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時間:2010-05-30 00:47來源:藍天飛行翻譯 作者:admin
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of B1 denotes a forward (nose down) tilt of the cone, Fig. 1.11, while a positive value
of A1 denotes a sideways component of tilt in the direction of ψ = 90°. The blade tips
trace out the ‘base’ of the cone, which is often referred to as the tip path plane or as
the rotor disc, Fig. 1.11.
In steady flight the blade motion must be periodic and is therefore capable of
being expressed in a Fourier series as
β = a0 – a1 cos ψ – b1 sin ψ – a2 cos 2ψ – b2 sin 2ψ – … (1.13)
For the case in question,
a0 = γ θ0/8, a1 = – B1, b1 = A1
a2 = b2 = … etc. = 0
When the flight condition is steady, eqn 1.9 can always be solved by assuming the
Basic mechanics of rotor systems and helicopter flight 13
form of eqn 1.13, substituting in the flapping equation, and equating coefficients of
the trigonometric terms. This is a method we shall be forced to adopt when the
flapping equation contains periodic coefficients, as will be the case in forward flight.
In terms of eqn 1.13, a0 represents the coning angle and a1 and b1 represent
respectively, a backward and sideways tilt of the rotor disc, the sideways tilt being in
the direction of ψ = 90°. The higher harmonics a2, b2, a3,…, etc., which will have
non-zero values in forward flight, can be interpreted as distortions or a ‘crinkling’ of
the rotor cone. But although these harmonics can be calculated, the blade displacements
they represent are only of the same order as those of the elastic deflections which, so
far, have been neglected. Thus, it is inconsistent to calculate the higher harmonics of
the rigid blade mode of motion without including the other deflections of the blade.
Stewart1 has shown that the higher harmonics are usually about one tenth of the
values of those of the next order above.
Comparison of eqns 1.11 and 1.12 shows that the amplitude of the periodic flapping
is precisely the same as the applied cyclic feathering and that the flapping lags the
cyclic pitch by 90°. The phase angle is exactly what we might have expected, since
the aerodynamic flapping moment forces the blade at its undamped natural frequency
and, as is well known, the phase angle of a second order dynamic system at resonance
is 90° whatever the damping. Further, the fact that the amplitude of flapping is
exactly the same as the applied feathering has a simple physical explanation. Suppose
that initially no collective or cyclic pitch were applied; the blades would then trace
out a plane perpendicular to the rotor shaft. If cyclic pitch were then applied, and the
blades remained in the initial plane of rotation, they would experience a cyclic
variation of incidence and, hence, of aerodynamic moment. The moment would
cause the blades to flap and, since, as we have found, blade flapping motion is stable,
the blades must seek a new plane of rotation such that the flapping moment vanishes.
This is clearly a plane in which there is no cyclic feathering and it follows from
Fig. 1.12 that this plane makes the same angle to the shaft as the amplitude of the
cyclic pitch variation. It is also obvious that the effect of applying cyclic pitch is
precisely the same as if cyclic pitch had been absent but the shaft had been tilted
through the same angle. Tilting the rotor shaft or, more precisely, the rotor hub
plane, is the predominant method of controlling the rotor of an autogyro. Tilting
the shaft of a helicopter is impossible if it is driven by a fuselage mounted engine,
and the rotor must be controlled by cyclic feathering.
Tip path plane a0
B1
Fig. 1.11 Interpretation of flapping and feathering coefficients
14 Bramwell’s Helicopter Dynamics
The above discussion illustrates the phenomenon of the so-called ‘equivalence of
feathering and flapping’; the interpretation is a purely geometric one. If flapping and
feathering are purely sinusoidal, the amplitude of either depends entirely upon the
axis to which it is referred. In Fig. 1.12, aa′ is the shaft axis, bb′ is the axis perpendicular
to the blade chord, cc′ the axis perpendicular to the tip path plane. If Fig. 1.12 shows
the blade at its greatest pitch angle, bb′ is clearly the axis relative to which the cyclic
feathering vanishes and is called the no-feathering axis. Similarly cc′ is the axis of no
flapping.
Let a1s be the angle between the shaft and the tip path plane and B1 the angle
between the shaft and the no-feathering axis. Viewed from the no-feathering axis the
cyclic feathering is, by definition, zero but the angle of the tip path plane is a1s – B1.
On the other hand, viewed from the tip path plane, the flapping is zero but the
 
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本文鏈接地址:Bramwell’s Helicopter Dynamics(14)
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