STABILITY CRITERIA AND CHANGING STABILITY Pilot Induced ...

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STABILITY CRITERIAAND

CHANGING STABILITY

Pilot Induced Oscillations (PIO) Videos

F-4B Sageburner PIO (May 18, 1961): – Pilot J. L. Felson attempted high-speed, low-altitude record run– Pitch damper failure led to severe PIO– Destroyed airplane and killed pilot

NASA conducted flight research with F-8C (1972 – 1985) – 1st digital fly-by-wire flight control system w/o mechanical back up – Smaller, more reliable– In military aircraft, much less vulnerable to battle damage – Aircraft much more responsive to pilot control inputs– Result: More efficient, safer aircraft with improved performance and design

Problem

A conventional aircraft is in trimmed, level unaccelerated flight. The wing is generating 40,000 lb of lift and has a moment around the aerodynamic center of -20,000 ft-lb. The aerodynamic center of the wing is located at 0.25c, the center of gravity is located at 0.45c, the aircraft has a chord of 5 ft, and the symmetrical tail aerodynamic center is located 10 ft behind the center of gravity. What is the lift generated by the tail, and what is the weight of the aircraft?

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Conventional Airplane

Ltcg

Lw

Ma.c.

xcg

xac

xt

ΣMcg = 0 = Ma.c + Lw (xcg – xac) – Lt (xt)

Moments and Forces

• Trimmed Flight ΣMcg = 0

• Straight and Level, Unaccelerated Flight (S.L.U.F.) ΣF = 0– L = W T = D

Problem Solution

ΣMcg = 0 = Ma.c + Lw (xcg – xac) – Lt (xt)

0 = -20,000 + 40,000 (0.45c-0.25c) – Lt (10)

0 = -20,000 + 40,000 (0.20X5) – Lt (10)

Lt = 2,000 lbs

W = L = Lw + Lt = 40,000 + 2,000 = 42,000lbs

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x

yz

(LongitudinalAxis)

(Vertical Axis)(Lateral Axis)

m n

l

Aircraft Axis System

•Right hand rule•Positive moments

RUDDERELEVATORAILERONRotation classically caused by

N(+ NOSE RT)

M(+ PITCH UP)

L(+ RT WING DOWN)

Moment about Axis(+ IAW RT HAND RULE)

WVU

YAWPITCHROLLMotion about Axis

z(+ out belly)

y(+ out right wing)

x(+ out nose)

Axis

Great Summary!!!

x

y

z

xy

z

x

y

z

Note: Longitudinal stability and control can be studied independently, but Lateral/Directional stability and control is coupled (yaw causes roll / roll causes yaw).

Longitudinal StabilityOverview

Absolute Angle of AttackTail Incidence Angle and Tail Angle of AttackRestoring Moments Moment CoefficientLongitudinal Stability: Wing Effects and Tail Effects Stability and Balance CriteriaNeutral PointStatic MarginAltering Stability

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Absolute Angle of Attack, αa

α α αa L= − =0

Absolute Angle of Attack

The angle between the relative wind and an airfoil’s zero lift lineAn airfoil positioned at its zero lift angle of attack has an absolute angle of attack of zero

α α αa L= − =0

αα

α

α L=0zero lift line

chord line

V ∞

CL vs. α and CL vs. αa

Always at the OriginαL=0 depends on camber

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Zero Lift Line

Lt

V∞itαt

Tail Incidence Angle and Tail Angle of Attack

xt

V∞

α

+Macwing

Lw

xw

Tail incidence angle, it , is the angle betweenChord Line of the tail and Aircraft Zero-Lift-Line.Sometimes fixed—sometimes moveable.

(Tail leading edge down is Positive)αt = (αa −it)

CM

CMo

BIG PICTUREStability and Balance Criteria in SLUF

αa

trimα(Trim angle of attack)

CΜα

αVNE

αStall Steeper Slope = More stable (stronger restoring moment)

)x ,x( C accgM f=α

Cargo, fuel, stores..

Variable wing sweep, Supersonic effects

CMο = f (CMac, it)(Moment Coefficient at zero lift)

Flaps Stick, trim

Restoring Moments

Desired Restoring Moment (-Mcg )

Disturbance (+ ) αa

V

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Restoring Moments

Desired Restoring Moment (+Mcg )

Displacement (- )αaV

Non-Restoring Moment and Loss of Control

JAS-39 Grippen, Stockholm Airshow 8 Aug 1993– Manufacturer and customer knew large and rapid stick movements could

cause divergent Pilot Induced Oscillations– Considered likelihood of it actually happening insignificant, so all pilots

weren't informed – Red warning light too late in telling pilot control system saturated for him to

do anything about it

JAS-39 Grippen on Landing

Moment Coefficient

Recall how we summed moments about the center of gravity:

M M L x x c L lcg ac ac t t= + − −∑ ( )

We can define this moment in terms of a coefficient:

C CMqScM M

cgcg

≡ =

The variation of this coefficient with changes in absolute angleof attack is the key to longitudinal static stability

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Longitudinal StabilityOverview


Zero Lift Line

V∞

α

+Macwing

Lw

Moment Contribution from Wing

xw

Recall:Macwing < 0 (for + camber)

andLw = CL q S = CLαα q S

Summing the moments and dividing by qSc:

C = (CLα xw/c )α+ CMacwing

Mcg αPositive slope(+)

Negative (-)

C (from wing)M cg

Zero Lift Line

Lt

αt = α - it

V∞itαt

Contribution from the Tail

C = - α

+ it

Mcg

(CLαt St xt )S c

(CLαt St xt )S c

αNegative slope (-)

Positive (+)

C (from tail)M cg

Summing the moments and dividing by qSc:

xt

Lt = CLαt α q StSymmetric airfoilSt = tail area

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Contributions to stability -Summary

Required Tail Contribution

Wing Only Contribution

Result – Wing and Tail

Longitudinal Stability—Tail Effects

it > 0it = 0

it < 0Tail incidence angle, it , is the angle betweenChord Line of the tail and Aircraft Zero-Lift-Line.Sometimes fixed—sometimes moveable.

Tail leading edge down is Positive

Stability Criteria

We want the change in moment coefficient to be opposite the change in angle of attackLeads to criteria for longitudinal static stability:

∂

∂α

∂

∂α α

CNote

CCM

a

M

aM

cg cg< =0 ( : )

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Balance Criteria

We want the aircraft to trim at a positive angle of attackThis gives the balance criteria:

CMo> 0

In summary, the stability curve must have a negative slopeand a positive intercept if the aircraft is to havelongitudinal balance and static stability

Neutral Point

The Neutral Point (n.p.) represents the c.g. location such that CMα = 0. It is analogous to the aerodynamic center for the wing alone (CMcg = constant as α changes).

Xcg is the distance from the leading edge of the wing to the Center of GravityXn is the distance from the leading edge of the wing to the Neutral Point

Xcg W

X n

C. G. Effect on Stability

Neutral Point –Where CMα = 0

CMcg

αa

Center of Gravity moving aft and retrim

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Static Margin: Stability Criteria

Non-dimensional difference between Neutral Point (n.p.) and Center of Gravity (c.g.) where:

cgn xxSM −=

cxxandcxx cgcgnn // ==

If S.M. > 0 (c.g. ahead of the neutral point) - aircraft is stableIf S.M. = 0 (c.g. at the neutral point)If S.M. < 0 (c.g. behind the neutral point)

- aircraft is neutral- aircraft is unstable

- CMαCL α

=

Other examples

Badminton shuttlecockArrow

Stability vs.Maneuverability (Control)

Stable Aircraft—not very easy to move – Not very maneuverable– C-5, C-17, B-52, Passenger airplanes

Maneuverable Aircraft—very easy to move– Not very stable (unstable in many cases)– Require Flight Control Systems

» Stability Augmentation System (SAS)» Fly-by-wire FCS

– F-16, F-22

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Static Margin: Stability Criteria (2)

Typical values:– Transports &

Consumer AC: 0.05 to 0.20

Cessna 172Learjet 3 5Boeing 747

P-51 MustangF-106

F-16A (early)F-16CX-29

– Fighters: 0 to 0.05

– Fighters - FBW

.19

.13

.27

.05

.09

-.02.01

-.33

More Stable

More Maneuverable

Maneuverable with other benefits

Altering Longitudinal, Static Stability

Most parameters are fixed once the aircraft is builtC.G. can be moved– Cargo location– Fuel location– Weapons, Stores, etc.

it changes the trim angle of attack, αe

Variable Geometry wings—change cg, CLαWand

moment arm (xcg-xac)

Longitudinal StabilityRecap


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Back up slides

Zero Lift Line

Lt

V∞itαt

Total Airplane Moment

xt

V∞

α

+Macwing

Lw

xw

C = (CLαw - CLαt )α + CMacw+ CLαt itMcg

xtc

StS

xwc

xtc

StS

Wing WingTail Tail

CMαCM0

Changing the CG Location

C = (CLαw - CLαt )α + CMacw+ CLαt itMcg

xtc

StS

xwc

xtc

StS

CMαCM0

LtxtLw xw

Move cg Aft

xw xt

increasesdecreases

CMα

CM0

Increases (slope rotates CCW)

depends on trim

c.g.

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Benefits of Canard Trim

Lw

Lw

Lw

Lw

Lt

Lt

Lc

Lc

np

np

np

np

Conventional Fighter X-29

Subsonic Subsonic

Supersonic Supersonic

SM > 0 (small)

SM > 0 (large)

SM < 0

SM > 0 (small)

cgn xxSM −= Requires > wing lift

Longitudinal Stability:Wing Effects

Mcgwing= Mac + Lw (xcg – xacW

) = Lw xW

+ Mac

Note: This is an unstable situation(Positive slope)

CMcg

αaα trim

Wing Only CMα > 0 cg

Lw

Ma.c.

xcg

xac

Mcgwing = (CLαw q S xw) αa + Mac

CMcgwing = (CLαw xw /c) αa + CMac

xw

Longitudinal Stability—Wing Effects

Wing a.c. forward of c.g. is unstable

Decrease instability (lower CMα)

– ↓ (xcg – xac) Shorter Moment Arm or move c.g. forward

– ↓ SW Smaller Wing Area (hard)– ↓ CLαW Less Efficient Wing (hard)

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Ltcg

Lw

Ma.c.

xcg

xacxt

Mcg = – Lt (xt) = -(CLαt q St xt)αt

CMcg

αa

Positive Stability

Mcg = -(CLαt q St xt) (αa −it)Mcg = -(CLαt q St xt) αa + CLαt q St xt itCMcg = -(CLαt St /S xt /c) αa + CLαt St /S xt/c it


it > 0it = 0

it < 0Tail incidence angle, it , is the angle betweenChord Line of the tail and Aircraft Zero-Lift-Line.Sometimes fixed—sometimes moveable.

Tail leading edge down is Positive


Tail aft of cg is stablizingCanards are destabilizingIncrease stability (more negative CMα) by xt Longer moment arm

St Larger tail

CLαt ARt or et

or move tail out of downwash

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Contributions to stability -Summary

Required Tail Contribution

Wing Only Contribution

Result – Wing and Tail

Quiz

Which of the following diagrams indicates an aircraft that has met the longitudinal static stability requirement for conventional ("upright") flight but is NOT currently trimmed for balanced, straight, level, unaccelerated flight? What should you do to trim the aircraft?

it > 0

Changing Variables*

C M cg

αα

C M cg

CMo _____CMα _____αtrim _____Vtrim _____

CMo _____CMα _____αtrim _____Vtrim _____

Slow Down (c. g. constant) Move c. g. Aft

* In Handout Package

No change

STABILITY CRITERIA AND CHANGING STABILITY Pilot Induced ...

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