C hc bay 2 Lateral Motion (Stick Fixed) Ng Khnh Hiu 1 C hc bay 2
Lateral Motion (Stick Fixed) Ng Khnh Hiu 2 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. C hc bay 2 Lateral Motion (Stick Fixed) Ng Khnh
Hiu 3 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: xa xaRolling moments = I p = I pL L||c c
A + A Ac cwhere 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 Ap is equal to , we can
rewrite Eq. (1) as follows:| Apxaaa p paxpLIL 1 pp =where , L
LLILLc c =A +A A = c c=(2) C hc bay 2 Lateral Motion (Stick Fixed)
Ng Khnh Hiu 4 Pure rolling motion (2/4) - The solution to Eq. (2)
for a step change in the aileron angle is: ( )( )t aapLp t 1 e LA =
A(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). ( )apl xass a ap l o xC
QSb/ILp= L C b 2u QSb/I A = A- The steady-state roll rate can be
obtained from Eq. (3), by assuming that time t is large enough that
e-t/ t is essentially 0: aplssao lCp b2u C= A(4) C hc bay 2 Lateral
Motion (Stick Fixed) Ng Khnh Hiu 5 Pure rolling motion (3/4) p al
l2 2xC 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: paa ao2 2op l xo p2 l x ss apb0.0385 (s)2u1Q u 4636.0125
(N/m ) 2b 1L C QSb/I 1.312 (/s) 0.7622 (s)2u LLL C QSb/I 4.6632 (/s
)p 0.31 (rad/s)17.76 (deg/s)L== == = = == = = A = =C hc bay 2
Lateral Motion (Stick Fixed) Ng Khnh Hiu 6 Pure rolling motion
(4/4) Example problem 1: (2/2)p(deg/sec)Time (sec)ssop b0.0122u=If
we fit the solution to the differential equation of motion to the
response we can obtain values for Loa and Lp, in turn, Cloa 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. p al l2 2xC 0.779 (/rad)C 0.17 (/rad)S 945 (ft
)b 96 (ft)I 273000 (slug.ft )= == = =C hc bay 2 Lateral Motion
(Stick Fixed) Ng Khnh Hiu 7 Roll control reversal - The aileron
control power per degree, (pb/2uo)/oa, 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. lrev2l m2kCUc
C C= wherek: the torsional stiffness of the wing, Clo, Clo: the
lift coefficients with respect to a change in angle of attack and
aileron angle, Cmo: the moment coefficient with respect to a change
in angle of attack. (5) C hc bay 2 Lateral Motion (Stick Fixed) Ng
Khnh Hiu 8 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 zrrYawing moments
= I orN I N N N NwhereN r r A = Ac c c cA = A + A+ A+ Ac c cc(6) r
A = A A = A A = Arr rN N + N N | |A A A = A |\ .(7) rrr z z z zN N
r N N where N ;N ; N ; NI I I Ic c c c c c c c= = = =n rUndamped
natural frequency: NNDamping ratio: 2 N== C hc bay 2 Lateral Motion
(Stick Fixed) Ng Khnh Hiu 9 Pure yawing motion (2/4) In the case of
the pure yawing motion, the frequency of oscillation is a function
of the airplanes static stability (weathercock or directional
stability) and the damping ratio is a function of the aerodynamic
damping derivative. rrn nn o2w w2zC 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 A|(0) = 0, Ar(0) = 0. C hc bay 2 Lateral Motion (Stick Fixed)
Ng Khnh Hiu 10 Pure yawing motion (3/4) Solution Ex. 3: (1/2) a.
The yawing moment equation rewritten in state-space form. where N 0
=rr r(7) N + N N A A A = Arr r rr= N r N + N A = AA A A Ainthe
state-space form:x Ax B = +rr r00 1 + N N Nrr ( (( A A( ( = A (( (
(A A( b.The characteristic equation and eigenvalues for the system.
212 0.7602 4.5504 0I A 0 0.38008 2.09904i+ + = = = rr r2 nzr no z2
nzQSbN =C 4.5504 (/s )Ib QSbN C 0.7602 (/s)2u IQSbN C 4.6145 (/s
)I== = = = C hc bay 2 Lateral Motion (Stick Fixed) Ng Khnh Hiu 11
Pure yawing motion (4/4) Solution Ex. 3: (2/2) c. rThe damping
ratio:N0.17822 N = =d.The response of the airplane to a 5o rudder
input. n The undamped natural frequency: N 2.1332 (rad/s) = =2nThe
damped natural frequency: 1 2.099 (rad/s) = =halvenThe time for
halving the amplitude:0.693 t 1.82 (s) = =( )halvenThe number of
cycles for halving the amplitude: N cycles 0.110 0.61 (cycle) = =
Ar AAmplitude degrees or degrees/second Time - seconds C hc bay 2
Lateral Motion (Stick Fixed) Ng Khnh Hiu 12 Lateral-directional
equation of motion (1/)
