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时间:2010-05-30 00:47来源:蓝天飞行翻译 作者:admin
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μ =0.05
–2
–3
Flight dynamics and control 175
It is interesting to note that increasing the rotor hub ‘stiffness’ (hub moment per
unit rotor tilt) is equivalent to decreasing the moment of inertia. It happens that
increasing the hub moment five times in the longitudinal case produces almost the
same effect as changing from the longitudinal to lateral moments of inertia, and we
see that the cubic equations for these cases are numerically very similar.
Let us decrease the moment of inertia so that it approaches zero. It can easily be
seen that the cubic characteristic equation, eqn 5.133, then degenerates to the quadratic
λ2 – yvλ – (lv/lp)wc = 0
whose roots are
λ = [ + ( / ) ] 12
1
4
2
yv± √ yv lv l p wc
This represents a lightly damped oscillation whose period, ignoring the yv term, is
T = 2πtˆ√(lp /lvwc )
This, in our notation, is Hohenemser’s formula1 for the period of oscillation in
hovering flight. For the longitudinal case the period is
T = 2πtˆ√(–mq/muwc )
For the case considered,
Time of oscillation (seconds)
Exact Hohenemser
Longitudinal (4% offset) 17.5 14.2
Longitudinal (hingeless) 15.4 15.1
Lateral (4% offset) 14.8 14.2
Lateral (hingeless) 15.2 15.0
Thus, for high control power or low inertia, Hohenemser’s formula gives quite
accurate results.
The stability in forward flight with the hingeless rotor, Fig. 5.16, shows that,
unlike the longitudinal case, the lateral stability is generally improved compared with
the aircraft with 4 per cent offset hinges.
5.8 Autostabilisation
We have seen that the helicopter is unstable both laterally and longitudinally in
hovering flight and that the longitudinal instability becomes worse with increase of
forward speed, particularly when the rotor has hingeless blades. We have also seen
that a tailplane is really effective only in the upper half of the speed range and that,
since the unstable characteristics of the rotor also deteriorate with speed, the tailplane
may be incapable of making the machine completely stable.
176 Bramwell’s Helicopter Dynamics
Although adequate control power is usually available to correct disturbances, an
unstable aircraft will require continuous correction and will be tiring to fly for long
periods, even in quite calm weather. Furthermore, in some conditions, such as flying
on instruments, an unstable aircraft could be quite dangerous and it is clearly desirable
to provide some form of artificial stabilisation to make good the inherent deficiencies.
The stabilisation devices in common use fall into two categories:
(i) a mechanical/gyro device which is an integral part of the rotor system, as used on
Bell and Hiller helicopters;
(ii) automatic flight control systems using feedback control based on signals from
sensors such as attitude or rate gyros.
5.8.1 Mechanical/gyro devices
The simple Bell stabilising bar embodies the essentials of all mechanical/gyro devices.
The Bell bar is basically a bar pivoted to the rotor shaft, Fig. 5.17, and provided with
viscous damping. The bar behaves like a gyroscope with lag damping.
The bar is linked to the blade so that a tilt of the bar relative to the shaft causes a
change of pitch of the rotor blade.
Since the bar can pivot relative to the shaft, its equation of motion is precisely the
same form as that of the blade under the excitation of gyroscopic and inertia moments,
eqn 1.16. Thus, if θbar is the angular displacement of the bar, its equation of motion
is
θ˙˙bar fθ˙bar θ ψ ˙ ψ
2
+ (2/T) + Ω bar = – 2Ωq sin + q cos
in which the viscous damping is conveniently represented by the ‘following’ time Tf,
or the time taken for a sudden displacement of the bar to diminish by 63 per cent. The
last term of eqn 1.16 is negligibly small in practice and so the equation above can
also be written
d2θbar/dψ2 + (2/TfΩ)dθbar/dψ + θbar = – 2 ˆ q sin ψ (5.134)
Let cl be the linkage ratio such that
θ = clθbar (5.135)
θ
Control
Damper
Bar
θbar
Fig. 5.17 Bell stabilising bar
Flight dynamics and control 177
so that, if cl is unity, one degree of the tilt of the bar produces one degree of blade
pitch. Since the bar does not affect the collective pitch, eqn 5.135 can be written in
the form
clθbar = – A1 sin ψ – B1 cos ψ (5.136)
where the cyclic pitch amplitudes A1 and B1 are functions of time.
 
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