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t
a
w
h
B B
B
1
= 1 = + 1 + – D 2 2
c
1
D c
1 c
ρ 1
α μ
Ω
∂
∂
∂
∂
∂
ˆ ∂
(5.147)
Similarly
zB1 = – ∂tc/∂B1 = – uzw (5.148)
Flight dynamics and control 181
The moment derivative is
′
m hx lz C
a
B 1 B 1B m w
1
1 1 1 s = – + – 1 + μ ∂
∂ ˆ
(5.149)
which can also be written as
′
m l ha t
B
t h C
a
w
h
h
B 1s m B
c
1
c
1 c
1
1 s
= –( – ) – ( + ) 1 + + D
∂
∂
∂
∂
∂
∂
μ ˆ
(5.150)
Numerical examples show that the terms in hcD can be neglected. The force and
moment derivatives for our example helicopter have been calculated and are shown
in Fig. 5.19.
5.9.2 The θ0 control derivatives
Differentiating eqn 3.33 with respect to θ0 gives
∂
∂
∂
∂
tc a
0
2
0
=
4
2
3
(1 + 3 /2) + θ
μ λ
θ
But λ = μαnf – λ
i
therefore
∂λ/∂θ0 = – ∂λi/∂θ0
since μ and αnf are independent of θ0.
Taking as before
λi c λ
2
i
= st /2√(Vˆ + 2 )
Fig. 5.19 Control derivatives (cyclic pitch)
0.6
0.5
0.4
0.3
0.2
0.1
0 0.1 0.2 0.3 μ
0.12
0.1
0.08
0.06
0.04
0.02
xB1
– 1 mB
zB1
zB1
xB1
and
– 1 mB
182 Bramwell’s Helicopter Dynamics
we have
∂
∂
∂
∂
∂
∂
λ
θ λ θ
λ
λ
λ
θ
i
0 2
i
2
c
0
c i
2
i
2 3/2
i
0
=
2 ( + )
–
2( + )
s
V
t st
√ˆ Vˆ
= i –
c
c
0
i
4 i
0
λ
θ
λ
t θ
∂ t
∂
∂
∂ v
or
(1 + ) = i
4 i
0
i
c
c
0
v ∂
∂
∂
∂
λ
θ
λ
t θ
t
(5.151)
giving finally
∂
∂
t a
a t
c
0
2
i c i
4 =
6
1 + 3 /2
θ 1 + ( /4 )(1 + )
μ
λ
⋅
v
(5.152)
For ˆV = = 0 μ we have approximately
∂
∂
t a
as
c
0
i
i
= 4
3
θ 8 +
λ
⋅ λ (5.153)
and for μ > 0.08
∂
∂
t
a
as
c
0
2
= 4
3
1 + 3 /2
θ 8 +
μ μ
μ (5.154)
From
a1
0
2 =
2 (4 /3 + )
1 – /2
μ θ λ
μ
∂
∂
∂
∂
a1
0
2
0
=
2
1 – /2
4
3
+ θ
μ
μ
λ
θ
=
2
1 – /2
4
3
– 2
i
0
μ
μ
λ
θ
∂
∂
Since ∂a1/∂θ0 is obviously zero when μ = 0, it is sufficient to consider only the
simpler case μ > 0.08. Then from eqns 5.151 and 5.152 we find
∂
∂
λ
θ
μ
μ
i
0
2
= 2
3
1 + 3 /2
8 +
as
as
(5.155)
giving
∂
∂
a as
as
1
0
2
12
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Bramwell’s Helicopter Dynamics(93)
