曝光臺 注意防騙 網(wǎng)曝天貓店富美金盛家居專營店坑蒙拐騙欺詐消費者

*If the centre of lift coincides with the centre of percussion of the blade the initial hinge reaction
will be zero.
It is sometimes said that, because helicopter blades can flap or bend, the rotor is
effectively self-alleviating with regard to the gust loads. There is a grain of truth in
this as far as the initial load is concerned because, while the blades are accelerating,
the reaction at the hinge will usually be less than the sudden change of lift*. But we
know that the blade flapping response is extremely rapid and can be regarded as
practically complete within about 1
4 second, so that after that time the lift load will
be fully transmitted to the hub and airframe. This extremely short time interval is
insignificant compared with the total response time, and justifies the assumption that
the blade flapping motion can be ignored. Furthermore, the rather academic example
of a sharp-edged gust represents the worst case; the more realistic ‘ramp’ type of gust
would give the blades even more time to reach equilibrium before the maximum gust
velocity is reached.
It is interesting to compare the initial acceleration response of the helicopter with
that of the fixed wing aircraft. If wg is the initial gust velocity, the normal acceleration
is given by
1 2 3
t seconds
e = 0.04
Hingeless
1
0.5
0
–0.5
Normal acceleration (g units)
Fig. 5.28 Effect of tailplane on gust response
192 Bramwell’s Helicopter Dynamics
ng
Z w
W g
w =
–
/
g
for the helicopter and for the fixed-wing aircraft.
For the helicopter,
n
z sA Rw
w sA R
z w
w R
w w = – = – g
c
2 2
g
c
ρ
ρ
Ω
Ω Ω
and for the fixed-wing aircraft
n z SW
= – wρ Vwg
For the helicopter we can take zw = –2aμ/(8μ + as), and for the fixed wing aircraft
z a w = – , 12
where a in the latter case is the lift slope of the complete aircraft.
Let us compare our example helicopter (hinged or hingeless) with a fixed wing
aircraft whose wing loading W/S is 1450 N/m2 and whose lift slope is 4. Then, if wg
is 10.5 m/s (standard gust velocity), the initial acceleration for both aircraft is as
shown in Fig. 5.29.
That the normal acceleration of the helicopter is practically constant is easily seen
physically since, for even quite high forward speeds, the rotor lift is dominated by the
(constant) rotational speed of the rotor, so the effect of a given vertical gust is almost
independent of speed. On the other hand, the response of the fixed wing aircraft
increases with speed.
5.9.6 The longitudinal characteristic quartic coefficients E1 and C1
The coefficient E1. Consider a helicopter changing from one trimmed speed to
another, the collective pitch remaining fixed. For small changes of speed from level
flight, the resulting angle of climb or descent will be quite small, corresponding to
the ‘gliding flight’ of classical fixed wing stability theory.
Fixed wing
W/S = 1430 N/m2
Helicopter
W/A = 270 N/m2
2.0
1.5
1.0
0.5
n
0 20 40 60 80 100 m/s
Speed
Fig. 5.29 Comparison of helicopter and fixed wing aircraft response to vertical gust
Flight dynamics and control 193
The pitching moment under these conditions will be a function of the tip speed
ratio μ, the (non-dimensional) vertical velocity w, and the longitudinal cyclic pitch
B1. Since the helicopter is supposed to be in trim, we must have Cm(μ, w, B1) = 0 at
all speeds. Therefore
d
d
= +
d
d
+
d
d
= 0
1
C C C 1
w
w C
B
m m m m B
μ μ μ μ
∂
∂
∂
∂
∂
∂
Since the thrust will be practically constant, we also have
d
d
= +
d
d
+
d
d
c c c c = 0
1
t t t 1
w
w t
B
B
μ μ μ μ
∂
∂
∂
∂
∂
∂
In terms of the aerodynamic derivatives, these two equations can be written
approximately as
m m
w
m
B
u + w B
d
d
+
d
d
= 0 1
1
μ μ
and
z z
w
z
B
u + w B
d
d
+
d
d
= 0 1
1
μ μ
Eliminating dw/dμ gives
d
d
=
–
–
　
中國航空網(wǎng) www.k6050.com
航空翻譯 www.aviation.cn
本文鏈接地址：Bramwell’s Helicopter Dynamics(98)