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

the derivatives. Since tcD and tc are almost identical, it is convenient to use eqn 3.33
for the relation between the thrust coefficient and the inflow ratio; i.e. starting from
t a
c 0
= 2
4
2
3
(1 + 3 /2) + θ μ λ 
 
 
(3.33)
where λ is referred to the no-feathering axis, we differentiate with respect to μ to
obtain
Flight dynamics and control 151
(4/a)∂tc /∂μ = 2μθ0 + ∂λ/∂μ (5.56)
Now, for small αnf
λ = μαnf – λ
i
where λ
i is the mean ‘momentum’ inflow ratio, therefore
∂λ/∂μ = αnf – ∂λi/∂μ (5.57)
since αnf remains constant with changes of μ.
In non-dimensional form, eqn 3.1 can be written (for small αD)
λi = stc /2(V + λ)
2
i
ˆ 2 1/2
so that
∂
∂
∂
∂
∂
∂
λ
μ λ μ λ
λ λ
μ
i
2
i
2 1/2
c c
2
i
= 2 3/2 i
2( + )
–
2( + )
s +
V
t st
V
V i
ˆ ˆ
ˆ 
 
 
since ∂/∂μ ≈ ∂/∂ ˆV , or
∂
∂
∂
∂
∂
∂
λ
μ
λ
μ
λ λ λ
μ
i i
c
c i
3
2
c
2
i
4
2
c
2
= – i
4
–
4
t
t V
s t s t
ˆ
(5.58)
Further, the induced velocity in hovering flight, or ‘thrust velocity’ v0 (Chapter 2)
is related to the thrust coefficient by
v0
2 12
c
= st Ω2R2
so that eqn 5.58 can also be written as
∂λ
∂μ
λ ∂
μ
λ
μ
i i
c
c
i
3
i
= – – 4 i
t
t
V ∂
∂
∂ v v (5.59)
where V = V/v0 and v i = vi/v0 , and v i can be taken from the chart, Fig. 3.2.
Then, from eqns 5.56, 5.57 and 5.59, we find,
∂
∂
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
t a
v
v
(5.60)
For μ > 0.08 (See sections 3.2 and 3.14)
λ i cμ i
≈ st /2 and v 4  1
and eqn 5.60 can be written
∂
∂
λ
μ
μθ α μ
μ
i = 0 nf c
2 + – 4 /
1 + 8
t a
/as
(5.61)
The incidence, αnf can be written alternatively as αD – a1.
With ∂λc/∂μ known, it is now possible to obtain ∂tc/∂μ. Eliminating ∂λi/∂μ from
eqn 5.59 using eqn 5.60 gives
152 Bramwell’s Helicopter Dynamics
∂
∂
t a V
a t
c 0 D 1 i
3
i
4
i c i
4 =
2 + – + /(1 + )
μ 4/ + ( / )/(1 + )
μθ α
λ
v v
v
(5.62)
and for μ > 0.08 this simplifies to
∂
∂
t a
as
c a st
0 D 1 c
= 2
2
8 +
(2 + – + /2 ) μ
μ
μ μθ α μ (5.63)
We note that ∂λi/∂μ and ∂tc/∂μ are both zero in hovering flight.
To find ∂a1/∂μ it is convenient to use the expression for a1 with λ referred to the
no-feathering axis, eqn 3.56
a1
0
2 =
2 (4 /3 + )
1 – /2
μ θ λ
μ
(3.56)
Differentiating with respect to μ and rearranging gives
∂
∂
a1 a1 ∂
2 = –
2
1 – /2
μ μ d
μ
μ
λμ
(5.64)
When μ is zero, eqn 5.64 reduces to
∂a1/∂μ = 8θ0/3 + 2λ (5.65)
To calculate ∂hcD∂/ μ we differentiate eqn 3.64; numerical examples show, however,
that only the profile drag term is of any importance, and we have simply
∂hc ∂
1
D 4 /μ = δ (5.66)
When considering derivatives with respect to the vertical velocity w, the component
　
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