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–( + c cos c)
1
1
x m m w
z
w q w x
B
B
˙ τ
1.2
1.0
0.8
0.6
0.4
0.2
0
0.05
0.04
0.03
0.02
0.01
0.2 0.3 μ
0.1
– 0 zθ
c.g. on shaft (l =0)
– 0 zθ
and
mθ′0
mθ′0
xθ 0
Fig. 5.20 Control derivatives (collective pitch)
xθ 0
Flight dynamics and control 185
U m w m w
z
x
z x w
m
w w x
B
B
w w
B
B
0 = c sin c – c cos c 1 + ( cos c – sin c ) c
1
1
1
τ τ τ τ
W x m z
x
z
V
m
u q u z
B
B
B
B
2 = – – + 1 +
1
1
1
ˆ
W xm zm Vm
x
z
Vx w
m
u q u q w z
B
B
u
B
B
1 = – ( – ) 1 – ( + c sin c)
1
1
1
ˆ ˆ τ
W mw mw
x
z
z w x w
m
u u z
B
B
u u
B
B
0 = c cos c – c sin c 1 + ( c cos c – c sin c)
1
1
1
τ τ τ τ
H m
z
w m
B
B
2 = 1 + 1
1
˙
H x z m zm x
m
m x m
z
u w u u w m
B
B
w w w
B
B
1 = – – + ( + ) 1 + ( – )
1
1
1
˙ ˙
H xz x z zm z m
x
m
x m x m
z
u w w u u w w u m
B
B
w u u w
B
B
0 = – + ( – ) 1 + ( – )
1
1
1
Equations 5.168, 5.169, and 5.170 are the transfer functions of the variables u, w,
and θ in relation to the control input B1.
Let us use the equations to find the normal acceleration due to a sudden increase
of cyclic pitch B1. The increment of acceleration can be expressed as ‘g’ units and
ng V
t
w
t
=
d
d
–
d
d
θ
or, in non-dimensional form, taking ˆV = , μ as
n
w
w
= 1 d
d
–
d
c d
μ θ
τ τ


(5.171)
The Laplace transform of eqn 5.171 is
n
p
w
= ( – w)
c
μθ
and for the input of cyclic pitch
B1 = B1/p
so that
n
B
w B
w
B
= 1 –
c 1 1
μ θ
 

 
(5.172)
Usually the solution of Δ = 0 consists of two real roots and a complex pair, and the
inverse of eqn 5.172 then takes the form
186 Bramwell’s Helicopter Dynamics
n = F + Ge1 + He 2 + e re [C cos + S sin ]
im im
λ τ λ τ λ τ λ τ λ τ (5.173)
with similar solutions for u, w, and θ. For the less usual solution of two complex
roots, the inverse of n takes the form
n = F + e re1 (C cos + S sin ) + e (C cos + S sin )
1 1
re2
1 im 1 im 2 im2 2 im2
λ τ λ τ λ τ λ τ λ τ λ τ
In the above two equations, F, G, H, C, C1, etc. are constants that are determined
by the initial conditions.
The normal acceleration in response to a step input of longitudinal cyclic pitch has
been calculated for our example helicopter (e = 0.04), for μ = 0.3, and is shown by
the full line of Fig. 5.21. Now, in a manouevre of this kind it would not be expected
that the stick would be held fixed for more than about three seconds, because the pilot
would want to retrim the aircraft into a new steady state. During such a relatively
short time interval, the forward speed would be expected to change very little and it
would seem reasonable to omit the speed terms, i.e. the u- and X-force derivatives,
from the response equations. The result of doing this is shown by the broken line of
Fig. 5.21, and it can be seen that the error in omitting these terms is quite small. The
simplification to the response equations is, however, considerable. The normal
acceleration can then be written as
n
w
w
= 1 d
d
–
d
c d
μ θ
τ τ


= – 1 ( + )
c
w 1 1
zBB zww (5.174)
from, the simplified eqn 5.163.
Then, for a step input of cyclic pitch,
　
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本文鏈接地址：Bramwell’s Helicopter Dynamics(95)