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Flight Dynamics and Aircraft Performance Lecture 1: Introduction – Longitudinal Static Stability G. Dimitriadis University of Liege
Flight Dynamics and Aircraft
Performance
Lecture 1:
Introduction – Longitudinal
Static Stability
G. Dimitriadis
University of Liege
What is it about?
Introduction
• The study of the mechanics and dynamics of flight is the means by which :
– We can design an airplane to accomplish efficiently a specific task
– We can make the task of the pilot easier by ensuring good handling qualities
– We can avoid unwanted or unexpected phenomena that can be encountered in flight
Reference material
• Lecture Notes
• Flight Dynamics Principles, M.V. Cook, Arnold, 1997
• Fundamentals of Airplane Flight Mechanics, David G. Hull, Berlin, Heidelberg : Springer-Verlag Berlin Heidelberg, 2007, http://dx.doi.org/10.1007/978-3-540-46573-7
Aircraft description
Pilot Flight Control
System Airplane Response Task
The pilot has direct control only of the Flight Control
System. However, he can tailor his inputs to the FCS by
observing the airplane’s response while always keeping an eye on the task at hand.
Control Surfaces
• Aircraft control is accomplished through
control surfaces and power
– Ailerons
– Elevators
– Rudder
– Throttle
• Control deflections were first developed by
the Wright brothers from watching birds
Wright Flyer The Flyer did not have
separate control surfaces.
The trailing edges of the windtips could be bent by
a system of cables
Modern control surfaces
Elevator
Rudder
Aileron
Elevon
(elevator+aileron)
Rudderon
(rudder+aileron)
Other devices
Flaps
Spoilers
•Combinations of control surfaces and other devices: flaperons,
spoilerons, decelerons (aileron and airbrake)
•Vectored thrust
Airbreak
Mathematical Model
Input
Aileron
Elevator Rudder
Throttle
Aircraft
equations of
motion
Output
Displacement
Velocity Acceleration
Flight Condition
Atmospheric Condition
Aircraft degrees of freedom Six degrees of
freedom:
3 displacements
x: horizontal motion
y: side motion
z: vertical motion
3 rotations
x: roll
y: pitch
z: yaw
x
z y
v
v: resultant linear velocity, cg: centre of gravity
: resultant angular velocity
cg
Aircraft frames of reference
• There are many possible coordinate systems: – Inertial (immobile and far away)
– Earth-fixed (rotates with the earth’s surface)
– Vehicle carried vertical frame (fixed on aircraft cg, vertical axis parallel to gravity)
– Air-trajectory (fixed on aircraft cg, parallel to the direction of motion of the aircraft)
– Body-fixed (fixed on aircraft cg, parallel to a geometric datum line on the aircraft)
– Stability axes (fixed on aircraft cg, parallel to a reference flight condition)
– Others
Airplane geometry
cg
c
c
c /4
s = b /2
c /4
lTlt
y
c(y)
Airplane references (1)
• Standard mean chord (smc)
• Mean aerodynamic chord (mac)
• Wing area
• Aspect Ratio
c = c 2 y( )s
s
dy / c y( )s
s
dy
c = c y( )s
s
dy / dys
s
AR = b2 /S
S = bc
Asymmetric Aircraft
Blohm und Voss 141
Ruttan Bumerang
Blohm und Voss 237
Wing geometry example
R
/3
Given this wing calculate:
1. The Standard Mean Chord
2. The Mean
Aerodynamic Chord
3. The wing area
4. The Aspect Ratio
5. The Mean
Aerodynamic Centre
R=0.8, =30o
Airplane references (2)
• Centre of gravity (cg)
• Tailplane area (ST)
• Tail moment arm (lT)
• Tail volume ratio: A measure of the
aerodynamic effectiveness of the
tailplane V T =
ST lTSc
Airplane references (3)
• Fin moment arm (lF)
• Fin volume ratio
c /4 c /4cg
lF
lf
V F =SF lF
Sc
Aerodynamic Reference Centres
• Centre of pressure (cp): The point at which the resultant aerodynamic force F acts. There is no aerodynamic moment around the cp.
• Half-chord: The point at which the aerodynamic force due to camber, Fc, acts
• Quarter-chord (or aerodynamic centre): The point at which the aerodynamic force due to angle of attack, Fa, acts. The aerodynamic moment around the quarter-chord, M0, is constant with angle of attack
Airfoil with centres
ac
Dc
cp
D Da
Lc L La
Fc F Fa
L
D M0
c /4c /2 hnc
c
Camber line
V0
By placing all of
the lift and drag
on the aerodynamic
centre we move
the lift and drag
due to camber
from the half-chord to the
quarter chord.
This is
balanced by the
moment M0
Static Stability
• Most aircraft (apart from high performance fighters) are statically stable
• Static stability implies:
– All the forces and moments around the aircraft’s cg at a fixed flight condition and attitude are balanced
– After any small perturbation in flight attitude the aircraft returns to its equilibrium position
• The equilibrium position is usually called the trim position and is adjusted using the trim tabs
Pitching moment equation
cg
mg
c
ac
LT
M0 MT
Lw
lT hc
h0c
h denotes the cg
position
Equilibrium equations
• Steady level flight is assumed
• Thrust balances drag and they both
pass by the cg
• Force equilibrium:
• Pitching moment around cg equilibrium:
Lw + LT mg = 0
M = M0 + Lw (h h0)c LT lT + MT = 0
(nose up moment is taken to be positive)
Stable or Unstable?
• An equilibrium point can be stable, unstable or neutrally stable
• A stable equilibrium point is characterized by
• A more general condition (takes into account compressibility effects) is
M = 0 and dM
d< 0
M = 0 and dM
dL< 0 or Cm = 0 and
dCm
dCL
< 0
Degree of stability
Trim point
1
2
3
4
1: Very stable
2: Stable
3: Neutrally stable
4: Unstable
Cm
Pitching moment stability (1)
• The pitching moment equation can be
written as
• Where
• And the tailplane is assumed to be
symmetric so that MT=0
Cm = Cm0+ CLw
(h h0) CLTV T = 0
Cm =M
12
V02Sc
, CLw=
Lw
12
V02S
, CLT=
LT
12
V02ST
Pitching moment stability (2)
• For static stability or,
approximately,
• Then
• Since M0 is a constant.
• Unfortunately, the derivative of the tail
lift with respect to the wing lift is
unknown
dCm /dCL < 0dCm /dCLw < 0
dCm /dCLw= (h h0) V TdCLT
/dCLw
Wing-tail flow geometry
V0 V0
T T
Tailplane
Elevator
Trim tab
Wing
• The downwash effect of the wing deflects the free stream flow seen by the tailplane by an angle .
• Total angle of attack of tail: T= - + T
• The total lift on the tailplane is given by:
CLT= 0 + a1 T + a2 + a3
Thin Airfoil Theory
• Thin Airfoil Theory is the main tool for
modelling incompressible aerodynamic
forces on 2D wing sections.
• It shows that cl=2 (A0+A1/2), where
• and z is the tail’s camber line
Aerodynamics example
• Prove that for a tailplane
CLT= 0 + a1 T + a2 + a3
Pitching moment stability (3)
• For small disturbances the downwash
angle is a linear function of wing
incidence :
• Wing lift is also a linear function of :
• So that
=d
d
CLw= a or = CLw
/a
T =CLw
a1
d
d
+ T
Pitching moment stability (4)
• The tail lift coefficient can the be written as
• The pitching moment equation becomes
• And the derivative of the pitching moment coefficient becomes
• since T is a constant
CLT= CLw
a1a1
d
d
+ a1 T + a2 + a3
dCm
dCLw
= (h h0) V Ta1a1
d
d
+ a2
d
dCLw
+ a3d
dCLw
Cm = Cm0+ CLw
(h h0) V T CLw
a1a1
d
d
+ a2 + a3 + a1 T
Elevator angle to trim
• In order to trim the aircraft, Cm=0
• The elevator angle required to achieve
this is given by
• i.e. it depends only on lift coefficient and
trim tab angle for a chosen angle of
attack
• Elevator and trim tab are
interchangeable (up to a point)
=1
V T a2Cm0 + CLw
(h h0)( )CLw
a
a1a2
1
d
d
a3a2
a1a2
T
Controls fixed stability
• Assume that the aircraft has reached
trim position and the controls are locked
• What will happen if there is a small
perturbation to the aircraft’s position
(due to a gust, say)?
• The pitching moment equation becomes
dCm
dCLw
= (h h0) V Ta1a1
d
d
Stability margin • Define the controls fixed stability margin as
• And the controls fixed neutral point as
• A stable aircraft has positive stability margin.
The more positive, the more stable.
• If the cg position (h) is ahead of the neutral
point (hn) the aircraft will by definition be
stable
• Too much stability can be a bad thing!
Kn =dCm
dCLw
Kn = hn h, so that hn = h0 + V Ta1
a1
d
d
Stability margin (2)
• Certification authorities specify that
at all times
• Of course, the stability margin can change:
– If fuel is used up
– If payload is released:
• Bombs
• Missiles
• External fuel tanks
• Paratroopers
• Anything else you can dump from a plane
Kn 0.05c
Elevator angle to trim
• Setting the tab angle to zero ( =0), it can be shown that the elevator angle to trim characteristic is given by
• It is therefore proportional to the controls fixed stability margin.
• Therefore, the stability margin can be obtained using measurements of the elevator angle to trim at various flight conditions
d
dCLw
=1
V T a2(hn h) =
1
V T a2Kn
Controls Fixed Example
• Calculate the controls fixed stability margin and neutral point
• Is the aircraft stable?
These values of
elevator angle to trim
were obtained from a Handley Page
Jetstream aircraft
Controls Free Stability
• Pilots don’t want to hold the controls throughout the flight.
• The trim tab can be adjusted such that, if the elevator is allowed to float freely, it will at an angle corresponding to the desired trim condition.
• This is sometimes called a hands-off trim condition.
• Therefore the pilot can take his hands off the elevator control and the aircraft will remain in trim.
Elevator Hinge Moment (1)
• The pitching moment equation is (earlier slides)
• The elevator angle, , is unknown and must be eliminated from the equation
• Consider the elevator hinge moment
Cm = Cm0+ CLw
(h h0) V T CLw
a1a1
d
d
+ a2 + a3 + a1 T
V0
T T
Tailplane
Elevator
Trim tab
H Elevator hinge
Elevator Hinge Moment (2) • Since the elevator is free to rotate, the
elevator hinge moment must be equal to zero.
• Assuming small displacement, the elevator hinge moment is a linear function of total angle of attack, elevator angle and trim tab angle, exactly like the lift. Therefore:
• Where b1, b2 and b3 are known constants
• Substituting for T and solving for the elevator angle gives
CH = b1 T + b2 + b3
=1
b2CH
CLw
a
b1b21
d
d
b3b2
b1b2
T
Controls Free Stability Margin
• This is an expression for that can be substituted into the pitching moment equation.
• Differentiating the latter with respect to wing lift coefficient gives
• Define the Controls Free Stability Margin, K'n, such that
• With h'n being the controls free neutral point
dCm
dCLw
= h h0( ) V Ta1a1
d
d
1
a2b1a1b2
Kn =dCm
dCLw
= hn h
Controls Free Neutral Point
• The controls free neutral point is then
• Using the expression for the controls
fixed neutral point gives
h n = h0 + V Ta1a1
d
d
1
a2b1a1b2
hn = h0 + V Ta1a1
d
d
h n = hn V Ta2b1ab2
1d
d
Discussion
• As with the controls fixed stability margin, the controls free stability margin is positive when the aircraft is stable.
• Similarly, the centre of gravity position must be ahead of the controls free neutral point if the aircraft is to be stable.
• Usually, the constants of the elevator and tab are such that h'n>hn.
• An aircraft that is stable controls fixed will usually be also stable controls free
Hands-off trim positions
• Assume that the aircraft is trimmed at a hands-off position (elevator is free)
• If the pilot changes elevator angle and then releases the control stick, the aircraft will return to the old trim position.
• In order to adopt a new hands off trim position the pilot must first move the elevator to the desired angle and then adjust the trim tab.
• The correct tab angle is the one that requires zero control force. The pilot can then take his hands off the control stick.
Tab angle to trim
• At a hands off trim position
• Differentiating with respect to wing lift coefficient but allowing to vary gives
• Therefore, the controls free stability margin can be estimated by measuring the tab angle to trim at various lift coefficients
Cm = Cm0+ CLw
(h h0) V T CLw
a1a1
d
d
1
a2b1a1b2
+ a3 1
a2b3a3b2
+ a1 T 1
a2b1a1b2
= 0
d
dCLw
=1
a3V T 1a2b3a3b2
( h n h) =1
a3V T 1a2b3a3b2
K n
Control force to trim • The control force to trim is the parameter
more relevant to the pilot.
• Consider an aircraft at a hands-off trim position. If the pilot moves the stick to assume a new position
• This time is constant but CH is non-zero since the pilot is applying a force to the control stick
• Differentiate with respect to lift to obtain
Cm = Cm0+ CLw
(h h0) V T CLw
a1a1
d
d
1
a2b1a1b2
+ a3 1
a2b3a3b2
+ a1 T 1
a2b1a1b2
+
a2b2
CH
= 0
dCH
dCLw
=1
V Ta2b2
( h n h) =1
V Ta2b2
K n
Measuring Controls Free Stability
• Therefore, the measurement of the controls free stability margin and controls free neutral point can be performed by measuring the elevator hinge moment required to trim around a given trim position
• The elevator hinge moment can be obtained from the mechanical gearing between the control stick and the elevator, g , as well as the control force applied, F .
F = g H =1
2V 2S c g CH
Controls free example
• Calculate the controls free stability margin and neutral point
• Is the aircraft stable?
These values of hinge
moment coefficient to
trim were obtained from a Handley Page
Jetstream aircraft
Summary of Longitudinal
Stability
cg
c
hc
h0c
ac
hnc
h nc
K nc
Knc
Longitudinal stability when cg ahead of ac
ac cg
Lw
LT
All the equations that have been derived up to now still
hold. The only difference is in the signs of h and T.
Notice that this configuration is inherently stable. An
increase in angle of attack causes an increase in lift (i.e.
nose down moment) and a decrees in tail downforce (again a nose down moment).
