曝光臺 注意防騙 網曝天貓店富美金盛家居專營店坑蒙拐騙欺詐消費者

0.7 0.8 0.9 1.0
k = 1.4
1.2
1.0
0.8
0.6
0.4
Component in phase with α
0.2
0.1
0.04
0.04
0.1
1.0
0.4
0.2
1.4
1.8
Component in quadrature with α
L
2πρV 2α
M
πρV 2α
Fig. 6.14 Amplitude and phase of lift and moment of aerofoil oscillating in pitch
V
–1 x
η
α
1
ξ
0
Fig. 6.15 Shed vortex elements
In particular, for a given increment of time dt, the change of circulation is dΓ and
a vortex element – Vγ dt is shed at the trailing edge. Since γ is a function of both time
and distance, we can write
dΓ/dt = –Vγ (l, t) (6.12)
and since the total amount of vorticity shed in the wake must be equal to the total
circulation about the aerofoil, we also have the relation
Γ( ) = – ( , )d
1
t b t
∞ ∫ γ ξ ξ (6.13)
From thin aerofoil theory7, the chordwise vorticity distribution can be expressed
as
γ( ) = 0 tan ( cosη) + sin (cosη)
12
–1
=1
x A A n –1
n n Σ
∞
(6.14)
Γ
γ dη
γ dξ
Rotor aerodynamics in forward flight 207
The assumed form, eqn 6.14, automatically satisfies the Kutta–Zhukowsky condition
at the trailing edge. The induced velocity at x due to a vortex element on the aerofoil
at η is
d ( ) =
( , ) d
2 ( – ) v x
t
x
γ η η
π η (6.15)
Substituting the series eqn 6.14 for γ and integrating across the chord from –1 to +
1, gives
v( ) = – – cos (cos ) 12
0
1
12
x A An n –1 Σ
∞
η
where we make use of the well known result
0
π φ
φ η
φ π η
∫ η
cos
cos – cos (cos )
d =
sin (cos )
–1 sin (cos )
–1
–1
n n
(6.16)
Let us suppose that, in addition to its forward velocity V, the aerofoil has a vertical
(heaving) velocity of ˙z at a point about which the aerofoil is also rotating, Fig. 6.16.
The relative air velocity normal to the chord at x due to the motion is
u(x) = αV˙ + z˙ – α˙b(x – x)
The induced velocity at x due to a vortex element in the wake is
d ( ) =
d
2 ( – )
w x
x
γ ξ
π ξ
and the velocity at x induced by the entire wake is
w x
x
( ) =
d
1 2 ( – )
∞ ∫ γ ξ
π ξ
Since there can be no flow perpendicular to the chord, we must have
u(x) + v(x) + w(x) = 0
or
– – cos (cos ) + [ + – ( – )] +
d
2 ( – )
= 0 12
0
12
=1
–1
1
A A n V z b x x
n n x Σ
∞ ∞ η α α ∫ γ ξ
π ξ
˙ ˙
(6.17)
α
0
–x
– x
α˙
z˙
Fig. 6.16 Aerofoil oscillating in heave
208 Bramwell’s Helicopter Dynamics
Now eqn 6.16 can also be expressed as
0
π φ
ξ φ
φ π ξ ξ
∫ ξ√
√
cos
– cos
d =
[ – ( – 1)]
( – 1)
2
2
n n
so that the Fourier coefficients An occurring in eqn 6.17 are easily found to be
A = 1
d
( – 1)
+ 2( + – )
A = 2
( – 1)
– 1 d + 2
A = 2
– ( – 1)
( – 1)
d
0
1
2
1
1
2
1
2
2
π
γ ξ
ξ
α α
π
ξ
ξ
γ ξ α
π
ξ ξ
ξ
γ ξ
∞
∞
∞
∫
∫
∫
√
√

 

 
√
√


V z bx
b
n
˙ ˙
˙






(6.18)
The linearised Bernoulli equation gives for the pressure difference between the
upper and lower surfaces
p p p
t
V
l – u = 2 + x
　
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本文鏈接地址：Bramwell’s Helicopter Dynamics(105)