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(aerodynamic) loading.
7.3.3 Solution of the forced response equation
We shall now consider the solution of the forced response equation, eqn 7.86 above.
For a rigid blade, φ1 can be identified with the flapping angle β and the right hand
side with the aerodynamic moment MA. As mentioned before, the orthogonal properties
of the normal modes decouple the inertia and stiffness terms so that the terms on the
left hand side of eqn 7.86 are independent of the other modes. Unfortunately, the
aerodynamic loading contains all the modes, since the complete blade motion is
required to determine the aerodynamic incidence. This means that the right hand side
is a function of φ1, φ2, …, etc., so that the mode equations cannot be solved independently.
We are therefore faced with the problem of integrating these equations in the most
efficient way.
The usual procedure is to assume a starting value for the blade motion– for example,
we may take the classical rigid blade flapping – and compute the aerodynamic loading
and the integral on the right hand side of eqn 7.86 for the modes being considered.
The value of the integral is then imagined to remain constant over a small azimuth
variation Δψ and the corresponding changes of φn and dφn/dψ are computed from
eqn 7.86. The loading integral is then re-calculated and φn and dφn/dψ are computed
for the next azimuth step Δψ. These stepwise integrations are continued until the
required values of φn converge to within an acceptable limit. The integrations may
extend over several resolutions before satisfactory convergence has been achieved.
Part of the process of convergence concerns the transient blade motion, because of
the inevitability of some error between the assumed blade motion, taken as the
starting value, and the true motion; and part will be due to the inherent features in the
integration process. There must, of course, be as many integrations as there are
modes taken to represent the blade motion.
Of the methods of integration, the most common is probably the fourth order
Runge-Kutta, but it appears that convergence difficulties arise if the step lengths are
too small, particularly for the higher modes.
A simpler and yet more accurate method is to take advantage of the fact that, if the
right hand side of eqn 7.86 is kept constant, the equation can be solved exactly over
the step length. Let us denote the right hand side of eqn 7.86 by fi, where fi is the
constant value corresponding to the ith step. Then eqn 7.86 can be written as
d2φn/dψ2 + λn2φn = φin (7.96)
To avoid confusion, we shall drop the suffix n from φn and let φ represent any mode
displacement. Let φ
i and ′ φ
i be the values of φ and dφ /dψ at the start of the ith step;
then eqn 7.96 can be written as
d2φ/dψ2 + λ2φ = fi
whose solution is easily found to be
274 Bramwell’s Helicopter Dynamics
φ
λ
λψ φ λψ φ
= (1 – cos ) + cos + λ sin λψ 2
fi
i
i′
If Δψ is the step change of ψ, the values of φ and dφ/dψ at the start of the (i + 1)th
step are
φ ′ ′
λ
λ ψ φ λ ψ φ
i λ λ ψ
i
i
f i
+1 2 = (1 – cos Δ ) + cos Δ + sin Δ (7.97)
and
φ′ λ λ ψ λφ λ ψ φ′λ ψ i
i
i i
f
+1 = sin Δ – sin Δ + cos Δ (7.98)
These are the initial values at the start of the next interval, at the end of which
φ
λ
λ ψ φ λ ψ
φ
i λ λ ψ
i
i
f i
+2
+1
2 +1
+1 = (1 – cos Δ) + cos Δ + ′ sin Δ (7.99)
Eliminating φi′+1 and φi′ by means of eqns 7.97 and 7.98 leads to the recurrence
relationship
φ φ λψ φ
λ ψ
λ i+2 = 2 i+1 cos – i – 2 fi +1 fi
1 – cos
Δ ( + )
Δ
(7.100)
connecting the mode displacement with the values at the two previous intervals and
the corresponding loading integrals.
Wilkinson and Shilladay16 have obtained the same result by appealing to sampled
data theory and using the Z-transform. The values of fi can be represented as discrete
input functions occurring at regular intervals, as in Fig. 7.25.
The use of constant values of fi over the interval is referred to as a zero-order hold,
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Bramwell’s Helicopter Dynamics(138)
