bai 5 - lateral motion stick fixed 2008

Cơ học bay 2 – Lateral Motion (Stick Fixed)

Ngô Khánh Hiếu

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Ngô Khánh Hiếu

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Introduction

- The stick fixed lateral motion of an airplane disturbed from its equilibrium state is a complicated combination of rolling, yawing, and sideslipping motions.

- Three potential lateral dynamic instabilities are of interest to the airplane designer: directional divergence, spiral divergence, and so-called Dutch roll oscillation.

Directional divergence: occurs when the airplane lacks the directional or weather-cock stability (directional static stability). If disturbed from its equilibrium state such an airplane will tend to rotate to ever-increasing angles of sideslip, and it will fly a curved path at large sideslip angles. Obviously, such a motion can be avoided by proper design of the vertical tail surface to ensure directional stability.

Spiral divergence: is a non-oscillatory divergence motion that can occur when directional stability is large and lateral stability is small. When disturbed from equilibrium, the airplane enters a gradual spiraling motion. The spiral becomes tighter and steeper as time proceeds and can result in a high-speed spiral dive if corrective action is not taken.

Dutch roll oscillation: is a coupled lateral-directional oscillation. This motion is characterized by a combination of rolling and yawing oscillations that have the same frequency but are out of phase with each other. The period can be on the order of 3 to 15 seconds.


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Pure rolling motion (1/4)

- Consider a wind-tunnel model free to roll about its x axis, the equation of motion for this model of a pure rolling motion is:

x

a xa

Rolling moments = I

δ p = Iδ pL L

φ

φ∂ ∂⇒ ∆ + ∆ ∆

∂ ∂

∑

where the contributions of the rolling moment in the left hand side are due to the deflection of the ailerons and the roll-damping.

(1)

- Because the roll rate ∆p is equal to , we can rewrite Eq. (1) as follows: φ∆

pxδa

aap p

δax

pLIL 1τ p p = δ where τ ,

δL L LI

L

L

∂ ∂ =∆ + ∆ − ∆ = − ∂ ∂ =

(2)


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- The solution to Eq. (2) for a step change in the aileron angle is:

( ) ( )tτδaa

p

Lp t 1 eδL

−∆ = − − ∆ (3)

- (4) for full aileron deflection can be used for sizing the aileron. The minimum requirement for this ratio is a function of the class of airplane under consideration (ex., Cargo or transport airplanes: 0.07; Fighter airplanes: 0.09).

( )δa

p

l xδass a a

p l o x

C QSb/ILp = δ δL C b 2u QSb/I

− ∆ = − ∆

- The steady-state roll rate can be obtained from Eq. (3), by assuming that time t is large enough that e-t/ τ is essentially 0:

δa

p

lssa

o l

Cp b δ2u C

= − ∆ (4)


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pδal l

2 2x

C 0.285 (/rad) C 0.039 (/rad)

S 18 (m ) b 6.7 (m) I 4676 (kg.m )

= − =

= = =

Example problem 1: (1/2)

Calculate the roll response of the F104A to a 5o step change in aileron deflection. Assume the airplane is flying at sea level with a velocity of 87 m/s. the F104A has the following aerodynamic and geometric characteristics:

Solution:

p

aaδa

o

2 2o

p l xo p

δ2δ l x ss a

p

b 0.0385 (s)2u

1Qρu 4636.0125 (N/m ) 2

b 1L C QSb/I 1.312 (/s) τ 0.7622 (s)2u L

LL C QSb/I 4.6632 (/s ) pδ 0.31 (rad/s ) 17.76 (deg/s)

L

=

= =

= = − ⇒ = − =

= = ⇒ = − ∆ = =


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Example problem 1: (2/2) p(deg/sec)

Time (sec)

ss

o

p b 0.0122u

=

If we fit the solution to the differential equation of motion to the response we can obtain values for Lδa and Lp, in turn, Clδa and Clp. The technique of extracting aerodynamic data from the measured response is often called the inverse problem or parameter identification.

Example problem 2:

Calculate the roll response of the De Havilland Canada airplane to a 5o step change in aileron deflection. Assume the airplane is flying at sea level with a velocity of 87 m/s.

pδal l

2 2x

C 0.779 (/rad) C 0.17 (/rad)

S 945 (ft ) b 96 (ft) I 273000 (slug.ft )

= − =

= = =


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Roll control reversal

- The aileron control power per degree, (pb/2uo)/δa, is essentially a constant when the speed is low. But at high speeds, it decreases until a point is reached where roll control is lost. This point is called the aileron reversal speed.

- The loss and ultimate reversal of aileron control is due to the elasticity of the wing.

δ

α δ

lrev 2

l m

2kCU

ρc C C= −

where k: the torsional stiffness of the wing, Clα, Clδ: the lift coefficients with respect to a

change in angle of attack and aileron angle,

Cmα: the moment coefficient with respect to a change in angle of attack.

(5)


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Pure yawing motion (1/4)

- Examine the motion of an airplane constrained so that it can perform only a simple yawing motion. The equation of motion can be written as follows:

- Because the center of gravity is constrained, we have:

x z

rr

Yawing moments = Iψ or N I ψ

N N N Nwhere Nβ β r δβ r δβ

∆ = ∆

∂ ∂ ∂ ∂∆ = ∆ + ∆ + ∆ + ∆

∂ ∂ ∂∂

∑

(6)

ψ β ψ β ψ r∆ = −∆ ∆ = −∆ ∆ = ∆

rrβ δ rβ

ψ N N ψ + N ψ N δ

∆ − − ∆ ∆ = ∆

(7)

rr

rβ δβz z z z

NδN r Nβ N βwhere N ; N ; N ; NI I I I

∂ ∂∂ ∂ ∂ ∂ ∂ ∂= = = =

nβ

r

β

Undamped natural frequency: ω N

NDamping ratio: 2 N

ξ

=

= −


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In the case of the pure yawing motion, the frequency of oscillation is a function of the airplane’s static stability (weathercock or directional stability) and the damping ratio is a function of the aerodynamic damping derivative.

β r

δr

n n

n o

2w w

2z

C 0.071 (/rad), C 0.125 (/rad)

C 0.072 (/rad), u 176 (ft/s)

S 184 (ft ), b 33.4 (ft)

I 3530 (slug.ft )

= = −

= − =

= =

=

Example problem 3:

Suppose an airplane is constrained to a pure yawing motion, using the data for the general aviation airplane in Appendix B, determine the following:

a. The yawing moment equation rewritten in state-space form.

b. The characteristic equation and eigenvalues for the system.

c. The damping ratio, and undamped natural frequency.

d. The response of the airplane to a 5o rudder input. Assume the initial conditions are ∆β(0) = 0, ∆r(0) = 0.


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Solution Ex. 3: (1/2)

a. The yawing moment equation rewritten in state-space form.

βwhere N 0=

rrβ δ r(7) ψ N ψ + N ψ N δ⇒ ∆ − ∆ ∆ = ∆

rrβ δ r

ψ r

r = N r Nψ + N δ

∆ = ∆

∆ ∆ − ∆ ∆

in the state-space form: x Ax Bη= +

rr

δβ r

0 0 1ψ ψ+δ

NN N rr

∆ ∆ = ∆ − ∆ ∆

b. The characteristic equation and eigenvalues for the system.

2

12

λ 0.7602λ 4.5504 0λI A 0 λ 0.38008 2.09904i

+ + =− = ⇒ = − ±

β

r

rδr

2β n

z

r no z

2δ n

z

QSbN =C 4.5504 (/s )Ib QSbN C 0.7602 (/s)

2u IQSbN C 4.6145 (/s )

I

=

= = −

= = −


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Solution Ex. 3: (2/2)

c.

r

β

The damping ratio:N 0.1782

2 Nξ = − =

d. The response of the airplane to a 5o rudder input.

nβ

The undamped natural frequency:

ω N 2.1332 (rad/s)= =

2n

The damped natural frequency:

ω ω 1 2.099 (rad/s)ξ= − =

halven

The time for halving the amplitude:0.693 t 1.82 (s)

ωξ= =

( )halven

The number of cycles for halving the amplitude:ω N cycles 0.110 0.61 (cycle)ωξ

= =

ψ∆

r∆

Ampl

itude

deg

rees

or d

egre

es/s

econ

d

Time - seconds


Ngô Khánh Hiếu

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Lateral-directional equation of motion (1/)

bai 5 - lateral motion stick fixed 2008

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