#### Transcript Part 2

```Steady, Level Forward Flight
I . Introductory Remarks
1
The Problems are Many..
Transonic Flow on
Noise
Thrust
Aeroelastic
Response
Shock
Waves
Aerodynamics
  0
  90
Main Rotor / Tail Rotor
/ Fuselage
Flow Interference
Tip Vortices
interactions
V
  180
  270
Dynamic Stall on
2
The Dynamic Pressure varies
V
  180 
Vtip  R

Retreating Side
Vtip  R  V
  90
  270 
Vtip  R  V
R
Reverse
Flow Region
  0
Vtip  R
3
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.
4
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.
5
II. Performance
6
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
7
Force Balance in Hover
Thrust
Drag
Rotor Disk
Drag
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
Sankar
2002 summed up.
8
Force Balance in Forward Flight
Thrust, T
Vehicle Drag, D
Flight Direction
Weight, W
9
Simplified Picture of Force Balance
T
aTPP
Rotor Disk, referred to
As Tip Path Plane
(Defined later)
c.g.
Flight Direction
D
T sin a TPP  D
T cos a TPP  W
W
10
Recall the Momentum Model
V
V+v
V+2v
11
Glauert’s Conceptual model
Freestream, V∞
Freestream, V∞
Induced velocity, v
Freestream, V∞
2v
12
Total Velocity at the Rotor Disk
V∞cosαTPP
V∞sinαTPP
Induced Velocity, v
Total Velocity 
V cos a TPP   V sin a TPP  v 
2
2
13
Relationship between
Thrust and Velocities
In the case of hover and climb, recall
T = 2 r A (V+v) v
Induced Velocity
Total velocity
Glauert used the same analogy in forward flight.
14
In forward flight..
T  2 rAv V cos aTPP   V sin aTPP  v
2
2
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 aTPP
It is convenient to non-dimensionalize all quantities.
15
Non-Dimensional Forms
T
rAR 2
Edgewise Freestream Component V cos a TPP V


Tip Speed
R
R
CT 
Non - dimensinal inflow ratio, i 
v
R
Glauert equation in non - dimensiona l
form becomes
C T  2i    tan a TPP  i 
2
2
16
Approximate Form at
High Speed Forward Flight
If advance ratio  is higher tha n 0.2,
and if tip path plane angle is small,
 far exceeds inflow ratio i so that
C T  2i  2   tan a TPP  i 
2
 2 i
CT
i 
2
In practice, advance ratio  seldom exceeds 0.4,
because of limitations associated with forward speed.
17
Variation of Non-Dimensional Inflow
CT
i 
2
i
CT  2i  2   tan a TPP   i2
2

Notice that inflow velocity rapidly decreases with advance ratio.
18
Power Consumption in
Forward Flight
We can compute " Ideal Power" from Glauert' s theory as Thrust
times normal velocity at the rotor disk.
The actual power will include blade profile power,
due to viscous drag on the blade. We will compute it later.
P  T V sin a TPP  v   Profile Power
Recall Tsin a TPP  D
Thus,
P  Tv  DV  Profile Power
Induced Power
Parasite Power
19
Power Consumption in Level Flight
P  Tv  DV  Blade Profile Power
The induced power Tv decreases with advance ratio  , as discussed earlier.
1
1

Parasite Power  DV   rV2C D S  V  rV3C D S
2
2

The parasite power increases as the cube of the velocity (or advance ratio  )
and dominates power consumptio n in high speed forward flight.
Here C D is vehicle parasite drag coefficien t, and S is the reference area C D
is based on.
Because there is no agreement on a common reference area, it is customary to
supply the product C D S .
This product is called f , the equivalent flat plate area.
1
Parasite Power  rV3 f
2
20
Power
Induced Power Consumption in
Forward Flight
Induced Power, Tv
V∞
21
Power
Parasite Power Consumption in
Forward Flight
V∞
22
Profile Power Consumption in
Forward Flight
We will later derive :
Profile Power  rAR 
Power
3
C d 0
8
1  3 
2
where
Cd 0  Average Drag Coefficien t of the Airfoil
V∞
23
Power
Power Consumption in Forward
Flight
V∞
24
Non-Dimensional Expressions for
Contributions to Power
P
rAR 3
C P  C P ,i  C P , parasite  C P ,0
Recall : C P 
Pi  Tv
Tv
T
v


 CT i
3
2
rAR  rAR  R
1
r
V3 f
1
1 f 3
Pparasite  rV3 f
C P , parasite  2


3
2
rAR  2 A
C P,i 
We will later show from blade element th eory that C P,0 
C d , 0
8
1  3 
2
1 f 3 C d , 0

Thus, C P  CT i 
 
1  3 2 
2A
8
25
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.
26
Empirical Corrections
Power Coefficent (Uncorrect ed) 
1 f 3 Cd 0
2
CT i 
 
1  3
2A
8


Profile Power
Induced power Parasite Power
Power Coefficent (Corrected ) 
1 f 3 C d 0
2
CT i 
 
1  4.6
8 2002

1.15

27
Excess Power Determines Ability to
Climb Rate of Climb= Excess Power/W
Excess Power
Power
Available Power
V∞
28
Forward Flight
29
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.
forces are created in the plane of the rotor. In order to avoid
pitch.
• From an aerodynamic perspective, lead-lag motion can
be neglected. Pitching and flapping motions must be
included.
30
31
Articulated Rotor
32
Flap Hinge
http://www.unicopter.com/0941.html#delta3
http://huizen.dds.nl/
~w-p/bookaut/scbprts.htm
33
Pitch Horn
http://www.unicopter.com/0941.html#delta3
34
35
36
37
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.
38
Hub Plane
z
Y-axis is perpendicular to
Fuselage symmetry plane
And perpendicular to z
=180 deg
Y
Z axis is normal to shaft
=270 deg
X-axis runs along fuselage symmetry plane
And is normal to Z
=90 deg
=0 deg
X
39
Tip Path Plane
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
X
azimuth angle =90 and =270 deg.
Z is perpendicular to TPP.
Z
Does not appear to be
Flapping..
40
Tip Path Plane
2Rb1c
b1c
b0+b1c
Rb0-b1c)
Rb0+b1c)
X
b0-b1c
41
No-Feathering Plane (NFP) or
Control Plane
Z
Swash Plate
Pilot Input
X
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
42
pitch up or down).
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
• 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.
43
Forward Flight
Response
44
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.
45
Schematic of Forces and Moments
dL
r
b
dCentrifugal
Force
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.
46
Velocity normal to
is
r+V∞sin
V∞
r

X
47
Moment at the Hinge due to
Aerodynamic Forces
From blade element theory, the lift force dL =


1
2
rc r  V sin   Cl dr
2
Moment arm = r cosb ~ r
1
2
Counterclockwise moment due to lift = rcr  V sin   rCl dr
2
Integrating over all such strips,
Total counterclockwise moment =
r R
1
2
r 0 2 rcr  V sin   rCl dr
48
Moment due to Centrifugal Forces
The centrifugal force acting on this strip = r  dm   2 rdm
2
Where “dm” is the mass of this strip.
This force acts horizontally.
The moment arm = r sinb ~ r b
Clockwise moment due to centrifugal forces =
r
 r bdm
2 2
Integrating over all such strips, total clockwise moment =
r R
2 2
2

r
b
dm

I

b

r 0
49
Moment at the hinge due to Inertial
forces
Small segment of mass dm
With acceleration rb
r
b
Flap Hinge
b is positive if blade is flapping up
 
Associated moment at the hinge = r rb dm
Integrate over all such segments:
Resulting clockwise moment at the hinge=
Ib
50
At equilibrium..
Ib  I 2 b 
Tip
2
1
rcCl r  V sin   rdr

2
Root
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 is
Lakshmi Sankar 2002
51
How does the blade dynamics behave when there is a
forcing function component on the right hand side of the
form Asint, and a damping term on the left hand side of
form c db/dt ?
To find out let us solve the equation:
2



Ib  cb  I b  A sin t
To solve this equation, we will assume a solution of form:
b  B sin t  C cos t
b 

A
A
cos t 
sin t  
2
c
c

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.
52
What happens when the pilot tilts
the swash plate back?
reaches 90 degrees
azimuth, pitches up.
• Lift goes up instantly.
occurs 90 degrees later
(recall the phase lag).
• The front part of the rotor
disk tilts up.
• The exact opposite
• 270 degree azimuth.
deg
deg
Z
X
53
What happens when the pilot tilts
the swash plate back?
TPP
T
Highest position
At =180 deg
lowest position
At =0 deg
NFP
SwashPlate
The tip path plane tilts back. The thrust points
backwards. The vehicle will tend to decelerate.
54
What happens when the pilot tilts
the swash plate to his/her right?
• Blade at = 180 deg
pitches up. Lift goes up.
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.
T
TPP
NFP
Helicopter viewed from
Aft of the pilot
55
Ib  I 2 b 
2
Tip
1
rcCl r  V sin   rdr

2
Root
has the general solution of the form
b  b 0  b1c cos  b1s sin  
b 2c cos 2  b 2 s sin 2   
First and higher harmonics
Coning Angle
=t
56
b1c determines the fore-aft tilt of
TPP
b1c
TPP
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.
If TPP is our reference coordinate system, what will b1c be? Ans: Zero.
57
b1s determines the lateral tilt of TPP
b1s
Y
Helicopter viewed from the back of the pilot
58
Pilot Input
 75% R   0  1c cos  1s sin 
Collective
Lateral control
Longitudinal control
The pitch is always, by convention
Specified at 75%R.
The pitch at all the other radial locations may be found
If we know the linear twist distribution.
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) by
Tilting the swash plate fore or aft as discussed earlier.
Lateral control means tilt the swash plate (and the TPP) laterally.
59
Longitudinal control
1s
b1c
X
1s
NFP
Note that 1s+ b1c is independent of the
Sankar
2002 are measured.
coordinate system in which
these
angles
60
Lateral Control
Line parallel to NFP
1c
TPP
b1s
NFP
1c
Y
Note that 1c-b1s is independent of the coordinate system in which 1c
And b1s were measured.
61
As a result..
b1c  1s  b1c
b1s  1c  b1s
NFP
 1s TPP
NFP
  1c TPP
62
In the future..
• We will always see b1c+1s appear in pair.
• We will always see b1s-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 (b1c or b1s).
• One degree of pitch= One degree of flap.
63
Forward Flight
Angle of Attack of the Airfoil
And
64
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
• Anhedral and dihedral effects due to coning of
• The first two bullets are self-evident. Let us look
at the other contributors to the angle of attack.
65
r db/dt
The blade section flaps up at a
velocity rb .
The airfoil thinks it is experienci ng
a downwash of equal magnitude.
66
Angle of Attack caused by the
Inflow and Freestream
Rotor Disk
V∞sina+v
a
V∞
r+V∞cosa
67
Anhedral/Dihedral Effect
V∞
V∞sinb
V∞sinb
V∞cosaV∞
See an upwash equal to V∞sinb
See an upwash equal to
V∞sinb
Blade at any  will see an upwash equal to
- V∞sinb cos 
68
Summing them all up..
a

UP
UT= r+V∞cosa
U P 
UP
a    tan     
UT
U T 
where,
U P  V sin a  v  rb  V cos sin b
1
69
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(a) = Cos(b)= Cos() ~ 1
• sin(a) ~ a, sin(b) ~ b, sin() ~ 
70
Angle of Attack
a effecive
UP 
U P U T  U P
   
   arctan 

UT
UT
 UT 
   0   tw
r
  1c cos   1s sin 
R
U T  r  V cos a s sin   r  V sin 
U P  R  rb  V cos a s b cos
 R  rb  Vb cos
s
Subscript s: All angles are in the shaft plane
71
Angle of attack (continued)
U T  U P
r


 r  V sin     0   tw  1c cos  1s sin  
R


 Rs  rb  Vb cos


db  db   d 
db

  
b
 

dt  d   dt 
d
 b1s cos  b1c sin  
72
Angle of Attack (Continued)
After some minor algebra,
U T  U P  r 0  1c  b1s  cos  1s  b1c sin  
V 0 sin   V 1c  b1s  cos sin  
V 1s  b1c sin 2   Vb 0 cos 
Va TPP  v
Notice b1c+1s appears in pair, as pointed out earlier.
Also 1c-b1s appears in pairs.
One degree of pitching is equivalent to one degree of
Flapping.
73
Relationship between aTPP and as
as
b1c
aTPP=as+b1c
74
Angle of Attack (Concluded)
r 0  1c  b1s  cos  1s  b1c sin  
V sin   V   b  cos sin 

U  U P
1  0
1c
1s

a effective  T


UT
U T  V 1s  b1c sin 2   Vb 0 cos


 Va TPP  v

75
• 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.
choices on the way the force coefficients
are computed.
76
Some choices for Computing
Sectional Loads as a function of a
• In analytical work, it is customary to use Cl=aa,
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 of a program (that we will show later),
these loads are corrected for compressibility
using Prandtl-Glauert Rule.
• They are also corrected for sweep or yaw
effects, in some instances.
77
Prandtl-Glauert Rule
V
a

UP
UT= r+V∞cosa
ψ
X
Compute Mach number=M= UT/a∞
Cl ,corrected 
Cl ,incompressible
1 M
2
78
Corrections for sweep or yaw angle
of the velocity
• The angle of attack is modified.
 amodified=a cosL
 L is the local sweep angle of the blade ¼
L
chord line.
Y
X
79
More accurate ways of computing
• In more sophisticated computer programs (e.g.
industry codes), a table look up is used to
compute lift, drag, and pitching moments from
stored airfoil database.
• In some cases, corrections are made for
stall.
• It is important to correct for dynamic stall effects
in high speed forward flight to get the vibration
levels of the vehicle correct.
80
Calculation of Sectional Forces
• After Cl, Cd, Cm are
found, one can find
the lift, drag, and
pitching moments per
unit span.
normal to, and along
the total velocity, and
must be rotated
appropriately.
L’=Lift per
foot of
span
a

D’
UT= r+V∞cosa
1
rU 2 cCl
2
1
 rc(U T2  U P2 )aa eff
2
1
rcaU T U T  U P 
2
L 
81
Forward Flight
To get
82
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.
83
Assumptions for
Analytical Integration
•
•
•
•
•
c= constant (untapered rotor)
v = constant (uniform inflow)
Cd = constant
Linearly twisted rotor
No cut out, no tip losses.
84
a
UP

UT= r+V∞cosa
85
Effective Angle of Attack
As discussed earlier,
a eff
UP
 
UT
U T  r  V sin 
U P  V sin a TPP  v  V b cos
 RTPP  V b cos
L 

1
1
1
rcClU 2  rcaU T2a eff  rca U T2  U T U P
2
2
2

86
Some algebra first..

r

2
r  V sin    0  tw  1c cos  1s sin  
1


L  rca
R



2
 r  V sin  R  V b cos 

3


r2
2 2
2
2
2 r


r

2
V

r

sin



V
sin





2
V


sin



0
0 
tw

tw
 0

R
R


 r V 2 sin 2    2 r 2 cos  2rV  sin  cos 

1c
 1c
 tw R 

 2

1
2
 rca V 1c sin  cos 

2
 2 r 2 sin   2rV  sin 2   V 2 sin 3 

1s
 1s
 1s


  2 RTPP r  V RTPP sin   rV b cos  V2 b cos sin 





Notice that we have first, second, and third harmonics present!
These fluctuations will be felt by the passengers/pilots as
87
Thrust
 1
T  b
 2
R
 



L
dr
d

0  0  

2
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,
Computer codes will do the integrations numerically, without
Sankar
88
to2002
make.
Anlaytical Integration of Thrust
We can interchange the order of integration.
Integrate with respect to ψ first. Use the formulas such as
2
2
2
0
0
0
2
sin

d


cos

d


sin


  cosd  0
2
 sin cosd  0
0
2
2
0
0
2
2
sin

d


cos

 d  
89
Result of Azimuthal Integration
0 2


2 2
 0  r  2 V



2
3
1
1
r 

2 r
2
Ld  rca   tw
 V  tw

2 0
2
R
R 


2
 rV1s   Rr TPP 


90
Normalize
3
3


 tw 2 
R
R
2
2
2
0
 0
 V R   tw
 V R  

1
3
2
4
4

T  rabc 
2
2
2

R
R

2

V




TPP
  1s TPP 2

2
1s TPP TPP 
3 2   tw
2
CT 
1   

 1    

2 3  2  4
2
2 
a  0 


Note that we will get the hover expressions back
 is set to zero.
91
Torque and Power
• We next look at how to compute the
instantaneous torque and power on a
• 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.
92
Profile Drag
r
D’
1
2
D  rU T cCd , 0
2
where
U T  r  V sin 
We will assume chord c and drag coefficient Cd0 are constant.
93
Integration of Profile Torque
 1
Q0  b 
 2
Non-dimensionalize:
Final result:
R

0 0 rDdr 

2
CQ , 0
Q

2
r R  AR
CQ,0 
Cd ,0
8
1   
2
94
Profile Power
 1
Q0  b 
 2
R

0 0 rDdr 

2
Non-dimensionalize:
Final result:
 1
P0  b 
 2
CP ,0
R

0 0 U T Ddr 

2
P0

3
r R  A
CP , 0 
Cd ,0
8
1  3 
2
95
Induced Drag
L’
a
Di

UP
UT= r+V∞cosa
UP 1
2 UP
Di  L
 rcClU T
UT 2
UT

1
rcClU TU P
2
96
Induced Torque and Power
2 R


0 0 rDi dr  d
2 R

1 
Pi 
  U T Di dr  d

2 0  0

1
Qi 
2
Performing the analytical integration,
CPi  i CT
This is a familiar result. Induced Power = Thrust times Induced Velocity!
97
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 Yforce.
• These forces will have inviscid components, and
viscous components.
98
Origin of In-Plane Forces
L’sinb=L’b
L’
L’b
Y
b
X
One source of in-plane forces is the tilting of the
Thrust due to the blade coning angle.
99
Origin of In-Plane Forces-II
V∞
Skin friction forces
A component of the free-stream flows along the blade,
isSankar
usually
Lakshmi
2002 neglected.
100
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), and
Y- direction (Y-Force).
Engineers are interested in both the instantaneous values (which determine
101
Vibration 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 nondimensionalized the same way thrust is nondimensionalized.
• Many text books (e.g. Leishman, Prouty) give
exact expressions for these coefficients.
102
Closed Form Expressions
C H TPP
 TPP 
 tw  
 2  0  2  

 

Cd 0 a 1c b 0
TPP 


 1s


4
2  6
4

 b 02



4


3

2  1c

 4 b 0  0  3  tw   4 TPP 



C a 
YTPP 

3
2  1s b 0
2
 6 1  3  2 b 0 TPP 


103
Forward Flight
Dynamics
104
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.
105
Flapping Dynamics
Ib  I 2 b 
Tip
2
1
rcCl r  V sin   rdr

2
Root
r 0  1c  b1s  cos  1s  b1c sin  
V sin   V   b  cos sin 

U  U P
a  0
1c
1s

Cl  aa effective  a T


UT
U T  V 1s  b1c sin 2   Vb 0 cos


 Va TPP  v

106
Solution Process
Assume the blade flapping dynamics equation
Ib  I 2 b 
Tip
2
1
rcCl r  V sin   rdr

2
Root
has the general solution of the form
b  b 0  b1c cos  b1s sin  
b 2 c cos 2  b 2 s sin 2   
Plug in the solution on both the left and right sides. The right side
Can 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 the
similar term on right.
107
Final Form
2
 80% R
b 1c   1s 
TPP

2
b0   
1  
 tw 


8
60
6
6


8 
3

   75% R  TPP 
3 
4

b 1c   1s 
3 2
1 
2
4
 b 0
b 1s   1c  3
1 2
1 
2


108
Level Flight
Calculation of Trim Conditions
Including
Fuselage Aerodynamics
109
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.
110
Horizontal Force Balance
HM
HT
DF
V∞
Total Drag= Fuselage Drag (DF) + H-force on main rotor (HM)
+ H-force on the tail rotor (HT)
111
Vertical Force Balance
LF
GW
Vertical Force = GW- Lift generated by the fuselage, LF
112
Tip Path Plane Angle
T
aTPP
V∞
T cos a TPP  GW  LF
T sin a TPP  DF  H M  H T
a TPP
 DF  H M  H T 
 tan 

 GW  LF 
1
113
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 database in computer codes.
114
Fuselage Angle of Attack
• Extracted from
– Tip path angle
– Downwash felt by the fuselage from the main
rotor
– Shaft inclination angle.
115
Shaft Inclination Angle
is
Fuselage Reference Axis
116
Relationship between Tip Path
Plane Angle of Attack and Shaft
Angle of Attack
V
∞
as
TPP
b1c
Shaft Plane
aTPP
aTPP  a s  b1c
117
Angle of Attack of the Fuselage
angle of attack.
• Subtract b1c 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
a F  a TPP  b1c  is 
V
118
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 b1c.
• Find tip path plane angle of attack.
• Recompute fuselage angle of attack.
• When iterations have converged, find main and tail rotor
power needed.
119
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.
120
Descent
HM
HT
LF
Wsinχ
χ
W cosχ
Rate of descent
V∞
DF
 descent ve locity 
  tan 

V


1
121
Tip Path Plane Angle in Descent
T cos a TPP  GW cos   LF
T sin a TPP  DF  H M  H T  GW sin 
a TPP
 DF  H M  H T  GW sin  
 tan 

GW
cos


L
F


1
122
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