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TD
HD
Z
V
X
αD
αD
δα1
Disturbed flight
TD + δTD
HD + δHD
V + u
w
Steady flight
Fig. 5.5 Longitudinal forces and flapping in disturbed flight
5.4 Calculation of the derivatives
The main contributions to the incremental forces and moments arise from the rotor,
as discussed at the beginning of this chapter. We have also assumed, with good
justification, that the rotor forces and moments, like those of the fixed wing aircraft,
depend only on the instantaneous values of speed, incidence, and rate of pitch. To
calculate the derivatives, then, it is necessary only to resolve the rotor forces onto the
chosen axes and perform straightforward differentiation on the expressions for force
and flapping in Chapter 3. Some of the fuselage contributions can be calculated with
fair confidence, particularly those of the tailplane.
As far as the rotor calculations are concerned it does not matter whether tip path
plane (disc) axes or no-feathering axes are used but, as we found in Chapter 3, the Hforce
is usually very small when the rotor forces are expressed in terms of either set
of axes, and this is an aid in the physical appreciation of the problem. We shall
therefore use disc axes in general in our analysis but occasionally use no-feathering
axes when it makes the differentiation easier.
Referring to Fig. 5.5, let αD be the disc incidence in steady flight. In disturbed
flight the longitudinal flapping increases by amount δa1 and the incidence of the wind
axes is α ≈ w/V. Resolving the thrust and in-plane force along the wind axes, we get
δX = –(TD + δTD) sin (αD + δa1) – (HD + δHD) cos (αD + δa1) – TD sin αD
– HD cos αD
≈ – TDδa1 – δTDαD – δHD
since αD and δa1 are small angles.
Also
δZ = – (TD + δTD) cos (αD + δa1) + (HD + δHD) sin (αD + δa1)
+ TD cos αD – HD sin αD
≈ – δTD + δHDαD + HDδa1
≈ – δTD
148 Bramwell’s Helicopter Dynamics
since the terms in HD are very small.
Then
X
X
u
T
a
u
T
u
H
u = = – D – – u
1
D
∂ D
∂
∂
∂
∂
∂
∂
∂
α (5.35)
Z
Z
u
T
u = ∂ = – uD
∂
∂
∂ (5.36)
X
X
w
T
a
w
T
w
H
w = = – D – – w
1
D
∂ D D
∂
∂
∂
∂
∂
∂
∂
α (5.37)
Z
Z
w
T
w = ∂ = – wD
∂
∂
∂ (5.38)
X
X
q
T
a
q
T
q
H
q = = – D – – q
1
D
∂ D D
∂
∂
∂
∂
∂
∂
∂
α (5.39)
Z
Z
q
T
q = ∂ = – qD
∂
∂
∂ (5.40)
Now, since the disc makes a small angle to the x axis, we can write
d
d
= 1 d
d
1 d
d
1 d
u ΩR uˆ ΩR D ΩRd ≈ μ ≈ μ
Then
x
X
sA R
t
a t h
u
= u = – c – –
1
D
c cD
ρ μ
α
Ω μ μ
∂
∂
∂
∂
∂
∂ (5.41)
and similarly
z
t
u = – ∂ c
∂μ (5.42)
x t
a
w
t
w
h
w w = – c – –
1
D
∂ c cD
∂
∂
∂
∂
ˆ ˆ ∂ ˆ α (5.43)
z
t
w w = – ∂ c
∂ ˆ
(5.44)
x t
a
q
t
q
h
q = – c – – q
1
D
c cD ∂
∂
∂
∂
∂
ˆ ˆ ∂ ˆ α (5.45)
z
t
q = – ∂ qc
∂ ˆ
(5.46)
To calculate the moment derivatives, we see from Fig. 5.6 that the moment of the
rotor forces δMr about the c.g. is
δMr = (–δX cos αs + δZ sin αs)lR + (δZ cos αs + δX sin αs)hR
　
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