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1
D
c cD ∂
∂
∂
∂
∂
ˆ ˆ ∂ ˆ α (5.43)
z
t
w w = – ∂ c
∂ ˆ (5.44)
Since ∂tc/∂ ˆ q is identically zero, zq = 0 and
x t
a
q
h
q = – c – q
1 cD ∂
∂
∂
ˆ ∂ˆ
but xq is negligible as a force derivative, although it should be included in the
moment derivative mq′ .
′ m hx lz C
a
u = – 1 u + 1 u + m + (mu)
1
s f
∂
∂μ (5.50)
= –( – 1s) + + + + ( )
c
c
1 c 1
f
D
s l ha
t
h t
a h
C
a
m mu
∂
∂
∂
∂
∂
∂
∂
μ μ μ ∂μ
(5.53)
′ m hx lz C
a
w
w = – 1 w + 1 w + m + (mw)
1
s f
∂
∂ ˆ
(5.51)
= –( – 1s) + + + + ( )
c
c
1 c 1
f
D
s l ha
t
w
h t
a
w
h
w
C
a
w
m mw
∂
∂
∂
∂
∂
∂
∂
ˆ ˆ ˆ ∂ ˆ
(5.54)
′ m hx C
a
q
q = – 1 q + m + (mq )
1
s f
∂
∂ ˆ
(5.52)
= c + + + ( )
1 c 1
f
D
s h t
a
q
h
q
C
a
q
m mq
∂
∂
∂
∂
∂
ˆ ˆ ∂ˆ
(5.55)
The tailplane derivatives, which may be the only contributions capable of being
calculated for the fuselage, are
( )T = – T + –
12
T
i i
T m V C a u L ′
μ λ
μ
λ
μ
∂
∂ (5.90)
( )T = – 1 –
12
T T
i m aV w ′
μ λ
μ
∂
∂ (5.91)
Flight dynamics and control 159
( )T = –
12
mq ′ μaTVTlT (5.92)
( )T = –
12
T T T
m aV l i
w˙ wˆ ′
∂
∂
λ
(5.93),
Other derivative terms which are used in the above are
∂
∂
∂
∂
∂
∂
λμ
α λ
μ
α α λ
μ = nf – = – –
i
D 1
i (5.57)
∂
∂
λ
μ
μθ α λ
λ
i 0 nf c i i
3
c i i
4 =
2 + – (4 / )
1 + (4/ )( / )(1 + )
t a V
a t
v
v
(5.60)
∂
∂
t V
a t
c 0 nf i
3
i
4
i c i
4 =
2 + + /(1 + )
4/ + ( / )/(1 + )
(= 0 when = 0) μ
μθ α
λ
μ
v v
v
(5.62)
∂
∂
t a
as
c st
0 nf
c
2
2
8 +
2 + +
2
(for > 0.08) μ
μ
μ μθ α
μ
≈ μ
(5.63)
∂
∂
∂
∂
a1 a1
2 = –
2
1 – /2
μ μ
μ
μ
λ
μ
⋅ (5.64)
∂
∂
hc 1
4
D = μ
δ (5.66)
∂
∂
t
w
a
a t
c
i c i
4 =
4
1
ˆ 1 + ( /4) / + ⋅
λ v
(5.74)
∂
∂
t
w
a
as
c i
i
=
2
ˆ 16 +
λ
λ when μ = 0 (5.78)
∂
∂
t
w as
c
2
ˆ 8 + ≈ μ
μ (for μ > 0.08) (5.79)
∂
∂
a
w a t
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本文链接地址:
Bramwell’s Helicopter Dynamics(82)