Spacecraft and Aircraft Dynamics - Lecture 4 ...control.asu.edu/Classes/MMAE441/Aircraft/441Lecture4.pdf · Aircraft Dynamics Lecture 4 In this lecture, we will discuss Airfoils:

Spacecraft and Aircraft Dynamics

Matthew M. PeetIllinois Institute of Technology

Lecture 4: Contributions to Longitudinal Stability

Aircraft DynamicsLecture 4

In this lecture, we will discuss

Airfoils:

• The wing Contribution to pitching momentI Rotate Lift and Drag vectorsI Calculate moment about CG ( ~M = ~r × ~F )I Use approximations to simplify expression.

M. Peet Lecture 4: 2 / 26

Two Confusing Figures

Confusion: For an airfoil, angle ofattack is measured to the zero-lift-line.

• Thus CM0 = 0 for an un-inclinedairfoil.

Confusion: We assume that theaerodynamic center is on the FRL.

• Thus as measure from the CG,

~rac =

Xcg −Xac

0Zcg

.

• If there is any confusion on aproblem, ask me to clarify.

M. Peet Lecture 4: 3 / 26

Next Subject: Wing ContributionLift and Drag

The wing is a lifting surface which produces both Lift and Drag.• Lift is perpendicular to free-stream velocity.

I If αFRL > 0, then lift vector pitches forward (nose-down direction) in thebody-fixed frame.

• Drag is parallel to free-stream velocity.I If αFRL > 0, then drag also rotates in the nose-down direction.

To determine the contributions of Lift and drag in the body-fixed frame, theseforces must be rotated by the angle of attack and any additional winginclination.

M. Peet Lecture 4: Wing Contribution 4 / 26

Rotating the Lift and DragRotation Matrices

Lift and Drag vectors rotate with angle of attack and wing inclination.

Example: Given a vector ~v =

x

y

z

and a pitch-up rotation, θ,

~v′ =

cos θ 0 sin θ0 1 0

− sin θ 0 cos θ

x

y

z

=

x cos θ + z sin θy

−x sin θ + z cos θ

The matrix is called a rotation matrix.


Rotating VectorsRotation Matrices

Rotation matrices can be used to calculate the effect of ANY rotation.

Roll Angle φ :

~v′ = R1(φ)~v

Pitch Angle θ :

~v′ = R2(θ)~v

Yaw Angle ψ :

~v′ = R3(ψ)~v

Remember to use the right-hand rule to determine what is a positive rotation.


Rotating VectorsRotation Matrices

The rotation matrices are (for reference):

Roll Rotation (φ) :

R1(φ)

=

1 0 00 cosφ − sinφ0 sinφ cosφ

Pitch Rotation (θ):

R2(θ)

=

cos θ 0 sin θ0 1 0

− sin θ 0 cos θ

Yaw Rotation (ψ):

R3(ψ)

=

cosψ − sinψ 0sinψ cosψ 00 0 1


Rotating VectorsRotation Matrices: Multiple Rotations

Rotation matrices, can be used to calculate a sequence of rotations:

Roll-Pitch-Yaw:~vRPY = R3(ψ)R2(θ)R1(φ)~v

Note the order of multiplication is critical.

~vRPY =

(

R3(ψ)

(

R2(θ)(

R1(φ)~v)

1

)

2

)

3


Rotating VectorsRotation Matrices: Example

Consider a pure lift force of 10MN after a pitch up of 10 deg and a yaw of20 deg.

~L =

00

−10

~LPY =

cos 20 − sin 20 0sin 20 cos 20 00 0 1

cos 10 0 sin 100 1 0

− sin 10 0 cos 10

00

−10

=

0.9397 −0.3420 00.3420 0.9397 0

0 0 1.0000

0.9848 0 0.17360 1.0000 0

−0.1736 0 0.9848

00

−10

=

0.9254 −0.3420 0.16320.3368 0.9397 0.0594−0.1736 0 0.9848

00

−10

=

−1.6318−0.5939−9.8481


Rotating VectorsRotation Matrices: Example

Compare this to the same rotations in the reverse order (Yaw, then Pitch)

~LY P =

cos 10 0 sin 100 1 0

− sin 10 0 cos 10

cos 20 − sin 20 0sin 20 cos 20 00 0 1

00

−10

=

−1.73650

−9.8481

Which is still in the x− z plane!!! Why?

Compare to Pitch, Yaw value:

~LPY =

−1.6318−0.5939−9.8481



The magnitude of lift and drag are given by:

Lwing = CL,wQS and Dwing = CD,wQS

The two important angles are angle of attack of the aircraft, αFRL and• inclination iw of the wing. Most wings are rotated up - this increases theangle of attack of the wing.

CL can be determined as

CL = CLααw = CLα(αFRL + iw)

where αw = αFRL + iw is the angle of the wing with the free-stream.M. Peet Lecture 4: Wing Contribution 11 / 26


CD is usually determined using CDα and CD0 from the tables.

CD = CDααw + CD0


Wing ContributionMoment Contribution

The wing makes three contributions to the moment about the Center of Gravity (CG)of the aircraft.

• The lift force, multiplied by “lever arm”

• The drag force, multiplied by a shorter “lever arm”

• The moment exerted on the wing itself (due to CM0)

As a reference point, we choose the leading edge of the wing.

• Xac is the distance from the leading edge to the aerodynamic center of the wing

• Xcg is the distance from the leading edge to the center of mass of the aircraft

• Zcg is the height of the CG above the Fuselage Reference Line (FRL)


Wing ContributionMoment Contribution

The moment produced by a force about point is

~M = ~r × ~F

For example, ~F =

Fx

0Fz

and ~r =

rx0rz

, which implies

~M = ~r × ~F = det

x y z

rx 0 rzFx 0 Fz

= −(rxFz − Fxrz)y.


Wing Contribution: Simple CaseZero Angle of Attack

Consider when α = 0. Then the force in the body-fixed frame, the lift and drag forcesare

Fw,α=0 =

−Dw

0−Lw

, applied at ~rw =

Xcg −Xac

0Zcg

Note that we assume the aerodynamic center is on the FRL. Now we have

~Mw,α=0 = M0,w y + ~r × ~F = Mwy + det

x y z

Xcg −Xac 0 Zcg

−Dw 0 −Lw

= (M0,w + Lw(Xcg −Xac)−DwZcg) y.


Wing Contribution: Simple CaseZero Angle of Attack

Thus the pitching moment produced by the lift and Drag at zero angle of attack

(αFRL = 0) is

Mw,α=0 =M0,w + Lw(Xcg −Xac)−DwZcg.

Converting to non-dimensional form, the pitching moment is

CMw,αFRL=0 = CM,w + CL,w(Xcg −Xac)− CD,wZcg.


Wing ContributionMoment Contribution: Angle of Attack

• Because Lift is perpendicular to free-stream and Drag is parallel, a positive angleof attack will rotate ~F in the negative pitch direction (nose down-direction).

• In the body-fixed frame, these forces become

~Fw =

cos(−α) 0 sin(−α)0 1 0

− sin(−α) 0 cos(−α)

−Dw

0−Lw

=

−Dw cosα+ Lw sinα

0−Dw sinα− Lw cosα

where α = αFRL and recall sin(−α) = − sinα, cos(−α) = cosα

Now we can calculate the moment contribution as before from

~Mw = M0,w + ~rw ×~Fw


Wing ContributionMoment Contribution: Angle of Attack

This contribution is readily calculated as

~Mw = ~M0,w + ~rw × ~Fw

That is,

Mw =M0,w + FxZcg − Fz(Xcg −Xac)

=M0,w + (Lw sinα−Dw cosα)Zcg

+ (Dw sinα+ Lw cosα) (Xcg −Xac)

or, in non-dimensional form, divide by QSc to get

CM,w = CM,0,w + (CL,w sinα− CD,w cosα)Zcg

c

+ (CD,w sinα+ CL,w cosα) (Xcg

c−Xac

c)

Recall that

CL,w = CLααw = CLα(αFRL + iw) and CD = CDααw + CD0

At this point we decide the equation are too complicated. We need to makesome Approximations.


Wing Contribution: SimplificationSmall Angle Approximations

Our moment equation is now

CM,w = CM,0,w + (CL,w sinα−CD,w cosα)Zcg

c

+ (CD,w sinα+CL,w cosα)

(

Xcg

c−

Xac

c

)

But this is nonlinear and I promised linear equations. Therefore we use theSmall Angle Approximations to linearize. If α ∼= 0, then

sinα ∼= α cosα ∼= 1


Wing Contribution: SimplificationOther Approximations

Replacing sinα with α and cosα with 1, we get

CM,w∼= CM,0,w + (CL,wα− CD,w)

Zcg

c+ (CD,wα+ CL,w)

(

Xcg

c−Xac

c

)

.

The equation is linear, but still complicated.

We now make 2 more assumptions

1. Zcg is small compared to Xcg −Xac.I CD,wZcg +CL,w (Xcg −Xac) ∼= CL,w (Xcg −Xac)

2. α is small.I CD,wα+CL,w

∼= CL,w

What we have left is

CM,w∼= CM,0,w + CL,w

(

Xcg

c−Xac

c

)

.


Wing Contribution: ApproximationsExceptions


Wing Contribution: ApproximationsExceptions


Wing Contribution: Continued

We now include the expression for Lift:

CL,w = CLα,wαw = CLα,w (αFRL + iw)

this yields

CM,w = CM,0,w + CLα,w (αFRL + iw)

(

Xcg

c−Xac

c

)

= CM,0,w + CLα,wiw

(

Xcg

c−Xac

c

)

+ CLα,wαFRL

(

Xcg

c−Xac

c

)

αFRL

Thus we haveCM,total = CM,total,0 + CM,total,ααFRL

where

CM,total,0 = CM,0,w + CLα,wiw

(

Xcg

c−Xac

c

)

CM,total,α = CLα,w

(

Xcg

c−Xac

c

)


Wing Contribution: Stability

Recall for nose-up steady flight, we need

CM,total,0 = CM,0,w + CLα,wiw

(

Xcg

c−Xac

c

)

≥ 0

and

CM,total,α = CLα,w

(

Xcg

c−Xac

c

)

≤ 0

Since CLα,w > 0, this can only be achieved when we have both

CG ahead of wing:

Xcg −Xac < 0

which is unusual.

Negative Camber:

CM,0,w > 0

which would be odd.

Thus we need a tail contribution.


Conclusion

In this lecture, we have

• Calculated the moment about the CG produced by Lift and Drag on aWing.

• Made small-angle approximations to get a linear expression.

• Neglected some Moment contributions.

• Developed impractical conditions for stability of a wing-only aircraft


Next Lecture

• Do the same thing for the tail.

• Discuss our new design variables.

• Discuss equilibrium and stability conditions.


Documents

Spacecraft and Aircraft Dynamics - Lecture 4 ...control.asu.edu/Classes/MMAE441/Aircraft/441Lecture4.pdf · Aircraft Dynamics Lecture 4 In this lecture, we will discuss Airfoils: