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AE 430 - Stability and Control ofAerospace Vehicles

Static/Dynamic Stability

Longitudinal StaticStability

We begin with the concept of Equilibrium (Trim).Equilibrium is a state of an object when it is at rest or insteady uniform motion, (i.e., with constant linear andangular momenta).

The resultant of all forces and moment about the CG mustboth be equal to zero.

Stability is defined as the ability of an aircraft to return to agiven equilibrium state after a disturbance (it is a property

of the equilibrium state) STATICALLY STABLE when

if it is disturbed from its equilibrium state by a smalldisplacement, then

the set of forces and moments so caused initially tend toreturn the aircraft to its original state

Static Stability

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Trimmed flight when all the forces and moments are balanced(trust = drag; lift = weight; pitching moment = 0; yawing moment = 0rolling moment = 0)

The steady flight condition may involve a steady acceleration e.g. acorrectly banked turn, or a steady dive or climb.

Pitch trim would be accomplished by deflecting the horizontalstabilizer, the elevator, or the elevator trim tab.

Trimmed state IS NOT NECESSARILY A STABLE STATE i.e. all the forces and moments may be balanced, but as soon

as the state is perturbed the aircraft departs from equilibrium.

Forces 0; Moments 0= =

0 for trimGM =

Trimmed Flight

(or steady unaccelerated flight)

Types of Stability

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Static Stability

Static stability of a body is anStatic stability of a body is an initialinitialtendency of that body totendency of that body to return toreturn to

its equilibrium stateits equilibrium stateafter aafter a

disturbance.disturbance.

Static longitudinal instabilityStatic longitudinal instabilityIn this case there is no tendency tono tendency to

return to equilibriumreturn to equilibrium

Any disturbance from equilibrium

leads to a larger disturbance, themotion is said to be divergent E

Neutral static stability is theNeutral static stability is the

boundary between stability andboundary between stability andinstability,instability, there is still no tendencyno tendency

to return to equilibriumto return to equilibrium, the motion is

therefore not stable

But, the motion does not diverge E

Energy is being dissipated

Positive damping

Energy is added to the systemNegative damping

Artificial damping is needed Stability Augmentation System SAS

Static Stability

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Dynamic Stability

DYNAMIC STABILITY characterizes the time history of motionDYNAMIC STABILITY characterizes the time history of motion

after a disturbance from equilibriumafter a disturbance from equilibriumAn aircraft is said to be dynamically stable if, after a

disturbance, it eventuallyreturns to its equilibrium state

and remains there

ABSOLUTE dynamic stability is not concerned withhow long this return takes

RELATIVE dynamic stability examines how long ittakes and what the behavior of that return motion is

To be dynamically stable, a system must first be staticallystable

A system can be dynamically unstable and be

statically stable -- but not vice versa

Dynamic Stability

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Dynamic Stability

Pilot-InducedOscillation

PIOs occur when a pilot over-controls an aircraft and a sustained oscillation results

Pilot-induced oscillations occur when the pilot of an aircraft inadvertently commands an often increasingseries of corrections in opposite directions, each an attempt to correct for the previous overcorrection withan overcorrection in the opposite direction. The physics of flight make such oscillations more probable forpilots than for automobile drivers. An attempt to cause the aircraft to climb, say by applying up elevator willalso result in a reduction in airspeed.

Another factor is the response rate of flight instruments in comparison to the response rate of the aircraftitself. An increase in power will not result in an immediate increase in airspeed. An increase in climb ratewill not show up immediately on the vertical speed indicator.

A pilot aiming for a 500 foot per minute descent, for example, may find himself descending too rapidly. Hebegins to apply up elevator until the vertical speed indicator shows 500 feet per minute. However, becausethe vertical speed indicator lags the actual vertical speed, he is actually descending at much less than 500feet per minute. He then begins applying down elevator until the vertical speed indicator reads 500 feet perminute, starting the cycle over. It's harder than it might seem to stabilize the vertical speed because theairspeed also constantly changes.

The most dangerous pilot-induced oscillations can occur during landing. A bit too much up elevator duringthe flare can result in the plane getting dangerously slow and threatening to stall. A natural reaction to thisis to push the nose down harder than one pulled it up, but then the pilot finds himself staring at the ground.An even larger amount of up elevator starts the cycle over again.

http://www.dfrc.nasa.gov/Gallery/Movie/F-8DFBW/HTML/EM-0044-01.html

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Statically Stable Response

noseup

nosedown

Balanced

positive pitchstiffness

(restoring moment)

Equilibriumpoint

Other necessarycondition to trim at

positive angle of attach,

m m Lm

L

dC dC dCC

d dC d

= = 0

m

L

dC

dC

Longitudinal Static Stability

Longitudinal static stability componentsLongitudinal static stability moments asa function of angle of attack. The curveis a composite of all the moment curvescaused by the different components ofthe airplane, (the wing, fuselage, tail,thrust, etc).

nose up (+)

nose down (-)

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Stable, neutral, and unstable static stability

DC-9. Note the contributions from the variouscomponents and the highly nonlinear post-stallcharacteristics

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There are different

degrees of stability

Some aircraft tend toreturn to equilibrium

faster

An aircraft can bestable at lower angles

of attack but may beunstable at higher

angles of attack

Wing Contribution

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WingContribution

( ) ( )

( ) ( )

cos sin

sin cos

w

w

w

cg

cg w w w cg ac w w w cg ac

w w w cg w w w cg ac

Moments M

M L i x x D i x x

L i z D i z M

=

= +

+ +

( ) ( )

( ) ( )

212

Dividing for :

cos sin

sin cos

w

w

cg cgac ac

m L w w D w ww wcg

cg cg

L w w D w w mw w ac

V Sc

x xx xC C i C i

c c c c

z zC i C i C

c c

= +

+ +

( ) ( )

( )

( )

( )0

cos 1; sin ;

negligible

;

w w

ww w w w

w w w w w w

L Dw w

cg cgacm L L w w mw wcg ac

cg

L w ww

cg cgac acm m L m m L L wwcg ac cg ac w

i i i

C C

x zxC C C i C

c c c

zC i

c

x xx xC C C C C C C

c c c c

= + +

= + = + +

0wL L L ww wC C C

= + Lift Coefficient

Well designed aircraft

Normal flight operation

Wing Contribution

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Nonlinear contributions

( )w w

cg cgac acm m L D w ww wcg ac

x xx xC C C C i

c c c c

= + +

2w

w w

L

D d

CC C

eAR= +

( ) cg

Lw w w Dw

zC i C

c

+

0wL L L ww wC C C

= + 2

w

Wind drag turn

Wing Contribution

0 0

0

w

w

cg acm m Lacw w

m m m wcg w wcg ac

m Lw w

x xC C C

c cC C C

x xC C

c c

= +

= +

=

To have a wing alone statically stable 0m wC < cg acx x

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Wing Contribution

Positive camber give nose-down pitching moment

Negatively cambered airfoil gives nose-up pitching moment and

cancels nose-down moment caused by lift and weight vectors

For straight-winged, tailless airplane, negative camber satisfies

conditions for stable, balanced flight

Not in general use

Dynamic characteristics poor

Drag and Clmax poor

Swept back wing with twisted tips

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Conventional and forward tail

arrangement

Tailless Aircraft

One example of

a tailless aircraftthat trims using a

positive Cm0airfoil section: theAeroVironment

Pathfinder, solar-powered aircraft

on a flight to over50,000 ft (15.2

km).

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Example # 1

For a given wing-body combination, the aerodynamic center lies 0.03

chord length ahead of the center of gravity. The moment coefficient

about the center of gravity is 0.0050, and the lift coefficient is 0.50.

Calculate the moment coefficient about the aerodynamic center.

, ,

, ,

,

,

,

( )

( )

0.005 0.5(0.03) 0.01

cg w ac w

ac w cg w

ac w

cg ac wM M Lw

cg ac w

M M Lw

M

x xC C C

c c

x xC C C

c cC

= +

=

= =

Example # 2

Consider a model of a wing-body shape mounted in a wind tunnel. The

flow conditions in the test section are standard sea-level propertieswith a velocity of 100 m/s. The wing area and chord are 1.5 m2 and

0.45 m, respectively.

Using the wind tunnel force and moment-measuring balance, the

moment about the center of gravity when the lift is zero is found to be

-12.4 N m.

When the model is pitched to another angle of attack, the lift and

moment about the center of gravity are measured to be 3675 N and20.67 N m, respectively.

Calculate the value of the moment coefficient about the aerodynamiccenter and the location of the aerodynamic center.

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Example # 2 Contd

,

, ,

2 2 2

,

1 10.225*100 6125 /

2 2

12.40.003

6125*1.5*0.45

0.003 at zero lift

cg w

cg w ac w

cg w

m

m m

q V N m

MC

q Sc

C C

= = =

= = =

= =

Example # 2 Contd

,

, ,

, ,

,

,

,

,

36750.4

6125*1.5

20.670.005

6125*1.5*0.45

( )

0.005 ( 0.003)

0.4

0.02

cg w

cg w ac w

cg w ac w

Lw

cg w

m

cg ac wm m Lw

m mcg ac w

Lw

cg ac w

LC

q S

MC

q Sc

x xC C C

c c

C Cx x

c c C

x x

c c

= = =

= = =

= +

= =

=