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N/CΩ2 = – a0a1 sin ψ
and, using the assumed form of β = a0 – a1 cos ψ, then
2β dβ/dψ + N/CΩ2 = a0a1 sin ψ – a1
2 sin 2ψ
where, again, we have taken B = C. The solution to eqn 1.20a is
ξ = – [a0a1/(1 – ε)] sin ψ + [ a1
2 /(4 – ε)] sin 2ψ
The second term is generally smaller than the first and, since, ε is very small, to
a fair approximation ξ can be written
Lag hinge axis
MA
Hub axis
a1
Fig. 1.14 Aerodynamic flapping moment component about lag hinge
a0
20 Bramwell’s Helicopter Dynamics
ξ = – a0 a1 sin ψ (1.22)
Thus, the flapping motion forces a lagging motion which lags the flapping motion
by 90°.
Equation 1.22 has a simple physical explanation. It can be seen from Fig. 1.15 that
the blade movement about the lag hinge, i.e. in the hub plane, is simply steady
motion of the blade in the tip path plane projected onto the hub plane. In other words,
the ξ motion calculated above corresponds to uniform motion in the tilted rotor cone,
and this is a result we should expect, for, since the flapping angle is constant relative
to the tip path plane, there can be no Coriolis moments in this plane and the blade
must rotate with constant angular velocity.
1.10 Feathering motion
For small pitch angles the blade feathering equation can be written
d2θ/dψ2 + θ = L/AΩ2 (1.23)
For free motion, L = 0, and the blade oscillates with shaft frequency. If θ is held
constant, corresponding to collective pitch application, there will be a moment AΩ2θ
trying to feather the blade into fine pitch. This moment is called the ‘feathering
moment’ and must be resisted by the feathering mechanism. The fact that the natural
a1
a0
ξ
Fig. 1.15 Blade lagging motion due to flap
Basic mechanics of rotor systems and helicopter flight 21
frequency of the feathering motion is exactly one cycle per revolution of the shaft –
exactly as required – means that, except to overcome friction, no forces are necessary
in the control links to maintain the motion.
The feathering moment AΩ2θ can be explained in terms of the centrifugal forces
acting on the blade. In Fig. 1.16, AA′ is a chord of the blade. Consider elementary
masses at the leading and trailing edges of the blade. The centrifugal forces acting on
these masses are inclined outwards and, therefore, have components in opposite
directions. But the centrifugal force directions both lie in planes perpendicular to the
shaft so that when the blade is pitched the opposite directed components exert a
couple tending to feather the blade into fine pitch. Integrating this moment in the
chordwise and spanwise directions can be shown to lead to AΩ2θ.
As with the flapping motion, the centrifugal moment acts like a spring giving a
frequency exactly equal to that of the shaft, but, again like the flapping motion, if the
feathering motion is viewed from a plane passing through the chord, the feathering
motion vanishes and the centrifugal moment in this plane also vanishes, Fig. 1.17.
1.11 Rotor forces and moments
So far we have derived the equations of blade flapping, lagging, and feathering and
have considered some simple cases of blade motion to illustrate some of its dynamic
properties. We now have to consider the effect of this blade motion on the helicopter
as a whole. We shall derive expressions for the forces and moments on the helicopter
A
A′
A
A′
Fig. 1.16 Feathering moment
Positive
incidence
Negative
incidence
Tilt of
shaft
Fig. 1.17 Feathering in tip path plane due to rotor tilt
Rotor
plane
22 Bramwell’s Helicopter Dynamics
and consider requirements for trimmed flight. These requirements will appear as the
control angles necessary to establish a given flight condition analogous to the static
stability analysis of the fixed wing aircraft.
In order to be able to write down the equations of motion of the helicopter in
steady and accelerated flight, it is necessary to calculate the forces exerted by the
blade on the hub. To do this we shall have to relate the motion expressed in terms of
axes fixed in the blade to axes fixed in the rotating hub and then to axes fixed in the
helicopter.
As before, let, i, j, k be the set of unit axes fixed in the blade. Let e1, e2, e3 be a
set of unit axes fixed in the rotating hub, Fig. 1.18.
When the blade is in its undeflected position, i.e. when there is no flapping or
　
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