曝光台 注意防骗
网曝天猫店富美金盛家居专营店坑蒙拐骗欺诈消费者
I /HelicaIPathofc.g
SpinAxis -- I
fc.g
Fig. 7.15 Schematicillustration of the effect of wing ffltin spin.
INERTIA COUPLING AND SPIN
~t.
I
SpinA,as -4
"
f
Fig. 7.16 Schematic illustration ofrotation about normal-to-wing chord.
653
We observe that a positive wing tilt Oy reduces the sideslip. When ey = y, the
sideslip is zero.
The sideslip plays animportant rolein the balance ofmoments. Usually, a certain
amount ofsideslip is always necessary to achieve a balance of all three components
of the moment Because the centrifugal force acting on all the components ofthe
airplane is directed radially outward and passes through the spin axis, it cannot
gena:ate any moment about the spin axis. Therefore, the resultant aerodynanuc
force must also pass through the spin axis as indicated in Fig. 7.12. Therefore,
the only way in which an airplane can have the right amount of wing tilt to adjust
the sideslip to the required value is through a rotation about the normal to the
chordline as shown in Fig. 7.16. Note that the normal to the wing chordline is
along the negative z-body axis. Hence, the airplane essentially rotates about its
z-body axis to gene:rate the required amount of wing tilt. Let X denote the angle
by whjich the aircraft is rotate7about the z-body axis. Then the angles X, Oy, and
ct are related by the following expression:r'
SillOy - -cos a- sin X (7.90)
To get a physical understanding of tlus relation, consider the two extreme cases
ofa = 0 and a - 90 deg. At a = 0, the spinning motion is all rolling because the
x-body axis co:incides with the vertical spin axis. The z-body axis is now in the
horizontalplane. So the rotation X about the z-body axis is numerically equalto Oy.
654 PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
For ce - 90 deg, the airplane is in a fiat spin, and the spinning motion is all yaw
about the z-body axis, which now coincides with the sp:in axis. Thus, any amount
of rotaLion about the z-body axis does not give any wing tilt because ffie wings are
in the horizontal plane for all values of X. Hence, the Oy = 0 for a ~ 90 deg.
The angular velocity vector S-2 now has the following components in the body
axes system:
p - s2 cos cr cos X
q - -c2 cos ct sin X
r -. 92 sina
7.5.1 Balance of Forces
.Arith U = V = W = O, Eqs. (7.74-7.76) reduce to
Fx = m(q W '- r V)
Fy = m(Ur - pW)
Fz = m(pV -Uq)
(7.91)
(7.92)
(7.93)
(7.94)
(7.95)
(7.96)
Substituting for U, V, W from Eqs. (7.80-7.82) for p, q, r from Eqs. (7.91-7.93)
and ignorin9 X (cOs x - 1, sin X - 0), we get
Fx - m C22R sincr
Fy =0
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