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larger than 10° would be regarded as extreme values – so that, referring to Fig. 1.21,
the approximate relations between the thrust and forces referred to the different axes
are
T
TD Resultant
H Tip path plane
HD
a1s
a1 B1
No-feathering
axis
Fig. 1.21 Thrust and H-force vectors
28 Bramwell’s Helicopter Dynamics
T ≈ TD ≈ Ts (1.39)
H ≈ HD + TDa1 ≈ Hs + TsB1 (1.40)
where B1 is the amplitude of the longitudinal cyclic pitch.
1.14 Longitudinal trim equations
The foregoing analysis in this chapter has provided the main features of rotor behaviour,
and demonstrated the dependence of rotor forces on the various different parameters
involved. By extending the analysis to include the helicopter fuselage whilst maintaining
the simplified rotor model, it is possible to derive the equations of equilibrium in
steady, uniform, trimmed flight. From this, the attitude of the fuselage may be
determined. These exercises are done for longitudinal flight in this and the following
section, and lateral control to trim is studied in the final section 1.16.
Referring to Fig. 1.22, and resolving forces in the vertical direction,
W + D sin τc = T cos (θ – B1) – H sin (θ – B1) (1.41)
and, resolving horizontally,
D cos τc = – T sin (θ – B1) – H cos (θ – B1) (1.42)
where θ is the angle between the vertical and shaft, positive nose up, and τc is the
angle of climb. Since θ and B1 are small angles, eqns 1.41 and 1.42 can be written
approximately as
W + D sin τc = T (1.43)
H + D cos τc = T(B1 – θ) (1.44)
Since D sin τc is very much smaller than W, then T ≈ W.
The origin of moments O is defined as the point on the shaft met by the perpendicular
from the c.g. If hR is the height of the hub above this point, and lR the distance of the
T
a1
B1 a1s
eR
Horizon
Fuselage axis Mf
Shaft
Horizon
θ
V
W
Fig. 1.22 Forces and moments in longitudinal plane
τc
lR O
MS
H
hR
Drag
Basic mechanics of rotor systems and helicopter flight 29
c.g. forward of this point, taking moments about O and making the small angle
assumption gives
– WlR – ThRB1 + HhR + Mf – Ms(B1 – a1) = 0 (1.45)
where Mf is the fuselage pitching moment and Ms = fbSeR is the centrifugal moment
per unit tilt of all the blades (eqn 1.38), and fb is a factor depending on the number
of blades.
Solving eqn 1.45 for B1 gives
B1 = (Mf – WlR + HhR + Msa1)/(ThR + Ms) (1.46)
Equation 1.46 gives the longitudinal cyclic pitch required for trim. For very small
or zero offset, we can put Ms = 0 and, since T ≈ W, we have
B1 = Mf/WhR – l/h + H/W (1.47)
and, if the fuselage pitching moment Mf = 0,
B1 = – l/h + H/W (1.48)
The denominator of the right-hand side of eqn 1.46 represents the control moment
for unit displacement of the rotor disc relative to the hub axis. The term ThR is the
moment due to the tilt of the thrust vector, which is the only control moment acting
if the hinges are centrally located (e = 0) or the rotor is of the see-saw type. Ms is the
centrifugal couple due to the flapping hinge offset, section 1.11. We shall see in
Chapter 4 that for a typical offset distance, e = 0.04 say, the total moment is more
than doubled by the offset hinge contribution.
The importance of the offset hinge moment is that it not only augments the control
power but is also independent of the thrust and can be designed to provide adequate
control power in those flight conditions where the thrust is temporarily reduced, e.g.
the ‘push-over’ manoeuvre or the transition from powered flight to autorotation.
Equation 1.48 has a simple physical interpretation: the longitudinal cyclic pitch
B1 must be such as to make the resultant rotor force pass through the helicopter’s
centre of gravity, Fig. 1.23; the figure shows that B1 + l/h = tan–1 (H/T) ≈ H/W, which
is eqn 1.48.
If Ms and Mf are not zero, the resultant force vector no longer passes through the
c.g. but must exert a moment about it in order to balance the Ms and Mf moments.
Resultant
T
H
N.F.A.
hR
c.g.
B1
Fig. 1.23 Resultant rotor force vector
lR
tan–1(HIT)
30 Bramwell’s Helicopter Dynamics
Expressed in terms of the tip path plane, eqn 1.47 becomes
B1 = a1 + HD/W – l/h + Mf /WhR (1.49)
Now HD, the force component in the plane of the rotor disc, is usually quite small,
　
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