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2
=
8 /3
1 – /2
1 –
(1 + 3 /2)
θ 8 +
μ
μ
μ
μ

 

 
(5.156)
Flight dynamics and control 183
Finally, differentiating eqn 3.64
∂
∂
∂
∂
∂
∂
hc a a a
0 0
D 1
0
D = D 0
8
–
θ θ 4
λ
θ
μλ θ 
 
 
 
 
=
8
1 + – 2 +
D
0
D
1
0
D 0
D
0
a a
∂ a
∂
∂
∂
∂
∂
λ
θ
λ
θ
μ λ θ λ
θ
 
 
 
 

 

 
(5.157)
Since
λD = μ(αnf + a1) – λ
i
∂
∂
∂
∂
∂
∂
λ
θ
μ
θ
λ
θ
D
0
1
0
i
0
= –
a
so that eqn 5.157 can be calculated in conjunction with eqns 5.155 and 5.156.
The force and moment derivatives are
x t
a t h
θ θ
α
0 θ θ
= – c – – D
1
0
D
c
0
c
0
∂
∂
∂
∂
∂
∂ (5.158)
z
t
θ 0 θ = – c
0
∂
∂ (5.159)
′m hx lz C
a
θ0 θ0 θ 0 ms θ = – 1 + 1 –
1
0
∂
∂ (5.160)
which can also be expressed as
′m l ha t
t h C
a
h
h
θ 0 θ ms θ θ
= –( – 1s ) + ( + ) + D
c
0
c
1
0
c
0
∂
∂
∂
∂
∂
∂ (5.161)
Again, the terms in hcD are negligibly small. The θ0 derivatives for our example
helicopter are shown in Fig. 5.20.
5.9.3 Control response in forward flight
As mentioned earlier, response to collective pitch variation will not be considered, so
the non-dimensional equations of motion reduce to
d /d – – + c cos c = 1 1 ( ) uτ xuu xww wθ τ xBBτ (5.162)
– + d /d – – d /d + c sin c = 1 1 ( ) zuu wτ zww Vˆθτ w θ τ zB B τ (5.163)
– – – d /d + d2/d 2 – d /d = ( )
1 1 muu mww mw˙ w τ θ τ mqθ τ mBB τ (5.164)
in which θ here is the fuselage pitch attitude change.
The cyclic pitch displacement B1 is a function of time, either as a prescribed
control input by the pilot or, as we discussed earlier in the section on autostabilisation,
184 Bramwell’s Helicopter Dynamics
related to some of the other variables through a control law. We have already considered
an example of the latter case in the section on stabilising bars.
Many of the control displacements which result in responses giving useful information
enable the above equations to be conveniently solved by the Laplace transformation.
Denoting the Laplace transforms of u, w, and θ by u, w and θ and supposing the
aircraft to be initially in trim, the transformed equations of motion become
( – ) – + c cos c = 1 1 p xu u xww wθ τ xBB (5.165)
– + ( – ) – ( – c sin c) = 1 1 zuu p zww Vp w zBB
ˆ τ θ (5.166)
– – ( + ) + ( 2 – ) =
1 1 muu pmw˙ mww p mqp θ mB B (5.167)
Solving for u, w and θ gives
u/B1 = xB /(p + U p + U P + U )/
3
2
2
1 1 0 Δ (5.168)
w/B1 = zB (p + W p + W p + W )/
3
2
2
1 1 0 Δ (5.169)
θ /1 = ( 2 + + )/
2
1 1 0 B mB H p Hp H Δ (5.170)
where
Δ = p4 + B1cp3 + C1p2 + D1p + E1 (the stability quartic)
and
U z m Vm x
z
w q w w x
B
B
2 = – ( + + ) + 1
1
ˆ ˙
U z m Vm m w Vx w
m
w q w w w x
B
B
1 = – + c sin c + ( – c cos c) 1
1
ˆ ˆ ˙ τ τ
　
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本文鏈接地址：Bramwell’s Helicopter Dynamics(94)