曝光台 注意防骗
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lagging, the blade axes coincide with the hub axes. Now, suppose the blade flaps
through angle β about e2, bringing the blade axes into a position whose unit vectors
are i1, j1, k1. The relationships between e1, e2, e3 and i1, j1, k1 are related through a
rotation matrix transformation as
i
j
k
e
e
e
1
1
1
1
2
3
=
cos 0 sin
0 1 0
– sin 0 cos
β β
β β
(1.24)
The blade now rotates about the lag axis through angle ξ, bringing the unit vectors
of the blade into their final positions i, j, k. The relationships between i, j, k and i1,
j1, k1, are, in matrix form
i
j
k
i
j
k
=
cos sin 0
– sin cos 0
0 0 1
1
1
1
ξ ξ
ξ ξ (1.25)
The relationships between e1, e2, e3 and i, j, k, are on multiplying the transformation
matrices in eqns l.24 and 1.25 together
Ω
e3
e2
k
j i β
ξ
ψ
Fig. 1.18 Deflected rotating blade
Basic mechanics of rotor systems and helicopter flight 23
i
j
k
e
e
e
=
cos cos sin cos sin
– sin cos cos – sin sin
– sin 0 cos
1
2
3
ξ β ξ ξ β
ξ β ξ ξ β
β β
(1.26)
The above relationships enable us to express quantities measured in one set of
axes in terms of another set, and we shall need them for calculating the forces and
moments on the helicopter.
Let the distance of the centre of gravity of the blade measured from the hinge be
rg. In terms of axes fixed in the blade the position vector of the c.g. is rg = rgi, and
in terms of hub axes the position vector is
rg + eRe1 = rg(cos β cos ξ e1 + sin ξ e2 + sin β cos ξ e3) + eRe1
Expressing the absolute acceleration ag of the c.g. as ag = a1e1 + a2e2 + a3e3, the
components, by applying the standard equations of the kinematics of a rigid body, are
found to be
a r
t
r
t
1 g r eR
2
2 g
2
= d g
d
(cos cos ) – 2 d
d
β ξ Ω (sin ξ) – Ω( cos β cos ξ + ) (1.27)
a r
t
r
t
2 g r
2
2 g
2
= d g
d
(sin ) + 2 d
d
ξ Ω (cos β cos ξ ) – Ω sin ξ (1.28)
a r
t 3 g
2
2 = d
d
(sin β cos ξ) (1.29)
Now, let the aerodynamic force on the blade be F and let R be the force exerted
by the hinge on the blade. If Mb is the blade mass, the equation of motion is
F + R = Mbag (1.30)
If F = F1e1 + F2e2 + F3e3 and R = R1e1 + R2e2 + R3e3,
R Ma F
R Ma F
R Ma F
1 b 1 1
2 b 2 2
3 b 3 3
= –
= –
= –
(1.31)
We now wish to resolve these rotating force components along axes fixed in the
helicopter. In order to comply with the usual stability axes, we take a set of unit axes
x, y, z, with the z axis pointing downwards along the negative direction of e3, Fig.
1.19.
If X, Y, Z are the hub force components along the fixed axes, then, remembering
that the force the blade exerts on the hub is –R,
X = R1 cos ψ – R2 sin ψ
Y = – R1 sin ψ – R2 cos ψ
Z = R3
These forces are time dependent, not only because of the sin ψ and cos ψ terms,
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Bramwell’s Helicopter Dynamics(19)
