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时间:2010-05-30 00:47来源:蓝天飞行翻译 作者:admin
曝光台 注意防骗 网曝天猫店富美金盛家居专营店坑蒙拐骗欺诈消费者

of flow through the rotor must be expressed as λ i – wˆ before differentiating. This
expression of the flow component is not valid if the flight path is steep, for then the
z axis, i.e. the w direction, makes a considerable angle to the rotor axis. However,
steep flight paths are possible in only a narrow range at low forward speed and
overall provide no practical restriction. The induced velocity ratio must then be
written
λi c λ
2
i
= st /2[Vˆ + ( – wˆ )2 ]1/2 (5.67)
where wˆ is made zero after differentiation.
Differentiating eqn 5.67 with respect to wˆ gives




λ ∂ ∂
λ
λ λ
λ
i
2
i
2 1/2
c c i i
2
i
2 3/2
=
2( + )

( / – 1)
ˆ ˆ ˆ 2[ + ( – ) ]
ˆ
w ˆ ˆ
s
V
t
w
st w
V w
= +
4
i 1 –
c
c i
4
2
c
2
λ λ λ i
t
t
w s t w



ˆ ∂ˆ
 
 
= i + 1 –
c
c
i
λ 4 λ i
t
t
w w



ˆ ∂ˆ v 
 
 
(5.68)
Flight dynamics and control 153
Writing λ = sin αnf – λi
ˆV






λ α α λ
ˆ
ˆ
w ˆ ˆ
V
w w
= cos nf –
nf i
But, except for steep flight paths, it is clear from Fig. 5.7 that the change of
incidence δαnf of the no-feathering axis due to the disturbance w is
δαnf = w/V
i.e. ∂α nf/∂wˆ = 1/Vˆ
Thus, providing αnf is not too large (so that cos αnf ≈ 1),
∂λ/∂ = 1 – ∂λi /∂ wˆ wˆ (5.69)
Since
λD = λ + μa1
the derivative of λD is
∂λD /∂wˆ = ∂λ/∂wˆ + μ∂a1/∂wˆ (5.70)
Differentiating eqn 3.33




t
w
a
w
c =
4
ˆ ˆ
λ
=
4
a 1 – i
w


λ
ˆ
 
 
(5.71)
Hence, from eqn 5.68,


λ λ
λ
i i c i
4
i c i
4 =
( /4) / +
wˆ 1 + ( /4) / +
a t
a t
v
v
(5.72)
so that


λ
wˆ a λ t
= 1
1 + ( /4) i / c + i
v 4
(5.73)
αnf
No-feathering axis
δαnf
V
w
Fig. 5.7 Change of incidence of no-feathering axis
154 Bramwell’s Helicopter Dynamics
and


t
w
a
a t
c
i c i
4 =
4
1
ˆ 1 + ( /4)λ / + v
(5.74)
Differentiating eqn 3.56 gives
∂a ∂
w w
1
d 2
=
2
ˆ 1 – /2 dˆ
μ
μ
λ
=
2
(1 – 2 /2)(1 + ( /4) / + )
i c i
4
μ
μ a λt v
(5.75)
Then, from eqn 5.70,


λ
λ
μ
μ
D
i c i
4
2
2 = 1
1 + ( /4) / +
1 + 3 /2
wˆ a t v 1 – /2
⋅ (5.76)
and so we have from eqn 3.64


h
w
a
a t
a
i
c
c i
4
12
1 0
D
2
D =
4(1 + ( / 4) / + )
– +
ˆ λ 1 – /2
μθ μλ
v μ

 

 
(5.77)
The derivatives 5.74, 5.75, and 5.77 can be simplified for most of the speed range.
For μ = 0, i = 1 and c = 2λi ,
v st 2 and we have
∂tc /∂wˆ = – zw = 2aλi/(16λi + as) (5.78)
and ∂a1/∂w = ∂hc /∂w = 0 D ˆ ˆ
For μ > 0.08, vi ≈ 0 and stc = 2μλi, and we have
 
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