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时间:2010-05-30 00:47来源:蓝天飞行翻译 作者:admin
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

is often so large that it is usually relieved by the provision of a lag hinge or equivalent
flexibility, as mentioned in section 1.2. This moment is the moment of the Coriolis
inertia forces acting in the in-plane direction.
More generally, if the rotor hub is pitching with angular velocity q, Fig. 1.8, the
angular velocity components of the blade are
{q sin ψ cos β + Ω sin β, q cos ψ – ˙β , –q sin ψ sin β + Ω cos β}
Ω
k
q
eR j
i
β
ψ
p
Fig. 1.8 Blade influenced by rotor hub pitching velocity q and rolling velocity p
Basic mechanics of rotor systems and helicopter flight 9
where ψ is the azimuth angle of the blade, defined as the angle between the blade
span and the rear centre line of the helicopter. The absolute accelerations of the hinge
point O are the centripetal acceleration Ω2eR acting radially inwards and eR( ˙ q cos ψ
– 2qΩ sin ψ) acting normal to the plane of the rotor hub.
Inserting these values into eqn A.1.15 and neglecting q2, which is usually very
small compared with Ω2, we finally obtain after some manipulation
β˙˙ + Ω2(1 + ε)β = MA/B – 2Ωq(1 + ε) sin ψ + q˙ (1 + ε) cos ψ (1.3)
The second term on the right is the gyroscopic inertia moment due to pitching
velocity, and the third term is due to the pitching acceleration.
We now find that the feathering moment L is
L = A(2Ωq cos ψ + ˙ q sin ψ)
which means that the pitching motion produces a moment tending to twist the blade.
The moment about the lag axis is hardly affected by the pitching motion, so that N
remains as before.
When the rotor hub is rolling with angular velocity p, Fig. 1.8, the equivalent
equation to 1.3 may be derived in like manner, and in this case the extra terms on the
right-hand side can be shown to be (1 + ε) (2Ωp cos ψ + ˙p sin ψ).
1.4 The equation of lagging motion
We assume the flapping angle to be zero and that the blade moves forward on the lag
hinge through angle ξ (Fig. 1.9). The angular velocity of the blade is (Ω + ˙ ξ )k and
a0 = – Ω2eR(cos ξi – sin ξ j). Then the third of Euler’s extended equations gives
Cξ˙˙ + MexgΩ2R2 sin ξ = N (1.4)
or, for small ξ,
ξ˙˙ + Ω2εξ = N/C (1.5)
where ε, in this case, is MbexgR2/C, eR being the drag-hinge offset distance.
It can easily be verified that if flapping motion is also included, the only important
term arising is the moment 2BΩβ ˙β calculated in the previous section. With a lag
Ω
eR
ξ
Ω + ˙ξ
Fig. 1.9 Blade lagging
10 Bramwell’s Helicopter Dynamics
hinge fitted, this moment can be regarded as an inertia moment and considered as
part of N. Then, if NA is taken as the aerodynamic lagging moment, together with any
artificial damping which may be, and usually is, added, eqn 1.5 can be written finally
as
˙˙ ξ + Ω2εξ – 2Ωβ ˙β = NA/C (since B ≈ C) (1.6)
The lagging motion produces no moment about the feathering axis, but the
instantaneous angular velocity Ω + ˙ ξ will affect the centrifugal and aerodynamic
flapping moments and may have to be taken into account when considering coupled
motion (see section 9.7).
1.5 Feathering motion
It is assumed that the flapping and lagging angles are zero and that the blade feathers
through angle θ (Fig. 1.10). The angular velocity components about i, j, k are
˙ θ
,
Ω
sin θ, Ω
cos θ. The first of Euler’s extended equations gives
A˙ θ + AΩ2 sin θ cos θ = L (1.7)
and, for the small feathering angles which normally occur, we can write
θ˙˙ + Ω2θ = L/A (1.8)
The second and third of Euler’s equations show that the feathering motion produces
no flapping moment but a lagging moment of –2A˙θΩ sin θ. This latter moment is
extremely small compared with the flapping Coriolis moment and can be neglected.
1.6 Flapping motion in hovering flight
The equation of blade flapping (1.2) is
d2β/dt2 + Ω2(1 + ε)β = MA/B
It is convenient to change the independent variable from time to blade azimuth
angle by means of ψ = Ωt. Then since
d/dt = Ωd/dψ and d2/dt2 = Ω2d2/dψ2
k
j
θ i
·θ
Fig. 1.10 Blade feathering
Basic mechanics of rotor systems and helicopter flight 11
 
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