322_3StaticLongStab

7/29/2019 322_3StaticLongStab

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Aero 322

AIRCRAFT STABILITY & CONTROL

for 73rd

E.C

Sqn Ldr Dr.Irtiza

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Lecture 3: Static Long Stab & ControlChapter 2, Section 2.3


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Longitudinal Static Stability

We need to have

A restoring moment generated after thedisturbance, that tries to bring back the ball to itsequilibrium

Equilibrium (or trim)

Resultant forces F=0

Resultant moments about cg M=0 IfF0 linear accelerations

IfM0 rotational accelerations

Consider Cm vs behaviour of two airplanes

Sign convention: Nose up moment: +ve

At trim point B: Cmcg=0 Disturbance increases to pt C

On airplane 1: restoring nose down (-ve) moment isgenerated tending to bring it back to pt B: staticallystable (1st requirement met)

Vice versa for airplane 2 2


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Longitudinal Static Stability

1st requirement for Static LongitudinalStability: dCm/d < 0

As there is a linear relationship betweenCL & , we can write: Cm = dCm/d = (dCm/dCL) (dCL/d)

Since (dCL/d) >0 (recall CL curve till stall)hence to have Cm < 0 we must have(dC

m/dC

L) < 0 or C

mCL< 0

2nd requirement: Consider two airplanes Both with dCm/d < 0

Physical Limitation: We can only trim anaircraft at a positive

This is met only if we have (y-intercept)

Cm0 > 0

Hence two requirements for staticlongitudinal stability are: Cm < 0 (or CmCL < 0 )

Cm0 > 0

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or CL


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Contribution of A/C components: Wing

Wing mean aerodynamic chord is shown

xac = distance from wing LE to a.c (a point wheremoments become independent of angle of attack)

xcg = distance from wing LE to cg

zcg = vertical displacement of cg

iw = wing installation angle (angle ofincidence)

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Sum of moments about cg:

M = Mcgw = Lwcos(w-iw)[xcg-xac] + Dwsin(w-iw)[xcg-xac] + Lwsin(w-iw)[zcg]

Dwcos(w-iw)[zcg] + Macw Dividing by V2Sc yields

Cmcgw = CLwcos(w-iw)[xcg-xac]/c + CDwsin(w-iw)[xcg-xac]/c + CLwsin(w-iw)[zcg]/c CDwcos(w-iw)[zcg]/c+ Cmacw

We assume that: angle of attack is small (>CD and zcg 0 then

Cmcgw = Cmacw + CLw[xcg-xac]/c

Since CLw = CL0w + CLww (from lift curve slope) then

Cmcgw = Cmacw + (CL0w+CLww)[xcg-xac]/c = Cmacw + CL0w[xcg-xac]/c + { CL w [xcg-xac]/c } w

To find Cm0w we put w = 0 to get

Cm0w = Cmacw + CL0w[xcg-xac]/c

To find Cmw we differentiate wrt w

Cm w = CL w [xcg-xac]/c


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Contribution of A/C components: Wing For stability we needed:

Cm < 0 (or CmCL < 0 )

Cm0 > 0

Hence if we want to derive static longitudinal stability from wing

alone (without horizontal tail e.g Mirage) we have to have both

requirements met from just the wing. Hence

Cm0w = Cmacw + CL0w[xcg-xac]/c > 0

means we need a -ve cambered wing airfoil

Cm w = CL w [xcg-xac]/c < 0 means, we need cg forward of a.c

This is not the case for all conventional aircraft with horizontal tail.

In fact wing for all such aircraft, has +ve cambered airfoil, as well as

cg is located aft of a.c. Hence wing contribution of all conventional

aircraft is UNSTABLE

We handle this instability by the use of horizontal tail5

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322_3StaticLongStab