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= 12
ρacΩ2R4
× ′ 
 
 

 

 
( + sin ) + –
d
d
– cos ( + sin ) d 2 θ μ ψ λ βψ
x x μ β ψx μ ψ x x
(3.45)
Expanding eqn 3.45 and integrating, assuming constant pitch angle, gives
MA ac R
12
= ρ Ω2 4
× ( ) ′

 
1 ∫
4
+ 2
3
sin + 12
sin + d – 1
4
d
d
– 1
3
0 cos
2 2
0
1
2 θ μ ψ μ ψ λ βψ
x x μ β ψ
+ sin d – 1
3
d
d
sin – 12
sin cos
0
1
2 μ ψ λ μ
βψ
∫ ′ ψ μ βψ ψ

 
x x (3.46)
It can easily be shown that for hovering flight the assumption of linear taper leads
to the same flapping moment as eqn 3.46, provided the pitch angle is taken as that at
0.8R (cf. 0.75R in the calculation of the thrust). This value will be changed slightly
when variable induced velocity is included, but it will be assumed that in calculations
involving a twisted blade it will be sufficiently accurate to use eqn 3.46 with the
collective pitch angle taken as that at 0.8R.
To evaluate the inflow integrals in eqn 3.46 we again use Mangler and Squire’s
induced velocity distribution (eqn 3.13), considering the first harmonic only. Then
0
1
2
0
1
2
nf i
0
1
0
2
i
0
1
1
∫ λ′x dx = ∫ x Vˆ sin α dx – 2λ∫ c x dx + 4λ cos ψ ∫ cx2 dx
= 1
3
sin – 15
4
nf i (1 – ) d
0
1
Vˆ α λ∫x4√ x2 x
– 15
64
i cos (9 – 4) d
1/2
0
1
π λ ν ψ ∫ x2 x3x
104 Bramwell’s Helicopter Dynamics
= 1
3
sin – 15
128
– 15
128
nf i i cos
1/2 ˆV α π λ π λ ν ψ
in which ν = (1 – sin αD)/(1 + sin αD).
To a reasonable approximation this integral may be expressed as
0
1
2 1/2
∫ λ′x dx = (λ – 1.1ν λ i cos ψ )/3
The second integral in eqn 3.46 is
0
1
0
1
nf i
0
1
∫ xλ′ dx =∫ xVˆ sin α dx – (15/4)λ∫ x3 √(1 – x2) dx
– 15
64
i cos (9 – 4) d
1/2
0
1
π λ ν ψ ∫ x2 x2 x
= sin – – 7
64
cos 12
nf
12
i i
Vˆ α λ π λ ν1/2 ψ
= ( – .69 cos ) 12
λ 0 λiν1/2 ψ
Hence,
MA ac R
2 4
0
= 1 2 2
8
1 + 8
3
sin + 2 sin ρ θ μ ψ μ ψ Ω ( ) 

+ 4
3
– 1.46 cos –
d
d
– 4
3
1/2 cos +2 sin
i λ ν λ ψ
βψ
μ β ψμ λψ
– 4
3
d
d
sin – 2 sin2 – 0.69 sin 2
i
1/2 μ βψ
ψ μ β ψ μλν ψ (3.47)
The flapping equation of a centrally hinged blade relative to a plane perpendicular
to the shaft axis has been found in Chapter 1 to be
d2βs/dψ2 + (1 + ε)βs = MA/BΩ2 (1.9)
As before, with β defined from the no-feathering plane, the equation can be shown
to be exactly similar in form, i.e.
d2β/dψ2 + (1 + ε)β = MA/BΩ2
Then, on using eqn 3.47 and rearranging, we obtain the differential equation of
flapping in the form
d
d
+
8
1 +
4
3
sin
d
d
+ 1 + +
4
3
cos + sin 2
2
2
2 β
ψ
γ μ ψ βψ
ε γ μ ψ μ ψ β 





 

 
8
Rotor aerodynamics and dynamics in forward flight 105
=
8
+
8
　
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本文鏈接地址：Bramwell’s Helicopter Dynamics(55)