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profile drag. Therefore as a good first approximation to hcD we can write
hc
1
D 4 = μδ (4.9)
and obtain, from eqn 4.8, the first approximation to αD.
The inflow ratio can now be calculated from
λ α λ D D i = sin – ˆV
= μD tan αD – λ
i
or, approximately,
λD = (μα)D – λ
i (4.10)
To obtain λ
i we can use the values of vi of Fig. 3.2 which applies when αD ≈ 0. Now
λi i
= v 0
vΩ
R
(4.11)
118 Bramwell’s Helicopter Dynamics
and
V V
R R
=
0
D
0
ˆΩ Ω
v ≈ μ v (4.12)
where v0 = √(T/2ρA) is the mean induced velocity in hovering flight or ‘thrust
velocity’, and μD = ˆV cos D α . Then, if v0 /ΩR is computed, V can be obtained from
eqn 4.12 and, with vi being read off from Fig. 3.2, λ
i can be finally calculated from
eqn 4.11.
With λD now known, or rather a first approximation to it, and taking tcD wc = ,
eqn 3.63 can be solved for the collective pitch angle θ0. These values of λD and θ0
enable a better approximation to hcD to be calculated from eqn 3.64 (or 3.65) and
new values of αD and λD to be obtained from eqns 4.8 and 4.10 respectively. In most
cases, however, the improved value of hcD makes very little difference to αD, as
indicated by the numerical example below, which also indicates the usefulness of the
disc axes for these calculations. Finally, the flapping coefficients can be calculated.
Example. To illustrate the procedure just described, we shall calculate the trim of a
four-bladed helicopter in level flight at sea level at tip speed ratio 0.3. The helicopter
is represented by the following data:
W = 45 000 N, solidity s = 0.05, R = 8 m, h = 0.25
δ = 0.013, ΩR = 208 m/s, SFP = 2.3 m2, b = 4, a = 5.7
Blade data: Mb = 74.7 kg; in terms of R, xg = 0.45, e = 0.04
It will be assumed that the fuselage pitching moment, Mf, including, possibly, a
tailplane, is zero.
The above data give
s A = 10 m2 , d = 0.23, w = 0.085, d = 0.0104 = 9.6 m/s
0 c
12
2
μ 0 v0
The first approximation to hcD is 1
4 μδ = 0.000 975, and hence
αD μ
12
2
= – ( 0 + c )/ c D d h w
= – 0.134 = – 7.67°
Now v0 /ΩR = 0.0462, so that V = μΩR/v 0 = 6.50. From Fig. 3.2 or eqn 3.2 we
find vi0 = 0.154, from which λ
i = v i0v0 / ΩR = 0.0071.
Then, from eqn 4.10,
λD = – 0.0473
Solving eqn 3.63 for θ0 gives
θ0 = 0.1824 = 10.5°
To obtain a better value of hcD we need an estimate of a1. From eqn 3.60, and
using the above values of θ0 and λD, we find
a1 = 0.104 = 5.93°
Trim and performance in axial and forward flight 119
and, using eqn 3.64 or 3.65, we calculate hcD , giving
hcD = 0.001 172
This value is about 20 per cent greater than the first approximation, 1
4μδ , but the
new value of λD using this revised value of hcD is
λD = –0.0479
which is close to one per cent of the first value. The small difference in λD using the
recalculated value of hcD is also found at other values of μ, suggesting that the
original approximation to hcD leads to sufficiently accurate values of λD and θ0. In
any case, it is very unlikely that the fuselage drag, upon which λD and αD depend
strongly, would be known accurately enough to justify ‘exact’ calculations of the
H-force.
From eqns 3.61, 3.57, and 3.66, we find
a0 = 0.066 = 3.78°, b1 = 0.037 = 2.1°
qc = 0.00579
The total power required at this speed is therefore 638 kW.
To calculate the cyclic pitch to trim, we write eqn 1.45 in terms of force components
related to disc axes, i.e.
–WfR – TDhRB1 + (HD + TDa1)hR + Mf – Ms(B1 – a1) = 0
Putting TD = W and solving for B1 gives
B1 = a1 + (Mf + HDhR – WfR)/(WhR + Ms) (4.13)
= 1 + ( f + cD – c )/( c + s) a Cm h h wf wh Cm (4.14)
where Cmf = Mf /ρsAΩ2R3
and Cms M sA R bSeR sA R = s / = /
2 3 12ρΩ ρΩ2 3
or Cms bM x e sAR = b g /2ρ
since S = Mbrg .
Ω2
The pitching moment Mf of the basic fuselage is not likely to be known very
accurately. The rather unstreamlined shape of most helicopter fuselages and the fact
that the fuselage lies in the complicated downwash pattern of the rotor makes it
　
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