Download - Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

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Page 1: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Flight Dynamics and Aircraft Performance

Lecture 8: Helicopters

G. Dimitriadis

University of Liege

Page 2: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Textbooks

  Bramwell’s Helicopter Dynamics, A. R. S. Bramwell, G. Done, D. Balmford, Butterworth-Heinemann, 2001   Basic Helicopter Aerodynamics, J. Seddon, BSP (Blackwell Scientific Publications) Professional, 1990   Principles of Helicopter Aerodynamics, J. G. Leishman, Cambridge University Press, 2000

Page 3: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Introduction   Helicopters can do all this: Hovering flight, reverse Forward flight, backflips etc

Page 4: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Not helicopters   Autogyros, gyrogliders etc are not helicopters.

Focke-Wulf Fw 61, 1936

Cierva C.4, 1923

Fa 330, Gyroglider (or rotorkite), 1943

Cierva C.19, 1932

Page 5: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Helicopters

Breguet-Dorand, Gyroplane Laboratoire,1935

Gyroplane de Breguet, 1907

Helicoptère de Paul Cornu, 1907

Pescara Helicopter No 3, 1924

Page 6: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Belgian First

  The first ever tandem rotor helicopter was built by Nicolas Florine.   It first flew in 1933 at the Laboratoire Aérotechnique de Belgique (now Von Karman Institute).

Page 7: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Modern helicopters Bell 204/205, 1955 Aerospatiale Alouette II, 1955 CH-47, Chinook, 1957

Kamov 50, 1982 Eurocopter Tiger, 1991 Mil Mi-26, Hind, 1977

Page 8: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Largest helicopter ever built   Mil-V12

Page 9: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

How does a helicopter fly?

  By accelerating downwards a column of air through the rotor.

The rotor creates a pressure difference Δp which accelerates flow through it. The velocity far upstream is 0, at the rotor vi and far downstream v∞.

Disk pi<p∞

p∞

p∞

pi+Δp>p∞

v=0

vi

v∞Flow field Pressure

field Velocity field

Page 10: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Pressure change

  Using Bernoulli’s equation on the upstream flow (assuming incompressibility) we have:

  On the downstream flow we have:

  So that

p∞ = pi +12ρvi

2

p∞ +12ρv∞

2 = Δp + pi +12ρvi

2

Δp =12ρv∞

2 (1)

Page 11: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Rotor thrust

  Mass flow through the rotor:

  Far downstream: the momentum flow, i.e. the momentum of the mass that flowed through the rotor is equal to:

  The thrust is the difference in momentum flow, i.e.

m = ρAvi

Jdownstream = m v∞ = ρAviv∞

T = Jdownstream − Jupstream = ρAviv∞ (2)

Page 12: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Airspeed at infinity   Noting that the pressure change across the rotor is a measure of the thrust,

  We can combine with equation (1) to show that   And that   Define: vi=induced velocity   Define w=T/A=disc loading   Define P=Tvi=induced power of the rotor

Δp =TA

= ρviv∞

v∞ = 2vi

T = 2ρAvi2

Page 13: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Thrust for vertical climb

  If the helicopter is climbing at speed Vc

The airspeed upstream is equal to Vc+vi. Donwstream it is equal to Vc+v∞. Bernoulli upstream: Bernoulli downstream:

p∞ +12ρ Vc + v∞( )2 = Δp + pi +

12ρ Vc + vi( )2

p∞ +12ρ Vc( )2 = pi +

12ρ Vc + vi( )2

Vc

Vc+vi

Vc+v∞

Page 14: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Thrust for vertical climb (2)

  The pressure change is therefore

  The thrust is given by:

  Combining with (3) gives v∞=2vi, i.e.

  The induced power is

T = ρA Vc + vi( ) Vc + v∞( ) − ρA Vc + vi( )Vc = ρA Vc + vi( )v∞

Pi = T Vc + vi( )

Δp =12ρv∞ 2Vc + v∞( ) (3)

T = 2ρA Vc + vi( )vi

Page 15: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Induced velocity and climb velocity

  Consider a hover case where the thrust is equal to Th, the power to Ph and the induced airspeed to vh.   Consider climbing flight at the same thrust, Th. The rotor climbs but also induces a velocity vi≠vh.   It is easy to see that   So that

vh2 = Vc + vi( )vi

vivh

= −Vc

2vh+

Vc

2vh

#

$ %

&

' (

2

+1

Page 16: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Induced power and climb power

  Therefore, we can write that   Leading to:

Pi = Th Vc + vi( )

PiPh

=Th Vc + vi( )

Thvh=

Vc

2vh+

Vc

2vh

"

# $

%

& '

2

+1

Page 17: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Real climbing rotor wake   The results

shown before assume that the wake is a column with a smooth and continuous vertical velocity distribution   A real wake is

much more complex

Page 18: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Real climbing rotor wake   Depending on

the rotation speed, climb speed, blade span and blade twist, the blade can produce: –  Lift near the tip

(the wake curls upwards)

–  Downforce near the root (the wake curls downwards)

Blade

Tip vortex

Inner vortex sheet

Page 19: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Real descending rotor wake

  When the helicopter is descending, the rotor descends into its own wake.

Page 20: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Descent   Climb is an easy case. The rotor wake lies

under the rotor and the rotor itself climbs into a smooth airflow.

  On the contrary, when the helicopter is descending, the rotor descends into its own wake.

  There are three different possibilities: –  Vortex ring flow: The rotor tips are caught inside

their own vortex rings. –  Turbulent wake state: The rate of descent is so

high that the rotor wake develops upwards but is quite turbulent.

–  Windmill brake state: The rate is even higher. The rotor wake develops upwards but is well defined.

Page 21: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Descent cases

Vortex ring flow, slow descent

Vortex ring flow, faster descent

Turbulent wake state

Windmill brake state

Page 22: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Vortex ring state   Denote by Vd the descent speed.   If Vd=O(vh), i.e the induced velocity in hover,

then some of the air recirculates around the rotor.

  Effectively, the rotor wake is squashed onto the rotor.

  The phenomenon leads to very high descent speeds and loss of stability.

  Recovery can be accomplished by moving the helicopter forward so that the rotor encounters clean air as its wake lies behind it.

Page 23: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Windmill brake state

  At much higher descent rates, i.e. Vd>>vh, the rotor wake develops upwards.   The wake is well defined.   The airflow decelerates on passing through the rotor.   The turbulent wake state lies between the vortex ring and windmill brake states. The rotor acts as a bluff body.

Page 24: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Safe descent   So how can a helicopter achieve a safe descent?   There are two methods:

– Descend very slowly so that Vd<<vh and the rotor wake effectively descends with the rotor.

– Descend with a forward velocity component so that the rotor wake lies behind the rotor.

  It must be said that near the ground the descent speed will be necessarily low.   Additionally, near the ground the helicopter can benefit from the ground effect.

Page 25: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Ground effect

  A helicopter hovering near the ground benefits from a large improvement in efficiency.   The vertical velocity of the wake on the ground must be equal to zero.   Therefore, the induced velocity of the rotor is very low. As P=Tvi, the power required to produce the same amount of thrust is much lower near the ground.

Page 26: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Induced velocity in ground effect

  Induced velocity in ground effect divided by induced velocity in free air.

r

h=height above ground r=distance from centre of rotor divided by R.

Page 27: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Blade Element Method   Blade Element Method (BEM), also known as strip theory in aeroelasticity.   It consists of estimating the aerodynamic forces on a small element of a blade, dy.

R

y

dy

c

ΩR

ψ

ψ=azimuth angle

Page 28: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Blade Element   The blade can have a pitch angle of θ. It also features an inflow angle φ=tan-1[(Vc+vi)/Ωy].   Its true angle of attack is given by α=θ-φ.

Page 29: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Blade element lift and drag

  The blade element lift and drag are given by:

  Where cl and cd come from the sectional characteristics of the blade element.   The thrust is given by:

  The in-plane torque is given by

dL =12ρU 2ccldy

dD =12ρU 2ccddy

dT = dLcosφ − dDsinφ

dQ = dLsinφ + dDcosφ( )y

Page 30: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Approximations

  The inflow angle is assumed to be small.   The drag coefficient is assumed to be much smaller than the lift coefficient.   Therefore:

dT ≈ dLU ≈ ΩydQ ≈ φdL + dD( )y

Page 31: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Non-dimensionalizations

  Define the following non-dimensional quantities:

  Also, for a rotor with N blades define the solidity factor as:

r = y /R =ΩyΩR

=UΩR

λ =Vc + viΩR

= rφ = inflow factor

dCT =dT

ρA ΩR( )2

dCQ =dQ

ρA ΩR( )2R

σ =blade areadisc area

=NcRπR2 =

NcπR

Page 32: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Total thrust and torque

  After non-dimensionalization, the blade element forces can be integrated over the blade span to yield:

  The rotor power requirement is given by P=ΩQ. Non-dimensionalising:

CT = dCT0

1

∫ dr =σ2

CLr2

0

1

∫ dr

CQ = dCQ0

1

∫ dr =σ2

φCL + CD( )r30

1

∫ dr

CP =P

ρA ΩR( )3= CQ

Page 33: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Thrust Approximation

  For attached flow, the lift coefficient of a blade element is given by

  where a is the lift curve slope. The thrust coefficient becomes

  So that, finally,

cl = aα = a θ − φ( )

CT =σ2

a θ − φ( )r20

1

∫ dr =σa2

θr2 − λr( )0

1

∫ dr =σa2

θ3−λ2

( )

* +

CT =σa2

θ3−λ2

& '

( )

(4)

Page 34: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Thrust in hover

  If the rotor is in hover, and

  Then, from (4),

  Which is a nonlinear equation relating pitch angle θ to thrust. It can be solved inversely as:

T = 2ρAvi2

CT = 2λ2

CT =σa2

θ3−12

CT

2%

& '

(

) *

θ =6σa

CT +32

CT

2

Page 35: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

About twist

  As shown earlier, helicopter blades produce little lift near the centre of the rotor because of the low linear speed.   Define the sectional lift as

  For the case where a=2π,

  Define

l =dLdy

=12ρU 2ccl =

12ρU 2ca θ − φ( )

lρ ΩR( )2c

= πr2 θ − φ( )

cl =l

ρ ΩR( )2c= πr2 θ − φ( )

Page 36: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Effect of twist

  Adding geometric twist to the blade can increase the sectional lift coefficient near the centre of the rotor.   This generally means increasing the twist towards the centre.   Consider two cases:

– Case θ=θ0. The pitch is constant over the blade.

– Case θ=θ1+θ2r. The pitch varies over the blade, i.e. there is geometric twist. For the pitch to be higher near centre of the rotor, θ2<0 and θ1>θ0.

Page 37: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Twist example

Keep in mind that this result was obtained using BEM. 3D effects near the wingtip have been ignored

Page 38: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Ideal twist   The ideal twist distribution is obtained when θr is

constant, i.e. θr=θ0.   This is a nonlinear twist that cannot be implemented at

the blade root but it is ideal because it corresponds to the minimum induced power.

Page 39: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Forward Flight   Forward flight

is different to vertical climb and hover!

  It creates a total thrust that is not centered on the rotor.

  This thrust causes a significant rolling moment on the rotor, making the helicopter impossible to fly.

Ω

Forward velocity V

V+ΩR

V-ΩR

V

Reversed velocity

Page 40: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Avoiding the rolling moment

  The way to cancel the rolling moment is to allow the blade to flap.   The additional lift of the advancing blade causes an upward flapping motion.   Similarly, the lower lift of the retreating blade causes a downward flapping motion.   Therefore, the rolling moment is not transmitted to the helicopter.

Page 41: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Flapping

  Flapping is a stable motion because flapping up causes the lift to drop and flapping down to increase

V+Ωr

β r

Advancing blade flaps upwards

α<θ

θ θ

Retreating blade flaps downwards

Ωr-V

β rα>θ

Page 42: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Corioli’s moments

  The flapping motion causes Corioli’s moments on the blades:

ΩR

(1-e)ΩRcosβ+ΩeR

The Corioli’s moment is due to the inequality of the tip speeds of the flapped and unflapped blades. It can cause a yawing moment on the helicopter

Page 43: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Lagging motion

  The way to avoid the yaw moment due to flapping is to allow the blade to lag:

Page 44: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Pitching (feathering)

  The rotor is not only the lifting surface but also the propulsion and main control system.   The main means of control of the rotor is the changing of the pitch of the blades (also known as feathering).   Pitch control can be either collective (all blades change pitch at the same time) or cyclic (the pitch change depends on whether the blade is advancing or retreating).

Page 45: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Westland Wessex hub

Flap hinge

Lag hinge

Pitch control

Pitch bearing

Page 46: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Westland Lynx Hingeless rotor: the blades are not hinged, they are solidly connected to the rotor hub. However, they have flexible elements near the root which allow flap and lag degrees of freedom, restrained by the stiffness of these elements.

Pitch bearings

Lag dampers

Flexible elements

Page 47: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Helicopter control   Control of the helicopter is handled almost exclusively by the rotor. There are two parameters of importance: – Magnitude of rotor thrust –  Line of action of rotor thrust

  Both of these parameters are controlled by rotor pitch. – Collective pitch increases the magnitude of the

thrust. – Cyclic pitch can change the line of action of

the thrust

Page 48: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Collective vs cyclic pitch The swashplate mechanism: -Lifting or lowering the swashplate increases or decreases collective pitch. -Tilting the swashplate introduces cyclic pitch. -In this case cyclic pitch is used to increase the angle of attack of the retreating blade.

Page 49: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Cyclic pitch   Cyclic pitch changes the pitch angle θ with

azimuth angle ψ.   This change is usually expressed as a first

order Fourier series:

  A1, the lateral cyclic coefficient, applies maximum/minimum pitch when the blades are at ψ=0o/ψ=180o. The blade response is phased by 90o, hence the lateral effect.

  B1, the longitudinal cyclic coefficient, applies maximum/minimum pitch when the blades are at ψ=90o/ψ=270o. Again, the blade response if phased by 90o.

θ ψ( ) = θ0 − A1 cosψ − B1 sinψ

Page 50: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Tip Path Plane   Using cyclic pitch it is possible to incline the rotor without inclining the rotor shaft.   The line of action of the thrust is perpendicular to the blade Tip Path Plane: T

Page 51: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Forward flight, Forward C.G

D

mg

C.G

TCase where the Centre of Gravity lies in front of the rotor shaft. In this case, the resultant of the weight and drag on the fuselage lies on the same line of action as the thrust.

Tip Path Plane

Page 52: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Forward flight, Aft C.G

D

mg

C.G

TTip Path Plane Case where the

Centre of Gravity lies aft of the rotor shaft. Again, the resultant of the weight and drag on the fuselage lies on the same line of action as the thrust. The pitch angle of the fuselage is much smaller than in the forward C.G. case.

Page 53: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Direct Head Moment

D

mg

C.G

TTip Path Plane

MfIn a more general case, the drag on the fuselage will also cause a fuselage pitching moment, Mf. This moment will be counteracted by the fact that the thrust and resultant of fuselage weight and drag are not colinear.

Page 54: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

How to start going forward   A hovering helicopter has no forward velocity.   The pilot uses cyclic pitch to tip the Tip Path

Plane forward and tilt the thrust vector forward.   The helicopter picks up forward speed.   The fuselage develops drag and pitches nose

down.   Now the rotor shaft is also pitched nose down;

there is no more need to apply cyclic pitch to the rotor.

T T T

W W WD D

Page 55: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Longitudinal stability in forward flight

  TD=rotor thrust perpendicular to TPP   HD=rotor drag parallel to TPP   τc=path angle to horizontal (climb rate)   αD=angle of attack of TPP

Page 56: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Longitudinal equilibrium equations

  Resolving forces horizontally and vertically gives:

  In forward flight the TPP is tipped forward so that αD+τc is usually small. The climb rate, τc, on the other hand, is not necessarily small. Hence:

TD cos αD + τ c( ) − HD sin αD + τ c( ) =W + Dsinτ cTD sin αD + τ c( ) + HD cos αD + τ c( ) = −Dcosτ c

TD ≈ T =W + Dsinτ cT αD + τ c( ) + HD = −Dcosτ c

Page 57: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Drag

  There are two main sources of drag: – Fuselage drag – Rotor drag

  Fuselage drag is usually calculated in terms of the so-called equivalent flat plate area   Rotor drag is subdivided into

– profile drag –  induced drag

Page 58: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Fuselage Drag   There are two source of fuselage drag:

– Parasite drag –  Interference drag

  Parasite drag has many sources:

  Interference drag is caused by the interaction of flow coming from these different components.

Page 59: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Parasite drag examples Define D=1/2ρV2SFP, SFP being the equivalent flat plate area, i.e. the area of a flat plate that has the same drag as the fuselage.

Page 60: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Rotor drag   The rotor drag is given by HD=Hp+Hi, where Hp is the profile drag and Hi the induced drag.   The profile drag is evaluated with respect to the drag of the chosen airfoil section and the angle of attack of the blade using blade element theory.   Making a polynomial approximation,

–  CHP=HP/ρA(ΩR)2=δ0+δ1αD+δ2αD2

  The induced drag can be assumed to be small for forward steady flight.

Page 61: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Longitudinal trim

  Therefore, the trim angle of attack of the TPP can be obtained from

  Such that:

  Where and T=W.

CT =W

ρA ΩR( )2+ CD sinτ c

CT αD + τ c( ) + CHD= −CD cosτ c

αD = −1

CT

12

ˆ V 2d0 cosτ c + CHD

% &

' ( − τ c

ˆ V = V /ΩR, d0 = SFP / A

Page 62: Flight Dynamics and Aircraft Performance - · PDF fileFlight Dynamics and Aircraft Performance Lecture 8: ... 1955 CH-47, Chinook, 1957 Mil Mi-26, Hind, ... blade. – Case θ=θ 1

Power required for forward flight

µ =V cosαD

ΩR

There is an optimum advance ratio, μ, requiring minimum power.

Maximum forward speed

Maximum climb rate