆β = −∆r +YβmV
∆β +g
V∆φ
∆r =NrIz
∆r +NβIz
∆β
The characteristic equation of the Dutch roll dynamics works out to
λ2 −(YβmV
+NrIz
)λ+
1
Iz
(YβmV
Nr +Nβ
)︸ ︷︷ ︸
ω2DR
= 0
2
Comparing with the standard spring-mass-damper system of the form x+cx+kx = 0, we note the following:
• The damping constant c is given byNrIz
+YβmV
. Generally, Nr as well as Yβ are negative as long as the
vertical tail is appropriately sized, and the Dutch roll mode damping is ideally not expected to presentany problems at low angles of attack.
• The stiffness depends on Nβ +YβNrmV
. Clearly, Nβ > 0 is a sufficient condition for ensuring that the
Dutch roll mode is stable, provided both Yβ < 0 and Nr < 0.
The Dutch roll mode manifests itself in the form of oscillations in yaw, as shown in Fig. 2. The yawingmotion is accompanied by a roll rate whose value is equal to the static residual in Eq. (3).
Figure 2: The Dutch roll mode manifests itself in the form of oscillations in yaw. Source:padpilot.eu
Just as in the case of the roll mode, the Dutch roll dynamics are not entirely self-contained. Therefore,the yaw rate and the sideslip converge to quasi-statically varying values ∆βs and ∆rs which depend on ∆φ,and are found by solving
YβmV
−1
NβIz
NrIz
∆βs
∆rs
=
−g
V∆φ
0
(4)
You should check that the static residuals equal
∆βs =−1
ω2DR
(g
V
NrIz
)∆φ
∆rs =1
ω2DR
(g
V
NβIz
)∆φ
where ω2DR is the natural frequency of the Dutch roll mode, given by the determinant of the matrix on the
LHS of Eq. (4):
ω2DR =
1
Iz
(YβmV
Nr +Nβ
)
3
3 Spiral Mode
This is the slowest of modes, and its dynamics are based entirely off the static residuals found above. Thespiral dynamics are given by
∆φ = ∆p ≡ ∆ps
where ∆ps is also found from the static residuals ∆rs and ∆βs:
∆ps = −LrLp
∆rs −LβLp
∆βs
=⇒ ∆φ =1
ω2DR
1
IzzLp(LβNr − LrNβ)
g
V∆φ
The stability condition for the spiral mode is:
LβNr − LrNβLp
< 0
For an aircraft with a stable roll mode, Lp < 0. Therefore, for spiral stability,
(LβNr − LrNβ) > 0
The sign of Lr is difficult to determine precisely. It gets a positive contribution from the tail and a negativecontribution from the wing. On the other hand, Nr < 0 for the most part; thus, it is important to haveLβ < 0 for stabilizing the spiral mode.
Figure 3 shows the response of an aircraft if the spiral mode is unstable. An instability in the spiral modeleads to an uncommanded turn which becomes progressively steeper with time. Depending on Lβ and Nβ ,the aircraft may either turn steeply with little sideslip, or get into a relatively gentler turn but with a largesidelsip.
Figure 3: An unstable spiral mode puts the aircraft into an uncommanded turn.Source: people.rit.edu/pnveme/EMEM682n/DynamicStab/roots lateral.html
4
4 Lateral-Directional Modes at Higher Angles of Attack
(a) Low α
(b) High α
Figure 4: The L-D modes of the F/A-18 HARV for the entire range of pre-stall α.
Figure 4 shows the lateral-directional modes of the F/A-18 HARV with increasing angle of attack. Thesemodes were plotted without assuming α0 = 0, and α0 was allowed to increase to near-stall values.
The roll mode in this aircraft is slower than in larger cargo and passenger aircraft. In addition, the rollmode tends to slow down with the angle of attack. This causes it to merge with the spiral mode and producean oscillatory mode called the lateral phugoid. Despite its name, the lateral phugoid is very much unlike itslongitudinal counterpart: it is well damped and is relatively faster, with a time constant of about 10 s.
The Dutch roll mode is quite poorly damped, with a coefficient of damping less than 0.2. The Dutch rolldamping can be improved by increasing the size of the vertical tail, which increases Nr as well as Yβ .
5
Chapter 9
Introduction to Manoeuvres and Aircraft Control
Aditya Paranjape
The course so far has covered aircraft performance and stability, and we have seen a hierarchical connec-tion between them.
Perforamance problems assume that the aircraft can attain and fly at the angle of attack and the bankangle that arise as solutions for various flight conditions such as maximum speed, fastest sustained turn, etc.
The problem of attaining and maintaining the desired values of angle of attack and other flight parameterswas addressed partially as a “trim and stability” problem. Among other things, we also saw that theperformance of a stable airframe may not be as good as that of an unstable airframe (e.g., recall thediscussion on the lift produced by stable and unstable aircraft). This leads designers to opt for airframesthat are inherently unstable, and these airframes need to be stabilized artificially using control systems tomake the aircraft safe and flight-worthy.
1 Manoeuvres and Pseudo Steady State Dynamics
We looked at level acceleration in Chapter as an example of non-equilibrium manoeuvres. The examplehad one special feature: every state that was not specified could be calculated quite easily because of thelongitudinal nature of the manoeuvre. In general, it is not as easy to do so that for general manoeuvres.
As an example, consider a wind axis roll manoeuvre, also called an a velocity vector roll (VVR). Inits simplest form, only the desired roll rate is specified with β = 0, while all other variables need to bedetermined from the equations of motion. Determining the exact time history of the state variables “byhand” is almost impossible, since V , θ and φ are time-varying.
One way to analyse such manoeuvres is to work with a restricted set of equations of motion. For instance,if the duration of the manoeuvre is very short (less than 5 seconds), then one may assume that the flightspeed is almost constant. Furthermore, if the manoeuvre takes place at reasonable high flight speeds, thenthe ratio (g/V ) can be ignored. A cursory look at the equations of motion shows that ignoring g/V comeswith a beneficial collateral consequence: the dynamics of the Euler angles θ and φ get decoupled from theother equations of motion. As such, we don’t expect these variables to be in equilibrium and their decouplingopens up the possibility of simplifying the analysis considerably. Let α0 denote the initial angle of attack,before the manoeuvre is commenced, and let δ0e the initial elevator deflection. This gives:
α = q − tanβ(p cosα+ r sinα)− 1
2ρV 2S(CL − C0
L)
β = p sinα− r cosα+Yββ + Yδaδa + Yδrδr
mV
p =Iy − IzIx
qr +Lpp+ Lrr + Lββ + Lδaδa + Lδrδr
Ix(1)
q =Iz − IxIy
rp+Mα(α− α0) + (Mq +Mα)q +Mδe(δe − δ0e)
Iy
r =Ix − IyIz
pq +Nrr +Nββ +Nδrδr
Iz
1
The equations stated in in Eq. (1) are called pseudo steady state (PSS) equations and are useful instudying manoeuvres such as VVR. In fact, the VVR is an equilibrium solution of the above equation.Prescribe α = α0 and q = β = 0. Then, for equilibrium, we need:
p = r = 0 :
[δaδr
]= −
[Lδa Lδr0 Nδr
]−1 [Lpp+ Lrr
Nrr
]=
−1
LδaNδr
[Nδr(Lp + Lrr)− LδrNrr
LδaNrr
]β = 0 : p sinα− r cosα+
Yδaδa + YδrδrmV
= 0 (2)
q = 0 : δe = δ0e −Iz − IxIy
rp
It is amply clear that Eq. (2) can be solved quite easily by hand. Choose p as the free variable. Next, findδa and δr in terms of p and r. Then, substitute into β = 0 to get a single equation connecting p and r. Inone go, this yields r, δa and δr. The elevator deflection is then found from q = 0.
The stability analysis that was performed for the complete equations of motion can be performed for thePSS equations as well, with the equilibrium defined appropriately. If level flight were to be analysed usingPSS, it is clear that we would be able to successfully isolate the short period, roll and Dutch roll modes quiteaccurately.
It is perhaps not surprising that the equilibrium solutions of the complete set of equations from Chapter 7are also equilibrium solutions of the PSS equations. However, the PSS equations have additional equilibriumsolutions which are not equilibrium solutions of the complete set of equations. The VVR is an example ofsuch solutions. Taking this idea a step further, one can conceive other subsets of the original dynamics andthe equilibrium solutions of these subsets can be viewed as manoeuvres in their own right. We will returnto this idea again later in this chapter.
2 Basic Elements of a Control System
An automatic control system is used on aircraft for several reasons:
• Reduce pilot workload by automating routine manoeuvres such as wings-level flight and turns.
• Improve the handling qualities by augmenting the eigenvalues of the flight dynamics.
• Eliminate pilot-induced oscillations by filtering the pilot inputs appropriately.
• Help the aircraft achieve the desired dynamical response, including for the purpose of creating controlsystem testbeds.
Figure 1: A-380 cockpit and a schematic of the typical flight control system. Source: Wikipedia
Figure 1 shows the schematic representation of an advanced flight control system that is typical of moderncommercial and combat aircraft. It consists of four primary components:
2
• Avionics: control column, autopilot console and the diplays.
• Flight control computer.
• Sensors: sense flight parameters and health monitoring
• Actuators: deflect control surfaces and actuate the engines
The flight control computer is the central node in this system - it provides position and rate commandsto the actuators based on the desired flight profile provided by the pilot (either through manual controlyoke deflection or through the autopilot) and the information it derives from the sensors. The control law isimplemented on the flight control computer as a software and can be of one of two types:
1. Completely closed-loop: there is no direct connection between the pilot’s controls and the actuators.Rather, the pilot’s stick movements are sent to the flight control computer, and the computer sendsthe appropriate commands to the actuators on the control surfaces and in the engines. This controlphilosophy is used, for example, on all Airbus aircraft starting with the Airbus A320. It eliminates theneed for mechanical linkages between the control column and the actuators, but renders the aircraftvulnerable to fatal crashes in the event of a computer malfunction.
2. Combination of closed-loop and open-loop: the control column has a direct link to the actuator position(typically via a power amplifier). However, stabilizing actuator commands are added on top of thepilot input by the flight control system. Most Boeing aircraft employ this control design philosophy.While this design offers the pilot the possibility of taking complete manual control of the aircraft, itrequires additional mechanical components to be built into the aircraft, thereby making the systemmore complex.
3 Types of Flight Control Problems
Consider the general nonlinear system
x = f(x, u), y = h(x), z = g(x)
where y denotes the variables which are to be controlled and z denotes the measured output signals. Notethat x ∈ Rn, while z ∈ Rm for some m ≤ n; i.e., not all states are necessarily measured.
Depending on the nature of the problem and the sensed variables, control problems can be classified asfollows.
• State versus output feedback problems: when z = g(x) = x, all of the aircraft states are availableto the flight computer in real time, and it can use them for control. This is called full state feedback. Incontrast, when z ⊂ x, i.e., only some of the state variables are measured, the resulting control problemis called an output feedback problem. Output feedback problems are generally harder to solve, and theflight computer will usually make for the limited information by using algorithms which help it predictthe state variables that are not measured.
• Regulation versus tracking problems: a tracking problem is one where the output y(t) is requiredto track a reference signal r(t). In such cases, it suffices to ensure that the remaining state variablesdo not assume unreasonable values; their exact values are not important subject to the qualificationof boundedness within a reasonable range. On the other hand, if the control objective is to ensurethat y(t) tends to zero asymptotically, then the control problem is referred to as a regulation problem.Regulation problems are generally easier to solve than tracking problems, although it is usually possibleto convert a tracking problem into a regulation problem by appropriate coordinate transformations.
The design of control systems which fall into either of these categories forms a complete subject in itsown right, and will not be pursued here. Rather, we will look into specific problems that control designersencounter, and which have their roots in flight dynamics and in the limitations of flight systems. We willassume that linear control techniques are employed for control design.
3
4 Time Scale Separation
The flight dynamics of most aircraft can be decomposed naturally into simpler blocks, each of which has a dis-tinct time-scale of its own. This decomposition is akin to the modal decomposition, except that longitudinal-lateral decoupling is avoided when aircraft are designed for high-rate manoeuvres, including turns and rolls.The decomposition used for the purpose of control follows a task-based hierarchy:
1. Fast rotational dynamics: p, q, r
2. Acceleration of the point mass: α, β, µ; also engine state η
3. Velocity states: V∞, γ, χ
4. Position variables: x, y, z
The logic behind this decomposition is that each stage tries to ensure that the slower stage prior to it tracksthe desired reference behaviour, unless the reference signal is meant for that stage itself. For instance, thevelocity commands Vc, γc, χc are chosen so that the aircraft flies along the desired path; it is the job of α, β,µ and the throttle η to ensure that V tracks Vc and so on. The actual control surface deflections are meantto ensure that the angular rates track the commanded values.
The benefits of using time scale separation is that it allows for a physically-intuitive control design which,naturally, leads to relatively simpler formulae for designing each component. The drawback of using timescale-separation in the above form is that it involves the underlying modal structure and associated couplingbetween the dynamics. For example, α and q obey a single time scale (short period dynamics), and ditto forβ and r (Dutch roll). This problem can be dealt with by merging the acceleration states and the rotationaldynamics into a single component - you should recongnize this component as the PSS dynamics seen earlier.
5 Nonlinearities
The flight dynamics are inherently nonlinear, and the nonlinearities are quite significant in the way theyinfluence stability and control properties of the aircraft. A complete discussion is beyond the scope of thiscourse, but here is a short list of some major effects:
1. The linearization depends strongly on the flight speed and the angle of attack. Therefore, controllershave to be designed for several equilibrium conditions and pieced together using an interpolation logic.
2. The “basin of attraction” of a stabilized equilibrium is a function of the equilibrium itself. Thisnecessitates tight bounds on the tolerable disturbances in certain flight regimes, or a richer grid ofcontrol design points in those regimes.
3. Asymmetric aerodynamics at high angle of attack: the aerodynamics of a perfectly symmetric airframeflying under symmetric conditions become asymmetric at high angles of attack. This point necessitatesa coupled longitudinal-lateral design at moderate and high angles of attack.
4. Departure phenomena: departure phenomena such as deep stall and spin arise due to the existenceof multiple equilibria for a given value of the control input. The only way to avoid departures is toprevent the aircraft from flying in the regime in which it is susceptible to departure. While departurescan be prevented using active control, this is not the preferred option because the basins of attractionof individual controllers tend to be quite small in the departure-prone regions.
6 Sensor and Actuator Limitations
Consider the system x = Ax+ Bu. When we design a control law for such systems by writing u = Kx, weimplicitly assume that the control actuator can physically produce any signal given to it. This is usually nottrue owing to some physical constraints of the actuator:
4
• Time-delay: in transmitting a signal; reaction time of the actuator
• Internal dynamics of an actuator
• Hysteresis: when an actuator is given a cyclic signal, its output is asymmetric and lags the cycle, asshown in Fig. 2
The internal dynamics of an actuator are usually neither easy to model perfectly nor easy to measure.Therefore, feedback of actuator states is ruled out.
Figure 2: Hysteresis. Green: expected cyclic path; Red: actual
As an example, consider the first order system
x = −2x+ u
If we desire that the system respond faster, we can add u = −kx. The closed-loop eigenvalue, given by−2 − k, can be moved as far left as desired. Now, suppose that system has an actuator with dynamicsu = −3u + 3uc, and suppose that we use the same feedback law as above. The closed-loop characteristicsystem is given by [
xu
]=
[−2 1−3k −3
] [xu
]The closed-loop eigenvalues are
−5±√
1− 12k
2, which suggests that the real part cannot be made any more
negative that −2.5. In other words, the convergence rate of the closed- loop is directly bounded by theactuator dynamics. Does this suggest why one should refrain from using the thrust to control anything butthe phugoid and the spiral modes (and that too if no other option is available)?
A similar problem occurs when the aircraft sensors have a finite bandwidth. Consider again the system
x = −2x+ u
This time, suppose that the state x is measured by a sensor with its internal dynamics, so that
u = −kxs, xs = −3xs + 3x
This creates the same sitation as that of the previous slide; once again, the closed-loop eigenvalues cannothave a real part any more negative that −2.5.
The discussion in this section leads us to conclude that the actuators and sensors should typically befaster than the mode that they are trying to control/sense. In particular, the limitations from actuators andsensors also limit the extent of instability that an airframe can be designed for.
5
