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24 Bramwell’s Helicopter Dynamics
but because the aerodynamic force will also be a function of the azimuth angle. For
performance and stability calculations we are interested in the time-averaged values.
Consider the average value of X taken over a complete revolution. We have
0
b
0
1
0
d = cos d – 1 cos d
τ τ τ
∫X t M∫a ψt∫F ψt
– b sin d + sin d
0
2
0
M a t F2 t
τ τ
∫ ψ ∫ ψ
where τ is a complete period.
Now, from eqn 1.27, the first term of a1 is rg d2(cos β cos ξ)/dt2. Integrating twice
by parts,
0
2
2
0
d
d
(cos cos ) cos d =
d
d
(cos cos ) cos
τ τ
∫ β ξ ψ β ξ ψ 
 
 
t
t
t
+ cos cos sin – cos cos cos d
0 0
Ω 
 

 
β ξ ψ Ω ∫ β ξ ψ
τ τ
t
If the other terms involving a1 and a2 are integrated in a similar way, we find that,
if the motion is periodic, i.e. if the flight condition is steady, all the terms in the
brackets vanish at the limits and all the remaining integrals cancel identically.
In other words, the mean values of all the inertia forces are zero and the only
forces which remain are the aerodynamic forces. The vanishing of the inertia forces
in steady unaccelerated flight might have been expected on physical grounds, but it
is a common mistake to believe that this is not necessarily the case; the reason for this
is that small angle approximations for the flapping and lagging angles are often made
when resolving the inertia forces, and considerable residuals may remain, particularly
as the centrifugal force is extremely large4.
The mean forces are therefore
(1/ ) d = = – (1/ ) cos d + (1/ ) sin d
0 0
1
0
τ τ ψ τ 2 ψ
τ τ τ ∫X t X ∫F t ∫ F t (1.32)
e3
y
e2
z
ψ
x
e1
Fig. 1.19 Blade and helicopter axes
Basic mechanics of rotor systems and helicopter flight 25
(1/ ) d = = (1/ ) sin d + (1/ ) cos d
0 0
1
0
τ τ ψ τ 2 ψ
τ τ τ ∫Y t Y ∫F t ∫F t (1.33)
(1/ ) d = = (1/ ) d
0 0
τ τ 3
τ τ ∫Z t Z ∫F t (1.34)
The force components X and Y in the plane of the hub give rise to pitching and
rolling moments about the helicopter’s centre of gravity. Further, the force at the
offset hinge, in the direction of the shaft, also exerts pitching and rolling moments.
This force has the same magnitude as Z, so that, if hR is the height of the hub above
the c.g. and eR is the distance of the hinge from the shaft axis, the average rolling
moment per blade on the helicopter is
L = YhR + eR Z sin dt
0 τ ψ
τ ∫
= + d
d
(sin cos ) sin d – sin d b g
2
0
2
2
0
YhR 3
M ex R
t
t eR F t τ β ξ ψ τ ψ
τ τ ∫ ∫
= –
2
sin cos sin d –
2
sin d b g
2 2
0
2
0
2
YhR 3
M ex R eR F
Ω π ∫ β ξ ψ ψ π∫ ψ ψ
π π
(1.35)
in which xgR = rg.
Similarly, the pitching moment M per blade is
M XhR
=– –M ex R eR F
2
sin cos cos d –
2
cos d b g
2 2
0
2
0
2
3
Ω π ∫ β ξ ψ ψ π∫ ψ ψ
π π
(1.36)
The first integrals in L and M are the inertia couples which arise when the plane
of a rotor with offset hinges is tilted relative to the shaft. If the flapping motion
relative to the shaft is βs = a0 – a1s cos ψ – b1s sin ψ, then, for small β and ξ, these
integrals are 12
MbexgΩ2R2b1s and 12
MbexgΩ2R2a1s respectively.
The second integrals in L and M are found to be very much smaller and can be
neglected, so that to a good approximation
L YhR M ex R b = + 12
b g
2 2
Ω 1s
= + 12
YhR SeRb1s (1.37)
and M XhR SeRa = – + 12
1s (1.38)
where S = MbxgΩ2R is the centrifugal force of the blade for zero offset, and
approximately so for small offset.
For the small hinge offsets that are usual, the second terms of eqns 1.37 and 1.38
　
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