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Cω˙ 3 – (A – B)ω1ω2 – Mb xgay = N (A.1.19)
If the hinge offset distance is eR and the rotor shaft rotates with angular velocity
Ω, then a0 is clearly of magnitude Ω2eR and is directed from the hinge to the shaft
axis.
A.2 The stability equations
The theory of Appendix A.1 can be applied to obtain the stability equations which are
discussed in detail in Chapter 5. We take an orthogonal set of axes, fixed in the
helicopter and whose origin is located at the helicopter’s centre of gravity. It can be
supposed, as is usual with the fixed wing aircraft, that the helicopter has a longitudinal
plane of symmetry, although this is rather less true of the single rotor helicopter
k
j i
Ω
c.g.
o
Fig. A.1.1 Blade pivot point
364 Bramwell’s Helicopter Dynamics
because of its tailrotor. The axes will be chosen so that the x and z directions lie in
the longitudinal plane with the y axis pointing to starboard, Fig. A.1.2.
The equations of motion of the helicopter, treating it as a rigid body, are
F = mdv/dt (A.2.1)
and T = dh/dt (A.2.2)
where F is the resultant external force on the helicopter and m its mass, and T is the
moment of this force.
We shall suppose that in trimmed flight there is no sideslip, so that the initial flight
velocity components along the x, y, z axes are U, 0, W. During disturbed flight the
increments of velocity will be denoted by u, v, w, so that the velocity vector v can be
written as
v = (U + u)i + vj + (W + w)k (A.2.3)
It will be supposed that the disturbance velocity components are small compared
with the steady components U and W. In trimmed flight the angular velocity of the
helicopter can be written as
= pi + qj + rk (A.2.4)
Then if X, Y, Z are the components of the force F, the force equations, from eqn
A.2.1 are
m[ ˙u + q(W + w) – vr] = X (A.2.5)
m[ ˙v + r(U + u) – p(W + w)] = Y (A.2.6)
m[ ˙w + pv – q(U + u)] = Z (A.2.7)
The angular momentum equation (eqn A.2.2) can be written
T = ∂h/∂t + × h (A.2.8)
y
x
z
Fig. A.1.2 Axis system for the helicopter
Appendices 365
and the components of momentum are, from eqn A.1.7,
h1 = Ap – Fq – Er
h2 = Bq – Dr – Fp
h3 = Cr – Ep – Dq
With T = Li + Mj + Nk, the expansion of eqn A.2.8 gives
A ˙p – (B – C)qr + D(r2 – q2) – E(pq + ˙r ) + F(pr – ˙q) = L (A.2.9)
B ˙q – (C – A)rp + E(p2 – r2) – F(qr + ˙p) + D(qp – ˙r ) = M (A.2.10)
C ˙r – (A – B)pq + F(q2 – p2) – D(rp + ˙q) + E(rq – ˙p) = N (A.2.11)
The two sets of equations A.2.5 to A.2.7 and A.2.9 to A.2.11 can be simplified
considerably by assuming that the disturbance velocity and angular velocity components
are so small that squares and products of them can be neglected. Further, if we
assume that the helicopter has a plane of symmetry, the products of inertia D and F
both vanish.
Then the force and moment equations simplify to
mu˙ + qW = X (A.2.12)
m˙v+ rU – pW = Y (A.2.13)
mw˙ – qU = Z (A.2.14)
and Ap˙ – Er˙ = L (A.2.15)
Bq˙ = M (A.2.16)
Cr˙ – Ep˙ = N (A.2.17)
In particular, if ‘wind axes’ are chosen, i.e. if the x axis is initially taken to lie
parallel to the flight direction, then W = 0 and the terms containing W vanish in eqns
A.2.12 and A.2.13.
A.3 Multiblade summations
In some helicopter problems it is necessary to calculate the total force or moment on
the helicopter by adding the contributions of the individual blades. In steady motion
these blade contributions are periodic, and a typical term for a given blade would be
An cos nψ. If there are b equally spaced blades, the contribution of the neighbouring
blades will be An cos n (ψ ± 2π/b) and the total effect of all the blades is therefore
An cos nψ + An cos n(ψ + 2π/b) + … + An cos n[ψ + 2π (b – 1)/b]
= cos ( + 2 / )
0
–1
An k= n k b
b
Σ ψ π
366 Bramwell’s Helicopter Dynamics
Let C n k b
k=
b
= cos ( + 2 / )
0
–1
Σ ψ π
and S n k b
k=
b
= sin ( + 2 / )
0
–1
Σ ψ π
so that
C S
k=
b
n kb n
k=
b
+ i = e = e e kn b
0
–1
i ( + 2 / ) i
0
–1
Σ ψ π ψΣ2π i/
The terms in the summation are a geometric series, and we easily find that
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