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

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

Moments and Forces

Trimmed Flight Mcg = 0

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

Problem Solution

Mcg = 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 rulePositive 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, a

a L= =0

Absolute Angle of Attack

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

a L= =0

L=0zero lift line

chord line

V

CL vs. and CL vs. a

Always at the OriginL=0 depends on camber

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Zero Lift Line

Lt

Vitt

Tail Incidence Angle and Tail Angle of Attack

xt

V

+Macwing

Lwxw

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

(Tail leading edge down is Positive)t = (a it)

CM

CMo

BIG PICTUREStability and Balance Criteria in SLUF

a

trim(Trim angle of attack)

C

VNE

Stall Steeper Slope = More stable (stronger restoring moment)

)x ,x( C accgM f=

Cargo, fuel, stores..

Variable wing sweep, Supersonic effects

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

Flaps Stick, trim

Restoring Moments

Desired Restoring Moment (-Mcg )

Disturbance (+ ) a

V

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Restoring Moments

Desired Restoring Moment (+Mcg )

Displacement (- )aV

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

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

Zero Lift Line

V

+Macwing

Lw

Moment Contribution from Wing

xwRecall:

Macwing < 0 (for + camber)and

Lw = CL q S = CL q S

Summing the moments and dividing by qSc:

C = (CL xw/c )+ CMacwing

Mcg Positive slope(+)

Negative (-)

C (from wing)M cg

Zero Lift Line

Lt

t = - it

Vitt

Contribution from the Tail

C = -

+ it

Mcg(CLt St xt )

S c(CLt St xt )

S c

Negative slope (-)

Positive (+)

C (from tail)M cgSumming the moments and dividing by qSc:

xt

Lt = CLt q StSymmetric airfoilSt = tail area

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Contributions to stability -Summary

Required Tail Contribution

Wing Only Contribution

Result Wing and Tail

Longitudinal StabilityTail Effects

it > 0it = 0

it < 0Tail incidence angle, it , is the angle betweenChord Line of the tail and Aircraft Zero-Lift-Line.Sometimes fixedsometimes 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:

C

NoteC

CMa

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 CM = 0. It is analogous to the aerodynamic center for the wing alone (CMcg = constant as 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 CM = 0

CMcg

a

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

- CMCL

=

Other examples

Badminton shuttlecockArrow

Stability vs.Maneuverability (Control)

Stable Aircraftnot very easy to move Not very maneuverable C-5, C-17, B-52, Passenger airplanes

Maneuverable Aircraftvery 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, eVariable Geometry wingschange cg, CLW and moment arm (xcg-xac)

Longitudinal StabilityRecap

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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Back up slides

Zero Lift Line

Lt

Vitt

Total Airplane Moment

xt

V

+Macwing

Lwxw

C = (CLw - CLt ) + CMacw+ CLt itMcgxtc

StS

xwc

xtc

StS

Wing WingTail Tail

CM CM0

Changing the CG Location

C = (CLw - CLt ) + CMacw+ CLt itMcgxtc

StS

xwc

xtc

StS

CM CM0

LtxtLw xw

Move cg Aft

xw xt

increasesdecreases

CM

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

a trim

Wing Only CM > 0 cg

Lw

Ma.c.

xcg

xac

Mcgwing = (CLw q S xw) a + MacCMcgwing = (CLw xw /c) a + CMac

xw

Longitudinal StabilityWing Effects

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

Decrease instability (lower CM) (xcg xac) Shorter Moment Arm or move

c.g. forward SW Smaller Wing Area (hard) CLW Less Efficient Wing (hard)

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Longitudinal StabilityTail Effects

Ltcg

Lw

Ma.c.

xcg

xacxt

Mcg = Lt (xt) = -(CLt q St xt)t

CMcg

aPositive Stability

Mcg = -(CLt q St xt) (a it)Mcg = -(CLt q St xt) a + CLt q St xt itCMcg = -(CLt St /S xt /c) a + CLt St /S xt/c it

Longitudinal StabilityTail Effects

it > 0it = 0

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

Tail leading edge down is Positive

Longitudinal StabilityTail Effects

Tail aft of cg is stablizingCanards are destabilizingIncrease stability (more negative CM) by xt Longer moment arm St Larger tail CLt 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

C M cg

CMo _____CM _____trim _____Vtrim _____

CMo _____CM _____trim _____Vtrim _____

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

* In Handout Package

No change