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of W2/3 as might have been expected, which is an indication of a large amount of
separation drag. The drag curve of the much cleaner fixed wing aircraft is more
nearly proportional to W2/3.
A breakdown of the fuselage parasite drag is shown in the table below.
4
3
2
1
0 20 40 60 80 100 120 140 160 180 kN
XH51A Clean fixed wing aircraft
SA 341
UH1
(research)
Lynx (Navy)
S52 Lynx (Army) Experimental helicopters
Scout
UH1B
S61
Belvedere
S58
Super Frelon
Typical production
helicopters
EH101(Civil)
CH 53A
CH 47
HH (X)
EH101 (Mil)
CH 46
Equivalent flat plate area SFP m2
Gross weight
S67
Fig. 4.15 Parasite drag of helicopters
Component Percentage drag
Basic fuselage with protuberances 20 to 40
Landing gear or fairing 6 to 25
Rotor pylon and hub 35 to 50
Tailrotor and tail surfaces 5 to 15
The drag of the rotor pylon and hub represents a high proportion of the overall
drag, and this is therefore an area where drag reduction leads to considerable benefit;
hence the appearance of hub and pylon fairings on the larger and faster helicopters.
S55
Trim and performance in axial and forward flight 131
Interference drag plays a significant role, because on a helicopter there are a number
of separate aerodynamic ‘shapes’ in close proximity whose pressure distributions
and boundary layers can interact with each other. Hub and pylon fairings are designed
to minimise interference drag in addition to reducing the basic parasitic drag contribution
of these components, the upper cambered shape of the former being a result3.
Also, larger and faster helicopters tend to utilise retractable landing gear, which
leads to the lower figure in the above table.
4.3.2 Analytical estimation of performance
Except at very low speeds (when the disc incidence may not always be small) we can
put ˆV = μ and λ
i = stc/2μ; also, writing λc for ˆV sin τc, eqn 4.20 can be expressed
as
q k
st s A
sA
c t d
2 12
c
2
t t
c c
12
3
= 0
8
δ (1 + 3μ ) + (1 + ) μ 1 + + λ + μ
 

 


(4.21)
where qc is the torque coefficient corresponding to the given power.
The expression for qc can be used to calculate either the torque and power for a
given flight condition or, as described below, the maximum speed and rate of climb
for a given torque.
To find the maximum level speed (λc = 0) for a given power, i.e. given qc, we have
to solve the quartic in μ expressed by eqn 4.21. Now at high speed we note that the
induced power is small; therefore, neglecting this term and the term 3μ2 of the profile
drag, we find as a first approximation to μ, μ1 say,
μ δ
1
3 c t t
0
=
2[q – (1 + sA /sA)/8]
d
The value of qc corresponding to the maximum power (900 kW) is 0.00834. Then,
with the previously given values of δ and d0, we find
μ1
3 = 0.0561 or μ1 = 0.383
With this value of μ, we calculate the terms previously neglected to give the
second approximation μ2 as
μ 2
3 = 0.0468 or μ2 = 0.36
which is extremely close to the correct value. Thus, the iteration provides the required
maximum value in two steps.
To find the maximum rate of climb we must satisfy the condition
∂λc/∂μ = 0
This condition leads to
6 0 + [3 – 2(1 + ) ] 1 + = 0
4 3
c
d kst 2 t t
s A
sA μ δμ 


(4.22)
132 Bramwell’s Helicopter Dynamics
To solve this equation for μ we note from Fig. 4.14 that the blade profile drag
contributes little to the slope of the power curve (which led to eqn 4.22), so that for
a first aproximation μ1 we can neglect the second term of eqn 4.22 and obtain
μ1
4 c
2
0
= t t
(1 + )
3
1 +
k st
d
s A
sA


From our data we find
μ1
4 = 0.000642
or
μ1 = 0.159
For the second approximation, a value for the second term is calculated using μ1,
giving
μ 2
4 = 0.000642 – 0.000117 = 0.000525
or
μ2 = 0.152
This agrees with the value obtained graphically and, again, the iteration leads to a
satisfactory answer in two steps. This value of μ is substituted into eqn 4.21 and the
　
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