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

resultant flapping will be of the same magnitude but rotated through angle βss so that
the new sideways flapping will be
b1 b1 a1 b
2
1
2 1/2
+ δ = ( + ) sin (ψ0 – βss)
= ( + ) (sin cos – sin cos ) 1
2
1
2 1/2
a b ψ0 βss βss ψ0
= b1 – a1βss , for small βss .
Therefore
δb1 = – a1βss = – a1v/(V cos αnf)
and so
∂ b1/∂vˆ = – a1/μ (5.121)
where v ˆ = v/ΩR.
Flight dynamics and control 171
It should be noted that the sign of ∂ b1 /∂vˆ depends on the sense of rotation of the
rotor, and this should be allowed for in the derivation of any formulae; but a little
consideration shows that the rotor always tilts away from the component of relative
wind so that there should be no confusion about the appropriate sign of the forces and
moments due to sideways flapping. In this work the rotor is supposed to be rotating
in an anticlockwise direction when viewed from above, i.e. positive sideways flapping
is directed to starboard.
By the same reasoning, the side wind will always cause a sideways component of
the longitudinal in-plane H-force, and it is easy to see that
∂Y/∂v = – HD/V cos αD
remembering that we are considering forces in the plane of the disc.
The non-dimensional form is
∂ yc∂ hc
1
4 / vˆ = – /μ ≈ – δ
in which yc = Y/ρsAΩ2R2 and the inflow contribution to hc is ignored (see eqn 3.64).
The calculation of the rolling moment due to the rate of roll p follows a similar
procedure to that described for the pitching moment, and we find that the nondimensional
rolling moment derivative due to the rotor forces is
(lp)r = h1(tc + aλD/8)∂b1/∂pˆ (5.123)
in which ∂yc/δb1 ≡ ∂hc/∂a1 = aλD/8 from eqn 3.64. Also, if the extra terms due to ˆp
from eqn 1.16a are added to the right-hand side of eqn 3.48, and an examination is
made of the various harmonic terms (as in the pitching case), then it can be shown
that
∂b1∂p
/ ˆ = – 16/γ(1 + μ2 /2) (5.124)
It has been shown11 that the fuselage contribution to the side force can be expressed
approximately as
(yv)f = – 0.3μSB/sA (5.125)
where SB is the projected side area of the fuselage.
The non-dimensional forms of all the lateral derivatives are
y t
a
s
t
w
S
sA
B
v = – c – – – 0.3
1 1
4 t
c
μ δ ∂ μ
∂ ˆ
(5.126)
′ ′ l ht C
a
h s
t
w v = – ( 1 c + m ) –
1
t t
c
s μ
∂
∂ ˆ (5.127)
l′ ′
h t a C
h s
t
w p
m = – 16 [ ( + /8) + ]
1 + /2
– 1 c D
2 t
2
t
s c
γ
λ
μ
∂
∂ ˆ
(5.128)
′ ′′ l hl t
w r = t t
∂ c
∂ ˆ
(5.129)
172 Bramwell’s Helicopter Dynamics
′ ′ ′ n ls t
w
v = t t + (nv)
c
f
∂
∂ ˆ
(5.130)
′ ′′ ′ n hls t
w
p = t t t + (np)
c
f
∂
∂ ˆ
(5.131)
′ ′ ′ n l s
t
w
r = – t + (nr)
2
t
c
f
∂
∂ ˆ
(5.132)
where st = stAt(ΩR)t/sAΩR. But, usually, (ΩR)t
ΩR; hence st = stAt/sA.
The fuselage derivatives are not likely to be known accurately except, perhaps, the
contribution of the fin. In the example below, the fuselage moment derivatives have
been taken as zero.
5.7 The lateral stability characteristics
The lateral stability derivatives for the example helicopter have been calculated and
are shown in Fig. 5.15, taking
lt = 1.2 and ht = 0.1
and considering level flight, τc = 0.
The non-dimensional moments of inertia are taken to be
iA = 0.033, iC = 0.11, iE = 0
0.08
0.06
0.04
0.02
0
–0.02
–0.04
–0.06
–0.08
–0.10
0.1 0.2 0.3 μ 0.4
nv′
lr′ and np′
1
2 yv
nr′
l p′
lv′
Fig. 5.15 Lateral derivatives for typical single rotor helicopter
Derivative values
　
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