© Lakshmi Sankar 20021 Steady, Level Forward Flight I. Introductory Remarks.

© Lakshmi Sankar 2002 1

Steady, Level Forward Flight

I . Introductory Remarks


The Problems are Many..

ThrustAeroelasticResponse

0

270 180

90

Dynamic Stall onRetreating Blade

Blade-Tip Vortexinteractions

UnsteadyAerodynamicsTransonic Flow on

Advancing Blade

Main Rotor / Tail Rotor/ Fuselage

Flow Interference

V

NoiseShock Waves

Tip Vortices


The Dynamic Pressure varies Radially and Azimuthally

0

270

180

90

V

RVtip

RVtip

VRVtip

VRVtip

RReverseFlow Region

Advancing SideRetreating Side


Consequences of Forward Flight

• The dynamic pressure, and hence the air loads have high harmonic content. Above some speed, vibrations can limit safe operations.

• On the advancing side, high dynamic pressure will cause shock waves, and too high a lift (unbalanced). – To counter this, the blade may need to flap up (or pitch down) to

reduce the angle of attack.• Low dynamic pressure on the retreating side.• The blade may need to flap down or pitch up to increase

angle of attack on the retreating side. This can cause dynamic stall.

• Total lift decreases as the forward speed increases as a consequence of these effects, setting a upper limit on forward speed.


Forward Flight Analysis thus requires

• Performance Analysis – How much power is needed?

• Blade Dynamics and Control – What is the flapping dynamics? How does the pilot input alters the blade behavior? Is the rotor and the vehicle trimmed?

• Airload prediction over the entire rotor disk using blade element theory, which feeds into vibration analysis, aeroelastic studies, and acoustic analyses.

• We will look at some of these elements.


Steady, Level Forward Flight

II. Performance


Inflow Model

• To start this effort, we will need a very simple inflow model.

• A model proposed by Glauert is used.• This model is phenomenological, not

mathematically well founded.• It gives reasonable estimates of inflow velocity at

the rotor disk, and is a good starting point.• It also gives the correct results for an elliptically

loaded wing.


Force Balance in Hover

Thrust

Weight In hover, T= W

That is all!

No net drag, or side forces. The drag forces on the individual blades

Cancel each other out, when summed up.

Drag

Drag

Rotor Disk


Force Balance in Forward Flight

Flight Direction

Weight, W

Vehicle Drag, D

Thrust, T


Simplified Picture of Force Balance

T

W

D

c.g. Flight Direction

TPP

Rotor Disk, referred toAs Tip Path Plane(Defined later)

WT

DT

TPP

TPP

cos

sin


Recall the Momentum ModelV

V+v

V+2v


Glauert’s Conceptual model

Freestream, V∞ Freestream, V∞

Induced velocity, v

Freestream, V∞

2v


Total Velocity at the Rotor Disk

Freestream V∞

Induced Velocity, v

V∞cosαTPP

V∞sinαTPP

Total velocity

22 vsinVcosV

Velocity Total

TPPTPP


Relationship between Thrust and Velocities

In the case of hover and climb, recall

T = 2 A (V+v) v

Induced Velocity

Total velocity

Glauert used the same analogy in forward flight.


In forward flight..

22 vsincosv2 TPPTPP VVAT

This is a non-linear equation for induced velocity v, which must be iteratively solved for a given T, A, and tip path plane angle TPP

It is convenient to non-dimensionalize all quantities.


Non-Dimensional Forms

R

R

V

R

V

RA

TC

TPP

T

v ratio, inflow dimensinal-Non

ratio, advance called is

cos

Speed Tip

Component Freestream Edgewise

i

2

22T tan2C

becomes form

ldimensiona-nonin equation Glauert

iTPPi


Approximate Form atHigh Speed Forward Flight

2

2

tan2C

thatso ratio inflow exceedsfar

small, is angle planepath tipif and

0.2,n higher tha is ratio advance If

22T

i

Ti

i

iTPPi

C

In practice, advance ratio seldom exceeds 0.4, because of limitations associated with forward speed.


Variation of Non-Dimensional Inflow with Advance Ratio

222T tan2C iTPPi

i

2T

i

C

Notice that inflow velocity rapidly decreases with advance ratio.


Power Consumption in Forward Flight

later.it compute will Weblade. on the drag viscous todue

power, profile blade include power will actual The

disk.rotor at the velocity normal times

Thrust as theory sGlauert' from Power" Ideal" computecan We

Power ProfileDv

,

Tsin Recall

Power Profilevsin

TPP

VTP

Thus

D

VTP TPP

Induced Power Parasite Power


Power Consumption in Level Flight

fV

f

SC

SCVSCV

TP

D

DD

3

DD

32

2

1Power Parasite

area. plateflat equivalent the, called isproduct This

.product supply the

tocustomary isit area, referencecommon aon agreement no is thereBecause

on. based is

C area reference theis S and t,coefficien drag parasite vehicleis C Here

flight. forward speedhigh in n consumptiopower dominates and

) ratio advance(or velocity theof cube theas increasespower parasite The

2

1V

2

1DVPower Parasite

earlier. discussed as , ratio advance with decreases Tvpower induced The

Power Profile Blade DVv


Induced Power Consumption in Forward Flight

Induced Power, Tv

V∞

Pow

er


Parasite Power Consumption in Forward Flight

Par

asite

Pow

erV∞

Pow

er


Profile Power Consumption in Forward Flight

Blade Profile Power

V∞

Pow

er

Airfoil theoft Coefficien Drag Average

318

RAPower Profile

:derivelater willWe

0

203

d

d

C

where

C


Power Consumption in Forward Flight

Induced Power, TvP

aras

ite P

ower

Blade Profile Power

V∞

Pow

er Tota

l Pow

er


Non-Dimensional Expressions for Contributions to Power

20,3

20,P,0

33

3

,3

23iP,

0,,,

3P

3182

1 ,

318

Ceory that element th blade from showlater willWe

2

121

2

1

R

vvC

v

C :Recall

diTP

d

parasitePparasite

iT

i

PparasitePiPP

C

A

fCCThus

C

A

f

RA

fVCfVP

CRA

T

RA

T

TP

CCCC

RA

P


Empirical Corrections

• The performance theory above does not account for– Non-uniform inflow effects– Swirl losses– Tip Losses

• It also uses an average drag coefficient.

• To account for these, the power coefficient is empirically corrected.


Empirical Corrections

203 3182

1

ed)(Uncorrect CoefficentPower

diT

C

A

fC

Induced power Parasite PowerProfile Power

203 6.4182

1

)(Corrected CoefficentPower

diT

C

A

fC

1.15


Excess Power Determines Ability to Climb

Induced Power, Tv

Par

asite

Pow

erBlade Profile Power

V∞

Pow

er Tota

l Pow

er

Available Power

Exc

ess

Pow

er

Rate of Climb= Excess Power/W


Forward Flight

Blade Dynamics


Background

• Helicopter blades are attached to the rotor shaft with a series of hinges:– Flapping hinges (or a soft flex-beam) , that allow blades to freely

flap up or down. This ensures that lift is transferred to the shaft, but not the moments.

– Lead-Lag Hinges. When the blades rotate, and flap, Coriolis forces are created in the plane of the rotor. In order to avoid unwanted stresses at the blade root, lead-lag hinges are used.

– Pitch bearing/pitch-link/swash plate: Used to control the blade pitch.

• The blade loads are affected by the motion of the blades about these hinges.

• From an aerodynamic perspective, lead-lag motion can be neglected. Pitching and flapping motions must be included.



Articulated Rotor


Flap Hingehttp://www.unicopter.com/0941.html#delta3

http://huizen.dds.nl/~w-p/bookaut/scbprts.htm

http://huizen.dds.nl/

http://huizen.dds.nl/~w-p/bookaut/scbph1.gif


Pitch Hornhttp://www.unicopter.com/0941.html#delta3


Lead-Lag Motion

http://huizen.dds.nl/~w-p/bookaut/scbph2.gif




Coordinate Systems

• Before we start defining the blade motion, and the blade angular positions, it is necessary to define what is the coordinate system to use.

• Unfortunately, there are many possible coordinate systems. No unique choice.


Hub Planez

X

Y

Z axis is normal to shaft

X-axis runs along fuselage symmetry planeAnd is normal to Z

Y-axis is perpendicular toFuselage symmetry planeAnd perpendicular to z

=0 deg

=90 deg

=180 deg

=270 deg


This plane is defined by two straight lines.The first connects the blade tips at azimuth angle =0 and =180 deg.The second connects the blade tips at azimuth angle =90 and =270 deg.Z is perpendicular to TPP.In TPP, the bladeDoes not appear to beFlapping..

Tip Path Plane

X

Z

Blade is at =0 deg.


Tip Path Plane

X

0+1c

0-1c

R0+1c)R0-1c)

2R1c1c


No-Feathering Plane (NFP) or Control Plane

The pilot controls the blade pitch by applying a collective control (all blades pitch up or down by the same amount), or by a cyclic control (which involves tilting the swash-plate). Some of the pitch links move up, while others move down. The airfoils connected pitch up or down).

X

ZPitch link

Swash Plate

Pilot Input

Blade at =90 deg

Blade at =270 deg


Differences between Various Systems

• For an observer sitting on the tip path plane, the blade tips appear to be touching the plane all the time. There is no flapping motion in this coordinate system.

• For an observer sitting on the swash plate, the pitch links will appear to be stationary. There is no vertical up or down motion of the pitch links, and no pitching motion of the blades either.

• In the shaft plane, the blades will appear to pitch and flap, both.

• One can use any one of these coordinate systems for blade element theory. Some coordinate systems are easier to work with. For example, in the TPP we can set the flapping motion to zero.


Forward Flight

Blade Flapping Dynamics and Response


Background

• As seen earlier, blades are usually hinged near the root, to alleviate high bending moments at the root.

• This allows the blades to flap up and down.• Aerodynamic forces cause the blades to flap up.• Centrifugal forces causes the blades to flap down.• Inertial forces will arise, which oppose the direction of

acceleration.• In forward flight, an equilibrium position is achieved,

where the net moments at the hinge due to these three types of forces (aerodynamic, centrifugal, inertial) cancel out and go to zero.


Schematic of Forces and Moments

dL

dCentrifugal Force

r

We assume that the rotor is hinged at the root, for simplicity.This assumption is adequate for most aerodynamic calculations.Effects of hinge offset are discussed in many classical texts.


Velocity encountered by the Blade

V∞

X

r

Velocity normal toThe blade leading edgeisr+V∞sin


Moment at the Hinge due toAerodynamic Forces

From blade element theory, the lift force dL =

drCVrc l2sin

2

1

Moment arm = r cos ~ rCounterclockwise moment due to lift = drrCVrc l

2sin2

1

Integrating over all such strips,Total counterclockwise moment =

Rr

r

ldrrCVrc0

2sin2

1


Moment due to Centrifugal Forces

The centrifugal force acting on this strip = rdm

r

dmr 22

Where “dm” is the mass of this strip.This force acts horizontally. The moment arm = r sin ~ rClockwise moment due to centrifugal forces = dmr 22

Integrating over all such strips, total clockwise moment =

2

0

22

IdmrRr

r


Moment at the hinge due to Inertial forces

r

dmrr

Small segment of mass dmWith acceleration

is positive if blade is flapping up

Associated moment at the hinge =

r

Flap Hinge

Integrate over all such segments: Resulting clockwise moment at the hinge= I


At equilibrium..

rdrVrcCIITip

Root

l

2

2 sin2

1

Note that the left hand side of this ODE resembles a spring-mass system, with a natural frequency of .

We will later see that the right hand side forcing term has first harmonic (terms containing t), second, and higher Harmonic content.

The system is thus in resonance. Fortunately, there isAdequate aerodynamic damping.


How does the blade dynamics behave when there is a forcing function component on the right hand side of the form Asint, and a damping term on the left hand side of form c d/dt ?

To find out let us solve the equation:

tAIcI sin2

To solve this equation, we will assume a solution of form:

tCtB cossin

2sincos

tc

At

c

A

In other words, the blade response will be proportional to the amplitude A of the resonance force, but will lag the force by 90 degrees.


What happens when the pilot tilts the swash plate back?

• The blade, when it reaches 90 degrees azimuth, pitches up.

• Lift goes up instantly.• The blade response

occurs 90 degrees later (recall the phase lag).

• The front part of the rotor disk tilts up.

• The exact opposite happens with the blade at

• 270 degree azimuth.

X

Z

Blade at =270 deg

Blade at =90 deg


What happens when the pilot tilts the swash plate back?

Blade reaches itsHighest positionAt =180 deg

Blade reaches itslowest positionAt =0 deg

T

SwashPlate

The tip path plane tilts back. The thrust points backwards. The vehicle will tend to decelerate.

TPP

NFP


What happens when the pilot tilts the swash plate to his/her right?

• Blade at = 180 degpitches up. Lift goes up.

• Blade responds by flapping up, and reaches its maximum response 90 degrees later, at = 270 deg.

• The opposite occurs with the blade at = 0 deg.

• TPP tilts towards the pilot’s right.

• The vehicle will sideslip.

TPP

Helicopter viewed fromAft of the pilot

NFP

T


Blade Flapping Motion

2sin2cos

sincos

form theofsolution general thehas

sin2

1

equation dynamics flapping blade The

22

110

2

2

sc

sc

Tip

Root

l rdrVrcCII

Coning AngleFirst and higher harmonics

=t


1c determines the fore-aft tilt of TPP

1c

X

This angle is measured in one of the three coordinate systems (Shaft plane, TPP, or NFP) that we have chosen to work with.Some companies like to use TPP, others like NFP or shaft plane.

TPP

If TPP is our reference coordinate system, what will 1c be? Ans: Zero.


1s determines the lateral tilt of TPP

Y

1s

Helicopter viewed from the back of the pilot


Pilot Input

sincos 110%75 scR

The pitch is always, by conventionSpecified at 75%R.The pitch at all the other radial locations may be foundIf we know the linear twist distribution.

Collective Lateral controlLongitudinal control

The pilot applies a collective pitch by vertically raising the swash plate up or down. All the blades collectively, and equally pitch up or down.

The pilot applies longitudinal control (i.e. tilts the TPP fore and aft) byTilting the swash plate fore or aft as discussed earlier. Lateral control means tilt the swash plate (and the TPP) laterally.


Longitudinal control

Blade at 90 deg azimuth

NFP

Line Parallel to NFP

X1s

1c

Note that 1s+ 1c is independent of the coordinate system in which these angles are measured.

1s


Lateral Control

Y1c

NFP

Line parallel to NFP

1c

Blade at =180

1s

Note that 1c-1s is independent of the coordinate system in which 1c

And 1s were measured.

TPP


As a result..

TPPcNFPscs

TPPsNFPcsc

1111

1111


In the future..

• We will always see 1c+1s appear in pair.• We will always see 1s-1c appear in pair.• As far as the blade sections are

concerned, to them it does not matter if the aerodynamic loads on them are caused by one degree of pitch that the pilot inputs in the form of 1c or 1s, or by one degree of flapping (1c or 1s).

• One degree of pitch= One degree of flap.


Forward Flight

Angle of Attack of the Airfoil

And

Sectional Load Calculations


The angle of attack of an airfoil depends on

• Pilot input: collective and cyclic pitch• How the blade is twisted• Inflow due to freestream component, and

induced inflow• Velocity of the air normal to blade chord,

caused by the blade flapping• Anhedral and dihedral effects due to coning of

the blades.• The first two bullets are self-evident. Let us look

at the other contributors to the angle of attack.


Blade Flapping Effect

magnitude. equal ofdownwash a

ngexperienci isit thinksairfoil The

.rvelocity

aat up flapssection blade The

r d/dt


Angle of Attack caused by the Inflow and Freestream

V∞

Rotor Disk

Induced inflow, v

V∞sin+v

r+V∞cos

Edgew

ise co

mpo

nent

Norm

al Com

ponent


Anhedral/Dihedral Effect

V∞cosV∞

V∞sin

Blade at =180 willSee an upwash equal to V∞sin

V∞sinV∞

Blade at =0 willSee an upwash equal to V∞sin

Blade at any will see an upwash equal to- V∞sin cos


Summing them all up..

UT= r+V∞cos

UP

Zero lift line

sincosVrvsin

,

tan 1

VU

where

U

U

U

U

P

T

P

T

P


Small Angle of Attack Assumptions

• The angle of attack a (which is the angle between the freestream and the rotor disk) is small.

• The cyclic and collective pitch angles are all small.

• The coning and flapping angles are all small.

• Cos() = Cos()= Cos() ~ 1• sin() ~ sin() ~ sin() ~


Angle of Attack

T

PT

T

P

T

Peffecive U

UU

U

U

U

U

arctan

sincos 110 sctw R

r

sinsincos VrVrU sT

cos

coscos

VrR

VrRU

s

sP

Subscript s: All angles are in the shaft plane


Angle of attack (continued)

cos

sincossin 110

VrR

R

rVr

UU

s

sctw

PT

sincos 11 cs

d

d

dt

d

d

d

dt

d


Angle of Attack (Continued)After some minor algebra,

vcossin

sincossin

sincos

02

11

110

11110

TPPcs

sctw

cssctwPT

VVV

VR

rV

R

rrUU

Notice 1c+1s appears in pair, as pointed out earlier.

Also 1c-1s appears in pairs.

One degree of pitching is equivalent to one degree ofFlapping.


Relationship between TPP and s

TPP=s+1c

V∞

s 1c

TPP


Angle of Attack (Concluded)

v

cossin

sincossin

sincos

1

02

11

110

11110

TPP

cs

sctw

cssctw

TT

PTeffective

V

VV

VR

rV

R

rr

UU

UU


Calculation of Sectional Loads

• Once the angle of attack at a blade section is computed as shown in the previous slide, one can compute lift, drag, and pitching moment coefficients.

• This can be done in a number of ways. Modern rotorcraft performance codes (e.g. CAMRAD) give the user numerous choices on the way the force coefficients are computed.


Some choices for Computing Sectional Loads as a function of

• In analytical work, it is customary to use Cl=a, Cd=Cd0 = constant, and Cm= Cmo, a constant. Here “a” is the lift curve slope, close to 2.

• In simple computer based simulations using Excel or a program, these loads are corrected for compressibility using Prandtl-Glauert Rule.

• More sophisticated calculations will use C-81 tables, with corrections for the local sweep angle.


Prandtl-Glauert Rule

X

V∞

ψ

UT= r+V∞cos

UP

Zero lift line

Compute Mach number=M= UT/a∞

2

,,

1 M

CC ibleincompressl

correctedl


More accurate ways of computing loads

• In more sophisticated computer programs (e.g. CAMRAD-II, RCAS), a table look up is often used to compute lift, drag, and pitching moments from stored airfoil database.

• These tables only contain steady airfoil loads. Thus, the analysis is quasi-steady.

• In some cases, corrections are made for unsteady aerodynamic effects, and dynamic stall.

• It is important to correct for dynamic stall effects in high speed forward flight to get the vibration levels of the vehicle correct.


Calculation of Sectional Forces

• After Cl, Cd, Cm are found, one can find the lift, drag, and pitching moments per unit span.

• These loads are normal to, and along the total velocity, and must be rotated appropriately.

UT= r+V∞cos

Zero lift line

L’=Lift per foot ofspan

D’

PTT

effPT

l

UUcaU

aUUc

cCUL

2

1

)(2

12

1

22

2


Forward Flight

Integration of Sectional Loads

To get

Total loads


Background

• In the previous sections, we discussed how to compute the angle of attack of a typical blade element.

• We also discussed how to compute lift, drag, and pitching moment coefficients.

• We also discussed how to compute sectional lift and drag forces per unit span.

• We mentioned that these loads must be rotated to get components normal to, and along reference plane.

• In this section, we discuss how to integrate these loads.• In computer codes, these integrations are done

numerically.• Analytical integration under simplifying assumptions will

be given here to illustrate the process.


Assumptions for Analytical Integration

• c= constant (untapered rotor)

• v = constant (uniform inflow)

• Cd = constant

• Linearly twisted rotor

• No cut out, no tip losses.


Blade Section

UT= r+V∞cos

UP

Zero lift line


Effective Angle of Attack

As discussed earlier,T

Peff U

U

sin VrUT

cos

cossin

VR

VvVU

TPP

TPPP

PTTeffTl UUUcacaUUcCL 222

2

1

2

1

2

1


Some algebra first..

sincoscossin

sinsin2sin

cossin

cossin2cossin

sin2sinsin2

2

1

cossin

sincossin

2

1

22

31

2211

22

21

2

112222

23222

0022

0

1102

VrVRVrR

VrVr

V

rVrVR

rR

rV

R

rVrVr

ca

VRVr

R

rVr

caL

TPPTPP

sss

c

cctw

twtw

sctw

Notice that we have first, second, and third harmonics present!These fluctuations will be felt by the passengers/pilots as vibratory loads.


Thrust

2

0 02

1ddrLbT

R

Thrust is computed by integrating the lift radially to get instantaneous thrust force at the hub, then averaging the thrust force over the entire rotor disk,and multiplying the force per blade by the number of blades.

Computer codes will do the integrations numerically, withoutany of the assumptions we had to make.


Anlaytical Integration of ThrustWe can interchange the order of integration.

Integrate with respect to ψ first. Use the formulas such as

2

0

22

0

2

2

0

2

0

22

0

2

0

cossin

0dcossin

0cossincossin

dd

ddd


Result of Azimuthal Integration

TPPs

twtw

RrrV

R

rV

R

r

Vr

cadL

21

23

2

20220

2

0

2

2

1

2

1


Next perform radial integration and Normalize

22

44232

12

22

1

23

2203

20

RRV

RVR

RVR

abcT

TPPTPPs

twtw

221

42

31

321220 TPPTPPstw

T

aC

Note that we will get the hover expressions back if advance ratio is set to zero.


Torque and Power

• We next look at how to compute the instantaneous torque and power on a blade.

• These are azimuthally-averaged to get total torque and total power.

• It is simpler to look at profile and induced components of torque are power separately.


Profile Drag

D’

r

sin

2

10,

2

VrU

where

cCUD

T

dT

We will assume chord c and drag coefficient Cd0 are constant.


Integration of Profile Torque

2

0 0

0 2

1 R

drDrbQ

ARR

QCQ 20,

Non-dimensionalize:

Final result: 20,0, 1

8

d

Q

CC


Profile Power

2

0 0

0 2

1 R

drDrbQ

AR

PCP 3

00,

Non-dimensionalize:

Final result: 20,0, 31

8

d

P

CC

2

0 0

0 2

1 R

T drDUbP


Induced Drag

UT= r+V∞cos

UP

Zero lift line

L’

Di

PTl

T

PTl

T

Pi

UUcC

U

UUcC

U

ULD

2

1

2

1 2


Induced Torque and Power

2

0 0

2

0 0

2

1

2

1

ddrDUP

ddrrDQ

R

iTi

R

ii

Performing the analytical integration, TiP CCi

This is a familiar result. Induced Power = Thrust times Induced Velocity!


In-Plane Forces

• In addition to thrust, that act normal to the rotor disk (or along the z-axis in the coordinate selected by the user), the blade sections generate in-plane forces.

• These forces must be integrated to get net force along the x- axis. This is called the H-force.

• These forces must be integrated to get net forces along the Y- axis. This is called the Y-force.

• These forces will have inviscid components, and viscous components.


Origin of In-Plane Forces

L’

One source of in-plane forces is the tilting of theThrust due to the blade coning angle.

L’sin=L’

L’

X

Y


Origin of In-Plane Forces-II

V∞

A component of the free-stream flows along the blade, in the radial direction. This causes radial skin friction forces. This is hard to quantify, and is usually neglected.

Radial flow causes radialSkin friction forces


Origin of In-Plane Forces III

D=Di+D0

Sectional drag (which is made of inviscid induced drag, and viscous drag)Can give rise to components along the X- direction (H-force), andY- direction (Y-Force).

Engineers are interested in both the instantaneous values (which determineVibration levels), as well as azimuthal averages (which determine force balance).


Closed Form Expressions for CH and CY

• Under our assumptions of constant chord, linear twist, linear aerodynamics, and uniform inflow, these forces may be integrated radially, and averaged azimuthally.

• The H- forces and the Y- forces are non-dimensionalized the same way thrust is non-dimensionalized.

• Many text books (e.g. Leishman, Prouty) give exact expressions for these coefficients.


Closed Form Expressions

TPPs

TPPc

tw

aY

TPPs

c

twTPP

dH

TPP

TPP

C

aCC

0201

100

2

20

101

0

0

2

331

6

43

2

4

3

4

46

22

24


Forward Flight

Calculation of Blade Flapping Dynamics


Background

• In the previous sections, we developed expressions for sectional angle of attack, sectional loads, total thrust, torque, power, H-force and Y-force.

• These equations assumed that the blade flapping dynamics a priori.

• The blade flapping coefficients are determined by solving the ODE that covers flapping.


Flapping Dynamics

rdrVrcCIITip

Root

l

2

2 sin2

1

vV

VV

VV

r

U

a

U

UUaaC

TPP

cs

sc

cssc

TT

PTeffectivel

cossin

sincossin

sincos

02

11

110

11110


Solution Process

2sin2cos

sincos

form theofsolution general thehas

sin2

1

equation dynamics flapping blade theAssume

22

110

2

2

sc

sc

Tip

Root

l rdrVrcCII

Plug in the solution on both the left and right sides. The right sideCan be integrated analytically, subject to usual assumptions.

Equate coefficients on the left side and right, term by term.For example coefficient with sinψ on the left with thesimilar term on right.


Final Form

2

0

11

2

%75

11

112

2%800

2

11

3

42

31

4

3

3

8

66601

8

cs

TPPR

sc

scTPPtw

R


Level Flight

Calculation of Trim Conditions

Including

Fuselage Aerodynamics


Background

• By trim conditions we mean the operating conditions of the entire vehicle, including the main rotor, tail rotor, and the fuselage, needed to maintain steady level flight.

• The equations are all non-linear, algebraic, and coupled.

• An iterative procedure is therefore needed.


Horizontal Force Balance

V∞

DF

HM

HT

Total Drag= Fuselage Drag (DF) + H-force on main rotor (HM)+ H-force on the tail rotor (HT)


Vertical Force Balance

LF

GW

Vertical Force = GW- Lift generated by the fuselage, LF


Tip Path Plane Angle

V∞ TPP

T

F

TMFTPP

TMFTPP

FTPP

LGW

HHD

HHDT

LGWT

1tan

sin

cos


Fuselage Lift and Drag

• These are functions of the fuselage geometry, and its attitude (or angle of attack).

• This information is currently obtained from wind tunnel studies, and stored as a data-base in computer codes.


Fuselage Angle of Attack

• Extracted from – Tip path angle – Blade flapping dynamics – Downwash felt by the fuselage from the main

rotor – Shaft inclination angle.


Shaft Inclination Angle

Fuselage Reference Axis

is


Relationship between Tip Path Plane Angle of Attack and Shaft

Angle of AttackTPP

Shaft Plane

V∞

s 1c

TPP

csTPP 1


Angle of Attack of the Fuselage

• Start with tip path plane angle of attack.

• Subtract 1c to get shaft angle of attack

• Subtract the inclination of the shaft

• Subtract angle of attack reduction associated with the downwash from the rotor

V

v1 scTPPF i


Iterative process

• Assume angle of attack for fuselage (zero deg).• Find LF and DF from wind tunnel tables.• Compute needed T. during the first iteration, T is

approximately GW-LF. Use this info. to find main rotor torque, main rotor H-force, tail rotor thrust needed to counteract main rotor torque, and tail rotor H- force.

• From blade trim equations, find 1c.• Find tip path plane angle of attack.• Recompute fuselage angle of attack.• When iterations have converged, find main and tail rotor

power. Add them up. Add transmission losses to get total power needed.


Autorotation in Forward Flight

• The calculations described for steady level flight can be modified to handle autorotative descent in forward flight.

• Power needed is supplied by the time rate of loss in potential energy.


Descent

V∞

Wsinχ

W cosχ

Rate of descent

χ

V

locitydescent vetan 1

DF

HT

HM

LF


Tip Path Plane Angle in Descent

F

TMFTPP

TMFTPP

FTPP

LGW

GWHHD

GWHHDT

LGWT

cos

sintan

sinsin

coscos

1


Iterative Procedure

• The iterative procedure involves – assume a rate of descent– Iterate on fuselage angle of attack to achieve forces

to balance, as done previously in steady level flight.– Compute the power needed to operate= main rotor+

tail rotor+ transmission losses.– Equate this power needed with the power available

from loss of potential energy= GW * Rate of descent.– Iterate until power needed = power available

Date post:	23-Dec-2015
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© Lakshmi Sankar 20021 Steady, Level Forward Flight I. Introductory Remarks.

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