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No speed terms
Full solution
0 1 2 3 4
t seconds
2.5
2.0
1.5
1.0
0.5
– / 1nBo
Fig. 5.21 Time history of normal acceleration in response to sudden change of cyclic pitch
Flight dynamics and control 187
n
z B
w p
z
p
p Bp C
B
= – 1 + w
+
+ +
1 1
c
1
2
1 1
Γ
′ ′
(5.175)
where Γ1 = –mq + μmB1 zB1 / and B1′c and C1′ are the quartic coefficients B1c and C1
with the speed derivatives neglected.
The response calculations presented in Figs 5.22 to 5.24 have been made using
eqn 5.175.
Figure 5.22 compares the response of our example helicopter (e = 0.04) with a
hingeless helicopter which, as before, is assumed to have a hub moment five times
larger than that of the 4 per cent offset flapping hinge. It can be seen that the greater
control power and the increased instability result in a rapid increase of acceleration,
and the roots of p2 B p C
+ 1′c + 1′ = 0 indicate this growth to be rapidly divergent.
A peculiarity of helicopter normal acceleration is the sudden jump of acceleration
which occurs with the initial application of cyclic pitch. This is due to the force
produced by the sudden change of rotor disc incidence which is roughly proportional
to speed.
Fitting the tailplane considered earlier to both helicopters, Figs 5.23 and 5.24,
shows that the rate of increase of acceleration reaches a maximum and then begins to
decrease.
5.9.4 Control response in hovering flight
Another important flying quality of the helicopter is the rolling response in hovering
flight, particularly in relation to accurate manoeuvring near the ground. We shall
assume that we need consider only pure rolling, in which case the equations of
motion reduce simply to
d
d
–
d
d
=
2
2 1 1
φ
τ
φτ
lp lAA (5.178)
0 1 2 3 4
t seconds
2.5
2.0
1.5
1.0
0.5
– / 1nBo
e = 0.04
Hingeless
Fig. 5.22 Time history of normal acceleration for hingeless helicopter
188 Bramwell’s Helicopter Dynamics
0 1 2 3 4
t seconds
– / 1nBo
2.0
1.5
1.0
0.5
Tailless, e = 0.04
With tailplane
Fig. 5.23 Effect of tailplane on longitudinal response (articulated rotor)
0 1 2 3 4
t seconds
2.0
1.5
1.0
0.5
– / 1nBo
Hingeless, no tailplane
With tailplane
The rolling response to a sudden application of cyclic pitch A1 is
d
d
= – 1 (1 – e )
1
φτ
l τ
l
A A
p
l p
Numerical values for our example helicopter with 4 per cent hinges and of helicopters
with zero offset and with a hingeless rotor are shown in Fig. 5.25.
It can be seen that the high roll damping of the hingeless helicopter enables it to
reach a constant rate of roll within less than a second, whereas the helicopter with
zero flapping hinge offset has not reached a steady rate by even four seconds. It is
interesting to note that, for a given amount of cyclic pitch, the final rates of roll are
the same. This is because the control power, represented by lA1 , and the roll damping
lp vary in roughly the same proportion, and it is the ratio between them that determines
the final rate of roll. Pilots have described the response of the hingeless helicopter as
‘crisp’, whereas some would say of the helicopter with zero hinge offset that it ‘lags’
the application of control. The probable explanation for this is that the roll damping
of the latter helicopter is so weak that, roughly speaking, control application demands
Fig. 5.24 Effect of tailplane on longitudinal response (hingeless rotor)
Flight dynamics and control 189
roll acceleration which is, so to speak, two stages in advance of the roll angle which
is usually the required response. Thus, since a steady state is not reached in a fairly
short time, considerable anticipation is required of the pilot in order to check the
rolling motion and prevent overshooting the required bank angle. The control of the
hingeless helicopter is said to command roll rate because of the high damping.
An indication of the pilot’s action required to achieve a 20° angle of bank in 2
seconds is shown in Fig. 5.26 for the hingeless helicopter and for one with zero
flapping hinge offset. The roll manoeuvre demanded is assumed to be given by
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Bramwell’s Helicopter Dynamics(96)
