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∂
∂
∂
∂
φ φ 
 
 
(6.19)
In terms of the vorticity on the aerofoil, the velocity potential is
φ γ ( ) = 12
d
–1
x b x
x ∫
so that
∂
∂
φ γ
x
b = 12
and
∂
∂
∂
∂
φ γ
t
b
t
x
x
= d 12
–1 ∫
The lift on the aerofoil is
L = b (pl – pu)dx
–1
+1 ∫
Substituting for the pressure difference from eqn 6.19 and the vorticity from
eqn 6.14 leads to
L b bt
A A bt
A A VA A = ( – ) + ( + ) + ( + ) 12
0
12
2 0
12
1 0
12
ρπ ∂ 1
∂
∂
∂

(6.20)
The circulation about the aerofoil is
Γ = d = ( + )
–1
+1
0
12
b∫ γx bπA A1 (6.21)
Rotor aerodynamics in forward flight 209
From eqns 6.18 and 6.21, the first two terms in the bracket of eqn 6.20 can be
written as
12
0
12
2
1
2 2
2
( – ) + 1 =
1 – + ( – 1)
( – 1)
b d
t
A A
t
b
t
∂
∂
∂
∂
∂
π π∂
ξ ξ ξ
ξ
Γ ∞ γ ξ ∫ √
√
+ b( V + z + bx) + 1
t
α˙ ˙˙ α˙ π
∂
∂
Γ
= d – ( – 1) d
1 1
b 2
t
b
π ∂ γ ξ ξ π t ξ γξ
∂
∂
∂
∞ ∞ ∫ ∫√
– b( V + z + bx) + 1
t
α˙ ˙˙ α˙ π
∂
∂
Γ (6.22)
Now, each element of vorticity in the wake is carried backwards with velocity V,
so the vorticity must be of the form
γ (ξ, t) = f(t – bξ /V)
Since, for a given element of vorticity, t – bξ /V is constant, it follows that
dγ = γ d + γ d = 0
ξ
∂ ξ
∂
∂
t ∂
t
and d – d = 0 t bV
ξ
i.e.
∂
∂
∂
∂
γ γ
t ξ
Vb
= – (6.23)
in the wake.
Differentiating under the integral sign and making use of eqn 6.23, we find
∂
∂
∂
t ∂t
d – ( – 1) d
1 1
2
∞ ∞ ∫γ ξ ξ ∫√ ξ γ ξ
= – d + ( – 1) d
1 1
2 Vb
Vb
∞ ∞ ∫∂ ∫√
∂
∂
∂
γ
ξ
ξ ξ ξ γ
ξ
ξ
= – + d + ( – 1) 1
1
2
1
Vb
Vb
Vb
[γ ξ]∞ γ ξ [γ √ ξ ]
∞ ∞ ∫
–
( – 1)
d
1
2
Vb
∞ ∫ √
γ ξ
ξ
ξ
= – [ – ( 2 – 1)] ( , ) + d
1
Vb
t Vb
{ξ √ξ γξ } γ ξ ∞ ∞ ∫ 1
–
( – 1)
d
1
2
Vb
∞ ∫ √
γ ξ
ξ
ξ ,
on integrating by parts.
210 Bramwell’s Helicopter Dynamics
Now ξ – √(ξ2 – 1) = 1/[ξ + √(ξ2 – 1)], from which it follows that, since γ (ξ, t) is
finite, the term in the bracket vanishes at the upper limit. At the lower limit the term
in the bracket is equal to (V/b)γ (1, t). This term cancels identically with the last term
of the eqn 6.22, so that
12
0
12
2 0
12
b ( – ) + 1 + ( + 1)
t
A A
t
∂ V A A
∂
∂
π ∂
Γ
= d –
( – 1)
d + ( + – ) + ( + )
1 1
2 0
12
1
V V b V z bx V A A π γ ξ π
γξ
ξ
ξ α α
∞ ∞ ∫ ∫√
˙ ˙˙ ˙
= VA0 + Vbα˙ + b(α˙V + ˙z˙ – α˙˙bx)
on eliminating the integrals by means of eqn 6.18. The lift can then be written as
L = 2 Vb[V + z + b( – x)] + Vb
　
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本文鏈接地址：Bramwell’s Helicopter Dynamics(106)