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

Yk = R1k sin ψk + R2k cos ψk
Zk = – R3k
Now, in steady flight the forces R1k, R2k, R3k will be periodic and therefore expressible
in the form of a Fourier series as
R1k = P0 + P1 cos ψk + P2 cos 2ψk + P3 cos 3ψk + …
+ Q1 sin ψk + Q2 sin 2ψk + Q3 sin 3ψk + … (8.1)
R2k = S0 + S1 cos ψk + S2 cos 2ψk + S3 cos 3ψk + …
+ T1 sin ψk + T2 sin 2ψk + T3 sin 3ψk + … (8.2)
R3k = U0 + U1 cosψk + U2 cos 2ψk + U3 cos 3ψk + …
+ V1 sin ψk + V2 sin 2ψk + V3 sin 3ψk + … (8.3)
Considering the force components along the shaft first, the total force in the Z
direction is
Z R bU U n V n
k
b
k n k
b
n k n k
b
= – = – – cos – n sin k
=0
–1
3 0 =1 =0
–1
=1 =0
–1
Σ ΣΣ Σ Σ
∞ ∞
ψ ψ
From the results of Appendix A.3, the summations can be simplified to give
Z = – b[U0 + Ub cos bψ + U2b cos 2bψ + …
+ Vb sin bψ + V2b sin 2bψ + …] (8.4)
Thus, the only harmonics which remain are those which are multiples of the number
of blades. If, for example, the rotor has four blades, only harmonics of frequency 4Ω,
8Ω, …, etc. contribute to the total vertical force, apart from the steady load bU0. This
result assumes, of course, that all the blades are perfectly matched (track and balance).
If one or more blades are not matched, other frequencies Ω, 2Ω, …, etc. may arise.
Considering now the force components in the plane of the hub, the X force is of
the form
X P S
k
b
k k
b
= – cos + sin k
=0
–1
0 =0
–1
0 Σ ψ Σ ψ
– cos cos – sin cos
=1 =0
–1
=1 =0
–1
Σ Σ Σ Σ
∞ ∞
n k
b
n k k n k
b
P nψ ψ Qn nψk ψk
+ cos sin + sin sin
–1 – 0
–1
=1 =0
–1
Σ Σ Σ Σ
∞ ∝
n k
b
n k k n k
b
S nψ ψ Tn nψk ψk (8.5)
If the blades are perfectly matched, P0 and S0 are the same for all blades and the
first two terms vanish. Using the well known formulae for the products of sines and
cosines, the remaining sums can be written as
Rotor induced vibration 293
X P T n P T n
n k
b
n n k n k
b
n n k = – ( + ) cos ( + 1) – ( – ) cos ( – 1) 12
=1 =0
–1
12
=1 =0
–1
Σ Σ Σ Σ ∞ ∞
ψ ψ
– ( – ) sin ( + 1) – ( + ) sin ( – 1) 12
=1 =0
–1
12
=1 =0
–1
Σ Σ Σ Σ
∞ ∞
n k
b
n n k n k
b
Q S n ψ Qn Sn n ψk
(8.6)
Using the results of Appendix A.3, the first and third sums reduce to b cos mbψ and
b sin mbψ, respectively, when n + 1 = mb, i.e. when n = mb – 1, (m = 1, 2, 3, …), and
the second and fourth sums to b cos mbψ and b sin mbψ when n = mb + 1. Therefore,
X becomes
X b P T P T mb
m mb – mb mb mb = – [ + + – ]cos 12
= 1 1 – 1 + 1 + 1 Σ
∞
ψ
– [ – + + ]sin 12
b = 1Q 1 S – 1 Q + 1 S – 1 mb
m mb – mb mb mb Σ
∞
ψ (8.7)
Similarly
Y b S Q S Q mb
m mb – mb mb mb = [ – + + ]cos 12
= 1 1 – 1 + 1 + 1 Σ
∞
ψ
+ [ + – + ]sin 12
b = 1P 1 T – 1 P + 1 T + 1 mb
m mb – mb mb mb Σ
∞
ψ (8.8)
Thus, if a blade of a three-bladed rotor produces a force F = R2 cos 2ψ on the hub,
the corresponding X and Y force components are – R2 cos 2ψ cos ψ and R2 cos 2ψ
sin ψ respectively. By expressing the trigonometrical products as the sums of sines
and cosines, these components are – 12
R2 cos 3ψ and – 12
R2 cos ψ in the X direction
and 12
R2 sin 3ψ and – 12
R2 sin ψ in the Y direction. But since n + 1 = b, for this case,
the terms in 3ψ add, whereas the terms in ψ cancel. If, however, F = R3 cos 3ψ, the
X and Y components for the individual blade would consist of terms in 2ψ and 4ψ, but
for the complete three-bladed rotor these terms would vanish since neither (n + 1) nor
　
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本文鏈接地址：Bramwell’s Helicopter Dynamics(147)