Flight Dynamics and ControlLecture 4:

Lateral stability DerivativesG. Dimitriadis

University of Liege1

Previously on AERO0003-1• We developed linearized equations of

motion

m − X w 0 0

0 m − Z w( ) 0 0

0 − M w Iy 00 0 0 1

"

#

$$$$$

%

&

'''''

uwqθ

"

#

$$$$

%

&

''''

+

− Xu − Xw − Xq − mWe( ) mgcosθe− Zu − Zw − Zq + mUe( ) mgsinθe− Mu − Mw − Mq 00 0 −1 0

"

#

$$$$$

%

&

'''''

uwqθ

"

#

$$$$

%

&

''''

=

XηXτ

Zη ZτMη

Mτ

0 0

"

#

$$$$

%

&

''''

η

τ"

#$%

&'

m 0 0 0 00 Ix −Ixz 0 00 −Ixz Iz 0 00 0 0 1 00 0 0 0 1

"

#

$$$$$$

%

&

''''''

vprϕψ

"

#

$$$$$$

%

&

''''''

+

− Yv − Yp + mWe( ) − Yr − mUe( ) −mgcosθe −mgsinθe− Lv − Lp − Lr 0 0− Nv − Np − Nr 0 00 −1 0 0 00 0 −1 0 0

"

#

$$$$$$

%

&

''''''

vprϕ

ψ

"

#

$$$$$$

%

&

''''''

=

Yξ YςLξ LςNξ

Nς

0 00 0

"

#

$$$$$$

%

&

''''''

ξ

ς"

#$%

&'

Longitudinal direction

Lateral direction

Lateral stability derivatives• We have already derived expressions for the most important

longitudinal stability derivatives.• Here we will do the same thing for the lateral stability

derivatives.• This means the derivatives of the sideforce Y, rolling

moment L and yawing moment N with respect to perturbations v, p, r, x, z.

• The discussion is based on – Etkin and Reid, Dynamics of flight.– Roskam, Methods for estimating stability and control

derivatives of conventional subsonic airplanes.• Both of these works include results from:

– Hoak et al, USAF Stability and Control Datcom (Data compendium)

3

Sideslip angle• The angle of

sideslip is a lateral angle of attack.

• It should not be confused with the yaw angle.

cg

V0

y

x

V

U

b

β = sin−1 VV0

4

Equilibrium flight condition• Consider rectilinear flight at a constant sideslip

angle be:

• The perturbed sideslip angle is given by:

• Take the difference form:

• This means that derivatives with respect to b are in fact derivatives with respect to v.

V =Ve + v, βe = sin−1 VeV0

β = sin−1Ve + vV0

≈ sin−1 VeV0+vV0

= βe +vV0

dβ = d βe +vV0

!

"#

$

%&=

1V0dv

5

Sideforce• The wing, fuselage, propeller and fin all

contribute to the sideforce Y.• Wing sideforce contribution:

– Sidewash factor– Wing sideforce

• Fuselage sideforce contribution:– Sidewash factor– Fuselage lift

• Propeller sideforce contribution:– Sidewash factor– Propeller normal force

• Fin sideforce contribution:– Fin lift 6

Rolling moment• The wing, fin and fuselage contribute to the

rolling moment.• Wing rolling moment contribution:

– Wingtip vortex effect– Dihedral effect– Sweep effect

• Fuselage rolling moment contribution:– High wing or low wing

• Fin rolling moment contribution:– Fin lift

7

Yawing moment• The wing, fin and fuselage contribute to the yawing moment.• Wing yawing moment contribution:

– Sidewash factor– Angle of attack difference on two half-wings due to rolling:

• Asymmetric induced drag– Angle of attack difference on two half-wings due to yawing:

• Asymmetric lift and drag, particularly when the wing is swept• Fuselage yawing moment contribution:

– Sidewash factor– Fuselage lift

• Fin yawing moment contribution:– Fin lift

• Propeller yawing moment contribution:– Propeller normal force

8

Fin lift• A fin is an upright wing.

Depending on the geometry, it can be a half-wing.

• In general, the fin will have a symmetric airfoil and its geometric angle of attack will be zero.

• The fin lift is then

• Where c1 is the fin’s lift curve slope, c2 is the rudder curve slope, aF is the fin angle of attack and z is the rudder deflection angle.

CLF = c1αF + c2ζ

9

Lift curve slope• The lift curve slope for a wing is given by:

– Straight tapered wings:

– Swept wings:

• For half-wing fins, we can divide aF by two but the aspect ratio must be for a full wing.

CLα= 0.995

clαE + clα /πAR

, where E =1+ 2λAR 1+λ( )

βCLα=

2π

2βAR

+1

k2 cos2 Λβ

+2

βAR"

#$

%

&'

2

where tanΛβ =tanΛ1/2

β and k =

βclα2π

β = 1− M 2

10

Fin angle of attack

• The angle of attack of the fin depends on:– Sideslip angle b– Sidewash angle s– Roll velocity p– Yaw velocity r

• The last two motions introduce effective curvature and camber to the fin but these effects can be modelled as angle of attack modifications.

11

Sidewash angle• The wing and fuselage

deflect the flow seen by the tail.

• The deflection angle is known as the sidewashangle s.

• The angle of attack of the fin is:

• We assume that the sidewash is a linear function of sideslip, so that:

cg

V0

y

x

V

U

b

b

sb-s

LF

αF = −β +∂σ∂β

β = − 1− ∂σ∂β

#

$%

&

'(β

αF = −β +σ

12

Sidewash derivative approximation

• The sidewash derivative with respect to b can be approximated as:

• where Zw is the vertical distance from the wing root quarter-chord to the fuselage centreline, positive downward, and

• SF is the fin area, S is the wing area, A the aspect ratio and Lc/4 the sweep angle at the quarter-chord.

∂σ∂β

= −0.276+3.06 SFS

11+ cosΛc/4

+ 0.4 Zw

d+ 0.009A

d = average fuselage cross sectional area0.7854

13

Angle due to roll velocity• The roll velocity introduces a

linearly varying velocity distribution on the fin, pz.

• A mean value for this velocity distribution is pzF, where zF is an appropriate mean height of the fin.

• We can simplify by saying that zF is the height of the fin’s aerodynamic centre.

• Note that the roll motion also has an effect on the sidewash angle.

• Then the change in angle of attack due to the roll motion is

Δα p = −pzFV0

+∂σ∂p

p 14

Angle due to yaw velocity• The yaw velocity r

introduces an additional airspeed rlF perpendicular to the fin.

• Yawing also has an effect on the sidewash angle.

• Then, the angle of attack change due the yaw velocity is:

Δαr =rlFV0

+∂σ∂r

r 15

Sideforce due to fin lift• The total sideforce due to fin lift is then given

by:

• Substituting b=be+v/V0:

• Separating the stiffness from the damping terms we get:

Ya =12ρV0

2SFc1 − 1−∂σ∂β

#

$%

&

'(βe − 1−

∂σ∂β

#

$%

&

'(vV0−pzFV0

+∂σ∂p

p+ rlFV0

+∂σ∂r

r#

$%

&

'(+12ρV0

2SFc2ζ

Ya =Yae +12ρV0

2SFc1∂σ∂p

p+ ∂σ∂r

r"

#$

%

&'+12ρV0SFc1 − 1−

∂σ∂β

"

#$

%

&'v+ rlF − pzF

"

#$

%

&'+12ρV0

2SFc2ζ

Ya =12ρV0

2SFc1 − 1−∂σ∂β

#

$%

&

'(β −

pzFV0

+∂σ∂p

p+ rlFV0

+∂σ∂r

r#

$%

&

'(+12ρV0

2SFc2ζ

Y aerodynamic derivatives due to fin

• The aerodynamic derivatives of Ybecome:!Yv = −

12ρV0SFc1 1−

∂σ∂β

#

$%

&

'(

!Yp =12ρV0

2SFc1∂σ∂p

−12ρV0SFc1zF

!Yr =12ρV0

2SFc1∂σ∂r

+12ρV0SFc1lF

!Yζ =12ρV0

2SFc2

Yv = −SFSc1 1−

∂σ∂β

#

$%

&

'(

Yp =SFSc1 V0

∂σ∂p

− zF#

$%

&

'(

Yr =SFSc1 V0

∂σ∂r

+ lF"

#$

%

&'

Yζ =SFSc2

17

DATCOM values for sideforcederivatives due to fin

• The terms !"/!$ and !"/!% are difficult to calculate.• The DATCOM proposes the following value for the sideforce

derivative due to roll:

&' = 2&*+, cos 0 − 2, sin 0

5 = −26,6 78 1 − !"!:+, cos 0; − 2, sin 0;

5• where 0; is the equilibrium angle of attack.• Similarly, for the sideforce derivative due to yaw, the DATCOM

proposes

&< = −2&*+, sin 0 + 2, cos 0

5 = 2 6,6 78 1 − !"!:+, cos 0; − 2, sin 0;

5• Note that both of these derivatives are only the contributions

from the fin. Contributions from the body and wing will be given later.

• Furthermore, these contributions are generally small.

Rolling moment diagram

• The fin generates a rolling moment around the centre of gravity.

• The rolling moment is equal to the fin lift times the distance between the fin’s aerodynamic centre and the centre of gravity.

19

Rolling moment due to fin

• The rolling moment due to the fin is given simply by:

• Substituting for the fin angle of attack:

• Substituting b=be+v/V0:

La =12ρV0

2SF c1 − 1−∂σ∂β

#

$%

&

'(β −

pzFV0

+∂σ∂p

p+ rlFV0

+∂σ∂r

r#

$%

&

'(+ c2ζ

#

$%%

&

'((zF

La = Lae +12ρV0

2SFzFc1∂σ∂p

p+ ∂σ∂r

r"

#$

%

&'+12ρV0SFzFc1 − 1−

∂σ∂β

"

#$

%

&'v− pzF + rlF

"

#$

%

&'+12ρV0

2SFzFc2ζ

La =12ρV0

2SFCLFzF =12ρV0

2SF c1αF + c2ζ( ) zF

L aerodynamic derivatives due to fin

• The aerodynamic derivatives of Lbecome (recalling that ): !Lv = −

12ρV0SFzFc1 1−

∂σ∂β

#

$%

&

'(

!Lp =12ρV0

2SFzFc1∂σ∂p

−12ρV0SFc1zF

2

!Lζ =12ρV0

2SFzFc2

Lv = −VFzFlFc1 1−

∂σ∂β

#

$%

&

'(

Lp =VFc1zFlF

V0∂σ∂p

− zF#

$%

&

'(

Lζ =VFzFlFc2

VF =SFlFSb

!Lr =12ρV0

2SFzFc1∂σ∂r

+12ρV0SFzFc1lF Lr =VF

zFlFc1 V0

∂σ∂r

+ lF"

#$

%

&'

21

DATCOM values for rolling moment derivatives due to fin

• Again, the terms !"/!$ and !"/!% are difficult to calculate.• The DATCOM proposes the following value for the rolling

moment due to roll:

&' = 2*+,-./ = −21-1 23 1 − !"!5

,-./

• Similarly, for the rolling moment derivative due to yaw, the DATCOM proposes

&6 = 2 1-1 23 1 − !"!5,- sin : + <- cos : ,- cos :? − <- sin :?

/.• Note that both of these derivatives are only the contributions

from the fin. Contributions from the body and wing will be given later.

• Furthermore, these contributions are generally small.

Yawing moment diagram

cg

V0

y

x

V

U

b

b

sb-s

LF

NlF

• The fin lift causes a restoring yawing moment around the aircraft’s CG.

• Assuming that the sideslip angle is small, the moment is given by:

Na = LFlF23

Yawing moment due to fin

• The yawing moment due to the fin is given simply by:

• Substituting for the fin angle of attack:

• Substituting b=be+v/V0:

Na = −12ρV0

2SFCLFlF = −12ρV0

2SF c1αF + c2ζ( )lF

Na = −12ρV0

2SF c1 − 1−∂σ∂β

#

$%

&

'(β −

pzFV0

+∂σ∂p

p+ rlFV0

+∂σ∂r

r#

$%

&

'(+ c2ζ

#

$%%

&

'((lF

Na = Nae−12ρV0

2SFlFc1∂σ∂p

p+ ∂σ∂r

r#

$%

&

'(+12ρV0SFlFc1 1− ∂σ

∂β

#

$%

&

'(v+ pzF − rlF

#

$%

&

'(−12ρV0

2SFlFc2ζ

N aerodynamic derivatives due to fin

• The aerodynamic derivatives of Nbecome (recalling that ): !Nv =

12ρV0SFlFc1 1−

∂σ∂β

#

$%

&

'(

!Np = −12ρV0

2SFlFc1∂σ∂p

+12ρV0SFlFc1zF

!Nr = −12ρV0

2SFlFc1∂σ∂r

−12ρV0SFlF

2c1

!Nζ = −12ρV0

2SFlFc2

Nv =SFlFSb

c1 1−∂σ∂β

#

$%

&

'(=VFc1 1−

∂σ∂β

#

$%

&

'(

Np =VFc1 zF −V0∂σ∂p

#

$%

&

'(

Nr = −VFc1 V0∂σ∂r

+ lF#

$%

&

'(

Nζ = −VFc2ζ

VF =SFlFSb

25

DATCOM values for yawing moment derivatives due to fin

• Again, the terms !"/!$ and !"/!% are difficult to calculate.• The DATCOM proposes the following value for the yawing

moment due to roll:&' = )*

• Similarly, for the yawing moment derivative due to yaw, the DATCOM proposes

&* = −2-.- /0 1 − !"!23. sin 7 + 9. cos 7 <

=<• Note that both of these derivatives are only the contributions

from the fin. Contributions from the body and wing will be given later.

• Furthermore, these contributions are generally small.

Static stability in yaw

• Static stability in yaw is also known as weathercock or yaw stiffness.

• At static conditions, p=r=0 and b is a constant.

• The yawing moment N becomes:

• Non-dimensionalise with respect to 12 ρV02Sb

Na = −12ρV0

2SFlF −c1 1−∂σ∂β

#

$%

&

'(β + c2δ

#

$%

&

'(

CN = −SFlFSb

−c1 1−∂σ∂β

#

$%

&

'(β + c2δ

#

$%

&

'(= −VF −c1 1−

∂σ∂β

#

$%

&

'(β + c2δ

#

$%

&

'( 27

Static stability in yaw (2)• The aircraft is stable in yaw if

• where

• If the rudder is held fixed, we obtain the controls fixed yaw stability criterion:

• Recall that this is only the fin contribution to stability in yaw. The wing, body and propeller contributions must also be included in order to properly asses the stability of the aircraft.

∂CN

∂β< 0

∂CN

∂β=VF c1 1−

∂σ∂β

#

$%

&

'(− c2

∂δ∂β

#

$%

&

'(

∂CN

∂β=VFc1 1−

∂σ∂β

#

$%

&

'(< 0

28

Effect of sideslip on wing• Straight rectangular wing with AR=8 at a=5o and b=10o.

• The left wingtip vortex moves away from the wing.

• The right wingtip vortex moves towards the wing.

• Sideslip has a significant effect on downwash and therefore induced drag.

• The effect on the lift is small.• The sideforce is small.

x(m

)

10

15

20

25

30

35

0

5

10

y (m)

5

Wake after 1.9 s

0-5-1029

y (m)-4 -3 -2 -1 0 1 2 3 4

Sec

tionallift

(N/m

)

406080100120140160

y (m)-4 -3 -2 -1 0 1 2 3 4

Sec

tionalin

duce

ddra

g(N

/m

)

0

1

2

3

4

y (m)-4 -3 -2 -1 0 1 2 3 4

Sec

tionallift

(N/m

)

0

50

100

150

200

y (m)-4 -3 -2 -1 0 1 2 3 4

Sec

tionalin

duce

ddra

g(N

/m

)

0

1

2

3

4

y (m)-4 -3 -2 -1 0 1 2 3 4

Sec

tionallift

(N/m

)

-50

0

50

100

150

200

y (m)-4 -3 -2 -1 0 1 2 3 4

Sec

tionalin

duce

ddra

g(N

/m

)

0

1

2

3

4y (m)

-4 -3 -2 -1 0 1 2 3 4

Sec

tionallift

(N/m

)-100-50050100150200

y (m)-4 -3 -2 -1 0 1 2 3 4

Sec

tionalin

duce

ddra

g(N

/m

)

0

1

2

3

4

Increasing angle b

b=0o b=5o

b=10o b=15o

30

Discussion of effect of b on wings

• For planar wings, the main effect of sideslip is an asymmetry in drag. – The trailing wing has high drag and the leading wing

has low drag.– This asymmetry causes a destabilising yawing

moment.• There is a small amount of asymmetry in lift very

near the wingtips but it can be neglected.• The sideforce is also very low and can be

neglected.• The main lateral load contribution of a planar wing

in sideslip is then the yawing moment N. Even this moment can be small and neglected if there is no dihedral. 31

Sideslip and dihedral• Adding dihedral changes completely the

situation:– Same wing as before at b=10o but with 10o

dihedral.

35

30

25

20

x (m)

15

10

Wake after 1.9 s

5

0-10

-5

0

y (m)

5

10

420-2-4

z(m

)

y (m)-4 -3 -2 -1 0 1 2 3 4

Sec

tionallift

(N/m

)

-50

0

50

100

150

200

y (m)-4 -3 -2 -1 0 1 2 3 4

Sec

tionalin

duce

ddra

g(N

/m

)

-4

-2

0

2

4

6

Dihedral effect• In the presence of dihedral, the yawing moment

caused by sideslip changes direction:– The trailing wing now has low drag and the

leading wing high drag. – The yawing moment is stabilizing.

• On the other hand, the lift is now completely asymmetric.– The leading wing has high lift and the trailing wing

low lift.– A rolling moment is created.

• Anhedral has exactly the opposite effect:– The yawing moment is destabilising.– The rolling moment is of opposite direction.

33

Sideslip and sweep

• Sweep also has an effect:– Same wing as before at b=10o but with 30o

sweep, no dihedral.

y (m)-4 -3 -2 -1 0 1 2 3 4

Sec

tionallift

(N/m

)

100105110115120125130135

y (m)-4 -3 -2 -1 0 1 2 3 4

Sec

tionalin

duce

ddra

g(N

/m

)

-1012345

3530

2520

x (m)

15

Wake after 1.9 s

105

0-10

-5

0

y (m)

5

10

4

2

0

-2

-4

z(m

)

34

Sweep effect• Sweep does not affect the direction of the

yawing moment:– The trailing wing has high drag and the leading

wing low drag. – The yawing moment is destabilizing but weak.

• It does however create a rolling moment, albeit weaker than that created by dihedral.– The leading wing has high lift and the trailing wing

low lift.– A rolling moment is created.

• Forward swept wings do not produce a significant rolling moment.

35

Roll angle• Aircraft at a roll angle

are usually subjected to a restoring moment.

• The lift is inclined so the aircraft will start sideslipping.

• If the wing has dihedral, the sideslip will cause a stabilising rolling moment.

• The low wing is also the leading wing and it has high lift.

• This phenomenon is known as roll stiffness.

36

Roll angle and dihedral• Dihedral has an effect when the wing is

rolled, even if the sideslip angle is 0o:– Same wing as before at b=0o but with 15o

dihedral at a 10o roll angle.

x (m)

30

20

10

0-10-5

Wake after 1.9 s

0y (m)510

-4

-3

-2

-1

0

1

2

3

4

z(m

)

y (m)-4 -3 -2 -1 0 1 2 3 4

Sec

tionallift

(N/m

)

100110120130140150160170

y (m)-4 -3 -2 -1 0 1 2 3 4

Sec

tionalin

duce

ddra

g(N

/m

)

0

1

2

3

4

37

Dihedral effect on roll• Dihedral causes mainly a lift asymmetry

when the wing is rolled:– The high wing has low lift and the low wing

high lift. – The result is a restoring rolling moment.

• It also creates a yawing moment.– The high wing has low drag and the low wing

high drag.• The effect of dihedral is increased when

the aircraft starts sideslipping.• Anhedral has exactly the opposite effect.

38

Effect of fuselage on roll moment

• The position of the wing on the fuselage has a significant effect on roll stability.– High wing: stabilising moment– Low wing: destabilising moment.

• This effect is due to the fact that the fuselage diverts the flow around the wing.

alowahigh

ahighalow

39

Roll Control

• Roll control can be accomplished using the ailerons

δ1

δ2

The total aileron angle is given by δa=(δ1+δ2)/2

More lift, more drag

Less lift, less drag Adverse yaw

40

Aileron adverse yaw

• Increasing the lift also increases the drag and vice versa.

• When deflecting ailerons, there is a net yawing moment in an opposite direction to the rolling moment.– When rolling left (in order to turn left), there

is a yawing moment to the right– This can make turning very difficult,

especially for high aspect ratio wings41

Roll control by spoilers

• Another way of performing roll control is by deforming a spoiler on the wing towards which we want to turn. To turn left:

The spoiler decreases lift and increases drag. Now the yawing moment is in the same direction as the roll.

Proverse yaw

Frise Ailerons

Frise ailerons are especially designed to create very high profile drag when deflected upwards. When deflected downwards the profile drag is kept low. Thus, they alleviate or, even, eliminate adverse yaw

The idea is to counteract the higher lift induced drag of the down wing with higher profile drag on the up wing.

43

Differential Aileron Deflection

• The roll rate of the aircraft depends on the mean aileron deflection angle. The individual deflections δ1 and δ2 do not have to be equal.Differential deflection means that the up aileron is deflected by a lot while the down aileron is deflected by a little. 44

Effect of roll rate

• Roll rate creates significant moments:– Same wing as before at a=0o with 18o/s roll

rate (three times faster than a watch).

-10-5

y (m)

05

Wake after 3.9 s

100

10

20

30x (m)

40

50

60

70

-2

-4

0

2

4

80

z(m

)

y (m)-4 -3 -2 -1 0 1 2 3 4

Sec

tionallift

(N/m

)

-20020406080100

y (m)-4 -3 -2 -1 0 1 2 3 4

Sec

tionalin

duce

ddra

g(N

/m

)

-0.50

0.51

1.52

2.5

Roll rate effect

• The plunging wing has high lift while the climbing wing has low lift:– A significant rolling moment is generated,

known as roll damping• The plunging wing also has high drag

while the climbing wing has low drag:– A significant yawing moment is generated

46

Roll subsidence• The pilot deflects the ailerons in order to start

rolling:– The climbing wing has high lift, the plunging wing low

lift.• As the aircraft accelerates in roll, the lift of the

climbing wing decreases and that of the plunging wing increases.

• This means that the roll rate cannot accelerate to very high values:– As the roll rate increases, the rolling moment

decreases and eventually falls to zero.• This phenomenon is know as roll subsidence.

47

Wing and body contribution to Yv

• The wing contribution to Yv (non-dimensional) is given by:

• where G the wing dihedral.• The body contribution to Yv (non-dimensional) is given by:

• Where S0 is the cross-sectional area of the fuselage at the point x0 along the body.

• The values for S0 and Ki are obtained from the following figures.

Yv = −0.0001 Γ180π

per rad

Yv = −2KiS0S

per rad

48

x1 is the body station where first reaches its maximum negative value. Sx is the fuselage cross-sectional area as a function of x.

dSxdx

49

Wing contribution to Lv• The wing contribution to Lv (non-dimensional) is

given by:

CL is the wing lift coefficient, G the wing dihedral, A is the aspect ratio, q is the wing twist (negative for washout) and Lc/2, Lc/4 the sweep angle at the half-chord or quarter-chord.

• All the other factors are given by the following figures (l is the taper ratio).

Lv =180π

CL

Clβ

CL

!

"#

$

%&Λc/2

KMΛ+Clβ

CL

!

"#

$

%&A

(

)**

+

,--+Γ

Clβ

Γ

!

"#

$

%&KMΓ

+θ tanΛc/4

ΔClβ

θ tanΛc/4

!

"#

$

%&

012

32

452

62

50

Body contribution to Lv• The body contribution to Lv (non-

dimensional) is given by:

• where Zw is the vertical distance from the wing root quarter-chord to the fuselage centreline, b is the span and

Lv =180 Aπ

db−0.0005 d

b−2.4π180

Zw

b"#$

%&'

d = average fuselage cross sectional area0.7854

53

Wing and body contribution to Nv

• The wing contribution to Nv is neglected by the USAF DATCOM.

• The body contribution is given by

• Where S is the wing area, SBs is the fuselage side area and lB is the fuselage length.

• The factors KN and KRl are obtained from the following figures.

Nv = −180π

KNKRl

SBsSlBb

54

55

Wing and body contribution to Lp

• The wing and body contribution to Yp is considered negligible.

• The wing and body contribution to Lp (non-dimensional) is given by:

• where k is the ratio of the profile lift curve slope to 2p and M is the Mach number. Furthermore,

• The factor in brackets is given in the following figures.

Lp =1−M 2Clp

κ

"

#

$$

%

&

''

κ

1−M 2

Clp( )Γ

Clp( )Γ=0

Clp( )Γ

Clp( )Γ=0

=1− 2 Zw

b / 2sinΓ+3 Zw

b / 2#

$%

&

'(2

sin2 Γ

56

β = 1−M 2

57

Wing contribution to Np

• The body contribution to Np is negligible.• The wing contribution to Np (non-dimensional) is

given by:

• where a is the angle of attack, Lpw is the wing contribution to Lp. Furthermore,

Np = −Lpwtanα − −Lp tanα −

cnpCL

"

#$

%

&'CL=0,M

CL

"

#

$$

%

&

''+

Δcnpθ

"

#$

%

&'θ

cnpCL

!

"#

$

%&CL=0,M

=A+ 4cosΛc/4

AB+ 4cosΛc/4

AB+ 12AB+ cosΛc/4( ) tan2 Λc/4

A+ 12A+ cosΛc/4( ) tan2 Λc/4

cnpCL

!

"#

$

%&CL=0,M=0

cnpCL

!

"#

$

%&CL=0,M=0

= −16A+ 6 A+ cosΛc/4( )A+ 4cosΛc/4

h0 − h( ) tanΛc/4

A+tan2 Λc/4

12)

*+

,

-.

B = 1−M 2 cos2 Λc/4

58

Calculation of Δcnpθ

"

#$

%

&'

59

Wing contribution to Lr• The wing and body contributions to Yrare negligible.• The body contribution to Lr is negligible.• The wing contribution to Lr is given by

• whereLr =CL

clrCL

!

"#

$

%&CL=0,M

+ΔclrΓ

!

"#

$

%&Γ+

Δclrθ

!

"#

$

%&θ

ΔclrΓ

#

$%

&

'(=

112

πAsinΛc/4

A+ 4cosΛc/4

clrCL

#

$%

&

'(CL=0,M

=

1+A 1−B2( )

2B AB+ 2cosΛc/4( )+AB+ 2cosΛc/4

AB+ 4cosΛc/4

tan2 Λc/4

8

1+ A+ 2cosΛc/4

A+ 4cosΛc/4

tan2 Λc/4

8

clrCL

#

$%

&

'(CL=0,M=0

Wing contribution to Nr

• The body contribution to Nr is negligible.• The wing contribution to Nr is given by

• Where the values in brackets are given by the figures overleaf.

Nr =cnrCL2

!

"#

$

%&CL

2 +cnrCD0

!

"##

$

%&&CD0

62

xc= h0 − h 63

DATCOM Rudder derivatives• The factor !" for the rudder’s contribution

to the fin lift is difficult to calculate.• The DATCOM proposes the following

value for the sideforce rudder derivative:

#$ = !&'()*'()+

'()+,-,.

/0/

• where all the terms where already introduced in lecture 3 and concern the influence of the flaps. They can be calculated using the following figures.

Flap influence terms

DATCOM Rudder derivatives (2)

• DATCOM value for the rolling moment rudder derivative:

!" = $%&'()&'(*

+,(*-.-/ 010

21 345 &6781 59: &6/

• DATCOM value for the rolling moment rudder derivative:

;" = −$%&'()&'(*

+,(*-.-/ 010

21 59: &6=81 345 &6/

Propeller normal force• Propellers at an angle of

attack generate both thrust (along the axis) and a normal force (in the propeller’s plane)

• The normal force is given by:

• Where sp is the propeller disk area and

TNp

cg

b

V0

V

U

lpNp =

12ρV0

2spCNp

CNp= eβ 67

Propeller normal force curve slope

• The propeller normal force curve slope can be determined from: e = fC

Yψ 0'

Tc =T

ρV 2d 2

d is the propeller diameter

68

Propeller normal force curve slope (2)

Blade twist angle at 0.75R (in degrees)

Propeller normal force curve slope (3)

SFF = 525 b / d( )0.3 + b / d( )0.6!" #$+ 270 b / d( )0.9

70

• Procedure:– Determine Tc and then f.– Using the blade twist angle determine– Calculate Side Force Factor (SFF)– Determine the correct from SFF. – More info in

• Notes on the propeller and slipstream in relation to stability, H. S. Ribner, NACA Wartime Report L-25, 1944

Propeller normal force curve slope (4)

CYψ 0'

CYψ 0'

71

Discussion of propeller normal force

• The propeller normal force generates a yawing moment that is:– Destabilising if the propeller lies ahead of the CG

(puller propeller)– Stabilising if the propeller lies behind the CG

(pusher propeller)• From the f graph, the propeller normal force is

a nonlinear function of thrust, since e=e(T). • However, we can always linearize around a

trimmed thrust condition.

72

Handling the thrust• We can write the thrust as:

• where Te is the trimmed thrust and t is a small perturbation.

• The propeller normal force curve slope becomes:

• Note that we usually ignore propulsion derivatives (i.e. with respect to t) in the lateral equations.

T = Te +τ

e T( ) = e Te +τ( ) = e Te( )+ ∂e∂τ

τ

73

Propeller contributions to Y and N

• Contribution to Y:

• Substituting b=be+v/V0 and e(T):

• Contribution to N:

• Substituting b=be+v/V0 and e(T):

Ya = −Np = −12ρV0

2spe T( )β

Ya =Yae −12ρV0

2spβe∂e∂τ

τ −12ρV0spe Te( )v

Na = Nplp =12ρV0

2splpe T( )β

Na = Nae+12ρV0

2splpβe∂e∂τ

τ +12ρV0splpe Te( )v 74

Propeller aerodynamic derivatives

• The propeller contributions to the aerodynamic derivatives are:!Yv =

12ρV0spe Te( )

!Yτ = −12ρV0

2spβe∂e∂τ

!Nv =12ρV0splpe Te( )

!Nτ =12ρV0

2splpβe∂e∂τ

75

Yv =spSe Te( )

Nv =spSlpe Te( )

Yτ = !Yτ

Nτ =1b!Nτ =

12ρV0

2splpbβe∂e∂τ