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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

completely define the operating state of the rotor. To do this we assume that the
transient response is stable and that, as in previous sections, the forced blade motion
is periodic and can be expressed as
β = a0 – a1 cos ψ – b1 sin ψ – a2 cos 2ψ – b2 sin 2ψ – …
remembering that β is defined relative to the plane of no-feathering.
This expression is substituted into eqn 3.48 and, assuming that it represents a
solution, the coefficients of the terms in sin ψ, cos ψ, … on the left- and right-hand
sides of eqn 3.48 can be equated. If we consider only the constant term and the two
first harmonic terms, we obtain, after manipulation of terms such as sin ψ, sin 2ψ,
etc.,
Fig. 3.24 Blade flapping response at high tip speed ratios
μ =0.5
β
Time
μ=2.25
Rotor aerodynamics and dynamics in forward flight 107
a0 0
= 2
8(1 + )
(1 + ) + 4
3
γ
ε[θ μ λ] (3.51)
a1 b
0
= 2 2 1
2 (4 /3 + )
1 – /2
+ 8 .
– /2
μ θ λ
μ γ
ε
1 μ
(3.52)
b
a
1 a
0
1/2
i
= 2 2 1
4( + 1.1 )/3
1 + /2
– 8 .
+ /2
μ ν λ
μ γ
ε
1 μ
(3.53)
If Glauert’s formula, eqn 3.4, for the induced velocity had been used instead of
Mangler and Squire’s, the values of a0 and a1 would be unaltered but it would have
been found that
b
a K
1 a
0 i
= 2 2 1
4( + 1.7 )/3
1 + /2
– 8 .
+ /2
μ λ
μ γ
ε
μ
5
1
(3.54)
For typical values of flapping hinge offset the terms in ε are usually negligible
except, perhaps, in the equations for b1 (eqns 3.53 and 3.54) at high values of μ.
When ε = 0, the formulae reduce to their classical forms:
a0 0
= 2
8
(1 + ) + 4
3
γ[θ μ λ] (3.55)
a1
0
2 =
2 (4 /3 + )
1 – /2
μ θ λ
μ
(3.56)
b
a
1
0
1/2
i
2 =
4( + 1.1 )/3
1 + /2
μ ν λ
μ
(3.57)
The case of constant induced velocity is found by putting vi = vi0 in eqn 3.13,
whence it is seen that the term involving ν1/2 in eqns 3.53 and 3.57 is absent.
Alternatively, letting K = 0 in eqn 3.54 leads to the same result.
Higher order flapping coefficients can be obtained by considering higher harmonics
of the flapping motion, but it is found that the coefficients appear explicitly as the
solutions of an infinite chain of simultaneous equations and cannot be evaluated
easily. Stewart17 has extracted the coefficients up to the fourth harmonic and has
shown that their magnitudes decrease very rapidly with order of harmonic. As a
rough rule it was found that the magnitude of a coefficient was about one tenth of the
value of that of the next lower harmonic. The calculation of the higher harmonics, as
has been mentioned in Chapter 1, is somewhat academic since the effects of the
higher modes of blade bending are at least as great as these harmonics of the first
(rigid) blade mode.
However, it should be noted here that the higher harmonic terms do become
important when considering rotor induced vibration (see Chapter 8). The simple rigid
blade flapping model used in the current analysis may be seen to give rise to first and
108 Bramwell’s Helicopter Dynamics
second harmonic forcing terms on the right-hand side of eqn 3.48 (or 1 Ω and 2 Ω
terms, to use the terminology adopted in Chapter 8), but in a more general case with
a flexible blade flapping model, forcing terms are generated over a wide range of
harmonics.
3.13 Force and torque coefficients referred to disc axes
It is useful to obtain expressions for the force and torque coefficients when referred
to the tip path plane axis (or rotor-disc axis) instead of the no-feathering axis. In
Chapter 1, section 1.13, we saw that, since the angle between these two axes is the
small flapping angle a1, we have the relationships
TD ≈ T
HD ≈ H – Ta1
For the tip-speed and inflow ratios we also have
μ α μ D D = cos ˆV ≈
λD = Vˆ sin αD – λ i ≈ λ + μa,
where TD, HD, μD, and λD are referred to the tip path plane.
Substituting for λ in eqns 3.55 to 3.57 for the flapping coefficients, we have for a1
 
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