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The development of a satisfactory feathering mechanism was the last link in the
creation of a successful helicopter and it enabled the rotor to be controlled without
tilting the hub or shaft, as had been possible with the free-wheeling autogyro rotor.
A brief description of the feathering mechanism is given below.
The principal feature of the feathering system is the swash plate mechanism, Fig.
1.13. This consists of two plates, of which the lower plate does not rotate with the
shaft but can be tilted in any direction by the pilot’s cyclic control. The upper plate
rotates with the shaft but is constrained to remain parallel to the lower plate. It can
be seen that if the swash plates are tilted the blade chord remains parallel to the swash
plate and, as the blade rotates with the shaft, cyclic feathering takes place relative to
18 Bramwell’s Helicopter Dynamics
the plane perpendicular to the shaft. The swash plate is, of course, a plane of nofeathering,
and the axis through the centre of the hub and perpendicular to the swash
plates is the no-feathering axis. Figure 1.5(b) shows a typical swash plate control
mechanism.
Other feathering mechanisms have been employed such as that in Fig. 1.6(a), but
the one described above is used a majority of helicopters.
Collective (constant) pitch is applied by the collective lever which effectively
raises or lowers the swash plate without introducing further tilt; this alters the pitch
angle of all the blades by the same amount.
1.8 Lagging motion
The lagging-motion equation, eqn 1.6, is
ξ˙˙ + Ω2εξ – 2Ωββ˙ = N/C
or
d2ξ /dψ2 + εξ – 2β dβ /dψ = N/CΩ2 (1.20)
where, as explained in section 1.4, the term 2Ωβ ˙β represents the Coriolis moment
due to blade flapping.
In finding the free lagging motion, we assume the flapping motion to be absent
and take N to be the aerodynamic (drag) moment of the blade about the lag hinge.
Let dΩ2 be the drag of the blade when it rotates at steady angular velocity Ω. The
lagging motion increases the instantaneous angular velocity of the blade to Ω + ˙ξ
and the drag can be assumed to be d(Ω + ˙ξ)2. Since ˙ξ is small compared with Ω, the
drag is approximately dΩ2 + 2dΩ ˙ξ
. If RD is the distance of the centre of drag of the
blade from the hub, and assuming the lag hinge offset to be small,
N = – dRD(Ω2 + 2Ω ˙ ξ )
But dRDΩ3 is the power, P say, required to drive one blade. Therefore
N = –(P/Ω)(1 + 2˙ ξ /Ω)
so that eqn 1.20 can be written
d2ξ/dψ2 + (2P/CΩ3)dξ/dψ + εξ = –P/CΩ3 (1.21)
Equation 1.21 is the equation of damped harmonic motion about a steady value
ξ = – P/CΩ3ε. Typical values of P/CΩ3 and ε are 0.006 and 0.075 respectively, giving
a steady value of ξ of about 412
°. The frequency of oscillation is 0.27Ω and the
damping is only about 2 per cent of critical. The much lower damping of the lagging
mode is due to the fact that the blade motion in this case is governed by the changes
of drag, and not of incidence. This low natural damping is usually augmented by
hydraulic or elastomeric damping to avoid potential instability problems.
Basic mechanics of rotor systems and helicopter flight 19
1.9 Lagging motion due to flapping
It will be assumed that the lag hinges are parallel to the rotor shaft so that ξ represents
a change of blade angle in the plane of the hub, as in the previous section. The
flapping motion must also be taken relative to this plane, and we will assume it takes
the form β = a0 – a1 cos ψ. In some later work we will have to distinguish flapping
relative to the shaft from that relative to the no-feathering axis by writing βs = a0s +
a1s cos ψ.
Regarding the Coriolis moment due to flapping as a forcing function, and ignoring
the damping, we rewrite eqn 1.20 as
d2ξ/dψ2 + εξ = 2β dβ/dψ + N/CΩ2 (1.20a)
and we have to consider the meaning of the moment N in this case. Since the blade
lift is perpendicular to the flow direction, the axis of the aerodynamic flapping
moment must lie in the tip path plane, because this is the plane in which the blades
move steadily. Thus there must be a component of the flapping moment about the lag
hinge, and it is clear from Fig. 1.14 that this component is – MAa1 sin ψ and is equal
to N.
But, we have seen that, for first harmonic flapping, and assuming zero flapping
hinge offset, MA = BΩ2a0, as explained in section 1.6.2. Therefore
　
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