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7/30/2019 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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! 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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! 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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! 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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! 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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! 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#

%&

()

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