曝光臺 注意防騙 網曝天貓店富美金盛家居專營店坑蒙拐騙欺詐消費者

vr =
× r
where
is the angular velocity of the body and axes, so that, from eqn A.1.1
h = Σr × (m
× r)
= Σm
(r · r) – Σmr(
· r) (A.1.6)
Now let us write
r = xi + yj + z k

= ω1i + ω2 j + ω3k
where i, j, k are a set of orthogonal unit vectors fixed in the body. If we also write
h = h1i + h2 j + h3 k
we find on expanding eqn A.1.6 that the the components of angular momentum are
given by
h A F E
h B D F
h C E D
1 1 2
2 2 3
3 3 1
= – –
= – –
= – –
ω ω ω
ω ω ω
ω ω ω
3
1
2

 
 
(A.1.7)
where A, B, C, D, E, F are the moments and products of inertia defined by
362 Bramwell’s Helicopter Dynamics
A = Σm(y2 + z2), B = Σm(x2 + z2) , C = Σm(x2 + y2)
D = Σmyz, E = Σmxz, F = Σmxy
The relations given by eqn 1.12 can be expressed conveniently in matrix form as
h
h
h
A F E
– F B D
– E – D C
1
2
3
1
2
3
=
– –
–


















ω
ω
ω
where the square matrix is sometimes referred to as the inertia tensor.
The inertia tensor can be regarded as an operator which transforms the angular
velocity vector into the angular-momentum vector.
It is always possible to choose axes through any point in the body such that the
products of inertia D, E, F vanish. These axes are called principal axes, in which case
h = Aω1i + Bω2 j + Cω3k (A.1.8)
If the origin of the axes is a fixed point or the centre of gravity, we find that
dh/dt = dH/dt = T (A.1.9)
since, in both cases, rg × a0 = 0 and either v0 = 0 or v0 = vg so that v0 × vg = 0.
Since, in general, the axes will be moving
dh/dt = ∂h/∂t +
× h (A.1.10)
and taking the special case of rotation about a fixed point and referring the motion to
principal axes, eqns A.1.8, A.1.9 and A.1.10 give
Aω˙1 – (B – C)ω2ω 3 = L (A.1.11)
Bω˙2 – (C – A)ω3ω1 = M (A.1.12)
Cω˙3 – (A – B)ω1ω 2 = N (A.1.13)
where L, M, N are the components of the torque T.
Equations A.1.11, A.1.12 and A.1.13 are known as Euler’s dynamical equations
and can easily be remembered from their cyclic form.
A.1.2. Euler’s equations for a rigid blade
In dealing with blade motion, however, we often wish to regard the blade as a rigid
body moving about a hinge system which is offset from the rotor axis. To simplify
matters we can assume that the hinge system is effectively concentrated at a point.
Thus, the blade moves about a point which is not fixed, Fig. A.1.1, and Euler’s
equations no longer apply.
On the other hand, it would be inconvenient to take the centre of gravity as origin
for, in calculating the torque, we would have to consider the unknown reactions at the
hinge. We can, however, extend Euler’s equations by making use of eqn A.1.3. Let us
write the position vector of the c.g. relative to the hinge as
Appendices 363
rg = xgi + yg j + zgk
and the absolute acceleration of the hinge as
a0 = axi + ay j + az k
Then the term rg × a0 in eqn A.1.3 becomes
rg × a0 = (ygaz – zgay)i + (zgax – xgaz)j + (xgay – ygax)k
and, taking principal axes as before, eqn A.1.3 can be expanded to give
Aω˙ 1 – (B – C)ω2ω3 – Mb(ygaz – zgay) = L (A.1.14)
Bω˙ 2 – (C – A)ω3ω1 – Mb(zgax – xgaz) = M (A.1.15)
Cω˙ 3 – (A – B)ω1ω2 – Mb(xgay – ygax) = N (A.1.16)
in which Mb is the blade mass. Equations A.1.14, A.1.15 and A.1.16 are referred to
as the ‘extended’ Euler equations. Actually, the centre of gravity of the blade can be
considered to lie practically on the i axis, so that we can write rg = xgi and the above
equations can be simplified to
Aω˙ 1 – (B – C)ω2ω3 = L (A.1.17)
Bω˙ 2 – (C – A)ω3ω1 + Mb xgaz = M (A.1.18)
　
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