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Transcript of Wight Dynamic
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Flight Dynamics and
Aircraft Performance
Lecture 1:
Introduction Longitudinal
Static Stability
G. Dimitriadis
University of Liege
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What is it about?
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Reference material
! Lecture Notes! Flight Dynamics Principles, M.V. Cook,
Arnold, 1997
! Fundamentals of Airplane FlightMechanics, David G. Hull, Berlin,Heidelberg : Springer-Verlag BerlinHeidelberg, 2007,
http://dx.doi.org/10.1007/978-3-540-46573-7
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Aircraft description
Pilot Flight ControlSystem 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 airplanes response while always keeping aneye on the task at hand.
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Control Surfaces
! Aircraft control is accomplished throughcontrol surfaces and power
!Ailerons! Elevators! Rudder! Throttle
! Control deflections were first developed bythe Wright brothers from watching birds
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Wright FlyerThe Flyer did not have
separate control surfaces.
The trailing edges of thewindtips could be bent by
a system of cables
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Modern control surfaces
Elevator
Rudder
Aileron
Elevon
(elevator+aileron)
Rudderon
(rudder+aileron)
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Other devices
Flaps
Spoilers
!Combinations of control surfaces and other devices: flaperons,spoilerons, decelerons (aileron and airbrake)
!Vectored thrust
Airbreak
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Mathematical Model
InputAileron
ElevatorRudder
Throttle
Aircraft
equations of
motion
Output
Displacement
VelocityAcceleration
Flight Condition
Atmospheric Condition
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Aircraft degrees of freedomSix degrees of
freedom:
3 displacements
x: horizontal motion
y: side motion
z: vertical motion
3 rotations
x: roll
y: pitch
z: yaw
x
zy
v !
v: resultant linear velocity, cg: centre of gravity
!: resultant angular velocity
cg
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Aircraft frames of reference
! There are many possible coordinate systems:! Inertial (immobile and far away)! Earth-fixed (rotates with the earths surface)! Vehicle carried vertical frame (fixed on aircraft cg,
vertical axis parallel to gravity)
!Air-trajectory (fixed on aircraft cg, parallel to thedirection of motion of the aircraft)
! Body-fixed (fixed on aircraft cg, parallel to ageometric datum line on the aircraft)
! Stability axes (fixed on aircraft cg, parallel to areference flight condition)
! Others
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Airplane geometry
cg
c
c
c /4
s = b /2
c /4
lT
lt
y
c(y)
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Airplane references (1)
! Standard mean chord (smc)! Mean aerodynamic chord (mac)! Wing area!Aspect Ratio
c = c2y( )
!s
s
" dy / c y( )!s
s
" dy
c = c y( )!s
s
" dy / dy!s
s
"
AR = b2/S
S=
bc
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Asymmetric Aircraft
Blohm und Voss 141
Ruttan Bumerang
Blohm und Voss 237
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Wing geometry example
!
R
!/3
Given this wingcalculate:
1.! The StandardMean Chord
2.! The MeanAerodynamicChord
3.! The wing area4.! The Aspect Ratio5.! The Mean
AerodynamicCentre
R=0.8, !=30o
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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
tailplaneV
T=STlT
Sc
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Airplane references (3)
! Fin moment arm (lF)! Fin volume ratio
c /4
c /4cg
lF
lf
VF=
SFlF
Sc
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Aerodynamic Reference Centres
! Centre of pressure (cp): The point at whichthe resultant aerodynamic forceFacts. Thereis no aerodynamic moment around the cp.
! Half-chord: The point at which theaerodynamic force due to camber,Fc, acts
! Quarter-chord (or aerodynamic centre): Thepoint at which the aerodynamic force due toangle of attack,F
a
, acts. The aerodynamicmoment around the quarter-chord,M0, isconstant with angle of attack
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Airfoil with centres
ac
Dc
cp
D Da
LcL La
FcF Fa
L
DM0
c /4
c /2
hnc
c
Camber line
V0
By placing all of
the lift and drag
on theaerodynamic
centre we movethe lift and drag
due to camber
from the half-chord to the
quarter chord.This is
balanced by the
momentM0
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Static Stability
! Most aircraft (apart from high performancefighters) are statically stable
! Static stability implies:!All the forces and moments around the aircrafts
cg at a fixed flight condition and attitude arebalanced
!After any small perturbation in flight attitude theaircraft returns to its equilibrium position
! The equilibrium position is usually called thetrim position and is adjusted using the trimtabs
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Pitching moment equation
cg
mg
c
ac
LT
M0MT
Lw
lT hc
h0c
h denotes the cgposition
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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 !LTlT +MT = 0
(nose up moment is taken to be positive)
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Degree of stability
Trim point
1
2
3
4
1: Very stable
2: Stable
3: Neutrally stable
4: Unstable
Cm
!
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Pitching moment stability (1)
! The pitching moment equation can bewritten as
! Where
!And the tailplane is assumed to besymmetric so thatMT=0
Cm=C
m0+C
Lw (h ! h
0)!C
LTVT=
0
Cm
=
M
1
2
!V0
2Sc
, CL
w
=
Lw
1
2
!V0
2S
, CL
T
=
LT
1
2
!V0
2ST
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Pitching moment stability (2)
! For static stability or,approximately,
! Then! SinceM0 is a constant.!
Unfortunately, the derivative of the taillift with respect to the wing lift is
unknown
dCm/dC
L< 0
dCm/dC
Lw< 0
dCm/dC
Lw
= (h ! h0)!VTdCLT
/dCL
w
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Wing-tail flow geometry
"V0
V0"
#
"T$T
$
%$
Tailplane
Elevator
Trim tab
Wing
! The downwash effect of the wing deflects thefree stream flow seen by the tailplane by anangle ".
! Total angle of attack of tail: "T="-#+$T
! The total lift on the tailplane is given by:C
LT
=!0+ a
1!
T+ a
2" + a
3#"
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Thin Airfoil Theory
! Thin Airfoil Theory is the main tool formodelling incompressible aerodynamic
forces on 2D wing sections.! It shows that cl=2&(A0+A1/2), where
! andzis the tails camber line
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Aerodynamics example
! Prove that for a tailplaneC
LT
=!0+ a
1!
T+ a
2" + a
3#"
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Pitching moment stability (3)
! For small disturbances the downwashangle is a linear function of wing
incidence ":
! Wing lift is also a linear function of":! So that
!= d!
d""
CL
w
= a! or ! =CL
w
/a
!T=
CLw
a1"
d#
d!
%&
()+*T
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Pitching moment stability (4)
! The tail lift coefficient can the be writtenas
! The pitching moment equation becomes!And the derivative of the pitching
moment coefficient becomes
! since "T is a constant
CL
T
=CL
w
a1
a1!
d"
d#
%&
()+ a1*T + a2* + a3+*
dC
m
dCL
w
= (h ! h0)!VTa1
a1! d"d#$
%& '
()+ a2 d*
dCL
w
+ a3
d+*
dCL
w
%&&
())
Cm=C
m0+C
Lw(h ! h0) !VT CLw
a1
a1!
d"
d#
$%&
'()+ a2* + a3+* + a1*T
%&
()
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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 andtrim tab angle for a chosen angle of
attack
! Elevator and trim tab areinterchangeable (up to a point)
! =1
VTa2
Cm0
+ CL
w
(h " h0)( )"CLw
a
a1
a2
$%
'( 1"
d)
d*
#$%
&'("
a3
a2
+! "a1
a2
!T
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Controls fixed stability
!Assume that the aircraft has reachedtrim position and the controls are locked
! What will happen if there is a smallperturbation to the aircrafts position(due to a gust, say)?
! The pitching moment equation becomesdC
m
dCL
w
= (h ! h0)!VTa1
a1!
d"
d#
%&
()
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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 neutralpoint (hn) the aircraft will by definition bestable
! Too much stability can be a bad thing!
Kn= !
dCm
dCLw
Kn = hn ! h, so that hn = h0 +VTa1
a 1!d"
d#
%& ()
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Stability margin (2)
! Certification authorities specify thatat 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
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Elevator angle to trim
! Setting the tab angle to zero (#"=0), it can beshown that the elevator angle to trimcharacteristic is given by
! It is therefore proportional to the controls fixedstability margin.
! Therefore, the stability margin can beobtained using measurements of the elevatorangle to trim at various flight conditions
d!dC
Lw
="
1VTa2
(hn " h) = " 1VTa2
Kn
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Controls Fixed Example
! Calculate the controls fixed stability marginand neutral point
! Is the aircraft stable?These values of
elevator angle to trim
were obtained from aHandley Page
Jetstream aircraft
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Controls Free Stability
! Pilots dont want to hold the controlsthroughout the flight.
! The trim tab can be adjusted such that, if theelevator is allowed to float freely, it will at anangle corresponding to the desired trimcondition.
! This is sometimes called a hands-off trimcondition.
! Therefore the pilot can take his hands off theelevator control and the aircraft will remain intrim.
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Elevator Hinge Moment (1)
! The pitching moment equation is (earlierslides)
! The elevator angle, ", is unknown andmust be eliminated from the equation
! Consider the elevator hinge moment
Cm=C
m0+C
Lw
(h ! h0) !V
TCL
w
a1
a1!
d"
d#
$
%&
'
()+ a2* + a3+* + a1*T
%&
()
V0!
$
!T"T
"
#"
Tailplane
Elevator
Trim tab
HElevator hinge
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Elevator Hinge Moment (2)! Since the elevator is free to rotate, the
elevator hinge moment must be equal to zero.
! Assuming small displacement, the elevatorhinge moment is a linear function of total
angle of attack, elevator angle and trim tabangle, exactly like the lift. Therefore:
! Where b1, b2 and b3 are known constants! Substituting for!Tand solving for the elevatorangle gives
CH= b
1!
T+ b
2" + b
3#"
! =1
b2
CH
"CLw
a
b1
b2
1"d#
d$
&'
)*"
b3
b2
+! "b1
b2
!T
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Controls Free Stability Margin
! This is an expression for" that can besubstituted into the pitching momentequation.
! Differentiating the latter with respect to winglift coefficient gives
!Define the Controls Free Stability Margin,K'n,such that
! With h'n being the controls free neutral point
dCm
dCL
w
= h ! h0( ) !VT
a1
a1!
d"
d#
$%&
'()1!
a2b1
a1b2
%&
()
!Kn= "
dCm
dCLw
= !hn" h
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Controls Free Neutral Point
! The controls free neutral point is then
! Using the expression for the controlsfixed neutral point gives
!hn= h
0+V
T
a1
a1"
d#
d$
%&'
()*1"
a2b1
a1b2
&'
)*
hn= h
0+V
T
a1
a1!
d"
d#
%&
()
!
hn=
hn "VTa2b1
ab21"
d#
d$
&' )*
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Discussion
! As with the controls fixed stability margin, thecontrols free stability margin is positive whenthe aircraft is stable.
! Similarly, the centre of gravity position mustbe ahead of the controls free neutral point ifthe aircraft is to be stable.
! Usually, the constants of the elevator and tabare such that h'n>hn.
! An aircraft that is stable controls fixed willusually be also stable controls free
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Hands-off trim positions
! Assume that the aircraft is trimmed at ahands-off position (elevator is free)
! If the pilot changes elevator angle and thenreleases the control stick, the aircraft willreturn to the old trim position.
! In order to adopt a new hands off trim positionthe pilot must first move the elevator to thedesired angle and then adjust the trim tab.
! The correct tab angle is the one that requireszero control force. The pilot can then take hishands off the control stick.
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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 margincan be estimated by measuring the tab angleto trim at various lift coefficients
Cm=C
m0+C
Lw(h ! h0)!VT CLw
a1
a1!
d"
d#
$%&
'()1!
a2b1
a1b2
$
%&
'
()+ a3*+ 1!
a2b3
a3b2
$
%&
'
()+ a1+T 1!
a2b1
a1b2
$
%&
'
()
%&
()= 0
d!"
dCLw
=
#1
a3VT 1#a2b3
a3b2
$
%&
'
()
( *hn# h) =
#1
a3VT 1#a2b3
a3b2
$
%&
'
()
*Kn
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Control force to trim! The control force to trim is the parameter
more relevant to the pilot.
! Consider an aircraft at a hands-off trimposition. If the pilot moves the stick to
assume a new position
! This time #" is constant but CH is non-zerosince the pilot is applying a force to thecontrol stick
! Differentiate with respect to lift to obtain
Cm=C
m0+C
Lw
(h ! h0)!VT CLw
a1
a1!
d"
d#
$%&
'() 1!
a2b1
a1b2
$
%&
'
()+ a3*+ 1!
a2b3
a3b2
$
%&
'
()+ a1+T 1!
a2b1
a1b2
$
%&
'
()+
a2
b2
CH
%&
()= 0
dCH
dCLw
=
!1
VT
a2
b2
( "hn! h) =
!1
VT
a2
b2
"Kn
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Measuring Controls Free Stability
! Therefore, the measurement of the controlsfree stability margin and controls free neutralpoint can be performed by measuring the
elevator hinge moment required to trimaround a given trim position
! The elevator hinge moment can be obtainedfrom the mechanical gearing between thecontrol stick and the elevator,g", as well as
the control force applied,F".
F!= g
!H=
1
2"V
2S!c!g!CH
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Controls free example
! Calculate the controls free stability marginand neutral point
! Is the aircraft stable?These values of hinge
moment coefficient to
trim were obtained froma Handley Page
Jetstream aircraft
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Summary of Longitudinal
Stability
cg
c
hc
h0c
ac
hnc
!hnc
!Knc
Knc
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Longitudinal stability when
cg ahead of ac
ac cg
Lw
LT
Is this configuration stable, neutrally stable or unstable?
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Answer
! All the equations that have been derived up to now stillhold. The only difference is in the signs ofh and "T.
! Notice that this configuration is inherently stable. Anincrease 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).
! The angle "Tdoes not appear in the equation for theneutral point.
! Therefore, the neutral point is where its always been,behind the aerodynamic centre. Therefore, the stabilitymargin is enormous. The aircraft is too stable!
Kn= h
n! h, so that h
n= h0 +VT
a1
a1!
d"
d#
%&
()